Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is $f(x)= \frac{1}{x} - \frac{1}{e^x-1}$ monotonic? We have $\displaystyle f(x)= \frac{1}{x} - \frac{1}{e^x-1}$, additionally $f(0)=\frac{1}{2}$. Determine whether $f(x)$ is monotonic.
I tried to do this by checking if $f'(x)<0$, however it does not look very helpful. I don't know if it's tricky, or I am just blind to ... | Observe that
$$f(x) = \frac{1}{x} - \frac{1}{2} \coth \frac{x}{2} + \frac{1}{2}.$$
Hence
$$f'(x) = -\frac{1}{x^2} + \frac{1}{4} \mathrm{cosech}^2 \frac{x}{2}.$$
So it is sufficient to prove that $4\sinh^2 \frac{x}{2} \ge x^2$ for $x \ge 0$. This follows from $\sinh y \ge y$ for $y \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Tangent line problems Problem 1
Find common tangent to the curve:
$y+x^2=-4$ and $x^2+y^2=4$.
My idea:
Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit different... | Another solution for problem 3:
Let solve the more general problem:
Let $p(x)$ be a polynomial of degree $n\ge 4$. Find bitangents to the
curve $y=p(x)$.
Let $y-p(x_0)=p'(x_0)(x-x_0)$ be the equation of the tangent at the point $(x_0,p(x_0))$. It will be a bitangent if the abscissae equation for the intersection po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Derivation and tangent problem set Problem 1
On the curve $y=\frac{1}{1+x^2}$ find a point in which tangent line is parallel to the horizontal axis.
My idea:
Let's find $y'$.
$$y'=\frac{-2x}{(1+x^2)^2}$$
If we want a tangent line to be parallel to x-axis it must look like this $t...y=kx+l$ such that $k=0$.
Let's fin... | The solutions to problems 1) and 2) are correct.
For problem 3) note that you are searching the straight lines of equation $(y-3)=m(x-1)$ that are tangent to the curve $3x^2-6x+2y^2-3=0$. This means that you want the values of $m$ such that the system:
$$
\begin{cases}
(y-3)=m(x-1)\\
3x^2-6x+2y^2-3=0
\end{cases}
$$
has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $N$ with $N^2=10^4M+N$ I'm a high school student in France and I participated in a math olympiad, and there was a question which I found impossible to solve. Maybe they are here some people who can help me:
We take a number $N$ which can be written by numbers $a,b,c$ and $d$ in
the same order.
so $N=a(10^3)+b(1... | $$N^2-N=10^4 \cdot M$$
$$N(N-1)=2^4\cdot 5^4\cdot M$$
$N(N-1)=16\cdot 625\cdot M$
$\gcd(N,N-1)=1$
Then 1)$$N=2^4k; N-1=5^4l, k,l \in \mathbb N$$ or 2)$$N-1=2^4k; N=5^4l, k,l \in \mathbb N$$
1) $10^4\le N-1=5^4l \le 10^5 \Rightarrow 2\le l \le 15$ and $16|N$
2)$10^4\le N=5^4l \le 10^5 \Rightarrow 2\le l \le 15$ and $16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to prove that there are no positive integer solutions $(x, y)$ to the equation $x^2 - y^2 = 1$ Prove the following:
Theorem. There are no positive integer solutions $(x, y)$ to the equation
$x^2-y^2=1$.
| Here is another take.
$x^2-y^2=1$ implies $x^2=y^2+1$.
$x^2=y^2+1 > y^2$ implies $x>y$, that is, $x\ge y+1$.
But then $x^2 \ge (y+1)^2 = y^2+2y+1 \ge y^2+2+1 > y^2+1$, since $y \ge 1$.
Thus, $x^2 > y^2+1$, contradiction.
Another, more succinct way to express this is:
$$
y^2 < y^2+1 < (y+1)^2
$$
Thus, $y^2+1$ is strict... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $\sin y+\cos y=\frac{1}{2}$ then find $\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y}$
If $$\sin y+\cos y=\frac{1}{2} \tag{1}$$ Then find $$x=\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y} \tag{2}$$
Given that $$\sin y=\frac{1}{2}-\cos y$$
Squaring both sides we get
$$8\cos^2y-4\cos y-3=0$$
Hence
$$\cos y=\frac... |
Can i get any better approach
How about the following way?
Let $s=\sin y,c=\cos y$.
Squaring the both sides of $s+c=1/2$ gives
$$s^2+2sc+c^2=\frac 14\quad\Rightarrow\quad sc=-\frac 38.$$
Hence,
$$\begin{align}\frac{s^3}{c^2}+\frac{c^3}{s^2}&=\frac{s^5+c^5}{(sc)^2}\\\\&=\frac{(s^2+c^2)(s^3+c^3)-s^2c^2(s+c)}{(sc)^2}\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Seeking combinatorial or group theoretic proof for permutation identity While working on another problem, I found the following combinatorial equality, but I got it analytically, and I'm curious to find a counting argument.
Fix $n$ a positive integer. For $n_1\leq n_2\leq \cdots \leq n_k$ with $\sum n_i=n$, let $s_{n_1... | Permutations with odd cycles only and cycle count marked have
the species
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_{=3}(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_{=5}(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_{=7}(\mathcal{Z})
+ \cdots).$$
which yields the EGF
$$G(z,u) =
\exp\left(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Generating function for the partition function Could someone explain what is the reasoning behind the following equality? Or maybe direct me to a proof of the following equality?
$$\sum_{n=0}^{\infty}p(n)x^n = \prod_{k=1}^{\infty}(1-x^k)^{-1}$$ where $\sum_{n=0}^{\infty}p(n)x^n$ is the generating function for the parti... | For the basic generating functions in term of $(1 - x^k) ^ {-1}$, we can write them out as a infinite sum.
$(1 - x) ^ {-1}$ = $1 + x + x^2 + x^3 + ....$
$(1 - x^2) ^ {-1}$ = $1 + x^2 + x^4 + x^6....$
$(1 - x^3) ^ {-1}$ = $1 + x^3 + x^6 + x^9....$
and so on.
So $\prod(1 - x^k) ^ {-1}$ is just an infinite product of
($1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the quadratic equation $x^2+(2-\tan\theta)x-(1+\tan\theta)=0$ has two integral roots, If the quadratic equation $x^2+(2-\tan\theta)x-(1+\tan\theta)=0$ has two integral roots,then find the sum of all possible values of $\theta$ in interval $(0,2\pi).$
The given quadratic equation is $x^2+(2-\tan\theta)x-(1+\tan\thet... | Let $t=\tan \theta$. We need that $t^2+8$ be a perfect square, so $$t^2+8=s^2\qquad(*)$$
When $\theta\in (0,2\pi)$, the value $t$ can be all real but if $t$ and $s$ must be integers we have $s=t+h$ where $h$ is a positive integer. It follows $$2ht+h^2=8\qquad (**)$$ An obvious solution of $(*)$ is $t=1$ so for $(**)$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Fewest steps to show $\sin^2(s) = \frac{1}{2} - \frac{1}{2} \cos(2s)$? Is there a method for getting from this statement
$
\sin^2(s)
$
to this statement
$
\frac{1}{2}\ -\ \frac{1}{2}\cos(2s)
$
in fewer steps than the following?
$\begin{align}
\cos(2s) &= \cos^2(s) - \sin^2(s) \\
\frac{1}{2} \cos(2s) &= \frac{1}{2} \cos... | You can certainly eliminate several of the steps. I might write it as
$$\cos 2s = \cos^2 s - \sin^2 s\tag{dbl angle formula}$$
$$\cos 2s = 1-2\sin^2 s\tag{$\cos^2$ in terms of $\sin^2$}$$
$$1-\cos 2s = 2 \sin^2 s\tag{subtr both sides from $1$}$$
$$\tfrac12 - \tfrac12\cos 2s = \sin^2 s\tag{halve both sides}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\... | Let $\cos\theta=c,\sin\theta=s$. Then,
$$\begin{align}&s(2+3s+5sc-7c^2+5c)-(1-c)(2s+3+5c^2+7cs+8c)\\&=5c^3+5cs^2-5c+3c^2+3s^2-3\\&=5c(c^2+s^2-1)+3(c^2+s^2-1)\\&=0\end{align}$$
from which the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| \begin{align*}
\frac{1}{n(n+3)} &= \frac{1}{3n}-\frac{1}{3(n+3)} \\
\sum_{n=1}^{\infty} \frac{1}{n(n+3)} &=
\sum_{n=1}^{\infty}
\left[ \frac{1}{3n}-\frac{1}{3(n+3)} \right] \\
&=\lim_{N\to \infty}
\left[
\sum_{n=1}^{N} \frac{1}{3n}-\sum_{n=1}^{N} \frac{1}{3(n+3)}
\right] \\
&=\sum_{n=1}^{3} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Determine whether $x^2 - 14x + 30 \equiv 0$ mod 1615 is solvable. If so, find its solutions... Determine whether $x^2 - 14x + 30 \equiv 0\pmod{1615}$ is solvable. If so, find its solutions.
I assume the best way to solve this is via Chinese Remainder Theorem, but first i would have to break down the mod and if there is... | $$x^2-14x+30\equiv 0\pmod{5\cdot 17\cdot 19}$$
$$\iff (x-7)^2\equiv 19\pmod{5\cdot 17\cdot 19}$$
$$\iff \begin{cases}(x-7)^2\equiv 19\equiv 2^2\pmod{5}\\(x-7)^2\equiv 19\equiv 6^2\pmod{17}\\(x-7)^2\equiv 19\equiv 0^2\pmod{19}\end{cases}$$
$$\iff \begin{cases}x-7\equiv \pm 2\pmod{5}\\x-7\equiv \pm 6\pmod{17}\\x-7\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that $\gcd(80,8a^2+1)=1$ Show that $\gcd(80,8a^2+1)=1$
Let $\gcd(80,8a^2+1)=d$, then we have: $d|80a^2+10,80a^2\Rightarrow\ d|10$
So $d=1\ or\ 2\ or\ 5\ or\ 10$
Obviously $d$ can't be $2\ or\ 10$,but how can we show $d$ can't be 5??
|
$\gcd(80,8a^2+1)$
Factors of $80$ are $8$ and $10$.So,if $\gcd\neq0$ then $8a^2+1$ must have $8$(or $2$ if you consider prime factors) or $10$ as factor.But clearly $8a^2+1$ cannot have $8$(or $2$) or $10$ as factors.Hence they have no common factors.hence,$\gcd=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}?$ For positive real numbers $a, b, c$, show that
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}.$$
I don't know how to solve this at all. Can you pro... | We have to prove that $\implies$
$\sum_{cyclic}\frac{a^2}{b}\geq{\frac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ca}}$
$\implies (ab+bc+ca)(\sum_{cyclic}\frac{a^2}{b})\geq(a+b+c)(a^2+b^2+c^2)$
Now expand both sides and many terms are cancelled out.Then we get the
inequality $\implies(\sum_{cyclic}\frac{ca^3}{b})\geq(ac^2+ba^2+cb^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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False Proof that $\sqrt{4}$ is Irrational Everyone with any basic knowledge of number theory knows the classic proof of the irrationality of $\sqrt{2}$. Curious about generalizations using elementary methods, I looked up the irrationality of $\sqrt{3}$, and found the following:
Say $ \sqrt{3} $ is rational. Then $\sq... | Just because $a^2$ is divisible by $4$, that doesn't mean $a$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
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You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning? You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning?
What I tried was
\begin{align*}
P(X=6) & = \frac{5}{36}\\
P(X=8) & = \frac{5}{... | Everything other than $6$, $7$, or $8$ is irrelevant. So we can imagine that we are repeating an experiment where the probability of $6$ is $5/16$, as is the probability of $8$, while $7$ has probability $6/16$.
Condition on the result of the first toss. Suppose it is a $6$. Then $6$'s become irrelevant, and the race ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$? How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$?
So I think we expand to $x^2$ since the lowest term for $\ln(1+x)$ is $x$
Let $u=\arctan{(x)}$
$\lim\limits_{x\to 0}\frac{e^{\arctan{(x)... | Let's try to simplify the expression first. We have
\begin{align}
L &= \lim_{x \to 0}\frac{e^{\arctan x} - xe^{\pi x} - 1}{(\log(1 + x))^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - xe^{\pi x} - 1}{\left(\dfrac{\log(1 + x)}{x}\right)^{2}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x + \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Approximate roots of nonlinear equation (non-integer polynomial) In case of pulsating bubble arising from underwater explosion, bubble radius satisfies the following equation.
$x^3\dot{x}^{2} + x^3 + \frac{k}{x^{3(\gamma-1)}} = 1$
The minimum and maximum bubble radii can be obtained by substituting $\dot{x}=0$ in the a... | Case $x$ maximum and $k$ small :
$x^3$ is large compared to $\frac{k}{x^{3(\gamma-1)}}$
$x^3=1-\frac{k}{x^{3(\gamma-1)}} \quad\to\quad x=\left(1-\frac{k}{x^{3(\gamma-1)}}\right)^{1/3} $
First approximate: $x\simeq 1 \quad\to\quad x\simeq \left(1-\frac{k}{1}\right)^{1/3} \simeq 1-\frac{k}{3}$
Second approximate: $x\sime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving the Diophantine Equation $x^2 - y! = 2001$ and $x^2 - y! = 2016$ I had recently faced a problem:
Solve the Diophantine Equation $x^2 - y! = 2001$.
Solving it was quite easy. You show how $\forall y \ge 6$, $9|y!$ and since $3$ divides the RHS, it must divide the LHS and if $3|x^2 \implies 9|x^2$ and so the LH... | Since $2016= 2^5 \cdot 3^2 \cdot 7$, we can use the same argument with $2$ or $7$ instead of $3$.
With $2$ we have:
If $y \ge 8$, then $2^5 \mid y!$ and so $2^5 \mid x^2$. This implies that $2^3 \mid x$ and, since $2^6 \mid y!$, we'd have $2^6 \mid 2016$, which is false.
Therefore, we only have to test $y \le 7$. The o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Let $\textbf{v} = (1, 1, 1, 1)$. Find a basis for... Let $\textbf{v}=(1,1,1,1)$. Find a basis for $\{\textbf{u}\in\Bbb{R}^4\ |\ \textbf{u}\cdot\textbf{v}=0\}$
How can I do this? In particular, I do not understand $\textbf{u}\cdot\textbf{v}=0.$
| Let
$$u=\begin{pmatrix}x\\y\\z\\w\end{pmatrix}$$
Then,
$$u\cdot v=0\Rightarrow x+y+z+w=0$$
$$w=-x-y-z$$
$$u=\begin{pmatrix}x\\y\\z\\-x-y-z\end{pmatrix}=x\begin{pmatrix}1\\0\\0\\-1\end{pmatrix}+y\begin{pmatrix}0\\1\\0\\-1\end{pmatrix}+z\begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$$
The set
$$\left\{\begin{pmatrix}1\\0\\0\\-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to simplify $y = \sin(\frac{\arcsin(x)}{n}), n≥1$? $y = \sin(\frac{\arcsin(x)}{n}), n≥1$
I know that:
$\lim \limits_{x \to 0} \frac{x}{y} = n$
But I can't figure out what the curve of $x/y$ practically represents. Is there an obvious simple solution?
| Let $\theta = \arcsin x$ and we have to assume that $|x| \lt \dfrac{1}{\sqrt 2}$.
Then $\tan \theta = \dfrac{x}{\sqrt{1-x^2}}$ and $|\tan \theta| \lt 1$.
\begin{align}
\cos \dfrac{\theta}{n} + i \sin \dfrac{\theta}{n}
&= (\cos \theta + i \sin \theta)^{\frac 1n} \\
&= (\cos \theta)^{\frac 1n}(1 + i \tan \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods.... | You already have a proof using AM-GM and Muirhead. For a pure AM-GM proof, note that $\frac12 \left(\frac{a}b+\frac{b}a\right) \geqslant 1$ and so on, so half the LHS takes care of the $3$ on RHS.
For the rest, note you can cyclically sum the (weighted) AM-GMs:
$$\frac5{18} \left(\frac{a}b+\frac{a}c\right)+\frac2{18}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$ Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.
I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$.
My reasonin... | Let's see what we can say about
comparing
$s_1(k)
=\sum_{n=k}^{\infty} \frac1{n^a}
$
with
$s_2(k)
=\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c
$
for large enough $k$,
where
$a > 1$ and $b > 1$
so the sums converge.
Using the integral approximation,
$s_1(k)
=\sum_{n=k}^{\infty} \frac1{n^a}
\approx \int_k^{\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$ Show that $$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$$ where $a,b>0$. I'm not sure how to simplify this. Any solutions or hints are greatly appreciated.
| You should take the partial fraction of $\dfrac{1}{(x^2+a^2)(x^2+b^2)}$.
This way, it is easier for you to integrate the expression and you should expect the $\arctan$ function in the numerator.
Recall:
$\dfrac{1}{(x^2+a^2)(x^2+b^2)}=\dfrac{Ax+B}{x^2+a^2}+\dfrac{Cx+D}{x^2+b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve limit $\lim_{x\to 0} \frac{x - \sin (x)}{(x \sin (x))^{3/2}}$ Hi I must solve the next limit $\infty - $\infty$ usibg L'Hopital and Taylor series.
$\lim\limits_{x \to 0} \frac{x - \sin (x)}{(x \sin (x))^{3/2} }$
I tried to eliminate the root with multiplying by
$1 = \frac{(x \sin(x))^{\frac{1}{2}})}{(x \sin ... | You have with Taylor series around $0$ : $$\sin(x)=x-\frac{x^3}{6}+o(x^3)$$
$$\lim\limits_{x \to 0} \frac{x - \sin (x)}{(x \sin (x))^{3/2} }=\lim\limits_{x \to 0} \frac{x - x + \frac{x^3}{6}+o(x^3)}{(x (x+o(x))^{3/2} }=\lim\limits_{x \to 0} \frac{\frac{x^3}{6}+o(x^3)}{x^3 +o(x^3) }= \frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0$ Prove that
$$6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0$$
Expression can be verified for different values of $A$ such as $\frac\pi4,\frac\pi2$ etc. But to prove it for general value of A?
| Set $x=\sin^2A$ and $y=\cos^2A$. Then $x+y=1$ and
\begin{align}
1&=(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\\
&=x^5+y^5+5xy(x^3+y^3)+10x^2y^2(x+y)\\
&=x^5+y^5+5xy(x+y)(x^2-xy+y^2)+10x^2y^2\\
&=x^5+y^5+5xy(x^2-xy+y^2)+10x^2y^2\\
&=x^5+y^5+5xy((x+y)^2-3xy)+10x^2y^2\\
&=x^5+y^5+5xy(1-3xy)+10x^2y^2\\
&=x^5+y^5-5x^2y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the value of $ab+ 2cb+\sqrt3 ac$? Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$, $b^2+c^2=9$ and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find... | What a prowess, @Piquito ! Here is another way of treating this issue.
Its "pros": it is systematic. Its "cons": it necessitates a computer algebra system (I used Mathematica).
Let us give the following numbers to the 3 equations:
$$\begin{cases}a^2+\sqrt 3ab+b^2&=&25 \ \ &(1)\\b^2+c^2&=&9 \ \ \ \ \ &(2) \\a^2+ac+c^2&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Let $f(x)=x^5$. For $x_1>0$, let $p_1=(x_1,f(x_1))$.Draw a tangent at the point $p_1$ Let $f(x)=x^5$. For $x_1>0$, let $p_1=(x_1,f(x_1))$. Draw a tangent at the point $p_1$ and let it meet the graph again at point $p_2$. Then draw a tangent at $p_2$ and so on . Show that , the ratio $\frac{A(\bigtriangleup p_np_{n+1}p... | Expanding on my comment above, consider any point $P(a, a^5)$ other than the origin on the curve $f(x)$. The slope of the tangent at $P$ is $5a^4$. Hence the point of intersection is $Q(b, b^5)$, where $a \neq b$ and
$$5a^4 = \frac{b^5-a^5}{b-a} \implies b^3 + 2ab^2+3a^2b+4a^3 = 0$$
$$\implies r^3+2r^2+3r+4 = 0$$
whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What's the solution to this puzzle? I saw this on Instagram with no solution and was wondering what the answer is. I got $33$.
$$1+4=5$$
$$2+5=12$$
$$3+6=21$$
$$8+11=?$$
| 96
For each line add the two numbers and store it in D. Add D to the C from the previous step.
A+B=C D
1+4=5 5 (4+1 gives 5)
2+5=12 7 (5+2 is 7. Add 7 to 5 = 12 from previous step)
3+6=21 9 (6+3 is 9. Add 9 to 12 = 21 from previous step)
4+7=32 11 (4+7 is 11. Add 11 to 21 = 32 from previo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{... | The numerator is
$$x^2-\frac{x^6}2-x^2\Bigl(1-\frac{x^4}2+\frac{x^8}{24}+o(x^8)\Bigr)=-\frac{x^{10}}{24}+o(x^{10}),$$
hence
$$\frac{x^2-\cfrac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}\sim_0\frac{-\dfrac{x^{10}}{24}}{x^{10}}=-\frac1{24}.$$
*Editors Note: Sorry, can't leave comments yet. When you factor in the $-x^2$ you'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$? How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$?
where $_{0}F_{1}$ is the hypergeometric series?
| One may recall the definition of the generalized hypergeometric function
$$
{}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z): = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}
$$ which gives
$$
{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A Chinese Exam Question which is.....quite hard Let $f(x)=x^2-2x-3$, and $x_n$ be some sequence.
$x_1=2$, $x_n =$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n, f(x_n))$.
Find an expression for $x_n$
Find $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.
| The slope of the line connecting points $(4,5)$ and $(x_n,y_n)$ is $m_n=\dfrac{5-y_n}{4-x_n}$
So the equation of the line is
$\begin{equation}
y-5=\dfrac{5-y_n}{4-x_n}(x-4)
\end{equation}$
$x_{n+1}$ will be the point on this line for which $y=0$, so (omitting the algebra)
$\begin{eqnarray}
x_{n+1}&=\dfrac{4x_n-4y_n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve the equation on the interval $\; 0 \le \theta \lt 2\pi $ Hey I have two questions for math exchange! Let me list them first and show you what I have tried. By the way one can not use a calculator on the test review!
Solve the equations on the interval $\; 0 \le \theta \lt 2\pi $:
*
*$2 \cos^2\theta - 3... | *
*$2\cos^2\theta-3\cos\theta+1=0$:
Use the change of variables $x=\cos\theta$, in order to write the equation as
$$
2x^2-3x+1=0
$$
Solving for $x$ yields
$$
x=1\quad or\quad x=\frac{1}{2}
$$
so in terms of $\theta$:
$$
\theta=0\quad or \quad \theta=\frac{\pi}{3}\quad or \quad \theta=-\frac{\pi}{3}
$$
*$\frac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expressing the roots of a cubic as polynomials in one root All roots of $8x^3-6x+1$ are real. (*)
The discriminant of $8x^3-6x+1$ is $5184=72^2$ and so the splitting field of $8x^3-6x+1$ has degree $3$.
Therefore, all three roots can be expressed as polynomials in any one given root.
Indeed, if $a$ is a root, then the ... | Based on the answers in this question, I'm able to finish the general case.
Let $x^3+px+q$ be a cubic with rational coefficients and having real splitting field of degree $3$. Let its roots be $a,b,c$. Then
$$b+c=-a, \qquad bc=-\dfrac{q}{a}$$
and $b,c$ are the roots of the quadratic
$$
h(x)=(x-b)(x-c)=x^2+ax-\dfrac{q}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
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Convert Integral Rectangular to Polar How can convert this problem
$$
\int_0^2 \int_x^\sqrt{8-x^2} \left(x^2+y^2\right)^{3/2} dydx
$$
I convert limits and funtion to polar cordinates as follows:
$$
\begin{split}
r^2 &= x^2+y^2\\
x = 2 &\to r \cos \theta = 2\\
x = 2 &\to r \cos \theta = 0\\
x = y &\to r \cos \theta ... | Do the usual and draw the region, and check that (fill in details)
$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\;\;,\;\;\frac\pi4\le t\le \frac\pi2\;,\;\;0\le r\le 2\sqrt2$$
so your integral becomes (don't forget the Jacobian!)
$$\int_0^{2\sqrt2}\int_{\pi/4}^{\pi/2}r^4\,dtdr=\frac\pi4\int_0^{2\sqrt2}r^4\,dr=\frac\pi{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding A 1-form on $R^2 - {\{(0,0)}\}$ I want to find a 1-form on $R^2 - {\{(0,0)}\}$ such that $w(Y) = 0$ and $w(X) = 1$. Here, $$X = -y\frac{\partial }{\partial x}
+ x\frac{\partial}{\partial y}\ \text{and}\ Y = x\frac{\partial }{\partial x}
+ y\frac{\partial}{\partial y}.$$
I let $w = adx+bdy$ be a 1-f... | Let $r = x^2 + y^2$. Note that if the values are $a = \frac{x}{r}$ and $b = \frac{y}{r}$, then we will have: $w = \frac{x}{r}$ $dx$ + $\frac{y}{r}$ $dy$. Then, the following follows: $ w(X) = \frac{1}{r} (-xy +xy)$ = $0$,$ w(Y) = \frac{1}{r} (x^2+y^2)$ = $\frac{1}{r} r = 1.$ This contradicts the conditions given ($w(X)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$. Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$.
My method:
If $n = 2$, $2^{n + 1} \le 3^n$ then $2^3 \le 3^2$ is $8 \le 9$, which holds for $n = 2$.
$2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$. Then
$2 \cdot 2 \... | Formal proof by induction.
First, show that this is true for $n=2$:
$2^{2+1}<3^2$
Second, assume that this is true for $n$:
$2^{n+1}<3^n$
Third, prove that this is true for $n+1$:
$2^{n+2}=$
$2\cdot\color\red{2^{n+1}}<$
$2\cdot\color\red{3^n}<$
$3\cdot3^n=$
$3^{n+1}$
Please note that the assumption is used only in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\sum_{r=1}^{n} \frac{r}{2^r}$ as $n\to \infty$
Consider the sequence $u_{n}=\sum\limits_{r=1}^{n} \frac{r}{2^r}$ with $n \ge 1$.
Then the limit of $u_{n}$ as $ n \rightarrow \infty$ is ?
I actually treated each term as an AGP and got $$u_{n}=2-\frac{1}{2^{r+1}}-\frac{n}{2^{r+1}}$$
But how to get the limi... | This is a well-known result due to Nicole Oresme
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=2
\end{align}
Here is the proof
Consider the function
\begin{align*}
f(z)=\frac{1}{1-z}
\end{align*}
The power series expansion of $f(z)$ is
\begin{align}
f(z)=\sum_{n=1}^{\infty}z^n
\end{align}
Now take the derivative of $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x²}-1}{\sqrt{1+x}-\sqrt{1-x}}$ First I need to prove that this limit; $\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}$ converges, then I have to find its limit. Now I don't know how to prove that it converges (these epsilon proofs are still somethin... | The standard way is to transform the limit into
$$
\lim _{x\to 0}
\frac{(\sqrt{1+x+x^2}-1)(\sqrt{1+x+x^2}+1)}
{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}
\cdot
\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
$$
that becomes
$$
\lim_{x\to0}
\frac{x+x^2}{2x}
\cdot
\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| $$\frac{2x-1}{x+2}=\frac{2(x+2)-5}{x+2}=\frac{2(x+2)}{x+2}-\frac{5}{x+2}=2-\frac{5}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating the integral $\int \frac{x^2+x}{(e^x+x+1)^2}dx$
Evaluate $$\int \frac{x^2+x}{(e^x+x+1)^2}dx$$
I tried converting in the form of Quotient rule(seeing the square in the denominator), neither am I able to make the denominators' derivative in the numerator. Some hints would be great. Thanks.
| Let $\displaystyle u=\frac{x^2+x}{xe^x}=(x+1)e^{-x}$ and $\displaystyle dv=\frac{xe^x}{(e^x+x+1)^2}dx,\;$ so $\displaystyle du=-xe^{-x}dx$ and $\displaystyle v=\frac{e^x}{e^x+x+1}$.
Then $\displaystyle\int\frac{x^2+x}{(e^x+x+1)^2}dx=(x+1)e^{-x}\cdot\frac{e^x}{e^x+x+1}-\int-\frac{x}{e^x+x+1}dx=\frac{x+1}{e^x+x+1}+\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$. All of its vertices satisfy $x^2+y^2+z^2=1$
Find the remaining two vertices with respect to $a$ and $b$.
| To solve everything from scratch will result in a too long answer,
so we assume these important properties to be known:
Image 1: Tetraeder in cube construction. (Source)
*
*The tetraeder in the task has the internal arm length $1$, meaning
the distance from origin to top vertex. Then the edge length between
two ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to calculate limit? I'm puzzled with this limit. The answer is -0.5, but how to get it? $\lim_\limits{x \to \infty}1-x+\sqrt{\frac{x^3}{x+3}}$
| I've tried to multiply by conjugate not only part with a variable, but all the expression, and this way I calculate it without replace and L'Hospital's Rule.
$\lim_\limits{x \to \infty} 1-x+\sqrt{\frac{x^3}{x+3}}=\lim_\limits{x \to \infty}\frac{\frac{x^3}{x+3}-(x-1)^2}{\sqrt{\frac{x^3}{x+3}}+x-1} = \lim_\limits{x \to \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Karatsuba multiplication for $x = 123 , y = 100$ Given the numbers $x = 123$ and $y = 100$ how to apply the Karatsuba algorithm to multiply these numbers ?
The formula is
xy=10^n(ac)+10^n/2(ad+bc)+bd
As I understand $n = 3$ (number of digits) and I tried writing the numbers as
x = 10*12+3 , y = 10*10 +0 thus a = 12 ... | $xy=123\cdot 100$
$(12\cdot 10+3)(10\cdot 10+0)$
$xy=10^n(ac)+10^n/2(ad+bc)+bd$, where $n=2
a=12, b=3, c=10, d=0$
subcategory 1:
$ac=12\cdot 10 = 120$
$bd=3\cdot 0 = 0$
$ad+bc= (12+3)(10+0)=150-120-0 = 30..$
$120\cdot 10^2 + 30\cdot 10^1 + 0$
subcategory 2:
$12\cdot 10$
$a=1,
b=2,
c=1,
d=0$
$ac=1$
$bd=0$
$ad+bc=(a+b)(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$ Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$.
This is the last part of a problem that I am working on. So far, we hav... | \begin{align*}
\sum_1^{\infty} \frac 1 {n^3} &= \sum_{\text{even}} \frac 1 {n^3} + s \\
&= \sum_1^{\infty} \frac{1}{(2k)^3} + s\\
&= \frac 1 8 \sum_1^{\infty} \frac 1 {k^3} + s
\end{align*}
Rearrange for the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove this inequality in complex domain (5) Let $z_{1},z_{2},z_{3}\in C$
show that
$$|z_{1}+z_{2}+z_{3}|^2+|(z_{1}-z_{2})(z_{1}-z_{3})|+|(z_{2}-z_{3})(z_{2}-z_{1})|
+|(z_{3}-z_{1})(z_{3}-z_{2})|\le 3(|z_{1}|^2+|z_{2}|^2+|z_{3}|^2)$$
Iif prove following inequality,Use Cauchy-Schwarz inequality we have
$$(|z_{1}+z_{2}+z... | Let $O$, $A$, $B$ and $C$ be points in the plain and $M$ is a gravity canter of $\Delta ABC$.
Hence,
$3(OA^2+OB^2+OC^2)=9OM^2+AB^2+AC^2+BC^2\geq$
$\geq9OM^2+AB\cdot AC+AB\cdot BC+AC\cdot BC$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Weird differential equation - Jacobi? In class we had differential equations of the type
$$y'=\frac{\left(Ax+By \right)y+ \alpha x + \beta y}{\left(Ax+By \right) x+ax+by},$$
where $A,\alpha,a,B,\beta,b$ are constants. The names of the constants were chosen in a way, so that the 3 different constants for each variable l... | For $y'=\dfrac{(x-y)y-x-y}{(x-y)x+x+y}$ ,
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{(x-xu)xu-x-xu}{(x-xu)x+x+xu}$
$x\dfrac{du}{dx}=\dfrac{(1-u)xu-1-u}{(1-u)x+1+u}-u$
$x\dfrac{du}{dx}=\dfrac{(x-1)u-xu^2-1}{(1-x)u+x+1}-u$
$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int _0^{\frac{\pi }{2}}\:\frac{\sqrt[3]{\sin^2\left(x\right)}}{\sqrt[3]{\sin^2\left(x\right)}+\sqrt[3]{\cos^2\left(x\right)}}dx$ how i evaluate this integral with integration by substitution $x=\frac{\pi}{2}-t$
$$\int _0^{\frac{\pi }{2}}\:\frac{\sqrt[3]{\sin^2\left(x\right)}}{\sqrt[3]{\sin^2\left(x\right)... | We have $$\begin{align} I=\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx=
& \int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}-\sqrt[3]{\cos^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Odd binomial sum equality has only trivial solution? Suppose $$\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2} = \sum_{k\ {\rm odd}}^m {m \choose k} 2^{(k-1)/2} 3^{(m-k)/2}.$$
Does $m=n=1$?
Clearly $m \leq n$, and for every $n$ there is at most one $m$.
| Here we find a closed expression for the series and show that a solution with $m$ even is not possible.
We use the convention $\binom{n}{k}=0$ if $0\leq n < k$ and start with the left-hand side. We obtain
\begin{align*}
\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2}&= \sum_{k=0}^n \binom{n}{2k+1} 2^k\tag{1}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing that $\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$ Infinite product
$F_{n}:=[1,1,2,3,5,8,\cdots]$ and
$L_{n}:=[1,3,4,7,\cdots]$
for $n=1,2,3,\cdots$ respectively.
$\frac{1+\sqrt5}{2}=\phi$
Show that,
$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{... | Use $$F_{4k+2}+1=F_{2k+2}L_{2k}\tag{1}$$ with $k=2^{n-1}$. Also note $$\prod_{k=1}^{n}L_{2^{k}}=\prod_{k=1}^{n}\phi^{2^k}+(-\phi)^{-2^{k}}=\frac{\phi^{2^{k+1}}-(-\phi)^{-2^{k+1}}}{\phi^2-(-\phi)^{-2}}=\frac{F_{2^{k+1}}}{F_{2}}\tag{2} $$ From $\text{(1)}$, note that $$\left(1+\frac{1}{F_{6}}\right)\cdot\left(1+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Number theory contradiction I was solving this question
How many pairs of integers $(a,b)$ that satisfies the following conditions:
*
*$1\le a,b\le 42$
*$a^9\equiv b^7\pmod{43}$
and I ran into a contradiction:
We know by Fermat's Little Theorem that $a^{42} \equiv 1 \pmod{43}$ and $b^{42} \equiv 1 \pmod{43}$. Thu... | 3 is a cyclical generator for $Z_{43}$
Every number in between 1 and 42 can be expressed as $3^n$
$a= 3^n, b = 3^m\\
a^9 \equiv 3^{9n} \mod 43\\
b^7 \equiv 3^{7m} \mod 43$
the only numbers that overlap are $3^{21} \equiv 42$ and $3^{42} \equiv 1.$
$a^9\equiv b^7\equiv 1 \mod 43$
$a = 1, 4, 16,21, 41,35, 11,$ and $b = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding an antiderivative $\newcommand{\dx}{\,\mathrm dx}$
I need to find the following:
$$\int \sqrt{\frac{1-x}{1+x}} \cdot \frac 1x \dx$$
Firstly, it has to be that $x\in (-1, 0)\cup (0,1]$. From this, it is implied that $1+x>0$. Also, it holds that $1-x \ge 0$.
What I did is that I multiplied both numerator and d... | Let $x=\cos2y\implies0\le2y\le\pi$
$\sqrt{\dfrac{1-x}{1+x}}=\tan y$
$\implies2\arctan\sqrt{\dfrac{1-x}{1+x}}=2y=\arccos x=\dfrac\pi2-\arcsin x$
But why don't we try with
$$\sqrt{\dfrac{1-x}{1+x}}=u\implies x=\dfrac{1-u^2}{1+u^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the definite integral $\int_0^3 \sqrt{9- x^2} \, dx$ I have been having a problem with the following definite integral:
$$\int_0^3 \sqrt{9- x^2} \, dx
$$
I am only familiar with u-substitution and am positive that it can be done with only that. Any help would be appreciated.
| Here's a quick method: If $y = \sqrt{9-x^2}$ then $y^2 = 9-x^2$ so $x^2+y^2=9$. If you know that $x^2 + y^2 = 9$ is the equation of a circle of radius $3$ centered at $(0,0)$, that means $y^2=9-x^2$ is also an equation of that circle, as is $y = \pm\sqrt{9-x^2}$. So $y = \sqrt{9-x^2}$ (without $\text{“}\pm\text{''}$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Writing answer for $2\cos x+\sin2x=0$ $$2\cos x + \sin 2x = 0\implies
\cos x = 0 \; \&\; \sin x = -1$$
So, the solution my book provides is
$x = π/2 + 2nπ\; \&\; x = 3π/2 + 2nπ$
Why is $3π/2$ (for $\cos$) not included in the general solution set?
|
$$2\cos x+\sin2x=0\\2 \cos x+ 2 \sin x \cos x=0 \\ \to 2 \cos x (1+\sin x)=0\\ \to \\ sinx=-1 ,cosx=0$$
With respect to figure $$\begin{cases}cos=0 & x = \pm \frac{\pi}{2}+2k\pi\\sinx=-1 & x=-\frac{\pi}{2}+2k\pi\end{cases} $$ and $$x=\pm \frac{\pi}{2} +2k\pi$$ include $x=\frac{3\pi}{2} ,\frac{-3\pi}{2},\frac{\pi}{2},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$ Let $x,y,z>0$. Prove that
$$\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$$
On the left-hand side we have arithmetic and harmonic means, while on the right geometric mean. The AM-GM-HM inequality stat... | Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(u)\geq0,$ where
$$f(u)=u+\frac{w^3}{v^2}-\frac{5w}{\sqrt[3]{16}}.$$
Id est, it's enough to prove our inequality for the minimal value of $u$, which happens for equality case of two variables.
Since our inequality is symmetric and homogeneous,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $\sin^{-1}x + \sin^{-1}y=\sin^{-1}(x \sqrt{1-y^{2}}+ y \sqrt{1-x^{2}})$, then what is the area represented by the locus of point $(x, y)$? If $\sin^{-1}x + \sin^{-1}y=\sin^{-1}(x \sqrt{1-y^{2}}+ y \sqrt{1-x^{2}})$, then what is the area represented by the locus of point $(x, y)$? I'm totally blank about this questio... | $$\alpha=sin^{-1}x\to sin\alpha=x \,\,\,\,\,and\,\,cos\,\alpha=\sqrt{1-sin^2 \alpha}=\sqrt{1-x^2}$$
$$\beta=sin^{-1}y\to sin\beta=y \,\,\,\,\,and\,\,cos\,\beta=\sqrt{1-sin^2\beta}=\sqrt{1-y^2}$$
let $t=sin^{-1}x+sin^{-1}y=\alpha+\beta$ , then
$$sin(t)=sin(\alpha+\beta)=sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum and minimum values of $\sin^2\theta+\sin^2\phi$ when $\theta+\phi=\alpha$ Find the maximum and minimum values of $\sin^2\theta+\sin^2\phi$ when $\theta+\phi=\alpha$(a constant).
$\theta+\phi=\alpha\implies\phi=\alpha-\theta$
$\sin^2\theta+\sin^2\phi=\sin^2\theta+\sin^2(\alpha-\theta)$
Let $f(\theta)=\... | Write
\begin{eqnarray}
\sin^2 \phi + \sin^2 \theta &=& \sin^2 \phi + 1-\cos^2 \theta \\
&=& 1+(\sin \phi - \cos \theta) (\sin \phi + \cos \theta) \\
&=& 1+(\sin \phi - \sin ( { \pi \over 2} - \theta)) (\sin \phi + \sin ( { \pi \over 2} - \theta)) \\
&=& 1+4 \cos ({ \phi -\theta + { \pi \over 2}\over 2} ) \sin ({ \ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Solve first order partial diferential equation Consider the Cauchy problem:
$$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
It is easy to show that the characteristic equations are given by:
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}... | Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{du}{dt}=u^2$ , we have $\dfrac{1}{u}=-t+f(y_0)=\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| Hint: Use the fact that $$(2^{3^n})^3 + 1 = \left(2^{3^n} +1\right) \left((2^{3^n})^2- 2^{3^n} + 1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Which is the explanation of the identity $\sum_{k=0}^n {n \choose k}k^2 = 2^{n-2}n(n+1)$? A friend of mine put me a problem some time ago, and after trying to do it, I finally surrendered. I looked for the answer online so maybe I could guess why is it like it is, but I just can't understand anything about it. Could so... | By definition,
$$\binom nk=\frac{n!}{k!(n-k)!},$$ so that
$$\binom nkk=\frac{n!}{(k-1)!(n-k)!}==\frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}=\binom {n-1}{k-1}n.$$
This trick allows you to "absorb" the factor $k$ inside the binomial number.
The trick doesn't work with $k^2$ because the denominator doesn't have a second factor $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
For the general ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ,show that the midpoints of the chords lie on a straight line.
Question: A collection of parallel chords connect pairs of points on an ellipse, as shown
For the general ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, show tha... | $$\Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + a^2(c^2-b^2) = 0 $$
By the midpoint of a parabola, the $x$-value of the midpoint of the intersection of the line and the ellipse will be at
$$\frac{-2mca^2}{2(b^2+a^2m^2)} = \frac{-mca^2}{b^2+a^2m^2}$$
Note that the $y$ value of the midpoint is linearly dependent on the $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
What is the sum of the prime factors of $2^{16}-1$? I know $2^{10}=1024$ and $2^6=64$, but it seems they are not very helpful in solving this problem. There must be a trick to solve the problem in an easy way.
What is the sum of the prime factors of $2^{16}-1$?
| Using $a^2-1=(a-1)(a+1)$ we have:
$$2^{16}-1=(2^8-1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=15\cdot 17\cdot 257=3\cdot 5\cdot 17\cdot 257$$
So, the answer is $3+5+17+257=282$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply... | $$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{(x+3)\color{red}{(x+1)}-(x-2)\color{red}{(x-1)}}{(x-1)(x+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Recurrence Relation with two parameters and Summation This is a recurrence relation with two parameters which came up in a problem I was trying to solve.
Given
$$\begin{align}&A_n=pB_{n-1};\qquad &&B_n=q(A_{n-1}+B_{n-1})\\
&A_4=p; \qquad &&B_4=q; \\
&p+q=1;&&(0< p,q< 1)\end{align}$$
evaluate
$$\sum_{n=4}^\infty... | From your own calculations, notice that you may write
$$\begin{pmatrix} A_n \\ B_n \end{pmatrix} = \begin{pmatrix} 0 & p \\ q & q \end{pmatrix} \begin{pmatrix} A_{n-1} \\ B_{n-1} \end{pmatrix} = \begin{pmatrix} 0 & p \\ q & q \end{pmatrix}^2 \begin{pmatrix} A_{n-2} \\ B_{n-2} \end{pmatrix} = \dots = \begin{pmatrix} 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a ... | Hint: $$(x+y)^3=x^3+y^3+3x^2y+3xy^2=x^3+y^3+3xy(x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Evaluate $\int_{1/10}^{1/2}\left(\frac{\sin{x}-\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}\right)^2dx$ How can I evaluate the following integral?
$$\int_{1/10}^{1/2}\left(\frac{\sin{x}-\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}\right)^2dx$$
I notice that the numerator and the denominator are very similar. So m... | Using product-to-sum formula, we have
$$
\begin{aligned}
2 \cos x(\sin x-\sin 3 x+\sin 5 x) = & \sin 2 x-(\sin 4 x+\sin 2 x)+(\sin 6 x+\sin 4 x) \\
= & \sin 6 x \cdots (1)
\end{aligned}
$$
and
$$
\begin{aligned}
2 \sin x(\cos x+\cos 3 x+\cos 5 x) = & \sin 2 x+(\sin 4 x-\sin 2 x)+(\sin 6 x-\sin 4 x) \\
= & \sin 6 x \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$ and verify that $P$ is a stationary point. A curve is defined by the equation $$2y+e^{2x}y^2=x^2+\frac{2}{e}$$
Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$
\begin{align}
2y+e^{2x}y^2 & = x^2+\frac{2}{e} \\
2\frac{dy}{dx}+2e^{2x}y^2+2e^{2x}... | You have a mistake in the final part because:
$$
e^2e^{-2}=1
$$
so
$$
1-e^2e^{-2}=0
$$
and
$$
\frac{1-e^{2}e^{-2}}{1+e^{2}e^{-1}}=\frac{0}{1+e}=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of k when $f(x) = x^3 -4x^2 + 6x + k$ If $ \ \ a ,b \ \ $and $\ \ c \ \ $ are roots of $ \ \ f(x) = x^3 -4x^2 + 6x + k \ \ $
and $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ \ \ $then find the value of $ \ k $
I have to tell you honestly that I cannot solve this ques... | HINTS to find $k^2$: Let $x=a^2+b^2$, $y=b^2+c^2$, $z=c^2+a^2$. Then $xy+yz+zx=(a^2+b^2+c^2)^2$ and you also know $xyz$ and $x+y+z$. Thus you know everything about $x,y,z$. Now $8a^2b^2c^2=(x+y-z)(x-y+z)(-x+y+z)$ and the R.H.S. is a symmetric polynomial in $x,y,z$, so you can find $a^2b^2c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proof of Leibniz $\pi$ formula I found the following proof online for Leibniz's formula for $\pi$:
$$\frac{1}{1-y}=1+y+y^2+y^3+\ldots$$
Substitute $y=-x^2$:
$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots$$
Integrate both sides:
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$
Now plug in $1$ for $x$:
$$\frac... | Yes, you need an extra step. You need to know this: If
$$
\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots
$$
holds for $-1 < x < 1$ and the series on the right side converges at $x=1$, then the equation also holds at $x=1$.
Arthur beat me to doing it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:
Suppose there is some point at which the line touches the curve, $a$, whi... | There's a simpler way to solve: take a line through $(2,0)$ with variable slope $t$:
$$y=t(x-2),$$
and write the equation for the intersection points with the curvbe has a double root:
$$t(x-2)=\frac{x^2}{x-1}\iff x^2=t(x-1)(x-2)\iff (1-t)x^2+3tx-2t=0.$$
There's a double root if and only if
$$\Delta=9t^2+8t(1-t)=t(t+8)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1... | The integral function is non-negative over the integration range and behaves like $\frac{1}{x^3}$ for large values of $x$, hence it is integrable for sure. Partial fraction decomposition gives:
$$\frac{2}{(x-1)x(x+1)} = \frac{1}{x(x-1)}-\frac{1}{x(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{2}{x}+\frac{1}{x+1}\right)\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Convert $16x^2+56x-80=0$ to the form of $(\text{something})^2=D$ Sorry about this very basic question, I want to convert the
equation $16x^2+56x-80=0$ to the form $(\text{something})^2=D$, I know that the answer is $(4x+7)^2-129$, but how can I convert this without usnig calculator
| You are "completing the square". The firs thing I would do is factor "16" out of the first two terms: $16(x^2+ \frac{7}{2}x)- 80= 0$. The reason I did that I that I know that a "perfect square" is of the form $(x+ a)^2= x^2+ 2ax+ a^2$. Compare that to $x^2+ \frac{7}{2}x$. The first two terms will be the same if $2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Help in proving an inequality Show that $$a^4 + b^4\ge\frac{1}{8}$$ if $a+b=1.$
| $\text {let: } k = \frac {(b-a)}{2}\\
\text {then: } b = \frac {(b+a)}{2} + \frac {(b-a)}{2} = \frac 12 + k\\
a = \frac {(b+a)}{2} - \frac {(b-a)}{2} = \frac 12 - k$
$a^4 + b^4\\
(\frac 12 - k)^4 + (\frac 12 + k)^2\\
[(\frac 12)^4 - 4(\frac 12)^3k + 6(\frac 12)^2k^2 -4(\frac 12)k^3 + k^4] + [(\frac 12)^4 + 4(\frac 12)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$
I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here
| Take two vectors: $v_1 = (a,b)$ and $v_2 = (\sin x, \cos x)$. Their scalar product is $(v_1, v_2) = |v_1||v_2|cos(\phi) = a \sin x + b \cos x$ where $\phi$ is the angle between $v_1$ and $v_2$, but $|v_1| = \sqrt{a^2 + b^2}, |v_2| = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
} |
A certain unique rotation matrix One can find that the matrix
$A=\begin{bmatrix}
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
\end{bmatrix} $
is at the same time $3D$... | The three row vectors of the matrix define a triorthonormal frame, so that the tips of the vectors belong to a unit sphere. The sum of entries being a constant corresponds to a plane in the first octant, $x+y+z=c$, which intersects the sphere along a (small) circle.
You can take any three points on that circle that for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Evaluate an increasing sum of binomial coefficients: $\sum_{k=1}^nk\binom{m+k}{m+1}$ I've been working on a problem and got to a point where I need the closed form of
$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$
I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this:
$$\sum_{k=1}^nk... | You can prove this by induction.
Here is the induction step:
$$
\begin{align*}
\sum_{k=1}^{n+1} k \binom{m+k}{m+1} &= \frac{n((m+2)n+1)}{(m+2)(m+3)}\binom{m+n+1}{m+1} + (n+1)\binom{m+n+1}{m+1} \\
&=\frac{(m+n+2)(m(n+1)+2n+3)}{(m+2)(m+3)} \binom{m+n+1}{m+1} \\
&=\frac{(n+1)(m(n+1)+2n+3)}{(m+2)(m+3)} \cdot \frac{(m+n+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof. I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^... | Using your hint, here is one way to go:
$$1^2 + 3^2 +\cdots +(2n+1)^2 = (1^2 + 2^2 + 3^2 + 4^2+\cdots +(2n+1)^2) - (2^2 + 4^2 + 6^2+\cdots+ (2n)^2)\\
(1^2 + 2^2 + 3^2 + 4^2+\cdots+ (2n+1)^2) - 4(1^2 + 2^2 + 3^2+\cdots+ n^2)$$
$$\frac{(2n+1)(2n+2)(4n+3)}{6} - \frac{4(n)(n+1)(2n+1)}{6}$$
Alternatively, you could use a pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Q27 from AMC 2012(Senior): Five consecutive integers that sum to a perfect square, and the three middle terms sum to a perfect cube.
Five consecutive integers $p,q,r,s,t$,each less than $10000$, produce a sum which is a perfect square, while the sum of $q,r,s$ is a perfect cube.What is the value of $ \sqrt{p+q+r+s+t}$... | Since $\frac{3}{5}x^2=y^3$ is a cube, the number of factors of 5 in $x^2$ must be even and one more than a multiple of 3. So choices are $4$ and $10$. Also the number of factors of 3 in $x^2$ must be even and one less than a multiple of $3$, so choices are $2$ and $8$. Now, $\frac{x^2}{5} \le 10000$, so $x$ must be a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Does this wrong cancellation of $B$ work for $\overline{AB}/ \overline{BC}=A/C$? My teacher says that wrong cancellation of $B$ for the fraction$$\frac{\overline{AB}}{\overline{BC}}=\frac{A}{C}$$ will work for some numbers. I see some trivial cases when $A=B=C$, but are there more of them?
$A$, $B$ and $C$ are some di... | Interesting question! I've solved this before! Let $\overline{AB}=10a+b$, $\overline{BC}=10b+c$, we get$$\frac{10a+b}{10b+c}=\frac{a}{c}$$ and $$b(10a-c)=9ac$$Note that $a$, $b$ and $c$ must be nonzero digits. $c=a$ is our trivial case, which leads to $a=b=c$. For $a \neq c$ let $d=c-a$. Its clear that $d$ cannot take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Fake proof that $\frac{e^x-1}{e^x+1}=e^x$, via integrating $\operatorname{sech} x$ in two ways We start with the integral:
$$\int \text{sech}(x)dx$$
Method 1
\begin{align}
\int \text{sech}(x)dx & = \int\frac{2}{e^x+e^{-x}}dx \\
&= \int\frac{2e^x}{e^{2x}+1}dx
\end{align}
Using the substitution $u=e^x$,
\begin{align}
... | Because $2\arctan\left(\frac{e^x-1}{e^x+1}\right)=2\left(\arctan(e^x)-\frac\pi4\right)=2\arctan(e^x)+C'$.
The results differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Choosing a value so a line and circle intersect a one, two, and no points Let l be a line and C be a circle.
$y=x+d$, where $d$ is to be determined.
$C=x^2+y^2=4$
*
*Pick a value for $d$ so that l and C intersect at one point.
*Pick a value for $d$ so that l and C intersect at two distinct points.
*Pick a value... | For intersection points both equations
$$
y = x + d \\
x^2 + y^2 = 2^2
$$
must hold. Inserting the first into the second we get
$$
y = x + d
$$
and
$$
2^2 = x^2 + (x+ d)^2 = 2 x^2 + 2xd + d^2 \iff \\
2 = x^2 + x d + d^2/2 = (x + d/2)^2 - d^2/4 + d^2/2 \iff \\
2 - d^2/4 = (x + d/2)^2 \iff \\
x = \frac{-d \pm \sqrt{8 - d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$
Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ without using a calculator.
I rearraged to get $4 \cdot 1976^4-1 > 10^{-19.76} \cdot 16 \cdot 1976^7$ and so we have to prove that $4 \cdot 1976^4-1-10^{-19.76} \cdot 16 ... | Seems really simple - am I missing something?
$a = 4\times 1976^3 < 4\times 2000^3 = 32\times 10^9$
$\implies 1/a > \frac{1}{32}10^{-9} > 3\times 10^{-11}$
$b= 16\times 1976^7 > 10^{22}$ and $1/b <10^{-22}$
$\implies \frac{1}{a} - \frac{1}{b} > 3 \times 10^{-11} - 10^{-22} >10^{-19.76}$ (howsoever you interpret that e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Number of positive unequal integer solutions of $x+y+z+w=20$ What is the number of positive different integer solutions of $x+y+z+w=20$, where $x,y,z,w$ are all different and positive?
It would be nice if coding is not used. I am given the answer $552$.
| Just for comparison with Brian Scott's solution, here is the longhand count solution.
As is observed in several places above, it is sufficient to find the number of solutions with $x<y<z<w$ and multiply by 24, because each such solution can be permuted to give 24 solutions to the original question.
Next note that $1+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Evaluate $\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}$ What's the value of $$\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}?$$
I've tried the AM-GM inequality with no luck.
Also tried to right the inequality as:
$$\left... | Clearly the expression is $a_{n}^{1/n}$ where $a_{n} = \dfrac{(2n + 1)!}{n!(n + 1)!2^{n}}$ and hence $$\frac{a_{n + 1}}{a_{n}} = \frac{(2n + 3)!}{2^{n + 1}(n + 1)!(n + 2)!}\cdot\frac{2^{n}n!(n + 1)!}{(2n + 1)!} = \frac{(2n + 3)}{(n + 2)} \to 2$$ so desired limit is also $2$. We have used the following standard result:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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eliminate the parameters Given:
$x = \frac{1}{2} \cos(\theta)$ and $y = 2\sin(\theta)$
Part a)
solving the first one for theta:
1) multiply both sides by $2$:
$$2x = \cos(\theta)$$
2) divide both sides by $\cos (\arccos)$:
$$\cos^{-1}(2x) = \theta$$
Part b)
Now plugging this into the second equation for theta:
1) plug... | This is an idea:
$$\begin{cases}x=\frac12\cos\theta\\{}\\y=2\sin\theta\end{cases}\implies \cos^2\theta+\sin^2\theta=4x^2+\left(\frac y2\right)^2=1$$
and you have an ellipse
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Closed form of the series.
Write the following series in closed form
$$\frac{1}{a-1}+\frac{2}{a-2}+\frac{2}{a-3}+\frac{1}{a-4}+\frac{1}{a-5}+\frac{2}{a-6}+\frac{2}{a-7}+\frac{1}{a-8}+\frac{1}{a-9}+\frac{2}{a-10}+\frac{2}{a-11}+\cdots+\frac{1}{a-4p+3}+\frac{2}{a-4p+2}+\frac{2}{a-4p+1}+\frac{1}{a-4p},$$
where $a$ is... | You have splitted the series in $4$ subseries. We see that for each sub series $(a-j)(a-i)=4,\forall j>i$. So we get the following series for every sub series:
$$(1):\sum_{i=0}^{p-1} \frac{1}{a-(4i+1)}\\(2):\sum_{i=0}^{p-1}\frac{2}{a-(4i+2)}\\(3):\sum_{i=0}^{p-1}\frac{2}{a-(4i+3)}\\(4):\sum_{i=0}^{p}\frac{1}{a-4i}.$$ T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve the equation $2^{x^2 + x} + \log_2x=2^{x+1}$
Solve the equation $2^{x^2 + x} + \log_2x=2^{x+1}$ where $x$ is real.
I tried to use derivatives, without success.
It's obvious that $x=1$ is solution.
Also, if $x \gt 1$ then $2^{x^2 + x} \gt 2^{x+1}$ and $\log_2x \gt 0$ therefore the equation has no solutions $x \... | It's clear that $x=1$ is a solution. Let's look at two cases: $x>1$ and $x<1$.
If $x>1$, then $\log_2x>0$ and $x^2+x>x+1$, which implies that $2^{x^2+x}>2^{x+1}$. It's impossible.
If $0<x<1$, then $\log_2x<0$ and $x^2+x<x+1$, which implies that $2^{x^2+x}>2^{x+1}$. It's impossible.
In summary, we only have one solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $\lambda$ and $\theta$ such that it validates this matricial equation Find $\lambda$ and $\theta$ such that it validates the matricial equation
$$ \left( \begin{array}{cc}
1 & 2 \\
2 & 3
\end{array} \right)
%
\left( \begin{array}{cc}
\cos \theta \\
\sin \theta
\end{array} \right) = \lambda\left( \begin{array}{cc... | For given $\theta$ this is an eigenvalue problem.
The characteristic equation of your matrix is $p(\lambda) = (1-\lambda)(3 - \lambda) -4 = \lambda^2 - 4\lambda -1$
And so the eigenvalues are
$ \lambda = \dfrac{4 \pm \sqrt{16+4}}{2} = 2 \pm \sqrt{5}$
It can be shown (though I used computer algebra) that the eigenvect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.
So I used the identity $$(1+x)^n=1 + \binom{n}{1}x ... | Note that $$\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$$
Therefore,
\begin{align}
\sum_{k=0}^{n}\frac{1}{k+1}{n\choose k}&=\frac{1}{n+1}\sum_{k=0}^{n}\frac{n+1}{k+1}{n\choose k}
\\&=\frac{1}{n+1}\sum_{k=0}^{n}{{n+1}\choose {k+1}}
\\&=\frac{1}{n+1}\left [\sum_{k=0}^{n+1}{{n+1}\choose k}-1\right ]
\\&=\frac{1}{n+1}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \ln \tan x\,\,dx= - \frac{2G}{5}$ Context: This question
asks to calculate a definite integral which turns out to be equal to $\displaystyle 4 \, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
4 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right),$ where $\text{Ti}_2(x) = \... | We can use the same technique of the linked answer for proving the claim. We first prove that $$2\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx=\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx.\tag{1}$$Let $$I=\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx$$ using the identity
$$\tan\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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How to use the generalized binomial theorem to produce the power series of $(1-x)^{1/2}$ I am trying to see how to get from $\sqrt{1-x}$ to the power series $\displaystyle\sum_{m=0}^\infty\frac{-1}{2m-1}\,{2m \choose m}\,\frac{x^m}{4^m}$, ideally using the generalized binomial theorem.
I see that the Taylor expansion e... | Let $f(x)=(1-x)^{1/2}$. Then, differentiating we have
$$\begin{align}
f^{(1)}(x)&=(-1)\left(\frac12\right) (1-x)^{-1/2}\\\\
f^{(2)}(x)&=(-1)^2\left(\frac12\right) \left(-\frac12\right) (1-x)^{-3/2} \\\\
f^{(3)}(x)&=(-1)^3\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right) (1-x)^{-5/2}\\\\
f^{(4)}(x)&=(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved?
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$
This is my attempt. It is too big, I am sure there is... |
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
$=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$
Then, y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$'a'$ for which roots of one equation lie between roots of other equation. If the range of values of $'a'$ for which the roots of the equation $x^2-2x-a^2+1=0$ lie between the roots of the equation $x^2-2(a+1)x+a(a-1)=0$ is $(p,q)$, find the value of $(q+\frac{1}{p^2})$
Could someone give me slight hint as I can't unde... | $x^2-2x-a^2+1=0$. Quadratic equation yields $x = \frac{2\pm \sqrt{4 - 4(a^+ 1)}}{2}$
$x^2-2(a+1)x+a(a-1)=0$. Q.E. yields $x = \frac{2(a+1) \pm \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}$
We are told $\frac{2(a+1) - \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}< \frac{2- \sqrt{4 - 4(a^+ 1)}}{2}$ and $\frac{2+ \sqrt{4 - 4(a^+ 1)}}{2} <\frac{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit of $\frac{\sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor }$ at $x=-1$ $$f(x):=\begin{cases}\frac{ \sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor } & \lfloor x+1\rfloor \ne0 \\
0 & \lfloor x+1\rfloor=0 \ \end{cases}$$
Then at $x=-1$ find the limit
My work
$$\lfloor x+1\rfloor=0$$
$$... | Read carefully the definition: the function is defined at the right of $-1$.
Make your life simpler and change $x+1$ into $x$, so you need the limits at $0$ of
$$
g(x)=\begin{cases}
\dfrac{\sin\lfloor x\rfloor}{\lfloor x\rfloor} & \text{if $\lfloor x\rfloor\ne0$}
\\[6px]
0 & \text{if $\lfloor x\rfloor=0$}
\end{cases}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
| We use a classical infinite descent argument.
Note that the right-hand side is even, so the left-hand side must be. It follows that two of $x$, $y$, $z$ are odd and the third even, or all three are even.
But two odd and one even is impossible, for then the right-hand side is divisible by $4$ and the left-hand side is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find $f(x)$ satisfying the functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ A function $f(x)$ satisfies the functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ for all real $x$. Then $f(x)$ is given by.
My work
$$x^2{f(x)} +f(1- x) =2x -x^4$$
Replacing $x$ $by$ $1- x$
$$(1-x)^2{f(1 - x)} +f(x) =2x -x^4 + 4x^3 -6x^2 +... | It is easier to solve a much more general problem:
Consider the functional equation
a(x) f(x) + b(x) f (1-x) = g(x).
Call a^* (x) = a(1-x) etc. and the equation becomes the two equations.
a f + b f^* = g
a^* f^* + b^* f = g^*. which is a 2-by-2 linear system:
Solving for f by inverting a 2-by-2 matrix gives
f(x) = (1/ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $10^{340} < \frac{5^{496}}{1985}$
Prove that $10^{340} < \dfrac{5^{496}}{1985}$.
I said since $2^{13} < 10^{4}$, we see that $5 = \dfrac{10}{2} > 10^{\frac{9}{13}}$ and so $10^{340} < \dfrac{10^{343.38}}{1985} <\dfrac{5^{496}}{1985}$. But then I have to estimate $\log_{10}{2} < 0.31$, which is pretty comp... | $$2^{340}\cdot 5^{340} \lt \dfrac{5^{496}}{1985}$$
is equivalent to
$$2^{340}\lt \frac{5^{496-340}}{5\times 397}=\frac{5^{155}}{397}$$
which is equivalent to
$$397\cdot 2^{340}\lt 5^{155}$$
Since $397\lt 1024=2^{10}$, it is sufficient to prove that
$$2^{350}\lt 5^{155}$$
which is equivalent to
$$2^{70}\lt 5\cdot 5^{30... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
which $X$ can satisfy $XSX^2=D$? In the question, $X$, $S$, and $D$ are symmetric positive definite (SPD) matrices.
Just for an example, $X=S^{-\frac{1}{2}}(S^{\frac{1}{2}}DS^{\frac{1}{2}})^{\frac{1}{2}}S^{-\frac{1}{2}}$ is the solution to $XSX=D$.
$X^2=XX$, $X^{\frac{1}{2}}X^{\frac{1}{2}}=X$, and $X^{-\frac{1}{2}}X^... | If $X$ is a solution, $SX=X^{-1}DX^{-1}$ must be positive definite. Therefore $X$ must commute with $S$. But then $XSX^2=D$ gives $X^3=S^{-1}D$. So, a solution exists if and only if $S$ commutes with $D$ and $X=S^{-1/3}D^{1/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$ How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?
| In general, it holds that $$\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}=n-\frac{1}{8n^2}+O\left(\frac1{n^3}\right)$$ and that $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac1{n^3}\right)\,$$ for all $n\geq 1$. Hence, their difference is $$1-\frac1{4n^2}+O\left(\frac1{n^3}\right)\,.$$ In pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 5
} |
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.