Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \fra... | If $T(n) = \frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}$, then
$$T(n+1) - T(n) = \frac{1}{2^{n+1}}$$
hence if $T(n) = \frac{2^n-1}{2^n}$, then
$$T(n+1) = T(n) + \frac{1}{2^{n+1}} = \frac{2^n-1}{2^n}+\frac{1}{2^{n+1}} = \frac{2(2^n-1)}{2^{n+1}}+\frac{1}{2^{n+1}} = \frac{2^{n+1}-2+1}{2^{n+1}},$$
giving the desired formul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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How to approach an integral over $g(\cos(t))$ from $0$ to $2\pi$, where $g(x)$ is nasty? For notational convenience, let $f(t) = a^2 + 2 a b \cdot \cos(t) + b^2$, where $a,b$ are both positive real constants and $t$ will be the variable of integration, which is supposed to be carried out from $t=0$ to $t=2 \pi$. I want... | Because of symmetry, we will take the integral from $t=0$ to $t=\pi$ and double it. Let $f(t)=u$.
It follows that
$$du=-2abSIN(t)dt$$
Converting the limits ($t=0$ & $t=\pi$) from t to u using the equation $u=a^2+2abCOS(t)+b^2$ we get $$lower limit=a^2+2ab+b^2$$
$$upper limit=a^2-2ab+b^2$$
Solving for cos(t), we get $$c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find $\int{\frac{\sqrt{x^2+1}}{x+2}dx}$ I have got such integral $$\int{\frac{\sqrt{x^2+1}}{x+2}dx}$$ and with Maple I got something like this:
$$\int\frac{1}{2} + \frac{1+3u^2+4u^3}{-2u^2+2u^4-8u^3}du$$ And I want to know how to achive this changes.
I tried to use WolframAlpha, but there is scarier solution. Th... | *
*We can use the Euler substitution $t=\sqrt{x^{2}+1}-x$ to obtain a rational fraction in terms of $t$
$$\begin{eqnarray*}
I =\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }\mathrm{d}t.
\end{eqnarray*}$$
*Since the integrand is a rational fraction, ... | {
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"url": "https://math.stackexchange.com/questions/145066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate square root of $i \Leftrightarrow z^2=i$ Let $z = r(\cos\theta+i\sin\theta)$. In my notes there was this example to calculate the square roots of $i$. What was done was:
$z = r(\cos\theta+i\sin\theta)\\z^2 = r^2(\cos(2\theta)+i\sin(2\theta))\\=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\\Leftrightarrow r=1 \ \ ... |
How did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?
This whole problem is easiest to do via Euler's identity: $$e^{i\theta} = \cos(\theta) + i\sin(\theta).$$ Now, any complex number $z$ can be
expressed as $z=re^{i\theta}$ where $r$ is some real
number and $0 \leq \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 1... | If $n\equiv 0\pmod 3$ Ok. If $n\equiv 1\pmod 3$, we have
\begin{equation}
n^{2} + 8 \equiv 1^{2} + 2\equiv 0\pmod 3.
\end{equation}
If $n \equiv 2\pmod 3$ we have
\begin{equation}
n^{2} + 8 \equiv 2^{2} + 2\equiv 6 \equiv 0\pmod 3.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 3
} |
Solving $|x-2| + |x-5|=3$
Possible Duplicate:
How could we solve $x$, in $|x+1|-|1-x|=2$?
How should I solve:
$|x-2| + |x-5|=3$
Please suggest a way that I could use in other problems of this genre too
Any help to solve this problem would be greatly appreciated.
Thank you,
| Well, there are a couple of ways.
Method 1. By cases.
One is to consider cases: note that
$$\begin{align*}
|x-2|&=\left\{\begin{array}{ll}
x-2 & \text{if }x\geq 2\\
2-x &\text{if }x\lt 2
\end{array}\right.\\
|x-5|&=\left\{\begin{array}{ll}
x-5 &\text{if }x\geq 5\\
5-x &\text{if }x\lt 5
\end{array}\right.
\end{align*}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is $f(x)={(x^2+2x-48)}/{x^2}$ increasing? decreasing? The question is to find where the graph is increasing or decreasing.
The original function is $f(x)={(x^2+2x-48)}/{x^2}$
I know I need to find the prime of this function and I think it is this after using the quotient rule:
$2(-x^2-x+49)/x^3$
Finally, in orde... | You computation of derivative is incorrect. It is easier to separate out each term and compute the derivative.
$$f(x) = \dfrac{x^2 + 2x - 48}{x^2} = 1 + \dfrac2{x} - \dfrac{48}{x^2}$$
Hence, $$f'(x) = 0 - \dfrac2{x^2} + \dfrac{2 \times 48}{x^3} = \dfrac{96-2x}{x^3}$$
Now setting, $f'(x) = 0$ gives us $96-2x = 0 \implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$... | I find it easier to go from polar to rectangular.
$r=2a\cos\theta+2b\sin\theta$; $r^2=2ar\cos\theta+2br\sin\theta$; $x^2+y^2=2ax+2by$; $x^2-2ax+y^2-2by=0$; $(x-a)^2+(y-b)^2=a^2+b^2$; voila, circle of radius $\sqrt{a^2+b^2}$ centered at $(a,b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.
If $n = 1 + m$, where $m$ is the product of four co... | Product of 4 consecutive numbers can be shown as
$$\begin{align}(x+1)(x+2)(x+3)(x+4)
&=(x+1)(x+4)(x+2)(x+3) \\
&=(x^2+5x+4) (x^2+5x+6) \\
&=\underset{A-B}{(x^2+5x+5 -1)} \underset{A+B}{(x^2+5x+5 +1)} \\
&=\underset{A^2-B^2}{(x^2+5x+5)^2 - 1}\end{align}$$
Where $(x^2+5x+5)^2$ is a perfect square.
Proved... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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"answer_id": 15
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The Fibonacci sum $\sum_{n=0}^\infty \frac{1}{F_{2^n}}$ generalized The evaluation,
$$\sum_{n=0}^\infty \frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}=\left(\frac{1-\sqrt{5}}{2}\right)^3+\left(\frac{1+\sqrt{5}}{2}\right)^2$$
was recently asked in a post by Chris here.
I like generalizations, and it turns out this is not a uniq... | Your conjecture is indeed right. Before proving your conjecture, let us obtain an intermediate result first. Let us prove the following claim first.
CLAIM:
If we have a sequence given by the recurrence,
$$a_{n+2} = ba_{n+1} + a_n,$$ with $a_0 =0 $ and $a_1 = 1$, we then have
$$\boxed{\color{blue}{\displaystyle \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
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$1 +1$ is $0$ ?
Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?
So:
$$
\begin{align}
1+1 &= 1 + \sqrt{1} \\
&= 1 + \sqrt{1 \times 1} \\
&= 1 + \sqrt{-1 \times -1} \\
&= 1 + \sqrt{-1} \times \sqrt{-1} \\
&= 1 + i \times i \\
&= 1 + (... | We have that $\sqrt{-1} \times \sqrt{-1} = (\sqrt{-1})^{2} = -1$ but $\sqrt{-1 \times -1} = \sqrt{1} = 1$. So $\sqrt{-1 \times -1} \neq \sqrt{-1} \times \sqrt{-1}$ which is the error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How to derive ellipse matrix for general ellipse in homogenous coordinates So lets say we have an ellipse with axes a and b and the rotation angle $\phi$ and center at $(0,0)$.
Now I apply the rotation to $x^2/a^2+y^2/b^2=1$ getting
$$x' = x\cos(\phi) + y\sin(\phi)$$
$$y' = y\cos(\phi) + x\sin(\phi)$$
$$x^2(b^2cos(\phi... | In general, an ellipse in a general position $[h,k]$ (what I needed) is implicitly given as
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
where $a$ and $b$ are semiaxes. Rotating the $x$ and $y$ coords yields (you're missing a $-$ sign in the first line)
$$x' = x\cos(\phi) - y\sin(\phi)$$
$$y' = x\sin(\phi) + y\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac... | One possible proof is to use the fact that for triangular $n\times n$ matrix $A$ it holds
$$\det(A) = \prod_{i=1}^n a_{i,i}$$
An example
$$\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \... | {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$ I don't know how to find an explicit form for this sum, anyone can help me?
$$\sum_{k=-\infty}^{\infty}
{1 \over \left\vert\,x - k\,x_{\atop{ \small 0}}\,\right\vert}
$$
Here are the calculations I made, but don't bring me anywhere:
(original image)
$$\begi... | As noted by some users, the series below is the one for the case
$$\sum_{k=-\infty}^{+\infty}\frac{1}{z-k}$$
i.e, there are no absolute values.
I scanned too fast but the last thing you have is this
$$\pi \cot(\pi z)=\frac 1 z+2z \sum_{n=1}^\infty \frac{1}{z^2-n^2} $$
One option is to use
$$\frac{{\sin \pi z}}{{\pi z}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| Expand $\frac{1}{1+x^{4}}$ into partial fractions. For this purpose you need
to factorize the polynomial in the denominator. You can proceed by writing it as a product of four linear terms
\begin{equation*}
x^{4}+1=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})
\end{equation*}
where $x_{1},x_{2},x_{3},x_{4}$ are its complex root... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 6
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What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$
Here is my attempt:
Letting
$$f(z)=\frac{1}{az^2+bz+c}$$
The poles of $f(z)$ are located at
$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$z_1 = \frac{-b-\sqrt{b^2-4ac}}... | Take $a=1,\ b=3, \ c=2$, then $z_0=-2, \ z_1=-1$, and so you have to compute $\cot(-\pi)$ and $\cot(-2\pi)$ which make no sense. However
$$
\sum_{n=0}^\infty\frac{1}{n^2+3n+2}=\sum_{n=0}^\infty(\frac{1}{n+1}-\frac{1}{n+2})
=\lim_{m\to \infty}(1-\frac{1}{m+2})=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 0
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Whats the sum of the length of all the sides of a triangle? You are given triangles with integer sides and one angle fixed at 120 degrees. If the length of the longest side is 28 and product of the remaining to sides is 240, what is the sum of all sides of the triangle?
I have tried to solve it using the formula given ... | Since the longest side must be opposite the largest angle in the triangle, the side which is 28 units is opposite the angle of degree measure of 120. Let us name the remaining sides $a$ and $b$. Using the cosine rule we get:
$$28^2=a^2+b^2-2ab\cos(120)$$
$$784=a^2+b^2-2ab\left(\frac{-1}{2}\right)$$
$$784=a^2+b^2+ab$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Error Term in Passing from Summation to Integral I encountered the following in a paper and do not understand how the error term is being bounded. In what follows, $n$ and $k$ are large integer constants.
$$ \sum_{i=0}^{n-1} \ln\left(1 - \frac{i}{k}\right) = \int_0^n
\ln\left(1 - \frac{x}{k}\right) dx \pm e(n,k) $$ w... | Note that $\ln \left( 1- \dfrac{x}k \right)$ is a decreasing function of $x$ for all $x < k$. Hence, $$\int_a^{a+1} \ln \left( 1 -\dfrac{x}k \right) dx \leq \ln \left( 1 - \dfrac{a}k\right) \leq \int_{a-1}^{a} \ln \left( 1 -\dfrac{x}k \right) dx $$
Hence, taking $a=0$ to $a=n-1$ and adding them up, we get that
$$\int_0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polynomials identity, factoring $x^{2^n}-1$ There is a proof that I can't solve.
Show that for any integer $k$, the following identity holds:
$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{k-1}})=1+x+x^2+x^3+\cdots+x^{2^k-1}$$
Thanks for your help.
| We work with the left-hand side.
$$(1+x)(1+x^2)=(1+x)(1)+(1+x)(x^2)=1+x+x^2+x^3.$$
Thus
$$\begin{align}
(1+x)(1+x^2)(1+x^4)&=(1+x+x^2+x^3)(1+x^4)\\
&= (1+x+x^2+x^3)(1)+(1+x+x^2+x^3)(x^4)\\
&=1+x+x^2+x^3+x^4+x^5+x^6+x^7\end{align}.$$
Continue. We will be multiplying $1+x+\cdots+x^7$ by $1+x^8$. Multiplying $1+x+\cdots+... | {
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"url": "https://math.stackexchange.com/questions/163323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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linear algebra: inverse of a matrix The inverse of the matrix
$A=\left( \matrix{1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} }\right)$
is
$A^{-1}=\left( \matrix{ 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 } \right)$.
Then, perhaps the matri... | Here you will find your answer and many other things about Hilbert matrices : http://www.jstor.org/stable/2975779
| {
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"url": "https://math.stackexchange.com/questions/165244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Squares of the form $x^2+y^2+xy$ How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?
Thanks !
| I have shown here that:
All coprime triples $(a,b,c)$ so that $a^2 + ab + b^2 = c^2$ can be
enumerated, without duplication, by taking two positive integers
$m \ge n$, where $3$ does not divide $n$, and either $mn$ is odd and
$\gcd(m,n) = 1$, or $8$ divides $mn$ and $\gcd(m,n) = 2$, and by setting
$$
\begin{align}
a&=m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Baby Rudin chapter 3 problem 12. Suppose $a_n > 0$ and $\sum a_n$ converges. Put $r_n = \sum_{m=n}^{\infty} a_m$.
1) Show that $\frac{a_m}{r_m} + ... + \frac{a_n}{r_n} > 1 - \frac{r_n}{r_m}$, if $m < n$, and deduce that $\sum \frac{a_n}{r_n}$ diverges.
2) Show that $\frac{a_n}{\sqrt[]r_n} < 2(\sqrt[]{r_n} - \sqrt[]{r_{... | For problem 1, note that
$$\frac{a_i}{r_i}=\frac{r_i-r_{i+1}}{r_i}$$
which gives us
$$\frac{a_m}{r_m} + ... + \frac{a_n}{r_n}=\frac{r_m-r_{m+1}}{r_m}+\cdots+\frac{r_n-r_{n+1}}{r_n}>\frac{r_m-r_{m+1}+\cdots+r_n-r_{n+1}}{r_m}$$
and nice things happen when you cancel terms in the numerator.
For problem 2, note that
$$\f... | {
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"timestamp": "2023-03-29T00:00:00",
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Verifying some trigonometric identities: $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$ Prove the following:
46. $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
I got as far as
Right Side: $\tan\theta\sin\theta$ to $\dfrac{\sin\theta}{\cos\the... | There is no "cross cancelling". You are subtracting the fractions, not multiplying them.
$$\begin{align*}
\frac{\csc\theta}{\cot\theta} - \frac{\cot\theta}{\csc\theta} & = \frac{\csc^2\theta - \cot^2\theta}{\cot\theta\csc\theta}\\
&= \frac{\quad\frac{1}{\sin^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}\quad}{\frac{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$ I need to evaluate this integral: $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$.
I've tried $t=\log(x+1)$, $t=x+1$, but to no avail. I've noticed that:
$\int_0^1 \frac{\log(x+1)}{1+x^2} dx = \int_0^1\log(x+1) \arctan'(x)dx =\left. \log(x+1)\arctan(x) \right|_{x=0}^{x=1} - ... | $\displaystyle A=\int_0^1\dfrac{\Big(\log(1+x)\Big)^2}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$
$\displaystyle A=\int_0^1\dfrac{\left(\log\left(\dfrac{2}{1+x}\right)\right)^2}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2-\log(1+x)\Big)^2}{1+x^2}dx$
$\displaystyle A=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "20",
"answer_count": 6,
"answer_id": 2
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Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how ... | $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$
$=\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2-(a^3+a^2) & b-a & 1-1 \\ c^3+c^2-(a^3+a^2) & c-a &1-1\end{pmatrix} $ (applying $R_2'=R_2-R_1\ and\ R_3'=R_3-R_1$)
$=(b-a)(c-a)
\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^2+a^2+ab+b+a... | {
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Integers of the form $a^2+b^2+c^3+d^3$ It's easy$^*$ to prove that if $n=3^{6m}(3k \pm 1)$ where $(m,k) \in \mathbb{N} \times \mathbb{Z}$, then $n=a^2+b^2+c^3+d^3$ with $(a,b,c,d) \in \mathbb{Z}^4$.
But how to prove that this is true if $n=3k$?
Thanks,
W
$^*$ Because $3k+1=0^2+(3k+8)^2+(k+1)^3+(-k-4)^3$, $3k+2=1^2+(3... | You do not need the final $d^3.$ Every integer is the sum of two squares and a cube, as long as we do not restrict the $\pm$ sign on the cube.
TYPESET FOR LEGIBILITY:
Solution by Andrew Adler:
$$ 2x+1 = (x^3 - 3 x^2 + x)^2 +(x^2 - x - 1)^2 -(x^2 - 2x)^3 $$
$$ 4x+2 = (2x^3 - 2 x^2 - x)^2 +(2x^3 -4x^2 - x + 1)^2 -(2x^... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int_\gamma \frac{\cos(z)}{z^3+2z^2} \ dz$ Let $\gamma(t)=e^{it},t \in [0,2\pi]$. We take a look at:
$$\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz$$
We let $$f(z)=\frac{\cos(z)}{z^2+2z}$$ and have
$$\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz=\int_\gamma \frac{f(z)}{z}dz$$
The problem is that $\lim_{z\rightarrow 0}... | $$\frac{\cos z}{z^2(z+2)}=\frac{1}{z^2}\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...\right)\left(\frac{1}{2}\frac{1}{1+\frac{z}{2}}\right)=$$
$$\frac{1}{2z^2}\left(1-\frac{z}{2}+\frac{z^2}{4}-...\right)\left(1-\frac{z^2}{4}+...\right)=\frac{1}{2z^2}-\frac{1}{4z}+...\Longrightarrow Res_{z=0}\left(\frac{\cos z}{z^2}\right)=-... | {
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"source": "stackexchange",
"question_score": "1",
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Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
| Remember how to complete the square:
$$Ax^2+Bx=A\left(x+\frac{B}{2A}\right)^2-\frac{B^2}{4A^2}$$
So now
$$ax^2+bx+c=0 ---- \text{complete square}$$
$$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}=-c$$
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$
$$x_{1,2}+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$
$$x_{1,2}=... | {
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"url": "https://math.stackexchange.com/questions/176439",
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"source": "stackexchange",
"question_score": "4",
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What is a good technique for solving polynomials? Say for example:
$6x^{3}-17x^{2}-4x+3=0$
I sort of look at it and don't know where to start, other than just guessing what the first one would be and trying to do from there. Is there a good technique for approaching such polynomials?
| $f(x)= 6x^3 -17x^2 -4x +3$
Always observe the polynomials behavior for different values of x
$f(-1)=-16$
$f(0)=3$
$f(1)=-12$
Here we see that there are two roots of $f(x)=0$ between $-1$ and $1$
Now the constant part of the term $6x^3$ is $6=(1)(2)(3)$
Check for $x=\frac{-1}{3}, \frac{1}{3},\frac{-1}{2}, \frac{1}{2}$
$... | {
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"source": "stackexchange",
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Solving $5^n > 4,000,000$ without a calculator
If $n$ is an integer and $5^n > 4,000,000.$ What is the least possible value of $n$? (answer: $10$)
How could I find the value of $n$ without using a calculator ?
| By logarithm rules:
$$5^{n}>4\cdot10^{6}\iff n>\log_{5}2^{2}2^{6}5^{6}=\log_{5}2^{8}+\log_{5}5^{6}=\log_{5}2^{8}+6=\log_{5}256+6$$
Since these are relatively small numbers I assume it is ok to write
: $5^{3}=125$ thus clearly $3<\log_{5}256<4$ hence the minimal $n$
that satisfies this inequality is $4+6=10$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 8
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Prove $\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$ How would I prove the following?
$$\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$$
I do not know how to do do the problem I do know $\sin(3x)$ can be $\sin(2x+x)$ and such yet I am not sure how to commence.
| Using $\sin 2\theta=2\sin\theta\cos\theta$ and $\sin^2\theta=(1-\cos2\theta)/2$ we get
$$\cos^2x ~\sin^3x=(\cos x\sin x)^2\sin x=\left(\frac{\sin 2x}{2}\right)^2\sin x=\frac{1}{4}\frac{1-\cos 4x}{2}\sin x \tag{$\circ$}$$
With the sum rule we have $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha.$ Therefo... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Determine the nature of a critical point (Lagrange multipliers)
Let $F \colon \mathbb R^2 \to \mathbb R$ be the function
$$
F(x,y):=xye^x + ye^y - e^x+1
$$
and denote with $C$ the set of zeroes of $F$, i.e. $C:=\{(x,y) \in \mathbb R^2 : F(x,y)=0\}$. Let also $f \colon \mathbb R^2 \to \mathbb R$ be a function whi... | In order to find the Hessian, we need to know the change in $f$ to second order in the deviation from $(0,0)$.
Expanding $F(x,y)$ around this point, we have up to second order
$$ F(x,y) = \nabla F(0,0)\begin{pmatrix} x \\y\end{pmatrix}
+ \tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_F \begin{pmatrix} x \\y\end{pmatrix}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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$1867k =\ldots 1992$, $k\in\mathbb{Z}^+$. Find the minimum value of k. The number $1867$ is multiplied by a positive integer $k$. The last four digits of the product are $1992$. Determine the minimum value of $k$.
$1867k =\ldots 1992$
| Hint $\ $ Let $\rm\, k = \ldots\!dcba\,$ have undetermined digits. Multiplying and comparing digits yields
$$\rm\begin{eqnarray} 1867\,(\ldots\!dcba)\ &= &\rm\ (7+6\cdot10+8\cdot 10^2 + 10^3)\,(a + b\cdot 10 + c\cdot 10^2+ d\cdot 10^3 + (\cdots)\, 10^4)\\
& =&\rm\ 7a + (6a\!+\!7b)\, 10 + (8a\!+\!6b\!+\!7c)\, 10^2 + (a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Integral: $\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t$ How to prove the following:
$$\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t=\frac{\pi}{2}\left(\ln(2)+\frac{\pi}{2} i\right)$$
| Let's evaluate the real and imaginary parts of the integral separately. Using $\Re(\ln(t+1)) = \ln(|t+1|)$ and $\Im(\ln(t+1)) = \pi [t < -1]$, where $[t<1]$ is the Iverson bracket:
$$ \begin{eqnarray}
\Re \int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t &=& \int_{-\infty}^\infty \frac{\ln(|t+1|)}{t^2+1} \mat... | {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the... | Hint $\rm\,\ n\,$ odd $\rm\:\:\! \Rightarrow\ 2\:|\:\color{#90f}{n^2 + 5}$
and, furthermore $\ \:\!\rm 8\:|\:\color{#0a0}{n^2}-\color{#c00}1,\ $ by $\ \rm mod\ 8\!:\ \color{#0a0}{odd^2} \equiv \{\pm 1,\pm 3\}^2\equiv \color{#c00}1$
multiplying $\rm \Rightarrow 2\cdot 8\:\!\:\!|\:\!(n^2\!-1)\:\!(\color{#90f}{n^2\!+\!5... | {
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"url": "https://math.stackexchange.com/questions/187033",
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"source": "stackexchange",
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I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, ... | $S_{2n}+4S_{n}=n(2n+1)^2$
$S_{2n}=S_{n}+S_{(n+1,2n)}$ -------(A)
$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(2n)^2$
$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(n+n)^2$
$S_{(n+1,2n)}=n(n)^2+(1^2+2^2+3^2+\cdots+n^2)+(2n)(1+2+3+\cdots+n)$
$S_{(n+1,2n)}=n^3+S_{n}+(2n)\frac{n(n+1)}{2}$
$S_{(n+1,2n)}=n^3+S_{n}+n^2(n+1)$
... | {
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"url": "https://math.stackexchange.com/questions/188712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.10 Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\displaystyle \frac{a}{\sqr... | We have over $(a,b,c)$:
$$\displaystyle LHS := \sum_{cyc} \frac{a}{\sqrt{a+2b}} = \frac{\sum_{cyc}\sqrt{a^2(b+2c)(c+2a)}}{\sqrt{(a+2b)(b+2c)(c+2a)}}$$
Using CS:
$$\displaystyle LHS \leq \sqrt{\frac{\left(a^2(b+2c)+b^2(c+2a)+c^2(a+2b)\right)\left(3(a+b+c)\right)}{(a+2b)(b+2c)(c+2a)}} \\
= \sqrt{3}\sqrt{\frac{a^2(b+2c)+... | {
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"source": "stackexchange",
"question_score": "10",
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Show $ \int_0^\infty\left(1-x\sin\frac 1 x\right)dx = \frac\pi 4 $ How to show that
$$
\int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=\frac{\pi}{4}
$$
?
| Another way is to use Laplace transform to evaluate this improper integral. In fact, let $f(x)=x-\sin x$ and then $F(s)=\frac{1}{s^2}-\frac{1}{s^2+1}$. Thus
\begin{eqnarray*}
\mathcal{L}\big\{\frac{f(x)}{x}\big\}&=&\int_s^\infty F(s)ds=-\frac{\pi}{2}+\frac{1}{s}+\arctan s, \\
\mathcal{L}\big\{\frac{f(x)}{x^2}\big\}&=&... | {
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"source": "stackexchange",
"question_score": "20",
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Compute this limit of series: $\lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 $ Compute this limit of series:
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 $
I used the definition of the definite integral
$\displaystyle \int_{a}^{b}f(x)dx=\lim_{n\rightarrow \infty} S_{n}$
w... | By induction, you can prove that
$\displaystyle \sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$
Thus
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{n^3}\sum_{k = 1}^{n - 1} k^2 = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2(n - 1) + 1)}{6} = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the sum function of $\sum_{n=0}^{\infty}\frac{n(n-2)}{n+1}x^{n-1}$ series summation:
$$\sum_{n=0}^{\infty}\frac{n(n-2)}{n+1}x^{n-1}$$
where $-1 <x <1$
is there a convinient function that sums the above series?
(unsure but this may be an expanded taylor series?)
| Write that as
$$\sum\limits_{n = 0}^\infty {\frac{{{n^2}}}{{n + 1}}{x^{n - 1}}} - \sum\limits_{n = 0}^\infty {\frac{{2n}}{{n + 1}}{x^{n - 1}}} $$
Now think about primitives and derivatives.
$$\eqalign{
& \sum\limits_{n = 0}^\infty {\frac{2}{{n + 1}}n{x^{n - 1}}} = f'\left( x \right) = \frac{d}{{dx}}\left[ {\sum\... | {
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"source": "stackexchange",
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Prove $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$ Please help me for proving this inequality $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$
| A more complicated approach is to prove the result by induction. Let
$$S(n)=\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}.$$
We want to show that $S(n)\lt 1-\dfrac{1}{n}$ for every integer $n \ge 2$.
The result is clearly true when $n=2$. We show that for any $k\ge 2$, if the result is true when $n=k$, it is true... | {
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"url": "https://math.stackexchange.com/questions/193261",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$ Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})]$, in the region $|x|<1$.
| \begin{align}
& \lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})] \\
&= \lim_{n\to\infty}\frac{(1-x)(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}... | {
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"url": "https://math.stackexchange.com/questions/193762",
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"source": "stackexchange",
"question_score": "1",
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Two unrelated equations, $w^2 = -\frac{15}{4} - 2i$ and $z^2 - (3-4i)z + (2-4i) = 0$? I am to solve the equation $z^2 - (3-4i)z + (2-4i) = 0$, and also have to help me that $w^2 = -\frac{15}{4} - 2i$.
I can ofcourse find $w$, but I fail to see how this helps me in solving $z^2 - (3-4i)z + (2-4i) = 0$. What am I missin... | $z=\frac{3-4i±\sqrt{(3-4i)^2-4\cdot 1\cdot (2-4i)}}{2}$
Now, $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i=1^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
So, $z=\frac{3-4i±(1-4i)}{2}=1$ or $2-4i$
$4w^2=-15-8i=(1)^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
$2w=±(1-4i)$
Alternatively, let $-\frac{15}{4}-2i=(a-ib)^2$
So, $-\frac{15}{4}-2i=a^2-... | {
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"url": "https://math.stackexchange.com/questions/195125",
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"source": "stackexchange",
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Inequality $(1+\frac1k)^k \leq 3$ How can I elegantly show that: $(1 + \frac{1}{k})^k \leq 3$ For instance I could use the fact that this is an increasing function and then take $\lim_{ k\to \infty}$ and say that it equals $e$ and therefore is always less than $3$
*
*Is this sufficient?
*What is a better wording th... | Using binomial theorem, $(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$
Now, consider ${k\choose r}\frac{1}{k^r}=\frac{k!}{r!(k-r)!}\frac{1}{k^r}=\frac{(1)(1-\frac{1}{k})(1-\frac{2}{k})...(1-\frac{r-1}{k})}{r!}\lt \frac{1}{r!}$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Inequality under condition $a+b+c=0$ I don't know how to prove that the following inequality holds (under condition $a+b+c=0$):
$$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$
| This one made me struggle so much that I was close to go crazy. Therefore let me post my solution to have a relief from this burden..
First we have $$2a^2=\frac43a^2+\frac23a^2=\frac43a^2+\frac23(b+c)^2\leq \frac43(a^2+b^2+c^2);$$ where the last inequality follows from the arithmetic-quadratic mean.
Analogously $$\begi... | {
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"source": "stackexchange",
"question_score": "5",
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Relations between the roots of a cubic polynomial How do I solve the last two of these problems?
The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\... | Let $a=\frac1{(1+\alpha)}$ etc,
so, $a,b,c$ are the roots of $6t^3-7t^2+3t-1=0$
$\implies a+b+c=\frac 7 6, ab+bc+ca=\frac 3 6=\frac 12$ and $abc=\frac1 6$
So, $$a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$$
$$=(a+b+c)((a+b+c)^2-3(ab+bc+ca))+3abc$$
$$=(\frac 7 6)((\frac 7 6)^2-3(\frac 12))+... | {
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Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$... | There's a nice trick:
$$\sin x + \sqrt 3 \cos x = 1 \\\\ = 2 \left(\frac{1}{2}\sin x + \frac{\sqrt 3}{2} \cos x\right) \\\\ = 2\left(\cos\left(\frac{\pi}{3} + 2k\pi\right)\sin x + \sin\left(\frac{\pi}{3} + 2k\pi\right)\cos x \right)\\\\= 2\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1.$$
When is $$\sin\left(x + \frac{\... | {
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How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
| $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\lim_{x\to 1} \frac{\sqrt{3+x}-2}{x-1}\cdot \lim_{x\to 1} \frac{x-1}{\sqrt[3]{7+x}-2}$$ $$=\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}$$
Use $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$ ... | {
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substitution in a non linear differential equation and to get a nicer form well I had this equation at the begining
$$
i \frac{\partial u}{\partial{z}} + \frac{1}{2 k_0} \frac{\partial^2 u}{\partial x^2} +\frac{1}{2}k_0 n_1 F(z) x^2 u-\frac{i[g(z) -\alpha(z)]}{2}u + k_0 n_2|u|^2 u = 0,
$$
If I substitute $X=x/w_0$, $Z=... | Lets see. Taking the change of variables $X = \frac{x}{w_0}$, $Z = \frac{z}{L_D}$ and $U = \frac{u}{C}$, we have
\begin{align}
\frac{\partial}{\partial x} &= \frac{\partial X}{\partial x}\frac{\partial}{\partial X} = \frac{1}{w_0}\frac{\partial}{\partial X},\\
\frac{\partial^2}{\partial x^2} &= \frac{1}{w_0^2}\frac{\pa... | {
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Check my workings: Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition. Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.
Precalculations
My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that
$$0<|x+... | You made a mistake here:
$$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|=|3(x+2)^2-4(x+2)|$$
It should be $\,8\,$ instead $\,4\,$ in the RHS. All the rest you did is fine, fixing this little mistake.
I show you now how'd I do it:
$$|3x^2+4x-2-2|=|3(x+2)^2-8(x+2)|=$$
$$|x+2|\,|3x-2|\stackrel{\text{for}\,|x+2|<0.5\Longrightarrow |3x-2... | {
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How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
| Consider the substitution $x=4\sin y$. Then $dx=4\cos y\ dy$, and
$$\frac 1{(16-x^2)^2} = \frac 1{(16\cos^2 y)^2} = \frac 1{(4\cos y)^4} $$
So,
$$\int\frac 1{(16-x^2)^2}dx = \int\frac{4\cos y}{(4\cos y)^4} dy$$
So, finally you need $\displaystyle\int\frac1{\cos^3}$.
| {
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Finding the derivative of $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ Find the derivative
How do I tackle this? My answer is totally different from the correction model, but I have tried for half an hour to show my answer in lateX but I don't know how to, it's too complicated, so, can someon... | Use the quotient rule:
\begin{align}
h'(x) & = \frac{(x^2+2)\frac{d}{dx}(2\sqrt{x}) - 2\sqrt{x}\frac{d}{dx}(x^2+2)}{(x^2+2)^2} \\[10pt]
& = \frac{(x^2+2)\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot 2x}{(x^2+2)^2}.
\end{align}
Now clear out fractions by multiplying the top and bottom both by $\sqrt{x}$:
$$
\frac{(x^2+2)-2x\cdot2x}... | {
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what is the limit of the sequence $x_0=a$, $x_1=b$ define $$x_{n+1}=\left(1-\frac{1}{2n}\right)x_n+ \frac{1}{2n} x_{n-1}$$
we need to find the limit of $\{x_n\}_n$ as $n\rightarrow \infty$.
Let $l$ be this limit , I tried taking the limit on both sides getting $l=l$ as $n\rightarrow \infty$.
| As $x_{n+1}=(1-\frac{1}{2n})x_{n}+\frac{1}{2n}x_{n-1}$, we can get that:
$$\begin{align*}
x_{n+1}-x_{n}&=-\frac{1}{2n}(x_{n}-x_{n-1})\\
x_{n}-x_{n-1}&=-\frac{1}{2(n-1)}(x_{n-1}-x_{n-2})\\
&\vdots\\
x_{2}-x_{1}&=-\frac{1}{2}(x_{1}-x_{0})\\
\end{align*}$$
then,by computation $$x_{n+1}-x_{n}=\frac{(-\frac{1}{2})^{n}}{n!}(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Analytic Geometry question I can't solve An isosceles triangle $ABC$ has 2 given vertices, $A(3,2)$ and $C (7,14$). The slope of AB is $\dfrac{1}{2}$. What are the coordinates of B?
I could figure out that line AB = $\dfrac{1}{2}x + \dfrac{1}{2} $
I found that the length of AC = is $\sqrt{160}$
But I haven't got a clue... | Put $B(x,y)$. I solve your problem with assume the triangle $ABC$ isosceles at $A$.
Because $AB =AC$, then $AB =\sqrt{160}$ or
$$x^2+y^2-6x-4y-147 = 0.$$
The coordinates of the point $B$ are solutions of the system
$$x^2+y^2-6x-4y-147 = 0, \quad y = \dfrac12x + \dfrac12.$$
We get $B(3 - 8\sqrt{2}, 2(1-2\sqrt{2})$ or... | {
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Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
| Yet another way is to start with:
$$
\sum_{0 \le k \le n} z^k = \frac{1 - z^{n + 1}}{1 - z}
$$
differentiate thrice:
$$
\frac{\mathrm{d}}{\mathrm{d} z}
\left( z \frac{\mathrm{d}}{\mathrm{d} z}
\left(
z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1 - z^{n + 1}}{1 - z}
\right)
\right)
= \frac{1 + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 5
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Proof of an inequality about $\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$ I've encountered an inequality pertaining to the following expression:
$\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$, where $z$ is a complex number.
After writing $z$ as $x + iy$ we have the inequality when $y \gt 1$ and $|x|... | As
$$1\leq y\leq|x+iy|\leq{y\over2}+y={3\over2} y$$
we have
$${1\over |x+iy|}\leq 1$$
and
$$|n^2+y^2-x^2-2ixy|\geq |n^2+y^2-x^2|\geq\Bigl(1-{1\over8}\Bigr)(n^2+y^2)\qquad(n\geq1)\ .$$
It follows that
$$\left|{1\over z}+\sum_{n=1}^\infty{2z\over z^2-n^2}\right|\leq 1+\sum_{n=1}^\infty {3y \over{7\over8}(n^2+y^2)}=1+{24\... | {
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"source": "stackexchange",
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How to find constants a and b in this function? $\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax+b}-2}x = \frac 5{12}$
How do you solve for constants $a$ and $b$?
| You can use $(x^3-y^3)=(x-y)(x^2+xy+y^2)$ with $x=\sqrt[3]{ax+b},y=2$ and multiply the term $\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)$ into nominator and denominator of the limit: $$\displaystyle \lim_{x\to0} \frac{\big(\sqrt[3]{ax+b}-2\big)\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}{x\big(\sqrt[3]{(ax+b)^2}... | {
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Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
| Assuming $f$ is sufficiently smooth, repeated application of the fundamental theorem of the calculus gives (finite Taylor expansion)
$$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \int_0^x (x-t)^2 \frac{(f'''(t)-f'''(0))}{2!}\, dt$$
Using the fact that $\sin' = \cos, \cos' = -\sin$, we can then exp... | {
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If I add a constant $c$ to each fraction's numerator and denominator in a sequence of fractions, how is the sequence affected? Given a sorted ascending sequence of fractions, if I add a constant $c$ to each fraction's numerator and denominator, how is the sequence affected?
For example, if I have a sequence in ascendin... | You need to check that $\frac{a}{b} < \frac{x}{y} \implies \frac{a+c}{b+c} < \frac{x+c}{y+c}$. The latter statement is equivalent to $(a+c)(y+c) < (x+c)(b+c)$ and this to $ay + ac + yc < xb + xc + bc$ as $ay < xb$ is given, you only need to see that $a-b < x-y$. This need not be true.
Consider $\frac{a}{b} = \frac{1}{2... | {
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Finding $\int\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x$. In Arnold's Mathematical Methods of Classical Mechanics he uses,
$$\int_0^1\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x = \left\{\frac{\pi}{2-b} : 0\leqslant b < 2\right\}$$ but he doesn't explain how to get it.
Via Mathematica (wolfram alpha works also), $$\int\frac{1}{\sq... | Write it in the following way:
$$\int_0^1\frac{dx}{\sqrt{x^b-x^2}}=\int_0^1 dx\,\frac{x^{-b/2}}{\sqrt{1-x^{2-b}}}.$$
Now make a trigonometric substitution $x^{1-\tfrac{b}{2}}=\sin\theta$ so that $\left(1-\frac{b}{2}\right)x^{-b/2}dx=\cos\theta\,d\theta$ and the integral becomes (ignore the bounds for a moment)
$$=\int ... | {
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How to find the roots of $x³-2$? I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\sqrt[3] 2$ and $\sqrt[3] {2}e^{\frac{2\pi}{3}i}$ but I don't why.
The first one is easy to find, but the another two roots?
I need help
Thank you
| If $\omega^3 = 1$ and $x^3 = 2$ then $(\omega x)^3 = \omega^3 x^3 = 2$.
Possible values of $\omega$ are $e^{\frac{1}{3}2 i \pi}$, $e^{\frac{2}{3}2 i \pi}$ and $e^{\frac{3}{3}2 i \pi}$. This is because $1 = e^{2 i \pi} = (e^{\frac{1}{k} 2 i \pi})^k$.
So the solutions of $x^3 - 2 = 0$ are $e^{\frac{1}{3}2 i \pi} \sqrt[3]... | {
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A geometry problem proposed at national olympiad. Let $ABC$ be a triangle with $\angle ABC=90^{\circ}$. We have: $$1) BH \perp AC;
$$ $$2)AD \text{ the bisector of } \angle{A} \text{ and } AD\cap BH=\{Q\},D\in BC;$$
$$3) CE \text{ the bisector of } \angle C \text{ and } CE \cap BH =\{P\},E \in AB;
$$
$$4) CE \cap AD =... | See the figure below:
Let $K$ be a point such that $K \in BQ$ and $\angle QKD$ is a right angle.
Using similarity and angle bisector theorem we get:
$$\frac{AH}{HC}=\frac{c^2}{a^2} \quad(1)$$
and
$$\frac{DB}{DC}=\frac{c}{b}. \quad(2)$$
From equation $(2)$ we conclude that
$$\frac{DB}{BC}=\frac{c}{b+c}. \quad(3)$$
Note... | {
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If the order of $a$ modulo $p$ is 6, is it true that the order of $(1-a)$ modulo $p$ is also 6? Given a prime $p$ and an integer $a$. If the order of $a$ modulo $p$ is 6, is it true that the order of $(1-a)$ modulo $p$ is also 6? If so prove it, if not give a counterexample.
It seems right. So I aimed to show that
$(1... | Note that $a$ has order $6$ iff $a$ is a solution of the congruence $x^6-1\equiv 0\pmod{p}$, but does not have order $\lt 6$.
We can factor $x^6-1$ as
$$x^6-1=(x^3-1)(x^3+1)=(x^2-1)(x^2+x+1)(x^2-x+1).$$
If $a^3-1\equiv 0\pmod p$ or $a+1\equiv 0\pmod{p}$, then $a$ has order $\lt 6$. So any element of order $6$ must be ... | {
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Structure of $x^2 + xy + y^2 = z^2$ integer quadratic form The pythagorean triples $x^2 + y^2 = z^2$ can be solved in integers using rational parameterization of solutions to $x^2 + y^2 = 1$.
It goes through $(1,0)$, then consider the line $y = -k (x - 1)$ so that $x^2 + k^2(x-1)^2 = 1$
We get $(1+ k^2 )x^2 - 2k^2 x + ... | Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a... | {
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Integer Solutions to $x^2+y^2=5z^2$ I'm looking for a formula to generate all solutions $x$, $y$, $z$ for $x^2 + y^2 = 5z^2$.
Any advice?
| Ok, so I am assuming rational solutions. This method can yield a parametrization of all integer solutions without too much work.
Note that $(1,2)$ lie on the circle $x^2 + y^2 = 5$. Let $x_0 = 1, y_0 = 2$. Now suppose $x^2 + y^2 = 5$. Let $m = x - x_0, n = y - y_0$, then we have $$m^2 + 2mx_0 + x_0^2 + n^2 + 2ny_0 + y_... | {
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"source": "stackexchange",
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Proof that a sum $x$ is found in$ [1, n (n + 1) / 2]$ squence In the subset sum existance problem, we have a sequence of integers. We are given an integer $x$ that we should look for a possible subsequence that sum to $x$. For example, if we have the sequence $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $x = 39$ then we ha... | Here is an alternative approach: consider the case for $n = 5$.
Then $n(n+1)/2 = 5(6)/2 = 15$.
Observe that we can write the numbers $1, \ldots, 15$ as follows:
$1, 2, 3, 4, 5,$
$1+5, 2+5, 3+5, 4+5,$
$1+4+5, 2+4+5, 3+4+5,$
$1+3+4+5, 2+3+4+5,$
$1+2+3+4+5$.
The number of elements summed for entries in the above rows, res... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the indefinite integral $\int x^n \sin x\,dx$ $\newcommand{\term}[3]{
\sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!}
}$
I am trying to prove that for $n \in\mathbb N$,
$$
\int x^n \sin x \, dx
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{(n-1)}{k}{n-2k-1}
$$
I started with differentiatio... | I finally got my proof with differentiation finished, too.
If
$$f(x) = \sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k}{n!\over(n-2k)!}\cos x+\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^kx^{n-2k-1}{n!\over(n-2k-1)!}\sin x$$
with $n\in \Bbb N$.
then
$$\begin{align}f'(x) &= \left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}... | {
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find $f(x)$ when $3f(x-6)-2f(x-9)=x^2-54$ I can easily show that with the assumption $f$ is a polynomial $f(x)=x^2$. But without that assumption how can I prove that $f(x)=x^2$???. I have tried many change of variables $x=u+k$ but to no result. I am lost here
| Let $f(x)$ be polynomial of $x$.
Let $n\ge 3,$ be the smallest power of $x$ whose coefficient$(a)$ may be $\ne 0$.
The coefficient of $x^n$ in $3f(x−6)−2f(x−9)$ is $3a-2a=a$
But the coefficient of $x^n$ where $n\ge 3$ in $x^2−54$ is $0\implies a=0$
So, $f(x)$ can not contain any higher powers $(\ge 3)$ of $x.$
So, $f(x... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq... | For the second and the third question,a generalized version is presented here:
Let $a,b,c>0$ and $a+b+c=3$,then
$$f(a,b,c)=(\sum\frac{ab}{c})+\lambda abc\geq3+\lambda$$
for $1\leq \lambda\leq9/4$,where the constant $9/4$ is optimal.
It is easy to see that 9/4 is optimal(a simple comparation between $f(1,1,1)$ and $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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The Length of a Bisector How can I prove that :
$$b_{A}=\frac{1}{b+c} \cdot \sqrt{bc\left[(b+c)^2-a^2\right]} ?$$ where $b_{A}$ is the length of the bisector from $A$ .
Thanks :)
| Let $P$ be the point where the bisector of $\angle A$ meets $BC$. Let $x=BP$ and $y=PC$. Then $x+y=a$. Moreover, by a standard theorem on angle bisectors, we have
$\dfrac{x}{c}=\dfrac{y}{b}$, or equivalently $bx=cy$.
We have two linear equations in two unknowns. Solve for $x$ and $y$. We obtain
$$x=\frac{ac}{b+c}\qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left(t + \frac{\pi}{3}\right)$
The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left( t + \frac{\pi}{3}\right)$
(a) At what time(s) in the interval $[0,2\pi]$ do the particles meet?
(b) What is the farthe... | (a)
For coincidence, $\sin(t+\frac{\pi}3)=\sin t$
So, $t+\frac{\pi}3=n\pi+(-1)^nt$ where $n$ is any intgere.
If $n=2m$(even), $t+\frac{\pi}3=2m\pi+t\implies 2m\pi=\frac{\pi}3$ which is impossible.
If $n=2m+1$(odd), $t+\frac{\pi}3=(2m+1)\pi-t\implies t=(6m+2)\frac{\pi}6=\frac{(3m+1)\pi}3$
Putting $m=0,t=\frac{\pi}3$
Pu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $x$ such that $\arctan(3/2)+\dots=\arctan x$ Find $x$ such that
$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$
| \begin{align}
& {} \quad \tan(a+b+c+d) \\
& = \frac{\tan a+\tan b+\tan c+\tan d\ \overbrace{ - \tan a\tan b\tan c - \cdots}^\text{4 terms}}{1-\ \underbrace{\tan a \tan b- \cdots}_\text{6 terms} +\tan a\tan b\tan c\tan d}
\end{align}
Therefore
\begin{align}
& {} \quad \tan\Big(\arctan(3/2) + \arctan(5/4) + \arctan(-5/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Combinatorics: how many ways are there to form 100 with 1, 2 and 5
Possible Duplicate:
Making Change for a Dollar (and other number partitioning problems)
dollar notes in denominations of 1, 2 and 5. How many ways are there to form exactly $100 using just multiples of these notes?
| $x+2y+5z=100$
$0\le x\le100\implies 0\le 100-2y-5z\le100\implies 0\le 2y+5z\le 100$
$\implies 0\le y\le \lfloor \frac{100-5z}2\rfloor$
As $0\le 5z\le 100, 0\le z\le 20 $
If $z$ is odd $=2a+1, 0\le y\le \lfloor \frac{100-5(2a+1)}2\rfloor=47-5a, 47-5a+1=48-5a$ values of $z$
So, $0\le 2a+1\le 20\implies 0\le a\le9 $
If $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite series and logarithm Is it true that:
$$\log_e 2 = \frac12 + \frac {1}{1\cdot2\cdot3} + \frac {1}{3\cdot4\cdot5}+ \frac{1}{5\cdot6\cdot7}+ \ldots$$
It was one of my homeworks . Thanks!
| $$a_n = \dfrac1{(2n-1)2n(2n+1)} = \dfrac12 \left( \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1}\right)$$
and $$a_0 = \dfrac12$$
Hence,
\begin{align}
\sum_{k=0}^n a_k & = \dfrac12 + \dfrac12 \left(\dfrac11 - \dfrac12 - \dfrac12 + \dfrac13 + \dfrac13 - \dfrac14 - \dfrac14 + \dfrac15 + \cdots + \dfrac1{2n-1} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$? I was tackling through an olympiad practice book when I saw one of these problems:
If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ - \ 1 \over\sqrt{x}}$?
The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that ... | When you know the answer, it is often easier...
Indeed, begin with $(x-1)/\sqrt{x} = 2\sqrt{2}$
It is equivalent to $x-1 = 2\sqrt{2x}$
Equivalent to $(x-1)^2 = 8x$
Equivalent to $x^2-2x+1 = 8x$
Equivalent to $x^2-10x+1 = 0$
Equivalent to $(x-a)(x-b) = 0$ where $a = 5+2\sqrt{6}$ and $b = 5-2\sqrt{6}$
Equivalent to {$x=a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I wa... | Here is another proof using the Reverse Rearrangment Inequality.
Observe that with $ a \leq b \leq c$, we have $ bc \leq ca \leq ab $ and $a^2 \leq b^2 \leq c^2$. Hence, applying the RRI on these similarly ordered sequences, we conclude that
$ ( a^2 + bc ) ( b^2 + ca) ( c^2 + ab) \leq ( a^2 + ab)( b^2 + bc)(c^2 + ca) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
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Generalized permutation and combination: number of solutions How many solutions are there to the equation: $$x_1 + x_2 + x_3 + x_4 + x_5= 21\;,$$
where $x_i$ is a non-negative integer such that
$$0 \le x_1 \le 3;\; 1 \le x_2 < 4;\text{ and }x_3 \ge 15\;?$$
| This should be the coefficient of $a^{21}$ in
$(1+a+a^2+a^3)(a+a^2+a^3)(a^{15}+a^{16}+a^{17}+a^{18}+...)(1+a+a^2+a^3+a^4+...)^2$
where $a$ is some real $0<a<1$
= coefficient of $a^5$ in
$(1+a+a^2+a^3)(1+a+a^2)(1+a+a^2+a^3+a^4+...)^3$
= coefficient of $a^5$ in
$(1-a^4)(1-a^3)(1-a)^{-6}$
= coefficient of $a^5$ in
$(1-a^4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| $\gcd(a+b, a^2-ab+b^2) = \gcd(a+b, (a+b)^2-3ab) = \gcd(a+b, 3ab)$ by the Euclidean algorithm. If the gcd was $d \ne 1 ,3$, then $d \mid a$ or $d \mid b$ in $3ab$ but then from $a+b$, $d$ would divide the other. Thus, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
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How to find the Laplace transform of $\frac{1-\cos(t)}{t^2}$? $$ f(t)=\frac{1-\cos(t)}{t^2} $$ $$ F(S)= ? $$
| Let
$$F(s) = \int_{0}^{\infty} f(t) \, e^{-st} \, dt = \int_{0}^{\infty} \frac{1-\cos t}{t^2} e^{-st} \, dt. $$
The function $f(t)$ satisfies the bound $ f(t) = O(1 \wedge t^{-2})$, thus it is absolutely integrable and we can apply Leibniz's integral to obtain
$$ F''(s) = \int_{0}^{\infty} (1-\cos t) \, e^{-st} \, dt =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
| Cancelation:
$$
\begin{array}{cccccccccccccccc}
x & (x^3 & + & x^2 y & + & xy^2 & + & y^3) \\
& & -y & (x^3 & + & x^2 y & + & xy^2 & + & y^3) \\[25pt]
= & x^4 & + & x^3 y & + & x^2y^2 & + & xy^2 \\
& & - & x^3y & - & x^2y^2 & - & xy^2 & - & y^3 \\[25pt]
= & x^4 & & & & & & & - & y^4
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/260362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
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How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$? How do I integrate this guy? I've been stuck on this for hours..
$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$
| Let $x = \sin^2y$.
(It seems we've started like @Sasha here and like @sos440.)
Then
$$\begin{eqnarray*}
I &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x(1-x)} \\
&=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
+ \underbrace{\frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{1-x}}_{x\to 1-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
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Manipulation of a binomial coefficient In obtaining a formula for the Catalan numbers I have got the expression $-\frac{1}{2}\binom{1/2}{n}(-4)^n$. All my efforts to show that this simplifies to $\frac{1}{n}\binom{2n-2}{n-1}$ have not succeeded. Is there some mistake in the original expression, and if not, how do I sim... | First we show that
$$\binom{1/2}{n}={1\over n!}(1/2-1)(1/2-2)...(1/2-(n-1))=$$$$={1\over n!}(-1/2)(-3/2)...(-(2n-3)/2)=$$
$$={1\over n!}{(-1)^{n-1}\over2^n}1\cdot3\cdots(2n-3)\frac{2\cdot4\cdot6...(2n-4)}{2^{n-2}(n-2)!}=$$
$$={1\over n!}{(-1)^{n-1}\over2^{2n-2}}\frac{(2n-3)!}{(n-2)!}\frac{(2n-2)(2n-1)2n}{(n-1)n\cdot2^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Convergence of a spiral in $\mathbb{C}$ Does the series
$$\sum_{k=0}^{\infty}\frac{i^k}{k!}$$converge, and if so, what is the value of it?
| Alternately (and equivalently to several of the other answers), if you didn't know the Euler formula but did know the power series for sin and cos, you could reason as follows:
$i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=i^0=1$. Therefore, the numerators of the power series are periodic of period 4; what's more, they split off... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Finding Laurent Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ at $z=2$. I need to find the Laurent Series for $z=2$, $z=i$, and $1<|z|<2$ for the following function:
$$ f(z)= \frac{z^2 -2z +5}{(z-2)(z^2+1)}.$$
I was thinking if I could get something like $$\frac{1}{1 - (z-2)}$$ I could use geometric series.
Like in... | *
*$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)}=\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}=\frac{1}{z-2}-\frac{2}{z^2+1}=\frac{1}{z-2}-\frac{2}{(z-2)^2+4z-3}=\frac{1}{z-2}-\frac{2}{(z-2)^2+4(z-2)+5}=\frac{1}{z-2}-\frac{2}{5}[1+\{\frac{4}5(z-2)+\frac{1}5(z-2)^2\}]^{-1}.$
Hence etc.
*
*$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)}=\frac... | {
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"timestamp": "2023-03-29T00:00:00",
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$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} =
-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$
| Hint: Use partial fractions,
$$\sum_{n=2}^{\infty} \frac {2}{(n^3-n)3^n} = \sum_{n=2}^\infty \frac {1}{3^n} \left( \frac {1}{n-1} - \frac {2}{n} + \frac {1}{n+1}\right)$$
Now, shift the indexing up/down 1 as necessary. Remember to check the power of 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proof of tangent half identity
Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$
I was unable to find any proofs of the above formula online. Thanks!
| There are several ways to proceed, apart from what Ayman gave
Approach 1: If you know the tangent t-formulas: For $t = \tan \frac {\theta} {2}$,
$$ \sin \theta = \frac {2t}{1+t^2}, \cos \theta = \frac {1-t^2}{1+t^2}$$
Substitute these into the equation and simplify.
Approach 2: (my preference)If you know that $\tan \f... | {
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"source": "stackexchange",
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Help figuring out all the alternative solutions to the integrals of sine and cosine I always worry a lot when doing integrals with trigonometric functions because there's always many ways to write the final answer. I am trying to figure out the general pattern for the various different solutions.
The integral
$$ \int ... | Putting $\cos x=z,dz=-\sin xdx$
$$ \int \sin^3\left(x\right)\cos^2\left(x\right)\, \mathrm{d}x=\int (1-z^2)z^2(-dz)=\int z^4 dz-\int z^2 dz=\frac{z^5}5-\frac{z^3}3+c=\frac{\cos^5x}5-\frac{\cos^3x}3+C $$
(i)
$$\frac{\cos^5x}5-\frac{\cos^3x}3=\frac{\cos^3x}{15}(3\cos^2x-5)=\frac{\cos^3x}{30}(6\cos^2x-10)=\frac{\cos^3x}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| This is wrong
If you are familiar with majorization, observe that the function is schur concave. Thus, its maximum occurs at a point where all variables are equal, and since that point exists in the constraint set (i.e. $a,b,c\geq 0$ and $a^2+b^2+c^2=3$), $a=b=c=1$ is the maxima. Thus, the inequality comes.
| {
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"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 8,
"answer_id": 7
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Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$
Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$
I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{... | The identity $k^2-1=(k+1)(k-1)$ shows that
$$
\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{k^2-1}{k^2}=\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2,
$$
and the value of limit should follow.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
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Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as:
$$\frac{x^4}{x(y^2+z^2)}+\fr... | By Cauchy-Schwarz you want to show that $\sum x(y^2+z^2) \leq 6$. But this is equivalent to
$$\begin{eqnarray}\sum x(3-x^2) \leq 6
\\ \Leftrightarrow 3\sum x \leq 6 + \sum x^3\end{eqnarray}$$
This then follows from $x^3 + 1 + 1 \ge 3x$ (AM-GM) and add up.
| {
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"url": "https://math.stackexchange.com/questions/290844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Missing and parasite roots in the trigonometric equation. I have this equation:
$$\boxed{\cos(2x) - \cos(8x) + \cos(6x) = 1}$$
RIGHT
And its right solution from the textbook is:
$$
\begin{align}
\cos(2x)+\cos(6x)&=1+\cos(8x)\\\\
2\cos(4x)*\cos(2x)&=2\cos^2(4x)\\\\
\cos(4x)*(\cos(2x)-\cos(4x))&=0\\\\
\cos(4x)*2\sin(3x)... | Just minor slippage. For example, from $\cos 2x=1$ you concluded that $2x=\pi k$. That is false, we need $k$ to be even: if for example $2x=\pi$, then $\cos 2x=-1$.
I don't understand your concern with $\frac{\pi}{3}$. It works.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Is this function differentiable at $(0,0)$? Let $f: \mathbb R^2 \to \mathbb R$ be the function
$$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$
where $(x,y) \neq (0,0)$ and $f(0,0)=0$.
Is $f$ differentiable at $(0,0)$ and if so, how can I prove it?
| We will prove that $f$ is differentiable at $(0,0)$ with derivative $df_{(0,0)}=0$.
Using $|\sin z|\leq |z|$, we have
$$
\frac{|f(x,y)|}{\sqrt{x^2+y^2}}\leq \frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}+\frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}.
$$
Now using Cauchy-Schwarz, we find
$$
\frac{|x|^3|x+y|}{(x^2+y^... | {
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"url": "https://math.stackexchange.com/questions/293984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The number of ways to put numbers in squares If one put the numbers from $1$ to $7$ in the squares in stead of letters in which the sum of cells is in descending order like shown in the figure .Every number should be used once or twice or never .Also the order of numbers in every oblique row its sum appear always in ... | If the inequalities are strict, $Cxy$ can only be $431$ or $521$ Starting with $521$, we must have $b=6, A=7$ and $AbCD$ fails. So we must have $Cxy=431$. Then $cxz=432$ or $531$ If it is $531$ we must have $abcd=7651$, but then $A=7$ and again $AbCD$ fails. If $b=6, A=7$ and again $AbCD$ fails, so $b=5$. Then $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inclusion Exclusion principle question What is the number of surjective (onto) functions
from the set [3] to the set [3].
| We first count the complement. There are $2^5$ functions that "miss" $1$, also $2^5$ that miss $2$, and $2^5$ that miss $3$. Add. We get $3\cdot 2^5$.
But we have double-counted the functions that miss both $1$ and $2$, also the functions that miss $2$ and $3$, also the functions that miss $3$ and $1$. There is $1$ (o... | {
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"url": "https://math.stackexchange.com/questions/298477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expression as a product of disjoint cycles Let $\alpha = (9312)(496)(37215) \in S_n, n \ge 9$. Express $\alpha$ as a product of disjoint cycles.
I know this is probably a really easy question, but my professor didn't elaborate on how to exactly do this and neither does my assigned text. If anyone could elaborate on the... | Forgive me if the answer is a little sloppy, this is my first answer on stack exchange.
We begin by writing in the following format,
(37215)=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
5&1&7&4&3&6&2&8&9
\end{pmatrix}
Now, if we take the cycle $(496)$ composed of this, we get
(496)o (37215)=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/299646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defi... | Introduce the notation $[a_0]=\sqrt{a_0}$, and $[a_0,a_1]=\sqrt{a_0+\sqrt{a_1}}$, and so on, including infinite lists:
$$[a_0,a_1,a_2,...]=\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}=\sqrt{a_0+[a_1,a_2,\ldots]}.$$
Generally $[a_0,a_1,\ldots]^2 = a_0 + [a_1,a_2,\ldots]$, so for constant-term lists we have a closed-for... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"answer_id": 1
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$k=2^n + 1$ is prime $\rightarrow n=2^m$ I am struggling with this proof. I want to prove the contrapositive, $n=2^ab \rightarrow 2^n + 1$ is composite.
My professor gave me a hint,
$n=2^ab$, $b=2r+1 \ge 3$ $\rightarrow$ $2^{2^a}+1 | 2^n+1$. I truly don't know what I am doing and need lots of explanation.
| Recall that if $b$ is odd, then
$$(x+1) \vert x^{b}+1$$
This can be seen immediately from the remainder theorem, since $(-1)^b + 1 = -1 + 1 = 0$. Equivalently, for odd $b$, we have
$$x^b+1 = (x+1)(1-x+x^2-x^3 \pm \cdots - x^{b-2} + x^{b-1})$$
Hence, if we write $n = 2^a \cdot b$, such that $2^a \Vert n$ i.e. $b$ is odd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Calculating the following limit: $\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}} $ I am trying to calculate this limit:
$$
\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}
$$
I've tried using conjugate of both denominator and numerator but I can't get the right result.
| $$\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$$
$$=\frac{(1+\sqrt{x+1})\{x^2+1-(x+1)\}}{(\sqrt{x^2+1}+\sqrt{x+1})(1-(x+1))}$$
$$=\frac{(1+\sqrt{x+1})\{x(x-1)\}}{(\sqrt{x^2+1}+\sqrt{x+1})(-x)}$$
$$=\frac{(1+\sqrt{x+1})(1-x)}{(\sqrt{x^2+1}+\sqrt{x+1})}\text { if } x\ne0$$
As $x\to0,x\ne0$
So, $$\lim_{x\to0}\frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/305497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Are there rigorous mathematical definitions for these waves? My friend linked this .gif to me tonight, and asked me if I knew of any equations that might model these bottom two waves (the blue and green waves). Unfortunately, I am not far enough in my education to recognize if any such model exists. Are these waves mod... | So, for the fist one obviously, it's sine function.
$$
h(t) = r \sin \omega t
$$
As for the second one,
$$
h(t) = \frac a2 \cdot \left \{ \begin{array}{lcc}
\tan \left (-\frac \pi 4 + \omega t \right )& \text{if} & 0 \le t \le \frac T4 \\
1 & \text{if} & \frac T4 \le t \le \frac T2 \\
\tan \left ( \frac {3\pi}4 + \omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$ How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?
My attempt:
$$\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2} = \frac{1}{2}\int_{0}^{2\pi} \frac{d\theta}{(2+\cos\theta)^2}$$
To find the singularity, I solve:
$ (2+\cos\theta... | Trigonometric substitution:
$$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1}dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}\Longrightarrow$$
$$\int\limits_0^\pi\frac{d\theta}{(2+\cos\theta)^2}=\int\limits_0^\infty\frac{2\,dx}{1+x^2}\frac{1}{\left(2+\frac{1-x^2}{1+x^2}\right)^2}=2\int\limits_0^\infty\frac{x^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
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