Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding value of $x$ in an A.P. If $1$ , $\log_{9}(3^{x+1} + 2)$, $\log_{3}(4⋅3^{x}-1)$ are in A.P. , then $x$ equals ?
| $$2\log_{9}(3^{x+1} + 2)=1+\log_{3}(4⋅3^{x}-1)$$
We have
$$\log_{3}(3^{x+1} + 2)=\log_{3}(4⋅3^{x+1}-3)$$
therefore
$$3^{x+1} + 2=4⋅3^{x+1}-3\implies 3^{x+2}=5$$
$$x=-2+\log_{3}(5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding the generating function of a recurrence relation in dependence of a variable Given this inhomogeneous linear recurrence relation of 2nd order :
$F_n = F_{n-2} + a$ for $n \geq 2$
with $F_1 = 1$ and $F_0 = 0$
How do I find the generating function of this recurrence relation in dependence of the variable a? I t... | You can split this 2nd order recurrence relation into two 1st order recurrence relations. Define $b_0=0$ and $b_n=b_{n-1}+a$ for $n\geq 1$. In the same way define $c_0=1$ and $c_{n}=c_{n-1}+a$ for $n\geq 1$. Now we get
$F_{2n}=b_n$ and $F_{2n+1}=c_n$, the first couple of terms of the generating function look like
$$
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of substitutions in solving equations I am currently working on this problem, I am asked to solve the following
$x^2 - 4 - x\sqrt{x^3 + 3x} = 7$.
I am able to manipulate the above to obtain
$x^5 - x^4 +3x^3 + 22x^2 - 121 = 0$.
The problem asks us to use a substitution but I am unable to figure out what to substitut... | The original equation may be written $$x^2=11+x\sqrt{x^3+3x}\,.$$ For the square root to be defined means $x^3+3x\geqslant0$, which implies $x\geqslant 0$. Now, if $x\leqslant1$, then $11+x\sqrt{x^3+3x}\geqslant11>x^2$; on the other hand, if $x>1$, then $11+x\sqrt{x^3+3x}>x\sqrt{x^3}>x^2$. Either way, our equation can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for ... | Condition A: The first two rolls musn't be both 6:
$$A=1-(\frac{1}{6}\times\frac{1}{6})=\frac{35}{36}$$
Condition B: The third roll mustn't be 6:
$$B=1-\frac{1}{6}=\frac{5}{6}$$
Condition C: The last two rolls must be 6:
$$C=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$
Probability of A, B and C being true at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How many different 4 letter words can be selected from the word ADVANCED? My attempt : $A-2 , D -2 , V - 1, N -1 , C -1 , E -1 $
$XXXX$ words $=0 $
$XXXY$ words $=0 $
$XXYY$ words $= \binom{2}{2}\times \frac{4!}{2!2!} = 6$
$XXYZ$ words $= \binom{2}{1}\times \binom{5}{2} \times \frac{4!}{2!} =240$
$XYZW$ words $= \bin... | $606$ is correct.
A g.f. way is to find the coefficient of $x^4$ in $4!(1+x)^4(1+x+\frac{x^2}{2!})^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)=l$, then what is the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$? If the function
$$\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)$$ exists and has a value equal to $l$ then what will be the value of $\frac{1}{a... | You can write the limit as
$$
\lim_{x\to0}\frac{1+bx-(1+ax)\sqrt{1+x}}{x^3}\frac{1}{\sqrt{1+x}(1+bx)}
$$
Since the second factor has limit $1$, we can disregard it. Use Taylor expansion up to order $3$:
$$
\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)
$$
Then the numerator is
$$
1+bx-1-\frac{1}{2}x+\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of algebraic expression.
If $0\leq x_{i}\leq 1\;\forall i\in \left\{1,2,3,4,5,6,7,8,9,10\right\},$ and $\displaystyle \sum^{10}_{i=1} x^2_{i}=9$
Then $\max$ and $\min$ value of $\displaystyle \sum^{10}_{i=1} x_{i}$
$\bf{My\; Try::}$
Using Cauchy-Schwarz Inequality
$$\left(x^2_{1}+x^2_{2}+.......+x^2_{10... | Clearly the minimum values is $9$ (since $a_i\geq a_i^2$), we can obtain it by taking $9$ values of $a_i$ to be $1$ and the rest to be $0$.
We flip the question around:
Given non negative reals $a_1,a_2\dots a_n$ with $a_1+a_2+\dots + a_n=A$
What is the minimum possible value of $a_1^2+a_2^2+\dots + a_n^2$?
Since the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solutions of the equation $x^3-y^3=z!-18$ What are the solutions of the equation $x^3-y^3=z!-18$? Here $x,y,z$ are non-negative integers. I have tried brute force but is there a better method?
| We can write equation in this form: $(x-y)^3+3xy(x-y)=z!-18$.
For $z\leq 2$:
If $x$ and $y$ give different residues modulo 3, then $z\in \{0,1,2\}$ (otherwise LS would be 1 or 2 modulo 3 and RS would be divisible by 3).
It can be easily check that only solution is $z=2$, $x=-2$, $y=2$.
For $z\geq 3$:
From previous par... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to ${2^n}{-1}$. Please explain in quotations!
"Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to $2^n-1$." In a very simple case, for $n = 3, 1 + 2 + 4 = 7 = 8 - 1$.
| Since we don't know what $1+2+4+\cdots+2^{n-1}$ is yet, we'll call that number $S$. So we have $S=1+\cdots+2^{n-1}$. Multiplying by $2$ on both sides, we get
$$
2S=2+4+8+\cdots+2^{n-1}+2^n
$$
Here comes the trick: calculate $2S-S$:
$$
\begin{align}
2S-S=&2+4+8+\cdots+2^{n-1}+2^n\\-(1+&2+4+8+\cdots+2^{n-1})
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a tricky composition of two piecewise functions I have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:
$$f(x) =
\begin{cases}
2x+1, & \text{if $x \le 0$} \\
x^2, & \text{if $x > 0$}
\end{cases}$$
$$g(x) =
\begin{cases}
-x, & \text{if $x ... | $$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases} $$
Therefore
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases} $$
So now we need to know when $f(x) < 2$ and when $f(x) \ge 2$.
$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases} $$
Let's look at one piece of $f$ a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$
Find $$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}.$$
My attempt:
$$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{3}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{1}{\frac{n-1}{3}}\right)^{2n+3}$$
Now ... | $$\begin{align}\left(\frac{n+2}{n-1}\right)^{2n+3}&= \left(1+\frac{3}{n-1}\right)^{2n+3}\\ &= \left(1+\frac{3}{n-1}\right)^{2(n-1)+5}\\ &= \left(\left(1+\frac{3}{n-1}\right)^{n-1}\right)^2\left(1+\frac{3}{n-1}\right)^5\end{align}$$
If you like, you can replace $n-1$ with $n$ as $n\rightarrow \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$
Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$
Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?
| We may even look out for integer solutions of $x^3+y^3+z^3=8$ alone. There is a huge literature on the sum of three cubes, and several data bases available. The equation has infinitely many solutions, for example
$$
\displaystyle (18t^3 + 2)^3 + (18t^4)^3 + (-18t^4 - 6t)^3 = 8
$$
for all integers $t$.
Among other solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to factorize the polynomial $a^6+8a^3+27$?
I would like to factorize $a^6+8a^3+27$.
I got different answers but one of the answers is
$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$
Can someone tell me how to get this answer? Thanks.
| Here's a slightly sneaky way to arrive at the factorization.
Let $P(a)=a^6+8a^3+27$. Then
$$P(10)=1008027=3\cdot3\cdot31\cdot3613=93\cdot3\cdot3613=(10^2-10+3)\cdot3\cdot3613$$
so we might guess that $a^2-a+3$ is a factor of $a^6+8a^3+27$. If it is, then the other factor would have to take the form $a^4+a^3+\cdots+9$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Sum of $a_0+a_1+\cdots + a_n$ If $a_0=2 , a_1=5 $and for $n>1 $
$$a_n=5a_{n-1}-6a_{n-2}$$
then
$$a_0+a_1+\cdots+a_n=?$$
I know telescoping series but
$$a_n=5(a_{n-1}-a_{n-2})-a_{n-2}$$
Thanks for your help.
Mani
| Clearly
$$ a_n-2a_{n-1}=3(a_{n-1}-2a_{n-2}). $$
So
$$ a_n-2a_{n-1}=3^{n-1}(a_{1}-2a_{0})=3^{n-1}. \tag{1}$$
Similarly
$$ a_n-3a_{n-1}=2^{n-1}(a_{1}-2a_{0})=-2^{n-1}. \tag{2}$$
(1)(2) give
$$ a_n=3^n+2^n.$$
So
$$ a_0+a_1+a_2+\cdots+a_n=(1+3+3^2+\cdots+3^n)+(1+2+2^2+\cdots+3^n)=\frac12\cdot(3^{n+1}-1)+(2^{n+1}-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\cos^n x + \sin^n x =1 $ the solutions of this equation as a function of the value of $n$??
\begin{align}
\cos^n x + \sin^n x =1
\end{align}
I already found the solution if n is odd,
| If $\cos x\sin x\not=0$, then $|\cos x|=\sqrt{1-\sin^2x}$ and $|\sin x|=\sqrt{1-\cos^2x}$ are both strictly less than $1$, which implies $\cos^nx+\sin^nx\lt\cos^2x+\sin^2x=1$ for $n\gt2$. For $n=1$, $\cos x+\sin x=1$ implies $\cos^2x+2\cos x\sin x+\sin^2x=1$, which implies $\cos x\sin x=0$. In sum, when $n$ is a pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?
| It suffices to compute:
$$\int \frac{y}{y^3 - r^3}dy$$
(putting $r = - \sqrt[3]{\frac{b}{a} }$). Let's assume that $a, b > 0$.
After decomposing into partial fractions, i.e.:
$$\frac{y}{y^3 - r^3} = \frac1{3r} \left( \frac1{y - r} - \frac{y - r}{y^2 + ry + r^2} \right)$$
we find:
$$\int\frac{y}{y^3 -r^3} dy = \frac1{3r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the value of an expression using substitution How to find the value of $x^{4000} + \frac{1}{x^{4000}}$ if $x+\frac{1}{x}=1$.
I think binomial theorem will be useful in this.
Here's my proceedings:
$$x+\frac{1}{x}=1$$
Raising both sides to the power of 4000
$$(x+\frac{1}{x})^{4000} = 1^{4000}$$
$${4000 \choose... | Hint: Our number $x$ is a primitive sixth root of unity. Taking high powers of it is exceptionally easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| Let $$I = \int\frac{x^2}{(x\sin x+\cos x)(x\cos x-\sin x)}dx$$
Now Put $x=2y\;,$ Then $dx = 2dy$
So $$I = \int \frac{2y^2}{(2y\sin 2y+\cos 2y)(2y\cos 2y-\sin 2y)}dy$$
So $$I = \int\frac{2y^2}{(y^2-1)\sin 2y+2y \cos 2y}dy$$
$$\bullet\; (y^2-1)\sin 2y+2y\cos 2y = (y^2+1)\left[\cos 2y\cdot \frac{2y}{1+y^2}+\sin 2y\cdot \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums?
How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums?
My Try: I have Solved It using Limit as a Su... | For $0 \le k \le n$, let $t_k = \frac{k}{n}$.
Divide the interval $[0,1]$ into $n$ sub-intervals $[t_{k-1},t_k]$ for $1 \le k \le n$ and apply MVT to $\tan^{-1} x$ on the intervals. We find for each $k$, there is a $x_k \in ( t_{k-1}, t_k )$ such that
$$\tan^{-1}t_k - \tan^{-1} t_{k-1} = \frac{t_k-t_{k-1}}{1+x_k^2}$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Show $1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$, where $z=\cos x+i \sin x$ Let $z=\cos x+i \sin x$. Show that
$$1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$$
| That identity has a rather simple geometric interpretation. Consider the following diagram:
where $x$ is the red angle and
\begin{align}
z &= \cos x + i\cdot\sin x, \\
m &= \cos \frac{x}{2} + i\cdot\sin \frac{x}{2}, \\
n &= m \cdot \cos\frac{x}{2}.
\end{align}
In other words, $z$ is any point with $|z| = 1$, and $x$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: C... | By writing the integrand function as its Taylor series centered at $x=0$ and performing termwise integration we get:
$$ I=\int_{0}^{1}\frac{dx}{1+x^6}=\sum_{k\geq 0}\frac{(-1)^k}{6k+1}=\sum_{n\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+7}\right).\tag{1} $$
The last series can now be computed through the discrete Fourier ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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If $f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
If $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
I tried to replace $x$ by $x+1, x-1$ in the equality,to get something like $f(x + k)=f(x)$ but without success.
Any help is appreciated.... | From the given equation, $$f(x+4)=\sqrt{3}f(x+3)-f(x+2),\\\sqrt{3}f(x+3)=3f(x+2)-\sqrt{3}f(x+1),\\f(x+2)=\sqrt{3}f(x+1)-f(x).$$ Adding those, we obtain that $$f(x+4)=f(x+2)-f(x),$$ and also $$f(x+6)=f(x+4)-f(x+2),$$ therefore $f(x+6)=-f(x)$. Hence, $$f(x+12)=-f(x+6)=f(x),$$ therefore $f$ is periodic with period $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
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Check convergence and find the sum $\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}$
$$\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}$$
I have starting an overview about series, the book starts with geometric series and emphasizing that for each series there is a corresponding infinite sequence.
For convergence I can look at the pa... | Note that
$$ \frac{1}{9n^2 + 3n - 2} = \frac{1}{(3n + 2)(3n - 1)} = \frac{1}{3} \left( \frac{1}{3n - 1} - \frac{1}{3n + 2} \right) $$
for all natural numbers $n$.
The partial sums of the sum that you are interested in are then given by
$$ \sum_{n=1}^N \frac{1}{9n^2 + 3n - 2} = \frac{1}{3} \sum_{n=1}^N \left( \frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of $n$ terms and infinite terms of series
The sum of $n$ terms of the series $$\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....$$
And also calcula... |
This series looks like
\begin{align*}
\frac{1}{2}&+\sum_{r=2}^\infty\frac{1}{r!}\left(\frac{1}{2}\right)^r(2r-3)!!\tag{1}\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}(2r-1)!!\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{(2r)!!}\tag{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the value of the integral: $\int_{0}^{\frac{\pi}{4}} \frac{1}{3\cos^2\theta + \sin^2\theta}d\theta$
Using the substitution $t = \tan\theta$ , find the value of the integral:
$$\int_{0}^{\pi/4} \cfrac{1}{3\cos^2\theta + \sin^2\theta}d\theta$$
I isolated $\theta$ in $t = \tan\theta$, and then substituted the exp... | You could also use that
$\displaystyle\int_{0}^{\pi/4} \cfrac{1}{3\cos^2\theta + \sin^2\theta}d\theta=\int_0^{\pi/4}\frac{\frac{1}{\cos^2\theta}}{\frac{3\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}}d\theta=\int_0^{\pi/4}\frac{\sec^2\theta}{3+\tan^2\theta} d\theta=\int_0^1\frac{1}{3+t^2}dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Stirling formula, O-Notation In lecture we proof that
: $ \sqrt{2 \pi n} (\frac{n}{e})^n< n! \le \sqrt{2 \pi n} (\frac{n}{e})^n e^{\frac{1}{12n}}$
But how we came form this to the formula $n!= \sqrt{2 \pi n} (\frac{n}{e})^n (1+ O(1/n)) $ with $0< O(1/n) \le \frac{1}{12n}$
I can rewirte the second formula as that $\li... | You have proved that, as $n\to \infty$,
$$
\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n< n! \le \sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^ne^{\large \frac{1}{12n}}\tag1
$$ from wich one deduces that, as $n \to \infty$,
$$
1< \frac{n!}{\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n} \le e^{\large \frac{1}{12n}} \tag2
$$ by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Combinatorial identity $\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}$ How to prove the combinatorial identity?
$$\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}$$
I took $\ln$ of the left hand side
$$\sum_{j=0}^n \ln(j^{n+1}) - \ln((j!)^2)$$
but not going anywhere from here. any help is welcome
| We have, by product operator $\prod$ properties
$$
\prod_{j=1}^{n}{ n \choose j}
=
\prod_{j=1}^{n}{ n! \over (n-j)!j! }
=
{\prod_{j=1}^n n! \over \prod_{j=1}^n (n-j)!\cdot j!}
=
{(n!)^n \over \prod_{j=1}^n (n-j)!\prod_{j=1}^nj!}
$$
The result can be obtained by induction on n. The proposition $P(n):\prod_{j=1}^{n}\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the reciprocal of the following infinite product I hae to evaluate the reciprocal of the following product to infinity
$$\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19}{2 \cdot 2 \cdot 6 \cdot 6 \cdot 10 \cdot 10 \cdot 14 \cdot 14 \cdot 18 \cdot 18}\cdot\ldots $$
I am gues... | Since $\frac{1\cdot 3}{2\cdot 2}=1-\frac{1}{2^2}$, $\frac{5\cdot 7}{6\cdot 6}=1-\frac{1}{6^2}$ and so on, we want to compute:
$$ P = \prod_{n\geq 1}\left(1-\frac{1}{(4n-2)^2}\right)^{-1} $$
while the Weierstrass product for the sine function $\frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)$ give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$ Solve differential equation
$$2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$$
I know that we have to use some smart substitution here, so that the equation becomes linear.
The only thing I came up with is a smart guessed particular solution: $y = x^2$. If we plug this function in, w... | $$
\begin{cases}
3y''y + 3(y')^2 - 2x^2 = 0, \\
y(0) = 1, \\
y'(0) = 0.
\end{cases}
$$
$$y''y+y'^2=(y'y)'\quad\to\quad 3(y'y)'=2x^2$$
$$3y'y=\frac{2}{3}x^3+c_1$$
$y'(0)=0\quad\to\quad c_1=0$
$$y'y=\frac{2}{9}x^3$$
$$2y'y=(y^2)'=\frac{4}{9}x^3$$
$$y^2=\frac{1}{9}x^4+c_2$$
$y(0)=1\quad\to\quad c_2=1$
$$y=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Derivation of Variance of Discrete Uniform Distribution over custom interval I'm trying to prove that the variance of a discrete uniform distribution is equal to $\cfrac{(b-a+1)^2-1}{12}$. I've looked at other proofs, and it makes sense to me that in the case where the distribution starts at 1 and goes to n, the varian... | So much time has passed, but at least now we know it can be done: https://en.wikibooks.org/wiki/Statistics/Distributions/Discrete_Uniform
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help solving $1 < \frac{x + 3}{x - 2} < 2$ I worked a lot of inequalities here in MSE and that greatly helped me.
I've seen a similar inequality [here] [1], but the one I have today is significantly different, in that I'll end up with a division by 0, which is not possible.
[1] [Simple inequality
$$1 < \frac{x + 3}{x -... | $$\frac { 1 }{ x-2 } -\frac { 1 }{ 5 } <0\\ \frac { 5-x+2 }{ 5\left( x-2 \right) } <0\\ \frac { 7-x }{ \left( x-2 \right) } <0\\ \frac { \left( 7-x \right) \left( x-2 \right) }{ { \left( x-2 \right) }^{ 2 } } <0\\ \left( 7-x \right) \left( x-2 \right) <0\\ x\in \left( -\infty ;2 \right) \cup \left( 7;+\infty \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find The Last Two Digits Of $9^{8^7}$ Find the last two digits of $9^{8^7}$.
I tried finding a secure pattern for the last two digits of powers of $9$ but that didn't work. Any answers?
| Alternatively
By FLT
$3^{\phi(100) = 40}= 9^{20} \equiv 1 \mod 100$
$9^{8^7} = 9^{2^{21}} \equiv 9^{2^{21} \mod 20} \mod 100$
$2^{\phi(5)=4} \equiv 1 \mod 5$ so $2^{21=4*5 + 1} \equiv 2 \mod 5$
$2^{21} \equiv 0 \mod 4$ os $2^{21} \equiv 2 + 5k \mod 20$ where $4|2 + 5k$.
i.e. $2^{21} \equiv 12 \mod 20$.
so $9^{2^{21}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_0^1 x(1+x^2)^n dx = \dfrac{1}{2n+2}$ $\int_0^1 x(1+x^2)^n dx = \dfrac{1}{2n+2}$
How is this calculated?
This is an interesting case because as a consequence of this, $\int_0^1 n^2x(1+x^2)^n dx$ converges to infinity with n to infinity, but the integrant converges point wise to 0.
(see Rudin "Real and c... | We use substitution rule with $1 - x^2 = t$. We have
\begin{align}
&dt = -2x dx \nonumber \\
&x dx = - \frac{1}{2} dt \nonumber
\end{align}
\begin{align}
&x = 0, \quad t = 1 - 0 = 1 \nonumber \\
&x = 1, \quad t = 0 \nonumber
\end{align}
Pulling it all into one single expression.
$$
\int_{0=x}^{1=x} x(1-x^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Squaring Infinite Series Expansion Of e^x $Fact$:$$\lim\limits_{n \to \infty}\frac{x^0}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!}=e^x$$
so
$$\lim\limits_{n \to \infty}\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}=e$$
also
$$\lim\limits_{n \to \infty}e^2=\frac{2^0}{0!}+... | Let $$f(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\tag{1}$$ then it is easy to see that the series on right is absolutely convergent for all values of $z \in \mathbb{C}$ and hence the function $f(z)$ is well defined for all complex values of $z$. Using Cauchy's product rule for multiplication of infinite series we see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find large power of a non-diagonalisable matrix
If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$.
The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresp... | Generalized eigenvector $\;u_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\;$for $\;\lambda=1\;$ :
$$Au_1=1\cdot u_1+v_1\iff \begin{pmatrix}a\\a+c\\b\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}+\begin{pmatrix}0\\1\\1\end{pmatrix}\implies \begin{cases}b=c+1\\{}\\a+c=b+1\end{cases}\implies \begin{cases}a=2\\{}\\b=1\\{}\\c=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares:
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$
Let's consider $a,b>0$ for now.
$$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{... | It was already shown that
$$
I_1(p)=\int_0^\infty \frac{dx}{\sqrt{(x^3+1)(x^3+p)}}=\frac{2 \pi}{3 \sqrt{3}} {_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right).
$$
By transformation 2.11(5) from Erdelyi, Higher transcendental functions (put $z=\frac{1-\sqrt{p}}{1+\sqrt{p}}$)
$$
{_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| You can solve this without much thinking.
By expanding the product, we find $m(m+1)(m+2)(m+3) = m^4 + 6m^3 + 11m^2 + 6m$. That's a bit bigger than $(m^2)^2 = m^4$. $(m^2+c)^2 = (m^4 + 2m^2c + c^2)$ is still too small because there is no term $m^3$. $(m^2+cm)^2 = m^4 + 2cm^3+c^2m^2$ looks better, especially if we let c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 3
} |
Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $... | Hint:
Let $\dfrac{\sin A}{\sqrt3}=\dfrac{\sin B}2=p$ and $\dfrac{\cos A}{\sqrt5}=\dfrac{\cos B}3=q$
$\implies(\sqrt3p)^2+(\sqrt5q)^2=1$
and $(2p)^2+(3q)^2=1$
Solve for $p^2,q^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$ Let $a_1, a_2,....,a_n, b_1, b_2,...,b_n$, let $\frac{b_1}{a_1} = max \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ ,
$\frac{b_n}{a_n} = min \{\frac{b_i}{a_i}, i=1,2, \cdots n ... | Note that
\begin{align*}
&\ \sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2 - \left(\sum_{i=1}^na_ib_i \right)^2-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2\left(\sum_{i=1}^na_i^2 \right)^2\\
=&\ \sum_{i,j=1}^n\left[a_i^2b_j^2- a_ib_ia_jb_j-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} ( \text{agm} (\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} ))^{-1}$ The following definite integral turns out to be expressible as the Arithmetic-Geometric Mean: $$I_4(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2... | Since $K(k)=\frac{\pi}{2} ~{}_2F_1(1/2,1/2;1;k^2)$ and it was shown in the question that
$$
I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right),
$$
the first question is equivalent to proving the equality
$$
\frac{\sqrt{2} }{b \sqrt{b^2+1}}{_2F_1\left(\frac{1}{2},\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational? If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational ?
Is it a "If and only if " statement or "only if " statement ?
For $a,b,c \in \mathbb Q$ , I think it is a "if and only if" statement . Am I correct ?
I can prove that if $a+b+c=0$ and $a,b,c \in \mathb... | If $a+b+c= 0 \implies x = 1$ is a root and is a rational number, and the other root is $x = \dfrac{c}{a}$ also a rational number since $a, c \in \mathbb{Q}$. To see a counter example for the other part, take $a = c = 3, b = 10$, then $a+b+c = 16 \neq 0$, yet the equation $3x^2+10x+3 = 0$ has rational roots $x = -3, -\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rationalizing denominator with cube roots Rationalize the denominator of $$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}$$ and simplify.
I already have an answer. I just want to compare answers with others. Maybe someone has different solutions? Also, I really disagree with the answer found at the back of the questio... | Use $x+x^2+x^3=x\cdot(x^3-1)/(x-1)$ with $x=\sqrt[3]{4}$.
Hence
$$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}=\frac{6(\sqrt[3]{4}-1)}{\sqrt[3]{4}(4-1)}=2-2^{1-2/3}=2-\sqrt[3]{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-... | For #3 use the technique of #2: For $n\in \mathbb N \cup \{0\}$ we have $$a^{2n+3}+b^{2n+3}=a^2(a^{2n+1}+b^{2n+1})+(-a^2+b^2)b^{2n+1}=$$ $$=a^2(a^{2n+1}+b^{2n+1})+(a+b)(b-a)b^{2n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A proof of the identity $ \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)} $. I have to prove that
$$
\forall n \in \mathbb{N}_{0}, ~ \forall x \in \mathbb{R} \setminus \mathbb{N}_{0}:
\qquad
\sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k}
= \frac{n!}{(x + 0) (x + ... | An induction on $n$ will work. Let
$$f(x,n)=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\;,$$
and for the induction step suppose that
$$f(x,n)=\frac{n!}{x(x+1)\ldots(x+n)}\;.$$
Then
$$\begin{align*}
\frac{(n+1)!}{x(x+1)\ldots(x+n+1)}&=\frac{n+1}{x+n+1}f(x,n)\\
&=\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
The integral $\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy$ What is $$\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy?$$ I split it as $\frac{y^{2}}{(y^2+1)^{0.5}} + \sqrt{y^2+1}.$ Now I substituted $y^{2}=u $ thus $2y\,dy=du$ so we get $0.5 \sqrt{\frac{u}{u + 1}} + 0.5 \sqrt{\frac{1 + u}{u}}$ but now what to do? Another idea was doi... | Let $\displaystyle y=\frac{1}{2}\left(t-\frac{1}{t}\right),\;dy=\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$,
so that $\displaystyle\sqrt{y^2+1}=\frac{1}{2}\left(t+\frac{1}{t}\right),\;\;t=y+\sqrt{y^2+1},\;\;\frac{1}{t}=\sqrt{y^2+1}-y.$
Then $\displaystyle\int \frac{2 (2y^2+1)}{(y^2+1)^{0.5}} dy=2\int\frac{\frac{1}{2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to factorise this expression $ x^2-y^2-x+y$ This part can be factorised as $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?
:)
| This is a far less elegant method than the other answers but I'll keep it up for posterity: set it up as a quadratic in $x$.
$$x^2-x+(y-y^2)=0$$
$$x=\frac{1\pm\sqrt{1+4(y^2-y)}}{2}$$
$$\left(x-\frac{1+\sqrt{1+4(y^2-y)}}{2}\right)\left(x-\frac{1-\sqrt{1+4(y^2-y)}}{2}\right)$$
$$\frac{1}{4}\left(2x-1+\sqrt{(2y-1)^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Solution of an Inequality Show that $(ab + bc + ca)(ab^{-1}+ bc^{-1} + ca^{-1}) \geq (a + b + c)^{2}.$
We, at first, change the given expression $(ab + bc + ca)(ab^{-1} + bc^{-1} +ca^{-1})$ into $(ab^2 + bc^2 + ca^2 )(1/a + 1/b + 1/c)$ . Now we will show that this term $\geq (a + b+ c)^{2}.$ Again by Chebyshev's inequa... | It follows from the Cauchy-Schwarz inequality:
$$(ab + bc + ca)(ab^{-1} + bc^{-1} + ca^{-1}) \ge \left(\sqrt{ab}\cdot \sqrt{ab^{-1}} + \sqrt{bc}\cdot \sqrt{bc^{-1}} + \sqrt{ca}\cdot \sqrt{ca^{-1}}\right)^2 = \cdots$$
Now finish the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help m... | For the LHS, use $\cos(120^\circ +A)=\cos 120^\circ \cos A - \sin 120^\circ \sin A$
and $\cos(240^\circ+A)=\cos 240^\circ \cos A - \sin 240^\circ \sin A$
Then $\cos^3(120^\circ+A)=\left(\cos 120^\circ \cos A - \sin 120^\circ \sin A \right)^3$
and $\cos^3(240^\circ+A)=\left( \cos 240^\circ \cos A - \sin 240^\circ \sin A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 3
} |
Number of $75\leq n\leq 300$ not divisible by any of $9,12,15$ - check work I need to find the number of $75\leq n\leq 300$ not divisible by any of $9,12,15$.
I thought about to solve this for $1\leq n\leq 300$ and $1\leq n\leq 74$ and then subtract. For each of them I thought to use inclusion-exclusion. I just want to... | Let $f(n)$ denote the amount of integers in the range $[1,n]$ not divisible by $9$ nor $12$ nor $15$:
$f(n)=n-\left\lfloor\frac{n}{9}\right\rfloor-\left\lfloor\frac{n}{12}\right\rfloor-\left\lfloor\frac{n}{15}\right\rfloor+\left\lfloor\frac{n}{lcm(9,12)}\right\rfloor+\left\lfloor\frac{n}{lcm(9,15)}\right\rfloor+\left\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words.
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$
$\begin{vmatrix}
n_{11} + 1 & n_{12} ... | Since $\texttt{rank}(A) = 1$ $A = u v^T$ for some $n \times 1$ matrices $u$ and $v$ and consequently $\texttt{trace}(A) = v^Tu.$
We have $$\begin{pmatrix} I & 0 \\ v^T & 1 \end{pmatrix} \begin{pmatrix}I & u \\ -v^T & 1 \end{pmatrix} = \begin{pmatrix} I & u \\ 0 & 1 + v^Tu \end{pmatrix}. \tag{1}$$
And we also have $$\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Convergence/divergence of $\sum_{k=1}^\infty\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}$ A problem asks me to determine if the series
$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$
converges or div... | By Stirling approximation, we have that
$$a_k=\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}=\frac{4^k\cdot (k!)^2}{(2k)!}=\frac{4^k}{\binom{2k}{k}}\sim \sqrt{\pi k}.$$
Hence the series $\sum_{k\geq 1} a_k$ diverges.
More generally the series $\sum_{k\geq 1} \frac{a_k}{k^r}$ c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{... | The triangle condition is equivalent to $0<a<1/2$, $0<b<1/2$ and $0<c<1/2$. Assume $c=\min(a, b, c)$, then
$$\sqrt{a^2+b^2} + \sqrt{b^2+c^2} + \sqrt{a^2+c^2} \le \sqrt{a^2+b^2} + \sqrt{b^2+bc} + \sqrt{a^2+ac} < \sqrt{a^2 + b^2} + (b+c/2) + (a+c/2) = \sqrt{a^2+b^2} + 1 < \sqrt{(1/2)^2+(1/2)^2} + 1 = 1+\frac{\sqrt{2}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$? Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$?
I tried graphing but it didn't help. Could anyone answer? Thanks!
| \begin{align}
f(a)
&= \dfrac{a+13}{3a+7} \\
&= \dfrac 13 \dfrac{3a+39}{3a+7} \\
&= \dfrac 13 \left(1 + \dfrac{32}{3a+7} \right) \\
\end{align}
This implies that $f(a)$ is strictly decreasing for positive integers.
Hence the maximum value must be $f(1) = \dfrac 75$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle b_{... | For the second recurrence
$$
b_{\,n + 1} = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{b_{\,n} }}
$$
we have
$$
\left\{ \begin{gathered}
b_{\,0} = 1/2 \hfill \\
b_{\,1} = \frac{2}
{{b_{\,0} }} = 4 \hfill \\
b_{\,n + 1} = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do t... | Let $f(x)=\log\left(\frac{x}{\sqrt{x^2+1}}\right)$ and
$$ g(t)=f(\sinh t) = \log\tanh t = \log\sinh t-\log\cosh t. $$
We have
$$ \frac{d}{dt}g(t) = \frac{\cosh t}{\sinh t}-\frac{\sinh t}{\cosh t} = \frac{1}{\sinh(t)\cosh(t)}$$
but the LHS also equals $\cosh(t)\,f'(\sinh t)$, hence
$$ f'(\sinh t) = \frac{1}{\sinh(t)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Solving system of $9$ linear equations in $9$ variables I have a system of $9$ linear equations in $9$ variables:
\begin{array}{rl}
-c_{1}x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} - c_{2}x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} - c_{3}... | obviously $x_i = 0$ is a solution. If you want a non-trivial solution
Subtract any two lines and you get $(c_i+1)x_i = (c_j+1)x_j = 0$
If any $c_j = -1$ we have all $(c_i + 1)x_i = 0$ so if $c_i \ne -1$ then $x_i = 0$ and if $c_i = -1$ then $x_i$ can be anything at all.
If none of the $c_i = -1$ then let $(c_i + 1)x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
As $\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)}$, is this is a function of only $x$ or both $x$ and $n$? $$\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)}$$
Now, the LHS seems to be a function of only $x$, whereas the RHS seems to be a func... | This is equality ,not a function
$$\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$\frac1{(1-x)^2} = \frac{d}{dx}\{x+x^2+x^3 +\cdots x^{n}\} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$x+x^2+x^3 +\cdots x^{n}=\frac{x(1-x^n)}{1-x}$$
$$\frac1{(1-x)^2} = \frac{d}{dx}\left\{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$ \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$ $$I=\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$$
I can't think of a substitution to solve this problem, by parts won't work here. Can anyone tell how should I solve this problem?
| Another way you might be interested .
Notice that $$\frac{d}{dx}\left(\frac{1}{x+\sqrt x+1}\right)=\frac{-(1+\frac{1}{2\sqrt x})}{(x+\sqrt x+1)^2}$$
Put then $$\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx=\frac{A+B\sqrt x+Cx}{x+\sqrt x+1}$$ and take the derivative in both sides. You get at once $C=0$ and the sy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find the generating function for $c_r = \sum^r_{i=1}i^2$ Find the generating function of
where $c_0 = 0, c_r = \sum^r_{i=1}i^2$.
Hence show that
$\sum^r_{i=1}i^2 = C^{r+1}_3 + C^{r+2}_3$
Attempt:
$c_r = \sum^r_{i=1}i^2$
= $x + 4x^2 + 9x^3 + ... + r^2x^r$
= $x(1 + 4x + 9x^2 + ... + r^2x^{r-1})$
= $x(\frac{1}{1-2x})$
How... | We are looking for a generating function
\begin{align*}
\sum_{r=0}^\infty c_r x^r=\sum_{r=0}^\infty \left(\sum_{i=1}^r i^2\right) x^r
\end{align*}
Note that Cauchy multiplication of a series with $\frac{1}{1-x}$ transforms the coefficient $a_r$ of a series to $\sum_{i=0}^r a_r$.
\begin{align*}
\frac{1}{1-x}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Rational solutions to a quadratic-type equation. For any fixed rational $t$, are there any nonzero rational solutions to the equation
$$y^{2} = (1-t^2)x^2 + \frac{1}{4} \text{ ?} $$
My attempt: consider the line $L$ whose gradient is $m$ and passes through the point $(0, \frac{1}{2})$. Then equation of $L$ is $y=mx + 1... | To simplify, rewrite the original equation as $(2y)^2+4(t^2-1)x^2=1$ and substitute $2y$ with $2mx+1$:
\begin{align}&(2mx+1)^2+4(t^2-1)=4(m^2+t^2-1)x^2+4mx+1=1\\
\iff & 4x\bigl((m^2+t^2-1)x+m\bigr)=0,
\end{align}
so the non-zero solution is
$$x=\frac{m}{1-m^2-t^2},\qquad y=\frac{m^2}{1-m^2-t^2}+\frac12=\frac{1+m^2-t^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solving equation in three variables please help me understand how the following equation with 3 variables and power of 2 is solved and what solution approach is the quickest.
$$3y^2 - 3 = 0$$
$$4x - 3z^2 = 0$$
$$-6xz+ 6z = 0 $$
| $$
\begin{cases}
3y^2-3=0\\
4x-3z^2=0\\
6z-6xz=0
\end{cases}\Longleftrightarrow
\begin{cases}
3y^2-3=0\\
x=\frac{3z^2}{4}\\
6z(1-x)=0
\end{cases}\Longleftrightarrow
\begin{cases}
3y^2-3=0\\
x=\frac{3z^2}{4}\\
6z\left(1-\frac{3z^2}{4}\right)=0
\end{cases}
$$
So, when you solve:
$$6z\left(1-\frac{3z^2}{4}\right)=0$$
We g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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For what value c is $f(x) = \frac{x^2+5x-14}{x^3-x^2-x-2}$ at 2 continuous? Im pretty new to continuinity and it seems for me the function can as easily be repaired with $c = 2$ so that $g(x) = \begin{cases}
\frac{x^2+5x-14}{x^3-x^2-x-2}, & x > 2\\
0, & x = 2 \\ \frac{x^2+5x-14}{x^3-x^2-x-2}, & x < 2
\end{cases} $
, be... | Instead of plugging $2$ directly into the expression, first factor the numerator and denominator as
$$f(x) = \frac{x^2 + 5x - 14}{x^3 - x^2 - x - 2} = \frac{(x + 7)(x - 2)}{(x - 2)(x^2 + x + 1)} = \frac{x + 7}{x^2 + x + 1}.$$
Then substituting $x = 2$ tells you that
$$\lim_{x \to 2}f(x) = \frac{2 + 7}{2^2 + 2 + 1} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$
I don't know how to begin to solve this problem
| Also we can use the following reasoning.
Since $\prod\limits_{cyc}(a+b)-(a+b+c+d)(abc+abd+acd+bcd)=(ac-bd)^2\geq0$,
it remains to prove that
$$(a+b+c+d)(abc+abd+acd+bcd)\geq\prod\limits_{cyc}(a+b+c-d)$$
which is also true by BW.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer
Given a sequence $(a_n)$, with $a_1 = 4$ and $a_{n+1} = a_n^2-2$ for all $n \in \mathbb{N}$, prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer.
We must ... | We see that $3a_n^2(a_n^2-4)$ must be a perfect square divisible by $4$. We notice that $a_n \equiv 2 \pmod{4}$ and so $a_n^2$ is a multiple of $4$, similarly with $a_n-2$ and $a_n+2$. Thus we just need $3a_n^2(a_n^2-4)$ to be a perfect square. To prove this, we prove the following: $$a_n^2(a_n^2-4) = 12(a_1 \cdots a_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solutions of $\sin2x-\sin x>0$ with $x\in[0,2\pi]$ What are the solutions of this equation with $x\in[0,2\pi]$?
$$\sin2x-\sin x>0$$
I took this to
$$(\sin x)(2\cos x-1)>0$$
Now I need both terms to be the same sign. Can you please help me solve this?
| Using Prosthaphaeresis Formula,
$$\sin2x-\sin x=2\sin\dfrac x2\cos\dfrac{3x}2$$
Check for $x=0,2\pi$
Else for $0<x<2\pi,\sin\dfrac x2>0$
So we need $$\cos\dfrac{3x}2>0$$ which is possible
if $2m\pi\le\dfrac{3x}2<2m\pi+\dfrac\pi2\iff\dfrac{4m\pi}3\le x<\dfrac{(4m+1)\pi}3$
or if $2n\pi+\dfrac{3\pi}2<\dfrac{3x}2\le2n\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to set up equation to find the unknown values if a limit exist.
For what values of the constants $a$ and $b$ does the following limit exist?
$$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$
for this question, $$f(x) = \frac{-(x+3)(\sqrt{ax+b}-2)}x,x<-3$$
$$f(x) = \frac{(x+3)(\sqrt{ax+b}-2)}x,x>=-3$$
Firstly, I foun... | Note that we can write
$$\begin{align}
\frac{|x+3|(\sqrt{ax+b}-2)}{x}&=\left(\frac{|x+3|(\sqrt{ax+b}-2)}{x}\right)\left(\frac{(\sqrt{ax+b}+2)}{(\sqrt{ax+b}+2)}\right)\\\\
&=\frac{|x+3|(ax+b-4)}{x(\sqrt{ax+b}+2)}\\\\
&=\left(\frac{|x+3|}{(\sqrt{ax+b}+2)}\right)\left(a+\frac{b-4}{x}\right)\tag 1
\end{align}$$
The first p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows:
Cecil has a 15-coin collection. Four coins are quarters, seven... | You computed the number of ordered selections in which Cecil selects no quarters. However, we are interested in the number of subsets, so the selections are not ordered.
Cecil selects no quarters.
The number of ways Cecil can select five coins from the eleven coins that are not quarters is
$$\binom{11}{5} = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding the other factors of the polynomial and $K$ If $(2x-1)$ is a factor of the polynomial $$p(x)=2x^3-5x^2-kx+3,$$ find $k$ and the other two factors of $p(x)$.
| by using the long division
$$\frac{2x^3-5x^2-kx+3}{2x-1}=x^2-2x-\frac{k+2}{2}+\frac{3-\frac{k+2}{2}}{2x-1}$$
the remainder or last term must be zero
or
$$3-\frac{k+2}{2}=0$$
hence $$k=4$$
the other factors
$$x^2-2x-3=(x-3)(x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$ Let $a,b,c$ and $n$ be natural numbers and $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=1$.
Does it possible to find all tuples $(a,b,c,n)$ such that:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$
| A linear diophantine equation
$$ax+by=c$$
where $c$ is divisible by $d=gcd(a,b)$, has solution
$$a=a_{0}+\frac{b}{d}\cdot k$$
$$b=b_{0}-\frac{a}{d}\cdot k$$
However, your equation is not linear, since
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=n$$
$$\frac{\left(a+b\right)ab+\left(b+c\right)bc+\left(a+c\right)ac}{abc}=... | {
"language": "en",
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"answer_count": 3,
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A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\t... | Create two zeroes in the first row ($C_2 \to C_2-\color{red}{4}C_1$ and $C_3 \to C_3-\color{blue}{9}C_1$) and expand:
$$\begin{vmatrix}
1 & \color{red}{4} & \color{blue}{9} \\
4 & 9 & 16 \\
9 & 16 & 25 \\
\end{vmatrix}=
\begin{vmatrix}
1 & 0 & 0 \\
4 & 9-16 & 16-36 \\
9 & 16-36 & 25-81 \\
\end{vmatrix} = \begin{vmatr... | {
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"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$, prove that:...... If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$, prove that: $\cos 3A+\cos 3B+ \cos 3C=3\cos(A+B+C)$.
My Attempt;
Here,
$$e^{iA}=\cos A+i\sin A$$
$$e^{iB}=\cos B+i\sin B$$
$$e^{iC}=\cos C+i\sin C$$
Then,
$$e^{iA}+e^{iB}+e^{iC}=0$$
Now, what should I d... | let $a = e^{iA}, b = e^{iB}, c = e^{iC}$. Also note that $a+b+c = 0$
$$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca) = 0$$
$$a^3 + b^3 + c^3 = 3abc$$
$$\mathrm{cis}\ 3A + \mathrm{cis}\ 3B + \mathrm{cis}\ 3C = 3\ \mathrm{cis}\ (a+b+c)$$
equating the real parts will give you the required expression:
$$\... | {
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Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.
Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$
I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
M... | When doing a proof by induction, I suggest you say so up front.
Label the base case and the inductive hypothesis. Here is your template.
Prove:
$4^3 + 4^4 + 4^5 \cdots 4^n = 4(4^n - 16)/3$
Proof by induction.
Base case:
$4^3 = 4(4^3-16)/6$
Inductive hypothesis:
Suppose, $4^3 + 4^4 + 4^5 \cdots 4^n = 4(4^n - 16)/3$
We ... | {
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Any other cases similar to $\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac 12 (3+\sqrt[4]{5}+\sqrt{5}+\sqrt[4]{125})$ Ramanujan gave many curious identities, one of which was $$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}=\frac 12(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125})\tag1$$... | The first identity of (1) can not be generalized, namely $5$ is the only integer satisfying (1). To see this, let
$$ \sqrt[4]{\frac{a+b\sqrt[4]{p}}{a-b\sqrt[4]{p}}}=\frac{\sqrt[4]{p}+q}{\sqrt[4]{p}-q} \tag{3}$$
where $p,q$ are integers such that $\sqrt[4]{p},\sqrt{p}$ are not integers and $a,b,$ are rationals. Then (3)... | {
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"answer_count": 1,
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$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $ I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.
| Assuming that you know about the polygamma functions, first write
$$(36n^2-1)^2=(6n+1)^2(6n-1)^2$$ and perform a first partial fraction decomposition
$$\frac 2{(36n^2-1)^2}=\frac{1}{2 (6 n+1)}-\frac{1}{2 (6 n-1)}+\frac{1}{2 (6 n+1)^2}+\frac{1}{2 (6 n-1)^2}$$
Computing the partial sum
$$\Sigma_p=\sum_{n=1}^p\frac 2{(36n... | {
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"answer_count": 2,
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trouble with double/iterated integration of $\int^1_0[\int^1_0v(u+v^2)^4du]dv$ I have:
For $\int^1_0v(u+v^2)^4du$:
u substitution (using x instead since there's a u in there already) with $x=(u+v^2), dx/du=1$
$v\int^1_0x^4=v\frac{1}{5}x^5=v\frac{(u+v^2)^5}{5}-\frac{v(0+v^2)^5}{5}|^1_0=\frac{(v+v^3)^5}{5}-\frac{v^{15}}{... | Note that $v(1+v^2)^5\ne (v+v^3)^5$. Rather, we have upon integrating with respect to $u$,
$$ \int_0^1 v(u+v^2)^4\,du=\frac v5 \left( (1+v^2)^5-v^{10} \right)$$
Now, note that the integration over $v$ is facilitated by the substitution $w=1+v^2$. This is left as an exercise for the reader.
| {
"language": "en",
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"question_score": "1",
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Solve the system of inequalities Solve the system:
$$|3x+2|\geq4|x-1|$$
$$\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$$
So for $|3x+2|\geq4|x-1|$ I got the solution $x=6$ and $x=2/7$ and Wolframalpha agrees with me. I'm having troubles writing the final solution for $\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$.
$1)$ $x^{2}+x-2\leq0... | Consider $$f(x)=\frac{x^{2}+x-2}{2+3x-2x^{2}}=\frac{(x-1)(x+2)}{-(2x+1)(x-2)}$$ It follows $$f(x)=\frac{-(x-1)(x+2)}{(2x+1)(x-2)}\le0\iff\frac{(x-1)(x+2)}{(2x+1)(x-2)}\ge0$$
One has at once the solution set for $f(x)\le 0$ is $$S_1=(-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)$$
Now consider $$g(x)=\left|\frac{3x+2}{x-1}... | {
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Show $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$
Positive real numbers $a,b,c$ satisfy $abc=1$. Prove
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
I tried AM–GM, Cauchy–Schwarz and Jensen's but they all failed.
| Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.
Hence, by C-S and Rearrangement we obtain $$\sum\limits_{cyc}\frac{1}{a^2+a+1}=\sum\limits_{cyc}\frac{y^2}{x^2+xy+y^2}=\sum\limits_{cyc}\frac{y^2(y+z)^2}{(x^2+xy+y^2)(y+z)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)... | {
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"source": "stackexchange",
"question_score": "2",
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value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$
$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-... | $$(5^x)^2+(a+2)5^x-(a+3)=(5^x+a+3)(5^x-1)<0$$
If $5^x-1>0,1<5^x<-(a+3)$ which is possible if $1<-(a+3)\iff a<-4$
If $5^x-1<0,-(a+3)<5^x<1$ which is possible if $1>-(a+3)\iff a>-4$
So, we need $a\ne4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Taylor Series of trigonometric function I searched quite a bit online but could only find the MACLAURIN SERIES of $\sin x$ and $\cos x$:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots$$
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+\cdots$$
Can anyone explain how we can expre... | Observe
\begin{align}
\sin x =&\ \sin(x-a+a) = \sin(x-a) \cos a +\cos(x-a) \sin a\\
=&\ \cos a \sum^\infty_{n=0} (-1)^n\frac{(x-a)^{2n+1}}{(2n+1)!} + \sin a\sum^\infty_{n=0}(-1)^n\frac{(x-a)^{2n}}{(2n)!}.
\end{align}
I will leave it as an exercise for the reader to figure out the Taylor expansion for $\cos x$ at $x=a$.... | {
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Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| $P(x)=1+(ax^2+bx+c)(x-1)^3=1-(c+bx+ax^2)(1-3x+3x^2-x^3)$
$=1-c+x(3c-b)+x^2(-a+3b-3c)+x^3(\cdots)$
We need $1-c=3c-b=-a+3b-3c=0$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $
Prove
$$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My... | Let $S_n$ be the sum $\sum_{k=1}^n\frac{1}{\sin(2^kx)}$.
we want prove by induction that
$\forall n\geq1 \;\; S_n=cot(x)-cot(2^nx)$.
the formula is true for $n=1$.
let $n\geq1 : S_n true$.
we must prove that
$S_{n+1}-S_n=cot(2^nx)-cot(2^{n+1})$.
$cot(2^nx)-cot(2^{n+1}x)=$
$\frac{\cos(2^nx)}{\sin(2^nx)}-\frac{\cos(2^{n+... | {
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Solve for $x$ where $0\leq x\leq 360$ Solve
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot \sin 2x \cdot \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x \cdot \cos x ). (4 \sin x \cdot \cos x \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I pr... | Using $\sin3B=3\sin B-4\sin^3B$
$$32\sin^3x\cdot\cos^2x\cdot\cos2x=\sin x(3-4\sin^2x)$$
If $\sin x=0,x=180^\circ n$ where $n$ is any integer
Else using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$32\cdot\dfrac{1-\cos2x}2\cdot\dfrac{1+\cos2x}2\cdot\cos2x=3-2(1-\cos2x)$$
$$\iff8(c-c^3)=1+2c\iff4c^3-3c=-\dfrac12$$
$$\implies\cos6x... | {
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"answer_count": 4,
"answer_id": 3
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real values of $a$ for which the range of function $ f(x) = \frac{x-1}{1-x^2-a}$ does not contain value from $\left[-1,1\right]$
All real values of $a$ for which the range of function $\displaystyle f(x) = \frac{x-1}{1-x^2-a}$ does not
contain any value from $\left[-1,1\right]$
$\bf{My\; Try::}$ Let $\displaystyle y ... | We want to find $a$ such that $|f(x)|=\bigg|\frac{x-1}{1-x^2-a}\bigg|\gt 1$ for every $x$ in the domain.
*
*If $a\not=0$, then $f(1)=\frac{0}{-a}=0$.
*If $a=0$, then $f(-2)=\frac{-2-1}{1-4-0}=1$.
Therefore, there is no such $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How many #8 digits can you build?
How many different 8-digit numbers can be formed using two 1s, two 2s, two 3s, and two 4s such that no two
adjacent digits are the same?
So we have $0 -> 9 = 10$ options for the digits but with the constraints we have something different.
Cases:
$1$ _ _
$2$ _ _
$3$ _ _
$4$ _ _
W... | Use inclusion/exclusion principle:
*
*Include the total number of sequences, which is $\frac{8!}{2!2!2!2!}$
*Exclude the number of sequences containing $11$, which is $\frac{7!}{1!2!2!2!}$
*Exclude the number of sequences containing $22$, which is $\frac{7!}{1!2!2!2!}$
*Exclude the number of sequences containing ... | {
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How to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$. I'm trying to figure out how to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$.
Hints?
| HINT:
Complete the square, which can be done by noting that:
$$6x^2 - 20x + 23 = 6(x^2 - \frac{10}{3}x + \frac{25}{9}) + \frac{19}{3}$$
Then multiply $(x^2 - \frac{10}{3}x + \frac{25}{9})$ by $9$ and divide $6$ by $9$.
| {
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Find this diophantine equation intgers $x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$ Find the solutions of the diophantine equation
$$x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$$
I have found $(x,y)=(\pm 1,0),(0,\pm 1)$: how to find all solutions?
Thanks
| $$\overbrace{(x+y)^4}^A-4\overbrace{xy^2(3x+2y)}^B=1$$
As $x,y$ are integers, a possible solution is $A=1, B=0$ i.e.
$$A=1\Rightarrow x+y=\pm 1\quad\cdots (1)\\
B=0\Rightarrow xy^2=0 \Rightarrow x=0 \text{ or } y=0\quad\cdots (2a)\\
\text{or}\\
3x+2y=0\Rightarrow 3x=-2y\quad\cdots (2b)$$
From $(1),(2a)$,
$$(x,y)=(0... | {
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Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rig... | There's no need for induction, the answer is immediate with modular arithmetic!
$$ 5\cdot10^n+10^{n-1}+3\equiv 5\cdot1^n+1^{n-1}+3=5+4=9\equiv0\pmod{9}$$
| {
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"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Calculate $ \lim\limits_{n \to \infty} \sum\limits_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $
Let $a \in (1, \infty)$. Calculate $$ \lim\limits_{n \to \infty} \sum_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $$
Here's what I tried:
Let $ x_n = \sum_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $. Let $a_n, b_n$ so that $a_n < x_n < b_n$. If $... | Let $a>\sqrt2$, we have that
\begin{align}
\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}&=\frac a{2a+2}+\sum_{k=0}^{\infty} \frac{2^k}{a^{2k+1}}(1+a^{-1-2k})^{-1}\\
&=\frac a{2a+2}+\sum_{k=0}^{\infty} \frac{2^k}{a^{2k+1}}\sum_{n=0}^{\infty}(-a^{-1-2k})^n
\end{align}
Applying Fubini's theorem,
\begin{align}
\sum_{k=0}^... | {
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Evaluate $\int_{|z|=4}\frac{\sin z}{z(z-2i)}\ dz$
Evaluate $$\int_{|z|=4}\frac{\sin z}{z(z-2i)}\ dz$$
The two singularities are included in inside the disk $|z|<4$, so I broke the integral in a sum of two integrals.
The first one:
$$\int \frac{\frac{\sin z}{z}}{z-2i}\ dz = 2i\pi f_1(2i)$$
where $f_1 = \frac{\sin z}{z... | Both methods work, but there is a small error in the second.
It is true that
$$
\frac{1}{z(z-2i)} = \frac{A}{z} + \frac{B}{z-2i}
$$
for $A = i/2$ and $B = -i/2$, but the error comes in the next line where your friend essentially says
$$
\frac{\sin z}{z(z-2i)} = \frac{A}{z} + \frac{B}{z-2i}
\newcommand{\contour}{{|z|=4}... | {
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Fibonacci Identity with Binomial Coefficients A friend showed me this cool trick: Take any row of Pascal's triangle (say, $n = 7$):
$$1, 7, 21, 35, 35, 21, 7, 1$$
Leave out every other number, starting with the first one:
$$7, 35, 21, 1$$
Then these are backwards base-5 "digits", so calculate:
$$7 + 35 \cdot 5 + 21 \cd... | Starting with Binet's Formula:
$$F_n = \frac{1}{\sqrt{5}} \left( \left(\frac{1 + \sqrt{5}}{2} \right)^n - \left(\frac{1 - \sqrt{5}}{2} \right)^n \right) $$
Expand using Binomial Formula:
$$F_n = \frac{1}{2^n\sqrt{5}} \left( \sum_{k=0}^n \left({n \choose k} \left( \sqrt{5} \right)^k\right) - \sum_{k=0}^n \left({n \choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$? The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ ... | Note that $(-1)^{\frac{1}{7}} = -1$
Also note that $(-1)^{\frac{9}{7}} = -(-1)^{\frac{2}{7}},\ (-1)^{\frac{11}{7}} = -(-1)^{\frac{4}{7}},\ (-1)^{\frac{13}{7}} = -(-1)^{\frac{6}{7}}$
You can then rewrite your addition as
$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
| Ignoring big theorems,
$$ f(n) = 1^4+2^4+\ldots+n^4 $$
is quite blatantly a polynomial in $n$ with degree $5$. It follows that it is enough to compute $f(0),f(1),\ldots,f(5)$ to recover the coefficients of such a polynomial by Lagrange interpolation:
$$ f(x) = \sum_{k=0}^{5}\,f(k)\!\!\!\!\prod_{\substack{j\in\{0,\ldots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the dimensions of the rectangle that will give the minimum perimeter. a farmer wants to make a rectangular paddock with an area of $ 4000 m^2$ However, fencing costs are high and she wants the paddock to have a minimum perimeter.
I have found the perimeter:
$$x\cdot y = 4000\\
y = \frac{4000}{x}$$
$$\begin{ali... | To minimise $2(x+y)$ subject to $xy=A$ note that: $$(x+y)^2=4xy+(x-y)^2=4A+(x-y)^2$$
Now since $x$ and $y$ are both positive, the minimum value of $2(x+y)$ occurs when $4(x+y)^2$ is a minimum and hence when $(x+y)^2$ is a minimum.
Since $(x-y)^2$ is non-negative, the minimum occurs when $x-y=0$ ie when you have a squar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Finding sum of infinite series $1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $ So the question is 'express the power series $$1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $$
in closed form'.
Now we are allowed to assume the power series of $e^x$ and also we derived the power series for $\cosh x$ usin... | We evaluate the power series $e^{\zeta x}$, noting that $\zeta^2 = -(1+\zeta)$ and $\zeta^3 = 1$, and get
$$
e^{\zeta x} = 1 + \zeta x + \frac{\zeta^2x^2}{2!} + \frac{\zeta^3x^3}{3!} + \frac{\zeta^4 x^4}{4!} + \cdots\\
= 1 + \zeta x - \frac{(1+\zeta)x^2}{2!} + \frac{x^3}{3!} + \frac{\zeta x^4}{4!} -\cdots
$$
while eval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Explain how and why the cancellation of $ 6 $’s in $ \dfrac{16}{64} $ to get $ \dfrac{1}{4} $ is a fallacious statement. Based on what we know from elementary and middle school teachers, most of us know that 16/64 correctly equals 1/4 because 16/64 is simplified with a common divisibility of 16. However, there is anoth... | The way someone might have justified that:
"Just remember that
$$\require {cancel}\frac {10}{20} = \frac{1\cancel {0}}{2\cancel {0}} = \frac{1}{2} $$
Therefore
$$\require {cancel}\frac{16}{64} = \frac{1\cancel {6}}{\cancel {6}4} = \frac{1}{4} $$
$\blacksquare$"
There is actually a problem on project euler regarding thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Sum of Maclaurin series Find the sum of the infinite series
\begin{equation}
\sum_{n=2}^\infty\frac{7n(n-1)}{3^{n-2}}
\end{equation}
I think it probably has something to do with a known Maclaurin series, but cannot for the life of me see which one.. Any hints would be appreciated!
Edit: Using your hints, I was able to ... | Hint: Notice
$$ \sum x^n = \frac{ 1}{1- x} $$
$$ \sum n x^{n-1} = \frac{1}{(1-x)^2} $$
$$ \sum n (n-1) x^{n-2} = \frac{2 }{(1-x)^3} $$
they all converge for $|x| < 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
An differential equation$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$ How to solve this ODE $$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$$
I tried to find its integral factor, but failed.
Many thanks for your help.
| Hint:
Let $t=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=2x\dfrac{dy}{dt}$
$\therefore(x^2+y^2+3)2x\dfrac{dy}{dt}=2x\left(2y-\dfrac{x^2}{y}\right)$
$(x^2+y^2+3)\dfrac{dy}{dt}=2y-\dfrac{x^2}{y}$
$(t+y^2+3)\dfrac{dy}{dt}=2y-\dfrac{t}{y}$
Let $u=y^2$ ,
Then $\dfrac{du}{dt}=2y\dfrac{dy}{dt}$
$\therefore\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \fr... | Please check the following hint. Firstly we have
$$
R := \sin A \sin C = \sin A \sin \left( \pi - (B+A) \right) = \sin A \sin \left( \frac{2\pi}{3} - A \right)
$$
thus,
\begin{align}
R
&= \sin A \left( \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \right)
= \frac{\sqrt{3}}{4} \sin 2A + \frac{1}{4} \left( 1 - \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
For $f(x)=4x^3+3x^2-x-1$, The Range of values $\frac{f(x_1)-f(x_2)}{x_1-x_2}$ can take is- My Attempt :-
$f'(x)=12x^2+6x-1$ where $f'(x) \ge \frac{-7}{4}$. So (I think) from LMVT we can directly say that
$$\frac{f(b)-f(a)}{b-a} \ge \frac{-7}{4}$$ But the answer Given is $$\frac{f(b)-f(a)}{b-a} > \frac{-7}{4}$$
So my ... | The mean value theorem does not tell you that every $c$ has a corresponding $a$ and $b$ with $\frac{f(a)-f(b)}{a-b}=f'(c)$. It tells you that given $a$ and $b$ you can find a $c$, but it doesn't say given $c$ you can find $a$ and $b$. In your case,
$\dfrac{f(a)-f(b)}{a-b}=4(a^2+ab+b^2)+3(a+b)-1=\dfrac{4(2a+2b+1)^2+(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as... | There are infinitely many. The complete rational solution to
$$a^2+b^2+c^2=1$$
is given by
$$\left(\frac{p^2-q^2-r^2}s\right)^k+\left(\frac{2pq}s\right)^k+\left(\frac{2pr}s\right)^k=1\tag1$$
where $s=p^2+q^2+r^2$ and $k=2$. But eq $(1)$ is also satisfied for $k=1$ if
$$p=\frac{q^2+r^2}{q+r}$$
For example, if $q=1,\,r=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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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.