Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
In a cyclic $\square ABCD$, $BC, CD$ and $DA$ are three tangents of such a circle that its center is on the side $AB$. Proving that $AD + BC = AB$
In a cyclic quadrilateral $ABCD$, $BC, CD$ and $DA$ are three tangents of a circle. The center of the circle is located on the side $AB$. Prove that $$AD + BC = AB$$
Attem... | Let $\mathcal{C}$ be a circle wichi is tangent to $BC,CD$ and $DA$ and let it center be $O$. Then $O$ must lie on angle bisector of angle $\angle ADC$ and angle $\angle DCB$. So these angle bisectors meet on $AB$.
Now let $E$ be on $AB$ so that $AE=AD$. We have to prove $BC=BE$ i.e. triangle $BCE$ is isosceles.
If $\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did ... | So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ \Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ \Rightarrow (x+3)^{2}=0 $
$ \therefore x=-3$ ($x$-ordinate of the point o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove value of $\sin(15°)$ I'm doing the following exercise: prove that
$$
\sin(15°)=\frac{1}{2\sqrt{2+\sqrt{3}}}
$$
I'm using this formula:
$$
\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)
$$
to get:
$$
\sin(45-30)=\sin(45)\cos(30)-\cos(45)\sin(30) \\
=\frac{1}{\sqrt2}\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\frac{1}{2} \\
=\fra... | They are equivalent. Since $$(1 + \sqrt{3})^2 = 1 + 2 \sqrt{3} + 3 = 2(2 + \sqrt{3}),$$ it follows that $$\sqrt{2 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{2}}.$$ Then $$\frac{1}{2 \sqrt{2 + \sqrt{3}}} = \frac{\sqrt{2}}{2 (1 + \sqrt{3})} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}.$$
It is also worth noting that we can obtain t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Finding expression for variable x in an equation (use Lambert function?) Let $a$, $b$ and $c$ be constants. How can one find an expression for variable $x$ in the following equation? $$\frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c})$$
From my research, I am being suggested to use the Lam... | $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\e{\mathrm{e}}$
\begin{align}
\frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c})
\tag{1}\label{1}
\end{align}
Indeed, the solution to equation \eqref{1}
can be expressed in terms of the Lambert W function.
The usual approach... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$... | Look at finite sums first (such that we do not subtract two diverging series):
$$
\sum_{n=1}^N\frac1{5n(5n-1)} =\\
\sum_{n=1}^N\frac{-1}{5n} + \frac{1}{5n-1} = \\
- \frac15 \sum_{n=1}^N(\frac{1}{n} - \frac{1}{n-1/5}) = \\
- \frac15 (\sum_{n=1}^N\frac{1}{n} - \sum_{n=0}^{N-1} \frac{1}{n+4/5} )= \\
- \frac15 (\sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach... | If $b=a$, then the last row is $\begin{bmatrix}0&0&-a&1-2a\end{bmatrix}$. In which case you can use the second row to kill entries in this row and then get echelon form $$ \begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&0 &0 &(1-3a) \end{bmatrix}$$
Now for rank to be $2$, you need $a=1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Evaluating $\int_0^{\pi/2} \log \left| \sin^2 x - a \right|$ where $a\in [0,1]$.
How to evaluate
$$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
$$
where $a\in[0,1]$ ?
I think of this problem as a generalization of the following proposition
$$
\displaystyle\int_0^{\pi/2} \log \left(\sin x\rig... | Using the symmetries and developing the $\sin^2$ term, we can express
\begin{align}
I&=\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \sin^2 x - a \right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1-2a}{2}-\frac{1}{2}\cos 2x\right|\,dx
%&=-\frac{\pi}{2}\ln 2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Help Solving Recurrence Relation: $a_n = n^3a_{n-1} + (n!)^3$ I'm trying to find an explicit formula for the following recurrence relation:
$a_0 = 1$, $\forall n \ge 1: a_n = n^3a_{n-1} + (n!)^3$
So far though, my attempts have been unsuccessful.
My Attempt
Let $A(x) = a_n\frac{x^n}{(n!)^3}$. Multiply by $\frac{x^n... | One simple way to find these sequences (so simple it is almost cheating!) is to generate the first few terms and search for the resulting sequence in the Online Encyclopedia of Integer Sequences (OEIS). In this case you have:
$$\begin{equation} \begin{aligned}
a_0 &= 1, \\[6pt]
a_1 &= 1^3 a_0 + (1!)^3 = 1^3 \cdot 1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Ways to simplify arctan() in integral results? Lately when I was computing $$\int\frac{\mathrm{d}x}{1+\sqrt{1-2x-x^2}},$$I got the result$$-2\arctan{\frac{\sqrt{1-2x-x^2}-1}{x}}-\ln\left(1-\frac{\sqrt{1-2x-x^2}-1}{x}\right)+\mathbf{C}.$$However, Wolfram|Alpha gives me its result in a different form, which may be writte... | Note that
$$
1-2x-x^2=2-(1+2x+x^2)=2-(1+x)^2
$$
so you can do the substitution $u\sqrt{2}=1+x$ and the integral becomes
$$
\int\frac{\sqrt{2}}{1+\sqrt{2}\sqrt{1-u^2}}\,du
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a ... | Hope it will help!
WLOG $ b\ge a\ge c\Rightarrow c\le 1\Rightarrow 1-c\ge 0$
We have: $$f\left(a,b,c\right)=\left(a-1\right)^3+\left(b-1\right)^3+\left(c-1\right)^3$$
And $$f\left(\frac{a+b}{2};\frac{a+b}{2};c\right)=2\left(\frac{a+b}{2}-1\right)^3+\left(c-1\right)^3$$
Note that:$$f\left(a,b,c\right)-f\left(\frac{a+b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\sum_{cyc}\frac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$
If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$ \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$
Here's what I did.
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{a(b^2 + c^2 + bc)}{(b^2c + c^2b + abc)^2} = \fr... | Another way:
By C-S
$$\sum_{cyc}\frac{a}{a+b^4+c^4}-1=\sum_{cyc}\frac{a(a^3+2)}{(a+b^4+c^4)(a^3+1+1)}-1\leq$$
$$\leq\sum_{cyc}\frac{a^4+2a}{(a^2+b^2+c^2)^2}-1
=-\frac{2\sum\limits_{cyc}(a^2b^2-a^2bc)}{(a^2+b^2+c^2)^2}=-\frac{\sum\limits_{cyc}c^2(a-b)^2}{(a^2+b^2+c^2)^2}\leq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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inequality related to square the sum of any two sides of a triangle with respect to square of other side Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.
My approach
As we know that the sum of any two sides ... | The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing f... | One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
which is the number of ways to put $3$ or fewer indistinguishable balls... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Check if vectors are the the fundamental set of solutions $$\begin{cases} x_1+2x_2+x_3-x_4+x_5=0 \\ 2x_1+x_2-x_3+2x_4-x_5=0 \\ x_1+5x_2+4x_3-5x_4+4x_5=0 \\ 4x_1+5x_2+x_3+x_5=0 \end{cases}$$
$$\begin{cases} \vec{a}_1=(-5,4,3,3,-3) \\ \vec{a}_2=(-7,5,-3,6,6) \\ \vec{a}_3=(5,-4,6,-3,-6) \end{cases}$$
Solution:
\begin{bmat... | $$
\left(
\begin{array}{rrrr}
1 & 0 & 0 &0 \\
-1 & 0 & 1 &0 \\
3 &-1 &-1 &0 \\
-5 & 0 & 1& 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1& 2 &1& -1 & 1\\
2& 1 &-1 & 2 &-1\\
1 &5 & 4 &-5 & 4\\
4 &5& 1 & 0& 1\\
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
1& 2 &1& -1 & 1\\
0& 3 &3 & -4 &3\\
0 &0 & 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the surface area of this function when rotated around the y axis. Stuck Say I have this equation:
$$y = \frac{1}{3} x^{\frac{3}{2}}$$ and the limits of the integration provided are $0 \leq x \leq 12$
So since we're rotating around the y axis, I'm going to change the limits of integration to use y.
$$0 \leq y \leq ... | Yes, you did make a mistake in the very beginning, when you kinda decided to switch to integration with respect to $y$ … but you didn't actually switch. Note that your integral
$$2\pi\int_0^{8\sqrt{3}}x\sqrt{1+\frac{1}{4}x}\,dx$$
is with respect to $x$, but you're using the $y$-values from $0\le y\le8\sqrt{3}$ as your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow ... | METHODOLOGY $1$: Feynman's Trick
Let $I(a,k)$ be the integral given by
$$I(a,k)=\int_0^\pi \log(a+k\cos(x))\,dx\tag1$$
Differentiating $(1)$ with respect to $a$ reveals
$$\begin{align}
\frac{\partial I(a,k)}{\partial a}&=\int_0^\pi \frac{1}{a+k\cos(x)}\,dx\\\\
&=\frac{\pi}{\sqrt{a^2-k^2}}\tag2
\end{align}$$
Integrating... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
The generalised Catalan Numbers and Borel's Triangle I am currently reading "Counting with Borel’s Triangle" (https://arxiv.org/abs/1804.01597), and am very confused on a stated formula.
We know:
$C_{n,k}=\frac{n-k+1}{n+1}{n+k \choose n}$
$C(x) = \sum_{n=0}^{\infty}\frac{1}{n+1}{2n \choose n}x^{n}=\frac{1-\sqrt{1-4x}}{... | We show the validity of the equality chain
\begin{align*}
\sum_{i\geq0}C(2;i)x^i=\frac{1+2xC(2x)}{1+x}=\frac{1}{1-xC(2x)}=\frac{4}{3+\sqrt{1-8x}}
\end{align*}
with $$C(x)=\frac{1-\sqrt{1-4x}}{2x}$$ the generating function of the Catalan numbers and the generalized Catalan numbers $C(2;i)$ stored in OEIS as A064... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all positive integers $n$ such that $\frac{3^n-1}{2^n}$ is an integer. I need to find all positive integers $n$ such that $\frac{3^n-1}{2^n}$ is an integer. So far I only found $n=1$, $n=2$ and $n=4$ and solutions to this problem? Is there a way to prove using modular arithmetic that there are no more solutions (... | If $m$ is odd then the highest power of $2$ that divides $3^m-1$ is $1$ and dividing $3^m+1$ is $2$. If $n=2^k\cdot m $ where $m$ is odd then factorize the numerator as $ (3^m -1)(3^m +1)(3^{2m} +1)...(3^{2^{k-1}m} +1 )$ .From the third term onwards highest power of $2$ dividing each factor is $1$ and hence highest pow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum_{n} t_n x^n = \frac{x^k}{(1-x^2)^k(1-x)} $
Let $t_n$ be a number of sequences
$$ 1 \le a_1 < a_2 < ... < a_k \le n $$
such that $a_{2i}$ is even and $a_{2i+1}$ is odd.
Prove that $$\sum_{n} t_n x^n = \frac{x^k}{(1-x^2)^k(1-x)} $$
My attempt. I want to use there enumerators and I have 4 cases.... | We have that
$$f(x):=\frac{x^k}{(1-x^2)^k(1-x)}=\frac{1}{1-x}\left(\frac{x}{1-x^2}\right)^k
=\frac{1}{1-x}\left(\sum_{j=0}^{\infty}x^{2j+1}\right)^k.$$
Therefore the coefficient $t_n=[x^n]f(x)$ is the number of ways we can write a positive integer $\leq n$ as the sum of $k$ odd numbers.
Now note that $a_1,a_2-a_1,\dots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Help verifying proof to proving that for all natural numbers $\sqrt{1} \leq$ than the sum Prove that for all natural numbers $n$,
$$\sqrt{n} \le 1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Solution: We must prove that $1 + \frac 1 {\sqrt{2}}, + \frac 1 {\sqrt{3}} +\,\cdots\, + \... | You are right. The inductive proof needs to establish the link between
$$\sqrt n\le1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}$$
and
$$\sqrt{n+1}-\frac1{\sqrt{n+1}}\le1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Obviously,
$$\sqrt{n+1}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplify trigonometric expression using trigonometric identities I have the trigonometric expression: $$2\sin x +2\sin \left(\frac{\pi} {3} -x\right) $$
and it should simplified in: $$\sin x + \sqrt 3 \cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?
| Using the formula for $\sin (\alpha - \beta)$ you obtain
\begin{align}
2&\sin x +2\sin \left(\frac{\pi} {3} -x\right)\\
= 2&\sin x +2\left[\sin \left(\frac{\pi} {3}\right) \cos x - \cos\left(\frac{\pi} {3}\right) \sin x\right]\\
= 2&\sin x + 2\left[{\sqrt 3 \over 2} \cos x - \frac 1 2 \sin x\right]\\[1ex]
= 2&\sin x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
A tricky integral involving hyperbolic functions Can anyone suggest a method for solving the integral below? I've tried numerous things but have had no luck yet. To be honest I'm not sure an analytical solution actually exists.
$$I=\int\cosh(2x)\sqrt{[\sinh(x)]^{-2/3}+[\cosh(x)]^{-2/3}}\,\textrm{d}x.$$
Thanks.
Here is ... | Thanks to Yuri for his working. Very helpful. I believe I have arrived at the solution now
\begin{align*}
I&=\int\cosh(2x)\sqrt{[\sinh(x)]^{-2/3}+[\cosh(x)]^{-2/3}}\,\textrm{d}x
\\
&=\int\cosh(2x)\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1/6}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
\\
&=\frac{6}{5}\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Solve $\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$
Find $$\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$$
Looking at the numerator, combined with the surd, you can get $$\int\frac{e^x(1+x)\left(\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right)}{1-x^2}\mathrm dx$$Then this begins to look like the quotient ... | Hint:
$$\dfrac{1+1-x^2}{(1-x)^{3/2}(1+x)^{1/2}}=\dfrac1{...}+f(x)$$
where $f(x)=\dfrac{\sqrt{1+x}}{\sqrt{1-x}},$
$f'(x)=?$
Recall $\dfrac{d(e^xf(x))}{dx}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$
The solution proposes that $x... | Suppose there is a larger $k$. Then we can apply Viete's root jump and 'descend' the roots infinitely, which is impossible since there is no infinite descending chain in the naturals. This is the crux of the whole article you linked to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Upper bound estimate $ab^3\leq c(a^2+b^2)^2$. Find optimal, or good, constants $c$ I am trying to find a good upper bound estimate for the expression $ab^3$, where $a,b\in\mathbb R$, and it should be of the form $ab^3\leq c (a^2+b^2)^2, c\in\mathbb R.$
(The reason for the latter is, because in my application of this, $... | By AM-GM
$$(a^2+b^2)^2=\left(a^2+3\cdot\frac{b^2}{3}\right)^2\geq\left(4\sqrt[4]{a^2\left(\frac{b^2}{3}\right)^3}\right)^2=\frac{16}{3\sqrt3}|ab^3|\geq\frac{16}{3\sqrt3}ab^3.$$
The equality occurs for $a=\frac{b}{\sqrt3},$ which says that $c=\frac{3\sqrt3}{16}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Eigenvalues of two symmetric $4\times 4$ matrices: why is one negative of the other? Consider the following symmetric matrix:
$$
M_0 =
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & 4 & 3 \\
2 & 4 & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
and a very similar matrix:
$$
M_1 =
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & -4 & 3 \... | I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
\left[\begin{array}{rrrr}
0 & 1 & 2 & 0 \\
1 & 0 & a & 3 \\
2 & a & 0 & 1 \\
0 & 3 & 1 & 0
\end{array}\right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
\begin{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Basic combinations logic doubt in probability
"If $3$ students are chosen at random from a class with $6$ girls and $4$ boys, what is the probability that all $3$ students chosen will be girls?"
$\left(\dfrac{6}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)$
So why can't we use that logic to answer thi... | The first question solved in the second method looks as follows:
$$\frac{{6\choose 3}}{{10\choose 3}}=\frac{6\cdot 5\cdot 4}{10\cdot 9\cdot 8}$$
Interpretation: There are ${6\choose 3}$ ways to choose $3$ girls out of $6$ and there are ${10\choose 3}$ ways to choose $3$ students out of $10$, hence the probability is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Understanding Mathematical Induction problems I have problem to understand this 3 formulas, I am new in this type of problems.
I have to solve this problems by induction.
\begin{align}
\sum_{j = 1}^{j = n} j^3 &= \left(\frac{n(n + 1)}{2}\right) ^ 2 &\text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j + 1) &= \frac{1}{3... | The general procedure for a proof by induction is to first show that the base case satisfies that proposition. Then you assume that your proposition holds for some $n$ and show that from their you can get $n+1$ to work.
I'll work the first the one and let you try the others using the same procedure.
We need to show t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$
Find the degree of splitting field of $f=X^4+2X^2+2$ over $\mathbf{Q}$.
By Eisenstein, $f$ is irreducible. By setting $Y=X^2$, we can solve for the roots: $Y=-1\pm i \iff X=\sqrt[4]{2}e^{a\pi i/8}$, $a\in\{3,5,11,13\}$. Clearly $f$ splits in $\mathbf{Q}(\sqr... | The OP asked in a comment
is there some easy way to show that $−1+i$ is not a square in $\mathbb Q(\zeta_8)$?
Let $a,b,c,d \in \mathbb Q$ and $\zeta_8 = e^{\tfrac14\pi i}$.
Consider the number
$$\tag 1 \nu = a + b e^{\tfrac14\pi i}+ c e^{\tfrac12\pi i} + d e^{\tfrac34\pi i}$$
Then
$\quad \nu^2 = a^2 + c^2 e^{i \pi} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let $P (x )=x^4+ax^3+bx+c=0$ and have real coefficient and have all real roots . Prove that $ab \leq 0$ Let $P (x )=x^4+ax^3+bx+c=0$ and have real coefficient
and have all real roots . Prove that $ab \leq 0$
First Let the roots of this polynomial (call it P(x)) be $q,r,s,t$
By Vieta's, $a=-(q+r+s+t)$ and $b=-(qrs+rst+q... | Let $p$, $q$, $r$ and $s$ be roots of our equation and
$$p+q+r+s=4u,$$
$$pq+pr+ps+qr+qs+rs=6v^2,$$ $$pqr+pqs+prs+qrs=4w^3$$ and $$pqrs=t^4,$$ where $v^2$ and $t^4$ can be negative.
Thus, $p$, $q$, $r$ and $s$ are roots of the equation:
$$x^4-4ux^4+6v^2x^2-4w^3x+t^4=0,$$ which says that by the Rolle's theorem the equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two posit... | COMMENT.-We have $$2^ky-1=a(y^2-x^2)\\y-1=b(2^k-1)$$ where the given solution gives the identities $2^{2k}=2^{2k}$ and the equivalent $2^k=2^k$, not properly a system of two independent equations.
Suppose now a true (independent) system
The first equation gives a quadratic in $y$
$$ay^2+(-2^k)y+(-ax^2+1)=0$$ and the di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+... +\frac{x_n}{n^b} $ is convergent Let $a\ge 0$ and $(x_n) _{n\ge 1}$ be a sequence of real numbers. Prove that if the sequence $\left(\frac{x_1+x_2+...+x_n}{n^a} \right)_{n\ge 1}$ is bounded, then the sequence $(y_n) _{n\ge 1}$, $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+...... | Let's do an Abel transformation. Denoting by $S_k = x_1 + ... + x_k$ for all $k \in \mathbb{N}$, you have
$$\sum_{n=1}^N \frac{x_n}{n^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}-S_n}{(n+1)^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}}{(n+1)^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} $$
$$= x_1 +\sum_{n=2}^{N} \frac{S_n}{n^b} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align... | Hint: We get $$\theta=\frac{1}{9}\arctan(\frac{3}{4})$$ so we get
$$3\csc\left(\frac{1}{3}\arctan(\frac{3}{4})\right)-4\sec\left(\frac{1}{3}\arctan(\frac{3}{4})\right)=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$
Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$
So I started by combining the two fractions, which gave me:
$$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)... | $$\frac{2\sin\theta\cos\theta}{1-\cos^2\theta} =\frac{2\sin\theta\cos\theta}{\sin^2\theta} = \frac{2\cos\theta}{\sin \theta} = 2\cot \theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$ Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of
$$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$
To find the minimum of $P$ I rewrite it as
$$P=2\left( ab+... | I would use that $$6=\frac{3}{2}(a+b+c)^2$$ and $$2=\frac{1}{4}(a+b+c)^3$$
Using this substitution we get the term
$$1/8\, \left( {a}^{2}+6\,ab+6\,ac+{b}^{2}+6\,bc+{c}^{2} \right)
\left( {a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}+2\,abc+3\,a{c}^{2}
+{b}^{3}+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the coefficient of $x^{32}$ in $(x^3 +x^4 +x^5 +x^6 +x^7)^7$ I don't understand the explanation in the book and why the final answer looks the way it does. I know I am supposed to factor and it will equal to $(x^3(1+x+\cdots+x^4))^7$. But after that, I am really confused about what happens. Can someone explain thi... | A strategy for solving problems like this one is to try to rewrite the expression as the quotient of a polynomial in $x$ and some power of $1-x$. You can then use either Newton’s generalized binomial theorem or the identity $${1\over(1-s)^{k+1}} = \sum_{n=0}^\infty \binom{n+k}k s^n$$ to expand the denominator as a powe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifying $\sin\frac{11\pi}{12}\sin\frac{29\pi}{12}-\cos\frac{13\pi}{12}\cos\frac{41\pi}{12}$. Why do I get the wrong answer? Can someone explain why I get wrong answer in simplifying this expression?
$$\sin\frac{11\pi}{12}\sin\frac{29\pi}{12}-\cos\frac{13\pi}{12}\cos\frac{41\pi}{12}$$
If we rewrite the expression wi... | No, you don't get $\frac12$. You get $0$ too, because\begin{align}\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)-\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{41\pi}{12}\right)&=\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)-\cos\left(\frac{11\pi}{12}\right)\cos\left(3\pi+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $a^3+b^3+3ab=1$ with $(a,b)\in \Bbb{Z}^2$
Solve the following equation for $(a,b)\in \Bbb{Z}^2$:
$$a^3+b^3+3ab=1$$
I tried all of the standard techniques I know. I tried modular arithmetic:
$$a^3+b^3+3ab\equiv 1 \pmod{3} $$
$$a^3+b^3\equiv 1 \pmod{3} $$
Now by Fermat's Little Theorem:
$$a^2 a+b^2 b\equiv 1 \p... | The item worth memorizing is
$$ x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( x^2 + y^2 + z^2 - yz - zx - xy \right) $$
where the quadratic form is positive semidefinite because
$$ \left( x^2 + y^2 + z^2 - yz - zx - xy \right) = \frac{1}{2} \left( (y-z)^2 + (z-x)^2 + (x-y)^2 \right) $$
If you then take $z=-1$ you get
$$ x^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(x)f(y)+f(xy)\le -\frac{1}{4},\forall x,y\in[0,1)$, show that $f(x)=-\frac{1}{2}$ Let $f:[0,1) \to \mathbb{R}$ be a function such that
$$f(x)f(y)+f(xy)\le -\dfrac{1}{4} \quad \forall\, x,y\in[0,1).$$
Show that
$$f(x)=-\dfrac{1}{2} \quad \forall\, x \in[0,1).$$
I have proved that $f(0)=-\dfrac{1}{2}$: if $x=y=0$, ... | Plugging in $y=0$ gives $f(x) \ge -\frac{1}{2}$ for each $x$.
Let $y=x$ to get $f(x)^2+f(x^2) \le -\frac{1}{4}$. This implies $f(x^2) \le -\frac{1}{4}$ for each $x$, and so $f(x) \le -\frac{1}{4}$ for each $x$. But then $f(x)^2+f(x^2) \le -\frac{1}{4}$ implies $f(x^2) \le -\frac{1}{4}-(\frac{1}{4})^2 = -\frac{5}{16}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force.... | Beginning with pairs $(1,0)$ and then $(8,3),$ all the other soltuions with positive variables satisfy
$$ x_{n+2} = 16 x_{n+1} - x_n \; , \; $$
$$ y_{n+2} = 16 y_{n+1} - y_n \; . \; $$
The $x_n$ begin
$$ 1, 8, 127, 2024, 32257, 514088, 8193151, $$
The $y_n$ begin
$$ 0, 3, 48, 765, 12192, 194307, 3096720, $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of the surface for the curve rotated about the x-axis
$y = \tan(x), 0\leq x \leq \frac{\pi}{3}$
I am struggling with constructing an integral for this formula.
Since the curve is rotated about the x-axis, I think this is the best way to setup the integral. Since the circumference is $2\pi r$, it makes s... | Hint: Do a change of variable $\cos x=u$, then $du=-\sin x$, so $$S=2\pi\int_{1/2}^1\frac{\sqrt{1+x^2}}{x^2}du$$
Note: Since the derivative will need to be squared, $$S=2\pi\int_{1/2}^1\frac{\sqrt{1+u^4}}{u^3}du$$
Here are some hints how to solve the rest: Make substitution $v=u^2$, $dv=2udu$ and you get an integral$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $P(n+1)$ for a polynomial P
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.
My attempt:
I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of de... | Your solution is fine.$$Q(-1) = \lambda (-1)(-2)\ldots (-(n+1))=\lambda (-1)^{n+1}(n+1)!=1$$
$$\lambda = \frac{(-1)^{n+1}}{(n+1)!}$$
$$Q(x)=\frac{(-1)^{n+1}x(x-1)\ldots (x-n)}{(n+1)!}=(x+1)P(x)-x$$
Let $x=n+1$
$$\frac{(-1)^{n+1}(n+1)n\ldots 1}{(n+1)!}=(n+2)P(n+1)-(n+1)$$
Hence $$P(n+1) = \frac{n+1 + (-1)^{n+1}}{n+2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$ Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$
I entered the equation in desmos.com and no roots were lying below or x=0 lines , hence all roots are imaginary.
Using Descartes rule for f(x) 4 sign change occurs , hence 4,... | To expand on Dr. Sonnhard Graubner's answer, let $f(x) = 54x^4-36x^2+15x^2-6x+1$. Then:
$$54x^4-36x^3+18x^2-6x+1 > f(x)$$
$$\Rightarrow (3x-1)^2(6x^2+1)$$
by the rational root test.
The smaller function has only one root: $x = \frac{1}{3}$, and there $f''(x) = 648x^2-216x+30$ is positive. Therefore, for all real $x$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of lattice points inside a circle
The number of lattice points inside the circle $x^2+y^2=a^2$ can not be
Options $(a)\; 202\;\;\; (b)\; 203\;\;\; (c)\; 204\;\;\; (d)\; 205$
Try: i have an idea of number of integer points on the circle $x^2+y^2=a^2$
Let $x,y\in\{4n,4n+1,4n+2,4n+3\}$
But no idea how to find num... | According to Gauss's circle problem, all choices cannot be ($r$ is radius, $N(r)$ is the number of lattice points):
$$\begin{array}{c|c|c}
r&0&1&2&3&4&5&6&7&8&9&10&11&12\\
\hline
N(r)&1&5&13&29&49&81&113&149&197&253&317&377&441
\end{array}$$
See the graph to verify the numbers $N(8)=197$ and $N(9)=253$:
$\hspace{5cm}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$.
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \}... | Note that$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Integrating $\int_0^{\infty} \frac{\log x}{(x + a)^2 + b^2} \operatorname d\!x$ I'm trying to show that $\int_0^{\infty} \frac{\log x}{(x + a)^2 + b^2} \operatorname d\!x = \frac{1}{b}\arctan \sqrt{a^2 + b^2}.$ However, I am a bit confused applying the key hole "method." I consider $f(z) = \frac{\log^2(z)}{(z+a)^2+b^2}... | without complex analysis$$I = \int^{\infty}_{0}\frac{\ln x}{(x+a)^2+b^2}dx$$
set $\displaystyle x = \frac{a^2+b^2}{y}$ and $\displaystyle dx=-\frac{a^2+b^2}{y^2}dy$
$$I = \int^{\infty}_{0}\frac{\ln(a^2+b^2)-\ln(y)}{(y+a)^2+b^2}dy$$
$$I = \ln(a^2+b^2)\int^{\infty}_{0}\frac{1}{(y+a)^2+b^2}dy-I$$
$$2I=\frac{\ln(a^2+b^2)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to write $[(2+\sqrt{3})^n + (2-\sqrt{3})^n + (2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1}]/6$ to the form $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$). We know that $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ is an integer (See here).
However, we want to write the formula
\begin{align}
&\frac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \frac{3-\sqr... | Call your formula $f(n)$. Since $2\pm\sqrt{3}$ are the roots of $x^2-4x+1$, it's easy to show that $f(n)$ is determined by $f(0)=1$, $f(1)=3$ and the recursive relation $f(n+1)=4f(n)-f(n-1)$.
After looking at @J.W.Tanner comment, I found out that we actually have
\begin{align*}
f(2n)&=f(n)^2+2\left(\sum_{0\leq k\leq n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Power series differential check $(x-1)y'' + y' = 0$ I'm a bit stuck on how to solve this:
$$(x-1)y'' + y' = 0 $$
so assuming y is a solution in this form:
$$\sum_{n=0}^\infty C_nx^n$$
$$\sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
subbing in and distributing:
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-... | Calling $Y = y' = \sum_{k=0}^{\infty}a_k x^k$ we have in $(x-1)Y'+Y = 0$
$$
(x-1)\sum_{k=1}^{\infty}k a_k x^{k-1}+\sum_{k=0}^{\infty}a_k x^k = a_0-a_1+\sum_{k=1}^{\infty}(k+1)a_k x^k-\sum_{k=1}^{\infty}(k+1)a_{k+1} x^k = 0
$$
hence $a_k - a_{k+1} = 0$ or $Y = a_0\sum_{k=0}^{\infty}x^k$ but $Y = y'$ so finally
$$
y = \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using De Moivre's law to compute $(-\sqrt3+i)^{2/3}$ Question:
If $z=-\sqrt{3}+i$, then $z^{2 / 3} = ?$
My work (which is wrong but I am not sure why):
We can write $z = r(\cos\theta + i\sin\theta)$
$$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$$
$$\theta = \tan^{-1} \left(\frac{-1}{\sqrt{3}}\right) = \frac{5\pi}... | You forgot $i$ when writing $z = r(\cos \theta + i \sin \theta)$. Arctangent was computed correctly, so
$$
z = 2\left(\cos(\frac{5}{6}\pi) + i \sin(\frac{5}{6}\pi)\right),
$$
$$
z^{2/3} = 2^{2/3}\left(\cos(\frac{5}{9}\pi) + i \sin(\frac{5}{9}\pi)\right).
$$
Now some trigonometry:
$$
\cos(\frac{5}{9}\pi) = \cos(\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
limit $\sqrt{-n^4+4n^2+4}-in^2 = -2i$? What is the limit of the following complex sequence? I get a different result than Wolfram Alpha. My approach was:
$$\sqrt{-n^4+4n^2+4}-in^2 = \sqrt{i^2n^4+4n^2+4}-in^2 = in^2\sqrt{1-\frac{1}{4n^2}-\frac{4}{n^4}}-in^2 \xrightarrow{n\xrightarrow{}\infty} 0$$
Wolfram Alpha says the ... | Let's forget about those pesky $i$s and consider the limit of
$$n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}-n^2$$
as $n\to\infty$. This equals
$$\frac{n^4\left(1-\frac4{n^2}-\frac4{n^4}\right)-n^4}{n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}+n^2}
=\frac{-4n^2-4}{n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}+n^2}
=\frac{-4-4/n^2}{\sqrt{1-\frac4{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&... | $c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$
$\iff c^4+s^2=c^2+s^4=P$(say) where $c=\cos x,s=\sin x$
$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-\dfrac{\sin^22x}2=2-\dfrac{1-\cos4x}4=?$
using $\sin2x=2\sin x\cos x,\cos2y=2-2\sin^2y$
We need $$\dfrac1P+\dfrac2P=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^... | For positive $n$, let $a=\sqrt[3]{n^3+3n^2}$, and let $b=\sqrt{n^2+2n}$.
Note that $a < n+1$ and
\begin{align*}
3n +1 &= (n+1)^3-a^3\\[4pt]
&= \bigl((n+1)-a\bigr)\bigl((n+1)^2+a(n+1)+a^2)\\[4pt]
&> \bigl((n+1)-a\bigr)(n+1)^2\\[4pt]
\end{align*}
hence $0 < n+1-a < {\large{\frac{3n+1}{(n+1)^2}}}$, which implies $\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
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Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$. If $f(x), g(x)$ are defined as following on $[0 , \infty)$,
$$\tag 1 f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$$
$$\tag 2 g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + ... | A simple proof:
Clearly:
$$
n^3+x^3\ge \frac{1}{4}(n+x)^3 \tag{1}
$$
and hence
$$
0\le \frac{nx}{n^3+x^3} \le \frac{4nx}{(n+x)^3}\le\frac{4(n+x)x}{(n+x)^3}=\frac{4x}{(n+x)^2}.
$$
Hence
$$
\sum_{n=1}^\infty\frac{nx}{n^3+x^3}\le
\sum_{n=1}^\infty\frac{4x}{(n+x)^2}=\frac{4x}{(1+x)^2}+\sum_{n=2}^\infty\frac{4x}{(n+x)^2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$ The inequality:
$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$$
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an e... | Since squaring is monotonic, this is equivalent to
$$
a^2 + b^2 + c^2 + d^2 + 2\sqrt{a^2+c^2}\sqrt{b^2+d^2} \ge a^2+b^2+c^2+d^2 + 2(ab + cd)
$$
which is in turn equivalent to
$$
ab + cd \le \sqrt{a^2+c^2}\sqrt{b^2+d^2},
$$
the Cauchy-Schwarz inequality.
Putting it in vector form is even more elegant. For $\mathbf{a} = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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In diophantine $3b^2=a^2$ where $a$ and $b$ are coprime, does $3|a$? Integers $a$ and $b$ are co-prime and $3\cdot b^2=a^2$.
$3\cdot b^2=a^2$, implies $a^2$ is divisible by 3 since, $3b^2$ is divisible by 3.
Is $a$ divisible by 3?
| Yep; in general, if $p|ab$ then $p|a$ or $p|b$ ($p$ is a prime). In your case, $3|a^2$ so $3|a$ or $3|a$; thus $a$ is divisible by $3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Volume of cylinder cut by two planes I try to find volume of cylinder $x^2 + y^2=9$ cut by inclined plane $z = x + y - 4$ and plane $z=1$
I should solve it in Cartesian, so my idea is to rotate the plane $z=x + y - 4$, and to split the cylinder in two symmetrical portions, which allows using only the portion over $x$ a... | $V = \int^3_{x =-3}\int^{\sqrt{9-x^2}}_{y = -\sqrt{9-x^2}}\int^{1}_{z = x+y-4}dzdydx=\int^3_{x =-3}\int^{\sqrt{9-x^2}}_{y = -\sqrt{9-x^2}} \bigg( 1 - x - y +4\bigg)dydx $
$V = \int^3_{x =-3}\bigg(5y-yx-\frac{y^2}{2}\bigg)^{\sqrt{9-x^2}}_{-\sqrt{9-x^2}}dx =\int^3_{x =-3} \big[(5-x)(2\sqrt{9-x^2}) - \frac{1}{2}(9-x^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(... | Note: $1-x^2\ge 0 \Rightarrow -1\le x\le 1$.
Square both sides and denote $1-x^2=t^2, -1\le t\le 1$:
$$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} \Rightarrow \\
(1+\sqrt{1-x^2})(2+6x^2+2\sqrt{(1-x^2)^3}=5-x^2+4\sqrt{1-x^2} \Rightarrow \\
(1+t)(2+6(1-t^2)+2t^3)=4+t^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
} |
Sec. 6 in G.F. Simmons' INTRO TO TOPOLOGY & MODERN ANALYSIS: A formula for the general terms of these two sequences please? The set $\mathbb{Q}^+$ of all the positive rational numbers is given by
$$ \mathbf{Q}^+ = \left\{ \ \frac{p}{q} \ \colon \ p \in \mathbf{N}, q \in \mathbf{N} \ \right\}.$$
This set is countably in... | Such formulae necessarily involve the Euler's totient function, because the number of positive fractions $p/q$ (in their lower terms, of course) such that $p+q=n$ is precisely $\varphi(n)$. Then, for the first sequence, the $k$-th term can be obtained this way:
Let $n=\max\{t\in\Bbb N:\sum_{j=1}^t\varphi(j)\le k\}$. In... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the amplitude of a complex number
If
$$z = (1-\cos{\theta}) + i \sin{\theta},$$
then the amplitude of $z$ is:
$$a) \frac{\pi}{4}-\frac{\theta}{2}$$ $$b)
\frac{\pi}{2}-\frac{\theta}{2}$$ $$c) \frac{\pi}{2}-\theta$$ $$d)
\frac{\pi}{2}-\frac{\theta}{4}$$ where 0<$\theta$ <$\pi$.
I tried :
$x= 1-\cos{\theta}, y= ... | $\sin\theta = 2\sin\frac{\theta}{2}\cdot \cos\frac{\theta}{2}$
$1-\cos\theta = 2\sin^2\frac{\theta}{2}$
So, $\tan\phi = \frac{\sin\theta}{1-\cos\theta} = \frac{2\sin\frac{\theta}{2}\cdot \cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$
$\tan\phi = \cot\frac{\theta}{2} = \tan(\frac{\pi}{2} - \frac{\theta}{2})$
$\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How do I solve this log equation for x? $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$ $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$
I only managed to solve up to this step: $\left(2+x\right)^{\frac{3}{2}}\sqrt{2-... | You are almost there.
$$\left(2+x\right)^{\frac{3}{2}}\sqrt{2-x}=\sqrt{4-x^2}\cdot125$$
$$\implies (2+x)\sqrt{2+x}\cdot\sqrt{2-x} = 125\sqrt{(2+x)(2-x)}$$
Here, notice that $x = 2$ and $x = -2$ are not the solutions for the original equation because $\log_50$ is not defined. Therefore, we can divide both sides by $\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Simultaneous equation of trig functions If I have the two following equations:
$$\begin{align}
A &= F\sin(x)+ G\cos(x)\\
B &= G\sin(x)+ F\cos(x)
\end{align}$$
and $A$,$B$,$F$,$G$ and are all constants, what is the easiest method of calculating $x$?
I've had a look at Cramers rule and the $R\cos(x-\alpha)$ rule. I need ... | If $F\ne G$, then subtracting the second equation from the first gives
\begin{eqnarray}
(F-G)\sin x-(F-G)\cos x&=&A-B\\\
\sin x-\cos x&=&\frac{A-B}{F-G}\\\
\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\right)&=&\frac{A-B}{F-G}\\\
\sqrt{2}\left(\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}\right)&=&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find $ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$ How to compute the limit
$$ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$$
Does exist an explicit formula for finding the numerator?
| The required limit is $1/(n+2)$. There probably does not exist a closed formula for the numerator. However, the numerator can be estimated by use of an integral, as follows.
Note that
$$\frac{k^{n+2}}{n+2}=\int_0^{k}x^{n+1}\,\mathrm dx < \sum_{x=0}^kx^{n+1}<\int_1^{k+1}x^{n+1}\,\mathrm dx=\frac{(k+1)^{n+2}}{n+2}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expr... | We can use a trick to express the function as something easier to deal with. We claim that we can write the function as $(d \cos \theta + e \sin \theta)^2 + f$ or $-(d \cos \theta - e \sin \theta)^2 + f$ for some constants $d, e, f$. From here, converting $d \cos \theta + e \sin \theta$ into a single trigonometric func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Why is: $1+2+2^2+\cdots+2^{n-2} = 2^{n-1}-1$? I know that this might be elementary, but I don't know how to prove this equality, some hint?
I tried to extract the common two:
$$ 1 + 2^1+2^2 + \cdots + 2^{n-2} = 1 + 2(1+2+\cdots+2^{n-3}) $$
But this didn't help
| The sum $1+2+2^2+\cdots+2^{n-2} $ doesn't change if we multiply it by $1$ i.e. $2-1$:
\begin{eqnarray}1+2+2^2+\cdots+2^{n-2} &=& (1+2+2^2+\cdots+2^{n-2})(2-1)\\
&=& (\color{red}{2+2^2+2^3+\cdots+2^{n-2}}+2^{n-1})-(1+\color{red}{2+2^2+2^3+\cdots+2^{n-2}})\\
&=&2^{n-1}-1
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1... | $\sum_{i=0}^{k\cdot m}c_ix^i=(1+x^1+x^2+...+x^m)^{k}$ is the generating function of the number of weak integer compositions (integer compositions with repetitions of $0$) of integer $i$ with $k$ parts where all parts are lower equal to $m$.
Unfortunately, it is not yet in OEIS.
$k,m>0$
Their coefficients are:
$$c_i=\su... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
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When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$ When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$, gives $\sum\limits_{n=1}^{\infty} \left(\frac{nx^{n-1}}{2^n}\right)$. so, $n =0$ , becomes $n = 1$. Then, if we were to differentiate $\sum\limits_{n=1}^{\infty} ... | No, because the reason the sum lower bound goes from 0 to 1 is because the derivative kills the first term in the sum. When we differentiate the sum starting from $n=1$, the first term is not killed by the derivative and hence you shouldn't drop it from the sum.
To deal with the sum
$$
F(x)=\sum_{n=1}^\infty n^2 x^n 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$.
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$
Before ... | $$\frac{xy}{z^2(x+y)}+\frac{yz}{x^2(z+y)}+\frac{xz}{y^2(x+z)}$$
$$=\frac{2x^2y^2}{z(x+y)}+\frac{2y^2z^2}{x(z+y)}+\frac{2x^2z^2}{y(x+z)}$$
$$\geq 2\frac{(xy+yz+xz)^2}{2(xy+yz+zx)}=xy+yz+xz$$
Using Titu's Lemma which is a variant of the Cauchy-Schwarz inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Polar decomposition of a general matrix How can I calculate the polar decomposition for a general matrix? For example for this simple one:
$$
\begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix}
$$
I know how to calculate it for a matrix with numbers, via eigenvalues, eigenvectors. I have been searching f... | So first let $A = \begin{pmatrix}a&-b\\b&a\end{pmatrix}=QS$.
Then, we have $A^{T}A=\begin{pmatrix}a&b\\-b&a\end{pmatrix}\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}a^2+b^2&0\\0&a^2+b^2\end{pmatrix}$.
The Eigenvalue(s) of this matrix is $a^2+b^2$, sometimes they're different but not too hard to handle.
The Eige... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How can i get the least $n $ such that $17^n \equiv 1 \mod(100$)? When I solve the problem:
$17^{2018}\equiv r \pmod{100} $
used Euler theorem since $\gcd (17,100)=1$ and so
$\phi(100)=40$
and so
$17^{40}\equiv 1 \pmod{100}$
But i also found that: $17^{20}\equiv 1 \pmod{100}$
How can i get the least n such that $17^{... | Euler's phi function, $\varphi(n)$, is a multiplicative function for which, if $a$ and $N$ are relatively prime, then $a^{\varphi(N)} \equiv 1 \pmod N$. In particular
$$\varphi(100) = \varphi(4)\varphi(25) = (4-2)(25-5)=40$$
So $17^{40} \equiv 1 \pmod{100}$. The smallest $n$ such that $17^n \equiv 1 \pmod{100}$ must th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Solve for integer $m,n$: $2^m = 3^n + 5$
Solve for integer $m$ and $n:$ $$2^m = 3^n + 5$$
My Attempt:
Easy to guess two solutions namely $(3,1)$ and $(5,3)$.
Also easy to see that a solution will exist iff $m > 0$ and $n > 0$.
Rewriting it as $2^m - 2 = 3^n + 3$ we get $2^m = 2 \mod 3 \Rightarrow m = 1 \mod 2$ and $3... | Rename $m\to x$ and $n\to y$
We see $x\geq 3$, $y\geq 1$.
Modulu 3 implies $x$ is odd. For $x\leq 5$ we get only $(3,1)$, $(5,3)$.
Say $x\geq 6$, then
$$3^y\equiv -5\;({\rm mod}\; 64)$$ It is not difficult to see
$$3^{11}\equiv -5\;({\rm mod}\; 64)$$
so $3^{y-11}\equiv 1\;({\rm mod}\; 64)$. Let
$r=ord_{64}(3)$, then s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$ Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$.
I tried Euler and hyperbolic substitutions, but both lead to complicated calculations and yet WolframAlpha is able to generate quite simple form of integral.
| Your integral is equivalent to
$$ \frac{1}{2} \underbrace{\int \frac{2x+1}{\sqrt{x^2 + x + 1}} \mathrm dx}_{=:I_1} - \frac{1}{2} \underbrace{\int \frac{\mathrm d x}{\sqrt{x^2 + x + 1}}}_{=: I_2}.$$
Note that $I_1$ can be easily solved with the substitution $u = x^2 + x + 1$.
For $I_2$ write
$$I_2 = \int \frac{ \mathrm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$
Solve system of congruences
$$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$
Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? ... | $\left.\begin{align}\rm Squaring &\ \ k^{\large 3}\equiv -l^{\large 3}\,\Rightarrow\, k^{\large 6}\,\equiv\, l^{\large 6}\\
\rm\&\ cubing &\ \ k^{\large 2}\equiv -l^{\large 2}\,\Rightarrow\, k^{\large 6}\!\equiv -l^{\large 6}\end{align}\right\} $ $\Rightarrow\, 2l^{\large 6}\equiv 0\,\Rightarrow\, l\equiv 0\ $ (in an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find K such that the line is tangent to $y=-\frac{3}{4}x+4$ The function is $f(x)= \frac{k}{x}$ and the line is $y =-\frac{3}{4}x+3$
So the derivative of $f(x)$ should equal $-\frac{3}{4}$
If I take $k$ to be a constant I can remove it from the fraction like so
$$f'(x)=k \frac{d}{dx} \frac{1}{x}$$
to get $-\frac{k}{x^... | From the given equation we obtain
$$-\frac{3}{4}x+3=\frac{k}{x}$$ so we get
$$-\frac{3}{4}x^2+3x-k=0$$ solving for $x$ we get
$$x=2\pm \frac{2}{3}\sqrt{9-3k}$$ since the line is a tangent line, we get
$$9-3k=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2... | $$\tan\frac{2\pi}{37}+4\left(-\sin\frac{4\pi}{37}+\sin\frac{14\pi}{37}+\sin\frac{16\pi}{37}+\sin\frac{24\pi}{37}-\sin\frac{28\pi}{37}+\sin\frac{32\pi}{37}-\sin\frac{36\pi}{37}\right)=\sqrt{37+6\sqrt{37}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
How would the most general $2 \times 2$ normal matrix look like?
How would the most general $2 \times 2$ normal matrix look like?
The normal matrix satisfy equation: $A^*A=AA^*$ where $A^*$ denotes
conjugate transpose.
I was thinking about the matrix:
$$
\begin{pmatrix}
a & -b \\
b & a \\
\end{pma... | Let $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$ For $A$ to be normal we require that $$\begin{bmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \overline{a} & \overline{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Summation closed form I have the following sum,
$$\sum_{j=0}^{\lfloor\frac{i+n-1}{n+2}\rfloor}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-1}{n-1}+\sum_{j=0}^{\lfloor\frac{i+n-2}{n+2}\rfloor}2(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-2}{n-1}+\sum_{j=0}^{\lfloor\frac{i+n-3}{n+2}\rfloor}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-3}{n-1... | Note that we can better write each term as
$$
\eqalign{
& S(k,n)=\sum\limits_{j = 0}^{\left\lfloor {{{i + n - k} \over {n + 2}}} \right\rfloor } {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
n - 1 \cr} \right)} = \cr
& = \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Area of a Polygon in a Polygon I'm dealing with a regular polygon with 7 corners. In this polygon is another polygon defined by connecting one point with the two opposite points of the same polygon.
I made a small sketch of the polygons
I need to calculate the ratio of the area of the big and of small polygon. I know... | Easy to see that $P_1Q_2P_3P_2$ and $P_1P_7Q_1P_2$ are parallelograms, which says that
$$P_1P_7=Q_1P_2=P_1P_2=P_1Q_2=P_2P_3.$$
Now, let $P_1P_4\cap P_6P_2=\{A\}$.
Thus, from $\Delta Q_1Q_2A$ we obtain:
$$Q_1Q_2=2AQ_2\sin\frac{\measuredangle Q_1AQ_2}{2}=2AQ_2\sin\frac{3\pi}{14}.$$
Also, from $\Delta P_1P_2A$ we obtain:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $3 \le \sum_{cyc}a\sqrt{b^3 + 1} \le \sum_{cyc}ab^2 + 3$ where $a, b, c \ge 0$ and $a + b + c = 3$.
$a$, $b$ and $c$ are non-negatives such that $a + b + c = 3$. Prove that $$\large 3 \le a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \le \frac{ab^2 + bc^2 + ca^2}{2} + 3$$
This problem is adapted from... | We have that $a, b, c \ge 0 \iff \sqrt{a^3 + 1} - 1, \sqrt{b^3 + 1} - 1, \sqrt{c^3 + 1} - 1 \ge 0$
$$\implies c\left(\sqrt{a^3 + 1} - 1\right) + a\left(\sqrt{b^3 + 1} - 1\right) + b\left(\sqrt{c^3 + 1} - 1\right) \ge 0$$
$$\iff a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \ge a + b + c = 3$$
The equality sign oc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
I have provided a (dumbfounding) solution down be... | We have that $\left\{ \begin{align} 5 &\mid x^2 - 2xy - y\\ 5 &\mid xy - 2y^2 - x \end{align} \right.$$\iff \left\{ \begin{align} 5 &\mid (x^2 - 2xy - y) + (xy - 2y^2 - x)\\ 5 &\mid (x^2 - 2xy - y) - (xy - 2y^2 - x) \end{align} \right.$
$\iff \left\{ \begin{align} 5 &\mid (x + y)(x - 2y - 1)\\ 5 &\mid (x - y)(x - 2y + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
| It's clear that the inequality
$$
\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\ge 0
$$holds for a triple $(a,b,c)$ if and only if it holds for the triple $(ta,tb,tc)$, for any $t > 0$.
Since $a,b,c > 0$, we can assume, by an appropriate scaling, that$\;a+b+c=1$.
With that assumption, it su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find $n \in \mathbb{Z}$ that maximizes $ (n+10)/(n^3+10)$ $(n+10)/(n^3+10)=(n+10)/(n^3+1000-990)$
I have thought about something like this. But am not sure about how to proceed with this. So I would like some help.
| You have that for $n=1$ that $11/11=1$ so you want to find $$\frac{n+10}{n^3+10}> 1$$
Otherwise the maximum value is $1$.
If $n^3+10>0$ we can multiply both sides of inequality to get
$$n+10> n^3+10\\n> n^3\\0> n(n-1)(n+1)$$
This happens when $n< -1$ or $0< n< 1$ The only integer satisfying $n^3+10>0$ and $n< -1$ is $-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the definite integral $\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx$ Writing:
Integrate[ArcTan[(a Cos[x] + b Sin[x])^2], {x, 0, 2 Pi}, Assumptions -> a^2 + b^2 > 0]
$$\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx,$$
where $a $ and $b $ are real numbers.
I get:
2 Pi ArcTan[Sqrt[1/2 (... | *
*We have $a\cos x+b\sin x=r\cos(x-\phi)$, where $$r=\sqrt{a^2+b^2},\quad\cos\phi=a/r,\quad\sin\phi=b/r,$$ and we can simply replace $x-\phi$ by $x$ in the integrand (because of its $2\pi$-periodicity). Denoting $c=(a^2+b^2)/2$, we see that the given integral is equal to $$\int_{0}^{2\pi}\arctan(2c\cos^2 x)\,dx=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, becaus... | First of all, we should check for $x \ge \frac{1}{2}$ and $x < \frac{1}{2}$ because in this step:
$$\sqrt{x^2+1}> 2x-1$$
we are squaring both sides. So sign of $2x-1$ may change the direction of inequality (precisely when $|2x-1| > \sqrt{x^2+1}$).
Now, for the case where $x < \frac{1}{2}$, we also have a solution for
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$ The sum of series is
$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$
I used to solve this problem as its $n$th term
$${1\over (2n+1)(2n+3)}$$
Now how can I proceed??
| This is a telescoping series:
$$
\frac{1}{(2n+1)(2n+3)}=\frac{1}{2}\frac{(2n+3)-(2n+1)}{(2n+1)(2n+3)}=\frac{1}{2}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)
$$
So:
$$
\sum_{n=1}^N\frac{1}{(2n+1)(2n+3)}=\frac{1}{2}\sum_{n=0}^N\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)=\frac{1}{2}\left(\frac{1}{2\cdot 1+1}-\frac{1}{2N+3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^... | Instead of comparing $\frac{1}{(n+1)^2} + \frac{1}{n+1}$ and $\frac{1}{n}$, we can compare $\frac{1}{(n+1)^2}$ and $\frac{1}{n}-\frac{1}{n+1}$. Then, what we have is
$$\frac{1}{(n+1)^2} = \frac{1}{(n+1)(n+1)} < \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \implies \frac{1}{(n+1)^2} < \frac{1}{n}-\frac{1}{n+1}$$
$$\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Help calculating a sum with exponential function So I have a sum to calculate:
$$\sum_{n=1}^{\infty} \frac{n-2}{n2^{n-1}+2^{n+1}}$$
So I first did this:
$$\sum_{n=1}^{\infty} \frac{n-2}{2^{n-1}(n+4)}$$
So first i thought to make this into a function of some sort and use derivation and integration to make it into someth... | Let's see if we can split this up:
$$\dfrac{n-2}{2^{n-1}(n+4)} = \dfrac{(n+4)-6}{2^{n-1}(n+4)} = \dfrac{1}{2^{n-1}}-\dfrac{6}{2^{n-1}(n+4)}$$
So, this sum becomes:
$$\sum_{n\ge 1} \dfrac{n-2}{2^{n-1}(n+4)} = 2-\sum_{n\ge 1} \dfrac{6}{2^{n-1}(n+4)}$$
Now, let's work with this summation:
$$\sum_{n\ge 1} \dfrac{6}{2^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the n... | Adapting the Extended Euclidean Algorithm, as implemented in this answer, to polynomial division and rotating to go down the page rather than across:
$$
\begin{array}{c|cc|c}
\color{#C00}{x^2}&\color{#C00}{1}&0\\
\color{#090}{x^2-2x+1}&0&\color{#090}{1}\\
2x-1&1&-1&1\\
\color{#C90}{\frac14}&\color{#C00}{-\frac12x+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation I'm trying to calculate the improper Fresnel integral $\int\limits_0^\infty\sin(x^2)dx$ calculation.
It uses several substitutions. There's one substitution that is not clear for me.
I could not understand how to get the right side from the left one. What ... | There's a neat trick to evaluate the integral
$$S_n(t)=\int_0^\infty \sin(tx^n)dx.$$
First, take the Laplace transform:
$$\begin{align}
\mathcal{L}\{S_n(t)\}(s)&=\int_0^\infty e^{-st}S_n(t)dt\\
&=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dxdt\\
&=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dtdx\\
&=\int_0^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Compute $\lim_{(x,y) \to (0,0)} \frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2}$
Does the limit $$\lim_{(x,y) \to (0,0)} \left( \frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2} \right) $$ exist?
I think it does and it's equal to $1$, but I don't know how to prove it. I tried to use Taylor expansion of $\cos(x)$ and $\cos(... | Note that, since $\cos(t)=1-\frac{t^2}{2}+O(t^4)$ as $t\to 0$, we have that
$$\frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2}=\frac{x^2 - 2+y^2+O(y^4) + 2}{y^2 - 2+x^2+O(x^4) + 2}=\frac{r^2+O(y^4)}{r^2+O(x^4)}=\frac{1+\frac{O(y^4)}{r^2}}{1+\frac{O(x^4)}{r^2}}$$
where $r^2=x^2+y^2$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute $ \int_0^1\frac{\ln^2(1+x)}{1+x^2}\, dx$ How to prove
$$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$
Where $ \operatorname{Li}_3(x)$ is the the trilogarithm function and $G$ is the Catalan constant.
Variant approaches are appreciated.
| I proved here in Eq $(1)$:
\begin{align}
\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx&=\Im\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}+\overset{\text{IBP}}{\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}\ dx}\\
&=\Im\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}-\int_0^1\frac{\ln x\tan^{-1}x}{1+x}\ dx-\int_0^1\frac{\ln(1+x)\tan^{-1}x}{x}\ dx\\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$ I have to simplify the following expression:
$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$
Answer: $\sqrt{a+b}-\sqrt{a-b}$
I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)
| You can write $$\frac{(a+b)^2-(a^2-b^2)}{\sqrt{a+b}(a+b+\sqrt{a^2+b^2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx$ using $\sinh x$
How to solve definite Integral
$$\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$
with hyperbolic function $\sinh(x)$?
Below is using $\tan x$ to do it. I thought it is same with this because $\sinh^2(x)+1=\cosh^... | \begin{align}
&\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}\overset{x\to 1/x}{dx}\\
=&\ \frac12\int_0^\infty \frac{1+x}{\sqrt{x^2+1}(x^2+x+1)} \overset{x=\sinh t}{dx}\\
=& \ \frac12\int_0^\infty \frac{1+\sinh t}{\cosh^2 t+\sinh t}dt
= \frac12 \int_0^\infty
\frac{\text{sech}\,t(\text{sech}\,t +\tanh t)}{1+\text{sech}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the vector $x ∈ \mathbb{R^3}$ that achieves $max||x||_1$ subject to $||x||_2 = 1$? I'm trying to answer the questions "What is the vector $x ∈ \mathbb{R^3}$
that achieves $max||x||_1$ subject to $||x||_2 = 1$?" and "What is the vector x ∈ $R^3$ that achieves $max||x||_∞$ subject to $||x||_2 = 1$?
I think the fi... | For the solution of your first question, you should also add the follwoing points: $\left(\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right), \left(\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right), \left(\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right), \left(-\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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On the Sum of Some of the Factors of $2310$
Given 3 natural numbers a,b,c such that $a\times b\times c=2310$. Find the sum of $\sum a+b+c$
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
So now since $a,b,c$ are symmetric , their sum should be the same , and I should get , $\sum a+b+c= ... | Consider $f(x)=2^5(x+\dfrac{2}{2})(x+\dfrac{3}{2})(x+\dfrac{5}{2})(x+\dfrac{7}{2})(x+\dfrac{11}{2})$ Your sum equals $3f(1)=3[2^5(1+\dfrac{2}{2})(1+\dfrac{3}{2})(1+\dfrac{5}{2})(1+\dfrac{7}{2})(1+\dfrac{11}{2})]=3(2+2)(2+3)(2+5)(2+7)(2+11)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Taylor series at infinity I'm required to make a Taylor series expansion of a function $f(x) = \arctan(x)$ at $x = +\infty$. In order to do this I introduce new variable $z = \frac{1}{x}$, so that $x \to +\infty$ is the same as $z \to +0$. Thus I can expand $f(z)$ at $z = 0$: $$f(z) = z - \frac{z^3}{3}+\frac{z^5}{5}-..... | Your expansion is not correct. You should find
$$f(z) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^2} - \frac{1}{5 x^5} + \mathcal{O}\left(\left(\frac{1}{x}\right)^7\right),$$
which is a perfectly good expansion in the small parameter $\frac{1}{x}$ for $x \rightarrow \infty$. See also Taylor expansion at infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Commutator subgroup of Heisenberg group. Dears,
Let $H$ be Heisenberg group, a group of $3\times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $\Bbb R$ above the main diagonal.
Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.
My question is -... | $$\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&-x&xz-y\\ 0&1&-z\\ 0&0&1\end{pmatrix}$$
Therefore \begin{align}ABA^{-1}B^{-1}&=\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&s&t\\ 0&1&u\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-x&xz-y\\ 0&1&-z\\ 0&0&1\end{pmatrix}\begin{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $\cot(A-B)=3$ and $\cot(A+B)=2$, then evaluate $\tan A + \frac{1}{\cot^2A}+\tan^3A+\frac{1}{\cot^4A}+\cdots$
Given:
$$\cot (A-B) = 3 \qquad \cot (A+B) = 2$$
Evaluate:
$$\tan A + \frac{1}{\cot^2 A} + \tan^3 A + \frac{1}{\cot^4 A} + \cdots $$
Attempts:
I think that the given information is used to get the value of $... | Let $\tan{A}=x$ and $\tan{B}=y$.
Thus, we obtain the following system.
$$\frac{x-y}{1+xy}=\frac{1}{3}$$ and
$$\frac{x+y}{1-xy}=\frac{1}{2}$$ or
$$3(x-y)=1+xy$$ and
$$2(x+y)=1-xy,$$ which gives
$$5x-y=2$$ or $$y=5x-2.$$
Now, since $$2(x+y)=1-xy,$$ we obtain:
$$2(x+5x-2)=1-x(5x-2)$$ or
$$12x-4=1-5x^2+2x$$ or
$$5x^2+10x-5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$ Prove the following for all real $x$
i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$
ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$
For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant,... | Let's denote with $\{ x \}$ the fractional part:
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}
$$
Then I would suggest that you first master the addition
$$
\eqalign{
& \left\lfloor {x + y} \right\rfloor = \left\lfloor {\left\lfloor x \right\rfloor + \left\{ x \right\} + \left\lfloor y \right\rfloor +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute the integral $\int\limits_0^1 \frac{3x}{\sqrt{4-3x^2}} dx $? I am struggling to compute the following equation.
\begin{equation}
\displaystyle\int_0^1 \dfrac{3x}{\sqrt{4-3x^2}} dx
\end{equation}
We are expected to use u-substitution, but I'm stuck and don't know how to proceed.
I don't know how to compute ... | In general, we would like to simplify the equation by use of u-substitution.
That being said, it would be smart to choose a value for $u$ that simplifies the equation to one that we know how to solve.
There are many options for $u$ in this case.
@Allawonder pointed out that $u=4-3x^{2}$ might be a good place to start, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t ... | I think if you take your linear map $\varphi$ and write it as a matrix in terms of the basis you provided, then it becomes
$$\begin{pmatrix}2 & -2&0 &0\\ -1 &1&1&-1\\-1&1&1&-1\\ 0&0&-2&2\end{pmatrix}$$
The column space of this linear map is spanned by two vectors, namely $(2,-1,-1,0)$ and $(0,1,1,-2)$. Rewriting these ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.