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Solution to two non polynomial equation If
$$x+\frac1x=-1$$
Find the value of
$$x^{99}+\frac{1}{x^{99}}$$
Is there any formal (traditional) way to solve these problems using only elementary algebra or high school math
| $\displaystyle (x^n + \frac 1 {x^n})(x + \frac 1 x) = x^{n+1} + x^{n-1} + \frac 1 {x^{n-1}} + \frac 1 {x^{n+1}} \\
\displaystyle \Rightarrow x^{n+1} + \frac 1 {x^{n+1}} = -(x^n + \frac 1 {x^n}) - (x^{n-1} + \frac 1 {x^{n-1}})$
If we denote $x^n + \frac 1 {x^n}$ by $f(n)$ then we have just shown that
$f(n+1) = -f(n) - f(n-1)$
Since you know that $f(0) = 2$ and $f(1)=-1$, this recursion allows you to find $f(n)$ for any $n >0$. If you work out $f(n)$ for $n=2,3,4 \dots$ you won't have to go very far before you see a pattern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Generating function of binary string with minimum lengths for 0s and 1s a) Show that the generating function by length for binary strings where every block of 0s has length at least 2, each block of ones has length at least 3 is:
$$\frac{(1-x+x^3)(1-x+x^2)}{1-2x+x^2-x^5}$$
b) Give a recurrence relation and enough initial conditions to determine coefficients of power series.
So for a), I came up with the block decomposition $((0^*00)^*(1^*111)^*)^*$ and found the generating function, using the fact that $0\leadsto x$, $1\leadsto x$, and $a^*\leadsto\frac{1}{1-a}$ where a is some binary string:
$$\frac{(1-x)^2}{(1-x-x^2)(1-x-x^3)}$$
which, clearly, does not equal the expected result. Could someone clear up for me where I went wrong?
Also, for b), how would you find a recurrence relation, since the degree of the numerator and denominator are the same, so there would be no general $a_n$ term.
Thanks in advance for any help!
| Let $a_n$ be the number of strings that start with $0$, and let $b_n$ be the number of strings that start with $1$. Then $a_0=b_0=1$, $a_1=0$, and, by conditioning on the length $k$ of the current run, we see that
\begin{align}
a_n &= \sum_{k=2}^n b_{n-k} &&\text{for $n \ge 2$}\\
b_n &= \sum_{k=1}^n a_{n-k} &&\text{for $n \ge 1$}.
\end{align}
Let $A(z)=\sum_{n=0}^\infty a_n z^n$ and $B(z)=\sum_{n=0}^\infty b_n z^n$. Then the recurrence relations imply
\begin{align}
A(z)-a_0 -a_1z &=\frac{z^2}{1-z} B(z) \\
B(z)-b_0 &=\frac{z}{1-z} A(z)
\end{align}
Solving for $A(z)$ and $B(z)$ yields
\begin{align}
A(z) &= \frac{1-2z+2z^2-z^3}{1-2z+z^2-z^3}\\
B(z) &= \frac{1-z}{1-2z+z^2-z^3}
\end{align}
So the desired generating function (subtracting the $1z^0=1$ term for the empty string that is otherwise counted twice) is
$$A(z)+B(z)-1=\frac{1-z+z^2}{1-2z+z^2-z^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$X$ a random variable, where $-1\leq X\leq \frac{1}{2}, \mathbb E[X]=0$. What is the maximal value of $\mathbb E[X^2]$? $X$ a random variable, where $$-1\leq X\leq \frac{1}{2}, \mathbb E[X]=0$$
Find the maximal value of $\mathbb E[X^2]$.
I came across a similar question, asking if the following inequality is always true $$\mathbb E[X^2]\leq \frac{1}{4}$$
The answer is no, for example:
$$\mathbb P(X=-1) = \frac{1}{3},\mathbb P(X=\frac{1}{2}) = \frac{2}{3}\implies\\ \mathbb E[X^2]= ((-1)^2)\cdot(\frac{1}{3})+(\frac{1}{2})^2\cdot(\frac{2}{3})=\\ \frac{1}{3} +\frac{2}{3\cdot 4} =\frac{1}{2}$$
I was wondering how can one find the maximum, for any discrete or continuos random variable.
Remark: I'm not sure this question has a "nice" solution.
| For the sake of generality say that $X\in [a,b]$ a.s. (this implies $a\leq 0$).
Since $X=\frac{b-X}{b-a}a + \frac{X-a}{b-a}b$ and the square function is convex, $X^2\leq \frac{b-X}{b-a}a^2 + \frac{X-a}{b-a}b^2$ and taking expectations, $$E(X^2)\leq \frac{b}{b-a}a^2 - \frac{a}{b-a}b^2 = -ab$$
This bound is tight: it is attained only for $X\sim \frac{b}{b-a} \delta_{a}+\frac{-a}{b-a} \delta_{b}$.
In the special case where $a=-1$ and $b=\frac 12$ this yields $E(X^2)\leq \frac 12$, with equality iff $X\sim \frac 13 \delta_{-1}+\frac 23 \delta_{1/2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove $\sin x + \arcsin x > 2x$ using Maclaurin series My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..
Prove that the inequality sin x + arcsin x > 2x holds for all values of x such
that 0 < x ≤ 1.
I know that the Maclaurin series of
sin(x) = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ...
arcsin(x) = x + $\frac{1}{2}\cdot\frac{x^3}{3}$ + ($\frac{1}{2}\cdot\frac{3}{4}$)$\cdot\frac{x^5}{5}$ + ...
However, I do not know how to prove this using there series...Could anyone have some ideas?
Thank you!
| For $0 < x \le 1$ we have
$$
\sin(x) + \arcsin(x) = 2 x + \sum_{n=2}^\infty \left(\frac{(-1)^n}{(2n+1)!} + \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} \right) x^{2n+1}
$$
because the $x^3$ terms cancel. Therefore it suffices to show that
$$
\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} > \frac{1}{(2n+1)!}
$$
for $n \ge 2$. This is equivalent to
$$
\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \cdot (2n)! > 1
$$
or
$$
\bigl( 1 \cdot 3 \cdots (2n-1) \bigr)^2 > 1 \, ,
$$
which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \sin{x}\ +\ \frac{1}{2}\sin{2x}\ +\ \frac{1}{3}\sin{3x}\ +\ \frac{1}{4}\sin{4x}\ =\ \frac{2}{3}\left(\cos{x}+1\right)\left(\sin^5x\ +\ 4\right)$ SORRY IF MY TITTLE IS UNCLEAR WITH ONLY MATH FUNCTIONS. IT'S MORE THAN 150 CHARACTERS
*
*This is my math problem
$$ \sin {x}\ +\ \frac{1}{2}\sin {2x}\ +\ \frac{1}{3}\sin {3x}\ +\ \frac{1}{4}\sin {4x}\ =\ \frac{2}{3}\left(\cos {x}+1\right)\left(\sin ^5x\ +\ 4\right) \left(*\right)
$$
*
*This is my effort
$$ \left(*\right) <=> 12\sin {x\left(1+\cos x\right)\ +\ 12\sin {x}\ -\ 16\sin ^3x\ +\ 12\sin {\left(x\right)\cos {\left(x\right)}\cos {\left(2x\right)}}}\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)
$$
$$ <=> 12\sin {x\left(1+\cos x\right)\ + 4\sin {x\left(2\cos {\left(2x\right)\left(1+\cos {x}\right)\\ +\ \left(1+\cos {x}\right)\left(2\cos ^2x\ -\ 2\cos {x\ +\ 1}\right)}\right)}}\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)
$$
$$ <=> 4\sin {x\left(1+\cos {x}\right)\left(6\cos ^2x\ -\ 2\cos {x\ +\ 2}\right)\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)}
$$
Now
$$\cos {x} = -1
$$
or
$$ 6\sin {x}\cos ^2x\ -\ 2\cos {x}\sin {x}\ +\ 2\sin {x}\ -\ 2\sin ^5x\ -\ 8\ =\ 0
$$
To here I tried many ways like converting all to \sin , group somes together but it didn't work. Please give me some HINTS
| It remains to solve the following equation.
$$\sin{x}(3\cos^2x-\cos{x}+1)=\sin^5x+4.$$
We'll prove that
$$\sin{x}(3\cos^2x-\cos{x}+1)<\sin^5x+4,$$ which says that this equation has no real roots.
Indeed,
$$\sin{x}(3\cos^2x-\cos{x}+1)=\sin{x}(3(1-\sin^2x)+1)-\sin{x}\cos{x}\leq$$
$$\leq\sin{x}(4-3\sin^2x)+\frac{\sin^2x+\cos^2x}{2}=4\sin{x}-3\sin^3x+\frac{1}{2}.$$
Now, let $\sin{x}=a$ and since $3\cos^2x-\cos{x}+1>0$ and $\sin^5x+4>0,$ we see that $a>0$
and we need to prove that
$$a^5+3a^3-4a+3.5>0,$$ which is true by AM-GM:
$$a^5+3a^3-4a+3.5>3a^3+2\cdot1.75-4a\geq\left(3\sqrt[3]{3\cdot1.75^2}-4\right)a>0.$$
I used the following AM-GM:
for non-negatives $x$, $y$ and $z$ we have:
$$\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$$ or
$$x+y+z\geq3\sqrt[3]{xyz}.$$
Here $x=3a^3$ and $y=z=1.75.$
| {
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"url": "https://math.stackexchange.com/questions/3536511",
"timestamp": "2023-03-29T00:00:00",
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All loops of order less than $5$ are groups I am studying loop theory and found that there are no non-trivial loops of order less than $5$. Can anybody help me in this, with some reference of a book or a research article?
A set $Q$ with a binary operation $\circ $ is called a loop if $e \in Q$, where $e$ is the identity element and both the equations $x \circ a = b$ and $a \circ x = b$ has unique solutions, for any $a, b \in Q$.
By trivial loop I mean it is a group.
| Without loss of generality we denote the loop's elements as $0,1,2,3$ with $0$ the identity and draw the Cayley table:
$$\begin{array}
00&1&2&3\\
1&&&\\
2&&&\\
3&&&\end{array}$$
By the Latin square property of quasigroups (shared with loops) there are only two ways to fill in the two remaining $1$ entries:
$$\begin{array}
00&1&2&3\\
1&&&\\
2&&1&\\
3&&&1\end{array}\qquad\begin{array}
00&1&2&3\\
1&&&\\
2&&&1\\
3&&1&\end{array}$$
The left possibility has only one way to fill in the $2$s and the $3$ by the Latin square property, which in turn forces the $0$s, giving a complete Cayley table:
$$\begin{array}
00&1&2&3\\
1&0&3&2\\
2&3&1&0\\
3&2&0&1\end{array}$$
But this is isomorphic to $\mathbb Z_4$ with $2$ as a generator.
Turning to the other option, we have two ways to place the $2$s while obeying the Latin square property:
$$\begin{array}
00&1&2&3\\
1&&&2\\
2&&&1\\
3&2&1&\end{array}\qquad\begin{array}
00&1&2&3\\
1&2&&\\
2&&&1\\
3&&1&2\end{array}$$
The second option forces the rest of the Cayley table:
$$\begin{array}
00&1&2&3\\
1&2&3&0\\
2&3&0&1\\
3&0&1&2\end{array}$$
which is the familiar presentation of $\mathbb Z_4$, generated by $1$. Once again, we have two possibilities for placing the $3$s in the last unfilled table above, and now both options force the remaining entries:
$$\begin{array}
00&1&2&3\\
1&0&3&2\\
2&3&0&1\\
3&2&1&0\end{array}\qquad\begin{array}
00&1&2&3\\
1&3&0&2\\
2&0&3&1\\
3&2&1&0\end{array}$$
The right table describes $\mathbb Z_4$, generated by $1$, and the left table describes the Klein four-group $\mathbb Z_2^2$. Hence
all loops of order $4$ are groups.
It is easy to check that all loops of order $1,2,3$ are groups by the same method, and I leave this as an exercise.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which arose from the deduction that $\frac{1}{4}\sqrt{\frac{256\sin^{4}40^{\circ}-80\sin^{2}40^{\circ}+12-\ 8\sqrt{3}\sin40^{\circ}}{\left(16\sin^{4}40^{\circ}-4\sin^{2}40^{\circ}+1\right)}}=\cos50^{\circ}$. Despite the apparent simplicity of the relationship, it seems quite tricky to prove. I managed to prove it by solving the equation as a quadratic in $(\sin\frac{\pi}{9})$ and then using the identity $\sqrt{\sec^2 x-1}=|\tan x|$, the double angle formulae and finally that $\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x$ can be written in the form $\sin\left(x+\frac{2\pi}{3}\right)$.
But it seems like quite a neat problem. So, does anyone have a better way of proving it?
| Using
$$\sin \frac{\pi}{3}=\frac{\sqrt 3}{2},~\cos \frac{\pi}{3}=\frac{1}{2},~\sec \frac{\pi}{3}=\frac{1}{\cos \frac{\pi}{3}}=2$$
see this.
and the triple angle forumla for $\sec$
$$2=\frac{\sec^3\frac{\pi}{9}}{4-3\sec^2\frac{\pi}{9}}=\frac{\sec\frac{\pi}{9}}{4\cos^2\frac{\pi}{9}-3}$$
$$\Leftrightarrow 2\left(4\cos^2\frac{\pi}{9}-3\right) = \sec\frac{\pi}{9}$$
squaring both sides and dividing by $4$
$$\Leftrightarrow \left(4\cos^2\frac{\pi}{9}-3\right)^2 = \\ 1-8\sin^2\frac{\pi}{9}+16\sin^4\frac{\pi}{9} = \frac{1}{4}\sec^2\frac{\pi}{9}$$
then using the triple angle formula for $\sin$
$$1-8\sin^2\frac{\pi}{9}+4\sin\frac{\pi}{9}\left(3\sin\frac{\pi}{9}-\sin\frac{\pi}{3}\right) = \\ 4\sin^2\frac{\pi}{9}-4\sin\frac{\pi}{3}\sin\frac{\pi}{9}+1 = \frac{1}{4}\sec^2\frac{\pi}{9}$$
$$\Leftrightarrow 4\sin^2\frac{\pi}{9}-2\sqrt 3\sin\frac{\pi}{9}+1 = \frac{1}{4}\sec^2\frac{\pi}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $e^{z+w}=e^ze^w$ using the Taylor expansion for $ e^z.$ For all $z,w ∈ \Bbb{C}$, using the power series definition for $e^z$, prove $e^{z+w}=e^ze^w$.
| Proved it the following way:
Claim: $e^{z+w}=e^ze^w$, where $z\;and\;w\in \Bbb{C}$.
Proof:
Let $e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$
and $e^w=1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots$ .
Now, for RHS:
$e^ze^w=\big(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\big)\big(1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots\big)$
$=1+w+\frac{w^2}{2!}+\frac{w^2}{3!}+z+zw+\frac{zw^2}{2!}+\frac{zw^3}{3!}+\cdots$
$=1+(z+w)+(\frac{z^2}{2!}+zw+\frac{w^2}{2!})+(\frac{z^3}{3!}+\frac{z^2w}{2!}+\frac{zw^2}{2!}+\frac{w^2}{3!})\cdots$
$=1+(z+w)+(\frac{z^2}{2!}+\frac{2zw}{2!}+\frac{w^2}{2!})+(\frac{z^3}{3!}+\frac{3z^2w}{3!}+\frac{3zw^2}{3!}+\frac{w^2}{3!})\cdots$
$=1+(z+w)+\frac{(z+w)^2}{2!}+\frac{(z+w)^3}{3!}+\cdots$
$=e^{z+w} \;QED.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542281",
"timestamp": "2023-03-29T00:00:00",
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How to find the stationary point under constraints analitically? I am working with the optimization problem from the paper, eq.(5)
$$\max_{X=(x_1, x_2, \ldots, x_{n+1})} f(X)=(A-B\sum_{i=1}^n \frac{1}{x_i})\times x_{n+1}$$
subject
to $$x_{n+1}=1-2k\sum_{i=1}^n x_i,$$ $$x_i \geq 0, \quad i = 1,2,\ldots, n+1.$$
Here $A \gg B > 0 \in \mathbb{R}$, $k \in \mathbb{Z}^+$, and $f(X)>1$.
From the paper I know the stationary point $$X^*=(\underbrace{\sqrt{\frac{B}{2kA}}, \ldots, \sqrt{\frac{B}{2kA}}}_{n\text{ times}}, 1 − 2 kn\sqrt{\frac{B}{2kA}}
)$$
and the optimal value $$f(X^*)=(\sqrt{A} − n \sqrt{ 2 \cdot k \cdot B})^2.$$
Question. How to find the stationary point under constraints analytically? Is it possible for $n=3, k=10$ case?
Attempt.
I have tried to use the Lagrange multiplier:
$$F(X, \lambda) = x_{n+1}(A-B\sum \frac{1}{x_i}) + \lambda(x_{n+1} - 1 + 2k\sum x_i)=0$$ and found the partical derivatives and have the $(n+2)$ system with $n+2$ variables:
\begin{cases}
F'_{x_i}(X, \lambda)= x_{n+1}\frac{B}{x_i^2} +2 \lambda k x_i=0, \quad i=1,2,..., n, \\
F'_{x_{n+1}}(X, \lambda)= A - B\sum \frac{1}{x_i}+\lambda=0, \\
F'_{\lambda}(X, \lambda)=x_{n+1} -1+2k\sum x_i =0.
\end{cases}
My problem now is how to express $x_i$, $i=1,2,..., n$, and $x_{n+1}$ through $\lambda$ and find roots.
I know the stationary point and have found the A & Q.
I will use the notation $\sum_{i=1}^n x_i := n \cdot x$ $\sum_{i=1}^n \frac{1}{x_i} := \frac{n} {x}$, and $x_{n+1}:=y$. Then the system will be
$$y\frac{B}{x^2}+2\lambda k x=0, \tag{2.1}$$
$$\lambda=B\frac{n}{x}-A, \tag{2.2}$$
$$y=1-2knx. \tag{2.3}$$
Put $(2.2)$ and $(2.3)$ in $(1.1)$:
$$( 1-2knx )\frac{B}{x^2}+2 (B\frac{n}{x}-A ) k x=0, \tag{3.1}$$
Multiple both sides $(3.1)$ on $x^2$:
$$( 1-2knx )B+2 (B\frac{n}{x}-A ) k x^3=0, \tag{4.1}$$
Open brackets and collect tems:
$$2kAx^3-2nBkx^2+2nBkx-B=0, \tag{5.1}$$
divide both sides on $2kA$:
$$x^3 - n\frac{B}{A}x^2 + n \frac{B}{A}x-\frac{1}{2k}\frac{B}{A}=0. \tag{6.1}$$
One can see the equation of power $3$, I am looking for a root $x \in \mathbb{R}$. I think the equation $(6.1)$
should has a simple real root and complex pair.
| EDIT
In the paper, they look for a set of probabilities $p_k\in (0, 1), k = 1, 2, \cdots, s$ and $p_{s+1}
= 1 - 2n \sum_{k=1}^s p_k \in [0, 1]$
such that condition (5) is satisfied, i.e. $p_{s+1}(A - B\sum_{k=1}^s \frac{1}{p_k}) > 1$.
(I put some images at the end.)
Condition (5) requires $A - B \sum_{k=1}^s \frac{1}{p_k} > 0$ and $p_{s+1} = 1 - 2n\sum_{k=1}^s p_k> 0$
which results in $\frac{A}{B} \cdot \frac{1}{2n}
> \sum_{k=1}^s \frac{1}{p_k} \cdot \sum_{k=1}^s p_k \ge s^2$,
or $A - 2ns^2 B > 0$.
As a result, in that paper, they solve the optimization problem
under the condition $A - 2ns^2 B > 0$ and $p_k\in (0, 1), k=1, 2, \cdots, s$ and $p_{s+1}\in (0, 1)$.
Under the condition $A - 2ns^2 B > 0$,
$p_1 = p_2 = \cdots = p_s = \sqrt{\frac{B}{2nA}}$
satisfy $1 -2 n \sum_{k=1}^s p_k > 0$ and hence the solution.
In the OP, $p_k$ is replaced with $x_i$, $s$ is replaced with $n$, $n$ is replaced with $k$. (I think the notation of the paper should be used.)
Using the notation of the OP, assuming that $A - 2kn^2 B > 0$, we can solve the optimization problem as follows.
With $x_i > 0, \forall i$ and $1-2k \sum_{i=1}^n x_i \ge 0$, we have
\begin{align}
\Big(A- B\sum_{i=1}^n \frac{1}{x_i}\Big)x_{n+1}
&= \Big(A- B\sum_{i=1}^n \frac{1}{x_i}\Big)\Big(1-2k \sum_{i=1}^n x_i\Big)\\
&\le \Big(A- B\frac{n^2}{\sum_{i=1}^n x_i}\Big)\Big(1-2k \sum_{i=1}^n x_i\Big) \tag{1}\\
&= A - B\frac{n^2}{\sum_{i=1}^n x_i}
- 2k A \sum_{i=1}^n x_i + 2kn^2B\\
&\le A - 2\sqrt{B\frac{n^2}{\sum_{i=1}^n x_i}\cdot 2k A \sum_{i=1}^n x_i} + 2kn^2 B \tag{2}\\
&= A - 2\sqrt{2kn^2 AB} + 2kn^2 B\\
&= (\sqrt{A} - n\sqrt{2kB})^2
\end{align}
with equality if and only if $x_1 = x_2 = \cdots = x_n = \sqrt{\frac{B}{2kA}}$,
and $p_{n+1} = 1 - 2kn\sqrt{\frac{B}{2kA}}$ (note: $1 - 2kn\sqrt{\frac{B}{2kA}} > 0$ since $A - 2kn^2 B > 0$).
Explanation: in (1), we have used Cauchy-Bunyakovsky-Schwarz inequality
to obtain $\sum_{i=1}^n \frac{1}{x_i} \ge \frac{n^2}{\sum_{i=1}^n x_i}$
with equality if and only if $x_1 = x_2 = \cdots = x_n$;
in (2), we have used $a + b \ge 2\sqrt{ab}$
with equality if and only if
$B\frac{n^2}{\sum_{i=1}^n x_i} = 2k A \sum_{i=1}^n x_i$
or $\sum_{i=1}^n x_i = \sqrt{\frac{Bn^2}{2kA}}$.
Some images from the paper:
image 1:
image 2:
image 3:
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How many integer solutions are there for $x+y+z+w=25$, if $x\geq 1, y \geq 2, z\geq 3, w\geq 4$?
How many integer solutions are there for $x+y+z+w=25$, if $x\geq 1, y \geq 2, z\geq 3, w\geq 4$?
Based off how to solve these problems from my last question I believe I'm meant to take $x^\prime =x-1, y^\prime=y-2,z^\prime=z-3,w^\prime=w-4$
then substituting these values $x=x^\prime +1,y=y^\prime +2,z=z^\prime +3,w=w^\prime +4$
I get $x^\prime+y^\prime+z^\prime+w^\prime=35$
Which I use the formula, ${n+k-1\choose k-1}$ to solve with $n=35,k=4$
Gives me ${35+4-1\choose 4-1}={38\choose 3}=8436$
Is this correct? If it is why does this substitution not change the problem?
| Your answer is incorrect. As a sanity check, notice that the number of integer solutions of the equation
$$x + y + z + w = 25$$
satisfying $x \geq 1$, $y \geq 2$, $z \geq 3$, $w \geq 4$ should be smaller than the number of solutions of the equation in the nonnegative integers since we cannot substitute $0$ for $x$, $0$ or $1$ for $y$, $0$, $1$, or $2$ for $z$, or $0$, $1$, $2$, or $3$ for $w$.
Your strategy is correct, but you made an error when you did your calculations.
\begin{align*}
x + y + z + w & = 25\\
x' + 1 + y' + 2 + z' + 3 + w' + 4 & = 25\\
x' + y' + z' + w' & = 15
\end{align*}
which is an equation in the nonnegative integers with
$$\binom{15 + 4 - 1}{4 - 1} = \binom{18}{3}$$
solutions, which is smaller than the $\binom{25 + 4 - 1}{4 - 1} = \binom{28}{3}$ solutions in the nonnegative integers, as we would expect.
Notice that by making these substitutions, we are first putting one object in the first box, two objects in the second box, three objects in the third box, and four objects in the fourth box, then distributing the remaining $25 - (1 + 2 + 3 + 4) = 25 - 10 = 15$ objects to the four boxes without restriction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$ In working to prove that
$$1+\cos\theta+\cos(2\theta)+\dots+\cos(n\theta)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{1}$$
I have shown
$$\begin{align}
1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta) &=\Re\left(\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}\right) \\[6pt]
&=\frac{1-\cos(\theta)+\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)} \\[6pt]
&=\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)} \tag{2}
\end{align}$$
but I am unsure how to proceed from here and get the last term of $(2)$ to match the last term of $(1)$:
$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{3}$$
I have read this post "How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?", but I cannot seem to convert from the form their answer is in to my form.
If possible, I would like to avoid using too many identities as this is an exercise in my complex analysis book.
| To prove
$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$$
examine the LHS
$$LHS = \frac{\cos(n\theta)-\cos(n\theta)\cos\theta+\sin(n\theta)\sin\theta}{2-2\cos\theta}$$
$$= \frac{\cos(n\theta)(1-\cos\theta)+\sin(n\theta)\sin\theta}{2(1-\cos\theta)}$$
Then, use $1-\cos\theta=2\sin^2\frac\theta2$ and $\sin\theta = 2\sin\frac\theta2 \sin\frac\theta2$ to simplify,
$$LHS= \frac{\cos(n\theta)\sin\frac\theta2+\sin(n\theta)\cos\frac\theta2}{2\sin\frac\theta2}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin\frac\theta2}=RHS$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expand a function in a Fourier Series Expand the function
$$ f(x) = \sin^6{x} \cdot \cos^3{x} $$
in a Fourier series on the segment $[-\pi, \pi]$.
I think we should start from the fact that this function is even, but then the $ a_n $ coefficient turns out to be very big
| If you know complex numbers:
Let
$$z:=\cos(x)+i\sin(x)$$
Then
$$\frac{1}{x}=\cos(x)-i\sin(x) \\
\cos(x)=\frac{1}{2}(z+\frac{1}{z})\\
\sin(x)=\frac{1}{2i}(z-\frac{1}{z})
$$
$$f(x) = \sin^6{x} \cdot \cos^3{x} = \frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^6\\
= \frac{-1}{2^9}(z^3+3z+3\frac{1}{z}+\frac{1}{z^3})(z^6-6z^4+15z^2 -20+15\frac{1}{z^2} -6 \frac{1}{z^4}+\frac{1}{z^6})\\
$$
Now open the brackets and use
$$z^n+\frac{1}{z^n}=2 \cos(nx) \\
z^n-\frac{1}{z^n}=2i \sin(nx)$$
Note A faster computation is to observe that
$$\frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^6=\frac{-1}{2^9}(z+\frac{1}{z})^3(z-\frac{1}{z})^3(z-\frac{1}{z})^3\\
=\frac{-1}{2^9}(z^2-\frac{1}{z^2})^3(z-\frac{1}{z})^3\\
=\frac{-1}{2^9}(z^6-3z^2+3\frac{1}{z^2}-\frac{1}{z^6})(z^3-3z+3\frac{1}{z}-\frac{1}{z^3})\\
$$
| {
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"source": "stackexchange",
"question_score": "2",
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Determine the value of the sum to infinity of $(u_{r+1} + \frac{1}{2^r})$ Determine the value of $$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$.
In earlier parts of the question, it is given $\displaystyle u_r= \frac{2}{(r+1)(r+3)}$, which when expressed in partial fractions is $\displaystyle\frac{1}{r+1}-\frac{1}{r+3}$.
Also previously, the question asks to show that $$\sum_{r=1}^n u_r = \frac{5}{6}-\frac {1}{n+2} -\frac{1}{n+3}$$ which I managed to.
I have also determined the values of $$\sum_{r=1}^\infty u_r$$ which is $$\sum_{r=1}^\infty u_r= \frac{5}{6}$$
But I don’t know how to solve for
$$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$
My attempts so far:
$\displaystyle u_{r+1} =\frac{2}{(r+1+1)(r+1+3)}$
$=\displaystyle\frac{2}{(r+2)(r+4)}$
Let $\displaystyle \frac{2}{(r+2)(r+4)}\equiv \frac{A}{(r+2)(r+4)}$
$2\equiv A(r+4)+B(r+2)$
Let r $=-4,B=-1$
Let $r=-2,A=1$
So, is $\displaystyle u_{r+1} \equiv \frac {1}{r+2}- \frac{1}{r+4}$?
Assuming that it is,$$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$
= $$\sum_{r=1}^\infty\Bigl(\frac{1}{r+2}+\frac{1}{r+4} + \frac{1}{2^r}\Bigr)$$
$=\frac{1}{3}-\require{cancel}\cancel{\frac{1}{5}}+\frac{1}{2}$
$\frac{1}{4}-\require{cancel}\cancel{\frac{1}{6}}+\frac{1}{4}$
$\require{cancel}\cancel{\frac{1}{5}}-\cancel{\frac{1}{7}}+\frac{1}{8}$
$\require{cancel}\cancel{\frac{1}{6}}-\cancel{\frac{1}{8}}+\frac{1}{16}$
.....
I don't know how to cancel the middle terms, how should this question be solved?
| You have
$$\begin{equation}\begin{aligned}
\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr) & = \sum_{r=1}^\infty u_{r+1} + \sum_{r=1}^\infty\frac{1}{2^r} \\
& = (\sum_{r=1}^\infty u_{r} - u_1) + 1 \\
& = \frac{5}{6} -\frac{1}{4} + 1 \\
& = \frac{10}{12} - \frac{3}{12} + \frac{12}{12} \\
& = \frac{19}{12}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I used $\sum_{r=1}^\infty\frac{1}{2^r}$ is the sum of an infinite geometric series, with $a = \frac{1}{2}$ being the first term and $r = \frac{1}{2}$ being the common ratio, so the sum is $\frac{a}{1-r} = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Term of rotation matrix entries equals 1 - proof concept?! I derived some stuff and it is happening that i come to the following expression:
$\frac{r_{13}^2 + r_{23}^2}{(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23} - r_{13}r_{22})^2}$
that must equal 1 for all first two rows of a rotation matrix (orthogonal matrix).
$$ \begin{matrix}
r_{11} & r_{12} & r_{13} & \\
\end{matrix} $$
$$ \begin{matrix}
r_{21} & r_{22} & r_{23} & \\
\end{matrix} $$
I am confused how to proof this! I see a bit of a cross product in the denominator but can't identify the relation combined with the numerator.
| I did this just by using the facts that the rows and columns have length one and are mutually orthogonal.
$$\begin{align}
&(r_{11}r_{23}-r_{13}r_{21})^2+(r_{12}r_{23}-r_{13}r_{22})^2\\
&=r_{11}^2r_{23}^2+r_{13}^2r_{21}^2-2r_{11}r_{21}r_{13}r_{23}+
r_{12}^2r_{23}^2+r_{13}^2r_{22}^2-
2r_{12}r_{22}r_{13}r_{23}\\
&=r_{23}^2(r_{11}^2+r_{12}^2)+r_{13}^2(r_{21}^2+r_{22}^2)
-2r_{13}r_{23}(r_{11}r_{21}+r_{12}r_{22})\\
&=r_{23}^2(1-r_{13}^2)+r_{13}^2(1-r_{23}^2)-2r_{13}r_{23}(-r_{13}r_{23})\\
&=r_{13}^2+r_{23}^2-2r_{13}^2r_{23}^2+2r_{13}^2r_{23}^2\\
&=r_{23}^2+r_{13}^2
\end{align}$$
I see that an answer has been posted since I started typing this (I'm awfully slow), but I'm going to post it anyway, since it's a bit different.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.
| You can't, because you need to work with the interplay of the sine functions. Concretely, using Taylor approximations (and collecting the errors together),
\begin{align}
\sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x}
&=\frac1{x+2}-\frac1{6(x+2)^3}-2\left(\frac1{x+1}-\frac1{6(x+1)^3} \right)\\ \ \\
&\ \ \ \ \ \ \ \ \ \ \ \ +\frac1x-\frac1{6x^3}+o(\frac1{x^5})\\ \ \\
&=\frac2{x(x+1)(x+2)}-\frac1{6(x+2)^3}+\frac2{6(x+1)^3}\\ \ \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac1{6x^3}+o(\frac1{x^5}).
\end{align}
Then
\begin{align}
x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
&=\frac2{(1+\tfrac2x)(1+\tfrac2x)}-\frac1{6(1+\tfrac2x)^3}\\ \ \\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac1{3(1+\tfrac1x)^3)}-\frac1{6x^3}+o(\tfrac1{x^2})\\ \ \\
&\xrightarrow[\vphantom{x_A}x\to\infty]{}2-\tfrac16+\tfrac13-\tfrac16=2.
\end{align}
\
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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The remainder when $( \sum_{k=1}^5 { ^{20} \mathrm C (2k-1) } )^6$ is divided by 11, is : How do I approach this? Any help is appreciated.
| Let $\displaystyle S =\binom{20}{1}+\binom{20}{3}+\binom{20}{5}+\binom{20}{7} +\binom{20}{9}\cdots (1)$
$\displaystyle S =\binom{20}{9}+\binom{20}{7}+\binom{20}{5}+\binom{20}{3}+\binom{20}{1}$
Using $\displaystyle \binom{n}{r}=\binom{n}{n-r}$
$\displaystyle S =\binom{20}{11}+\binom{20}{13}+\binom{20}{15}+\binom{20}{17}+\binom{20}{19}\cdots (2)$
Add $(1)$ and $(2)$
$\displaystyle 2S =\sum^{10}_{k=1}\binom{20}{2k-1}=2^{19}$
$$\displaystyle S =2^{18}$$
We have to calculate remainder when $\bigg(2^{18}\bigg)^6=2^{108}$ is divided by $11$
Now $2^{3}\cdot \bigg(2^{5}\bigg)^{21}=8\cdot (32)^{21}$
$=-8\cdot (1-33)^{21}=-8+M(11)=3-11+M(11)=3+M'(11)$
So remainder is $3$ when $2^{108}$ is divided by $11$
Where $M(11)$ and $M'(11)$ is Multiple of $11$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to use Chinese Remainder Theorem A cubic polynomial $f(x)=ax^3+bx^2+cx+d$ gives remainders $-3x+3$ and $4x-1$ when divided by $x^2+x+1$ and $x^2+2x-4$. Find the value of $a,b,c,d$.
I know it’s easy but i wanna use Chinese Remainder Theorem(and Euclidean Algorithm) to solve it.
A hint or a detailed answer would be much appreciated
| EDIT: I just noticed my answer below doesn't meet OP's expectations, as I'm not explicitely using the Chinese theorem. I haven't read the question carefully enough...
Reduction of $f(x)$ modulo $x^2+x+1$ can be computed by successively replacing all instances of $x^2$ in $f$ by $-x-1$, until we obtain an expression of degree $<2$. The process reduces $f(x)$ to $a(-x-1)x +b(-x-1) +cx +d=-ax^2+(c-a-b)x+d-b$, which in turns gives $ -a(-x-1)+(c-a-b)x+d-b = (c-b)x+a+d-b$.
The same method with $x^2+2x-4$ tells us that $f(x)$ reduces to $(c-2b+8a)x-8a+d$.
These two reductions must be respectively $-3x+3$ and $4x-1$. By very definition of equality between polynomials (that is, we can identify the coefficients), it reduces to the following system with $4$ equations and $4$ unknowns, which I let your solve by yourself:
$$\left\{
\begin{array}{}
-3&=c-b\\
3&=a+d-b\\
4&=c-2b+8a\\
-1&=-8a+d\\
\end{array}
\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A non-probabilistic take on $\lim_{n\to \infty }\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}$ I saw this question here and I tried it on my own. I don't have a background in Probability theory, so I tried to employ algebra only to evaluate this limit.
$$\lim_{n\to \infty}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}$$
My Attempt:
$$\begin{aligned}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}&=\frac{1}{5^n}\sum_{k=0}^{4n} {k+n-1\choose n-1} \left(\frac{4}{5}\right)^{k}\\ &=\frac{1}{5^n}\left(\text{coeff. of }x^{n-1} \text{ in }\sum_{k=0}^{4n}(1+x)^{k+n-1}\left(\frac{4}{5}\right)^k\right)\\ &=\frac{1}{5^n}\left[\text{coeff. of }x^{n-1} \text{ in } \left((1+x)^{n-1}\frac{\left(\frac{4}{5}(1+x)\right)^{4n+1}-1}{\left(\frac{4}{5}(1+x)\right)-1}\right)\right]\end{aligned}$$
I don't know how to proceed because computing the coefficient in the above term is a really long task. Any hints are appreciated. Thanks
| $$\begin{aligned}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}&=\frac{1}{5^n}\sum_{k=0}^{4n} {k+n-1\choose n-1} \left(\frac{4}{5}\right)^{k}\\ &=\frac{1}{5^n}\left(\text{coeff. of }x^{n-1} \text{ in }\sum_{k=0}^{4n}(1+x)^{k+n-1}\left(\frac{4}{5}\right)^k\right)\end{aligned}$$
therefore what you re after is
=$\frac{1}{5^n} \sum_{k=0}^{4n}{n-1\choose k+n-1}\left(\frac{4}{5}\right)^k $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_{0}^{2\pi} \frac{\mathrm{e}^{-i k a \cos\phi \sin\theta}}{1+\cos\phi \sin\theta}\,\mathrm d\phi$ How to simplify the following equation:
$$\int_{0}^{2\pi} \frac{\mathrm{e}^{-i k a \cos\phi \sin\theta}}{1+\cos\phi \sin\theta}\,\mathrm d\phi$$
where $k$, $a$ and $\theta$ can be regarded as the constants, and $0<\theta<\frac{\pi}{2}$. Call $\sin\theta=b$ for brevity.
Attempt:
As Von Neumann pointed out, which is really helpful, using the residues theorem,
$$\frac{2}{ib}\int_{|z|=1} \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{z^2 + \frac{2}{b}z + 1}\ \text{d}z$$
Where does the pole locate?
$$z_{1,\ 2} = -\frac{1}{b}\pm \sqrt{\frac{1}{b^2}-1}$$
noted that: $\frac{1}{b}=\frac{1}{\sin(\theta)}>1$, then only one pole within the circle is considered,where
$$z_{1} = -\frac{1}{b}+ \sqrt{\frac{1}{b^2}-1}$$
Using the residues theorem, I got the result
$$\frac{2\pi}{\cos\theta}\exp(ika)$$
Then I programmed the original integral and the derived result, there are discrepancies. My question is that should we consider the numerator when using residues theorem? That is
$$\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)$$ because using the pole $z_{1}$ we can derive that
$$\cos\phi =\frac{z^2+1}{2z}=\frac{\left(-\frac{1}{b}+\sqrt{\frac{1}{b^2}-1}\right)^2+1}{2\left(-\frac{1}{b}+\sqrt{\frac{1}{b^2}-1}\right)} = -\frac{1}{b}=-\frac{1}{\sin\theta}$$
As it is shown that $$\cos\phi=-\frac{1}{\sin\theta}$$ which conflicts with each other. Could someone help? Thank you very much in advance.
| Methinks we can appreciate it carefully with residues theorem.
First of all we step into the complex plane with the usual rules for sine and cosines, that is:
$$\mathcal{J} = \phi \to e^{i\psi} = z ~~~~~~~ \cos\phi \to \ \frac{1}{2}\left(z + \frac{1}{z}\right) ~~~~~~~ \text{d}\phi = \frac{\text{d}z}{iz}$$
We will also call $\sin\theta = b$ for brevity.
The integral becomes (arrange it a bit)
$$\frac{1}{2i}\int_{|z|=1} \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{1 + \frac{b}{2}\frac{z^2+1}{z}}\ \frac{\text{d}z}{z}$$
That is,
$$\frac{1}{ib}\int_{|z|=1} \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{z^2 + \frac{2}{b}z + 1}\ \text{d}z$$
Poles occurs at $$z_{1,\ 2} = -\frac{1}{b}\pm \sqrt{\frac{1}{b^2}-1}$$
Now by residues theorem:
$$\mathcal{J} = 2\pi i \frac{1}{ib} \sum_{z_k} \lim_{z\to z_k} (z-z_k)f(z)|_{z = z_k}$$
Where
$$f(z) = \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{(z-z_1)(z - z_2)}$$
Can you proceed?
Doing the calculation you shall obtain:
$$\frac{2\pi}{b}\left(\frac{e^{-2 i a \sqrt{\frac{1}{b^2}-1} b k}}{2 \sqrt{\frac{1}{b^2}-1}} -\frac{e^{2 i a \sqrt{\frac{1}{b^2}-1} b k}}{2 \sqrt{\frac{1}{b^2}-1}}\right) = -\frac{2\pi}{b}\frac{i \sin \left(2 a \sqrt{\frac{1}{b^2}-1} b k\right)}{\sqrt{\frac{1}{b^2}-1}} = -\frac{2\pi}{b}\frac{i b \sin \left(2 a \sqrt{1-b^2} k\right)}{\sqrt{1-b^2}}$$
Eventually:
$$-2\pi i\frac{ \sin \left(2 a \sqrt{1-b^2} k\right)}{\sqrt{1-b^2}}$$
Where $b = \sin\theta$.
I will check this again later, for the sake of correctness.
| {
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Solving $(4-x^2)y''+2y=0$ by series I know that there are other ways to solve this, but I want to solve the ODE $(4-x^2)y''+2y=0$ by series close to $0$, that is, I'm looking for solutions of the form $y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}.$
Substituing in the ODE, I got:
$$(8a_{2}+2a_{0})+(24a_{3}+2a_{1})x+\sum_{n=2}^{\infty}[4(n+2)(n+1)a_{n+2}+2a_{n}-n(n-1)a_{n}]x^{n}=0\implies \\
a_2=-\frac{a_0}{4};\quad a_3=-\frac{a_{1}}{12}; \quad a_{n+2}=\frac{(n-2)a_{n}}{4(n+2)}. $$
But this recorrence relation give me a problem: $a_4=0\implies a_{2n}=0$ for all $n>2 $.
My question is: Did I made a mistake? Or this solution considers, for even $n$, only the terms $a_2$ and $a_0$, that is, $y_{1}(x)=a_{0}\left(1-\frac{1}{4}x^{2}\right)$?
| Since you correctly obtained a solution $y_1(x)=x^2-4$ (I set $a_0=-4$ in your notation), here is a way to find all solutions. Suppose that a general solution $y(x)$ takes the form $y_1(x)z(x)=(x^2-4)z(x)$. We get
$$y'(x)=y_1(x)z'(x)+y_1'(x)z(x)=(x^2-4)z'(x)+2xz(x)$$
and
$$y''(x)=y_1(x)z''(x)+2y_1'(x)z'(x)+y''_1(x)z(x)=(x^2-4)z''(x)+4xz'(x)+2z(x).$$
From the original ODE $-(x^2-4)y''(x)+2y(x)=0$, so
$$-(x^2-4)\big((x^2-4)z''(x)+4xz'(x)+2z(x)\big)+2\big((x^2-4)z(x)\big)=0.$$
That is
$$-(x^2-4)z''(x)-4xz'(x)=-\big((x^2-4)z''(x)+4xz'(x)+2z(x)\big)+2z(x)=0.$$
Hence if $u(x)=z'(x)$, then
$$(x^2-4)u'(x)+4xu(x)=0$$
or
$$\frac{u'(x)}{u(x)}=-\frac{4x}{x^2-4}=-\frac{2}{x-2}-\frac{2}{x+2}.$$
This implies
$$\ln u(x)=-2\ln|x-2|-2\ln|x+2|+c'=-\ln\big((x-2)^2(x+2)^2\big)+c'$$
for some constant $c'$. That is, if $c=e^{c'}$, we have
$$z'(x)=u(x)=\frac{c}{(x-2)^2(x+2)^2}=\frac{c}{32}\left(\frac{1}{x+2}+\frac{2}{(x+2)^2}-\frac{1}{x-2}+\frac{2}{(x-2)^2}\right).$$
That is, if $b=-c/32$ and $a$ is a constant, then
$$z(x)=a-b\left(\ln |x+2|-\frac{2}{x+2}-\ln|x-2|-\frac{2}{x-2}\right)=a+b\left(\frac{4x}{x^2-4}-\ln\left|\frac{x+2}{x-2}\right|\right).$$
So
$$y(x)=y_1(x)z(x)=ay_1(x)+by_2(x),$$
where
$$y_2(x)=4x-(x^2-4)\ln\left|\frac{x+2}{x-2}\right|.$$
It is possible to find $y_2(x)$ through the series method. Note that
$$y_2(x)=\sum_{n=0}^\infty b_nx^n$$
where $b_0=0$, $b_1=8$, and $b_n=\frac{n-4}{4n}b_{n-2}$ for $n\ge3$. So $b_{2n}=0$ for all integers $n\ge0$. And for $n\ge 1$,
$$b_{2n+1}=8\left(\frac{2n-3}{4(2n+1)}\right)\left(\frac{2n-5}{4(2n-1)}\right)\cdots \left(\frac{1}{4\cdot 5}\right)\left(\frac{-1}{4\cdot 3}\right)=-\frac{1}{2^{2n-3}(2n+1)(2n-1)}.$$ That is
$$y_2(x)=-\sum_{n=0}^\infty\frac{x^{2n+1}}{2^{2n-3}(2n+1)(2n-1)}=-16\sum_{n=0}^\infty\frac{1}{4n^2-1}\left(\frac{x}{2}\right)^{2n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}$ I need to proof that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}
\end{align}
is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the residual theorem:
I started with $\gamma: [0,\pi] \to \mathbb{C}, t \to e^{2it}$. Since $\cos(t) = \frac{1}{2}\left(e^{it}+e^{-it}\right)$ and $\gamma'(t) = 2ie^{2it}$ we find that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(e^{it}+e^{-it}\right)} \cdot \frac{2ie^{2it}}{2ie^{2it}}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t
\end{align}
I did this with the aim to use
\begin{align}
\int_{\gamma} f(z) \mathrm{d}z = \int_{a}^{b} (f\circ \gamma)(t)\gamma'(t) \mathrm{d}t,
\end{align}
so we find
\begin{align}
\int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t = \int_{\gamma} \frac{-i}{6z+2z\sqrt{z}+2\sqrt{z}} \mathrm{d}z
\end{align}
At this point I don't know how to continue. Can anyone help?
| We proved here that
$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
set $a=3$ and $b=2$ then integrate both sides we have
$$\int_0^\pi\frac{dt}{3+2\cos t}=\int_0^\pi\frac{dt}{\sqrt{5}}+\frac{2}{\sqrt{5}}\sum_{n=1}^\infty\left(\frac{\sqrt{5}-3}{2}\right)^n\underbrace{\int_0^\pi\cos(nt)\ dt}_{0}=\frac{\pi}{\sqrt{5}}$$
| {
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"source": "stackexchange",
"question_score": "1",
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Prove that $p(x)=x^4-x+\frac{1}{2}$ has no real roots.
What is the simplest way to prove that the polynomial
$p(x)=x^4-x+\frac{1}{2}$ has no real roots?
I did with Sturm's theorem:
$p_0(x)=x^4-x+\frac{1}{2}$
$p_1(x)=4x^3-1$
$p_2(x)=\frac34x-1$
$p_3(x)=-\frac{229}{27}$
The signs for $-\infty$ are $+,-,-,-$ and for $\infty$ are $+,+,+,-$. In the end $1-1=0$ real roots.
Can it be done faster?
| You can use AM-GM to show that there are no real roots. If $x< 0$, then $$x^4-x+\frac12> x^4+\frac12 > \frac12>0.$$
If $x\ge0$, then
$$x^4-x+\frac12 =\left(x^4+\frac1{4}+\frac1{8}+\frac1{8}\right)-x\geq 4\cdot\sqrt[4]{x^4\cdot \frac14\cdot\frac18\cdot\frac18}-x=x-x=0,$$
but the equality case doesn't occur because $\frac14\ne\frac18$.
You can even find the minimum value of $x^4-x+\frac12$ using AM-GM. Note that when $x\ge 0$,
\begin{align}x^4-x+\frac12 &=\left(x^4+3\cdot\frac1{\sqrt[3]{4^4}}\right)-x+\left(\frac12-\frac{3}{\sqrt[3]{2^8}}\right)
\\&\geq 4\cdot\sqrt[4]{x^4\cdot \left(\frac1{\sqrt[3]{4^4}}\right)^3}-x+\left(\frac12-\frac{3}{\sqrt[3]{2^8}}\right)
\\&=x-x+\left(\frac12-\frac{3}{\sqrt[3]{2^8}}\right)=\frac12-\frac{3}{\sqrt[3]{2^8}}.
\end{align}
Note that the minimum value $\frac12-\frac3{\sqrt[3]{2^8}}\approx 0.02753$ is achieved if and only if $x=\frac{1}{\sqrt[3]{4}}\approx 0.62996$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Closed-form expression of a series involving combinatorics How to find the closed-form expression of the function $$F(x,y)=\sum_{n=0}^\infty\sum_{m=0}^\infty {{m+n}\choose {n}}x^ny^m=1+x+y+x^2+y^2+2xy+...?$$
| The inner sum is
$$\sum_{m=0}^\infty \binom{m+n}{n} y^m=\frac{1}{(1-y)^{n+1}}.$$
So
\begin{align}
F(x,y)&=\sum_{n=0}^\infty \frac{1}{(1-y)^{n+1}} x^n \\
&= \frac{1}{1-y}\sum_{n=0}^\infty \left(\frac{x}{1-y}\right)^n \\
&= \frac{1}{1-y}\cdot\frac{1}{1-x/(1-y)} \\
&= \frac{1}{1-x-y}.
\end{align}
| {
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"source": "stackexchange",
"question_score": "2",
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Number of possible integers less than 100000 such that the digits 4,5,6 appear in that order I have been stuck on this problem for some time and I am not sure how to approach it:
"How many positive integers less than 100,000 have digits containing 4,5,6 in that particular order?" (by this it means that the digits "456" must appear one after another in the number)
I am thinking that I would multiply 10X10X3X2X1 since the two of the five digit spaces can hold any digit between 0-9 and the other 3 must have 456 however I do not think this considers the order of the digits or the fact that the smallest possible integer would be 456. (if that makes any sense)
How would I modify this so that the order of the numbers matter?
What would the answer be to this?
thanks in advance!
| We can represent any nonnegative integer less than $100000$ as a five-digit string by appending leading zeros as necessary. For instance, we can represent $456$ as $00456$ and $4956$ as $04956$.
There are ten possible positions in which the first $4$, first $5$, and first $6$ could appear:
$456\square\square$
$45\square 6 \square$
$45 \square \square 6$
$4 \square 56 \square$
$4 \square 5 \square 6$
$4 \square \square 56$
$\square 456 \square$
$\square 45 \square 6$
$\square 4 \square 56$
$\square \square 456$
Any position before the first $4$ can be filled in $7$ ways since it cannot be a $4$, $5$, or $6$.
Any position after the first $4$ but before the first $5$ can be filled in $8$ ways since it cannot be a $5$ or $6$.
Any position after the first $5$ but before the first $6$ can be filled in $9$ ways since it cannot be a $6$.
Any position after the first $6$ can be filled in $10$ ways.
Thus, the number of nonnegative integers less than $100000$ in which the digits $4$, $5$, and $6$ appear in that order is
$$1 \cdot 1 \cdot 1 \cdot 10 \cdot 10 + 1 \cdot 1 \cdot 9 \cdot 1 \cdot 10 + 1 \cdot 1 \cdot 9 \cdot 9 \cdot 1 + 1 \cdot 8 \cdot 1 \cdot 1 \cdot 10 + 1 \cdot 8 \cdot 1 \cdot 9 \cdot 1 + 1 \cdot 8 \cdot 8 \cdot 1 \cdot 1 + 7 \cdot 1 \cdot 1 \cdot 1 \cdot 10 + 7 \cdot 1 \cdot 1 \cdot 9 \cdot 1 + 7 \cdot 1 \cdot 8 \cdot 1 \cdot 1 + 7 \cdot 7 \cdot 1 \cdot 1 \cdot 1$$
If you make the assumption that the digits $4$, $5$, and $6$ each appear exactly once, you obtain Rezha Adrian Tanuharja's answer. This is because there would be only seven choices for each of the other two positions.
| {
"language": "en",
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"source": "stackexchange",
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} |
Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$ Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $x=y=z.$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi - c.$$ This follows that \begin{align*} \cos(a+b)=\cos(\pi - c) & \implies \cos a \cos b - \sin a \sin b = - \cos c \\ & \implies xy-\sqrt{1-x^2} \sqrt{1-y^2}=z.\end{align*} Now i am not able to proceed from here. Please help me to solve this.
| With your substitution you have $a,b,c\in [0,2\pi],\ a+b+c=\pi$ and
$$\cos a+\cos b+\cos c=\frac{3}{2}$$
We have that:
$$\cos c=1-2\sin^2 \frac{c}{2},\ \cos a+\cos b=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}$$
Therefore:
$$\cos a+\cos b+\cos c=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}+\cos c\\
=2\sin \frac{c}{2}\cos \frac{a-b}{2}+1-2\sin^2\frac{c}{2}\leq 2\sin \frac{c}{2}+1-2\sin^2\frac{c}{2}$$
Therefore:
$$\frac{3}{2}\leq 2\sin \frac{c}{2}+1-2\sin^2\frac{c}{2}\Rightarrow 2\left(\sin \frac{c}{2}-\frac{1}{2}\right)^2\leq 0$$
This implies $c=\frac{\pi}{3}$. Similarly we can prove that $a=b=\frac{\pi}{3}$, which gives $x=y=z=\frac{1}{2}$.
| {
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"source": "stackexchange",
"question_score": "1",
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Prove that $1-3^x+5^x-7^x = 0 \Leftrightarrow x=0$ Problem :
Prove that equation $$1-3^x+5^x-7^x = 0$$
has unique root $x=0$.
Since $\lim_{x\to-\infty}(1-3^x+5^x-7^x) = 1$ and $\lim_{x\to\infty}(1-3^x+5^x-7^x) = -\infty$,
I tried to show $x \to 1-3^x+5^x-7^x$ is decreasing function, but it wasn't easy to determine its derivative is negative.
Thanks for any help.
| $$1-3^x+5^x-7^x = 0$$
$$\rightarrow0\geq 1-3^x=7^x-5^x \geq 0 $$
$$\rightarrow0= 1-3^x=7^x-5^x = 0 $$
$$\rightarrow x=0$$
EDIT
If $x<0$ we'd get $$\rightarrow 0\leq 1-3^x=7^x-5^x \leq 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate using differentiation under the sign of integration: $\int_{0}^{\pi} \frac {\ln (1+a\cos (x))}{\cos (x)} dx$ Evaluate by using the rule of differentiation under the sign of integration $\int_{0}^{\pi} \dfrac {\ln (1+a\cos (x))}{\cos (x)} \textrm {dx}$.
My Attempt:
Given integral is $\int_{0}^{\pi} \dfrac {\ln(1+a\cos (x))}{\cos (x)}$. Here $a$ is the parameter so let
$$F(a)=\int_{0}^{\pi} \dfrac {\ln (1+a\cos (x))}{\cos (x)} \textrm {dx}$$
Differentiating both sides w.r.t $a$
$$\dfrac {dF(a)}{da} = \dfrac {d}{da} \int_{0}^{\pi} \dfrac {\ln (1+a\cos (x))}{\cos (x)} \textrm {dx}$$
By Leibnitz Theorem:
$$\dfrac {dF(a)}{da} = \int_{0}^{\pi} \dfrac {1}{1+a\cos (x)} \times \dfrac {1}{\cos (x)} \times \cos (x)
\textrm {dx}$$
$$\dfrac {dF(a)}{da}=\int_{0}^{\pi} \dfrac {dx}{1+a\cos (x)} \textrm {dx}$$
Now writing $\cos (x)= \dfrac {1-\tan^{2} (\dfrac {x}{2})}{1+\tan^2 (\dfrac {x}{2})}$ and proceeding with integration becomes quite cumbersome to carry on. Is there any way to simplify with some easy steps?
| Let $a^2<1$
$$I(a)=\int_{0}^{\pi} \frac{\ln 1+ a \cos x)}{\cos x} dx~~~~(*)$$
D.w.r.t.$a$ on both sides, to get
$$\frac{dI}{da}=\int_{0}^{\pi} \frac{\cos x}{(1+a\cos x)\cos x} dx=\int_{0}^{\pi} \frac{dx}{1+a \cos x}=J(a)~~~~(1)$$
Use $$\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx~~~(2)$$
Then $$J(a)=\int_{0}^{\pi} \frac{dx}{1-a \cos x}~~~(3)$$Adding (1) and (3), we get
$$2J(a)=2\int_{0}^{\pi} \frac{dx}{1-a^2\cos^2x}= 4 \int_{0}^{\pi/2} \frac{dx}{1-a^2 \cos^2x}=4\int_{0}^{\pi/2} \frac{\sec^2 x dx}{\tan^2 x+(1-a^2)}.$$
Let $\tan x=u$, then
$$2\frac{dI}{da}=4 \int \frac{du}{u^2+(\sqrt{1-a^2})^2}=\frac{4}{\sqrt{1-a^2}}
\tan^{-1}(u/\sqrt{1-a^2})|_{0}^{\infty}=\frac{2\pi}{\sqrt{1-a^2}}$$
So $$\frac{dI}{da}=\frac{\pi}{\sqrt{1-a^2}} \implies I(a) =\pi \int \frac{da}{\sqrt{1-a^2}}+C \implies I(a)= \sin^{-1}{a}+C \implies I(0)=C$$
From (*), we have $I(0)=0 \implies C=0$. Hence
$$I=\pi \sin^{-1} a, ~a^2<1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve indefinite integral $ \int \frac{x^4}{\sqrt{x^2+4x+5}}dx $ I need to solve the next problem:
$$
\int \frac{x^4}{\sqrt{x^2+4x+5}}dx
$$
I know the correct answer is
$$
(\frac{x^3}{4}-\frac{7x^2}{6}+\frac{95x}{24}-\frac{145}{12})*\sqrt{x^2+4x+5}\space+\space\frac{35}{8}\ln{(x+2+\sqrt{x^2+4x+5})}+C
$$
still, I cannot find the solution.
I've already tried substituting $u=x+2$ and that has't given any satisfying result
| With $x+2=\sinh t$ to rewrite the integral as,
$$
I=\int \frac{x^4}{\sqrt{x^2+4x+5}}dx=\int (\sinh t-2)^4dt$$
$$=\int (\sinh^4 t-8\sinh^3 t + 24\sinh^2 t - 32\sinh t+16)dt
\tag 1$$
where,
$$\begin{array}
& & \int \sinh t dt = \cosh t \\
& \int \sinh^2 t dt = \frac12\int (\cosh 2t-1)dt=\frac14\sinh 2t -\frac12t \\
& \int \sinh^3 t dt = \int (\cosh^2 t-1)d(\cosh t)=\frac13\cosh^3 t -\cosh t \\
& \int \sinh^4 t dt = \frac14\int (\cosh 2t-1)^2dt
=\int ( \frac18\cosh 4t -\frac12\cosh 2t+\frac38) dt \\
& \>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>=\frac1{32}\sinh 4t -\frac14\sinh 2t+\frac38t \end{array}$$
Plug above integrals back into (1) to get,
$$I= \frac1{32}\sinh 4t -\frac83\cosh^3 t+\frac{23}4\sinh 2t -24\cosh t + \frac{35}8t+C$$
$$= \cosh t\left(\frac18\sinh t(2\sinh^2 t+1) -\frac83(\sinh^2+1) +\frac{23}2\sinh t -24\right) + \frac{35}8t+C$$
Substitute back $\sinh t = x+2$ and $\cosh t = \sqrt{1+\sinh^2 t} = \sqrt{x^2+4x+5}$ to obtain,
$$
I = \sqrt{x^2+4x+5}\left(\frac{x^3}{4}-\frac{7x^2}{6}+\frac{95x}{24}-\frac{145}{12}\right)\space+\space\frac{35}{8}\sinh^{-1}(x+2)+C
$$
Furthermore, $\sinh^{-1}(x+2)=\ln{(x+2+\sqrt{x^2+4x+5})}$ by applying the identity $\sinh^{-1}u=\ln(u+\sqrt{1+u^2})$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Greatest common divisor of $(x+1)^{4n+3} + x^{2n}$ and $x^3-1$. I have to find the greatest common divisor of
$$(x+1)^{4n+3} + x^{2n}$$
and
$$x^3-1$$
I know I can express the second polynomial as:
$$x^3-1 = (x-1)(x^2+x+1)$$
So I would have to check if the first polynomial is divisible by $(x^3-1)$, $(x^2+x+1)$ or $(x-1)$ and if it is not divisible by any of those, then the two polynomials do not have a common divisor except for $1$. But I don't know how I can divide the polynomial
$$(x+1)^{4n+3} + x^{2n}$$
by those $3$ other polynomials and therefore can't check the greatest common divisor.
| Let $Q(x) = (x+1)^{4n+3} + x^{2n}$. Then $Q(1) = 2^{4n+3} + 1 \neq 0$, hence $x-1$ does not divide $Q(x)$.
We have $$Q(x) = (x+1)^{4n+3} + x^{2n} = (x+1)[(x+1)^2]^{2n+1} + x^{2n} =$$
$$= (x+1)[(x^2+x+1) + x]^{2n+1}+ x^{2n} = $$
$$=(x^2+x+1)P(x) + (x+1)x^{2n+1} + x^{2n} = (x^2+x+1)(P(x) + x^{2n}),$$
hence $x^2+x+1$ divides $Q(x)$.
| {
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Concurrency of normals If $\theta_1$, $\theta_2$, and $\theta_3$ are the eccentric angles of three points on the hyperbola $x^2/a^2 - y^2/b^2 = 1$ such that
$\sin (\theta_1+\theta_2) +
\sin (\theta_2+\theta_3) +
\sin (\theta_3+\theta_1) = 0$ then prove that the normals at these points are concurrent.
P.S. My attempt was to use the same idea as for an ellipse here, but then I realised that it wouldn't do for quite obvious reasons.
Edit 1: As a clarification, do NOT use the hyperbola $xy = c^2$ to answer this question. You may use ONLY the hyperbola I mentioned above.
Edit 2: Here are my efforts so far....
Assume that the equation of normal in parametric form is
$$\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2$$
Then, taking $\tan \frac{\theta}{2} = t$ and using the familiar trigonometric identities gives a biquadratic equation in $t$ as:
$$ bkt^4 + 2(a^2 + b^2 + ah)t^3 + 2(a^2 + b^2 - ah)t - bk = 0$$
where we have substituted $(h,k)$ for $(x,y)$ as the point of intersection of the four normals at $\theta_i$ ($i= 1,2,3,4$) as the roots of the above equation.
Observe that this gives
$$\theta_1 + \theta_2 + \theta_3 + \theta_4 = 2n\pi, n \in \mathbb Z$$
but it doesn't get me near the condition at all. I cannot 'eliminate' one root to get a cubic having three roots. I need some help, particularly in eliminating one root. Any efforts towards this could be welcome.
Other attempts I tried include complex numbers and using the fact that if we have a repeated root, then the equation will have its slope zero at this particular root.
Edit 3: In reply to Blue's comment, the point of intersection comes out to be
$$P \equiv
\left(
\frac{(a^2 + b^2)\sec \theta_1 \sec \theta_2 \cos
\left(
\frac{\theta_1-\theta_2}{2}
\right)}
{a\cos
\left(
\frac{\theta_1 + \theta_2}{2}
\right)},
-\frac{(a^2 + b^2)\tan \theta_1 \tan \theta_2 \tan
\left(
\frac{\theta_1 + \theta_2}{2}
\right)}{b}
\right)$$
On substituting in the third equation, we get
$$
\sin^2 \theta_1 \cos^2 \theta_2 \sin \theta_3 \cos \theta_3-
\cos^2 \theta_1 \sin^2 \theta_2 \sin \theta_3 \cos \theta_3+
\\ \quad
\sin^2 \theta_1 \sin \theta_2 \cos \theta_2 \sin \theta_3 \cos \theta_3+
\sin \theta_1 \cos^2 \theta_1 \sin ^2 \theta_2 \sin \theta_3-
\\ \quad
\sin^2 \theta_1 \sin \theta_2 \cos^2 \theta_2 \cos \theta_3-
\sin \theta_1 \cos \theta_1 \sin^2 \theta_2 \cos \theta_2 \cos \theta_3
\\ =
\sin^2 \theta_1 \cos \theta_1 \cos^2 \theta_2 \cos \theta_3+
\sin \theta_1 \cos^2 \theta_1 \sin \theta_2 \cos \theta_3 \cos \theta_3-
\\ \quad
\sin \theta_1 \cos \theta_1 \sin \theta_2 \cos^2 \theta_2 \cos \theta_3-
\cos^2 \theta_1 \sin^2 \theta_2 \cos \theta_2 \cos \theta_3$$
Now I am unable to simplify. Please note that this has been homogenised by using the fact that $\sin^2 \theta + \cos^2 \theta = 1$. Yet I am unable to get to the hypothesis.
| Start with $$\frac{ax}{\text{sec}\ \theta_1} + \frac{by}{\text{tan}\ \theta_1} = a^2 + b^2$$
$$\frac{ax}{\text{sec}\ \theta_2} + \frac{by}{\text{tan}\ \theta_2} = a^2 + b^2$$
$$\frac{ax}{\text{sec}\ \theta_3} + \frac{by}{\text{tan}\ \theta_3} = a^2 + b^2$$
When are these three straight lines concurrent?
Equate the following determinant to $0$ and simplify!
\begin{vmatrix}
\frac{a}{\sec \theta_1} & \frac{b}{\tan \theta_1} & a^2 + b^2 \\
\frac{a}{\sec \theta_2} & \frac{b}{\tan \theta_2} & a^2 + b^2 \\
\frac{a}{\sec \theta_3} & \frac{b}{\tan \theta_3} & a^2 + b^2 \notag
\end{vmatrix}
Note:
In other words, you need to equate the following to $0$ and simplify:
\begin{vmatrix}
\frac{1}{\sec \theta_1} & \frac{1}{\tan \theta_1} & 1 \\
\frac{1}{\sec \theta_2} & \frac{1}{\tan \theta_2} & 1 \\
\frac{1}{\sec \theta_3} & \frac{1}{\tan \theta_3} & 1 \notag
\end{vmatrix}
A few more steps:
$1.$ Multiply $R_1$ by $\sin \theta_1$, $R_2$ by $\sin \theta_2$ and $R_3$ by $\sin \theta_3$
$2.$ Try to show that the above simplifies to
$-2 \sin \frac{\theta_1 - \theta_2}{2} \sin \frac{\theta_2 - \theta_3}{2} \sin \frac{\theta_3 - \theta_1}{2}[\sin (\theta_1 + \theta_2) + \sin (\theta_2 + \theta_3) + \sin (\theta_3 + \theta_1)] = 0$
$3.$ Argue that $\sin (\theta_1 + \theta_2) + \sin (\theta_2 + \theta_3) + \sin (\theta_3 + \theta_1) = 0$
In fact, you need to prove the converse of this result. That is also true and you can write the proof "bottom-up"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int_0^\infty \frac{\log(2+x^2)}{4+x^2}\,\mathrm dx$ Evaluate the integral
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
--
I started by stating that the integral from 0 to infinity should be the same as half the integral from $-\infty$ to $\infty$, that is:
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx = \frac{1}{2}\int_{-\infty}^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
and then by stating that this must be equal to
$$\pi i \cdot \sum(\text{residues in upper plane})$$
noting that there is a singularity at $z=2i$ in the upper half plane, and an "issue" (I don't know if it's strictly a singularity) with the log function at $z=\sqrt{2}i$.
The residue at $z=2i$ is easily dealt with, but when I try to deal with the log issue, I can't make any headway. I decided to make my branch cut between $z=\sqrt{2}i$ and $z=-\sqrt{2}i$, and form a contour that goes around this cut and the point, but it doesn't seem to be working out for me.
Suggestions would be appreciated!
| $$I=\int_{0}^{\infty} \frac{\log(2+x^2)}{4+x^2}dx=\int_{0}^{\infty} \frac{\log(a^2+x^2)}{4+x^2}dx=I(a)$$
$$\implies \frac{dI}{da}=2a \int_{0}^{\infty} \frac{dx}{(a^2+x^2)(4+x^2)]}=
\frac{2a}{4-a^2}\int_{0}^{\infty} \left( \frac{1}{a^2+x^2}-\frac{1}{4+x^2}\right) dx$$
$$\frac{dI}{da}=\frac{2a}{4-a^2}[\frac{\pi}{2a}-\frac{\pi}{4}]=\frac{\pi}{2(a+2)} \implies I(a)=\frac{\pi}{2} \ln (a+2)+C~~~~(*)$$
$$I(0)=2\int_{0}^{\infty} \frac{\ln x}{x^2+4}dx=J~~~~(1)$$
Let $x=1/t \implies dx=-dt/t^2$ in $J$, the
$$J=2\int_{0}^{\infty} \frac{-\ln t}{t^2+4}dt=2\int_{0}^{\infty} \frac{-\ln x}{x^2+4}dx~~~~(2)$$
Adding (1) and (2), we get $J=0=I(0)$, from (*) we get $C=-\frac{\pi}{2}\ln 2$
Thus, $$I(a)=\frac{\pi}{2} \ln (a+2)-\frac{\pi}{2} \ln 2= \frac{\pi}{2}\ln\frac{a+2}{2}=\frac{\pi}{2}\ln (1+\frac{1}{\sqrt{2}}),~ as ~a=\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How do I establish the differentiability of this function by definition? Given this function
$f(x) = (x^2 + 1)^{\sin(x)} $
How would I establish its differentiability over the entire function? I understand how to establish differentiability at a point $c$, by assessing the limit of the different quotient at $x = c$. But how would I establish the differentiability of the entire function?
| You shouldn't use the definition of derivative, instead you should use a tricky technique:
$\displaystyle f(x)=(1+x^2)^{\sin x} \implies \ln(f(x)) = \ln\left( (1+x^2)^{\sin x} \right) = \sin x \ln(1+x^2)$
$\displaystyle \implies \frac{d}{dx} \ln(f(x)) = \frac{d}{dx} \sin x \ln(1+x^2) \implies \frac{f'(x)}{f(x)} = \cos x \ln(1+x^2) + \sin x \frac{2x}{1+x^2}$
$\displaystyle f'(x)= f(x) \left( \cos x \ln(1+x^2) + \sin x \frac{2x}{1+x^2} \right) = (1+x^2)^{\sin x} \left( \cos x \ln(1+x^2) + \sin x \frac{2x}{1+x^2} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
| By my previous post it's enough to prove that:
$$x^2\ln{x}+x+e^x-3x^2>0,$$ where $x$ is a root of the equation $$x^2-x+e^x(x-2)=0.$$
The last equality gives $$e^x=\frac{x^2-x}{2-x}$$ or
$g(x)=0,$ where $$g(x)=x-\ln\frac{x^2-x}{2-x}$$ and since $$\frac{x^2-x}{2-x}>0,$$ we obtain $$1<x<2.$$
Now, $$g'(x)=\left(x-\ln\frac{x^2-x}{2-x}\right)'=-\frac{x^3-4x^2+6x-2}{x(2-x)(x-1)}=-\frac{x(x-2)^2+2(x-1)}{x(2-x)(x-1)}<0,$$
which says that $g$ decreases.
Also, $$g(1.7672)=0.0052...>0,$$ which says that the root of the equation $x^2-x+e^x(x-2)=0$ is greater than $1.7672$.
Id est, it's enough to prove that $$x^2\ln{x}+x+e^x-3x^2>0,$$ where $e^x=\frac{x^2-x}{2-x}$ and $1.7672<x<2$ for which it's enough to prove that:
$$x^2\ln{x}+x+\frac{x^2-x}{2-x}-3x^2>0$$ or $f(x)>0$, where
$$f(x)=\ln{x}-\frac{6x-3x^2-1}{x(2-x)}.$$
But $$f'(x)=\frac{x^3-4x^2+6x-2}{(x-2)^2x^2}=\frac{x(2-x)^2+2(x-1)}{x^2(2-x)^2}>0,$$
which says that $f$ increases.
Thus, $$f(x)>f(1.7672)=0.000096...>0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to obtain a formula for $f(z)$ given this recurrence I am trying to figure out how to derive a formula for $f(z)$ that is a function of $z$ and maybe $k \in \mathbb{N}$:
$$
f(z) = 1+z f \bigg(\frac{z}{1+z} \bigg)
$$
As an attempt, I tried a change of variable $z=\frac{1}{x}$, and I get:
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}f\bigg(\frac{1/x}{1+1/x}\bigg)
$$
which evaluates to the following after simplifying the terms inside $f$:
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}f\bigg(\frac{1}{1+x}\bigg)
$$
Now, given the above, is the following telescoping operation valid?
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}\bigg[ 1 + \frac{1}{x} + f\bigg(\frac{1}{2+x}\bigg) \bigg] = 1 + \frac{1}{x} + \frac{1}{(x+1)^2} + ... + \frac{1}{(x+k-3)^{k-2}} + \frac{1}{(x+k-2)^{k-1}}f\bigg(\frac{1}{k+x}\bigg)
$$
Also, how do I transform the above to a solution for $f(z)$?
| In the last step you plug $1+x$ in the place of $x$, hence you should have
$$ f\left( \frac{1}{x}\right) = 1 + \frac{1}{x}f\left( \frac{1}{1+x}\right) = $$
$$ 1 + \frac{1}{x}\left[ 1 + \frac{1}{1+x}f\left( \frac{1}{2+x} \right)\right]$$
So I think the telescopic series you achieved isn't correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\int \sin^2x\cos^4xdx$ $$\int \sin^2x\cos^4xdx$$
I tried
$$I = \int (1-\cos^2x)\cos^4xdx = \int \frac{\sec^2x-1}{\sec^6x}dx = \int \frac{\tan^2x}{\sec^6x}dx$$
Take $\tan x = t \implies \sec^2xdx = dt$
$$I = \int \frac{t^2}{(t^2+1)^4}dt$$
And I could not proceed further from here.
| With even powers of the trig functions use the half-angle identities until you have a an odd power.
$\sin^2 x = \frac 12 (1-\cos 2x)\\
\cos^2 x = \frac 12 (1+\cos 2x)$
$(\sin^2 x)(\cos^4 x) = \frac {1}{8}(1-\cos 2x)(1+cos 2x)(1+\cos 2x)\\
\frac18(1-\cos^2 2x)(1+\cos 2x)\\
\frac 18 (1 +\cos 2x - \cos^2 2x - \cos^3 2x)\\
\frac 18 (1 +\cos 2x - \frac 12 (1+\cos 4x) - \cos 2x(1-\sin^2 2x))\\
\frac 1{16} (1 - \cos 4x - 2\sin^2 2x\cos 2x)$
And each of those terms is straight forward to integrate.
Alternatively, you can use some complex analysis and say:
$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\cos x = \frac {e^{ix} + e^{-ix}}{2}$
$(\sin^2 x)(\cos^4 x) = \frac {(e^{2ix} -2 + e^{-2ix})(e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix})}{-64}\\
\frac {e^{6ix} + 2e^{4ix} - e^{2ix} -4 - e^{-2ix} +2 e^{-4ix} + e^{-6ix}}{-64}\\
\frac {-\cos 6x - 2\cos 4x + \cos 2x + 2}{32}$
And again not to bad to integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$
This is the answer the assignment was looking for.
| Firstly, $$\sin\theta=\sqrt{1-\cos^2\theta}$$ is wrong. Try $\theta=-\frac{\pi}{2}$.
But for $\theta\in(0,\pi)$ we see that $-3<3\cos\theta<3$ and $x=3\cos\theta$ gets any value from $(-3,3)$.
Also, for these values of $\theta$ we obtain a right formula: $$\sin\theta=\sqrt{1-\cos^2\theta}.$$
Secondly, $$\cos^3\theta=\frac{1}{4}(\cos3\theta+3\cos\theta)$$ and we can evaluate the integral shorter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many times must I apply L’Hopital? I have this limit:
$$\lim _{x\to 0}\left(\frac{e^{x^2}+2\cos \left(x\right)-3}{x\sin \left(x^3\right)}\right)=\left(\frac 00\right)=\lim _{x\to 0}\frac{\frac{d}{dx}\left(e^{x^2}+2\cos \left(x\right)-3\right)}{\frac{d}{dx}\left(x\sin \left(x^3\right)\right)}$$
$$\lim_{x\to0}\frac{2e^{x^2}x-2\sin \left(x\right)}{\frac{d}{dx}\left(x\right)\sin \left(x^3\right)+\frac{d}{dx}\left(\sin \left(x^3\right)\right)x}=\lim_{x\to0}\frac{2e^{x^2}x-2\sin \left(x\right)}{\sin \left(x^3\right)+3x^3\cos \left(x^3\right)+\sin \left(x^3\right)}$$
but yet we have $(0/0)$. If I apply L’Hopital again, I obtain
$$=\lim_{x\to0}\frac{2\left(2e^{x^2}x^2+e^{x^2}\right)-2\cos \left(x\right)}{15x^2\cos \left(x^3\right)-9x^5\sin \left(x^3\right)}$$
again giving $(0/0)$. But if I apply L’Hopital a thousand times I'll go on tilt. What is the best solution in these cases? With the main limits or applying upper bonds?
| You have one too many copies of the $\sin x^3$ in your denominator after the first derivative, so let's take a step back. If you're prepared to use Taylor series,$$\begin{align}e^{x^2}+2\cos x-3&=1+x^2+\frac12x^4+2-x^2+\frac{1}{12}x^4-3+o(x^4)\\&\sim\frac{7}{12}x^4,\\x\sin x^3&\sim x^4,\end{align}$$so the limit is $\frac{7}{12}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{1/x}$ $$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=?$$
I have tried a lot but I am stuck when I solve this by using this hints. $a^x=\exp(\ln(a^x))=\exp(x\ln a)$, so then $a=\frac{4^x+5^x}{4}$. The above expression becomes
$$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=\lim_{x\to\infty}\exp\left(\ln\left(\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}\right)\right)=\exp\left(\lim_{x\to\infty}\frac{1}{x}\ln\left(\frac{4^x+5^x}{4}\right)\right).$$
Now here I am stuck that how I can deal this indeterminate expression?
| Do you have to use this hint?
You could simplify your task if you simplify the given expression first:
$$\frac{4^x+5^x}{4}=\frac{1}{4}\cdot(4^x+5^x)=\frac{1}{4}\cdot5^x\cdot\left(\left(\frac{4}{5}\right)^x+1\right),$$
therefore
$$\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=\left(\frac{1}{4}\cdot5^x\cdot\left(\left(\frac{4}{5}\right)^x+1\right)\right)^{\frac{1}{x}}=\frac{1}{4^{\frac{1}{x}}}\cdot5\cdot\left(\left(\frac{4}{5}\right)^x+1\right)^{\frac{1}{x}}.$$
From this, the limit should be pretty evident.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587623",
"timestamp": "2023-03-29T00:00:00",
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Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle \sum_{k=1}^{n+1}\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
which can be summarised to:
$ \displaystyle \sum_{k=1}^{n}\frac{1}{(5k + 1) (5k + 6)} + \frac{1}{(5(n+1) + 1) (5(n+1) + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
Using the Induction Hyptohesesis this turns to:
$ ( \frac{1}{30} - \frac{1}{5(5n + 6)} ) + \frac{1}{(5n + 6 ) (5n+6+5)} $
when we let s = 5n +6 we get:
$ \frac{1}{30} - \frac{1}{5s} + \frac{1}{(s) (s+5)} = \frac{1}{30} - \frac{1}{5(s + 5)} $
substracting both sides by $\frac{1}{30}$ yields:
$- \frac{1}{5s} + \frac{1}{(s) (s+5)} = - \frac{1}{5(s + 5)}$ and this is where I'm stuck. I'm working on this for more than 6h but my mathematical foundation is too weak to get a viable solution. Please help me.
| By multiplying with $ 5(+5)$ we clear the denominators:
$- \frac{1(s+5)}{5s(s+5)} + \frac{5}{5s (s+5)} = - \frac{1s}{5s(s + 5)}$
$-s = -s$
Which proves the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that: $\sum_{k=0}^{\infty}\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\frac{\pi}{4}$ I was observing this paper and by some experimental chance I got this result, but I can not prove it.
$$\sum_{k=0}^{\infty}\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\frac{\pi}{4}\tag1$$
I tried to factorise $4k^4+12k^3+13k^2+6k+5$ but it is not factorable.
I guess that $$\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\arctan(x)+\arctan(y)$$
Does anyone knows how to prove $(1)?$
| Let $ u = \frac{2}{ k^2 + 2k + 1}, v = \frac{2}{ 4k^2 + 4k + 1}$. Then,
$$ \arctan (u) - \arctan (v) = \arctan ( \frac{ u - v} { 1 + uv} ) = \arctan ( \frac{6k^2 + 4k } { 4k^4 + 12k^3 + 13k^2 + 6k + 5 }). $$
Then use $\sum_{k=1} \arctan \frac{2}{k^2} = \frac{3 \pi } { 4} $ (2.14) and $ \sum_{k=0} \arctan \frac{2}{(2k+1)^2 } = \frac{ \pi}{2}$ (2.8) as given in the paper, to justify
$\sum_{k=0} \arctan ( \frac{6k^2 + 4k } { 4k^4 + 12k^3 + 13k^2 + 6k + 5 }) $
$= \sum_{k=0} \arctan \frac{2}{(k+1)^2} - \arctan \frac{2}{(2k+1)^2} $
$ = \sum_{k=1} \arctan\frac{2}{k^2} - \sum_{k=0} \arctan \frac{2}{ (2k+1)^2 } $
$ = \frac{ 3 \pi }{4} - \frac{\pi}{2} $
$= \frac{ \pi}{4} .$
Note: The 2 identities given in the paper can again be proved via telescoping sums.
2.8 is done via 1 telescoping sum ($u=2k, v = 2k+2)$.
2.14 is done via 2 telescoping sums.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Reduction of order on a 2nd order LI DE The equation and $y_1(x)$ are $$y_1(x)=x^{\frac{-1}{2}}\cos(x) $$
$$x^2y''+xy'+(x^2-\frac{1}{4})y=0$$
Using the formula that solves for $y_2(x)$
(the one with an integration factor over $y_1(x)$ squared, all inside an integral multiplied by $y_1(x)$)
We get $$ x^{-1/2}\cos(x) \int \frac{\frac{1}{x}}{x^{1/4}\cos^2(x)}dx $$
Now from here what I thought to do was to rationalize the integrand. There by making the equation easier for integration.
Just focusing on the integral I got
$$\int \frac{1}{x^4\sqrt{x} *cos^2(x)} dx$$
Now it only seems to me that the integration has only gotten more complicated. Could this be done by parts or u-sub?
| We are given
$$x^2y''+xy'+\left(x^2-\dfrac{1}{4}\right)y=0 \tag 1$$
We are also given a solution as
$$y_1(x)=x^{-1/2}\cos(x)$$
Using Reduction of Order, a second solution is given by
$$y_2(x) = y_1(x) v(x) = x^{-1/2} \cos(x) v(x)$$
The first derivative is
$$y_2'(x) = \frac{\cos (x) v'(x)}{\sqrt{x}}-\frac{v(x) \cos (x)}{2 x^{3/2}}-\frac{v(x) \sin (x)}{\sqrt{x}}$$
The second derivative is
$$y_2''(x) = -\frac{\cos (x) v'(x)}{x^{3/2}}-\frac{2 \sin (x) v'(x)}{\sqrt{x}}+\frac{\cos (x) v''(x)}{\sqrt{x}}+\frac{v(x) \sin (x)}{x^{3/2}}+\frac{3 v(x) \cos (x)}{4 x^{5/2}}-\frac{v(x) \cos (x)}{\sqrt{x}}$$
Substituting this into the ODE
$x^2 y_2'' + x y_2'+\left(x^2-\dfrac{1}{4}\right)y_2$
$= x^2\left(-\dfrac{\cos (x) v'(x)}{x^{3/2}}-\dfrac{2 \sin (x) v'(x)}{\sqrt{x}}+\dfrac{\cos (x) v''(x)}{\sqrt{x}}+\dfrac{v(x) \sin (x)}{x^{3/2}}+\dfrac{3 v(x) \cos (x)}{4 x^{5/2}}-\dfrac{v(x) \cos (x)}{\sqrt{x}}\right) + x \left(\dfrac{\cos (x) v'(x)}{\sqrt{x}}-\dfrac{v(x) \cos (x)}{2 x^{3/2}}-\dfrac{v(x) \sin (x)}{\sqrt{x}}\right) + \left(x^2-\dfrac{1}{4}\right)x^{-1/2} \cos(x) v(x) = 0$
Simplifying, we get
$$x^{3/2} \left(\cos (x) v''(x)-2 \sin (x) v'(x)\right) = 0$$
Now solve for $v(x)$ and then find $y_2(x)$.
Can you proceed?
The final result should be
$$y_2(x) = \dfrac{\cos (x) (c_1 \tan (x)+c_2)}{\sqrt{x}}$$
You can substitute this into $(1)$ and verify it satisfies the ODE.
| {
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A five digit number minus a four digit number equals $33333$. What are the two numbers, if you are only allowed to use the numbers $1-9$
Question: A five digit number minus a four digit number equals $33333$. What are the two numbers, if you are only allowed to use the numbers $1-9$ once? More precisely,
\begin{align}
& & A_1 \, A_2 \, A_3 \, A_4 \, A_5 \\
& - & A_6 \, A_7 \, A_8 \, A_9 \\
& & \hline 3\,\,\,\,\,3 \,\,\,\,\,3 \,\,\,\,\,\,3 \,\,\,\,\,\,3\\
& & \hline
\end{align}
where $A_1,A_2,A_3,A_4,A_5, A_6, A_7, A_8, A_9 \in \{1,2,3,4,5,6,7,8,9\}$ and they form a pairwise distinct set.
For me, I would guess $A_1=3$ or $A_1 = 4.$ But that is all I got.
I am interested to know its thought process.
| The sum of the digits $1$ through $9$ is odd. They contribute to the parity of the digit sum of the result no matter which row they’re in. The digit sum of the result is odd. Thus there must be an even number of borrowings.
A column that causes borrowing must have a $7$, $8$ or $9$ in the bottom row, so we cannot have four borrowings.
On the other hand, if there were no borrowing at all, the possible pairs in a column would be $9-6-3$, $8-5-2$ and $7-4-1$, but we can use at most one from each of these three groups.
It follows that there are exactly two borrowings. Thus the difference between the digit sums of the rows must be $5\cdot3-2\cdot9=-3$, and since the sum of all digits is $\frac{9(9+1)}2=45$, the top row must sum to $21$ and the bottom row to $24$.
We need to have exactly two of $7$, $8$ and $9$ in the bottom row to cause the two borrowings.
It can’t be $7$ and $8$ because then $7$ would have to be subtracted from $1$ and $8$ from $2$, so the two borrowing columns would have to be the two lending columns.
If it were $8$ and $9$, that would leave a sum of $7$ for the bottom row, so that could be $3,4$ or $2,5$ or $1,6$. It can’t be $3,4$ because one of those needs to be $A_1$; it can’t be $2,5$ because $5$ would need to be subtracted from $8$ or $9$; and it can’t be $1,6$ because $6$ would need to be subtracted from $9$.
Thus $7$ and $9$ are in the bottom row. That leaves a sum of $8$ for the bottom row, which could be $3,5$ or $2,6$. But it can’t be $2,6$, again because $6$ would need to be subtracted from $9$.
Thus we have $3,5,7,9$ in the bottom row and $1,2,4,6,8$ in the top row. So $4$ must be $A_1$, $7$ must be subtracted from $1$, $9$ from $2$, $3$ from $6$ and $5$ from $8$. Thus the lenders must be $4$ and $1$, so the top row must start $412$. That leaves two possibilities for the order of the last two columns, so there are two solutions:
41286 41268
-7953 and -7935
----- -----
33333 33333
The solutions are confirmed by this Java code. (Full disclosure: I initially made a mistake in the proof and wrote the code to find it, so I knew the solution before I completed the proof.)
| {
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Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)^{20}$ and then working with binomial expansion.
| There is a thing called multinomial theorem which is a slight generalization of the binomial theorem.
First of all, note $(x^2+x^7+x^9)^{20}= x^{40}(1+x^5+x^7)^{20}.$ So, basically, now will find the coefficients of $x^{17}.$ Can you continue?
Note that $(1)^{17} (x^5)^2(x^7)^1$ gives us $x^{17}.$ The coffocient of $x^{17}$ is $\dfrac{20!}{17! 2! 1!}$.
The idea of binomial/multinomial theorem is purely combinatorial.
Suppose we have $(x+y+z)^{20}.$ Consider the term $x^{10}y^{6}z^{4}.$ Note that the coefficient of this term is $\dfrac{20!}{10! 6! 4!}$ The combinatorial interpretation is that you can create/form the $x^{10}y^{6}z^{4}$ $\dfrac{20!}{10! 6! 4!}$ ways. This should you give you some idea to think more about the problem. I hope it helps.
| {
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Examples of triangles, which related ellipses are perfectly packed with circles. Ellipse can be
perfectly packed with $n$ circles
if
\begin{align}
b&=a\,\sin\frac{\pi}{2\,n}
\quad
\text{or equivalently, }\quad
e=\cos\frac{\pi}{2\,n}
,
\end{align}
where $a,b$ are the major and minor semi-axis
of the ellipse and $e=\sqrt{1-\frac{b^2}{a^2}}$ is its eccentricity.
Consider a triangle and any ellipse, naturally associated with it,
for example, Steiner circumellipse/inellipse, Marden inellipse,
Brocard inellipse,
Lemoine inellipse,
ellipse with the circumcenter and incenter as the foci
and $r+R$ as the major axis, or any other ellipse
you can come up with,
which can be consistently associated with the triangle.
The question is: provide the example(s) of triangle(s)
for which the associates ellipse(s) can be perfectly packed with circles.
Let's say that the max number of packed circles is 12,
unless you can find some especially interesting case with more circles.
For example, the Steiner incircle for the famous $3-4-5$ right triangle
can not be perfectly packed.
The example of the right triangle with the Marden inellipse,
perfectly packed with six circles is given in the self-answer below.
|
This is example of the
ellipse, perfectly packed with three circles,
inscribed into the equilateral triangle $ABC$.
Let the center of the ellipse be $M=0$
and its semi-axes defined as
\begin{align}
s_a=|DF_1|=|DF_2|&=1
,\\
s_b=|MD|=|ME|&=\sin\frac\pi{2\cdot3}=\frac12
,
\end{align}
locations of the top and bottom points are
\begin{align}
D&=(0,-\tfrac12)
,\quad
E=(0,\tfrac12)
,\\
F_1&=(-\tfrac{\sqrt3}2,0)
,\quad
F_2=(\tfrac{\sqrt3}2,0)
.
\end{align}
\begin{align}
\text{Then the equation of the ellipse is }
\quad
x^2+2\,y^2&=0
\end{align}
and for the upper arc we have
\begin{align}
y(x)&=\tfrac12\,\sqrt{1-x^2}
,\\
y'(x)&=-\tfrac12\,\frac{x}{\sqrt{1-x^2}}
,
\end{align}
so ve can find the point $K$, tangent to
the circumscribed equilateral $\triangle ABC$:
\begin{align}
-\tfrac12\,\frac{x}{\sqrt{1-x^2}}
&=\tan\tfrac\pi6=\tfrac{\sqrt3}3
,\\
x&=-\tfrac{2}{13}\sqrt{39}
,\\
y(x)&=\tfrac{1}{26}\sqrt{13}
.
\end{align}
So, the tangential points are
\begin{align}
K&=(-\tfrac{2}{13}\sqrt{39},\tfrac{1}{26}\sqrt{13} )
,\quad
L=(\tfrac{2}{13}\sqrt{39},\tfrac{1}{26}\sqrt{13} )
,
\end{align}
and the location of the vertices of $\triangle ABC$
can be easily found as
\begin{align}
A&=(-\tfrac16\,\sqrt3\,(\sqrt{13}+1), -\tfrac12)
,\quad
B=(\tfrac16\,\sqrt3\,(\sqrt{13}+1), -\tfrac12)
,\\
C&=(0,\tfrac12\,\sqrt{13})
,
\end{align}
the side length of the triangle is thus
\begin{align}
|AB|=|BC|=|CA|=a&=\tfrac13\,\sqrt3\,(\sqrt{13}+1)
,
\end{align}
and the tangential points $D,K,L$ divide the side segments
as follows:
\begin{align}
|AK|=|BL|&=\sqrt{\tfrac2{39}(7+\sqrt{13})}
,\quad
|CK|=|CL|=\tfrac4{13}\,\sqrt{39}
.
\end{align}
This ellipse is neither Steiner nor Mandart inellipse,
it is essentially a
[![generalized Steiner inellipse]][Linfield1920],
for which the foci are
the roots of the derivative of the rational
function
\begin{align}
f(z)&=(z-A)^u (z-B)^v (z-C)^w
,
\end{align}
where $A,B,C$ are the coordinates of the vertices of the triangle,
and the tangent points $L,K,D$ divide the segments
$BC,CA$ and $AB$
as $v:w,w:u$ and $u:v$, respectively.
In this case
\begin{align}
u=v&=\tfrac1{12}\,(\sqrt{13}-1)
,\\
w&=\tfrac16\,(7-\sqrt{13})
.
\end{align}
Semi-axes of such ellipse, expressed in terms of the side length
$a$ of the equilateral triangle, are
\begin{align}
s_a&=\frac{a\sqrt3}{12}\,(\sqrt{13}-1)
,\quad
s_b=\frac{a\sqrt3}{24}\,(\sqrt{13}-1)
.
\end{align}
Similarly, for another orientation,
we have
\begin{align}
s_a&=\frac{a\sqrt3}{3}\,(\sqrt{7}-2)
,\quad
s_b=\frac{a\sqrt3}{6}\,(\sqrt{7}-2)
\end{align}
and
\begin{align}
u&=\tfrac{11}3-\tfrac43\,\sqrt7
,\quad
v=\tfrac23\,\sqrt7-\tfrac43
,\quad
w=v
.
\end{align}
Reference
[Linfield1920]: Ben-Zion Linfield.
“On the relation of the roots and poles of a rational function to the roots of its
derivative”.
In: Bulletin of the American Mathematical Society 27.1 (1920), pp. 17-21
| {
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Find the inverse of $f(x)= 3x^2 - 3x -11$, where $x>2$
Find the inverse of $f(x)= 3x^2 - 3x -11$, where $x > 2$
The answer guide says the answer is $f^{-1}(x)=\frac{1}{2} + \sqrt{ \frac{x}{3}} +\frac{47}{12}$,
but i had $\frac{1}{2} + \sqrt{\frac{x}{3} +\frac{47}{12}}$ instead.
How do i get the correct answer?
| To find the innverse of $f(x)$, you have to swap $x$ and $y$, obtaining:
$$x=f(y)=3y^2-3y-11=\left(\sqrt{3}y-\frac{\sqrt{3}}{2}\right)^2-11-\frac{3}{4}$$
From here you solve for $y$, having:
$$x+11+\frac{3}{4}=3\left(y-\frac{1}{2}\right)^2$$
And so:
$$\sqrt{\frac{x}{3}+\frac{47}{12}}=y-\frac{1}{2} \leftrightarrow y = \sqrt{\frac{x}{3}+\frac{47}{12}}+\frac{1}{2}$$
Your answer is correct. There is certanly a typing error in the answer given.
| {
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Prove $a_t \leq \frac{c}{t}$ We have the recursive equation $$a_t^2 = (1 - a_t)a_{t-1}^2$$ $$a_0 = 1$$
Prove that there exists constant $c$ such that $a_t \leq \frac{c}{t}$, and find the smallest such $c$.
I was able to get that $a_t^2 = (1 - a_t)(1 - a_{t-1})\dots(1 - a_1)$, but I don't know how to proceed after that, and I'm not sure if I'm going in the right direction.
| It turn out that
$c \ge 2$ will work.
$a_t^2
= (1 - a_t)a_{t-1}^2
$.
$a_1^2
= 1-a_1$
so
$a_1
=\dfrac{-1\pm\sqrt{5}}{2}
=\dfrac{-1+\sqrt{5}}{2}
$.
$a_t^2+a_ta_{t-1}^2-a_{t-1}^2
= 0
$
so
$\begin{array}\\
a_t
&=\dfrac{-a_{t-1}^2+\sqrt{a_{t-1}^4+4a_{t-1}^2}}{2}\\
&=\dfrac{-a_{t-1}^2+a_{t-1}\sqrt{a_{t-1}^2+4}}{2}\\
&=a_{t-1}\dfrac{-a_{t-1}+\sqrt{a_{t-1}^2+4}}{2}\\
&=a_{t-1}\dfrac{-a_{t-1}+\sqrt{a_{t-1}^2+4}}{2}\dfrac{a_{t-1}+\sqrt{a_{t-1}^2+4}}{a_{t-1}+\sqrt{a_{t-1}^2+4}}\\
&=a_{t-1}\dfrac{2}{a_{t-1}+\sqrt{a_{t-1}^2+4}}\\
&=\dfrac{2}{1+\sqrt{1+4/a_{t-1}^2}}\\
\text{if}
&a_{t-1} \le \dfrac{c}{t-1}\\
a_t
&\le\dfrac{2}{1+\sqrt{1+4/(c/(t-1))^2}}\\
&=\dfrac{2}{1+\sqrt{1+4(t-1)^2/c^2}}\\
&\lt\dfrac{2}{1+\sqrt{4(t-1)^2/c^2}}\\
&=\dfrac{2}{1+2(t-1)/c}\\
\end{array}
$
so we want
$\dfrac{2}{1+2(t-1)/c}
\le \dfrac{c}{t}
$
or
$\dfrac{2c}{c+2(t-1)}
\le \dfrac{c}{t}
$
or
$2t
\le c+2(t-1)
=2t+c-2
$
or
$c \ge 2$.
Therefore
$c \ge 2$
will work.
| {
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if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity. I was given a statement that if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity.
The roots of $x^n−1$ in $\mathbb C$ which are not also roots of $x^m −1$ for some $1 ≤ m ≤ n$ are called the primitive n’th complex roots of unity
So if this example works with $n=6$ I would get the following:
We have $x^6−1 = (x−1)(x^5+x^4+x^3+x^2+x+1)$, $x^6−1 = (x^2−1)(x^4+x^2+1)$ and $x^6 −1 = (x^3 −1)(x^3 + 1)$. The roots of $x−1$, $x^2 −1$, $x^3 −1$ are $1,−1,\frac {−1} 2 ± \frac{{\sqrt3}} 2 i$. Thus the remaining two roots of $x^6−1$, namely, $ω^1 = \frac {1} 2 + \frac{{\sqrt3}} 2 i$ and $ω^5 = \frac {1} 2 - \frac{{\sqrt3}} 2 i $ are the primitive 6’th complex roots of unity.
Is it correct to prove the above statement by just pointing out that if we set $\omega = \frac {−1} 2 ± \frac{{\sqrt3}} 2 i$ that $-\omega = \frac {1} 2 - \frac{{\sqrt3}} 2 i$ and $-\omega = \frac {1} 2 + \frac{{\sqrt3}} 2 i$ which are both sixth roots of unity.
Thanks in advance.
| $\omega$ a primitive third root of unity implies $\omega=e^{(2\pi i k)/3},\,k=1,2$. And $-1=e^{\pi i}$. Thus $-\omega=e^{\pi i}e^{(2\pi i k)/3}=e^{(6+4k)\pi i/6}=e^{(2\pi i k)/6}\,,k=1,5$.
| {
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Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.
| By Lagrange, we minimize
$$x^2+y^2+\lambda(x^4+y^4+3xy-2).$$
The derivatives on $x$ and $y$ yield
$$2x+\lambda(4x^3+3y)=2y+\lambda(4y^3+3x)=0,$$ and after elimination of $\lambda$,
$$2x(4y^3+3x)=2y(4x^3+3y)$$ or
$$(8xy-6)(y^2-x^2)=0.$$
The solutions $y=\pm x$ imply
$$2x^4\pm3x^2-2=0,$$ giving
$$x^2+y^2=2x^2=\frac{\mp3\pm5}2$$ and only the positive solutions $1$ and $4$ are valid.
Next, with $xy=\dfrac34$,
$$x^4+y^4+3xy-2=(x^2+y^2)^2-2x^2y^2+3xy-2=(x^2+y^2)^2-\frac98+\frac94-2=0$$
and
$$x^2+y^2=\sqrt{\frac78}.$$
Anyway, we would have $(x-y)^2=x^2+y^2-2xy<0$, which is not possible.
Final answers: the solution points are $\pm\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$ and $\pm(\sqrt2,-\sqrt2).$
| {
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Show $\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$ Show that
$$\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$$
My Attempt:
Let $$I=\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
Using $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x)dx$ we can write:
$$I=\int_{0}^{\pi} \frac {(\pi - x)dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} $$
$$I=\int_{0}^{\pi} \frac {\pi dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - \int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}$$
$$I=2\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - I$$
$$I=\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
| $$I=\int_{0}^{\pi} \frac{x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(1)$$
Apply $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$.
$$I=\int_{0}^{\pi} \frac{(\pi-x) dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(2)$$
Add (1) and (2)
$$2I=\pi\int_{0}^{\pi} \frac{ dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(3)$$
Use $\int_{0}^{2a} f(x) dx=\int_{0}^{a} f(x) dx, ~if ~f(2a-x)=f(x)$
$$I=\pi \int_{0}^{\pi/2} \frac{dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(4)$$
Let $$J(a,b)=\int_{0}^{\pi/2} \frac{dx}{(a^2 \sin^2 x+ b^2 \cos^2x)}=\int_{0}^{\pi/2}
\frac {\sec^2 x dx}{b^2+a^2\tan^2 x} =\frac{\pi}{2ab}~~~(5)$$
D. (5) w.r.t. $a$ to get
$$\frac{J(a,b)}{da}=\int_{0}^{\pi/2} \frac{-2a \sin^2 xdx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=-\frac{\pi}{2a^2b}~~~~(6)$$
$$\implies \int_{0}^{\pi/2} \frac{\sin^2x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=\frac{\pi}{4a^3b}~~~(7)$$
Similarly by D.w.r.t. $b$, we can get
$$\implies \int_{0}^{\pi/2} \frac{\cos^2x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=\frac{\pi}{4a^3b}~~~(8)$$
Adding (7) and (8), we get from (4)
$$I=\frac{\pi^2}{4}\frac{a^2+b^2}{a^3b^2}.$$
| {
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Evaluating Series Integral How do I show that $$\int_0^\infty\frac{(\ln x)^2dx}{1+x^2}$$=$$4(1-1/3^3+1/5^3-1/7^3...)$$
I expanded the integral to $$\int_0^\infty(\ln x)^2(1-x^2+x^4...)dx$$ using the power series for $$\frac{1}{1+x^2}$$ but I'm not sure how to continue from here.
| Starting with Integrand's observation in the comments, we have $$
\begin{array}{rcl} \displaystyle I = 2\int_{0}^{1}{\displaystyle\frac{\log^{2}{x}}{1+x^{2}}\,\mathrm{d}x}=2 \int_0^1 \log^2{x} \sum_{k \ge 0} (-1)^k x^{2k}\,\mathrm{d}x = 2\sum_{k \ge 0} (-1)^k\int_0^1 x^{2k}\log^2{x}\,\mathrm{d}x\end{array}
$$
Now, consider $\displaystyle f(\alpha) = \int_0^1 x^{2\alpha}\,\mathrm{dx} = \frac{1}{1+2\alpha}$. Then finding $f''(\alpha)$ for both sides gives:
$$\displaystyle \int_0^1 x^{2\alpha} \log^2{x}\,\mathrm{dx} = \frac{2}{(1+2\alpha)^3}$$
Therefore:
$$I = 4 \sum_{k \ge 0} \frac{(-1)^k}{(2k+1)^3} = 4 \cdot \frac{\pi^3}{32} = \frac{\pi^3}{8}. $$
| {
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There is a number, the second digit of which is smaller than its first digit by 4, and if the number There is a number, the second digit of which is smaller than its first digit by 4, and if the number was divided by the digit's sum, the remainder would be 7.
Actually I know the answer is 623
I found it by using computer program which checks the conditions for all numbers but I wanted to know if there is a mathematical way to slove this problem.
| One-digit case is impossible, since $4\not \equiv7 \mod 4$
Two-digit case: write number as $10(a+4)+a$.
$$10(a+4)+a \equiv 7 \mod (2a+4)$$
$$11a\equiv -33 \mod 2a+4$$
$$a+3 \equiv 0 \mod 2a+4$$
$$2(a+3) \equiv 2 \not \equiv 0 \mod 2a+4$$
Therefore, two-digit is impossible.
Three-digit case: Write number as $100(a+4)+10a+b$.
$$100(a+4)+10a+b\equiv 7 \mod (2a+b+4)$$
$$110a+b+400 \equiv 7 \mod (2a+b+4)$$
$$108a+396\equiv7 \mod (2a+b+4)$$
$$108a+389\equiv 0\mod (2a+b+4)$$
When $a=1$,
$$497\equiv 0 \mod b+6$$
Since $497=7\times 71$, $b+6=7$. However, we "do not want that", since the modulo must be greater than 7. "Remainder = 7" $\implies$ "Modulo > 7"
When $a=2$,
$$605 \equiv 0 \mod b+8$$
$605 = 5 \times 11^2$. Therefore, we can take $b+8=11 \implies b=3$.
Therefore our final answer is $\fbox{623}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $n\ge 3$ determine all real solutions of the system of $n$ equations.
Question: For $n\ge 3$ determine all real solutions of the system of $n$ equations: $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
My approach: It is given that $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
Define $$S_n:=x_1+x_2+\cdots+x_n.$$
This implies that $$x_1+x_2+\cdots+x_{n-1}+x_n=x_n+\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}++x_i+x_{i+1}+\cdots+x_n=x_i+\frac{1}{x_i}\\ \cdots \\ x_1+x_2+\cdots+x_{n-1}+x_n=x_1+\frac{1}{x_1}.$$
Therefore, we have $$x_j+\frac{1}{x_j}=S_n, \forall 1\le j\le n.$$
Now for any $1\le j\le n,$ we have $$x_j+\frac{1}{x_j}=x_n+\frac{1}{x_n}\\ \implies \frac{1}{x_j}-\frac{1}{x_n}=x_n-x_j\\ \implies \frac{x_n-x_j}{x_nx_j}=x_n-x_j\\ \implies (x_n-x_j)\left(\frac{1}{x_nx_j}-1\right)=0\\ \implies x_j=x_n \text{ or } x_j=\frac{1}{x_n}.$$
Now let us have $i (0\le i\le n-1)$ of the $(n-1)$ numbers $x_j, 1\le j\le n-1$ such that $$x_j=x_n$$ and the rest $(n-1-i)$ numbers such that $$x_j=\frac{1}{x_n}.$$
Therefore, since $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\\implies i.x_n+(n-1-i).\frac{1}{x_n}=\frac{1}{x_n}\\ \implies i.x_n^2+(n-2-i)=0\\\implies i.x_n^2=2+i-n.$$
Now since $x_n\in\mathbb{R}$ and $x_n\neq 0\implies x_n^2>0.$ Now since $i\ge 0\implies i.x_n^2\ge 0 \implies 2+i-n\ge 0\implies i\ge n-2.$ Therefore $i=n-2,n-1$.
Now when $i=n-2$, we have $(n-2)x_n^2=0.$ Now since it is given that $n\ge 3\implies n-2\ge 1>0$. Therefore $x_n^2=0\implies x_n=0$. But $x_n\neq 0$. Therefore $i\neq n-2$.
Now when $i=n-1$, we have $(n-1)x_n^2=1\implies x_n^2=\frac{1}{n-1}\implies x_n=\pm\frac{1}{\sqrt{n-1}}.$ Therefore we have $$x_1=x_2=\cdots=x_n=\pm\frac{1}{\sqrt{n-1}}.$$
Therefore the required set of solutions are $$(x_1,x_2,\cdots,x_n)=\left(\frac{1}{\sqrt{n-1}},\frac{1}{\sqrt{n-1}},\cdots,\frac{1}{\sqrt{n-1}}\right),\left(-\frac{1}{\sqrt{n-1}},-\frac{1}{\sqrt{n-1}},\cdots,-\frac{1}{\sqrt{n-1}}\right).$$
Can someone check if my solution is correct or not? And if correct, is there a more better and efficient solution than this?
| We have $$x_1+x_2+\cdots+x_{n-1}-\frac{1}{x_n}=0\\x_1+x_2+\cdots-\frac{1}{x_{n-1}}+x_n=0\\.........................\\.........................\\-\frac{1}{x_1}+x_2+\cdots+x_{n-1}+x_n=0 $$ It follows
$$\sum_{i=1}^{i=n}\left((n-1)x_i-\frac{1}{x_i}\right)=\sum_{i=1}^{i=n}\frac{(n-1)x_i^2-1}{x_i}=0 $$
Then a clear solution is given by $(n-1)x_i^2-1=0$ for all $i$.
Thus the two solutions given by the O.P.
| {
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"timestamp": "2023-03-29T00:00:00",
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If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)? I have a question that goes:
If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?
So what I tried was I know that since the line is normal to the 2nd circle, so it must pass through the center of the second circle which is $(-1,2)$.
So from that I got that $$-a+2b+c=0$$
But I cant really find any other equations here that would help, I tried differentiation the curves but I dont have the point of contact so can't really do anything there.
I also know that the tangent to the circle $x^2+y^2+2gx+2fy+c = 0$ at $(a,b)$ is $ax+by+(a+x)g+(b+y)f +c = 0$
I don't know how to proceed, can someone help?
| By sketching a graph in mind, we see that the line $ax+by+c=0$ is just a line that is passing through $(-1,2)$ (the center of the circle $x^2+y^2+2x-4y+1=0$) and is tangent to the circle $x^2+y^2-2x=3/5$.
Therefore we looking for tangent lines from point $(-1,2)$ to the circle $x^2+y^2-2x=3/5$.
The line passing through $(-1,2)$ with slop $m$ is
$$y=m(x+1)+2.$$
The intersects points of the line and the circle $x^2+y^2-2x=3/5$ is found by
$$x^2+(m(x+1)+2)^2-2x=3/5$$
simplifying:
$$(1+m^2)x^2+[2m(m+2)-2]x+(m+2)^2-3/5=0\tag{*}$$
Since we need the line $y=m(x+1)+2$ to touch the circle, then the discriminant of $(*)$ must be zero:
$$(2m(m+2)-2)^2-4(1+m^2)((m+2)^2-3/5)=0$$
which simplifies to
$$\frac{-4}{5}(m+3)(m+\frac{1}{3})=0$$
i.e. $m=-3$ or $m=-1/3$. Thus the tangent lines are
$$y=-3(x+1)+2\quad\text{and}\quad y=-\frac13(x+1)+2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadrilateral/Geometry: Find the area of all the triangles in a trapezoid
I know the area of $BDE$ is $4$ but can't find out the area of $ABE$.
To solve $BDE$:
Since triangle $ACD$ and $CDB$ share the same base and height and knowing their shared area $CDE$ is 3, we know $BDE$ is 4.
Question: How do you solve for $ABE$?
| The area of the triangle ADC is equivalent to area of triangle ACE plus area of triangle CDE $$\frac{1}{2} \times AC \times DC = 4+3$$ $$AC \times DC = 2(4+3).$$ Thus, $AC \times DC = 14.$
Now for area of triangle of BDC, it is the area of BDE plus area of CDE $$\frac{1}{2}\times AC\times DC = \text{area of triangle BDE} + 3$$ $$\frac{1}{2}\times 14 = \text{area of triangle BDE} + 3.$$ Thus, area of triangle BDE is 4.
Area of triangle CDE is the vertical distance of E to DC $\times DC$ such that it is point Y. So, $$3= \frac{1}{2} EY \times DC \text{ so } EY \times DC = 6$$
Furthermore, $$\frac{EY\times DC}{AC\times DC}=\frac{3}{7}.$$ Now, area of ABE is $\frac{1}{2} \times \text{ vertical line from E to arbitrary N}$ such that $$=\frac{1}{2}\times EN\times AB = \frac{1}{2}\times(AC-EY)\times AB$$ $$=\frac{1}{2}\times(AC-\frac{3}{7}AC)\times AB$$ $$=\frac{2}{7}AC\times AB.$$ Now, area of triangle ABC is area of ACE plus area of ABE so $$\frac{1}{2}\times AC\times AB = 4 + \text{area of ABE}$$ $$\frac{1}{2}\times\frac{7}{2}\times\frac{2}{7} \times AC \times AB = 4 + \text{area of ABE}$$ $$\frac{7}{4} \times \text{area of triangle ABE} = 4 + \text{area of ABE}$$ $$\frac{7}{4} \times \text{area of ABE} - \text{area of ABE} = 4$$ $$\frac{3}{4}\times \text{area of ABE} =4.$$ So, $$\text{area of triangle ABE} = \frac{16}{3}.$$ And we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Different solutions with different results for an inequality Find m such that the following inequality:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
is always true for $\forall x \in R$.
1st solution:
1st case
$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$
$$<=>x^2+2x-m-1>0$$
$$\Leftrightarrow 1^2+(m+1)< 0$$
$$\Leftrightarrow m<- 2$$
2nd case
$$4x-2m-\frac{1}{2}< -(-x^2 + 2x +\frac{1}{2} -m)$$
$$\Leftrightarrow x^2-6x+3m>0$$
$$\Leftrightarrow 3^2-3m<0$$
$$\Leftrightarrow m>3$$
2nd solution:
The inequality is the same as:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
Since the left-hand side is always positive, in order for the inequality to be always true, $\frac{3}{2}-m$ has to be negative, or $m > \frac{3}{2}$
The 2 solutions give different answers, so I was quite confused
But I get more confused as Wolfram Alpha gives me the solution:
$$m > \sqrt{3} - \frac{1}{4} \text{ or } m < -\sqrt{3} - \frac{1}{4} $$
There's a high chance that Wolfram Alpha's solution is correct (after testing out some $m$ value). How do I approach their solution? (Or maybe if you believe that solution is wrong, then what's the exact solution to the problem?)
| Feedback on the First Solution
Your solution is somewhat correct but not complete. In fact the question asks us to find all $m$ that the given inequality holds for all $x \in \mathbb{R}$. Your decomposition into the cases is correct, but you then tried to find all $m$ that each of the cases holds for all $x \in \mathbb{R}$, which is not a necessity to find all such $m$'s because all $x \in \mathbb{R}$ need not satisfy each cases, separately; in other words, the "all $x \in \mathbb{R}$" must lie before the decomposition.
Clarifying the above explanation, consider, for example, $m=2$. According to your conclusion, $m=2$ must not be acceptable. However, $m=2$ satisfies the inequality in the following way, for example: for $x\in [0, \infty )$ the first case holds, and for $x\in (- \infty , 0)$ the second case holds. So $m=2$ satisfies the inequality for all $x \in \mathbb{R}$.
Feedback on the Second Solution
The inequality is the same as:$$(x-1)^2+|4x-2m-\frac{1}{2}| \gt \frac{3}{2}-m$$Since the left-hand side is always positive, in order for the inequality to be always true $\frac{3}{2}-m$ has to be negative.
Your argument is not correct because for some values of $m$ the right-hand side may be positive while the left-hand side may be always greater than the right-hand side for all $x \in \mathbb{R}$ (For example, consider $m=-2$).
The Correct Solution
When you face an absolute value inequality that using some absolute value properties to solve it may lead to some misleading results, it is better first to get rid of the absolute value. So, let us solve this problem in this way as follows.$$\left |4x-2m-\frac{1}{2} \right | =\begin{cases} 4x-2m-\frac{1}{2}, & \text{if } x \ge \frac{4m+1}{8} \\ -(4x-2m- \frac{1}{2}), & \text{if } x \lt \frac{4m+1}{8} \end{cases}.$$ So, the original problem is now translated into finding all $m$ for which the following inequalities hold:$$\begin{cases} 4x-2m-\frac{1}{2} \gt -x^2+2x+\frac{1}{2}-m , & \text{if } x \ge \frac{4m+1}{8} \\ -(4x-2m- \frac{1}{2}) \gt -x^2+2x+\frac{1}{2}-m, & \text{if } x \lt \frac{4m+1}{8} \end{cases}.$$Solving each inequality, we have$$\begin{cases} x^2+2x-m-1 \gt 0, & \text{if } x \ge \frac{4m+1}{8} \\ x^2-6x+3m \gt 0, & \text{if } x \lt \frac{4m+1}{8} \end{cases}.$$As we know, a quadratic $x^2+bx+c$ is positive for all $x \in \mathbb{R}-[r_1,r_2]$, where $r_1$ and $r_2$ are the roots of the quadratic. The roots of the first quadratic are$$r_1=-1-\sqrt{m+2}, \qquad r_2=-1+\sqrt{m+2}.$$Since the first inequlity concerns the $x$'s greater than $\frac{4m+1}{8}$, we need to consider the $x$'s greater than $r_2$ to satisfy the first inequality. So, all possible $m$'s for which the first inequality holds can be obtained as follows.$$\frac{4m+1}{8}= -1+\sqrt{m+2} \quad \Rightarrow \quad 16m^2+8m-47=0$$$$\Rightarrow \quad m_1=-\frac{1}{4}-\sqrt{3}, \qquad m_2=-\frac{1}{4}+\sqrt{3}$$Only $m=m_2$ is acceptable because there are $m \gt m_1$ (for example, $m=0$) not satisfying the first inequality.
A similar argument can be applied to the second inequality. The roots of the second quadratic are$$r'_1=3-\sqrt{9-3m}, \qquad r'_2=3+\sqrt{9-3m}.$$Since the second inequlity concerns the $x$'s less than $\frac{4m+1}{8}$, we need to consider the $x$'s less than $r'_1$ to satisfy the second inequality. So, all possible $m$'s for which the second inequality holds can be obtained as follows.$$\frac{4m+1}{8}=3-\sqrt{9-3m} \quad \Rightarrow \quad 16m^2+8m-47=0$$$$\Rightarrow \quad m'_1=-\frac{1}{4}-\sqrt{3}, \qquad m'_2=-\frac{1}{4}+\sqrt{3}$$Only $m=m'_1$ is acceptable because there are $m \lt m'_2$ (for example, $m=0$) not satisfying the second inequality.
Thus, the following are the possible values for $m$ satisfying the original inequality:$$m \in (-\infty , -\frac{1}{4}-\sqrt{3} ) \cup (-\frac{1}{4}+\sqrt{3} , \infty ).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An A level question about partial differentiation
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My answer almost matches the requirement, but I don't understand how the negative symbol got there.
| Using the product rule gives: $$d(f(u)g(v))=f'(u)g(v)du+f(u)g'(v)dv$$
Thus I like to work your equation like this:
$$d(2x^4+xy^3+y^4)=8x^3dx+y^3dx+3xy^2dy+4y^3dy=d(10)=0$$
Now we factorize to get the result:
$(8x^3+y^3)dx+(3xy^2+4y^3)dy=0\iff \dfrac{dy}{dx}=-\dfrac{8x^3+y^3}{3xy^2+4y^3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trouble with $4\times4$ matrix determinant $$
\begin{vmatrix}
1 & -6 & 7 & 5 \\
0 & 0 & 3 & 0 \\
3 & -2 & -8 & 6\\
2 & 0 & 5 & 4\\
\end{vmatrix}
$$
Clearly I want to expand along the second row yielding:
$((-1)^5)3$ times the following matrix
$$
\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 & 6 \\
2 & 0 & 4 \\
\end{vmatrix}
$$
and then breaks down into several smaller matrices:
2 times
$$
\begin{vmatrix}
-6 & 5 \\
-2 & 6 \\
\end{vmatrix}
$$
and 4 times
$$
\begin{vmatrix}
1 & -6 \\
3 & -2 \\
\end{vmatrix}
$$
which should come out to be $-3[(2(-36+10))+(4(-2+18))]$
$-3[(2(-16))+(4(16))]$
$-3(-32+64)=32 \times -3$
but the answer is -36 I don't know what went wrong?
| Note that, in your $3\times 3$ matrix, you could have subtracted twice the first column to the last, obtaining instantly the determinant:
$$\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 & 6 \\
2 & 0 & 4 \\
\end{vmatrix}=\begin{vmatrix}
1 & -6 & 3 \\
3 & -2 & 0 \\
2 & 0 & 0 \\
\end{vmatrix}=3\cdot 4\quad_\text{(expanding along the last column),}$$
whence the final determinant $-3\cdot 12=-36$.
| {
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Find the area between the graphs of $f(x)=e^{.25x}$ and $g(x)=\frac{-1}{x}$ between $x=1$ and $x=2$ 5.5
Can somebody verify this solution for me?
Find the area between the graphs of $f(x)=e^{.25x}$ and $g(x)=\frac{-1}{x}$ between $x=1$ and $x=2$
Since $f(x) > g(x)$ on $(1,2)$, the area between the graphs is:
$\int_1^2 e^{.25x} - (\frac{-1}{x})dx$
$=\int_1^2 e^{.25x} + (\frac{1}{x})dx$
$= \frac{e^{.25x}}{.25} + ln(|x|)|_1^2dx$
$= \frac{e^{.25(2)}}{.25} + ln(2) - \frac{e^{.25(1)}}{.25} - ln(1)$
$= \frac{e^{.25(2)}}{.25} + ln(2) - \frac{e^{.25(1)}}{.25} - ln(1)$
Since $ln(1)=0$, we get:
$= 4e^{.5} + ln(2) - 4e^{.25}$
| Well, area is an absolute thing. So we know that the total area $\mathcal{A}$ is given by:
$$\mathcal{A}=\int_1^2\exp\left(\frac{x}{4}\right)\space\text{d}x+\left|\int_1^2-\frac{1}{x}\space\text{d}x\right|=$$
$$\left[4\exp\left(\frac{x}{4}\right)\right]_1^2+\left|\left[-\ln\left|x\right|\right]_1^2\right|=$$
$$4\exp\left(\frac{2}{4}\right)-4\exp\left(\frac{1}{4}\right)+\left|-\ln\left|2\right|-\left(-\ln\left|1\right|\right)\right|=$$
$$4\exp\left(\frac{1}{2}\right)-4\exp\left(\frac{1}{4}\right)+\left|-\ln\left|2\right|-\left(-0\right)\right|=$$
$$4\exp\left(\frac{1}{2}\right)-4\exp\left(\frac{1}{4}\right)+\left|-\ln\left|2\right|\right|=$$
$$4\exp\left(\frac{1}{2}\right)-4\exp\left(\frac{1}{4}\right)+\ln\left(2\right)\approx2.15193\tag1$$
So, yes you're right.
| {
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Sum of squared Fresnel sine integral I'm trying to find the following sum:
$$ \sum_{n=0}^{\infty} \frac{S\left(\sqrt{2n}\right)^2}{n^3}$$
where $S(n)$ is the fresnel sine integral, however, I think I made a mistake somewhere.
To start, I considered using parseval's identity:
$$ 2\pi\sum_{n=-\infty}^{\infty} |c_n|^2 = \int_{-\pi}^{\pi} |f(x)|^2 \space dx$$
where $f(x)$ is:
$$ f(x) = \sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} $$
$c_n$ becomes:
$$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} \right ) e^{-inx}\space dx $$
This integral is complicated, so I plugged it into wolfram alpha and found that
$$ c_n = \frac{1}{2\pi} \left(-\sqrt{2\pi} \cdot \frac{S\left(\sqrt{2n}\right)}{n^{3/2}} \right)$$
so,
$$ |c_n|^2 = \frac{1}{4\pi^2} \left(2\pi \cdot \frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \frac{1}{2\pi} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right)$$
I think $|c_n|^2$ is finite for all n and is an even function of n. If this is true, then parseval's identity gives:
$$ 2\pi\sum_{n=-\infty}^{\infty} \frac{1}{2\pi} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx$$
and if $|c_n|^2$ is even then this expression becomes:
$$ 2\sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx$$
I believe that
$$ \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx = \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right)^2 \space dx$$
and if I plug in the second integral into wolfram alpha again, I find that (EDIT user Claude Leibovici correctly found that):
$$ \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right)^2 \space dx = \pi^2$$
So, in total I have:
$$ 2\sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \pi^2$$
or
$$ \sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \frac{\pi^2}{2}$$
The problem is that wolfram alpha suggests that the sum approaches .549, but my answer is ~4.93. Where did I make a mistake?
| This is not answer.
I am stuck with the problem but I have a few remarks
*
*I suppose that the summation starts at $n=1$ and not at $n=0$
*Using Wolfran Alpha (see here)
$$\int_{-\pi}^\pi\left(\frac{\sqrt{-i x}}{\sqrt{2}}+\frac{\sqrt{i x}}{\sqrt{2}}\right)^2\,dx= \pi^2$$
*Using Wolfran Alpha (see here)
$$\int_{-\pi}^\pi\left|\frac{\sqrt{-i x}}{\sqrt{2}}+\frac{\sqrt{i x}}{\sqrt{2}}\right|^2 \,dx= \pi^2$$
| {
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Probelm in solving $\sin\left(\frac{x}{2}\right) - \cos\left(\frac{3x}{2}\right) = 0$ While solving $\sin\left(\frac{x}{2}\right) - \cos\left(\frac{3x}{2}\right) = 0$, if I convert $\sin$ into $\cos$, I am getting answer (i.e. $\left(n + \frac{1}{4}\right)\pi$ and $\left(2n - \frac{1}{2}\right)\pi$. However, if I convert $\cos$ to $\sin$, I am not getting answer (i.e. $\left(\frac{1}{2} - n\right)\pi$ and $\left(\frac{n}{2} +\frac{1}4{}\right)\pi$).
Where am I wrong?
Is there any specific way to solve these equations?
| Write $\sin \frac x2=-\cos(\frac\pi2+\frac x2)$ and factorize with $\cos a+\cos b =2\cos\frac{a+b}2 \cos\frac{a-b}2$,
$$\sin\frac x2 - \cos\frac{3x}2
= -2\cos(x+\frac\pi4)\cos(\frac x2-\frac\pi4)=0$$
which leads to the solutions
$$x= \frac\pi4+n\pi,\> -\frac{\pi}2+2n\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622038",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Nice integral $\int_{0}^{\infty}\ln\Big(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\Big)\frac{1}{x}dx=-\frac{3\pi^2}{4}$ Last integral of the day :
$$\int_{0}^{\infty}\ln\Big(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\Big)\frac{1}{x}dx=-\frac{3\pi^2}{4}$$
I have tried integration by parts and some obvious substitution but I failed.
I have tried moreover the Feynmann trick without success .WA get an antiderivative .But I don't see how to get that . I'm not against complex integration if it's necessary but I would like a detailed answer in this case.
So if you have nice ideas...
Thanks a lot for your contributions !
| Let $\displaystyle I = \int_0^{\infty} \ln \left( \frac{x^3 - x^2 - x + 1}{x^3 + x^2 + x + 1}\right)\frac{1}{x} dx$. We can write,
\begin{align}
I & = \int_0^{\infty} \ln \left( \frac{(x-1)^2(x+1)}{(x+1)(x^2 + 1)}\right)\frac{1}{x} dx \\
& = \int_0^{1} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx + \int_1^{\infty} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx \\
& = \int_0^{1} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx + \int_0^{1} \ln \left( \frac{(1-u)^2}{u^2 + 1}\right)\frac{1}{u} du \\
& = 2\int_0^{1} \ln \left( \frac{(1 - x)^2}{x^2 + 1}\right)\frac{1}{x} dx \\
& = 4\int_0^{1} \frac{\ln \left( 1 - x\right)}{x} dx - 2\int_0^{1} \frac{\ln \left( 1+ x^2\right)}{x} dx \\
& = 4\int_0^{1} \frac{\ln \left( 1 - x\right)}{x} dx - \int_0^{1} \frac{\ln \left( 1+ t\right)}{t} dt
\end{align}
The third step uses the substitution $u = 1/x$ and the last one uses $t = x^2$.
Consider,
\begin{align}
I_1 & = \int_0^{1} \frac{\ln \left( 1 - x\right)}{x} dx \\
& = -\sum_{r = 1}^{\infty} \int_0^1 \frac{x^{r-1}}{r} dx \\
& = -\sum_{r = 1}^{\infty} \frac{1}{r^2} \\
& = -\frac{\pi^2}{6}
\end{align}
Similarly, we can write, $\displaystyle \int_0^{1} \frac{\ln \left( 1+ x\right)}{x} dx = \frac{\pi^2}{12}$.
Plugging the values, we get that $\displaystyle I = -\frac{3 \pi^2}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim\limits_{n\to\infty}\frac1{n^4}·\left[1·\sum\limits_{k=1}^nk+2·\sum\limits_{k=1}^{n-1}k+3·\sum\limits_{k=1}^{n-2}k+\cdots+n·1\right]$
Find the value of$$L=\lim_{n\to\infty}\frac1{n^4}\cdot\left[1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1\right].$$
My attempt is as follows:
$$S=\sum_{i=1}^{n}i\sum_{j=1}^{n-i+1}j$$
$$S=\sum_{i=1}^{n}i\cdot\dfrac{(n-i+1)(n-i+2)}{2}$$
$$S=\dfrac{1}{2}\sum_{i=1}^{n}i\cdot(n^2-ni+2n-in+i^2-2i+n-i+2)$$
$$S=\dfrac{1}{2}\sum_{i=1}^{n}i\cdot(n^2+i^2-2ni-3i+3n+2)$$
$$S=\dfrac{1}{2}\left(\dfrac{n^3(n+1)}{2}+\dfrac{n^2(n+1)^2}{4}-\dfrac{n(n+1)(2n+1)(2n+3)}{6}+\dfrac{n(n+1)(3n+2)}{2}\right)$$
$$S=\dfrac{n(n+1)}{4}\left(n^2+\dfrac{n(n+1)}{2}-\dfrac{(2n+1)(2n+3)}{3}+3n+2\right)$$
$$S=\dfrac{n(n+1)}{4}\left(\dfrac{6n^2+3n(n+1)-2(2n+1)(2n+3)+6(3n+2)}{6}\right)$$
$$S=\dfrac{n(n+1)}{4}\left(\dfrac{9n^2+3n-2(4n^2+8n+3)+18n+12}{6}\right)$$
$$S=\dfrac{n(n+1)}{4}\left(\dfrac{n^2+5n+6}{6}\right)$$
$$S=\dfrac{n(n+1)(n+2)(n+3)}{24}$$
$$L=\lim_{n\to\infty}\frac1{24}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)\left(1+\dfrac{3}{n}\right)}$$
hence answer is $\dfrac{1}{24}$.
But is there any shorter way to do this as in the last we got $S=\dfrac1{24}n(n+1)(n+2)(n+3)$, which is a nice closed expression.
| WE can write it as $$L=\lim_{n \to \infty} \frac{1}{n^4}\sum_{j=1}^{n} j \sum_{k=1}^{n-j+1} k =\lim_{n \to \infty} \frac{1}{n^4}\sum_{j=1}^{n} \frac{j(n-j+1)(n-j+2)}{2}$$ $$L = \lim_{n \to \infty} \frac{1}{2n^4}\sum_{j=1}^n(j^3-2nj^2+n^2j+....)$$
Let us use the asymptotic result when $n$ is very large: $\sum_{k=1}^n k^z \sim\frac{n^{z+1}}{z+1}$
we get $$L=\lim_{n \to \infty} \frac{1}{2n^4} (\frac{n^4}{4}-2\frac{n^4}{3}+\frac{n^4}{2})=\frac{1}{24}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
Solving the line for $y$ we get $y=\frac{4}{3}x+\frac{20}{3}$
Substituting this to the equation of the circle we get $(x-2)^2+(\frac{4}{3}x+\frac{20}{3}-1)^2=2^2$, but solving this for $x$ ended up with no real roots. What am I missing here?
| What you would do is draw a perpendicular from the centre of the circle to that line.
That perpendicular passes through $C=(2,1)$ and is (obviously) perpendicular to $y=\dfrac{4}{3}x+\dfrac{20}{3}$
Line has a gradient of $-\dfrac{3}{4}$
$\begin{align} \dfrac{y-1}{x-2} &= -\dfrac{3}{4} \\ y&=-\dfrac{3}{4}x +\dfrac{5}{2}\end{align}$
The two lines intersect at $P=(-2, 4)$ which you can verify by solving this:
$\dfrac{4}{3}x+\dfrac{20}{3}=-\dfrac{3}{4}x +\dfrac{5}{2}$
The distance $PC$ is:
$\sqrt{(2-(-2))^2+(4-1)^2}=5$
So the shortest distance to the circumference is $PC$ minus the radius of the circle which is:
$5-2=\boxed{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Simple factorization problem I cannot find the solution! This is not a homework assignment.
$11x^2+13x-7$
| Do you know about completing squares? Write your expression as $$11\left(x^2+\frac{13}{11}x\right)-7.$$ The goal is to write this as a difference of squares, which is easy to factorise. To this end, complete the square on the expression in brackets, which becomes $$x^2+2x\frac{13}{22}+\left(\frac{13}{22}\right)^2-\left(\frac{13}{22}\right)^2=\left(x+\frac{13}{22}\right)^2-\left(\frac{13}{22}\right)^2.$$
Thus our expression becomes $$11\left(x+\frac{13}{22}\right)^2-11\left(\frac{13}{22}\right)^2-7=\left(\sqrt {11}\left(x+\frac{13}{22}\right)\right)^2-\frac{169}{44}-7=\left(\sqrt {11}\left(x+\frac{13}{22}\right)\right)^2-\left(\sqrt{\frac{477}{44}}\right)^2,$$ from where you should now be able to continue.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $\sum^N_{n=0}\frac12\leq\sum^N_{n=0}\sum^{2^{n+1}-1}_{r=2^n}\frac1r\leq\sum^N_{n=0}1=\frac{N+1}2\leq\sum^{2^{N+1}-1}_{r=1}\frac1r\leq N+1$?
Consider the following inequalities:
$$\frac{1}{2} \leq \sum^{2^{n+1}-1}_{r=2^n}\frac{1}{r}\leq 1 \tag{1}$$
Upon summing over $(1)$ from $n=0$ to $n=N$, we obtain
$$\sum^N_{n=0}\frac{1}{2}\leq \sum^N_{n=0}\sum^{2^{n+1}-1}_{r=2^n}\frac{1}{r}\leq\sum^N_{n=0}1 \tag{2}$$
or equivalently
$$\frac{N+1}{2}\leq\sum^{2^{N+1}-1}_{r=1}\frac{1}{r}\leq N+1 \tag{3}$$
But I do not see how one can proceed from the double sum in $(2)$ to the single sum in $(3)$?
I tried to make sense of $(3)$ by directly expanding the summation terms in $(2)$:
$$S=\sum^N_{n=0}\sum^{2^{n+1}-1}_{r=2^n}\frac{1}{r}=\sum^N_{n=0}\left(\frac{1}{2^n}+\frac{1}{2^n+1}+...+\frac{1}{2^{n+1}-2}+\frac{1}{2^{n+1}-1} \right) \tag{4}$$
However, further expansion followed by applying the sum to each term of $(4)$ individually led to the following predicaments:
$1.$ Sums like
$$\sum^N_{n=0}\left(\frac{1}{2^{n+1}-4}\right) \tag{5}$$
contain one or more terms that are undefined and so do not make any sense.
*Can we circumvent this by ignoring terms that appear before and including the 'trouble-causing' term, e.g. in the case of $(5)$
$$\sum^N_{n=0}\left(\frac{1}{2^{n+1}-4}\right) \to \sum^N_{n=2}\left(\frac{1}{2^{n+1}-4}\right) \tag{6}$$ ?
$2.$ There are repeated terms generated by $(4)$ and they do not cancel,and are therefore inconsistent with $(3)$?
e.g. $1$ appears twice due to $\sum^N_{n=0} \left(\frac{1}{2^n}\right)$ and $\sum^N_{n=0} \left(\frac{1}{2^n+1} \right)$
Can someone please explain where my conceptual errors lie?
| In fact it is a $\underline{\textbf{telescopic sum}}$, let $ N\in\mathbb{N} $, denoting for any $ n\in\mathbb{N} $, $ S_{n}=\sum\limits_{k=1}^{2^{n}-1}{\frac{1}{k}} $, we have : \begin{aligned}\sum_{n=0}^{N}{\sum_{r=2^{n}}^{2^{n+1}-1}{\frac{1}{r}}}&=\sum_{n=0}^{N}{\left(\sum_{r=1}^{2^{n+1}-1}{\frac{1}{r}}-\sum_{r=1}^{2^{n}-1}{\frac{1}{r}}\right)}\\ &=\sum_{n=0}^{N}{\left(S_{n+1}-S_{n}\right)}\\ &=S_{N+1}-S_{0}\\\sum_{n=0}^{N}{\sum_{r=2^{n}}^{2^{n+1}-1}{\frac{1}{r}}}&=\sum_{n=0}^{2^{N+1}-1}{\frac{1}{r}}\end{aligned}
| {
"language": "en",
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"source": "stackexchange",
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}} = -6$.
I can show my attempt but it's pointless as the result i achieved $+\infty$, which is very wrong. is there an strategy trying to calculate a limit without l'hospital rule.
| The expression tends to $$\frac{2\cdot1-\sqrt{1^2+3}}{\sqrt{1+3}+\sqrt{2\cdot1+2}}=0.$$
You may not use L'Hospital here.
(Or is there a typo in the question ?)
Update:
$$\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}=\frac{\sqrt{x+3}+\sqrt{2x+2}}{2x+\sqrt{x^2+3}}\cdot \frac{4x^2-x^2-3}{x+3-2x-2}\to\frac{4}{4}(-6).$$
| {
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"source": "stackexchange",
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Showing that $\sum_{n=1}^\infty n^{-1}\left(1+\frac{1}{2}+...\frac{1}{n}\right)^{-1}$ is divergent
How to show that $\sum_{n=1}^\infty \frac{1}{n\left(1+\frac{1}{2}+...\frac{1}{n}\right)}$ diverges?
I used Ratio test for this problem and this is the result: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1-\frac{1}{(n+1)+\frac{(n+1)}{2}+...+1}\right)= 1$$
Then I thought using abel or dirichlet test. But I couldn't solve it.
| First note that
$$
\sum\limits_{k = 1}^n {\frac{1}{k}} \le 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t} } } = 1 + \int_1^n {\frac{{dt}}{t}} = 1 + \log n,
$$
for all $n\geq 1$. Then for all $N\geq 3$,
\begin{align*}
& \sum\limits_{n = 1}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}} > \sum\limits_{n = 3}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}} \ge \sum\limits_{n = 3}^N {\frac{1}{{n(\log n + 1)}}} \\ & \ge \sum\limits_{n = 3}^N {\frac{1}{{n(\log n + \log n)}}} = \frac{1}{2}\sum\limits_{n = 3}^N {\frac{1}{{n\log n}}}
\\ &
\ge \frac{1}{2}\sum\limits_{n = 3}^N {\int_{n}^{n+1} {\frac{{dt}}{{t\log t}}} } = \frac{1}{2}\int_3^{N+1} {\frac{{dt}}{{t\log t}}} \\ & = \frac{{\log \log( N+1) - \log \log 3}}{2} .
\end{align*}
This shows that the series diverges.
| {
"language": "en",
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"source": "stackexchange",
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Find an ellipse tangent to triangle given centre position of the ellipse I have a fully defined triangle, $PT_1T_2$, of side lengths $a$, $b$, & $c$ with corresponding opposite angles $A$, $B$, & $C$. I want to find an ellipse that is tangent to the lines $PT_1$ and $PT_2$ at points $T_1$ and $T_2$ respectively.
As there are many ellipses that satisfy this condition, I impose that the centre point, $O$, of the ellipse is at a distance $d$ from point $P$ and that the line $OP$ is at an angle $D$ from the line $PT_1$, where $0<D<A$.
Given these conditions, can I find the focus of the ellipse and the angle of the major axis relative to the line $PO$? What are the equations for these?
| I like the more-geometric answers, but I'm somewhat committed to this analytic approach, so let's continue ...
Place $O$ at the origin, and let $P = (-d,0)$. As @Aretino helpfully observes in a comment, $\overleftrightarrow{OD}$ must bisect the chord $\overline{T_1T_2}$, so define $M$ as midpoint of the chord, with $|PM|=m$, so that $M=(-d+m,0)$; also, define $\theta = \angle OMT_1 = \beta+\delta$. Then, with $a := |T_1T_2|$, we can write
$$T_1 = M + \frac{a}2(\cos\theta,\sin\theta) \qquad T_2 = M - \frac{a}2(\cos\theta,\sin\theta) \tag{1}$$
Since $O$ is the center of the ellipse, we can define, say, $T_3=-T_1$ to get a third point on the ellipse. Since $T_1$ and $T_2$ are points of tangency, we can tease them into double-points by defining
$$T_1' = T_1 + t_1(P-T_1) \qquad T_2' = T_2 + t_2(P-T_2) \tag{2}$$
for "infinitesimally small" $t_1$ and $t_2$ that we can treat as non-zero or zero as to our advantage.
With five points on the conic, we can use a determinant to get its equation. (See, for instance, this answer.) Using a computer algebra system such as Mathematica to expand the determinant, we obtain factors of $t_1$ and $t_2$ that we divide-out (because they're non-zero), then set the remaining instances of these values to $0$ (because they're not non-zero), so that the equation becomes ...
$$\begin{align}
0 &= x^2 a^2 \sin^2\theta
- 2 a^2 x y \cos\theta \sin\theta
+ y^2\left( 4 m (d-m) + a^2 \cos^2\theta \right) \\
&- a^2d\sin^2\theta \left( d - m \right)
\end{align}\tag{3}$$
From here, we can consult, say, this answer for recipes that express metric properties of a conic in terms of the coefficients of the general second-degree polynomial. We find that the angle $\phi$ that the major axis makes with the $x$ axis satisfies
$$\tan2\phi = \frac{a^2 \sin 2\theta}{4 (d - m) m + a^2 \cos 2\theta} \tag{4}$$
and the major and minor radii of the ellipse are given by
$$r_{\pm}^2 = \frac{d}{8m}\left( a^2 + 4m (d - m) \pm \sqrt{ a^4 + 16 m^2 (d - m)^2 + 8 a^2 m (d - m) \cos 2\theta )}\right) \tag{5}$$
To rewrite in terms of the problem's stated parameters, the Law of Cosines gives
$$a^2 = b^2 + c^2 - 2 b c\cos\alpha \tag{6}$$
where $\alpha := \angle T_1PT_2 = 180^\circ - \beta - \gamma$. We can use Stewart's Theorem to show
$$m^2 = \frac14\left(-a^2+2b^2+2c^2\right) = \frac14\left( b^2 + c^2 + 2b c \cos\alpha\right) \tag{7}$$
Also, since $|\triangle PT_1T_2| = \frac12b c \sin\alpha = \frac12am\sin\theta$, we can find
$$\begin{align}
\sin^2\theta &= \frac{4b^2 c^2 \sin^2\alpha}{(b^2+c^2-2b c\cos\alpha)(b^2+c^2+2b c \cos\alpha)} = \frac{4b^2c^2\sin^2\alpha}{(b^2-c^2)^2+4b^2c^2\sin^2\alpha} \tag{8} \\[4pt]
\cos^2\theta &= \frac{(b^2-c^2)^2}{(b^2-c^2)^2+4b^2c^2\sin^2\alpha} \tag{9} \\[4pt]
\cos2\theta &= \frac{(b^2-c^2)^2-4b^2c^2\sin^2\alpha}{(b^2-c^2)^2+4b^2c^2\sin^2\alpha} \tag{10}
\end{align}$$
Expressions $(4)$ and $(5)$ don't seem to get appreciably better by substituting-in from these expressions and simplifying (the arbitrary $d$ gets in the way), so I'll leave things here. $\square$
| {
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"question_score": "2",
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Closed form of the recursive function $F(1):=1,\;F(n):=\sum_{k=1}^{n-1}-F(k)\sin\left(\pi/2^{n-k+1}\right)$ Suppose that $F$ is defined via the recurrence relation
$$F(1)=1, \qquad F(n)=\sum_{k=1}^{n-1}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr)$$ What is $F(N)$? I don't have any idea how to solve this problem. Only one thing that I've noticed is that: $$ 0=\sum_{k=1}^{n}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr).$$
Edit:
From the comment section:
I was trying to rewrite $\sin\left(\frac{\pi}{2^n}\right)$ by the formula for a double argument and I've ended up with $$\sin\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2}\right)}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)}=\frac{1}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)},n\gt 1,$$
but it doesn't seem to help.
$F$ is used in another formula. It should be true for most of the functions
$$g(x)=\sum_{n=1}^\infty \left(\sin\left(x2^{n-1}\right)\sum_{k=1}^n\left(F(k)g\left(\frac{\pi}{2^{n-k+1}}\right)\right)\right),\;x\in\left(0,\frac{\pi}{2}\right)$$
First four values of $F$ are:
\begin{align*}F(1)&=1\\F(2)&=-\sin\left(\frac{\pi}{4}\right)\\F(3)&=-\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{4}\right)\\F(4)&=-\sin\left(\frac{\pi}{16}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)-\sin^3\left(\frac{\pi}{4}\right)\end{align*}
These terms look like if they created some pattern, but the fifth term which is
\begin{align*}F(5)=-\sin\left(\frac{\pi}{32}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{16}\right)-3\sin^2\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)+\sin^4\left(\frac{\pi}{4}\right)\end{align*}
meses the pattern up.
| By putting
$$
G(n) = F(n + 1)
$$
we can rewrite the recurrence as
$$ \bbox[lightyellow] {
\sum\limits_{k = 0}^n {G(k)\sin \left( {{{\pi /2} \over {2^{\,n - k} }}} \right)}
= \sum\limits_{k = 0}^n {G(n - k)\sin \left( {{{\pi /2} \over {2^{\,k} }}} \right)} = \left[ {0 = n} \right]
} \tag{1}$$
where $[P]$ denotes the Iverson bracket,
here equivalent to the Kronecker delta.
The recurrence can be unfolded to give
$$
\left\{ \matrix{
G(0) = 1 \hfill \cr
G(n) = - {1 \over {\sin \left( {\pi /2} \right)}}
\sum\limits_{k = 1}^n {G(n - k)\sin \left( {{{\pi /2} \over {2^{\,k} }}} \right)} \hfill \cr} \right.
$$
Recursion (1) clearly indicates that we are dealing with a convolution, in particular with a multiplicative inversion (aka. reversion) in terms
of power series.
So if we put
$$
H(x,y) = \sum\limits_{0\, \le \,n} {\sin \left( {{x \over {2^{\,n} }}} \right)y^{\,n} }
$$
then we readily have the ogf for $G(n)$ as
$$
\eqalign{
& \sum\limits_{0\, \le \,n} {G\left( n \right)y^{\,n} } = {1 \over {H(\pi /2,y)}}
= {1 \over {\sum\limits_{0\, \le \,n} {\sin \left( {{\pi \over {2^{\,n + 1} }}} \right)y^{\,n} } }} = \cr
& = 1 - {{\sqrt 2 } \over 2}y + \left( {{{1 - \sqrt {2 - \sqrt 2 } } \over 2}} \right)y^{\,2} + \cr
& - \left( {\sin \left( {{\pi \over {16}}} \right)
+ {{\sqrt 2 \left( {1 - 2\sqrt {2 - \sqrt 2 } } \right)} \over 4}} \right)y^{\,3}
+ O\left( {y^{\,4} } \right) \cr}
$$
To try and find a closed form for $G$ let's try one of the various reversion approaches, which hing
on viewing identity (1) as a system of linear equations with $G(n)$ as unknowns.
$$ \bbox[lightyellow] {
\left( {\matrix{
{\sin \left( {x/2^{\,0} } \right)} & 0 & 0 & 0 \cr
{\sin \left( {x/2^{\,1} } \right)} & {\sin \left( {x/2^{\,0} } \right)} & 0 & 0 \cr
{\sin \left( {x/2^{\,2} } \right)} & {\sin \left( {x/2^{\,1} } \right)} & {\sin \left( {x/2^{\,0} } \right)} & 0 \cr
\vdots & \ddots & \ddots & \ddots \cr
} } \right)\left( {\matrix{
{G(0)} \cr
{G(1)} \cr
{G(2)} \cr
\vdots \cr
} } \right) = \left( {\matrix{
1 \cr
0 \cr
0 \cr
\vdots \cr
} } \right)
} \tag{2}$$
where the matrix is a lower triangular Toeplitz matrix.
However it doesn't look that these methods might lead to a closed expression for the $G(n))$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Definite Integral of $\int_{\frac{-1}{2}}^\frac{1}{2}\int_{\frac{-1}{2}}^\frac{1}{2} \sqrt{x^2+y^2} dxdy$ Here's my attempt at trying to evaluate the integral.
Let $x = y tan\theta$
$$\frac{dx}{d\theta} = \frac{y}{cos^2\theta}$$
$$dx = \frac{y}{cos^2\theta}d\theta$$
The new bounds of inner integral would be $\theta = tan^-(\frac{1}{2y}) $ and $\theta = tan^-(\frac{-1}{2y})$
$$\int_{\frac{-1}{2}}^\frac{1}{2}\int_{tan^-(\frac{-1}{2y})}^{tan^-(\frac{1}{2y})} y^2 sec^3{\theta} dy$$
Evaluating the innermost integral
$$\int_{tan^-(\frac{-1}{2y})}^{tan^-(\frac{1}{2y})} sec^3{\theta} dy = {\huge|} \frac{sec\theta tan\theta + ln|sec\theta + tan\theta|}{2}{\huge|}_{tan^-(\frac{-1}{2y})}^{tan^-(\frac{1}{2y})}$$
$$\begin{multline} = \left( \frac{sec(tan^-(\frac{1}{2y})) tan(tan^-(\frac{1}{2y})) + ln|sec(tan^-(\frac{1}{2y})) + tan(tan^-(\frac{1}{2y}))|}{2}\right)
- \\ \left( \frac{sec(tan^-(\frac{-1}{2y})) tan(tan^-(\frac{-1}{2y})) + ln|sec(tan^-(\frac{-1}{2y})) + tan(tan^-(\frac{-1}{2y}))|}{2}\right) \end{multline}$$
$$\begin{multline} = \left( \frac{\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} + 1}{2y}|}{2}\right)
- \left( \frac{-\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} - 1}{2y}|}{2}\right) \end{multline}$$
$$\begin{multline} = \frac{1}{2}\left[\left(\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} + 1}{2y}|\right)
- \left(-\frac{\sqrt{4y^2 + 1}}{4y^2} + ln|\frac{\sqrt{4y^2 + 1} - 1}{2y}|\right) \right] \end{multline}$$
$$\begin{multline} = \frac{1}{2}\left[\left(\frac{\sqrt{4y^2 + 1}}{2y^2} + ln|\sqrt{4y^2 + 1} + 1|
- ln|\sqrt{4y^2 + 1} - 1| \right) \right] \end{multline}$$
$$\begin{multline} = \frac{1}{2}\left[\left(\frac{\sqrt{4y^2 + 1}}{2y^2} + ln|\frac{\sqrt{4y^2 + 1} + 1}{\sqrt{4y^2 + 1} - 1}| \right) \right] \end{multline}$$
Evaluating the outermost integral
$$\begin{multline} \int_{\frac{-1}{2}}^\frac{1}{2}\frac{\sqrt{4y^2 + 1}}{4} + y^2 \frac{ln{\large|}\frac{\sqrt{4y^2 + 1} + 1}{\sqrt{4y^2 + 1} - 1}{\large|}}{2} dy \end{multline}$$
I am kinda stuck at this point, any help is greatly appreciated -:). I may have done something wrong in the steps above.
| We can use symmetry to say that
$$I = \int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{x^2+y^2}\:dxdy = 8 \int_{0}^{\frac{1}{2}}\int_{0}^{x}\sqrt{x^2+y^2}\:dydx$$
From here, we can use polar coordinates:
$$I = 8 \int_{0}^{\frac{\pi}{4}} \int_0^{\frac{1}{2}\sec\theta}r^2\:drd\theta = \frac{1}{3}\int_{0}^{\frac{\pi}{4}}\sec^3\theta\:d\theta$$
Now, we could use integration by parts and trig identities to solve this integral, but instead let's use the substitution $\tan \theta = \sinh \tau$:
$$I = \frac{1}{3}\int_0^{\sinh^{-1}(1)}\cosh^2 \tau\:d\tau = \frac{1}{6}\int_0^{\sinh^{-1}(1)} 1 + \cosh 2\tau \:d\tau$$
$$= \frac{1}{6}+\frac{1}{6}\sinh\tau\cosh\tau \Biggr|_0^{\sinh^{-1}(1)} = \frac{\sinh^{-1}(1)+\sqrt{2}}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$$\int_0^1\frac{\ln(x)\text{Li}_2(x)}{1-ax}dx=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$
multiply both sides by $\frac{a}{3}$ then replace $a$ by $1+i$ and consider the the real parts of both sides we have
$$\Re\ \text{Li}_4(1+i)=-\frac16\Re\ \text{Li}_2^2(1+i)+\frac23\zeta(2)\Re\ \text{Li}_2(1+i)+\frac13\Re \int_0^1\frac{(1+i)}{1-(1+i)x}\ln(x)\text{Li}_2(x)dx$$
For the integral, use $\Re\frac{1+i}{1-(1+i)x}=\frac{1-2x}{2x^2-2x+1}$ which gives $I$.
What I tried is subbing $1-2x=y$ which gives
$$I=\int_{-1}^1\frac{-y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\int_{-1}^1 f(y)dy=\underbrace{\int_{-1}^0 f(y)dy}_{y\to\ -y}+\int_{0}^1 f(y)dy$$
$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy-\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy$$
I think I made it more complicated. Any help would be appriciated.
| $$\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\operatorname{Li}_4\left(\frac12\right)-\frac{\pi^4}{30}+\frac{21}8\ln2\zeta(3)-\frac{\pi^2}{12}\ln^22$$
$$\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=\frac{C^2}{2}+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{701 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{5 \log ^4(2)}{128}-\frac{3}{64} \pi ^2 \log ^2(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx$ Evaluate $$\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx\,,$$ where $a$ being parameter. I am not able to solve this.
| Let $$I(a) = \int_0^{1}\frac{\arctan {ax}}{x\sqrt{1-x^2}}dx$$
Differentiating the integral with respect to $a$,
$$I'(a) = \int_0^{1}\frac{dx}{\sqrt{1-x^2}(1+a^2x^2)}$$
Let, $$u^2 = \frac{1-x^2}{1+a^2x^2} \implies x^2=\frac{1-u^2}{1+a^2u^2} \therefore xdx=-\frac{u(1+a^2)}{(1+a^2u^2)^2}du$$
Therefore, the integral reduces to,
$$I'(a)=\frac{1}{\sqrt{1+a^2}}\int_0^{1}\frac{du}{\sqrt{1-u^2}}$$
$$\therefore I'(a)=\frac{\pi}{2\sqrt{1+a^2}} \implies I(a)=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|+C$$ where C is an undetermined constant that can be found by putting $a=0$ in the original integral which evaluates to $0$. Therefore, $C=0$ and the integral is
$$I(a)=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Functional equation: $ f : \mathbb R ^ * \to \mathbb R $ with $ \frac 1 x f ( - x ) + f \left( \frac 1 x \right) = x $ I tried to solve this problem:
Determine all function $ f : \mathbb R ^ * \to \mathbb R $ such that $ \forall x \in \mathbb R ^ * $
$$ \frac 1 x f ( - x ) + f \left( \frac 1 x \right) = x \text . $$
Basically I tried the classical way to substitute so I got this :
Let $ P ( x ) $ be the assertion above.
*
*$ P ( 1 ) $: we obtain $ f ( - 1 ) + f ( 1 ) = 1 $.
All what I know about this function is this information!
I don't know if there's another technique to simplify it or reduce it. Some help please!
| Fix $a \neq 0$. Then plugging in respectively $x = a$ and $x = -\frac{1}{a}$, we get,
\begin{align*}
\frac{1}{a}f(-a) + f\left(\frac{1}{a}\right) &= a \\
-af\left(\frac{1}{a}\right) + f(-a) &= -\frac{1}{a}.
\end{align*}
This is a system of linear equations in terms of unknowns $x = f(-a)$ and $y = f\left(\frac{1}{a}\right)$, represented by the following augmented matrix
$$\left[\begin{array}{cc|c}\frac{1}{a} & 1 & a \\ 1 & -a & -\frac{1}{a}\end{array}\right].$$
Row reducing, we get
$$\left[\begin{array}{cc|c}1 & 0 & \frac{1}{2}\left(a^2 - \frac{1}{a}\right) \\ 0 & 1 & \frac{1}{2}\left(a + \frac{1}{a^2}\right)\end{array}\right].$$
In particular, this implies
$$f(-a) = \frac{1}{2}\left(a^2 - \frac{1}{a}\right) \implies f(x) = \frac{1}{2}\left(x^2 + \frac{1}{x}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I proceed to the results? Please help me to proceed. Thanks in advance.
| We need to prove that $$\sum_{cyc}\left(a^3-a^{\frac{7}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}\right)\geq0$$ and since $$(3,0,0)\succ\left(\frac{7}{3},\frac{1}{3},\frac{1}{3}\right),$$ it's true by Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\frac{\left|\sin(nx)\right|}{\left|\sin(x)\right|}\leq\frac{\pi}{2x}$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$,we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}+\frac{\pi^{2}}{8}\left(n^{2}-1\right)<\frac{\pi^{2}n^{2}}{4}.$$
But using mathematica I found this inequality can still be improved.
And after calculating some terms I found it seems that when $n\geq 2$ we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}.$$
But I cannot prove this.So is there any method to improve my result?Any help will be thanked.
| Let $ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx $
$$ a_{n}-a_{n-1}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos(2n-2)x-\cos 2nx}{\sin^2 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} dx $$
$$ a_{n}-a_{n-1}-(a_{n-1}-a_{n-2})=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x-\sin(2n-3)x}{\sin x} dx=2\int_{0}^{\frac{\pi}{2}} \cos(2n-2)x dx=0 $$
$$ a_{n}-2a_{n-1}+a_{n-2}=0 $$
So $ a_{n} $ is an arithmetic sequence,$ a_{n}=a+bn,a_{0}=0,a_{1}=\frac{\pi}{2} $
We get
$$ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n\pi}{2} $$
Thus
$$ \int_{0}^{\frac{\pi}{2}}x \frac{\sin^4 nx}{\sin^4 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \frac{\left| \sin nx \right |}{\sin x} \left| \sin nx \right | \frac{\sin^2 nx}{\sin^2 x} dx ＜ \int_{0}^{\frac{\pi}{2}} \frac{\pi}{2} \times n \times 1 \times \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n^2 \pi^2}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Any $(x, y, z)$ can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$? Please tell me whether there any $(x, y, z)$
which can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$ ?
No process or just solve it by calculator are both fine.
Thank you.
| We have
\begin{align}
&5x^2 + 2y^2 + 6z^2 - 6xy - 2xz + 2yz\\
= \ & 2y^2 + (-6x + 2z)y + 5x^2 - 2xz + 6z^2\\
= \ & 2\left(y + \frac{-3x+z}{2}\right)^2 - 2 \left(\frac{-3x+z}{2}\right)^2
+ 5x^2 - 2xz + 6z^2\\
= \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}x^2 + xz + \frac{11}{2}z^2 \\
= \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}(x + z)^2 - \frac{1}{2}z^2 +
\frac{11}{2}z^2\\
= \ & 2\left(y + \frac{-3x+z}{2}\right)^2 + \frac{1}{2}(x+z)^2 + 5z^2.
\end{align}
Thus, $5x^2 + 2y^2 + 6z^2 - 6xy - 2xz + 2yz$ is never negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that this inequality is true Show that $\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{999998}{999999} > \frac{1}{100}$.
I tried to take another multiplication $\frac{3}{5} \cdot \frac{6}{8} \cdot \frac{9}{11} \cdot ... \cdot \frac{999996}{999998}$ so that we would have their multiplications equal to $\frac{2}{999999}$. And if we assume that first one equals to $x$, second one equals $y$ we would have an inequality like $x \gt y$ and $x^2 \gt y \cdot x$ so that we can prove that $ x \gt \frac{1}{1000}$ but I can't make it for $\frac{1}{100}$.
| Let $$A=\frac{2}{3}\cdot\frac{5}{6}\cdot...\cdot\frac{999998}{999999},$$
$$B=\frac{1}{2}\cdot\frac{4}{5}\cdot...\cdot\frac{999997}{999998}$$ and $$C=\frac{3}{4}\frac{6}{7}\cdot...\cdot\frac{999999}{1000000}.$$
Thus, since $$\left(\frac{3n+2}{3n+3}\right)^2>\frac{3n+1}{3n+2}\cdot\frac{3n+3}{3n+4},$$ we obtain: $$A^2>BC,$$ which gives
$$A^3>ABC=\frac{1}{1000000}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Polynomial $x^3-2x^2-3x-4=0$ Let $\alpha,\beta,\gamma$ be three distinct roots of the polynomial $x^3-2x^2-3x-4=0$. Then find $$\frac{\alpha^6-\beta^6}{\alpha-\beta}+\frac{\beta^6-\gamma^6}{\beta-\gamma}+\frac{\gamma^6-\alpha^6}{\gamma-\alpha}.$$
I tried to solve with Vieta's theorem. We have
$$\begin{align}
\alpha+\beta+\gamma &= 2, \\
\alpha\beta+\beta\gamma+\gamma\alpha &= -3, \\
\alpha\beta\gamma &= 4.
\end{align}$$
For example, $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=10$ and similarly, we can find $\alpha^3+\beta^3+\gamma^3$...
But it has very long and messy solution. Can anyone help me?
| Let $\alpha+\beta+\gamma=3u$, $\alpha\beta+\alpha\gamma+\beta\gamma=3v^2$ and $\alpha\beta\gamma=w^3$.
Thus, $$\sum_{cyc}\frac{\alpha^6-\beta^6}{\alpha-\beta}=\sum_{cyc}(2\alpha^5+\alpha^4\beta+\alpha^4\gamma+\alpha^3\beta^2+\alpha^3\gamma^2)=$$
$$=2(243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3)+$$
$$+81u^3v^2-81uv^4-9u^2w^3+15v^2w^3+$$
$$+27uv^4-18u^2w^3-3v^2w^3=$$
$$=9(54u^5-81u^3v^2+7u^2w^3+24uv^4-2v^2w^3).$$
Now, use your work.
I got $608$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Implementation of Cauchy product on $\cos x\cdot \sin x$ I have a mistake that I can't find somewhere along the way.
Please help me find the place things go wrong.
Find the product of $\cos x$ and $\sin x$ as defined:
$$\cos(x) = \sum_{k=0}^{\infty} \frac{\alpha_k}{k!} x^k $$
$$\sin(x) = \sum_{k=0}^{\infty} \frac{-\alpha_{k+1}}{k!} x^k $$
with $$\alpha_k = \frac{i^k}{2} \left( 1 + (-1)^k \right) $$
My attempt, using the following tricks:
1) $$ \sum_{n=0}^{\infty} \frac{z^n}{n!} = e^z $$
2) $$ e^{z+w} = \sum_{n=0}^{\infty} \frac{z^n}{n!} \cdot \sum_{n=0}^{\infty} \frac{w^n}{n!} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{z^k}{k!} \frac{w^{n-k}}{(n-k)!} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} n! \cdot \frac{z^k}{k!} \frac{w^{n-k}}{(n-k)!} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \cdot z^k \cdot w^{n-k} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n\choose k} \cdot z^k \cdot w^{n-k} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} (z+w)^n = e^{z+w} $$
Here is what I did:
$$ \cos(x) \cdot \sin(x) = \sum_{k=0}^{\infty} \frac{\alpha_k}{k!} x^k \cdot \sum_{k=0}^{\infty} \frac{-\alpha_{k+1}}{k!} x^k = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{-\alpha_{k+1}}{k!} x^k \cdot \frac{\alpha_{n-k}}{(n-k)!} x^{n-k} = \\
\sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \left[ -\alpha_{k+1} \cdot x^k \cdot \alpha_{n-k} \cdot x^{n-k} \right] = \\
\sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[-1 \cdot \frac{i^{k+1}}{2} \left( 1 + (-1)^{k+1} \right) \cdot x^k \right] \cdot \left[ \frac{i^{n-k}}{2} \left( 1 + (-1)^{n-k} \right) \cdot x^{n-k} \right] = \\
\sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[-1 \cdot \frac{1}{2i} \cdot \left( i^k + (-i)^{k} \right) \cdot x^k \right] \cdot \left[ \frac{1}{2} \left( i^{n-k} + (-i)^{n-k} \right) \cdot x^{n-k} \right] = \\
-\frac{1}{2i} \cdot \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ \left( i^k + (-i)^{k} \right) \cdot x^k \right] \cdot \left[ \left( i^{n-k} + (-i)^{n-k} \right) \cdot x^{n-k} \right] = \\
-\frac{1}{2i} \cdot \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k + (-ix)^{k} \right] \cdot \left[ (ix)^{n-k} + (-ix)^{n-k} \right] = \\
-\frac{1}{2i} \cdot \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k (ix)^{n-k} + (ix)^k (-ix)^{n-k} + (-ix)^{k} (ix)^{n-k} + (-ix)^{k} (-ix)^{n-k}\right] = \\
-\frac{1}{2i} \cdot \frac{1}{2} \cdot \left[ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k (ix)^{n-k} \right] + \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k (-ix)^{n-k} \right] + \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (-ix)^{k} (ix)^{n-k} \right] + \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (-ix)^{k} (-ix)^{n-k} \right] \right] = \\
-\frac{1}{2i} \cdot \frac{1}{2} \cdot \left[ \sum_{n=0}^{\infty} \frac{1}{n!} (ix + ix)^n + \sum_{n=0}^{\infty} \frac{1}{n!} (ix-ix)^{n} + \sum_{n=0}^{\infty} \frac{1}{n!} (-ix+ix)^{n} + \sum_{n=0}^{\infty} \frac{1}{n!} (-ix-ix)^{n} \right] = \\
$$
What do I do about
$$ \sum_{n=0}^{\infty} \frac{1}{n!} (ix-ix)^{n} $$
and
$$ \sum_{n=0}^{\infty} \frac{1}{n!} (-ix+ix)^{n} $$
Do they converge to 1 or diverge?
How do I get the expected answer:
$$ \frac{1}{2} \sin(2x) $$
Thanks
| Note that $\alpha_{k}=0$ if $k$ is odd and $(-1)^{\frac{k}{2}}$ otherwise. Applying Cauchy product yields
\begin{align*}
\displaystyle \sum_{n=0}^{\infty} \displaystyle \sum_{k=0}^{n} \frac{(-1)^k x^{2k}}{(2k)!}\cdot \frac{(-1)^{n-k}x^{2n-2k+1}}{(2n-2k+1)!}=&\displaystyle \sum_{n=0}^{\infty} \displaystyle \sum_{k=0}^{n} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} \binom{2n+1}{2k} \\
=&\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \displaystyle \sum_{k=0}^{n} \binom{2n+1}{2k} \\
=&\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \cdot 4^{n} \\
=&\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}2^{2n} \\
=&\frac{1}{2} \cdot \displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}2^{2n+1} \\
=&\frac{1}{2} \cdot \displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{(2x)^{2n+1}}{(2n+1)!}\\
=&\frac{\operatorname{sin}(2x)}{2}.
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/3677168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone? Observe that $x_1=1$ and $x_2=\dfrac{1}{\sqrt{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)>\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=1$.
Thus, $x_2>x_1$. In general, we also have $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)>\dfrac{1}{\sqrt{n}}\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\ldots+\dfrac{1}{\sqrt{n}}\right)=1$.
Thus, $x_n\geq 1$ for all $n\in \mathbb{N}$. Also, we have,
$x_{n+1}=\dfrac{1}{\sqrt{n+1}}\left(\sqrt{n}x_n+\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}$.
Is it true that $x_{n+1}>x_n$?
Edit :
Thanks to the solution provided by a co-user The73SuperBug. Proving $x_{n+1}-x_n>0$ is equivalent to proving $x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$. This is explained below :
\begin{equation}
\begin{aligned}
&x_{n+1}-x_n>0\\
\Leftrightarrow & \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}-x_n>0\\
\Leftrightarrow & \left(1-\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)x_n-\dfrac{1}{n+1}<0\\
\Leftrightarrow & x_n<\dfrac{1}{(n+1)\left(1-\dfrac{\sqrt{n}}
{\sqrt{n+1}}\right)}\\
\Leftrightarrow & x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}.
\end{aligned}
\end{equation}
| The sequence is indeed increasing. Using what you have left off we need to prove: $x_{n+1} - x_n > 0\iff ...x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$. We prove this by induction on $n \ge 1$. Clearly $x_1 = 1 < 1+ \sqrt{\frac{1}{2}}$. Assume $x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$, we show: $x_{n+1} < 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$. Using the recursive formula you had above: $x_{n+1} = \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}< \dfrac{\sqrt{n}}{\sqrt{n+1}}\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)+\dfrac{1}{n+1}= 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}< 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$ which is clear because $n(n+2) < (n+1)^2$. Thus by induction $x_n < 1 +\dfrac{\sqrt{n}}{\sqrt{n+1}}$ and in turn implies $x_{n+1} > x_n, \forall n \ge 1$. Thus the sequence is increasing.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving limits of functions using first principles Prove using first principles that $\lim_{x \to 2}$ ($\frac{x}{1+x}$) = $\frac{2}{3}$
I know that you need to use a $\delta$-$\varepsilon$ proof where you fix $\varepsilon > 0$ and find $\delta > 0$ such that $0<|x - 2|<\delta$ $\implies$ $|\frac{x}{1+x}$ - $\frac{2}{3}|$ < $\varepsilon$
I got $|x-2| < 3\varepsilon|1+x|$ from $|\frac{x}{1+x} - \frac{2}{3}| < \varepsilon$. I know how you finish off the proof once you have found a $\delta$. The part I am not quite sure about is if I can use $\delta$ = 3$\varepsilon$|1+x| directly or if I need to get rid of the x somehow
| You need to "get rid of the $x$" by solving the inequality. Start fixing $\epsilon>0$ and consider
$$
\left|\frac{x}{1+x}-\frac{2}{3}\right|<\epsilon.
$$
As soon as $0<\epsilon<\frac 13$, by solving the previous inequality we get $\frac{2-3\epsilon}{1+3\epsilon}<x<\frac{2+3\epsilon}{1-3\epsilon}$. We would like to find $\delta>0$ such that this interval in $x$ is contained in $(2-\delta,2+\delta)$. Note that
$$
\frac{2-3\epsilon}{1+3\epsilon}=2-\frac{9\epsilon}{1+3\epsilon}\quad\text{and}\quad \frac{2+3\epsilon}{1-3\epsilon}=2+\frac{9\epsilon}{1-3\epsilon}.
$$
Since $\frac{9\epsilon}{1+3\epsilon}<\frac{9\epsilon}{1-3\epsilon}$, we can choose $\delta=\frac{9\epsilon}{1+3\epsilon}$ and the definition of limit is satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$ For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$
$$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$
My solution$:$
Let $x=a^2,\,y=b^2,\,z=c^2$ then $a+b+c=1,$ we need to prove$:$ $$\sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq 1\Leftrightarrow \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq a+b+c$$
By AM-GM$:$ $$\text{LHS} = \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a \sqrt{2a^2(b^2+c^2)}} \geqq \sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)} \geqq a+b+c$$
Last inequality is true by SOS$:$
$$\sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)}-a-b-c=\sum\limits_{cyc} {\frac {{c}^{2} \left( a-b \right) ^{2} \left( a+b \right) \left( 2\, {a}^{2}+ab+2\,{b}^{2}+{c}^{2} \right) }{a \left( 2\,{a}^{2}+{b}^{2}+{c }^{2} \right) b \left( {a}^{2}+2\,{b}^{2}+{c}^{2} \right) }} \geqq 0$$
PS:Is there any another solution for original inequality or the last inequality$?$
Thank you!
| Let $x\geq y\geq z$.
Thus, by C-S we obtain: $$\sum_{cyc}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}=\sum_{cyc}\frac{x^2-xy-xz+yz}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{xy+xz}{\sqrt{2x^2(y+z)}}=$$
$$=\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{\sqrt{2(y+z)}}{2}\geq$$
$$\geq \frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(y-x)(y-z)}{\sqrt{2y^2(x+z)}}+\frac{1}{2}\sum_{cyc}(\sqrt{y}+\sqrt{z})=$$
$$=\frac{x-y}{\sqrt2}\left(\frac{x-z}{x\sqrt{y+z}}-\frac{y-z}{y\sqrt{x+z}}\right)+1\geq$$
$$\geq \frac{x-y}{\sqrt2}\left(\frac{\frac{x}{y}(y-z)}{x\sqrt{y+z}}-\frac{y-z}{y\sqrt{x+z}}\right)+1=$$
$$=\frac{(x-y)(y-z)}{y\sqrt2}\left(\frac{1}{\sqrt{y+z}}-\frac{1}{\sqrt{x+z}}\right)+1\geq1.$$
| {
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"url": "https://math.stackexchange.com/questions/3684495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the determinant $\begin{vmatrix} y+z&z&y\\z&z+x&x\\y&x&x+y\end {vmatrix}$ Performing the operation $R_1\rightarrow R_1-R_2-R_3$
$$\begin{vmatrix} 0&-2x&-2x \\ y&z+x&x \\ z & x&x+y \end{vmatrix}$$
Pulling $-2x$ out and performing $C_2\rightarrow C_2-C_3$
$$-2x\begin{vmatrix} 0&0&1\\ y&z&x \\ z&-y&x+y \end{vmatrix}$$
$$=-2x(-y^2-z^2)$$
However, this is not given in the options, which are
$4xyz$, $xyz$, $xy$, $x^3+y^3$
What am I doing wrong?
| There is an error in your solution. The third row is not the same in the first step. The first step should be
$$\begin{vmatrix} 0&-2x&-2x \\ z&z+x&x \\ y & x&x+y \end{vmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Tough polynomial form problem Find all real values of $a$ for which the equation $(x^2 + a)^2 + a = x$
has four real roots.
I played around a bit with a graphing calculator and suspect that $(-\infty, -\frac{3}{4})$ are solutions, but I'm not sure if that's all.
| We have the cheeky difference of squares factorization :
$$
(x^2+a)^2 + a-x = (\color{red}{x^2+a})^2 -\color{blue}{x}^2 +x^2+a-x \\ = (\color{red}{x^2+a}-\color{blue}{x})(\color{red}{x^2+a}+\color{blue}x)+x^2+a-x = (x^2+x+a+1)(x^2-x+a)
$$
which has four real roots if and only if both the factors have two real roots each, which is if and only if their discriminants are non-negative. Find these, solve the simple inequalities for $a$ and combine the ranges.
| {
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"url": "https://math.stackexchange.com/questions/3689720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another series involving $\log (3)$ I will show that
$$\sum_{n = 0}^\infty \left (\frac{1}{6n + 1} + \frac{1}{6n + 3} + \frac{1}{6n + 5} - \frac{1}{2n + 1} \right ) = \frac{1}{2} \log (3).$$
My question is can this result be shown more simply then the approach given below? Perhaps using Riemann sums?
Denote the series by $S$ and let $S_n$ be its $n$th partial sum.
\begin{align}
S_n &= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 3} + \frac{1}{6k + 5} - \frac{1}{2k + 1} \right )\\
&= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 2} + \frac{1}{6k + 3} + \frac{1}{6k + 4} + \frac{1}{6k + 5} + \frac{1}{6k + 6} \right )\\
& \quad - \sum_{k = 0}^n \left (\frac{1}{2k + 1} + \frac{1}{2k + 2} \right ) - \frac{1}{2} \sum_{k = 0}^n \left (\frac{1}{3k + 1} + \frac{1}{3k + 2} + \frac{1}{3k + 3} \right )\\
& \qquad + \frac{1}{2} \sum_{k = 0}^n \frac{1}{k + 1}\\
&= H_{6n + 3} - H_{2n + 2} - \frac{1}{2} H_{3n + 3} + \frac{1}{2} H_{n + 1}.
\end{align}
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n \frac{1}{k}$. Since $H_n = \log (n) + \gamma + o(1)$ where $\gamma$ is the Euler-Mascheroni constant we see that
$$S_n = \log (6n) - \log (2n) - \frac{1}{2} \log (3n) + \frac{1}{2} \log (n) + o(1) = \frac{1}{2} \log (3) + o(1).$$
Thus
$$S = \lim_{n \to \infty} S_n = \frac{1}{2} \log (3).$$
| Your sum is$$\sum_{n\ge0}\int_0^1x^{6n}(1-2x^2+x^4)dx=\int_0^1\dfrac{(1-x^2)^2}{1-x^6}dx=\int_0^1\dfrac{1-x^2}{1+x^2+x^4}dx,$$where the first $=$ uses monotone convergence. Since$$1+x^2+x^4=(1+x^2)^2-x^2=\prod_\pm(1\pm x+x^2),$$you can show as an exercise that this integral is$$\left[\dfrac12\ln\dfrac{1+x+x^2}{1-x+x^2}\right]_0^1=\dfrac12\ln3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\lim (f(x) + 1/f(x)) = 2 $ prove that $\lim_{x \to 0} f(x) =1 $
Let $f:(-a,a) \setminus \{ 0 \} \to (0 , \infty) $ and assume $\lim_{x
\to 0} \left( f(x) + \dfrac{1}{f(x) } \right) = 2$. Prove using the
definition of limit that $\lim_{x \to 0} f(x) = 1$
Attempt:
Let $L = \lim_{x \to 0} f(x) $. Let $\epsilon > 0$ be given. If we can find some $\delta > 0$ with $|x| < \delta $ such that $|f(x) - 1 | < \epsilon $ then we will be done.
We know that since $f(x) > 0$, then $\lim 1/f(x) $ is defined. In fact, applying the limit to hypothesis, we end up with
$$ L+ \dfrac{1}{L} = 2 $$
and certainly $L=1$ as desired. I am having difficulties making this proof formal in $\delta-\epsilon$ language. Can someone assist me?
| First, note that:
$$a + \frac{1}{a} - 2 = \frac{a^2 - 2a + 1}{a} = \frac{(a - 1)^2}{a}.$$
So, roughly speaking, given we can make $\left|f(x) + \frac{1}{f(x)} - 2\right|$ as small as we like, we should be able to make $\frac{(f(x) - 1)^2}{|f(x)|}$ as small as we like. There are two ways for this to happen: either $f(x)$ is exceptionally close to $1$ or $|f(x)|$ is extremely, unreasonably large. However, in the latter case, $|f(x)|$ being very large means that $(f(x) - 1)^2$ is even larger, which will prevent the whole fraction from being small, so there really was only one option.
Let's formalise this. First, we need to eliminate the possibility of $|f(x)|$ becoming overly large, so let's try to show that $f(x)$ is bounded on some neighbourhood of $x = 0$.
I'm going to pick, fairly arbitrarily, $\varepsilon = 1$, and apply it to the limit definition of $f(x) + 1/f(x) \to 2$. Then, there exists some $\delta_0 > 0$ such that
\begin{align*}
0 < |x| < \delta_0 &\implies \left|f(x) + \frac{1}{f(x)} - 2\right| = \frac{(f(x) - 1)^2}{|f(x)|} < 1 \\
&\implies f(x)^2 - 2f(x) + 1 < |f(x)| \\
&\implies f(x)^2 + (\pm 1 - 2)f(x) + 1 < 0,
\end{align*}
where the $\pm$ depends on the sign of $f(x)$. In either case, we have a convex parabola, and the solutions to the above inequalities will be bounded. That is, there exists some $M$ such that $|f(x)| \le M$ for all $x \in (-\delta_0, \delta_0) \setminus \{0\}$. Since $f(x) \neq 0$ for at least some $x$ near $0$, we know $M > 0$.
Now we prove the limit. Suppose $\varepsilon > 0$. We have, for $0 < |x| < \delta_0$,
\begin{align*}
|f(x) - 1| < \varepsilon &\impliedby (f(x) - 1)^2 < \varepsilon^2 \\
&\impliedby \frac{(f(x) - 1)^2}{M} < \frac{\varepsilon^2}{M} \\
&\impliedby \frac{(f(x) - 1)^2}{|f(x)|} < \frac{\varepsilon^2}{M} \\
&\impliedby \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M}.
\end{align*}
Using the definition of the known limit, there is some $\delta_1$ such that
$$0 < |x| < \delta_1 \implies \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M},$$
and hence
$$0 < |x| < \min\{\delta_0, \delta_1\} \implies |f(x) - 1| < \varepsilon,$$
completing the proof.
| {
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Find all positive integers which are representable uniquely as $\frac{x^2+y}{xy+1}$ with $x,y$ positive integers. $\textbf{Question:}$ Find all positive integers,which are representable uniquely as
$$\frac{x^2+y}{xy+1}\,,$$
where $x$ and $y$ are positive integers.
I think this question maybe has something to do with vieta jumping.
I also found that for such $x$ and $y$ to exist $y < x^2$ must hold.
| Suppose that $k$ be a positive integer such that there exists a pair $(x,y)\in\mathbb{Z}_{>0}^2$ for which
$$\frac{x^2+y}{xy+1}=k\,.\tag{#}$$
Then, $t=x$ is a root to the quadratic polynomial
$$q(t):=t^2-(ky)t+(y-k)\,.$$
Note that $$t=ky-x=\frac{y-k}{x}$$ is also a root of $q(t)$.
If $y-k\leq 0$, then $ky-x\leq 0$ and $y\leq k$, so that $x\geq ky\geq y^2$. Now, observe that
$$ky^2=\frac{x^2y^2+y^3}{xy+1}=xy-1+\frac{y^3+1}{xy+1}\,$$
with
$$\frac{y^3+1}{xy+1}\leq \frac{y^3+1}{y^2\cdot y+1}=1\,.$$
Since $ky^2$ is a positive integer, we must have $x=ky$ and $y=k$, implying $(x,y)=(k^2,k)$.
We now suppose that $y>k$. The discriminant of $q(t)$ is given by
$$(ky)^2-4(y-k)=z^2\,,$$
for some nonnegitive integer $z$. Observe that
$$(ky)^2>z^2\geq (ky-2)^2\,.$$
That is, $z^2=(ky-2)^2$ or $z^2=(ky-1)^2$. The case $z^2=(ky-2)^2$ implies $k=1$, while the case $z^2=(ky-1)^2$ is impossible since $z^2$ must have the same parity as $(ky)^2$. Therefore, $k=1$ and thus,
$$q(t)=t^2-yt+(y-1)=\big(t-(y-1)\big)(t-1)\,,$$
implying $x=y-1$ or $x=1$.
In summary, the solutions $(x,y)\in\mathbb{Z}_{>0}^2$ to (#)$ are the following.
*
*If $k=1$, then the solutions are $(x,y)=(1,m)$ and $(x,y)=(n,n+1)$, where $m$ and $n$ are positive integers with $n\geq 2$.
*If $k>1$, then there is a unique solution $(x,y)=(k^2,k)$.
Furthermore, all solutions $(x,y)\in\mathbb{Z}^2$ to (#) are the following.
*
*If $k\leq -2$, then the solutions are $(x,y)=(0,k)$, $(x,y)=(k^2,k)$, and $(x,y)=(-k-1,-1)$.
*If $k=-1$, then the solutions are $(x,y)=(0,-1)$, $(x,y)=(-3,5)$, and $(x,y)=(-2,5)$.
*If $k=0$, then the solutions are $(x,y)=(n,-n^2)$, where $n$ is an integer such that $n\neq1$.
*If $k=1$, then the solutions are $(x,y)=(1,m)$ and $(x,y)=(n,n+1)$, where $m$ and $n$ are integers with $m\neq -1$ and $n\neq 1$.
*If $k\geq 2$ is not a perfect square, then the solutions are $(x,y)=(-k-1,-1)$, $(x,y)=(0,k)$, and $(x,y)=(k^2,k)$.
*If $k\geq 4$ is a perfect square, then the solutions are $(x,y)=(-k-1,-1)$, $(x,y)=(\pm\sqrt{k},0)$, $(x,y)=(0,k)$, and $(x,y)=(k^2,k)$.
However, if we solve the Diophantine equation $$x^2+y=k(xy+1)$$ for $(k,x,y)\in\mathbb{Z}^3$, then there is an extra solution for each $k\in\mathbb{Z}$: $(x,y)=(1,-1)$.
| {
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Find $\lim_{n \to \infty} n^2 \int_{n}^{5n}\frac{x^3}{1+x^6}dx$ Question:Find the limit $\lim_{n \to \infty} n^2 \int_{n}^{5n}\frac{x^3}{1+x^6}dx$
I tried to convert it into $\frac{0}{0}$ indeterminate form,then applying
L'Hospital's rule but the expressions in numerator are not nice to integrate.I do not know other way to solve this limit.
Can anybody help me out!
| Since you already received good answers for the limit itself, let me show how we could have the partial terms.
$$\frac{x^3}{x^6+1}=\frac{x^3}{(x^3-i)(x^3+i)}$$ Using partial fraction decomposition
$$\frac{x^3}{x^6+1}=\frac{x-2 i}{6 \left(x^2-i x-1\right)}+\frac{x+2 i}{6 \left(x^2+i
x-1\right)}-\frac{1}{6 (x-i)}-\frac{1}{6 (x+i)}$$ and the integration does not make much problems. Skipping the steps and recombining to have a compact result
$$12\int\frac{x^3}{x^6+1} dx=-2 \log \left(x^2+1\right)+\log \left(x^2-\sqrt{3}
x+1\right)+\log \left(x^2+\sqrt{3} x+1\right)-$$ $$2 \sqrt{3} \tan
^{-1}\left(\sqrt{3}-2 x\right)-2 \sqrt{3} \tan ^{-1}\left(2
x+\sqrt{3}\right)$$ Now, computing
$$I_k=\int_n^{kn}\frac{x^3}{x^6+1} dx \qquad \text{with} \qquad k >1$$ and expanding the result as series for large values of $n$
$$I_k=\sum_{p=0}^\infty (-1)^p\frac{1- k^{-(2+6p)}}{(2+6p)\,n^{2+6p}}$$
$$n^2 \,I_k=\sum_{p=0}^\infty (-1)^p\frac{1- k^{-(2+6p)}}{(2+6p)\,n^{6p}}$$
$$\lim_{n \to \infty} n^2 \int_{n}^{kn}\frac{x^3}{1+x^6}dx=\frac{1}{2} \left(1-\frac{1}{k^2}\right)$$ and the asymptotics
$$n^2 \,I_k=\frac{1}{2} \left(1-\frac{1}{k^2}\right)-\frac 18\left(1-\frac{1}{k^8}\right)\frac 1 {n^6}+O\left(\frac{1}{n^{12}}\right)$$
| {
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Sine Parametric function exercise Find the biggest negative value of $a$ , for which the maximum of
$f(x) =sin(24x+\frac{πa}{100})$ is at $x_0=π$
The answer is $a=-150$, but I don't understand the solving way. I would appreciate if you'd help me please
| Recall that the general solution of $\sin\theta = 1$ is:
$$
\theta = \frac{\pi}{2} + 2\pi n, \text{where }n \in \mathbb Z
$$
So the general solution for finding all maximum values of $f(x)$ is:
\begin{align*}
24x + \frac{\pi a}{100} &= \frac{\pi}{2} + 2\pi n, \text{where }n \in \mathbb Z \\
24x &= \frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n, \text{where }n \in \mathbb Z \\
x &= \frac{1}{24} \left(\frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n \right), \text{where }n \in \mathbb Z \\
\end{align*}
In particular, we know that there is some $n_0 \in \mathbb Z$ such that $x_0 = \pi$, so:
\begin{align*}
\pi &= \frac{1}{24} \left(\frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n_0 \right) \\
24\pi &= \frac{\pi}{2} - \frac{\pi a}{100} + 2\pi n_0 \\
\frac{\pi a}{100} &= \frac{\pi}{2} - 24\pi + 2\pi n_0 \\
a &= -2350 + 200 n_0 \\
\end{align*}
Since $a < 0$, we know that $n_0 < \frac{2350}{200} = 11.75$. Rounding down to the nearest integer, we find that $n_0 = 11$ so that $a = -2350 + 200(11) = -150$, as desired. $~~\blacksquare$
| {
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How may one solve problems over expressions like $(2+px)^6$ without the binomial theorem? A friend of mine posed a problem on a mathematics discord server.
The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.
I immediately thought of employing the binomial theorem, as what was required, actually. But, I decided to do another method, which led me to the question of how we can solve such problems without using the binomial theorem.
How can we solve this question without the binomial theorem or newton's method?
My Attempt:
Lemma: $$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc.$$
Note that $$(2+px)^6=(2+px)^2(2+px)^2(2+px)^2$$ $$=((px)^2+4px+4)((px)^2+4px+4)((px)^2+4px+4)$$ Therefore, in the lemma, substitute $x\mapsto (px)^2$ and $a,b,c\mapsto 4px+4$. It follows $$(2+px)^6=(px)^6+12(px+1)(px)^4+48(px+1)^2(px)^2+64(px+1)^3.$$ We can ignore the first two terms since they contain no strict $x^2$ coefficient. Thus, upon expanding $$48(px+1)^2(px)^2+64(px+1)^3$$ it follows the coefficient of $x^2$ is $240p^2$. $$240p^2=60\tag{$p>0$}$$ $\therefore p=\frac 12$.
| This implicitly uses the Binomial Theorem (edit: no it does not), and calculus, but anyway...
Let
$$p(x)=(2+px)^6=a_0+a_1x+60x^2+\mathcal{O}(x^3).$$
Differentiate twice:
$$\begin{align}
p'(x)&=6(2+px)^5\cdot p=a_1+60\cdot 2x+\mathcal{O}(x^2)
\\ \Rightarrow p''(x)&=6p\cdot 5\cdot(2+px)^4\cdot p=120+\mathcal{O}(x)
\end{align}$$
Evaluate at $x=0$:
$$30p^2\cdot 2^4=120\Rightarrow p^2=\frac{1}{4}\Rightarrow p\underset{p>0}{=}+\sqrt{\frac14}=\frac12.$$
| {
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1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac{1}{\sqrt{x^2+1}}\cdot C$$
But when I try to calculate $y'$, I get a strange equation. I think I had to be wrong somewhere. I will be grateful for your help.
| You must have asked Wolfram Alpha to solve the homogeneous equation, i.e. with $0$ instead of $x$ on the right side. According to the standard method for solving first-order equations, your integrating factor is
$$ \eqalign{\mu(x) &= \exp \left(\int \frac{x\; dx}{1+x^2}\right) \cr
&= \exp\left(\frac{1}{2} \log(1+x^2)\right)\cr
&= \sqrt{1+x^2}}$$
and then the general solution is
$$ \eqalign{y &= \frac{1}{\sqrt{1+x^2}} \left(\int x \sqrt{1+x^2} \; dx + C \right) \cr
&= \frac{1}{\sqrt{1+x^2}} \left( \frac{(1+x^2)^{3/2}}{3} + C \right)\cr
&= \frac{1+x^2}{3} + \frac{C}{\sqrt{1+x^2}}}$$
| {
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Does this $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ converges ?and what about its bounds? I want to evaluate this sum $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ such that I want to know what is the value of that series if it is converge , I have started by the evaluation of the series of the term :$\frac{(-1)^n}{\sqrt n+(-1)^n}$ using the following idea:
\begin{eqnarray*}
\frac{(-1)^n}{\sqrt n+(-1)^n}&=&\frac{(-1)^n}{\sqrt n}\times\frac{1}{1+\frac{(-1)^n}{\sqrt n}}\\
&=&\frac{(-1)^n}{\sqrt n}\left(1-\frac{(-1)^n}{\sqrt n}+\frac{1}{n}+o\left(\frac 1n\right)\right)\\
&=&\frac{(-1)^n}{\sqrt n}-\frac{1}{n}+\frac{(-1)^n}{n\sqrt n}+o\left(\frac{1}{n\sqrt n}\right).
\end{eqnarray*}
The series of general term $\frac{(-1)^n}{\sqrt n+(-1)^n}$ is sum of two convergents series (Altern series ) with harmonic series which it is diverge which gives a diverge series , but really I can't use the same idea to check whethere $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ converge or not ? and what is its value however the nested radical which montioned in denominator of that series follow the quadratic equation $x^2=n+x$ which means $x=\frac{1+\sqrt{1+4n}}{2}$ or $x=\frac{1-\sqrt{1+4n}}{2}$ then we have two possibility for the titled sum for $x\geq 0$ probably yield to divergent series , but rea really its difficult for me to assure that is a convergent series ?and what about its bounds ? any help ?
| As you noticed, $$y_n = \sqrt{n + \sqrt{n+\sqrt{n+...}}} = \frac{1+\sqrt{1+4n}}{2}$$ by using the fact that $y_n = \sqrt{n+y_n}$.
Thus the sum simplifies to $$\sum_{n=1}^{\infty} \frac{(-1)^n}{\frac{1+\sqrt{1+4n}}{2} + (-1)^n}$$
This can be split up into even $n$ and odd $n$ as $$\sum_{n=1}^{\infty} \left(\frac{1}{\frac{1+\sqrt{1+8n}}{2} + 1} + \frac{-1}{\frac{1+\sqrt{-3+8n}}{2} -1}\right)$$
Expanding this yields $$\frac{1}{3}-\frac{2}{-1+\sqrt{5}}+\sum_{n=2}^{\infty}\frac{5+2\sqrt{-3+8n}-\sqrt{1+8n}+2n\sqrt{1+8n}-2n\sqrt{-3+8n}}{\left(2-2n\right)\left(2-4n\right)}-\sum_{n=2}^{\infty}\frac{8n}{\left(2-2n\right)\left(2-4n\right)}$$
Although the first sum converges, the second sum diverges by the limit comparison test with $\frac{1}{n}$.
| {
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Diagonalization of a quadratic form, such as $\lambda_i=\{-1,1,0\}$ I am given a quadratic form
$$Q(x) = x_1^2 + 5 x_2^2 + 3 x_3^2 - 8 x_1 x_3 + 8 x_2 x_3$$
and what I need is to diagonalization of a quadratic form, such as $\lambda_i=\{-1,1,0\}$ using the Lagrange method (could not find the term for that in English).
So I am trying to extract square terms and end up with a canonical quadratic form such as $Q(y)=3y_1^2-\frac{13}{3}y_2^2+y_3^2$ where I use the following substitutions
$$
\begin{cases} y_1 = -\frac{4}{3}x_1+\frac{4}{3}x_2+x_3 \\ y_2 =x_1-\frac{16}{13}x_2 \\ y_3 = 9x_2 \end{cases}
$$
Which gives us a transformation of the type $Yx=y$ where the matrix Y is:
$$
\begin{pmatrix} -\frac{4}{3} & \frac{4}{3} & 1 \\ 1 & -\frac{16}{13} & 0 \\ 0 & 9 & 0 \end{pmatrix}
$$
Then we make the following substitutions:
$$
\begin{cases} z_1 = -\frac{1}{\sqrt3}y_1 \\ z_2 =\sqrt\frac{3}{13}y_2 \\ z_3 = y_3 \end{cases}
$$
Which gives us a matrix $Z$, which we use for transformation $y=Zz$.
As a result we have equation $Yx=y$ and $y=Zz$, so the matrix of transformation from the initial quadratic form to the diagonal form with $\lambda=\{-1,1,0\}$ will be the matrix $Y^{-1}Z$
Inverse matrixes give
$$
Y^{-1}=\begin{pmatrix} -1 & 1 & \frac{16}{117} \\ 0 &0& -\frac{1}{9} \\ 0 & \frac{4}{3} & \frac{4}{117} \end{pmatrix}
$$
$$
Z=\begin{pmatrix} \sqrt3 & 0 & 0 \\ 0 &\sqrt\frac{13}{3}& 0 \\ 0 & 0 & 1 \end{pmatrix}
$$
Giving me as the end result
$$
\begin{pmatrix} -\sqrt3 & \sqrt\frac{13}{3} & \frac{16}{117} \\ 0 &0& \frac{1}{9} \\ 0 & \sqrt\frac{208}{27} & \frac{4}{117} \end{pmatrix}
$$
Is my reasoning correct and if so, is there a better way to do this task?
| Your method seems to be less efficient than a faster method I know; I can't vouch for the accuracy of your method if you don't give a final answer.
The correct way to do this is to write
$$\mathbf v = \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix},$$
and then write the quadratic form as
$$Q(x) = \mathbf v^\intercal A \mathbf v,$$
for an appropriate symmetric matrix $A$, which in this case is
$$\begin{pmatrix}1&0&-4\\0&5&4\\-4&4&3\end{pmatrix}.$$
Since this is symmetric, we can diagonalise it, writing $A = PDP^{-1}$, where $P$ is the matrix of eigenvectors, and $D$ is a diagonal matrix of eigenvalues. Multiplying $P$ and $P^{-1}$ beside $\mathbf v$ as well will give you your required form.
| {
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For $\pi<\alpha<\frac{3\pi}{2}$ what is the value of $\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$? I solved part of it this way:
$$\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$$
$$= \sqrt{4\sin^4\alpha + 4\sin^2\alpha \cos^2\alpha} + 2\cos\left(\frac{\pi}{2} - \alpha\right) + 2$$
because $\sin 2\alpha = 2\sin\alpha\cos\alpha$ and $\cos2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1$.
The expression then reduces to:
$$\sqrt{4\sin^2\alpha(\sin^2 \alpha + \cos^2\alpha)} + 2\sin\alpha + 2$$
since $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha$. Now, because $\sin^2 \alpha + \cos^2\alpha = 1$, the expression becomes:
$$\sqrt{4\sin^2\alpha} + 2\sin\alpha + 2$$
Keep in mind that $\pi < \alpha < \frac{3\pi}{2}$. How do you proceed? I always seem to be left with extraneous sines of $\alpha$, while according to my textbook, the answer is $2$.
| Hint: For that range of $\alpha$, $\sin\alpha\leq 0).
Also note that for all real numbers $x$, one has that $\sqrt{x^2}=|x|$.
| {
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Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?
| How about using calculus?
$$f(x)=\frac{x+2}{(x+1)^2}=\frac{1}{x+1}+\frac{1}{(x+1)^2} \\ f’(x) =\frac{-1}{(x+1)^2}-\frac{2}{(x+1)^3} =0 \\ \implies x=-3$$ which you can see clearly corresponds to a minimum. We see $f(-3) =-\frac 14$. Now, $\lim_{x\to\pm\infty}f(x)=0$ and also $\lim_{x\to -1} f(x) =+\infty$ which means that $-\frac 14$ is the global minimum. There is no upper bound and hence the range is $$\left[-\frac 14,\infty\right)$$
| {
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Proving: $\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac1{x^2}\right)=\frac{\pi ^2}3$ without L'Hospital Evaluating
$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.
| Using Maclaurin expansion we get:
$$
\frac{\sin\pi x}{\pi x}
= \frac{\pi x - \frac16 (\pi x)^3 + O(x^5)}{\pi x}
= 1 - \frac{\pi^2 x^2}{6} + O(x^4)
\\
\frac{\pi x}{\sin\pi x} = 1 + \frac{\pi^2 x^2}{6} + O(x^4)
\\
\left(\frac{\pi x}{\sin\pi x}\right)^2 = 1 + \frac{\pi^2 x^2}{3} + O(x^4)
\\
\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}
= \frac{1}{x^2}\left[\left(\frac{\pi x}{\sin\pi x}\right)^2-1\right]
= \frac{\pi^2}{3} + O(x^2)
\to \frac{\pi^2}{3}
$$
| {
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Modulus operation to find unknown If the $5$ digit number $538xy$ is divisible by $3,7$ and $11,$ find $x$ and $y$ .
How to solve this problem with the help of modulus operator ?
I was checking the divisibility for 11, 3:
$5-3+8-x+y = a ⋅ 11$ and $5+3+8+x+y = b⋅3$ and I am getting more unknowns ..
| This is a solution with divisibility properties only:
$$11\mid 538xy \implies 11\mid y-x+8-3+5 = y-x+10$$
so $$11\mid y-x-1\implies y-x-1=0$$
Also $$3\mid 538xy \implies 3\mid y+x+8+3+5 = y+x+16$$ so $$3\mid y+x+1 = 2(x+1)\implies x\in\{2,5,8\}$$
Now try for each pair $(x,y)\in \{(2,3),(5,6),(8,9)\}$ if $7\mid 538xy$...
| {
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How to make inverse to work here? I have this equation:
$$\sqrt{5 - x} = 5 - x^2$$
My current approach is - I note that if I will let: $f(x) = \sqrt{5 - x}, g(x) = 5 - x^2$ then I will have $f(g(x)) = g(f(x)) = x$ Or, in other words, $f(x) = g^{-1}(x)$ (they're inverse) which means that if they intersect, then they must do so on the line y = x . This in turn means that the original equation is same as:
$$x = \sqrt{5 - x} = 5 - x^2$$
Which is of course much easier to solve. Since we have 5 - x under a square root we note that x should not be greater than 5, but also since square root is non-negative, right side should be non-negative as well, thus |x| cannot be greater than $\sqrt{5}$. With this, we can go and solve the quadratic equation:
$x^2 + x - 5 = 0$ this gives two solutions, $x = \frac{-1\pm\sqrt{21}}{2}$
and only one satisfies the condition for $|x|\le\sqrt{5}$ so we conclude $x = \frac{-1+\sqrt{21}}{2}$
Done deal! But.. if I verify this, it turns out this answer is not complete. Take a look:
Clearly there should be one more solution for this. I also plotted $h(x) = -\sqrt{5 - x}$ because that will be the inverse for the negative half of the $g(x) = 5 - x^2$ (the solution to the quadratic which we discarded earlier is the one which solves $-\sqrt{5 - x} = 5 - x^2$ which is also confirmed by $y=x$ passing through that point).
My questions:
*
*Where's my mistake?
*How to make this work with inverse functions (if possible at all)?
| There are two issues in your argument:
*
*the function $g$ is not injective, hence not invertible;
*for a bijective function, we have $f(x)=x\implies f(x)=f^{-1}(x)$, but $f(x)=f^{-1}(x)\implies f(x)=x$ doesn't holds, in general.
First note that the equation $\sqrt{5 - x} = 5 - x^2$ is equivalent to
$$\left\{\begin{array}{l}|x|\leq 5\\(x^2-5)^2+x-5=0\end{array}\right.$$
As you noted, every root of $x^2+x-5$ is a root of $(x^2-5)^2+x-5$.
Consequently, the polynomial $(x^2-5)^2+x-5$ is divisible by $x^2+x-5$, indeed we have
$$(x^2-5)^2+x-5=(x^2+x-5)(x^2-x-4)$$
Hence the third solution of $\sqrt{5 - x} = 5 - x^2$ is a root of $x^2-x-4$.
| {
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Find the maximum and minimum of $(1/x-1)(1/y-1)(1/z-1)$ if $x+y+z=1$ I get more confused when I try to solve this. For my first approach, I'm using normal AM-GM inequality:
\begin{equation}
(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)=\frac{(1-x)(1-y)(1-z)}{xyz} \\=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1
\end{equation}
By $x+y+z=1\geq \sqrt[3]{xyz}$ $\Rightarrow \frac{1}{27} \geq xyz$ we get
\begin{equation}
(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\geq 8
\end{equation}
So this is the minimum value. I do some research for how to find the maximum and come across an interesting Theorem " Lagrange Multipliers " than I'm using this to solve the problem. I'm not pretty good at Latex so I'm going for a shortcut. So we have $f(x,y,z)= (\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ and $g(x)=x+y+z-1$. After derivated I get:
\begin{equation}
\frac{-1+z+y-yz}{x^2yz}=\frac{-1+x+z-xz}{xy^2z}=\frac{-1+x+y-xy}{xyz^2}=\lambda
\end{equation}
Solving this I get $x=y=1$, $z=-1$ and so on...this kind of solution give 0 for the answer. And $x=y=z=\frac{1}{3}$ which give 8. But that mean 8 is the maximum!? Please help
| Actually the expression $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)$$ can be made arbitrarily large with the constraint $(x+y+z)=1$
Proof: Let $N>0$ be any arbitrary large real number. Choose $\varepsilon\in(0,1)$ such that $$\frac{1}{\varepsilon}>(N+1)$$ Then let $x=\varepsilon,y=z=\frac{1-\varepsilon}{2}$. Then clearly $$x,y,z>0,(x+y+z)=1\quad\text{and}\quad\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)=\left(\frac{1}{\varepsilon}-1\right)\left(\frac{2}{1-\varepsilon}-1\right)^2>\left(\frac{1}{\varepsilon}-1\right)(2-1)^2=\left(\frac{1}{\varepsilon}-1\right)>N$$ $$\tag*{$\left[\text{since $\frac{2}{1-\varepsilon}>2$}\right]$}$$
Hence the expression $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)$$
can be made arbitrarily large and hence attains no maximum in the set $$\mathcal{S}=\{(x,y,z):x,y,z>0;(x+y+z)=1\}$$ So $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\in[8,\infty)\;\forall\;(x,y,z)\in\mathcal{S}$$
$\tag*{$\square$}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power series approximation for $\ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ to calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $ Problem
Approximate $f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ and then calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $
My attempt
Let
$$f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)}) \iff $$
$$f(x) = (1+x)\ln(1+x) + (1-x)\ln(1-x) \quad $$
We know that the basic Taylor series for $\ln(1+x)$ is
$$ \ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} \quad (1)$$
As far as $\ln(1-x)$ is concerned
$$y(x) = \ln(1-x) \iff y'(x) = \frac{-1}{1-x} = - \sum_{n=0}^\infty x^n \text{ (geometric series)} \iff$$
$$y(x) = \int -\sum_{n=0}^\infty x^n = - \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \quad (2)$$
Therefore from $f(x), (1), (2)$ we have:
$$ f(x) = (1+x)\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - (1-x)\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \iff$$
$$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$
Why I hesitate
It all makes sense to me up to this point. But the exercise has a follow up sub-question that requires to find:
$$ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $$
I am pretty sure that this sum is somehow connected with the previous power series that we've found, but I can't find a way to calculate it, so I assume that I have made a mistake.
Any ideas?
| Alternative approach for the second part:
$$\begin{eqnarray*} S &=& \sum_{n\geq 1}\frac{1}{n(2n+1)}=2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n\geq 1}\int_{0}^{1}\left(x^{2n-1}-x^{2n}\right)\,dx\\&=&2\int_{0}^{1}\sum_{n\geq 1}\left(x^{2n-1}-x^{2n}\right)\,dx = 2\int_{0}^{1}\frac{x-x^2}{1-x^2}\,dx = 2\int_{0}^{1}\frac{x}{1+x}\,dx \\ &=&2\int_{0}^{1}\left(1-\frac{1}{x+1}\right)\,dx = 2\left[x-\log(x+1)\right]_{0}^{1} = \color{red}{2(1-\log 2)}. \end{eqnarray*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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