Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solution to two non polynomial equation If
$$x+\frac1x=-1$$
Find the value of
$$x^{99}+\frac{1}{x^{99}}$$
Is there any formal (traditional) way to solve these problems using only elementary algebra or high school math
| $\displaystyle (x^n + \frac 1 {x^n})(x + \frac 1 x) = x^{n+1} + x^{n-1} + \frac 1 {x^{n-1}} + \frac 1 {x^{n+1}} \\
\displaystyle \Rightarrow x^{n+1} + \frac 1 {x^{n+1}} = -(x^n + \frac 1 {x^n}) - (x^{n-1} + \frac 1 {x^{n-1}})$
If we denote $x^n + \frac 1 {x^n}$ by $f(n)$ then we have just shown that
$f(n+1) = -f(n) - f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Generating function of binary string with minimum lengths for 0s and 1s a) Show that the generating function by length for binary strings where every block of 0s has length at least 2, each block of ones has length at least 3 is:
$$\frac{(1-x+x^3)(1-x+x^2)}{1-2x+x^2-x^5}$$
b) Give a recurrence relation and enough initi... | Let $a_n$ be the number of strings that start with $0$, and let $b_n$ be the number of strings that start with $1$. Then $a_0=b_0=1$, $a_1=0$, and, by conditioning on the length $k$ of the current run, we see that
\begin{align}
a_n &= \sum_{k=2}^n b_{n-k} &&\text{for $n \ge 2$}\\
b_n &= \sum_{k=1}^n a_{n-k} &&\text{fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$X$ a random variable, where $-1\leq X\leq \frac{1}{2}, \mathbb E[X]=0$. What is the maximal value of $\mathbb E[X^2]$? $X$ a random variable, where $$-1\leq X\leq \frac{1}{2}, \mathbb E[X]=0$$
Find the maximal value of $\mathbb E[X^2]$.
I came across a similar question, asking if the following inequality is alway... | For the sake of generality say that $X\in [a,b]$ a.s. (this implies $a\leq 0$).
Since $X=\frac{b-X}{b-a}a + \frac{X-a}{b-a}b$ and the square function is convex, $X^2\leq \frac{b-X}{b-a}a^2 + \frac{X-a}{b-a}b^2$ and taking expectations, $$E(X^2)\leq \frac{b}{b-a}a^2 - \frac{a}{b-a}b^2 = -ab$$
This bound is tight: it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\sin x + \arcsin x > 2x$ using Maclaurin series My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..
Prove that the inequality sin x + arcsin x > 2x holds for all values of x such
that 0 < x ≤ 1.
I know that the Maclaurin series of
sin(x) = x - $\fra... | For $0 < x \le 1$ we have
$$
\sin(x) + \arcsin(x) = 2 x + \sum_{n=2}^\infty \left(\frac{(-1)^n}{(2n+1)!} + \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} \right) x^{2n+1}
$$
because the $x^3$ terms cancel. Therefore it suffices to show that
$$
\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$ \sin{x}\ +\ \frac{1}{2}\sin{2x}\ +\ \frac{1}{3}\sin{3x}\ +\ \frac{1}{4}\sin{4x}\ =\ \frac{2}{3}\left(\cos{x}+1\right)\left(\sin^5x\ +\ 4\right)$ SORRY IF MY TITTLE IS UNCLEAR WITH ONLY MATH FUNCTIONS. IT'S MORE THAN 150 CHARACTERS
*
*This is my math problem
$$ \sin {x}\ +\ \frac{1}{2}\sin {2x}\ +\ \frac{1}{3}\si... | It remains to solve the following equation.
$$\sin{x}(3\cos^2x-\cos{x}+1)=\sin^5x+4.$$
We'll prove that
$$\sin{x}(3\cos^2x-\cos{x}+1)<\sin^5x+4,$$ which says that this equation has no real roots.
Indeed,
$$\sin{x}(3\cos^2x-\cos{x}+1)=\sin{x}(3(1-\sin^2x)+1)-\sin{x}\cos{x}\leq$$
$$\leq\sin{x}(4-3\sin^2x)+\frac{\sin^2x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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All loops of order less than $5$ are groups I am studying loop theory and found that there are no non-trivial loops of order less than $5$. Can anybody help me in this, with some reference of a book or a research article?
A set $Q$ with a binary operation $\circ $ is called a loop if $e \in Q$, where $e$ is the identit... | Without loss of generality we denote the loop's elements as $0,1,2,3$ with $0$ the identity and draw the Cayley table:
$$\begin{array}
00&1&2&3\\
1&&&\\
2&&&\\
3&&&\end{array}$$
By the Latin square property of quasigroups (shared with loops) there are only two ways to fill in the two remaining $1$ entries:
$$\begin{arr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which ar... | Using
$$\sin \frac{\pi}{3}=\frac{\sqrt 3}{2},~\cos \frac{\pi}{3}=\frac{1}{2},~\sec \frac{\pi}{3}=\frac{1}{\cos \frac{\pi}{3}}=2$$
see this.
and the triple angle forumla for $\sec$
$$2=\frac{\sec^3\frac{\pi}{9}}{4-3\sec^2\frac{\pi}{9}}=\frac{\sec\frac{\pi}{9}}{4\cos^2\frac{\pi}{9}-3}$$
$$\Leftrightarrow 2\left(4\cos^2\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove $e^{z+w}=e^ze^w$ using the Taylor expansion for $ e^z.$ For all $z,w ∈ \Bbb{C}$, using the power series definition for $e^z$, prove $e^{z+w}=e^ze^w$.
| Proved it the following way:
Claim: $e^{z+w}=e^ze^w$, where $z\;and\;w\in \Bbb{C}$.
Proof:
Let $e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$
and $e^w=1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots$ .
Now, for RHS:
$e^ze^w=\big(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\big)\big(1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots\big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the stationary point under constraints analitically? I am working with the optimization problem from the paper, eq.(5)
$$\max_{X=(x_1, x_2, \ldots, x_{n+1})} f(X)=(A-B\sum_{i=1}^n \frac{1}{x_i})\times x_{n+1}$$
subject
to $$x_{n+1}=1-2k\sum_{i=1}^n x_i,$$ $$x_i \geq 0, \quad i = 1,2,\ldots, n+1.$$
Here $A \... | EDIT
In the paper, they look for a set of probabilities $p_k\in (0, 1), k = 1, 2, \cdots, s$ and $p_{s+1}
= 1 - 2n \sum_{k=1}^s p_k \in [0, 1]$
such that condition (5) is satisfied, i.e. $p_{s+1}(A - B\sum_{k=1}^s \frac{1}{p_k}) > 1$.
(I put some images at the end.)
Condition (5) requires $A - B \sum_{k=1}^s \frac{1}{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How many integer solutions are there for $x+y+z+w=25$, if $x\geq 1, y \geq 2, z\geq 3, w\geq 4$?
How many integer solutions are there for $x+y+z+w=25$, if $x\geq 1, y \geq 2, z\geq 3, w\geq 4$?
Based off how to solve these problems from my last question I believe I'm meant to take $x^\prime =x-1, y^\prime=y-2,z^\prim... | Your answer is incorrect. As a sanity check, notice that the number of integer solutions of the equation
$$x + y + z + w = 25$$
satisfying $x \geq 1$, $y \geq 2$, $z \geq 3$, $w \geq 4$ should be smaller than the number of solutions of the equation in the nonnegative integers since we cannot substitute $0$ for $x$, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$ In working to prove that
$$1+\cos\theta+\cos(2\theta)+\dots+\cos(n\theta)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{1}$$
I have shown
$$\begin{align}
1+\cos(\theta)+\cos(2\theta)... | To prove
$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$$
examine the LHS
$$LHS = \frac{\cos(n\theta)-\cos(n\theta)\cos\theta+\sin(n\theta)\sin\theta}{2-2\cos\theta}$$
$$= \frac{\cos(n\theta)(1-\cos\theta)+\sin(n\theta)\sin\theta}{2(1-\cos\theta)}$$
Then, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Expand a function in a Fourier Series Expand the function
$$ f(x) = \sin^6{x} \cdot \cos^3{x} $$
in a Fourier series on the segment $[-\pi, \pi]$.
I think we should start from the fact that this function is even, but then the $ a_n $ coefficient turns out to be very big
| If you know complex numbers:
Let
$$z:=\cos(x)+i\sin(x)$$
Then
$$\frac{1}{x}=\cos(x)-i\sin(x) \\
\cos(x)=\frac{1}{2}(z+\frac{1}{z})\\
\sin(x)=\frac{1}{2i}(z-\frac{1}{z})
$$
$$f(x) = \sin^6{x} \cdot \cos^3{x} = \frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^6\\
= \frac{-1}{2^9}(z^3+3z+3\frac{1}{z}+\frac{1}{z^3})(z^6-6z^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Determine the value of the sum to infinity of $(u_{r+1} + \frac{1}{2^r})$ Determine the value of $$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$.
In earlier parts of the question, it is given $\displaystyle u_r= \frac{2}{(r+1)(r+3)}$, which when expressed in partial fractions is $\displaystyle\frac{1}{r+1}-\fr... | You have
$$\begin{equation}\begin{aligned}
\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr) & = \sum_{r=1}^\infty u_{r+1} + \sum_{r=1}^\infty\frac{1}{2^r} \\
& = (\sum_{r=1}^\infty u_{r} - u_1) + 1 \\
& = \frac{5}{6} -\frac{1}{4} + 1 \\
& = \frac{10}{12} - \frac{3}{12} + \frac{12}{12} \\
& = \frac{19}{12}
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Term of rotation matrix entries equals 1 - proof concept?! I derived some stuff and it is happening that i come to the following expression:
$\frac{r_{13}^2 + r_{23}^2}{(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23} - r_{13}r_{22})^2}$
that must equal 1 for all first two rows of a rotation matrix (orthogonal matrix).... | I did this just by using the facts that the rows and columns have length one and are mutually orthogonal.
$$\begin{align}
&(r_{11}r_{23}-r_{13}r_{21})^2+(r_{12}r_{23}-r_{13}r_{22})^2\\
&=r_{11}^2r_{23}^2+r_{13}^2r_{21}^2-2r_{11}r_{21}r_{13}r_{23}+
r_{12}^2r_{23}^2+r_{13}^2r_{22}^2-
2r_{12}r_{22}r_{13}r_{23}\\
&=r_{23}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ ... | You can't, because you need to work with the interplay of the sine functions. Concretely, using Taylor approximations (and collecting the errors together),
\begin{align}
\sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x}
&=\frac1{x+2}-\frac1{6(x+2)^3}-2\left(\frac1{x+1}-\frac1{6(x+1)^3} \right)\\ \ \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
The remainder when $( \sum_{k=1}^5 { ^{20} \mathrm C (2k-1) } )^6$ is divided by 11, is : How do I approach this? Any help is appreciated.
| Let $\displaystyle S =\binom{20}{1}+\binom{20}{3}+\binom{20}{5}+\binom{20}{7} +\binom{20}{9}\cdots (1)$
$\displaystyle S =\binom{20}{9}+\binom{20}{7}+\binom{20}{5}+\binom{20}{3}+\binom{20}{1}$
Using $\displaystyle \binom{n}{r}=\binom{n}{n-r}$
$\displaystyle S =\binom{20}{11}+\binom{20}{13}+\binom{20}{15}+\binom{20}{17}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to use Chinese Remainder Theorem A cubic polynomial $f(x)=ax^3+bx^2+cx+d$ gives remainders $-3x+3$ and $4x-1$ when divided by $x^2+x+1$ and $x^2+2x-4$. Find the value of $a,b,c,d$.
I know it’s easy but i wanna use Chinese Remainder Theorem(and Euclidean Algorithm) to solve it.
A hint or a detailed answer would be ... | EDIT: I just noticed my answer below doesn't meet OP's expectations, as I'm not explicitely using the Chinese theorem. I haven't read the question carefully enough...
Reduction of $f(x)$ modulo $x^2+x+1$ can be computed by successively replacing all instances of $x^2$ in $f$ by $-x-1$, until we obtain an expression ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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A non-probabilistic take on $\lim_{n\to \infty }\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}$ I saw this question here and I tried it on my own. I don't have a background in Probability theory, so I tried to employ algebra only to evaluate this limit.
$$\lim_{n\to \infty}\sum... | $$\begin{aligned}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}&=\frac{1}{5^n}\sum_{k=0}^{4n} {k+n-1\choose n-1} \left(\frac{4}{5}\right)^{k}\\ &=\frac{1}{5^n}\left(\text{coeff. of }x^{n-1} \text{ in }\sum_{k=0}^{4n}(1+x)^{k+n-1}\left(\frac{4}{5}\right)^k\right)\end{aligned}$$
t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_{0}^{2\pi} \frac{\mathrm{e}^{-i k a \cos\phi \sin\theta}}{1+\cos\phi \sin\theta}\,\mathrm d\phi$ How to simplify the following equation:
$$\int_{0}^{2\pi} \frac{\mathrm{e}^{-i k a \cos\phi \sin\theta}}{1+\cos\phi \sin\theta}\,\mathrm d\phi$$
where $k$, $a$ and $\theta$ can be regarded as the constant... | Methinks we can appreciate it carefully with residues theorem.
First of all we step into the complex plane with the usual rules for sine and cosines, that is:
$$\mathcal{J} = \phi \to e^{i\psi} = z ~~~~~~~ \cos\phi \to \ \frac{1}{2}\left(z + \frac{1}{z}\right) ~~~~~~~ \text{d}\phi = \frac{\text{d}z}{iz}$$
We will also... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $(4-x^2)y''+2y=0$ by series I know that there are other ways to solve this, but I want to solve the ODE $(4-x^2)y''+2y=0$ by series close to $0$, that is, I'm looking for solutions of the form $y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}.$
Substituing in the ODE, I got:
$$(8a_{2}+2a_{0})+(24a_{3}+2a_{1})x+\sum_{n=2}^{\i... | Since you correctly obtained a solution $y_1(x)=x^2-4$ (I set $a_0=-4$ in your notation), here is a way to find all solutions. Suppose that a general solution $y(x)$ takes the form $y_1(x)z(x)=(x^2-4)z(x)$. We get
$$y'(x)=y_1(x)z'(x)+y_1'(x)z(x)=(x^2-4)z'(x)+2xz(x)$$
and
$$y''(x)=y_1(x)z''(x)+2y_1'(x)z'(x)+y''_1(x)z(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}$ I need to proof that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}
\end{align}
is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the re... | We proved here that
$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
set $a=3$ and $b=2$ then integrate both sides we have
$$\int_0^\pi\frac{dt}{3+2\cos t}=\int_0^\pi\frac{dt}{\sqrt{5}}+\frac{2}{\sqrt{5}}\sum_{n=1}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $p(x)=x^4-x+\frac{1}{2}$ has no real roots.
What is the simplest way to prove that the polynomial
$p(x)=x^4-x+\frac{1}{2}$ has no real roots?
I did with Sturm's theorem:
$p_0(x)=x^4-x+\frac{1}{2}$
$p_1(x)=4x^3-1$
$p_2(x)=\frac34x-1$
$p_3(x)=-\frac{229}{27}$
The signs for $-\infty$ are $+,-,-,-$ and for $... | You can use AM-GM to show that there are no real roots. If $x< 0$, then $$x^4-x+\frac12> x^4+\frac12 > \frac12>0.$$
If $x\ge0$, then
$$x^4-x+\frac12 =\left(x^4+\frac1{4}+\frac1{8}+\frac1{8}\right)-x\geq 4\cdot\sqrt[4]{x^4\cdot \frac14\cdot\frac18\cdot\frac18}-x=x-x=0,$$
but the equality case doesn't occur because $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Closed-form expression of a series involving combinatorics How to find the closed-form expression of the function $$F(x,y)=\sum_{n=0}^\infty\sum_{m=0}^\infty {{m+n}\choose {n}}x^ny^m=1+x+y+x^2+y^2+2xy+...?$$
| The inner sum is
$$\sum_{m=0}^\infty \binom{m+n}{n} y^m=\frac{1}{(1-y)^{n+1}}.$$
So
\begin{align}
F(x,y)&=\sum_{n=0}^\infty \frac{1}{(1-y)^{n+1}} x^n \\
&= \frac{1}{1-y}\sum_{n=0}^\infty \left(\frac{x}{1-y}\right)^n \\
&= \frac{1}{1-y}\cdot\frac{1}{1-x/(1-y)} \\
&= \frac{1}{1-x-y}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Number of possible integers less than 100000 such that the digits 4,5,6 appear in that order I have been stuck on this problem for some time and I am not sure how to approach it:
"How many positive integers less than 100,000 have digits containing 4,5,6 in that particular order?" (by this it means that the digits "456"... | We can represent any nonnegative integer less than $100000$ as a five-digit string by appending leading zeros as necessary. For instance, we can represent $456$ as $00456$ and $4956$ as $04956$.
There are ten possible positions in which the first $4$, first $5$, and first $6$ could appear:
$456\square\square$
$45\squa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$ Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $x=y=z.$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi... | With your substitution you have $a,b,c\in [0,2\pi],\ a+b+c=\pi$ and
$$\cos a+\cos b+\cos c=\frac{3}{2}$$
We have that:
$$\cos c=1-2\sin^2 \frac{c}{2},\ \cos a+\cos b=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}$$
Therefore:
$$\cos a+\cos b+\cos c=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}+\cos c\\
=2\sin \frac{c}{2}\cos \frac{a-b}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $1-3^x+5^x-7^x = 0 \Leftrightarrow x=0$ Problem :
Prove that equation $$1-3^x+5^x-7^x = 0$$
has unique root $x=0$.
Since $\lim_{x\to-\infty}(1-3^x+5^x-7^x) = 1$ and $\lim_{x\to\infty}(1-3^x+5^x-7^x) = -\infty$,
I tried to show $x \to 1-3^x+5^x-7^x$ is decreasing function, but it wasn't easy to determine it... | $$1-3^x+5^x-7^x = 0$$
$$\rightarrow0\geq 1-3^x=7^x-5^x \geq 0 $$
$$\rightarrow0= 1-3^x=7^x-5^x = 0 $$
$$\rightarrow x=0$$
EDIT
If $x<0$ we'd get $$\rightarrow 0\leq 1-3^x=7^x-5^x \leq 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate using differentiation under the sign of integration: $\int_{0}^{\pi} \frac {\ln (1+a\cos (x))}{\cos (x)} dx$ Evaluate by using the rule of differentiation under the sign of integration $\int_{0}^{\pi} \dfrac {\ln (1+a\cos (x))}{\cos (x)} \textrm {dx}$.
My Attempt:
Given integral is $\int_{0}^{\pi} \dfrac {\ln(... | Let $a^2<1$
$$I(a)=\int_{0}^{\pi} \frac{\ln 1+ a \cos x)}{\cos x} dx~~~~(*)$$
D.w.r.t.$a$ on both sides, to get
$$\frac{dI}{da}=\int_{0}^{\pi} \frac{\cos x}{(1+a\cos x)\cos x} dx=\int_{0}^{\pi} \frac{dx}{1+a \cos x}=J(a)~~~~(1)$$
Use $$\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx~~~(2)$$
Then $$J(a)=\int_{0}^{\pi} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve indefinite integral $ \int \frac{x^4}{\sqrt{x^2+4x+5}}dx $ I need to solve the next problem:
$$
\int \frac{x^4}{\sqrt{x^2+4x+5}}dx
$$
I know the correct answer is
$$
(\frac{x^3}{4}-\frac{7x^2}{6}+\frac{95x}{24}-\frac{145}{12})*\sqrt{x^2+4x+5}\space+\space\frac{35}{8}\ln{(x+2+\sqrt{x^2+4x+5})}+C
$$
still, I canno... | With $x+2=\sinh t$ to rewrite the integral as,
$$
I=\int \frac{x^4}{\sqrt{x^2+4x+5}}dx=\int (\sinh t-2)^4dt$$
$$=\int (\sinh^4 t-8\sinh^3 t + 24\sinh^2 t - 32\sinh t+16)dt
\tag 1$$
where,
$$\begin{array}
& & \int \sinh t dt = \cosh t \\
& \int \sinh^2 t dt = \frac12\int (\cosh 2t-1)dt=\frac14\sinh 2t -\frac12t \\
& \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Greatest common divisor of $(x+1)^{4n+3} + x^{2n}$ and $x^3-1$. I have to find the greatest common divisor of
$$(x+1)^{4n+3} + x^{2n}$$
and
$$x^3-1$$
I know I can express the second polynomial as:
$$x^3-1 = (x-1)(x^2+x+1)$$
So I would have to check if the first polynomial is divisible by $(x^3-1)$, $(x^2+x+1)$ or $(x-... | Let $Q(x) = (x+1)^{4n+3} + x^{2n}$. Then $Q(1) = 2^{4n+3} + 1 \neq 0$, hence $x-1$ does not divide $Q(x)$.
We have $$Q(x) = (x+1)^{4n+3} + x^{2n} = (x+1)[(x+1)^2]^{2n+1} + x^{2n} =$$
$$= (x+1)[(x^2+x+1) + x]^{2n+1}+ x^{2n} = $$
$$=(x^2+x+1)P(x) + (x+1)x^{2n+1} + x^{2n} = (x^2+x+1)(P(x) + x^{2n}),$$
hence $x^2+x+1$ div... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Concurrency of normals If $\theta_1$, $\theta_2$, and $\theta_3$ are the eccentric angles of three points on the hyperbola $x^2/a^2 - y^2/b^2 = 1$ such that
$\sin (\theta_1+\theta_2) +
\sin (\theta_2+\theta_3) +
\sin (\theta_3+\theta_1) = 0$ then prove that the normals at these points are concurrent.
P.S. My attempt ... | Start with $$\frac{ax}{\text{sec}\ \theta_1} + \frac{by}{\text{tan}\ \theta_1} = a^2 + b^2$$
$$\frac{ax}{\text{sec}\ \theta_2} + \frac{by}{\text{tan}\ \theta_2} = a^2 + b^2$$
$$\frac{ax}{\text{sec}\ \theta_3} + \frac{by}{\text{tan}\ \theta_3} = a^2 + b^2$$
When are these three straight lines concurrent?
Equate the foll... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int_0^\infty \frac{\log(2+x^2)}{4+x^2}\,\mathrm dx$ Evaluate the integral
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
--
I started by stating that the integral from 0 to infinity should be the same as half the integral from $-\infty$ to $\infty$, that is:
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx =... | $$I=\int_{0}^{\infty} \frac{\log(2+x^2)}{4+x^2}dx=\int_{0}^{\infty} \frac{\log(a^2+x^2)}{4+x^2}dx=I(a)$$
$$\implies \frac{dI}{da}=2a \int_{0}^{\infty} \frac{dx}{(a^2+x^2)(4+x^2)]}=
\frac{2a}{4-a^2}\int_{0}^{\infty} \left( \frac{1}{a^2+x^2}-\frac{1}{4+x^2}\right) dx$$
$$\frac{dI}{da}=\frac{2a}{4-a^2}[\frac{\pi}{2a}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I establish the differentiability of this function by definition? Given this function
$f(x) = (x^2 + 1)^{\sin(x)} $
How would I establish its differentiability over the entire function? I understand how to establish differentiability at a point $c$, by assessing the limit of the different quotient at $x = c$. Bu... | You shouldn't use the definition of derivative, instead you should use a tricky technique:
$\displaystyle f(x)=(1+x^2)^{\sin x} \implies \ln(f(x)) = \ln\left( (1+x^2)^{\sin x} \right) = \sin x \ln(1+x^2)$
$\displaystyle \implies \frac{d}{dx} \ln(f(x)) = \frac{d}{dx} \sin x \ln(1+x^2) \implies \frac{f'(x)}{f(x)} = \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
| By my previous post it's enough to prove that:
$$x^2\ln{x}+x+e^x-3x^2>0,$$ where $x$ is a root of the equation $$x^2-x+e^x(x-2)=0.$$
The last equality gives $$e^x=\frac{x^2-x}{2-x}$$ or
$g(x)=0,$ where $$g(x)=x-\ln\frac{x^2-x}{2-x}$$ and since $$\frac{x^2-x}{2-x}>0,$$ we obtain $$1<x<2.$$
Now, $$g'(x)=\left(x-\ln\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to obtain a formula for $f(z)$ given this recurrence I am trying to figure out how to derive a formula for $f(z)$ that is a function of $z$ and maybe $k \in \mathbb{N}$:
$$
f(z) = 1+z f \bigg(\frac{z}{1+z} \bigg)
$$
As an attempt, I tried a change of variable $z=\frac{1}{x}$, and I get:
$$
f\bigg(\frac{1}{x}\bigg) ... | In the last step you plug $1+x$ in the place of $x$, hence you should have
$$ f\left( \frac{1}{x}\right) = 1 + \frac{1}{x}f\left( \frac{1}{1+x}\right) = $$
$$ 1 + \frac{1}{x}\left[ 1 + \frac{1}{1+x}f\left( \frac{1}{2+x} \right)\right]$$
So I think the telescopic series you achieved isn't correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int \sin^2x\cos^4xdx$ $$\int \sin^2x\cos^4xdx$$
I tried
$$I = \int (1-\cos^2x)\cos^4xdx = \int \frac{\sec^2x-1}{\sec^6x}dx = \int \frac{\tan^2x}{\sec^6x}dx$$
Take $\tan x = t \implies \sec^2xdx = dt$
$$I = \int \frac{t^2}{(t^2+1)^4}dt$$
And I could not proceed further from here.
| With even powers of the trig functions use the half-angle identities until you have a an odd power.
$\sin^2 x = \frac 12 (1-\cos 2x)\\
\cos^2 x = \frac 12 (1+\cos 2x)$
$(\sin^2 x)(\cos^4 x) = \frac {1}{8}(1-\cos 2x)(1+cos 2x)(1+\cos 2x)\\
\frac18(1-\cos^2 2x)(1+\cos 2x)\\
\frac 18 (1 +\cos 2x - \cos^2 2x - \cos^3 2x)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\c... | Firstly, $$\sin\theta=\sqrt{1-\cos^2\theta}$$ is wrong. Try $\theta=-\frac{\pi}{2}$.
But for $\theta\in(0,\pi)$ we see that $-3<3\cos\theta<3$ and $x=3\cos\theta$ gets any value from $(-3,3)$.
Also, for these values of $\theta$ we obtain a right formula: $$\sin\theta=\sqrt{1-\cos^2\theta}.$$
Secondly, $$\cos^3\theta=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How many times must I apply L’Hopital? I have this limit:
$$\lim _{x\to 0}\left(\frac{e^{x^2}+2\cos \left(x\right)-3}{x\sin \left(x^3\right)}\right)=\left(\frac 00\right)=\lim _{x\to 0}\frac{\frac{d}{dx}\left(e^{x^2}+2\cos \left(x\right)-3\right)}{\frac{d}{dx}\left(x\sin \left(x^3\right)\right)}$$
$$\lim_{x\to0}\frac{2... | You have one too many copies of the $\sin x^3$ in your denominator after the first derivative, so let's take a step back. If you're prepared to use Taylor series,$$\begin{align}e^{x^2}+2\cos x-3&=1+x^2+\frac12x^4+2-x^2+\frac{1}{12}x^4-3+o(x^4)\\&\sim\frac{7}{12}x^4,\\x\sin x^3&\sim x^4,\end{align}$$so the limit is $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{1/x}$ $$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=?$$
I have tried a lot but I am stuck when I solve this by using this hints. $a^x=\exp(\ln(a^x))=\exp(x\ln a)$, so then $a=\frac{4^x+5^x}{4}$. The above expression becomes
$$\lim_{x\to\inft... | Do you have to use this hint?
You could simplify your task if you simplify the given expression first:
$$\frac{4^x+5^x}{4}=\frac{1}{4}\cdot(4^x+5^x)=\frac{1}{4}\cdot5^x\cdot\left(\left(\frac{4}{5}\right)^x+1\right),$$
therefore
$$\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=\left(\frac{1}{4}\cdot5^x\cdot\left(\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle... | By multiplying with $ 5(+5)$ we clear the denominators:
$- \frac{1(s+5)}{5s(s+5)} + \frac{5}{5s (s+5)} = - \frac{1s}{5s(s + 5)}$
$-s = -s$
Which proves the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Show that: $\sum_{k=0}^{\infty}\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\frac{\pi}{4}$ I was observing this paper and by some experimental chance I got this result, but I can not prove it.
$$\sum_{k=0}^{\infty}\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\frac{\pi}{4}\tag1$$
I tried to f... | Let $ u = \frac{2}{ k^2 + 2k + 1}, v = \frac{2}{ 4k^2 + 4k + 1}$. Then,
$$ \arctan (u) - \arctan (v) = \arctan ( \frac{ u - v} { 1 + uv} ) = \arctan ( \frac{6k^2 + 4k } { 4k^4 + 12k^3 + 13k^2 + 6k + 5 }). $$
Then use $\sum_{k=1} \arctan \frac{2}{k^2} = \frac{3 \pi } { 4} $ (2.14) and $ \sum_{k=0} \arctan \frac{2}{(2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Reduction of order on a 2nd order LI DE The equation and $y_1(x)$ are $$y_1(x)=x^{\frac{-1}{2}}\cos(x) $$
$$x^2y''+xy'+(x^2-\frac{1}{4})y=0$$
Using the formula that solves for $y_2(x)$
(the one with an integration factor over $y_1(x)$ squared, all inside an integral multiplied by $y_1(x)$)
We get $$ x^{-1/2}\cos(x) \i... | We are given
$$x^2y''+xy'+\left(x^2-\dfrac{1}{4}\right)y=0 \tag 1$$
We are also given a solution as
$$y_1(x)=x^{-1/2}\cos(x)$$
Using Reduction of Order, a second solution is given by
$$y_2(x) = y_1(x) v(x) = x^{-1/2} \cos(x) v(x)$$
The first derivative is
$$y_2'(x) = \frac{\cos (x) v'(x)}{\sqrt{x}}-\frac{v(x) \cos (x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A five digit number minus a four digit number equals $33333$. What are the two numbers, if you are only allowed to use the numbers $1-9$
Question: A five digit number minus a four digit number equals $33333$. What are the two numbers, if you are only allowed to use the numbers $1-9$ once? More precisely,
\begin{ali... | The sum of the digits $1$ through $9$ is odd. They contribute to the parity of the digit sum of the result no matter which row they’re in. The digit sum of the result is odd. Thus there must be an even number of borrowings.
A column that causes borrowing must have a $7$, $8$ or $9$ in the bottom row, so we cannot have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)... | There is a thing called multinomial theorem which is a slight generalization of the binomial theorem.
First of all, note $(x^2+x^7+x^9)^{20}= x^{40}(1+x^5+x^7)^{20}.$ So, basically, now will find the coefficients of $x^{17}.$ Can you continue?
Note that $(1)^{17} (x^5)^2(x^7)^1$ gives us $x^{17}.$ The coffocient of $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Examples of triangles, which related ellipses are perfectly packed with circles. Ellipse can be
perfectly packed with $n$ circles
if
\begin{align}
b&=a\,\sin\frac{\pi}{2\,n}
\quad
\text{or equivalently, }\quad
e=\cos\frac{\pi}{2\,n}
,
\end{align}
where $a,b$ are the major and minor semi-axis
of the ellipse and $e=\... |
This is example of the
ellipse, perfectly packed with three circles,
inscribed into the equilateral triangle $ABC$.
Let the center of the ellipse be $M=0$
and its semi-axes defined as
\begin{align}
s_a=|DF_1|=|DF_2|&=1
,\\
s_b=|MD|=|ME|&=\sin\frac\pi{2\cdot3}=\frac12
,
\end{align}
locations of the top and bottom poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of $f(x)= 3x^2 - 3x -11$, where $x>2$
Find the inverse of $f(x)= 3x^2 - 3x -11$, where $x > 2$
The answer guide says the answer is $f^{-1}(x)=\frac{1}{2} + \sqrt{ \frac{x}{3}} +\frac{47}{12}$,
but i had $\frac{1}{2} + \sqrt{\frac{x}{3} +\frac{47}{12}}$ instead.
How do i get the correct answer?
| To find the innverse of $f(x)$, you have to swap $x$ and $y$, obtaining:
$$x=f(y)=3y^2-3y-11=\left(\sqrt{3}y-\frac{\sqrt{3}}{2}\right)^2-11-\frac{3}{4}$$
From here you solve for $y$, having:
$$x+11+\frac{3}{4}=3\left(y-\frac{1}{2}\right)^2$$
And so:
$$\sqrt{\frac{x}{3}+\frac{47}{12}}=y-\frac{1}{2} \leftrightarrow y = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a_t \leq \frac{c}{t}$ We have the recursive equation $$a_t^2 = (1 - a_t)a_{t-1}^2$$ $$a_0 = 1$$
Prove that there exists constant $c$ such that $a_t \leq \frac{c}{t}$, and find the smallest such $c$.
I was able to get that $a_t^2 = (1 - a_t)(1 - a_{t-1})\dots(1 - a_1)$, but I don't know how to proceed after that,... | It turn out that
$c \ge 2$ will work.
$a_t^2
= (1 - a_t)a_{t-1}^2
$.
$a_1^2
= 1-a_1$
so
$a_1
=\dfrac{-1\pm\sqrt{5}}{2}
=\dfrac{-1+\sqrt{5}}{2}
$.
$a_t^2+a_ta_{t-1}^2-a_{t-1}^2
= 0
$
so
$\begin{array}\\
a_t
&=\dfrac{-a_{t-1}^2+\sqrt{a_{t-1}^4+4a_{t-1}^2}}{2}\\
&=\dfrac{-a_{t-1}^2+a_{t-1}\sqrt{a_{t-1}^2+4}}{2}\\
&=a_{t-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity. I was given a statement that if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity.
The roots of $x^n−1$ in $\mathbb C$ which are not also roots of $x^m −1$ for some $1 ≤ m ≤ n$ are... | $\omega$ a primitive third root of unity implies $\omega=e^{(2\pi i k)/3},\,k=1,2$. And $-1=e^{\pi i}$. Thus $-\omega=e^{\pi i}e^{(2\pi i k)/3}=e^{(6+4k)\pi i/6}=e^{(2\pi i k)/6}\,,k=1,5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function woul... | By Lagrange, we minimize
$$x^2+y^2+\lambda(x^4+y^4+3xy-2).$$
The derivatives on $x$ and $y$ yield
$$2x+\lambda(4x^3+3y)=2y+\lambda(4y^3+3x)=0,$$ and after elimination of $\lambda$,
$$2x(4y^3+3x)=2y(4x^3+3y)$$ or
$$(8xy-6)(y^2-x^2)=0.$$
The solutions $y=\pm x$ imply
$$2x^4\pm3x^2-2=0,$$ giving
$$x^2+y^2=2x^2=\frac{\mp3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show $\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$ Show that
$$\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$$
My Attempt:
Let $$I=\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
Using $\int_{a}^{b} f(x) dx=\int... | $$I=\int_{0}^{\pi} \frac{x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(1)$$
Apply $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$.
$$I=\int_{0}^{\pi} \frac{(\pi-x) dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(2)$$
Add (1) and (2)
$$2I=\pi\int_{0}^{\pi} \frac{ dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(3)$$
Use $\int_{0}^{2a} f(x) dx=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating Series Integral How do I show that $$\int_0^\infty\frac{(\ln x)^2dx}{1+x^2}$$=$$4(1-1/3^3+1/5^3-1/7^3...)$$
I expanded the integral to $$\int_0^\infty(\ln x)^2(1-x^2+x^4...)dx$$ using the power series for $$\frac{1}{1+x^2}$$ but I'm not sure how to continue from here.
| Starting with Integrand's observation in the comments, we have $$
\begin{array}{rcl} \displaystyle I = 2\int_{0}^{1}{\displaystyle\frac{\log^{2}{x}}{1+x^{2}}\,\mathrm{d}x}=2 \int_0^1 \log^2{x} \sum_{k \ge 0} (-1)^k x^{2k}\,\mathrm{d}x = 2\sum_{k \ge 0} (-1)^k\int_0^1 x^{2k}\log^2{x}\,\mathrm{d}x\end{array}
$$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
There is a number, the second digit of which is smaller than its first digit by 4, and if the number There is a number, the second digit of which is smaller than its first digit by 4, and if the number was divided by the digit's sum, the remainder would be 7.
Actually I know the answer is 623
I found it by using comput... | One-digit case is impossible, since $4\not \equiv7 \mod 4$
Two-digit case: write number as $10(a+4)+a$.
$$10(a+4)+a \equiv 7 \mod (2a+4)$$
$$11a\equiv -33 \mod 2a+4$$
$$a+3 \equiv 0 \mod 2a+4$$
$$2(a+3) \equiv 2 \not \equiv 0 \mod 2a+4$$
Therefore, two-digit is impossible.
Three-digit case: Write number as $100(a+4)+10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For $n\ge 3$ determine all real solutions of the system of $n$ equations.
Question: For $n\ge 3$ determine all real solutions of the system of $n$ equations: $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
... | We have $$x_1+x_2+\cdots+x_{n-1}-\frac{1}{x_n}=0\\x_1+x_2+\cdots-\frac{1}{x_{n-1}}+x_n=0\\.........................\\.........................\\-\frac{1}{x_1}+x_2+\cdots+x_{n-1}+x_n=0 $$ It follows
$$\sum_{i=1}^{i=n}\left((n-1)x_i-\frac{1}{x_i}\right)=\sum_{i=1}^{i=n}\frac{(n-1)x_i^2-1}{x_i}=0 $$
Then a clear solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)? I have a question that goes:
If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?
So what I tried was I know that since th... | By sketching a graph in mind, we see that the line $ax+by+c=0$ is just a line that is passing through $(-1,2)$ (the center of the circle $x^2+y^2+2x-4y+1=0$) and is tangent to the circle $x^2+y^2-2x=3/5$.
Therefore we looking for tangent lines from point $(-1,2)$ to the circle $x^2+y^2-2x=3/5$.
The line passing through... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Quadrilateral/Geometry: Find the area of all the triangles in a trapezoid
I know the area of $BDE$ is $4$ but can't find out the area of $ABE$.
To solve $BDE$:
Since triangle $ACD$ and $CDB$ share the same base and height and knowing their shared area $CDE$ is 3, we know $BDE$ is 4.
Question: How do you solve for $... | The area of the triangle ADC is equivalent to area of triangle ACE plus area of triangle CDE $$\frac{1}{2} \times AC \times DC = 4+3$$ $$AC \times DC = 2(4+3).$$ Thus, $AC \times DC = 14.$
Now for area of triangle of BDC, it is the area of BDE plus area of CDE $$\frac{1}{2}\times AC\times DC = \text{area of triangle B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Different solutions with different results for an inequality Find m such that the following inequality:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
is always true for $\forall x \in R$.
1st solution:
1st case
$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$
$$<=>x^2+2x-m-1>0$$
$$\Leftrightarrow 1^2... | Feedback on the First Solution
Your solution is somewhat correct but not complete. In fact the question asks us to find all $m$ that the given inequality holds for all $x \in \mathbb{R}$. Your decomposition into the cases is correct, but you then tried to find all $m$ that each of the cases holds for all $x \in \mathbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An A level question about partial differentiation
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My ans... | Using the product rule gives: $$d(f(u)g(v))=f'(u)g(v)du+f(u)g'(v)dv$$
Thus I like to work your equation like this:
$$d(2x^4+xy^3+y^4)=8x^3dx+y^3dx+3xy^2dy+4y^3dy=d(10)=0$$
Now we factorize to get the result:
$(8x^3+y^3)dx+(3xy^2+4y^3)dy=0\iff \dfrac{dy}{dx}=-\dfrac{8x^3+y^3}{3xy^2+4y^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Trouble with $4\times4$ matrix determinant $$
\begin{vmatrix}
1 & -6 & 7 & 5 \\
0 & 0 & 3 & 0 \\
3 & -2 & -8 & 6\\
2 & 0 & 5 & 4\\
\end{vmatrix}
$$
Clearly I want to expand along the second row yielding:
$((-1)^5)3$ times the following matrix
$$
\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 ... | Note that, in your $3\times 3$ matrix, you could have subtracted twice the first column to the last, obtaining instantly the determinant:
$$\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 & 6 \\
2 & 0 & 4 \\
\end{vmatrix}=\begin{vmatrix}
1 & -6 & 3 \\
3 & -2 & 0 \\
2 & 0 & 0 \\
\end{vmatrix}=3\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area between the graphs of $f(x)=e^{.25x}$ and $g(x)=\frac{-1}{x}$ between $x=1$ and $x=2$ 5.5
Can somebody verify this solution for me?
Find the area between the graphs of $f(x)=e^{.25x}$ and $g(x)=\frac{-1}{x}$ between $x=1$ and $x=2$
Since $f(x) > g(x)$ on $(1,2)$, the area between the graphs is:
$\int_1^2 ... | Well, area is an absolute thing. So we know that the total area $\mathcal{A}$ is given by:
$$\mathcal{A}=\int_1^2\exp\left(\frac{x}{4}\right)\space\text{d}x+\left|\int_1^2-\frac{1}{x}\space\text{d}x\right|=$$
$$\left[4\exp\left(\frac{x}{4}\right)\right]_1^2+\left|\left[-\ln\left|x\right|\right]_1^2\right|=$$
$$4\exp\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of squared Fresnel sine integral I'm trying to find the following sum:
$$ \sum_{n=0}^{\infty} \frac{S\left(\sqrt{2n}\right)^2}{n^3}$$
where $S(n)$ is the fresnel sine integral, however, I think I made a mistake somewhere.
To start, I considered using parseval's identity:
$$ 2\pi\sum_{n=-\infty}^{\infty} |c_n|^2 = \... | This is not answer.
I am stuck with the problem but I have a few remarks
*
*I suppose that the summation starts at $n=1$ and not at $n=0$
*Using Wolfran Alpha (see here)
$$\int_{-\pi}^\pi\left(\frac{\sqrt{-i x}}{\sqrt{2}}+\frac{\sqrt{i x}}{\sqrt{2}}\right)^2\,dx= \pi^2$$
*Using Wolfran Alpha (see here)
$$\int_{-\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Probelm in solving $\sin\left(\frac{x}{2}\right) - \cos\left(\frac{3x}{2}\right) = 0$ While solving $\sin\left(\frac{x}{2}\right) - \cos\left(\frac{3x}{2}\right) = 0$, if I convert $\sin$ into $\cos$, I am getting answer (i.e. $\left(n + \frac{1}{4}\right)\pi$ and $\left(2n - \frac{1}{2}\right)\pi$. However, if I conve... | Write $\sin \frac x2=-\cos(\frac\pi2+\frac x2)$ and factorize with $\cos a+\cos b =2\cos\frac{a+b}2 \cos\frac{a-b}2$,
$$\sin\frac x2 - \cos\frac{3x}2
= -2\cos(x+\frac\pi4)\cos(\frac x2-\frac\pi4)=0$$
which leads to the solutions
$$x= \frac\pi4+n\pi,\> -\frac{\pi}2+2n\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Nice integral $\int_{0}^{\infty}\ln\Big(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\Big)\frac{1}{x}dx=-\frac{3\pi^2}{4}$ Last integral of the day :
$$\int_{0}^{\infty}\ln\Big(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\Big)\frac{1}{x}dx=-\frac{3\pi^2}{4}$$
I have tried integration by parts and some obvious substitution but I failed.
I have tr... | Let $\displaystyle I = \int_0^{\infty} \ln \left( \frac{x^3 - x^2 - x + 1}{x^3 + x^2 + x + 1}\right)\frac{1}{x} dx$. We can write,
\begin{align}
I & = \int_0^{\infty} \ln \left( \frac{(x-1)^2(x+1)}{(x+1)(x^2 + 1)}\right)\frac{1}{x} dx \\
& = \int_0^{1} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx + \int_1^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluating $\lim\limits_{n\to\infty}\frac1{n^4}·\left[1·\sum\limits_{k=1}^nk+2·\sum\limits_{k=1}^{n-1}k+3·\sum\limits_{k=1}^{n-2}k+\cdots+n·1\right]$
Find the value of$$L=\lim_{n\to\infty}\frac1{n^4}\cdot\left[1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1\right].$$
My attempt is a... | WE can write it as $$L=\lim_{n \to \infty} \frac{1}{n^4}\sum_{j=1}^{n} j \sum_{k=1}^{n-j+1} k =\lim_{n \to \infty} \frac{1}{n^4}\sum_{j=1}^{n} \frac{j(n-j+1)(n-j+2)}{2}$$ $$L = \lim_{n \to \infty} \frac{1}{2n^4}\sum_{j=1}^n(j^3-2nj^2+n^2j+....)$$
Let us use the asymptotic result when $n$ is very large: $\sum_{k=1}^n k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
S... | What you would do is draw a perpendicular from the centre of the circle to that line.
That perpendicular passes through $C=(2,1)$ and is (obviously) perpendicular to $y=\dfrac{4}{3}x+\dfrac{20}{3}$
Line has a gradient of $-\dfrac{3}{4}$
$\begin{align} \dfrac{y-1}{x-2} &= -\dfrac{3}{4} \\ y&=-\dfrac{3}{4}x +\dfrac{5}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
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Simple factorization problem I cannot find the solution! This is not a homework assignment.
$11x^2+13x-7$
| Do you know about completing squares? Write your expression as $$11\left(x^2+\frac{13}{11}x\right)-7.$$ The goal is to write this as a difference of squares, which is easy to factorise. To this end, complete the square on the expression in brackets, which becomes $$x^2+2x\frac{13}{22}+\left(\frac{13}{22}\right)^2-\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Why does $\sum^N_{n=0}\frac12\leq\sum^N_{n=0}\sum^{2^{n+1}-1}_{r=2^n}\frac1r\leq\sum^N_{n=0}1=\frac{N+1}2\leq\sum^{2^{N+1}-1}_{r=1}\frac1r\leq N+1$?
Consider the following inequalities:
$$\frac{1}{2} \leq \sum^{2^{n+1}-1}_{r=2^n}\frac{1}{r}\leq 1 \tag{1}$$
Upon summing over $(1)$ from $n=0$ to $n=N$, we obtain
$$\sum^... | In fact it is a $\underline{\textbf{telescopic sum}}$, let $ N\in\mathbb{N} $, denoting for any $ n\in\mathbb{N} $, $ S_{n}=\sum\limits_{k=1}^{2^{n}-1}{\frac{1}{k}} $, we have : \begin{aligned}\sum_{n=0}^{N}{\sum_{r=2^{n}}^{2^{n+1}-1}{\frac{1}{r}}}&=\sum_{n=0}^{N}{\left(\sum_{r=1}^{2^{n+1}-1}{\frac{1}{r}}-\sum_{r=1}^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{... | The expression tends to $$\frac{2\cdot1-\sqrt{1^2+3}}{\sqrt{1+3}+\sqrt{2\cdot1+2}}=0.$$
You may not use L'Hospital here.
(Or is there a typo in the question ?)
Update:
$$\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}=\frac{\sqrt{x+3}+\sqrt{2x+2}}{2x+\sqrt{x^2+3}}\cdot \frac{4x^2-x^2-3}{x+3-2x-2}\to\frac{4}{4}(-6).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Showing that $\sum_{n=1}^\infty n^{-1}\left(1+\frac{1}{2}+...\frac{1}{n}\right)^{-1}$ is divergent
How to show that $\sum_{n=1}^\infty \frac{1}{n\left(1+\frac{1}{2}+...\frac{1}{n}\right)}$ diverges?
I used Ratio test for this problem and this is the result: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1-\frac{... | First note that
$$
\sum\limits_{k = 1}^n {\frac{1}{k}} \le 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t} } } = 1 + \int_1^n {\frac{{dt}}{t}} = 1 + \log n,
$$
for all $n\geq 1$. Then for all $N\geq 3$,
\begin{align*}
& \sum\limits_{n = 1}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}} > \sum\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find an ellipse tangent to triangle given centre position of the ellipse I have a fully defined triangle, $PT_1T_2$, of side lengths $a$, $b$, & $c$ with corresponding opposite angles $A$, $B$, & $C$. I want to find an ellipse that is tangent to the lines $PT_1$ and $PT_2$ at points $T_1$ and $T_2$ respectively.
As the... | I like the more-geometric answers, but I'm somewhat committed to this analytic approach, so let's continue ...
Place $O$ at the origin, and let $P = (-d,0)$. As @Aretino helpfully observes in a comment, $\overleftrightarrow{OD}$ must bisect the chord $\overline{T_1T_2}$, so define $M$ as midpoint of the chord, with $|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Closed form of the recursive function $F(1):=1,\;F(n):=\sum_{k=1}^{n-1}-F(k)\sin\left(\pi/2^{n-k+1}\right)$ Suppose that $F$ is defined via the recurrence relation
$$F(1)=1, \qquad F(n)=\sum_{k=1}^{n-1}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr)$$ What is $F(N)$? I don't have any idea how to solve this problem. Only o... | By putting
$$
G(n) = F(n + 1)
$$
we can rewrite the recurrence as
$$ \bbox[lightyellow] {
\sum\limits_{k = 0}^n {G(k)\sin \left( {{{\pi /2} \over {2^{\,n - k} }}} \right)}
= \sum\limits_{k = 0}^n {G(n - k)\sin \left( {{{\pi /2} \over {2^{\,k} }}} \right)} = \left[ {0 = n} \right]
} \tag{1}$$
where $[P]$ denotes t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Definite Integral of $\int_{\frac{-1}{2}}^\frac{1}{2}\int_{\frac{-1}{2}}^\frac{1}{2} \sqrt{x^2+y^2} dxdy$ Here's my attempt at trying to evaluate the integral.
Let $x = y tan\theta$
$$\frac{dx}{d\theta} = \frac{y}{cos^2\theta}$$
$$dx = \frac{y}{cos^2\theta}d\theta$$
The new bounds of inner integral would be $\theta = t... | We can use symmetry to say that
$$I = \int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{x^2+y^2}\:dxdy = 8 \int_{0}^{\frac{1}{2}}\int_{0}^{x}\sqrt{x^2+y^2}\:dydx$$
From here, we can use polar coordinates:
$$I = 8 \int_{0}^{\frac{\pi}{4}} \int_0^{\frac{1}{2}\sec\theta}r^2\:drd\theta = \frac{1}{3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$... | $$\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\operatorname{Li}_4\left(\frac12\right)-\frac{\pi^4}{30}+\frac{21}8\ln2\zeta(3)-\frac{\pi^2}{12}\ln^22$$
$$\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=\frac{C^2}{2}+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx$ Evaluate $$\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx\,,$$ where $a$ being parameter. I am not able to solve this.
| Let $$I(a) = \int_0^{1}\frac{\arctan {ax}}{x\sqrt{1-x^2}}dx$$
Differentiating the integral with respect to $a$,
$$I'(a) = \int_0^{1}\frac{dx}{\sqrt{1-x^2}(1+a^2x^2)}$$
Let, $$u^2 = \frac{1-x^2}{1+a^2x^2} \implies x^2=\frac{1-u^2}{1+a^2u^2} \therefore xdx=-\frac{u(1+a^2)}{(1+a^2u^2)^2}du$$
Therefore, the integral reduce... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Functional equation: $ f : \mathbb R ^ * \to \mathbb R $ with $ \frac 1 x f ( - x ) + f \left( \frac 1 x \right) = x $ I tried to solve this problem:
Determine all function $ f : \mathbb R ^ * \to \mathbb R $ such that $ \forall x \in \mathbb R ^ * $
$$ \frac 1 x f ( - x ) + f \left( \frac 1 x \right) = x \text . $$
... | Fix $a \neq 0$. Then plugging in respectively $x = a$ and $x = -\frac{1}{a}$, we get,
\begin{align*}
\frac{1}{a}f(-a) + f\left(\frac{1}{a}\right) &= a \\
-af\left(\frac{1}{a}\right) + f(-a) &= -\frac{1}{a}.
\end{align*}
This is a system of linear equations in terms of unknowns $x = f(-a)$ and $y = f\left(\frac{1}{a}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I proceed to the results? Please help me to proceed. Thanks in advance.
| We need to prove that $$\sum_{cyc}\left(a^3-a^{\frac{7}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}\right)\geq0$$ and since $$(3,0,0)\succ\left(\frac{7}{3},\frac{1}{3},\frac{1}{3}\right),$$ it's true by Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\... | Let $ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx $
$$ a_{n}-a_{n-1}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos(2n-2)x-\cos 2nx}{\sin^2 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} dx $$
$$ a_{n}-a_{n-1}-(a_{n-1}-a_{n-2})=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x-\sin(2n-3)x}{\sin x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Any $(x, y, z)$ can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$? Please tell me whether there any $(x, y, z)$
which can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$ ?
No process or just solve it by calculator are both fine.
Thank you.
| We have
\begin{align}
&5x^2 + 2y^2 + 6z^2 - 6xy - 2xz + 2yz\\
= \ & 2y^2 + (-6x + 2z)y + 5x^2 - 2xz + 6z^2\\
= \ & 2\left(y + \frac{-3x+z}{2}\right)^2 - 2 \left(\frac{-3x+z}{2}\right)^2
+ 5x^2 - 2xz + 6z^2\\
= \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}x^2 + xz + \frac{11}{2}z^2 \\
= \ & 2\left(y + \frac{-3x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that this inequality is true Show that $\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{999998}{999999} > \frac{1}{100}$.
I tried to take another multiplication $\frac{3}{5} \cdot \frac{6}{8} \cdot \frac{9}{11} \cdot ... \cdot \frac{999996}{999998}$ so that we would have their multiplications... | Let $$A=\frac{2}{3}\cdot\frac{5}{6}\cdot...\cdot\frac{999998}{999999},$$
$$B=\frac{1}{2}\cdot\frac{4}{5}\cdot...\cdot\frac{999997}{999998}$$ and $$C=\frac{3}{4}\frac{6}{7}\cdot...\cdot\frac{999999}{1000000}.$$
Thus, since $$\left(\frac{3n+2}{3n+3}\right)^2>\frac{3n+1}{3n+2}\cdot\frac{3n+3}{3n+4},$$ we obtain: $$A^2>BC,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Polynomial $x^3-2x^2-3x-4=0$ Let $\alpha,\beta,\gamma$ be three distinct roots of the polynomial $x^3-2x^2-3x-4=0$. Then find $$\frac{\alpha^6-\beta^6}{\alpha-\beta}+\frac{\beta^6-\gamma^6}{\beta-\gamma}+\frac{\gamma^6-\alpha^6}{\gamma-\alpha}.$$
I tried to solve with Vieta's theorem. We have
$$\begin{align}
\alpha+\b... | Let $\alpha+\beta+\gamma=3u$, $\alpha\beta+\alpha\gamma+\beta\gamma=3v^2$ and $\alpha\beta\gamma=w^3$.
Thus, $$\sum_{cyc}\frac{\alpha^6-\beta^6}{\alpha-\beta}=\sum_{cyc}(2\alpha^5+\alpha^4\beta+\alpha^4\gamma+\alpha^3\beta^2+\alpha^3\gamma^2)=$$
$$=2(243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3)+$$
$$+81u^3v^2-81uv^4-9u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Implementation of Cauchy product on $\cos x\cdot \sin x$ I have a mistake that I can't find somewhere along the way.
Please help me find the place things go wrong.
Find the product of $\cos x$ and $\sin x$ as defined:
$$\cos(x) = \sum_{k=0}^{\infty} \frac{\alpha_k}{k!} x^k $$
$$\sin(x) = \sum_{k=0}^{\infty} \frac{-\alp... | Note that $\alpha_{k}=0$ if $k$ is odd and $(-1)^{\frac{k}{2}}$ otherwise. Applying Cauchy product yields
\begin{align*}
\displaystyle \sum_{n=0}^{\infty} \displaystyle \sum_{k=0}^{n} \frac{(-1)^k x^{2k}}{(2k)!}\cdot \frac{(-1)^{n-k}x^{2n-2k+1}}{(2n-2k+1)!}=&\displaystyle \sum_{n=0}^{\infty} \displaystyle \sum_{k=0}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone? Observe that $x_1=1$ and $x_2=\dfrac{1}{\sqrt{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)>\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=1$.
Thus, $x_2>x_1$. In genera... | The sequence is indeed increasing. Using what you have left off we need to prove: $x_{n+1} - x_n > 0\iff ...x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$. We prove this by induction on $n \ge 1$. Clearly $x_1 = 1 < 1+ \sqrt{\frac{1}{2}}$. Assume $x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$, we show: $x_{n+1} < 1+\dfrac{\sqrt{n+1}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Proving limits of functions using first principles Prove using first principles that $\lim_{x \to 2}$ ($\frac{x}{1+x}$) = $\frac{2}{3}$
I know that you need to use a $\delta$-$\varepsilon$ proof where you fix $\varepsilon > 0$ and find $\delta > 0$ such that $0<|x - 2|<\delta$ $\implies$ $|\frac{x}{1+x}$ - $\frac{2}{3}... | You need to "get rid of the $x$" by solving the inequality. Start fixing $\epsilon>0$ and consider
$$
\left|\frac{x}{1+x}-\frac{2}{3}\right|<\epsilon.
$$
As soon as $0<\epsilon<\frac 13$, by solving the previous inequality we get $\frac{2-3\epsilon}{1+3\epsilon}<x<\frac{2+3\epsilon}{1-3\epsilon}$. We would like to find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$ For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$
$$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$
My solution$:$
Let $x=a^2,\,y=b^2,\,... | Let $x\geq y\geq z$.
Thus, by C-S we obtain: $$\sum_{cyc}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}=\sum_{cyc}\frac{x^2-xy-xz+yz}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{xy+xz}{\sqrt{2x^2(y+z)}}=$$
$$=\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{\sqrt{2(y+z)}}{2}\geq$$
$$\geq \frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the determinant $\begin{vmatrix} y+z&z&y\\z&z+x&x\\y&x&x+y\end {vmatrix}$ Performing the operation $R_1\rightarrow R_1-R_2-R_3$
$$\begin{vmatrix} 0&-2x&-2x \\ y&z+x&x \\ z & x&x+y \end{vmatrix}$$
Pulling $-2x$ out and performing $C_2\rightarrow C_2-C_3$
$$-2x\begin{vmatrix} 0&0&1\\ y&z&x \\ z&-y&x+y \end{vmatr... | There is an error in your solution. The third row is not the same in the first step. The first step should be
$$\begin{vmatrix} 0&-2x&-2x \\ z&z+x&x \\ y & x&x+y \end{vmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Tough polynomial form problem Find all real values of $a$ for which the equation $(x^2 + a)^2 + a = x$
has four real roots.
I played around a bit with a graphing calculator and suspect that $(-\infty, -\frac{3}{4})$ are solutions, but I'm not sure if that's all.
| We have the cheeky difference of squares factorization :
$$
(x^2+a)^2 + a-x = (\color{red}{x^2+a})^2 -\color{blue}{x}^2 +x^2+a-x \\ = (\color{red}{x^2+a}-\color{blue}{x})(\color{red}{x^2+a}+\color{blue}x)+x^2+a-x = (x^2+x+a+1)(x^2-x+a)
$$
which has four real roots if and only if both the factors have two real roots eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Another series involving $\log (3)$ I will show that
$$\sum_{n = 0}^\infty \left (\frac{1}{6n + 1} + \frac{1}{6n + 3} + \frac{1}{6n + 5} - \frac{1}{2n + 1} \right ) = \frac{1}{2} \log (3).$$
My question is can this result be shown more simply then the approach given below? Perhaps using Riemann sums?
Denote the serie... | Your sum is$$\sum_{n\ge0}\int_0^1x^{6n}(1-2x^2+x^4)dx=\int_0^1\dfrac{(1-x^2)^2}{1-x^6}dx=\int_0^1\dfrac{1-x^2}{1+x^2+x^4}dx,$$where the first $=$ uses monotone convergence. Since$$1+x^2+x^4=(1+x^2)^2-x^2=\prod_\pm(1\pm x+x^2),$$you can show as an exercise that this integral is$$\left[\dfrac12\ln\dfrac{1+x+x^2}{1-x+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $\lim (f(x) + 1/f(x)) = 2 $ prove that $\lim_{x \to 0} f(x) =1 $
Let $f:(-a,a) \setminus \{ 0 \} \to (0 , \infty) $ and assume $\lim_{x
\to 0} \left( f(x) + \dfrac{1}{f(x) } \right) = 2$. Prove using the
definition of limit that $\lim_{x \to 0} f(x) = 1$
Attempt:
Let $L = \lim_{x \to 0} f(x) $. Let $\epsilon > 0$... | First, note that:
$$a + \frac{1}{a} - 2 = \frac{a^2 - 2a + 1}{a} = \frac{(a - 1)^2}{a}.$$
So, roughly speaking, given we can make $\left|f(x) + \frac{1}{f(x)} - 2\right|$ as small as we like, we should be able to make $\frac{(f(x) - 1)^2}{|f(x)|}$ as small as we like. There are two ways for this to happen: either $f(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find all positive integers which are representable uniquely as $\frac{x^2+y}{xy+1}$ with $x,y$ positive integers. $\textbf{Question:}$ Find all positive integers,which are representable uniquely as
$$\frac{x^2+y}{xy+1}\,,$$
where $x$ and $y$ are positive integers.
I think this question maybe has something to do with vi... | Suppose that $k$ be a positive integer such that there exists a pair $(x,y)\in\mathbb{Z}_{>0}^2$ for which
$$\frac{x^2+y}{xy+1}=k\,.\tag{#}$$
Then, $t=x$ is a root to the quadratic polynomial
$$q(t):=t^2-(ky)t+(y-k)\,.$$
Note that $$t=ky-x=\frac{y-k}{x}$$ is also a root of $q(t)$.
If $y-k\leq 0$, then $ky-x\leq 0$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n \to \infty} n^2 \int_{n}^{5n}\frac{x^3}{1+x^6}dx$ Question:Find the limit $\lim_{n \to \infty} n^2 \int_{n}^{5n}\frac{x^3}{1+x^6}dx$
I tried to convert it into $\frac{0}{0}$ indeterminate form,then applying
L'Hospital's rule but the expressions in numerator are not nice to integrate.I do not know other w... | Since you already received good answers for the limit itself, let me show how we could have the partial terms.
$$\frac{x^3}{x^6+1}=\frac{x^3}{(x^3-i)(x^3+i)}$$ Using partial fraction decomposition
$$\frac{x^3}{x^6+1}=\frac{x-2 i}{6 \left(x^2-i x-1\right)}+\frac{x+2 i}{6 \left(x^2+i
x-1\right)}-\frac{1}{6 (x-i)}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sine Parametric function exercise Find the biggest negative value of $a$ , for which the maximum of
$f(x) =sin(24x+\frac{πa}{100})$ is at $x_0=π$
The answer is $a=-150$, but I don't understand the solving way. I would appreciate if you'd help me please
| Recall that the general solution of $\sin\theta = 1$ is:
$$
\theta = \frac{\pi}{2} + 2\pi n, \text{where }n \in \mathbb Z
$$
So the general solution for finding all maximum values of $f(x)$ is:
\begin{align*}
24x + \frac{\pi a}{100} &= \frac{\pi}{2} + 2\pi n, \text{where }n \in \mathbb Z \\
24x &= \frac{\pi}{2} - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How may one solve problems over expressions like $(2+px)^6$ without the binomial theorem? A friend of mine posed a problem on a mathematics discord server.
The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.
I immediately thought of employing the bino... | This implicitly uses the Binomial Theorem (edit: no it does not), and calculus, but anyway...
Let
$$p(x)=(2+px)^6=a_0+a_1x+60x^2+\mathcal{O}(x^3).$$
Differentiate twice:
$$\begin{align}
p'(x)&=6(2+px)^5\cdot p=a_1+60\cdot 2x+\mathcal{O}(x^2)
\\ \Rightarrow p''(x)&=6p\cdot 5\cdot(2+px)^4\cdot p=120+\mathcal{O}(x)
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac... | You must have asked Wolfram Alpha to solve the homogeneous equation, i.e. with $0$ instead of $x$ on the right side. According to the standard method for solving first-order equations, your integrating factor is
$$ \eqalign{\mu(x) &= \exp \left(\int \frac{x\; dx}{1+x^2}\right) \cr
&= \exp\left(\frac{1}{2} \log(1+x^2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does this $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ converges ?and what about its bounds? I want to evaluate this sum $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ such that I want to know what is the value of that series if it is converge , I have started... | As you noticed, $$y_n = \sqrt{n + \sqrt{n+\sqrt{n+...}}} = \frac{1+\sqrt{1+4n}}{2}$$ by using the fact that $y_n = \sqrt{n+y_n}$.
Thus the sum simplifies to $$\sum_{n=1}^{\infty} \frac{(-1)^n}{\frac{1+\sqrt{1+4n}}{2} + (-1)^n}$$
This can be split up into even $n$ and odd $n$ as $$\sum_{n=1}^{\infty} \left(\frac{1}{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Diagonalization of a quadratic form, such as $\lambda_i=\{-1,1,0\}$ I am given a quadratic form
$$Q(x) = x_1^2 + 5 x_2^2 + 3 x_3^2 - 8 x_1 x_3 + 8 x_2 x_3$$
and what I need is to diagonalization of a quadratic form, such as $\lambda_i=\{-1,1,0\}$ using the Lagrange method (could not find the term for that in English)... | Your method seems to be less efficient than a faster method I know; I can't vouch for the accuracy of your method if you don't give a final answer.
The correct way to do this is to write
$$\mathbf v = \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix},$$
and then write the quadratic form as
$$Q(x) = \mathbf v^\intercal A \mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $\pi<\alpha<\frac{3\pi}{2}$ what is the value of $\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$? I solved part of it this way:
$$\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$$
$$= \sqrt{4\sin^4\alpha + 4\sin^2\alpha \cos... | Hint: For that range of $\alpha$, $\sin\alpha\leq 0).
Also note that for all real numbers $x$, one has that $\sqrt{x^2}=|x|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find t... | How about using calculus?
$$f(x)=\frac{x+2}{(x+1)^2}=\frac{1}{x+1}+\frac{1}{(x+1)^2} \\ f’(x) =\frac{-1}{(x+1)^2}-\frac{2}{(x+1)^3} =0 \\ \implies x=-3$$ which you can see clearly corresponds to a minimum. We see $f(-3) =-\frac 14$. Now, $\lim_{x\to\pm\infty}f(x)=0$ and also $\lim_{x\to -1} f(x) =+\infty$ which means t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Proving: $\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac1{x^2}\right)=\frac{\pi ^2}3$ without L'Hospital Evaluating
$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to ... | Using Maclaurin expansion we get:
$$
\frac{\sin\pi x}{\pi x}
= \frac{\pi x - \frac16 (\pi x)^3 + O(x^5)}{\pi x}
= 1 - \frac{\pi^2 x^2}{6} + O(x^4)
\\
\frac{\pi x}{\sin\pi x} = 1 + \frac{\pi^2 x^2}{6} + O(x^4)
\\
\left(\frac{\pi x}{\sin\pi x}\right)^2 = 1 + \frac{\pi^2 x^2}{3} + O(x^4)
\\
\frac{\pi ^2}{\sin ^2\pi x}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Modulus operation to find unknown If the $5$ digit number $538xy$ is divisible by $3,7$ and $11,$ find $x$ and $y$ .
How to solve this problem with the help of modulus operator ?
I was checking the divisibility for 11, 3:
$5-3+8-x+y = a ⋅ 11$ and $5+3+8+x+y = b⋅3$ and I am getting more unknowns ..
| This is a solution with divisibility properties only:
$$11\mid 538xy \implies 11\mid y-x+8-3+5 = y-x+10$$
so $$11\mid y-x-1\implies y-x-1=0$$
Also $$3\mid 538xy \implies 3\mid y+x+8+3+5 = y+x+16$$ so $$3\mid y+x+1 = 2(x+1)\implies x\in\{2,5,8\}$$
Now try for each pair $(x,y)\in \{(2,3),(5,6),(8,9)\}$ if $7\mid 538xy$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to make inverse to work here? I have this equation:
$$\sqrt{5 - x} = 5 - x^2$$
My current approach is - I note that if I will let: $f(x) = \sqrt{5 - x}, g(x) = 5 - x^2$ then I will have $f(g(x)) = g(f(x)) = x$ Or, in other words, $f(x) = g^{-1}(x)$ (they're inverse) which means that if they intersect, then they mus... | There are two issues in your argument:
*
*the function $g$ is not injective, hence not invertible;
*for a bijective function, we have $f(x)=x\implies f(x)=f^{-1}(x)$, but $f(x)=f^{-1}(x)\implies f(x)=x$ doesn't holds, in general.
First note that the equation $\sqrt{5 - x} = 5 - x^2$ is equivalent to
$$\left\{\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the maximum and minimum of $(1/x-1)(1/y-1)(1/z-1)$ if $x+y+z=1$ I get more confused when I try to solve this. For my first approach, I'm using normal AM-GM inequality:
\begin{equation}
(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)=\frac{(1-x)(1-y)(1-z)}{xyz} \\=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1
\end{equation... | Actually the expression $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)$$ can be made arbitrarily large with the constraint $(x+y+z)=1$
Proof: Let $N>0$ be any arbitrary large real number. Choose $\varepsilon\in(0,1)$ such that $$\frac{1}{\varepsilon}>(N+1)$$ Then let $x=\varepsilon,y=z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Power series approximation for $\ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ to calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $ Problem
Approximate $f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ and then calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $
My attempt
Let
$$f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)}) ... | Alternative approach for the second part:
$$\begin{eqnarray*} S &=& \sum_{n\geq 1}\frac{1}{n(2n+1)}=2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n\geq 1}\int_{0}^{1}\left(x^{2n-1}-x^{2n}\right)\,dx\\&=&2\int_{0}^{1}\sum_{n\geq 1}\left(x^{2n-1}-x^{2n}\right)\,dx = 2\int_{0}^{1}\frac{x-x^2}{1-x^2}\,dx =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.