Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$?
Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$?
I can calculate the answer using a calculator, however I suspect to do so may be missing the point of the question.
The problem appears in a book immediately after a section called 'Rules for square roots'with $\sqrt{ab} = \sqrt{a}.\sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ as the given rules.
| We want to find the sign of $\sqrt{1001} - \sqrt{1000} - \dfrac{1}{10}$. Multiplying by the obviously positive $\sqrt{1001} + \sqrt{1000} + \dfrac{1}{10}$ we get $(\sqrt{1001})^2 - (\sqrt{1000} + \dfrac{1}{10})^2 = 1001 - 1000 - \dfrac{2\sqrt{1000}}{10} - \dfrac{1}{100} = \dfrac{99}{100} - \dfrac{2\sqrt{1000}}{10}$
Since $\sqrt{1000} > 30$ this quantity is obviously negative. So $\dfrac{1}{10}$ is larger.
| {
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"url": "https://math.stackexchange.com/questions/763272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
} |
Find the value of $\sin(B-A)$. If $A$ is an acute angle whose tangent is $\frac{15}{8}$ and $B$ is and obtuse angle whose sine is $\frac{12}{13}$, find $\sin (B-A)$.
[Without calculators]
I suppose I gotta use this formula: $\sin B \cos A - \cos B \sin A$
Before that they asked me to find $\tan 2A$ which is $\frac{-240}{161}$ and $\cos 2B$ which is $\frac{-119}{169}$
But I can't understand how to proceed....help please!
| Label a right triangle with one acute angle $A$. Since $\tan A = \dfrac{15}{8}$ you can label the opposite side with length $15$ and the adjacent side with length $8$. Then the hypotenuse has length $17$ so that $\sin A = \dfrac{15}{17}$ and $\cos A = \dfrac{8}{17}$.
For $B$, use $\sin^2 B + \cos^2 B = 1$. Since $B$ is obtuse, its cosine is negative. Thus $\cos B = - \dfrac{5}{13}$. Now use the formula for $\sin(B-A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Question about a sum Why is it that
$$\sum_{k=1}^n(n+k+1)(n+1)=\frac{3}{2}\sum_{k=1}^n3k^2+k.$$
I cannot understand it. This is not homework, I am just a little interested in this!
| The first sum can be written as$$\sum_{k=1}^n(n+k+1)(n+1)=(n+1)\sum_{k=1}^n(n+k+1)=(n+1)n(n+1)+(n+1)\sum_{k=1}^nk\\=n(n+1)^2+\frac{n(n+1)^2}{2}=\frac{3n(n+1)^2}{2}$$
instead the second one as
$$\frac{3}{2}\sum_{k=1}^n3k^2+k=\frac{3}{2}\left(\sum_{k=1}^n3k^2+\sum_{k=1}^nk\right)=\frac{3}{2}\left(3\sum_{k=1}^nk^2+\sum_{k=1}^nk\right)=\frac{3}{2}\left(3\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)=\frac{3}{2}\left(\frac{n(n+1)}{2}(2n+2)\right)=\frac{3}{2}\left(\frac{n(n+1)}{2}2(n+1)\right)=\frac{3n(n+1)^2}{2}$$
which equals the first one!
| {
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"url": "https://math.stackexchange.com/questions/764579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Prove inequality $(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$ I am trying to prove the following inequality:
$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$
for all $a>0, b>0$.
Does anyone know how to prove it?
Thanks a lot in advance!
| Inequality that is another form of a well-known inequalities:
For $0<y<x$
$$ \sqrt{xy}<\frac{x-y}{\ln x-\ln y} < \frac{x+y}{2}. (1)$$
For $a>b$ inequality of enunciation transform so:
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(a-b)(\ln a-\ln b)<=>$$
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})(\ln a-\ln b)<=>$$
$$\sqrt{a} - \sqrt{b} < \frac{1}{2}(\sqrt{a}+\sqrt{b})(\ln \sqrt{a}-\ln \sqrt{b})<=>$$
$$\frac{\sqrt{a} - \sqrt{b} }{\ln \sqrt{a}-\ln \sqrt{b}}< \frac{\sqrt{a}+\sqrt{b}}{2}$$
which is obtained from (1) for $x=\sqrt{a}, y=\sqrt{b}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
} |
non-homogeneous recurrence relations The question is:
$a_{n}=12a_{n-2}+16a_{n-3}+9\cdot 4^{n}+81n$with $a_{0}=0 ,a_{1}=1 ,a_{2}=98$
I tried to deal with the particular solution first by:
$A4^{n}+Bn=12[A4^{n-2}+B(n-2)]+16[A4^{n-3}+B(n-3)]+9\cdot 4^{n}+81n$.
However, turns out the coefficient of $4^{n}$being 0 which means 9=0 which is clearly wrong.
Could anyone please show me where did I do wrong in the working? Many thanks.
| Go for generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, rewrite the recurrence without index subtractions:
$$
a_{n + 3} = 12 a_{n + 1} + 16 a_n + 576 \cdot 4^n + 81 n + 243
$$
Multiply by $z^n$, sum over $n \ge 0$, and recognize a few sums to get:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
= 12 \frac{A(z) - a_0}{z} + 16 A(z)
+ 576 \frac{1}{1 - 4 z}
+ 81 \frac{z}{(1 - z)^2}
+ 243 \frac{1}{1 - z}
$$
This gives the formidable:
\begin{align}
A(z)
&= \frac{z + 92 z^2 + 240 z^3 - 1408 z^4 + 832 z^5}
{1 - 6 z - 3 z^2 + 52 z^3 - 12 z^4 - 96 z^5 + 64 z^6} \\
&= \frac{14}{1 + 2 z} - \frac{6}{(1 + 2 z)^2}
- \frac{5}{1 - z}
+ \frac{3}{(1 - z)^3}
- \frac{4}{1 - 4 z}
+ \frac{4}{(1 - 4 z)^2}
\end{align}
Geometric series and the generalized binomial theorem, in particular:
$$
\binom{-m}{k} = (-1)^k \binom{m + k - 1}{m - 1}
$$
finish this off:
\begin{align}
a_n
&= 14 \cdot (-2)^n
- 6 \cdot \binom{n + 1}{2} \cdot (-2)^n
- 5
+ 3 \cdot \binom{n + 2}{2}
- 4 \cdot 4^n
+ 4 \cdot \binom{n + 1}{1} \cdot 4^n \\
&= (8 - 6 n) \cdot (-2)^n
+ \frac{2 n^2 + 9 n - 4}{2}
+ n \cdot 4^{n + 1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION.
This is from my other thread click here to view it
Everything you see below is me finding the axis of rotation by solving $Kv=v$. Just to show you how much working it requires:
Noting that the axis of rotation consists of vectors that remain unmoved. That is a vector $v$ satisfying $Kv = v$. Or, $Kv - Iv=0$ where $I$ is the $3\times3$ identity matrix. For matrix $K$ after solving the homogeneous equations given by $(K-I)v=0$ and showing the working:
$(K-I)v=0$
So
$$K-I=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}-\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}=\
\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}$$
therefore
$$\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}v=0$$
writing out the components for $v$ gives
$$\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=0$$
Multiplying out gives three equations
$-x+z=0$
$-x-y=0$
$-y-z=0$
Since
$$
v=\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
Here's the solution parametrically in terms of $x$
\begin{align*}
z&= x\\
y&=-x\\
x&=x
\end{align*}
Hence the axis of rotation is given by the line
$$
\begin{bmatrix}
x\\-x\\x
\end{bmatrix}=x\begin{bmatrix}1\\-1\\1\end{bmatrix}\quad x\in\Bbb R
$$
That is, the axis of rotation is
$$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
As you can see this was a lot of work so i would be so grateful if someone could please explain in simple english how to get the answer: $$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
by using Inspection?
Many thanks to all that helped so far particularly Brian Fitzpatrick in the last thread
| Edit. For any $3\times3$ rotation matrix $K$, if it is not symmetric, you can read off the rotation axis directly from its skew-symmetric part $W=K-K^T$. More specifically, the rotation axis is parallel to $(w_{32}, w_{13}, w_{21})^T=-(w_{23}, w_{31}, w_{12})^T$.
Reason: any anticlockwise rotation for an angle $\theta$ about a unit vector $\mathbf u=(x,y,z)^T$ can be put into axis-angle form:
$$
K=\begin{bmatrix}
\cos\theta + x^2(1-\cos\theta) &xy(1-\cos\theta) - z\sin\theta &xz(1-\cos\theta) + y\sin\theta\\
yx(1-\cos\theta) + z\sin\theta &\cos\theta + y^2(1-\cos\theta) & yz(1-\cos\theta) - x\sin\theta\\
zx(1-\cos\theta) - y\sin\theta &zy(1-\cos\theta) + x\sin\theta & \cos\theta + z^2(1-\cos\theta)
\end{bmatrix}.
$$
When it is not symmetric, $\sin\theta\ne0$. Hence the skew-symmetric part of $K$ is given by
$$
W=K-K^T=2\sin\theta\,\begin{bmatrix}
0&-z&y\\
z&0&-x\\
-y&x&0
\end{bmatrix}.
$$
Thus $(w_{32}, w_{13}, w_{21})=2(\sin\theta)(x,y,z)=-(w_{23}, w_{31}, w_{12})$.
In your case,
$$
W = K-K^T =
\begin{pmatrix}
\ast & \ast & 1\\
-1 & \ast & \ast\\
\ast & -1 & \ast
\end{pmatrix}.
$$
Therefore the rotation axis is parallel to $(-1,1,-1)^T$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Definite integral problem I was solving a definite integral problem which was reduced to :
$$\int^{1}_{0} \frac{\ln(1+t)}{t} dt$$
I couldn't solve it and when I saw the solution, the answer was simply given as $\frac{\pi^2}{12}$, and claimed that this is an identity.
Can anybody give me a proof of this identity?
| \begin{align*}
\int_0^1 \frac{\log(x+1)}{x} \, dx &= \int_0^1 \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{n-1}}{n}dx\\
&= \sum_{n=1}^\infty (-1)^{n+1}\int_0^1 \frac{x^{n-1}}{n}dx\\
&= \sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^2}\\
&= \frac{\pi^2}{12}\\
\end{align*}
To calculate that sum, let us assume that the value of the following series
$$S_n = 1 + \frac 1 {2^2} + \frac 1 {3^2} + \dots = \frac{\pi^2}{6}$$
Now if we consider only the even values,
\begin{align*}
S_{2n} &= \frac 1 {2^2} + \frac 1 {4^2} + \frac 1 {6^2} + \dots \\
&= \frac{1}{2^2 \cdot 1} + \frac 1 {2^2\cdot 2^2} + \frac{1}{2^2 \cdot 3^2} + \dots \\
&= \frac{1}{4} S_n\\
&= \frac{\pi^2}{24}\\
\end{align*}
To get the value of our series, we take $S_n - 2 S_{2n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\sec \psi}{\tan \psi+1+\sec \psi}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi - 1}{\tan \psi+\sec \psi+1}d\psi$
Now Multiply both Numerator and Denominator by $\left(\tan \psi+\sec \psi-1\right)$
So $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi+1} \times \frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi-1}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\left(\tan \psi+\sec \psi-1\right)^2}{\left(\tan \psi+\sec \psi\right)^2-1}d\psi$
Now I did not understand how can i solve after that
Help required
Thanks.
| Solution:
Let $$t=x+1+\sqrt{1+x^2}\to 1+x^2=(t-1-x)^2\to x=\frac{t^2-2t}{2t-2}\to dx=\frac{t^2-2t+2}{2(t-1)^2}dt$$
We have $$x-1+\sqrt{1+x^2}=t-2$$
And $$\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}=\frac{t-2}{t}$$
Hence $$I=\int_{0}^{1}\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}dx=\int_{2}^{2+\sqrt{2}}\frac{t-2}{t}\frac{t^2-2t+2}{2(t-1)^2}dt$$
$$=\int_{2}^{2+\sqrt{2}}\left[\frac{1}{2}-\frac{2}{t}+\frac{1}{t-1}-\frac{1}{2(t-1)^2}\right]dt=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/768387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
| Hint $\ {\rm mod}\ 3\!:\ x\equiv 2\equiv -1\,\Rightarrow\, 1+2x+4x^2\equiv 1-2+4\equiv 3\equiv 0.\,$
Generally for polynomial $\ p(x),\,\ \ p(-1) = p_0-p_1+p_2-p_3+\cdots = $ alternating coef sum.
Alternatively $\,\color{#c00}{4\equiv1}\,\Rightarrow\, \color{#c00}{4x^2}\!+2x+1\equiv \color{#c00}{x^2}\!+2x+1\equiv (x\!+\!1)^2\,$ which has root $\,x\equiv-1\equiv 2.$
There are also many other ways to proceed, and all provide good exercise in modular arithmetic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find equation of a circle given a sector A sector of a semi-circle which is $60^\circ$ has an area of $\frac{3\pi}{2}$ units squared.
Show that the curved section is a function of the form $f(x)= \sqrt{9-x^2}$ with domain $[0,\frac{3\sqrt{3}}{2}]$
| Area of sector
\begin{align}
A&=\frac{60^\circ}{360^\circ}\pi r^2\\
\frac{3\pi}{2}&=\frac{1}{6}\pi r^2\\
r&=3
\end{align}
Equation of circle with center on the origin
\begin{align}
x^2+y^2&=r^2\\
x^2+y^2&=3^2\\
y&=\sqrt{9-x^2}
\end{align}
then use parametric equation (polar coordinate) $x=r\cos\theta$ and $y=r\sin\theta$ to determine the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factoring in terms of Irreducibles Factor the polynomial $x^5 + 2x^3 + 3x^2 + 1$ as a product of irreducible polynomials in $\mathbb{Z}_5[x]$.
My thoughts: I know what the definition of an irreducible function is but as far as methods to find the correct answer as easy as possible is difficult to understand.
| You found the linear factors already, leaving the quartic $x^4 + 3x^3 + x^2 + x + 3$. The only possibility is that this factors into two quadratics, thus:
$$
(x^2+ax+b)(x^2+cx+d)\equiv x^4 + 3x^3 + x^2 + x + 3\pmod5
$$
which you can expand as
$$
x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd\equiv x^4 + 3x^3 + x^2 + x + 3\pmod5
$$
which you can solve as 5 simultaneous equations.
The equations are:
$$1\equiv1,\ a+c\equiv3,\ ac+b+d\equiv1,\ ad+bc\equiv1,\ bd\equiv3$$
You can drop the first one, which is already solved:
$$a+c\equiv3,\ ac+b+d\equiv1,\ ad+bc\equiv1,\ bd\equiv3$$
Now there are only two ways to get 3 mod 5 as a product of two numbers: 1 times 3 or 2 times 4 (in either order, of course). Since the problem is symmetrical with respect to switching (a,b) and (c,d), you can assume without loss of generality that $b\le d$ if you like, so there are two possibilities: $b=1,d=3$ and $b=2,d=4$. The first:
$$a+c\equiv3,\ ac+4\equiv1,\ 3a+c\equiv1,\ 3\equiv3$$
and the second:
$$a+c\equiv3,\ ac+1\equiv1,\ 4a+2c\equiv1,\ 3\equiv3$$
Simplifying both you get
$$a+c\equiv3,\ ac\equiv2,\ 3a+c\equiv1$$
and
$$a+c\equiv3,\ ac\equiv0,\ 4a+2c\equiv1$$
Solve both, using $c\equiv3-a$, and see if you can find any solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$? How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$?
I tried by using integration by parts, but always got sucked. Should be very easy...
| Let $u=\sqrt{x}$ and $x=u^2$, $\frac{\sqrt{x}}{1-\sqrt{x}}=-1+\frac{1}{1-\sqrt{x}}=-1+\frac{1}{1-u}$, then
\begin{align}
\int{\frac{\sqrt{x}}{1-\sqrt{x}}dx}&=\int{\left(-1+\frac{1}{1-u}\right)(2udu)}\\
&=-\int{2udu}+2\int{\frac{u}{1-u}du} \\
&=-u^2+2\int{\left(-1+\frac{1}{1-u}\right)du} \\
&=-u^2-2u+2\ln|1-u|+C \\
&=-x-2\sqrt{x}+2\ln|1-\sqrt{x}|+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Non homogeneous Recurrence relation problem So here i have this non homogeneous recurrence relation i need to solve:
$$a_{n}=12a_{n-2}+16a_{n-3}+9\cdot 4^{n}+81n,$$
where $a_{0}=0$, $a_{1}=1$ $a_{2}=98$. I'm confused at the homogeneous relation part of this relation. What's the characteristic of this type of homogeneous relation? I am planning to solve this like solving a second order recurrence, but is there any simpler way to deal with this?
| Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence with no index subtractions:
$$
a_{n + 3}
= 12 a_{n + 1} + 16 a_n
+ 576 \cdot 4^n + 81 n + 243
$$
The resulting recurrence is now valid for $n \ge 0$ ( no terms for negative index allowed originally).
Multiply by $z^n$, sum over $n \ge 0$, recognize some sums:
\begin{align}
\sum_{n \ge 0} a_{n + r} z^n
&= \frac{A(z) - a_0 - a_1 z - \ldots - a_{r - 1} z^{r - 1}}{z^r} \\
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2}
\end{align}
and get:
$$
\frac{A(z) - z - 98 z^2}{z^3}
= 12 \frac{A(z)}{z} + 16 A(z)
+ 576 \frac{1}{1 - 4 z} + 81 \frac{z}{(1 - z)^2} + 243 \frac{1}{1 - z}
$$
Written as partial fractions:
$$
A(z)
= \frac{1}{(1 - 4 z)^2}
- \frac{1}{1 - 4 z}
+ \frac{6}{(1 + 2 z)^2}
+ \frac{14}{1 + 2 z}
- \frac{3}{(1 - z)^2}
- \frac{5}{1 - z}
$$
Use geometric series and the generalized binomial theorem to read off the coefficients:
$$
\binom{-m}{k} = (-1)^k \binom{m + k - 1}{m - 1}
$$
Note in particular that:
$$
\binom{-2}{n} = (-1)^n (n + 1)
$$
So:
\begin{align}
a_n
&= (n + 1) \cdot 4^n - 4^n
+ 6 (n + 1) \cdot (-2)^n + 14 \cdot (-2)^n
- 3 (n + 1) - 5 \\
&= n \cdot 4^n + (6 n + 20) \cdot (-2)^n - 3 n - 8
\end{align}
| {
"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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Finding the value of one-sided limits and greatest integer function.
$$
\lim_{x \to 0} \frac{a}{x} \left\lfloor\frac{x}{b} \right\rfloor
$$
The $\lfloor \rfloor$ stands for the greatest integer function.
I have calculated and the left-hand limit is coming as (ab). But, I have doubt in the right-hand limit. I did this problem by sandwich-theorem. Can, anyone help me to find the right-hand limit correctly?
| I assume that
$a > 0$
and $b > 0$.
If $0 < x < b$,
$\lfloor \frac{x}{b} \rfloor
= 0$.
Therefore,
for
$0 < x < b$,
$\frac{a}{x}\lfloor \frac{x}{b} \rfloor
= 0$.
Therefore,
$\lim_{x \to 0^{+}}\frac{a}{x}\lfloor \frac{x}{b} \rfloor
= 0$.
The expression is not defined at $x = 0$.
If $0 > x > -b$,
$\lfloor \frac{x}{b} \rfloor
= -1$.
Therefore,
for
$0 > x > -b$,
$\frac{a}{x}\lfloor \frac{x}{b} \rfloor
= -\frac{a}{x}$.
This is not defined
as $x \to 0^{-}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
finding square $k$ such that $k\mid 2^{k-1} -1$ Can we find a positive square integer $k>1$, which satisfies $k\mid 2^{k-1} -1$ ?
If yes, what are such $k>1$ values? Here $k = n^2$ and $n$ is some positive integer. If we cannot find such square integer $k>1$, then how to disprove the statement $k\mid 2^{k-1} -1$?
High regards and advanced thanks!
Richard Sieman
| Given that $n^2|(2^{n^2-1}-1)$.
Now it is obvious that $(2^r-1)|(2^{n^2-1}-1)$, where $r|(n^2-1)$, hence,
$2^r-1=n^2$ for some positive integer $r$. Now $r$ cannot be even, (otherwise, $2^{2k}-n^2=1$ for $r=2k$, and factorising yield $2^k-n=1$ and $2^k+n=1$ for which $n=0$, or $2^k-n=-1$ and $2^k+n=-1$ for which $n=1$) so let $r=2k+1$, hence, $2^r-1=n^2$ becomes; $2^{2k+1}-1=n^2$ or
$n^2-2(2^k)^2=-1$ which is Pell's equation for which all solutions are given by; $n+(2^k){\sqrt2} \equiv (1+{\sqrt2})^{2t+1}$ for all non-negative integers $t$, $r=(2k+1)|(n^2-1)$, and can be used to narrow your search
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving simultaneous congruences with the Chinese remainder theorem Solve the following system of congruences using the Chinese remainder theorem:
$$\begin{align*}
2x &\equiv 3 \pmod{7} \\
x &\equiv 4 \pmod{6} \\
5x &\equiv 50 \pmod{55}
\end{align*}
$$
I was a little confused how to reduce the congruences into a form where the Chinese remainder theorem is applicable.
| The congruence $2x\equiv 3\pmod{7}$ holds if and only if $x\equiv 5\pmod{7}$. And $5x\equiv 50\pmod{55}$ if and only if $x\equiv 10\pmod{11}$. Now the problem is in "standard" form.
Remark: To get the first result, we can multiply both sides of $2x\equiv 3\pmod{7}$ by the inverse of $2$ modulo $7$. Note that $4$ is the inverse to $2$. Or more simply we replace the $3$ by $10$, and divide by $2$.
The second result comes from the general fact that if $k\ne 0$, then $ka\equiv kb\pmod{km}$ if and only if $a\equiv b\pmod{m}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many ways can you build a football/soccer team? I'm reading a book on the history of soccer. There are several ways to position a team, for example, four defenders, four midfielders and two forwards is the common 4-4-2. How many ways can you build a team, given that there are 10 players that move because the keeper is always in the same position?
I made some schemes:
10
1-9
1-1-8
1-1-1-7
1-1-1-1-6
1-1-1-1-1-5
1-1-1-1-1-1-4
1-1 1-1-1-1-1-3
1-1-1-1-1-1-1-1-2
1-1-1-1-1-1-1-1-1-1
1-2-7
1-2-1-6
1-2-1-1-5
1-2-1-1-1-4
1-2-1-1-1-1-3
1-2-1-1-1-1-1-2
1-21-1-1-1-1-1-1
1-3-6
1-3-1-5
Then I thought of another scheme, one that would group all possible combinations of diagrams from the number of lines that are willing players on the court.
L1 10
L2 1-9,9-1,2-8,8-2,3-7,7-3,6-4,4-6,5-5
L3
1-1-8,1-8-1,8-1-1,1-2-7,1-7-2, 2-7-1,2-1-7,7-2-1, 7-1-2,1-3-6,1-6-3 .....
I can not find the formula to translate this, if it is correct.
In looking for an answer I found this site. Could you guide me to a book or article to solve my problem?
| You are looking for the number of compositions of 10, this is $2^{10 - 1} = 512$.
The proof of this fact is similar to stars-and-bars: To divide $n$ into $k$ nonempty pieces is to lay down $n$ stars, and place separators (bars) in $k - 1$ of the $n - 1$ positions between stars, i.e, $\binom{n - 1}{k - 1}$ ways. Adding over all possible numbers of divisions is:
$$
\sum_{1 \le k \le n} \binom{n - 1}{k - 1}
= \sum_{0 \le k \le n - 1} \binom{n - 1}{k}
= 2^{n - 1}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ but that didn't help me much . After this I obtained integrals of the form
$$
\int_0^1 \log \big[\big(x\pm\sqrt{1-x^2}\big)^2\big] \frac{x\, dx}{1-x^2}
$$
which I am not sure how to handle. Thanks.
| So here is a solution which follows the logic of both @gar's and @xpaul's but avoids the issue of poles.
Let $I(a)$ be defined by
$$ I(a) = \int_{0}^{1} \frac{2x}{1-x^{2}} \log \left| \frac{x + \sqrt{1-x^{2}}}{x + a\sqrt{1-x^{2}}} \right| \, dx. $$
The value we want to calculate is $I(-1)$. With the substitution $t = x/\sqrt{1-x^{2}}$ (this substitution is essentially the same as in @Felix's answer), it follows that
$$ I(a)
= \int_{0}^{\infty} \frac{2t}{1+t^{2}} \log \left| \frac{1+t}{a+t} \right| \, dt
= \Re \int_{0}^{\infty} \frac{2t}{1+t^{2}} \log \left( \frac{1+t}{a+t} \right) \, dt, $$
where the branch of logarithm is chosen to avoid the upper half-plane. Then by DCT,
$$ I(a)
= \lim_{\epsilon \downarrow 0} \Re \int_{i\epsilon}^{\infty+i\epsilon} \frac{2t}{1+t^{2}} \log \left( \frac{1+t}{a+t} \right) \, dt =: \lim_{\epsilon \downarrow 0} I_{\epsilon}(a). $$
Now, for $0 < \epsilon \ll 1$, we have
\begin{align*}
I_{\epsilon}'(a)
&= - \Re \int_{i\epsilon}^{\infty+i\epsilon} \frac{2t}{(1+t^{2})(t+a)} \, dt \\
&= - \frac{2a}{1+a^{2}} \Re \int_{i\epsilon}^{\infty+i\epsilon} \left( \frac{1}{a} \frac{1}{1+t^{2}} + \frac{t}{1+t^{2}} - \frac{1}{t + a} \right) \, dt \\
&= - \frac{2a}{1+a^{2}} \Re \left( \frac{\pi}{2a} - \frac{1}{a}\arctan(i\epsilon) + \log(a+i\epsilon) - \log\sqrt{1-\epsilon^{2}} \right) \\
&= - \frac{2a}{1+a^{2}} \left( \frac{\pi}{2a} + \log\sqrt{a^{2}+\epsilon^{2}} - \log\sqrt{1-\epsilon^{2}} \right).
\end{align*}
This gives, together with the initial condition $I_{\epsilon}(1) = 0$, that
$$ I_{\epsilon}(a) = \int_{a}^{1} \frac{2s}{1+s^{2}} \left( \frac{\pi}{2s} + \log\sqrt{s^{2}+\epsilon^{2}} - \log\sqrt{1-\epsilon^{2}} \right) \, ds. $$
Taking $\epsilon \downarrow$, again by DCT, we get
$$ I(a) = \int_{a}^{1} \frac{\pi + 2s\log|s|}{1+s^{2}} \, ds. $$
Plugging $a = -1$ we finally obtain the desired answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question. Part (a) was the following:
Write the sum: $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2)$ using summation notation.
It was simple enough : $\sum k(k+2)$.
For this question, the base case $(n=1)$ holds, as $1\cdot(1+2) = 3 = (1\cdot2\cdot9)/6$.
Induction step: Assume the above holds for all $n = k$, prove that it holds for all $n = k+1$
I'm a bit lost from here, help?
| Note that
$$
k(k + 2) = k^2 + 2k = k^2 - k + 3k = 2{k \choose 2} + 3{k \choose 1}
$$
Definition 1: Let $\{x_n\}$ a sequence. $\Delta x_n = x_{n+1} - x_n$.
Definition 2: If $\Delta X_n = x_n$, then $X_n = \Delta^{-1}x_n + C$ with $C \in \mathbb{R}$.
Proposition 1: $\Delta^{-1}$ is linear and
$$
\Delta \biggl(\sum_{k =n_0}^{n-1}x_k \biggr) = x_n \quad \text{and} \quad \Delta {n \choose p} = {n \choose p - 1}
$$
Proof. Exercise.
Proposition 2: $\sum_{k =m}^{n-1}x_k = \Delta^{-1}x_n\biggl|_{m}^{n} = X_n - X_m$.
Proof. Exercise.
Proposition 3: $\Delta^{-1}{n \choose p} = {n \choose p+1} + C$.
Proof. Exercise.
Thus,
$$
\sum_{k=1}^{n}k(k+2) = \sum_{k=1}^{n}\biggl[2{k \choose 2} + 3{k \choose 1}\biggr]
= \biggl[2{k \choose 3} + 3{k \choose 2}\biggr]_{1}^{n+1}
$$
$$
= 2{n+1 \choose 3} + 3{n+1 \choose 2} = \dfrac{1}{3}(n+1)n(n-1) + \dfrac{3}{2}(n+1)n
$$
$$
=n(n+1)\biggl[\dfrac{n}{3} - \dfrac{1}{3} + \dfrac{3}{2}\biggr] = \dfrac{n(n+1)(2n+7)}{6}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding the image of a disk Find the image of the disk $D= \{ z \in \mathbb{C}: |z|<1 \}$ under the mapping $z \to w=\frac{1}{z-1}$.
My attempt:
First I graphed the boundary.
When $z=i, f(z)=-i/2-1/2$
When $z=-i, f(z)=i/2-1/2$
When $z=-1, f(z)=-1/2$
When $z=1, f(z)$ is undefined.
Then I graphed a few other points, $z=-1,0,\frac{1}{2}$ and saw that everything is being mapped to the left hand side of the the vertical line $x<-\frac{1}{2}$, is this statement true?
| Say, $z=x+iy$, then $f(z)=f(x+iy)$ becomes $$\frac{x-1}{(x-1)^2+y^2}-\frac{i y}{(x-1)^2+y^2}$$
Note that in $D$, $x$ and $y$ are between $0$ and $1$.
Now considering the real part we have $(x-1)^2+y^2 = x^2 +1 -2x +y^2 < 2-2x$. It follows $$\frac{1}{x^2 +1 -2x +y^2} > \frac{1}{2-2x}$$ which implies $$\frac{x-1}{x^2 +1 -2x +y^2} < \frac{x-1}{2-2x} $$ since $x-1$ is negative. It now follows that the real part is less than $ \displaystyle -\frac {1} {2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How did my book simplify this? How did my book go from:
$\frac{4}{5}=\frac{x}{30}$
to
$\frac{4}{1}=\frac{x}{6}$
I understand that I could have cross multiplied it in the first place but what I don't understand is why my book changes the denominators without changing the numerators when simplifying the problem.
| You can rewrite 30 as $5\times 6$:
$$\frac{4}{5}=\frac{x}{5\times 6}\;\; \text{ multiply both sides by 5 } \;\;5\times \frac{4}{5}=5\times \frac{x}{5 \times 6}$$
if you then do some rearrangement you can get:
$$ \frac{5}{5}\times \frac{4}{1}=\frac{5}{5}\times \frac{x}{6}$$
Then replacing $5/5$ with 1 gives your result:
$$ \frac{4}{1}=\frac{x}{6} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve $\sin(2\theta) -\tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi $ I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$
to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$
My solution:
$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $
so $\frac{2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)}{1 + \tan^2(\theta)} = 0$
so $ 2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)= 0$
$ \implies \tan^3(\theta) + \tan(\theta) = 0$ $\implies \tan(\theta) [\tan^2(\theta) + 1] = 0$ $\implies \tan(\theta) = 0 \;\textrm{or}\; \tan^2(\theta) + 1 =0 $ since $\theta$ must be real.
Then we solve $\tan(\theta) = 0$ $\implies$ $\theta = n\pi, \ \ $ $ n \in Z \ \ $ so $\theta = \pi,2\pi$
| You're missing a solution
$$ \sin(2x) = 2 \cos x \sin x = \tan x = \frac{\sin x}{\cos x} \implies \cos^2x = \frac{1}{2} $$
$$\therefore \cos x = \pm \frac{1}{\sqrt{2}} \quad or \quad \sin x = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
calculating the sum of a series I was wondering if there is a way to use complex-analysis to solve the sum of the following series? (just like you can use it to solve integrals of some kinds (using fourier's transform for integrals).
$$\sum^{\infty}_{n=0}\frac{1}{n^2+25}$$
and
$$\sum^{\infty}_{n=0}\frac{(-1)^n}{n^2+25}$$
And if not, how to solve them otherwise?
thanks in advance!
| Solution via Fourier's series:
Note that (see here)
$$
\int \cos {nx} ~ \cosh {bx} dx = \dfrac{1}{n^2+b^2}
\left(n \sin nx \cosh bx + b \cos nx \sinh bx \right) + C
$$
(form is very close to given $\pm\dfrac{1}{n^2+25}$).
Now consider function
$$
f(x)=\cosh {5x}, \qquad x\in [-\pi,\pi];
$$
and make it $2\pi$-periodic (so, $f(x)$ is continuous function).
Fourier series of this even function will have form
$$
f(x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{nx},
$$
where
coefficients $a_n$ are:
$$
a_n = \dfrac{2}{\pi}\int_{0}^{\pi} \cosh {5x} ~ \cos nx \; dx
=
\dfrac{10 \cos {n\pi} \sinh{5\pi}}{\pi(n^2+25)} = \dfrac{10\sinh{5\pi}}{\pi} \cdot \dfrac{(-1)^n }{n^2+25}.
$$
As $f(x)$ is continuous, then
$$
\dfrac{a_0}{2}+f(0)
= \sum\limits_{n=0}^{\infty} a_n \cos (n \cdot 0)
= \sum\limits_{n=0}^{\infty} a_n
= \dfrac{10\sinh{5\pi}}{\pi} \sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{n^2+25},
$$
$$
\dfrac{a_0}{2} + f(\pi)
= \sum\limits_{n=0}^{\infty} a_n \cos n\pi
= \sum\limits_{n=0}^{\infty}(-1)^n a_n
= \dfrac{10\sinh{5\pi}}{\pi} \sum\limits_{n=0}^{\infty} \dfrac{1}{n^2+25}.
$$
Note that
$~~\dfrac{a_0}{2}+f(0) = \dfrac{\sinh 5\pi}{5\pi}+1$,
$~~\dfrac{a_0}{2}+f(\pi) = \dfrac{\sinh 5\pi}{5\pi}+\cosh{5\pi}$,
so
$$
\sum_{n=0}^\infty \dfrac{(-1)^n}{n^2+25} = \dfrac{\pi\left(\frac{\sinh 5\pi}{5\pi}+1\right)}{10\sinh 5\pi}
= \dfrac{1}{50}+\dfrac{\pi}{10}\dfrac{1}{\sinh 5\pi},
$$
$$
\sum_{n=0}^\infty \dfrac{1}{n^2+25} = \dfrac{\pi\left(\frac{\sinh 5\pi}{5\pi}+\cosh{5\pi}\right)}{10\sinh 5\pi}
= \dfrac{1}{50}+\dfrac{\pi}{10}\dfrac{\cosh 5\pi}{\sinh 5\pi}.
$$
| {
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"url": "https://math.stackexchange.com/questions/792817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the sum $\sum\limits_{n=1}^{\infty}\frac{1}{n!(n^4+n^2+1)}$ Find this sum
$$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}$$
My try: since
$$n^4+n^2+1=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$
then
$$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}=\sum_{n=1}^{\infty}\dfrac{1}{2n\cdot n!}\left(\dfrac{1}{n(n-1)+1}-\dfrac{1}{n(n+1)+1}\right)$$
then I can't go further.
| Let $S$ denote the sum to be evaluated. For every $n$,
$$
\frac2{n^4+n^2+1}=\frac{n+1}{n(n+1)+1}-\frac{n-1}{(n-1)n+1},
$$
hence
$$
2S=\sum_{n\geqslant1}\frac{n+1}{n(n+1)+1}\frac1{n!}-\sum_{k\geqslant1}\frac{k-1}{(k-1)k+1}\frac1{k!}.
$$
The RHS is almost a telescoping series. To wit, the change of variable $n=k-1$ in the last sum and the fact that its $k=1$ term is zero yield
$$
2S=\sum_{n\geqslant1}\frac{n+1}{n(n+1)+1}\frac1{n!}-\frac{n}{n(n+1)+1}\frac1{(n+1)!},
$$
that is,
$$
2S=\sum_{n\geqslant1}\frac{(n+1)^2-n}{n(n+1)+1}\frac1{(n+1)!}\stackrel{k=n+1}{=}\sum_{k\geqslant2}\frac1{k!}=\mathrm e-\frac1{0!}-\frac1{1!},
$$
and finally,
$$
S=\frac12\mathrm e-1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \log\left(\frac{1+ x}{1-x}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}
=\frac{\pi^2}{2}.
$$
I tried to write
$$
I=\int_0^1 \frac{\log(1+ax)}{x\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1-ax)}{x\sqrt{1-x^2}}dx
$$
If we work with one of these integrals we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1} a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx-\sum_{n=1}^\infty \frac{a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx,
$$
simplifying this I get an infinite sum of Gamma functions. which i'm not sure how to relate to the $\arcsin$ Thanks.
| It's easier to take the derivative of both sides according to $a$, than to perform the integration:
$$\int_0^1 \frac{2}{1-a^2x^2}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{\sqrt{1-a^2}}$$
The left hand side giving:
$$2 \text{arctanh}\left(\frac{\sqrt{a^2-1}}{\sqrt{1 - x^2}}x\right)\frac{1}{\sqrt{a^2-1}}+C$$
Which, when applying the limits, gives:
$$\frac{\pi}{\sqrt{1-a^2}}+C$$
as desired. Now all that needs to be done is compare at a single point to prove that the $C=0$, which you already have done.
| {
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How should I solve integrals of this type? The general form of the integral I want to solve is:
$$ \int e^{bx}\sin(ax) dx$$
Euler's formula has a nice connection, but the i makes it too complicated.
Doing it by parts doesn't seem to get me anywhere.
Do you have any tips for how to begin solving this?
| A third (or maybe fourth, depending on how you're counting) method (related to the one in David's answer) which uses a bit of linear algebra is as follows:
\begin{align*}
\left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right]' &= \left[
\begin{array}{c}
ae^{ax}\cos(bx) - be^{ax}\sin(bx)\\
ae^{ax}\sin(bx) + be^{ax}\cos(bx)
\end{array}
\right]\\
&= \left[
\begin{array}{cc}
a & -b\\
b & a
\end{array}
\right] \left[
\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}
\right]
\end{align*}
Using the adjoint method, and the fact that the integral is (up to a constant) the functional inverse of the derivative on the space of smooth functions, we have:
\begin{align*}
\int \left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right] \text{d}x &= \frac{1}{a^2+b^2} \left[
\begin{array}{cc}
a & b\\
-b & a
\end{array}
\right] \left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right] + \left[\begin{array}{c}
C_1\\
C_2
\end{array}\right]\\
&= \left[\begin{array}{c}
\frac{1}{a^2+b^2}\left(ae^{ax}\cos(bx) + be^{ax}\sin(bx)\right) + C_1\\
\frac{1}{a^2+b^2}\left(ae^{ax}\sin(bx)-be^{ax}\cos(bx)\right) + C_2
\end{array}\right]
\end{align*}
Hence we have:
\begin{align*}
\int e^{ax}\cos(bx) \,\text{d}x &= \frac{1}{a^2+b^2}\left(ae^{ax}\cos(bx) + be^{ax}\sin(bx)\right) + C\\
\int e^{ax}\sin(bx) \, \text{d}x &= \frac{1}{a^2+b^2}\left(ae^{ax}\sin(bx)-be^{ax}\cos(bx)\right) + C
\end{align*}
Why the downvote? This method works quite well computing integrals of this type.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$ Hi I am trying to prove this
$$
I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\,
\frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x
=-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\left(\alpha\right),\qquad \alpha^2<1.
$$
What an amazing result this is! I tried to write
$$
I=\int_0^{\pi/4} \log \sin x\frac{\cos 2x}{1+\alpha^2\sin^2 2x}-\int_0^{\pi/4}\log \cos x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx
$$
and played around enough here to realize it probably isn't the best idea.
Now back to the original integral I, we can possibly change variables $y=\tan x$ and
re-writing the original integral to obtain
$$
\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+{\alpha^2}\big(1-\cos^2 (2x)\big)}dx=\int_0^1 \log y \frac{1-y^2}{1+y^2}\frac{1}{1+{\alpha^2}\big(1-(\frac{1-y^2}{1+y^2})^2\big)}\frac{dy}{1+y^2}.
$$
Simplifying this we have
$$
I=\int_0^1\log y \frac{1-y^2}{1+y^2}\frac{(1+y^2)^2}{(1+y^2)^2+4\alpha^2y^2}\frac{dy}{1+y^2}=\int_0^1\log y \frac{1-y^2}{(1+y^2)^2+4\alpha^2y^2}dy
$$
Another change of variables $y=e^{-t}$ and we have
$$
I=-\int_0^\infty \frac{t(1-e^{-2t})}{(1+e^{-2t})^2+4\alpha^2 e^{-2t}} e^{-t}dt
$$
but this is where I am stuck...How can we calculate I? Thanks.
| Integrate by parts; then you get that
$$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$
The first term on the RHS is zero. To evaluate the integral, expand the arctan into a Taylor series and get
$$I(\alpha) = -\frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \alpha^{2 k} \int_0^{\pi/2} du \, \sin^{2 k}{u} = -\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k}$$
A little manipulation leads us to
$$\alpha I'(\alpha) +I(\alpha) = -\frac{\pi}{4} \sum_{k=0}^{\infty} (-1)^k \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k} = -\frac{\pi}{4} \frac1{\sqrt{1+\alpha^2}}$$
The LHS is just $[\alpha I(\alpha)]'$, so the solution is
$$I(\alpha) = -\frac{\pi}{4} \frac{\operatorname{arcsinh}(\alpha)}{\alpha} $$
| {
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} |
Please solve this recurrence relation question for $8a_na_{n+1}-16a_{n+1}+2a_n+5=0$ Suppose $a_1=1$ and $$8a_na_{n+1}-16a_{n+1}+2a_n+5=0,\forall n\geq1,$$Please help to sort out the general form of $a_n$.
Here are the first a few values of the series. Not sure if they are useful as they seem quite random to me.
$$1,{7\over8},{3\over4},{13\over20},{7\over12},\dots$$
Thanks.
| Let $p_n$, $q_n$ be two sequences to be determined such that $\displaystyle\;a_n = \frac{p_n}{q_n}\;$. In terms of $p_n$, $q_n$, the recurrence relation for $a_n$ can be rewritten as
$$a_{n+1} = \frac{2a_n+5}{-8a_n+16}
\quad\iff\quad\frac{p_{n+1}}{q_{n+1}} = \frac{2p_n+5q_n}{-8p_n+16q_n}\tag{*1}
$$
If we scale $p_n, q_n$ to make them satisfy the linear recurrence relation
$$\begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix}
= \begin{bmatrix}2 & 5\\-8 & 16\end{bmatrix} \begin{bmatrix}p_{n}\\q_{n}\end{bmatrix}
\tag{*2}
$$
then any solution of $(*2)$ will lead to a solution of $(*1)$. It is easy to check the square matrix in $(*2)$ has eigenvalues $6$ and $12$ with corresponding eigenvectors
$\begin{bmatrix}5\\4\end{bmatrix}$ and $\begin{bmatrix}1\\2\end{bmatrix}$. The general
solution of $(*2)$ has the form:
$$
\begin{bmatrix}p_{n}\\q_{n}\end{bmatrix}
= A \times 6^n \begin{bmatrix}5\\4\end{bmatrix} + B \times 12^n \begin{bmatrix}1\\2\end{bmatrix}$$
for suitable chosen constants $A$ and $B$. By inspection, we can reproduce $a_1 = 1$ by setting $A = \frac16$ and $B = \frac{1}{12}$.
This leads to a solution of the original recurrence relation subject to $a_1 = 1$:
$$a_n = \frac{p_n}{q_n} = \frac{5 \times 6^{n-1} + 12^{n-1}}{4 \times 6^{n-1} + 2\times 12^{n-1}} = \frac{2^{n-1} + 5}{2^n + 4}$$
| {
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Find all real solutions of $6^x+1=8^x-27^{x-1}$
Find all real solutions of $6^x+1=8^x-27^{x-1}$.
Things I tried:
We want solutions of
$$2^x3^x+1 = (2^x)^3-\frac{(3^x)^3}{27}.$$ Write $a=2^x$ and $b=3^x$. This gives
$$ab+1 = a^3-\frac{b^3}{27}$$
or
$$27ab+27=27a^3-b^3$$
How to continue?
| As you did, let $a=2^x, b=3^x$. We have $$ab+1=a^3-\frac{b^3}{27}$$
Note the standard identity $$u^3+v^3+w^3-3uvw=(u+v+w)\left(\frac{(u-v)^2+(v-w)^2+(w-u)^2}{2}\right)$$
Take $u=a, v=-\frac{b}{3}, w=-1$. Then $u^3+v^3+w^3-3uvw=a^3-\frac{b^3}{27}-1-ab=0$.
Thus $u+v+w=0$ or $(u-v)^2+(v-w)^2+(w-u)^2=0$. In the latter case, we get $u=v=w$ so $a=-\frac{b}{3}=-1$, so $a=-1, b=3$. But clearly $2^x=a \not =-1$, a contradiction.
Thus $u+v+w=0$.
Thus $a-\frac{b}{3}-1=0$, i.e. $2^x=3^{x-1}+1$.
Write $x=y+1$, and rewrite as $1^y+3^y=2^y+2^y$. Finally use the theorem I prove here. (Quoted below)
Theorem: Let $a, b, c, d$ be real numbers such that $0<a<b \leq c<d$. Then the equation $$a^x+d^x=b^x+c^x$$ has
*
*Exactly two solutions, $x=0$ and $x=t>0$ for some $t$, if $ad-bc<0$
*Exactly two solutions, $x=0$ and $x=t<0$ for some $t$, if $ad-bc>0$
*Exactly one solution, $x=0$, if $ad-bc=0$
with $(a, b, c, d)=(1, 2, 2, 3)$, $ad-bc=-1<0$, and $y$ in place of $x$.
Thus there are exactly two solutions for $y$, given by $y=0$ and $y=t>0$, some $t$. By inspection, $y=1$ is a solution. Thus all real solutions for $y$ are $y=0, 1$.
Thus all real solutions for $x$ are $x=1, 2$.
| {
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Find the last column of a matrix. Find the matrix. $A\left[\begin{matrix} 1 & 0 \\ 0 & 0 \\ 1 & 1 \end{matrix} \right]$ = $\left[\begin{matrix} 2 & 3 \\ -1 & 0 \\ 5 & -7 \\ 0 & 6 \end{matrix} \right]$
(1)Find the last column of $A$.
(2)Find the matrix $A$.
I figured that the size of $A$ has to be $4 \times 3$. Therefore there are 4 entries in any column of $A$.
The answer to (1) is: $\left[\begin{matrix} 3 \\ 0 \\ -7 \\ 6 \end{matrix} \right]$
and the answer to (2) is $\left[\begin{matrix} -1 & * & 3 \\ -1 & * & 0 \\ 12 & * & -7 \\ -6 & * & 6 \end{matrix} \right]$ where * is any number.
However I have no idea how to get this answer.
| Multiplying $A$ by $\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}$ on the right sums the first column of $A$ with its last column.
Multiplying $A$ by $\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ on the right maintains the last column.
Using the last column and the first line of this answer you're able to obtain the first column of the product.
It also follows that the second column of $A$ is irrelevant.
This can and should all be confirmed by writing $A=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ a_{41} & a_{42} & a_{42}\\\end{bmatrix}$ and carrying out the computation.
| {
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Cylindrical coordinates: $\iiint (x^2 + y^2 + z^2) dxdydz$ Show that
$$ I= \iiint_S (x^2 + y^2 + z^2) dxdydz = \frac{2^{10} a^5 k}{75} \left(1 + \frac{k^2}{3} \right), a>0, k>0$$
where $S$ is the region bounded by the cilinder $x^2 + y^2 = 2ax$ and the upper and lower sections of the cone $z^2 = k^2(x^2 + y^2)$.
My attempt:
$$
I = \int_{-\pi/2}^{\pi/2} \int_0^a \int_{-kr}^{kr} r(r^2 + z^2) dz dr d\theta
$$
using cylindrical coordinates, but I couldn't find the given value of this integral.
Thank in advance!
| Using Cylinderical coordinates, Mathematica evaluates the following integral as
$$\int_{-\pi/2}^{\pi/2} \int_0^{2 a \cos \theta } \int_{-kr}^{kr} (r^2 + z^2) r dz dr d\theta = \frac{1024}{225} a^5 k \left(k^2+3\right)$$
I suppose you made error in the bounds of $r$ since center of cylinder is $(a, 0)$ and polar equation of circle with center $(a, 0)$ with radius $a$ is $r = 2a \cos(\theta)$
Also on rectangular coordinates the following code on Mathematica produces same value
Integrate[
x^2 + y^2 + z^2, {x, 0, 2 a}, {y, -Sqrt[2 a x - x^2],
Sqrt[2 a x - x^2]}, {z, -Sqrt[k^2 (x^2 + y^2)],
Sqrt[k^2 (x^2 + y^2)]}, Assumptions :> {a > 0, k > 0}]
| {
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"source": "stackexchange",
"question_score": "2",
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Probability Question: Drawing balls from an urn Question:
An urn contains 12 balls where 3 balls colored red, 3 balls colored blue, 3 balls colored green, and 3 balls colored white. If we draw 4 balls in succession without replacement, what are the probabilities that
*
*the first is not red, the second is not blue, the third is not green, the fourth is not white and all four balls are different color?
*the first is not red, the second is not blue, the third is not green, the fourth is not white, and the drawn balls may have same color?
I can answer question no. 1 using derangement and the answer is $!4*\frac{3}{12}*\frac{3}{11}*\frac{3}{10}*\frac{3}{9}=\frac{27}{440}$, but how to answer question no. 2? Any help would be appreciated. Thanks in advance.
| For the second problem, we first find the probability of the complement, that the first is red or the second is blue or the third is green or the fourth is white.
We use Inclusion/Exclusion.
The probability the first is red is $\frac{3}{12}$, with similar expressions for the others. If we add these $\frac{3}{12}$, we get a first approximation $\binom{4}{1}\cdot\frac{3}{12}$.
This first approximation, double counts, for example, the cases where first is red and second is blue. The probability of this is $\frac{3}{12}\cdot\frac{3}{11}$. Subtract all $\binom{4}{2}$ instances. Our second approximation is $\binom{4}{1}\cdot\frac{3}{12}-\binom{4}{2}\cdot\frac{3}{12}\cdot\frac{3}{11}$.
But we have subtracted too much, for example the cases where first is red, second is blue, and third is green. So we add back $\binom{4}{3}\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}$. Finally, we subtract the probability of red then blue then green then white.
So the probability of the complement is
$$\binom{4}{1}\cdot\frac{3}{12}- \binom{4}{2}\cdot\frac{3}{12}\cdot\frac{3}{11}+\binom{4}{3}\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}-\binom{4}{4}\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}\cdot\frac{3}{9}.$$
| {
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prove that $2^{15} - 2^3 $ divides $ a^{15} - a^3$ Prove that $$2^{15} - 2^3 $$ divides $$ a^{15} - a^3$$ for any integer $a$.
Hint: $$ 2^{15} - 2^3 = 5\cdot7\cdot8\cdot9\cdot13$$
| We can factor:
$$a^{15}-a^3=a^3(a-1)(a+1)(a^2+1)(a^2+a+1)(a^2-a+1)(a^4-a^2+1)$$
Now, we just need to prove that this is divisible by $2^3\cdot3^2\cdot5\cdot7\cdot13$.
If $a$ is odd, $(a-1)(a+1)(a^2+1)$ is the product of three even numbers, and is divisible by 8. If $a$ is even, $a^3$ is divisible by 8.
If $a$ is a multiple of 3, $a^3$ is divisible by $9$. Otherwise, one of $(a-1)$ and $(a+1)$ is divisible by 3, as is one of $(a^2-a+1)$ and $(a^2+a+1)$, and therefore $a^2(a-1)(a+1)(a^2-a+1)(a^2+a+1)$ is divisible by 9.
If $a$ is a multiple of 5, obviously so is our product. If not, $a^2$ is either 1 more or one less than a multiple of 5, so $a(a-1)(a+1)(a^2+1)$ is divisible by 5.
Consider $a\mod7$. If it's 0, then $a$ is divisible by 7. If it's 1, $(a-1)$ is divisible by 7. If it's 2 or 4, $(a^2+a+1)$ is divisible by 7. If it's 3 or 5, $(a^2-a+1)$ is divisible by 7. If it's 6, $(a+1)$ is divisible by 7. Regardless, our product is divisible by 7.
Consider $a\mod13$. If it's 0, then $a$ is divisible by 13. If it's 1, $(a-1)$ is divisible by 13. If it's 2 or 6 or 7 or 11, $(a^4-a^2+1)$ is divisible by 13. If it's 3 or 9, $(a^2+a+1)$ is divisible by 13. If it's 4 or 10, $(a^2-a+1)$ is divisible by 13. If it's 5 or 8, $(a^2+1)$ is divisible by 13. If it's 12, $(a+1)$ is divisible by 13. Regardless, our product is divisible by 13, and we're done.
| {
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"source": "stackexchange",
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"answer_id": 5
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Show that $2 < e^{1/(n+1)} + e^{-1/n}$ I'm trying to show $2 < e^{1/(n+1)} + e^{-1/n}$.
I can show that $ 2 < e^{1/n} + e^{-1/(n+1)}$ since
$$2 \leq 2\cosh\left(\frac{1}{n}\right) = e^{1/n} + e^{-1/n} < e^{1/n} + e^{-1/(n+1)}$$
but I'm still having trouble with the other inequality. I though using $\cosh$ again might help but I can't get anywhere.
| well, since
$$ \left(\frac1{4! n^4}-\frac1{5!n^5}\right)+\cdots+\left(\frac1{(4+2k)!n^{4+2k}} -\frac1{(4+2k+1)!n^{4+2k+1}}\right)>0 $$
and
$$ \left(\frac1{4! n^4}-\frac1{5!n^5}\right)+\cdots+\left(\frac1{(4+2k)!n^{4+k}} -\frac1{(4+2k+1)!n^{4+2k}}\right) +\frac1{(4+2k+2)!n^{4+2k+2}} >0 $$
then,
$$\sum_{k\geq4}\frac{(-1)^k}{k!n^k}>0$$
we have (or the lemma below)
$$e^{-\frac1n}=1-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\sum_{k\geq4}\frac{(-1)^k}{k!n^k}\gt 1-\frac1n+\frac1{2n^2}-\frac1{6n^3}$$
it is obvious that
$$e^{\frac1{1+n}}\gt 1+\frac1{1+n}+\frac1{2{(1+n)^2}}+\frac1{6{(1+n)^3}}+ \frac1{24{(1+n)^4}}$$
if $n>2$, it is easy to check that
$$e^{-\frac1n}+e^{\frac1{1+n}}\gt 2-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\frac1{1+n}+\frac1{2{(1+n)^2}}+\frac1{6{(1+n)^3}}+ \frac1{24{(1+n)^4}}\\ >2$$
In fact
$$n^3(n+1)^3\left(-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\frac1{1+n}+\frac1{2{(1+n)^2}}+\frac1{6{(1+n)^3}}+ \frac1{24{(1+n)^4}}\right)=\frac{n^3}{24(n+1)}-\frac16=\frac{n(n^2-4)-4}{24(n+1)}\gt0$$
edit:
Lemma
$x\ne0$, $k$ is a odd number, then
$$e^x\gt 1+x+\frac{x^2}{2!}+\dotsb+ \frac{x^k}{k!}$$
Proof $\qquad$ according to Taylor formula, there exists $\theta(0\lt\theta\lt1)$ such that
$$e^x= 1+x+\frac{x^2}{2!}+\dotsb+ \frac{x^k}{k!}+\frac{e^{\theta x}}{(k+1)!}x^{k+1}$$
| {
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Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
Do you know that series? Thank you.
| Realize the last series you ask about as a geometric series...
$$
\frac{1}{z^2+1} = \frac{1}{z^2}\;\frac{1}{1-(-1/z^2)} =
\frac{1}{z^2} - \frac{1}{z^4} +\frac{1}{z^6} - \frac{1}{z^8}+\dots
$$
| {
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Partial Fractions Integration Question $$\int\frac{x^5+x-1}{x^3 +1} dx$$
Have tried everything ... polynomial long division, partial fractions, trig substitution etc...
Not for an assignment, so if a complete solution could be provided that'd be much appreaciated
| First divide the denominator into the numerator to get
$$\frac{x^5+x-1}{x^3+1}=x^2 -\frac{x^2-x+1}{x^3+1}$$
Now we factor the denominator, $$x^3+1=(x+1)(x^2-x+1)$$
Note that we have cancellation so we get
$$\int x^2-\frac{1}{x+1} dx=\frac{x^3}{3}-\ln |x+1|+C$$
| {
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calculate the Integral $ \int \ln\left(1+2a\cos x+a^2\right)dx\;,$ where $a\in \mathbb{R}.$ How can we calculate the Integral $\displaystyle \int \ln\left(1+2a\cos x+a^2\right)dx\;,$ where $a\in \mathbb{R}.$
$\bf{My\; Try::}$ Using $\displaystyle \cos (x) = \frac{e^{ix}+e^{-ix}}{2},$ we get ...
\begin{align}\displaystyle \int \ln(1+2a\cos x+a^2)dx &= \int \ln\left(1+2a\cdot \left(\frac{e^{ix}+e^{-ix}}{2}\right)+a^2\right)dx\\ &= \int \ln(1+a\cdot (e^{ix}+e^{-ix})+a^2)dx\\&=\int \ln \left((1+ae^{ix})\cdot (1+ae^{-ix})\right)dx\\
&= \int \ln(1+ae^{ix})dx+\int \ln(1+ae^{-ix})dx\end{align}
Now How can I solve after that
Help me
Thanks
| The integral
\begin{align}
I_{+} = \int \ln(1+a e^{ix}) \ dx
\end{align}
can be evaluated by making the substitution $t = e^{ix}$ which leads to
\begin{align}
I_{+} = - i \int \frac{\ln(1+at)}{t} \ dt = \left. i Li_{2}(-a t) \right|_{t = e^{ix}} = i \ Li_{2}(-a e^{ix}),
\end{align}
where $Li_{2}(x)$ is the dilogarithm function. In a similar manor
\begin{align}
I_{-} = \int \ln(1 + a e^{-ix}) \ dx = - i \ Li_{2}(-a e^{-ix}).
\end{align}
As stated by the author of the problem the integral in question is:
\begin{align}
\int \ln(1+2a\cos x+a^2)dx &= \int \ln\left(1+2a\cdot \left(\frac{e^{ix}+e^{-ix}}{2}\right)+a^2\right)dx\\
&= \int \ln(1+a\cdot (e^{ix}+e^{-ix})+a^2)dx\\
&=\int \ln \left((1+ae^{ix})\cdot (1+ae^{-ix})\right)dx\\
&= \int \ln(1+ae^{ix})dx+\int \ln(1+ae^{-ix})dx \\
&= i \left[ Li_{2}(-a e^{ix}) - \ Li_{2}(-a e^{-ix}) \right].
\end{align}
In series form the dilogarithm is
\begin{align}
Li_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}}
\end{align}
and leads to
\begin{align}
\int \ln(1+2a\cos x+a^2)dx &= - 2 \ \sum_{n=1}^{\infty} \frac{(-1)^{n} \ a^{n} \ \sin(nx)}{n^{2}}
\end{align}
which is valid for $|a| \leq 1$. For the case $a > 1$ the logarithm in the integral is sen as
\begin{align}
\ln(1+2a\cos x+a^2) = 2 \ln(a) + \ln\left(1 + \frac{2}{a} \cos(x) + \frac{1}{a^{2}} \right)
\end{align}
and leads to
\begin{align}
\int \ln(1+ 2a \cos(x) + a^{2}) dx = 2 \ln(a) \ x - 2 \ \sum_{n=1}^{\infty} \frac{(-1)^{n} \ \sin(nx)}{a^{n} \ n^{2}}.
\end{align}
As an example of the use of this integral consider the case when the limits are 0 and $\pi/2$. The integral becomes
\begin{align}
\int_{0}^{\pi/2} \ln(1+2a\cos x+a^2)dx = 2a \ \sum_{n=0}^{\infty} \frac{ (-1)^{n} \ a^{2n} }{(2n+1)^{2}}.
\end{align}
For the case of $a = 1$ this leads to the known integral value
\begin{align}
\int_{0}^{\pi/4} \ln(\cos(x)) \ dx = \frac{\textbf{G}}{2} - \frac{\pi}{2} \ \ln(2)
\end{align}
where $\textbf{G}$ is Catalan's constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying fraction with square root as denominator I'm trying to find the integral of:
$$\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}$$
but I first need to simplify it so I tried dividing by the $\sqrt{x}$ for each of the numbers on the top like so:
$$\dfrac {2\sqrt{x}}{\sqrt{x}}$$
and did the same for the others. For the one above it was easy to see that it just simplifies to $2$. But I am unsure how to do the same for the others for instance $\dfrac {-3x}{\sqrt{x}}$. I know to $-\sqrt{x}$ but i don't know what $-3x - \sqrt{x}$ would come out with?
| HINT :
Rewrite
$$
\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}=\dfrac {2\sqrt{x} - 3\sqrt{x}\sqrt{x} + x\cdot x}{\sqrt{x}}=\dfrac {2\sqrt{x} - 3\sqrt{x}\sqrt{x} + \sqrt{x}\sqrt{x}\cdot x}{\sqrt{x}}
$$
or
$$
\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}=\dfrac {2x^{\large\frac12} - 3x^1 + x^2}{x^{\large\frac12}},
$$
where $x=\sqrt{x}\sqrt{x}$ and $\sqrt{x}=x^{\large\frac12}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Congruency by completing the square $x^2+x+1\equiv 0\mod 49$
We have the ring isomorphism $\mathbb{Z}/49\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$.
Consider $x^2+x+1\equiv 0\mod 7$
I usually solve these polynomial congruencies using the 'complete the square' method. How would I go about doing that in this case?
EDIT:
$x^2+x+1\equiv x^2-6x+8\equiv(x-3)^2\equiv 1\mod 7$
I get solutions $x=4$ and $x=2$ $\mod 7$.
Now I tried to use Hemel's lifting, so,
We have $f(x)=x^2+x+1$ thus $f'(x)=2x+1$.
In the case of $x\equiv 2\mod 7$, we have $f(2)=7$, $f'(2)=5$; as $7\nmid f'(2)$, we can lift to a unique solutions by solving $f'(2)\equiv -\frac{f(2)}{7}\mod 7$, i.e. $5t\equiv -1\mod 7$.
Clearly $t\equiv 4\mod 7$, leading to the solution $x\equiv 2+4\cdot 7\equiv 30\mod 49$.
In the case of $x\equiv 4\mod 7$, we have $f(4)=21$, $f'(4)=9$; as $7\nmid f'(4)$, we can lift to a unique solutions by solving $f'(4)t\equiv -\frac{f(4)}{7}\mod 7$ or $9t\equiv -\frac{9}{7}\mod 7$...and this is where I'm stuck, assuming I'm even doing it correctly.
| A start: Equivalently, we solve $4x^2+4x+4\equiv 0\pmod{49}$, that is, $(2x+1)^2+3\equiv 0\pmod{49}$.
| {
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How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$ How can I determine these limits:
$$ a) \lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$$
$$ b) \lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$$
Note I cannot use l'Hospital. But I know that $\lim_{n\rightarrow 0} \frac{\sin n}{n} = 1$ and $\lim_{n\rightarrow 0} \frac{\cos n -1}{n} = 0$
I already found the limit of $\lim_{n\rightarrow 0} \frac{\sin(3n) \sin(2n)}{n^2}$:
$$\lim_{n\rightarrow 0} \frac{\sin(3n) \sin(2n)}{n^2}
= \lim_{n\rightarrow 0} \frac{\sin(3n)}{n} \lim_{n\rightarrow 0} \frac{ \sin(2n)}{n} =
3 \lim_{n\rightarrow 0} \frac{\sin(3n)}{3n} 2\lim_{n\rightarrow 0} \frac{ \sin(2n)}{2n} = 3 \cdot 2 = 6$$
Can I use a similar approach here?
| For (a), you can use a very similar approach. Since $\cos(x)$ is continuous on $\mathbb{R}$, you can pull the limit inside, and you can rewrite the limit as:
\begin{align*}
\lim_{n\to 0} \cos\left(\frac{\pi}{n}\sin(n)\cos(n)\right) &= \lim_{n\to 0} \cos\left(\frac{\sin(n)}{n}\pi\cos(n) \right)\\
&= \cos\left(\lim_{n\to 0} \frac{\sin(n)}{n} \pi \cos(n) \right)
\end{align*}
For (b), try
\begin{align*}
\lim_{n\to 0} \frac{n}{\sin(2n) - \cos\left(\frac{n}{2}\right) +1}
&= \lim_{n\to 0} \frac{1}{\frac{\sin(2n)}{n} - \frac{\cos\left(\frac{n}{2}\right) -1}{n}}\\
&= \lim_{n\to 0} \frac{1}{2\frac{\sin(2n)}{2n} - \frac{1}{2}\frac{\cos\left(\frac{n}{2}\right) -1}{\frac{n}{2}}}
\end{align*}
(You will want to make sure you understand the steps in these equalities. The first is dividing top and bottom by $n$, but the second is manipulating each term in the denominator to make them look like the limits you already know.)
And from here, your previous approach will prove useful.
| {
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"source": "stackexchange",
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Math Olympiad Algebraic Question Comprising Square Roots
If $m$ and $n$ are positive real numbers satisfying the equation $$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
find the value of $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$$
I came across this question in a Math Olympiad Competition and had no idea how to solve it. Can anyone help? Thanks.
| We have
$$
\begin{align*}
m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n&=3\\
m+4\sqrt{mn}+4n-2\sqrt{m}-4\sqrt{n}-3&=0\\
(\sqrt{m}+2\sqrt{n})^2-2(\sqrt{m}+2\sqrt{n})-3&=0\\
(\sqrt{m}+2\sqrt{n}-3)(\sqrt{m}+2\sqrt{n}+1)&=0
\end{align*}
$$
which gives
$$
\sqrt{m}+2\sqrt{n}=3\,\,\text{or}\,\,\sqrt{m}+2\sqrt{n}=-1.
$$
We can disregard the second solution as $m$ and $n$ are real so
$$
\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}=\frac{3+2014}{4-3}=2017.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Because $(a-b)(a+b)(a^2+b^2)=a^4-b^4$, and $\lim_{x\to 7}a=\lim_{x\to 7}b=2$, we rewrite as
$$\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}=\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{(x+9)-16}(2+2)(2^2+2^2)$$
$$=32\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}$$
Because $(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)=a^6-b^6$, and also $\lim_{x\to 7}a=\lim_{x\to 7}b=3$, we rewrite as
$$32\lim_{x\to 7}\frac{(x+2)^3-(x+20)^2}{x-7}\frac{1}{(3+3)(3^2-3\cdot 3+3^2)(3^2+3\cdot 3+3^2)}$$
$$=\frac{32}{1458}\lim_{x\to 7}\frac{x^3+5x^2-28x-392}{x-7}=\frac{32}{1458}\lim_{x\to 7}x^2+12x+56=\frac{32}{1458}(189)=\frac{112}{27}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| You should note that you get
$$\sqrt{x^2 + 2x + 2} = \sqrt{(x+1)^2 + 1} \approx (x+1)$$
as you are squaring i.e. $(x+1)^2$ is not approximately equal to $x^2$ since $(x+1)^2 - x^2 \approx 2x$ you can only ignore the constant.
| {
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Evaluating e using limits What algebraic operations can I use on the $RHS$ to show $RHS = LHS$
$$e=\lim_{k\to\infty}\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)^k$$
| Since$$\left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)^k=e^{k \log \left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)} $$
it's sufficient to show that
$$\lim_{k \to \infty} k \log \left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)=1 .$$
We can rewrite this expression as
$$\frac{\log \left(\frac{2+\sqrt{3+9k^2}}{3k-1} \right)}{\frac{1}{k}}, $$
which approaches the indeterminate form $0/0$ as $k \to \infty$. We thus apply L'hôpital's rule, and obtain the equivalent limit
$$\lim_{k \to \infty} \frac{ \frac{3k-1}{2+\sqrt{3+9k^2}} \frac{d}{dk} \left[ \frac{2+\sqrt{3+9k^2}}{3k-1} \right] }{-\frac{1}{k^2}}=\lim_{k \to \infty} -\frac{3 k^2 \left(\frac{2 k}{\sqrt{k^2+\frac{1}{3}}}+1\right)}{1-9 k^2}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument:
$P_{k+1} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)\cdot(2k+1)}{1\cdot2\cdot3\cdot\ldots\cdot k \cdot (k+1)} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)}{1\cdot2\cdot3\cdot\ldots\cdot k}\cdot \frac{2k+1}{k+1} \leq 2^k \cdot\frac{2k+1}{k+1} \leq 2^{k+1}\cdot \frac{2k+1}{k+1}$ using $P_k$ and that these are all positive numbers.
Now dividing through by $\frac{2k+1}{k+1}$ would give the desired result.
I just feel like I've done something wrong here. Can anyone tell me if I've overlooked anything?
| All you need is to prove that: $\dfrac{2k+1}{k+1} \leq 2$
| {
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Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$ Let $n$ be a positive whole number. Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$, without using a computer.
$n$ must be greater than 3 and less than $2^6=64$, so using a computer it is easy to verify that $n$ must be 6,7,14,21 or 42.
| Using Fermat's Little Theorem,
$$a^7\equiv a\pmod7$$
$$a^7-a=a(a^6-1)=a(a-1)\{a^5+a^4+\cdots+a+1\}$$ where by Little Theorem, $\displaystyle2|a(a-1)$
$$a^7-a=a(a^6-1)=a(a^2-1)(a^4+a^2+1)=(a^3-a)(a^4+a^2+1)$$ where by Little Theorem, $\displaystyle3|(a^3-a)$
So, $\displaystyle a^7-a$ will be divisible by lcm$(2,3,7)$
| {
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Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int xu^{-1}$$
$$=\int \left(\dfrac{u}{2}-\dfrac{1}{2}\right)u^{-1}\text{du}=\int\left[\dfrac{1}{2}-\dfrac{1}{2}u^{-1}\right]\text{du}$$
$$=\dfrac{1}{2}x-\dfrac{1}{2} \ln \left|2x+1\right|$$
Looking at the answer in the back, this is wrong.
The answer is $x \ln(2x+1)-x+\dfrac{1}{2}\ln(2x+1)+c$.
What have I done wrong?
| Your issue is in what follows this step:
\begin{align*}
x\ln(2x+1) - \int \frac{2x}{2x+1}\,\text{dx} &= x\ln(2x+1) - \int \frac{2x+1-1}{2x+1}\,\text{dx}\\
&= x\ln(2x+1) - \left(\int \frac{2x+1}{2x+1}\,\text{d}x - \int \frac{1}{2x+1}\,\text{d}x\right)\\
&= x\ln(2x+1) - \left(\int \,\text{d}x - \int \frac{1}{2x+1}\,\text{d}x\right)\\
&= x\ln(2x+1) - \left(x - \frac{1}{2}\ln(2x+1)\right) + C\\
&= x\ln(2x+1) - x + \frac{1}{2}\ln(2x+1) + C
\end{align*}
(The second to last equality follows from a u-substitution, using $u = 2x+1$. Technically, it seems that the answer should be
$$x\ln(2x+1)-x+\frac{1}{2}\ln|2x+1|+C$$
but perhaps there is some reason for them to be able to drop that absolute value.)
We can also compute this integral using substitution: Use $u = 2x+1$, $2x = u-1$, $du = 2dx$, to compute it as follows:
\begin{align*}x\ln(2x+1) - \int \frac{2x}{2x+1}\,\text{d}x &= x\ln(2x+1) - \frac{1}{2}\int \frac{u-1}{u}\,\text{d}u\\
&= x\ln(2x+1) - \frac{1}{2}\int\,\text{d}u + \frac{1}{2}\int \frac{1}{u}\,\text{d}u\\
&= x\ln(2x+1) - \frac{u}{2} + \frac{1}{2}\ln|u| + C\\
&= x\ln(2x+1) - \frac{2x+1}{2} + \frac{1}{2}\ln|2x+1| + C\\
&= x\ln(2x+1) - x - \frac{1}{2} + \frac{1}{2}\ln|2x+1|+C\\
&= x\ln(2x+1) - x + \frac{1}{2}\ln|2x+1|+C'
\end{align*}
(the last equality is just a matter of absorbing the constant $\frac{1}{2}$ into the integration constant $C + \frac 12 = C'$)
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.
| Notice $$
\begin{align}
k^3
&= \color{red}{(k^3 - k)} + \color{blue}{k}\\
&= \color{red}{(k-1)k(k+1)} + \color{blue}{k}\\
&= \frac{\color{red}{(k-1)k(k+1)}(k+2)-(k-2)\color{red}{(k-1)k(k+1)}}{4}
+ \frac{\color{blue}{k}(k+1) - (k-1)\color{blue}{k}}{2}
\end{align}$$
is telescoping. We have
$$\begin{align}
\sum_{k=0}^n k^3
&= \frac{(n-1)n(n+1)(n+2)}{4} + \frac{n(n+1)}{2}\\
&= \frac{n(n+1)}{2}\left(\frac{(n-1)(n+2)}{2} + 1\right)\\
&= \left(\frac{n(n+1)}{2}\right)^2
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help w/ Partial Fraction Decomposition I need some help figuring out how to decompose $\displaystyle\frac{1}{x^4+1}$ into partial fractions.
This is what I have done so far: $$\frac{1}{x^4+1} = \frac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}$$
From there, I do not know how to decompose it into partial fractions, or if I even set it up correctly.
| I'm guessing that this problem is homework so I'll provide a step forward from where you're stuck to see if it'll help you.
Currently, you have this factored into two distinct irreducible quadratics, each of which gets a term in the form $\frac{Ax+B}{ax^2+bx+c}$. Therefore, you should set up the equation to be:
$$\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}=\frac{Ax+B}{x^2-\sqrt{2}x+1}+\frac{Cx+D}{x^2+\sqrt{2}x+1}$$
Continue by determining the values of A, B, C, and D you need.
I tried that, and I could not get valid equations. Assuming I computed everything correctly, I got the following equations: B+D=0 and B+D=1, which is invalid.
Those aren't the equations you're looking for. If you multiply both sides by $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$, you will get:
$$1=(Ax+B)(x^2+\sqrt{2}x+1)+(Cx+D)(x^2-\sqrt{2}x+1)$$
Multiply out to get
$$1=Ax^3+A\sqrt{2}x^2+Ax+Bx^2+B\sqrt{2}x+B+Cx^3-C\sqrt{2}x^2+Cx+Dx^2-D\sqrt{2}x+D$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How is $2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$ calculated? $$2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$$
I cant figure out why this is $\frac64$. I try to use telescopic series without success.
| Ok well first of all $\frac{6}{4}=\frac{3}{2}$.
The series telescopes as has been pointed out, using that formula we have
$$\left(\frac{1}{1}-\frac{1}{3}\right)+
\left(\frac{1}{2}-\frac{1}{4}\right)+
\left(\frac{1}{3}-\frac{1}{5}\right)+
\left(\frac{1}{4}-\frac{1}{6}\right)+\cdots =1+\frac{1}{2}=\frac{3}{2}$$
| {
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Value of the given limit I need to calculate the value of :
$$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{2n}{\frac{r}{\sqrt{n^2+r^2}}}$$
I had been trying to use Cesàro summation but somehow, I might be messing up.
The options are : $$\sqrt{5}+1,\sqrt{5}-1,\sqrt{2}-1,\sqrt{2}+1$$
| Hint: As a Riemann Sum:
$$
\lim_{n\to\infty}\sum_{r=1}^{2n}\frac{r/n}{\sqrt{1+r^2/n^2}}\frac1n=\int_0^2\frac{x}{\sqrt{1+x^2}}\mathrm{d}x\tag{1}
$$
We can also use the fact that
$$
\sqrt{n^2+r^2}-\sqrt{n^2+(r-1)^2}=\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\tag{2}
$$
and
$$
\begin{align}
&\left|\,\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}-\frac{r}{\sqrt{n^2+r^2}}\,\right|\\[6pt]
&=\small\left|\,\frac{r}{\sqrt{n^2+r^2}}\frac{\sqrt{n^2+r^2}-\sqrt{n^2+(r-1)^2}}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}-\frac1{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\,\right|\\[6pt]
&=\small\left|\,\frac{r}{\sqrt{n^2+r^2}}\frac{2r-1}{\left(\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}\right)^2}-\frac1{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\,\right|\\[6pt]
&\le\frac{2n}{4n^2}+\frac1{2n}\\[12pt]
&\le\frac1n\tag{3}
\end{align}
$$
Therefore, using $(3)$ gives
$$
\left|\,\frac1n\sum_{r=1}^{2n}\left[\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}-\frac{r}{\sqrt{n^2+r^2}}\right]\,\right|\le\frac2n\tag{4}
$$
and using $(2)$ yields
$$
\frac1n\sum_{r=1}^{2n}\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}=\sqrt5-1\tag{5}
$$
Finally, $(4)$ and $(5)$ say
$$
\left|\,\frac1n\sum_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}-(\sqrt5-1)\,\right|\le\frac2n\tag{6}
$$
Therefore, $(6)$ gives
$$
\lim_{n\to\infty}\frac1n\sum_{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}=\sqrt5-1\tag{7}
$$
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} |
Find a generating function for $\sum_{k=0}^{n} k^2$
Find a generating function for $\sum_{k=0}^{n} k^2$
I know my solution is wrong, but why?
My solution:
If $F(x)$ generates $\sum_{k=0}^{n} k^2$ then $F(x)(1-x)$ generates $k^2$.
$\frac{x}{(1-x)^4}: \left\{ 0,1,4,9,16,25... \right\}$
$\frac{x}{(1-x)^3}: \left\{ 0,1,3,5,7,9... \right\}$
$\frac{x}{(1-x)^2}: \left\{ 0,1,2,2,2,2... \right\}$
$\frac{x}{1-x}: \left\{ 0,1,1,1,1,1... \right\}$
$x: \left\{ 0,1,0,0,0,0... \right\}$
So, $F(x)=\frac{x}{(1-x)^5}$ generates $\sum_{k=0}^{n} k^2$
| The differences of $\{0,1,2,2,2,2,\dots\}$ are $\{0,1,1,0,0,\dotsc\}$, not $\{0,1,1,1,1,\dotsc\}$.
So you should get
$$\begin{align}
x+x^2 &: \{0,1,1,0,0,\dotsc\}\\
\frac{x+x^2}{1-x} &: \{ 0,1,2,2,2,\dotsc\}\\
\frac{x+x^2}{(1-x)^2} &: \{ 0,1,3,5,7,\dotsc\}\\
\frac{x(1+x)}{(1-x)^3} &: \{ 0,1,4,9,\dotsc\}
\end{align}$$
and
$$\frac{x(1+x)}{(1-x)^4}$$
as the generating function of $\sum k^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/832858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that limit $\lim_{n\to\infty}\sqrt{4+\frac{1}{n^2}}+\sqrt{4+\frac{2}{n^2}}+\cdots+ \sqrt{4+\frac{n}{n^2}}-2n=\frac{1}{8}$
Let $$a_{n}=\sqrt{4+\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n^2}}+\cdots+
\sqrt{4+\dfrac{n}{n^2}}-2n,$$
show that
$$\lim_{n\to\infty}a_{n}=\dfrac{1}{8}$$
My attempt:
Since
$$\sqrt{4+\dfrac{i}{n^2}}-2=\dfrac{\dfrac{i}{n^2}}{\sqrt{4+\dfrac{i}{n^2}}+2}=\dfrac{1}{\sqrt{4n^2+i}+2}\dfrac{i}{n}$$
then I can't work it. Thank you.
|
My solution: Note
$$\sqrt{4+\dfrac{k}{n^2}}<\sqrt{2^2+\dfrac{k}{n^2}+(\dfrac{k}{4n^2})^2}=2+\dfrac{k}{4n^2}$$
$$\sqrt{4+\dfrac{k}{n^2}}=\sqrt{4+\dfrac{k-1}{n^2}+\dfrac{1}{n^2}}>\sqrt{2^2+\dfrac{k-1}{n^2}
+(\dfrac{k-1}{4n^2})^2}=2+\dfrac{k-1}{4n^2}$$
and
$$\lim_{n\to\infty}\dfrac{k-1}{4n^2}=\lim_{n\to\infty}\dfrac{k}{4n^2}=\dfrac{1}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Limit by diagonalisation I want to compute the limit of the sequence $x_0=a$, $x_1=b$ and $x_n=\frac{x_{n-1}+x_{n-2}}{2}$.
I did it by setting:
$\begin{pmatrix}
x_{n+1}\\ x_{n}
\end{pmatrix}=$
$\begin{pmatrix}
\frac{1}{2} &\frac{1}{2}\\
1 &0\\
\end{pmatrix}$
$\begin{pmatrix}
x_{n}\\ x_{n-1}
\end{pmatrix}$
and calculating, that $$\lim\limits_{n\to\infty}\begin{pmatrix}
\frac{1}{2} &\frac{1}{2}\\
1 &0\\
\end{pmatrix}^n=\begin{pmatrix}
\frac{2}{3} &\frac{1}{3}\\
\frac{2}{3} &\frac{1}{3}\\
\end{pmatrix}$$
And thus $\lim\limits_{n\to\infty}x_n=\frac{2}{3}b+\frac{1}{3}a$.
But this can't be true, because the result should be symmetric in $a$ and $b$. Where is my mistake?
| $x_2 = \dfrac{a + b}{2}$
$x_3 = \dfrac{a+3b}{4}$
It is already not symmetric in $a$ and $b$. There is no error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Consider convergence of series: $\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$ Consider convergence of series:
$$\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$$
My tried:
We have
$$\sum_{n=1}^{\infty }\sin(\pi (2+\sqrt{3})^{n})=\sum_{n=1}^{\infty}\sin\left(\pi[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}]-\pi(2-\sqrt{3})^{n}\right)\, (*)$$
Because $$(2+\sqrt{3})^{n}=\sum_{k=0}^{n}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$
$$(2-\sqrt{3})^{n}=\sum_{k=0}^{n}(-1)^{k}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$
Hence $$(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}=\left\{\begin{matrix} 0&,k=2l+1 \\ m\in N&,k=2l \end{matrix}\right.$$
$$\Rightarrow (1)=\sum_{n=1}^{\infty}\sin\left(m\pi-\pi(2-\sqrt{3})^{n}\right)=\sum_{n=1}^{\infty}(-1)^{m+1}\sin\frac{\pi}{(2+\sqrt{3})^{n}}<\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$$
$\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$ converge
Hence series is converge.
True or False?
| A simple way to show convergence is to note that $$(2+\sqrt{3})^2=7+4\sqrt{3}=2*7-(2- \sqrt{3})^2$$
$$(2+\sqrt{3})^3=26+15\sqrt{3}=2*26-(2-\sqrt{3})^3$$
$$(2+\sqrt{3})^4=97+56\sqrt{3}=2*97-(2-\sqrt{3})^4$$
and in general
$$(2+\sqrt{3})^n=2k-(2-\sqrt{3})^n$$
with k integer.
The summation then reduces to $$-\sum_{n=1}^{\infty}\sin\left[\pi\left(2-\sqrt{3}\right)^n\right]$$
where the terms clearly tend to zero as n tends to infinity. Convergence can therefore easily be shown using the comparison test, observing that $$\sum_{n=1}^{\infty}\sin\left[\pi\left(2-\sqrt{3}\right)^n\right]<\sum_{n=1}^{\infty}\sin[\pi(1/2)^n]<\sum_{n=1}^{\infty}\pi(1/2)^n=\pi$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\
&= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\
&= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\
&= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= 2\sin2\theta\cos\theta.}$$
| Use the sum of angles identities for the $\sin$ function:-
$$\sin(3\theta)=\sin(2\theta+\theta)=\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta)$$
$$\sin(\theta)=\sin(2\theta-\theta)=\sin(2\theta)\cos(\theta)-\cos(2\theta)\sin(\theta)$$
Adding both equations results in
$$\sin(3\theta)+\sin(\theta)=2\sin(2\theta)\cos(\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If $T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$. Determine $||T||$. Define $T : \mathbb{R}^2 → \mathbb{R}$ ( $\mathbb{R}^2$ & $\mathbb{R}$ being
equipped with the Euclidean norm) by
$T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$.
Determine $||T||$.
My thoughts:-
We know that
$ \qquad \left\|{T}\right\| = \sup \left\{{\left|{Th}\right|: \left\|{h}\right\| = 1}\right\}$
here $h=(x,y)$ and $||h||=\sqrt{x^2+y^2}=1$
So we need to find out the maximum value of $2x+y$ with the condition that $x^2+y^2=1$.
Am I right?
| Yes, you are right. To do so, note that
$$ T'(x,y) = \begin{pmatrix} 2 & 1 \end{pmatrix} $$
and for $F(x,y) = x^2 + y^2 - 1$ we have
$$ F'(x,y) = \begin{pmatrix} 2x & 2y \end{pmatrix} $$
We hence have to solve
\begin{align*}
2 &= 2x \lambda\\
1 &= 2y\lambda\\
x^2 + y^2 &= 1
\end{align*}
So $x = 1/\lambda$, $y = 1/2\lambda$, hence $y = \frac x2$. Plugging this into the third equation gives
$$ x^2 + \frac{x^2}4 = 1 \iff x^2 = \frac 45 \iff x = \pm \frac 2{\sqrt 5} $$
This gives $y = \pm \frac 1{\sqrt 5}$. Now evaluate $T$ at the two points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/840418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Given a sequence $ (a_{n})_{n \in \mathbb{N}} $, find $ \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $. I would appreciate it if somebody could help me with the following problem:
If the sequence $ (a_{n})_{n \in \mathbb{N}} $ satisfies
$$
a_{1} = 1
\quad \text{and} \quad
\forall n \in \mathbb{N}, n \ge 2: \quad
a_{n} = \frac{(a_{1} + 2) (a_{2} + 2) \cdots (a_{n-1} + 2)}{2^{n}},
$$
find $ \displaystyle \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $.
| For $n \ge 2$, we have: $a_{n+1} = \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)(a_n+2)}{2^{n+1}}$ $= \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)}{2^n}\cdot \dfrac{a_n+2}{2} = \dfrac{a_n(a_n+2)}{2} = \dfrac{1}{2}a_n^2+a_n$.
Thus, $\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}} = \dfrac{a_{n+1}-a_n}{a_na_{n+1}} = \dfrac{\tfrac{1}{2}a_n^2}{a_n \cdot \tfrac{1}{2}a_n(a_n+2)} = \dfrac{1}{a_n+2}$.
Therefore, $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{a_n+2} = \dfrac{1}{a_1+2} + \sum_{n=2}^{\infty}\left[\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}\right] = \dfrac{1}{a_1+2} + \dfrac{1}{a_2} = \dfrac{1}{1+2} + \dfrac{1}{\tfrac{3}{4}} = \boxed{\dfrac{5}{3}}$,
where we have used the fact that the sum is telescoping and $a_n \to \infty$ as $n \to \infty$.
EDIT: We can show that $a_n \to \infty$ without assuming that the series converges.
Note that $\dfrac{a_{n+1}+1}{2} = \dfrac{1}{4}a_n^2+\dfrac{1}{2}a_n+\dfrac{1}{2} = \left(\dfrac{a_n+1}{2}\right)^2+\dfrac{1}{4} \ge \left(\dfrac{a_n+1}{2}\right)^2$ for all $n \ge 2$.
Since $a_3 = \dfrac{33}{32}$, by induction we have $\dfrac{a_n+1}{2} \ge \left(\dfrac{a_3+1}{2}\right)^{2^{n-3}} = \left(\dfrac{65}{64}\right)^{2^{n-3}}$ for all $n \ge 3$.
Clearly, $\displaystyle\lim_{n\to\infty}\left(\dfrac{65}{64}\right)^{2^{n-3}} = +\infty$. Therefore, $\displaystyle\lim_{n\to\infty}a_n = +\infty$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Trigonometric Triangle Equality $A, B, C$ are the angles of a triangle then $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is
always greater than what integral value.
| Here $A,B,C$ are the angles of a $\triangle.$ So Here $$\displaystyle 0<\frac{A}{2},\frac{B}{2},\frac{C}{2}<\frac{\pi}{2}$$.
Now Using $\bf{A.M\geq G.M}\;,$ We Get
So we get $$\displaystyle \frac{\tan^2 \left(\frac{A}{2}\right)+\tan^2 \left(\frac{B}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{A}{2}\right)\cdot \tan^2 \left(\frac{B}{2}\right)} = \tan\left(\frac{A}{2}\right)\cdot \tan \left(\frac{B}{2}\right)\color{red}\checkmark$$
Similarly $$\displaystyle \frac{\tan^2 \left(\frac{B}{2}\right)+\tan^2 \left(\frac{C}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{B}{2}\right)\cdot \tan^2 \left(\frac{C}{2}\right)} = \tan \left(\frac{B}{2}\right)\cdot \tan \left(\frac{C}{2}\right)\color{red}\checkmark$$
Similarly $$\displaystyle \frac{\tan^2 \left(\frac{C}{2}\right)+\tan^2 \left(\frac{A}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{C}{2}\right)\cdot \tan^2 \left(\frac{A}{2}\right)} = \tan \left(\frac{C}{2}\right)\cdot \tan \left(\frac{A}{2}\right)\color{red}\checkmark$$
Now Add all Three, We Get
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)\color{blue}\checkmark\color{blue}\checkmark$$
Now Here $$\displaystyle \frac{A}{2}+\frac{B}{2}=\frac{C}{2}\Rightarrow \tan\left(\frac{A}{2}+\frac{B}{2}\right)=\tan \frac{C}{2}$$
So We Get $$\displaystyle \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)=1$$
Put into $\color{blue}\checkmark\color{blue}\checkmark\;,$ We Get
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Inequality for harmonic means Prove that for real numbers $a_1 ,a_2 ,...,a_n >0$ the following inequality holds
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$
| Upper Bound
By Cauchy-Schwarz, we have
$$
\begin{align}
\left(\sum_{j=1}^ka_j\right)\left(\sum_{j=1}^k\frac{j^2}{a_j}\right)
&\ge\left(\sum_{j=1}^kj\right)^2\\[3pt]
&=\frac{k^2(k+1)^2}{4}\tag{1}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j}
&\le\sum_{k=1}^n\frac4{k(k+1)^2}\sum_{j=1}^k\frac{j^2}{a_j}\\
&=\sum_{j=1}^n\frac{j^2}{a_j}\sum_{k=j}^n\frac4{k(k+1)^2}\\
&\le\sum_{j=1}^n\frac{j^2}{a_j}\sum_{k=j}^n2\left(\frac1{k^2}-\frac1{(k+1)^2}\right)\\
&\le\sum_{j=1}^n\frac{j^2}{a_j}\frac2{j^2}\\
&=2\sum_{j=1}^n\frac1{a_j}\tag{2}
\end{align}
$$
Thus, the ratio is at most $2$.
Sharpness
Set $a_k=k^\beta$ for $0\lt\beta\lt1$. First
$$
\begin{align}
\sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j}
&=\sum_{k=1}^n\frac{k}{\frac1{1+\beta}k^{\beta+1}+O(k^\beta)}\\
&=(1+\beta)\sum_{k=1}^n\frac1{k^\beta+O(k^{\beta-1})}\\
&=\frac{1+\beta}{1-\beta}n^{1-\beta}+O(1)\\[3pt]
&\sim\frac{1+\beta}{1-\beta}n^{1-\beta}\tag{3}
\end{align}
$$
Next
$$
\begin{align}
\sum_{k=1}^n\frac1{a_k}
&=\sum_{k=1}^n\frac1{k^\beta}\\
&=\frac1{1-\beta}n^{1-\beta}+O(1)\\[3pt]
&\sim\frac1{1-\beta}n^{1-\beta}\tag{4}
\end{align}
$$
Thus, as $\beta\to1^-$, the ratio tends to $2$ for large $n$. Therefore, $2$ is sharp.
| {
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"url": "https://math.stackexchange.com/questions/844995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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limit $f(x)=((x-7)(x+4))^{1/2}$ I have to calculate the Asymptotes in the infinity(and minus infinity) for this function:
$f(x)=((x-7)(x+4))^{1/2}$
I know that
$\lim_{x \to\infty} f(x)/x= 1$
And I get into trouble with:
$\lim_{x \to\infty} f(x)-x$
which is
$\lim_{x \to\infty} ((x-7)(x+4))^{1/2}-x = \lim_{x \to\infty} \sqrt{x^2-3x-28}-x$
wolfram alpha says it is $-3/2$ but i don't get why.... please help me with that, thanks
| $$
\sqrt{x^2-3x-28} - x = \frac{(\sqrt{x^2-3x-28} - x)(\sqrt{x^2-3x-28} + x)}{(\sqrt{x^2-3x-28} + x)}\\
= \frac{x^2-3x-28-x^2}{(\sqrt{x^2-3x-28} + x)}
= \frac{-3x-28}{x(\sqrt{1-3/x-28/x^2} + 1)} \to \frac {-3}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Apostol (6.22.10): Finding $\int \frac{\arcsin x}{x^2} dx$ I'm trying to solve another integral from Apostol (Chapter 6, Section 6.22, Question 10) which says to show the following:
$$
\int \frac {\arcsin x}{x^2}dx = \log|{\frac {1-\sqrt{1-x^2}}{x}}| - \frac {\arcsin x}{x} +C
$$
I have tried the following:
With $\int \frac {\arcsin x}{x^2}dx$ I started using integration by parts, choosing $u = \arcsin x $ and $dv = 1/x$ which lead to
$$
\int \frac {\arcsin x}{x^2}dx = -\frac{\arcsin x}{x} + \int \frac{dx}{x\sqrt{1-x^2}}
$$
$$
= -\frac{\arcsin x}{x} + \int \frac{x.dx}{x^2.sqrt{1-x^2}}
$$
With the Integral on the RHS, I set $u^2 = 1 - x^2$ which implied $x^2 = 1- u^2$ and $xdx = -u. du$. This meant
$$
\int \frac{xdx}{x^2\sqrt{1-x^2}} = \int \frac{-u du}{u (1-u^2)} = \int \frac{-du}{ 1-u^2} = -\int \frac{du}{1-u^2}
$$
For the integral $\int \frac{du}{1-u^2}$, I set $u = \sin \theta$ which meant $du = \cos\theta. d\theta$ which implied
$$
\int \frac{du}{1-u^2} = \int \frac{\cos\theta}{1 -\sin^2\theta}d\theta = \int\sec\theta\,d\theta = \log|\sec \theta + \tan \theta| = \log|\frac{1+\sin \theta}{\cos \theta}|
$$
Now, since $u = \sin \theta$ in the last substitution and $\sin^2\theta + \cos^2\theta = 1$ it follows that $\cos\theta = \sqrt{1-u^2}$ which means
$$
\log|\frac{1+\sin \theta}{\cos \theta}| = \log |\frac{1+u}{\sqrt{1-u^2}}|
$$
$$
= \log|\frac{1+u}{\sqrt{1-u}\sqrt{1+u}}|
$$
$$
= \log|\frac{\sqrt{1+u}}{\sqrt{1-u}}| = \frac{1}{2}\log|\frac{1+u}{1-u}|
$$
That is
$$
\int \frac{du}{1-u^2} =\frac{1}{2}\log|\frac{1+u}{1-u}|
$$
Now, in the first substitution $u^2 = 1 - x^2$ implies $u = \sqrt{1-x^2}$ which means
$$
\frac{1}{2}\log|\frac{1+u}{1-u}| = \frac{1}{2}\log|\frac{1+ \sqrt{1-x^2}}{1 - \sqrt{1-x^2}}|
$$
$$
= \frac{1}{2}\log|\frac{(1+\sqrt{1-x^2})(1+\sqrt{1-x^2})}{(1 - \sqrt{1-x^2})(1+\sqrt{1-x^2})}|
$$
$$
= \frac{1}{2}\log|\frac{2 - x^2 +2\sqrt{1-x^2}}{x^2}|
$$
$$
= \log|\frac{1+\sqrt{1-x^2}}{x}|
$$
That is
$$
\int \frac{xdx}{x^2\sqrt{1-x^2}} = - \log|\frac{1+\sqrt{1-x^2}}{x}|
$$
with respect to the original integral means
$$
\int \frac {\arcsin x}{x^2}dx = -\frac{\arcsin x}{x} - \log|\frac{1+\sqrt{1-x^2}}{x}| + C
$$
which is not the answer as stated in Apostol's book. I've probably made some error in the calculations which I just haven't been able to find, so if someone can point out the error in my calculations that would be appreciated
| A much more computationally straightforward approach follows. The integrand suggests the substitution $$\theta = \sin^{-1} x, \quad x = \sin \theta, \quad dx = \cos \theta \, d\theta,$$ giving $$I = \int \frac{\sin^{-1} x}{x^2} \, dx = \int \theta \csc \theta \cot \theta \, d\theta.$$ Then integration by parts with the choice $u = \theta$, $du = d\theta$, $dv = \csc\theta \cot\theta \, d\theta$, $v = -\csc \theta$, yields $$\begin{align*} I &= -\theta \csc \theta + \int \csc \theta \, d\theta \\ &= -\theta \csc \theta + \int \frac{\csc^2 \theta + \csc\theta \cot\theta}{\csc \theta + \cot \theta} \, d\theta.\end{align*}$$ Now with the choice $t = \csc\theta + \cot\theta$, $dt = -(\csc^2 \theta + \csc\theta \cot\theta) \, d\theta$, we obtain $$\begin{align*}I &= -\theta \csc\theta - \log|\csc\theta + \cot\theta| + C \\ &= -\frac{\sin^{-1} x}{x} - \log \left| \frac{1 + \sqrt{1-x^2}}{x} \right| + C.\end{align*}$$ Robert Israel's answer then shows the equivalence of this antiderivative to the claimed result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Jordan Canonical Form of $A^{-1}$
Let $A$ be a $5 \times 5$ invertible matrix over $\mathbb{R}$. Suppose
$c_A(x)=(x+1)^3(x+2)^2$ and $m_A(x)=(x+1)(x+2)^2$, what is a Jordan
Canonical Form of $A^{-1}$?
Is it possible to obtain a JCF of $A^{-1}$ from that of $A$, or we have to explicitly solve for $c_{A^{-1}}(x)$ and $m_{A^{-1}}(x)$ in the first place?
| Its easy to see that
$$A=\begin{pmatrix}
-2&1&0&0&0\\
0&-2&0&0&0\\
0&0&-1&0&0\\
0&0&0&-1&0\\
0&0&0&0&-1\\
\end{pmatrix}$$
so we see
$$A^{-1}=\begin{pmatrix}
-\frac{1}{2}&-\frac{1}{4}&0&0&0\\
0&-\frac{1}{2}&0&0&0\\
0&0&-1&0&0\\
0&0&0&-1&0\\
0&0&0&0&-1\\
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real numbers such $ x+y+z = 0 $.
prove : $ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $
Please think and write your solutions! ; )
| Let $x,y,z$ be the roots of $t^3-Qt - P=0$, $Q= -(xy+yz+zx), P = xyz$.
$0 = (x+y+z)^2 = -2Q + x^2+y^2+z^2 \implies x^2+y^2+z^2 = 2Q$.
You have $t^3 = P + Qt$ and when you replace with $x,y,z$ and sum:
$$
x^3+y^3+z^3 = 3P + Q(x+y+z) = 3P
$$
You also get the identities:
$$
x^4+y^4+z^4 = Q(x^2+y^2+z^2) = 2Q^2
\\
x^{n+3}+y^{n+3}+z^{n+3} = P(x^n+y^n+z^n) + Q(x^{n+1}+y^{n+1}+z^{n+1})
$$
Now return to the problem. Define $S_n = x^n+y^n+z^n$.
$$
LHS = \frac 1{49}S_7^2
= \frac 1{49}(PS_4+ QS_5)^2
= \frac 1{49}(2PQ^2 + Q(PS_2 + QS_3))^2\\
= \frac 1{49}(2PQ^2 + 2PQ^2 + 3PQ^2 )^2
= P^2Q^4
$$
$$
RHS = \frac1{50}S_5^2S_4
= \frac1{50}(PS_2+QS_3)^2 \times 2Q^2\\
= \frac1{50}(2PQ +3PQ)^2 \times 2Q^2
= P^2Q^4 = LHS
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
How to get the Normal line? My book proposed to me to find the the Normal lines to the curve that pass through the origin.The answer must be the intersection points between them.
The curve: $\dfrac{2}{1+x^2}$
My first idea is to get the derivative: $f'(x) = $ $\frac{-4x}{x^4 + 2x^2 + 1}$, then get the perpendicular of $f'(x)$, resulting in
$M = \dfrac{-1}{f'(x)}$ = $\dfrac{x^4 + 2x^2 + 1}{4x}$.
The segment that pass through the origin with the angular coeficient $M$ is:
$g(x) = M * (x - x0) + y0$; to $x0$ = 0 and $y0$ = 0.
So, $g(x) = \dfrac{x^4 + 2x^2 + 1}{4}$
Finally to get the common points between $g(x)$ and $f(x)$, I have the equality: $f(x) = g(x)$.
$\dfrac{x^4 + 2x^2 + 1}{4}$ = $\dfrac{2}{x^2 + 1}$
Resulting in two real solutions: $x = -1$ and $x = 1$ and another four complex roots. If you look carefully, we can see a "hidden" solution at x = 0.
Is it the right way to procced to get the points !?
| If $\displaystyle x=\tan\theta, y=\frac2{1+x^2}=2\cos^2\theta$
So, $\displaystyle\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}=-4\tan\theta\cos^4\theta=-4\sin\theta\cos^3\theta$
So, the equation of the normal at $(\tan\theta,2\cos^2\theta)$
$$\frac{y-2\cos^2\theta}{x-\tan\theta}=-\frac1{-4\sin\theta\cos^3\theta}$$
$$\implies(y-2\cos^2\theta)4\sin\theta\cos^4\theta=x-\sin\theta \ \ \ \ (1)$$
Now as the normal passes through the origin, set $x=y=0$ to find the required values of $\theta$
$$\implies(0-2\cos^2\theta)4\sin\theta\cos^4\theta=0-\sin\theta\iff\sin\theta(8\cos^6\theta-1)=0$$
If $\displaystyle\sin\theta=0\implies\cos^2\theta=1, (1)$ becomes $\displaystyle y-2=x$
If $\displaystyle8\cos^6\theta-1=0,$ the only real value of $\displaystyle\cos^2\theta$ is $\displaystyle\frac12\implies\sin\theta=\pm\frac1{\sqrt2}$
Put the values of $\displaystyle\cos^2\theta,\sin\theta$ in $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.
So $\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4$.
Now how can I find the max. and min. values of $|z|$?
Help me please.
Thanks
| Let $z=\color{red}{\sqrt{2}}re^{i\theta}$, then $\left|z+\frac{2}{z}\right|=2$ reduces to $$r^2+\frac{1}{r^2}=2-2\cos{2\theta} \tag{1}$$
Observe that $(1)$(considering it as a quadratic in $r^2$) will have two real roots of the form $\alpha$ and $\frac{1}{\alpha}$ only if $\cos{2\theta} \le 0$. Thus if the maximum value is $\alpha$, the minimum would be $\frac{1}{\alpha}$.
Without loss of generality assume maximum(of $r^2$) is $\ge 1$, notice that $r^2+\frac{1}{r^2}$ is an increasing function when $r\ge 1$, hence maximum is attained when R.H.S is maximum, i.e. when $\cos{2\theta}=-1$. Solving for $r$, get $r=\sqrt{2+\sqrt{3}}\quad,\quad\frac{1}{\sqrt{2+\sqrt{3}}}$
Maximum value of $|z|=\sqrt{2}\cdot \sqrt{2+\sqrt{3}}=1+\sqrt{3}$ and minimum value of $|z|=\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/853992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$
How do I find the sum of this series?
$$\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$$
Hints in the right direction would be appreciated.
| From the identity $ \displaystyle \left(\sin(x)\right)^3 = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3$ you get $ \displaystyle \left(\sin(x)\right)^3 = \frac{3}{4}\sin(x)-\frac{1}{4}\sin(3x) $
In particular : $ \displaystyle \frac{\sin^3(3^n)}{3^n} = \frac{3}{4}\left(\frac{\sin(3^n)}{3^n}-\frac{\sin(3^{n+1})}{3^{n+1}}\right)$
So we have a telescopic partial sum : $$ \sum_{n=0}^N \frac{\sin^3(3^n)}{3^n} = \frac{3}{4} \sin(1)-\frac{3}{4} \frac{\sin(3^{N+1})}{3^{N+1}} $$
So since $ \displaystyle \lim_{x \rightarrow +\infty} \frac{\sin(x)}{x} = 0 $ by taking the limit as $ N \rightarrow +\infty $ gives : $$ \sum_{n=0}^{+\infty} \frac{\sin^3(3^n)}{3^n} = \frac{3}{4} \sin(1) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Show that the following series is less than $4 \pi^2 / 3$. Show: For any $k = 0,1,2,...$,
$$ \sum_{i=0}^{i=k} \frac{(k+1)^2}{(i+1)^2 (k-i+1)^2} \leq \frac{4 \pi^2}{3}. $$
| Hint:
1) From $(\frac{a+b}{2})^2\le\frac{a^2+b^2}{2}$ verify that $(k+1)^2 < 2[(i+1)^2+(k-i+1)^2]$,
2) $\sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\lim\limits_{x \to+\infty} \frac{4x-3}{2x+5}$ and $\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$ Hello guys help me with these:
1.) $\lim \limits_{x \to +\infty} frac{4x-3}{2x+5}$
$=\lim \frac{4-\frac{3}{x}}{2+\frac{5}{x}}$
$= \lim \frac{4-3\frac{1}{x}}{2+5\frac{1}{x}}$
=$\frac{\lim 4- \lim 3 \lim \frac{1}{x}}{\lim 2+ \lim 5 \lim\frac {1}{x}}$ // lim of $\frac{1}{x}$=0
=$\frac{4 - 3 * 0}{2 + 5 * 0}$
=$\frac{4 - 0}{2 + 0}$
=$\frac{4}{2}
=2$
lim of the function as $x$ approaches positive infinity is $2$.
Is it correct?
If then what should I do if the given is:
$\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$
| When you evaluate a limit approaching either positive or negative infinity, you could easily evaluate it by dividing top and bottom by $x$ which has the highest power.
$$
\lim_{x \rightarrow \infty} \frac{x}{x^2}
$$
Given / $x^2$:
$$
\lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{1} = \lim_{x \rightarrow \infty} \frac{\frac{1}{\infty}}{1} = 0 \because \frac{1}{\infty} = 0
$$
When $x$ goes to infinity only the highest power x matters. It is because how fast the value of each x changes is quite different from other's.
To take an instance:
$f(x) = x, g(x) = x^2$
When $x = 2$, $f(2) = 2$ and $g(2) = 4$
After a while, when $x = 20$, $f(20) = 20$, $g(20) = 400$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$ First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.
I'm trying to solve the following limit:
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}
$$
This limit is $0/0$ if evaluated directly, so I tried multiplying by the conjugate of the denominator:
$$
\begin{align}
& = \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)}\\
& = \lim_{x\to\pi/6}\frac{(2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{3} + 3)} \\
& = \lim_{x\to\pi/6}\frac{2\sin{x} - 1}{2\sqrt{3}\cos{x} + 3}\\
& = \frac{2(1/2) - 1}{2\sqrt{3}\frac{\sqrt{3}}{2} + 3}\\
& = \frac{0}{6}\\
& = 0
\end{align}
$$
But according to WolframAlpha this is incorrect, and the limit should be 1. What have I done wrong?
Also, as I have only just begun I am unfamiliar with L'Hopital's rule.
| Another way is to use change of variables. Let $y = x - \pi/6$.
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3} = \lim_{y \to 0}\frac{1 - 2\sin{(y + \pi/6)}}{2\sqrt{3}\cos{(y+\pi/6)} - 3} = L
$$
But,
$$
1 - 2\sin(y + \pi/6) = (1 - \cos y) - \sqrt{3}\sin y
$$
and
$$
2\sqrt{3}\cos{(y+\pi/6)} - 3 = 2\sqrt{3}\bigg(\frac{\sqrt{3}}{2}\cos y - \frac{1}{2}\sin y\biggr) - 3
$$
$$
= 3\cos y - \sqrt{3}\sin y - 3 = -3(1 - \cos y) - \sqrt{3}\sin y
$$
Thus,
$$
L = \lim_{y \to 0}\dfrac{\dfrac{1 - \cos y}{y} - \dfrac{\sqrt{3}\sin y}{y}}{-3\dfrac{1 - \cos y}{y} - \sqrt{3}\dfrac{\sin y}{y}} = 1
$$
I used the fact that
$$
\lim_{y \to 0}\frac{1 - \cos y}{y} = 0 \quad \text{and} \quad \lim_{y \to 0}\frac{\sin y}{y} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
A closed form of the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$ I have found a closed form of the following new series involving non-linear harmonic numbers.
Proposition. $$\sum_{n=1}^{\infty} \dfrac{H_n^2-(\gamma + \ln n)^2}{n} = \dfrac{5}{3}\zeta(3)-\dfrac{2}{3}\gamma^3-2\gamma \gamma_{1}-\gamma_{2} $$
where \begin{align} & H_{n}: =\sum_{k=1}^{n}\frac{1}{k} \\ &\gamma: =\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{1}{k}-\ln n\right) \\ & \gamma_{1}:=\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{\ln k}{k}-\frac{1}{2}\ln^2 n\right)\\& \gamma_{2}: =\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{\ln^2 k}{k}-\frac{1}{3}\ln^3 n\right), \end{align}
$\gamma_1, \gamma_2$ being Stieltjes constants.
What tool would use to prove it?
|
Here is a nice solution from R. Tauraso, for those interested in this problem:
Solution proposed by Roberto Tauraso, Roma, Italy.
The Multiple Harmonic Sum is defined by $$H_{n}(s_{1}, \dots, s_{l}) == \sum_{0 < k_{1} < k_{2} < \dots < k_{l} \leq n}\prod_{i = 1}^{l}\frac{1}{k_{i}^{s_{i}}}$$ Then, by the known properties of the stuffle product,
\begin{align}
H_{n}^{2}(1) &= 2H_{n}(1, 1) + H_{n}(2)\notag\\
H_{n}^{3}(1) &= 6H_{n}(1, 1, 1) + 3H_{n}(1, 2) + 3H_{n}(2, 1) + H_{n}(3)\notag
\end{align}
Therefore, for $N > 0$,
\begin{align}
\sum_{n = 1}^{N}\frac{H_{n}^{2}}{n} &= 2\sum_{n = 1}^{N}\frac{H_{n}(1, 1)}{n} + \sum_{n = 1}^{N}\frac{H_{n}(2)}{n}\notag\\
&= 2H_{N}(1, 1, 1) + 2H_{N}(1, 2) + H_{N}(3) + H_{N}(2, 1)\notag\\
&= \frac{1}{3}H_{N}^{3}(1) + \frac{2}{3}H_{N}(3) + H_{N}(1, 2)\notag\\
&= \frac{1}{3}H_{N}^{3}(1) + \frac{2}{3}\zeta(3) + \zeta(2, 1) + o(1)\notag\\
&= \frac{1}{3}H_{N}^{3}(1) + \frac{5}{3}\zeta(3) + o(1)\tag{1}
\end{align}
because $\zeta(2, 1) = \zeta(3)$ (see for example Thirty-two Goldbach Variations by J. M. Borwein and D. M. Bradley).
Moreover, $H_{N}(1) = \ln N + \gamma + O(1/N)$ implies that
\begin{align}
\frac{1}{3}H_{N}^{3}(1) - \sum_{n = 1}^{N}\frac{(\gamma + \ln n)^{2}}{n} &= \frac{1}{3}H_{N}^{3}(1) - \gamma^{2}H_{N}(1) - 2\gamma\sum_{n = 1}^{N}\frac{\ln n}{n} - \sum_{n = 1}^{N}\frac{\ln^{2}n}{n}\notag\\
&= \frac{\ln^{3}N}{3} + \frac{\gamma^{3}}{3} + \gamma\ln^{2}N + \gamma^{2}\ln N - \gamma^{2}(\ln N + \gamma)\notag\\
&\,\,\,\,\,\,\,\, -2\gamma\left(\gamma_{1} + \frac{\ln^{2}N}{2}\right) - \left(\gamma_{2} + \frac{\ln^{3}N}{3}\right) + o(1)\notag\\
&= -\frac{2}{3}\gamma^{3} - 2\gamma\gamma_{1} - \gamma_{2} + o(1)\tag{2}
\end{align}
By adding together $(1)$ and $(2)$, and by taking the limit as $N \to \infty$, we get the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 1,
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} |
$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down. Otherwise known as $12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}$
How do you simplify this multi level fractional radical expression into $4\sqrt{6}$.
| Here's yet another approach: $\frac1{\sqrt{2}}$ is often (equivalently) written as $\frac{\sqrt{2}}2$ (an equivalency that can be easily seen by multiplying both numerator and denominator by $\sqrt{2}$). Using this form turns the original formula into $12\cdot\dfrac{\frac{\sqrt{2}}2}{\frac{\sqrt{3}}2}$ $= 12\cdot\dfrac{\sqrt{2}}{\sqrt{3}}$ (by multiplying numerator and denominator by an easy factor of two). Now, again 'clear the radical' in the denominator by multiplying by $1=\dfrac{\sqrt{3}}{\sqrt{3}}$: $12\cdot\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}$ $= 12\cdot\dfrac{\sqrt{6}}{3}$ $=4\sqrt{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Another integral for $\pi$ Here is a new integral for $\pi$.
$$\int_{0}^{1}\sqrt{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} = \pi $$
where $\left\{x\right\}$ denotes the fractional part of $x$.
Do you have any proof?
| Let me suggest a general approach.
We start with a theorem.
Theorem. (O. Oloa) Let $f$ be a piecewise continuous function on $(0,1)$ verifying $ \displaystyle \int_{0}^{1} \frac{ \left|f (x) \right| }{x} \: \mathrm{d}x <\infty.$
Then $$ \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} =
\int_{0}^{1} f(x) \frac{ \mathrm{d}x}{x} \qquad (*) $$ where $\left\{x\right\}$ denotes the fractional part of $x$.
Proof. One has \begin{align*} \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} &=
\sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}}
f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} \\
&= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left(\left\{ u \right\}\right) \: \frac{\mathrm{d} u}{u(u-1)} \\
&= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left(u-k\right) \: \frac{\mathrm{d} u}{u(u-1)} \\
&= \sum_{k=1}^{\infty} \int_{0}^{1} f \left(v\right) \: \frac{\mathrm{d} v}{(v+k)(v+k-1)} \\
&= \int_{0}^{1} f \left(v\right) \sum_{k=1}^{\infty} \frac{1}{(v+k)(v+k-1)} \mathrm{d} v \\
&= \int_{0}^{1} f(v) \frac{ \mathrm{d}v}{v},\end{align*} where the permutation between the infinite sum and the integration is allowed by the dominated convergence theorem: let $v \in (0,1)$, as $N$ tends to $\infty$, \begin{align}
\displaystyle f \left(v\right) \sum_{k=1}^{N} \frac{1}{(v+k)(v+k-1)} & = f(v) \sum_{k=1}^{N}\left(\frac{1}{v+k-1}-\frac{1}{v+k}\right)
= \frac{f(v)}{v}-\frac{f(v)}{v+N} \longrightarrow \frac{f(v)}{v} \nonumber \end{align}
and one has \begin{align}
\displaystyle \left|f \left(v\right) \sum_{k=1}^{N} \frac{1}{(v+k)(v+k-1)}\right|
= \left| \frac{f(v)}{v}-\frac{f(v)}{v+N}\right|
= \frac{\left| f(v) \right|}{v}-\frac{\left| f(v) \right|}{v+N} \leq \frac{ \left|f (v) \right| }{v} \nonumber
\end{align} the latter function being such that $ \displaystyle \int_{0}^{1} \frac{ \left|f (v) \right| }{v} \: \mathrm{d}v <\infty$ by hypothesis. $\square$
Now, to get our result one may apply $(*)$ with $\displaystyle f(x)=\sqrt{\frac{x}{1-x}}$. Clearly $f$ satisfies the hypotheses of our Theorem and $$
\displaystyle \int_{0}^{1} \left| f(x) \right| \: \frac{ \mathrm{d}x}{x} =
\int_{0}^{1} \left( \frac{x}{1-x}\right)^{1/2} \frac{ \mathrm{d}x}{x}
= 2 \int_{0}^{1} \frac{1}{\sqrt{1-u^2}} \: \mathrm{d}u
= \pi. $$
Similarly, inserting successively \begin{align} \displaystyle f(x):= \sqrt{\frac{x}{1-a^2 x}} \, ,\qquad 0 < a \leq 1 \nonumber \end{align} and
\begin{align}\displaystyle f(x):=\sqrt[n]{\frac{x}{1-x}} \, ,\qquad n=2,3, \cdots\nonumber \end{align}
into $(*)$ gives some generalizations.
Proposition 1. Let $ 0 < a \leq 1$. Then \begin{align}
\displaystyle \int_{0}^{1} \sqrt{\frac{\left\{1/x\right\}}{1-a^2 \left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x}
& = \frac{2\arcsin a}{a} \end{align} where $\left\{x\right\}$ denotes the fractional part of $x$.
The initial integral is obtained with $a=1$.
Proposition 2. Let $n=2,3, \cdots .$ Then \begin{align}
\displaystyle \int_{0}^{1} \sqrt[n]{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x}
& = \frac{\pi}{\sin(\pi/n)} \end{align} where $\left\{x\right\}$ denotes the fractional part of $x$.
The initial integral is obtained with $n=2$.
The (new) identity $(*)$ may be seen as a dual of the Gauss famous invariance result in ergodic theory:
Theorem. Let $f \in L^1(0,1)$. Then
$$ \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{x+1} =
\int_{0}^{1} f(x) \frac{ \mathrm{d}x}{x+1}
$$
where $\left\{x\right\}$ denotes the fractional part of $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$ The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.
Ramanujan established many such denesting of radicals such as $$\sqrt{\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}} = \sqrt[5]{1 + \sqrt[5]{2} + \sqrt[5]{8}} = \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}}\tag {2}$$$$\sqrt[3]{\sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}\tag {3}$$$$\sqrt[4]{\frac{3 + 2\sqrt[4]{5}}{3 - 2\sqrt[4]{5}}} = \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\tag{4}$$$$\sqrt[\color{red}6]{7\sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}\tag{5}$$$$\sqrt[6]{4\sqrt[3]{\frac{2}{3}} - 5\sqrt[3]{\frac{1}{3}}} = \sqrt[3]{\frac{4}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{1}{9}}\tag{6}$$
$$\sqrt[8]{1\pm\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2}\,\frac{\sqrt[4]{5}\pm 1}{\sqrt{2}}\tag{7}$$
with the last one found in Ramanujan's Notebooks, Vol 5, p. 300. Most of these radical expressions are units (a unit is an algebraic integer $\alpha$ such that $\alpha\beta = 1$ where $\beta$ is another algebraic integer).
For me the only way to establish these identities is to raise each side of the equation to an appropriate power using brute force algebra and then check the equality. However for higher powers (for example equation $(2)$ above) this seems very difficult.
Is there any underlying structure in these powers of units which gives rise to such identities or these are mere strange cases which were noticed by Ramanujan who used to play with all sorts of numbers as a sort of hobby? I believe (though not certain) that perhaps Ramanujan did have some idea of such structure which leads to some really nice relationships between units and their powers. I wonder if there is any sound theory of such relationships which can be exploited to give many such identities between nested and denested radicals.
| We also have the following identity,
$$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$
For $m=n=1$ we get $(1)$.
For $m=4$ and $n=1$ we get $(5)$.
| {
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"source": "stackexchange",
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Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?
Again the sequence has a group of three terms it repeats: Add one to first term to get second term, then double second term to get third term.
How do you compute the $n$th term of such sequences directly, without iterating through all preceding terms?
| A more general approach to solving for the $n$th term of such sequences uses matrix multiplication. Suppose the even terms are nonzero constant $r\neq 1$ times the preceding odd terms, while the odd terms are constant $d$ plus the preceding even terms. We have:
$$ \begin{pmatrix} 1 & 0 & d \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} a_{2n} \\ a_{2n-1} \\ 1 \end{pmatrix} =
\begin{pmatrix} a_{2n+1} \\ a_{2n} \\ 1 \end{pmatrix} $$
$$ \begin{pmatrix} r & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} a_{2n+1} \\ a_{2n} \\ 1 \end{pmatrix} =
\begin{pmatrix} a_{2n+2} \\ a_{2n+1} \\ 1 \end{pmatrix} $$
Combining these by matrix multiplication gives the double step:
$$ \begin{pmatrix} r & 0 & rd \\ 1 & 0 & d \\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} a_{2n} \\ a_{2n-1} \\ 1 \end{pmatrix} =
\begin{pmatrix} a_{2n+2} \\ a_{2n+1} \\ 1 \end{pmatrix} $$
The problem is then reduced to finding a closed form for natural powers of the matrix:
$$ A = \begin{pmatrix} r & 0 & rd \\ 1 & 0 & d \\ 0 & 0 & 1 \end{pmatrix} $$
which can be done by diagonalization, since $A$ has three distinct eigenvalues $0,1,r$.
Represent $A$ with respect to the corresponding basis of eigenvectors:
$$ \left\{ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} -rd \\ - d \\ r-1 \end{pmatrix},
\begin{pmatrix} r \\ 1 \\ 0 \end{pmatrix} \right\} $$
and the resulting similarity transformation diagonalizes $A$, say $A = S D S^{-1}$ where $D= \operatorname{diag}(0,1,r)$. Thus, assuming an initial value $a_1$ and $a_2 = ra_1$:
$$ A^n \begin{pmatrix} a_2 \\ a_1 \\ 1 \end{pmatrix} =
S D^n S^{-1} \begin{pmatrix} ra_1 \\ a_1 \\ 1 \end{pmatrix} =
\begin{pmatrix} a_{2n+2} \\ a_{2n+1} \\ 1 \end{pmatrix} $$
The powers of $D$ are explicitly $D^n = \operatorname{diag}(0,1,r^n)$, so this gives a direct expression for any terms in the sequence starting from $a_1$:
$$ a_{2n+1} = r^n a_1 + \frac{r^n -1}{r-1} d $$
$$ a_{2n+2} = r a_{2n+1} = r^{n+1} a_1 + \frac{r^n -1}{r-1} rd $$
taking advantage of the calculation DanielV carried out in the Comment below.
This matrix multiplication technique can be modified to handle more general mixtures of arithmetic and geometric rules.
| {
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"source": "stackexchange",
"question_score": "5",
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Show that {$a_n$} is convergent and find sup{$a_n| n \in Z_+ $} $a_1 = 1$ and $ a_{n+1} = \frac{4+3a_n}{3+2a_n} ; \forall n \in Z_+$
Show that {$a_n$} is convergent, find its limit and find sup{$a_n| n \in Z_+ $} if exists.
I found the limit as follows -
Let$$\lim_{n \to \infty} a_n = L$$
$$\lim_{n +1\to \infty} a_n+1 = L$$
$$L = \frac{4+3L}{3+2L}$$ which gives$ L= \sqrt{2}$
$$\therefore \lim_{n \to \infty} a_n = \sqrt{2}$$
${a_n} $ is convergent ( $\sqrt{2} \in \mathbb R$)
Is there a fault in this? Can I straightaway say that $\sup\{a_n| n \in Z_+ \} =\sqrt{2} $ ?
| $a_{n+1} - a_n = \dfrac{4+3a_n}{3+2a_n} - a_n = \dfrac{4-2a_n^2}{3+2a_n}$. So consider $f(x) = \dfrac{4+3x}{3+2x}$, we have: $f'(x) = \dfrac{1}{(3+2x)^2} > 0$, thus $f$ increases strictly. Next assume by induction that : $0 < a_n < \sqrt{2}$, then: $a_{n+1} = f(a_n) < f(\sqrt{2}) = \dfrac{4+3\sqrt{2}}{3+2\sqrt{2}} = \sqrt{2}$. Thus by induction $a_n < \sqrt{2}$, $\forall n$. Also this implies: $a_{n+1} - a_n = \dfrac{4-2a_n^2}{3+2a_n} > 0$. Thus: $0 < a_n < a_{n+1} < \sqrt{2}$, $\forall n$. Thus: $a_n$ increases and is bounded hence convergent to $L$. You solved $L = \sqrt{2}$, and this should imply that $\sqrt{2} = L = \sup\{a_n:n\in \mathbb{N}\}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the probability of each outcome when a biased die is rolled, if rolling a 2 or 4 is three times as likely as rolling each of the other$\dots$ Question:Find the probability of each outcome when a biased die is rolled, if rolling a $2$ or $4$ is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4.
My Attempt:
Let $O_1, O_2, \dots, O_6$ be the out come of dies.
Since rolling $O_2$ or $O_4$ is twice as likely, then
$$P(O_2) = P(O_4) = 3P(O_1) = 3P(O_3) = \dots = 3P(O_6)$$
Thus,
$$P(O_1) + P(O_2) + P(O_3) + P(O_4) + P(O_5) + P(O_6) = 1$$
$$P(O_1) + 3P(O_1) + P(O_1) + 3P(O_1) + P(O_1) + P(O_1) = 1$$
$$10P(O_1) = 1$$
$$P(O_1) = \dfrac{1}{10}$$
Therefore,
$$P(O_1) = P(O_3) = P(O_5) = P(O_6) = \dfrac{1}{10}$$ and,
$$P(O_2) = P(O_4) = \dfrac{3}{10}$$
Problem:
The book gave a different answer such that,
$$P(O_1) = P(O_3) = P(O_5) = P(O_6) = \dfrac{1}{16}$$ and,
$$P(O_2) = P(O_4) = \dfrac{6}{16}$$
Frankly, is $P(O_2) = P(O_4) = 6P(O_1)$ now? I think I'm missing something fundamental here.
| Your confusion stems from the obscure description of the question at hand.
The actual meaning of the question is apparently $\displaystyle P(2,4) = 3 \cdot P(1,3,5,6)$.
Therefore, $\displaystyle P(2,4) = \frac{12}{16} = 3 \cdot \frac{4}{16} = 3 \cdot P(1,3,5,6)$ implies the following:
*
*$\displaystyle P(2) = \frac{1}{2} \cdot \frac{12}{16} = \frac{6}{16}$
*$\displaystyle P(4) = \frac{1}{2} \cdot \frac{12}{16} = \frac{6}{16}$
*$\displaystyle P(1) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$
*$\displaystyle P(3) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$
*$\displaystyle P(5) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$
*$\displaystyle P(6) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$
In other words, you should change this attempt:
*
*$P(2) = P(4) = 3P(1) = 3P(3) = 3P(5) = 3P(6)$
To this attempt:
*
*$P(2) + P(4) = 3P(1) + 3P(3) + 3P(5) + 3P(6)$
Where:
*
*$P(2) = P(4)$
*$P(1) = P(3) = P(5) = P(6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calculate the series without using Wolfram Alpha? Please help me. Grazie!
| Let
$$\begin{align}f(x)&=\sum_{n=0}^\infty(2n+1)x^{2n+1}\\&=x\sum_{n=0}^\infty(2n+1)x^{2n}\\&=x\frac{d}{dx}\left(\sum_{n=0}^\infty x^{2n+1}\right)\\&=x\frac{d}{dx}\left( \frac{x}{1-x^2}\right)\\&=x\frac{x^2+1}{(1-x^2)^2}\end{align}$$
and notice that the desired sum is $f\left(\frac15\right)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
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Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$ Please help me to evaluate this integral:
$$
I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\,
\,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1
$$
Mathematica could not evaluate it in a closed form. A numerical integration returned
$$I \approx 4.25314982536869548103063\ldots\,,\tag2$$
but neither
WolframAlpha nor ISC+ could find a plausible closed form for this.
| \begin{align}
-\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
&=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\
&=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\
&=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}{x^4-2x^2-1}{\rm d}x\\
\end{align}
For simplicity's sake, let $\displaystyle\frac{1}{x^4-2x^2-1}=\sum^4_{k=1}\frac{c_k}{x-r_k}$. The integral becomes
$$I=-4\sum^4_{k=1}\int^1_0\frac{c_k\ln(1-x^4)-c_k\ln{2}-2c_k\ln{x}}{x-r_k}{\rm d}x$$
The second integral is
\begin{align}
-c_k\ln{2}\int^1_0\frac{1}{x-r_k}{\rm d}x=-c_k\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)
\end{align}
The third integral is
\begin{align}
-2c_k\int^1_0\frac{\ln{x}}{x-r_k}{\rm d}x
&=2c_k\int^{1/r_k}_0\frac{\ln(r_kx)}{1-x}{\rm d}x\\
&=-2c_k{\rm Li}_2\left(\frac{1}{r_k}\right)
\end{align}
Pluck these results back in.
$$I=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)-\int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\right]$$
The remaining integral is
\begin{align}
&\ \ \ \ \ \int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\\
&=\sum_{j=1,-1,i,-i}\int^1_0\frac{\ln(1+jx)}{x-r_k}{\rm d}x\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\ln\left(1+\frac{j\lambda}{y}+jr_k\right)\frac{{\rm d}y}{y}\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\left[\ln\left(1+\frac{1+jr_k}{j\lambda}y\right)-\ln\left(\frac{y}{j\lambda}\right)\right]\frac{{\rm d}y}{y}\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{1+jr_k}{j-jr_k}}_{-\frac{1+jr_k}{jr_k}}\frac{\ln(1+y)}{y}-\frac{1}{y}\ln\left(\frac{y}{1+jr_k}\right){\rm d}y\\
&=-\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]
\end{align}
Final Result:
\begin{align}
\color\purple{\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)+\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]\right]}
\end{align}
where
\begin{align}
\frac{c_1}{x-r_1}&=\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x-\sqrt{1+\sqrt{2}}\right)}\\
\frac{c_2}{x-r_2}&=-\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x+\sqrt{1+\sqrt{2}}\right)}\\
\frac{c_3}{x-r_3}&=-\frac{1}{4\sqrt{2}\left(x+i\sqrt{\sqrt{2}-1}\right)}\\
\frac{c_4}{x-r_4}&=-\frac{1}{4\sqrt{2}\left(x-i\sqrt{\sqrt{2}-1}\right)}\\
\end{align}
| {
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prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$
Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then:
$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$
yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc} \frac {a^3} {b+c+d})\times(\sum\limits_{cyc} a(b+c+d)) \geq (\sum\limits_{cyc} a^2)^2$
so my problem is simplified to proving this:
$\frac {(\sum\limits_{cyc} a^2)^2} {(\sum\limits_{cyc} a(b+c+d))} \geq \frac {1} {3}$
$3 \times (\sum\limits_{cyc} a^2)^2 \geq \sum\limits_{cyc} a(b+c+d)$
$3 \times (a^2+b^2+c^2+d^2)^2 \geq 2(ab+ac+ad+bc+bd+cd)$
someone said to me if i play around with AM-GM it could be solved and i'm almost there
my idea is this right now:
prove $(a^2+b^2+c^2+d^2) \geq ab+bc+cd+da=1$ proved(with help of Jineon Baek hint)
prove $3(a^2+b^2+c^2+d^2) \geq 2(ab+ac+ad+bc+bd+cd)$ proved(with help of Jineon Baek hint)
| To finish your idea, just add up all inequalities of type $a^2+b^2 \geq 2ab$.
| {
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"source": "stackexchange",
"question_score": "4",
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Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| Another way :
\begin{align}
\int\frac{1}{1+\color{blue}{\sin x}+\color{red}{\cos x}}\ dx&=\int\frac{1}{\sin^2\frac x2+\cos^2\frac x2+\color{blue}{2\sin\frac x2\cos\frac x2}+\color{red}{\cos^2\frac x2-\sin^2\frac x2}}\ dx\\
&=\int\frac{1}{2\cos\frac x2\left(\sin\frac x2+\cos\frac x2\right)}\ dx\\
&=\int\frac{1}{2\cos^2\frac x2\left(\tan\frac x2+1\right)}\ dx\\
&=\int\frac{\frac12\sec^2\frac x2}{1+\tan\frac x2}\ dx\\
&=\int\frac{d\left(\tan\frac x2\right)}{1+\tan\frac x2}\\
&=\ln\left|1+\tan\frac x2\right|+C,
\end{align}
then use trigonometric identity: $\displaystyle\color{blue}{\tan\frac x2=\frac{\sin\frac x2}{\cos\frac x2}}$ and will, of course, obtain the desired answer.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2} \right )$. I tried all identities I know but I have no idea how to proceed.
| It shouldn't be that hard cause you have this two.
$\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})}+\frac{\cos (\frac{B}{2})}{\sin (\frac{B}{2})}+\frac{\cos (\frac{C}{2})}{\sin (\frac{C}{2})}=\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})}\times \frac{\cos (\frac{B}{2})}{\sin (\frac{B}{2})}\times \frac{\cos (\frac{C}{2})}{\sin (\frac{C}{2})}$
$A+B+C=180$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Find limit of a fraction without derivatives: $\lim\limits_{h\rightarrow0} \frac{\sqrt[3]{x+h} - \sqrt[3]{x}}h$ How to find limit of this?
\begin{equation}
\lim_{h \rightarrow 0} \frac {\sqrt[3]{x+h} - \sqrt[3]{x}} {h}.
\end{equation}
I see here the derivative definition, so I understand in "derivative mind" that answer has to be \begin{equation} \frac{1}{3} x^{-\frac{2}{3}}, \end{equation}but in the book this expression goes before derivatives, so it must be another approach to find this limit.
| Rationalise the numerator by the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$
$\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}$
$=\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}{\frac { \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3
}}{ \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3}}}
$
$=\lim _{h\rightarrow 0}{\frac {(x+h)-x}{h \left( \left( x+h \right) ^{2/3}+
\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3} \right) }}$
$=\lim _{h\rightarrow 0}{\frac {1}{ \left( x+h \right) ^{2/3}+
\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3} }}$
$={\frac {1}{ \left( x \right) ^{2/3}+
\sqrt [3]{x}\sqrt [3]{x}+{x}^{2/3} }}$ by plugging in $h=0$
$={\frac{1}{3x^{2/3}}}$
=$\frac{1}{3}x^{-2/3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Verification of the Stokes theorem for the surface that is a part of a cone
Let $S$ consist of the part of the cone $z=(x^2+y^2)^{1/2}$ for $x^2+y^2\leq9$ and suppose $${\bf A}=(-y,x,-xyz).$$ Verify that Stokes theorem is satisfied for this choice of $\bf A$ and $S$.
In the solution to this question in order to work out the surface integral you can project on to $z=3$ and evaluate over the region $x^2+y^2\leq9$. I was just wondering whether or not you could do the same but this time project onto the plane $z=0$?
| By Stoke's theorem,
$$\int_S (\nabla \times \mathbf{A}) \cdot \mathbf{n} \,d\sigma = \oint_C \mathbf{A} \cdot d\mathbf{r},$$
where $C = \{(x,y,z):x^2+y^2=9,z =3\}$ oriented clockwise.
With $\mathbf{A} = -y\mathbf{i}+x\mathbf{j}+-xyz\mathbf{k},$ the line integral is
$$\eqalign{
\oint_C \mathbf{A} \cdot d\mathbf{r}&=-\int_C(-ydx+xdy)\\&=-\int_{0}^{2\pi}[(-3 \sin \theta)( -3 \sin \theta) + (3 \cos \theta)( 3 \cos \theta)]\,d \theta\\&=-18\pi.
}$$
Now we can do the surface integral by projecting on the plane $z=0$.
We have
$$\nabla \times \mathbf{A}= -xz\mathbf{i}+yz\mathbf{j}+2\mathbf{k}.$$
The surface can be expressed as $z=g(x,y)=\sqrt{x^2+y^2}$ over the region $\{(x,y):0 \leq x^2+y^2 \leq 9\}$ in the $x$-$y$ plane. The differential element of area is
$$d\sigma = \sqrt{1+ \Big{(}\frac{\partial g}{\partial x}\Big{)}^2+\Big{(}\frac{\partial g}{\partial y}\Big{)}^2 }dxdy$$
and the outer unit normal vector is
$$\mathbf{n}=\frac{ \Big{(}\frac{\partial g}{\partial x}\Big{)}\mathbf{i}+\Big{(}\frac{\partial g}{\partial y}\Big{)}\mathbf{j}-\mathbf{k} }{\sqrt{1+ \Big{(}\frac{\partial g}{\partial x}\Big{)}^2+\Big{(}\frac{\partial g}{\partial y}\Big{)}^2 }},$$
where
$$\frac{\partial g}{\partial x}= \frac{x}{\sqrt{x^2+y^2}},\,\frac{\partial g}{\partial y}= \frac{y}{\sqrt{x^2+y^2}}$$
The surface integral reduces to
$$\eqalign{
\int_S (\nabla \times \mathbf{A}) \cdot \mathbf{n} \,d\sigma &= \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\left[-xz\frac{\partial g}{\partial x}+yz\frac{\partial g}{\partial y}-2\right]\,dx\,dy\\
&=\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\left(-x^2+y^2-2\right)\,dx\,dy\\
&=4\int_{0}^{3} \left[\int_{0}^{\sqrt{9-x^2}}\left(-x^2+y^2-2\right)\,dy\right]\,dx\\
&=4\int_{0}^{\pi/2} \left[\int_{0}^{3}\left(-r^2 \cos^2 \theta+r^2\sin^2 \theta-2\right)\,rdr\right]\,d \theta\\
&=4\int_{0}^{\pi/2} \left[\int_{0}^{3}\left(-r^2 \cos 2 \theta - 2\right)\,rdr\right]\,d \theta\\
&=4\int_{0}^{\pi/2} \left(-\frac{81}{4} \cos 2 \theta - 9\right)\,d \theta\\
&=\left.4\left(-\frac{81}{8} \sin 2 \theta - 9\theta\right)\right|_0^{\pi/2}\\
&=-18\pi
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find depth of a half-filled parabolic cross-section Given a cross-section of an object that is parabolic in shape, how do you find the depth of the object when it is "half full".
A full example given in an exam:
A long trough whose cross-section is parabolic is $1\frac{1}{2}$ metres wide at the top and $2$ metres deep. Find the depth of water when it is half-full.
| Model the cross-section as a parabola centred at the origin. Then we know that:
$$
y = ax^2
$$
for some constant $a$. But from the given dimensions, we also know that the parabola contains the point $(3/4, 2)$. Substituting, we find that:
$$
2 = a(3/4)^2 \iff 2 = \frac{9a}{16} \iff a = \frac{32}{9}
$$
Now the area of the cross-section when full is given by:
$$
A = 2\int_0^{3/4} (2 - \tfrac{32}{9}x^2) \, dx = 2 \left[2x - \frac{32}{27}x^3 \right]_0^{3/4} = 2\left[\frac{3}{2} - \frac{1}{2} \right] = 2
$$
Now suppose that the half-width of the cross-section when half-full is $x = k$ so that its depth when half-full is $\frac{32}{9}k^2$. Then it follows that:
\begin{align*}
\frac{A}{2} &= 2\int_0^{k} (\tfrac{32}{9}k^2 - \tfrac{32}{9}x^2) \, dx \\
\frac{9A}{128} &= \int_0^{k} (k^2 - x^2) \, dx \\
\frac{9(2)}{128} &= \left[k^2x - \frac{x^3}{3} \right]_0^k \\
\frac{9}{64} &= \frac{2k^3}{3} - 0 \\
\frac{27}{128} &= k^3 \\
k &= \frac{3}{4\sqrt[3]{2}}
\end{align*}
Hence, we conclude that the desired depth is:
$$
\frac{32}{9}k^2 = \frac{32}{9}\cdot \frac{9}{16\sqrt[3]{4}} = \frac{2}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \sqrt[3]{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$ How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$
Wolfram Alpha agrees numerically.
I tried replacing $\sin p$ by $\dfrac{e^{ip}-e^{-ip}}{2i}$ and similarly for $\cos p$, but in vain.
| The following is an approach using contour integration.
$ $
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{3}(\pi x^{2}) \cos(4x^{2})}{x^{5}} \ dx &= \frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3} \ dt \\ &= \frac{1}{4} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) \cos(4t)}{t^{3}} \ dt \\ &= \frac{1}{4} \text{Re} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{16} \text{Re} \int_{-\infty}^{\infty} \frac{[3 \sin (\pi t) - \sin(3\pi t)]e^{4it}}{t^{3}} \ dt\\ &= \frac{1}{32} \text{Re} \frac{1}{i} \int_{-\infty}^{\infty} \frac{(3e^{\pi i t} - 3e^{ -\pi i t} - e^{3 \pi i t} + e^{-3 \pi i t})e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{(4+\pi) i t} - 3 e^{(4- \pi) i t} - e^{(4+3 \pi) i t} + e^{(4-3 \pi) i t}}{(t-i \epsilon)^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} 2 \pi i \ \text{Res} \left[ \frac{3e^{(4+\pi)iz} - 3 e^{(4- \pi)iz} - e^{(4+ 3 \pi)iz}}{(z- i \epsilon)^{3}},i \epsilon \right] \\ &= \frac{\pi }{16} \text{Re} \lim_{\epsilon \to 0^{+}} \frac{1}{2!} \lim_{z \to i \epsilon} \frac{d^{2}}{dz^{2}} \left( 3e^{(4+ \pi)iz} - 3 e^{(4-\pi)iz} - e^{(4+ 3\pi)iz}\right) \\ &= \frac{\pi}{32} \left(-3( 4+ \pi)^{2} + 3 (4- \pi)^{2} +(4 + 3 \pi)^{2}\right) \\ &= \frac{\pi}{32} \left(3 \pi - 4 \right)^{2} \end{align}$$
EDIT:
Notice that on line 6, $4 - 3 \pi <0$.
So by Jordan's lemma, $\displaystyle \int_{-\infty}^{\infty}\frac{e^{(4- 3\pi)it}}{(z-i \epsilon)^{3}} dt$ would need to be broken off and evaluated separately by closing the contour with the lower half of $|z|=R$ as opposed to the upper half of $|z|=R$.
But since the pole is in the upper half-plane, $\displaystyle \int_{-\infty}^{\infty} \frac{e^{(4- 3\pi)it}}{(t-i \epsilon)^{3}} \ dt $ evaluates to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How do you prove this integral involving the Glaisher–Kinkelin constant According to wikipedia on the page Glaisher–Kinkelin constant
$$\int_0^{1/2} \ln\Gamma(x) dx=\frac32\ln \text{A}+\frac5{24}\ln 2+\frac14\ln\pi$$
I got interested in how you possibly could prove something like that, but couldn't find any citations about it on the wiki page.
| In general, $$ \begin{align} \int_{0}^{z} \log \Gamma(x) \ dx &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z) \\ &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} + \ln A - \frac{1}{12} + \zeta^{'}(-1,z) \end{align}$$
where $\zeta(a,z)$ is the Hurwitz zeta function and where I used the fact that $\zeta'(-1) = \frac{1}{12} - \log A$.
You can find a derivation here.
Then $$ \int_{0}^{1/2} \log \Gamma(x) \ dx = \frac{\log 2 \pi}{4} + \frac{1}{24} + \log A + \zeta' \left( -1, \frac{1}{2} \right) .$$
But using the identity $$\zeta \left( z, \frac{1}{2} \right) = (2^z-1) \zeta(z) \ , \tag{1}$$
we have
$$\zeta' \left( z, \frac{1}{2} \right) = \log(2) 2^{z} \zeta(z) +(2^z-1) \zeta'(z) . $$
And replacing $z$ with $-1$, $$\begin{align} \zeta' \left( -1, \frac{1}{2} \right) &= \frac{\log 2}{2} \zeta(-1) - \frac{1}{2} \zeta'(-1) \\ &= - \frac{\log 2}{24} + \frac{\log A}{2} - \frac{1}{24} \end{align}$$
where I used the fact that $\zeta(-1) = -\frac{1}{12}$.
Therefore,
$$ \begin{align} \int_{0}^{1/2}\log\Gamma (x) \ dx &= \frac{\log 2 \pi}{4} - \frac{\log 2}{24} + \frac{3}{2} \log A \\ &= \frac{3}{2} \log A + \frac{5}{24} \log 2 + \frac{\log \pi}{4} . \end{align}$$
$ $
$(1)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (11)
EDIT:
A proof showing that $\zeta'(-1) = \frac{1}{12} - \ln A$ can be found here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$ in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$
How can I integrate this expressions?
| Another idea (just reducing it to another form):
Let $$I=6\int \frac{1}{x^2+6x+3} dx=6\int \frac{1}{(x+3)^2-6} dx=\int \frac{1}{(\frac{1}{\sqrt{6}}(x+3))^2-1} dx$$.
Now let $$\frac{1}{\sqrt{6}}(x+3)=\cosh a$$, hence using $$\cosh^2 a - 1 = \sinh ^2 a$$ and $$\frac{1}{\sqrt{6}} = \sinh a \frac{da}{dx}\Leftrightarrow dx = da \sinh a \sqrt{6}$$ we get $$I=\int \frac{1}{\sinh ^2 a} da \sinh a\sqrt{6} = \sqrt{6} \int \frac{1}{\sinh a} da$$.
EDIT: Can someone please show me how to write bigger LaTeX?
EDIT2: Neat!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have done so far:
Using Cauchy inequalities I can write:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})(b+c+a) \geq (a+b+c)^2$$
$a+b+c=1$,So:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq 1$$
| lemma:
$$\sum_{cyc}\dfrac{a^2}{b}\ge\dfrac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ac}$$
Proof:
$$\Longleftrightarrow (ab+bc+ac)(\sum_{cyc}\dfrac{a^2}{b})\ge (a+b+c)(a^2+b^2+c^2)$$
$$\Longleftrightarrow a^3+b^3+c^3+a^2c+c^2b+b^2a+\sum_{cyc}\dfrac{a^3c}{b}\ge a^3+b^3+c^3+\sum_{sym }a^2b$$
$$\Longleftrightarrow\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge ac^2+cb^2+ba^2$$
By AM-GM,we have
$$\dfrac{a^3c}{b}+\dfrac{b^3a}{c}\ge 2a^2b$$
$$\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge 2b^2c$$
$$\dfrac{a^3c}{b}+\dfrac{c^3b}{a}\ge 2c^2a$$
if we prove this
$$\dfrac{a+b+c}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Deciding whether a number is rational (2 examples) 1) Prove that number irrational $\sqrt{7-\sqrt{2}}$
I created a polynomial $x=\sqrt{7-\sqrt{2}}$ so
$P(x)=x^4-14x^2+47$ and since $47$ is prime we check $P(x)$ for $ {1,-1,47,-47}$ and since all of them are $P(x)\neq0$ it means our number is irrational.
Is my prof OK ?
2) Decide if the number $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational or irrational. I don't know how to tackle this one. I'd be grateful for hints
| For 2).
If $p=\sqrt{\sqrt 5+3}+\sqrt{\sqrt 5-2}$ is a rational number, then
$$\begin{align}p-\sqrt{\sqrt5+3}=\sqrt{\sqrt5-2}&\Rightarrow p^2-2p\sqrt{\sqrt5+3}+\sqrt5+3=\sqrt5-2\\&\Rightarrow 2p\sqrt{\sqrt5+3}=p^2+5\\&\Rightarrow 4p^2(\sqrt 5+3)=(p^2+5)^2\\&\Rightarrow \sqrt5=\frac{(p^2+5)^2}{4p^2}-3\end{align}$$
implies that $\sqrt 5$ is a rational number. This is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Prove $\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$ Question:
Prove or disprove
$$I=\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$$
I know
\begin{align}\frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=&\left(1+\dfrac{1}{2^{2^n}}\right)\left(1+\dfrac{1}{2^{2^n}+2}\right)\\&\cdots\left(1+\dfrac{1}{2^{
2^n+1}}\right)\end{align}
so
$$\lim_{n\to\infty}\left(1+\dfrac{1}{2^{2^n}}\right)\left(1+\dfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\dfrac{1}{2^{
2^n+1}}\right)=\sqrt{2}\ ?$$
I feel this result is very surprising. This problem comes from Chris's sis.
and I use wolfram,limit wofl can't find it
I often use this theta function and is this true? Thank you.
| The logarithm of the expression under the limit can be rewritten as
$$\sum_{k=0}^{2^{2^n-1}}\ln\left(1+\frac{1}{2^{2^n}+2k}\right)=\sum_{k=0}^{2^{2^n-1}}\frac{1}{2^{2^n}+2k}+O\left(2^{-2^n}\right).$$
Denoting $N=2^{2^n-1}$, it is easy to see that the limit of the logarithm can be computed as the limit of a Riemann sum:
$$\frac{1}{N}\sum_{k=0}^N\frac{1}{2\left(1+\frac{k}{N}\right)}\stackrel{N\rightarrow\infty} \longrightarrow \frac12\int_0^1\frac{dx}{1+x}=\ln\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)} dx$ =? After reading articles on differentiation under the integral sign, I hit this post from MIT OCW, where after introducing the power tool, it challenges reader to do
$$\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)} dx$$
Obviously I have no clue where to start. Could any one give a hint?
| This is a possible way to evaluate the integral. Partial fraction decomposition and the double angle formula yield
$$\int^\infty_0\frac{\sin^2{x}}{x^2(1+x^2)}dx=\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{x^2}dx-\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{1+x^2}dx$$
The first integral can be evaluated in many ways, differentiation under the integral sign is one of them. I prefer to proceed with a simple fact that follows from the definition of the gamma function.
$$\int^{\infty}_0t^{n-1}e^{-xt} \ dt=\frac{\Gamma(n)}{x^n}$$
Hence the first integral is
\begin{align}
\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{x^2}dx
&=\frac{1}{2}\int^\infty_0(1-\cos{2x})\int^\infty_0te^{-xt} \ dt \ dx\\
&=\frac{1}{2}\int^\infty_0t\int^\infty_0e^{-xt}(1-\cos{2x}) \ dx \ dt\\
&=\int^\infty_0\left(\int^\infty_0e^{-xt}\sin{2x} \ dx\right)dt\\
&=\int^\infty_0\frac{2}{t^2+4}dt\\
&=\frac{\pi}{2}\\
\end{align}
The second integral can be broken up further and evaluated using the residue theorem.
\begin{align}
\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{1+x^2}dx
&=\frac{\pi}{4}-\frac{1}{4}\Re\oint_{\Gamma}\frac{e^{2iz}}{1+z^2}dz\\
&=\frac{\pi}{4}-\frac{1}{2}\Re\left(\pi i\operatorname{Res}(f,i)\right)\\
&=\frac{\pi}{4}-\frac{1}{2}\Re\left(\pi i\frac{e^{-2}}{2i}\right)\\
&=\frac{\pi}{4}-\frac{\pi}{4e^2}
\end{align}
Hence
$$\int^\infty_0\frac{\sin^2{x}}{x^2(1+x^2)}dx=\frac{\pi}{4}\left(1+e^{-2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find $\int_{0}^{\pi/2} \log ({1+\cos x}) dx$ using real-variable methods? How do you find the value of this integral, using real methods?
$$I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx$$
The answer is $2C-\dfrac{\pi}{2}\log {2}$ where $C$ is Catalan's constant.
| \begin{align}J&=\int_0^{\frac{\pi}{2}}\ln(1+\cos x)dx\\
K&=\int_0^{\frac{\pi}{2}}\ln(1-\cos x)dx\\
J+K&=2\int_0^{\frac{\pi}{2}}\ln(\sin x)dx\\
&=\int_0^{\frac{\pi}{2}}\ln(\sin x)dx+\underbrace{\int_0^{\frac{\pi}{2}}\ln(\sin x)dx}_{u=\frac{\pi}{2}-x}\\
&=\int_0^{\frac{\pi}{2}}\ln(\sin u\cos u)du\\
&=\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin(2u)}{2}\right)du\\
&\overset{z=2u}=\frac{1}{2}\int_0^\pi \ln\left(\frac{\sin z}{2}\right)dz \\
&=\frac{1}{2}\int_0^\pi \ln\left(\sin z\right)dz-\frac{1}{2}\pi\ln 2 \\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}} \ln\left(\sin z\right)dz+\frac{1}{2}\underbrace{\int_{\frac{\pi}{2}}^\pi \ln\left(\sin z\right)dz}_{t=\pi-z}-\frac{1}{2}\pi\ln 2\\
&=\frac{1}{2}(J+K)-\frac{1}{2}\pi\ln 2\\
J+K&=\boxed{-\pi\ln 2}\\
K-J&=\int_0^{\frac{\pi}{2}}\ln\left(\frac{1-\cos x}{1+\cos x}\right)dx\\
&\overset{u=\sqrt{\frac{1-\cos x}{1+\cos x}}}=4\int_0^1 \frac{\ln u}{1+u^2}du=-4\text{G}\\
J&=\frac{1}{2}\Big((J+K)-(K-J)\Big)=\frac{1}{2}(-\pi\ln 2+4\text{G})=\boxed{2\text{G}-\frac{1}{2}\pi\ln 2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
limit of trigonometric series Given
$$a_n = \frac 1{n^2}\sum_{k=1}^n {(2k+1)\sin\left(\frac{k^2 \pi}{n^2} \right)}$$
find $\lim_{n \to \infty} a_n$.
My try: Because $k = 1,2,\dots, n$, $$0\le\frac {k^2}{n^2} \le 1$$
hence
$$0\le\frac {k^2}{n^2} \pi \le \pi$$
so
$$0 \le \sin \left( \frac {k^2}{n^2} \pi\right) \le 1.$$
That means $a_n$ is a non-negative sequence.
$$0 \le \frac 1{n^2}\sum_{k=1}^n {(2k+1)\sin\left (\frac{k^2 \pi}{n^2} \right)} \le
\frac 1{n^2}\sum_{k=1}^n {(2k+1)} $$
$$
\frac 1{n^2} n(2n+1) = 2 + \frac 1n \to 2
$$
so, by the comparison rule, the given series converges. The problem is the "sandwich" is not sufficient.
| Consider that:
$$a_n = \frac{1}{n^2}\sum_{k=1}^{n}\sin\left(\frac{k^2\pi}{n^2}\right)+\frac{2}{n}\sum_{k=1}^{n}\frac{k}{n}\sin\left(\frac{k^2\pi}{n^2}\right),\tag{1}$$
where the first term on the right is $O(1/n)$ while the second one is a Riemann sum converging to the integral:
$$\int_{0}^{1}2x\sin(\pi x^2)\,dx=\int_{0}^{1}\sin(\pi y)\,dy=\frac{2}{\pi}.$$
Due to the error term of the trapezoidal rule, the difference between the last integral and the rightmost term in $(1)$ is just $O(1/n^2)$, hence:
$$\lim_{n\to+\infty}a_n=\frac{2}{\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
log-trig integral with sin, cos, and tan Here is another log-trig integral you may find challenging/fun. Or not :)
$$\int_{0}^{\frac{\pi}{2}}\ln(1+\sin(x))\ln(1+\cos(x))\tan(x)dx=\frac{\pi^{2}}{8}\ln(2)-\frac{5}{16}\zeta(3)$$
| The integral equals
$$
\int_0^1 dx\, \ln \left( 1 + \sqrt{1-x^2} \right) \frac{\ln(1+x)}{x}
\\= \left. -\text{Li}_2(-x) \ln \left( 1 + \sqrt{1-x^2} \right) \right\lvert_0^1 + \int_0^1 dx\, \text{Li}_2(-x)\frac{1}{x} \left[1 - \frac{1}{\sqrt{1-x^2}} \right]
\\=\text{Li}_3(-1)-\int_0^1 dx\, \frac{\text{Li}_2(-x)}{x\sqrt{1-x^2}}.
$$
Now, $\text{Li}_3(-1) = -\beta(3) = -\left(1 - 2^{1-3}\right) \zeta(3) = -\frac{3}{4} \zeta(3)$. In the second term, write $\text{Li}_2(-x)$ as a series and interchange summation and integration:
$$
-\int_0^1 dx\, \frac{\text{Li}_2(-x)}{x\sqrt{1-x^2}}
= \sum_{k\geq 1} \frac{(-1)^{k-1}}{k^2} \int_0^1 dx\,x^{k-1} \left(1-x^2\right)^{-1/2}
= \sum_{k\geq 1} \frac{(-1)^{k-1}}{2k^2} B\left(\frac{k}{2},\frac{1}{2}\right).$$
We calculate this sum, following Cody's suggestion to start with
$$
\sum_{k\geq 1} (-1)^{k-1} B\left(\frac{k}{2},\frac{1}{2}\right) x^{k-1} = \frac{\pi - 2\arcsin x}{\sqrt{1-x^2}}.
$$
Integrating once from $0$ to $x$ yields
$$
\sum_{k\geq 1} \frac{(-1)^{k-1}}{k} B\left(\frac{k}{2},\frac{1}{2}\right) x^{k}
= \pi \arcsin x -\arcsin^2 x.
$$
Dividing by $x$ and integrating from $0$ to $1$ yields
$$
\sum_{k\geq 1} \frac{(-1)^{k-1}}{k^2} B\left(\frac{k}{2},\frac{1}{2}\right)
= \int_0^{\pi/2}dx\, \frac{\pi x \cos x}{\sin x} - \frac{x^2 \cos x}{\sin x}
\\= -\int_0^{\pi/2}dx\, \pi \ln \sin x-2x\ln \sin x = \frac{\pi^2}{2} \ln 2 + 2 \int_0^{\pi/2}dx\, x\ln \sin x.
$$
The latter integral can be calculated as follows: use the identity $\ln \sin x = -\ln 2 -\sum_{k\geq0} \cos(2 k x)$ and interchange summation and integration. This gives the value $$2\int_0^{\pi/2}dx\, x\ln \sin x = -\dfrac{\pi^2}{4}\ln 2 + \beta(3) = -\dfrac{\pi^2}{4}\ln 2 + \dfrac{7}{8} \zeta(3).$$
Adding everything up yields the correct value of the integral:
$$
-\frac{3}{4} \zeta(3) + \frac{1}{2} \left[\frac{\pi^2}{2} \ln 2 -\dfrac{\pi^2}{4}\ln 2 + \dfrac{7}{8} \zeta(3)\right] = \frac{\pi^2}{8} \ln 2 - \frac{5}{16} \zeta(3).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Simplifying a trigonometric identity Simplify $1 + \tan^2x$
My attempt:
$$\begin{align}1 + \tan^2x&\\
&= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\
&= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\
&= \frac{\sin^2x}{\cos^2x}\\
&= \tan^2x\end{align}$$
The correct answer, however..is $sec^2x$ Wherever I went wrong, please show.
| 4 th line of your solution is clearly wrong. In denominator their must be
lcm of [(cosx)^2 * (cosx)^2] which is (cosx)^2
and in 5th line you dont have used well known identity (cosx)^2 + (sinx)^2 = 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find a real numbers $a,b$ such $a^n+b^n$ is rational Question:
prove or disprove :there exsit real numbers $a,b$ such follow two condition:
(1):$a+b$ is irrational
(2): for any postive integer $n\ge 2$, then $a^n+b^n$ is rational.
I have know if
$n=2k$ case is true,because I let $a=\sqrt{2}+1,b=\sqrt{2}-1$,so
$$a^{2k}+b^{2k}=(\sqrt{2}+1)^{2k}+(\sqrt{2}-1)^{2k}\in Q$$
But for $n=2k+1$,I can't find a example.(if you can't find,can you prove when$n=2k+1$,there can't exsit?) Thank you for help
| Suppose that $a,b$ are real numbers such that $a^n+b^n$ is rational for all integers $n \ge 2$.
Since $a^2+b^2$ and $a^4+b^4$ are rational, $\dfrac{(a^2+b^2)^2-(a^4+b^4)}{2} = a^2b^2$ is also rational.
Then, since $(a^5+b^5)-(a^2+b^2)(a^3+b^3)+a^2b^2(a+b) = 0$, we have that
$\dfrac{(a^2+b^2)(a^3+b^3)-(a^5+b^5)}{a^2b^2} = a+b$ is rational, as desired.
EDIT: The above is only valid if $a^2b^2 \neq 0$, but the case where $a = 0$ or $b = 0$ is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve the inequality $x^2>10$ using square roots? Solve the inequality: $$x^2>10$$
How am I supposed to do this? It doesn't make sense when I take into account that if $x^2=10$ then $x=+\sqrt{10}$ and $x=-\sqrt{10}$
But how am I supposed to apply this to an inequality, I would get $x>\sqrt{10}$ and $x>-\sqrt{10}$
But for some reason this just doesn't make sense to me. Can someone explain it to me mathematically, instead of just having to memorize these kinds of things?
| Here is how I think about problems like this:
If $x^2 > 10$, then
$x^2 - 10 > 0$
$(x + \sqrt {10})(x - \sqrt {10}) > 0 $
$(x + \sqrt {10})$ and $(x - \sqrt {10})$ are either both positive or both negative.
Thus, when they are both positive and we divide both sides by one of the terms:
$x + \sqrt {10} > 0$ and $x - \sqrt {10} > 0$
This simplifies to:
$x > \sqrt {10}$
When they are both negative and we divide both sides by one of the terms:
$x + \sqrt {10} < 0$ and $x - \sqrt {10} < 0$
This simplifies to:
$x < -\sqrt {10}$
The final solution is:
$x < -\sqrt {10}$ or $x > \sqrt {10}$
Conversely,
if $x^2 < 10$, then
$x^2 - 10 < 0$
$(x - \sqrt{10})(x + \sqrt{10}) < 0$
We know that one of the terms, $(x - \sqrt{10})$ or $(x + \sqrt{10})$, is negative.
Since $(x - \sqrt{10})$ is always smaller, we know that it is the negative term.
Thus, when we divide both sides by $(x - \sqrt{10})$ we get:
$x + \sqrt{10} > 0$
This simplifies to:
$x > -\sqrt{10}$
when we divide both sides by $(x + \sqrt{10})$ we get:
$x - \sqrt{10} < 0$
This simplifies to:
$x < \sqrt{10}$
The final solution is:
$-\sqrt {10} < x < \sqrt {10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 9,
"answer_id": 1
} |
Where did the negative answer come from? The question is to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$
$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+x$$
$$x^2-x-2=0$$
$$(x-2)(x+1)=0$$
$$x=2,-1$$
because $x$ is positive $x=2$ is the answer. but where did the $x=-1$ come from ?
| $x=\sqrt{2+\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}_{\text{$x$}}}$, so we get $x=\sqrt{2+x}$.
Now there is only one solution. If we square both sides, we add the case $-x=\sqrt{2+x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 1
} |
Finding the sum to infinity Question:
Find the sum to infinity for the following series
$$1, -\frac{1}{2}, \frac{1}{2^2}, -\frac{1}{2^3},\cdots$$
What would be the technique used to find such a sum?
| Write the sum of the $n$ and $(n+1)^{th}$ terms as
$$
\frac{1}{2^n} - \frac{1}{2^{n+1}}
= \frac{1}{2^{n+1}}
$$
then sum them up to get
$$
\sum_{n \in 2 \mathbb{N} }^\infty \frac{1}{2^{n+1}}
= \sum_{k=0}^\infty \frac{1}{2^{2k+1}}
= \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k}
= 2/3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.