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Height of a cube edge from the floor
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?
I'm sorry for my bad English, but I try to illustrate it as follows.
The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?
|
Let $p = \angle ACA'$, $q = \angle ECA$ (in the original figure), where $A,C,A'$ are the same as in Math Lover's answer.
Then $\tan p = \frac{1}{\sqrt 7}, p = \tan^{-1} \frac{1}{\sqrt 7}$ and $\tan q = \frac{2}{2 \sqrt 2} = \frac{1}{\sqrt 2}, q = \tan^{-1} \frac{1}{\sqrt2}$. Now using the arctangent addition formula:
$$p + q = \tan^{-1} \frac{1/\sqrt7 + 1/\sqrt2}{1 - (1/\sqrt7)(1/\sqrt2)} = \tan^{-1} \frac{\sqrt{2} + \sqrt{7}}{\sqrt{14}-1} = \tan^{-1} \frac{\sqrt{2\cdot2\cdot7}+\sqrt{2\cdot7\cdot7}+\sqrt{2}+\sqrt{7}}{13} = \frac{1}{13} \left(8 \sqrt2 + 3 \sqrt7 \right).$$
So $\sin(p+q) = EN / EC \Rightarrow EN = 2\sqrt{3}a \sin(p+q)$. Now since $\sin(\tan^{-1} x) = \frac{x}{\sqrt{x^2+1}}$ (draw a triangle), then we have the monstrous task of simplifying this expression:
$$\sin(p+q) = \frac{(1/13) \left(8 \sqrt2 + 3 \sqrt7 \right)}{\sqrt{(1/169)(191+48\sqrt{14})+1}}$$
and after incorporating the $1$, denesting the square root, and dividing, I can confirm that this eventually gives $EN = a(1 + \sqrt{7/2})$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$ $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$
I have no idea which story to form, I thought about $n$ students assuming n=4
picking 4 students for a committee with 3 roles. since each students has 3 options.
then on the other side.. $2^{0}\binom{4}{0} + 2^{1}\binom{4}{1} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{3}+2^{4}\binom{4}{4}$ = $2^{0}\binom{4}{4} + 2^{1}\binom{4}{3} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{1}+2^{4}\binom{4}{0}$
but im just stuck here...
|
Define $A_n$ to be the set of strings of length $n$ made of $\{0, 1, 2\}$. It is clear that $|A_n| = 3^n$. We will prove that $|A_n| = \sum_{k=0}^n 2^{k}\binom{n}{k}$.
For $k=0, 1, \dots, n$ partition $A_n$ into $k+1$ disjoint sets $B_{n,k}$ each consisting of strings that contain exactly $k$ $2$s. It is clear that
$$
A_n = \bigcup_{k=0}^n B_{n,k}
$$
and so
$$
|A_n| = \sum_{k=0}^n |B_{n,k}|\tag1.
$$
Each element in $B_{n,k}$ is uniquely determined by the choice of $k$ positions on which to place a $2$ and a binary string on the remaining $n-k$ positions. Therefore, $|B_{n,k}| = 2^{n-k}\binom{n}{k}$. Substituting into $(1)$ and using $\binom{n}{k} = \binom{n}{n-k}$, we get
$$
|A_n| = \sum_{k=0}^n 2^{n-k}\binom{n}{k} = \sum_{k=0}^n 2^{k}\binom{n}{k}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3951876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$
*$\left(1- \frac {x+1}{4x+3}\right)^x$
*$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$
*$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$
*$e^{\infty} = \infty$
answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!
|
Being
$$\lim_{x \to \infty} \left(\frac{3 x + 2}{4 x + 3}\right)^{x} = \lim_{x \to \infty} e^{\ln{\left(\left(\frac{3 x + 2}{4 x + 3}\right)^{x} \right)}}= e^{\lim_{x \to \infty} x \ln{\left(\frac{3 x + 2}{4 x + 3} \right)}}=e^{\lim_{x \to \infty} x \color{blue}{\ln{\left(\lim_{x \to \infty} \frac{3 x + 2}{4 x + 3} \right)}}}$$
If you multiply and divide by $x$ the ratio $$\frac{3 x + 2}{4 x + 3}=\frac{\frac{3 x + 2}x}{\frac{4 x + 3}x}=\frac{3+\frac{2}x}{4+\frac{3}x}$$
you have
$$\lim_{x \to \infty} \frac{3+\frac{2}x}{4+\frac{3}x}=\frac 34$$
But if we remember that $\ln \frac 34<0$, the limit for $x\to \infty$ is $-\infty$ (to the exponent of the Napier's number) i.e.
$$\to e^{-\infty}=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$ The exercise in my textbook states
You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$
Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$
I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx$$
$$\int^1_0{x(1-2x^4)^{n-1}}dx + \int^1_0{-2x^4(1-2x^4)^{n-1}}dx$$
$$I_n=I_{n-1} + \int^1_0{-2x^5(1-2x^4)^{n-1}}dx$$
Then setting up the second integral for integration by parts
$$ I_n=I_{n-1} + \frac{1}{2}\int^1_0{x^2*-4x^3(1-2x^4)^{n-1}}dx $$
Differentiating $x^2$ and integrating $-4x^3(1-2x^4)^{n-1}$
$$I_n=I_{n-1} + \frac{1}{2}\left(\left[\frac{x^2(1-2x^4)}{n}\right]^1_0 -\frac{2}{n}\int^1_0{x(1-2x^4)^n}dx\right)$$
Evaluating $\left[\frac{x^2(1-2x^4)}{n}\right]^1_0$
$$I_n=I_{n-1} + \frac{1}{2}\left(\frac{(-1)^n}{n} -\frac{2}{n}I_n \right)$$
Collecting like terms
$$I_n=I_{n-1} +\frac{(-1)^n}{2n} -\frac{1}{n}I_n$$
$$I_n=\frac{1}{1+\frac{1}{n}}I_{n-1} + \frac{(-1)^n}{2n(1+\frac{1}{n})}$$
Simplifying the fractions
$$I_n=\frac{n}{1+n}I_{n-1} + \frac{(-1)^n}{2n+2}$$
However this is not equivelant to the expression in the question, have I messed up somewhere or missed something out?
|
Let $J = \int_{0}^{1}x^{5} (1 - 2x^{4})^{n-1} dx$.
Integration by parts gives: $$I_n=\int^1_0{x(1-2x^4)^n}dx = \frac{(-1)^{n}}{2} + 4nJ$$.
Separating the integral as you have done gives:
$$I_n=\int^1_0{x(1-2x^4)^n}dx =\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx=I_{n-1} -2 J$$
Solving for $J$ in the first and substitute it into the second gives the desired reduction formula for $I_{n}$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find minimal polynomial of given matrix I want to find the minimal polynomial of this matrix
$$\left[\begin{array}{ccccc}
2 & 0 & 0 & 0 & 0\\
4 & 2 & 0 & 0 & 0\\
1 & 4 & 2 & 0 & 0\\
0 & 0 & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 3
\end{array}\right]$$
I found it charastic polynomial $\chi_{A}=\left(2-x\right)^{3}\left(3-x\right)^{2}$
I also know that the minimal polynomial irreducible factors have the same irreducible factors as charastic polynomial so
$$m_{A}\left(x\right)\in\left\{ \left(2-x\right)^{3}\left(3-x\right)^{2},\left(2-x\right)^{3}\left(3-x\right),\left(2-x\right)^{2}\left(3-x\right),\left(2-x\right)\left(3-x\right),\left(2-x\right)\left(3-x\right)^{2}\right\} $$
but I don't know how to continue from here
|
The minimal polynomial is $(X-2)^3(X-3)$. This a triangular matrix with a $2\times 2$ diagonal block whose eigenvalue is $3$ thus whose minimal polynomial is $X-3$. There exists a block which is a $3\times 3$ matrix which has $2$ on the diagonal. The minimal polynomial of this block is $(X-2)^3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating the residues What are the residues at $s=0,\pm 1$ for the following function
$$I=\frac{1}{e^{2\pi is}-1}\frac{e^{isu}}{s^2(s^2-1)}?$$
Now it appears to me that, two pieces, $e^{2\pi is}$ and $s^2(s^2-1)$, diverge in the denominator. What are the residues then?
|
Since you asked only for the residues and not how to get them:
$$\text{res}\left(I,0\right)=-\frac{i \left(3 u^2-6 \pi u+2 \pi ^2-6\right)}{12 \pi }$$
$$\text{res}\left(I,1\right)=\frac{e^{i u} (2 u+5 i-2 \pi )}{8 \pi }$$
$$\text{res}\left(I,-1\right)=\frac{e^{-i u} (-2 u+5 i+2 \pi )}{8 \pi }$$
To get the first residue, we note that $s=0$ is a $4$th order pole of the function $I(s)$, then we must compute
$$\text{res} (I,0)={\frac {1}{3!}\lim _{s\to 0}{\frac {d^3}{ds^3}}}\left((s-0)^{4}I(s)\right)$$
which is quite "time consuming" :) (I did it with Mathematica)
$$\frac{1}{3!}\underset{s\to 0}{\text{lim}}\left(\frac{\left(-\frac{i e^{i s u} u^3}{-1+e^{2 i \pi s}}+\frac{6 e^{i u s+2 i \pi s} i \pi u^2}{\left(-1+e^{2 i \pi s}\right)^2}+3 e^{i s u} \left(\frac{4 e^{2 i \pi s} \pi ^2}{\left(-1+e^{2 i \pi s}\right)^2}-\frac{8 e^{4 i \pi s} \pi ^2}{\left(-1+e^{2 i \pi s}\right)^3}\right) i u+e^{i s u} \left(\frac{48 i e^{6 i \pi s} \pi ^3}{\left(-1+e^{2 i \pi s}\right)^4}-\frac{48 i e^{4 i \pi s} \pi ^3}{\left(-1+e^{2 i \pi s}\right)^3}+\frac{8 i e^{2 i \pi s} \pi ^3}{\left(-1+e^{2 i \pi s}\right)^2}\right)\right) s^2}{s^2-1}+3 \left(-\frac{e^{i s u} u^2}{-1+e^{2 i \pi s}}+\frac{4 e^{i u s+2 i \pi s} \pi u}{\left(-1+e^{2 i \pi s}\right)^2}+e^{i s u} \left(\frac{4 e^{2 i \pi s} \pi ^2}{\left(-1+e^{2 i \pi s}\right)^2}-\frac{8 e^{4 i \pi s} \pi ^2}{\left(-1+e^{2 i \pi s}\right)^3}\right)\right) \left(\frac{2 s}{s^2-1}-\frac{2 s^3}{\left(s^2-1\right)^2}\right)+\frac{e^{i s u} \left(\left(\frac{24 s}{\left(s^2-1\right)^3}-\frac{48 s^3}{\left(s^2-1\right)^4}\right) s^2+6 \left(\frac{8 s^2}{\left(s^2-1\right)^3}-\frac{2}{\left(s^2-1\right)^2}\right) s-\frac{12 s}{\left(s^2-1\right)^2}\right)}{-1+e^{2 i \pi s}}+3 \left(-\frac{2 i e^{i u s+2 i \pi s} \pi }{\left(-1+e^{2 i \pi s}\right)^2}+\frac{i e^{i s u} u}{-1+e^{2 i \pi s}}\right) \left(\left(\frac{8 s^2}{\left(s^2-1\right)^3}-\frac{2}{\left(s^2-1\right)^2}\right) s^2-\frac{8 s^2}{\left(s^2-1\right)^2}+\frac{2}{s^2-1}\right)\right)=\frac{i \left(-3 u^2+6 \pi u-2 \pi ^2+6\right)}{12 \pi }$$
$s=1;\;s=-1$ are a double poles
$$\text{res}\left(I,1\right)=\underset{s\to 1}{\text{lim}}\frac{dI}{ds}(s-1)^2$$
and
$$\text{res}\left(I,-1\right)=\underset{s\to 1}{\text{lim}}\frac{dI}{ds}(s+1)^2$$
It is useful L'Hopital rule in the last two limits
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$ $$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$
I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ where $a_1,a_2,a_3\in \{a,b,c,d,e\}$.
|
Notice that for all $x,y$ we have $$x^2+xy+y^2 = {1\over 2}(x^2+(x+y)^2+y^2)\geq 0$$
Now we have:
$$3(a-b) = (b-e)\color{red}{\Big(} \underbrace{b+c+d}_x )^2+\underbrace{(b+c+d)(c+d+e)}_{xy}+(
\underbrace{ c+d+e}_y)^2\color{red}{\Big)} $$
so $${\rm sign} (a-b) = {\rm sign} (b-e) $$ and similary for:
\begin{align} {\rm sign} (b-c) &= {\rm sign} (c-a)\;\;\;\;(2) \\
{\rm sign} (c-d) &= {\rm sign} (d-b)\;\;\;\;(3) \\
{\rm sign} (d-e) &= {\rm sign} (e-c) \;\;\;\;(4)\\
{\rm sign} (e-a) &= {\rm sign} (a-d) \;\;\;\;(5)\\
\end{align}
*
*If $a>b$ then $b>e$ so $a>e$ and from $(5)$ we have $d>a$ so $d>e$ and now from $(4)$ we have $e>c$ and now $d>c$ so from $(2)$ we have $b>d$ so $b>c$ and now from $(2)$ we have $c>a$. So we have $$a>b>d>e>c>a$$ A contradiction.
*If $b>a$ we again get a contradiction, so $a=b=c=d=e=:x$ and all we need is to solve $x=9x^3$...
|
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"timestamp": "2023-03-29T00:00:00",
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|
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.
Solution
We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$
Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$
Hence, the answer is $31a^4$.
The original exam
|
This answer provides some motivation for a high school student learning the binomial theorem and problem solving techniques; we also get a 'feeling' for homogeneous polynomials.
Employing the binomial theorem we naturally write out (including the quadratic),
$\tag 1 x + y \color\red{ = a} $
$\tag 2 x^2 + y^2 = (x+y)^2- (2xy) $
$\tag 3 x^3 + y^3 = (x+y)^3 - (3 x^2 y + 3 x y^2) \color\red {= 10a^3}$
$\tag 4 x^4 + y^4 = (x+y)^4 - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $
Using $\text{(1)}$,
$\tag 1 x + y \color\red{ = a} $
$\tag 2 x^2 + y^2 = \color\red {(a)^2} - (2xy) $
$\tag 3 x^3 + y^3 = \color\red {(a)^3} - (3 x^2 y + 3 x y^2) \color\red {= 10a^3}$
$\tag 4 x^4 + y^4 = \color\red {(a)^4} - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $
It looks likes we have a good shot at writing $xy$ in terms of $a^2$ and from $\text{(3)}$,
$\quad \text{(3)} \implies -3xy(x+y) = 9a^3 \land \text{(1)} \implies xy = -3a^2$
Checkpoint,
$\tag 1 x + y \color\red{ = a} $
$\tag 2 x^2 + y^2 = \color\red {7a^2}$
$\tag 2 xy = \color\red{ -3\, a^2}$
$\tag 3 x^3 + y^3 \color\red {= 10a^3}$
$\tag 4 x^4 + y^4 = \color\red {a^4} - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $
Always hopeful, we tackle the remaining part,
$\quad - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) =$
$\quad -2xy(2x^2+3xy+2y^2) =$
$\quad -2xy\bigr(2(x^2+y^2) + 3xy\bigr) =$
$\quad 6a^2\bigr(14a^2 -9a^2\bigr) = 30a^4$
and so
$\tag{ANS} x^4 + y^4 = 31a^4$
Bonus Section
Keep going?
$\quad x^5 + y^5 = (x+y)^5 -(5 x^4y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 ) = $
$\quad \quad \quad \quad \quad \quad a^5 -xy(5 x^3 + 10 x^2 y^1 + 10 x^1 y^2 + 5 y^3 ) =$
$\quad \quad \quad \quad \quad \quad a^5 +3a^2(50 a^3 + 10 x^2 y^1 + 10 x^1 y^2 ) =$
$\quad \quad \quad \quad \quad \quad a^5 +3a^2\bigr(50 a^3 + 10 xy(x + y )\bigr) =$
$\quad \quad \quad \quad \quad \quad a^5 +3a^2(50 a^3 -30a^3) =$
$\quad \quad \quad \quad \quad \quad 61a^5$
More?
$\quad x^6 + y^6 = \,?$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC As title,
I could only figure out that
$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$
But couldn't get how to derive $2(a+b+c)$.
|
Not sure if there is a more straightforward solution but in meantime,
$3 \sin\frac{A}{2} = 4 \sin\frac{B}{2} = 5 \sin\frac{C}{2} = r$
$\frac{C}{2} = 90^0 - \frac{A+B}{2}$
So, $\cos \frac{A+B}{2} = \frac{r}{5}$
Expanding and writing values in terms of $r$ and squaring both sides, we simplify it to a cubic
$\frac{\sqrt {9-r^2} \times \sqrt {16-r^2}}{12} - \frac{r^2}{12} = \frac{r}{5}$
$25(9-r^2)(16-r^2) = (5r^2 + 12r)^2$
$120r^3 + 769r^2 - 3600 = 0$.
As $0 \lt r \lt 3$, we get $r \approx 1.9$ (with help from WolframAlpha)
From here on, we get sides as $\approx 5.842, 6.947, 8.145$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Summation of integration terms If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then
(A) $I>\ln99$
(B) $I<\ln99$
(C) $I<\frac{49}{50}$
(D) $I>\frac{49}{50}$
The official answer is B and D
My approach is as follow
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $$
$$\frac{A}{x} + \frac{B}{{x + 1}} = \frac{{A\left( {x + 1} \right) + Bx}}{{x\left( {x + 1} \right)}} = \frac{{x\left( {A + B} \right) + A}}{{x\left( {x + 1} \right)}}$$
$$A = k + 1\text{ and } B = - \left( {k + 1} \right)$$
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\left( {\frac{{k + 1}}{x} - \frac{{k + 1}}{{x + 1}}} \right)} dx} \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\int\limits_k^{k + 1} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx} $$
$$I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left. {\ln\frac{x}{{x + 1}}} \right|_k^{k + 1}} \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{k + 1}}{{k + 2}} - \ln\frac{k}{{k + 1}}} \right)}$$ $$\Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{{{\left( {k + 1} \right)}^2}}}{{k\left( {k + 2} \right)}}} \right)} $$
Not able to approach from here
|
Hint: $$\sum_1^{98} (k+1)\ln\frac{(k+1)^2}{k(k+2)} \\ = \sum_1^{98} (k+1) \left( \ln\frac{k+1}{k} - \ln \frac{k+2}{k+1} \right) \\ = 2(\ln 2-\ln\frac 32) + 3(\ln \frac 32 -\ln \frac 43) +4(\ln \frac 43 -\ln \frac 54) +\dots +99(\ln \frac{99}{98} -\ln \frac{100}{99}) \\ =2\ln 2 + \big( \ln\frac 32 +\ln \frac 43 +\dots +\ln \frac{99}{98} \big) -99\ln\frac{99}{100} \\ =2\ln 2+\ln\frac{99}{2} -99\ln\frac{99}{100}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$
This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$
I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.
|
With a u-substitution:
$\int x^3\sqrt{x^2-9} \ dx\\
\int \frac 12 x^2\sqrt{x^2-9} (2x\ dx)\\
u = x^2 - 9\\
x^2 = u+9\\
du = 2x\ dx\\
\frac 12\int (u+9)\sqrt{u}\ du\\
\frac 12(\frac 23 (9u^\frac 32) + \frac 25 u^\frac 52)+ C\\
3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$
With a trig substitution:
$\int x^3\sqrt{x^2-9} \ dx\\
x = 3\sec\theta\\
dx = 3\sec\theta\tan\theta\\
3^5\int \sec^4\theta\tan^2\theta\ d\theta\\
3^5\int (\tan^2\theta\ + \tan^4\theta)\sec^2\theta d\theta\\
3^5 (\frac 13\tan^3\theta\ + \frac 15\tan^5\theta) + C\\
3^5 (\frac 13 (\frac {x^2}{9}-1)^\frac 32 + \frac 15(\frac {x^2}{9}-1)^\frac 52) + C\\
3^5 (\frac 13 \frac {(x^2-9)^\frac 32}{3^3} + \frac 15\frac {(x^2-9)^\frac 52}{3^5}) + C\\
3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$
If you want to use the substitution
$\theta = \sec^{-1} \frac {x}{3}\\ dx = \frac {3}{x\sqrt{x^2-9}}$
$\int x^3\sqrt{x^2-9} \ dx\\
\int \frac 13 x^4(x^2-9)\left(\frac {3}{x\sqrt{x^2-9}}\ dx\right)\\
\frac 13 \int (3^4\sec^4 \theta) (3^2\tan^2\theta)\ d\theta\\
$
And continue as above.
|
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|
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$
|
The series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}=1+\frac{1}{3}+\frac{1\cdot 2}{3\cdot 5}+\frac{1\cdot2\cdot3}{3\cdot5\cdot7}+\cdot\cdot\cdot$
causes $2-\frac{\pi}{2}$ to have nice properties with the given sequence $a_{n+1}= \frac{2n+1}{n}a_n-\frac{1}{n}.$
$a_1 = 2-\frac{\pi}{2}=1-\frac{1}{3}-\frac{1\cdot 2}{3\cdot 5}-\frac{1\cdot2\cdot3}{3\cdot5\cdot7}-\cdot\cdot\cdot$
$a_2=1-\frac{2}{5}-\frac{2\cdot 3}{5\cdot7}-\cdot \cdot \cdot$
$\vdots$
$a_n=1-\frac{n}{2n+1}-\frac{n(n+1)}{(2n+1)(2n+3)}-\cdot \cdot \cdot$
from which $a_n >1-\frac{1}{2}-\frac{1}{2^2}-\cdot \cdot \cdot = 1-\sum\limits_{i=1}^\infty 2^{-i}=1-1=0.$
Note that if $a_n$ is a decreasing sequence implies $a_n \leq \frac{1}{n+1}$
$a_{n+1}\leq a_n \Longleftrightarrow$
$\frac{2n+1}{n}a_n -\frac{1}{n} \leq a_n \Longleftrightarrow$
$(2n+1)a_n-1\leq na_n \Longleftrightarrow$
$(n+1)a_n\leq 1 \Longleftrightarrow$
$a_n\leq \frac{1}{n+1}$
and the first inequality
$a_{n+1}=1-\frac{n+1}{2n+3}-\frac{(n+1)(n+2)}{(2n+3)(2n+5)}-\cdot \cdot \cdot \leq a_n =1-\frac{n}{2n+1}- \frac{n(n+1)}{(2n+1)(2n+3)}-\cdot \cdot \cdot$ does hold because for each term
$\frac{(n)(n+1)(n+2)\cdot\cdot\cdot(n+m)}{(2n+1)(2n+3)(2n+5)\cdot \cdot \cdot (2n+2m+1)}\leq \frac{(n+1)(n+2)\cdot\cdot\cdot(n+m)(n+m+1)}{(2n+3)(2n+5)\cdot \cdot \cdot (2n+2m+1)(2n+2m+3)}$
and cancelling like terms it is true that
$\frac{n}{2n+1} \leq \frac{n+m+1}{2n+2m+3}$ for $m\geq0.$
Although induction is tempting, I think all the properties of the sequence are captured in the series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}$.
Note that I proved also that $a_n$ is bounded below and decreasing, and thus convergent. Instead of looking at the inequalities $0<a_n\leq \frac{1}{n+1}$ as a possible induction proof, you can look at them as squeeze inequalities, i.e. $a_n \rightarrow 0$.
|
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|
Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions
Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.
My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\sqrt{2}}(x)=\frac{\ln(x)}{\ln(\sqrt{2})}$$
then, we can re-write the equation as $$\left(\frac{\ln(x)}{\ln(2)}\right)^{2}-\left( \frac{\ln(x)}{\ln(\sqrt{2})}\right)^{2}=a-\sqrt{a+\frac{\ln(x)}{\ln(2)}}$$
Let $t:=\ln(x)$ so the equation above we can re-write in terms of $t$ as $$\left(\frac{1}{\ln(2)}t\right)^{2}-\left( \frac{1}{\ln(\sqrt{2})}t\right)^{2}=a-\sqrt{a+\frac{1}{\ln(2)}t}$$
Then, the idea that I was thinking is squaring both sides of the equation, but there appears an equation of degree 4 that is difficult to solve. How could I continue from there? or maybe a simpler approach?
New approach: Using the point out of @Ross Millikan and since that $$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt{2})}=2\log_{2}(x)$$
and let $t:=\log_{2}(x)$ so the equation can be we-write in terms of $t$ as $$t^{2}-2t=a-\sqrt{a+t} \implies t^{2}-2t\color{blue}{+1}=a-\sqrt{a+t}\color{blue}{+1} \implies (t-1)^{2}=a+1-\sqrt{a+t}$$
or maybe $$ t^{2}-2t=a-\sqrt{a+t} \overset{y^{2}=a+t}{\implies} (y^{2}-a)^{2}-2(y^{2}-a)=a-y^{2}$$
solving a little the expression above we can see that $$y^{4}-2y^{2}a+a^{2}-2y^{2}+2a-a+y^{2}=0$$
how can I solve this?
|
Somewhat oddly, this becomes easier by first eliminating $a$. Using your substitution for $t$, let $b=\sqrt{a+t}$. Then $a=b^2-t$ and the equation becomes
$$t^2 - 2t = b^2 - t - b$$
Or $$
0 = t^2 - b^2 - (t - b)
= (t-b)(t+b-1)
$$
At this point it might be tempting to get back to $t$ by replacing $b$ with $\sqrt{a+t}$. Don't do it. When solving equations with radicals, it's almost always better to solve for the value of the radical instead of solving for the original variable. The reason is that constraint on $b$ is just $b\ge0$, which is very convenient. So keep solving for $b$ by getting rid of $t$ with $t = b^2-a$.
$$
\begin{cases}
(b^2 - b - a)(b^2 + b - a - 1)=0\\[2ex]
b\ge0
\end{cases}
$$
The 4 solutions for $2b$ and the requirement for $b\ge0$ are:
*
*$1-\sqrt{1+4a}$, where $-1\le4a\le0$
*$1+\sqrt{1+4a}$, where $-1\le4a$
*$-1+\sqrt{5+4a}$, where $-4\le4a$
*$-1-\sqrt{5+4a}$, never non-negative
Whenever root 1 is valid, so are roots 2 and 3. So, aside from degenerate roots, the only way to just have two roots is with roots 2 and 3 but not 1. So, $a>0$
Degenerate Roots
It is important to remember that roots can be degenerate. While we established that for $-1\le4a\le0$, there are 3 roots, we cannot eliminate that range just yet. What if there's some value of $a$ in that range where 2 of the 3 roots are equal.
There are two possibilities
*
*$b_1 = b_2$
Then, $4a = -1$ and the third root $b_3$ equals the other two so we will have a triple degenerate root and only 1 distinct root.
*$b_3 = b_1$ or $b_3=b_2$
Then $-1+\sqrt{5+4a} = 1 \pm\sqrt{1+4a}$
But, since $$(\sqrt{5+4a}\pm\sqrt{1+4a})(\sqrt{5+4a}\mp\sqrt{1+4a})=4$$
If $\sqrt{5+4a}\pm\sqrt{1+4a} = 2$, then $\sqrt{5+4a}\mp\sqrt{1+4a} = 2$ from which it follows that $\sqrt{1+4a}=0$ and again we would end up with a triple degenerate root.
This leaves us with $a>0$ as the solution
|
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|
Difficult Cauchy Problem Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.
I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.
I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.
|
By Cauchy-Schwarz,
\begin{align}
\left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right)(a+b+c) & \geq \left(\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab}\right) ^2 \\
& = \left (\dfrac{a^2+b^2+c^2}{abc} \right)^2 \\
& = \left (\dfrac{3abc}{abc} \right)^2 \\
& = 9. \\
\end{align}
Thus, dividing both sides of the preceding inequality by $a+b+c$, we have that
$\left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right) \geq \dfrac{9}{a+b+c}$, and we are done.
|
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|
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$
$$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$
$$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$
$$= (250\sqrt{2009} - 1259)^5$$
However, the answer given is $32$, so there could have been more simplifications.
As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?
|
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
I see no reason for elegance. Since one of the factors in the numerator is $1$, computing $a^3$ is is straightforward.
$$a^3 = \left(\frac{1}{8}\right) \times
\left[
1 + 3\sqrt{2009} + 3(2009) + 2009\sqrt{2009}
\right]$$
$$=~
\left(\frac{1}{8}\right) \times
\left[6028 + 2012\sqrt{2009}\right]
~=~
\frac{1507 + 503\sqrt{2009}}{2}.
$$
Therefore,
$$(a^3 - 503a - 500)$$
$$=~ \frac{1507 + 503\sqrt{2009}}{2} ~-~
\frac{503 + 503\sqrt{2009}}{2} - \frac{1000}{2}
~=~ \frac{4}{2} \implies $$
$$(a^3 - 503a - 500)^5 = 2^5 = 32.$$
|
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|
Prove specially rearranged alternating harmonic series converges to $\frac 12 \ln{\frac{4p}{q}}$ By Leibnitz's test the alternating series is convergent.
$\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac 1n
=\frac 12 \ln{\frac{4×1}{1}}$
\begin{align}
& \left(1-\frac 12- \frac 14\right)+\left(\frac 13-\frac 16-\frac 18\right)+...\\
& =\frac 12 \ln {\frac{4\cdot 1}{2}}
\end{align}
Let $\sum \sigma$ is a rearrangement of the alternating series.$\sum t$ is obtained by grouping it's terms in such a manner that each group contains $p$ positive terms and $q$ negative terms. Prove that $\sum t=\frac 12\ln {\frac{4p}{q}}$
|
$t_{(p+q)n}$
\begin{align}
& =\left(1+\frac 13+...+\frac{1}{2p-1}-\frac 12-\frac 14-...-\frac{1}{2q}\right)+\left(\frac{1}{2p+1}+...+\frac{1}{4p-1}-\frac{1}{2q+2}-...-\frac{1}{4q}\right)+...+\left(\frac{1}{2p(n-1)+1}+...+\frac{1}{2pn-1}-\frac{1}{2q(n-1)+2}-...-\frac{1}{2qn}\right)\\
&=\left(1+\frac 13+...+\frac{1}{2pn-1}\right)-\frac 12\left(1+\frac 12+\frac 13+...+\frac{1}{qn}\right)\\
&=\left(1+\frac 12+\frac 13+...+\frac{1}{2pn}\right)-\left(\frac 12+\frac 14+...+\frac{1}{2pn}\right)-\frac 12(\gamma_{qn}+\ln{qn})\\
&=(\gamma_{2pn}+\ln{2pn})-\frac 12(\gamma_{pn}+\ln{pn})-\frac 12(\gamma_{qn}+\ln{qn}) \\
& =\left(\frac 12\ln 2^2+\frac 12\ln\frac pq\right)+\left(\gamma_{2pn}-\frac 12\gamma_{pn}-\frac 12\gamma_{q n}\right)
\end{align}
$\therefore \displaystyle \lim_{n \to \infty}t_{(p+q)n}=\frac 12\ln{\frac{4p}{q}} $
|
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|
$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster? I have to calculate this integral:
\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}
My attempt:
Using, trigonometric substitution
\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow \ x&=a \tan \theta\\ \Longrightarrow \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta
\end{align}
Thus,
\begin{align}
\int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}}
\end{align}
My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help
|
Integrate by parts
\begin{align}
I & =\int x^{2}\sqrt{a^{2}+x^{2}}\,dx
=\int \frac x3d(a^2+x^2)^{3/2} \\
&=\frac13x( a^2+x^2)^{3/2}-\frac13I-\frac{a^2}3\underset{=J}{\int \sqrt{a^2+x^2}dx} \\
&=\frac14x( a^2+x^2)^{3/2}-\frac{a^2}4J\tag 1
\end{align}
Integrate $J$ by parts again
\begin{align}
J & =\int \sqrt{a^{2}+x^{2}}\,dx=x\sqrt{ a^2+x^2}-J-\int \frac{a^2}{\sqrt{a^2+x^2}}dx\\
&=\frac12 x\sqrt{ a^2+x^2}-\frac{a^2}2\sinh^{-1}\frac xa \tag2
\end{align}
Plug (2) into (1) to obtain
$$I= \frac14x( a^2+x^2)^{3/2}- \frac{a^2}8 x\sqrt{a^2+x^2} - \frac{a^4}8\sinh^{-1}\frac xa +C
$$
|
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|
Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.
|
From your solution,
\begin{gather*}
x\cos \theta =y\sin \theta \\
y=x\cot \theta \\
\end{gather*}
Substitute the corresponding value of y into any of the given two equations, and you should be able to get the values of x any y both.
Let's see what we get by substituting it into the first equation:
\begin{gather*}
2x=y\tan \theta +\sin \theta \\
2x=x+\sin \theta \\
x=\sin \theta \\
y=x\cot \theta =\cos \theta \\
\\
\end{gather*}
Can you now prove what is required?
|
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|
Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $ find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$
Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $
So I can express $b=-a(\alpha+\beta)$
$c=a.\alpha\beta$
Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$
I want to know whether there is any different approach other than this method?
Any hint is higly valued. thank you!
|
This is somewhat along the lines of the approach Quanto takes, but causes the roots to emerge in a different fashion.
Dividing the given quadratic equation $ \ ac·x^2 \ - \ b(c+a)·x \ + \ (c+a)^2 \ = \ 0 \ \ $ through by $ \ ac \ $ produces
$$ x^2 \ - \ \frac{b(c+a)}{ac}·x \ + \ \frac{(c+a)^2}{ac} \ = \ 0 \ \ \Rightarrow \ \ x^2 \ - \ \frac{b}{a} · \left(1+\frac{a}{c}\right)·x \ + \ \frac{c}{a} · \left(1+\frac{a}{c}\right)^2 \ = \ 0 \ \ . $$
Applying the Viete relations you cited, we can then express this as
$$ x^2 \ + \ (\alpha + \beta) · \left(1+\frac{1}{\alpha·\beta}\right)·x \ + \ (\alpha·\beta) · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ = \ 0 \ \ . $$
Now it is a fairly familiar fact (which is simple enough to prove, though it is not often expressed this way) that if $ \ r \ $ and $ \ s \ $ are the roots of a quadratic equation $ \ x^2 + mx + n \ = \ 0 \ \ , $ then $ \ r + s \ = \ -m \ $ and $ \ r - s \ = \ \sqrt{\Delta} \ = \ \sqrt{m^2 - 4n} \ \ . $ For the equation at hand, we thus have
$$ r \ + \ s \ = \ - \left(\alpha \ + \ \beta \ + \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ , $$ $$ \Delta \ = \ (\alpha + \beta)^2 · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ - \ \ 4 · (\alpha·\beta) · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ \ = \ \ (\alpha - \beta)^2 · \left(1+\frac{1}{\alpha·\beta}\right)^2 $$
$$ \Rightarrow \ \ \sqrt{\Delta} \ \ = \ \ (\alpha - \beta) · \left(1+\frac{1}{\alpha·\beta}\right) \ \ = \ \ \left(\alpha \ - \ \beta \ + \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \right) \ \ . $$
Consequently,
$$ r \ \ = \ \ \frac12 · \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ + \ \alpha \ - \ \beta \ + \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\beta \ + \ \frac{1}{\alpha} \right) $$
and similarly,
$$ s \ \ = \ \ \frac12 · \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ - \ \alpha \ + \ \beta \ - \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\alpha \ + \ \frac{1}{\beta} \right) \ \ . $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3978334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
if $a_{n+1}=xa_{n}+ya_{n-1}$ then have this results :$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}$？ it is well known:
if $F_{1}=1,F_{2}=1$,and $F_{n+1}=F_{n}+F_{n-1}$.then we have
$$F_{m+n}=F_{m}F_{n+1}+F_{m-1}F_{n}$$
prove 1
and I have know this :if $a_{0}=0,a_{1}=1$ and $x,y$ be give postive integers,and $$a_{n+1}=xa_{n}+ya_{n-1}$$
then we have
$$a_{m+n}=a_{m}a_{n+1}+ya_{m-1}a_{n}$$
prove 2
My question is
if the sequence such $a,b,x,y$ be give postive integers,and such that $a_{1}=a,a_{2}=b,a,b\in N^{+}$,and $$a_{n+1}=xa_{n}+ya_{n-1}$$,then is also have this Similar results
$$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}?$$
if there exist,Find the $X,Y$?Thanks
|
So, the answer is indeed, yes, such a relation exists (with some caveats), and we can prove it via induction, just as in the 1st answer you have linked in the question.
MAJOR EDITS: My original answer was incorrect, as pointed out by @mathlove. Here is a revised answer.
First, let us prove the induction step. For it, assuming the relation is true upto $(m,n)$, we have to prove it for $(m+1,n)$ and $(m,n+1)$, just as mentioned in the $1^{st}$ answer.
For $(m+1,n)$, we have
\begin{align}
a_{m+1+n} &= xa_{m+n}+ya_{(m-1)+n} \\
&= x(Xa_ma_{n+1}+Ya_{m-1}a_n) + y(Xa_{m-1}a_{n+1}+Ya_{m-2}a_n) \\
&= Xa_{n+1}(xa_m+ya_{m-1}) + Ya_n(xa_{m-1}+ya_{m-2}) \\
&= Xa_{m+1}a_{n+1} + Ya_ma_n
\end{align}
Similarly we can do so for $(m,n+1)$. Thus, the inductive step is proved.
For the base case, as the induction relation involves $m-1$ and $n$, the lowest values that these can have is $1$ (As $a_0$ is not defined). So, the base case for the induction must be for $(m,n) = (2,1)$.
For the base case to hold, the following relation must hold -
\begin{align}
a_{2+1} &= Xa_2a_{1+1} + Ya_{2-1}a_1 \\
\implies a_3 &= Xb^2 + Ya^2
\end{align}
From the given recursion relation, we get
\begin{align}
a_3 &= xa_2 + ya_1 \\
&= xb + ya
\end{align}
Hence, the given relation holds iff $X,Y$ satisfy
$$Xb^2 + Ya^2 = xb + ya$$
Addendum: I added this as this was part of my original answer, and I wanted to show where I originally went wrong.
Now, let us determine what restrictions $X$ and $Y$ must have, if such a result was to exist.
Let $m=2$. Then, the induction gives
\begin{align}
a_{n+2} &= Xa_2a_{n+1}+Ya_1a_n \\
&= (Xb) a_{n+1} + (Ya) a_n
\end{align}
Now, we have already been given that $a_{n+2}=xa_{n+1}+ya_{n}$. Thus, we get
\begin{align}
(Xb) a_{n+1} + (Ya) a_n &= xa_{n+1}+ya_{n}\\
\implies (Xb-x)a_{n+1} &= (y-Ya)a_n
\end{align}
Now, if either $X = \frac{x}{b}$ or $Y = \frac{y}{a}$, then this relation shows that the other must be zero as well. Hence, we have either $(X,Y) = \left(\frac{x}{b}, \frac{y}{a}\right)$, or
\begin{align}
a_{n+1} &= \frac{y-Ya}{Xb-x}a_n \\
\implies a_n &= \left(\frac{y-Ya}{Xb-x}\right)^{n-1} a_1
\end{align}
Note: We can easily see that in both the cases, $Xb^2 + Ya^2 = xb + ya$ holds (as it should).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$
I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:
$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$
So that:
$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$
Then i tried to separate this last integral by partial fractions, my result on this was:
$$\frac {1} {(x+a)(x^2 + 1)} = \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1}\right)$$
And the integral reduces to:
$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$
Then this last expression evaluates to:
$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$
Then integrating from 0 to 1 with respect to $a$ we will get:
$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$
(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)
But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.
Do i need some other approach? Or did i made any mistake?
Any help is appreciated.
|
$$f(x)=\frac{1}{(x+a)(x^2+1)}=$$
$$\frac{A}{x+a}+\frac{Bx+C}{x^2+1}$$
we find that $$A=\frac{1}{a^2+1}$$
and
$$Bi+C=\frac{1}{a+i}=\frac{a-i}{a^2+1}$$
So,
$$B=-\frac{1}{a^2+1}\;,\;C=\frac{a}{a^2+1}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-x-y$ and replace:
I get LHS = $x^2y+xy^2+2(x+y)-x^2-y^2-3xy-1$ but don't know how to proceed.
Thank you for your help.
|
As $0 \leq x, y, z \leq 1, (1-x), (1-y), (1-z) \geq 0$
Using A.M - G.M,
$(1-x) + (1-y) = z \geq 2 \sqrt{(1-x)(1-y)} \ $ (as $ \ x + y + z = 2$)
Similarly for $x, y$.
Multiplying them, we have
$xyz \geq 8 (1-x)(1-y)(1-z)$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $
How to go further
|
An alternative way: By the rearrangement inequality we have $$\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\geq \frac{a}{\sqrt{a}}+\frac{a}{\sqrt{a}}+\frac{b}{\sqrt{b}}+\frac{b}{\sqrt{b}}+\frac{c}{\sqrt{c}}+\frac{c}{\sqrt{c}}=2(\sqrt{a}+\sqrt{b}+\sqrt{c})$$ Now proceed by AM-GM as in the answer by Albus Dumbledore.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Equivalence class for the following relation A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$
Show that $\sim$ is equivalence relation (I have done this part)
Determine the equivalence class $[-1]_\sim$
Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.
I am facing difficulty in second and third part of the question.
For the second one I think it is $[-1]_\sim=\{-1,1\}$ as $-1\sim x: (-1)^4-x^2 =x^4-(-1)^2 $ so both $-1$ and $1$ are proving the equality.
I think the third statement is true but how can I prove it?
Please help.
|
Rearrange: $a^4 - b^4 = b^2 - a^2$
But $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$ (DOTS)
So $(a^2 - b^2)(a^2 + b^2) = b^2 - a^2$
(1) Assume $a^2 - b^2 \neq 0$
Then divide both sides by $a^2 - b^2$
$a^2 + b^2 = -1$
Since $a^2$ and $b^2$ are both positive in $\Bbb R$, this is a contradiction.
(2) Therefore $a^2 - b^2 = 0$, or $a^2 = b^2$.
Therefore $b = \pm a$ and each equivalence class of a comprises $\{a, -a \}$.
The equivalence class of $-1$ is $\{-1, 1 \}$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Given a directional derivative to find a point Let $ f(x,y,z)$ be differentiable, and assume that
$$ f(x, y, x^2 + y) = 3x - y $$
(for all values of $x,y$).
Also, given the direct-derivative of the point $A=(0, 12, 12)$ and the direction vector is $(1, 0, 1)$ is equal to $3$, I need to find the gradient of $f$ at $A$.
I have no idea how to even start, because $f$ is not given explicitly, only the derivative — but again, not explicitly for $f(x,y,z)$, but for this weird combination $f(x, y, x^2 + y)$.
What should I do?
|
Using the chain rule:
$$\frac{ \partial}{ \partial x} f(x,y,x^2 + y) = f_x(x,y,x^2 + y) \cdot \frac{ \partial}{ \partial x} (x) + f_y (x,y, x^2 + y) \cdot \frac{ \partial }{ \partial x}(y) + f_z(x,y,x^2 + y) \cdot \frac{\partial }{\partial x} (x^2 + y) = f_x(x,y,x^2 + y) + 2x \cdot f_z(x,y,x^2 + y) = \frac{\partial}{ \partial x}(3x -y) = 3
f_x(x,y,x^2 +y) = 3- 2x f_z(x,y,x^2+y) $$
Notice that $A$ fits perfectly to the $x,y, x^2 + y$ pattern ($0^2 + 12 = 12 = A_z$) Thus:
$$ f_x(0,12,12) = 3- 2 \cdot 0 \cdot f_z(x,y,x^2+y) = 3$$
So we have one third of the gradient at the point $A$.
Using the chain rule again but now differentiating on $y$:
$$ f_y(x,y,x^2 + y) = f_x(x,y,x^2 + y) \cdot (x)_y + f_y(x,y,x^2 + y) \cdot (y)_y + f_z(x,y,x^2 + y) \cdot (x^2 +y)_y \\= f_y(x, y,x^2 +y) + f_z(x,y,x^2 + y) = (3x -y)_y = -1 \\ \Rightarrow \color{red}{f_y(0,12,12) + f_z(0,12,12) = -1} $$
Now using the directional derivative:
First we need to normalize the direction vector, which is given by $$ \frac{1}{||(1,0,1)||} \cdot (1,0,1) = ( \frac{1}{\sqrt{2}} , 0, \frac{1}{\sqrt{2}})$$
Now using the fact that the function is differentiable, you don't need to actually calculate the derivative by definition, we can just use this known formula (which is equal to $3$ according to the question):
$$ \nabla f(0,12,12) \cdot ( \frac{1}{\sqrt{2}} , 0, \frac{1}{\sqrt{2}}) = 3$$
By the definition of the gradient at point $A$, it is equal to all the vector of all partial derivatives at that point:
$$ \nabla f(0,12,12) = \left ( f_x(0,12,12), f_y(0,12,12) , f_z(0,12,12) \right ) $$
Regular dot multiplication gives us:
$$ f_x(0,12,12) \cdot \frac{1}{\sqrt{2}} + 0 \cdot f_y(0,12,12) + f_z(0,12,12) \cdot \frac{1}{ \sqrt{2}} = 3$$
Now, we already know what is the value of $f_x(0,12,12)$ and it is equal to $3$. Substituting back gives us:
$$f_z(0,12,12) \cdot \frac{1}{\sqrt{2}} = 3 - \frac{3}{\sqrt{2}} \\ f_z(0,12,12) = 3 \sqrt{2} - 3$$
Recall the red equation, we can substitute $f_z$ back and get:
$$ f_y(0,12,12) = -1 -f_z(0,12,12) = -1 -3 \sqrt{2} + 3 = 2 - 3 \sqrt{2} $$
Thus we can finally say that the answer is:
$$ \nabla f(0,12,12) = (3, 2-3\sqrt{2}, 3 \sqrt{2} -3 ) $$
And rest.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3985814",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Expanding Dawson's integral in a series of Hermite functions We want to show that
$$F(x)=\sqrt{\pi } \sum
_{n=0}^{\infty }
\frac{(-1)^n H_{2
n+1}(x)}{2^{3 n+3} \Gamma
\left(n+\frac{3}{2}\right)}$$
where $F(x)$ is the Dawson Integral ($F(x)=\exp \left(-x^2\right)
\int_0^x \exp
\left(u^2\right) \, du$) and $H_n(x)$ is the $n$ th Hermite polynomial.
Since $F(x)$ is odd, we can write $F(x)=\sum _{n=0}^{\infty }
c_{2 n+1} H_{2 n+1}(x).$ where
$$c_{2 n+1}=\frac{\int_{-\infty
}^{\infty } \exp
\left(-x^2\right) F(x) H_{2
n+1}(x) \, dx}{2^{2 n+1} (2
n+1)! \sqrt{\pi }}$$
Employing the generating function for the Hermite polynomials we can write
$$\int_{-\infty }^{\infty } \exp
\left(-(x-t)^2\right) F(x)
\, dx=\sum _{n=0}^{\infty }
\frac{t^n \int_{-\infty
}^{\infty } F(x) \exp
\left(-x^2\right) H_n(x) \,
dx}{n!}$$
How do we proceed to get the coefficient of $t^n$ ($n$ odd) on the left hand side ?
|
We will proceed to evaluate the integral appearing above in $c_{2 n+1}$.
The Dawson Integral can be expressed in the form
$$F(x)=\int_0^{\infty } \exp
\left(-y^2\right) \sin (2 x
y) \, dy$$
Hence,
$$\int_{-\infty }^{\infty } \exp
\left(-x^2\right) F(x) H_{2
n+1}(x) \,
dx=\int_0^{\infty } \exp
\left(-y^2\right)
\int_{-\infty }^{\infty }
\exp \left(-x^2\right) H_{2
n+1}(x) \sin (2 x y) \, dx
\, dy$$
To evaluate the $x$-integral on the right hand side we invoke the generating function for the Hermite polynomials.
$$\exp \left(-x^2\right) \sin (2
y x) \exp \left(2 x
t-t^2\right)=\sum
_{n=0}^{\infty }
\frac{H_n(x) t^n \exp
\left(-x^2\right) \sin (2 y
x)}{n!}$$
Integrating both sides with respect to $x$ from $-\infty$ to $\infty$ we obtain
$$\int_{-\infty }^{\infty } \exp
\left(-(x-t)^2\right) \sin
(2 y x) \, dx=\sum
_{n=0}^{\infty } \frac{t^n
\int_{-\infty }^{\infty }
H_n(x) \exp
\left(-x^2\right) \sin (2 y
x) \, dx}{n!}$$
The integral on the left is readily evaluated as $\sqrt{\pi } \exp
\left(-y^2\right) \sin (2 t
x y)$. Expanding this expression in a power series in $t$ we obtain the realtion
$$\sqrt{\pi } \exp
\left(-y^2\right) \sum
_{n=0}^{\infty }
\frac{(-1)^n (2 y t)^{2
n+1}}{(2 n+1)!}=\sum
_{n=0}^{\infty } \frac{t^{2
n+1} \int_{-\infty
}^{\infty } H_{2 n+1}(x)
\exp \left(-x^2\right) \sin
(2 y x) \, dx}{(2 n+1)!}$$
where we have recognized that the integral vanishes for even values of $n$.
Equating coefficient of like powers of $t$ we have
$$\int_{-\infty }^{\infty } H_{2
n+1}(x) \exp
\left(-x^2\right) \sin (2 y
x) \, dx=(-1)^n \sqrt{\pi }
\exp \left(-y^2\right) (2
y)^{2 n+1}$$
We then obtain
$$\int_{-\infty }^{\infty } \exp
\left(-x^2\right) F(x) H_{2
n+1}(x) \, dx=(-1)^n
\sqrt{\pi } 2^{2 n+1}
\int_0^{\infty } \exp
\left(-2 y^2\right) y^{2
n+1} \, dy$$
The integral on the right hand side evaluates to $(-1)^n \sqrt{\pi } 2^{n-1} n!$
Finally, we find the desired coefficient
$$c_{2 n+1}=\frac{(-1)^n
n!}{2^{n+2} (2 n+1)!}$$
which can be shown to be equivalent to that given in the original problem statement
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to compute a series whose terms are a rational function times an exponential function? How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}
I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n = \left(\frac{11}{14n-7}-\frac{1}{n+3}\right)\left(-\frac{1}{3}\right)^n\text{.}
\end{equation}
But I don't know what to do next.
|
$$\frac{n+10}{2n^2+5n-3}=\frac{3}{2 n-1}-\frac{1}{n+3}$$
Consider the first sum
$$S_1=\sum_{n=1}^\infty\frac{x^n}{2 n-1}=y\sum_{n=1}^\infty\frac{y^{2n-1}}{2 n-1}=y\tanh ^{-1}(y)=\sqrt{x} \tanh ^{-1}\left(\sqrt{x}\right)$$ With $x=-\frac 13$ we then have
$$S_1=-\frac{\pi }{6 \sqrt{3}}$$
For the second sum
$$S_2=\sum_{n=1}^\infty\frac{x^n}{ n+3}=\frac 1 {x^3}\sum_{n=1}^\infty\frac{x^{n+3}}{ n+3}=-\frac 1 {x^3}\left(\log(1-x)+x+\frac {x^2}2 +\frac {x^3}3\right)$$
With $x=-\frac 13$ we then have
$$S_2=\frac{47}{6}-27\log \left(\frac{4}{3}\right)$$
For the total
$$\frac{47}{6}-\frac{\pi }{2 \sqrt{3}}-27\log \left(\frac{4}{3}\right)$$
|
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|
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$
To find the domain:
$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$
$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$
$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$
$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$
The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$
But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.
Thanks
|
As IV_ said, your inverse function should be $y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$.
Your original function $ f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ is equal to a sine function squared. If you graph f(x), you can see that its biggest value is $1$, and it achieves its smallest value at $x=(-3\pi+1)/2$ because the domain is restricted. Calculating the smallest $y$-value, we get $f((-3\pi+1)/2)$, which is approximately $0.382$ according to the graph. Thus, the range of $f(x)$ is $[0.382,1]$.
So, your function maps $x$-values in its domain to the range $[0.382,1]$. An inverse function reverses this process, so it maps $[0.382,1]$ back to the domain of the original function. That means that the domain of your inverse function is $[0.382,1]$.
Notice how we didn't have to find the domain of the original function to find the domain of the inverse?
|
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|
Why is this definite integral divergent? The integral given is
$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx$$
The steps I have taken to solve it are first applying the double angle formula for the cosine function
$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ (2\cos^2(2x)-1)}dx $$
Then simplifying
$$\int^{\frac{\pi}{3}}_0 \frac{5}{2\cos^2(2x)}dx \iff \frac{5}{2}\int^{\frac{\pi}{3}}_0\sec^2(2x) dx$$
$$ \left[\frac{5}{2}\tan(2x) \right]^{\frac{\pi}{3}}_0 = \frac{-5 \sqrt{2}}{4}$$
However trying to verify this on multiple sources such as integral-calculator.com and wolframalpha have returned no value for the integral and state that this is divergent.
Does this integral really have no finite value over this interval? I'm pretty sure I have not made a method mistake so if there is infact no finite solution to this integral then what does this value of $\frac{-5\sqrt{2}}{4}$ represent?
|
You properly found that
$$\int\frac{5}{1+ \cos(4x)}dx=\frac{5}{4} \tan (2 x)$$ Now, write
$$I=\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx=\int^{\frac{\pi}{4}-\epsilon}_0 \frac{5}{1+ \cos(4x)}dx+\int_{\frac{\pi}{4}+\epsilon}^{\frac \pi3} \frac{5}{1+ \cos(4x)}dx$$ $$I=\frac{5}{4} \cot (2 \epsilon )+\frac{5}{4} \cot (2 \epsilon )-\frac{5 \sqrt{3}}{4}=\frac{5}{2} \cot (2 \epsilon )-\frac{5 \sqrt{3}}{4}$$
Now, what happens when $\epsilon \to 0$ ?
|
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|
coordinate form of an affine transformation I am having trouble with this seemingly simple question: Write the standard coordinate form of an affine transformation in $\mathbb{A}^{2}(\mathbb{R})$ that maps
the point (1, −2) to the point (0, 10), and the lines $10x_{1} − 4x_{2} = 1$ and $3x_{1} − 3x_{2} = −7$ to the lines $x_{1} − 2x_{2} = −3$ and $x_{1} − x_{2} = 6$, respectively.
Now as I understand for the point transformation
$\begin{pmatrix}0\\10\\0\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12} & b_{1}\\a_{21} & a_{22} & b_{2}\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1\\-2\\0\end{pmatrix}$ and I need to find this matrix of transformation, this matrix equation gives me two equations with 6 unknowns. So I guess 4 more equations will be coming from the transformations of the two lines. Now here is where I am having trouble, I don't know how to get those equations. Any kind of hints would be really helpful. Thanks in advance.
|
Here is a different method. Let $A$ be the looked for matrix.
First of all, let us determine the intersection of the "initial" axes and of the "final" axes. It is rather easy to find that the first two axes intersect at $(14/9,35/9)$ and the second pair of axes at $(15,9)$.
Therefore, $(15,9,1)$ is the image of $(14/9,35/9,1)$ by $A$.
Besides, directing vectors of the initial axes are
$$\binom{2}{5} \ \ \text{and} \ \ \binom{1}{1}$$
and directing vectors of the final axis are:
$$s\binom{2}{1} \ \ \text{and} \ \ t\binom{1}{1}$$
(parameters $s$ and $t$ are there in order to take into account the change of scale)
With all these informations, we can say that
$$A : \begin{pmatrix}2\\5\\0\end{pmatrix} \to \begin{pmatrix}2s\\1s\\0\end{pmatrix}, \ \begin{pmatrix}1\\1\\0\end{pmatrix} \to \begin{pmatrix}1t\\1t\\0\end{pmatrix}, \ \begin{pmatrix}14/9\\35/9\\1\end{pmatrix}\to
\begin{pmatrix}15\\9\\1\end{pmatrix} $$
Otherwise said :
$$A \times \underbrace{\begin{pmatrix}2&1&14/9\\5&1&35/9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{initial affine frame}\\B\end{array}} = \underbrace{\begin{pmatrix}2s&1t&15\\1s&1t&9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{final affine frame}\\C\end{array}} $$
which is equivalent to :
$$A=C \times B^{-1} = \dfrac{1}{3}\begin{pmatrix}5t-2s&2s-2t&45-14s/3\\
5t-s&s-2t& 27-7s/3\\
0& 0& 3\end{pmatrix}\tag{1}$$
Now, the last step is expressing that
$$A\begin{pmatrix}1\\-2\\1\end{pmatrix}=\begin{pmatrix}0\\10\\1\end{pmatrix}$$
which yields a system of two equations in two unknowns $s$ and $t$, finally giving :
$$s=9 \ \ \text{and} \ \ t=17/3$$
that have to be plugged into (1) to give the final answer:
$$A=\begin{pmatrix} 31/9&20/9&1\\ 58/9&-7/9& 2\\0&0&1\end{pmatrix}$$
Remark: the interest of this method is that all the computations can be done using a Computer Algebra System ; here is what I have done with Matlab:
syms x1 x2 s t
[X1,X2]=solve(10*x1-4*x2==0,...
3*x1-3*x2==-7,...
x1,x2)
B=[2, 1, X1;
5, 1, X2;
0, 0, 1];
[X1,X2]=solve(x1-2*x2==-3,...
x1 -x2==6,...
x1,x2)
C=[2*s, t, X1 ;
s, t, X2;
0, 0, 1];
A=C*inv(B);
[S,T]=solve(A*[1,-2,1]'==[0,10,1]',s,t);
subs(A,{s,t},{S,T})
|
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|
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$ I came across this integral
$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$
According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.
A promising way might be the following
\begin{align*}
\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3} \right )^2 + \frac{8}{3}}} \\
&= \cdots
\end{align*}
A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )
So, any ideas how to evaluate it?
P.S: Are there techniques available for these type of problems?
|
This is an elliptic integral of the first kind, so there is no analytical way to evaluate it.
More on Elliptical Integrals: https://en.wikipedia.org/wiki/Elliptic_integral
In any case, we might try to give your integral a numerical result because it's bounded between $0$ and $1$.
We can expand the whole integrand in Taylor-MacLaurin series. Notice that the very first term of the expansion would just be $f(0) = \frac{1}{\sqrt{3}}$, so:
Order zero
$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}}$$
Hence
$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} \approx 0.57735(...)$$
Let's go on with another term.
Second Order
Going on with the expansion we find:
$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} + O(x^4)$$
Whence
$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\int_0^1 x^2\ \text{d}x = \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\cdot \frac{1}{3} = \frac{8}{9 \sqrt{3}} \approx 0.5132(...)$$
We can go on with the terms.
Numerical Exact Result
Via the help of W. Mathematica, the numerical inteegration gives the result
$$0.500539(...)$$
Which is rather near to our last result, as you can see.
Let's do better.
Fourth Order
$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} - \frac{x^4}{3 \sqrt{3}} + O\left(x^6\right)$$
So again
$$\int_0^1 f(x)\ \text{d}x \approx \frac{8}{9 \sqrt{3}} - \int_0^1 \frac{x^4}{3 \sqrt{3}} = \frac{37}{45 \sqrt{3}} \approx 0.47471(...) $$
It starts to oscillate around $0.500$. You can go on with the terms, what you will have to integrate are just polynomials.
A Big Order
Say we go on to order ten:
$$\frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}+O\left(x^{12}\right)$$
Whence
$$\int_0^1 f(x)\ \text{d}x \approx \int_0^1 \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}\ \text{d}x = \frac{238984}{280665 \sqrt{3}} \approx 0.491609(...)$$
Got it?
|
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|
How to find all polynomials that follows the equation $P(P'(x)) = P'(P(x))$ I want to find all polynomials that has the following property
$$P(P'(x)) = P'(P(x))$$
where P(x) is a polynomial.
Can you please tell me a way to solve this problem?
|
Let $ax^{n}$ be the leading coefficient, then the leading coefficient of both sides of the equation are equal giving:
$a(anx^{n-1})^{n} = an(ax^{n})^{n-1}$
So $a^{n+1}n^{n} = a^{n}n$
So $a= \frac{1}{n^{n-1}}$
So one such family is $\frac{x^{n}}{n^{n-1}}$
Now consider the constant coefficient K and the linear coefficient L:
$P'(K) = P(L)$
So if $P(x) = Lx+K$ is linear then, LK=K
So linear functions of the form
$P(x) = x+k$
Work.
$P(x) = ax^{2}+Lx+K$
Has $a=\frac{1}{2}$ and $\frac{L^{2}}{2}+L^{2} + K = K + L$
So K is free, and Either L=0 or $L = \frac{2}{3}$.
$L = \frac{2}{3}$ doesn't work when plugging back into $P'(P(x)) = P(P'(x))$
So we get $\frac{x^{2}}{2}+k$.
At third order and higher P and P' are both non-linear which forces K to be a solution of a polynomial of order higher than 2 and so K can only take on so many values. It is likely that the first family given will describe all of the remaining polynomials. We will demonstrate for the third order polynomials.
$\frac{x^{3}}{9} + bx^{2} + Lx + K$
Looking at 5th order terms:
$\frac{b}{27} = \frac{2b}{81}$
Which forces $b=0$.
Then looking at second order terms:
$\frac{L^{2}}{3} = 0$
which forces $L=0$.
$$\frac{K^{2}}{3} + 2bK + L = \frac{L^{3}}{9} + bL^{2} + L^{2} + K$$
So $\frac{K^{2}}{3} - K = 0$.
So K=3 or K=0.
$\frac{1}{3}(\frac{x^{3}}{9} + K )^2 = \frac{1}{9}(\frac{x^{2}}{3})^3 + K$
Looking at the 3rd order term forces K=0.
So $\frac{x^{3}}{9}$ is the only 3rd order solution.
|
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|
Is the answer to this summation correct? Consider the summation below:
$$\sum_{k=1}^\infty(2k+1)x^{2k}$$
The problem asks to find what the mentioned summation is equal to. The solution provided in the book starts like the following:
$$x^3+x^5+x^7+...=\frac{x^3}{1-x^2}$$
and then takes the derivative of both sides and uses that to find the solution.
My question is shouldn't $x^3+x^5+x^7+...=\frac{x^3(1-x^{n+1})}{1-x^2}$?
How does the book assume that $x^3+x^5+x^7+...$ equals $\frac{x^3}{1-x^2}$ when we're not told whether $\lvert x\rvert<1$ or not?
|
Let's make some order.
As @Crostul and @cosmo5 pointed out in the comments we clearly have to suppose $|x| < 1$ otherways the series diverges.
Then using only some known properties of geometric series:
$$ x^3 + x^5 + \dots + x^{3+2n} = \sum_{n=0}^N x^{3 + 2j} = x^3 \cdot \sum_{n=0}^N (x^2)^j = x^3 \cdot \frac{1-(x^2)^{N+1}}{1-x^2} = \frac{x^{3}(1-x^{2(N+1)})}{1-x^2}$$
Since we have supposed $|x| < 1$ we can state
$$ \sum_{n=0}^\infty x^{3 + 2n} = \frac{x^{3}}{1-x^2} \qquad \qquad (1)$$
Now you can take what suggest your book [see somewhere why you can do the $\star$ passage!]
Differentiating the left part in $(1)$ we get:
$$ \frac{d}{dx}\sum_{n=0}^\infty x^{3 + 2n} \stackrel{\star}{=} \sum_{n=0}^\infty \frac{d}{dx} \left(x^{3 + 2n} \right)= \sum_{n=0}^\infty (3+2n)x^{2n+2} = 3x^2 + \sum_{n=1}^\infty (3+2n)x^{2n+2}$$
Differentiating the right part in $(1)$ we get:
$$ \frac{d}{dx} \left( \frac{x^{3}}{1-x^2} \right)= \frac{x^2(3 - x^2)}{(1-x^2)^2}$$
Dividing both results for $x^2$ we get:
$$ 3 + \sum_{n=1}^\infty (3+2n)x^{2n} = \frac{(3 - x^2)}{(1-x^2)^2}$$
Split
$$\sum_{n=1}^\infty (3+2n)x^{2n} = \sum_{n=1}^\infty (1+2n)x^{2n} + 2\sum_{n=1}^\infty x^{2n}$$
and you get:
$$\begin{align}\sum_{n=1}^\infty (1+2n)x^{2n} & = \frac{(3 - x^2)}{(1-x^2)^2} -3 - 2\sum_{n=1}^\infty x^{2n} \\[10pt]
& = \frac{(3 - x^2)}{(1-x^2)^2} - 3- \frac{2x^2}{1-x^2} \\[10pt]
& = \frac{3x^2 - x^4}{(1-x^2)^2}
\end{align}$$
|
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|
Integrals of the type $\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}$ I'm interested in understanding the sequence of substitutions needed to obtain
\begin{align}
\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}
=\frac{\tan ^{-1}\left(\frac{b x}{a \sqrt{a^2+b^2+x^2}}\right)}{a b}\, ,
\end{align}
with $a,b>0$.
The obvious substitution
\begin{align}
x=\sqrt{a^2+b^2}\tan\xi
\end{align}
gets rid of the square root but complicates the $a^2+x^2$ factor:
\begin{align}
\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}
= \int \frac{d\xi}{ (a^2+ (a^2+b^2)\tan^2\xi)\cos\xi} \tag{1}
\end{align}
which doesn't seem to be immediately helpful.
The substitution $a^2+x^2=b^2\tan^2\xi$ makes things worse since
$dx$ will convert to $b^2 \tan\xi d\xi/(\cos^2\xi\sqrt{b^2\tan^2\xi-a^2})$
with no realistic chance of eliminating the square root.
That would leave me with Euler's substitution
\begin{align}
\sqrt{a^2+b^2+x^2}=x+t
\end{align}
but before I get there I wonder if there's a more "obvious" way.
|
Your first choice is a better start than it seems, because Bioche's rules advise continuing with$$u=\sin\xi=\frac{\sqrt{a^2+b^2}\tan\xi}{\sqrt{a^2+b^2}\sec\xi}=\frac{x}{\sqrt{a^2+b^2+x^2}}.$$Then$$\frac{d\xi}{(a^2+ (a^2+b^2)\tan^2\xi)\cos\xi}=\frac{\cos\xi d\xi}{a^2\cos^2\xi+(a^2+b^2)\sin^2\xi}=\frac{du}{a^2+b^2u^2}.$$
|
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|
Show $\cos\frac\gamma2=\sqrt{\frac{p(p-c)}{ab}}$
Show that the equality $$\cos\dfrac{\gamma}{2}=\sqrt{\dfrac{p(p-c)}{ab}}$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and $\measuredangle ACB=\gamma$.
I have just proved that $$l_c=\dfrac{2ab\cos\dfrac{\gamma}{2}}{a+b}$$
Is this a well known formula for angle bisectors?
Can we use it somehow?
Thank you in advance!
|
Form half-angle trig-identity
$$\cos\gamma=2\cos^2\frac{\gamma}{2}-1\iff \cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos\gamma}{2}}$$
Substituting the value from cosine formula: $\cos \gamma=\frac{a^2+b^2-c^2}{2ab}$
$$\cos \frac{\gamma}{2}=\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}$$
$$=\sqrt{\frac{(a+b)^2-c^2}{4ab}}$$
$$=\sqrt{\frac{(a+b+c)(a+b-c)}{4ab}}$$
$$=\sqrt{\frac{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b+c}{2}-c\right)}{ab}}$$
$$=\sqrt{\frac{p(p-c)}{ab}}$$
|
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|
Largest consecutive integers with no prime factors except $2$, $3$ or $5$? The number $180$ has a special property. Its prime factors are only $2$, $3$, and $5$. However the number $220$ does not have this special property because one of its prime factors is $11$.
In the first one-hundred natural numbers, the numbers with this property are plentiful. They cover almost half of that range. However, into the thousands, or tens of thousands, they get increasingly rare, and you might go hundreds of integers without finding one. As far as I know, $80$ and $81$ are the two largest consecutive numbers that both have this property — $80 = 2^4 \times 5$ and $81 = 3^4$. I have tested numbers up to one million, and have not found another pair of numbers like this.
But, is it possible to mathematically prove this? Has anyone ever tried to do it before? All evidence points to the fact that $80$ and $81$ are the largest pair of integers with this property, but has it ever been proven before?
|
Only one of the two consecutive numbers can be divisible by $2$. Similarly, only one is divisible by $3$ and only one is divisible by $5$. Depending on how we pair the prime powers we therefore have six possibilities.
The cases $p^Aq^B=r^C-1$ can all be solved in the same way.
$2^A3^B=5^C-1 $
If $A\ge4$ then modulo $16$ we have $C\equiv 0\pmod 4$ and then $5^C-1$ has a factor of $13$.
If $B\ge2$ then modulo $9$ we have $C\equiv 0\pmod 6$ and then $5^C-1$ has a factor of $31$.
The remaining values of $A,B$ give the solutions $\{4,5\},\{24,25\}$.
$2^A5^B=3^C-1 $
If $A\ge5$ then modulo $32$ we have $C\equiv 0\pmod 8$ and then $3^C-1$ has a factor of $41$.
If $B\ge2$ then modulo $25$ we have $C\equiv 0\pmod {20}$ and then $3^C-1$ has a factor of $11$.
The remaining values of $A,B$ give the solutions $\{2,3\},\{8,9\},\{80,81\}$.
$3^A5^B=2^C-1 $
If $A\ge2$ then modulo $9$ we have $C\equiv 0\pmod 6$ and then $2^C-1$ has a factor of $7$.
If $B\ge2$ then modulo $25$ we have $C\equiv 0\pmod {20}$ and then $2^C-1$ has a factor of $31$.
The remaining values of $A,B$ give the solutions $\{1,2\},\{3,4\},\{15,16\}$.
The cases $p^Aq^B=r^C+1$.
$2^A\times3^B=5^C+ 1$
Modulo $4$ we have $2^A\times3^B\equiv 2\pmod 4$. Therefore $A=1$ and $2\times3^B=5^C+ 1.$
If $B\ge2$ then modulo $9$ we have $C\equiv 3\pmod {6}$ and then $5^C+1$ has a factor of $7$.
The remaining values of $B$ give the solution $\{5,6\}$.
$2^A\times 5^B=3^C+ 1$
Modulo $4$ we have $2^A\times3^B\equiv 2\pmod 4$. Therefore $A=1$ and $2\times5^B=3^C+ 1.$
If $A\ge3$ then modulo $8$ we have $3^C\equiv -1\pmod {8}$ which is impossible.
If $B\ge2$ then modulo $25$ we have $C\equiv 10\pmod {20}$ and then $3^C+1$ has a factor of $1181$.
The remaining values of $A,B$ give the new solution $\{9,10\}$.
$3^A\times 5^B=2^C+ 1$
The cases $AB=0$ have been covered in previous equations. So we can suppose $AB>0$ when the equation is impossible modulo $15$.
|
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|
a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3 A random variable has the density $f\left(x\right)=a+bx^{2}$ with $0<x<1$. Determine and b so that its mean will be 2/3.
I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\right)=\int_{0}^{1}ax^{2}dx+\int_{0}^{1}bx^{4}dx$$
$$P\left(0<X<1\right)=a\int_{0}^{1}x^{2}dx+b\int_{0}^{1}x^{4}dx=a\left(\frac{x^{3}}{3}\right)_{0}^{1}+b\left(\frac{x^{5}}{5}\right)_{0}^{1}$$
$$P\left(0<X<1\right)=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=\frac{2}{3}=\frac{5a+3b}{15}$$
$$10=5a+3b$$
I'm assuming x as a continuous random variable
In this case shouldn't be a=0 b=1?
|
$$E[X]=\int_0^1 xf(x)dx=\int_0^1(ax+bx^3)dx=a\frac{x^2}{2}+b\frac{x^4}{4}\bigg|_0^1=\frac{2a+b}{4}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$
Given $x,y,z$ are positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$
My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1} \right)$$
$$x^2+1 \geq 2x \Rightarrow \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}\right)\leq \frac{1}{2}\left(x^2y+y^2z+z^2x\right)=\frac{3}{2}$$
$$P \geq x^3y+y^3z+z^3x - \frac{3}{2}$$
I was stuck when I was finding the minium of $x^3y+y^3z+z^3x$
Help me, thank you so much
|
Too long for a comment.
Considering the minimization problem
$$
\max_{x,y,z} -f(x,y,z)\ \ \text{s. t.}\ \ g(x,y,z) = 0
$$
with
$$
\cases{
f(x,y,z) = \frac{x^5 y}{x^2+1}+\frac{x z^5}{z^2+1}+\frac{y^5 z}{y^2+1}\\
g(x,y,z) = x^2 y + y^2 z + z^2 x - 3
}
$$
and according to the notes, the point $x=y=z=1,\lambda=\frac 56$ is a stationary point on the lagrangian
$$
L=-f+\lambda g
$$
This point can be qualified using the bordered hessian which is
$$
H = \left(
\begin{array}{cccc}
0 & 2 x y+z^2 & x^2+2 y z & 2 x z+y^2 \\
2 x y+z^2 & -\frac{2 x^3 \left(3 x^4+9 x^2+10\right) y}{\left(x^2+1\right)^3} & -\frac{x^4 \left(3 x^2+5\right)}{\left(x^2+1\right)^2} & -\frac{z^4 \left(3
z^2+5\right)}{\left(z^2+1\right)^2} \\
x^2+2 y z & -\frac{x^4 \left(3 x^2+5\right)}{\left(x^2+1\right)^2} & -\frac{2 y^3 \left(3 y^4+9 y^2+10\right) z}{\left(y^2+1\right)^3} & -\frac{y^4 \left(3
y^2+5\right)}{\left(y^2+1\right)^2} \\
2 x z+y^2 & -\frac{z^4 \left(3 z^2+5\right)}{\left(z^2+1\right)^2} & -\frac{y^4 \left(3 y^2+5\right)}{\left(y^2+1\right)^2} & -\frac{2 x z^3 \left(3 z^4+9
z^2+10\right)}{\left(z^2+1\right)^3} \\
\end{array}
\right)
$$
and at the point of interest we have
$$
H(1,1,1) = \left(
\begin{array}{cccc}
0 & 3 & 3 & 3 \\
3 & -\frac{11}{2} & -2 & -2 \\
3 & -2 & -\frac{11}{2} & -2 \\
3 & -2 & -2 & -\frac{11}{2} \\
\end{array}
\right)
$$
now according to the notes, $n = 3, m = 1$ so we need to consider the two minors $H_3$ and $H_4$. Following we have $\det(H_3) = 63 > 0$ and $\det(H_4) = -\frac{1323}{4} < 0$. The signs are $\text{sign}(\det(H_3)) = (-1)^{m+1} = 1$ and $\text{sign}(\det(H_4))$ should alternate as negative.
This characterizes the point $(x,y,z)=(1,1,1)$ as a local minimum for $f$ constrained.
NOTE
Taking from the restriction $x$ as
$$
x = \phi(y,z) = \frac{\sqrt{z^4-4 y^3 z+12 y}-z^2}{2 y}
$$
and substituting into $f(x,y,z)$ we obtain a surface
$$
u = f(\phi(y,z),y,z)
$$
which can be depicted from the following plot. In blue the relative minimum point with coordinates $u=\frac 32, y=1,z=1$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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|
Famous or common mathematical identities that yield $1$ To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:
$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe
$$
-e^{i\pi} = 1 \tag{2}
$$
Are there other famous (as in commonly used) identities that yield $1$ in particular?
|
Limits
$$\lim_{x\to0}\frac{\sin x}{x}=1$$
This is actually a special case of the more general identity: If $f(0)=0$ and $f'(0)=1$, then
$$\lim_{x\to0}\frac{f(x)}{x}=1$$
which includes things like
$$\lim_{x\to0}\frac{\arctan x}{x}=1$$
Series
Any geometric series (which converges) will do as long as the first term, $a$, is equal to $1-r$, where $r$ is the common ratio. For example:
$$\begin{align}1&=\frac{1}{3}+\frac{1}{3}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^2+\frac{1}{3}\left(\frac{2}{3}\right)^3+\cdots\\
&=\frac{1}{4}+\frac{1}{4}\left(\frac{3}{4}\right)+\frac{1}{4}\left(\frac{3}{4}\right)^2+\frac{1}{4}\left(\frac{3}{4}\right)^3+\cdots\\
&=\sum_{k=1}^\infty (1-r)r^{k-1},~~~~~~~\lvert r\rvert <1\end{align}$$
Another quite interesting series:
$$\sum_{k=1}^\infty \frac{k}{(k+1)!}=1$$
I can also list many integrals, such as
$$\begin{align}
\int_0^\frac{\pi}{2}\cos x~dx=1
\end{align}$$
And just for fun:
$$0!=1$$
I will try to add some more interesting identities if I can remember/come across them.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$
But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.
|
$$x^3 + y^3 + 3 x^2 y^2 = x^3 y^3\tag{1}$$
$$\frac{x+y}{xy}\tag{2}$$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=0$$
can be factorized$^1$ as
$$(x+y-xy) \left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)=0$$
$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$ has solution$^2$ given by $x=y=-1$ we have $$\frac{x+y}{xy}=-2$$, the solutions of $(1)$ are only those of $x+y-xy=0$ that is $x+y=xy$ and finally
$$\frac{x+y}{xy}=1$$
$(^1)$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=(x+y)^3-3x^2y-3xy^2+ 3 x^2 y^2 - x^3 y^3=$$
$$=\left[(x+y)^3- x^3 y^3\right]+\left[-3x^2y-3xy^2+ 3 x^2 y^2 \right]=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)\right]-3 x y (x+y-x y)=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)-3xy\right]=(x+y-xy)\left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)$$
$(^2)$
$$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$$
set $y=kx$. We get
$$k^2 x^4+k^2 x^3+k^2 x^2+k x^3-k x^2+x^2=0\to x^2 \left(k^2 x^2+k^2 x+k x+k^2-k+1\right)=0$$
As $x\ne 0$ we have
$$k^2 x^2+k^2 x+k x+k^2-k+1=0$$
the discriminant of this equation is $D=-k^4+2 k^3-k^2=-k^2(k-1)^2$
$D\ge 0\to k=1 $ (solution $k=0$ is discarded)
if $k=1$ we have $x^2+2x+1=0\to x=-1$ thus the solutions are
$$x=-1;\;y=-1$$
|
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|
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.
|
At the core of contest problems like
Given $x^n+1/x^n=p$, evaluate $x^m+1/x^m$
Given $x^n+y^n=p$, evaluate $x^m+y^m$
for most likely, integers $m,n$, lies the recurrence relation
$$x^{n+1}+y^{n+1}=(x+y)(x^n+y^n)-xy(x^{n-1}+y^{n-1})$$
Our question seems a bit generalized but in the same spirit, the following recurrence relation can be formed :
$$ax^{n+1}+by^{n+1}=(x+y)(ax^{n}+by^{n})-xy(ax^{n-1}+by^{n-1})$$
Thus the problem reduces to finding the linear coefficients $(x,y)$ or $(x+y,xy)$.
To ease the calculation, we do the following :
$$217=79(x+y)-23xy \tag{1}$$
$$691=217(x+y)-79xy \tag{2}$$
Lets do $3\times(1)-(2)$. We get
$$-40=20(x+y)+10xy$$
$$\Rightarrow (x+2)(y+2)=0$$
Thus one of $x,y$ is $-2$ but not both. (Why?) Putting $x=-2$ in $(1)$ gives $y=3$. Our recurrence relation is
$$ax^{n+1}+by^{n+1}=(ax^{n}+by^{n})+6(ax^{n-1}+by^{n-1})$$
which means this ladder goes up in following manner :
$$79+6\cdot 23=271$$
$$271+6\cdot 79=691$$
$$691+6\cdot 217=1993$$
Thus we have observed a systematic method to tackle these type of problems.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$
I can use Werner and prostaferesis formulas and I find the identity $(1)$.
But if we suppose of not to use these formulas I have done a try writing:
$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$
$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$
$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$
Then I have abandoned because I should do long computations and I think that is not the right way.
Is there any trick without to use the prostapheresis or Werner's formulas?
|
I think that we can establish the identity without too much work; it certainly isn't short though. Your statement is equivalent to $$\cos^2 4x-\cos^2 2x=-\sin 6x\sin2x$$We'll start with left hand side; my strategy will be to make everything in terms of $\cos 2x$. I will make use of the following identities:
$$\begin{align}
\sin^2\theta&\equiv1-\cos^2\theta\\
\cos2\theta&\equiv2\cos^2\theta-1\\
\sin3\theta&\equiv3\sin\theta-4\sin^3\theta
\end{align}$$
Here we go:
$$\begin{align}
\cos^2 4x-\cos^2 2x&\equiv(2\cos^2 2x-1)^2-\cos^22x\\
&\equiv4\cos^42x-5\cos^22x+1
\end{align}$$
Now for the right hand side:
$$\begin{align}
-\sin6x\sin2x&\equiv-\sin2x(3\sin2x-4\sin^3 2x)\\
&\equiv4\sin^42x-3\sin^2x\\
&\equiv4(1-\cos^22x)^2-3(1-\cos^22x)\\
&\equiv4(\cos^4 2x-2\cos^2x+1)-3+3\cos^22x\\
&\equiv4\cos^42x-5\cos^22x+1
\end{align}$$
Hence we have $$\cos^2 4x-\cos^2 2x\equiv-\sin 6x\sin2x$$ as required.
I personally would also be very interested in any shortcut or trick that could be used here instead; if you find one please let me know.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$. $x,y,z \in \mathbb R$
My attempt:
Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.
Maybe I’m wrong.
|
First, we prove for all positive $x,y,z$.
If all of $x,y,z\geq1$ we have
$$x^2\geq x ,\\ y^2\geq y \\ z^2\geq z$$ $$⇒x^2+y^2+z^2\geq x+y+z \geq xyz\\$$
If at least one of $x,y,z$ is less than $1$ then take (WLOG) $x\geq y\geq z>0$ & $1>z>0$
$$\frac{x}{y}\geq1 $$
$$⇒\frac{x}{yz}\geq1 $$
$$⇒\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\geq 1$$
$$⇒x^2+y^2+z^2\geq xyz$$
If two of $x,y,z$ is negative (say $x$ and $y$)
we have $$-x-y+z ≥ x+y+z ≥xyz=(-x)(-y)z$$
$$⇒-x-y+z ≥(-x)(-y)z$$ Here $-x,-y,z$ are positive numbers. So from the earlier result we can say,
$$(-x)^2+(-y)^2+z^2\geq (-x)(-y)z$$ $$⇒x^2+y^2+z^2\geq xyz$$
The inequality is obvious when one of $x,y,z$ is $0$ or if one of $x,y,z$ is negative or if all of $x,y,z$ is negative.
|
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|
Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$ I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connection between the summation and either of the two closed-form expressions.
$$
\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}
$$
Some specific cases written out for some simple visualization: ($n=0$ through $n=3$)
$$
\begin{align}
\binom{0}{0} 2^{0} &= 1\binom{0}{0} = \frac{1}{2}\binom{2}{1} \\
\binom{0}{0} 2^{2} + \binom{2}{1} 2^{0} &= 3\binom{2}{1} = \frac{2}{2}\binom{4}{2} \\
\binom{0}{0} 2^{4} + \binom{2}{1} 2^{2} + \binom{4}{2} 2^{0} &= 5\binom{4}{2} = \frac{3}{2}\binom{6}{3}\\
\binom{0}{0} 2^{6} + \binom{2}{1} 2^{4} + \binom{4}{2} 2^{2} + \binom{6}{3} 2^{0} &= 7\binom{6}{3} = \frac{4}{2}\binom{8}{4}
\end{align}
$$
EDIT: I should specify that I have already found an inductive proof, but it is not satisfying to me. I am specifically looking for a bijective/double-counting proof ideally.
|
Proving the equality between the two closed form is easier:
$$\begin{align*}
(2n+1)\binom{2n}n &=(2n+1)\frac{(2n)!}{n!n!}\\
&= \frac{(2n+1)!}{n!n!}\\
&= \frac{(n+1)^2}{(2n+2)}\cdot\frac{(2n+2)!}{(n+1)!(n+1)!}\\
&= \frac{n+1}{2}\binom{2(n+1)}{n+1}
\end{align*}$$
By induction, consider the $n+1$ case of the summation form and the second closed-form,
$$\begin{align*}
\sum_{k=0}^{n+1} \binom{2k}{k}2^{2(n+1)-2k} &= 2^2\sum_{k=0}^{n}\binom{2k}{k}2^{2n-2k} + \binom{2(n+1)}{n+1}2^0\\
&= 2^2\cdot \underbrace{\frac{n+1}{2} \binom{2(n+1)}{n+1)}}_{\text{by induction hypothesis}} + \binom{2(n+1)}{n+1}2^0\\
&= [2(n+1)+1]\binom{2(n+1)}{n+1}
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$ Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.
|
$\textbf{Solution.}$ (Khang Nguyen) We have $$(3-a)(3-b)(3-c)\ge 0 \rightarrow abc+9(a+b+c)\le 36$$
Assume that $$(b-1)(c-1)\ge 0 \rightarrow 36\ge 8a+(9+a)(b+c)\Rightarrow b+c\le \dfrac{(36-8a)}{9+a}$$
Therefore
\begin{align*}
\text{LHS}&=\sqrt{2a+b+c+2\sqrt{(a+b)(a+c)}}+\sqrt{b+c}\\&
\le \sqrt{2a+\dfrac{36-8a}{9+a}+2\sqrt{a^2+3}}+\sqrt{\dfrac{36-8a}{9+a}}\\&\le 3+\sqrt{3}.
\end{align*}
Let $\dfrac{36-8a}{9+a}=x^2\, \left(1\le x\le 2\right),$ after simplify it becomes:
$$ \left( {x}^{2}+8 \right) \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{2
}+8}\sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10
{x}^{2}+72}$$
Or $$\sqrt{x^2+8} \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10
{x}^{2}+72}$$
Since $x\le 2\rightarrow \sqrt{3}+3-x\ge \sqrt{3}+3-2>0\rightarrow \text{VT}>0.$ By squaring both sides, we need to prove
$$-2 \left( \sqrt {3}+3 \right) \left( \sqrt {3}{x}^{2}+{x}^{3}-6{x
}^{2}-10\,\sqrt {3}+8\,x+6 \right) \ge 4\sqrt {3}\sqrt {7\,{x}^{4}-50\,
{x}^{2}+124}$$
It's easy to prove $\text{LHS}>0$ and from here the inequality is equivalent to
$$24\left( 2+\sqrt {3} \right) \left( {x}^{2}+8 \right) \left( 2
\sqrt {3}x+{x}^{2}+8\sqrt {3}-9x-10 \right) \left( x-2 \right)
\left( x-1 \right) \ge 0,$$
One more time, we can check that this inequality is true for all $1 \le x \le 2$ and done!
Equality holds when $(a,b,c)=(3,1,0)$ and permutations.
See also here.
|
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|
Calculating $\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx$ without using complex numbers One can find the antiderivate with help of the partial fraction method introducing complex numbers:
$$\frac{x\ln(x+1)}{x^2+1}=\frac{1}{2}\left(\frac{\ln(x+1)}{x-i}+\frac{\ln(x+1)}{x+i}\right).$$
The result is a complicated function containing dilogarithms and logarithms with complex arguments. The value of the definite integral turns out to be
$$\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx = \frac{1}{96}(\pi^2+12\ln^2(2)).$$
Is there a way to find that value more directly - without introducing complex numbers?
|
Rewrite the integral as
$$I=\int_0^1 \frac{x\log (x+1)}{x^2+1}dx =\int_0^1\int_0^1\frac{x^2\:dydx}{(yx+1)(x^2+1)}$$
A partial fraction and integration order reverse later we get
$$\int_0^1\int_0^1 \frac{xy}{(y^2+1)(x^2+1)}-\frac{1}{(y^2+1)(x^2+1)} +\frac{1}{xy+1}-\frac{y^2}{(y^2+1)(xy+1)}\:dxdy = \int_0^1\frac{y}{2(y^2+1)}\log(2) -\frac{\pi}{4}\frac{1}{y^2+1}+ \frac{\log(1+y)}{y}-\frac{y\log(y+1)}{y^2+1}\:dy$$
The first two terms simplify easily and the last term is the integral we started with. The third term is given by Taylor series
$$\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^1x^{n-1}dx = \sum_{n=1}^\infty \frac{(-1)^n}{n^2} =\frac{\pi^2}{12}$$
thus the equation simplifies to
$$\implies I = \frac{\log^22}{4}-\frac{\pi^2}{16}+\frac{\pi^2}{12}- I$$
therefore $I = \frac{1}{96}(\pi^2 +12\log^22)$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4030962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Kernel, Basis and Column Space If a matrix $A$ has row reduced form
$A=$$\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$
then the kernel of the linear map $T : \mathbb{R}^4 → \mathbb{R}^3$ defined by $T(x) = Ax$ has basis $$(−3,1,0,0), (−3, 0, 1, 1).$$
How do I prove or disprove the claim regarding the kernel?
Any help is appreciated guys. Also, how can I show that columns 2 and 4 forms a basis for the column space?
|
The reduced matrix is $A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$ and if we want to find the generator of the Kernel we have to solve the system $A\underline x=\underline0$:
$$A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}\underline x=\underline 0\implies \begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \end{pmatrix}\begin{pmatrix}x\\y\\z\\w \end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies$$
$$\begin{cases}x+3y+3w=0\\z-w=0 \end{cases}\implies z=w=t\in\mathbb R,y=s\in\mathbb R,x=-3t-3s.$$
So we can state that $\operatorname{Ker}(F)=\Big\langle\begin{pmatrix}-3\\1\\0\\0 \end{pmatrix},\begin{pmatrix}-3\\0\\1\\1 \end{pmatrix} \Big\rangle$.
|
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"source": "stackexchange",
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|
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
|
\begin{align}
\sqrt{x} - \sqrt{x-1}
&= \dfrac{(\sqrt{x} - \sqrt{x-1})(\sqrt{x} + \sqrt{x-1})}{\sqrt{x} + \sqrt{x-1})} \\
&= \dfrac{x-(x-1)}{\sqrt{x} + \sqrt{x-1}} \\
&= \dfrac{1}{\sqrt{x} + \sqrt{x-1}}
\end{align}
\begin{align}
\sqrt{x+1} - \sqrt{x}
&= \dfrac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x})} \\
&= \dfrac{(x+1)-x}{\sqrt{x+1} + \sqrt{x}} \\
&= \dfrac{1}{\sqrt{x+1} + \sqrt{x}}
\end{align}
So
\begin{align}
\sqrt{x+1} &> \sqrt{x-1} \\
\sqrt{x+1} + \sqrt x &> \sqrt x + \sqrt{x-1} \\
\left(\dfrac{1}{\sqrt{x+1} + \sqrt x}\right)
&< \left(\dfrac{1}{\sqrt x + \sqrt{x-1}}\right) \\
\sqrt{x+1} - \sqrt x &< \sqrt x - \sqrt{x-1} \\
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $f(x) = x^2 + 3$ is continuous at $x=3$ Prove that $f(x) = x^2 + 3$ is continuous at $x=3$.
I have tried using $\delta = \sqrt{\epsilon + 9} - 3$.
I tried to split $|x^2-9| = |(x-3)(x+3)|$ and tried to make $x+3$ in terms of $\delta$.
But I get $\delta^2 + 6\delta$.
I don't really get what I'm doing wrong or right so I need some help with finding the right $\delta$.
Also, I cannot really understand the goal of the proof, I read on elsewhere that I need to make $\delta$ as small as possible?
|
You need to solve the equation
$$0<|x-3|<\delta\implies|(x^2+3)-(3^2+3)|<\epsilon$$ for $\delta$.
As the function $x^2+3$ is monotonic around $x=3$, you can solve the equation
$$|x^2-9|=\epsilon$$ and use any $\delta$ such that $(3-\delta,3+\delta)$ is wholly contained between the two solutions, $x=\sqrt{9\pm\epsilon}.$ (Other possibilities are $x=-\sqrt{9\pm\epsilon}$, but this does not straddle $x=3$.)
The largest value of $\delta$ is thus
$$\min(\sqrt{9+\epsilon}-3,3-\sqrt{9-\epsilon})$$
and this is always $\sqrt{9+\epsilon}-3$.
Check:
$$-\delta<x-3<\delta\implies3-\delta<x<3+\delta\implies x^2<(6-\sqrt{9+\epsilon})^2,(\sqrt{9+\epsilon})^2.$$
But
$$(6-\sqrt{9+\epsilon})^2=36-12\sqrt{9+\epsilon}+9+\epsilon\ge 9-\epsilon$$
because
$$36+2\epsilon\ge 12\sqrt{9+\epsilon}$$
$$1296+144\epsilon+4\epsilon^2\ge 1296+144\epsilon$$
and $$\epsilon^2>0.$$
|
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|
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}
It checks out numerically, for $n=1,\dots,7$ that
\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}
so \eqref{0} seems to follow the pattern
\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}
for some function $f$.
Items \eqref{5} and \eqref{7} look promising
as they agree to $f(n)=2n\cot^2(\tfrac\pi{n})$,
but the other fail on that.
Edit:
Also, it looks like
\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}
and
\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}
with the same $f$.
Edit
Thanks to the great answer by @Quanto,
the function $f$ can be defined as
\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}
$\endgroup$
|
Not a complete answer, but an elaboration of what I said in the comments.
To evaluate $$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$$
Write it as $$\int_1^\infty\frac{\ln\left(x^{-2n}+1\right)+\ln\left(x^{2n}\right)}{1+x^2} dx = \int_1^\infty\frac{\ln\left(x^{-2n}+1\right)}{1+x^2} dx+2n\int_1^\infty\frac{\ln\left(x\right)}{1+x^2} dx$$
The second integral is a standard integral for Catalan's constant. To solve the first, make the substitution $x \to \frac{1}{x}$ to get $$\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx + 2nG$$
So if you know the value of any of the integrals with bounds $(0,\infty)$ or $(1,\infty)$ or $(0,1)$, you could find the other two.
The series for $\ln(1+x)$ converges for $|x|<1$, so expand it to get $$\int_0^1\frac{\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}x^{2nk}}{1+x^2}dx = \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\frac{x^{2nk}}{1+x^{2}}dx$$
Then you can expand $\frac{1}{1+x^2}$ to get $$\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}x^{2nk}\sum_{m=0}^{\infty}\left(-1\right)^{m}x^{2m}dx = \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\sum_{m=0}^{\infty}(-1)^m\int_{0}^{1}x^{2nk}x^{2m}dx$$
Finally, evaluate the inner integral to get
$$\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\sum_{m=0}^{\infty}(-1)^m\frac{1}{2nk+2m+1}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Question on Rudin Theorem 3.31 Proof On page 65, we have:
$e-s_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\dots < \frac{1}{(n+1)!}(1 + \frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots)=\frac{1}{n!n}$
Given $s_n = \sum_{k=0}^{n}{\frac{1}{k!}}$.
What I don't understand is, where does $\frac{1}{n!n}$ come from?
|
Note that since for $ |a| < 1$ (Geometric series formula)
$$
\sum_{i=0}^{\infty}a^i = \frac{1}{1-a},
$$
choosing $a := \frac{1}{n+1}$, on the right hand side it follows that
$$
\frac{1}{(n+1)!}\left(\sum_{i=0}^{\infty}\left(\frac{1}{n+1}\right)^i\right) = \frac{1}{(n+1)!}\left(\frac{1}{1 - \frac{1}{n+1}}\right) = \frac{1}{n!n}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:
*
*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.
*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2z^2 = 3xyz
$$
*Since (from 1.) the left hand side (LHS) is strictly positive, then $xyz \gt 0$
*Exploring the LHS of the equation from 1.:
$$
2x^2y^2+2x^2z^2+2y^2z^2 = (xy-xz)^2+(xy-yz)^2+(xz-yz)^2+2xyz(x+y+z)
$$
therefore
$$
x^2y^2+x^2z^2+y^2z^2 \geq xyz(x+y+z)
$$
*Replacing LHS of the last inequation by $3xyz$ (according to 2.) gives:
$$3xyz \geq xyz(x+y+z)$$ and since (from 3.) $xyz$ is positive, we can divide both sides by $xyz$, obtaining: $$x+y+z \leq 3$$
*$xyz$ can be positive in two cases:
Case I. $x, y, z$ are all positive. In this case $x \geq 1$, $y \geq 1$, $z \geq 1$,
so
$$
x+y+z \geq 3
$$
but according to 5. this must be an equation. Therefore all inequations for $x$, $y$ and $z$ must turn into equations, so $x=y=z=1$.
Case II. One of $x, y, z$ is positive, two others are negative. Without loss of generality we can assume $x \gt 0$, $y \lt 0$, $z \lt 0$
...
|
Hint: If $(x,y,z)$ is a solution, then show that so is $(x,-y,-z)$ and other cyclical permutations. This then severely restricts the values of $|x|, |y|, |z|$ can take after you have derived the inequality:
$$x+y+z\leq 3. $$
|
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|
limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$ How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhere.
$$\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}} $$
|
Assume WLOG $x\geq0$. Thus, $$\frac{\sqrt{x+2}-\sqrt{x+1}}{\sqrt{x+1}-\sqrt x}\cdot\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+1}+\sqrt x}$$$$=(\sqrt{x+1}+\sqrt x)(\sqrt{x+2}-\sqrt{x+1})\cdot\frac{\sqrt{x+2}+\sqrt{x+1}}{\sqrt{x+2}+\sqrt{x+1}}$$$$=\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}=1-\frac{\sqrt{x+2}-\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}$$
|
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"url": "https://math.stackexchange.com/questions/4043952",
"timestamp": "2023-03-29T00:00:00",
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The value of the second-order Eulerian polynomials at x = -1/2. Recently, the second-order Eulerian polynomials
$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n $
have been discussed on MSE [ a ,
b ].
$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n =
\sum_{k=0}^n \left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle \, x^k $$
Here $\left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle $
are the second-order Eulerian numbers A340556.
The values of these polynomials at $ x = \frac12 $ generate a sequence
that represents the solution of Schröder's fourth problem
(see MSE and
A000311).
That's a pretty nice result that suggests the question:
Do the values of the polynomials at $ x = -\frac12$ (times $ 2^n $) also have
a combinatorial meaning? That this might indeed be the case is a conjecture
suggested by many examples of combinatorial polynomials. The reader can get
an impression for himself while browsing through the OEIS.
Polynomials
P(x)
2^nP(1/2)
2^nP(-1/2)
Laguerre
A021009
A103194
A000262
Motzkin
A064189
A330796
A000244
BigSchröder
A080247
A065096
A239204
StirlingSet
A048993
A005493
A000110
StirlingCycle
A132393
A000254
A000774
Eulerian 1st
A173018
A180119
A001710
However, this question may not be easy to answer, so we ask a more concrete question.
The sequence $ 2^n \left\langle \!\left\langle - 1/2 \right\rangle \!\right\rangle_n $
can also be generated without reference to the second-order Eulerian polynomials by
series reversion.
$$ 2^n \left\langle\!\!\left\langle - \frac{1}{2} \right\rangle\!\!\right\rangle_n
= (n + 1)!\, [x^{n+1}]\, \text{Reversion}\left(\frac{6x + \exp(3x) - 1}{9}\right) \quad (n \ge 0). $$
Can someone confirm this equation?
Addendum:
The general form of the reversion is described by:
$$ \left\langle\! \left\langle x \right\rangle\! \right\rangle_n =
(n+1)!\, (1-t)^{2n + 1}\, [x^{n+1}]\,
\operatorname{Reversion}_{x}(x + t - t \exp(x)) \quad (n \ge 0) $$
|
Here are some comments. Seeking to invert
$$-\frac{1}{9} + \frac{2}{3} x + \frac{1}{9} \exp(3x) = z$$
we consult Wikipedia on
LambertW
to find that the closed form solution to
$$x = a + b \exp(cx)$$
is given by
$$x = a - \frac{1}{c} W(-bc \exp(ac)).$$
We thus write
$$x = \frac{3}{2} z + \frac{1}{6} - \frac{1}{6} \exp(3x).$$
to obtain
$$\frac{1}{6} + \frac{3}{2} z
- \frac{1}{3} W((1/2)\times \exp((9/2)z+1/2))
\\ = \frac{1}{6} + \frac{3}{2} z
- \frac{1}{3} W(\exp((1+9z)/2)/2).$$
We have
$$W(z) = \sum_{m\ge 1} (-1)^{m-1} m^{m-1} \frac{z^m}{m!}$$
We then obtain for $n\ge 1$
$$-\frac{1}{3} (n+1)! [x^{n+1}]
\sum_{m\ge 1} (-1)^{m-1} \frac{m^{m-1}}{m!}
\frac{\exp(m/2)}{2^m} \exp(9mx/2)
\\ = \frac{1}{3}
\sum_{m\ge 1} (-1)^{m} \frac{m^{m+n}}{m!}
\frac{\exp(m/2)}{2^{m+n+1}} 9^{n+1}
\\ = \frac{3^{2n+1}}{2^{n+1}}
\sum_{m\ge 1} (-1)^{m} \frac{m^{m+n}}{m!}
\frac{\exp(m/2)}{2^{m}}
.$$
Using the notation from the cited post we get for our closed form
$$\frac{3^{2n+1}}{2^{n+1}} Q_n(-\exp(1/2)/2).$$
Now with $T(-\exp(1/2)/2) = -1/2$ we get with the cited identity
$$\frac{3^{2n+1}}{2^{n+1}}
\frac{1}{(3/2)^{2n+1}}
\sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^{k}
\\ = 2^n \sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^{k}.$$
This is the claim. Note that with $n=0$ we get $Q_0(-\exp(1/2)/2) - 1$
from the series for a total of $\frac{3}{2} +
\frac{3}{2} (Q_0(-\exp(1/2)/2)-1) = \frac{3}{2} +
\frac{3}{2} \frac{1}{3/2} - \frac{3}{2} = 1$ which also agrees with
the polynomial.
Concerning the addendum. Inverting $x+t-t\exp(x)$ with respect to
$x$ we obtain
$$-W(- t \exp ( -t + z )) -t + z.$$
We then get for the proposed closed form with $n\ge 1$
$$- (1-t)^{2n+1} (n+1)! [x^{n+1}]
\sum_{m\ge 1} (-1)^{m-1} \frac{m^{m-1}}{m!}
(-1)^m t^m \exp(-tm) \exp(mx)
\\ = (1-t)^{2n+1}
\sum_{m\ge 1} \frac{m^{m+n}}{m!}
t^m \exp(-tm).$$
This is $(1-t)^{2n+1} Q_n(t\exp(-t)).$ Now we have $T(t\exp(-t)) = t$
so we obtain
$$(1-t)^{2n+1} \frac{1}{(1-t)^{2n+1}}
\sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle t^k
\\ = \sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle t^k$$
as claimed. We get for $n=0$, that the coefficient on the singleton $z$
is $(1-t)$ for a total of $1-t+ (1-t) (Q_0(t\exp(-t))-1) = 1-t + (1-t)
t/(1-t) = 1$ which is the correct value.
|
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|
Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection Clearly, the total number of subsets possible is $2^5$
For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.
Now there are far too many cases (IMO) for me to manually curate, so is their a shorter way, complementary probability perhaps?
Even in complimentary method there seem to be a lot of subsets, but here is what I got
$$\binom 51(\binom 41 +\binom 42 +\binom 43 +\binom 44) + \binom 52 (\binom 31 + \binom 32 +\binom 33).....+\binom 54 (\binom 11)$$
$$=5(2^4-1) +10 (2^3-1) +10(2^2-1) +5 (2-1)$$
$$=180$$
Which comes out to be greater than $2^5$, which isn’t possible. Where am I going wrong?
Edit: My reasoning
I found the cases where one set has an element, and then the other set does not have that element, although I now realise I missed the cases where they have 1 element common or 3 or 4 or 5, which effectively makes my attempt useless.
|
Let's say instead of choosing two subsets from a single set, we choose subsets from the two sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5\}$, so number of subsets that can be chosen from $A$ are $2^{5}$ and similarly total number of subsets chosen from $B$ are $2^5$.
So, in a total of $2^5 \cdot 2^5=2^{10}$ ways, we can choose the subsets from $A$ and $B$.
Now, we need to find the cases, where there are exactly $2$ elements common to them.
If there are two elements in subset choosen from $A$, which can be done in $\displaystyle \binom{5}{2}$ ways, for two elements to be common, we can choose subsets in $B$ in $\displaystyle \binom{2}{2}$ ways, and for the rest of $3$ the elements, they can be chosen in $2^3$ ways.
If there are three elements in subset choosen from $A$, which can be done in $\displaystyle \binom{5}{3}$ ways, so choose $2$ elements from those three choosen for $A$, that is, $\displaystyle\binom{3}{2}$ and for rest choose them in $2^2$ ways.
If there four elements in subset choosen from $A$, there are $\displaystyle \binom{5}{4}$ ways. Choose two from those $4$, i.e $\displaystyle\binom{4}{2}$, choose rest in $2^1$ ways.
If all five elements are choosen in $A$, choose any two elements in $B$, for which there are $\displaystyle\binom{5}{2}$ ways.
So we have, probability that there is exactly one element common is
$$\dfrac{\binom{5}{2}\binom{2}{2}2^3+\binom{5}{3}\binom{3}{2}2^2+\binom{5}{4}\binom{4}{2}2^1+\binom{5}{5}\binom{5}{2}2^0}{2^{10}}=\dfrac{135}{512}$$
|
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|
Solve $\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $ Solve
$$\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $$
under degrees (not radians).
There is a well known identity
$$4 \sin x \sin(60 ^{\circ} −x) \sin(60 ^{\circ} +x) = \sin3x$$
but it doesn't quite help here. I can guess a solution $x = 6^{\circ} $, but I would like a more satisfactory solution other than guessing ...
|
Apply $\sin(60-x)=\frac{\sqrt3}2\cos x -\frac12\sin x$ and
rearrange the equation to isolate $x$
\begin{align}
\cot x &=\frac{1+2 \sin6\sin18} {2\sqrt3\sin 6\sin 18}
\end{align}
Then, substitute $ \sin 18 =\frac{\sin36}{2\cos18}
= \frac{\sin72}{4\cos18\cos36}
= \frac1{4\cos36}$
to obtain
\begin{align}
\cot x &=\frac{2\cos36+\sin6} {\sqrt3\sin 6}
=\frac{\cos36+(\sin54 +\sin6)} {\sqrt3\sin 6}\\
&=\frac{\cos36+\cos24} {\sqrt3\sin 6}
= \frac{2\cos30\cos6} {\sqrt3\sin 6}= \cot 6
\end{align}
Thus, $x = 6^\circ+ 180^\circ n$.
|
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|
Question about the exponential generating function for $T_n=T_{n-1}+(n-1)\cdot T_{n-2}$ So I came across this step when finding an expontential generating function using this recurrence relation,
$$
T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}
$$
Since $T_{0}=T_{1}=1$
For $n=0$
$$
\sum_{n \geq 0} \frac{T_{0} x^{0}}{0 !}=1
$$
For $n=1$
$$
\sum_{n \geq 1} \frac{T_{1} x^{1}}{1 !}=x
$$
Therefore,
$$
f(x)=\sum_{n \geq 0} \frac{T_{n} x^{n}}{n !}=1+x+\sum_{n \geq 2} \frac{T_{n} x^{n}}{n !}
$$
From the recurrence relation, $T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}$. Hence,
$$
\Rightarrow 1+x+\sum_{n \geq 2} \frac{T_{n-1}+(n-1) T_{n-2}}{n !} x^{n}
$$
Distributing the terms using the commutative law of addition, we get,
$$
\begin{gathered}
\therefore 1+x+\sum_{n \geq 2} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n} \\
\Rightarrow 1+\sum_{n \geq 1} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n}
\end{gathered}
$$
Where did the ' $x$ ' go in the last step above? And how did the lower limit change from $n \geq 2$ to $n \geq 1$ ?
Could anyone help explain? Any help would be highly appreciated.
|
Just compare the two lines, focusing on
$$\sum_{n\geq 1} \frac{T_{n-1}}{n!}x^n = \color{red}{\underbrace{\frac{T_{1-1}}{1!}x^1}_{n=1}} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$
$$x + \sum_{n\geq 2} \frac{T_{n-1}}{n!}x^n = \color{red}{x} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$
so it all comes down to $$\frac{T_{1-1}}{1!}x^1 \stackrel{?}{=} x$$ and this is true since $T_0 = 1$.
|
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|
Leap year probability A year in the 2020s (i.e., from 2020 to 2029 inclusive) is selected uniformly at random, a month
is selected uniformly at random from that year, and a day of the month is selected uniformly at
random from that month.
(a) What is the (exact) probability that the day is the 29th of February?
->I use 3/10 * 1/12 * 1/29 [not very sure]
(b) Given that the day is the 29th, what is the probability that the month is February?
For this one, I know I should use Baye's rule, but cannot proceed further. Is anyone willing to help?
|
Your answer to a is correct. You must choose a leap year, and then you must choose February, and then you must choose the 29th.
For part (b), $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$ Setting $A = $ "February" and $B = $ "29th", we find $P(A) = 1/12$, regardless of the year chosen.
Finding $P(B)$ is the sum of all the ways it can happen. $$\frac{1}{10}\cdot \frac{1}{12} \cdot \frac{1}{31}$$ is the probability of getting January 29th, 2020. Likewise there are another $112$ possibilities. We just need to group them. $$\frac{7}{10}\cdot \frac{1}{12}\cdot [\frac{1}{31} + \frac{1}{31} + \frac{1}{30} + \frac{1}{31} + ...]$$ has eleven terms, $7$ of which are $\frac{1}{31}$ and $4$ of which are $\frac{1}{30}$, giving us $$\frac{7}{10}\cdot \frac{1}{12}\cdot [\frac{7}{31} + \frac{4}{30}]$$ gives the probability for non-leap years. Add to that $$\frac{3}{10} \cdot \frac{1}{12} \cdot [\frac{7}{31} + \frac{4}{30} + \frac{1}{29}]$$ to get the total for $P(B)$.
|
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|
If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$
But the actual answer is $2.$
Am I missing something ?
|
After completing the squares you can use the inequality between the arithmetic and geometric mean (AM-GM) in the form $a^2+b^2 \geq 2ab$:
Hence,
$$\frac{(2(x+1))^2+3^2}{6(x+1)}\stackrel{AM-GM}{\geq}\frac{2\cdot 2(x+1)\cdot 3}{6(x+1)}=2$$
Equality is reached for $2(x+1)=3 \Leftrightarrow x = \frac 12$.
|
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|
Finding the Closed Form of a Recursion I wanted to find the closed form of the recursion: $$a_n = \frac{1}{n+1} + \frac{n-1}{n+1} a_{n-1}$$ where $a_0 = 0$.
So far, my progress has been to multiply both sides by $\frac{n(n+1)}{2}$ which gives $$\frac{n(n+1)}{2} a_n = \frac{n}{2} + \frac{n(n-1)}{2} a_{n-1}.$$
Now, if we let $b_n = \frac{n(n+1)}{2} a_n$, the recursion becomes: $$b_n = \frac{n}{2} + b_{n-1}.$$
How do I continue on from here? I feel like I have to figure out a closed form for $b_n$ and use that to compute $a_n$, but I can't seem to figure out a way to do so.
Any help would be greatly appreciated.
|
I think all of $a_n = \frac{1}{2}$
You can easily prove it using induction also.
$a_1 = \frac{1}{2}$
Assume $a_m = \frac{1}{2}$
To prove $a_{m+1} = \frac{1}{2}$
$a_{m+1} = \frac{1}{m+2} + \frac{m+1-1}{m+2} a_m$
$a_{m+1} = \frac{1}{m+2} + \frac{m}{m+2} \frac{1}{2}$
$a_{m+1} = \frac{1}{2}$
Since the relation is true for $n=1$ and assuming the result for $a_m$, we can prove it for $a_{m+1}$, hence the result is true by induction.
Also we can get the same from your equation,
$b_n = \frac{n}{2} +b_{n-1}$
$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$
$b_n = \frac{n}{2} +\frac{n-1}{2}+ \dots + \frac{n-k+1}{2} + b_{n-k}$
Substitute k = n
$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$
$b_n = \frac{n}{2} +\frac{n-1}{2}+ \dots + \frac{1}{2} + b_{0}$
$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$
$b_n = \frac{n + n-1 +n-2 + \dots +1}{2}+0$
$b_n = \frac{n(n+1)}{4}$
$\frac{n(n+1)}{2} a_n = \frac{n(n+1)}{4}$
$a_n =\frac{1}{2}$
|
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|
How to go about finding whether this limit exists: $\lim_{x \to 0} x [[\frac{1}{x}]]$? I am trying to solve few challenge questions on Real Analysis from Kaczor and Nowak's Problems in Mathematical Analysis, to become more proficient and stimulate thinking. I'd like someone to (a) verify if my proof is correct (b) is there a way to rigorously show that the limit does not exist?
Note. $[[x]]$ is the greatest integer less than or equal to $x$ for all $x \in \mathbf{R}$.
Find the limits or state that they do not exist.
Problem 1.1.1 (b) $\lim_{x \to 0}x\cdot[[\frac{1}{x}]]$
Proof.
Consider the sequences $a_n:=\frac{1}{n}$ and $b_n=-\frac{1}{n}$. Both these sequences converge to zero.
The corresponding image sequences are,
\begin{align*}
(f(a_n)) &= \frac{1}{1}\cdot 1, \frac{1}{2} \cdot 2, \frac{1}{3}\cdot 3, \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}
\begin{align*}
(f(b_n)) &= -\frac{1}{1}\cdot (-1), -\frac{1}{2} \cdot (-2), -\frac{1}{3}\cdot (-3), \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}
Also, consider the sequence $c_n:=\frac{1}{\sqrt{n}}$. The sequence $(c_n)$ also converges to zero.
\begin{align*}
(f(c_n)) &= \frac{1}{\sqrt{1}}\cdot [[\sqrt{1}]], \frac{1}{\sqrt{2}} \cdot [[\sqrt{2}]], \frac{1}{\sqrt{3}}\cdot [[\sqrt{3}]], \frac{1}{\sqrt{4}} \cdot [[\sqrt{4}]], \ldots \\
&= 1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},1,\frac{2}{\sqrt{5}},\frac{2}{\sqrt{6}},\frac{2}{\sqrt{7}},\frac{2}{\sqrt{8}},1,\ldots
\end{align*}
Thus, the image sequence oscillates between $0$ and $1$ and is not convergent. This violates the definition of functional limits. We require that, for all sequences $(x_n)$ in the domain of the function $f$, such that $(x_n) \to a$, $x_n \ne a$ for all $n \in \mathbf{N}$, the image sequence $f(x_n)$ converges to $L$. Then, $\lim_{x \to a}f(x) = L$. So, the above limit does not exist.
|
Hint: $\frac 1x-1\lt [[\frac 1x]]\le 1/x $ for all $x\ne 0$
By Squeeze theorem, $\lim_{x\to 0^+} x[[\frac 1x]]=1=\lim_{x\to 0^-} x[[\frac 1x]]$
|
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|
Evaluating $\sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}$ Question :
My attempt:
Let $a=17399172$
$$\begin{align}
&\sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(\ln\left(e^{C_j^k}\right)\right)\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(C_j^k\right)\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(2^k\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k} \\
\end{align}$$
Can you give me some hints what to do afterwards? I'll try by myself.
|
This is not a full answer but I will elaborate on @Dr.WolfgangHintze comments. We have
$$
\begin{align}
S
&= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}\\
&= -\sum_{k=0}^{a-1}(1+H_{k+1})\frac{(\frac{1}{1-e})^{k+1}}{k+1} \\
&= -\int_0^{\frac{1}{1-e}}\sum_{k=0}^{a-1}(1+H_{k+1})t^k\,\mathrm dt\\
&=: -\int_0^{\frac{1}{1-e}}S^\ast(t)\,\mathrm dt.
\end{align}
$$
Now write
$$
\begin{align}
S^\ast(t)
&= \sum_{k=0}^{a-1}(1+H_{k+1})t^k\\
&= \frac{1-t^a}{1-t}+\sum_{k=0}^{a-1}H_{k+1}t^k\\
&= \frac{1-t^a}{1-t}+\sum_{k=0}^\infty H_{k+1}t^k-\sum_{k=a}^\infty H_{k+1}t^k\\
&= \frac{1-t^a}{1-t}+\sum_{k=0}^\infty H_{k+1}t^k-\sum_{k=0}^\infty H_{a+k+1}t^{a+k}.
\end{align}
$$
Working with the recurrence relations for the harmonic numbers as well as the Cauchy product we find
$$
\begin{align}
\sum_{k=0}^\infty H_{a+k+1}t^{a+k}
&=t^a\sum_{k=0}^\infty (H_a+\sum_{\ell=1}^{k+1}\frac{1}{a+\ell})t^k\\
&=t^a\left(\frac{H_a}{1-t}+\sum_{k=0}^\infty\sum_{\ell=0}^k\frac{1}{a+1+\ell}t^k\right)\\
&=\frac{t^a}{1-t}\left(H_a+\sum_{k=0}^\infty\frac{1}{a+1+k}t^k\right)\\
&=\frac{t^a}{1-t}\left(H_a+\sum_{k=0}^\infty\frac{(1)_k\Gamma(a+1+k)}{\Gamma(a+2+k)}\frac{t^k}{k!}\right)\\
&=\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right).
\end{align}
$$
Setting $a=0$ gives the closed form for the other series term and so
$$
\begin{align}
S^\ast(t)
&= \frac{1-t^a}{1-t}-\frac{\log(1-t)}{1-t}-\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right).
\end{align}
$$
Therefore,
$$
\begin{align}
S
&= -\int_0^{\frac{1}{1-e}} \frac{1-t^a}{1-t}-\frac{\log(1-t)}{1-t}-\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right)\,\mathrm dt.
\end{align}
$$
Most of the integrals are easy to evaluate with the exception of the integral involving the hypergeometric function which will take a little extra work.
|
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|
Big O for error terms The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the example give here we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.
1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$
Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?
2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?
I appreciate your help! :)
|
Always remember that $\mathcal{O}(g)$ is a set. So it is more appropriate to say that when $x\to a, f\in \mathcal{O}(g)$ which means that there exists an $\epsilon\gt 0 $ for which you can find a $\delta\gt 0$ such that $0\le |f(x)|\le \epsilon |g(x)|$ for all $0\lt|x-a|\lt \delta$.
Your question 1): Why $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$
Answer:
Note that $\lim_{x\to 0}\Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|=\frac 16$ hence by limit definition there exists a $\delta\gt 0$ such that
$\begin{align}
0\lt|x|\lt \delta& \implies \Big|\Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|-\frac 16\Big|\lt 1\\
&\implies \Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|\lt \frac 76\\
& \implies \Big|\frac{x^3}{3!}+\frac{x^4}{4!}+...\Big|\le \frac 76 |x^3|\\
&\implies \frac{x^3}{3!}+\frac{x^4}{4!}+...\in \mathcal{O}(x^3) \end{align} $
Similarly, you can prove the second one.
|
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|
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
|
One more option that apparently hasn't been mentioned yet: in your original question you mentioned being stuck on the term
$$\sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
This can be rewritten as
$$\frac{1}{2}\sum_{n=1}^{ \infty } \left(-\frac{1}{3}\right)^n$$
which can also be evaluated as a geometric series.
|
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|
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
|
Consider the function $\displaystyle f(x) = \frac{1-x^2}{(1+x^2)^2} $. The Maclaurin series of $f$ is given by:
$$\displaystyle \sum_{k \ge 0} [k(-1)^k x^{2k} - k(-1)^k x^{2k-2}]$$
The integral is equal to $\displaystyle \int_0^1 f(x)\log{x}\,\mathrm{d}x $. Using the series expansion you should get
$$\displaystyle I = \sum_{k \ge 0} \bigg(\frac{k(-1)^k}{(2k-1)^2} + \frac{k(-1)^{k+1}}{(2k+1)^2}\bigg) = \sum_{k \ge 0} \frac{(-1)^{k+1}}{(2k+1)} = -\frac{\pi}{4}.$$
|
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|
Show that a number belongs or does not belong to a Cantor set I have trouble showing that $\frac{10}{31}$ and $\frac{45}{69}$ belong/do not belong to the Cantor set. I tried to do the following
$$\sum_{n = x}^\infty \frac{2}{3^n} = \frac{2}{3^x} + \frac{2}{3^{(x + 1)}} + \frac{2}{3^{(x + 2)}} + \dots = \frac{2}{3^x} \left[ 1 + \frac{1}{3^1} + \frac{1}{3^2} + \dots \right] = \frac{2}{3^x} \sum_{k = 0}^\infty \frac{1}{x^k} = \frac{2}{3^x} \times \frac{1}{1- \frac{1}{3}} = \frac{1}{3^{x- 1}}$$
and then plug different values of $x = 1, 2, \dots$, but I could not seem to get these numbers. Does it mean they do not belong to the Cantor set? Please help if you can. Thank you so much.
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Neither one is in the Cantor set. $\dfrac{10}{31} = 0.02220101110012..._{3}$, and $\dfrac{45}{69} = \dfrac{15}{23} = 0.12212110220122..._{3}$. Both has $1$ in their ternary presentation.
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|
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
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You can crudely bound $1+\frac{1}{2} + \cdots + \frac{1}{n}$ by
$$\sum_{k=1}^n \frac{1}{k} \le \sum_{k=1}^{\lceil\sqrt{n}\rceil} \frac{1}{k} + \sum_{k=\lceil \sqrt{n} \rceil}^n \frac{1}{k} \le \lceil \sqrt{n}\rceil \cdot 1 + n \cdot \frac{1}{\sqrt{n}} \le 3\sqrt{n},$$
so that when you divide this by $n$, the expression is bounded by $3 \frac{\sqrt{n}}{n} \to 0$.
|
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|
Equation of 2nd order kernel of arc-cosine kernel In the paper "Kernel method for deep learning" (Cho and Saul, NIPS 2009), a part $J_n(\theta)$ of definition of the arc-cosine kernel is given as
$$J_n(\theta) = (-1)^n (\sin \theta)^{2n+1} \left( \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \right)^n \left( \frac{\pi - \theta}{\sin \theta} \right).$$
For $n=2$, eq.(7) in the paper shows
$$J_2(\theta) = 3 \sin \theta \cos \theta + (\pi - \theta) (1 + 2 \cos^2 \theta).$$
How to derive this equation?
In my calculation, $J_2(\theta)$ becomes $J_2(\theta) = 2 \sin \theta \cos \theta + (\pi - \theta) (1 + \cos^2 \theta)$.
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Most likely you made a computational mistake somewhere.
\begin{align}
J_2(\theta) &= (-1)^2 (\sin\theta)^5 \left( \frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)^2\left( \frac{\pi - \theta}{\sin \theta}\right)\\
&= (\sin^5 \theta) \left( \frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)\frac1{\sin \theta} \frac{-\sin \theta -[\cos \theta] (\pi - \theta) }{\sin^2 \theta} \\
&= -\sin^4 \theta \frac{\partial }{\partial \theta } \left( \frac{\sin \theta + [\cos \theta] (\pi - \theta)}{\sin^3 \theta}\right)\\
&= - \sin^4 \theta \left(\frac{\sin^3 \theta (\cos \theta -\cos \theta -[\sin \theta] (\pi - \theta)) - (\sin \theta + [\cos \theta] (\pi - \theta))(3\sin^2 \theta \cos \theta)}{\sin^6 \theta} \right)\\
&= -\frac1{\sin^2 \theta} \left([ -\sin^4 \theta] (\pi-\theta) - (\sin \theta + [\cos \theta] (\pi - \theta))(3\sin^2 \theta \cos \theta) \right)\\
&= [\sin^2 \theta] (\pi - \theta) + (\sin \theta + [\cos \theta](\pi-\theta))3\cos \theta)\\
&= 3 \sin \theta \cos \theta + (2\cos ^2 \theta + 1)(\pi - \theta)
\end{align}
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How would I integrate $\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$? $$\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$$
I have known this equals to $\frac{\pi^2}{16}$, but I cannot prove it.
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Note
\begin{align}
I=&\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx\\
= &\int_0^1 \frac{\frac12\ln(1-x^2)+\ln(1+ \frac x{\sqrt{1-x^2}})}{x}\,dx=\frac12I_1+I_2\tag1
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln(1-x^2)}{x}\,dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln(1-x)}{x}\,dx\\
&= \int_0^1 \frac{\ln(1-x)}{x}\,dx+ \int_0^1 \frac{\ln(1+x)}{x}\,dx
= -\int_0^1 \frac{\ln(1+x)}{x}\,dx\\
I_2 &= \int_0^1 \frac{\ln \left(1+\frac x{\sqrt{1-x^2}}\right)}x
\overset{x=\frac t{\sqrt{1+t^2}}}=\int_0^1\frac{\ln(1+t)}{t(1+t^2)}dt+ \int_1^\infty \underset{t\to1/t}{\frac{\ln(1+t)}{t(1+t^2)}dt}\\
&=\int_0^1 \frac{\ln(1+t)}t dt - \frac14 \int_0^1 \frac{\ln t}{1+t}dt
=\frac54 \int_0^1 \frac{\ln(1+t)}t dt
\end{align}
Substitute $I_1$ and $I_2$ into (1) to obtain
$$I= \frac34 \int_0^1 \frac{\ln(1+t)}t dt =\frac34\cdot \frac{\pi^2}{12}=\frac{\pi^2}{16}
$$
$\int_0^1 \frac{\ln(1+t)}t dt=\frac{\pi^2}{12}$
|
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|
Abundant products of iterations of Euler's totient function Let $a_0(n) = n$ and $a_{i+1}(n) = \varphi(a_i(n))$ for $i\geq 0$, where $\varphi(n)$ is Euler's totient function (the number of positive integers less than or equal to $n$ and coprime with $n$). Denote $f(n) = \prod_{k=0}^{\infty}a_k(n)$.
Determine all positive integers $n$ such that the sum of positive divisors of $f(n)$ is strictly
greater than $2f(n)$.
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Let's observe that:
\begin{align*}
a_0(n) &= n \,, \\
a_1(n) &= \varphi(n) \,, \\
a_2(n) &= \varphi(a_1(n)) = \varphi(\varphi(n)) = \varphi^{(2)}(n)\,, \\
... \\
a_k(n) &= \varphi^{(k)}(n)\,,
\end{align*}
where $\varphi^{(k)}(n)$ is $k$ times composition of the totient function. This gives us the function $f(n)$ as
\begin{align*}
f(n) = n\prod_{k=1}^{\infty}\varphi^{(k)}(n) \,.
\end{align*}
Now it is intuitively clear why $\varphi^{(k+1)}(n)\leq\varphi^{(k)}(n)$; that is, we expect the totient function to decrease with each iteration (until it becomes $1$). This means that for a finite number $n$, we have a finite number of elements in the product of $f(n)$.
The problem asks us to find all numbers $n$ that give us $2f(n)<\sigma(f(n))$, where
\begin{align*}
\sigma(f(n)) = \sum_{d|f(n)}d\,,
\end{align*}
is called the divisor function and is the sum of all divisors of $f(n)$.
Numbers that satisfy $2f(n)<\sigma(f(n))$ are called abundant numbers. However, we are going to look for numbers $2f(n)>\sigma(f(n))$, where $n$ is referred to as deficient numbers. Deficient numbers contain all odd numbers with distinct primes and all even numbers that are powers of $2$. Why we choose to look at this case will become apparent soon.
The totient function can be written as
\begin{align*}
\varphi(n) = n\prod_{p|n}\bigg(\frac{p-1}{p}\bigg)\,,
\end{align*}
where $p|n$ means all primes that divide $n$.
The only number that is even and prime is $2$. This means that numbers that are made up of primes greater than $2$ will always give $\varphi(n)$ as an even number. This means that for all $n>2$ we have that $f(n)$ will always be an even number so we do not need to worry about odd deficient numbers.
However, for $n = 2^{m}$ we have
\begin{align*}
\varphi(2^{m}) = 2^{m}\bigg(\frac{2-1}{2}\bigg) = 2^{m-1}\,.
\end{align*}
Thus $f(2^{m})$ is
\begin{align*}
f(2^{m}) = 2^{m+(m-1)+(m-2)+...+1} = 2^{m(m+1)/2}
\end{align*}
and will be a deficient number because $f(2^{m})$ is a power of $2$ (as stated before).
We need to also exclude perfect numbers that satisfy $\sigma(f(n)) = 2f(n)$. We already worked out that $f(n)$ is always even and we do not need to worry about odd perfect numbers (still an open problem in number theory). Even perfect numbers are of the form $2^{p-1}(2^{p}-1)$, where $p$ is a prime and $2^{p}-1$ is also a prime (Mersenne prime). We need $n = 2^{p} - 1$ because this is the only term that is odd. Now if we apply the totient function $\varphi(n) = n - 1 = 2^{p}-2$ (because $n$ is prime and will be always coprime with all numbers up to itself), we expect $\varphi(n) = 2^{p-1}$ and this gives us the equation
\begin{align*}
2^{p} - 2 &= 2^{p-1} \\
2^{p-1} - 1 &= 2^{p-2} \,.
\end{align*}
The LHS is odd while RHS is even. So only $p = 2$ is valid and we get $n = 3$ and $f(3)$ will be a perfect number.
In conclusion, the numbers that do not satisfy the inequality are $3$ and $2^{m}:m\in\mathbb{N}$. Any other number satisfies the inequality.
|
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|
Find all natural solutions that satisfy $2^ + 3^ = ^2$ It looks like an easy question but I couldn't find a way to solve it. I found (0,1,2),(3,0,3),(4,2,5) by trial and error and I'm kinda sure they are the only answers but I'm not sure how to prove it.
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If $y=0$, we have $2^x=z^2-1=(z+1)(z-1)$, for which the only solution is $x=z=3$. That is, we have $z+1=2^a$ and $z-1=2^b$ with $a+b=x$. But then $2=(z+1)-(z-1)=2^a-2^b$, for which the only solution is $a=2$ and $b=1$, from which $x=z=3$ follows.
If $y\ge1$, then $2^x$ is a square mod $3$, which implies $x$ is even. Writing $x=2w$, we have
$$3^y=z^2-2^{2w}=(z+2^w)(z-2^w)$$
which implies $2^{w+1}=3^y-1$. (That is, $z+2^w$ and $z-2^w$ must each be a power of $3$ whose exponents' sum is $y$, so their difference, which is $2\cdot2^w$, is the difference of those two powers of $3$. But $2^{w+1}$ is not divisible by $3$, so the smaller exponent must be $0$.) Now if $y$ is odd, then $3^y\equiv3$ mod $8$, which implies $2^{w+1}\equiv2$ mod $8$, so that $w$ can only be $0$, which gives us the solution $(x,y,z)=(0,1,2)$ and no other solutions with $y$ odd. On the other hand, if $y$ is even, say $y=2v$, then, writing $z'=3^v$, we have $2^{w+1}=z'^2-1$, for which we saw in the first paragraph there is only the solution $w+1=z'=3$, giving us the solution $(x,y,z)=(4,2,5)$ and no other solutions with $y$ even.
So that's it: the equation $2^x+3^y=z^2$ has the three solutions $(3,0,3)$, $(0,1,2)$, and $(4,2,5)$ and no others.
|
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|
Use binomial series for $\sqrt{1+x}$ to find the binomial series for $\frac{1}{\sqrt{1+x}}$ The question specifically asks me to use the binomial series for $\sqrt{1+x}$ to find the series for $\frac{1}{\sqrt{1+x}}$.
First, I need to find the derivative of the binomial series $\sqrt {1+x}$
I first constructed the binomial series based on the formula $$1+rx+\frac {r(r-1)x^2}{2!} + ... + \frac{r(r-1)...(r-n+1)x^n}{n!}$$
I got $$1+\frac {1}{2} x - \frac{1}{2!}\frac{1}{2^{2}}x^2 + ... + \frac{(-1)^{n+1}}{n!}\frac{1\cdot 3\cdot 5...(2n-3)}{2^n}x^n$$
$$1 + \Sigma_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}\frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^n}x^n$$
When I take the derivative, the first few terms becomes:
$$0 + \frac{1}{2} - \frac{1}{1!}\frac{1}{2^{2}}x + \frac{1}{2!}\frac{1\cdot3}{2^3}x^2 ... $$
So $$\frac{d}{dx}\sqrt{1+x} \,=\, \frac{1}{2} + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^{n+1}}x^{n}$$
And I also know that $$\frac{d}{dx}\sqrt{1+x} = \frac{1}{2\sqrt{1+x}}$$
Therefore, $$\frac{1}{\sqrt{1+x}} = 2\frac{d}{dx} \sqrt{1+x}$$
So multiply the series I found by 2 gives me $$2\cdot (\frac{1}{2} + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^{n+1}}x^{n}) = 1 + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{n!} \frac{1\cdot 3 \cdot 5...\cdot (2n-3)}{2^{n}}x^{n}$$
However, my answer was wrong, and instead of $$1\cdot 3 \cdot 5...\cdot (2n-3)$$, it should be $$1\cdot 3 \cdot 5...\cdot (2n-1)$$
Why is that?
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The expanded form of $\frac{d}{dx}\sqrt{1+x}$ is correct, the compact one is not. You could do $$\frac{d}{dx}\left(1+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n!}\frac{1\cdot3\cdots(2n-3)}{2^n}x^n\right)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!}\frac{1\cdot3\cdots(2n-3)}{2^n}x^{n-1}$$ without expanding (at $n=1$ we have an empty product in both cases, which is assumed to be $1$).
Now, to "shift the index", put $n=m+1$ everywhere, and you get $$\sum_{m=\color{red}{0}}^\infty\frac{(-1)^m}{m!}\frac{1\cdot3\cdots(2m-1)}{2^{m+1}}x^m.$$
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A Difficult Area Problem involving a Circle and a Square A few days ago, I encountered the following problem:
After a little bit of thinking, I managed to come up with the following solution:
*
*Rotate the square $90^\circ$ clockwise and let the new bottom left corner of the square be $(0,0)$.
*The circle inscribed in the square is hence centered at $(5,5)$ with a radius of $5$. The circle equation thus becomes $(x-5)^{2} + (y-5)^{2} = 25 \Rightarrow y = 5 + \sqrt{25 - (x-5)^{2}}$ in the first quadrant.
*Similarly for the quarter circle, the equation becomes $y = \sqrt{100-x^2}$.
The graph hence looks like this:
My intention is to find the shaded area in the above graph. To do so, first I find $X$ by equating $5 + \sqrt{25 - (x-5)^{2}} = \sqrt{100-x^2} \Rightarrow x=\frac{25 - 5\sqrt{7}}{4}$.
From this, I calculate the area of the shaded region as follows:
$$\text{Area} = (10 \cdot \frac{25 - 5\sqrt{7}}{4} - \int_0^\frac{25 - 5\sqrt{7}}{4} \sqrt{100-x^2} \,\mathrm{d}x) + (10 \cdot (5 - \frac{25 - 5\sqrt{7}}{4}) - \int_\frac{25 - 5\sqrt{7}}{4}^5 5 + \sqrt{25 - (x-5)^{2}} \,\mathrm{d}x) \approx 0.7285$$
Now, the diagram looks like this:
From here, I figured out the shaded area as follows:
$$\text{Area} \approx 10^{2} - \frac{\pi(10^{2})}{4} - (\frac{10^{2} - \pi(5^{2})}{4} + 2 \times 0.7285) \approx \boxed{14.6 \: \text{cm}^{2}}$$
While I did figure out the correct solution, I find my approach to be rather lengthy. I was wondering if there is a quicker, simpler and more concise method (that probably does not require Calculus) that one can use and I would highly appreciate any answers pertaining to the same.
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Here is an alternate solution,
We assume right bottom vertex to be the origin then, equations of circles are
Circle S: $x^2+y^2 = 100$
Circle T: $(x+5)^2+(y-5)^2 = 25$
Solving both equations, intersection points are $A \left(\frac{5}{4}\left(\sqrt7-5\right), \frac{5}{4}\left(\sqrt7+5\right)\right)$ and $B\left(-\frac{5}{4}\left(\sqrt7+5\right), -\frac{5}{4}\left(\sqrt7-5\right)\right)$
So length of chord $AB$ at intersection is $\frac{5 \sqrt7}{\sqrt2}$.
We note that this chord $AB$ is common chord of both circles.
Angle subtended by chord at the center is given by,
At $P$, $\angle APB = \alpha = 2 \arcsin \left(\frac{\sqrt7}{2\sqrt2}\right)$
At $O$, $\angle AOB = \beta = 2 \arcsin \left(\frac{\sqrt7}{4\sqrt2}\right)$
Shaded area is difference of area of two circular segments of this chord, which are $ATB$ and $ASB$.
$ = \left(circular \ sector \ PATB - \triangle PAB\right) - \left(circular \ sector \ OASB - \triangle OAB\right)$
$ = \left(25 \times \frac{\alpha}{2} - \frac{25 \sqrt7}{8}\right) - \left(100 \times \frac{\beta}{2} - \frac{125 \sqrt7}{8}\right) \approx 14.638 $
EDIT: to find $AB$ without coordinate geometry, we know that $OP = 5 \sqrt2$. If perp from $P$ to $AB$ is $x$ then,
$\left(5\sqrt2 + x\right)^2 + \left(\frac{AB}{2}\right)^2 = 10^2$ and $x^2 + \left(\frac{AB}{2}\right)^2 = 5^2$. Solving them, we get value of $AB$.
|
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Find the number of three-digit numbers in which exactly one digit $3$ is used? Find the number of three-digit numbers in which exactly one digit $3$ is used?
The number is of one of the forms $$\_\text{ }\_\text{ }3 \\ \_\text{ } 3 \text{ } \_ \\ 3\text{ }\_ \text{ }\_$$ There are $V_9^2-V_8^1=9\times8-8=64$ possibilities for each of the first $2$ forms, and $V_9^2$ for the third. This makes $2\times64+72=200$ possibilities in total. The given answer in my book is $225$. What am I missing?
|
We have $9$ digits other than $0$, but $0$ can't be the first digit. So for both the first and second, we have $8 \cdot 9=72$ possibilities each. For the last one we have $9 \cdot9=81$ possibilities.
Alternatively, there are $9 \times 10 \times 10 = 900$ three-digit numbers. $8 \times 9 \times 9 = 648$ of them have no $3$s as the first digit cannot be $0$ or $3$, and $1$ number has all three $3$s. For the numbers with two $3$s, they must be in the form _ 33, 3 _ 3, 33 _, which makes $8 + 9 + 9 = 26$ possibilities.
Thus there are $900 - 648 - 1 - 26 = 225$ three-digit numbers with exactly one $3$.
|
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Prove $\{a_n\}$ converges.
Suppose $a_1,a_2>0$ and
$a_{n+2}=2+\dfrac{1}{a_{n+1}^2}+\dfrac{1}{a_n^2}(n\ge 1)$. Prove
$\{a_n\}$ converges.
First, we may show $\{a_n\}$ is bounded for $n\ge 3$, since $$2 \le a_{n+2}\le 2+\frac{1}{2^2}+\frac{1}{2^2}=\frac{5}{2},~~~~~~ \forall n \ge 1.$$
But how to go on?
|
Another approach: Let $L \approx 2.3593$ be the unique solution of $L = 2 + 2/L^2$ in the interval $[2, 2.5]$, and $b_n = a_n - L$. We want to show that $b_n \to 0$.
The recursion formula becomes
$$
b_{n+2} = \frac{1}{(L+b_{n+1})^2} + \frac{1}{(L+b_{n})^2} - \frac{2}{L^2} \, .
$$
We estimate
$$
\left| \frac{1}{(L+b_{n})^2} - \frac{1}{L^2}\right| = \frac{|b_n|(2L+b_n)}{L^2(L+b_n)^2} \le \frac{5}{16} |b_n|
$$
for $n \ge 3$, so that
$$
|b_{n+2} | \le \frac{5}{16} (|b_{n+1}| + |b_n|) \, .
$$
Then
$$
2 |b_{n+2} | + |b_{n+1} | \le \frac{13}{8}|b_{n+1} | + \frac{5}{8}|b_{n}|
\le \frac{13}{16} \left(2 |b_{n+1} | + |b_{n} | \right) \, .
$$
This shows that $2 |b_{n+1} | + |b_{n} |$ decreases geometrically to zero. It follows that $b_n \to 0$, as desired.
|
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irrational integral $ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$ I have to solve this irrational integral $$ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$$
It seems that the most convenient way to operate is doing the substitution
$$ x= \frac{t^2}{3-2t}$$
according to the rule,
obtaining the integral:
$$ \int \frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}\, dx$$
Then $$\frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}= \frac{A}{t-2}+\frac{B}{t+3}+\frac{C}{3-2t}+\frac{D}{(3-2t)^2}$$
then I found the coefficient $A= \frac{12}{5},B= -\frac{12}{5}, C= \frac{15}{2},D= -\frac{9}{2}$
In my book the integral on which to operate is:
$$ \int -2\frac{(t^2-3t)(3+t-t^2)}{(t^2-7t+6)(3-2t)^2}\, dx$$
that is different from my
$$ \int \frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}\, dx$$
Perhaps I made mistakes in the first passages and I checked lots of times my calculations. Can someone indicate where I'm making mistakes?
|
Note that the substitution $x=\frac{t^2}{3-2t}$ leads to
$ \sqrt{x^2+3x}=\frac{3t-t^2}{3-2t}, \>\>\>\>\> dx = -\frac{2(t^2-3t)}{(3-2t)^2}dt
$
and
$\int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}dx
= -\int \frac{1+\frac{3t-t^2}{3-2t}}{2-\frac{3t-t^2}{3-2t}}\, \frac{2(t^2-3t)}{(3-2t)^2}dt
= -\int \frac{3+t-t^2}{t^2-7t+6}\, \frac{2(t^2-3t)}{(3-2t)^2}dt
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$ Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$
I tried substituting $x=\tan(t)$ in order to get away with square root. ($\:dx=\frac{1}{\cos^2(t)}dt\:$)
$\sqrt{x^2+1}=\frac{1}{\cos(t)}\:\:$ and $\:\:x^4-1=\frac{\sin^4(t)-\cos^4(t)}{\cos^4(t)}$
Now after putting both into main Integral and by simplifying I have :
$$
\int\frac{\sin(t)\cdot\cos^2(t)}{\sin^4(t)-\cos^4(t)} \, dt=\text{?}
$$
Now need a bit help if possible.
Thank you in advance :)
|
Let $\sqrt{x^2+1}=y, dy=\dfrac x{\sqrt{x^2+1}}$
$$\implies x^2=y^2-1$$
$$\int\dfrac{dy}{(y^2-1)^2-1}=\int\dfrac{dy}{y^2(y^2-2)}$$
Now use,
$$\dfrac2{y^2(y^2-2)}=\dfrac{y^2-(y^2-2)}{y^2(y^2-2)}=\dfrac1{y^2-2}-\dfrac1{y^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the series generated by the sequence diverges Consider the positive sequence $\{x_n\}_{n\ge 1}$ given by
$$\begin{cases}
x_1&=1,\\
x_{n+1}^2+x_{n+1}x_n^2-x_n^2&=0,\ \forall n\ge 1.
\end{cases}$$
More specifically, the second equation together with condition that the sequence is positive-valued mean that $$ x_{n+1}=\dfrac{-x_n^2+\sqrt{x_n^4+4x_n^2}}{2}>0.$$
Prove that $\sum_{n=1}^\infty x_n=+\infty$.
With some ideas from Kaind, it can be seen that
\begin{align*}
x_{n+1}< x_n\Longleftrightarrow -x_n^2+\sqrt{x_n^4+4x_n^2}< 2x_n\Longleftrightarrow x_n^4+4x_n^2 < (x_n^2+2x_n)^2=x_n^4+4x_n^3+4x_n^2.
\end{align*}
Since the latter inequality is true, we have $x_{n+1}< x_n$ for all $n,$ which means $x_n$ is decreasing. Since the sequence is bounded from below by $0,$ its limit exists. Assume the limit is $L,$ we have $L^2+L^3-L^2=0,$ or equivalently, $L=0.$
Thus $x_n\downarrow 0$. It remains to show that $\sum_{n=1}^\infty x_n=+\infty$ and I am still stucking here.
|
This is the answer from one of my friends.
We will prove that $x_n\ge 1/n$ by induction. Firstly, notice that
\begin{align*}
x_{n+1}=\dfrac{-x_n^2+\sqrt{x_n^4+4x_n^2}}{2}=\dfrac{4x_n^2}{2(x_n^2+\sqrt{x_n^4+4x_n^2})}=\dfrac{2}{1+\sqrt{1+\dfrac{4}{x_n^2}}}.
\end{align*}
Consider the function $f(x)=\dfrac{2}{1+\sqrt{1+\dfrac{4}{x^2}}}$, we have $x_{n+1}=f(x_n).$ Since $f$ is increasing, supposing $x_n\ge 1/n$, we have
\begin{align*}
x_{n+1}=f(x_n)\ge f(1/n)=\dfrac{2}{1+\sqrt{1+4n^2}}.
\end{align*}
It remains to show that $\dfrac{2}{1+\sqrt{1+4n^2}}\ge \dfrac{1}{n+1}$ which is equivalent to
\begin{align*}
2n+2\ge 1+\sqrt{1+4n^2}\Longleftrightarrow 2n+1\ge \sqrt{1+4n^2}.
\end{align*}
Squaring both side gives us a always-true expression $4n\ge 0$, and thus $x_{n+1}\ge \dfrac{1}{n+1}.$ Together with $x_1\ge 1$, by induction we have $x_n\ge 1/n$ for all $n.$ Thus $\sum_{n=1}^\infty x_n=\infty.$
|
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|
Prove that $\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$ Prove that
$$\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$$
Here I am trying the following
\begin{align*}
\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} &= \prod_{i\geq 1} \left( \sum_{n \geq 0} x^{n}y^{n(2i-1)} \right)\\
&= \sum_{m\geq 0} \sum_{n \geq 0} p(2m-1,n)x^{n}y^{2m-1}\\
&= \sum_{n\geq 0}\left( \sum_{m \geq 0} p(2m-1,n)y^{2m-1}\right)x^{n} \\
&= \sum_{n\geq 0} \frac{y^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}x^{n}\\
\end{align*}
However, I am not sure if this is totally correct.
|
We obtain
\begin{align*}
\color{blue}{\prod_{k=1}^\infty\frac{1}{1-xy^{2k-1}}}&=\prod_{k=0}^\infty\frac{1}{1-xy^{2k+1}}\\
&=\prod_{k=0}^{\infty}\frac{1}{1-\left(xy\right)\left(y^{2}\right)^k}\\
&=\frac{1}{\left(xy;y^2\right)_{\infty}}\tag{1}\\
&=\sum_{n=0}^{\infty}\frac{1}{\left(y^2;y^2\right)_n}(xy)^n\tag{2}\\
&=\sum_{n=0}^\infty\frac{1}{\prod_{k=0}^{n-1}\left(1-\left(y^2\right)^{k+1}\right)}\left(xy\right)^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\frac{1}{\prod_{k=1}^{n}\left(1-y^{2k}\right)}\left(xy\right)^n}\\
\end{align*}
Comment:
*
*In (1) we use the $q$-Pochhammer symbol
\begin{align*}
(a;q)_{n}&=\prod_{k=0}^{n-1}\left(1-aq^k\right)\\
(a;q)_{\infty}&=\prod_{k=0}^{\infty}\left(1-aq^k\right)\qquad\qquad |a|<1,|q|<1
\end{align*}
*In (2) we use the identity
\begin{align*}
\color{blue}{\frac{(at;q)_{\infty}}{(t;q)_{\infty}}=\sum_{n=0}^\infty\frac{(a;q)_n}{(q;q)_n}t^n\qquad\qquad |q|<1,|t|<1}\tag{3}
\end{align*}
where we set $a=0$. This is Theorem 2.1 in The Theory of Partitions by G. E. Andrews. We follow the proof there and write (3) as
\begin{align*}
\prod_{n=0}^\infty\frac{1-atq^n}{1-tq^n}=\sum_{n=0}^\infty\prod_{k=0}^{n-1}\frac{1-aq^k}{1-q^{k+1}}t^n
\end{align*}
The left-hand side is a function $F(t)$ which we want to expand as generating function
\begin{align*}
F(t)=\prod_{n=0}^\infty \frac{1-atq^n}{1-tq^n}=\sum_{n=0}^\infty A_n t^n\tag{4}
\end{align*}
in the unknown $A_n=A_n(a,q)$. Multiplying (4) with $1-t$ gives
\begin{align*}
(1-t)F(t)&=(1-at)\prod_{n=1}^\infty \frac{1-atq^n}{1-tq^n}=(1-at)\prod_{n=0}^\infty\frac{1-atq^{n+1}}{1-tq^{n+1}}\\
&= (1-at)F(tq)\\
&=(1-at)\sum_{n=0}^\infty A_nt^nq^n\tag{5}
\end{align*}
Denoting with $[t^n]$ the coefficient of $t^n$ in a series and making coefficient comparison in (5) gives
\begin{align*}
[t^n](1-t)F(t)&=A_n-A_{n-1}\\
&=[t^n](1-at)F(tq)\\
&=q^nA_n-aq^{n-1}A_{n-1}
\end{align*}
and
\begin{align*}
A_n(1-q^n)&=A_{n-1}\left(1-aq^{n-1}\right)\tag{6}\\
A_0&=1
\end{align*}
follows. From the recurrence relation (6) we obtain for $n\geq 1$:
\begin{align*}
\color{blue}{A_n}&=\frac{1-aq^{n-1}}{1-q^n}A_{n-1}\\
&=\frac{\left(1-aq^{n-1}\right)\left(1-aq^{n-2}\right)}{\left(1-q^n\right)\left(1-q^{n-1}\right)}A_{n-2}\\
&=\prod_{k=0}^{n-1}\frac{1-aq^k}{1-q^{k+1}}A_0\\
&\,\,\color{blue}{=\frac{(a;q)_n}{(q;q)_n}}
\end{align*}
and the claim (3) follows.
|
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|
Finding the intersection of $x= y^2 + y - 2$ and $y = -x^2 - \frac32x + 1 $ Goodmorning, I am struggling in finding the points of intersection of the following parabolas:
$$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$$
I know that these two can be solved either algebraically or with matrices, but at the moment the algebraic solution is the one that I'm going for.
How can I solve this system without getting lost in complex and boring calculations? (I'm asking specifically if there are some tricks to solve this more quickly end elegantly)
|
$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$
The first one can be written as,
$x = (y-1) (y+2)$ ...(i)
The second one is $y = -x^2 - \frac32x + 1 \implies y-1 = - x^2 - \frac{3}{2}x$ ...(ii)
Plugging in $y-1$ from (ii) into (i),
$x = (- x^2 - \frac{3}{2}x) (- x^2 - \frac{3}{2}x + 3)$
$x = x (x + \frac{3}{2}) (x(x + \frac{3}{2})-3)$
$x = 0$ is a solution. If $x \ne 0$,
$x(x+\frac{3}{2})^2 - 3 (x + \frac{3}{2}) = 1$
$a^3 - \frac{3}{2} a^2 - 3 a - 1 = 0 \ $ where $a = x + \frac{3}{2}$
$a (a - 2)(a+\frac{1}{2}) - 2 (a+\frac{1}{2}) = 0$
$(a + \frac{1}{2})(a^2 - 2 a - 2) = 0$
So $a = -\frac{1}{2}, a = 1 \pm \sqrt3$
$\implies x = - 2, x = -\frac{1}{2} \pm \sqrt3$
So all in all, we got all $4$ values of $x$ at intersections,
$x = 0, - 2, -\frac{1}{2} \pm \sqrt3$. Find corresponding $y$ values.
|
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|
Calculating a limit. Is WolframAlpha wrong or am I wrong? What I'm trying to solve:
$$\lim _{x\to -\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}$$
What I put into WolframAlpha:
(sqrt(x^2+14)+x)/(sqrt(x^2-2)+x)
My result: $1$, which I get by simply dividing bot the numerator and the denominator by $x$, then letting it go towards $-\infty$
$$\frac{\frac{\left(\sqrt{x^2+14}+x\right)}{x}}{\frac{\left(\sqrt{x^2-2}+x\right)}{x}}=\frac{\left(\sqrt{\frac{x^2}{x^2}+\frac{14}{x^2}}+\frac{x}{x}\right)}{\left(\sqrt{\frac{x^2}{x^2}-\frac{2}{x^2}}+\frac{x}{x}\right)}=\frac{\left(\sqrt{1+\frac{14}{x^2}}+1\right)}{\left(\sqrt{1-\frac{2}{x^2}}+1\right)}\:=\:\frac{\left(\sqrt{1}+1\right)}{\left(\sqrt{1}+1\right)}\:=\:\frac22 \ = \ 1$$
WolframAlpha's result: $-7$. It has a long, complicated 25 step solution.
Is the solution $1$, $-7$ or neither?
Edit: of course, I set it $x$ to go towards $-\infty$ in WolframAlpha too
|
The limit is as $x$ goes to $-\infty,$ so you can't just divide by $x.$ Both the numerator and the denominator have the form $\infty -\infty.$ WolframAlpha is correct.
|
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|
Solve for integers $x, y$ and $z$: $x^2 + y^2 = z^3.$
Solve for integers $x, y$ and $z$:
$x^2 + y^2 = z^3.$
I tried manipulating by adding and subtracting $2xy$ , but it didn't give me any other information, except the fact that $z^3 - 2xy$ and $z^3+2xy$ are perfect squares.
This doesn't give us much information to work on. I don't know if my steps are correct, I do not know how to approach this problem.
Any help would be appreciated.
|
Let $C$, $D$, $S$ and $T$ be integers, and define
\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}
Then a routine verification shows that $x^2+y^2=z^3$. I will show that every solution is of this form. Moreover, if we require $S$ and $T$ to be coprime and nonnegative, every solution will have precisely one such representation, making this a proper parametrization.
Let $x$, $y$ and $z$ be integers such that
$$x^2+y^2=z^3.$$
First note that $x$ and $y$ are not both odd, as otherwise we get a contradiction by reducing mod $8$.
Let $d:=\gcd(x,y)$ and let $a$ and $b$ be integers such that $d=ab^3$ and $a$ is cube-free. Then $d^2=a^2b^6$ divides $z^3$ and hence $a$ divides $z$. Writing $x=au$, $y=av$ and $z=aw$ we see that
$$a^3w^3=z^3=x^2+y^2=(ab^3u)^2+(ab^3v)^2=a^2b^6(u^2+v^2),$$
from which it follows that $b^2$ divides $w$ because $a$ is cube-free. So writing $x=ab^3X$, $y=ab^3Y$, $z=ab^2Z$ shows that
$$X^2+Y^2=aZ^3,$$
where $X$ and $Y$ are coprime. Factoring in $\Bbb{Z}[i]$ then shows that
$$aZ^3=(X+Yi)(X-Yi),$$
where $X$ and $Y$ are coprime and not both odd, so the two factors are coprime. Then
$$X+Yi=(A+Bi)(U+Vi)^3,$$
for some integers $A$, $B$, $U$ and $V$ such that $\gcd(A,B)=\gcd(U,V)=1$ and $A^2+B^2=a$ and $U^2+V^2=Z$. Then
\begin{eqnarray*}
X&=&AU^3-3BU^2V-3AUV^2+BV^3,\\
Y&=&BU^3+3AU^2V-3BUV^2-AV^3,
\end{eqnarray*}
and hence for $C=bA$ and $D=bB$ we find that
\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}
In particular, parametrizations given in the other answers and comments correspond to $(C,D,S,T)=$
$$(1,0,a,b),\qquad(k,k,1,0),\qquad(1,k,1,0),\qquad(a,b,1,0).$$
|
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|
$ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
&= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{2+2 x+4 \sqrt{x+1}} \cdot d x\\&=\int \frac{1}{1+x+2 \sqrt{x+1}} \cdot d x\\
&\quad +\int \frac{\sqrt{x+1}}{2(1+x)+4 \sqrt{x+1}}\cdot d x \\
&\quad -\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot d x
\end{split}
$$
From this last line i can evaluate the first two antiderivatives but the last one $$\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot dx$$ seems a bit hard for me two evaluate, i don't find the good substitution. I am opened to any suggestion. Thanks in advance!
|
Too large for a comment in the other answer.
Let us first try to identify the correct substitution
For real calculus, we need $x+1\ge0\text{ and }3-x\ge0\iff-1\le x\le3$
$$\iff-\dfrac12-\dfrac12\le\dfrac{x-1}2\le\dfrac32-\dfrac12$$
WLOG $\dfrac{x-1}2=\cos2t, 0\le2t\le\pi\implies x=2\cos2t+1$
$\sqrt{x+1}=+(2\cos t),\sqrt{3-x}=+(2\sin t)$ as $0\le t\le\dfrac\pi2$
Finally use $(\sin t+\cos t+1)(\sin t+\cos t-1)=(\sin t+\cos t)^2-1=?$
|
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|
Find the radius of convergence of $\sum_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}$ I need to find the radius of convergence of
$$\sum\limits_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}$$.
I came up with the following solution:
Since $\cos{\frac{1}{3^n}\sim1-\frac{1}{2(3^n)^2}}$, then
$$\ln\left(\cos{\frac{1}{3^n}}\right)\sim\ln\left(1-\frac{1}{2(3^n)^2}\right)\sim-\frac{1}{2(3^n)^2}$$
Therefore,
$$\sum_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}\sim\sum_{n=0}^{\infty}{-\frac{x^n}{2(3^n)^2}}=-\sum_{n=0}^{\infty}{\frac{x^n}{2(3^n)^2}},$$ where we can assess easily what the radius is using the ratio test:
$$R = \lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
Is the solution correct? If not, could you suggest an alternative solution, please?
|
As a few commentors have pointed out: Yes, this looks good. Ratio test is always going to be your friend on questions such as these. Let's go ahead and work it out so you're sure you got the right final answer:
\begin{align}
\frac{a_{n+1}}{a_n} &= \frac{{\frac{x^{n+1}}{2(3^{n+1})^2}}}{{\frac{x^n}{2(3^n)^2}}}\\
&= {\frac{x^{n+1}}{2(3^{n+1})^2}}\cdot\frac{2(3^n)^2}{x^n}\\
&= {\frac{x}{(3^{n+1})^2}}\cdot\frac{(3^n)^2}{1}\\
&= {\frac{x}{3^2}}\cdot\frac{1}{1}\\
&= {\frac{x}{9}}\\
\end{align}
so we have that radius of convergence is $-9 \leqslant x \leqslant 9$.
|
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|
Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$ Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$
First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^2-6x+15)\\x^4-12x^3+15x^2-10x^3+120x-150x+15x^2-180x+225=\\=4x^3-24x^2+60x$$ which is an equation I can't solve. I tried to simplify the LHS by $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{(x^2-6x+15)-4x}{x^2-6x+15}=1-\dfrac{4x}{x^2-6x+15}$$ but this isn't helpful at all. Any help would be appreciated! :) Thank you in advance!
|
If you take out $x$ (clearly $x\ne 0$) we get: $$\dfrac{x(x-10+{15\over x})}{x(x-6+{15\over x})}=\dfrac{4x}{x(x-12+{15\over x})}.$$
Cancel $x$ and let $t=x+ {15\over x}$, then we have $${t-10\over t-6} = {4\over t-12}$$ or $$t^2-22t+120 = 4t-24$$
and so on...
|
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|
Using polar coordintaes to evaluate $\int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy$ I want evaluate the following integral using polar coordinates.
$$ \int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy $$ over the positive quadrant of the circle $$x^2+y^2=1$$
I used the substitution $x=r\cos\theta$, $y=r\sin\theta$ and reduced the integral to
$$ \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} {\sqrt {\frac {1-r^2} {1+r^2}}}r drd\theta$$
Then, I made the substitution $u=r^2$, giving me, $2rdr=du$ and got
$$\int {\sqrt{\frac{1-u}{1+u}}}du$$
I then split the integrand into partial fractions $-1 + {\frac{2}{1+u}}$
How do I now proceed? Can I reduce it to the form $\int \sqrt{x^2-a^2}\ dx$?
|
Integrate by parts as follows
$$\int {\sqrt{\frac{1-u}{1+u}}}du
= \int {\sqrt{\frac{1-u}{1+u}}}d(1+u)
=\sqrt{1-u^2}+\int \frac{du}{\sqrt{1-u^2}}
$$
|
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|
Study the series of $\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$ with respect to $a>0$ I have to study the series $\sum a_n$ with $a_n=\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$, with respect to $a>0$.
I have thought to use the asymptotic criterion for series.
In particular I can observe that:
$\sin{x}\sim x-\frac{x^3}{3!}$ and so: $\sin{(\sqrt[3]{x})}\sim \sqrt[3]{x}- \frac{x}{3!}$ for $x\to 0$.
Now this means that: $$\int_0^{\frac{1}{n^a}}\sqrt[3]{x}- \frac{x}{3!}\,dx=[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}]\rvert_0^{\frac{1}{n^a}}-[\frac{x^{2}}{2\cdot 3!}]\rvert_0^{\frac{1}{n^a}}=\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}$$ Now since $\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}$ it self is asymptotic to $\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$ then I can say:
$$a_n\sim \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}\sim \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$$
Now $\sum_{n=1}^{\infty} \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$ is convergent for $\frac{4}{3}a>1\iff a>\frac{3}{4}$ and so from asymptote criterion also the original series converges for this value of $a$.
I need a check of my attempt and in case there is something wrong can you tell me where and how can I correct this?
(I can't use the Lebesgue theore since it is not something that I know)
|
Here is a simpler proof of the equivalent: let $f(t) = \frac{\sin(t)}{t}$. It is well known that $|f(t)|\le 1$ for all $t\in{\mathbb R}$ and that $f(0)=1$. The substitution $x = \frac{y}{n^a}$ in the integral gives
\begin{equation}
a_n = \left(\frac{1}{n^a}\right)^{4/3}\int_0^1 f\left(\left(\frac{y}{n^a}\right)^{1/3}\right) y^{1/3} d y
\end{equation}
By Lebesgue's dominated convergence theorem, the above integral converges to
$\int_0^1 y^{1/3} d y = \frac{3}{4}$ when $n$ tends to $\infty$, hence
\begin{equation}
a_n \sim \frac{3}{4}\left(\frac{1}{n^a}\right)^{4/3}
\end{equation}
|
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|
Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$ Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$
And Prove it by induction.
Attempt:
$$A(x)= \sum_{k=0}^n (k+1)(n-k+1)$$
$$\sum_{n=0}^\infty(\sum_{k=0}^n (k+1)(n-k+1))x^n$$
The series of differences is :$\{(n+1),2(n),3(n-1),4(n-2)…,1,…\}$
The generating functions of this series is : $(1-x)A(x)$
The series of differences is :$\{(n+1),(n-1),(n-3),(n-5),…,2-n,…\}$
The generating functions of this series is : $(1-x)^2A(x)$
The series of differences is :$\{(n+1),-2,-2,-2,…,-2,…\}$
The generating functions of this series is : $(1-x)A^3(x)$
The series of differences is :$\{(n+1),-(n+3),0,0,…,0,…\}$
The generating functions of this series is : $(1-x)^4A(x)$
On the other hand, the generating functions $(n+1)+(-n-3)x$,
Therefore: $(1-x)^4A(x)=(n+1)+(-n-3)x$
$$A( x) =\frac{( n+1) +( -n-3) x}{( 1-x)^{4}} =(( n+1) +( -n-3) x) \cdot \sum _{n=0}^{\infty }\binom{n+4-1}{4-1} x^{n}$$
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\sum _{n=0}^{\infty }( n+1) \cdot \binom{n+4-1}{4-1} x^{n} +\sum _{n=0}^{\infty }( -n-3) \cdot \binom{n+4-1}{4-1} x^{n+1}$$
Prove in with induction :
\begin{aligned}
( n+1) \cdot \binom{n+3}{3} +( -n-3)\binom{n+2}{3} & =\frac{( n+3)( n+2)( n+1)^{2} n}{3!} +\frac{( n+2)( n+1)( n-1) n( -n-3)}{3!}\\
& =\frac{( n+3)( n+2)( n+1)^{2} -( n+2)( n+1)( n+3) n}{6}\\
& =\frac{( n+3)( n+2)( n+1)}{6}\\
& =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right)
\end{aligned}
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right)$$
induction basis: $n=0$
$$\displaystyle \sum _{k=0}^{n}( 0+1)( 0-0+1) =1=\frac{1}{6}\left( 0^{3} +6\cdotp 0^{2} +11\cdotp 0+6\right)$$
For $n+1$:
\begin{aligned}
\sum _{k=0}^{n+1}( k+1)( n-k+1) & =\sum _{k=0}^{n}( k+1)( n-k+1) +( n+2)(( n+1) -( n+1) +1)\\
& =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right) +( n+2)\\
\end{aligned}
If anyone can enlighten me, and find my mistakes I appreciate that.
Unfortunately, I did not succeed.
|
Unclear if this is off-topic. The OP specifically requires that generating functions be used.
Other responses have covered this constraint, so I'll show an alternative approach.
Known (or routinely proven by induction) that
*
*$\displaystyle \sum_{i = 1}^n i = \left(\frac{1}{2}\right)(n+1)(n).$
*$\displaystyle \sum_{i = 1}^n i^2 = \left(\frac{1}{6}\right) (n+1)(2n+1)(n).$
Since $(n - k + 1) = (n + 2) - (k + 1)$
$\displaystyle \sum_{k = 0}^n (k+1) (n - k + 1)$ can be re-expressed as
$$\sum_{k = 0}^n (k + 1) (n + 2) -
\sum_{k = 0}^n [(k + 1) (k + 1)]. \tag{1}$$
The first summation from expression (1) above is
$$ (n+2) \sum_{k=1}^{n+1} k = (n+2) \left(\frac{1}{2}\right) (n+1)(n+2). \tag{2}$$
The second summation from expression (1) above is
$$\sum_{k=1}^{n+1} k^2 = \left(\frac{1}{6}\right) (n+2)(2n+3)(n+1). \tag{3}$$
Taking common factors between expressions (2) and (3) above, you have that the overall computation is
$$\left(\frac{1}{6}\right)(n+1)(n+2) ~~[3(n+2) - (2n + 3)] $$
$$= ~~ \left(\frac{1}{6}\right)(n+1)(n+2)(n+3) = \binom{n+3}{3}.$$
|
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|
An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.
I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
with $0<\beta<\pi$. Hence $\sin \beta=\sqrt{1-5/25}=2\sqrt5/5$. But $\overline{AB}=\overline{BC}\cos \beta$ and the area
$$\mathcal A(\triangle ABC)=\frac12(\overline{BC}\cos \beta)\cdot (\overline{BC}\sin \beta)=\frac12\overline{BC}^2 \tag 1$$
We know that:
$$\overline{AH}+\overline{CH}+\overline{HB}=7, \quad \text{with}\quad \overline{CH}+\overline{HB}=\overline{BC}$$
Hence
$$\overline{AH}+\overline{BC}=7 \iff \overline{BC}=7-\overline{AH}$$
and $\overline{AH}=\overline{AB}\cos\gamma=\overline{BC}\cos\gamma\cos\beta$.
Putting this last identity to the condition $\overline{BC}=7-\overline{AH}$ I will have:
$$\overline{BC}=7-\overline{BC}\cos\gamma\cos\beta \iff \overline{BC}(1+\cos\gamma\cos\beta)=7$$
After,
$$\overline{BC}=\frac{7}{1+\cos\gamma\cos\beta}$$
with $\cos \gamma=\cos(\pi/2-\cos\beta)=\sin\beta=2\sqrt5/5$.
Definitively
$$\overline{BC}^2=\frac{49}{\left(1+\frac 25\right)^2}=\frac{49}{\frac{49}{25}}$$
and
$$\mathcal{A}(\triangle ABC)=\frac15\overline{BC}^2=5 \tag 2$$
I'm very tired and often don't find immediate solutions to problems. I had thought of using Euclid's second theorem by placing
$\overline{AH}=x$ and $\overline{BC}=y$, and $x+y=7$. After I will have $x^2=\overline{BH}\cdot \overline{CH}$. But I have left this path.
Can this possible alternative be used or is there another more immediate way?
|
$\cos\beta=\dfrac{\sqrt5}{5}$ and $\sin\beta=\dfrac{2\sqrt5}{5}.$ Let $\overline{AH}=x,$ then you can obtain values $$\overline{AB}= \dfrac{\sqrt{5}x}{2},\qquad \overline{BC}= \dfrac{5x}{2},\qquad \overline{AC}=\sqrt{5}x$$ and $$\overline{AH}=x,\qquad \overline{BH}=\dfrac{x}{2},\qquad \overline{CH}=2x$$ using simple trigonometry. Now, compute the value of $x$ via $\overline{AH}+\overline{BH}+\overline{CH}=7$ and use it to compute the area.
|
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|
Given 3 points. What's the normal to the plane that contains these 3 points?
Given are these points $$v_1=\begin{pmatrix} 1\\ 2\\ 3
\end{pmatrix}, v_2=\begin{pmatrix} 2\\ 3\\ 3 \end{pmatrix},
v_3=\begin{pmatrix} 2\\ 4\\ 4 \end{pmatrix}$$
Determine the normal to the plane which contains these 3 points.
My problem with this is I'm not sure what is really meant with "normal". Is it just the normal-vector or I will also need some parameter multiplied with the normal-vector?
So far I've done that but I'm not sure if that is all what is asked?
First calculate both direction vectors:
$$\begin{pmatrix}
2\\
3\\
3
\end{pmatrix}-
\begin{pmatrix}
1\\
2\\
3
\end{pmatrix}=
\begin{pmatrix}
1\\
1\\
0
\end{pmatrix} \text{ and }
\begin{pmatrix}
2\\
4\\
4
\end{pmatrix}-
\begin{pmatrix}
1\\
2\\
3
\end{pmatrix}=
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}$$
Now cross-product of these two vectors gives the normal vector:
$$\begin{pmatrix}
1\\
1\\
0
\end{pmatrix}
\times
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}=
\begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$$
But now is this really an answer to the question? Or I need some parametric form of it? Please help :/
|
“The” normal to a plane typically really does refer to any vector that is normal to the plane. In other words, your suggested answer (or any non-zero scalar multiple of it) is perfectly fine.
No need to express the normal generally as $$k\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}\quad(k\in\mathbb R\setminus\{0\}).$$
|
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|
Evaluating limits using Taylor expansions The limits are
$$\lim_{x\to 0}(\frac{\cos{x}-e^{-x^2/2}}{x^4})$$
$$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}$$
Probably wrong things that I've tried
$\lim_{x\to 0}\frac{\cos{x}-e^{-x^2/2}}{x^4}=\lim_{x\to 0}(\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^5)-1+\frac{x^2}{2}+\frac{x^4}{4}+o(x^4)}{x^4})$
$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}=\lim_{x\to 0}\frac{(1+x+\frac{x^2}{2}+o(x^2))(x+o(x^2))-x-x^2)}{x^3}=\lim_{x\to 0}\frac{\frac{x^3}{2}+o(x^2)}{x^3}=1/2$
Could you please help me understand how these kinds of limits can be computed I keep getting all of them wrong and if this goes on for a little bit longer I may have a panic attack.
|
$$\forall\;x\in\Bbb R\;,\;\;\;e^x=\sum_{n=0}^\infty\frac{ x^n}{n!}=\implies e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}{4\cdot2!}-\ldots$$
and since
$$\cos x=1-\frac{x^2}2+\frac{x^4}{4!}-\ldots$$
we get
$$\cos x-e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}{24}-\ldots-\left(1-\frac{x^2}2+\frac{x^4}8-\ldots\right)=-\frac{x^4}{12}+\ldots$$
and thus
$$\lim_{x\to0}\frac{\cos x-e^{-x^2/2}}{x^4}=-\frac1{12}$$
The parts with "$\;\ldots\;$" mean there are powers of $\;x\;$ higher than $\;4\;$ , so at the limit those will cancel with $\;x^4\;$ in the denominator and all that part will tend to zero. Try to make this formal in your proof of the value of this limit, and attempt the other one in a simlar fashion.
|
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|
Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS
&= \tan 20° + 2\tan 50° \\
&= \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \tan 20° - \tan^2 20°)}{√3 - \tan 20°}
\end{align}
$
Why are the two sides not equal despite being expressed in the same terms? Can someone offer some help?
Much to my surprise, my friend just expanded tan70° and cross-multiplied the terms to prove it.
|
You can use the identity $2 \cot 2 A = \cot A - \tan A$ which makes the job easy.
$\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$
$= \tan 20° + \cot 20° - \tan 20^0 = \cot 20^0 = \tan 70^0$
Otherwise proceed from $\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$ as below -
$\displaystyle \tan 20^0 + 2 \cot 40^0 = \tan 20° + 2 \cdot \frac{\cos^2 20^0 - \sin^2 20^0}{2 \cos 20^0 \sin 20^0}$
$= \tan 20^0 + \cot 20^0 - \tan 20^0 = \tan 70^0$
|
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|
Finding distance from Local maximum of $f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant
What is the distance of the local maximum of the function
$f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant?
$1)1\qquad\qquad2)\sqrt2\qquad\qquad3)2\qquad\qquad3)2\sqrt2$
It is a question from timed exam so the fastest answers are the best.
My approach:
First I realized the function defined for $x\in[0,4]$. to find local maximum I put derivative equal to zero :
$$f'(x)=1+\frac{4-2x}{2\sqrt{4x-x^2}}=0$$
$$x-2=\sqrt{4x-x^2}$$
$$2x^2-8x+4=0\rightarrow x^2-4x+2=0\rightarrow x=2\pm\sqrt2$$
From here I noticed that the quadratic $-x^2+4x$ is symmetric along $x=2$ line so after plugging $x=2\pm\sqrt2$ in the $\sqrt{4x-x^2}$, hence $f(x)$ is maximum for the value $x=2+\sqrt2$ and $f(2+\sqrt2)=2+\sqrt2+\sqrt{2}=2+2\sqrt2$
Now we should find the distance from the point $(2+\sqrt2,2+2\sqrt2)$ from the line $y=x$. Now by recognizing that $(2+\sqrt2,2+\sqrt2)$ is a point on the line $y=x$, I guess the distance might be $1$ and because it is the least value in the choices then I conclude it is the answer.
Is it possible to solve this problem quickly with other approaches?
|
$x^2 - 4x + 4 = (x-2)^2$ is a perfect square, so rewrite $f(x)$ as $x + \sqrt{4 - (x-2)^2}$. We want to get $\sqrt{4 - 4 \cos^2 u} = 2 \sin u$ in the given domain, which leads to the parametrisation:
$$(x,y) = (2 + 2 \cos t, 2 + 2 \cos t + 2 \sin t).$$
Now $2 + 2 \cos t + 2 \sin t = 2(1 + \sqrt{2} \cos(t - \pi/4))$, which attains a maximum at $t = \frac{\pi}{4}$, thus:
$$x = 2 + 2 \cos \pi/4 = 2 + \sqrt{2}, y = 2(1 + \sqrt{2}).$$
|
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|
Given $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ can we conclude that $x^n+y^n+z^n\in\mathbb{Q}$ for all $n\in\mathbb{N}$. Is there an explicit form? I had seen a fun problem that is exactly the problem in the question, except it was a specific case of this. Turns out, if $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ then $x^5+y^5+z^5 = 6$. I put this equation into Wolfram Alpha for different values of $n$. For instance, $x^8+y^8+z^8=\frac{51}{72}$. $x^{11}+y^{11}+z^{11}=\frac{11117}{216}.$
Firstly, is there an explicit expression to evaluate $x^n+y^n+z^n$? And, if not, are we able to prove that this is always a rational number, at least? WolframAlpha is unable to calculate an explicit form for $x^n+y^n+z^n$.
|
None of the calculations we need are mysterious, but one might need some experience dealing with symmetric functions and their relations to know what to look for.
We denote $x^n+y^n+z^n$ by $p_n$. You can verify the following inductive formula:
$p_n = (x+y+z)p_{n-1}-(xy+yz+zx)p_{n-2}+(xyz)p_{n-3}$
We can calculate the coefficients of the $p_i$ above as follows:
$x+y+z = 1$
$xy + yz + zx = \frac{(x+y+z)^2 -(x^2+y^2+z^2)}{2} = \frac{1^2-2}{2} = -\frac{1}{2}$
$xyz = \frac{(x^3+y^3+z^3)-(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{3} = \frac{3-1 \cdot (2-(-\frac{1}{2}))}{3} = \frac{1}{6}$
Thus our inductive formula becomes
$p_n = p_{n-1}+\frac{1}{2}p_{n-2}+\frac{1}{6}p_{n-3}$
It is clear from this recursive equation that all of the $p_n$ will be rational numbers, since $p_1, p_2$ and $p_3$ are rational numbers. In fact, it even follows that the denominator of $p_n$ will be of the form $2^a3^b$ for some non-negative integers $a$ and $b$. For an explicit form we could look at roots of the characteristic/auxiliary polynomial of the corresponding recursive equation. In this case we get
$T^3-T^2-\frac{1}{2}T-\frac{1}{6}$
whose roots are not that nice. However, Wolfram Alpha gives an explicit formula. Then again, one could perhaps just solve the original system directly to find $x$, $y$ and $z$.
|
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|
Algebra trick clarification question I have the equation $x = \sqrt{3} + \sqrt{2} \,\,$ and the trick is that $\frac{1}{x} = \sqrt{3} - \sqrt{2}$. I don't see how the minus sign appears just by taking 1 over x?
|
\begin{align}
\frac1x &= \frac1{\sqrt{3}+\sqrt{2}}\\
&= \left(\frac1{\sqrt3 + \sqrt2} \right) \left( \frac{\sqrt{3}-\sqrt{2}}{\sqrt3-\sqrt2}\right)\\
&= \frac{\sqrt3-\sqrt2}{3-2}
\end{align}
In general if $x = \sqrt{a}+\sqrt{b} \ne 0, a \ne b$, then $\frac1x = \frac{\sqrt{a}-\sqrt{b}}{a-b}$
|
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|
What kind of transformation is this matrix refering to?
Given is the 3D transformation $$M_1=\begin{pmatrix} 0 & 1 & 0\\ 3 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
What kind of transformation is $M_1$?
I've used several different points \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}
and did operations with it $$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}=\begin{pmatrix} 0 & 1 & 0\\ 3 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$
By several test-inputs I know that the solution is always in the form of \begin{pmatrix}
y\\
3x\\
z
\end{pmatrix}
So basically the $z$ value never changes, but the $x$ value is swapped with the $y$ value and the $x$ value which now is the $y$ value got multiplied by $3$.
I'm really missing the name of the transformation here because it's neither a scaling for me, nor a rotation or a shearing... It's more like a mixture of several transformations (called affine transformation?)? Or there is not really a name for this? It's very confusing I'm sorry if it is :(
|
$$M_1=\begin{pmatrix} 0 & 1 & 0\\ 3 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
$$=\begin{pmatrix} 1 & 0 & 0\\ 0 &
3 & 0\\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 0 & 1 & 0\\ 1 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
Looks like a reflection in $y=x$ together with a scaling.
|
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|
Changing the order of integration to find the volume My goal is to set up the triple integral that will solve the volume $\iiint \ xyz \ dV $ if S if the region bounded by the cylinders $x^2 + y^2 = 25$ and $ x^2+ z^2 = 25$ and the 1st octant with dV = dxdydz.
With order dV = dzdydx, the value of the volume is $\frac{15625}{24}$. My attempt is to set up the triple integral with order dV = dxdydz in which it should have the same answer with the triple integral with order dV = dzdydx.
The S I have computed is S $\lbrace (x, y, z) \in \mathbb{R}^3: 0 \leq x \leq 5, 0 \leq y \leq \sqrt{25 - x^2}, 0 \leq z \leq \sqrt{25 - x^2} \rbrace$ this is for triple integral with order dV = dzdydx
The S I have computed is S $\lbrace (x, y, z) \in \mathbb{R}^3: 0 \leq x \leq \sqrt{25 - \frac{y^2}{2} - \frac{z^2}{2}}, 0 \leq y \leq \sqrt{50 - z^2}, 0 \leq z \leq \sqrt{50} \rbrace$ this is for triple integral with order dV = dxdydz
When I set up the triple integral, I found out that it should be $\frac{1}{2} \int_0^\sqrt{50} \int_0^\sqrt{50-z^2} \int_0^\sqrt{25 - \frac{y^2}{2} - \frac{z^2}{2}} xyz \ dxdydz$ and not simply $\int_0^\sqrt{50} \int_0^\sqrt{50-z^2} \int_0^\sqrt{25 - \frac{y^2}{2} - \frac{z^2}{2}} xyz \ dxdydz$. What is the principle behind this? Why I should divide the volume of this triple integral to 2?
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The second $S$ is incorrect. On one surface we have $0\le x\le\sqrt{25-y^2}$ and on the other $0\le x\le\sqrt{25-z^2}$, which one do you choose? Since we have to take the intersection of the regions enclosed by these surfaces, we take the intersection of these inequalities as well, i.e.$$\begin{align*}0\le x\le\sqrt{25-y^2}\text{ and }0\le x\le\sqrt{25-z^2}&\iff0\le x\le\min\{\sqrt{25-y^2},\sqrt{25-z^2}\}\\&
\iff0\le x\le\begin{cases}\sqrt{25-y^2},&0\le z<y\le5\\
\sqrt{25-z^2},&0\le y\le z\le5\end{cases}
\end{align*}$$
I have attached a visual of the region. The horizontal plane is the $xy$ plane and the vertical axis is $z$ axis. I have tried to show its cross-sections lying in $xy,yz,xz$ planes. You may see the cross-section in the $yz$ plane is the square $0\le y,z\le5$.
Thus we will have to split the integral according as $y>z$ or $y\le z$:$$I=\left[\int_{z=0}^{z=5}\int_{y=z}^{y=5}\int_{x=0}^{x=\sqrt{25-y^2}}+\int_{z=0}^{z=5}\int_{y=0}^{y=z}\int_{x=0}^{x=\sqrt{25-z^2}}\right]xyz~dx~dy~dz$$
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"timestamp": "2023-03-29T00:00:00",
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Centroid of the region bounded by y = f(x) and x = f(y) curve How do i find the centroid of the region if $x = f(y)$ is involved ?
For example : find the centroid of the region bounded by $y = x^2$ and $x = y^2$
And how do i find it if there is curve intersection that is below x axis ?
For example : find the centroid of the region bounded by $y = x^3$ and $x = y^2-1$
The formula I found in the book was
$$x = \frac 1{\mathcal{A}}\int_a^b x[f(x)-g(x)] \,dx$$
$$y = \frac 1{\mathcal{A}}\int_a^b \frac 12 (f(x)^2-g(x)^2) \,dx$$
$\mathcal{A}$ is the area of the region bounded by the curve.
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In this case you can first try to sketch the region, as shown below.
From the sketch we learn that it suffice to consider the first quadrant where $ x, y \geq 0 $.
Thus the equation $ x = y^2 $ can be converted to $ y = \sqrt{x} $.
Put $ f(x) = \sqrt{x} $, $ g(x) = x^2 $, $ a = 0 $, $ b = 1 $, and using the formulae you have given yield
$$ A = \int_0^1 \sqrt{x} - x^2 \operatorname{d}\!x = \left. \frac23 x^{3/2} - \frac13 x^3 \right|^1_0 = \frac13, $$
$$ x = \frac1A \int_0^1 x(\sqrt{x} - x^2) \operatorname{d}\!x = 3 \left[ \frac25 x^{5/2} - \frac14 x^4 \right]_0^1 = 0.45, $$
and
$$ y = \frac1A \int_0^1 \frac12 [(\sqrt{x})^2 - (x^2)^2]\operatorname{d}\!x = \frac32 \left[ \frac12 x^2 - \frac15 x^5 \right]_0^1 = 0.45. $$
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"timestamp": "2023-03-29T00:00:00",
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Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.
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And now, a geometric argument.
If any of the terms on the right hand side are nonpositive, the inequality follows trivially. If they are all positive, $a,b,c$ form the legs of a triangle:
By Heron's formula, the area of this triangle is
$$
\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}
$$
The area of the triangle is also half the sum of the area of the two rectangles. However, a rectangle with diagonal $d$ can't have more area than $d^2/2$, so the total area of the two rectangles is less than $(a^2 + b^2)/2$. Since the triangle has half this area, we have
$$
\frac{a^2 + b^2}{4} \ge \frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}
$$
Squaring both sides and clearing denominators then gives your inequality.
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Limit of the series $\sum_{k=1}^\infty \frac{n}{n^2+k^2}.$ I am trying to evaluate $$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$ Now I am aware that clearly $$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$ but I do not know what to do if each sum is already sent to infinity. Im taking a limit of limits. I suppose I could rewrite my limit as $$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}?$$ But I am unaware if this is helpful at all. Any hints would be appreciated. Obviously, Wolfram calculates this as $\frac{\pi}{2}$ but I am unaware of the steps and logic to get there.
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Denote $\displaystyle \mathcal{S}(n) = \sum_{k \ge 1} \frac{n}{n^2+k^2}$. We have
\begin{aligned} \mathcal{S}(n) & = \sum_{k\geq 1}\frac{n}{ n^2+k^2} \\& = \frac{1}{2}\sum_{k\geq 1}\frac{2n }{ n^2+k^2} \\& = \frac{1}{2}\sum_{k\geq 1} \frac{d}{dn}\log\left(1+\frac{n^2}{k^2}\right) \\& = \frac{1}{2}\frac{d}{dn} \log \prod_{k\geq 1}\left(1+\frac{n^2}{k^2}\right) \\& = \frac{1}{2}\frac{d}{dn} \bigg[\log \pi n\prod_{k\geq 1}\left(1+\frac{n^2}{k^2}\right)- \log(\pi n)\bigg] \\& = \frac{1}{2}\frac{d}{dn} \log \pi n \prod_{k\geq 1}\left(1+\frac{n^2}{k^2}\right)-\frac{1}{2 n}\end{aligned}
Weierstrass factorization for hyperbolic sine is:
$$\sinh \pi z=\pi z \prod_{k=1}^\infty\left(1+\frac{z^2}{k^2}\right)$$
Therefore \begin{aligned} \mathcal{S}(n) & = \frac{1}{2} \frac{d}{dn} \log \sinh \pi n -\frac{1}{2 n} \\&= \frac{1}{2}\pi \coth(n \pi)-\frac{1}{2 n } \end{aligned}
And finally taking the limit as $n \to \infty$
\begin{aligned} \lim_{n \to \infty} \mathcal{S}(n) &= \lim_{n \to \infty} \bigg(\frac{1}{2}\pi \coth(n \pi)-\frac{1}{2 n }\bigg) \\& = \frac{\pi}{2}.\end{aligned}
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Subsets and Splits
Fractions in Questions and Answers
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