Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$ Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$
$y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,\frac{1}{4})$.I dont know how to sol... | $p(x,y)=(x,y)+(1,3)=(x+1,y+3), \quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix
$\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}$ which is $\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}\pmatrix{x \\ y \\ 1}=\pmatrix{x+1 \\ y+3 \\ 1}$
so for $y=(x-1)^2$ it is
$\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}\pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Summation of series and Taylor series are giving different results $A = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots$
$B = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots$
$A + B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1... | As others have already pointed out, the series $\sum_{n=1}^\infty \frac1n$ and $\sum_{n=1}^\infty \frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.
Another way to analyze this is to look at the partial sums $A_N=\sum_{n=1}^{N} \frac{1}{2n}$ and $B_N=\sum_{n=1}^N \frac{1}{2n-1}$.
While $A_N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $p>1$ that ${\int\limits^p_1}\frac{1}{x}\,\mathrm{d}x={\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x$ Find $p>1$ that $${\displaystyle\int\limits^p_1}\dfrac{1}{x}\,\mathrm{d}x={\displaystyle\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x$$
\begin{align*}
&{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x=\ln\left(... | Find $p>1$ that
$$
{\displaystyle\int\limits^p_1}\dfrac{1}{x}\,\mathrm{d}x={\displaystyle\int\limits^p_1}\ln\left(x\right)\,\mathrm{d}x
$$
\begin{align*}
&{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x=\ln\left(\mid x \mid \right) && \vert \ \text{general integr}
\end{align*}
$F_1(x)=\ln\left({\mid x \mid} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$ The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.
Prove $$\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)... | We have $\log\left(\frac{1+x^2}{(1-x)^2}\right)=\log(1-x^4)-\log(1-x^2)-2\log(1-x)$, hence by integration by parts
$$ I_1 = \frac{3\pi^3}{32}-2\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx\qquad \chi(n)=\left\{\begin{array}{rcl}1&\text{if}&n\equiv 1\pmod{2}\\ 2 & \text{if} & n\equiv 2\pmod{4}\\ 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 4,
"answer_id": 3
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Denote $f(x)=\int_x^{x+1}\cos t^2 {\rm d}t.$Prove $\lim\limits_{x \to +\infty}f(x)=0.$ Problem
Denote $$f(x)=\int_x^{x+1}\cos t^2 {\rm d}t.$$Prove $\lim\limits_{x \to +\infty}f(x)=0.$
Proof
Assume $x>0$. Making a substitution $t=\sqrt{u}$，we have ${\rm d}t=\dfrac{1}{2\sqrt{u}}{\rm d}u.$ Therefore,
\begin{align*}
f(x)&=... | Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $\cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.
Another Proof
Likewise, assume $x>0$. Making a substitution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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power series for $\frac1{x^2+x+1}$ So I am trying to find the sequence $(a_n)_{n\geq0}$ such that
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{... | Another way of doing this is to let the roots of $1+x+x^2=0$ be $a$ and $b$ so that $1+x+x^2=(1-ax)(1-bx)$ with $ab=1$ and $a+b=-1$. Then use partial fractions $$\frac 1{1+x+x^2}=\frac A{1-ax}+\frac B{1-bx}$$ where clearing fractions gives $$1=A(1-bx)+B(1-ax)$$
Set $a=\frac 1b$ to obtain $1=B(1-\frac ab)$ so that $B=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring $3^6-3^3 +1$ $3^6-3^3 +1$ factors?, 37 and 19, but how to do it using factoring, $3^3(3^3-1)+1$, can't somehow put the 1 inside
| Viewing this as a quadratic in $3^3$ leads you to try to factorise $x^2-x+1$ or $x^6-x^3+1$, neither of which splits into anything useful. However, you can pull out a factor of $9$ from $3^6$ and a factor of $3$ from $3^2$ to obtain
$$3^6-3^3+1 = 9 \cdot 3^4 - 3 \cdot 3^2 + 1$$
And the quartic $9x^4-3x^2+1$ does factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For what values of n (where n is a natural number) is this statement true: $3^n - n - 1 ≡ 0\pmod5$ I'm clueless as where to start with this, $3^n$ seems to be of period 4, where $ - n - 1$ is of period 5.
| $\, n = 4q\!+\!r,\ 0\le r\le 3.\,$ $\bmod 5\!:\ \color{#0a0}1 \equiv 3^{\large n}\!-n\equiv 3^{\large r} \overbrace{(3^{\large 4})^{\large q}}^{\large 3^{\Large 4} \equiv\ 1}\!-4q\!-\!r\equiv \color{#0a0}{3^{\large r}\!+q-r}$
$\,r=0\,\Rightarrow\ \color{#0a0}{1\equiv 3^{\large 0}\!+q-0}\ $ so $\,\color{#c00}{q\equiv 0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Find other angles
Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Then find the other angles
My Attempt
Let $a=\sqrt{3}+1$, $b=\sqrt{3}-1$ and $C=60$
$$
c^2=a^2+b^2-2.a.b... | Playing around with the angle sum and the Law of Sines ($|\angle A|>|\angle B|$):
$|\angle A|+|\angle B|=120°$
$(\sin A)/(\sin B)=(\sqrt{3}+1)/(\sqrt{3}-1)=2+\sqrt{3}$
Put $|\angle A|=120°-|\angle B|$ and apply the formula for the wine of a difference:
$((\sqrt{3}/2)(\cos B)-(1/2)(\sin B))/(\sin B)=2+\sqrt{3}$
$(\cos B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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A Series for $\pi$
Question: Can we show that $$\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}=\frac{\pi}{8} $$ ?
According to wolfram alpha this result is true.
Just amateur curiosity, not sure of a starting point to show if this true.
| Note that after cancellation we have
$$\frac{(2n-3)!!}{(2n+3)!!}=\frac{1}{(2n-1)(2n+1)(2n+3)}$$
Taking partial fractions gives
$$
\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{8}\left(-\frac{1}{2n-1}+\frac{2}{2n+1}-\frac{1}{2n+3}\right)
$$
Now, the standard arctangent series for $\pi$ gives $\sum_{n=0}^\infty (-1)^n \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}... | Letting $h=\frac1x$:
$$\begin{array}{cl}
&\displaystyle \lim_{x \to \infty} x \sin \left( \sqrt{x^2+3} - \sqrt{x^2+2} \right) \\
=&\displaystyle \lim_{x \to \infty} x \sin \left( \frac 1 {\sqrt{x^2+3} + \sqrt{x^2+2} } \right) \\
=&\displaystyle \lim_{h \to 0^+} \frac1h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my questio... | Actually your finite sum is equal to $$\sum_{r=2}^n\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}-\frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
Ho... | Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
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The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is? Question based on Permutations and Combinations
The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?
A) $2^{20} + \binom{20}{10}$
B) $2^{2... | If you choose $n$ of the unique object, then there are $20\choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$\sum_{n=0}^{10} {20\choose n}$$
By symmetry, we have
$$2\sum_{n=0}^{10} {20\choose n} = \sum_{n=0}^{20} {20\choose n} + {20\choose 10} = 2^{20} + {20\choose 10}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this simple equation $\frac{46}{y} + y = 25$? How do I solve this simple equation? $\frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
| $$\frac{46}{y}+y = 25$$
Here, assuming $y \neq 0$, you can multiply both sides of the equation by $y$, yielding
$$46+y^2 = 25y$$
$$y^2-25y+46 = 0$$
Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:
$$(y+y_1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear Transformation involving polynomial and matrix Determine if its linear the transformation
$f:\Re_{3} [x]\rightarrow M_{2x2}$
such as
$f(ax^3+bx^2+cx+d) = \begin{bmatrix}
a-c & 0 \\
0&b+d
\end{bmatrix}$
for any $ax^3+bx^2+cx+d \in \Re _{3} [x]$
Im having trouble associating the polynomial with the resulting ... | That looks fine. Another way to go is this:
Consider the standard basis $$\bigl\{A_{1,1},A_{2,1},A_{1,2},A_{2,2}\bigr\}$$ for $M_{2\times 2},$ where $A_{i,j}$ is the $2\times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis $\{x^3,x^2,x,1\}$ for $\mathfrak{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Determinant of a Hankel matrix with sequence (1, 2, 3,..., n, 1, 2,..., n-1) I'm looking for a closed form of the determinant of matrices like
$\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\\ 3& 1 &2\end{bmatrix}$
or
$\begin{bmatrix}1 & 2 & 3 &4\\2 & 3 &4 & 1\\ 3&4& 1 &2\\ 4& 1&2&3\end{bmatrix}$,
that means Henkel matrices of ... | These matrices are actually circulant matrices, whose determinant is well known.
For the matrices at hand, the determinant is
$
\Delta_n=\prod _{j=0}^{n-1}f(\omega _{j})
$,
where $f(z)=1+2z+3z^{2}+\dots +nz^{n-1}$ and $w_j$ is the $j$-th root of unity.
We have $f(\omega_0)=f(1)=1+2+3+\cdots +n = \dfrac{n(n+1)}{2}$.
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a value of $t$ above which these inequalities aren't jointly satisfiable Let $\{a_{1}, ...., a_{8}\} \in [0, 1]$ be such that $\sum \limits_{0 < i \leq 8} a_{i} = 1$ and let $t \in (\frac{1}{2}, 1]$. Is there a value of $t$ above which the following inequalities aren't jointly satisfiable? I know that they ar... | Let’s start a quest for a largest $t<1$ for which there exist feasible $a_1,\dots, a_8$ that satisfy all six inequalities. In this quest we can drop the condition $\sum a_i=1$, because if non-negative numbers $a_i$ satisfy the inequalities for some $t$ then numbers $a’_i=\frac {a_i}{\sum a_j}$ also satisfy the inequal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n}\cos\left(\frac{n\pi}{3}\right)=-\frac{\pi^2}{36}$ I learnt on www.pi314.net that $$\sum_{n=1}^{\infty}\frac{H_n}{n}\cos\left(\frac{n\pi}{3}\right)=-\frac{\pi^2}{36}$$
This result is hard to verify using Wolfram Alpha since the series converges very slowly.
I do not kno... | Perhaps it is in the extraction of the values you are having trouble with. If so, here is one rather (tedious and laborious) way to do it.
From Eq. (590) in the link, one has
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \operatorname{Li}_2 (x) + \frac{1}{2} \ln^2 (1 - x).$$
Setting $x = e^{i\pi /3}$ we see that
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Inequality. $\sum_{cyc}(\frac{1}{a+b+\sqrt{2a+2c}})^3 \le \frac{8}{9}$ Problem. When $a, b, c>0, a, b, c \in \Bbb R, 16(a+b+c)\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, Prove that
$$\sum_{cyc}(\frac{1}{a+b+\sqrt{2a+2c}})^3 \le \frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that
$$4(x+y+z) \ge -2\... | Let $a=\frac{x}{4},$ $b=\frac{y}{4}$ and $c=\frac{z}{4}.$
Thus, the condition gives $$x+y+z\geq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ or
$$1\leq\frac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$\sum_{cyc}\frac{1}{\left(x+y+2\sqrt{2(x+z)}\right)^3}\leq\frac{1}{72}.$$
Now, by AM-GM
$$\left(x+y+2\sqrt{2(x+z)}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. Th... | Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the inequality $\frac{2}{x} > 3 x$ I've been slowly losing my mind on how to do this question, any help would be greatly appreciated.
Solve the inequality $$\frac{2}{x} > 3 x.$$
| To solve the equation
$$
\frac{2}{x} > 3x
$$
we could multiply be $x$ to obtain
$$
2 > 3x^2.
$$
But you have to be careful!
If $x < 0$, the inequality is reversed.
In other words:
$$
\frac{2}{x} > 3x
\iff \begin{cases}
2 < 3x^2 & x < 0 \\
2 > 3x^2 & x > 0.
\end{cases}
$$
Notice the equation isn't defined for $x = 0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove that a semigroup satisfying $a^pb^q=ba$ is commutative
Let $(S, \cdot)$ be a semigroup. There are natural numbers $p,q \geq 2$ such that $a^pb^q=ba$ for all $a,b \in S$. Prove that $S$ is commutative.
I wrote
$$\begin{align}
a^{p+1}b^{q+1} &=b^{(q+1)p}a^{(p+1)q} \\
&=b^{p}\cdot(b^q)^p \cdot (a^p)^q\cdot a^q \... | Let $a\in S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $d\mid p+q-2$. But we also have $$a^3=\color{red}{a^2}\cdot \color{blue}{a}=\color{blue}{a^p}\color{red}{(a^2)^q}=a^{p+2q}$$ which means $d\mid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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derivative with quotient rule and summations I want to differentiate this with respect to $\eta$:
$$C(\eta) = \frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\left(1-\frac{\eta}{\eta_c}\right)}$$
Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, ... | Here's how you could differentiate this particular function:
$$
C'(\eta) = \left[\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\left(1-\frac{\eta}{\eta_c}\right)}\right]'=
\left[\frac{1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}}{3\left(1-\frac{\eta}{\eta_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating the probability of obtaining exactly four distinct values when a die is rolled six times Please could someone help me with determining the probability of getting $4$ distinct numbers (no order in the outcome e.g. 1,2,3,4 or 4,5,6,2 etc) or $5$ distinct from rolling a die $6$ times. So far I was able to cal... | Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $\binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$
My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is... | Note that
$$a^2+b^2+c^2-\frac{1}{3}(a+b+c)^2=\frac{1}{3}((a-b)^2+(b-c)^2+(a-c)^2)$$
(this is useful if you read Mark Bennet's hint).
So it remains to prove that
$$-2(a-b+1)\leq \frac{1}{3}((a-b)^2+(b-c)^2+(a-c)^2).$$
Now you may use the real variables $x=a-b$, $y=b-c$ and $z=c-a$ (and you are right, the condition $a>1... | {
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In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$. Where's my error?
In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, point $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$.
I know the correct solution, but I'm more interested in where is the problem i... | So we have that $\frac{\sin \alpha}{\sin (180-\beta)} = \frac{\sin 20}{\sin 80}$.
The first thing we use is that $\alpha + \beta = 160$ from the triangle $ABD$. From here, $180 - \beta = 180 - (160-\alpha) = 20 + \alpha$.
Next, we note that:
$$
\frac{\sin 20}{\sin 80} = \frac{\sin 20}{\cos(90-80)} = \frac{\sin 20}{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I s... | Hint
First, let $n=\frac 1x$ and compose Taylor series around $x=0$; back to $n$, get successively
$$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}+O\left(\frac{1}{n^4}\right)$$
$$\sqrt{n^2+\sqrt{n^4+1}}=\sqrt{2} n+\frac{1}{4 \sqrt{2} n^3}+O\left(\frac{1}{n^5}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $a$ or $b \equiv 0 \pmod{5}$ if $a^2-b^2 \equiv 0 \pmod{5}$? I have this question, knowing the last in title can i define that at least a or b is divisible by 5?
| Well, ... no because $a^2 - a^2 \equiv 0$ but neither $a$ need not be equivalent to $0 \mod 5$.
Or for that matter $a^2 - (-a)^2 \equiv 0$ but if $a \ne \equiv 0 \pmod 5$ then $-a\equiv 5-a\not \equiv 0$ either.
You can exhaust all cases. $0^2 - 0^2, 1^2 - 1^2, 1^2 - 4^2, 2^2 - 2^2, 2^2 - 3^2, 3^2 - 3^2, 3^2 - 2^2, 4^... | {
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In square $ABCD$ with side $1$, points $E$, $P$, $F$ are the midpoints of $AD$, $CE$, $BP$. What is the area of $\triangle BFD$? I'm not sure if my solution to this Olympiad Geometry question is valid.
Let $\square ABCD$ be a square with side length $1$. Let $E$, $P$, $F$ be the midpoints of $AD$, $CE$, $BP$, respect... | The equation of the line $\overleftrightarrow{BD}$ is $x+y-1=0$. The coordinates of the point $F$ are $\left(\dfrac 34, \dfrac 38 \right)$. So the distance from the point $F$ to the line $\overleftrightarrow{BD}$ is
$$h = \dfrac{\left|\dfrac 34 + \dfrac 38 - 1 \right|}{\sqrt{1^2+1^2}}
= \dfrac{1}{8 \sqrt 2}$$
The a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximize the following equation under a constraint Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.
Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:
| Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
\begin{eqnarray} g(r,\theta)&=&(r\cos\theta)^4+(r\sin\theta)^4\\
&=&r^4(\cos^4\theta+\sin^4\theta)\\
\frac{dg}{dr}&=&4r^3(\cos^4\theta+\sin^4\theta)=0\\
\frac{dg}{... | {
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Mixed Fractions and Multiplication (with Variables) I stumbled over this expression: $3 \frac{1}{x^3}$.
How should you interpret something like that?
While you could see that as implicitit multiplication ($3 * \frac{1}{x^3}$),
you could also argue that $3 \frac{1}{x^3}$ is a mixed fraction ($3 + \frac{1}{x^3}$).
I thin... | When there is no operator between expressions the usual interpretation is multiplication, so $3\frac 1{x^3}=3\cdot \frac 1{x^3}=\frac 3{x^3}$
Mixed fractions are an exception to this, so $3\frac 13 \neq 3 \cdot \frac 13$. Instead it is $3 \frac 13=3+\frac 13$.
If there is any possibility of confusion one should sup... | {
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"timestamp": "2023-03-29T00:00:00",
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Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k... | For $x \in [0,1]$ let
$$ f(x) = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n} x^{2n} \, . $$
Using the power series of $\arcsin$ we find
$$ x \frac{\mathrm{d}}{\mathrm{d} x} x \frac{\mathrm{d}}{\mathrm{d} x} f(x) = 4 \frac{\mathrm{d}}{\mathrm{d} x} [\arcsin(x) - x] = 4 \left[\frac{1}{\sqrt{1-x^2}} - 1 \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Infinite series problem
The sum of $$\frac{2}{4-1}+\frac{2^2}{4^2-1}+\frac{2^4}{4^4-1}+\cdots \cdots $$
Try: write it as $$S = \sum^{\infty}_{r=0}\frac{2^{2^{r}}}{2^{2^{r+1}}-1}=\sum^{\infty}_{r=0}\frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
| You may prove by induction that $$\sum_{r=0}^n\frac{2^{2^r}}{2^{2^{r+1}}-1}=1-\frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$\sum_{r=0}^{k+1}\frac{2^{2^r}}{2^{2^{r+1}}... | {
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"timestamp": "2023-03-29T00:00:00",
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On the evalution of an infinite sum
I wish to show that
$$\sum_{n = 0}^\infty (-1)^n \left [\frac{2n + 1/2}{(2n + 1/2)^2 + 1} + \frac{2n + 3/2}{(2n + 3/2)^2 + 1} \right] = \frac{\pi}{\sqrt{2}} \frac{\cosh \left (\frac{\pi}{2} \right )}{\cosh (\pi)}.$$
The reason I wish to find such a sum is as follows.
The question... | $$
\begin{align}\newcommand{\Re}{\operatorname{Re}}
&\sum_{n=0}^\infty(-1)^n\left[\frac{2n+1/2}{(2n+1/2)^2+1}+\frac{2n+3/2}{(2n+3/2)^2+1}\right]\tag1\\
&=\frac12\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac{2n+1/2}{(2n+1/2)^2+1}+\frac{2n+3/2}{(2n+3/2)^2+1}\right]\tag2\\
&=\frac14\sum_{n\in\mathbb{Z}}(-1)^n\left[\frac{n+\frac1... | {
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"url": "https://math.stackexchange.com/questions/3088089",
"timestamp": "2023-03-29T00:00:00",
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Correctness of the induction step Prove by induction that for all $ n \geq 2$ that the following:
$\sum_{i = 1}^n \frac{i}{i+1} < \frac{n^2}{n+1}$
Looking at the $n+1$ step, is it safe to assume the following ?
$\sum_{i = 1}^{n+1} \frac{i}{i+1} < \frac{(n+1)^2}{n+2} \iff \frac{n+1}{n+2} \leq \frac{(n+1)^2}{n+2} - \frac... | Hint: You must show that $$\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{n}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
$$\frac{n^2}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$$
And we get
$$\frac{(n+1)^2}{n+2}-\frac{n+1}{n+2}-\frac{n^2}{n+1}=\frac{n}{(n+1) (n+2)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the complex numbers $z$ such that $w=\frac{2z-1}{2+iz}$ is real
Find the complex numbers $z$ such that $w=\dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one.
Trouble is that, by manipulating $w$, it just seems to get much worse.
A... | $$w=\frac{(2x-1)+iy}{(2-y)+ix}=\frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2\left(x^2-\frac... | {
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"timestamp": "2023-03-29T00:00:00",
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Rewriting $2x^2-12x+11$ by completing the square I want to rewrite $2x^2-12x+11$ in square form. According to my book, the answer is $2(x-3)^2-7$. However, I get $2(x-3)^2+2$ for some reason.
My Steps:
$2x^2-12x+11$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+9$
$2(x - 3)^2=-2$
$2(x-3)^2+2$
What am I doing wrong?
| As noted in the comments, in the fourth line, you have to add 18 to the right member and not 9.
$2x^2-12x+11=0$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+18$
$2(x - 3)^2=7$
$2(x-3)^2-7=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the minimum value of $9x^2+2y^2-8xy-6x+11$ For any $x, y$ in real, find the minimum value of $9x^2+2y^2-8xy-6x+11$
| Let $$9x^2+2y^2-8xy-6x+11=k$$
Thus, the the following quadratic equation of $y$
$$2y^2-8xy+9x^2-16x+11-k=0$$ has real solutions, which says
$$16x^2-2(9x^2-6x+11-k)\geq0$$ or
$$(x-3)^2\leq k-2,$$ which gives $$k\geq2.$$
The equality occurs for $x=3$ and $y=6,$ which says that $2$ a minimal value.
| {
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Is there any other way to establish this trig identity? $\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $ I needed to verify this trig identity: $$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $$
what I did is I worked on both sides individually:
LHS: $$\frac{\frac1{\cos(x)} + 1}{\frac {\sin(x)}{... | $$\begin{align*} \frac{\sec x + 1}{\tan x} &= \frac{1 + \cos x}{\sin x} = \frac{(1 + \cos x)(1 - \cos x)}{\sin x (1 - \cos x)} = \frac{1 - \cos^2 x}{\sin x (1 - \cos x)} = \frac{\sin^2 x}{\sin x (1 - \cos x)} \\ &= \frac{\sin x}{1 - \cos x}. \end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving equations involving square roots I am a student and I often encounter these type of equations:
$$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6$$
I usually solve these by taking one term ($\sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some method... | A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.
Given that
$$\begin{align}\sqrt{x^2 +(y-2)^2} + \sqrt{x^2+(y+2)^2} &= 6 &[1]\\ \end{align}$$
and that
$$\begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\\ \end{align}$$
we have (by $\frac{[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$a + b = c + c$ So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get:
(
| In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=\frac{a^2+b^2}{2}$$
$$X=\frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=\frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$
Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$
and that the remainder of the division of polynomial $P(x)$ by $x-3$ is $18$. Evaluate the constant term of polynomial $P(x-1)$.
All I cou... |
the remainder of the division of polynomial $P(x)$ by $x−3$ is $18$
implies:
$$P(x)=(x-3)R(x)+18 \Rightarrow \\
P(x+1)=(x-2)R(x+1)+18=(x^2 -4)Q(x)+3ax+6\\
P(2+1)=18=6a+6 \Rightarrow a=2.$$
Hence:
$$P(\color{red}x-1)=P(x-2+1)=((x-2)^2-4)Q(x-2)+6(x-2)+6 =\\
(x^2-4x)Q(x-2)+6x-6 \Rightarrow \\
P(\color{red}0-1)=-6.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mes... | [You can use the trick of observing that it's a telescoping sum, or you just blast ahead with a proof by induction. I think you're asking about the second approach, so here's an answer.]
It's like any other induction proof. You get to assume "P(k)", which means you get to assume
$$
\sum_{i=1}^{k} \frac{1}{i(i+1)} = \fr... | {
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \time... | Proof by induction that $\sum_{k=1}^{n} (2k-1)(n-k+1) = \sum_{k=1}^{n} k^2 \space \forall n \ge 1$:
Base case: For $n=1$ we have $1 \times 1 = 1$
Assume true for $n=k$.
For $n=k+1$:
$\sum_{k=1}^{n+1} (2k-1)((n+1)-k+1) \\
=\sum_{k=1}^{n+1} (2k-1)((n-k+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} (2k-1)((n-k+1) + (2(n... | {
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"url": "https://math.stackexchange.com/questions/3099670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
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determining if sequence has upper bound I am somewhat stuck in my calculations when determining if sequence has an upper bound.
The sequence $$x_n = \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n-1}+\frac{1}{2n}$$
Is equal to $$\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}})$$
And s... | By C-S $$\sum_{i=1}^n\frac{1}{n+i}=1+\sum_{i=1}^n\left(\frac{1}{n+i}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{i=1}^n\frac{i}{n+i}=$$
$$=1-\frac{1}{n}\sum_{i=1}^n\frac{i^2}{ni+i^2}\leq1-\frac{1}{n}\frac{\left(\sum\limits_{i=1}^ni\right)^2}{\sum\limits_{i=1}^n(ni+i^2)}=1-\frac{1}{n}\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $y=\frac {a+bz}{c+dz}$, $z=\frac{a+bx}{c+dx}$, $x=\frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$ I need to solve this problem and I don’t know how.
If $x, y, z$ are unequal and $y = \frac {a + bz}{c + dz}, z = \frac {a + bx}{c + dx}, x = \frac {a + by}{c + dy},$ then $ad + bc + b^2 + c^2 = 0$.
I see that I nee... | $$
\begin{bmatrix}
y\\
1
\end{bmatrix}
=
\alpha
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}
\begin{bmatrix}
z\\
1
\end{bmatrix}, \qquad
\begin{bmatrix}
z\\
1
\end{bmatrix}
=
\beta
\begin{bmatrix}
b & a\\
d & c
\end{bmatrix}
\begin{bmatrix}
x\\
1
\end{bmatrix}, \qquad
\begin{bmatrix}
x\\
1
\end{bmatrix}
=
\gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that planes $x+2y+3z=8$ and $2x+3y+4z=11$ intersect in a line coplanar with $\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$ Question is
Show that the line of intersection of the planes
$$ x + 2y + 3z = 8 \quad\text{and}\quad 2x + 3y + 4z = 11 $$
is coplanar with the line
$$\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}... | The line of intersection between the two planes can be represented parametrically. Let $z = t$:
$$
\left\{
\begin{array}{c}
x + 2y = 8 - 3t \\
2x + 3y = 11 - 4t \\
\end{array}
\right.
$$
Then $2 (1) - (2)$ gives $y = (16 - 6t) - (11 - 4t) = 5 - 2t$, and so $x = (8 - 3t) - 2(5 - 2t) = -2 + t$.
We have to check tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
PMF of throwing a die 4 times We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
*
*What is the probability that $X \ge 4$?
*Compute the PMF of $X$.
*Determine the mean and variance of $X$.
My attempt:
*
*$(1/2)^4$ because that means each of the $4$ rolls, you eith... | 1) is now correct after your edit based on Daniel's comment.
2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).
My approach to this question is to break it down into a bunch of simpler problems as such:
For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
computing the adjoint of operator $T$ on the space $P_2(\mathbb{R})$ Suppose that the inner product on $P_2(\mathbb{R})$ is defined by $$\langle f,g \rangle:= f(-1)g(-1)+f(0)g(0)+f(1)g(1).$$
Consider the operator $T \in B(P_2(\mathbb{R}))$ which is defined as $Tf=f'$, so the derivative of $f$. Find the adjoint of $T$.
... | Every $f\in P_2$ can be written as
$$
f=f(-1)\frac{1}{2}x(x-1)+f(0)(1-x^2)+f(1)\frac{1}{2}x(x+1)$$
Therefore,
$$
f'(x)=f(-1)(x-\frac{1}{2})-2f(0)x+f(1)(x+\frac{1}{2})
$$
which gives
\begin{align}
f'(-1)&=-\frac{3}{2}f(-1)+2f(0)-\frac{1}{2}f(1) \\
f'(0)&=-\frac{1}{2}f(-1)+\frac{1}{2}f(1) \\
f'(1)&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\lim_{n->\infty}\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{3}{n}}}{\sqrt[6]{n^2+\sin{\frac{2}{n}}}-\sqrt[3]{n}}?$ I tried to get rid off cube root as written below but still can not get throught the next steps. What should be the right step to take after the steps below? Did I start as I should or do I... | As said in comments, using Taylor series would make life much easier.
Start dividing all terms by $n^{\frac 13}$ and let $x=\frac 1n$. Now, use the well known Taylor series for the sine and the cosine. Continue with Taylor or with the binomial expansion.
Using three terms in the expansions, you should end with the limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\int_{0}^{1} x^2 \sqrt{1+x^2} dx$? I'm trying to calculate the following integral:
$\int_{0}^{1} x^2 \sqrt{1+x^2} dx$
I tried solving by parts but i'm getting nowhere close. I feel like some substitution will be good here, however neither $x=\cos(u)$ nor $x=\sin(u)$ get me anywhere.
| First let $x = \tan u$, we get
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
={\displaystyle\int}\sec^2\left(u\right)\tan^2\left(u\right)\sqrt{\tan^2\left(u\right)+1}\,\mathrm{d}u
\end{equation}
Using the fact that
$\tan^2 u + 1 =\sec^2 u$
we get
\begin{equation}
{\displaystyle\int}x^2\sqrt{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solution to a differential equation involving inseparable variables. What is the solution for the following DE
$\frac{dy}{dx} - \epsilon{y} = x$
Where $0 \leq x \leq 1$ and initial condition y = 1 when x = 0 and where $\epsilon$ is any positive parameter
| $$\frac{dy}{dx} - \epsilon y = x$$
$$\frac{dy}{dx} = x + \epsilon y$$
Let $v = x + \epsilon y$
$$\frac{dy}{dx} = v$$
Differentiating $v = x + \epsilon y$
$$\frac{dv}{dx} = 1 +\epsilon\frac{dy}{dx}$$
$$\frac{dy}{dx} = \frac{\frac{dv}{dx} - 1}{\epsilon}$$
Substituting $\frac{dy}{dx}$ we have $$ \frac{\frac{dv}{dx} - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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question about calculus I want to find the difference in volume between two balls, one with radius of x while the other with radius of $x+dx$. I find two answers, $\frac{4}{3}\pi(x+dx)^3-\frac{4}{3}\pi x^3$ and $4πx^2 dx$. Please tell me what I did wrong. Thanks.
Pardon my bad English.
| Your answers are the same if we treat higher powers of $dx$ as being equal to $0$.
$$\begin{align}
\frac{4}{3}\pi(x+dx)^3-\frac{4}{3}\pi x^3
&= \frac{4}{3}\pi(x^3+3x^2dx + 3x dx^2 + dx^3)-\frac{4}{3}\pi x^3 \\
&= \frac{4}{3}\pi(x^3+3x^2dx)-\frac{4}{3}\pi x^3 \\
&=\frac{4}{3}\pi x^3+4 \pi x^2dx-\frac{4}{3}\pi x^3 \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.
HINT: The answer is $-2\pm \sqrt5$.
What I have tried so far:
$$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$
$$\tan{\frac{x}{2}} = \frac{1-\sin^2... | Hint: Use that $$\frac{1}{\sin(x)}-\sin(x)=1/2\,{\frac {1+ \left( \tan \left( x/2 \right) \right) ^{2}}{\tan
\left( x/2 \right) }}-2\,{\frac {\tan \left( x/2 \right) }{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Sum of square of binomial coeffcient with positive and negative terms Finding $\displaystyle \binom{2n}{1}^2-2\binom{2n}{2}^2+3\binom{2n}{3}^2-\cdots \cdots -2n\binom{2n}{2n}^2.$
What I've tried:
$$(1-x)^{2n}=\binom{2n}{0}-\binom{2n}{1}x+\binom{2n}{2}x^2+\cdots \cdots +\binom{2n}{2n}x^{2n}$$
$$-2n(1-x)^{2n-1}=-\binom{2... | Starting from
$$\sum_{k=0}^{2n} (-1)^{k-1} k {2n\choose k}^2
= \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
k {2n\choose k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose k-1}
= 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose 2n-k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
[z^{2n-k}] (1+z)^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Show an identity involving a sum of arctangents of algebraic expressions
Show that
$$
2\tan^{-1}\frac{\sqrt{x^2+a^2} - x + b}{\sqrt{a^2-b^2}} + \tan^{-1}\frac{x\sqrt{a^2-b^2}}{b\sqrt{x^2+a^2} + a^2} + \tan^{-1}\frac{\sqrt{a^2-b^2}}{b} = n\pi .
$$
I tried using $$ x= a \tan \theta ,\; b= a \sin\phi,$$
but then calcu... | Hints:
*
*Write the equation as $\tan^{-1}\frac{x\sqrt{a^2-b^2}}{b\sqrt{x^2+a^2} + a^2} + \tan^{-1}\frac{\sqrt{a^2-b^2}}{b} = n\pi - 2\tan^{-1}\frac{\sqrt{x^2+a^2} - x + b}{\sqrt{a^2-b^2}} $
*Apply $\tan$ on both sides and use $\tan (n \pi - 2\alpha) = -\tan 2\alpha$
*Resolve the tangents on the left using $\tan \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration, get stuck at x=tan($\theta$) when calculating arc length I am learning to calculate the arc length by reading a textbook, and there is a question
However, I get stuck at calculating
$$\int^{\arctan{\sqrt15}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta$$ How can I co... | Think of the identities that you have available:
$$\begin{align}\cos^2\theta+\sin^2\theta&=1\\
1+\tan^2\theta&=\sec^2\theta\\
\cot^2\theta+1&=\csc^2\theta\\
\cosh^2\theta-\sinh^2\theta&=1\\
1-\tanh^2\theta&=\text{sech}^2\,\theta\\
\coth^2\theta-1&=\text{csch}^2\,\theta\end{align}$$
Of the $4$ that give you a formula fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Ellipse equations paradox We know that an ellipse can be plotted in cartesian coordinates using the following parametric function:
$$ ellipsePoint(\theta)=\left[\begin{array}{c}a \cdot \cos(\theta)\\b \cdot \sin(\theta)\end{array}\right], 0\leqslant \theta \lt 2\pi $$
I have rewritten it as an equation in polar coordin... | Switching to polar coordinates,
$$\rho=\sqrt{a^2\cos^2t+b^2\sin^2t},\\\tan\theta=\frac ba\tan t.$$
After elimination of $t$,
$$\rho=\sqrt{a^2\cos^2\left(\arctan\left(\frac ab\tan\theta\right)\right)+b^2\sin^2\left(\arctan\left(\frac ab\tan\theta\right)\right)}
\\=\sqrt{a^2\frac1{\dfrac{a^2}{b^2}\tan^2\theta+1}+b^2\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| Recall Taylor series : $$\sin^2 y =y^2+O(y^3) ,\ \ln\
(1+x^2)=x^2+O(x^3)
$$ so that there is $\delta_i$ s.t. $$ 1-\epsilon < \frac{y^2}{\sin^2y},\ \frac{x^2}{\ln\ (1+x^2)}<1+\epsilon $$ for $|y|<\delta_1$ and $|x|<\delta_2$ \begin{align*}\frac{x^2+y^2}{\sin^2y +\ln\ (1+x^2)} &=
\frac{x^2+y^2}{y^2 + O(y^3) + x^2 +O(x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How many square matrices are there with given columns, rows and diagonals Suppose we have $n\times n$ matrix that contains numbers from $1$ to $n^2$.
How many matrices are there that their $n$ columns, $n$ rows and $2n$ diagonals contain given numbers?
For example $3\times 3$ matrix:
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5... | I believe the previous answers have sufficiently addressed the $3\times3$ matrix. In this post I will consider the $4\times4$ matrix below.
\begin{pmatrix}
0&1&2&3\\
4&5&6&7\\
8&9&A&B\\
C&D&E&F
\end{pmatrix}
Consider first the groups containing $0$:
$$(0,1,2,3),(0,4,8,C),(0,5,A,F),(0,7,A,D)$$
We see that $A$ appears in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int \frac{dt}{(t^2-1)^2}$ I have this problem of evaluating the indefinite integral of a rational function
$$\int \frac{dt}{(t^2-1)^2}$$
and I'm a bit unsure about how to proceed. I could use partial fractions:
$$\frac{1}{(t+1)(t-1)(t+1)(t-1)}$$
$$\frac{1}{(t+1)^2(t-1)^2}$$
$$\frac{A}{t+1} + \frac{B}{t-1} + ... | $$\frac{1}{t^2-1} = \frac{1}{2}\left(\frac{1}{t-1} - \frac{1}{t+1}\right)\tag{1}$$
Square both sides:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} + \frac{1}{(t+1)^2}\right) - \frac{1}{2} \frac{1}{t^2-1}$$
The last term's partial fraction expansion is given by (1), therefore we have:
$$\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
show this inequality $\sum\frac{x}{2+xy}\ge\frac{1}{2}$
Let $x,y,z\ge 0$ such that $$x+y^2+z^3=1.$$
Show that
$$\dfrac{x}{2+xy}+\dfrac{y}{2+yz}+\dfrac{z}{2+zx}\ge\dfrac{1}{2}$$
I try do
$$\sum_{cyc}\dfrac{x}{2+xy}=\sum_{cyc}\dfrac{x^2}{2x+x^2y}\ge\dfrac{(x+y+z)^2}{(2x+2y+2z)+(x^2y+y^2z+z^2x)}$$
it have to prove
... | I don't know if this would help you but here is a hint for you.
$\left \{ \begin{aligned} x, y, z \ge 0\\ x + y^2 + z^3 = 1 \end{aligned} \right. \implies 0 \le x, y, z \le 1 \implies \left\{ \begin{aligned} x = x\\ y \ge y^2\\ z \ge z^3 \end{aligned} \right. \implies x + y + z \ge x + y^2 + z^3 = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$ Let $a,b,c$ be reals with $a+b+c=1$. Show that :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$
I have tried to something like this:
$$18a^4-24a^3+6a^2-12a+12\geq 0$$
$$18b^4-24b^3+6b^2-12b+12\geq 0$$
$$18c^4-24c^3+6c^2-12c+12\g... | Writing your inequality in the form $$\frac{18(a^4+b^4+c^4)}{a+b+c}+6(a^2+b^2+c^2)(a+b+c)+(a+b+c)^3-24(a^3+b^3+c^3)\geq 0$$ so we get that this is equivalent to
$$\frac{\left(a^2-4 a b-4 a c+b^2-4 b
c+c^2\right)^2}{a+b+c}\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$... | Option:
Normal of the plane: $(1,1,1)$,
Normalized(i.e of unit length): $(1/√3)(1,1,1)$
Line through $(8,0,-6)$:
$\vec r= (8,0,-6)+t (1/√3)(1,1,1)$.
Determine $t$ when line intersects plane:
$(8+t/√3)+t/√3+(-6+t/√3)=6;$
$3(t/√3)=4$;
$t= 4/√3.$
Distance = $4/√3$ (Why?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the angle in a quadrilateral
This is the picture, and we are aiming for the angle $x$
It's easy to see that $\angle DGA = \angle CGB = 100°$, $\angle CGD = \angle AGB = 80°$, $\angle CBG = 50°$, but now i'm missing $\angle GBA = ? $ and $\angle GAB = x$
Any hints?
| Let $y$ denote the angle $\angle GBA$. By the law of sines applied to the four triangles with vertex $G$, we obtain the relations
$$\frac{AB}{\sin 80^\circ}=\frac{BG}{\sin x}=\frac{AG}{\sin y}.$$
$$\frac{AD}{\sin 100^\circ}=\frac{AG}{\sin 40^\circ}=\frac{DG}{\sin 40^\circ},$$
$$\frac{BC}{\sin 100^\circ}=\frac{BG}{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Functional equation to determine $f(-1)$
There's a function $f$ satisfying:
$$f\left(\frac{1}{1−x}\right)+2\cdot f\left(\frac{x−1}{x}\right)=3x$$
Find the value of $f(−1)$.
I have no idea how to solve this one. If anyone could help me to understand it, I would be grateful.
| Substitute $1-\frac{1}{x}$ instead of $x$.
We obtain:
$$f\left(\frac{1}{1-1+\frac{1}{x}}\right)+2f\left(1-\frac{1}{1-\frac{1}{x}}\right)=3\left(1-\frac{1}{x}\right)$$ or
$$f(x)+2f\left(\frac{1}{1-x}\right)=3\left(1-\frac{1}{x}\right).$$
Now, substitute $1-\frac{1}{x}$ instead of $x$ in the last equation.
We obtain:
$$f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the integral $\int \frac{(\ln(x))^2}{x^3} \, dx$ $$\int \frac{(\ln(x))^2}{x^3} \, dx $$
Starting off with Integration by Parts
$$
\begin{align}
u = \ln(x)^2 &~~~ dv = x^{-3} \\\\
du = 2\ln(x)dx &~~~ v = \frac{x^{-2}}{-2}
\end{align}
$$
$$
\begin{align}
\int \frac{(\ln(x))^2}{x^3} &= (\ln(x))^2 \left( \frac{-x^{-2}... | Let's prove by induction $$\displaystyle I_p(n)=\int\dfrac{\ln(x)^n}{x^p}\mathop{dx}=\dfrac{P_n(\ln(x))}{x^{p-1}}$$ where $P_n$ is a polynomial of degree $n$.
We will assume in the following $p>1$, so we can carry on integration by parts.
$\displaystyle I_p(0)=\int \dfrac{\mathop{dx}}{x^p}=\dfrac{1-p}{x^{p-1}}$ thus $P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
3-variable symmetric inequality Given $a,b,c>0$ satisfying $a^2+b^2+c^2=3$. Prove that
$$2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+3(a+b+c)\geq 15.$$
I've tried to use the inequality $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{9}{a+b+c}$ as well as AM-HM and the condition $a+b+c \leq 3$ but it still d... | There is also the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, and $abc=w^3$.
Thus, by AM-GM $v\geq w$ and the condition gives $3u^2-2v^2=1,$ which gives $u=\sqrt{\frac{1+2v^2}{3}}.$
Thus, $$\sum_{cyc}\frac{1}{a}=\frac{ab+ac+bc}{abc}=\frac{3v^2}{w^3}\geq\frac{3}{v}.$$
Id est, it's enough to prove that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find some digits of $17!$
$17!$ is equal to $$35568x428096y00$$
Both $x$ and $y$, are digits. Find $x$ and $y$.
So, $$17!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13\times 17=(2^3\times 5^3)\times 2^{12}\times 3^6\times 7^2\times 11\times 13\times 17$$
If there`s a product of $(2\times 5)^3$
Then this nu... | HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "22",
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Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$. I am new to multivariable limits and I am having a hard time arguing this limit:
Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$.
I can work out the overarching context of the proof as foll... | Break down the problem
This function is a compositition of the multivariate affine function $(x, y) \mapsto x + y - 2$, and the univariate square root function $x \mapsto \sqrt{x}$. Both of these functions are continuous on their domains. Affine functions are very straightforward functions, and are easier to prove cont... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse image of an open set with a function Let $f: (\mathbb{R},|.|)\to (\mathbb{R},|.|)$ defined by $f(x)=\dfrac{1}{1+x^2}$
I want to find $f^{-1}(]a,b[)$
What I do:
$$f^{-1}(]a,b[)=\{y\in \mathbb{R}, \frac{1}{1+y^2}\in ]a,b[\}$$
that is : $$a<\frac{1}{1+y^2}<b$$
*
*If $a>0$ then $$\frac1b-1<y^2<\frac1a-1$$
af... | I think a good way to approach the problem is to think first about the function $f$:
First of all, since $x^{2} \geq 0$ then $1+x^{2} \geq 1$ which leads to $ f(x) = \frac{1}{1+x^{2}} \leq 1$. This means that if $a > 1$ then $f^{-1}(]a,b[) = \emptyset$.
We also have that $1+x^{2} \geq 1 > 0$ so $f(x) = \frac{1}{1+x^{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove $\tan1 < \frac{\pi}{2}$? Prove that the equation
$$\sin x \sin({\sin x}) = \frac{\pi}{2} \cos({\sin x})$$
has no real solutions.
Let $t=\sin x$, $-1\leq t\leq 1$. Then the expression above is equvalent to $t\sin t = \frac{\pi}{2} \cos t$. As the function $f(t)=t\sin t - \frac{\pi}{2} \cos t$ is even, and... | This is a Community Wiki repost of one of several deleted answers to a deleted question from last year: Compare $\arcsin (1)$ and $\tan (1)$.
[I have taken this opportunity to incorporate a simplification - but only a slight one - I made later in a comment.]
I would be pleased if we were allowed to revisit the other de... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The inductive proof of $\sum_{k=1}^n \frac k{k+1} \leq n - \frac1{n+1}$ is unclear Prove by induction of $n$
$$\sum_{k=1}^n \frac k{k+1} \leq n - \frac1{n+1}$$
\begin{align}\sum_1^{n+1}\frac k{k+1}&\leq n-\frac 1{n+1}+\frac{n+1}{n+2}\\&=n-\frac 1{n+1}+1-\frac 1{n+2}\\&=(n+1)-\frac{2(n+2)-1}{(n+1)(n+2)}\\&=(n+1)-\frac ... | Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
$$\sum_{k = 1}^{n} \frac{k}{k + 1} \leq n - \frac{1}{n + 1},$$
and want to end with the conclusion t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please... | Let $a, b, c\in [0, 1]$ satisfying all the hypothesis and $a=\max\{a, b, c\}>b$, it's clear that this expression must have a maximum. Set
$$
a'=\frac{a+b}2<a\\
b'=\frac{a+b}2>b
$$
then $a'+b'+c=1$ and
$$
a'b'+a'c+b'c=\frac{(a+b)^2}4+ab+bc=ab+bc+ac+\frac{(a-b)^2}{4}\\
2a'b'c=2c\frac{(a+b)^2}{4}=2abc+2c\frac{(a-b)^2}4\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Understanding a technique mentioned in 2016 IMO Shortlist The following question is there in the IMO shortlist 2016:
Let $a, b$ and $c$ be positive real numbers such that min $ (ab, bc, ca) > 1$. Prove that
$$((a^2 + 1)(b^2 + 1)(c^2 + 1))^{1/3} \leq \left(\frac{a+b+c}{3}\right)^2 + 1$$
In the shortlist document htt... | I understood the question being about the "mixing variables" technique, not about proving the lemma. While I haven't heard the term, the technique itself is familiar, though I could rarely apply it in an olympiad setting.
Using the lemma on $(a,b), (a,c)$ and $(b,c)$ and multiplying the result we get
$$(a^2+1)^2(b^2+1)... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $ with Taylor I have to calculate some limits and try to solve them in use of taylor.
$$
\lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right)
$$
In taylor pattern I have $x_0$ to put, but there $x_0$ is $\infty$ so I want to replace i... | You just have to write the Taylor expansion of $\ln \left( 1 + \frac{1}{x} \right)$ when $x$ tends to $+\infty$ :
$$x - x^2 \ln \left( 1 + \frac{1}{x} \right) = x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + o\left( \frac{1}{x^2} \right) \right) = \frac{1}{2} + o(1)$$
So the limit is equal to $\frac{1}{2}$.
| {
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How obtain $\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$ when integrating $\sqrt{\dfrac{x-1}{x+2}}, x > 1$ The book I'm using states two different answers, one being
$\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$
I have tried two different approaches.
Approach one was to rewrite the integral like this... | A slightly different substitution from yours gives the result stated in your book:
$$u=\sqrt{x+2}\implies x=u^2-2\implies\mathrm dx=2u\,\mathrm du$$
$$\begin{align*}
\int\sqrt{\frac{x-1}{x+2}}\,\mathrm dx&=\int\frac{\sqrt{u^2-3}}u(2u\,\mathrm du)\\[1ex]
&=2\int\sqrt{u^2-3}\,\mathrm du\\[1ex]
&=u\sqrt{u^2-3}-3\ln\left|\... | {
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Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 +... | You can also bring in the divisibility test for $11$ into the problem. First note that $700+400=1100$ and $ab+ba=11(a+b)$, forcing your sum $1a21$ to be a multiple of $11$.
By the divisibility test for $11$, then, $1-a+2-1$ must be a multiple if $11$, forcing $a=2$. The ones digit sum then must be $b+2=11(\ge 2)$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3145603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Pattern in power towers of 2 involving last digits We have
\begin{align}
2^{2^{2}} &\mod 10 = 6 \\
2^{2^{2^2}} &\mod 100 = 36 \\
2^{2^{2^{2^2}}} &\mod 1000 = 736 \\
2^{2^{2^{2^{2^{2}}}}} &\mod 10000 = 8736 \\
2^{2^{2^{2^{2^{2^2}}}}} &\mod 100000 = 48736
\end{align}
I think you get the point. Basically, it seems like ... | Let $x_0=4$ and $x_{n+1}=2^{x_n}$.
The claim is $x_{n+1}\equiv x_n\pmod{10^n}$.
By induction on $n$, we assume $x_{n+1}\equiv x_n\pmod{10^n}$ and we prove $x_{n+2}\equiv x_{n+1}\pmod{10^{n+1}}$.
We have
$$x_{n+2}-x_{n+1}=2^{x_{n+1}}-2^{x_n}=2^{x_n}(2^{x_{n+1}-x_n}-1)$$
Since $x_n\geq n+1$ we get $x_{n+2}\equiv x_{n+1}\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$y-xy' =2(x+yy'), y(1) =1 $ is..
Question: $y-xy' =2(x+yy'), y(1) =1 $ is..
Firstly, I do not even know what the question is asking for.
Secondly, Why does this question put $y(1)=1$? Isn't it obvious or is there any special meaning to it.
Surprisingly, the answer to it is also very unique.
The answer is: $\tan^{-... | How the equation can be solved
The equation can be rewritten as
$$
y' = \frac{y-2x}{2y+x}.
$$
Introduce $u = y/x$ so that $y = xu$ and $y' = xu' + u.$ This makes the equation take the form
$$
xu' + u = \frac{xu-2x}{2xu+x} = \frac{u-2}{2u+1}
$$
i.e.
$$
xu' = \frac{u-2}{2u+1} - u = \frac{(u-2)-u(2u+1)}{2u+1} = -2 \frac{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151552",
"timestamp": "2023-03-29T00:00:00",
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Prove that $a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that
$$a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$$
I found a point at which the equality is attended, say $a=0,b=1,c=2$. But I have no idea how to prove it. I tried to use the AM-GM in... | Start out as before:
$$
a\sqrt{b^{3}+1}+b\sqrt{c^{3}+1}+c\sqrt{a^{3}+1}\leq5
$$
$$
a\cdot\sqrt{\left(b+1\right)\left(b^{2}-b+1\right)}\leq a\cdot\frac{b^{2}+2}{2}
$$
$$
\frac{a\left(b^{2}+1\right)+b\left(c^{2}+1\right)+c\left(a^{2}+1\right)}{2}=3+\frac{ab^{2}+bc^{2}+ca^{2}}{2}
$$
Continue by lagrange multipliers:
$$
\L... | {
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How to prove this identity nicely?
Show that$$\sum_{1\le i<j\le n}\left((x_j-x_i)-(x_j-x_i)^2\right)=\left(\sum_{i=1}^n{x_i}\right)^2-n\sum_{i=1}^n{x_i^2}-\sum_{i=1}^n{(n-2i+1)x_i}\\=-n\sum_{i=1}^n\left(x_i-\frac1n\sum_{j=1}^n{x_j}+\frac{n-2i+1}{2n}\right)^2+\frac1{4n}{\sum_{i=1}^n(n-2i+1)^2}.$$
I wonder how this ide... | $\def\peq{\mathrel{\phantom{=}}{}}$For the first identity, because$$
\sum_{i < j} (x_j - x_i) = \sum_{k = 1}^n x_k \left( \sum_{l < k} 1 + \sum_{l > k} (-1) \right) = \sum_{k = 1}^n (2k - n - 1) x_k,
$$\begin{gather*}
\sum_{i < j} (x_j - x_i)^2 = \sum_{k = 1}^n x_k^2 \left( \sum_{l < k} 1 + \sum_{l > k} 1 \right) - 2 \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of the series $1+1/2 -1/3 -1/4 -1/5 +1/6 +1/7 - 1/8 -1/9 - 1/10 + 1/11 + 1/12-...$ I would like to prove that $1+1/2 -1/3 -1/4 -1/5 +1/6 +1/7 - 1/8 -1/9 - 1/10 + 1/11 + 1/12 -...$ converges or diverges
I first showed that $1-1/2 -1/3 +1/4+1/5-1/6-1/7+1/8+1/9-1/10...$ converges by writing it as $1+\Sigma(-1)... | Note that you can group all but the first two terms into groups of five terms of the form
$$
-\frac{1}{5k+3}-\frac{1}{5k+4}-\frac{1}{5k+5}+\frac{1}{5k+6} + \frac{1}{5k+7}
< -\frac{1}{5k+3}
$$
This means that the $5k+7$-th partial sum will be bounded above by
$$
1 + \frac{1}{2} - \frac{1}{3}-\frac{1}{8} - \frac{1}{13} ... | {
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Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.
Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.
I have been doing similar problems, but can't get my head around this one. I'd rather get a tip instead of the full answer. Thanks to anyone who contributes.
My Solution After Tips:
Let $u = x + y$ and $ v = x - ... | You can solve it directly as follows:
*
*$f(x+y,x-y) = xy+y^2 = (x+y)y$
Now, write
*
*$y = \frac{1}{2}((x+y)-(x-y))$
It follows:
$$f(x+y,x-y)= (x+y)\cdot \frac{1}{2}((x+y)-(x-y)) = \frac{1}{2}\left( (x+y)^2 - (x+y)(x-y)\right)$$ $$\Rightarrow f(x,y) = \frac{1}{2}(x^2-xy)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the functio... | Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation
$$
y^{\prime\prime} + y^\prime - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
\begin{align}\label{eq:1}\tag{1}
y^\prime(x) = 2Ax + B \quad \text{and} \quad y^{\prime\prime}(x) = 2A.
\end{align}
So, if $y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$
If $a,b,c\in\mathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is
what i try
Cauchy schwarz inequality
$$(a^2+(2b)^2+c^2)(2^2+\frac{1}{2^2}+3^2)\geq (2a+b+3c)^2$$
How do i solve it without Cauchy schwarz inequality Help me please
| Let $(a,b,c)=\left(\frac{160}{53},\frac{20}{53},\frac{240}{53}\right).$
Thus, we get a value $\frac{1600}{53}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$a^2+4b^2+c^2\geq\frac{1600}{53}\cdot\left(\frac{2a+b+3c}{20}\right)^2$$ or
$$37a^2-16(b+3c)a+208b^2-24bc+17c^2\geq0,$$ for which it's eno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.
Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.
Approach:
*
*I defined a function $f: \mathbb{R... | Consider the following:
$$2^{x}-2^{-x}-2^{x+1}=2^x-2^{-x}-2\cdot2^x= -2^x-2^{-x}=-2(2^{x-1}+2^{-x-1}) = - 2 \left(\frac{2^x+2^{-x}}{2}\right) $$
Hence
$$2^{x}-2^{-x}-2^{x+1} = - 2 \left(\frac{2^x+2^{-x}}{2}\right) = -2 \cosh(x \log 2)$$
Similarly
$$2^{x}-2^{-x}+2^{1-x}=2^x-2^{-x}+2\cdot2^{-x}= 2^x+2^{-x}=2(2^{x-1}+2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding a limit of a matrix raised to the $n$-th power There is a stochastic matrix given
$$\mathbb{P} = \begin{bmatrix} 0&\frac{1}{2}&\frac{1}{2}&0 \\ 0&1&0&0 \\ \frac{1}{2}&0&0&\frac{1}{2} \\ \frac{1}{3}&0&\frac{1}{3}&\frac{1}{3} \end{bmatrix}$$
I am to find $\mathbb{P}^n$. I know a theorem which says that if $\exist... | We have $P=SDS^{-1}$, where
$$
S=\pmatrix{
-1 & -1 & \frac{9}{16} & 1 \\
0 & 0 & 0 & 1 \\
1 & 0 & \frac{15}{16} & 1 \\
0 & 1 & 1 & 1 \\
}
,\quad
D=\pmatrix{
-\frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & \frac{5}{6} & 0 \\
0 & 0 & 0 & 1 \\
}
$$
Therefore, $P^\infty = S D^\infty S^{-1}$.
Since
$$
D^... | {
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"url": "https://math.stackexchange.com/questions/3170445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partial fraction decomposition of even function I need to do partial fraction decomposition of this function (to solve its integral):
$\frac{t^2}{t^4+4}$
Since $t^4+4=(t^2+2t+2)(t^2-2t+2)$ I would do:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}+C\frac{2t-2}{t^2-2t+2}+D\frac{1}{t^2-2t+2}$
And then cont... | Since the function is even we have that
$$A\frac{2t+2}{(t+1)^2}+B\frac{1}{(t+1)^2}+C\frac{2t-2}{(t-1)^2}+D\frac{1}{(t-1)^2}$$ is equal to
$$A\frac{-2t+2}{(-t+1)^2}+B\frac{1}{(-t+1)^2}-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}.$$ Reording terms we get
$$-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}-A\frac{2t-2}{(-t+1)^2}+B\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bounded recursive sequence - Proof by induction
Given the sequence $(x_n)$ defined by
\begin{cases}
x_1 &= 1\\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right),
\end{cases}
prove that $1 \leq x_n \leq \frac{3}{2}, \forall n \in N$.
I verified the base case for $n=1$ and $n=2$.
Assumed that the boundaries hold f... | For induction step we have to prove
$1\leqslant x\leqslant\frac32\Rightarrow 2\leqslant x+\frac2x\leqslant3$
that is $x^2-3x+2\leqslant0$ and $x^2-2x+2\geqslant0$ for $1\leqslant x\leqslant\frac32$
Can you finish?
| {
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"answer_id": 0
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Find the value of $\lambda$ in $\frac{3-\tan^2 {\pi\over 7}}{1-\tan^2 {\pi\over 7}}=\lambda\cos{\pi\over 7}$
Find the value of $\lambda$ in $$\dfrac{3-\tan^2 {\pi\over 7}}{1-\tan^2 {\pi\over 7}}=\lambda\cos{\pi\over 7}$$
The numerator looks similar to expansion of $\tan 3x$, so I tried this
$$\dfrac{3\tan {\pi\over ... | Let $7x=\pi$
$$\lambda\cos x=\dfrac{3-\tan^2x}{1-\tan^2x}=\dfrac{3\cos^2x-(1-\cos^2x)}{\cos^2x-(1-\cos^2x)}$$
$$\iff2\lambda\cos^3x-4\cos^2x-\lambda\cos x+1=0$$
From How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$,
$\cos\dfrac{(2n+1)\pi}7; n=0,1,2$ are the roots of $8t^3-4t^2-4t+1=0$
Comparing wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a closed form for coefficients in $x^{3n}=x_0\left(a_nx+b_n+\frac {c_n}{x}\right)$ Consider,
$$
x^3=x+1
$$
Let $x_0$ be a solution to the above equation. Now consider $x^{3n}$. For $n=2$ we have:
$$
x^6=(x+1)^2
$$
$$
=x^2+2x+1
$$
$$
=x\left(x+2+\frac {1}{x}\right)
$$
$$
=x_0\left(x+2+\frac {1}{x}\right)
$$
For ... | We derive generating functions for the recurrence relation:
\begin{align*}
a_{n+1}&=a_n+b_n\tag{1}\\
b_{n+1}&=a_n+b_n+c_n\qquad\qquad (n\geq 2)\tag{2}\\
c_{n+1}&=a_n+c_n\tag{3}\\
a_2&=1,b_2=2,c_2=1\\
\end{align*}
Let $A(x)=\sum_{n\geq 2} a_nx^n, B(x)=\sum_{n\geq 2} b_nx^n, C(x)=\sum_{n\geq 2} c_n x^n$.
We obtain from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How can I prove a curve lies on an ellipsoid? I am trying to prove a curve parameterized by
$$\mathbf{r} (t) = \cos(t) \, \mathbf{i} + \sqrt{2} \sin(t) \, \mathbf{j}-\sin(t) \, \mathbf{k}$$
lies on an ellipsoid. How do I do this?
| The general equation of an ellipsoid in $\Bbb R^3$ is
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1, \tag 1$
where
$abc \ne 0; \tag 2$
in order for the curve
$r(t) = \cos t \mathbf i + \sqrt 2 \sin t \mathbf j - \sin t \mathbf k \tag 3$
to lie on a surface (1), it's components
$x(t) = \cos t, \; y(t) = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving an algebraic binomial identity related to Bertrand's ballot theorem I am trying to answer the math.stackexchange question found here by developing all the theory from scratch and not using Bertrand's ballot theorem. My logic boils down to being able to prove the following identity:
$$ \frac{n-m+1}{n+1} \binom{n... | Well, we just need to simplify the right hand side:
\begin{align}
RHS &= \frac{n-1-m+1}{n} {n+m-1 \choose n-1} + \frac{n-m+2}{n+1} {n+m-1 \choose n} \\
&= \frac{n-m}{n} \frac{(n+m-1) !}{(n-1)! m!} + \frac{n-m+2}{n+1} \frac{(n+m-1) !}{n! (m-1)!} \\
&= \Big[\frac{n-m}{n+m} + \frac{(n-m+2)m}{(n+1)(n+m)}\Big] {n+m \choose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | $$3x+2y=36\tag1$$
$$5x+4y=64\rightarrow \rightarrow \rightarrow \rightarrow 3x+2y+3x+2y-x=64$$
$$36+36-x=64$$
$$x=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 5
} |
For positive $a$, $b$, $c$, $d$ with $a+b+c+d=4$, show that $\sum_{cyc} \left(\frac{1}{a^2}-a^2\right)\ge0$ From MOP 2012:
If $a,b,c,d>0$, $a+b+c+d=4$, show
$$\sum_{cyc} \left(\frac{1}{a^2}-a^2\right)\ge0$$
My attempt: Using the derivative and tangent line, I obtain
$$(x-1)(-x^3+7x^2-x-1)=0$$
Does using $n-1$ EV ... | EV works of course, but there is something nicer:
$$\sum_{cyc}\frac{1}{a^2}=\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)+\sum_{cyc}\frac{1}{ab}=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\sum_{cyc}\frac{1}{ab}=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2+\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $a,b$ are even integers I'm trying to prove the following theorem:
Let $a,b\in\mathbb{Z}$ . Then $$a^{2}+b^{2}\equiv0\pmod 4 \iff a \;\text{and}\; b\;\text{are even}$$
I always struggle to prove some number is odd or even. How to prove it?
I thought of using the $(a+b)^2=a^2+2ab+b^2$ formula but not sur... | $\Rightarrow$:
By the rules of modular arithmetic we have
\begin{align}
0
&= (a^2 + b^2) \bmod 4 \\
&= \left( \left( a^2 \bmod 4 \right) + \left( b^2 \bmod 4 \right) \right) \bmod 4
\end{align}
For each summand we have
\begin{align}
x^2 \bmod 4
&= \left( \left( x \bmod 4 \right) \left( x \bmod 4 \right) \right) \bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to solve $|a+b|+|a-b|=c$? It is intuitive that $a=\pm \frac{c}{2}$, with $-\frac{c}{2}\leq b\leq \frac{c}{2}$ or vice-versa are solutions to the problem. Can I get to these solutions without dividing the expression in all the cases (i.e. $a$ and $b$ being positive or negative and $|a|$ being greater than or less th... | There are several standard methods: intervals (cases), squaring, graphical.
Let's do squaring:
$$(|a+b|+|a-b|)^2=c^2 \iff \\
2a^2+2b^2+2|a^2-b^2|=c^2 \iff \\
(2|a^2-b^2|)^2=(c^2-2(a^2+b^2))^2 \iff \\
4a^2+4b^2-8a^2b^2=c^4-4(a^2+b^2)c^2+4a^2+4b^2+8a^2b^2 \iff \\
c^4-4(a^2+b^2)c^2+16a^2b^2=0 \Rightarrow \\
c^2=2(a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show With Induction that $2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$ Show with induction that
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$
My solution:
Base case 1: n = 1
LHS = $(1+1)\cdot2^{1-1} = 2$
RHS = $1\cdot2^{1}= 2$
Case 2: n = p
When... | The solution without induction:
$$2\cdot2^0+3\cdot2^1+...+(n+1)2^{n-1}=\frac{1}{2}\left(2\cdot2^1+3\cdot2^2+...+(n+1)2^n\right)=$$
$$=\frac{1}{2}\left(x^2+x^3+...+x^{n+1}\right)'_{x=2}=\frac{1}{2}\left(\frac{x^2(x^n-1)}{x-1}\right)'_{x=2}=$$
$$=\frac{1}{2}\left(\frac{\left((n+2)x^{n+1}-2x\right)(x-1)-\left(x^{n+2}-x^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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.