Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sum of fourth powers in terms of sum of squares The sum of the fourth powers of the first $n$ integers can be expressed as a multiple of the sum of squares of the first $n$ integers, i.e.
$$\begin{align}
\sum_{r=1}^n r^4&=\frac {n(n+1)(2n+1)(3n^3+3n-1)}{30}\\
&=\frac{3n^2+3n-1}5\cdot \frac {n(n+1)(2n+1)}6 \\
&=\frac{3n... | Statement we'd like to show:
$$
\sum^{n}_{r=1}r^4=\frac{3n^2+3n-1}{5}\sum^{n}_{r=1}r^2
$$
Or equivalently:
$$
5\sum^{n}_{r=0}r^4=(3n^2+3n-1)\sum^{n}_{r=0}r^2
$$
We may recall one of approaches how to get closed form of this expression and utilize it in a slightly different way.
$$
\sum^{n}_{r=0}r^5 + (n+1)^5=\sum^{n}_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Calculus Integral $\int \frac{dt}{2^{t} + 4}$
I try resolve this integral, with $u = 2^t + 4 $, but can´t ... plz
$$\int \frac{dt}{2^t + 4} = \frac{1}{\ln 2}\int\frac{du}{u(u-4)} = \text{??} $$
| $\displaystyle \int\dfrac{dt}{2^t+4}$
on u substitution, i.e., $2^t+4=u,\;$ we get-
$\dfrac{1}{\ell n2}\displaystyle \int\dfrac{du}{u(u-4)}\;....(1),\;$ as you already done.
Now, we need partial fraction to solve it further.
For this, $\dfrac{1}{u(u-4)}=\dfrac{A}{u}+\dfrac{B}{u-4}$
$=\dfrac{Au-4A+Bu}{u(u-4)}$
$\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Find the $n$th derivative of $f(x)=x\sin(x)\cos(2x)$ If it helps, it ask the value for $n=100$ and $x=\pi/2$.
I can't do it by induction because it has too many factors and trying to use an equality for $\cos(2x)$ didn't helped.
I don't see the relation in the derivatives.
| Observe
\begin{align}
x\sin x \cos 2x =&\ x\frac{e^{ix}-e^{-ix}}{2i} \frac{e^{2ix}+e^{-2ix}}{2} = \frac{x}{4i}\left(e^{3ix}-e^{ix}+e^{-ix}-e^{-3ix} \right)\\
=&\ \frac{1}{2}x\sin 3x-\frac{1}{2}x\sin x \\
=&\ \frac{1}{2}\left(x-\frac{\pi}{2}\right)\sin 3x -\frac{1}{2}\left(x-\frac{\pi}{2}\right)\sin x + \frac{\pi}{4}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of the sequence $x_{n+1}=\sqrt{3x_n}$
Let $x_{n+1}=\sqrt{3x_n}$ and $x_1=1$. Prove $x_n=3^{1-(\frac{1}{2^{n-1}})}$ for all $n$ and find the limit of $\{x_n\}$.
Notes: The first few terms of the sequence are $1,\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}} ...$ I do not know from this how to find t... | If you want to see where the formula comes from, notice the pattern:
\begin{eqnarray}
x_1 &=& 1\\
x_2 &=& \sqrt{3} &=& 3^{\frac12}\\
x_3 &=& \sqrt{3\sqrt{3}} &=& \bigl(3\cdot3^{\frac12}\bigr)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 &=& 3^{\frac12 + \frac14}\\
x_4 &=& \sqrt{3\sqrt{3\sqrt{3}}} &=& \left(3\bigl(3\cdot3^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$ Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$
My Attempt:
$LHS=\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{(\sin\frac{x}{2}-\cos\frac{x}{2})(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\sin\fra... | You have correctly proved they are not equal.
WolframAlpha agrees with you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4... | $8 \mid x$ if and only if $x\equiv 0\mod 8$. Then you have $4k^2 + 4k - 4l^2 - 4l$ If $k$ is odd, $k^2$ is too. Then $4k^2 + 4k \equiv 0 \mod 8$ if $k$ is even then so is $k^2$ then $4k^2 + 4k \equiv 0\mod 8$. A similar argument for $l$ will finish your proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Necessary conditions for a Sudoku puzzle to have no repetitions Is it true that if a Sudoku puzzle has the following features there will be no repetitions in rows, columns and $3 \times 3$ subsquares?
*
*The sum of each row must be $45$
*The sum of each column must be $45$
*The sum of each $3 \times 3$ subsquare m... | If all the cells are distinct 1 through 9 then the sum is 1+2+....+9 =45. But there is utterly no reason earth to assume the converse, that is $a+b+.... +i = 45$ then they are all distinct.
For any $b,...,h =N$ we can have $a$ be any $1 \le a \le 45-N $ and $i = 45-N-a$. And we can determine values for the other rows... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960908",
"timestamp": "2023-03-29T00:00:00",
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Prove that for $n\ge 2$, $(n + 1)^n n! < (2n)! < 4^n (n!)^2 .$ Prove that for all real $n\ge 2$, the following inequalities hold: $$(n + 1)^n n! < (2n)! < 4^n (n!)^2 .$$
I have posted my proof below, if anyone has a shorter, or more interesting proof, please share :-)
| First note we can re-write the left hand side as:
$$(n+1)^n n!=1\cdot2\cdot3\cdots n\cdot \underbrace{(n+1)\cdot(n+1)\cdots(n+1)}_{n\text{ times}},$$
and the middle term as
$$(2n)!=1\cdot2\cdot3\cdots n\cdot(n+1)\cdot (n+2)\cdots (n+(n-1))\cdot 2n.$$
The first $n+1$ terms are equal, and the remaining terms can be compa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question on Partial Fraction Decomposition I'm supposed to evaluate the following integral by decomposing the integrand into a sum of partial fractions. Note, this question isn't about evaluating.
$$\int\frac{6x+6}{(x^2+1)(x-1)^3}dx$$
From my understanding the numerator should be equated to the following form:
$$\frac{... | By setting $f(x)=\frac{6x+6}{(x^2+1)(x-1)^3}$ we have:
$$ E = \lim_{x\to 1}(x-1)^3 f(x) = \lim_{x\to 1}\frac{6x+6}{x^2+1} = 6 \tag{E} $$
$$ D = \lim_{x\to 1}\frac{d}{dx}(x-1)^3 f(x) = -\lim_{x\to 1}\frac{6(x^2+2x-1)}{(x^2+1)^2}=-3\tag{D} $$
$$ C = \frac{1}{2}\lim_{x\to 1}\frac{d^2}{dx^2}(x-1)^3 f(x) = 6\lim_{x\to 1}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula Let $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real.
It can be shown using a distance formula that the minimal value of $f(x)$ is
$D = \sqrt{(a-c)^2+(|b|+|d|)^2}$.
Show ... | Let, w.l.o.g., $b > 0$ and $d > 0$.
Define
$$
x_0 = \frac{ad+bc}{b+d}
$$
Further, as already stated in the question, the minimum value of $f$ is
$D = \sqrt{(a-c)^2+(b+d)^2}$.
The following equality holds
$$
f(x) = \sqrt {(g(x))^2 + (h(x))^2}+\sqrt {(g(x))^2 + (D - h(x))^2}
$$
where
$$g(x) = \frac{b+d}{D} \; (x-x_0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $d$ must be a perfect square
Prove that if $a,b,c,d$ are integers such that $$(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2 = d,$$ then $d$ is a perfect square.
In order for $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2$ to be an integer, either $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ has to be the square root of a pos... | Theorem: Suppose that $a+2^{1/3}b+2^{2/3}c=0$ where $a,b,c\in\mathbb{Z}$. Then $a=b=c=0$.
Proof: Note that
$$
\begin{align}
&\left(a+2^{1/3}b+2^{2/3}c\right)\left(\left(a^2-2bc\right)+2^{1/3}\left(2c^2-ab\right)+2^{2/3}\left(b^2-ac\right)\right)\\
&=a^3+2b^3+4c^3-6abc\tag{1}
\end{align}
$$
Thus, if $a+2^{1/3}b+2^{2/3}c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve differentil equation if the begining values are given? How do I solve differentil equation if the begining values are given?
$\frac{dx}{dt}=x^2+5x$ with x(0)=-3. I need to find x(t).
$\int \frac{dx}{x^2+5x}=\int dt$
So when I put it on Symbollab I get that left side is $-\frac{2}{5}arctanh(\frac{2}{5}(x+... | If $y = \mbox{arc}\tanh x$, then $$x = \tanh y =\frac{e^y - e^{-y}}{e^y+e^{-y}}= \frac{e^{2y} - 1}{e^{2y}+1}.$$ So
$$ x(e^{2y}+1) = e^{2y}-1$$
or $$e^{2y}(x-1) = -x-1$$
So $$e^{2y} = \frac{1+x}{1-x}.$$ That is
$$2y = \log\left( \frac{1+x}{1-x}\right)$$
and
$$y= \frac{1}{2}(\log(1+x)-\log(1-x)).$$
So we have $\mbox{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Trouble with integral expressing $\mathbb E[X^2]$ where $X \sim N(0,1)$ Let $X \sim N(0,1)$
Then to find $E[X^2]$, we can do:
$$
\int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx
$$
Let's say we didn't know this is an odd function, and decided to take the integral by splitting it into 2 pieces:
$$
\fr... | Let
\begin{align}
I(a) &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \mathrm{d}x \\
&= \frac{1}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\
&= \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{a}} \mathrm{erf}(y) \Big|_{0}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{a}}
\end{align}
using the subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the 2016th power of a complex number Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac... | $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$
Lets simplify $\frac{-1 + i\sqrt 3}{1 + i}$
$\frac {1}{1+i}(-1 + i\sqrt 3)\\
\frac {1-i}{2}(-1 + i\sqrt 3)\\
(1-i)(-\frac12 + i\frac{\sqrt 3}2)$
Convert to polar:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)$
Now we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 5
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Continuity of this function in the origin Does anyone have any idea on how can I prove if this partial derivative is continuous at the origin:
$f(x,y) = \frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}$
Of course I know the definition: for every $\epsilon$ there is a $\delta$ if $ ||(x,y)||$ is less that $\delta$ than $|f(x,y)|$... | HINT:
Transforming to polar coordinates $(\rho,\phi)$, we have
$$\begin{align}
f(x,y)&=\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\\\\
&=\frac{\rho^4\sin(\phi)\cos(\phi)(\cos^2(\phi)+2\sin^2(\phi))}{\rho^3}
\end{align}$$
Then, the limit as $\rho\to0$ is trivially $0$. If $f(0,0)=0$, then $f$ is continuous at the origin.
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinatorial proof of $\sum_{1\le i\le n,\ 1\le j\le n}\min(i,j)=\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$ $$\sum_{1\le i\le n,\ 1\le j\le n}\min(i,j)=\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$$
Is there a link, if any, between these two identical sums?
| $$\begin{array}{rr|rrrrrrr}
\hline
\min(i,j)&&i\\
&&1&2&3&4&\cdots &(n-2)&(n-1)&n\\
\hline
j&1&1&1&1&1&\cdots &1&1&1\\
&2&1&2&2&2&\cdots &2&2&2\\
&3&1&2&3&3&\cdots &3&3&3\\
&4&1&2&3&4&\cdots &4&4&4\\
&\vdots&&&&&\ddots \\
&\vdots \\
&\vdots \\
&(n-2)&1&2&3&4&&\color{green}{(n-2)}&\color{green}{(n-2)}&\color{green}{(n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the roots of any cubic with trigonometric roots.
Question: How would you find the roots of a cubic polynomial whose roots are only expressive in trigonometric forms?
I'm really confused on how you would solve it. Some examples:$$x^3+x^2-2x-1=0\\x_1=2\cos\frac {2\pi}7,x_2=2\cos\frac {4\pi}7,x_3=2\cos\frac {8\... | For $x^3+x^2-10x-8 = 0$.
$(\Bbb Z/31 \Bbb Z)^*/\{\pm 1\}$ is cyclic of order $15$, so it has only one subgroup $H$ of index $3$.
$H = \{\pm 1; \pm 2; \pm 4; \pm 8; \pm 15\}$ (generated by $\pm 2$)
So, the roots should be integer combinations of the quantities
$\sum_{n \in kH} 2\cos(\frac {2n\pi}{31})$ where $kH \in G/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary number theory proof without contradiction Question: suppose $a, b, c$ are integers and $\gcd(a,b) = 1$. Prove that if $c|(a^2+b^2)$ then $\gcd(a,c) = gcd(b,c) = 1$.
Prove by contradiction (naive approach) : assume $p \in PRIMES$ and $\ne1$ such that: $p|a$ and $p|c$. Since $c|(a^2+b^2)$ then $p|(a^2+b^2) \Ri... | Suppose $a^2+b^2=kc$ and, as implied by $\text{gcd}(a,b)=1$, we have $ma+nb=1$ for some integers $m$ and $n$. Then
$$
1=m^2a^2+n^2b^2+2mnab.
$$
Consider the RHS above:
\begin{aligned}
1=\text{RHS}&=m^2(a^2+b^2)+(n^2-m^2)b^2+2mnab=c(km^2)+b[(n^2-m^2)b+2mna];\\
1=\text{RHS}&=n^2(a^2+b^2)+(m^2-n^2)a^2+2mnab=c(kn^2)+a[(m^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Discuss the inequality different methods inequality $\sum_{k=1}^{n}k^2y_{k}\ge (n+1)\sum_{k=1}^{n}ky_{k}$
Nice Problem: Let $n\ge 3$,and $y_{1},y_{2},\cdots,y_{n}$ be real numbers, and such that $$2y_{k+1}\le y_{k}+y_{k+2}.1<k\le n-2$$
and $\displaystyle\sum_{k=1}^{n}y_{k}=0$. Show that
$$\sum_{k=1}^{n}k^2y_{k}... | The inequality to prove is obviously equivalent to $\sum_{k=1}^{n} k(n-k+1)y_k \le 0$.
Question 1: Idea in the given proof is to use summation by parts to transform the above sum in $y_k$ into a sum in $\Delta_k^{2} \;y_k = y_{k+2}-2 y_{k+1} + y_k$ where the terms are known to be positive by convexity of the sequence. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Why is $\lim_{n \rightarrow \infty} \bigg( \bigg| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} \bigg| \bigg)=e$? I know that the following is the correct limit, but I have difficulties in seeing just why this is.
$$\lim_{n\to\infty}\left| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}}\right|=e$$
| Hint. By using the Taylor series expansion, as $x \to 0$, one has
$$
\log(1+x)=x-\frac{x}2+\frac{x^3}3+o(x^3)
$$ giving, as $n \to \infty$,
$$
\begin{align}
-n^2\log\left(1+\frac1n\right)&=-n+\frac12-\frac1{3n}+o\left(\frac1{n}\right)
\\\\
(n+1)^2\log\left(1+\frac1{n+1}\right)&=n+\frac12+\frac1{3n}+o\left(\frac1{n}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Limit involving a trigonometric function as $x\rightarrow +\infty$ I am probably supposed to use this identity: $$\lim_{x\to0} \frac{\sin(x)}{x} = 1$$
I have this mathematical problem:
$$\lim_{x\to+∞} (x^2-1) \sin(\frac{1}{x-1}) = \lim_{x\to+∞} (x+1)(x-1) \sin(\frac{1}{x-1}) = ???$$
Can I do this to the sinus limit?
$... | Let $X=\frac{1}{x-1}$ so that $x-1=\frac{1}{X}$ and $x+1=2+\frac{1}{X}$
We have $\lim_{x \to +\infty} (x^2-1) \sin(\frac{1}{x-1})=\lim_{X \to 0^+} (2+\frac{1}{X}) \frac{\sin(X)}{X}=+\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How was this calculated $(1-1/z)^{(\frac{-1}{3})}=(1+\frac{1}{3z}+\dots)$ *Background: a simplification step from an interesting longer problem here where the residue is needed so only the first two numbers are needed.
$$(1-\tfrac1z)^{-\frac13}=(1+\tfrac1{3z}+\cdots)$$
Was the likely step to compute the cube root: $(1+... | Considering $$A=\frac{1}{\sqrt[3]{1-\frac{1}{z}}}$$ let for simplicity $z=\frac 1x$ to make $$A=(1-x)^{-1/3}$$ and use either the generalized binomial theorem or Taylor series around $x=0$. This will give $$A=1+\frac{x}{3}+\frac{2 x^2}{9}+\frac{14 x^3}{81}+\frac{35 x^4}{243}+\frac{91
x^5}{729}+O\left(x^6\right)$$ Re... | {
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"url": "https://math.stackexchange.com/questions/1983825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it 'not mathematical' to compare the L.H.S. and R.H.S in such type of equations? $$x+\frac{1}{x}=25 + \frac{1}{25}$$
The solution is very simple. But the problem is whether my solution is correct or not. I did it by simply comparing the LHS and the RHS. Thus, I got $x=25$ or $\frac{1}{25}$. But my book does it in th... | The "longer" method is:
\begin{align}
x + \frac{1}{x} &= 25 + \frac{1}{25} \\
x^{2} + 1 &= \left(25 + \frac{1}{25} \right) x \\
x^{2} - \left(25 + \frac{1}{25} \right) x + 1 &= 0 \\
x &= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \sqrt{\left(25 + \frac{1}{25} \right)^2 - 4} \\
&= \frac{1}{2} \, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding if sentence is true using induction: $\sqrt1 + \sqrt2 +\dots+\sqrt{n} \le \frac23(n+1)\sqrt{n+1}$ I have to prove
$\sqrt{1} + \sqrt{2} +...+\sqrt{n} \le \frac{2}{3}*(n+1)\sqrt{n+1}$
by using math induction.
First step is to prove that it works for n = 1 , which is true.
Next step is to prove it for n + 1. We ... | To finish your proof by induction we could use the fact the following inequalities are equivalent to each other
\begin{align*}
\frac23(n+1)^{3/2}+\sqrt{n+1} &\le \frac23(n+2)^{3/2}\\
\frac32\sqrt{n+1} &\le (n+2)^{3/2} - (n+1)^{3/2}\\
\frac32\sqrt{n+1} &\le (\sqrt{n+2} - \sqrt{n+1}) (n+2 + \sqrt{(n+2)(n+1)} + n+1)\\
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to split 59 in $\mathbb{Q}(\sqrt{13})$ How to split 59 in $\mathbb{Q}(\sqrt{13})$
as i know that 59 is prime and we can write $(x+y\sqrt{13})(x-y\sqrt{13})=59$
$x^2-13y^2=59$ but i cant find the x,y
| Agree with Parcly Taxel.
In fact for squarefree $-50 \leq d \leq 50$, $59$ splits in $\Bbb{Q}[\sqrt{d}]$ only for \begin{align}
d &=-43 & 59 &= 4^2 + 43 \cdot 1^2 \\
d &=-34 & 59 &= 5^2 + 34 \cdot 1^2 \\
d &=-23 & 59 &= 6^2 + 23 \cdot 1^2 \\
d &=-11 & 59 &= \left(\frac{15}{2}\right)^2 + 11 \cdot \left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solve Recurrence Relation for Maximum and Minimum in an Array. I know that recursive equation of this algo is
$T\left ( n \right )=2T\left ( \frac{n}{2} \right )+2 $
Given that $T\left ( 1 \right )=0,T\left ( 2 \right )=1 $
and its solution is also given here, ijust want to clear my doubt where i am stuck at..
I solve... | Changing my base case from $T\left ( 1 \right )=0$ to $T\left ( 2 \right )=1 $ .my equation looks like
$\Rightarrow T\left ( n \right )=2^{k}*T\left ( \frac{n}{2^{k}} \right )+2^{k}+\cdots 2^{2}+2^{1}$
Taking $T\left ( 2 \right )=1 $
$\frac{n}{2^{k}}=2 \Rightarrow k=\log_{2}\frac{n}{2} $
And our equation becomes
$T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $ \int_{-\infty}^{\infty} \frac{e^{-ix}}{\sqrt{x+2i} + \sqrt{x+5i}}dx$ I want to compute following improper integral via residue theorem.
$ \int_{-\infty}^{\infty} \frac{e^{-ix}}{\sqrt{x+2i} + \sqrt{x+5i}}dx$
Note this is a exercise 4.57 in Krantz, complex analysis textbook. (Function theory of one complex ... | We take a branch of $\sqrt{z}$ as follows:$$
\sqrt{z}=\sqrt{r}e^{i\frac{\theta }{2}},\quad \quad z=re^{i\theta} \,\,\left(-\frac{\pi}{2}<\theta <\frac{3\pi}{2}\right)
.$$
Then $\sqrt{z}$ is analytic and single-valued in $\mathbb{C}\setminus [0,-i\infty)$.
We consider a contour $\Gamma$ consisting of the horizontal line... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How prove this $BC$ always passes through a fixed point with $\frac{x^2}{4}+y^2=1$
if the point $A(0,1)$ on the ellipse $\Gamma:$ $\dfrac{x^2}{4}+y^2=1$ and the circle $\tau:$ $(x+1)^2+y^2=r^2(0<r<1)$,if $AB,AC$ tangent the circle $\tau$ ,$B,C\in \Gamma$,show that the line $BC$ always passes through a fixed point
I... | When $r\rightarrow 1$, it's easy to see that $y$-axis will be a tangent line, which means that the fixed point should be on the $y$-axis. So later, we need to put $x=0$ in the equation of tangent line to find the fixed $y$-coordinate. Use the tangent conditions, get $\frac{|k-1|}{\sqrt{k^2+1}}=r$ (here, we get $k_1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rig... | Induction hypothesis:
$$9|5\times10^k+10^{k-1}+3$$
Thus,
$$9|10(5\times10^k+10^{k-1}+3)$$
$$9|5\times10^{k+1}+10^k+30$$
$$9|5\times10^{k+1}+10^k+3+27$$
But $9|27$. Hence,
$$9|5\times10^{k+1}+10^k+3$$
which is what had to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Symmetric ratio of binomial probabilities around the midpoint $\frac{m+1}{2}$ greater $\frac{p}{1-p}$ for $p>0.5$ I am trying to show the following:
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)}\geq \frac{p}{1-p}$,
with $n>m\geq \frac{n+1}{2... | Note that
\begin{align}
\sum_{i=(2m-n+1)/2}^{(m-1)/2} \binom{m}{i}p^i(1-p)^{m-i} &=~\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{m-i}p^{m-i}(1-p)^i \\
&=~\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{m-i}(1-p)^i
\end{align}
Therefore,
\begin{align}
&\frac{\Pr\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{\Pr\left(\tfrac{2m-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$f(x) = x^2$. For each positive integer $n$, let $P_n$ be the partition $P_n = \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n},1 \}$ of $[0,1]$. Let $f(x) = x^2$. For each positive integer $n$, let $P_n$ be the partition $P_n = \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n},1 \}$ of $[0,1]$.
Show that $S(P)=\frac{1... | For $j=1,2...n\;$, we have
$$x_j=j\frac{1}{n},$$
$$x_{j}-x_{j-1}=\frac{1}{n},$$
$$\displaystyle{M_j=sup_{x\in[x_{j-1},x_j]}f(x)=f(x_j)=x_j^2=\frac{j^2}{n^2}},$$
since $f$ is increasing.
and
$$S(P)=\sum_{j=1}^n(x_j-x_{j-1})M_j=\frac{1}{n^3}\sum_{j=1}^n j^2$$
$$=\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}$$
$$=\frac{1}{6n^2}(2n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine whether $\sum\limits_{n=1}^{\infty} \frac{ \sqrt{n} }{n^2+3}$ converges or diverges. - Please check my reasoning. Determine whether $\sum_{n=1}^{\infty} \dfrac{ \sqrt{n} }{n^2+3}$ converges or diverges.
I would appreciate it if someone could check my reasoning for this problem and indicate if I have made any ... | One may observe that, for $n>1$,
$$
0<\frac{\sqrt{n}}{n^2+3} < \frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}
$$ since $\displaystyle \sum\frac{1}{n^{3/2}}$ is convergent then the initial series $\displaystyle \sum\frac{\sqrt{n}}{n^2+3} $ is convergent by the comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Generalization of Leibniz formula for $\pi$ The well-known Leibniz formula for $\pi$ is
$$
1-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{\pi}{4}.
$$
Looking at some nonstandard techniques, I've happened upon the following formula:
\begin{equation} \label{armdark}
\begin{split}
\Big( \frac{1}{1} + \frac{1}{... | The idea is to use $\sum_{n=1}^\infty \exp(2\pi i nm/\ell)/n=-\log(1-\exp(2\pi i m/\ell))$ for any $m$ not divisible by $\ell$. From here you can find $\sum_{n=1}^\infty c_n/n$ for any $\ell$-periodic sequence $c_n$ such that $\sum_{n=1}^\ell c_n=0$, by finding $d_m$'s such that $c_n=\sum_{m=1}^\ell d_m\exp(2\pi i nm/\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Inequality based on AM/GM Inequality Find the greatest value of $x^3y^4$
If $2x+3y=7 $ and $x≥0, y≥0$.
(Probably based on weighted arithmetic and geometric mean)
| My usual attempt to generalize.
To find the max of
$x^m y^n$
subject to
$ax+by = c$
where
$m$ and $n$
are integers.
Following
Macavity's solution,
use the
AGM inequality
in the form
$\sum_{i=1}^p a_i
\ge p (\prod_{i=1}^p a_i)^{1/p}
$.
Then
$\begin{array}\\
c
&=m\frac{ax}{m}
+n\frac{by}{n}\\
&\ge (m+n)((\frac{ax}{m})^m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\left(0.5^x-0.25^x\right)^{\frac{1}{x}}$ is strictly increasing The problem is to prove that
$$\left(0.5^x-0.25^x\right)^{\frac{1}{x}}$$
is strictly increasing on $\mathbb{R}^+$. The problem I'm having is that once I found the derivative to be
$$\left(\dfrac{1}{2^x}-\dfrac{1}{4^x}\right)^\frac{1}{x}\left(\d... | Assume $a$ is a real number such that $a>1$. Then let
$$
f_a(x):=\left(1-\frac1{a^x}\right)^{1/x}, \qquad x>0.
$$ We have
$$
\begin{align}
f_a'(x)&=\left( -\frac1{x^2}\cdot \ln \left(1-\frac1{a^x}\right)+\frac1x \cdot \frac{\ln a \cdot \frac1{a^x}}{1-\frac1{a^x}} \right)\cdot\left(1-\frac1{a^x}\right)^{1/x}
\\\\&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a_{ij}=\max(i,j)$, calculate the determinant of $A$
If $A$ is an $n \times n$ real matrix and
$$a_{ij}=\max(i,j)$$
for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.
So, we know that
$$A=\begin{pmatrix}
1 & 2 & 3 & \dots & n\\
2 & 2 & 3 & \dots & n\\
3 & 3 & 3 & \dots & n\\
\vdots & \vdots & \vdots ... | Well you can prove this using induction too. The n = $1$ case is trivial. Let, $\det(A_{n}) = n(-1)^{n-1}$. We'll need to prove $$\det(A_{n+1}) = (-1)^{n}(n+1)$$Now $A_{n+1}$ looks like:-
$$A_{n+1}=\begin{pmatrix}A_{n}&P\\
P^{T}&n+1
\end{pmatrix}
$$
where $P^{T} = (n+1)(1,1,...1)^{T}$ $$\det(A_{n+1}) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Taylor's series problems Can you help please me solve this by using Taylor series?
I would be grateful if you can explain how did you solve it
$$\lim_{x\to 0}{\cosh{2x\over 2+x^4}+\cos{2x\over 2+x^4}-2e^{x^4\over 2}\over \tan\sqrt{1+x^4}-\tan\sqrt{1-x^4}}$$
https://i.stack.imgur.com/xUZA1.png
| Let's develop every term to the fifth order:
$$\cosh \frac{2x}{2+x^4} \approx 1+ \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^5)$$
$$\cos \frac{2x}{2 + x^4} \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^5)$$
$$2\ e^{x^4/2} \approx 2 + x^4 + \mathcal{O}(x^5)$$
$$\sqrt{1 + x^4} \approx 1 + \frac{x^4}{2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-2... | Your $(2)$ is incorrect. Instead, we should square both sides of $(1)$ to get
$$(\alpha-\beta)^2=(x-5)^2$$
$$\alpha^2-2\alpha\beta+\beta^2=(x-5)^2$$
$$2\alpha\beta=(x-5)^2-\alpha^2-\beta^2$$
Square both sides again
$$4\alpha^2\beta^2=\left[(x-5)^2-\alpha^2-\beta^2\right]^2$$
Now all radicals are removed and you should... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Derivative with quotient rule Trying to get $\dfrac{d}{dx}\left[\frac{20e^x}{(e^x+4)^2}\right]$.
After quotient rule: $$f'(x)=20\dfrac{e^x(e^{2x}+8e^x+16)-e^x(2e^x+8e^x)}{(e^x+4)^4}\\\\\\= 20\dfrac{e^{3x}+8e^{2x}+16e^x-2e^{2x}-8e^x}{(e^x+4)^4}\\\\=20\dfrac{e^{3x}+6e^{2x}+8e^x}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+4)(e^x+2)}... | $$
f'(x)=20\frac{e^x(e^{2x}+8e^x+16)-e^x(2e^{\color{red}{2x}}+8e^x)}{(e^x+4)^2}
$$
You also do $e^x\cdot 8e^{x}=8e^x$, which is wrong.
On the other hand, if you consider
$$
g(x)=\frac{x}{(x+4)^2}
$$
you have
$$
g'(x)=\frac{(x+4)^2-x\cdot 2(x+4)}{(x+4)^2}=
\frac{x+4-2x}{(x+4)^3}=-\frac{x-4}{(x+4)^3}
$$
Since
$$
f(x)=20g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}}$ How can I calculate the below limit?
$$
\lim\limits_{x\to \infty} \left( \mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}} \right)
$$
In fact I know should use the L’Hospital’s Rule, but I do not how to use it.
| Using the fact that $\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =1\\ \\ $ we can write
$$\lim _{ x\rightarrow \infty }{ \left( e^{ \sqrt { x+1 } }-e^{ \sqrt { x } } \right) } =\lim _{ x\rightarrow \infty }{ { e }^{ \sqrt { x } }\left( e^{ \sqrt { x+1 } -\sqrt { x } }-1 \right) } =\lim _{ x\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | Hint:
Apply $a^3+b^3 = (a+b)(a^2-ab+b^2)$ where you take $a = \cos^2 x$ and $b=\sin^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 3
} |
Prove that this is a valid formula using axioms of propositional logic The question is how to prove using basic axioms this expression:
$$(A \to B) \to ((\lnot C \to A) \to (\lnot C \to B))$$
I have the list of axioms, one of them looks like this: $A \to (B \to A)...$ But I don't understand how to apply this to my expr... | Start with a premised proof:
$$\begin{array} {rl}
\text{Premise} :& A \Rightarrow B
\\ \text{Premise} :& \lnot C \Rightarrow A
\\ \text{Premise} :& \lnot C
\\ \text{MP} :& A
\\ \text{MP} :& B
\end{array}$$
Then apply deduction theorem:
$$\begin{array} {rl}
\text{Ax 1} :& (A \Rightarrow B) \Rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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BMO1 2011 Problem 1: Find $n$ such that $n^2+20n+11$ is a perfect square
*
*Find all (positive or negative) integers
$n$ for which
$$n^2+20n+11$$
is a perfect square. Remember that you must justify that you have found
them all.
For the first part I did so:
$$n^2+20n+11=M^2 $$ where M is an integer.
That factors dow... | Note that
$$n^2+20n+11=(n+10)^2-89$$
so we have to find all the pairs of squares $a^2$, $b^2$ such that $a^2-b^2=89$. Of course, $a$ and $b$ are integers. We see that $|a|>|b|$.
Since $a^2-b^2=(a+b)(a-b)$ and $89$ is a prime number, then we have these possibilities:
*
*$a+b=89$ and $a-b=1$, that is, $a=45$, $b=44$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find all $n\in\mathbb N$ such that $\sqrt{n^2+8n-5}$ is an integer. $n^2+8n-5$ has to be a perfect square.
How to find all $n$?
| Note that $n^2 + 8x - 5 = (n + 4)^2 - 21$. Now this number will certainly be no square if it is bigger than $(n + 3)^2$ (since there are no squares strictly between $(n + 3)^2$ and $(n + 4)^2$), i.e. if
$$\begin{align*}
&n^2 + 8n - 5 > (n + 3)^2 \\
\iff& n^2 + 8n - 5 > n^2 + 6n + 9 \\
\iff& 2n > 14 \\
\iff& n > 7
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
determine the roots of the equation $7\sec^2{x}+2\tan{x}-6=2\sec^2{x}+2$ Determine the roots of the equation $7\sec^2{x}+2\tan{x}-6=2\sec^2{x}+2$, when $0≤x≤2π$. I don't know how to solve this equation algebraically. Please help?
| Using $1+tan^2=sec^2$,
$$7\sec^2(x)+2\tan(x)-6=2\sec^2(x)+2\iff $$
$$7+7\tan^2(x)+2\tan(x)-6=2+2\tan^2(x)+2\iff $$
$$5\tan^2(x)+2\tan(x)-3=0 \iff$$
$$(\tan(x)+1)(5\tan(x)-3)=0 \iff$$
$$\tan(x)=-1 \hbox{ or } \tan(x)=3/5.$$
Thus, in $[0,2\pi)$, $x=3\pi/4$ or $x=7\pi/4$ or $x=\tan^{-1}(3/5)$ or $x=\tan^{-1}(3/5)+\pi.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A nice infinite series: ${\sum\limits_{n=1}^\infty}\frac{1}{n!(n^4+n^2+1)}=\frac{e}{2}-1$ - looking for a more general method while clearing out some stuff I found an old proof I wrote for an recreational style problem a while back, here's the sum:
${\sum\limits_{n=1}^\infty}\frac{1}{n!(n^4+n^2+1)}=\frac{e}{2}-1$
I'll ... | Here is my heuristics understanding on why the summation is so special. Consider the function
$$ f(z) = \sum_{n=0}^{\infty} \frac{1}{z(z-1)\cdots(z-n)}. $$
Let $a \in \Bbb{C} \setminus \{0,1,2,\cdots\}$ and $\epsilon > 0$ be sufficiently small. Putting the matter of rigor aside, computing the following contour integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$... | Hint: Multiply the top and bottom of the first limit in order to rewrite it as
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3 (\sqrt{1+\tan(x)} + \sqrt{1 + \sin(x)})} =
\frac 12 \lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}
$$
Then we have
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3} =
\lim_{x \to 0} \frac 1{\cos(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove:$x^2+2xy+3y^2-6x-2y\ge-11\;\;\forall(x,y)\in\Bbb{R}$
Problem Statement:-
Prove that for all real values of $x$ and $y$
$$x^2+2xy+3y^2-6x-2y\ge-11$$
I have no idea how to approach this question all I could think on seeing it was tryin to find the linear factors, turns out that the determinant
$$\begin{vmatrix... | The matrix for the quadratic form is
$$
A=\begin{bmatrix}
1 & 1 & -3 \\
1 & 3 & -1 \\
-3 & -1 & 11
\end{bmatrix}
$$
A sufficient condition for the quadratic form to be positive semidefinite is that the leading principal minors but the last (that ought to be nonnegative) are positive. If also the full determinant is pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to prove that $\int_{-1}^{+1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$ I'am trying to prove that
$$\int_{-1}^{+1} (1-x^2)^n\:\mathrm{d}x = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$$
Here what I have done so far. We know that:
\begin{equation}
\sin^{2}x + \cos^{2}x = 1
\end{equation}
Let $x = \sin\alpha$ so $\mathrm{d}... | Consider the integral $\int_{0}^{+1} (1-x^2)^n dx$
Let $I(n)$ be the above integral do the substitution $x=\sin t$. Then $dx=\cos t dt$ and
$1-x^2=cos^2 t$.
Also $t=0$ for $x=0$ and $t=\pi/2$ for $x=1$ so
$I(n) = \int_0^{\pi/2} \cos^{2n+1} t \ dt = \int_0^{\pi/2} \cos^{2n} t \ \cos t \ dt$.
To integrate by parts do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
power series solution for $xy''+y'+xy=0$ I need to solve
$$xy''+y'+xy=0$$
at $x_0 =1$
so I supposed:
$$y = \sum_{n=0}^{\infty}a_n(x-1)^n\implies$$
$$y'= \sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}\implies$$
$$y'' = \sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}$$
Then I did:
$$(x-1+1)\sum_{n=2}^{\infty}n\cdot (n-... | Rewrite as $$(x-1)y''+(x-1)y+y+y'+y''=0 $$
The coefficient of $(x-1)^n$
*
*in $y$ is $a_n$
*in $(x-1)y$ is $a_{n-1} $ (with $a_{-1}=0$ understood)
*in $y'$ is $(n+1)a_{n+1}$
*in $y''$ is $(n+2)(n+1)a_{n+2}$
*in $(x-1)y''$ is $(n+1)na_{n+1}$
Thus you get the equation
$$(n+1)na_{n+1} + a_{n-1}+a_n+(n+1)a_{n+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What's wrong with the way I solved the complex equation $z^5=-16\overline{z} (z \in \mathbb{C})$
Solve the fallowing complex equation($z \in \mathbb{C})$: $z^5=-16\overline{z}$.
Here's how I tried to solve the equation:
$$ z^5=-16\overline{z} \rightarrow z^6=-16(x^2+y^2)$$
$$z= \varepsilon(\cos{\theta} +i\sin{\theta}... | $x$ and $y$ are unknown, so it makes no sense saying $z_0=16(x^2+y^2)e^{i\pi/6}$. Recall that $z=x+iy$, according to what you wrote.
A proper way would be first noting that $z=0$ is a solution, so we can henceforth assume $z\ne0$.
Write $z=re^{i\varphi}$; then the equation becomes
$$
r^5e^{5i\varphi}=-16re^{-i\varphi}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Different sites give different answers for this series sum For the series $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}$ wolframalpha gives answer $\displaystyle\frac{\pi^2}{64}$, but another site gives $\displaystyle\frac{\pi^2}{64}-\frac{1}{12}$: http://www.emathhelp.net/calculators/calculus-2/series-calc... | $$\begin{align}
S&=\sum_{n=0}^\infty\frac{1}{((2n+1)^2-4)^2}=\sum_{n=0}^\infty\frac{1}{(4n^2+4n-3)^2}=\sum_{n=0}^\infty\frac{1}{((2n-1)(2n+3))^2}\\
&=\frac{1}{4}\sum_{n=0}^\infty\frac{2n+3-(2n-1)}{((2n-1)(2n+3))^2}=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)^2(2n+3)}-\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.
For $x^2\equiv 1\pmod {7}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are ... | You have $x\equiv\pm1\mod7$ and $x\equiv\pm1\mod13$. For all the solutions, you have to consider the systems:
$$x\equiv1\mod7$$
$$x\equiv1\mod13$$
and
$$x\equiv-1\mod7$$
$$x\equiv1\mod13$$
and
$$x\equiv1\mod7$$
$$x\equiv-1\mod13$$
and
$$x\equiv-1\mod7$$
$$x\equiv-1\mod13$$
as each system will get you a valid answer. I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan... | The second solution $\sin(\theta) = -1$ doesn't work with the original equation because then $\tan$ and $\sec$ are both undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Closed form of recursively defined sequence Let $a_0, a_1 \in \mathbb R$. I am given the sequence $(a_n)_{n\in \mathbb N}$ recursively defined by $$a_n = \frac{2}{5} a_{n-2} + \frac{3}{5} a_{n-1}$$
for $n \geq 2$. I want to find an explicit term describing $a_n$, I tried to figure it out by writing out the first 3-4 te... | Given the recurrence
$$
a_{\,n} = \frac{3}
{5}a_{\,n - 1} + \frac{2}
{5}a_{\,n - 2} \quad \left| {\;a_{\,n\, < \,0} = 0} \right.
$$
in my opinion, the best way to solve it is through the generating function.
So let's write it as:
$$
a_{\,n} - \frac{3}
{5}a_{\,n - 1} - \frac{2}
{5}a_{\,n - 2} - \left[ {1 = n} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trouble with a substitution I'm struggling to show that
$$ \int_{-1}^{1} \frac{(1-x^2)^{1/2}}{1+x^2} dx $$
to
$$ -\pi + \int_{-\pi}^{\pi} (1+\cos^2(\theta))^{-1}d\theta$$
with $x=\cos(\theta)$
I'm aware I'm missing something obvious but I end up with a stray $\sin(\theta)$
| You can use $u=\tan(\frac{\theta}{2})$ in the calculation. In fact,
\begin{eqnarray}
\int_{-\pi}^{\pi}\frac{1}{1+\cos^2\theta}d\theta&=&\int_{-\pi}^{\pi}\frac{1}{1+\frac{1+\cos(2\theta)}{2}}d\theta\\
&=&\int_{-\pi}^{\pi}\frac{2}{3+\cos(2\theta)}d\theta\\
&=&\int_{-2\pi}^{2\pi}\frac{1}{3+\cos(\theta)}d\theta\\
&=&2\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding particular solution to 1D - wave equation given general solution. Given the general solution of the wave equation:
$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$
Find the particular solution given that:
$\displays... | $$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$$
$$\displaystyle u(x,t) = \sum_{n=0}^\infty \frac{b_n}{2}\left[ \cos\left[\pi\left(n+\frac{1}{2}\right)(x-ct) \right]-\cos\left[\pi\left(n+\frac{1}{2}\right) (x+ct)\right] \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Zeta Infinite Summation $\sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n)$ Let $Re\{s\}\gt0$ :
$$
\sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} = \sum_{n=1}^{\infty} \left( \frac{n}{n+1}\right) n^{-(s+1)} = \sum_{n=1}^{\infty} \left( 1 - \frac{1}{n+1}\right) n^{-(s+1)} = \zeta(s+1) - \sum_{n=1}^{\infty} \frac{n^{-(s+1)}}{n+1} = \\[6... | Let
$S(N)
=\sum_{n=1}^{N} \frac{n^{-s}}{n+1}
=\sum_{n=1}^{N} \frac{1}{n^s(n+1)}
$.
If $Re(s)> 0$,
$\lim_{N \to \infty} S(N)$
exists by comparison with
$\sum_{n=1}^{N} \frac{1}{n^{1+Re(s)}}
$.
Then,
since
$\frac1{1+x}
=\sum_{k=0}^{2m} (-1)^k x^k
-\frac{x^{2m+1}}{1+x}
$
$\begin{array}\\
S(N)
&=\sum_{n=1}^{N} \frac{1}{n^s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of digits after the decimal point of the order of $10$ in $(\frac{\Bbb{Z}}{p\Bbb{Z}})^*$ Let $p$ be a prime number and $a$ an integer such that $0<a<p.$
If I look at $(\frac{\Bbb{Z}}{p\Bbb{Z}})^*$, and the order of $10.$
I discover numerically that if the order of $10$ is even $2n$ then if we look at the list $(... | If the order of $10$ modulo $p$ is $r$, then we have
$$\frac{1}{p} = \frac{Q}{10^{r}-1} = \sum_{m = 1}^\infty \frac{Q}{(10^{r})^m},$$
where
$$Q = \sum_{k = 1}^r q_k \cdot 10^{r-k}.$$
If $r$ is even, $r = 2n$, we can write $Q = 10^n\cdot a + b$, where $0 \leqslant a,b < 10^n$, and then
$$\frac{1}{p} = \sum_{m = 1}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ Divergent? Did I do this right? $$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$$
$$a_n = \frac{1}{\sqrt{n^2+1}} <\frac{1}{\sqrt{n^2}} \le \frac{1}{n}, \forall n\ge1$$
Then by The Comparison Test with $b_n$, where $b_n$ is a divergent p-series = $\frac{1}{n}$ and $a_n \le b_n$,
... | This conclusion is not quite correct. Note that $n^2+1>n^2 \implies \frac{1}{\sqrt{n^2+1}}<\frac1n$ is correct. But $\frac{1}{n^2}\le \frac1n$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
Instead, we can assert that for $n\ge 1$, $n^2+1\le 2n^2$. Therefore, $\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{2}\,n}$. And si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
repeating decimals equation problem The repeating decimals $0.abab\overline{ab}$ and $.abcabc\overline{abc}$ satisfy
$0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37}$
where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.
| The numbers can be expressed as $0.\overline {ab}$ and $0.\overline {abc}$ and written as fractions fractions as ${ 0.\overline {ab} }= {ab\over99} $ and $ 0.\overline {abc} = { abc\over999} $:
$$ {ab\over99} + { abc\over999} = {33\over 37} $$
Let $h = ab $ and $k = c \implies abc = 10h + k$
where $0\leq h<100 $ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Binomial coefficients are odd if and only if $n = 2^k-1$
Let $n$ be an integer, $n > 0$. Prove that all the coefficients of the expansion of the Newtonian binomial $(a+b)^n$ are odd if and only if $n$ is of the form $2^k-1$.
Why in the solution below do they take $n > 8$? Also, I don't get the middle paragraph argume... | If $n=2n_1+1$, then the coefficients are:
$$\frac{2n_1+1}{1}, \frac{(2n_1+1)\cdot (2n_1)}{1\cdot 2}, \cdots , \frac{(2n_1+1)\cdot (2n_1)\cdots 1}{1\cdots (2n_1)}. $$
Removing all the odd factors from the numerator and denomiator we get the sequence:
$$ 1, \frac{2n_1}{2}, \frac{2n_1}{2},\frac{(2n_1)\cdot(2n_1-2)}{2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
I don't have any idea as to how to proceed with this questio... | Write $\ f = q\,g + r\ $ with $\,\deg r \le 2 = \deg g,\ $ by the Division Algorithm.
$\color{#0a0}5 \ = f(1)\ =\ r(1)\ \ $ by $\ g(1)\ =0\ $
$\color{#0a0}5\! =\! f(-2)\!=\! r(-2)\ $ by $\ g(-2)\!=\!0,\ $ so $\ r = \color{#0a0}5 + \color{#c00}c\,(x\!-\!1)(x\!+\!2)$
$\!\begin{align} 3\! =\! f(-1)\!\! &=\! r(-1)\ \, {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding an unknown length in a triangle. I have incurred a question and I am having a hard time with it. Please refer the image below
Here it is given that $AP:PC = 3:4$, $QM:MP = 3:2$ and $QB=12cm$. We have to find the length of $AB$. I have tried as many theorems I knew but was not able to conclude to the answer. Pl... | In following link, you may see the triangle drawn.
http://i.imgur.com/AUskpyR.png
Solution:
$\frac{AP}{PC} = \frac{3}{4}, \frac{QM}{MP} = \frac{3}{2} $ are given in the question. Lets draw a line from A to T through M.
$\frac{AP}{AC} \times \frac{CT}{TQ} \times \frac{QM}{MP} = 1$ by the menelaus theorem. When we plug ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$
I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know ho... | As $F_{n+2}=F_{n+1}+F_n$. One first solves the characteristic polynomial.
$$
X^2=X+1
$$
on gets two roots $r=\frac{1+\sqrt{5}}{2}$ and $\bar{r}=\frac{1-\sqrt{5}}{2}$.
One now solves $\alpha.r^n+\beta.\bar{r}^n=F_n$ for the two first indices.
The OP did not indicate his initialization so I suppose that $F_0=0;\ F_1=1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
$\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{\sin{4x}}{4}$ - solving a trigonometric equation The question is to solve the following equation:
$$ \sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{\sin{4x}}{4} $$
There is a tedious and mistake-prone way to do this, that is using trigonometric identities to write the equation in term... | $$\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{2\cdot\sin{2x}\cdot \cos 2x}{4}$$
$$2\cdot\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\sin{2x}\cdot \cos 2x $$
$$\sin 2x(2\cdot\sin{x}\cdot\sin{3x}-\cos 2x) =0$$
$$\sin 2x(-\cos 4x+\cos 2x-\cos 2x)=0 \Rightarrow \sin2x \cdot\cos4x=0$$
$$\sin 2x =0 \Rightarrow x=\frac{k\pi}{2}$$
$$\cos 4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $f(x)$ from a equation $f(x)+f(\frac{x-1}{x}) = x+1$ is given.
find $f(x)$.
I did tried to change it into another form buy substituting $x$ with $\frac{x}{x-1}$
The result were $f(\frac{1}{x})+f(\frac{x}{x-1})=\frac{2x-1}{x-1}$
I don't know what to do next.
| All you have to do is make the substitution again.
$$f(x)+f\bigg(\frac{x-1}{x}\bigg)=x+1$$
$$\implies f\bigg(\frac{x-1}{x}\bigg)+f\bigg(\frac{1}{1-x}\bigg)=\frac{x-1}{x}+1$$
$$\implies f\bigg(\frac{1}{1-x}\bigg)+f(x)=\frac{1}{1-x}+1$$
Now subtract the first and second equations to get
$$f(x)-f\bigg(\frac{1}{1-x}\bigg)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.
| More generally, the coefficient of $x^{n}$ is given by
$$[x^{n}](1+x^2(1-x))^8=[x^{n}]\sum_{k=0}^8\binom{8}{k}x^{2k}(1-x)^k
=(-1)^n\sum_{n/3\leq k\leq \min(n/2,8)}\binom{8}{k}\binom{k}{n-2k}.$$
For $n=10$ we get
$$\sum_{4\leq k\leq 5}\binom{8}{k}\binom{k}{10-2k}=\binom{8}{4}\binom{4}{2}+\binom{8}{5}\binom{5}{0}=476.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two... | Note that $a^2 + b^2 = c^2 + d^2$ is equivalent to $a^2 - c^2 = d^2 - b^2$, i.e. $(a-c)(a+c) = (d-b)(d+b)$. If we factor any odd number $m$ as $m = uv$, where $u$ and $v$ are both odd and $u < v$, we can write this as $m = (a-c)(a+c)$ where $a = (u+v)/2$ and $c = (v-u)/2$. So any odd number with more than one factori... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 13,
"answer_id": 10
} |
If $x:y = 7:3$ , then find the value of $\frac{y}{x-y}$ If $x:y = 7:3$ , then can I in order to find the value of $\frac{y}{x-y}$ replace $x$ and $y$ with $7$ and $3$ respectively ?
| $ \frac{x-y}{y}=\frac{x}{y}-1=\frac{7}{3}-1=\frac{4}{3}$. Hence, what you want $=\frac{3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Why is $2^b-1=2^{b-1}+2^{b-2}+...+1$? Can I get an intuitive explanation why the formula in the title holds?
I know that it works but I am not sure why
$2^b-1=2^{b-1}+2^{b-2}+...+1$
| Consider the number $N=2^{b-1}+2^{b-2}+\dots+2^2+2^1+2^0$. Now let's look at $N+1$:
$$N+1 = 2^{b-1}+2^{b-2}+\dots+2^2+2^1+2^0+2^0\\
=2^{b-1}+2^{b-2}+\dots+2^2+2^1+2^1\\
=2^{b-1}+2^{b-2}+\dots+2^2+2^2\\
\vdots\\
=2^{b-1}+2^{b-2}+2^{b-2}\\
=2^{b-1}+2^{b-1}\\
=2^b$$
(since $2^k+2^k=2^{k+1}$ for any $k$), so $N=2^b-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&... | Observe it's rather:
$$
\begin{align}
\color{blue}{2\sin x} - 2\sin^3 x + \color{blue}{\sin x} - 2\sin^3 x &=\color{blue}{2\sin x} + \color{blue}{\sin x}- 2\sin^3 x - 2\sin^3 x
\\\\&=3\sin x - 4\sin^3 x.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute $ \lim_{n\rightarrow \infty }\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k} $ Calculate $ \lim_{n\rightarrow \infty }\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k} $.
So far I found that $ \frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=\frac{(k-5)(k-4)(k-3)}{(k-2)(k-1)k(k+1)(k+2)}=\frac{((... | You did a big part of the job.
$k^{5}-5k^{3}+4k=(k-2)(k-1)k(k+1)(k+2)$
Now you can rewrite $$\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=\frac{A}{k-1}+\frac{B}{k}+\frac{C}{k+1}+\frac{D}{k-2}+\frac{E}{k+2}$$.
By multiplying the numerators and denominators of the RHS, and identifying the new numerator in terms of $A,B,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2... | Try taking out $2^n$ and $2^{n-1}$ from the denominator and the numerator. I got $$\frac{1}{2^{(n-1)^2}}.$$
If you can't get it let me know. I'll show the work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Factorize $x^4+16x-12$ over reals Factorize $x^4+16x-12$ over reals.
The factor is $x^4+16x-12=(x^2-2x+6)(x^2+2x-2)$
It can be factorized again but I am stuck in this step.If we want to add and then subtract we have a lot of thing to add and subtract.Another idea that I saw in books is writing as this:
$x^4+16x-12=(x^... | @JohnHughes explains how to find that this is not a cubic times linear factor. Once you figure that out, we know it's a quadratic times quadratic. To solve this, I'm going to go off the equation you had:
$$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$$
Expand the right side:
$$x^4+16x-12=x^4+(a+a')x^3+(b+b'+aa')x^2+(ab'+ba')x+bb'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find integral $\int\frac{\arcsin(x)}{x^{2}}dt$ Find integral $$\int\frac{\arcsin(x)}{x^{2}}dx$$
what I've done: $$\int\frac{\arcsin(x)}{x^{2}}dx=-\int\arcsin(x)d(\frac{1}{x})=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$ I got stuck with that
| Perform integration by parts to get that $\displaystyle \int \frac{\arcsin(x)}{x^2} dx$ = $\displaystyle \frac{-\arcsin(x)}{x} + \int \frac{1}{x\sqrt{1 -x^2}}dx$.
To solve this latter integral, let $x = \sin(\theta)$. Then $dx = \cos(\theta)d\theta,$ so the integral becomes $\displaystyle \int\frac{\cos(\theta)}{\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and ... | I don't think I've ever seen $x \frac{y}{z}$ used to mean $x + \frac{y}{z}$ except when $x$, $y$ and $z$ are literal integers (e.g. $2 \frac{3}{4}$). That's not to say it never happens, but it would be terribly confusing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 3
} |
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx $ Problem :
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx$.
My approach :
\begin{align}
&\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx \\
=& \ln(1^x +2^x +3^x +6^x) \frac{x^2}{2} - \int^1_{-1} \frac{1}{1^x+2^x+3^x+6^x}(2^x\log2+3^x \log3+6^x \log... | $$I=\int_{-1}^0f(x)dx+\int_0^1f(x)dx$$
$$=\int_0^1(f(-x)+f(x))dx$$
$$=\int_0^1(-f(x)+x\ln(6^x)+f(x))dx$$
$$=\ln(6)\int_0^1x^2dx$$
$$=\frac{\ln(6)}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding linear asymptote We have
$$ f(x) = (2x^2-x^3)^{1/3} $$
A linear asymptote is like: $$y = px+q$$ $$p \not= 0$$
We had this definition in our lecture:$$ p = \lim\limits_{x \rightarrow \infty}{\frac{f(x)}{x}}
$$
so
$$ p = \lim\limits_{x \rightarrow \infty}{\frac{(2x^2-x^3)^\frac{1}{3}}{x}} $$
$$ =\lim\limits... | I would use the generalized
binomial theorem:
$\begin{array}\\
f(x)
&= (2x^2-x^3)^{1/3}\\
&= (-x)(1-\frac{2}{x})^{1/3}\\
&= (-x)(1-\frac{2}{x}\frac13+(-\frac{2}{x})^2\frac13 (-\frac23)+O(\frac1{x^3}))\\
&= (-x)(1-\frac{2}{3x}-\frac{8}{9x^2}+O(\frac1{x^3}))\\
&= -x+\frac{2}{3}+\frac{8}{9x}+O(\frac1{x^2}))\\
\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find Delta for given limit of function I have : $$\lim_{x \to-2}\frac{3x+6}{x^3+8} = \frac{1}{4}$$
This is what i've come up with so far :
$$|f(x) - L| = \frac{3x+6}{x^3+8} -\frac{1}{4}| =|\frac{12x+24}{4x^3+32} -\frac{x^3+8}{4x^3+32}| = |\frac{-x^3+12x+16}{4x^3+32}|$$
I am looking for a $\delta(\epsilon)$ so that $|x ... | $$\left|\frac{3(x+2)}{x^3+8}-{1\over4}\right|<\epsilon\Longleftrightarrow\left|\frac{3(x+2)}{(x+2)(x^2-2x+4)}-{1\over4}\right|<\epsilon$$
$$x\ne-2\hspace{1cm}\left|\frac{3}{x^2-2x+4}-{1\over4}\right|<\epsilon$$ $$\left|\frac{-x^2+2x+8}{4(x^2-2x+4)}\right|<\epsilon\Longleftrightarrow-\epsilon<\frac{-x^2+2x+8}{4(x^2-2x+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alte... | Let $f(x)=(1-x)^{-n}$. Then by applying mean value theorem for $f$ we have,
$$\frac{f(x)-f(0)}{x-0}=f'(\zeta)$$(for some $0<\zeta<x<1$)
$$\frac{(1-x)^{-n}-(1-0)^n}{x-0}=(-n)(1-\zeta)^{-n-1}(-1)=\frac{n}{(1-\zeta)^{n+1}}>n$$
The above line is true as $1-\zeta<1\Rightarrow\frac{1}{1-\zeta}>1$
$$\Rightarrow (1-x)^{-n}-1>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
What fractions can fill a $N$ by $N$ matrix given that their sum is always 2? My problem is the following: I have a matrix $N$ by $N$ in size. I want to fill it with fractions of $1$ of increasing denominator in relation to their distance from the center of the matrix. The central value is always $1$.
$N$ is always odd... | Let's say we have a $2n+1 \times 2n+1$ matrix $A$ with a $1$ in the center, which is $A_{n+1,n+1}$, and $\frac{1}{d\cdot \text{taxicab}(x,y,n+1,n+1)}$ for $A_{x,y}$ for any cell that is not the center. We want all of this to sum to $2$, so without the $1$ in the center, we want the elements to sum to $1$.
In Python, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trignometric integration We have to solve the following integration.
$$\int\frac{\tan 2\theta}{\sqrt{\cos^6 \theta+\sin^6\theta}}\ d\theta$$
In this, I write the denominator as $\sqrt 1-3\sin^2\theta \cos^2\theta$
But now how to proceed?
| as you write $\displaystyle \sin^6 \theta+\sin^6 \theta = 1-3\sin^2 \theta \cos^2 \theta=\frac{1}{4}\left(4-3\sin^2 2 \theta\right)$
let $\displaystyle \mathcal{I} = \int\frac{\tan 2 \theta }{\sqrt{\sin^6 \theta+\sin^6 \theta}}d\theta = 2\int\frac{\sin 2 \theta }{\cos 2 \theta \sqrt{4-3\sin^2 2 \theta}}d\theta = 2\int\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve a matrix equation I need to find $X$ from
$$\begin{pmatrix}
1 & 2\\
-3 &-6
\end{pmatrix} X \begin{pmatrix}
1 &2 \\
-1 &-2
\end{pmatrix}=\begin{pmatrix}
2 &4 \\
-6 & -12
\end{pmatrix}$$
I wrote $X$ as
$$X=\begin{pmatrix}
a & b\\
c &d
\end{pmatrix}$$
and I got $a+2c-b-2d=2$ but I do not know what to do nex... | You are on good way. When you multiply matrices on left side, you finally get $$a-b+2c-2d = 2$$ which is a linear equation. Solutions are given with three parameters, and thus
$$ X = \begin{pmatrix}
a & b\\
c & \frac{a-b+2c-2}{2}
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal.
\begin{equation*}
A =
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
\end{equation*}
I know that a $n \times n$ matrix is called orthogonal ... | You are right. Now solve the system of equation in terms of $a$ and $b$. However, if $b$ is restricted to be an integer, the equation $$4 + b^2 = 1$$ does not have a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| Since $$(1 - \sqrt{5})^3 = 16 - 8 \sqrt{5}$$ and similarly $$(1 + \sqrt{5})^3 = 16 + 8 \sqrt{5}$$ it follows that $$(2 + \sqrt{5})^{1/3} + (2 - \sqrt{5})^{1/3} = \left(\frac {16+8\sqrt{5}} 8 \right)^{1/3} +\left(\frac {16-8\sqrt{5}} 8 \right)^{1/3} = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
} |
Probability event with $70$% success rate occurs three consecutive times for sample size $n$ It has been a long time since I've done probability so I am not sure which to do (if either are correct). Thank you for taking the time to look at my work.
Probability an event occurs is $70$%.
I'm looking for the probability o... |
Let's denote the event under consideration with $a$
\begin{align*}
P(X=a)=0.7
\end{align*}
and we denote the complementary event with $b$.
We are looking for the words of length $n$ and their probability of occurrence which do not contain three or more consecutive $a$'s. The result is $1$ minus this probability.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}... | WLOG, $a>0$. As there are singularities at $x=-a$ and $x=a$, the expressions can differ in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$, depending on the signs of $x+a$ and $x-a$.
Correct expressions are
$$\int\frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{a+x}{a-x}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Expected value and probability
The distribution function of a discrete random variable X is given
$$F_X(x)=\begin{cases} 0, &x<3\\ \frac{6}{11},& 3\leq x< 4 \\ \frac{7}{11}, & 4\leq x<5 \\ \frac{8}{11}, & 5\leq x<6 \\ 1, & 6\leq x \end{cases} $$
Let $A=(X=3)\cup (X=5)$. Calculate: $P(A)$ and $E(X)$
How I did it:
$... | The values that you got for the probabilities $P(X=x)$ are wrong. $P(2)=0$ and $P(3)=\frac6{11}$ and so on. If you correct this, then you get for the first one:\begin{align}P(X=3)&=P(X\le 3)-P(X<3)=F_X(3)=\frac6{11}\\[0.2cm]P(X=5)&=P(X\le 5)-P(X<5)=F_X(5)-F_X(5-)=\frac8{11}-\frac7{11}=\frac1{11}\end{align} Hence $P(A)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all $n$ such that $n \mid x_n$ where $x_n = x_{n-1} + \lfloor n^2 / 4 \rfloor$, $x_0 = 0$ $X_n$ is sequence such that $x_n=x_{n-1}+[\frac{n^2}{4}]$ and $x_0=0$. Find all positve integers $n$ for which $x_n$ is divisible by $n$.
[X] means integer part.
| $$x_{2k}=x_{2k-1}+k^2=x_{2k-2}+2k^2-k=2k^2-k+(2(k-1)^2-(k-1))+x_{2k-4}=\cdots=\sum_{n=1}^{k} 2n^2-\sum_{k=1}^{k} n=\frac{k(k+1)(2k+1)}{3}-\frac{k(k+1)}{2}=\frac{k(k+1)(4k+2-3)}{6}=\frac{k(k+1)(4k-1)}{6}$$
Also
$$a_{2k+1}=a_{2k}+k^2+k\\a_{2k+1}=\frac{k(k+1)(4k-1)}{6}+\frac{6k(k+1)}{6}=\frac{k(k+1)(4k+5)}{6}$$
Now splitt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Integrability of $f(t) =\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{\Gamma \left( \frac{it+1}{1.5} \right) }{\Gamma \frac{ it+1}{2} }$ Can we show that the following function is integrable
\begin{align}
f(t) =\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{\Gamma \left( \frac{it+1}{1.5} \right) }{\Gamm... | We have $$\left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right| = |(1+it)^{\frac{1+it}{6}}|= \left|\exp\left(\left(\frac{1+it}{6}\right)\ln(1+it)\right)\right|.$$ Remember that $|e^z|=e^{\Re(z)}$ and $$\Re\left(\left(\frac{1+it}{6}\right)\ln(1+it)\right)=\frac{\ln(\sqrt{1+t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n... | Your solution looks good.
Another approach could be:
$$a_n=\frac{3n^2+n}{(n+1)^2}=\frac{3(n+1)^2-5n-3}{(n+1)^2}=3-\left[\frac{5n}{(n+1)^2}+\frac{3}{(n+1)^2}\right]=3-\left[\frac{5}{n+2+\frac{1}{n}}+\frac{3}{(n+1)^2}\right]$$
$a_n$ is increasing so what can we conclude about $\frac{1}{a_n}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
I don't know how to start...
| Let's start by applying the quadratic formula.
We have: $(x^{2}+4x+8)^{2}+3x\hspace{1 mm}(x^{2}+4x+8)+2x^{2}=0$
$\Rightarrow x^{2}+4x+8=\dfrac{-3x\pm\sqrt{(3x)^{2}-4(1)(2x^{2})}}{2(1)}$
$\hspace{27.75 mm} =\dfrac{-3x\pm{x}}{2}$
$\hspace{27.75 mm} =-2x,-x$
Then,
$\Rightarrow x^{2}+4x+8=-2x$
$\Rightarrow x^{2}+6x+8=0$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\... | Note that this proof is a bit longer than expected, but it shows some line of thoughts....We have: If $x \le 0 \implies f(x) \le 0 < \dfrac{1}{2} \implies f(x) < \dfrac{1}{2}$. Thus if $x > 0 \implies f(x) = \dfrac{x-1}{x+\sqrt{x^2-x+1}}$. Here we can reason a little bit. If $0 < x \le 1 \implies f(x) \le 0 \implies f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$ Let $x_1,x_2,x_3,\cdots ,x_n (n\ge2)$ be real numbers greater than $1.$ Suppose that $|x_i-x_{i+1}|<1$ for $i=1,2,3,\cdots,(n-1)$.
Prove that $$\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_... | Set $d_i = x_{i+1} - x_i$ for $i = 1, \ldots, n-1$. Then
$$
\frac{x_n}{x_1} = 1 + \sum_{i=1}^{n-1} \frac{d_i}{x_1} \quad ,
\quad \frac{x_i}{x_{i+1}} = 1 - \frac{d_i}{x_{i+1}}
$$
and therefore
$$
\frac{x_n}{x_1} + \sum_{i=1}^{n-1} \frac{x_i}{x_{i+1}} =
n + \sum_{i=1}^{n-1} d_i \left( \frac{1}{x_1} - \frac{1}{x_{i+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Using AM-GM inequality Let $a,\ b,\ c$ be positive real numbers. Prove that:
$$\left(a+\dfrac{bc}{a}\right)\left(b+\dfrac{ca}{b}\right)\left(c+\dfrac{ab}{c}\right)\geq 4\sqrt[3]{\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)}\tag1$$
| By AM-GM $$(a^2+bc)(b^2+ac)=c(a^3+b^3)+ab(c^2+ab)\geq2\sqrt{abc(a^3+b^3)(c^2+ab)}.$$
Thus, $$\prod_{cyc}\left((a^2+bc)(b^2+ac)\right)\geq8\sqrt{a^3b^3c^3\prod_{cyc}(a^3+b^3)\prod_{cyc}(c^2+ab)}$$ or
$$\prod_{cyc}(a^2+bc)^3\geq64a^3b^3c^3\prod_{cyc}(a^3+b^3)$$ or
$$\prod_{cyc}\left(a+\frac{bc}{a}\right)\geq4\sqrt[3]{\pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\mathbb Q(\sqrt[3]{2})[Y]/\langle Y^2+Y+1\rangle$ is a splitting field of $X^3-2$ Show that $\mathbb Q(\sqrt[3]{2})[Y]/\langle Y^2+Y+1\rangle$ is a splitting field of $f(X) = X^3-2$ where $Y$ is an indeterminate over $\mathbb Q$
| Since $Y^2 + Y + 1$ is irreducible, the elements of $\mathbb{Q}(\sqrt[3]{2})(Y)/ \langle Y^2 + Y + 1 \rangle$
are of the form $ a+b \theta$, where $\theta$ is a root of $Y^2 + Y + 1$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$. The roots of $Y^2 + Y + 1$ are
$$ -\frac{1}{2}\pm i\frac{\sqrt{3}}{2}.$$
Choose the one with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute ${11 \choose 1} + {11 \choose 3} + ... + {11 \choose 11}$? I did this with brute force, and got 1024.
What is the faster method of solving this?
| Since $\binom{11}{1}=\binom{11}{10},\binom{11}{3}=\binom{11}{8}$ and so on,
$$ \binom{11}{1}+\binom{11}{3}+\ldots+\binom{11}{11} = \binom{11}{10}+\binom{11}{8}+\ldots+\binom{11}{0} $$
so both the terms are half the sum $\binom{11}{0}+\binom{11}{1}+\binom{11}{2}+\ldots+\binom{11}{11}=2^{11}$, i.e. $2^{10}=\color{red}{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Help to Prove that $\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}$ I need help on proving $(1)$.
$$I=\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}\tag1$$
This is what I have attempted;
Enforcing a sub: $u=\cot^2{x}$ then $du=-2\cot{... | We first write $I$ as
$$ I= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \left( \arctan(\cot^2 x) - \arctan(1) \right) \, dx. $$
Now using addition formulas for $\arctan$ and $\cos$, we have
$$ \arctan(\cot^2 x) - \arctan(1)
= \arctan\left(\frac{\cot^2 x - 1}{\cot^2 x + 1} \right)
= \arctan(\cos 2x). $$
Consequently we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Probability with two boxes and balls There are $10$ white and $10$ black balls in the first box, and $8$ white and $10$ black balls in the second box. Two balls are put from the first to the second box, then we choose one ball from the second box. What is the probability that the chosen ball from the second box is whit... | Case $1$: $2$ white balls are put from first box to second and one ball chosen from the second box is white
probability $=\dfrac{\dbinom{10}{2}}{\dbinom{20}{2}} \times
\dfrac{10}{20}$
Case $2$: $2$ black balls are put from first box to second and one ball chosen from the second box is white
probability $=\dfrac{\db... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
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.