Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Manipulation of Taylor expansion of $e^x$ I have an expression:
$$ f(x) = \sum\limits_{N=0}^\infty (N-x)^2 \frac{x^N}{N!}$$
I want to figure out whose Taylor Expansion this is.
I've found that I can separate out the above expression, and get the following:
$$f(x) = \sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} - 2 \sum\l... | Let's look at the mysterious sum. First, let's just write out some of the terms
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = 0 + x + \frac{4 x^2}{2!} + \cdots$$
so the first term is zero, let's drop it
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} =\sum\limits_{N=1}^\infty N^2 \frac{x^N}{N!}$$
and then shift the ind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Value of $L +\frac{153}{L}=$
If $\displaystyle L = \lim_{x\rightarrow 0}\bigg(\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}\bigg).$
Then value of $\displaystyle L +\frac{153}{L}=$
what i try
$$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}$$
from D L Hopital rule
$$L=\lim_{... | $$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}=\\
\lim_{x\rightarrow 0}\frac{\ln\left(1+\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)\right)}{x\frac{\ln(1+x)}{x}\cdot\frac{\ln(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}-1}\cdot\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve system of equations $ y + 3x = 4x^3$ and $x + 3y = 4y^3$ Problem: In the set of real numbers find all $x,y$ that satisfy
$ y + 3x = 4x^3,x + 3y = 4y^3$.
What I don't understand, is how do I know that a condition satisfies both equations(I will show later exactly).
The solution goes like this: By addition and subs... | Hint.
We have with $f(x) = 4x^3-3 x$
$$
y = f(x)\\
x = f(y)\\
$$
or
$$
y = f^2(y)\\
x = f^2(x)
$$
so the solutions are the fixed points
which are
$$
\left\{0,\pm 1, \pm\frac{\sqrt 2}{2},\frac 14\left(-1\pm\sqrt 5\right),\frac 14\left(1\pm\sqrt 5\right)\right\}
$$
NOTE
$$
f^2(x)-x = f(f(x)) -x = 8 (x-1) x (x+1) \left(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc}... | Observe that, in virtue of the QM-AM inequality, we obtain $$\sqrt{a^2+ab+b^2}\geqslant \frac{a+\sqrt{ab}+b}{\sqrt3}$$ Similarly $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geqslant \frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}$$ We thus want to prove $$\frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?
$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$
I have ... | We can prove by L'Hospital that
$$\lim_{x\to \infty}\frac{\ln\left(1+\frac1x\right)-\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}\right)}{-\frac1{4x^4}}=\lim_{x\to \infty}\frac{\frac1{x^5+x^4}}{\frac1{x^5}}=\lim_{x\to \infty}\frac{x^5}{x^5+x^4}=1 \\\iff \ln\left(1+\frac1x\right)=\frac1x-\frac1{2x^2}+\frac1{3x^3}+O\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Sketching the graph and finding the function by using the given. A function f that is defined and continuous for $x\neq 0$ satisfies the following conditions:
*
*f($2-{\sqrt 2}$) = $\sqrt[3]{1-1/\sqrt 2}$, f($\frac{2}{3}$) = $0$, f(2) = $\sqrt[3]{2}$, f($2+{\sqrt 2}$) = $\sqrt[3]{1+1/\sqrt 2}$
*$\lim_{x \to 0^-}f(... | First, since the left and right limits at zero are unequal (in fact, do not exist), $f$ is discontinuous at $0$, so that $d<0$. Since $f\left(\frac{2}{3}\right)=0$, the factor $(ax+b)^c$ must vanish there, so that $c>0$ and $ax+b = r(3x-2)$ for some constant $r$. So rewriting, we get
$$f(x) = r^c(3x-2)^c x^d.$$
Next, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$.
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.
Question from Jalil Hajimir
| You first fix $y, z$ and let $x > 0$ vary. Taking derivative with respect to $x$, dropping all those nonnegative terms such as $y^2+1$ to simplify notation, leads to
$$ \frac{d (OP\ full\ epxr)}{d x} \approx x - \frac{(x^2+1)}{x+y+z} = \frac{x(y+z) - 1}{x+y+z}, $$
where the $\approx$ means I dropped some positive term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\int \frac{dx}{x^3+x^2\sqrt{x^2-1}-x}$ solve:
$$\int \frac{\mathrm dx}{x^3+x^2\sqrt{x^2-1}-x}$$
I tried:
$$\begin{align}\int \frac{\mathrm dx}{x(x^2-1)+x^2\sqrt{x^2-1}}&=\int \frac{\mathrm dx}{x(\sqrt{x^2-1}\sqrt{x^2-1})+x^2\sqrt{x^2-1}}\\&=\int \frac{\mathrm dx}{x\sqrt{x^2-1}(\sqrt{x^2-1}+x)}\end{align}$$
$x=\sin t$
... | You can proceed from where you are like this, e.g.
\begin{eqnarray}
\mathcal I &=& \int\frac{dx}{x\sqrt{x^2-1}\left(\sqrt{x^2-1}+x\right)} =\\
&=&-\int\frac{\sqrt{x^2-1}-x}{x\sqrt{x^2-1}}dx=\\
&=&-\int\frac1xdx +\int\frac1{\sqrt{x^2-1}}dx=\\
&=&-\log|x|+\log\left(\sqrt{x^2-1}+x\right)+C
\end{eqnarray}
EDIT
For the seco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$ $\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$
I wanted to use L' Hopital's rule so I wrote the term as:
... | To simplify we have that
$$\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)=$$
$$\sin^2{x}\frac{\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}{\cos^2 x}$$
and since $t=\sin^2 x \to 1$ we can consider the limit
$$\lim_{t\to 1} \frac{\sqrt{2t^2+3t+4}-\sqrt{t^2+6t+2}}{1-t^2}$$
whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Rectangle inscribed on two parabolas but not parallel to the axis Let $f(x)=2x^2$ and $g(x)=36-x^2$. Let $R$ be the closed region between them. Draw a rectangle $ABCD$ such that $A$ and $B$ are points on $f(x)$, and $C$ and $D$ are points on $g(x)$, and the rectangle is not parallel to the axis, or prove that to be imp... | Brute force:
First, we relax the condition to finding any parallelogram with 2 vertices on each equation.
Let $ A = (a,2a^2), B = (a+b, 2a^2 + 4ab + 2b^2), D = (d, 36-d^2)$, which makes $ C = (d+b, 36-(d+b)^2 = 36 - d^2 + 4ab + 2b^2)$. This gives $ b=0$(rejected), $d =-2a- \frac{3b}{2}$.
Now, we to make the parallelogr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality... | Here I give a proof by using the standard pqr method.
Proof: Let $p = x+y+z$, $q = xy+yz+zx$ and $r = xyz$.
We will use the following facts (see [1], Facts N12 and N6):
(i) $q^3 + 9r^2 \ge 4pqr$.
(ii) $q^3 \ge 27r^2$.
We need to prove that
$$\frac{pq}{r} \ge 9 \sqrt{\frac{p^2-2q}{q}}$$
or
$$\frac{p^2q^2}{r^2} \ge 81 \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
$(...((x-2)^2-2)^2-...-2)^2\;\;\text{(n parentheses)}$ Find the coefficient next to $x^2$ in the development of a polynomial:
$$(...((x-2)^2-2)^2-...-2)^2\;\;\text{(n parentheses)}$$
I am not sure but I think we get the coefficient in the following way:
We have something like this quasi-recursive formula:
$$...+k_1\cd... | We have that
$$(x-2)^2=x^2-4x+4$$
$$((x-2)^2-2)^2=(x^2-4x+2)^2=\ldots+20x^2-16x+4$$
$$(((x-2)^2-2)^2-2)^2=(\ldots-16x+2)^2=\ldots 336x^2-64x+4$$
therefore we can guess that
*
*coefficient for $x$ are $-4,-16,-64,\ldots \implies k_2(n)=-(4^n)$
*coefficient for $x^2$ are $1,20,336,\ldots \implies k_1(n)=(k_2(n-1))^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to prove this $\sum_{i=1}^{n}(x_{i})^{S-x_{i}}>1?$ Question:
Let $x_{i} \in (0,1),i=1,2,\cdots,n$. Show that
$$
x_{1}^{S-x_{1}}+x_{2}^{S-x_{2}}+\cdots+x_{n}^{S-x_{n}}>1
$$
where $S=x_{1}+x_{2}+\cdots+x_{n}$.
I have proved when $n=2$,because it use this Bernoulli's inequality
$$
(1+x)^a\le 1+ax,0<a\le 1,x>-1
$... | Update
An ugly proof for $n=4$:
Due to symmetry, assume that $d = \max(a, b, c, d)$. Note that
$v\mapsto u^v$ is decreasing on $v > 0$ provided $u\in (0,1)$,
and that $x\mapsto x^y$ is convex on $x>0$ provided $y\ge 1$. We have
\begin{align}
a^{S-a} + b^{S-b} + c^{S-c} + d^{S-d} &\ge a^S + b^S + c^S + d^{S-d}\\
& \ge 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
True/false: $\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$ I have the following proposition about which I have to say whether it is true or false.
$\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$. $I$ is the identity matrix.
I tried brutal ways (taking... | You can also check that it is true via the real Jordan normal form.
First if we have a complex eigenvalue $\alpha + i \beta$ and one real eigenvalue $\lambda$, then $A$ is (after conjugation) of the form
$$ A = \begin{pmatrix} \alpha & \beta & 0 \\ -\beta & \alpha & 0 \\ 0 & 0 & \lambda
\end{pmatrix} $$
and thus
$$ det... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\f... | A way to do basically the same thing with less weird operator manipulation is to think of the procedure as using the integrating factor method on two first order ODEs. If you have say
$$y''+4y=x^2 e^{2ix}$$
then you can write it as
$$(D+2i)(D-2i)y=x^2 e^{2ix}.$$
Then you can let $u=(D-2i)y$ and solve $(D+2i)u=x^2 e^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
What is the expected length of the hypotenuse formed by bending a unit length randomly at a right angle? This is easy enough to simulate and find the answer is somewhere around .812. However I am not finding it so easy to solve the integral involved which I believe is...
$$\int_0^1 \sqrt{x^2+(1-x)^2} dx$$
I don't seem ... | $$\begin{align}I&=\int_0^1 \sqrt{x^2+(1-x)^2} dx\\
&=\int_0^1\sqrt{2x^2-2x+1}dx\\
&=\int_0^1\sqrt2\sqrt{\left(x-\frac12\right)^2+\frac14}dx\end{align}$$
Let $x-\frac12=\frac12\tan\theta$. Then, $dx=\frac12\sec^2\theta d\theta$.
$$\begin{align}I&=\frac{1}{2\sqrt2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}|\sec\theta|\sec^2\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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evaluation of Trigonometric limit
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$
What i try
Put $\displaystyle \frac{n\pi}{2}=x,$ when $n\rightarrow\infty,$ Then $x\rightarrow \infty$
$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos... | Hint:
$$\frac{1}{n^2}=\frac{1}{n\left(\color{red}n\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+\color{red}{\require{cancel}\cancel{n}}\sin^2 \frac{n\pi}{2}\right)}=\frac1n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle
Such that : $a+b+c=1$ Then
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$
My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$
Also :
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don... | Homogenizing the equation, we WTS
$$\frac{(a+b+c)^3}{2} -(a+b+c)(a^2+b^2+c^2)-4abc > 0$$
Expanding and factoring, we obtain
$$(a+b-c)(a-b+c)(-a+b+c) > 0$$
This is true via the triangle inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Does the denominator cancel? I recently asked this question: Simplify expression cross/dot products
The answer was straightforward, yet I can't apply the same to the expression below.
Can the following expression be simplified? In particular, I would like to remove the denominator:
$\frac{( b \times (a \times b)) \cdot... | Your numerator is a dot product of two vector triple products. The former is equal to $(b \cdot b)a - (b \cdot a)b = \|b\|^2a - (a \cdot b)b$ and the latter is equal to $(a \cdot a)b - (a \cdot b)a = \|a\|^2b - (a \cdot b)a$, so their dot product is \begin{align*}(\|b\|^2a - (a\cdot b)b)\cdot(\|a\|^2b - (a\cdot b)a) &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other way... | You can avoid squaring by comparing
$$\eqalign{
\sqrt{12}a&=\sqrt{84}+\sqrt{120}<10+11=21\ ,\cr
\sqrt{12}b&=\sqrt{36}+\sqrt{228}>6+15=21\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the fol... | We may assume without loss of generality that $m, n$ are both positive as this is equivalent to them both being negative and if only one of them is negative then this inequality is trivial.
We can then consider 2 cases:
Case 1: m< 2n
By your argument we need only show that $|\sqrt{3}n^2|+|mn|<5n^2$. This follows from $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
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Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$
Solve the system:
$$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
x^2-y^2=8\end{array}$$
First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do nex... | Rewrite the first equation as
$$x^2+y^2 - \dfrac{10}{3}xy = \frac13(x-3y)(3x-y)=0$$
which yields $x=3y$ and $x=\frac y3$. Plug them into $x^2-y^2=8$ to obtain the real solutions $(3,1)$ and $(-3,-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
using absolute function to translate the inequality How to use the absolute value function to translate each of the following statements into a single inequality.
(a) $\ x ∈ (-4,10) $
(b) $\ x ∈ (-\infty,2] \cup[9,\infty) $
I think in the first one the absolute value of $\ x$ should be greater than 4 and less than 10... | $x \in (-4,10)\implies$
$-4 < x < 10\implies$
$-4 + k < x + k < 10 + k$
If we have $M=|10+k| = |-4+k|$ (and presumably $-M= -4k < 0 < 10+k=M$) then we would have $-M < x + k < M$ so $|x + k| <M$.
So if $10+k = -(-4+k)=4-k$ then $2k = -6$ and $k -3$ and $M=10 -3=7$ and
$-4 -3 < x - 3< 10-3$
$-7 < x-3 < 7$
$|x-3| < 7$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\fr... | $\cos^4 x = \frac 18 (\cos 4x + 4\cos 2x + 3)\\
\sin^4 x = \frac 18 (\cos 4x - 4\cos 2x + 3)\\
\cos^4 x+ \sin^4 x = \frac 14(\cos 4x + 3)$
$\frac {4\cos 2x}{\cos 4x + 3}$
That looks a little better.
Lets say $\cos 4x = 1 - 2\sin^2 2x$
$\int \frac {4\cos 2x}{4 - 2\sin^2 2x}\ dx$
Now we can do a u-subtitution
$u = \sin 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots... | For each $n\in\mathbb N$, the first non-null term of the Taylor series of $x^n\sin^n(x)$ centered at $0$ is $x^{2n}$, whereas the first non-null term of the Taylor series of $x^n-\sin^n(x)$ centered at $0$ is $kx^{n+2}$, for some $k\neq0$. So, your limit will be finite and non zero if and only if $2n=n+2$. And this occ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?
| We can assume without loss of generality that $x,y\ge 0$ (why?)
So by AM-GM inequality:
$$x^2+y^2=x^2+\frac{y^2}3+ \frac{y^2}3+ \frac{y^2}3\ge4\sqrt[4]{\frac{1}{3^3}x^2y^6}=4\cdot3^{-\frac34}\sqrt{xy^3}.$$
By squaring:
$$(x^2+y^2)^2\geq16\cdot3^{-\frac32}xy^3\approx3.08 xy^3.$$
So we have the sharper bound $$\boxed{(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
solutions of $a+b=c^2 , a^2+c^2=b^2$ ; $a,b,c$ are natural numbers So it all started with a fun observation, $12+13=5^2$ and these are Pythagorean triplets($5,12,13$), so I thought are there more such numbers?
with brute force I was able to get $(24,25,7)$ and $(40,41,9)$.
Then I was able to find 3 families of solutio... | $$a^2+c^2=b^2$$
$$\implies a^2+a+b=b^2$$
$$\implies \Big(a+\frac{1}{2}\Big)^2= \Big(b-\frac{1}{2}\Big)^2$$
$$\implies a+\frac{1}{2}=b-\frac{1}{2}$$
$$\implies b=a+1$$
Hence,
$$2a+1=c^2$$
Therefore, $c$ is odd, let $c=2k+1$. Putting it in above equation, you get, $$a=2k^2+2k$$
$$\implies b=a+1=2k^2+2k+1$$
This is the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Finding residues in a complex integral $\int_0^\infty\frac{z^6\,\mathrm{d}z}{(z^4+1)^2}$ using Laurent series expansion I am trying to compute
$$\int_0^\infty\frac{z^6\,\mathrm{d}z}{(z^4+1)^2}.$$
I am getting stuck on computing the residues. I am only considering the residues when $z=\mathrm{e}^{\frac{\mathrm{i}\pi}4}... | Let's compute the residue by finding a few terms of the Laurent series for
$$
p(z) = \frac{z^6}{(z^4+1)^2}
$$
at $\omega = e^{i\pi/4} = \frac{1+i}{\sqrt{2}}$.
To expand in powers of $z-\omega$, I prefer to write $r=z-\omega$ and expand in powers of $r$.
$$
p(r+\omega) = \frac{(r+\omega)^6}{((r+\omega)^4+1)^2}
$$
Do thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$. Exercise 4.4, problem 24, page 144 of the book on differential equations by KKOP [here][1].
Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$, where $a,b,\omega$ are real constants, $a<b$. Consider separately t... | You first mistake is at the step
$$\begin{align}
\scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{\sqrt{b^2-a^2}}\left[\cos \left(\omega - \sqrt{b^2-a^2}x\right) - \cos \left(\omega + \sqrt{b^2-a^2}x\right)\right]\\
\scriptsize c_2'(x)&=\scriptsize e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I calculate the limit $\lim\limits_{n \to \infty} \frac1n\ln \left( \frac{2x^n}{x^n+1} \right)$. I have the following limit to find:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg)$$
Where $n \in \mathbb{N}^*$ and $x \in (0, \infty)$.
I almost got it. For $x > 1$, I observed th... | Since $\frac{d}{dn}x^n=x^n\ln x$, $\frac{d}{dn}\frac{2x^n}{x^n+1}=-2\frac{d}{dn}\frac{1}{x^n+1}=\frac{2x^n\ln x}{(x^n+1)^2}$ and $\frac{d}{dn}\ln\frac{2x^n}{x^n+1}=\frac{\ln x}{x^n+1}$. So you want $\lim_{n\to\infty}\frac{\ln x}{x^n+1}$, regardless of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron who... | Yes, because what you are really doing is finding the volume of a volume of a parallelepiped. You can also approach the problem this way:
height= |a|
and the area is given by
|b × c|
So, then you have Volume = height * area
Volume= |a||b × c|=|a ∙( b × c)|
For b: (0,2,0) – (0,0,0) = (0,2,0)
For c: (0,0,2) – (0,0,0) = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the det... | A determinant of the left matrix is equal to
$$\prod_{cyc}(a^2+b^2)+2a^2b^2c^2-\sum_{cyc}(a^2+b^2)a^2b^2=$$
$$=\sum_{cyc}\left(a^4b^2+a^4c^2+\frac{2}{3}a^2b^2c^2\right)+2a^2b^2c^2-\sum_{cyc}(a^4b^2+a^4c^2)=4a^2b^2c^2.$$
A determinant of the right matrix is equal to
$$0+2abc-0=2abc$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$ I tried substitution $x=\dfrac{1}{t}$, then $z=t^2$, tried rationalizing the denominator but haven't been able to pull it off. I can't think of any trig substitution either. I prefer an intuitive approach rather than an "plucked-out-of-thin-air" counter-intui... | Here is an elementary method without using hyperbolic functions:
$$\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$$
$$=\int \frac{1}{x^5(\frac{1}{x^4}+1)\sqrt{\sqrt{\frac{1}{x^4}+1}-1}}dx$$
Now substitute $t=\frac{1}{x^4}+1$, required integral becomes,
$$-\frac{1}{4}\int\frac{1}{t\sqrt{\sqrt {t}-1}}dt$$
Now substitute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Possible values of expression Real numbers $x,y,z$ are selected so, that $\frac{1}{|x^2 +2yz|}$, $\frac{1}{|y^2 +2xz|}$, $\frac{1}{|z^2 +2xy|}$ are sides of a triangle(they satisfy the triangle inequality). Determine the possible values of the expression $xy+xz+yz$.
It is easy to prove that all positive and negative re... | Suppose that $xy+yz+zx=0$. Then, we have the following identities
$$
\frac{1}{|x^2+2yz|}=\frac{1}{|x^2+2yz-xy-zx-yz|}=\frac{1}{|x-y|\cdot |x-z|},
\\
\frac{1}{|y^2+2zx|}=\frac{1}{|y^2+2zx-xy-zx-yz|}=\frac{1}{|x-y|\cdot |y-z|},
\\
\frac{1}{|x^2+2yz|}=\frac{1}{|z^2+2xy-xy-zx-yz|}=\frac{1}{|z-x|\cdot |y-z|}.
$$
Now, it's e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamm... | Expressing the two complex and conjugate roots as a function of the third root $\gamma$ ,we obtain
$\alpha=-\frac{2\gamma+1}{4}+I\frac{\sqrt{12\gamma^{2}+4\gamma+7}}{4}$
$\beta=\frac{2\gamma+1}{4}-I\frac{\sqrt{12\gamma^{2}+4\gamma+7}}{4}$
therefore
$\Big(\frac{1}{\gamma^{3}}+I\frac{(2\gamma-1)\sqrt{12\gamma^{2}+4\gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Mistake in evaluating $\int_{0}^{1}\frac{x^2\ln x}{\sqrt{1-x^2}}dx$ $$\int_{0}^{1}\dfrac{x^2\ln x}{\sqrt{1-x^2}}dx$$
My attempt is as follows:
$$x=\sin\theta\Rightarrow dx=\cos\theta d\theta$$
$$\int_{0}^{\frac{\pi}{2}}\sin^2\theta\ln (\sin\theta)d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1-\cos2\theta)\ln(\sin\theta)... | Actually，
\begin{align*}
\int_0^{\frac\pi2}\cos(2\theta)\ln(\sin\theta)d\theta&=
\frac12\int_0^{\frac\pi2}\ln(\sin\theta)d(\sin2\theta)\\
&=\frac12\sin2\theta\ln(\sin\theta)\bigg|_0^{\frac\pi2}-\frac12\int_0^{\frac\pi2}\sin2\theta\cdot\frac{\cos\theta}{\sin\theta}d\theta\\
&=-\frac12\int_0^{\frac\pi2}2\cos^2\theta d\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $a^{4b}+b^{4a}\geq \frac{1}{2}$ Inspired by a problem of Vasile Cirtoaje I propose this :
Let $a,b>0$ such that $a+b=1$ then we have :
$$a^{4b}+b^{4a}\geq \frac{1}{2}$$
I compute the derivative of $f(x)=x^{4(1-x)}+(1-x)^{4x}$ on $]0,1]$ we get :
$$ f'(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x)) + ... | Alternative Proof:
We need to prove that, for all $a$ in $(0, 1/2]$,
$$a^{4 - 4a} + (1 - a)^{4a} \ge 1/2.$$
We split into two cases:
Case I $\,\, a\in (0, 1/4)$:
We have
$$a^{4 - 4a} + (1 - a)^{4a}
\ge (1 - a)^{4a} \ge 1 - a > \frac12.$$
Case II $\,\,a \in [1/4, 1/2]$:
First, using Bernoulli inequality, we have
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of the given function- Find the minimum value of-
$$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$
I tried opening the brackets and trying to cancel the terms and using AM-GM.
| Hint.
Calling
$$
u = x^3+\frac{1}{x^3}\\
v = \left(x+\frac{1}{x}\right)^3
$$
we have
$$
\frac{v^2-u^2}{v+u}=v-u = 3\left(x+\frac{1}{x}\right)\ge 6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Extrema of $f(x,y) = xy\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$ From Demidovich
Find the extrema of the function of $z = xy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$
What I've done so far:
I've make $z_{x}'= 0$ and $z_{y}' = 0$, wich gives me:
$$\frac{\partial z}{\partial x} = y \sqrt{1 - \frac{x^2}{a^2} - \... | I suppose that $a,b>0$ are given parameters. The given function is then defined in the region
$$
R = \left\{\ (x,y)\ :\ \frac {x^2}{a^2}+ \frac {y^2}{b^2}\le 1\ \right\}\ .
$$
The four quadrants $I, II, \dots$ are cutting $R$ in four pieces, $R_I$, $R_{II}$, ... and it is enough to search for the local / global maxima... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$.
find $\alpha^2+\beta^2$
Both equations have a common root
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has equal roots
$$4b^2-20a=
0$$
$$b^2=... | $$a\alpha^2-2b\alpha+5=0\tag{1}$$
$$\alpha^2-2b\alpha-10=0\tag{2}$$
Subtracting $(1)$ and $(2)$
$$\alpha^2(a-1)+15=0$$
$$\alpha^2=\dfrac{15}{1-a}$$
From the first equation $$\alpha^2=\dfrac{5}{a}$$
$$\dfrac{15}{1-a}=\dfrac{5}{a}$$
$$3a=1-a$$
$$a=\dfrac{1}{4}$$
As first equation has equal roots
$$D=0$$
$$b^2-5a=0$$
$$b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the values for which $A^2 = I_2$, A is a matrix, with $A \neq I_2$ and $A \neq -I_2$ First I tried to find $A^2$ with
$$
A=\begin{bmatrix}
\alpha & \beta\\
\delta & \gamma\\
\end{bmatrix}
$$
I multiplied this by itself and got:
$$
\begin{bmatrix}
\alpha^2+\beta\delta& \beta(\alpha + \gamma)... | I found a much simpler way to solve this (probably the one intended by the authors).
Since
$$AA = I_2 \Leftrightarrow A = I_2.A^{-1} \Leftrightarrow A = A^{-1}$$
So,
$$A=\begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix} = A^{-1} = \det(A)^{-1}*\begin{bmatrix}
d & -b\\
-c & a\\
\end{bmatrix} $$
$\de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
What condition the roots of the polynomial meet Coefficients of the polynomial $f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$ meet the condition $a_3^3-4a_2a_3+8a_1=0.$ By substitution $x=y-\frac{a_3}{a_4}$ polynomial $f(x)$ is transformed into a biquadratic polynomial $g(y)=y^4+b_2y^2+b_0,$ where $b_2, b_0$ depend on the polyno... | It is given $f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$. When we substitute $x=y-\frac{a_3}{a_4}$, we get the following polynomial,
$$f(y)=a_4\left(y-\frac{a_3}{a_4}\right)^4+a_3\left(y-\frac{a_3}{a_4}\right)^3+a_2\left(y-\frac{a_3}{a_4}\right)^2+a_1\left(y-\frac{a_3}{a_4}\right)+a_0$$
Now, we are expanding each term of the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2... | Since our inequality is homogeneous, we can assume that $a+b+c=3.$
Thus, $$\sum_{cyc}\frac{a}{(b+c)^2}-\frac{9}{4(a+b+c)}=\sum_{cyc}\frac{a}{(3-a)^2}-\frac{3}{4}=\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)=$$
$$=\frac{1}{4}\sum_{cyc}\frac{-a^2+10a-9}{(3-a)^2}=\frac{1}{4}\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}=$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Find the volume of the solid rotating about $x=2$ $y=x^3, y=0, x=1$ rotated about $x=2$
Plugging in $x=1$ into $y=x^3$ we get 1 so our bounds are $[0,1]$
In order to account for rotating about the line $x=2$ subtract 2 from the function
$$\int_0^1 \pi [y^{\frac{1}{3}}-2]^2 dy$$
$$\pi \int_0^1 y^{\frac{2}{3}}-4y^{\frac{... | For any given $y$-value, what is the area of the corresponding annulus ("washer") cross section of our solid of rotation? The outer radius is $2 - y^{1/3}$, while the inner radius is $1$. So the area becomes
$$
\pi(2-y^{1/3})^2 - \pi\cdot 1^2 = \pi\left[(2-y^{1/3})^2 - 1\right]
$$
This is what you should integrate. All... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $\lim\limits_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1}=-1$? The context: p-adic convergence
I first ran into the formula:
$$
S = \lim_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} = -1
$$
in the context of $p-$adics, where we define a new norm, where for $x = 5^a \cdot b$, $5 \not \lver... | I think you're conflating a couple of things here. Let $S_n$ be the sequence of partial sums of the geometric series, $S_n=\sum_{i=0}^n4\cdot 5^i$. Then it is true for all $n$ that $S_n=5^{n+1}-1$ by the usual formula, so that if we look at the 'associated' sequence $\mathfrak{S}_n=S_n-5^{n+1}$, then $\mathfrak{S}_n$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to deal with negative area when evaluating a definite integral
Find the area bounded by the curves $y=2-x^2$ and $x+y=0$
$$
-x=2-x^2\implies x^2-x-2=(x-2)(x+1)=0
$$
My Attempt
$A_1:$ Area above the x-axis and $A_2:$ Area below the x-axis
$$
A_1=\int_{-1}^\sqrt{2}(2-x^2)dx-\int_{-1}^0(-x)dx=\Big[2x-\frac{x^3}{3}\... | The simplest way to do it I see is
$$\int_{-1}^2\left(2-x^2+x\right)dx=\left.2x-\frac {x^3}3+\frac{x^2}2\right|_{-1}^2\\
4-\frac {8}3+\frac 42-(-2)-\frac{1}3-\frac 12=\\
=\frac{9}2$$
This is the approach in your reference, but there is a typo in the upper limit of the second integral. Your $A_1$ is trying to get the a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Steady state probabilities of divergent Markov matrices I have a problem for which the Markov matrix turns out to be the following:
$$P = \begin{pmatrix}
0 & 0.5 & 0 & 0.5\\
0.5 & 0 & 0.5 & 0 \\
0 & 0.5 & 0 & 0.5\\
0.5 & 0 & 0.5 & 0\\
... | This Markov chain is periodic with period $2$, meaning that for any initial state $i$, the limiting probability $\lim_{n\to\infty}\mathbb P(X_n=j\mid X_0=i)$ does not exist. However, $\{X_n\}$ is irreducible, positive recurrent, and has finitely many states, so a unique stationary distribution $\pi$ exists. Since $P$ i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=... | The equation can also be factorized as follows-
$$\begin{align}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
&=\frac 12(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)\\
&=\frac 12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
\end{align}
$$
Hence the equation can only be true when either $a+b+c=0 $ or $a=b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\sum_{i = 1}^na_1^2 = 1$. Calculate the maximum value of $\sum_{cyc}|a_1 - a_2|$.
Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$ $(n \in \mathbb N, n \ge 3)$. Calculate the maximum value of $$\large |... | We have that $|x - y| = 2 \cdot \max(x, y) - (x + y)$
$$\implies \sum_{cyc}|a_1 - a_2| = 2 \cdot \left[\sum_{cyc}\max(a_1, a_2) - \sum_{i = 1}^na_1\right]$$
which could be rewritten as $$\sum_{cyc}|a_1 - a_2| = 2 \cdot \sum_{i = 1}^nx_ia_i$$ where $x_i \in \{-1, 0, 1\}, i = \overline{1, n}$ and $\displaystyle\sum_{i = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514654",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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example for calculation of the exterior derivative I don't understand how to calculate exterior derivatives. For the form
$$
\theta = \frac{x\, dy - y\, dx}{x^2 + y^2}
$$
I arrive at the following solution:
$$
\begin{align*}
d\theta
&= d\left(\frac{x}{x^2 + y^2}\right) \wedge dy
+ d\left(\frac{-y}{x^2 ... | If you apply the definition you have:
$$
\begin{align*}
d\left(\frac{x}{x^2 + y^2}\right) \wedge dy
&= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy
+ \partial_y\left(\frac{x}{x^2 + y^2}\right) dy \wedge dy \\
&= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy \\
&= \frac{-x^2 + y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3518302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Power Series Representation of $f(x) = \frac{x^3}{(2-x)^3}$ I am trying to represent the function, $f(x) = \frac{x^3}{(2-x)^3}$ as a power series and saw this on a different post.
$
{1\over (1-u)} =\sum_{n=0}^{\infty}u^{n}, \qquad|u|<1, \tag1
$
$-{1 \over (1-u)^2} =\sum_{n=1}^{\infty}nu^{n-1}, \qquad|u|<1,\tag2$
${2 \... | $$f(x)=\frac{x^3}{(2-x)^3}=\frac{x^3}{8}(1-\frac{x}{2})^{-3}=\frac{x^3}{8}[1-{-3 \choose 1}\frac{x}{2}+{-3 \choose 2} (\frac{x}{2})^2 -{-3 \choose 3}(\frac{x}{2})^3$$ $$+{-3 \choose k} (\frac{x}{2})^4 .....] ~~~~(1)$$
Using $${-p \choose k}=\frac{-p(-p-1)(-p-2)....(-p-k+1)}{k!},$$
we find that $${-3 \choose 1}=-3, {-3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagra... | By AM-GM and C-S we obtain:
$$\frac{\sqrt2(a+b)^2}{4}=\frac{\sqrt{2(a+b)}\sqrt{a+b}(a+b)^2}{4ab}\geq(\sqrt{a}+\sqrt{b})\sqrt{a+b}=$$
$$=\sqrt{a^2+ab}+\sqrt{b^2+ab}\geq\sqrt{a^2+\frac{4a^2b^2}{(a+b)^2}}+\sqrt{b^2+\frac{4a^2b^2}{(a+b)^2}}=$$
$$=\sqrt{a^2+4}+\sqrt{b^2+4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Line integral of vector field when Stokes' theorem cannot be applied directly
Define, on $\mathbb{R}^3 \setminus \{(0,0,z) \in \mathbb{R}^3 \ \mid \ z \in \mathbb{R} \}$, the vector fields $$G(x,y,z) = \left(\frac{x}{x^2+y^2}, \frac{2y}{x^2+y^2}, 2 \right) $$ and $$H(x,y,z) = \left(\frac{-y}{x^2+2y^2}, \frac{x}{x^2+2y... | If you don't want to use Stoke's theorem then integrate around the path itself.
The cylinder x+ y+ z= 2 and the paraboloid $z= x^2+ y^2$ intersect where $x+ y+ (x^2+ y^2)= 2$. Complete the squares in both variables: $x^2+ x+ \frac{1}{4}+ y^2+ y+ \frac{1}{4}= 2+ \frac{1}{4}+ \frac{1}{4}$, $(x+ \frac{1}{2})^2+ (y+ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum without computing the derivative. I am trying to find the minimum of the following function:
$$H(x)=\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})},\hspace{1cm}x>0$$
What I usually do to find extrema of a function is computing the derivative of the function and finding its ... | Observe that
$$H(x) = \frac{\left(\left(x+\frac{1}{x}\right)^3\right)^2 - \left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3 + \left(x^3+\frac{1}{x^3}\right)} = \left(x+\frac{1}{x}\right)^3 - \left(x^3+\frac{1}{x^3}\right) = 3\left(x+\frac{1}{x} \right) \geq 6$$
(with equality at $x=1$) where in the last s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Calculate $\displaystyle\sum_{k=1}^n \frac{k^2}{2^k}$ Calculate $\sum _{k=1}^n\:\frac{k^2}{2^k}$
I got $\sum _{k=1}^n\:\frac{k^2}{2^k} = (1+2+ \dots +k + \dots +n-1+n)(1+\frac{1}{2} + \dots +\frac{1}{2^{k-1}}+ \dots +\frac{1}{2}^{(n-1)}-\frac{n}{2^n})$. Not sure how to simplify further.
| Firstly,
$$\sum_{k=1}^n\ \frac k{2^k}=\sum_{k=1}^n\ \left(\frac k{2^{k-1}}-\frac {k+1}{2^k}+\frac 1{2^k}\right)=1-\frac {n+1}{2^n}+\sum_{k=1}^n\ \frac 1{2^k}=2-\frac {n+2}{2^n}$$ .
On the other hand,
$$\sum_{k=1}^n\ \frac {k^2}{2^k}=\sum_{k=1}^n\ \left(\frac {k(k+1)}{2^k}-\frac k{2^k}\right)=\sum_{k=1}^n\ \left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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For real $a$, $b$, $c$ in $(0,1)$, if $a+b+c=2$, then $\frac{a}{1-a}\times\frac{b}{1-b}\times\frac{c}{1-c}\geq 8$
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$
Prove that
$$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$
I tried by substituting $x$ for$(1-a)$ and similarly for o... | $$a+b-c=2-2c>0,$$ which says that $a$, $b$ and $c$ are sides-lengths of a triangle.
Now, let $a=y+z$, $b=x+z$ and $c=x+y.$
Thus, $x$, $y$ and $z$ are positives, $x+y+z=1$ and by AM-GM
$$\prod_{cyc}\frac{a}{1-a}=\prod_{cyc}\frac{y+z}{x}\geq\prod_{cyc}\frac{2\sqrt{yz}}{x}=8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0... | Because of the $b^3$ in numerator, I should rather start with
$$\frac{b^{3}}{1 - \theta(1-b)^{2}}=\frac{b+2}{\theta }+\frac{b (3 \theta +1)-2 \theta +2}{\theta \left(\theta(b-1)^2 -1\right)}$$
$$\frac{b^{3}}{1 - \theta(1-b)^{2}}=\frac{b+2}{\theta }+\frac{b (3 \theta +1)-2 \theta +2}{\theta \left(\theta( b-1)^2 -1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$ I need to find the number of divisors of $N = 3^55^77^9$ that are of the form $4n+1$, $n\geq 0$.
My try:
I noticed that $5$ itself is a number of the form $4n+1$ so all of its power satisfy the required condition, so number for the exponent of $5$ will be $7+1 = ... | A generating function approach:
Let $$f(x)=\left(1+x+\cdots+x^5\right)\left(1+x+\cdots+x^9\right)$$
Then the divisors of $3^57^9$ which are of the form $4n+1$ are the number of $3^a7^c$ with $0\leq a\leq 5, 0\leq c\leq 9$ and $a+c$ is even. This is the sum of the even coefficients of $f(x)$ which can be written as:
$$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore... | No, it is not, since what you got is not a power series (see what you get if you put $n=0$).
Use the fact that\begin{align}\frac1{x^2}&=\frac14+\left(\frac1{x^2}-\frac14\right)\\&=\frac14+\int_4^x-\frac1x\,\mathrm dx\\&=\frac14-\int_4^x\frac1{-2+(x+2)}\,\mathrm dx\end{align}and you will get that the answer is$$\frac1{x... | {
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"source": "stackexchange",
"question_score": "3",
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Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) ... | As your error has been pointed out, I am providing a different way to tackle the problem without introducing extra solutions.
From the given equation, we have $1=(1-\sin x)(1-\cos x)$, which is equivalent to
$$1=\Biggl(1-\cos\left(\frac{\pi}2-x\right)\Biggr)\left(2\sin^2 \frac{x}{2}\right)=4\sin^2\left(\frac{\pi}{4}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'... | You made a sign mistake here:
$$\frac{-1}{2}v'+\frac{2}{x}v = \frac{5}{x^2}$$
And also
$$\left (\frac 1 {y^2} \right )'=-2\frac {y'}{y^3}$$
$$\implies v'=-2\frac {y'}{y^3}$$
Another way:
$$x^2dy+ 2xydx - 5y^3dx=0$$
$$d(x^2y) - 5y^3dx=0$$
$$\frac {dx^2y}{(x^2y)^3} - 5\frac {dx}{x^6}=0$$
$$- \frac {1}{2(x^2y)^2} +\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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An exponential equation over positive real numbers $4^x+14^x+3^x=11^x+10^x$ Solve the following equation over the positive reals:
$$4^x+14^x+3^x=11^x+10^x.$$
By inspecting the graph, the solutions must be $x \in \{1,2\}$
I tried using inequalities like $3^x+4^x<5^x$ for $x>2$ but I couldn’t work it forward. Thanks in a... | We'll solve this equation for any real value of $x$.
Indeed, let $x>2$ and $x=1+t$.
Thus, $t>1$ and since $f(x)=x^t$ is a convex function, by Jensen we obtain:
$$4^x+3^x+14^x=(3\cdot3^t+8\cdot14^t)+\left(4\cdot4^t+6\cdot14^t\right)\geq$$
$$\geq11\left(\frac{3\cdot3+8\cdot14}{11}\right)^t+10\left(\frac{4\cdot4+6\cdot14... | {
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"url": "https://math.stackexchange.com/questions/3535196",
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"question_score": "2",
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Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.... | Hint
Let $a=b^2$ where $b\ge0$
WLOG any point on $$x^2+y^2=b^2$$ be $P(b\cos t,b\sin t)$
Now $$(b\cos t-4)^2+(b\sin t-2)^2=20+b^2-4b(2\cos t+\sin t)$$
Again $$-\sqrt{2^2+1^2}\le-(2\cos t+\sin t)\le\sqrt{2^2+1^2}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Finding the derivative of $N$ with respect to $t$ of $N=500(1-\frac{3}{(t^2+2)^2})$ Finding the derivative of $N$ with respect to $t$ of $N= 500\left(1-\frac{3}{(t^2+2)^2}\right)$
The first step to simplify things is to note that since $\frac{3}{(t^2+2)^2}=3(t^2+2)^{-2}$, we have
$$N= 500\left(1-\frac{3}{(t^2+2)^2}\rig... | Your answer is correct, and it is important initially to write out all the steps, just like you have done. Note that the derivative is positive or negative depending upon whether or not $t$ is, so from the derivative (and from the original function as well) it is found that $N$ has a minimum at $0$, where $N = 125$. Fu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $5\mid\frac {n^{r-1}-1}{n-1}$ and $r $ is even then $r=10k+6$ Let $5\mid\frac {n^{r-1}-1}{n-1}$ and $r $ is even. I'm trying to show $r=10k+6$.
Since
$ \frac {n^{r-1}-1}{n-1}=n^{r-2}+...+n+1$ and I couldn't find any $n $ for which
$5|n^2+n+1$
$5|n^6+n^5+n^4+n^3+n^2+n+1$
$5|n^8+n^7+n^6+n^5+n^4+n^3+n^2+n+1$
$5|n^{10... | First of all, let's take $a=r-1$, since it is a bit easier to calculate with it.
Define $l$ like mentioned above $n^a-1=5l(n-1)$. Looking this equation with modulo $n$ we will get:
$$-1\equiv -5l \mod(n) \Leftrightarrow n \mid 5l-1$$
From this we get that 5 can not divide n, since otherwise we would have $5\mid 5l-1$, ... | {
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"url": "https://math.stackexchange.com/questions/3537757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rationalizing the denominator of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$
Simplify
$$\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$$
I think this should be expressed without square roots at the denominator. I tried to multiply by conjugate.
| $$
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}=\\\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2+\sqrt{2+\sqrt{2}}}}=\\\frac{2 \cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}=\\
\frac{2\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}} \cdot \frac{2-\sqrt{2+\sqrt{2}}}{2-\sqrt{2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538395",
"timestamp": "2023-03-29T00:00:00",
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} |
Determine the value of $\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}}{\sqrt{2\sqrt {2\sqrt{2...}}}}$ Determine the value for
$$\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}}{\sqrt{2\sqrt {2\sqrt{2...}}}}$$
I think the formula $S_\infty =\frac {a}{1-r}$ should be used for this question but I don’t know how to find the fir... | Hint:
$$\sqrt{2\sqrt {2\sqrt{2...}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots}$$
and
$$\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}=4^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots}$$
and now you can use the geometric series formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Problem regarding unique solution of differential equation
A unique solution to the differential equation $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist
then choose the correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 ... | $$(y')^2-xy'=-y$$
Complete the square:
$$(y')^2-xy'+\frac {x^2} 4 =-y+\frac {x^2} 4 $$
$$(y'-\frac x 2 )^2=-y+\dfrac {x^2} 4 $$
The differential equation reduces to:
$$(z')^2=z$$
Where $z=-y+\dfrac {x^2} 4 $
Now you have to analyse what happens when $z <0,z=0,z>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Consider the linear function$ f : \mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ defined as follows: Consider the linear function $f : \mathbb{R}^{2\times2} → \mathbb{R}^{3\times2}$ defined as follows:
$$\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix} \in \mathbb{R}^{2\times2}\mapsto f\begin{pmatrix} r_1 & r... | Let's calculate the kernel first. We are searching for all matrices $R=\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}$ satisfying
$$f(R)=\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\ -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix} =0.
$$
So, from the last line of the above matrix we obtain $r_3=r_4=0$, and from th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n... | The $n$ roots of $z^n=-1$ are $n$ equidistant points on the unit circle. $|z-1|$ is simply the distance between one such root and $1$. Therefore, $|z-1|$ is minimized when $z=e^{\frac{\pi i}{n}}$. Therefore, $|z-1|\geq|e^{\frac{\pi i}{n}}-1|=2\sin{\frac{\pi}{2n}}$(Which can be found using geometry or algebra, as one wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Is there non real number of x that sufficient for this $11-\sqrt 7 x = 4x - 10$
Is there non real number of x that sufficient for $11-\sqrt 7 x = 4x - 10$ and $|\sqrt 7 x - 11 | = 4x - 10$
There is a solution which is real number.
| $|anything|$ is always a positive real number.
So $|anything| = 4x - 10$ will only have real solutions (if any solutions at all).
Also for complex numbers, $x$ we tend not to use the notation $\sqrt{x}$ as it is ambiguous.
I'm going to assume you meant $\sqrt{7x}$ (which is the square root of $7x$) and not $\sqrt{7}x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a differential-initial value equation: $y'(x) = \frac{y}{x+y^3}$, $ y(2)=2$
The equation and given values are
$$
\begin{split}
\frac{dy}{dx} &= \frac{y}{x+y^3}\\
y(2) &= 2
\end{split}
$$
At first I thought it was a separation of variables problem, but I then was told later on by a tutor that it could be sol... | The differential equation, $$\frac{dy}{dx}=\frac{y}{x+y^3}$$ is not homogenous so the substitution $y=ux$ is not helpful.
If you change your eqaution into $$\frac{dx}{dy}=\frac{x+y^3}{y}$$
then it is linear and you may solve it using integrating factor method.
$$\frac{dx}{dy}=\frac{x+y^3}{y}=(1/y)x+y^2$$
The integratio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
... | Go upto $(n+1)$ (not $n$)
You have
\begin{eqnarray*}
3 \sum_{i=1}^{n} i^2 + 3 \sum_{i=1}^{n} i + \sum_{i=1}^{n} 1=(n+1)^3-1.
\end{eqnarray*}
Now use
\begin{eqnarray*}
\sum_{i=1}^{n} 1&=&n \\
\sum_{i=1}^{n} i&=& \frac{n(n+1)}{2}. \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\... | Hint:
$a=x^2$
$b=-x$
$c=1$ $$Min{\frac{(a+b+c)^3}{a^3+b^3+c^3}}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Determine the point on the plane $4x-2y+z=1$ that is closest to the point $(-2, -1, 5)$ Determine the point on the plane $4x-2y+z=1$ that is closest to the point $(-2, -1, 5)$. This question is from Pauls's Online Math Notes. He starts by defining a distance function:
$z = 1 - 4x + 2y$
$d(x, y) = \sqrt{(x + 2)^2 + (y ... | Here is a simple solution using tools from middle school for the computation:
Denote $x,y,z$ the coordinates of the orthogonal projection of the point $(-2,-1,5)$
It satisfies the equations of proportionality:
$$\frac{x+2}4=\frac{y+1}{-2}=\frac{z-5}1$$
This common ratio is also equal to
$$\frac{4(x+2)-2(y+1)+1(z-5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
... | Hint:
$$\int\dfrac{x - 1}{x^2 + x + 1}\,\mathrm dx\equiv\int\dfrac{2x + 1}{2(x^2 + x + 1)} - \dfrac{3}{2(x^2 + x + 1)}\,\mathrm dx$$
Then, let $u = x^2 + x + 1\implies\mathrm du = 2x + 1\,\mathrm dx$. So,
$$\int\dfrac{2x + 1}{x^2 + x+ 1}\,\mathrm dx\equiv\int\dfrac1u\,\mathrm du.$$
Notice that $$\int\dfrac1{x^2 + x + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Minimum three variable expression $\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$ If $a,b,c \ge 0$ are distinct real numbers, find the minimum value of
$$\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$$
What I did: I used $a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2$, but no success. A friend of mine said the minimum... | Your friend was right!
Indeed, we can assume that $a<b<c$.
Thus, it's enough to prove that
$$(a+b+c)^3(c-a)(c-b)(b-a)\leq bc(c-b)(b+c)^3$$ or
$$a((b+c)(b^3+c^3)+(b+c)(2b^2+bc+2c^2)a-bca^2-2(b+c)a^3-a^4)\geq0,$$ which is obvious.
Also, $$(a^2+b^2+c^2)^3\geq(b^2+c^2)^3$$ and after assuming $c=xb$ it's enough to find $$\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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sum of binomial series with alternate terms
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$
what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displays... | With the following the purpose was to use slightly different functions
from what we saw in the answer by @MarkusScheuer. Start from
$$\sum_{k=0}^n (-4)^k {n+k\choose 2k}
= \sum_{k=0}^n (-4)^k [z^{n-k}] \frac{1}{(1-z)^{2k+1}}
\\ = [z^n] \frac{1}{1-z}
\sum_{k=0}^n (-4)^k \frac{z^k}{(1-z)^{2k}}.$$
Here the coefficient ext... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63}$. Why is this assumption necessary? Question:
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63} $. Why is this assumption necessary?
Proof:
Since $\gcd(a,3)=1$ $\Leftrightarrow a\equiv 1\pmod 3$ $\Leftrightarrow a^7\equiv 1\pmod3\equiv a\pmod3$
Then using Fer... | $\gcd(a,7) = 1$ is not necessary. If $\gcd(a,7)=7$ then $a^7\equiv a \equiv 0 \pmod 7$ and no theorem needed.
FLT says if $\gcd(a,7)=1$ then $a^6\equiv 1\pmod 7$ and from that we conclude that $a^7 \equiv a \pmod 7$ ALWAY; trivialy so if $7|a$--- and by FLT if $\gcd(a,7)=1$.
But we need $\gcd(a,3) = 1$.
Note: $3^7 \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{1}{(p+q)^3}\left(\frac{1}{p^3}+\frac{1}{q^3}\right)+...$
Simplify $\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)$ if $p\ne -q, p\ne0$ and $q\ne 0$.
$\dfrac{1}{(p+q)^3}\big(\dfrac... | Big hairy guns:
$\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)=$
$\frac {p^3 + q^3}{p^3q^3(p+q)^3} + 3\frac {p^2 + q^2}{p^2q^2(p+q)^4}+6\frac {p+q}{pq(p+q)^5}=$
$\frac {(p+q)^2(p^3+q^3) + 3(p+q)pq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiable function $f(x)=4x^7 −14x^4 +30x−17$ I am trying to prove that the function $f:\Bbb R→\Bbb R$, $f(x)=4x^7 −14x^4 +30x−17$,is injective. To do this I need to prove it is differentiable from first principles. I can then prove its derivative is strictly increasing to show it is injective. Any help on the pro... | The OP is asking for a first-principles derivation of the derivative of a specific polynomial. Here's one way to do it, using the algebraic identity $(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y^2+\cdots+xy^{n-2}+y^{n-1})$.
If $f(x)=4x^7-14x^4+30x-17$, then
$$\begin{align}
{f(x)-f(a)\over x-a}
&={4(x^7-a^7)-14(x^4-a^4)+30(x-a)\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve the following trigonometric equation for x
Solve $$ \sin^2x \tan x + \cot x \cos^2 x - \sin2x = 1 + \tan x + \cot x $$
I converted the whole equation in $\sin x$ and $\cos x$ and after rearranging a bit I got $$ (\sin x + \cos x)(1-\sin x \cos x) - 2 \sin^2x \cos ^2x = \sin x \cos x +1 $$
I supposed $\sin x+\c... | Let for the sake of writing less stuff $c=\cos x$ and $s = \sin x$.
Case 1: $cs \ne0$; we can multiply your equation by $cs$ without losing roots. We get
$$s^4 + c^4 - 2 s^2c^2 = sc + s^2+c^2$$
after that
$$(s^2+c^2)^2 - 4 s^2c^2 = sc + 1$$
or
$$4s^2c^2+sc=0.$$
Since $sc\ne0$, we get $4sc+1=0$, or, in terms of $x$, $ 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I show this series diverges? The series in question is $$\sum_{n=1}^\infty\frac{n^2+1}{2n^2+5}$$
If $a_n$ is the $n$'th term of this sum, then $a_n \rightarrow \frac{1}{2}$ as $n\rightarrow \infty$. I believe this implies that the series doesn't converge - since this is effectively the negation of the statement ... | Just for your curiosity.
As already said, the problem is over as soon as you showed that the term tends to anything which is not zero.
If fact, we can even approximate the partial sums
$$\frac{n^2+1}{2n^2+5}=\frac{1}{2}-\frac{3}{4 \left( n^2+\frac52\right)}$$
Now, use partial fraction decomposition
$$\frac{1}{ n^2+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inductive proof of inequality involving summation and nested exponents What is the best way to approach this problem?
$$\sum_{k=1}^{n} \frac{1}{2^{k^2}} \leq
1-\frac{1}{2^{n^2}}$$
So far I have proved this works for a base case where $n=1$
$$\sum_{k=1}^{1} \frac{1}{2^{k^2}} \leq
1-\frac{1}{2^{1^2}}$$
$$\frac{1}{2} \l... | You have the inductive case of $n = m$ giving
$$\sum_{k=1}^{m} \frac{1}{2^{k^2}} \leq 1-\frac{1}{2^{m^2}} \tag{1}\label{eq1A}$$
A common method to use induction with inequalities is to go from the inductive case, given above, to the case you're trying to prove, i.e., for $n = m + 1$, by changing the values so one side ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find sum of geometric-like series with binomial coefficients using complex analysis Studying analytic number theory, I stumbled across the problem of finding the sum of the series
$\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{1}{5}\right)^n$
A professor gave me the hint of "using basic complex analysis" but honestly, I ... | Following the hints we introduce
$${2n\choose n} =
\frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{(1+z)^{2n}}{z^{n+1}} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z}
\sum_{n\ge 0} \frac{(1+z)^{2n}}{5^n\times z^{n}} \; dz.$$
We must now determine $\varepsilon$ for the geometric series t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that if $a,b,c > 0$ and $a + b + c = 1$, we have: $\frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$ Prove that if $a,b,c > 0$ such that $a + b + c = 1$, then the following inequality holds:
$$S = \frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$$
Wh... | Because by AM-GM we have:
$$\frac{a^2}{a^3+5}\leq\frac{a}{4}$$
$$a^3+2\cdot2.5\geq3\sqrt[3]{a^3\cdot2.5^2}>4a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\... | Hint : First use: $a^2 + 4 \ge 4a$ for each term on the left. Then split the log and show next that the left is $\ge 0$. Equality is at $ x = y = z = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one e... | Let me continue TonyK's comment. If $\sqrt{2}=1.b_1b_2\ldots b_{n-1}11\ldots11b_{2n+1}\ldots$, then
$$
\sqrt{2}=\frac{k}{2^{n-1}}+\left(\frac{1}{2^n}+\ldots+\frac{1}{2^{2n}}\right)+r,
$$
where $k=(\overline{1b_1b_2\ldots b_{n-1}})_2$ is a positive integer and $r\in(0,\frac{1}{2^{2n}})$ (since $\sqrt{2}$ is not rationa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Probability problem; if sum and product are not divisible by number Out of set of numbers {10,11,...,100} three are being chosen randomly. Find these probabilities:
*
*At least one out of three are divisible by 6
*Exactly two are not divisible by 4
*Their sum is not divisible by 4
*Their product is not divisible ... | Well, I took the time to work out the hard part. Here is a table:
$$\begin{array}{cccc|c|c}0&1&2&3&\text{Number}&\text{Total}\\ \hline
23&22&23&23&&\\ \hline
3&0&0&0&\binom{23}{3}&1771\\
1&1&0&1&\binom{23}{1}\binom{22}{1}\binom{23}{1}&11638\\
1&0&2&0&\binom{23}{1}\binom{23}{2}&5819\\
0&2&1&0&\binom{22}{2}\binom{23}{1}&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Prove using induction that $1^2 + 3^2 + 5^2 + . . . + (2n + 1)^2 = (n + 1) ∗ (2n + 1) ∗ (2n + 3)/3$ My question reads as follows:
Prove using induction that
$$1^2 + 3^2 + 5^2 + . . . + (2n + 1)^2 = (n + 1) \times (2n + 1) \times (2n + 3)/3.$$
Firstly, I must prove the base case. For $n=0$, my LHS $= 1$ and my RHS $= 1... | $\frac {(k+1)(2k+1)(2k+3)}3 + (2k+3)^2 = \frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 \iff$
$\frac {(k+1)(2k+1)(2k+3)+ 3(2k+3)^2}3= \frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 \iff$
$(k+1)(2k+1)(2k+3)+ 3(2k+3)^2 = (k+2)(2(k+1)+1)(2(k+1)+3)\iff$
$(2k+3)[(k+1)(2k+1) + 3(2k+3)] = (k+2)(2k+3)(2k+5)\iff$
$[(k+1)(2k+1) + 3(2k+3)]=(k+2)(2k+5)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Differentiate $((x^{2}-1)f_{n-1}(x))^{(n)}$.
Differentiate $((x^{2}-1)f_{n-1}(x))^{(n)}$.
Using Leibniz rule, I obtain the following:
$\frac{x^{2}-1}{2^{n}n!}f_{n-1}^{(n)}(x)+\frac{x}{2^{n-1}(n-1)!}f_{n-1}^{(n-1)}(x)+\frac{1}{2^{n}(n-2)!}f_{n-1}^{(n-2)}(x)$
I am trying to show for Legendre polynomial $\phi_{n}(x)$ ... | In order to prove that
$$\frac{x^2-1}{n}P_n' = xP_n - P_{n-1}$$
a different route should be followed.
Two other formulas will be used in the derivation below:
\begin{equation}\label{1}
(n+1)P_{n+1} = (2n+1)xP_n-nP_{n-1}
\end{equation}
and
\begin{equation}\label{2}
P_{n+1}'-P_{n-1}' = (2n+1)P_n
\end{equation}
Here we go... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
| The trivial solution $n=0$ is seen by inspection.
For $n \ne 0$, $y^2=n^6+n^4+1 \Rightarrow (y+1)(y-1)=n^6+n^4$.
Now $(y+1)-(y-1)=2 \wedge \gcd{(y+1),(y-1)}=1,2$ so we must resolve $n^6+n^4$ into two factors that differ by $2$ and have at most one factor of $2$ in common.
The factors $n^4,(n^2+1)$ are relatively prime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:
Let $a,b,c$ be non-negative real numbers. Prove that
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
Do you have any hints to solve this inequality? Any hints would be fine.
I t... | First, note that
$$2(1-a+a^2)(1-b+b^2)=1+a^2b^2+(a-b)^2+(1-a)^2(1-b)^2 \geqslant 1+a^2b^2$$
Hence it is enough to show that
$$3(1+a^2b^2)(1-c+c^2) \geqslant 2(1+abc+a^2b^2c^2)$$
Considering the above as a quadratic in $c$, it is clear when $c=0$ this is true, so we just need to confirm the discriminant doesn't get posi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Calculus - limit calculate Calculate the limit not using L'Hospitale rule:
$\lim\limits _{x \to 2} \frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$
I tried multiplying for both conjugates but I got nothing. I think I have to make something with $\sqrt{x^3-4}\cdot \sqrt{x^2+12}$, but I don't know what to do.
| \begin{align}
\lim_{x\to2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}&=\lim_{x\to2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}\color{blue}{\cdot\frac{\sqrt{x^2+12}+4}{\sqrt{x^2+12}+4}\cdot\frac{2+\sqrt{x^3-4}}{2+\sqrt{x^3-4}}}\\
&=\lim_{x\to2}\frac{2+\sqrt{x^3-4}}{\sqrt{x^2+12}+4}\cdot\frac{(x^2+12)-(16)}{(4)-(x^3-4)}\\
&=\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this... | Let $u= \cos x$, $du = - \sin x \ dx$:
$$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
$$= -\int_{1}^{0} \frac{du \ (u^5)}{(1-2(1-u^2)u^2)^2}$$
$$= \int_{0}^{1} \frac{u^5}{(1-2u^2+2u^4)^2} \ du$$
Then substitute again: let $v = u^2, dv = 2u\ du$:
$$= \frac{1}{2} \int_{0}^{1} \frac{v^2 \ dv}{(1-2v+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$
takes all real values for $x\in R$ are:
My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My... | The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$
$\dfrac{ax^2+3x-4}{a+3x-4x^2}$ will take all the values for $x \in \mathbb R$ only if the denominator is not equal to zero!
So, For $a + 3x - 4x^2 \neq 0; \iff -3 \pm \sqrt{9 +16a} \neq 0$
We need to check all the poi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\arctan{x}+\arctan{y}$ from integration I was trying to derive the property
$$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$
for $x,y>0$ and $xy<1$ from the integral representation
$$
\arctan{x}=\int_0^x\frac{dt}{1+t^2}\,.
$$
I am aware of "more trigonometric" proofs, for instance using that $\tan{(\alpha+\beta)}... | We want show that
\begin{eqnarray*}
\int_x^{ \frac{x+y}{1-xy}} \frac{dt}{1+t^2} = \int_{0}^{y} \frac{du}{1+u^2}
\end{eqnarray*}
that's to say the LHS is actually independent of $x$.
The substitution
\begin{eqnarray*}
t=x+ \frac{u(1+x^2)}{1-ux}
\end{eqnarray*}
will do the trick.
The limits are easily checked and we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Inequality for Olympiad students Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that
$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}} \ge 6$
My attempts you can see here: https://scontent-xsp1-2.xx.fbcdn.net/v/t1.0-9/90231854_2782055341915789_3356982430379540480_n.jpg?_nc_cat=101&_nc_sid=8... | Another way (L.Hadassy, Y.Ilany).
If $a\geq b\geq c$ we have $$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3b+\frac{1}{b}}$$ because it's $$(a-b)(b-c)\geq0.$$
Thus,
$$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}.$$
If $a\geq c\geq b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solutions of the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ I have seen the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ in many places and the answer is $x=2$ which is obtained by making the substitution for $x$ inside the radical sign i.e $x=\sqrt{2+x}$ which renders the quadratic $x^2 -x -2=0$
having the so... | For $x=\sqrt{2 + \sqrt{2 +x}}$ you got the bi-quadratic $x^4-4x^2-x+2=0$ with solutions
$$x=-1,2, \frac{-1-\sqrt{5}}{2} , \frac{-1+\sqrt{5}}{2}.$$
Now, the first and third values ($x=-1, \frac{-1-\sqrt{5}}{2}$) can not be a solution because they are negative.
The fourth value is approximately $0.61803398875,$ but we k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $2003$ divides the numerator of $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335}$. I came up with this problem on a book in number theory :
Let $p$ and $q$ be natural numbers such that
$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335} = \frac{p}{q} \,.$$
Show ... | Suppose that $k>1$ is a positive integer such that $3k-1$ is a prime number. Then the numerator of the alternating series
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2k-1}$$
is divisible by $3k-1$. For example when $k=2$, we have $2k-1=3$ and $$1-\frac12+\frac13=\frac{5}{6}$$ with numerator divisible by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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