Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5... | Your equation $4$ is not right.
It would be $ q=-27p-141 $
Then you will equate equation $4$ and $3$
$$ -27p-141=p-1 $$
You will get $p$ as $-5$.
Substitute $p$ as $-5$ in equation $3$ and we will get $q$ as $-6$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute (if it exists) the one-sided limit of $\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $
Compute (if it exists) the one-sided limit of $$\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $$
I'm finding the question really confusing, especially the one-sided limit of... | Since$$|x|<1\implies\frac1{1+x}=1-x+x^2-x^3+\cdots,$$you have (again, if $|x|<1$)\begin{align}\frac1{(1+x)^2}&=-\left(\frac1{1+x}\right)'\\&=1-2x+3x^2-4x^3+\cdots\\&=\sum_{k=1}^\infty(-1)^{k+1}kx^{k-1}\end{align}and therefore$$|x|<1\implies\frac x{(1+x)^2}=-\sum_{k=0}^\infty(-1)^kkx^k.$$So,$$\lim_{x\to1^-}\sum_{k=0}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872694",
"timestamp": "2023-03-29T00:00:00",
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Recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n + n + 3$ with $a_{0} = 1$ and $a_{1} = 4$ This is a nonhomogeneous recurrence relation, so there is a homogeneous and a particular solution.
Homogenous:
$a_n - 4a_{n-1} + 3a_{n-2} = 0$
$r^2 - 4r + 3 = 0$
$(r - 3)(r - 1)$
$a_n^h = \alpha(3^n) + \beta(1^n)$
This is whe... | So we have
$$
a_{\,n} - 4a_{\,n - 1} + 3a_{\,n - 2} = 2^{\,n} + n + 3 = q(n)
$$
and the solutions to the homogeneous equations are
$$
3^{\,n} ,\;1
$$
The homogeneous equation has constant coefficients and
$$
q(n) = 2^{\,n} + \left( {n + 3} \right)
$$
is the sum of two terms of the form
$$
c^{\,n} \cdot {\rm polyn... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \zeta(2p)a^{2p}$ I'm trying to solve the problem 14.3.9 (Applications of Fourier Series) from Arfken's Mathematical Methods For Physicists:
a) Show that the fourier expansion of $\cos(ax)$ is:
\begin{equation}
\cos(ax) = \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} ... | We have \begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \sum_{p=0}^{\infty} \dfrac{a^2}{n^2}\left(\dfrac{a^2}{n^2}\right)^p
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$
Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$
Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq
\frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$
My Solution.
We write the inequality as
$$\big... | Suppose $a+b+c=1,$ let $p=a+b+c,\,q=ab+bc+ca,\,r=abc$ then $p \leqslant \frac{1}{3}.$ We write inequality as
$$\frac{5(1-2q+3r)}{q-r} \geqslant \frac{1-2q}{q}+10,$$
or
$$(1+23q)r+2q(2-9q) \geqslant 0. \quad (1)$$
If $0 < q \leqslant \frac{2}{9}$ then $(1)$ is true.
If $\frac{2}{9} < q \leqslant \frac{1}{4},$ from condi... | {
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| The Contradiction method also helps here.
Indeed, let $b+c>2,$ $b=kx$, $c=ky$, where $k>0$ and $x+y=2$.
Thus, $$k(x+y)>2,$$ which gives $k>1$ and
$$3=b^2+bc+c^2=k^2(x^2+xy+y^2)>x^2+xy+y^2,$$ which is a contradiction because we'll prove now that $$x^2+xy+y^2\geq3.$$
Indeed, we need to prove that:
$$x^2+xy+y^2\geq3\left(... | {
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How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equatio... | The telescoping summation helps: $$a_n=\frac{1}{4}a_{n-1}+\frac{1}{4}\left(\frac{2}{3}\right)^{n-1},$$
$$\frac{1}{4}a_{n-1}=\frac{1}{4^2}a_{n-2}+\frac{1}{4^2}\left(\frac{2}{3}\right)^{n-2},$$
$$\frac{1}{4^2}a_{n-2}=\frac{1}{4^3}a_{n-3}+\frac{1}{4^3}\left(\frac{2}{3}\right)^{n-3},$$
$$\cdot$$
$$\cdot$$
$$\cdot$$
$$\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \c... | By the binomial theorem, $$9^{n+1}+7^{2n}=(10-1)^{n+1}-(50-1)^n=(10a+(-1)^{n+1})-(50b+(-1)^n)=10a-50b$$ since $n$ and $n+1$ have different parities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ wh... | we have to prove $$7\sum \frac{1}{1+a}+2\sum_{cyc}\frac{b^2}{b+ab}\ge 69/4$$
but by $AM\ge HM$ $$7\sum \frac{1}{1+a}\ge 63/4$$
also using $ab+bc+ca\le {(a+b+c)}^2/3=1/3$
$$2\sum_{cyc} \frac{b^2}{b+ab}\ge 2\frac{(a+b+c)^2}{ab+bc+ca+a+b+c}\ge 6/4$$
The conclusion is now obvious
| {
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Upper and Lower semicontinuity and right-left continuity. Let $f: [0,1]\rightarrow\mathbb{R}$ be a function. Is there any relation between upper-lower semicontinuity of $f$ and right-left continuity of $f$?
I mean, is there any implication?
Thanks in advance.
| The only implication is that both left + right continuity and upper + lower semicontinuity are equivalent (and, of course, are equivalent to continuity). Otherwise, we can create a table of counterexamples. Here, we define several real functions of a variable $x$, which possess the given continuity properties at $0$:
$... | {
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Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots
Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.
Obviously as soon as you factor this equation to $$(4x^... | Rewrite it as a polynomial in $a$:
$$ a^3 x + a^2 (x^2 + 1) + a ( 4x^3 + 8x^2 ) + (4x^4 + 8x^3 + 4x^2 + 8x) = 0.$$
The factorization almost immediately jumps out by observation:
$$ (a^2 + 4x^2 + 8x ) ( ax + x^2 + 1) = 0 $$
The idea of changing the variable is a common trick. It can be helpful when factoring (e.g. appl... | {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos... | Using Principal values
$$-\dfrac\pi3\le\cos^{-1}x-\dfrac\pi3\le\pi-\dfrac\pi3 $$
$$\cos^{-1}x-\dfrac\pi3=\begin{cases} \cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) &\mbox{if } \cos^{-1}x-\dfrac\pi3\ge0\iff x\le\cos\dfrac\pi3 \\
-\cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) & \mbox{if } x>\cos... | {
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Prove $f(x)=x^2+3x$ is Continuous at $x=a$ I am attempting to give a $\delta,\epsilon$ proof that the following function, $f(x)=x^2+3x$ is continuous at $x=a$, where $a$ is any real number. I have the following:
Given any $\epsilon > 0$, find a $\delta > 0$ such that if $|x-a|<\delta$ then $|(x^2+3x)-((a^2)+3a)|<\epsi... | Use the sequential criterion for continuity at $a$. Let $(a_n)_{n=1}^\infty$ be a real sequence with limit $a$. Now, it suffices to show $\lim_{n\to\infty}f(a_n)=f(a)$.
$$\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a_n(a_n+3)=\lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}(a_n+3)=a(a+3)=f(a)$$
| {
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)... | Let $C$ be the solutions. Since $C = -C$ so we can look for solutions with non negative real part.
Note that $2 = |2i| = |z+i - (z-i)| \le |z+i|+|z-i|$.
Let $f(x) = |x+iy+i| + |x+iy-i|$ and note that $f$ is continuous everywhere, differentiable for $ x > 0$ and $f'(x) >0$ for such $x$. It follows from the mean value th... | {
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Prove using the $\epsilon-\delta$ definition that $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$ for any $a > 0$, $a \in \mathbb{R}$. Scratch Work
Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$. Then for every $\epsilon > 0$, we aim to find a $\delta > 0$ such that
$$
0 < |x-a| < \delta \Rightarrow \left|\sin \... | I agree with user837206’s comment: your post is correct.
I suggest some tweaks:
You start with “Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$.” That’s the statement you are trying to prove. It’s just some notes, not the actual proof, so it’s not technically wrong. But make sure you are clear about what you hav... | {
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Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $
Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b = 0$, $\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \... | As $\vec a \perp \vec b$ like $\hat i \perp \hat j$, it's strategic to represent
$$ \vec c = \frac{1}{4}\vec a - \frac{\mu}{4}\vec b$$
Remaining two conditions are,
$$ (\vec b - \vec a)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)=0$$
and
$$ (\vec b - \vec a)\cdot(\vec b - \vec a) = 2^2(\frac{1}{4}\vec a + \frac{4-\... | {
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"timestamp": "2023-03-29T00:00:00",
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A simple binomial inequality seeking its proof from the book? Let $k\in \mathbb{N}$ and $1\leq a \leq k$ be fixed. Imagine we have two groups $G_{1}, G_{2}$ of $k$ indistinguishable elements each forming two sequences with indices $1, \dots , k$. Choose the element with index $a$ in $G_{1}$ and the element with index $... | By expanding the binomials, it can be shown directly that
$$
\frac{f(k,a,b)}{f(k,a,b-1)} = \frac{\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}}{\binom{a+b-3}{a-1}\binom{2k-b-a+1}{k-a}} \\
= \frac{a+b-2}{b-1} \frac{k-b+1}{2k-a-b+1} = (1 + \frac{a-1}{b-1})(1 - \frac{k-a}{2k-a-b+1})
$$
Interpreting this ratio as a function of a c... | {
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Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.
(2) Assuming that the condition hold for $n=k.$
(3) Proving that it holds for $n=k+1$
$$3^{4(k+1... | An alternative strategy for this kind of problem is to consider the difference between consecutive terms.
Let $f(n)=3^{4n+2}+1$. Then $f(n+1)-f(n)=80 \cdot 3^{4 n + 2}$ is a multiple of $5$.
The claim follows by induction since $f(0)=10$ is a multiple of $5$.
(Actually, this proves that $f(n)$ is always a multiple of ... | {
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$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4... | Dividing both the numerator and the denominator by $\sqrt{n^3}$, you have
\begin{eqnarray}
&&\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}\\
&=&\frac{2}{\big(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}+2\big)\big( \sqrt{1-\frac1{n^2}}+1\big)}.
\end{eqnarray}
Now you can take the limit to ... | {
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"timestamp": "2023-03-29T00:00:00",
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Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can fro... | $2x+1|x^3-3x+2\iff 2x+1|8(x^3-3x+2)$
$\iff 2x+1|8(x^3-3x+2)-(2x+1)^3=-12x^2-30x+15$
$\iff 2x+1|3(2x+1)^2-12x^2-30x+15=-18x+18$
$\iff 2x+1|9(2x+1)-18x+18=27$
$\iff x\in \{-14,-5,-2,-1,0,1,4,13\}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose $x$ and $y$ are unequal real numbers
Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$
For the above question, I got $x$ not equal to $-1,0,1... | Tricky one.
Let $x+y = u$, $x-y=v$.
$$ xy=1 \Rightarrow u^2-v^2 =4$$
and cubing original equation :$$ \sqrt[3]{\dfrac{u}{v}} + \sqrt[3]{\dfrac{v}{u}} = u $$
gives
$$ \dfrac{u}{v} + \dfrac{v}{u} + 3u = u^3 $$
$$ \Rightarrow u^2 + v^2 + 3u^2v = u^4v$$
Substitute $u^2=4+v^2$ ,
$$(4+v^2) + v^2 + 3(4+v^2)v = (4+v^2)^2v $$
d... | {
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"timestamp": "2023-03-29T00:00:00",
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Distance between the curves from a particular point The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to
(A) 1
(B) 2
(C) 3
(4) 4
My appr... | The distance between the circle and $P(u,v)$ is the distance between $P$ and the centre minus the radius. So we have$$\left|\sqrt{(u+2)^2+v^2}-2\right|=\sqrt{(u-2)^2+v^2}$$Squaring,$$\begin{align*}&(u+2)^2+v^2+4-4\sqrt{(u+2)^2+v^2}=(u-2)^2+v^2\\\implies&2u+1=\sqrt{(u+2)^2+v^2}\\\implies&4u^2+4u+1=(u+2)^2+v^2\\\implies&... | {
"language": "en",
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Cubic Formula Doesn't Seem to Work for $x^3+3x^2-50x-52=0$ So I'll jump right into it. I've derived the following formula set for solving cubics of the form $ax^3+bx^2+cx+d=0$:
$$z_{k\pm}=\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}\pm\sqrt{\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+4b^3d}{108a^4}}}$$
$$x=z_{k\pm}-\frac{\left(\frac... | For negative radicands, the 6 $z$-values translates into 3 distinctive $x$-values as well.
Take the example $x^3+3x^2-50x-52=0$ given in the post, $z$‘s are evaluated as
$$z_{k\pm} =\sqrt[3]{\pm \sqrt{-\frac{53^3}{27}}}= \pm\sqrt{\frac{53}3}e^{i \frac{1+4\pi k}6},\>\>\>k=0,1,2$$
and the corresponding $x$’s are
$$x_{k\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3906005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1... | Consider the identity $$(2k+1)^2=4k^2+4k+1$$ which relates the sum of the odd squares to the sum of the squares.
The extra terms $4k+1$ are easily dealt with with a telescoping,
$$(k+1)^2-k^2=2k+1.$$
Altogether,
$$1^2+3^2+ \dots +(2n+1)^2=4\frac{n(n+1)(2n+1)}6+2(n+1)^2-1-n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| $5^{6n}-2^{2n}=125^{2n}-2^{2n}$
$=(125-2)\left(125^{2n-1}+125^{2n-2}2+125^{2n-3}2^2+\cdots+125^22^{2n-3}+125\cdot2^{2n-2}+2^{2n-1}\right)$
is divisible by $3$, because $125-2$ is.
Now substitute $m=n+1$ to show that $5^{6m+6}-2^{2m+2}$ is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n\leq \frac{x}{2}$ using induction I was asked to show that
For every $n\in\mathbb{N}$ and $x\in [0,1]$ the following is true:
\begin{equation*}
\left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n\leq \frac{x}{2}.
\end{equation*}
However, I found a mistake in my proof and... | Rearrange the induction hypothesis so it reads $\left(1-x+\dfrac {x^2}2\right)^k \le \dfrac x2 + (1-x)^k$.
Now the induction step can go as follows:
\begin{align}\left(1-x+\frac {x^2}2\right)^{k+1} - (1-x)^{k+1} &\le \left(1-x+\frac {x^2}2\right)\left(\frac x2 + (1-x)^k\right)-(1-x)^{k+1}\\
&=\frac x2-\frac {x^2}2+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $\alpha$ be a root of $(x^2-a)$ and $\beta$ be a root of $(x^2-b)$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$. QUESTION: Let $K$ be a field of characteristic different of $2$. Let $F$ be a splitting field for $(x^2-a)(x^2-b)\in K[x]$. Let $\alpha$ be a root of $x^2-a$ and $\beta$ be a root of ... | Once you know that $[ K( \alpha,\beta ) : K ]=4$, with basis $\{1,\alpha,\beta,\alpha\beta\}$, you can proceed as follows:
$$
\begin{pmatrix} 1 \\ \gamma \\ \gamma^2 \\ \gamma^3 \end{pmatrix}
=
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ a+b & 0 & 0 & 2 \\ 0 & a+3b & 3a+b & 0
\end{pmatrix}
\begin{pmatrix} 1 \\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\si... | The set $E$ is a ball of radius ${1\over2}$, centered at $\left({1\over2},0,0\right)$. The integrand
$$f(x,y,z):={yz\over x^2+y^2+z^2}$$
satisfies $f(x,-y,z)\equiv-f(x,y,z)$. This means that $f$ is odd with respect to the symmetry plane $y=0$ of $E$. It is then obvious that the requested integral has value $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\e... | With a couple of corrections, your answer seems consistent:
$z=\dfrac{1\pm i}2\implies a=\color{red}\pm\sqrt{\dfrac{1\pm i}2}$
$\pm i+4b^2=1\implies 4b^2=1\color{red}\mp i\implies b=\color{red}\pm\dfrac{\sqrt{1\color{red}\mp i}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How do I prove that if $n$ is greater or equal to $4$, then $(2n)!$ is greater than or equal to $10^n$? I used the method of induction to prove this.
For the basis step, when $n=4$, the statement is $[2(4)]!≥10^4$ which is the same as $40320≥10000$ which is true.
Next, for the inductive step, we show $\mathrm{S}_k$ imp... | With induction but more clear:
You have proved the base case, so let's continue.
Let $(2n)!\ge 10$. Then:
$$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!=(4n^2+6n+2)(2n)!\ge (4n^2+6n+2)\cdot 10^n$$
Clearly $4n^2+6n+2 \ge 10$ for $n \ge 1$ so:
$$(2(n+1))! \ge (4n^2+6n+2)\cdot 10^n \ge 10\cdot 10^n=10^{n+1}$$
$$(2(n+1))! \ge 10^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3931938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Recursive relation induction problem: $c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$ This is part of problem 18 of chapter 13 of Michael Spivak's Calculus:
Define $c_n = \int_{0}^{1}x^n dx$
(b) Prove: $$c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \c... | Note that $$\sum_{k \ge 0} a_{2k} = \sum_{k \ge 0} \frac{1+(-1)^k}{2}a_k.$$
Taking $a_k=\binom{p+1}{k}\frac{1}{k+1}$ yields
\begin{align}
\frac{1}{2^{p+1}}\sum_{k\ge 0} \binom{p+1}{2k}\frac{1}{2k+1}
&=\frac{1}{2^{p+1}}\sum_{k\ge 0} \frac{1+(-1)^k}{2} \binom{p+1}{k}\frac{1}{k+1} \\
&=\frac{1}{2^{p+1}(p+2)}\sum_{k\ge 0} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$
Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
{\rm d}t=\dfrac{\pi }{\sqrt{2}}$.
We can obtain
$$\lim_{x \to 1}\int_0^... | Finally I was able to handle this problem. It is just tedious calculation. Let $u=\frac{f(t)}{f(x)}$ and then
\begin{eqnarray}t&=&1-\frac{(1-x)(x-1+\sqrt{(1-x)^2+8u(1+x)}}{2u(1+x)},\\
dt&=&-\frac12\frac{ (x-1) \left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)}{u^2(1+x)\sqrt{(1-x)^2+8u(1+x)}}du.
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Nicomachus theorem proof - what did I do wrong? The exercise asked me to proof
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$...$
I formulate the equation as
(1)$$a^3 = \sum_{i=0}^{a-1} (a-1)a+1+2i$$
Proof base case:
$$1^3=\sum_{i=0}^{1-1} (1-1)(1)+1+2i=1$$
Proof (a+1) case:
(2) $$(a+1)^3=a^3+3a^2+3a+1$$
(3) $$(a+1)^3=\sum_{i=0}^{(a... | There are simpler ways to prove this relation.
The sum $\displaystyle S(n)=\sum\limits_{i=0}^n i=\frac{n(n+1)}{2}$ is well known and easy to prove, for instance
$\begin{align}2S(n)&=\bigg(1+2+\cdots+(n-1)+n\bigg)+\bigg(n+(n-1)+\cdots+2+1\bigg)\\&=\bigg((n+1)+(n+1)+\cdots+(n+1)\bigg)\\&=n(n+1)\end{align}$
Then notice $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that determinant is zero iff $x= 2, 3$
Prove that
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix}=0\Leftrightarrow x= 2, 3$$
I see that if determinant is zero, two lines must equal, in this problem, there are between line 1 and line 2, 3, but how can I prove there is no other sol... | Why not to expand the determinant and find all the solutions?
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix} =
\begin{vmatrix}
1 & x & x^{2}\\
0 & 2-x & 2^{2}-x^2\\
0 & 1 & 5
\end{vmatrix} =
\begin{vmatrix}
2-x & 2^{2}-x^2\\
1 & 5
\end{vmatrix}$$
$$=10-5x-(4-x^2) = x^2-5x+6=(x-2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Elliptic Curve Discriminant Here is my attempted approach to prove that the discriminant $\triangle = 4a^3+27b^2$ of an elliptic curve in the form of $y^2 = x^3 + ax +b$ is zero. I have a problem at the end which doesn't bring me to the expected conclusion. Below is the process, please let me know the error in the proc... | When $3x^2 + a \ne0$ and $2y = 0$, the curve is still nonsingular, its just that the slope of the tangent line is infinite.
The singularity occurs when
$$3x^2 + a = 0 \text{ and } 2y =0$$
simultaneously i.e.
$a = - 3x^2$ so $0 = x^3 + (-3x^2)x + b$
giving
$$ b = 2x^3$$
this is what leads to the equation
$$27b^2 + 4a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2} \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+... | Note that\begin{align}\int_1^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx&=-\int_1^0\frac{1+y^{-2}}{1+y^{-4}}\left(-\frac1{y^2}\right)\,\mathrm dy\\&=\int_0^1\frac{1+y^2}{1+y^4}\,\mathrm dy\end{align}and that therefore\begin{align}\int_0^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx&=\int_0^1\frac{1+x^2}{1+x^4}\,\mathrm dx+\int_1^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Apply product and chain rule to differentiate with respect to one variable I have the following:
$$
u = \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}} \phi\left(\frac{x}{t}\right)
$$
I need to find $u_t$
This equation was found by solving the characteristic equation:
$$
\frac{dx}{xt} = \frac{dt}{t^2} = \frac{du}{-\left(\frac{1}... | Your mistake is in the second term. It should be
$$\frac{x^2}{4t^{2}}\frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) = \frac{x^2}{4t^{2}}u$$
You just forgot the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Height of a cube edge from the floor
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?
I'm sorry for my bad English, but I try to illustr... | You can also use similar triangles. Let $P$ be the point on $MN$ directly below $A$, and let $Q$ be the point of intersection of $EN$ and line $AC$. Since $\angle CAM$ is a right angle, we have that
$$\triangle CAP \sim \triangle AMP \sim \triangle EAQ$$
$CA \,$ is clearly $\, 2a \sqrt{2}\, $ so $\, PC = a\sqrt{7}\,.$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$ $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$
I have no idea which story to form, I thought about $n$ students assuming n=4
picking 4 students for a committee with 3 roles. since each students has 3 options.
then on the other side.. $2^{0}\binom{4}{0} + 2... | The "binomial theorem" says that $(x+ y)^n= \sum_{i=0}^n \begin{pmatrix}n\\ i\end{pmatrix} x^i y^{n-i}$. If x= 2 and y= 1 that becomes $(2+ 1)^n= \sum_{i=0}^n\begin{pmatrix}n \\i \end{pmatrix} (2^i)(1^{n-i})$ or $3^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix} 2^i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{... | Now @Infinity_hunter has explained your error, note that $\frac34-\frac{3x+2}{4x+3}=\frac{1}{4(4x+3)}>0$, so $0<\left(\frac{3x+2}{4x+3}\right)^x<\left(\frac34\right)^x$ proves the limit is $0$ by squeezing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Proving inequality given some conditions I want to show that $-x^2y^2-y^4+4y^2 \geq 0$, if $x^4+2y^2 \leq 4$ and $-x^2y^2-y^4+4y^2 \leq 0$, if $x^4+2y^2 \geq 32$. Is there a factoring trick that helps to prove this explicitly?
| *
*For the first question you need to prove
$$
-2x^2-2y^2+8 \geq 0
$$
with the condition $-2y^2 \geq x^4 -4$. This condition implies $x^2 < 2$ since otherwise it can never be fulfilled. Combining the two gives the sharper $
-2x^2+x^4+4 \geq 0
$ or $(x^2-1)^2 +3\ge 0$, done.
*For the second question you need to prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other. Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.
I'm not sure how t... | $f(x) = x^2 +bx+ 9;\;g(x) = x^2 +ax+c$
The sum of the roots of $f(x)$ is $s=-b$ and their product is $p=9$
The sum the roots of $g(x)$ is $-a$ and their product is $c$
So we must have $a=-b;\;c=9$
$f(x)=x^2+bx+9;\;g(x)=x^2-bx+9$
$h(x)=f(x)+g(x)=2x^2+18=0\to x=\pm 3i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Probability to get 5 successes in at most 7 trials The night before the party, you plan to rehearse the song you are to sing and would like to sing it perfectly 5 times before sleeping. Since you are a great singer, the probability of you singing it perfectly on each attempt is 5/6. Suppose your attempts are mutually i... | If the problem is asking for the $5^{th}$ success in the $7^{th}$ trial then your approach is incorrect, continue reading for the correct solution. If the question is like the body "$5^{th}$ success in $7$ trials then what you did is correct in terms of choosing the binomial distribution, not the arithmetic, I think th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perf... | Here's another way. If $a+bq+cq^2=0$, then $a=-(bq+cq^2)$, hence $a^2=b^2q^2+2bcq^3+c^2q^4$. So if $q^3=p$, then we have two equations:
$$\begin{align}
a+bq+cq^2&=0\\
(a^2-2bcp)-c^2pq-b^2q^2&=0
\end{align}$$
Multiplying both sides of the first equation by $b^2$ and the both side of the second equation by $cp$, and then... | {
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"url": "https://math.stackexchange.com/questions/3957348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Two variables function limit In an exercise of a limit of a function of two variables in the solution I read this inequality:
$$ \frac{x^2y^2}{(x^2+y^2)^\frac{3}{2}} \le \frac{1}{2}\sqrt{x^2+y^2} $$
how did they arrive at this result?
| That inequality is equivalent to$$x^2y^2\leqslant\frac12\sqrt{x^2+y^2}(x^2+y^2)^{3/2}=\frac{(x^2+y^2)^2}2.$$Besides,$$(x^2+y^2)^2-2x^2y^2=x^4+y^4\geqslant0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$ $$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$
I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ w... |
Thanks to @CalvinLin for pinpointing that we cannot "assume WLOG" that the five variables are sorted as $a \le b \le c \le d \le e$.
We can still assume that the maximum element among them is $e$ $^{(*)}$. Then, by comparing $\color{red}{\text{red}}$ variables, we observe
*
*$$3d = (\color{red}e + a + b)^3 \ge (a +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank y... | One can use trigonometric functions to avoid squaring both sides to get a degree 4 polynomial. Clearly $x>1$. Let $x=\sec t$, $t\in(0,\frac{\pi}{2})$. Then the equation becomes
$$ \sec t+\csc t=\frac{35}{12} $$
or
$$ \frac{\sin t+\cos t}{\sin t\cos t}=\frac{35}{12}. $$
Squaring both sides gives
$$ \frac{1+\sin(2t)}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution... | A very nice question.
You can think in this way: if we denote
$$x+y = s\\
x^3 + y^3= t\\
x^4 + y^4 = u$$
$s$, $t$, $u$ depend on only two parameters $x$, $y$, so they have only two degrees of freedom. Therefore, there must be some relation between $s$, $t$, $u$. It can be obtained by eliminating $x$, $y$ from the above... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Summation of integration terms If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then
(A) $I>\ln99$
(B) $I<\ln99$
(C) $I<\frac{49}{50}$
(D) $I>\frac{49}{50}$
The official answer is B and D
My approach is as follow
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k +... | You have
$$\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx \le \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x} = \int_1^{99} \frac{dx}{x} = \ln 99$$
hence B is true. Also
$$\begin{aligned}\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx &\ge \sum_{k = 1}^{98} \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Intuition on the concept of bounding a sum For example, in calculating the following limit
$$\lim_{n\to\infty}\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}$$
I know (because my professor wrote) that one of the methods of solving this limit is via the sandwich rule since we know the lower and upper bound
$$\frac{3n}{\sqrt{n^2+3... | A single term of the series obeys the inequality $$\frac{1}{\sqrt{n^2 + 3n}} < \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{\sqrt{n^2 + 1}}$$ for any integer $k \in \{1, 2, \ldots, 3n\}$, but the sum does not. Instead, you have to sum each of the expressions accordingly:
$$\sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 3n}} < \sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions
Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.
My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\... | Consider the function $$f(x)=\left(\log_2 x\right)^2-\log_{\sqrt 2} x-a+\sqrt{a+\log_2 x},$$ defined for all $x\ge 2^{-a}.$ Then we find that $$f'(x)=\frac{1}{2x\log_e 2\cdot \sqrt{a+\log_2 x}}\left(4\log_2 x\sqrt{a+\log_2 x}-4\sqrt{a+\log_2 x}+1\right).$$ Letting $M=\log_2 x,$ we see that $f'(x)=0$ whenever $16(M+a)(M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many variables in the multinomial expansion have an exponent that's different from 2? Multinominal: $(a + b + c + d)^{10}$
Question: How many variables in the multinomial expansion have an exponent that's different from 2?
I want to solve it using generating functions.
So we can write: $t_1 + t_2 + t_3 + t_4 = 10$,... | You can rewrite $$(1+x+x^3+x^4...)^4=(1+x)^4(1+x^3+x^5+x^7+...)^4$$
$$=(x^4+4x^3+6x^2+4x+1)(1+x^3+x^5+x^7+x^9+...)^4$$
Now first look at the $x^4$ term: the only way to add exponent up to $10$ is $4+3+3=10$, therefore we have ${4\choose 2}=6$ ways.
Next look at $4x^3$ term, the only way is $3+7=10$, therefore we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966608",
"timestamp": "2023-03-29T00:00:00",
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Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$
Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.
I tried solving this question through substitution, e.g. $a_1=\... | The case $x_1=x_2=...=x_n=1$ is trivial, and for $n >2$, is the only case when equality holds. We prove this below.
Suppose the $x_i$s are not all $1$. Then, there exists $i,j \in \{1,2,..,n\}, i \neq j$, such that $x_i<1<x_j$. Replace the pair $(x_i \ ,x_j)$ with $(x_i'\ ,x_j')$ , such that:
$$x_i'=1, \ x_j'=x_ix_j.$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Difficult Cauchy Problem Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried... | Proposed inequality can be rewriten like this:$$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$$
By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^2$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we make it easy to solve quadratics without the Quadratic Formula? For example ,
$$5x^2 - 7x-2=0$$
Then, we only need to find is when
$$5x^2-7x=2$$
$$x(5x-7)=2$$
Since that when we get $2-2=0$.
Is there any way we can find them easily without solving quadratics using formula?
I am just getting very enthusiast... | Unfortunately, your strategy does not help that much because it is very difficult to solve
$$
x(5x-7)=2
$$
without simply returning to the original equation. This is because there is a $2$ on the RHS. The reason we want to be able to write a quadratic in the form
$$
(x-p)(x-q)=0
$$
is so that we can exploit the zero-pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the density function of $Y = X^2 - 1$ for a piecewise-linear r.v. $X$ A density function $f$, of a continuous random variable $X$, is given by:
$$
f(x) = \left\{
\begin{array}{lc}
\frac{1}{2} & \quad 0 < x < 1 \\
\frac{x}{3} & \quad 1 \leq x < 2 \\
\ 0 & \text{else}
... | First note that Y can take values between -1 and 3. If we find the cumulative probability for Y - let's call it $G(y) = p(Y \le y)$ - then its derivative will be the density function.
For $0 < x < 1$, $y = x^2 - 1$, so $-1 < y < 0$. For $y$ in this range,
$p(Y \le y) = p(X^2 - 1 \le y) = p(X \le \sqrt{y+1}) = \frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Proving an inequality with n positive real numbers Is there any alternative way of proving
$$\sum_{k=1}^n\frac{a_k-1}{a_k+1}\geq\frac{\prod_{k=1}^n a_k -1}{\prod_{k=1}^na_k+1}$$ for any positive real numbers $a_1,a_2,\cdots,a_n,$ greater than or equal to 1, other than the one given in one of the the answers; An ine... | We write the inequality as
$$\sum_{i=1}^n\frac{1}{a_i+1}\leq \frac{n-1}{2}+\frac{1}{a_1a_2\cdots a_n+1}. \quad (1)$$
For $n=1$ the inequality become
$$\frac{1}{a_1+1}+\frac{1}{a_2+1} \leqslant \frac{1}{2} + \frac{1}{a_1a_2+1}, \quad (2)$$
or
$$\frac{(a_1-1)(a_2-1)(a_1a_2-1)}{2(a_1+1)(a_2+1)(a_1a_2+1)} \geqslant 0.$$
Wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{200... | Your working leads to the answer. Here is the correction -
$a = \frac{1 + \sqrt{2009}}{2}$
$a^2 - 3 = \frac{999 + \sqrt{2009}}{2}$
$a (a^2 - 3) = 752 + 250 \sqrt {2009}$
$500 (a + 1) = 250 ( \sqrt {2009} + 3)$
$a(a^2-3) - 500(a+1) = 2$
So $(a^3 - 503a - 500)^5 = 32$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$ Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$
Using the Chinese Remainder Theorem, I obtained $x^2 \equiv 23 \pmod{7}$ and $x^2 \equiv 23 \pmod{11}$.
For $x^2 \equiv 23 \pmod{7}$, the answers are:
\begin{align}
x \equiv 3 \pmod{7} \\
x \equiv 4 ... | To get the given solution, combine congruences in your solution with the Chinese remainder theorem: $3\bmod7\land1\bmod11\iff45\mod77$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster? I have to calculate this integral:
\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}
My attempt:
Using, trigonometric substitution
\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow ... | $$x^2 \sqrt{a^2+x^2};\;\, x\to a \sinh u$$
$$dx=\cosh u\,du$$
$$\int x^2 \sqrt{a^2+x^2}\,dx=\int (a^2 \sinh ^2 u)(a \cosh u )\sqrt{a^2 \sinh ^2 u+a^2}\,du=$$
$$=a^4\int \sinh^2 u\cosh^2 u\,du=\frac{a^4}{4}\int\sinh^2 2u\,du=\frac{a^4}{8} \int (\cosh 4 u-1) \, du=$$
$$=\frac{1}{8} a^4 \left(\frac{1}{4} \sinh 4 u-u\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Write the equation $f(x,y)=0$ that describes the set $S$ Let $A=(1,1)$, $B=(1, -2)$ and $C=(5,-5)$.
Write the equation $f(x,y)=0$ with $f\in \mathbb{R}[x,y]$ that describes the set $S$ of points that have same distance of $A$ as to $BC$.
$$$$
I have done the following:
Let $P=(x,y)$ be a point of $S$.
The it must be $d... | We have
$$\sqrt{(x-1)^2+(y-1)^2}=\frac{|3x+4y+5|}{\sqrt{3^2+4^2}} \iff 25(x-1)^2+25(y-1)^2-(3x+4y+5)^2=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding maximum radius of inscribed circle I am working through a pure maths text book as a hobby. I am having difficulty with the second half of this problem:
The equal sides AB and AC of an isosceles triangle have length a, and the angle A is denoted by $2 \theta$. Show that the radius of the inscribed circle of the ... | HINT:
Let $$\frac{dr}{d\theta}=0$$
and then rewrite the numerator completely in terms of $\sin \theta$.
If you need any more help please don't hesitate to ask :)
EDIT:
As you correctly found, the numerator of the derivative is equal to
$$a(1-2\sin^2\theta-\sin^3\theta)$$
Hence, $$a(1-2\sin^2\theta-\sin^3\theta)=0$$
as... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $ find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$
Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\fra... | Let $r_1$ and $r_2$ be the roots of the equation
$$
a c x^{2}-b(c+a) x+(c+a)^{2}=0
$$
Then $$
\begin{aligned}
r_{1}+r_{2} &=\frac{b(c+a)}{a c}=\frac{b}{a}+\frac{b}{c}\\&=-(\alpha+\beta)-\frac{\alpha+\beta}{\alpha \beta} \\
&=-(\alpha+\beta)-\left(\frac{1}{\beta}+\frac{1}{\alpha}\right)\\&=-\left(\alpha+\frac{1}{\beta}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2] I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.
I tried multiple ways to underst... | The well ordering principle comes into play in trying to find the first $n$ where $n^3 > 3^n$.
We know $1^3 < 3^1$.
But can there be any $n^3 > 3^n$?
If so, there must be a first $n$ where $n^3 > 3^n$. But if $n$ is the first then it must be that $(n-1)^3 \le 3^{n-1}$.
Now hopefully we will be able to show $(n-1)^3 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} ... | Even if I think that @Tavish's solution is the most efficient, wht you did must work.
$$f(a) = \int_0^1 \frac{\log(x+a)} {x^2 + 1} \,dx$$
$$f'(a) = \frac{1} {a^2 + 1} \left(\log(a+1) - \log(a) - \ln(4)+ \frac{\pi}{4} a\right)$$ Now, you must use $a^2+1=(a+i)(a-i)$ to make
$$ \int f'(a)\,da=\frac{1}{8} \pi \log \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981454",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove a well-known inequality using Cauchy-Schwarz or AM-GM For $a;b>0$ and $ab \geq 1$ we have a well-known inequality:
$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2} \geq \dfrac{2}{1+ab}$
Which is equivalent to: $\dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)} \geq 0$ (true)
But is there a solution for this inequality by Cauchy-Schw... | Again not AM-GM or C-S (unless you count $a^2+b^2\geq2ab$) but I thought this was kinda sweet:
\begin{align*}
\frac{1}{1+a^2}+\frac{1}{1+b^2}=1-\frac{(ab)^2-1}{1+a^2+b^2+(ab)^2}&\geq1-\frac{(ab)^2-1}{1+2ab+(ab)^2}\\
&=1-\frac{ab-1}{ab+1}\\
&=\frac{2}{1+ab}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Simplified expression for centers of period-three Mandlebrot bulbs The Mandelbrot set contains three regions (two bulbs and a cardioid) with periods of three. These regions each contain a fixed point which is a root of the expression $x^3 + 2x^2 + x + 1$.
The fixed point at the center of the period-three cardioid has t... | Given that I know the real root of the function $x^3 + 2x^2 + x + 1$ is $-ρ^2$, I can factor that out to get $(x+ρ^2)(x^2 + (2 - ρ^2) x + (2 - ρ^6 + 3ρ^4 - 3ρ^2))$.
Using the quadratic formula on that second term, I get $\frac{ρ^2 ± \sqrt{-4 + 8ρ^2 - 11ρ^4 + 4ρ^6} - 2} {2}$.
Because $ρ^3 = ρ + 1$ and $ρ^4 = ρ^2 + ρ$, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab... | By AM-GM $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge 2(\sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} )=2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$
Now using $x^2+y^2+z^2\ge xy+yz+zx$ $$\begin{align*}2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 2(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Time period of a nonuniform oscillator I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the va... | Taking on board the comment of @Hans Ludmark, I believe I have found the answer but I'm not entirely sure if it is rigorous. Considering the discontinuity at $\theta=\pi$ I now re-examine the equation
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\fra... | Look at the plot below (first picture). The function can be inverted only in the interval where it is bijective.
Take the derivative
$$f'(x)=\frac{4}{3} \sin \left(\frac{1}{3} (2 x+1)\right) \cos \left(\frac{1}{3} (2 x+1)\right)=\frac{2}{3} \sin \left(\frac{2}{3} (2 x+1)\right)$$
$$f'(x)=0\to \sin \left(\frac{2}{3} (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$ I saw this twitter post that reads:
Find all the pairs of positive integers $(x,y)$ satisfying $$ x^2 + y^2 - 5xy + 5 = 0 . $$
I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's int... | This sort of thing has a clear description in terms of Conway's topograph; I find it more convenient to use equivalent form $u^2 + uv - 5 v^2.$ The outcome for the original problem is sequences (note that the rule deals with every other element). For instance,
$5\cdot 9 - 2 = 43$ and $5 \cdot 14 -3 = 67$
$$ x_{n+4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Area of shaded triangular region from absolute value functions So, I got this question from my teacher. I've tried solving it but to no avail. I cannot work out a method either, which can help me solve this.
So, it gives us two absolute value functions
$g(x) = 4|x-3|+3$
$f(x) = -6|x-3|+9$
There are two points, A & B as... | From your diagram, we can see that we must use $A,B$, and the vertex of $f(x)$ to calculate the area of our triangle. $A, B$ are the intersection points of $f(x)$ and $g(x)$.
Thus, we set $g(x) = f(x)$ to solve for our intersection points, and we get $4|x-3| +3 = -6|x-3|+9$. Simplifying, we get $10|x-3| = 6$, so $|x-3|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$. Why is this inequality true? The proof to the generalized harmonic series has the following inequality:
$$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$$
with $c > 1$ and $c = 1... | Group by packets of size $2^n$:
$$\sum_{k=1}^{+\infty}\frac 1 {k^c}=\sum_{n=0}^{+\infty}\left(\sum_{m=2^n}^{2^{n+1}-1}\frac 1 {m^c}\right)< \sum_{n=0}^{+\infty}\frac{2^n}{(2^n)^c}=\sum_{n=0}^{+\infty}\frac 1 {2^{\omega n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$
Given $x,y,z$ are positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$
My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{... | There's no minimum of $P$ at all. Consider an intersection of $x^2y+y^2z+z^2x = 3$ and $y = x^{-4}$, i.e. $x^{-2} + x^{-8}z + z^2x = 3$ or $x^9 z^2 + z + x^6 - 3x^8 = 0$. We can find from here that there exists a positive solution
$$
z = \dfrac{-1 + \sqrt{1+12x^{17}-4x^{15}}}{2x^9}.
$$
at least for $x$ big enough. More... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$ I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$
We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.
But I have a question: what should I do if the problem is for 4 variables, eg:
Let $a;b;... | If we fix $b$ and $d$ and write $a=x, c=s-x$ where $s=3-b-d\ge 0$ then
$$f(x) = ab^2+bc^2+cd^2+da^2 = b^2 x+b(s-x)^2+d^2(s-x)+dx^2, x \in [0,s]$$
is quadratic with leading coefficient $b+d \ge 0$ so its maximum is obtained at either $x=0 (a=0) $ or $x=s (c=0)$.
Similarly if we fix $a$ and $c$, then setting either $b=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Puzzling solution for $1/x = 3 - 2\sqrt{x}$ What is the solution for
$$\frac{1}{x} = 3 - 2\sqrt{x}$$
When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.
However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.
What am I missing?
| Rewrite the equation as
$$
2\sqrt{x}=3-\frac{1}{x}
$$
This shows that there is an implied condition, besides $x>0$ necessary in order that the equation makes sense: indeed you need
$$
3-\frac{1}{x}\ge0
$$
which becomes, taking into account that $x>0$, $3x-1\ge0$, so $x\ge1/3$.
Now you can safely square and get
$$
4x=9-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Smallest possible area for triangle I'm solving the following question:
The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?
This is my attempt at solving the problem:
Let $a$ be the le... |
Outline : We will write down the equation of tangent at arbitrary point $P$ of ellipse. Let it intersect the coordinate axes in $A$ and $B$. We identify the legs of the right triangle formed can be identified as $x$- and $y$- intercepts of the tangent line. So we find the smallest area of triangle minimizing the produ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{... | Hint:
Let $\dfrac{x+y}{xy}=k$
$$(xy)^3=3(xy)^2+(x+y)^3-3xy(x+y)$$
$$\iff(xy)^3(1-k^3)-3(xy)^2+3xy(kxy)=0$$
$$\iff(xy)^3(1-k^3)+3(xy)^2(k-1)=0$$
As $xy\ne0$ $$(xy)(1-k^3)=3(1-k)$$
Clearly, $k=1$ is a solution.
Otherwise, $$xy=\dfrac3{1+k+k^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unkno... | Hint:
$$(a+b)(ax^2+by^2)-(ax+by)^2=\cdots=ab(x-y)^2\ \ \ \ (1)$$
$$(a+b)(ax^3+by^3)-(ax+by)(ax^2+by^2)=\cdots=ab(x-y)^2(x+y)\ \ \ \ (2)$$
On division, we get $x+y$
$$(a+b)(ax^3+by^3)-(ax^2+by^2)^2=\cdots=abxy(x-y)^2\ \ \ \ (3)$$
$(1)/(3)$ will give $xy$
$$(a+b)(ax^4+by^4)-(ax+by)(ax^3+by^3)=\cdots=ab(x-y)^2((x+y)^2-xy)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$. $x,y,z \in \mathbb R$
My attempt:
Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.
Maybe I’m wr... | I guess you guys are too fast and / or assume too much prior knowledge. Let's digest it.
We need to be careful since all numbers are reals. Hence AM-GM and square root methods don't work.
First, note that if $xyz < 0$, then there is nothing to prove, since in this case $x^2 + y^2 + z^2 \geq 0 \geq xyz$ holds in all ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$ I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connec... | A Couple of "Common" Series
We have the Geometric Series
$$
\sum_{k=0}^\infty 4^kx^k=\frac1{1-4x}\tag1
$$
and the Generating Function for the Central Binomial Coefficients
$$
\sum_{k=0}^\infty\binom{2k}{k}x^k=\frac1{\sqrt{1-4x}}\tag2
$$
Equation $(2)$ is the Generalized Binomial Theorem applied to $\binom{-1/2}{k}=\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\inf... | I thought it might be instructive to present an approach that relies on straightforward contour integration. To that end, we now proceed.
Let $I$ be the integral given by
$$I=\int_0^\infty \frac{\log(x)}{x^3-1}\,dx$$
Now, moving to the complex plane, we analyze the contour integral $J$ given by
$$J=\oint_C \frac{\log^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
Bounding the difference between consecutive terms in the sequence $a_n=(1+1/n)^n$ I'm pretty sure the following estimate holds for $C>1$, but I can't prove it:
$$a_n:=\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1}\leq\frac{C}{n^2}$$
For context, I was looking for some such bound in order to prove that ... | Suppose that $p\geq 1$. Then
$$
\left( {1 + \frac{1}{p}} \right)^p = \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right) \ge \exp \left( {1 - \frac{1}{{2p}}} \right) \ge e\left( {1 - \frac{1}{{2p}}} \right)
$$
and
\begin{align*}
&\left( {1 + \frac{1}{p}} \right)^p \le \exp \left( {p\log \left( {1 + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
| Alternative approach that avoids Calculus:
square both sides.
LHS squared is $(2x-1) - 2\sqrt{x(x-1)}$
RHS squared is $(2x+1) - 2\sqrt{x(x+1)}$
So, taking LHS - RHS, the question reduces to whether
$-2 + 2\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 0.$
This is equivalent to asking whether
$\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
A Stirling number identity representing the second-order Eulerian numbers. Graham, Knuth, and Patashnik give in
CMath
a thorough introduction to the Stirling numbers.
On table 250 and table 251, they compile two pages of
Stirling number identities.
Of course, there are many more such identities. We give here one
which ... | We seek to show that with $0\le k\le n$ the following identity holds:
$$\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j} =
\sum_{j=0}^{n-k} (-1)^j {2n+1\choose j} {2n-k-j+1\brack n-k-j+1}.$$
We will start with the LHS. The chapter 6.2 on Eulerian Numbers
of Concrete Mathematics by Knuth et al. proposes the formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
computing the limit $\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$ I'm trying to compute the following limit and would greatly appreciate your heartening feedback on my solution.
The limit:
$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$
My steps in deriving the solution:
Preliminary identi... | Your proof is fine. Here's another:$$\sec\theta,\,\tan\theta\to\infty\implies\sec\theta+\tan\theta\to\infty\implies\sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta}\to0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
What did I do wrong with this integral problem?
Find the value of the following definite integral
$$\int_0^{\pi/2} \sqrt{1 - \sqrt{3}\sin 2x + 2\cos^2x} \, dx$$
*
*My attempt:
$$\int_0^{\pi/2} \sqrt{1 - \sqrt{3}\sin 2x + 2\cos^2x} \, dx$$
$$= \int_0^{\pi/2} \sqrt{2 - \sqrt{3}\sin 2x + \cos2x} \, dx$$
$$= \int_0^{\... | $$\int_0^{\pi/2}\sqrt{1-\sqrt{3}\sin(2x)+2\cos^2(x)}dx$$
then as you did we know that:
$$\cos(2x)=2\cos^2(x)-1\Rightarrow 2\cos^2(x)=\cos(2x)+1$$
so we have:
$$\int_0^{\pi/2}\sqrt{2-\sqrt{3}\sin(2x)+\cos(2x)}dx$$
try substituting $u=2x\Rightarrow dx=du/2$ so:
$$\int_0^\pi\sqrt{2-\sqrt{3}\sin(u)+\cos(u)}\frac{du}2$$
now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove an identity by induction I am working on the following problem:
$\text{For $n$ $\ge$ 1, Prove the following:}$
$n(1+x)^{n-1} = \sum_{k=1}^{n} k {n\choose k}x^{k-1} $
$\text{Deduce that: } \sum_{k=1}^{n}k{n \choose k}=n2^{n-1} $
I followed the proof by induction as follows:
$\text{Base Case: let n=1 , both ... | Well, $(n+1)(1+x)^n = (1+x)\frac{n+1}n\cdot n(1+x)^{n-1}= (1+x)\frac{n+1}n\cdot \sum\limits_{k=1}^nk{n\choose k}x^{k-1}=$
$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1}+ x\sum\limits_{k=1}^nk{n\choose k}x^{k-1})=$
$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1} + \sum\limits_{k=1}^nk{n\choose k}x^{k})=$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{ali... | The close-form result can be expressed as
$$\color{blue}{ \int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx = -nG+\frac\pi2 n \ln 2
+ \pi \sum_{k=1}^{[\frac n2]}\ln \cos\frac{(n+1-2k)\pi}{4n} }
$$
as shown below. Note that
\begin{align}
I_n =
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
\overset{x\to\frac1x} == \frac12 J_n - nG
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Solve in the set of integers $2^x +5^x = 3^x + 4^x$
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x + 4^x$$
With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+... | For $x\geq2$ we have $3^{x-2},4^{x-2}\leq5^{x-2}$, and so from $5^2=3^2+4^2$ it follows that
$$2^x+5^x>5^x=5^{x-2}(3^2+4^2)=5^{x-2}3^2+5^{x-2}4^2\geq3^x+4^x.$$
For $x\leq-2$ we have $3^{x+2},4^{x+2}\leq2^{x+2}$, and so from $2^{-2}>3^{-2}+4^{-2}$ it follows that
$$2^x+5^x>2^x=2^{x+2}(3^{-2}+4^{-2})=2^{x+2}3^{-2}+2^{x+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:
*
*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.
*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2... | ... then write $m = -y, n = -z$. Then $x,m,n$ are positive integers satisfying:
$$\frac{x(-m)}{-n} + \frac {x(-n)}{(-m)} + \frac {(-m)(-n)}{x} = 3$$
or
$$\frac {xm}{n} + \frac {xn}{m} + \frac {mn}{x} = 3$$
which is just our original equation. Using our previous analysis, $x=m=n=1$. Hence $y=z=-1$...
A shortcut to your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Challenge: find angle x in the figure via logical approach Background: this problem was from social app, but I cannot find original source. This is an interesting problem that I have not seen similar ones. It is a hard problem, cannot be solved with angle chasing. I worked out via analytic geometry programming, but ca... | Label the bottom two points (starting from left) with $A$, $B$, and the top points (starting from left) with $C$, $D$, $E$. Let $F$ be the intersection of $AD$ and $BC$. Let $\omega$ be the circumcircle of $ABF$. Let $G$ be the second common point of $\omega$ and $AE$. Let $H$ be the second common point of $BD$ and $\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4042623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $a,b\in \mathbb R^+$, $ |a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}}$ Prove $a+b\leq 2$
Question: If $a,b\in \mathbb R^+$, $|a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}},$ prove $a+b≤2$.
I figured out that $a+b\leq \frac {1}{\sqrt{a}}+\frac {1}{\sqrt{b}}$, but I am not sure how... | Just my 2 cents to the Albus' answer, without using power mean. We can assume $0<a\leq b$ ($a=0$ can be checked easily) then $$\frac{(a+b)^2}{4}\leq a^3+b^3 \iff \frac{\left(1+\frac{b}{a}\right)^3}{4}\leq 1+\left(\frac{b}{a}\right)^3 \tag{1}$$
Now
$$f(x)=\frac{\left(1+x\right)^3}{4} -1-x^3=(1+x)\left(\frac{(1+x)^2}{4}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two circles inscribed in a semicircle Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)
|
Let $EM=a$ be tangential to both circles at the point $E$ and the angle $\theta$. Then, the area $[ABE] = 33= a^2\sin\theta$ and
\begin{align}
[ABCD]&=153 =[AOD]+[BOC]=[COD] \\
& = \frac12Rr_1\cos\alpha + \frac12Rr_2\cos\beta + \frac12R^2\sin(\alpha+\beta)
\end{align}
Substitute $\sin\alpha = \frac{r_1}{R-r_1}$, $\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a formula for (0.5)²+(1)²+(1.5)²+(2)²+...+(25)² (1/4)(1³) + (1/9)(1³+2³) + (16)(1³+2³+3³) + ... + (1/2704)(1³+2³+3³+...+51³) = ?
Initially, I factored it as 1³ (1/2² + 1/3² + ... + 1/52²) + 2³ (1/3² + ... + 1/52²) + 3³ (1/4² +...+ 1/52²) + ... + 51³ (1/52²), but got stuck to split them as separate series.
Alte... | The sum can be expressed as:
$$
\frac1{2^2}\cdot 1^3+\frac1{3^2}\cdot (1^3+2^3)+\frac1{4^2}\cdot (1^3+2^3+3^3)+\cdots +\frac1{52^2}\cdot (1^3+2^3+\cdots +51^3)=$$
$$\sum_{i=2}^{52} \frac1{i^2} \sum_{j=1}^{i-1} j^3\stackrel{(1)}{=}\\
\sum_{i=2}^{52} \frac1{i^2}\cdot (\frac{(i-1)i}{2})^2=\sum_{i=2}^{52} \frac{(i-1)^2}4=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $a+b+c=3,$ show $a+b+c\geq ab+bc+ac$ by using AM-GM A am having issues with a problem that states: let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Show that $$a^bb^cc^a\le1$$
As this was an example problem, a solution path was provided but I am confused as to why the solution works. It writes:
$$1=... | Hint
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$
but, by $MA\ge MG$: $a^2+b^2\ge 2ab, a^2+c^2\ge 2ac, b^2+c^2\ge 2bc$, what gives you
$$a^2+b^2+c^2\ge ab+ac+bc.$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trying to find if this trigonometry problem is wrong or not I know That we have a triangle ABC with $$A=90° $$ $$ b+c=18 $$ $$ \cot(B)=4\sin^2(C) $$ $$ 0<B,C<90 $$
But , upon calculating i end up a = 18, b = 18, c = 0 which should not be possible
| $$\frac{c}{b}=\cot(B)=4\sin^2(C)=\frac{4c^2}{a^2}$$
$$a^2=4bc=b^2+c^2$$
Now use $c=18-b$ and get:
$$b^2-18b+54=0$$
$$b=9\pm3\sqrt{3}$$
Therefore, $a=6\sqrt{6}$; $b,c=9\pm3\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How/why $\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}$ turn into (negative fraction) $-\frac{x+a}{x²a²}$? I have that the solution of $\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)$ is $-\frac{x+a}{x²a²}$, but when I try to solve it I get (positive) $\frac{x+a}{x²a²}$.
The way I've been getting it is by making the denominato... | $$\frac{1}{x^2(x - a)} - \frac{1}{a^2(x - a)}$$
$$\frac{1}{(x-a)}\left(\frac{1}{x^2} - \frac{1}{a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{a^2 - x^2}{x^2a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{(a + x)(a - x)}{x^2a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{(a + x) -1(x - a)}{x^2a^2}\right)$$
$$\frac{-(a + x)}{x^2a^2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_... | The series can also be written as
$$
\begin{align}
\overbrace{\ \sum_{k=1}^\infty\frac1{3^k}\ }^{\substack{\text{each power}\\\text{of $3$}}}+\overbrace{\ \sum_{k=1}^\infty\frac1{9^k}\ }^{\substack{\text{even powers}\\\text{of $3$}}}
&=\frac{\frac13}{1-\frac13}+\frac{\frac19}{1-\frac19}\\
&=\frac12+\frac18\\[6pt]
&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | Integrate by parts
\begin{align}
\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx
&= \int_0^\infty \frac{\ln x}{2x}\>d(\frac{x^2}{x^2+1})\\
&\overset{ibp}= \frac12\int_0^\infty \underset{=\ 0}{\frac{\ln x}{x^2+1}dx }- \frac12\int_0^\infty \frac{1}{x^2+1}dx
= -\frac\pi4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.