Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and... | Alternatively, $(a+b)^2 = a^2+b^2+2ab \le a^2+b^2+(a^2+b^2) = 2(a^2+b^2) = 2\cdot 4 = 8\implies a+b \le \sqrt{8} < 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$? To prove $\lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$ exists, we prove that the sequence
$$f_n=\left(1+\frac{x}{n}\right)^n$$
is bounded and monotonically increasing toward that bound.
Proof Attempt:
We begin by showing $f_n=\left(1+\fra... | Let's observe that $$\left(1+\frac{x}{n}\right) ^n=1+x+\sum_{k=2}^n\frac{x^k}{k!}\left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right)=\sum_{k=0}^{n}a_k\frac{x^k}{k!}\tag{1}$$ where $$a_0=a_1=1,a_k=\left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right),k=2,\dots,n$$ Note that the coefficients $a_k$ are posit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ be a polynomial where a and b are real numbers and $f(x) = 0$ has two distinct negative roots and equal positive roots, fi... | Note that if $r\neq 0$ is a root of $f(x)=0$, then so is $\frac{1}{r}$. This is easily noted by the symmetry of the coefficients and can algebraically shown by the fact that
$$r^4f(\frac{1}{r})=f(r)$$
Hence, if the positive root is $r$ and the negative root of smaller magnitude is $s$, then our four roots are
$$r,\frac... | {
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"url": "https://math.stackexchange.com/questions/4232734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac... | Another similar approach: consider the Beta integral
$$\int_0^1 (1-x)^2 x^n d x = B(3, n+1) = \frac{\Gamma(3)\Gamma(n+1)}{\Gamma(3 + n+1)} = \frac{2! \cdot n!}{(n+3)!} = \frac{2}{(n+1)(n+2)(n+3)}$$
We get
$$\sum_{n\ge 0} \frac{1}{(4n+1)(4n+2)(4n+3)} = \frac{1}{2} \int_{0}^1 (1-x)^2 (\sum_{n\ge 0}x^{4n}) dx=\\ = \frac{1... | {
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"url": "https://math.stackexchange.com/questions/4232859",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
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Solving the equation $2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$
Solve the equation $$2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$$
Here is my work:
$$2x^2-5x+4=(4x-7)\sqrt{x-1}$$
$$2(x-1)^2-(x-1)+1=(4(x-1)-3)\sqrt{x-1}$$
Using the substitution $\sqrt{x-1}=t$,
$$2t^4-t^2+1=t(4t^2-3)$$
$$2t^4-4t^3-t^2+3t+1=0$$
Here I plugge... |
$2t^4-4t^3-t^2+3t+1=0$
Alt. hint: $\;\;\displaystyle
2t^4-4t^3\color{red}{+2t^2-2t^2}-t^2+3t+1 = \underbrace{2t^2(t-1)^2-3t(t-1)+1}_{\small{\displaystyle 2u^2-3u+1=(u-1)(2u-1)}}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:
$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \fr... | Start from the infinite product expansion of normalized sinc function:
$${\rm sinc}(z) =\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \left(1 - \frac{z^2}{n^2}\right)
$$
Substitute $z$ by $2z$, grouping factors in RHS in pairs and then divide it by expansion of ${\rm sinc}(z)$, we obtain an infinite product expansion o... | {
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"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Determine the minimum value of: $P=\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}+\frac{\sqrt{y^2+yz+z^2}}{(y+z)^2+1}+\frac{\sqrt{z^2+zx+x^2}}{(z+x)^2+1}.$ Given $x,y,z>0$ satisfy $x+y+z=\dfrac{3}{2}$.Determine the minimum value of:
$$P=\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}+\frac{\sqrt{y^2+yz+z^2}}{(y+z)^2+1}+\frac{\sqrt{z^2+zx+x^2... | The minimum does not exist.
Let $x\rightarrow\frac{3}{2}^-$ and $y=z\rightarrow0^+$.
Thus, we get a value $\frac{12}{13}.$
We'll prove that it's an infimum.
If it's given that $x\geq0$, $y\geq0$ and $z\geq0$ so for $(x,y,z)=\left(\frac{3}{2},0,0\right)$ we obtain a value $\frac{12}{13}$ again
and we'll prove that it's ... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\... | Substitution is a good way to solve these kinds of problems. In the original problem, the term $\sqrt{1-x^2}$ suggests the substitution $x=\sin\theta\ \text{or}\cos\theta$. Also, notice that $\sin\frac\pi 4=\cos\frac\pi 4=\frac{1}{\sqrt2}$. Now Substituting $x=\sin\theta$ where $\theta\in[-\frac\pi 2,\frac\pi 2]$, we g... | {
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"source": "stackexchange",
"question_score": "5",
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If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$?
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?
I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequali... | Let $k$ be the minimal value.
Thus, $k>0$ and the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ or $$c^2-k(a+b)c+10a^2+10b^2-kab\geq0$$ is true for any reals $a$,$b$ and $c$ (because after homogenization the condition $ab+ac+bc=1$ is not relevant already), which says $$k^2(a+b)^2-4(10a^2+10b^2-kab)\leq0,$$ which gives... | {
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"url": "https://math.stackexchange.com/questions/4243893",
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}... | Consider, $$S=\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots \tag{1}$$ Now multiply $S$ by 3, $$3S=1+\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots \tag{2}$$ $(2)-(1)$, $$2S=1$$ $$S=\frac12$$
Thus, for finite terms, $$\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots+\frac{1}{3^n}\lt\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Computing functions with discontinuity what is the best way to compute the value of the following function in a computer program, e.g., C/Fortran, in double precision?
$$
g(a,b)={{e^a/a - e^b/b}\over{a-b}} + {1\over{ab}}.
$$
The above function has removable discontinuities at a=0, b=0, and a=b. I also need to code a si... | The answer is the uppermost rightmost element of the matrix exponential
$$
\exp\begin{pmatrix} 0 & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{pmatrix}
$$
So you can use the Taylor series of $e^x$ and apply it to
$$
A= \begin{pmatrix} 0 & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{pmatrix}
$$
Proof:
We have
$$ A= SDS^{-1} =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$ with Popoviciu's inequality. I want to prove the inequality $$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$$, for positive reals. I have a hint that it is solved with Popoviciu'... | We dont really need Popoviciu's Inequality here,Basically use the following lemma.
If $x,y,z>0\in \mathbb{R}$ $$\dfrac{x}{y}+\dfrac{x}{z}\geq \dfrac{4x}{y+z}$$
Proof : Note that this is equivalent to $x\left(\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{4}{y+z}\right)\Longleftrightarrow\dfrac{x(y-z)^2}{yz(y+z)}\geq 0$ which is clea... | {
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"source": "stackexchange",
"question_score": "2",
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Doubt when something is some fraction less than some quantity. Why is $\frac{1}{4}$ less than X equals $\left(\text{X}-\frac{\text{X}}{4}\right)$ and not equals $\left(\text{X}-\frac{{1}}{4}\right)$ ?
For example: I was given the following problem to solve:
Q) One fourth less than 50% of 120 equals to ?
My solution:
50... | \begin{align*}
\dfrac{x}{4} &\lt x-\dfrac{x}{4}\Large{?}\\ \\
x-\dfrac{x}{4}
&=\dfrac{4x-x}{4}=\frac{3x}{4}\\ \\
\dfrac{x}{4} &\lt \frac{3x}{4}\\ \\
\therefore\space \dfrac{x}{4} &\lt x-\dfrac{x}{4}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $\int\frac{x+1}{x^2+x+1}dx$ Find
$\int\frac{x+1}{x^2+x+1}dx$
$\int \frac{x+1dx}{x^2+x+1}=\int \frac{x+1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
From here I don't know what to do.Write $(x+1)$ = $t$?
This does not work.Use partial integration?I don't think it will work here.
And I should complete square then find.
| \begin{align}
\int \frac{x+1}{x^2+x+1}\, dx &= \int \frac{x+\frac12 + \frac12}{x^2+x+1} \, dx \\
&= \int \frac{x+\frac12}{x^2+x+1} \, dx + \frac12 \int \frac1{x^2+x+1}\, dx \\
&=\frac12\ln |x^2+x+1| + \frac12 \int\frac1{(x+\frac12)^2 + \frac34} \, dx
\end{align}
I leave the rest as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\... | Just to add to the noise, this is how I would personally attack this problem. This is a matter of personal style and may not work for you.
$|\frac 7{x+4}| < 1$.
Either $\frac 7{x+4}$ is non-negative (positive or zero) or it is negative.
It is positive if $x+4$ is positive. It is negative if $x+4$ is negative. (And i... | {
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"question_score": "2",
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"answer_id": 7
} |
Find $s,t$ such that $\gcd(509,1177) = 509s+1177t$? How to find $s,t\in\mathbb{Z}$ such that $\gcd(509,1177) = 509s+1177t$?
I know that $509$ is prime and that $1177=107\cdot 11$ so we have $\gcd(509,1177)=1$. How can I proceed from here? Kind of system of equations in two variables but only one equation.
Actually I ca... | $$ \gcd( 1177, 509 ) = ??? $$
I prefer, for the "extended" part, writing it as a continued fraction, rather than "back substitution." The reason this is possible is that the cross product of consecutive convergents is always $\pm 1$
$$ \frac{ 1177 }{ 509 } = 2 + \frac{ 159 }{ 509 } $$
$$ \frac{ 509 }{ 159 } = 3 + ... | {
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"source": "stackexchange",
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For all diagonals $a$ of Pascal's triangle (figurate numbers), $\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2$? I am looking for the derivation of the closed form along any given diagonal $a$ of Pascal's triangle,
$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$
Numbered observations follow. As for the limit proposed in th... | Using algebra
$$f_n(x)=\sum_{k=a}^n {k\choose a}x^k=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}-x^{n+1} \binom{n+1}{a} \, _2F_1(1,n+2;n+2-a;x)$$ where appears the gaussian hypergeometric function.
$$g(x)=\sum_{k=a}^\infty {k\choose a}x^k=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}$$
So
$$f_n\left(\frac{1}{2}\right)=2-\... | {
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"source": "stackexchange",
"question_score": "4",
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Find $\int \frac{1}{(1+m\cos(\theta))^2} \mathrm d\theta$ where $0I tried Weistrass substitution, integration by parts and partial fractions for this integral, but it made the integral even worse. Actually I got this integral when I was working with the Kepler's second law.
When I did Weistrass substitution I got some... | This can be done via e.g. Mathematica. Try the substitution $w = \tan(\theta/2)$, so $$\textrm{d}w = \frac{1}{2} \sec^2\left(\frac{\theta}{2}\right)\mathrm{d}\theta$$ and also, $$\cos(\theta) = \frac{1-w^2}{1+w^2}$$ Then, using $\sec^{2}(\theta) = 1 + \tan^{2}(\theta)$, we have $\sec^{2}(\theta/2) = 1 + w^2$, and there... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Introduce $x$, $y$, $\alpha$ and $\delta$, so
$$x=\frac {a+b} 2$$
$$y=\frac {c+d} 2$$
$$\alpha=\frac {a-b} 2$$
$$\delta=\frac {c-d} 2$$
Now
$$a=x-\alpha$$
$$b=x+\alpha$$
$$c=y-\delta$$
$$d=y+\delta$$
Rewrite $a+b=c+d$ using $x$, $y$, $\alpha$ and $\delta$:
$$x-\alpha+x+\alpha=y-\delta+y+\delta$$
$$2x=2y$$
$$x=y$$
Simil... | {
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"url": "https://math.stackexchange.com/questions/4260123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
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How to find the probability of $P(Y=6)$ Anna writes down a random sequence created by the following process. She repeatedly rolls a fair
6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if
any), then she writes the new number down and then continues rolling. Otherwise, s... |
Question:
what is P(Y=6)
observe that the sequence's length can only be $X \in\{1,2,3,4,5,6\}$
$X=1$ when you roll immediately 6
$X=5$ when you roll 4 numbers in increasing order, and then 6. The only favourable sequences are, in this case
$1234$
$1235$
$1245$
$1345$
$2345$
...and 6 on the fifth roll. Observe that ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Minimum of $abc$ when $a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}.$
Let $a,b,c$ be positive real numbers with $abc=k$ such that the inequality $$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$$
holds for all $a,b,c$. Find the minimum value of $k$.
I found that $abc=2$ works. Here is my proof for $abc=2$... | Let $abc=k$.
Thus, after homogenization we need to prove that
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{k}}\right)\geq0,$$ which for $a=b=c$ gives $k\geq2$ and it's enough to prove that:
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)\geq0,$$ which you proved already by C-S.
Also, it's true by AM-GM and Muir... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Showing LI without determinant or Reduced row echelon form. Consider the $\mathbb{R}$-vector space $ P^3(\mathbb{R})$ of polynomials over $\mathbb{R}$ with degree at most 3. Define $$W= \{f \in P^3(\mathbb{R}): f^{'}(2)=f(1)=0\}$$
EDIT: Calculate $\dim(W)$. by finding a basis for W.
We need to find a basis for W the ea... | For a polynomial $f = ax^3 + bx^2 + cx +d$ in $P^3(\mathbb R)$ one has that $f(1) = a+b+c+d$ and $f'(2) = 12a+4b+c$, so
$$
f \in W \implies
\begin{cases} 12a+4b+c = 0 \\[1mm] \qquad \quad \& \\[0.5mm]
a+b+c+d = 0
\end{cases} \implies
\begin{cases} c = -12a-4b \\[1mm] \qquad \quad \& \\[0.5mm]
d = -a-b-c = 11a+3b ... | {
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"url": "https://math.stackexchange.com/questions/4266883",
"timestamp": "2023-03-29T00:00:00",
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Finding the maximum value of the following product.
Let $x,y,z$ be three reals which satisfies $x^2+y^2+z^2=1$. Find the maximum value of $$P = (x^2-yz)(y^2-zx)(z^2-xy).$$
I think expanding would be a bad idea. I tried to apply the Weirstrass product inequality but it may give the minimum not maximum value.
| Lagrange multiplier is the correct tool for solving this type of constrained optimization problem. I don't know your backgroud, but the following is a completely elementary approach (which I only found out with the help of Lagrange).
First it's not hard to show that the maximum (which exists as the sphere is compact) i... | {
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Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}... | $$\sum_{n=1}^{k}\frac{1}{(n)(n+1)(n+2)}$$
By partial fraction decomposition,
$$\frac{1}{(n)(n+1)(n+2)}=\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}$$
$$\sum_{n=1}^{k}\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}$$
By splitting the terms and taking the constants outside,
$$\frac{H_{k}}{2}-(H_{k+1}-1)+\frac{H_{k+2}-\frac{3}... | {
"language": "en",
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Is $g(x,y)= \frac{f(x^{2y+1},y)}{f(x,y)}$ always an integer? This question is similar to this other question:
Let $$ f(x,y):= \frac{x^y -1}{x+(-1)^y}$$
and $$ g(x,y):= \frac{f(x^{2y+1},y)}{f(x,y)}.$$
Let $y\ge1$ be an integer. Show that $g(x,y)$ is a polynomial function of $x$ with integer coefficients.
Moreover, it s... | If $y$ is even, then
$$g(x,y)=\frac{\frac{x^{y(2y+1)}-1}{x^{2y+1}+1}}{\frac{x^y-1}{x+1}}=\frac{(x^{y(2y+1)}-1)(x+1)}{(x^{2y+1}+1)(x^y-1)}.$$
If $y$ is odd, then
$$g(x,y)=\frac{\frac{x^{y(2y+1)}-1}{x^{2y+1}-1}}{\frac{x^y-1}{x-1}}=\frac{(x^{y(2y+1)}-1)(x-1)}{(x^{2y+1}-1)(x^y-1)}.$$
One has
$$x^t-1=\prod_{d\mid t}\Phi_d(x... | {
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Prove that $\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge 2$
Given $a+b+c+d=4, $ and $a,b,c,d$ are positive real numbers. Prove that $$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge 2$$
Can someone give me hints for this?
To show
$$\frac{a^2}{a+b^2ca}+\frac{b^2}{b+c^2... | It's enough to prove that:
$$a^2bd+b^2ca+c^2bd+d^2ac\leq4,$$
Now, use AM-GM twice.
| {
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Isn't my book using logarithm wrongly to differentiate? Problem:
Differentiate with respect to $x$: $\frac{e^{-3x}(3x+5)}{7x-1}$
My book's attempt:
Let,
$$y=\frac{e^{-3x}(3x+5)}{7x-1}$$
$$\ln(y)=\ln(\frac{e^{-3x}(3x+5)}{7x-1})\tag{1}$$
$$\text{rest of the math...}$$
Question:
*
*Isn't taking $\ln()$ unfounded in $(1)... | This is called "logarithmic differentiation". It can be used to make taking derivatives of expressions involving lots of products, quotients, and powers, easier by invoking the properties of the logarithm.
For example, say we want the derivative of
$$y = \frac{(x^2+1)^3\sqrt{x^3+2x}}{e^x(x+1)}.$$
Taking logarithms on b... | {
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In triangle. Prove that: $2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$ Problem: Given a,b,c are length of triangle. Prove that: $$2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Ha... | Let $a=y+z$, $b=x+a$ and $c=x+y$.
Thus,$x$, $y$ and $z$ are positives and we need to prove that:
$$4(x+y+z)\sum_{cyc}\frac{1}{x+y}-9\geq\sum_{cyc}\sqrt{17-\frac{16yz(2x+y+z)}{\prod\limits_{cyc}(x+y)}}$$ or
$$\sum_{cyc}(4x^3+7x^2y+7x^2z+6xyz)\geq\sum_{cyc}\sqrt{\prod_{cyc}(x+y)(17\prod_{cyc}(x+y)-16yz(2x+y+z))}.$$
Now, ... | {
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"source": "stackexchange",
"question_score": "2",
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Winning strategy for a game with cubic equation The following problem is from USSR $1990$:
The following equation with erased coefficients is written on a blackboard: $$x^3+\dots x^2+\dots x+\dots =0$$
Two players are playing a game. In one move the first player chooses a number and the second player puts it instead o... | In case 2, the idea is:
*
*If $N$ is a perfect square, then $x^3 - Nx$ factors as $x(x+\sqrt N)(x-\sqrt N)$. This means we can win by playing $-N$, then $0$ if $-N$ is chosen as the coefficient of $x$.
*If $N = pq(p+q)$, then the expansion of $(x+p)(x+q)(x-p-q)$ has an $x^2$ coefficient of $0$ and constant term of $... | {
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"source": "stackexchange",
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Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be writte... | You can also consider taking the $\ln$ from both sides:
\begin{align*}
y = (x + \sqrt{1 + x^{2}})^{3/2} & \Rightarrow \ln(y) = \frac{3\ln(x + \sqrt{1 + x^{2}})}{2}\\\\
& \Rightarrow \frac{y'}{y} = \frac{3}{2(x + \sqrt{1 + x^{2}})}\times\left(1 + \frac{x}{\sqrt{1 + x^{2}}}\right)\\\\
& \Rightarrow y' = \frac{3y}{\sqrt{1... | {
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"source": "stackexchange",
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Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*... | As you posted:
$$ \int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx \\ \sum_0^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}$$
You can rewrite it as the following using the definition of the central binomial coefficient:
$$\sum_0^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2} = \sum_0^{\infty} \frac{4^n}{\frac{(2n)!}{n!^2}(2n+1)^2}... | {
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"source": "stackexchange",
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What's the measure of height relative to $BC$ in the triangle below?. For reference:In the isosceles triangle $ABC$ ($AB = BC$) is marked
the interior point O whose distances to sides $AB, BC and AC$ are $2, 3$ and $5$ respectively,
If the sum of the height measurements of triangle $ABC$ is $30$, calculate the
measure ... | If altitude from $A$ and $C$ to the opposite sides is $h_1$ and from $B$, it is $h_2$, then
$2 h_1 + h_2 = 30 \tag1$
If $A$ is the area of $\triangle ABC$,
$ \displaystyle a = c = \frac{2A}{h_1}, b = \frac{2A}{h_2}$
Now note that $A = Area ~ \triangle OAB + Area ~ \triangle OBC + Area ~ \triangle OAC$
$ \displaystyle A... | {
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Calculate $E[|X^{k}|]$ when k is odd, $X$ is a random variable with standard normal distribution I want to calculate $E[|X^{k}|]$ when k is odd, $X$ is a random variable with standard normal distribution. My approach is as follows.
Since $E[X^{k}]=0$ when $k$ is odd. We have $$E[X^{k}] = 0 = \int_{-\infty}^{\infty}x^{k... | Let $k$ be a positive integer and $$I(k) = \int_{x=0}^\infty x^{2k-1} e^{-x^2/2} \, dx.$$ Then the substitution
$$u = x^2/2, \quad du = x \, dx$$ gives
$$I(k) = \int_{u=0}^\infty (2u)^{k-1} e^{-u} \, du = 2^{k-1} \int_{u=0}^\infty u^{k-1} e^{-u} \, du = 2^{k-1} \Gamma(k).$$ So we have
$$\operatorname{E}[|X^{2k-1}|] =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Substitution to evaluate $\int{\sqrt{\sin x}\cos^3x\ dx}$? I'm already solving more daunting exercises, but for some reason I can't tackle this one:
$$
\int{\sqrt{\sin x}\cos^3x\ dx}
$$
None of the "obvious" substitutions ($\sin x$, $\sqrt{\sin x}$, $\cos x$, $\cos^3x$) seem to make sense. I'm probably missing somethin... | Another trivial substitution is $u^2=\sin x$.
So we have
$$\cos x\,\mathrm{d}x=2u\,\mathrm{d}u$$
As a result,
\begin{align}
\int \sqrt{\sin x}\cos^3 \,\mathrm{d}x
&= \int \sqrt{\sin x}\, \cos^2 x \,\cos x \, \mathrm{d}x \\
&= \int \sqrt{u^2}\, (1-u^4) \,2u\, \mathrm{d}u \\
&= 2\int u^2(1-u^4)\,\mathrm{d}u\\
&= \frac{2... | {
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Demonstrate that $3ba^2 = a^3\sqrt3 +b^3$ in the triangle below For reference: In triangle ABC: $AC=BC=a, AB=b$ and $\angle C= 40^\circ$.
Demonstrating identity: $3ba^2 = a^3\sqrt3+b^3$
My progress:
Draw $CD \perp AB, D\in AB$ and $AF \perp CD,(F \in CD)$
$\triangle ABF \sim \triangle CBD \implies:
\frac{AF}{CD}=\fra... | Draw line segment $\small AD$ as in the figure.
$\small \triangle ACB\sim\triangle BAD\implies\dfrac{BD}b=\dfrac ba\implies BD=\dfrac{b^2}a$
Therefore, $\small CD=a-\dfrac{b^2}a$.
Now applying cosine rule to $\small \triangle ACD$ with the angle $\small 30^\circ$ gives the desired formula.
$\small \left(a-\dfrac{b^2... | {
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Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists I think this is the first time when I handle of non-linear differential equation.
$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }_{x > 0} \tag{1} $$
$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$
$$ \u... | This is a special case of a Bernoulli differential equation, wich has a well know method of solution by transforming in a linear equation.
| {
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In a room of $n \geq 7$ people, what is the probability that, for each day of the week, there is at least one person whose birthday falls on that day? In a room of $n \geq 7$ people, what is the probability that, for each day of the week, there is at least one person whose birthday falls on that day? Assume each day of... | Since there are seven possible days of the week on which each person could have been born, there are $7^n$ possible birthday assignments.
If at least one person is born on each day of the week, we must exclude those birthday assignments in which at least one day is missing. There are $\binom{7}{k}$ ways to exclude $k$... | {
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Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\fra... | The desired inequality is written as
$$\frac{2x}{\sqrt{1 + x^2}}
+ \frac{2y}{\sqrt{1 + y^2}} +
\frac{2z}{\sqrt{1 + z^2}}
+ 8 \cdot \frac{1 + x^2}{4x^2}
+ 8 \cdot \frac{1 + y^2}{4y^2}
+ 8 \cdot \frac{1 + z^2}{4z^2} \ge 27.$$
Using AM-GM, we have
\begin{align*}
\mathrm{LHS}
&\ge 27\,\sqrt[27]{\frac{2x}{\sqrt{1 + x^2}}... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $R_{ab} = \frac{1}{2} S g_{ab}$ (2 dimensional Ricci curvatures) In the title $R_{ab}$ is the usual Ricci curvature tensor and $S$ is the scalar curvature. As we are in a two dimensional Riemanian manifold with metric $g$, we can write $S= g^{11} R_{11} + g^{12} R_{12} + g^{21} R_{21} + g^{22} R_{22}$, where ... | You have to use that in dimension $2$, the curvature tensor has a very specific formula. Namely, it holds that $$R(X,Y)Z=K(g(Y,Z)X-g(X,Z)Y),$$where $K$ is the Gaussian curvature of the surface. Trace in $X$ to obtain $${\rm Ric}(Y,Z) = Kg(Y,Z).$$Now take the $g$-trace to obtain $S = 2K$. So $K=S/2$ and so ${\rm Ric} = ... | {
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How to turn into a product of $2$ factors the expression : $ x^4 -3x^2+1$? Source : Lebossé & Hémery, Algèbre et Analyse ( Classe de seconde , 1965).
The exercice requires : turn into a product of $2$ factors the expression $ x^4 -3 x^2 +1$.
The only way I see is to use the property : $P(x)$ being a polynomial , if $... | It is quite easy question. You said two factors.
$$x^4-3x^2+1=(x^4-2x^2+1)-x^2=(x^2-1)^2-x^2=(x^2+x-1)(x^2-x-1).$$
Is it your answer?
| {
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Olympiad Math Question - If the sum of the positive inverses of 4 positive integers equals $1.1$, what’s the lowest possible sum of the integers? I was going through some Olympiad pass papers and came across this question:
Given four different positive integers $A, B, C, D$ so that $\frac{1}{A}+\frac{1}{B}+ \frac{1}{C... | Without loss of generality, suppose $A < B < C < D$.
Notice that if $A \ge 3$, then $B \ge 4$, $C \ge 5$, $D \ge 6$, and then $\tfrac{1}{A}+\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} \le \tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5}+\tfrac{1}{6} = \tfrac{19}{20} < 1.1$. So we need $A = 1$ or $A = 2$.
Case 1: $A = 2$. Then we need... | {
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Gaussian integral $\int_{0}^{\infty} dx \, x \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right)$ I've been having trouble evaluating the Gaussian integral of the form
$$\int_{0}^{\infty} dx \, x \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right) \, ,$$
which can b... | As @rtem Alexandrov commented, a formal expansion around $x=0$ would perfectly work and would probably converge fast since
$$\sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2}=-2 c+\frac{b^2 c}{x^2}+\frac{b^2 c \left(4 c^2-3 b^2\right)}{4x^4}+O\left(\frac{1}{x^6}\right)$$
Expanded as series around $x=0$, we have $$\sqrt{(... | {
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Manipulating a polar equation I started with this:
$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
And via substitution, got this far:
$r=\left(\frac{144}{9-25\sin^{2}\theta}\right)^{.5}$
For the fact that Desmos plots these the same, I assume I'm right so far.
The goal in this section is to end with the form
$r=\frac{ep}{1-e\sin... | The right focus is $(5, 0)$. Shifting the hyperbola to the left by $5$ units results in the new equation:
$ \dfrac{(x + 5)^2}{16} - \dfrac{y^2}{9} = 1 $
Now using polar coordinates, but measuring the angle $\theta $ from the positive $y$ axis direction, then $x = -r \sin \theta , y = r \cos \theta $
Hence,
$ \dfrac{... | {
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Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and ... | $ \displaystyle 2x \cos \alpha - 3y \sin \alpha = 6 \implies y = \frac{2 \cot \alpha}{3} x - 2 \csc \alpha = mx + c$ where $m = \tan \theta = \dfrac{2 \cot \alpha}{3}$
Consider $0 \leq \alpha \leq 2\pi$ that gives all possible lines for the given equation. If $\alpha = \frac{\pi}{2}$ or $\frac{3 \pi}{2}$, the line is $... | {
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Showing that $b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$ For an ODE where the solution should be on the interval $[-\frac{1}{2},\frac{1}{2}]$ it has to be $b \leq \frac{b}{\frac{1}{4}+b^2}$.
The task is:
Let $f: [-\frac{1}{2}, \frac{1}{2}]\times[-b,b]\to \mathbb{R}$ with $f(x,y):=x^2+y^... | The "butterfly" (double cone) condition that you have to check is that
$$
Ma\le b,
$$
with $a,b$ the sizes of the box under consideration, $a=\frac12$, and $M=a^2+b^2=\frac14+b^2$ the maximum of the function value inside the box. Inserted this gives
$$
\frac12(\frac14+b^2)\le b\iff \frac14-2b+b^2\le 0.
$$
To compute a... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logar... | If we are allowed to use series expansions, then the problem is fairly trivial. We want to evaluate $$\lim_{x\to \infty} \frac{1 - x \log \left(1+ \frac1x\right)}{\log \left(1+ \frac1x\right)}$$
The series expansion of $\log \left(1+ \frac1x\right)$ is given by $$\log \left(1+ \frac1x\right) = \frac{1}{x} - \frac{1}{2x... | {
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What is $\lim_{n\to\infty} n (\sum_{k = 1}^{n} \frac{1}{\sqrt {n^2 + k}} - 1)$? I am trying to find this limit: $\lim_{n \to \infty} a_n^{b_n}$, where $a_n = \sum_{k = 1}^{n} \frac{1}{\sqrt{n^2+ k}}$ and $b_n = n$. I have managed to prove that $\lim_{n \to \infty} a_n = 1$, by using the squeeze lemma (bounding $a_n$ be... | We can rewrite the limit as
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{\sqrt{n^2+k}}-n = \lim_{n\to\infty}\sum_{k=1}^n \frac{n-\sqrt{n^2+k}}{\sqrt{n^2+k}} = \lim_{n\to\infty}\sum_{k=1}^n \frac{1-\sqrt{1+\frac{k}{n^2}}}{\sqrt{1+\frac{k}{n^2}}}$$
Since the numerator approaches zero we will rationalize and continue to simpl... | {
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"answer_id": 0
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Optimization Triangle I can't seem to finish this task. I found the derivative and the critical numbers, except when the derivative $= 0$. I don't understand how I can find it manually.
Task:
$2$ points A and C lie on a line $l$. a point B is $4$ away from line $l$. The sum of AB + BC is always equal to $10$. Find the... | From $$0 = \frac{x}{\sqrt{(x+4)(x-4)}} + \frac{x-10}{\sqrt{(14-x)(6-x)}},$$ simply multiply both sides by $\sqrt{(x+4)(x-4)} \cdot \sqrt{(14-x)(6-x)}$ to get
$$0 = x\sqrt{(14-x)(6-x)} + (x-10){\sqrt{(x+4)(x-4)}}.$$
Move one of the two terms to the left hand side, and then square both sides to get
$$x^2(14-x)(6-x) = (x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4321915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left... | We need more terms to obtain
$$\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}+o\left(x^{4}\right)\right)=-\frac{x^{2}}{2}-\frac{x^{4}}{12}+o\left(x^{4}\right)$$
and then by binomial expansion
$$\frac{1}{-\frac{x^{2}}{2}-\frac{x^{4}}{12}+o\left(x^{4}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, w... | Hint :
$$a^2 + ab + b^2 = \left( a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that there is a unit $u \in R$ such that $ub = bu = a$
Let $R$ be a ring with identity containing elements $a$ and $b$ with $ab = b$ and $b^2 = a$. Prove that there is a unit $u \in R$ such that $ub = bu = a$.
Source: Problem $15.3.6$, Algebra in Action: A Course in Groups, Rings, and Fields by Shahriar Shahriar... | We know that $b^3 = b$ and so $b^4 = b^2$. We may then compute that
$$
(b^2+b-1)^2 = b^4 + 2 b^3 - b^2 - 2 b + 1 = b^2 + 2b - b^2 - 2b + 1 = 1.
$$
Hence $b^2+b-1$ is a unit. Furthermore, we have
$$
(b^2+b-1)b = b(b^2+b-1) = b^3 + b^2 - b = b + b^2 - b = b^2,
$$
so it is the unit we are after.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
| The equation can be properly written as $y(x)+1+(x+1)y'(x)=0,$ which suggests the substitution $z(x)=(x+1)[y(x)+1].$ Hence $z'(x)=y(x)+1+(x+1)y'(x),$ implying $z'(x)=0.$ Since $y$ need not be differentiable at $-1$ to satisfy the equation everywhere else, we account for this, so we have that $(x+1)[y(x)+1]=A$ for every... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?
My approach:
We can see that the $n^{... | I tried to rewrite the sum:
$$\sum _{n=1}^k \frac{\left(\frac{1}{4}\right)^n (2 n)!}{n! (n+1)!}$$
computationally using the Mathematica Code:
Sum[Factorial[2*n]*(1/4)^n/(Factorial[n]*Factorial[n + 1]), {n, 1, k}]
As a result I got:
$$\frac{2^{-2 k-1} \left(-k (2 (k+1))!-2 (2 (k+1))!+2^{2 k+1} (k+1)! (k+2)!\right)}{(k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Find the area of region ABC. For reference: In figure $T$ and $K$ are points of tangency,
$MT = a$ and $KN = b$; calculate area of region $ABC$.
(Answer:$2\sqrt{ab}(\sqrt a+\sqrt b)^2$)
My progress:
$$S_{ABC} = p \cdot r = \frac{r \cdot (AB+BC+AC)}{2}\\
AC +2R = AB+BC\\
S_{ABC} = AG \cdot GC \qquad \text{(property)... | Here is a geometrical solution without much algebra.
For a right triangle, if you draw a line through the points of tangency of the incircle with the perpendicular sides, it does bisect the arcs of the circumcircle on both sides. In other words, $M$ and $N$ are midpoints of minor arcs $AB$ and $BC$ respectively. At th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving $\sqrt3\;\cos(x)= \sin\;(\frac {x}{2})$ I'm currently stuck on this trigonometric equation:
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
Here's what I've tried so far (2 methods):
Method n.1
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \;(1-\sin^2x)= \sin\;(\frac {x}{2})$$
$$\sqrt3-\sqrt3\;\sin^2x= \sin\;(... | We begin with $\sqrt{3}\cos(x)=\sin(\frac{x}{2})$. We can start by using our half-angle formula to replace $\sin(\frac{x}{2})$ with $\sqrt{\frac{1-\cos(x)}{2}}$. Making the substitution and squaring both sides, we now see that $3\cos^2 (x)=\frac{1-\cos(x)}{2}$. Multiplying both sides by $2$, we get $6\cos^2(x)=1-\cos(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4337047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots Find the range of $m$ so that the equation $(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots.
Either $(x^2+2mx+7m-12)=0$ or $(4x^2-4mx+5m-6)=0$
The discriminant of $(x^2+2mx+7m-12)=0$ is $4(m-3)(m-4)$
and the discriminant of $(4x^2-4mx+5m-6)... | We can also look at the "vertex form" of each quadratic factor of $ \ (x^2+2mx+7m-12)·(4x^2-4mx+5m-6) \ $ to see what they contribute to their product function.
The first factor can be written as
$$ f(x) \ \ = \ \ x^2 \ + \ 2mx \ + \ (7m - 12) \ \ = \ \ (x \ + \ m)^2 \ + \ (7m - 12) \ - \ m^2 $$ $$ = \ \ (x \ + \ m)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\ \sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converge for all $x\ ?$ Does $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converge for all $x\ ?$
It obviously converges for any $x\ $ of the form $\ 2^mk \pi\ $ where $\ m,k\in\mathbb{Z},\ $ but for any other values of $\ x\ $ the question is interes... | As it has been over a year and no answer/edit has been written to encompass the answer in the comments by @achille_hui, I'll present their answer here for posterity:
At $x=\frac{\pi}{7}$ and $n\geq 1$ we have
$$\sin\left(2^n\frac{\pi}{7}\right)=\sin\left(2\pi\frac{2^{n-1}}{7}\right)=\begin{cases}
\sin\left(2\pi\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Power series solution of $y'-y=x^2$ $y'-y=x^2$
I try to get a power series solution.
$y(x)=\sum_{n=0}^{\infty}a_{n}X^{n},y'(x)=\sum_{n=1}^{\infty}na_{n}X^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}$.
$\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}-\sum_{n=0}^{\infty}a_{n}X^{n}=X^{2}$
Then $ \space a_{n+1}=\frac{a_n}{n+1} \space ... | Your solution amounts to
$$
y - a_0 \left( 1 + x + \dfrac{x^2}{2} \right) = (2 + a_0) \left( \mathrm{e}^x - 1 - x - \dfrac{x^2}{2} \right) .
$$
Notice you've indeed made a mistake. You have $a_{n + 1} = \dfrac{a_n}{n + 1}$, so $a_5 = \dfrac{a_4}{5} = \dfrac{2 + a_0}{5!}$, $a_6 = \dfrac{a_5}{6} = \dfrac{2 + a_0}{6!}$, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does the theorem 2 of algebra of limits contradict in this Q? Q: $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$
Solution in my textbook using theorem 2 of algebra of limits I.e $\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}$ .
$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1} &=\li... | $x^{15}-1=(x-1)(1+x+x^2+...+x^{13}+x^{14}).$
$x^{10}-1=(x-1)(1+x+x^2+...+x^{8}+x^{9}).$
So if $x\ne 1$ then $$\frac {x^{15}-1}{x^{10}-1}=\frac {1+x+x^2+...+x^{13}+x^{14}}{1+x+x^2+...+x^{8}+x^{9}}$$ which obviously converges to $\frac {15}{10}$ as $x\to 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$ I want to expand function $$f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$$ for $1 < |z| < 2$.
My work so far
By partial fraction decomposition we can get that:
$$\frac{z - 4}{(z + 1)^2(z - 2)} = \frac{-3(z - 4)}{(z + 1)^2} + \frac{\frac 1 9 (z - 4)}{z - 2}$$
Now deriving L... | You want a series for $\frac1{(1+z)^2}$ that’s good outside the unit circle, don’t you? Thus a series in $1/z$. So write
$$
\frac1{(1+z)^2}=\frac1{z^2}\frac1{(1+1/z)^2}\,.
$$
If you find that confusing, expand $t^2\frac1{(1+t)^2}$ in powers of $t$, and substitute $1/z$ for $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A curious family of Chebyshev-like polynomials Consider the family $f_n(x)$ of functions of $x$ for $0\leq x\leq1$, each indexed by a variable $n \in \mathbb{N}$, described by the following equation:
$$f_n(x) = \sin^2\left(n \arcsin\left(\sqrt{x}\right)\right)$$
Evaluation in numerical algebra programs yields the follo... | Chebyshev polynomials of the second kind satisfy $\,U_{n-1}(\cos \theta)\,\sin \theta =\sin(n\theta)\,$.
With $\,\theta=\arcsin\left(\sqrt{x}\right)\,$ it follows that:
$$
\sqrt{x} \; U_{n-1}\left(\sqrt{1-x}\right) = \sin\left(n \arcsin(\sqrt{x})\right) \;\;\iff\;\; f_n(x) = x \cdot U_{n-1}^2\left(\sqrt{1-x}\right)
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Doubt regarding a differential equation. There's a problem in a book regarding exact differential equation :
Solve :
$$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}$$
The author further proceeds to rearrange above in the form $M\,dx+N\,dy=0$
where
$$M=x+\frac{a^2y}{x^2+y^2} ; N= y-\frac{a^2x}{x^2+y^2}$$
Which further ... | $$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}\tag 1$$
As you correctly found : Equation $(1)$ is an exact ODE.
You transformed Eq.$(1)$ into Eq.$(2)$:
$$(x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy =0\tag 2$$
Again you correctly found : Equation $(2)$ is not an exact ODE.
This is not contradictory because Eqs.$(1)$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Support for marginals of joint PDF I have two joint PDF-s and three questions below.
I have a random vector' $(U,V)'$ and it's PDF. I need to find the marginal PDF-s for both $U$ and $V$. I tried deploying the idea from this anwer, but I still have some questions.
This is the joint PDF:
$$
f_{U,V}(u,v)=\begin{cases}
... | Your work for $1$ and $2$ is mostly correct except some typos.
$ \displaystyle f_U(u) = \frac{12\sqrt{1-u^2}-6}{4\pi-3\sqrt{3}}, ~2 |u| \leq \sqrt3$
Also, $ \displaystyle f_V(v) = \frac{24\sqrt{1-v^2}}{4\pi-3\sqrt{3}}, 1/2 \leq v \leq 1$
So, $ \displaystyle E[U] = \int_{-\sqrt3/2}^{\sqrt3/2} u \cdot f_U(u) ~ du~,$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and... | Proving $\displaystyle\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$:
Set $a=z$ and $b=1-z$ in the Beta function:
$$\operatorname{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}\mathrm{d}x,$$
we obtain
\begin{gather}
\Gamma(z)\Gamma(1-z)=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find all positive integers s.t. $\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$ Find all positive integers $a, b, c, d$ such that :
$$\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$$
The original problem came from atomic electron transitions :
I would like to find out non-trivial positiv... | we take, $(a,b,c,d)=(pq,rq,st,wt)$
where, $r^2=p^2+q^2$ & $w^2=s^2+t^2$
And, $(r,p,q)=((u^2+v^2),(u^2-v^2),(2uv))$
$(w,s,t)=((m^2+n^2),(m^2-n^2),(2mn))$
We have Identity,
$(1/rp)^2=(1/pq)^2-(1/rq)^2$ ---(1)
$(1/ws)^2=(1/st)^2-(1/wt)^2$ ----(2)
we want, eqn (1)=eqn (2)
Hence, $pr=sw$
or
$(u^4-v^4)=(m^4-n^4)$ ----(3)
e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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Continuity of this function as $(x,y)$ tends to $(0,0)$
Here's a function in $x$ and $y$ defined piecewise as
$$f(x,y)=
\left\{\begin{array}{ll}
0 & (x,y)=(0,0)\\
\frac{x^2y}{x^4+y^2} & (x,y) \neq (0,0) \\
\end{array}\right.$$
Examine its continuity as the ordered pair tends to $(0,0)$.
Okay, so I first tried ... | Notice that :
$$f(t, t^2) = \dfrac{t^2 t^2}{t^4 + t^2} = \dfrac{t^4}{2 t^4} = \dfrac{1}{2} \underset{t \to 0}{\to} \dfrac{1}{2} \neq f(0, 0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the limit of $(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$. To find the limit of $(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$.
First notice that $(1+\frac{1}{n^2})$ is the smallest term in the product and $(1+\frac{n}{n^2})$ is the greatest. So: $ (1+\frac{1}{n^2})^n \leq (1+\frac{1}{n^2})(... | You have proved that $x-\frac{x^2}{2}\leq\ln(x+1)\leq x$
Now as @Gary said $$\prod_{i=1}^n(1+\frac{i}{n^2})=e^{\sum_{i=1}^n\ln(1+\frac{k}{n^2})}$$
From $(1)$ we get , $$\sum_{i=1}^n(\frac{i}{n^2}-\frac{i^2}{2n^4})\leq\sum_{i=1}^n\ln(1+\frac{i}{n^2})\leq\sum_{i=1}^n(\frac{i}{n^2})$$
$$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $\varphi$ is golden ratio, how do I prove this $\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$? Given $ \varphi$ is golden ratio, how do I prove this:
$ \displaystyle \tag*{}\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$
... | Hint
For $x \in \mathbb R$
$$\cos x =\frac{e^{ix}+e^{-ix}}{2}$$
Now let $$f(x)= \sum_{n=1}^\infty \frac{x^n}{n^2}.$$
We have $f(0)=0$ and $$f^\prime(x)=\sum_{n=1}^\infty \frac{x^{n-1}}{n}=-\frac{1}{x}\ln(1-x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int {\frac{{dx}}{{{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^2}}}} $ My first solution was:
$$\begin{array}{l}
\int {\dfrac{{dx}}{{{{(\sin \dfrac{x}{2} + \cos \dfrac{x}{2})}^2}}}} \\
= \int {\dfrac{{dx}}{{{{\sin }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}}}} \\
= \int {\d... | If $f(x) = \tan x - \sec x + C $
Then $f'(x) = \sec^2 x - \sec \tan x = \sec x ( \sec x - \tan x ) $
Multiplying top and bottom by $(\sec x + \tan x ) $
$f'(x) = \dfrac{\sec x}{\sec x + \tan x } $
Multiplying top and bottom by $\cos x $
$f'(x) = \dfrac{1}{1 + \sin x }$
So, your answer is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is (f, g, h) an orthonormal string for $\langle f,g \rangle = \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(x)dx? I'm working on this exercise about orthogonality related to Fourier series. Basically, the exercise is as follows:
Study the inner product space of the space $C([0,2\pi])$:
$$
\langle f,g\rangle=\frac{1}{2\pi}\int_{0}... | It seems like you are using the terms orthogonal and orthonormal interchangeably in your question. This is not correct. In fact, orthonormality is a stronger condition than orthogonality.
A set of vectors $\{v_i\}_{i\in I}$ in an inner product space is orthogonal if $\langle v_i,v_j\rangle=0$ for all $i\neq j$. That se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 ... | $$
P(x) = \prod_{d \mid 2n+1, d \neq 1} \Phi_d (x)\\
$$
So focus on getting a good decomposition for $\Phi_d (x)$.
$\Phi_d (x)$ is a product of $\phi(d)$ terms. $d$ is not even because it divides $2n+1$ and it is greater than $1$ by assumption. For the factor corresponding to $k$ there is also the factor corresponding ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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how to prove $A+A'B+A'B'C+A'B'C'D=A+B+C+D$ Prove the above relationship by using the Boolean definition. I tried $A+A'B=A+B$, but end up with $A+B+A'B'(C+D)$, how can I go next?
|
I tried $A+A'B=A+B$
Apply this to $\,A+A'B'C=A+B'C\,$ then again to $\,B+B'C=B+C\,$:
$$
\begin{align}
A+A'B+A'B'C &= \color{red}{A+B}+A'B'C
\\ &= \color{red}{A}+B+\color{red}{B'C}
\\ &= A+\color{red}{B+C}
\end{align}
$$
Repeat the steps to prove the equality with all four terms.
[ EDIT ] $\;$ Another way also using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the point... | Using polar coordinates
$(x, y) = (r \cos \theta, r \sin \theta)$
In the range of $ 0 \le \theta \le \dfrac{\pi}{4} $ we want
$ r \le \left( \dfrac{a}{2} - r \cos \theta \right ) $
Hence,
$ r \le \dfrac{ a } { 2 (1 + \cos \theta ) } $
The area enclosed by this polar curve
$ r(\theta) = \dfrac{ a } { 2 (1 + \cos \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx$ Let $f$ be a analytic function in the closed unit circle with its center
at the point $\alpha\in\mathbb{R}$, then:
\begin{equation*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+... | If we convert the integral into a contour integral, we get
$$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx &= \frac{1}{2}\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx \\ &= \int_{-\pi}^{\pi}\frac{f(\alpha+e^{ix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Distribution density of the sum of two independent, uniform ditributioned random variables Random variable $ξ_1$ and $ξ_2$ are independent and have uniform distribution on interval [-a;a]. What is the disribution density of random variable $\eta = 2ξ_1 + ξ_2^2$? Obviously, we should use
convolution $f_{\eta}(y) = \int\... | I will refer to $~ξ_1$ as $X$, $ξ_2$ as $Y$ and $\eta$ as $Z$.
$X, Y$ are independent and we have $X, Y \sim U(-a, a)$ and we need to find density of $Z = 2X + Y^2$. We can first find $F_z(z)$ and then differentiate to find the density.
We first note that $-2a \leq z \leq 2a + a^2$.
$i$) For $-2a \leq z \lt a^2 - 2a$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Weird Problem on Polynomial Roots The polynomial $f(x)=x^3-3x^2-4x+4$ has three real roots $r_1$, $r_2$, and $r_3$. Let $g(x)=x^3+ax^2+bx+c$ be the polynomial which has roots $s_1$, $s_2$, and $s_3$, where
\begin{align*}
s_1 &= r_1+r_2z+r_3z^2, \\
s_2 &= r_1z+r_2z^2+r_3, \\
s_3 &= r_1z^2+r_2+r_3z,
\end{align*}and $z=\d... |
Since the answer is g(1), I need to calculate
$1−s_1^3$.
Well, the real part of $1−s_1^3$ is to be calculated, thus we need to calculate $\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$ where $\overline{z}$ is the complex conjugate of z.
$\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$ $\implies$$\frac{ 2−(s_1^3 + \overline{s_1^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Tracing a circle by a sliding triangle An isosceles triangle with a unit length base is sliding on two lines which make an angle of $60^\circ$ between them. The third vertex traces a circle centered at the intersection of the two lines. What is the altitude of the triangle, and what is the radius of the circle ?
Wha... |
Let one line be horizontal and the other making an angle of $\phi$ with it.
I'll take the origin of the coordinate system at their intersection. Placing the triangle at a general orientation with its base making an angle of $\theta$ with the horizontal as shown above, we can write the following equation from the law ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Differentiability of $f(x,y)= \frac{xy^3}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $0$ for $(x,y)=(0,0)$.
Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ with\begin{align*} f:=\begin{cases} \frac{xy^3}{x^2+y^2},&(x,y)\neq (0,0)\\0, & (x,y)=(0,0). \end{cases} \end{align*}
Show that $f$ is differentiable at $(0,0)$.
My approa... | It looks correct, but it's easier to see that, if $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, then$$\left|\frac{xy^3}{(x^2+y^2)^{3/2}}\right|=\rho|\cos(\theta)\sin^3(\theta)|\leqslant\rho=\sqrt{x^2+y^2}.$$Therefore,$$\lim_{(x,y)\to(0,0)}\frac{xy^3}{(x^2+y^2)^{3/2}}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4393649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$f(x)$ and $g(x)$ are monic cubic polynomials, with $f(x)-g(x)=r$. If $f$ has roots $r+1$ and $r+7$, and $g$ has roots $r+3$ and $r+9$, then find $r$.
Let $f(x)$ and $g(x)$ be two monic cubic polynomials, and let $r$ be a real number. Two of the roots of $f(x)$ are $r+1$ and $r+7$. Two of the roots of $g(x)$ are $r + ... | Some properties of these two polynomials may be worth remarking upon first:
• because $ \ f(x) \ = \ g(x) + r \ \ , \ \ r \ $ real, the curves of the two functions never intersect;
• for the same reason, the quadratic $ \ (b) \ $ and linear $ \ (c) \ $ coefficients are identical ;
• from the given information about... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Is there an elementary method to evaluate the indefinite integral $\int \frac{1}{1+\sin ^{6} x} d x?$ Inspired by the post, I want to increase the power of $\sin x$ by $2$ to $6$,
$$
I=\int\frac{1}{1+\sin ^{6} x} d x.
$$
As usual, we multiply both the numerator and denominator by $\sec^6 x$ and get
$$
\begin{aligned}
I... | Thanks to @Claude Leibovici for his shortest solution. I want to add an elementary but a bit long one. $$
\begin{aligned}
\int \frac{d x}{1+\sin ^{4} x} &=\int \frac{\sec ^{4} x}{\sec ^{4} x + \tan ^{2} x} d x \\
\\
&=\int \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}+t^{4}} d t, \quad \textrm{ where } t =\tan x \\
&=\int \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$? What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$?
I know the answer is going to be $\frac{6}{5}$ because it will occur when there is equality between th... | Here's a "Only 2-variable AM-GM" approach.
(Per the comments: Since the expression is homogenous, I'm working with $a+b+c = 3 $ to simplify the working. There isn't a huge difference with working with $a+b+c = 2$, but the constants get uglier.
Also I'm restricting to positive values, since otherwise there is no minimum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $\sum_{n=0}^\infty \big( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\big)$ without power series or integration? In a calculus, I have to find a direct way to compute the sum
$$\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right).$$
I tried to reform a partial sum:
$$\sum_{n=0}^N a_n = \fra... | Update: This uses integration, which is something OP sought to avoid.
$$\begin{split}
\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right)
&=\sum_{n=0}^\infty \int_0^1x^{3n}(x^3-3x+2)dx\\
&= \int_0^1\frac{x^3-3x+2}{1-x^3}dx\\
&= \int_0^1 \left( \frac 3 {x+x^2+1}-1\right)dx\\
&= \left[ 2{\sqrt 3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Proving integral identity via Substitution my task is to prove that $\int_{0}^{\infty} \frac{t^a}{sinh(t)} \frac{x}{x^2+t^2} dt= x^a \int_{0}^{\infty} \frac{t^a}{sinh(xt)} \frac{1}{1+t^2} dt$
so starting with the integral on the right hand side and by substituting u=xt one gets
$x^a \int_0^{\infty} \frac{\frac{u^a}{x^a... | I'm going to assume that $x>0$, so
\begin{align}
x^a \int_{0}^{\infty} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt&=x^a \lim_{b\to+\infty}\int_{0}^{b} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt\qquad(\text{doing $u=xt$})\\
&=x^a \lim_{b\to+\infty}\int_{0}^{bx} \frac{\frac{u^a}{x^a}}{\sinh(u)} \frac{1}{1+ (\frac{u}{x})^2} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What technique can be used to prove this inequality? For $a,b\in(0,1)$, prove
$$\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\ge \frac{\sqrt{181}}{6}$$
where the equality holds for $a=\frac49$ and $b=\frac25$.
Attempted for a while, it seems no standard inequality tricks can... | If you're looking for 'standard inequalities', then check Minkowski's inequality,
$$\sqrt{\color{red}{a^2}+\color{blue}{b^2}} + \sqrt{\color{red}{(1-a)^2}+\color{blue}{\tfrac14}} + \sqrt{\color{red}{\tfrac49}+\color{blue}{(1-b)^2}} \geqslant \sqrt{\color{red}{(a+1-a+\tfrac23)^2} + \color{blue}{\left(b+\tfrac12+1-b \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral with elliptical coordinates Good evening everyone. I'm doing an integral of which I know the result but it comes out different to me.
Anyone able to tell me where am I wrong?
The result is $\frac{64}{27} \sqrt{3} \pi$.
The starting integral was a triple integral. Passing in elliptical coordinates I found:
$x =... | In short, you are missing the Jacobian of transformation in your integral. If you plug that in, rest of your work is correct.
If the projection of $z = x + 2$, intersected by the paraboloid surface, in xy-plane is $E$,
$x + 2 \geq 2 \sqrt{ (x^2 + y^2)} \implies 3x^2 - 4x + 4 y^2 \le 4$
$ \displaystyle E: ~\left(x - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that: $ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}\le abc+\frac{1}{2}\sqrt[3]{\frac{(a^{2}+bc)^{2}(b^{2}+ca)^{2}(c^{2}+ab)^{2}}{abc}}$
Let $a,b,c>0$. Prove that: $$a b \sqrt{a b}+b c \sqrt{b c}+c a \sqrt{c a} \leqslant a b c+\frac{1}{2} \sqrt[3]{\frac{\left(a^{2}+b c\right)^{2}\left(b^{2}+c a\right)^{2}\left(c^{2}+a b\r... | Using AM-GM, it suffices to prove that
$$ab\frac{a + b}{2} + bc\frac{b + c}{2} + ca\frac{c + a}{2}
\le abc + \frac12\sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}$$
or
$$a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2 - 2abc
\le \sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}.$$
Since the inequality is h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x$? Stuck by the integral
$$\int_{0}^{1} \sqrt{\frac{x}{1-x^{3}}} d x,$$
I finally solve the the problem using Beta Function.
Then I generalize the result to
$$
I(m,n)=\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x
$$
can be tackled by the Beta function by let... | $$\int \sqrt{\frac{x^{m}}{1-x^{n}}}\, dx=\frac 2{2+m}x^{\frac{m}{2}+1} \, _2F_1\left(1,\frac{m+n+2}{2 n};\frac{m+2n+2}{2
n};x^n\right)$$
$$\int_0^1 \sqrt{\frac{x^{m}}{1-x^{n}}}\, dx=\frac{\sqrt{\pi }\, \Gamma \left(\frac{m+2}{2 n}\right)}{n\, \Gamma
\left(\frac{m+n+2}{2 n}\right)} \quad \text{if} \quad \Re(n)>0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Application of squeeze theorem to $\displaystyle\lim_{x \rightarrow 0}x^4\cos(\frac{\pi}{x^2})+1$ Use squeeze theorem to evaluate
$$\lim_{x \rightarrow 0}x^4\cos(\frac{\pi}{x^2})+1$$
Can I sandwich it between $-x^4\cos(\frac{\pi}{x^2})+1$ and $x^4\cos(\frac{\pi}{x^2})+1$?
| You can squeeze using $1-x^4$ and $1+x^4$. First, show $1-x^4 \leq 1+x^4cos\left(\frac{\pi}{x^2}\right) \leq 1+x^4$. Then we can use the squeeze theorem to show $\displaystyle\lim_{x\rightarrow 0} 1+x^4cos\left(\frac{\pi}{x^2}\right) = 1$ as $\displaystyle\lim_{x\rightarrow 0} 1 + x^4 = 1 = \displaystyle\lim_{x\rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
... | There are arbitrary length examples with real numbers. Moreover, you can take almost arbitrary first $k-1$ numbers $a_1$, $a_2$, ... , $a_{k-1}$ and always exists $a_k$ such that $$(a_1+a_2+...+a_{k-1}+a_k)^2=a_1^2+a_2^2+...+a_{k-1}^2+a_k^2$$
Let's prove it: Mark partial sums as $a_1 +...+a_{k-1}=A \neq 0$ and $a_1^2+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 6,
"answer_id": 2
} |
$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that:... | Let $x = a + 2, \; y = b + 2, \; z = c + 2$
Hence, we need to prove that
$$\frac{4-x}{x} + \frac{4-y}{y} + \frac{4-z}{z} \geq 1$$
$$\implies \frac{4}{x} - \frac{x}{x} + \frac{4}{y} - \frac{y}{y} + \frac{4}{z} - \frac{z}{z} \geq 1$$
$$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} - 3 \geq 1$$
$$\implies \frac{4}{x} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\g... | I've wanted to give U a little help with the thinking here.
we have the base case: $1-\frac{1}{\sqrt{2}}<\frac{2}{2^{2}} \longrightarrow -\frac{1}{\sqrt{2}}\lt -\frac{1}{2} \\\\\text{and that's true because}\longrightarrow \sqrt{2}\lt 2$
let there be n such that: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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A problem related to coefficient of $\int{(t-a)^n}dt$ Let's make some antiderivative of $(t-a)^n$.
When $n=1$, we get $\frac{1}{2}t^2-at$ for the antiderivative (ignoring some constant). Now when we put $t=a$, we get $-\frac{1}{2}a^2$ and $-\frac{1}{2}$ is the coefficient.
When $n=3$, we get $\frac{1}{4}t^4-at^3+\frac{... | Setting the constant equal to $0$ and taking $t=a$ means you are computing :
$$I_n = \int_0^a (t-a)^n\text dt$$
A change of variable $x = t-a$ tells us that :
$$I_n = \int_0^a(-x)^n\text dx$$
And we can compute :
$$I_n = \frac{(-1)^n a^{n+1}}{n+1}$$
In particular when $n$ is odd, you have :
$$I_n = -\frac{a^{n+1}}{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why can't a recurring pattern be found by using the second term in sine graph? In the question
$\sin(x) = \sin (4x), 0 < x < \pi$
I can use the equation
$4x = \pi - x$ ,
$4x = 2\pi + x$ ,
$4x = 3\pi - x$ ,
to work out the solutions, which are $\pi/5, 2\pi/3, 3\pi/5$.
But $\sin(x) = \sin(\pi - x)$.
I am thinking that mu... | The sine of an angle in standard position (vertex at the origin, initial side along the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.
Clearly, $\sin\theta = \sin\varphi$ if $\theta = \varphi$. By symmetry, $\sin\theta = \sin\varphi$ if $\theta =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are,
$$x_1=\sqrt{2}+\sqrt{3}$$
$$x_2=\sqrt{2}-\sqrt{3}$$
$$x_3=-\s... | There are two issues here: First, you seem to have made a data entry error, which resulted in a non-integer answer. Second, $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ isn't integrally closed, so you're computing the discriminant of some non-maximal order of $\mathbb{Q}(\sqrt{2} + \sqrt{3})$.
The following SageMath code verifies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of ordered triples $(a,b,c)$ such that $a+b+c=2019$ **How many ordered triples $(a, b, c)$ are there satisfy the following criteria **
*
*$a,b,c$ are positive numbers
*$a+b+c=2019$
*a, b, c form increasing arithmetic progression
*a , b and c are suitable to be sides of triangle
My idea includes application... | $\color{blue}{1-)}$ $a ,b,c$ form increasing arithmetic progression , so $a<b<c$
Let say $b=a+x$ , and $c=b+x$ where $x$ is positive integer
*
*$c=a+x+x$
*$b=a+x$
$a+b+c =3a+3x$ where $a,x$ all are positive integers
$\color{blue}{2-)}$ $a , b$ and $c$ are suitable to be sides of triangle , then
*
*$a+b > c \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $-1 \le x \le 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method) If $-1 \leq x \leq 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method)
As stated in the problem, I can't use calculus. Therefore, I'm using things I've learnt so far instead:
One of the things I ... | We have
$$(x + \sqrt{1 - x^2})^2
+ (x - \sqrt{1 - x^2})^2
= 2x^2 + 2(1 - x^2) = 2 \tag{1}$$
which results in
$$x + \sqrt{1 - x^2} \le \sqrt 2.$$
(Note: Alternatively, we my use $(a + b)^2 \le 2(a^2 + b^2)$ to get
$(x + \sqrt{1 - x^2})^2 \le 2[x^2 + (1 - x^2)] = 2$.)
Also, when $x = 1/\sqrt 2$, we have $x + \sqrt{1 - x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to tackle $\int_{0}^{\frac{\pi}{2}}y \ln (1+\cos y)\,d y$? I recently encounter an integral problem consisting the integral
$$
I:=\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y,
$$
I tried to tackle $I$ using the double angle formula and the result of $$
\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y
$$
\begin{aligned}
... | Another solution, obtained by one integration by parts is
$$\int y \log (1+\cos (y))\,dy=$$ $$\frac{1}{2} y^2 \log (1+\cos (y))+2 i y \text{Li}_2\left(-e^{i y}\right)-2 \text{Li}_3\left(-e^{i y}\right)+\frac{i
y^3}{6}-y^2 \log \left(1+e^{i y}\right)$$ giving for the definte integral
$$\int_0^{\frac \pi 2} y \log (1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3... | The expression is the product of the roots of the quartic equation satisfied by $\,z = x^3 + 1\,$. Using polynomial resultants, the equation in $\,z\,$ is $\,\text{res}(x^4 - x^3 + 2 x^2 + x + 1, z - x^3 - 1, x)=0\,$ which results in $\,z^4 + 4 z^3 - 4 z^2 - 16 z + 16 = 0\,$, so the expression evaluates to $\,16\,$.
If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Generating Function Approach giving wrong combination count while normal brute force casework approach giving correct answer I am solving a problem: How many words are less than four letters long and contain only the letters A, B, C, D, and E? Here, 'word' refers to any string of letters.
My Solution 1 uses Generating ... | You will have to consider whether the e.g.f. (exponential generating function) approach is worth while here for such a short problem. However, the approach would stand you in good stead for more complex problems.
You will have to extract from the expansion of $(1+x+x^2/2!+x^3/3!)^5$,
(coefficient of $x$) + $2!$(coeffic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$A^2 - B^2$ is an invertible matrix and $A^5 = B^5$ and $A^3B^2 = A^2B^3$ . Then what is the determinant of the matrix $A^3 + B^3$? $A,B$ are two $3 \times 3$ real-matrices with following three properties.
*
*$A^2 - B^2$ is an invertible matrix
*$A^5 = B^5$
*$A^3B^2 = A^2B^3$
Then what is the determinant of the mat... | $A$ and $B$ are real $3\times3$ matrices, so their eigenvalue polynomials are cubic and hence have at least one real zero. Let $\lambda\in\Bbb{R}$ be an eigenvalue of $B$, and $x\in\Bbb{R}^3$ be the corresponding eigenvector. Then
$$
A^5x=B^5x=\lambda^5x.\tag{1}
$$
We also have
$$
\lambda^3A^2x=(A^2B^3)x=A^3B^2x=\lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.