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Evaluate $\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$
Evaluate $$\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$$
I tried to create a telescoping sum but I couldn't. The last step I could reach was turning the limit into $$\lim_{n\to \infty} \sum_{r=1}^n \left(\frac {1}{2^r(1-\tan (2^{-r+1})} -\frac {1}{2^r(1+\tan (2^{-r+1})}\right) $$
But couldn't proceed further. Also I thought about Riemann sums but it was a pure dead end. Any help would be greatly appreciated
| Solution 1. By the sine double-angle formula,
$$ \sin x = 2^n \sin(2^{-n}x) \prod_{k=1}^{n} \cos(2^{-k}x). $$
Now log-differentiating both sides gives
$$ \cot x = \frac{1}{2^n}\cot\left(\frac{x}{2^n}\right) - \sum_{k=1}^{n} \frac{1}{2^k}\tan\left(\frac{x}{2^k}\right). $$
Taking $n\to\infty$ and simplifying gives
$$ \sum_{k=1}^{\infty} \frac{1}{2^k}\tan\left(\frac{x}{2^k}\right) = \frac{1}{x} - \cot x. $$
Solution 2. The above computation suggests the type of telescoping that we have to create. Indeed, by the tangent double-angle formula
$$ \frac{1}{2}\cot\left(\frac{x}{2}\right) - \frac{1}{2}\tan\left(\frac{x}{2}\right) = \frac{1-\tan^2(x/2)}{2\tan(x/2)} = \cot x $$
and hence
\begin{align*}
\sum_{k=1}^{n} \frac{1}{2^k}\tan\left(\frac{x}{2^k}\right)
&= \sum_{k=1}^{n} \left( \frac{1}{2^k}\cot\left(\frac{x}{2^k}\right) - \frac{1}{2^{k-1}}\cot\left(\frac{x}{2^{k-1}}\right) \right) \\
&= \frac{1}{2^n}\cot\left(\frac{x}{2^n}\right) - \cot x.
\end{align*}
Taking $n\to\infty$ gives the same answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=(1,-\frac 52)$ Let $$T:\mathbb R^3\to \mathbb R^2\mid M_{B_1B_2}=\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}$$ a linear transformation and the basis $$B_1=\{(1,0,0),(0,-3,1),(0,0,-2)\},\qquad B_2=\{(2,0),(0,-1)\}.$$
Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=\left(1,-\dfrac 52\right).$
I did a similar exercise and the steps I did are the same:
$$\left(\begin{array}{cc:c}2&0&1\\0&-1&-5/2\end{array}\right)\sim \left(\begin{array}{cc:c}1&0&1/2\\0&1&5/2\end{array}\right).$$ Then
$$\left(\begin{array}{ccc:c}1&0&0&2\\0&-3&0&0\\0&1&-2&1\end{array}\right)\sim \left(\begin{array}{ccc:c}1&0&0&2\\0&1&0&0\\0&0&1&-1/2\end{array}\right)$$
so we need to do
$$\begin{array}{cccccc}
M_{B_1B_2}&\cdot &[v]_{B_1}&=&[T(v)]_{B_2}& \\
\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}&\cdot&\begin{pmatrix}2\\0\\-1/2\end{pmatrix}&=&\begin{pmatrix}\color{red}{1/2}\\5/2\end{pmatrix}&\\
&\begin{pmatrix}2a-1\\-5/2\end{pmatrix}&&=&\begin{pmatrix}\color{red}{1/2}\\5/2\end{pmatrix}&\Rightarrow \begin{cases}2a-1&=&\color{red}{1/2}\\-5/2&=&5/2\end{cases}
\end{array}$$
but $$\boxed{-\dfrac 52\neq \dfrac 52\quad\wedge\quad a=\dfrac 34}$$ and the answer says $\boxed{a=1}\;\text{(apparently without contradictions)}$.
So I came to an absurd and the value of $a$ gives me different, what can be said of the exercise? Is it wrong or am I wrong?
Thanks!
$\color{red}{\text{Edited in red.}}$
| Note that $$[(2,0,1)]_{B_1}=(2,0,\frac{-1}{2})_{B_1}$$ because $$(2,0,1)=2(1,0,0)+0(0,-3,1)+\frac{-1}{2}(0,0,-2)$$, thus, $$T(2,0,1)=(2a-1,\frac{-5}{2})$$ ,take a=1
| {
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How do I find the sum of the series -1^2-2^2+3^2+4^2-5^2… upto 4n terms? I tried by giving
$$ S = \sum_{k=0}^{n-1} \left((4k+3)^2+(4k+4)^2-(4k+1)^2-(4k+2)^2\right) $$
but I am stuck here. I have no idea what to do next. The answer in my book says 4n(n+1). How can I get it? I tried expanding (4k+1)^2, etc. and got $ \sum_{k=0}^{n-1} (8k+20) $, I tried to further expand this by taking it as $ 8\sum_{k=0}^{n-1}k + 20n = 8(n-1)(n)/2 + 20n = 4n^2+16n= 4n(n+4) $ which is not the right answer. What have I done wrong? What must I do now?
| Expand
\begin{align}
&(4k+3)^2-(4k+2)^2+(4k+4)^2-(4k+1)^2\\
\qquad&=(4k+3-4k-2)(4k+3+4k+2)+(4k+4-4k-1)(4k+4+4k+1)\\
\qquad&=8k+5+3((8k+5)\\
\qquad&=4(8k+5)\\
\qquad&=32k+20
\end{align}
So your sum is
$$
\sum_{k=0}^{n-1}(32k+20)=
32\frac{n(n-1)}{2}+20n=16n^2-12n
$$
| {
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Minimum value of $\dfrac{a+b+c}{b-a}$
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
Find the minimum value of $\dfrac{a+b+c}{b-a}$
Attempt:
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the expression $\equiv \dfrac{a+b+\frac {b^2}{4a}}{b-a}$
The answer given is $3$.
| Rewrite the equation as $x^2+Bx+C$. Now, $B^2\le 4C$ So, the minimum value of $C=\dfrac{B^2}{4}$. Now, $\dfrac {a+b+c}{b−a} =\dfrac {1+B+(B^2/4)}{B−1}$. Find the minimum value of function.
| {
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Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$? Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$?
So what I did was basically simplify the terms on the right $\mod 13$.
$2^n5^{2n+1}+3^{2n+1}7^{n+1} \mod 13$
$= 5\cdot 2^n\cdot 5^n\cdot 5^n+3\cdot 7\cdot 3^{n}\cdot 3^n7^n \mod 13$
$= 5\cdot 50^n+21\cdot 63^n \mod 13$
$= 5\cdot 50^n+21\cdot 63^n \mod 13$
$= 5\cdot 11^n+8\cdot 11^n \mod 13$
$= 13\cdot 11^n \mod 13$
$= 0\cdot 11^n \mod13$
$=0 \mod13$
Thus, for all $n$, $13$ will divide this term, and the statement is true. Can anyone tell me if what I have done is correct/wrong? Would really appreciate it!
| Yes it is correct and very nice. The only minor note, we can also conclude at that step of course $$...\equiv 13\cdot 11^n \mod 13$$
As an alternative you can try also by induction.
| {
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Minimum value of $\frac{pr}{q^2}$ in quadratic equation If $y^4-2y^2+4+3\cos(py^2+qy+r)=0$ has $2$ solutions and $p,q,r\in(2,5)$. Then minimum of $\displaystyle \frac{pr}{q^2}$ is.
solution I try
$$-3\cos(py^2+qy+r) =(y^2-1)^2+3\geq 3$$
$$\cos(py^2+qy+r)\leq -1\implies \cos(py^2+qy+r)=-1$$
$$py^2+qy+r=(4n+1)\pi\implies py^2+qy+r-(4n+1)\pi=0$$
for real and distinct roots
$$q^2-4p[r-(4n+1)\pi ]\geq 0$$
How do I find minimum of $\displaystyle \frac{p r}{q^2}$? Help me.
| Rewrite the equation as $(y^2 - 1)^2 + 3 (1 + \cos (py^2 + qy + r)) = 0$. Both terms on the LHS are manifestly non-negative, so the equation is solved only when both terms are zero. The only possible solutions are $y = \pm 1$, and both are obtained only when $p+q+r$ and $p-q+r$ are both odd multiples of $\pi$. As $2 < p-q+r < p+q+r < 15$, this is only possible if $p - q + r = \pi$ and $p + q + r = 3\pi$, which can be solved to give $q = \pi$ and $p + r = 2\pi$. The quantity to minimize is thus $$\frac{pr}{q^2} = \frac{p (2\pi - p)}{\pi^2}$$ which takes possible values $$\frac{10\pi - 25}{\pi^2} < \frac{pr}{q^2} \leq 1,$$ the maximum attained at $p = \pi$ and the infimum (not strictly a minimum, as the allowable interval for $p$ is open) obtained at the endpoint $p = 5$.
| {
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How to prove that in a field with order $4$, $x^2 = x +1$ for $x$ different from $0$ and $1$? For a field $\mathbb{F}$ with order 4, prove that elements in $\mathbb{F}$ (except for $0$ and $1$) satisfies the equation:
$$
x^2 = x +1
$$
My thought is as followings:
Set $G = \{0,1,a,b\}$.
We can prove $Char \mathbb{F} = 2$ and consider $ab$ since the field has no zero divisors then $ab \neq 0$ and if $ab = a$ then $b = 1$ but $b \neq 1$, which implies $ab = 1$. Then consider $a^2$, since $a \neq b$ then $a^2 \neq 1$, by the same reason $a^2 \neq a,0$ then $a^2 = b$, then use the same way to consider $a+1$ we can get the answer.
But does there exist a easier way?
| This is obviously not true for $x=0$ or $x=1$. But then the question looks a bit weird. Anyway,
Suppose that $x$ is neither $0$ nor $1$. We have $(x+1)(x^2 - x - 1) =(x+1)( x^2 - x + 1 ) = x^3 -1$. Since $F-\{0\}$ is cyclic of order $3$, $x^3 - 1 = 0$. Hence $x = -1 = 1$ or $x^2 = x+1$. So $x^2 = x+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?
I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.
For instance:
$\sin(2x) = 2 \sin(x) \cos(x)$
| Since $\cos 2x=\pm\frac{7}{25}$, $\tan x=\frac{\sin 2x}{1+\cos 2x}\in\left\{\frac{24}{32},\,\frac{24}{18}\right\}=\left\{\frac{3}{4},\,\frac{4}{3}\right\}$ so $\sin x=\pm\frac{\tan x}{\sqrt{1+\tan^2 x}}\in\pm\left\{\frac{3}{5},\,\frac{4}{5}\right\}=\left\{-\frac{4}{5},\,-\frac{3}{5},\,\frac{3}{5},\,\frac{4}{5}\right\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786868",
"timestamp": "2023-03-29T00:00:00",
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About an interesting improper integral $\int_{0}^{\infty} \frac{x dx}{1+x^6 \sin^2 x}$ Is there a non-negative $C^\infty$ function $f(x)$ which is not bounded in $[a, \infty)$, but whose improper integral $\int_{a}^{\infty} f(x) dx$ exists?
Yes!
$\frac{x}{1+x^6 \sin^2 x}$ is not bounded in $[0, \infty)$ and not negative, but
$$\int_{0}^{\infty} \frac{x dx}{1+x^6 \sin^2 x}$$
exists.
I found this example in a calculus book written by Teiji Takagi.
His proof of the above fact is the following:
$$\int_{n \pi}^{(n+1) \pi} \frac{x dx}{1+x^6 \sin^2 x} < (n+1) \pi \int_{0}^{\pi} \frac{x dx}{1+(n \pi)^6 \sin^2 x} < \frac{1}{n^2}$$
I cannot derive the following inequaltiy:
$$(n+1) \pi \int_{0}^{\pi} \frac{x dx}{1+(n \pi)^6 \sin^2 x} < \frac{1}{n^2}$$
Please derive it.
| Using the fact that
\begin{align}
\frac{2}{\pi}x \leq \sin x \ \ \text{ for } \ \ 0\leq x\leq \frac{\pi}{2}
\end{align}
and
\begin{align}
1-\frac{2}{\pi}\left(x-\frac{\pi}{2}\right) \leq \sin x \ \ \text{ for } \ \ \frac{\pi}{2}\leq x \leq \pi.
\end{align}
Hence we have
\begin{align}
\int^\pi_0 \frac{xdx}{1+(n\pi)^6\sin^2 x} =&\ \int^{\pi/2}_0 \frac{xdx}{1+(n\pi)^6\sin^2 x}+\int^\pi_{\pi/2} \frac{xdx}{1+(n\pi)^6\sin^2 x}\\
\leq& \int^{\pi/2}_0 \frac{xdx}{1+4n^6\pi^4 x^2}+\int^\pi_{\pi/2} \frac{xdx}{1+(n\pi)^6(1-\frac{2}{\pi}(x-\frac{\pi}{2}))^2}\\
=&\ \frac{\tan^{-1}(n\pi)^3}{2\pi n^3} \leq \frac{1}{4n^3}.
\end{align}
Hence everything else follows immediately.
| {
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Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$$\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1).$$
| Note that, $(a^2-1)(a^{2m-2}+a^{2m-4}+\dots+a^2+1)\\=a^2(a^{2m-2}+a^{2m-4}+\dots+a^2+1)-(a^{2m-2}+a^{2m-4}+\dots+a^2+1)\\=a^{2m}+a^{2m-2}+a^{2m-4}+\dots+a^2-(a^{2m-2}+a^{2m-4}+\dots+a^2+1)\\=a^{2m}-1$
| {
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Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown off by the $-1$ when it comes to the prime factorization part.
My current solution looks like this :
$$ \sqrt {11} -1 = \frac {a}{b}$$
$$ \sqrt {11} = \frac {a}{b} + 1$$
$$ \sqrt {11} = \frac {a+b}{b}$$
$$ 11 = \left(\frac {a+b}{b}\right)^2$$
$$ 11 = \frac {(a+b)^2}{b^2}$$
$$ 11 = \frac {a^2 + 2ab + b^2}{b^2}$$
$$ 11 b^2 = a^2 + 2ab +b ^2$$
$$ 10b^2 = a^2 + 2ab $$
At that point, is it acceptable to conclude that a² is a multiple of 11 even though we have a trailing $2ab$?
The required method is then to conclude using prime factorization that $a = 11k$ and replace all that in the formula above to also prove $b$, however, I am again stuck with the ending $2ab$.
Would it instead be correct to prove that $\sqrt{11}$ is rational using the usual method and that, by extension, $\sqrt{11} - 1$ is also rational?
Thank you
| Continuing where you stopped:
$$
10b^2 = a^2 + 2ab
$$
Write this as
$$
b(10b-2a) = a^2
$$
Therefore, $b$ divides $a^2$. Since $\gcd(a,b)=1$, the only possibility is $b=1$. But then $\sqrt{11}-1=a$ is an integer, which implies $\sqrt{11}$ is an integer, which it clearly isn't because $3^2 < 11 < 4^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the minimal polynomial of $2\sqrt[3]{3}+\sqrt[3]{4}$ in the most efficient way . How could one find the minimum polynomial of $2\sqrt[3]{3}+\sqrt[3]{4}$, withot using the method where we let $\alpha=2\sqrt[3]{3}+\sqrt[3]{4}$, and keep cubing until we get a polynomial for $\alpha.$
I know that method will work eventually, but the degree of the polynomial is 9 as the degree of the extension is 9 and so it takes far too long.
Any help would be greatly appreciated.
| Let $$x:= 2\sqrt[3]{3}+\sqrt[3]{4} = \sqrt[3]{24}+\sqrt[3]{4}$$
Then $$x^3 = 24+4+3(\sqrt[3]{24}\cdot\sqrt[3]{4})(\underbrace{\sqrt[3]{24}+\sqrt[3]{4}}_{=x})= 28+6x\sqrt[3]{12}$$
So $$216\cdot 12x^3 = (x^3-28)^3 =...$$
| {
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Solving Recurrence Relation for Series Solution of an ODE I am trying to solve the below problem:
Assume $y = \sum_{n=0}^{\infty}a_nx^n$ is a solution to $(x-1)y''-(x-3)y'-y=0$. Find $a_n$.
I took both derivatives of $y$, plug them into the equation, modify the indices until each series has the same $x^n$, and take terms out of the series such that they have the same index to bring all the terms under the same summation. I arrive at
$\sum_{n=0}^{\infty}[(n+1)(n)a_n-(n+1)(n+2)a_{n+2} - na_n+3(n+1)a_{n+1}-a_n]x^n + (-2a_2+3a_1-a_0)=0$
Therefore $$a_2 = \frac{3a_1}{2}-\frac{a_0}{2}$$
and
$$a_{n+2} = \frac{(n+3)a_{n+1}-a_n}{(n+2)}, n = 1,2,3...$$
I also find some terms as
$$a_3 = \frac{4a_2}{3}-a_1 = \left(\frac{4\cdot 3}{3\cdot 2} - 1 \right) a_1-\frac{4}{3\cdot 2}a_0$$
$$a_4 = \left(\frac{5\cdot 4\cdot 3}{4\cdot 3\cdot 2} - \frac{13}{8} \right) a_1-\left(\frac{5\cdot 4}{4\cdot 3\cdot 2} - \frac{1}{8} \right)a_0$$
$$a_5 = \left(\frac{6\cdot 5\cdot 4\cdot 3}{5\cdot 4\cdot 3\cdot 2} - \frac{53}{40} \right) a_1-\left(\frac{6\cdot 5\cdot 4}{5\cdot 4\cdot 3\cdot 2} - \frac{17}{60} \right)a_0$$
I can see a pattern for the first set of coefficients, but not the next. For example the first part for $a_1$ is $a_n = \frac{n+1}{2}$
Any help would be greatly appreciated. Thanks!
| After Leucippus's answer.
Since Leucippus identified the proper sequences, if
$$a_{n+2} = \frac{(n+3) \, a_{n+1} - a_{n}}{n+2}$$ the coefficients can write
$$a_n=(2 a_0 -a_1) +e \,(a_1 -a_0 )\, \frac{\Gamma (n+1,1)}{n!}$$ where appears the incomplete gamma function.
| {
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Solving $4^m-3^n=p^2$ for natural $(m,n,p)$. I know that such exponential equations, like the one in question, $4^m-3^n=p^2$ (where $m,n,p$ are natural) are usually solved using numerical methods, so I tried the following.
My attempt
I’ve split the search into parts; at least intuitively (and WolframAlpha confirms), the only solutions are $(0,0,0)$ and $(1,1,1)$. So, I’ve shown that these are the solutions for $m,n\le 2$ by direct calculation. Then, I analysed the last digits of the square difference in each case, first noting that:
$$l(4^k)=\begin{cases}
4, 2\mid k\\
6, 2\nmid k
\end{cases}$$
And that:
$$l(3^k)=\begin{cases}
3, k\equiv 1\: (\text{mod }4)\\
9, k\equiv 2\: (\text{mod }4)\\
7, k\equiv 3\: (\text{mod }4)\\
1, 4\mid k
\end{cases}$$
Where $l$ is the last-digit function. For $2\mid m$ and $n\equiv 1 \text{ or } 2\: (\text{mod }4)$ and for $2\nmid m$ and $k \equiv 3 \text{ or } 0\: (\text{mod }4)$, the last digits are $3$ or $7$, so $p$ cannot be a natural number.
However, I am not sure how I should go about the other $4$ cases. Any hints on how to continue with this proof, or does anybody know a simpler / nicer approach?
| Hint: Write like this
$$(2^m-p)(2^m+p) =3^n$$
Solution:
So $2^m-p = 3^a$ and $2^m+p = 3^b$ for some $a,b$ where $a+b=n$ and $b\geq a\geq 0$.
Now
$$ 2^{m+1}= 3^b+3^a = 3^a(3^{b-a}+1)\implies a=0 \;\;\;{\rm and}\;\;\; 2^{m+1}= 3^b+1$$
so
$$(-1)^{m+1}\equiv_3 1 \implies m+1=2k$$
so $$ (2^k-1)(2^k+1)=3^b$$ and we can repeat story $2^k-1 =3^x$ and $2^k+1=3^y$ for some $x,y$ where $x+y=k$ and $y\geq x\geq 0$.
So $$2 =3^y-3^x = 3^x(3^{y-x}-1)\implies x=0 \;\;\;{\rm and}\;\;\;3^y=3\implies y=1$$
So $k=1$ and $m=1$. So $n=1$ and $p= 1$.
If $0$ is also natural number then for $b=0$ we have $n=0$ and $m=0$ and $p=0$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2794280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_0^1x(\tan^{-1}x)^2~\textrm{d}x$
Evaluate $\int\limits_0^1x(\tan^{-1}x)^2~\textrm{d}x$
My Attempt
Let, $\tan^{-1}x=y\implies x=\tan y\implies dx=\sec^2y.dy=(1+\tan^2y)dy$
$$
\begin{align}
&\int\limits_0^1x(\tan^{-1}x)^2dx=\int\limits_0^{\pi/4}\tan y.y^2.(1+\tan^2y)dy\\
&=\int\limits_0^{\pi/4}\tan y.y^2dy+\int\limits_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-\int2y.\log|\sec y|dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-y^2.\log|\sec y|+\int\frac{\tan y\sec y}{\sec y}y^2dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
\end{align}
$$
How do I proceed further and solve the integration?
| Let's try integration by parts:
\begin{align}
\int x(\arctan x)^2\,dx
&=
\frac{x^2}{2}(\arctan x)^2-
\int\frac{x^2}{2}2\arctan x\frac{1}{1+x^2}\,dx
\\
&=
\frac{x^2}{2}(\arctan x)^2-
\int\frac{1+x^2-1}{1+x^2}\arctan x\,dx
\\
&=
\frac{x^2}{2}(\arctan x)^2-
\int\arctan x\,dx+
\int\frac{1}{1+x^2}\arctan x\,dx
\end{align}
Now note that
$$
\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx
$$
and that
$$
\int\frac{1}{1+x^2}\arctan x\,dx=
\frac{1}{2}(\arctan x)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$? I guess it's easy, but I still need help. The inequality is
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
If you set $t=2^x$, then it becomes
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
The set of solutions is
$$x \in (-\infty,-2) \cup \{1\}$$
which is not what I get.
My attempt
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
This is undefined for $x=-2$, because the right hand side becomes: $\frac{1}{2^{-2+2}-1}=\frac{1}{2^0-1}=\frac{1}{1-1}=\frac{1}{0}$. Therefore,
$$x \neq -2$$
Now, let's set $t=2^x$.
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
Multiply both sides with $(t+3)(4t-1)$.
We must split the inequality because we don't know is $(t+3)(4t-1)$ positive or negative. Question: what if it's zero? Should we consider that possibility too?
For $(t+3)(4t-1) > 0$:
$$4t-1 \geq t+3$$
$$3t \geq 4$$
$$t \geq \frac{4}{3}$$
$$2^x \geq \frac{4}{3}$$
$$x \geq \log_2 \left(\frac{4}{3}\right)$$
For $(t+3)(4t-1) < 0$:
$$4t-1 \leq t+3$$
$$3t \leq 4$$
$$t \leq \frac{4}{3}$$
$$2^x \leq \frac{4}{3}$$
$$x \leq \log_2 \left(\frac{4}{3}\right)$$
I think it's already obvious where and how I'm wrong, so I think I don't need to continue with this attempt. If I do, then please ask in the comment.
| Continue from your efforts, $2^x = t$ already implies $t \gt 0$.
$$\frac{1}{t+3}- \frac{1}{4t-1} \ge 0\\
\frac{3t-4}{(t+3)(4t-1)}\ge 0$$
Solving this inequality with the help of zero points of the three factors, gives $t\in (-3,1/4) \cup [4/3, \infty) $ but we also had $t\gt 0$. So final answer is $t \in (0,1/4) \cup [4/3, \infty)$.
So $x \in (-\infty, -2) \cup [2-\log_2(3), \infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?
| Presumably the intended problem is:
*
If $(x,y,z)$ is randomly chosen from the set of nonnegative integer triples $(x,y,z)$ satisfying $$x+y+z=10$$
what is the probability that $z$ is even?
Using that interpretation . . .
Suppose $x+y+z=10$, and $z$ is even.
Write $z=2c$.
Note that $x,y$ are either both even, or both odd.
If $x,y$ are both even, write $x=2a,\,y=2b$, and count the solutions to
$$2a+2b+2c=10$$
or equivalently,
$$a+b+c=5$$
If $x,y$ are both odd, write $x=2a+1,\,y=2b+1$, and count the solutions to
$$(2a+1)+(2b+1)+2c=10$$
or equivalently,
$$a+b+c=4$$
Then just sum those counts, and divide by $66$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Stuck with solving inequality to find the product of highest and lowest integer solutions The inequality in question is
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
This time our job is to find the product of highest and lowest integer solutions.
My attempt
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
$$\sqrt{((5+2\sqrt{6})^{x})^2}+\sqrt{((5-2\sqrt{6})^{x})^2}\leq98$$
$$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x\leq98$$
What next?
The solution is $-4$.
| We have
$$
(a+b)^2+(a-b)^2= 2a^2+2b^2 = 98
$$
with $a = 5$ and $b = 2\sqrt 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
You roll a die until you get a $5$, what is the expected value of the minimum value rolled? I am struggling to work out a simple way to answer this, and a rationale behind this approach using tail sum (as I do not understand):
$$E\left( x\right) =\sum ^{5}_{k=1}P\left( x\geq k\right) =\dfrac {1}{6}\sum ^{5}_{k=1}\left( \sum ^{\infty }_{i=0}\left( \dfrac {k}{6}\right) ^{i}\right) = \frac{137}{60}.$$
Does this always hold?
$$E\left( x\right)=\sum ^{n}_{k=1}kP\left( x= k\right) =\sum ^{n}_{k=1}P\left( x\geq k\right)$$
I have never seen this formula, but working through it I understand.
EDIT: I can get the right answer with a long winded method, calculating each probability separately, which I believe the Tails sum speeds up:
This is my long winded approach.
$$E\left( X_{\min }\right) = 5P\left( x= 5\right) +\ldots +1P(x=1)$$
$$=5\left( \dfrac {1}{6}\sum ^{\infty }_{i=0}\left( \dfrac {1}{6}\right) ^{i}\right) +4\left( \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {2}{6}\right) ^{i}- \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {1}{6}\right) ^{i}\right)+\ldots$$ $$+1\left( \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {5}{6}\right) ^{i}- \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {4}{6}\right) ^{i}\right),$$
where $\dfrac {1}{6}$ represents getting a $5$, so the first sum is all the possibilities of getting repeated $6$s and then a $5$, or just rolling a $5$. The next sum is all the possibilities of getting a $4$ or a $6$, then a $5$, minus all the possibilities of just getting a $6$ then a $5$. So it represents all the strings of just $4$s or $6$s before getting a $5$, so all the possibilities of minimum value being a $4$.
| I seem to get $\dfrac{137}{60}$, slightly more than $2$, which seems as plausible to me as your similar answer
More precisely: $$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15$$
I think you are saying that if only $6$s are thrown before the first $5$ then the minimum is $6$ to calculate
$$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac16+6 \times \frac1{30}$$
I am saying the minimum is $5$, since a $5$ is thrown to stop the game. This is what is causing the difference between $\frac{139}{60}$ and $\frac{137}{60}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Sum of coefficients of even powers of x in $(1+x)^5 (1+x^2)^5$? the exact question is asking the sum of coefficients of even powers of x in the expansion of $(1+x+x^2+x^3)^5$ and in the solution this expression is simplified to $(1+x)^5(1+x^2)^5$. The solution to this question says this expression is equal to $(1+5x+10x^2+10x^3+5x^4+x^5)(1+x^2)^5$, this I understood, then it says that required sum of coefficients is equal to $(1+10+5)2^5$. How did they get this?
| The second term $(1+x^2)^5$ will have only even powers.
Now you have to calculate the sum of coefficients,
$\Rightarrow$ $2^5=32$ as ${5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}=2^5$
And calculate the sum of all even powers in the first term i.e. $(1+x)^5$ as sum of 2 even numbers is an even number.
$\Rightarrow$ $(1+10+5)=16$
So answer is $16*32=512$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks.
| Cauchy Schwarz Inequality
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}=\frac{\sin^4 x}{\cos x\sin x}+\frac{\cos^4 x}{\sin x\cos x}\geq \frac{(\sin^2 x+\cos^2 x)^2}{2\sin x\cos x}$$
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\geq \frac{1}{\sin 2x}\geq 1$$
equality hold when $$\frac{\sin^2 x}{\sin x\cos x}=\frac{\cos^2 x}{\sin x\cos x}$$
that is $\displaystyle x=\frac{\pi}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Converting parametric $x = \sec \theta + \tan \theta$, $y = \csc\theta + \cot\theta$ to Cartesian form This question comes from Solomon C4 Paper K, Question 7b.
Consider the parametric equations:
$$\begin{align}
x &= \sec(\theta) + \tan(\theta) \\
y &= \csc(\theta) + \cot(\theta)
\end{align}$$
I would like to express them in Cartesian form. To check my answer, I have used the Desmos Graphing Calculator to draw the graph of the parametric equations ($-\pi<\theta<\pi$):
Firstly I find:
$$\begin{align}
x+\frac{1}{x} &= 2\sec(\theta) \\[4pt]
y+\frac{1}{y} &= 2\csc(\theta)
\end{align}$$
Dividing the first equation by the second:
$$\frac{x+\dfrac{1}{x}}{y+\dfrac{1}{y}}=\tan(\theta)$$
Using both the original parametric equation for $x$ and the one found allows me to simplify to:
$$y+\frac{1}{y} = \frac{2\left(x+\dfrac{1}{x}\right)}{x-\dfrac{1}{x}}$$
At that point, I decide to check with Desmos again:
This graph seems to include the one found from the parametric equations, but with an extra negative reciprocal curve.
At this point, I check the mark scheme for the question, and find that it simplifies to:
$$\begin{align}
\cos(\theta)&=\frac{2x}{x^2+1} \\[4pt]
\sin(\theta)&=\frac{2y}{y^2+1}
\end{align}$$
It then uses the identity: $\sin^2A+\cos^2A=1$ to form the following Cartesian equation:
$$\frac{4y^2}{(y^2+1)^2}+\frac{4x^2}{(x^2+1)^2}=1$$
Plugging that into Desmos returns:
This graph once again includes the original one, but with even more extra reciprocal curves. I think this is because of the squares in the identity used.
I then decide to muck around with various reciprocal equations until I find one that matches the first graph. I find:
$$y=\frac{2}{x-1}+1$$
At this point, I am very confused. How (if it is possible) can I get from my parametric equations to that final Cartesian equation? Where (if I am going wrong) am I (and the mark scheme) going wrong with the first two attempts? Do the first two Cartesian equations have solutions that are not solutions for the parametric equations?
| When in doubt, here are some heuristics that can sometimes help:
*
*Rewrite everything in terms of $\sin$ and $\cos$.
*Try putting things over a common denominator.
*Try computing $x+y$, $x^2$, $y^2$ and $xy$ and see if you can find relationships between them.
In this case, we have
$$x = \frac{1}{\cos\theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin\theta}{\cos \theta}$$
$$y = \frac{1}{\sin\theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos\theta}{\sin \theta}$$
Finding a common denominator and adding these we have
$$x + y = \frac{\sin\theta + \sin^2\theta + \cos\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1+ \sin\theta + \cos\theta}{\sin\theta\cos\theta}$$
while multiplication gives
$$xy = \frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta \cos\theta} = \frac{1+\sin\theta+\cos\theta+\sin\theta \cos\theta}{\sin\theta \cos\theta}$$
Now step back and look at these two results: hopefully you notice that they are almost identical, except for one extra term in the numerator of the second expression. In fact we have
$$xy - (x+y) = \frac{\sin\theta \cos\theta}{\sin\theta \cos\theta} = 1$$
so any point on the parametrized curve satisfies the equation
$$xy - x - y = 1$$
Now solve this for $y$ using elementary algebraic methods, and you arrive at the form $$y = \frac{x+1}{x-1}$$
which is equivalent to the equation you found by "mucking around".
As far as why the marking scheme leads to an equation that has "extraneous" solutions: this happens because part of their solution involved squaring both equations. To see a simpler example of this, suppose the equations were
$$ x = t + 5, y = t - 5. $$
The most straightforward way to combine these into one equation is to write $y = x - 10$. However, suppose instead you square both sides, getting:
$$x^2 = t^2 + 10t + 25, y^2 = t^2 - 10t + 25$$
Then one can observe that $x^2 - y^2 = 20t$. But also,
$$ x + y = 2t$$
so we can write $$x^2 - y^2 = 10(x + y)$$
This leads to the equations
$$x^2 - 10x - 10y - y^2 = 0$$
which factors into
$$(x - y - 10)(x + y) = 0$$
leading to two separate solutions:
$$y = x - 10 \textrm{ or } y = -x$$
So the graph of $x^2 - 10x - 10y - y^2 = 0$ consists of two lines, one of which is the one we actually want; the other one is a spurious solution introduced by the act of squaring the original parametric equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.
| Simple computer search gives:
\begin{array}{|c|c|} \hline
m & n \\ \hline
212 & 2 \\
213 & 3 \\
214 & 4 \\
215 & 5 \\
300 & 90 \\
301 & 91 \\
324 & 114 \\
325 & 115 \\ \hline
\end{array}
Further observation:
\begin{array}{|c|c|} \hline
n \equiv m \pmod{\lambda} & \lambda \\ \hline
2,3,4,5 & 210 \\
6,7 & 462 \\
8,9,10,11 & 858 \\
10,11 & 1430 \\
12,13 & 1326 \\
12,13 & 2210 \\
14,15,16,17 & 1938 \\
14,15 & 3230 \\
16,17 & 4522 \\
18,19 & 2622 \\
18,19 & 6118 \\
20,21,22,23 & 690 \\
20,21 & 1610 \\
20 & 3795 \\
24,25 & 870 \\ \hline
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$ I have become stuck while solving a trig identity. It is:
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$
I have simplified the left side as far as I can:
\begin{align}
\frac{1-2\sin(x)}{\sec(x)}
&=\frac{1-2\sin(x)}{1/\cos(x)}=(1-2\sin(x))\cos(x)\\
&=\cos(x)-2\sin(x)\cos(x)=\cos(x)-\sin(2x)
\end{align}
However, I'm not sure what to do on the right side. I know I can use a compound angle formula to break $\cos(3x)$ into $\cos(2x)\cos(x)-\sin(2x)\sin(x)$; however, I do not know where to go after that. My main problem is with the denominator of the right side, I can't figure out how to get rid of it, either by multiplying, or by using a trig identity. Any help in solving this identity would be greatly appreciated!
| The identity is equivalent to
$$
\cos3x=\cos x(1-4\sin^2x)
$$
(except for the values where the denominators vanish). The right-hand side can be rewritten as
$$
\cos x(\cos^2x+\sin^2x-4\sin^2x)=\cos^3x-3\cos x\sin^2x
$$
which is known to be the same as $\cos3x$: by De Moivre
\begin{align}
\cos 3x+i\sin3x
&=(\cos x+i\sin x)^3 \\
&=\cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x\\
&=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x)
\end{align}
Of course you can also use
\begin{align}
\cos3x
&=\cos(2x+x)\\
&=\cos2x\cos x-\sin2x\sin x\\
&=(\cos^2x-\sin^2x)\cos x-2\cos x\sin^2x\\
&=\cos^3x-3\cos x\sin^2x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Factoring a given quartic form Description
Below is our polynomial:
$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4$$
I've gone to almost end of it but,it seems there's a small -or conversely huge- fault in my solving which made $\mathbf{A}\:and\: \mathbf{B}$ imaginary numbers as I didn't find any two numbers which Could have the sum of $-8xy$ and product of $24x^2y^2$ OR in another sense , I can not find $\mathbf{A}\:and\: \mathbf{B}$.
My work
$$\begin{align}
& 3x^4-8x^3y+14x^2y^2-8xy^3+3y^4 \\
& = 3(x^4+y^4)-8xy(x^2+y^2)+14x^2y^2 \\
& = 3(x^4+y^4+2x^2y^2)-8xy(x^2+y^2)+8x^2y^2\\
& = 3(x^2+y^2)^2-8xy(x^2+y^2)+8x^2y^2\\
& = \frac{1}{3}\times[3^2(x^2+y^2)^2- 3\times8xy(x^2+y^2)+24x^2y^2] \\
& =\frac{1}{3}(3(x^2+y^2)+\mathbf{A})(3(x^2+y^2)+\mathbf{B})
\end{align}$$
| Assume by symmetry
$$(ax^2+bxy+cy^2)(cx^2+bxy+ay^2)=\\=acx^4+(ab+bc)x^3y+(a^2+b^2+c^2)x^2y^2+(ab+bc)xy^3+(ac)y^4$$
then we need
*
*$ac=3 \implies a=3\, c=1$
*$ab+bc=-8 \implies3b+b=-8\implies b=-2$
*$a^2+b^2+c^2=14 \implies 9+4+1=14$
thus
$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4=(3x^2-2xy+y^2)(x^2-2xy+3y^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all functions $ f : \mathbb R \to \mathbb R $ satisfying $ f ( x ) f ( y ) + f ( x y ) + f ( x ) + f ( y ) = f ( x + y ) + 2 x y $
Find all functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f ( x ) f ( y ) + f ( x y ) + f ( x ) + f ( y ) = f ( x + y ) + 2 x y $$
for all $ x , y \in \mathbb R $.
I tried the standard way: $ x = 0 $, $ x = y $, $ x = 1 $, $ \dots $, but without any success. I spent quite some time trying to solve it but didn't succeed.
I tried to reduce it to Cauchy's 1-4 equations but didn't succeed. In the course of it, I found interesting works of Aczél, Erdős and even Putnam, but they are not directly related, I guess.
Any idea? I am interested in this problem but I couldn't solve!
| "Quick" solution: let $g(x)=x-f(x)$. Then in terms of $g$, the functional equation can be rearranged to $$g(x+y)+g(x)g(y)=(1+x)g(y)+(1+y)g(x)+g(xy).$$ By my answer to this question, the solutions to this functional equation are $g(x)=0$, $g(x)=3x$, and $g(x)=x(x+1)$. We conclude that the solutions for $f$ are $f(x)=x$, $f(x)=-2x$, and $f(x)=-x^2$.
Alternatively, here's a direct solution which I was foolish enough to work out thoroughly before realizing that this equation reduces to one I've solved before. :)
It is easy to verify that $f(x)=x$, $f(x)=-2x$, and $f(x)=-x^2$ are solutions. I claim they are the only solutions.
First, setting $y=0$ gives $f(0)(f(x)+2)=0$ for all $x$, so either $f(x)=-2$ for all $x$ or $f(0)=0$. It is easily verified that $f(x)=-2$ does not work, so we must have $f(0)=0$.
Now let us write $a=f(1)$. Setting $x=y=2$ gives $f(2)^2+2f(2)=8$, so $f(2)=2$ or $f(2)=-4$. Setting $x=y=1$ gives $a^2+3a=f(2)+2$. If $f(2)=2$ we get either $a=1$ or $a=-4$. If $f(2)=-4$ we get either $a=-1$ or $a=-2$.
Setting $y=1$ we have $(a+2)f(x)+a=f(x+1)+2x$ and so $$f(x+1)=(a+2)f(x)-2x+a\tag{*}$$ for all $x$. Note that this recurrence can be used to find $f(x)$ for all integers $x$ from $a$. We now consider the possible values of $a$ one by one.
If $a=-2$, $(*)$ immediately gives $f(x+1)=-2(x+1)$ and so $f(x)=-2x$ for all $x$.
If $a=-4$, then we have $f(2)=2$ and $f(-1)=-1$ and $f(-2)=1/2$ from $(*)$. But then the functional equation fails with $x=2$ and $y=-1$. So this case is impossible.
If $a=-1$, we find that $f(x)=-x^2$ for all integers $x$. The recurrence $(*)$ has the form $$f(x+1)=f(x)-2x-1.$$ Now setting $y=-1$ in the original equation gives $f(-x)-1=f(x-1)-2x$. Replacing $x$ with its negative, we find $$f(-(x+1))=f(x)-2x-1.$$ Comparing the two centered equations above, we see $f(x+1)=f(-(x+1))$ so $f$ is an even function. Now setting $x=y$ and $x=-y$ in the original equation give $$f(x)^2+f(x^2)+2f(x)=f(2x)+2x^2$$ and $$f(x)^2+f(x^2)+2f(x)=-2x^2$$ respectively. Comparing these equations, we get $f(2x)=-4x^2$ and so $f(x)=-x^2$ for all $x$.
Finally, suppose $a=1$. We then have $f(x)=x$ for all integers $x$ and the recurrence $(*)$ is $$f(x+1)=3f(x)-2x+1.$$ Using the recurrence backwards we have $f(x-1)=\frac{f(x)+2x-3}{3}$ and so we get $$f(-x)=\frac{f(x)-4x}{3}.$$ Replacing $x$ with $-x$ we get $$f(x)=\frac{f(-x)+4x}{3}=\frac{\frac{f(x)-4x}{3}+4x}{3}=\frac{f(x)}{9}+\frac{8x}{9}.$$ Solving for $f(x)$ we conclude that $f(x)=x$ for all $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^{\infty}\frac{1}{(1+x^2)(3-\cos x)}dx$ Greetings I tried to evaluate $$I=\int_0^{\infty}\frac{1}{(1+x^2)(3-\cos x)}dx$$ Here is my try, it is abit longer, but I wrote it all so that I wont have a silly mistake. My main ideea was to expand into fourier series $$g(t)=\frac{1}{3-\cos t}$$ so I took(I am not sure if it's correct but that is how I learned) $$a_0=\frac{1}{\pi}\int_0^{2\pi}\frac{1}{3-\cos t}dt$$ $$a_n=\frac{1}{\pi}\int_0^{2\pi}\frac{\cos(nt)}{3-\cos t}dt$$$$b_n=\frac{1}{\pi}\int_0^{2\pi}\frac{\sin(nt)}{3-\cos t}dt$$ $$a_n+ib_n=\frac{1}{\pi}\int_0^{2\pi}\frac{e^{int} }{3-\cos t}dt$$ we let $$e^{it}=z \rightarrow dt=\frac{dz}{iz}\, ; |z|=1$$ And we can write $\cos t=\frac{z^2+1}{2z}$ $$a_n+ib_n=\frac{2}{i\pi}\int_{|z|=1} \frac{z^n}{-z^2+6z-1}dz$$our function $$g(z)= \frac{z^n}{-z^2+6z-1}$$ has in the interior of the circle $|z|=1$ simple poles at: $$-z^2+6z-1=0\rightarrow-(z-3)^2+8=0$$giving $$z_1=3-2\sqrt{2}$$ the other pole is not in the circle. So $$\frac{2}{i\pi}\int_{|z|=1} g(z)dz=2\pi i Res(g;z_1)$$ $$Res(g;3-2\sqrt 2)=\lim_{z\to 3-2\sqrt 2} \, \frac{z^n}{z-(3+2\sqrt 2)}=\frac{(3-2\sqrt 2)^n}{-4\sqrt 2}+i\cdot 0$$ Thus the $b_n$ term vanishes and $$a_n=\frac{(3-2\sqrt 2)^n}{-4\sqrt 2}\rightarrow a_0=-\frac{1}{4\sqrt 2}$$ so as a fourier series we have $$g(t)=-\frac{1}{8\sqrt 2}-\frac{1}{4\sqrt 2} \sum_{n=1}^{\infty}\frac{\cos(nt)}{(3+2\sqrt2)^n}$$ So we can rewrite the original integral as $$I=-\frac{1}{8\sqrt 2}\int_0^{\infty}\frac{1}{1+x^2}dx-\frac{1}{4\sqrt 2} \sum_{n=1}^{\infty}\frac{1}{(3+2\sqrt 2)^n}\int_0^{\infty}\frac{\cos(nx)}{1+x^2}dx$$ The first integral is simple, and the second one is well known around here to be $\frac{\pi}{2e^n}$ see for example: Integral evaluation $\int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx$ so I get $$I=-\frac{\pi}{8\sqrt 2}\left(\frac{1}{2}+\sum_{n=1}^{\infty} \frac{1}{(e(3+2\sqrt 2))^n}\right)=-\frac{\pi}{16\sqrt 2}\left(\frac{3 e+2e\sqrt 2+ 1}{3e+2e\sqrt 2 - 1}\right) $$
Now since wolfram fails to compute this I am not sure, also for sure there is a sign mistake because the integral is positive... Could you please correct my answer?
| By following the same approach given by Random Variable in Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$ , we have that if $1<|a|<e$ then
$$\begin{align} \frac{1}{2a}\int_{0}^{\infty} \frac{1}{(1+x^{2})(\frac{1+a^2}{2a}-\cos x )} \ dx &= \frac{1}{1-a^{2}} \int_{0}^{\infty} \Big(1+2 \sum_{k=1}^{\infty} a^{k} \cos(kx) \Big) \ \frac{1}{1+x^{2}} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{2}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \int_{0}^{\infty} \frac{\cos(kx)}{1+x^{2}} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \sum_{k=1}^{\infty} \Big(\frac{a}{e} \Big)^{k} \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}}\cdot \frac{a/e}{1-a/e} = \frac{\pi}{2} \frac{1}{1-a^{2}} \cdot\frac{e+a}{e-a} \end{align}$$
Since $\frac{1+a^2}{2a}=3$ for $a=3-2\sqrt{2}$ and $|a/e|<1$ we get
$$\int_{0}^{\infty} \frac{1}{(1+x^{2})(3-\cos x )} \ dx
=\frac{a\pi}{1-a^{2}} \cdot\frac{e+a}{e-a}=\frac{\sqrt{2}\pi}{8} \cdot\frac{e+3-2\sqrt{2}}{e-3+2\sqrt{2}}\approx 0.630190099$$
which is $-4$ times your result.
| {
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Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$. Could someone please explain a step in the following proof?
Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.
The above limit exists if for every $\varepsilon > 0$, there exists a real number $\delta > 0$ such that if $0<|x-1|<\delta$, then $\left |\frac{1}{x+1}-\frac{1}{2} \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$.
I understand that $|x-1|$ can be made small, but for $|x+1|$, why does the solution say the condition "if $|x-1| < 1$, $x+1\in (1,3)$"? Why and how is the condition $|x-1| <1$ made?
| So far we have $$\left |\frac{1}{x+1}-\frac{1}{2} \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$$ In order to achieve the last inequality, we have to control the denominator as well as the numerator.
By assuming $|x-1|<1$ we can control $|x+1|$, therefore the fraction, $ \frac{|x-1|}{2|x+1|}$ could be made less than $\epsilon$ by choosing a small enough $ \delta$ .
| {
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Find the number of ways to select $1,2,3,4,5,6,7,8,9,10$ fruits from a pile (using generating functions) Find the number of ways to select $1,2,3,4,5,6,7,8,9,10$ fruits from a pile of $3$ apples, $5$ oranges and $2$ bananas. (Use generating functions.)
Any tips?
| We encode zero up to
*
*three apples as: $\ \quad1+x+x^2+x^3=\frac{1-x^4}{1-x}$
*five oranges as: $\ \quad1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x}$
*two bananas as: $\quad 1+x+x^2=\frac{1-x^3}{1-x}$
Denoting with $[x^k]$ the coefficient of $x^k$ of a series, we are looking for
\begin{align*}
\color{blue}{[x^k]\frac{(1-x^4)(1-x^6)(1-x^3)}{(1-x)^3}}\qquad \qquad 1\leq k\leq 10
\end{align*}
We obtain
\begin{align*}
[x^k]&\frac{(1-x^4)(1-x^6)(1-x^3)}{(1-x)^3}\\
&=[x^k](1-x^4)(1-x^6)(1-x^3)\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\tag{1}\\
&=[x^k](1-x^3-x^4-x^6+x^7+x^9+x^{10})\sum_{j=0}^\infty\binom{j+2}{2}x^j\tag{2}\\
\end{align*}
Comment:
*
*In (1) we apply the binomial series expansion.
*In (2) we expand the factors up to powers of $x^{10}$ since higher powers do not contribute to $[x^k]$ for $1\leq k\leq 10$. We also apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
Example: We calculate from (2) the coefficient of $x^k$ for $k=7$.
\begin{align*}
\color{blue}{[x^7]}&\color{blue}{(1-x^3-x^4-x^6+x^7+x^9+x^{10})\sum_{j=0}^\infty\binom{j+2}{2}x^j}\\
&=\left([x^7]-[x^4]-[x^3]-[x^1]+[x^0]\right)\sum_{j=0}^\infty\binom{j+2}{2}x^j\\
&=\binom{9}{2}-\binom{6}{2}-\binom{5}{2}-\binom{3}{2}+\binom{2}{2}\\
&=36-15-10-3+1\\
&\,\,\color{blue}{=9}
\end{align*}
The $9$ solutions of seven fruits are
$$
\begin{array}{c|c|c}
\text{apples}&\text{oranges}&\text{bananas}\\
\hline
3&4&0\\
3&3&1\\
3&2&2\\
2&5&1\\
2&4&1\\
2&3&2\\
1&5&1\\
1&4&2\\
0&5&2\\
\end{array}
$$
| {
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Special properties of the number $146$ I'm a math teacher. Next week I'll give a special lecture about number theory curiosities. It will treat special properties of numbers — the famous story with Ramanujan, taxicab numbers, later numbers divisible by all their digits, etc.
I was given class number $146$ for the lecture and I think it would be fine to start with a special property of our class's number. Ramanujan would surely find something at once, but I can't. Do you see any special properties of $146$?
Here are some of my observations, but these properties are not very special:
*
*$146$ is a semiprime number (product of two distinct primes), while the reversal $641$ is prime.
*$146 = 4^3 + 4^3 + 3^2 + 3^2$.
Here is a very similar question, just to show what kind of question this is and what kind of answers I would like to see.
| The sum of the first $n$ cubes, starting from $1^3$, is given by the square of the $n$th Triangular number, $T_n$:
\begin{align}
1^3 &= & 1^2 &= & T_{1}^2\\
1^3+2^3 &= & 3^2 &= & T_{2}^2 \\
1^3+2^3+3^3 &= & 6^2 &= & T_{3}^2 \\
1^3+2^3+3^3+4^3 &= & 10^2 &= & T_{4}^2 \\
\end{align}
Noting this we can write the sum of the first $n$ squares of the Triangular numbers as:
\begin{align*}
\sum_{i=1}^{n}T_i^2&=T_1^2+T_2^2+\dotsb+T_n^2\tag{1}\\
&=n\times1^3+(n-1)\times2^3+\dotsb+1\times n^3\\
&=\sum_{k=1}^{n}(n-k+1)k^3\tag{2}
\end{align*}
Note, taking $n=4$ in $(1)$ gives
$$T_{1}^2+T_{2}^2+T_{3}^2+T_{4}^2=1^2+3^2+6^2+10^2=146$$
and taking $n=4$ in $(2)$ gives
$$4\times1^3+3\times2^3+2\times 3^3+1\times 4^3=146$$
As a consequence of $(1)$ another nice pattern emerges:
$$\frac{1}{4}\left[2^2(1^2+3^2)+4^2(3^2+5^2)\right]=146$$
| {
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For any prime number $n >5$, prove the final digit of $n^4$ is $1$ So I am struggling a bit with this question
$n$ is prime
we can ignore $2$ and $5$ as $n>5$
now if $n$ is prime
for the digits: $\{0,1,2,3,4,5,6,7,8,9\}$
$\{0,2,4,6,8\}$ can be discounted as $n$ cannot be even that
$5$ can be discounted as $n$ is not a multiple of $5$ either
therefore any prime must have the last digit $q$ such that $q\subset\{1,3,7,9\}$
For $n^4$, if $n=10p+q$
for a natural number $p$
I don't quite know how I can get to the desired result.
| Hint
$$n^4-1=(n^2-1)(n^2+1)=(n^2-1)(n^2-4+5)\\
=(n^2-1)(n^2-4)+5(n-1)(n+1)\\
=(n-2)(n-1)(n+1)(n+2)+5(n-1)(n+1)$$
Show that
$$10| (n-2)(n-1)(n+1)(n+2) \,, \mbox{ and }\\
10|5(n-1)(n+1)$$
| {
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Without relating the problem to polynomials, how can we solve this inequality? $a,b,c$ are real;
$a<b<c$ ;
$a+b+c=6$ ;
$ab+bc+ca = 9$.
Prove that $0<a<1<b<3<c<4$.
I solved this by treating $a,b,c$ as roots of $x^3 - 6x^2 + 9x + d$ for some $d$, but I have not been able to solve it in a more direct way.
| Partial solution:
Let $\{x,y,z\} =\{a,b,c\}$. Since $z=6-x-y$ we get a quadratic equation:
$$ x^2+x(y-6)+(y-3)^2 =0$$
and since this one has to have a real solution we have a discriminant nonnegative, so
$$ -3y(y-4)\geq 0 \implies y\in [0,4]$$
and the same we can say for $x$ and $z$.
Now can any of $x,y,z\in\{0,4\}$?
Say $z=0$ then we get $$6=x+y >2\sqrt{xy} = 2\cdot 3 =6$$
a contradiction.
Say $z=4$ then we get $x+y=2$ and so $xy=1$ so $$2=x+y >2\sqrt{xy} = 2\cdot 1 =2$$
a contradiction again.
So if we return to $a,b,c$ we get: $0<a<b<c<4$.
Now if $b\geq 3$ then $c>3$ and thus $a+b+c>6$ a contradiction.
If $b\leq 1$ then we get $ a<1$ and so $a+b+c<6$ a contradiction again. So $$1<b<3$$
We are left to prove $a<1$ and $c>3$....
| {
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olympic mathematics For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2,
\ldots$ by:
$$a_{n+1} = \begin{cases} \sqrt{a_n}, & \mbox{if } \sqrt{a_n}\mbox{ is an integer,} \\a_n+3, &\mbox{otherwise,}\end{cases}$$
Determine all values of $a_0$ for which there is a number $A$ such that $a_n = A$ for infinitely many values
of $n$.
I have figured that if there exists an $i$ such that $a_i
\ \mathrm{mod}\, 3 = 2$, the elements after $a_i$ are all different.
I think there are many ways to solve this question so i would appreciate any answers.
i think the answer is $a_0\ \mathrm{mod}\, 3 = 0$
but i can't prove it
examples: 7,10,13,16,4,2 doesn't work
3,6,9,3 works
please tell me if i can improve my question.
| There is no integer $a$ so that $a^2 \equiv 2 \mod 3$ (as $0^2 \equiv 0 \mod 3$, $1^2 \equiv 1 \mod 3$ and $2^2 \equiv 1 \mod 3$ so if $a_0\equiv 2 \mod 3$ then by induction $a_{i+1} = a_i + 3 \equiv a_0 \equiv 2 \mod 3$ and the sequence never repeats.
There are infinitely many squares that are multiples of $3$. The square roots of these are multiples of $3$. So $a_0 = 3k$ will eventually lead to a $a_n = (3m)^k$ so $a_{n+1} = 3m < a_n$ so $a_n$ will repeat infinitely. So any multiple of $3$ will be such a starting point.
(Obviously if a sequence repeats once it will, by induction, repeat infinitely many times.
This just means we must consider $a_0 \equiv 1 \mod 3$.
If $a_0 = 1$ then $a_1 = \sqrt 1 = 1$ so $a_0$ is such a starting point.
But $a_0 = 4 \implies a_1 = 2$ so $4$ is not a valid starting point.
If $a_0 = 7; a_1 = 10;a_2 = 13; a_3 = 16; a_4 = 4$ so $7$ iss not a valid starting point. Is any $1 + 3k$ other than $1$?
Well, it can't be less than $25$ as that will lead to $25$ which leads us to $5\equiv 2 \mod 3$. We need to be lead to a $(3k + 1)^2 $ that leads to $3k + 1$ that does not lead to and $3k + 1 < (3m + 2)^2 < (3k+1)^2$.
Not sure any such number exists. $49$ for example is such that $7 < 16 < 49$ is a failure.
But if $(3k+1)^2> 49$ then $k > 2$ and $(3k-2)^2 = 9k^2 - 12k + 4 \ge 18k-12k +4 = 6k+ 4 > 3k + 1$ so $3k + 1 < (3k-2)^2 < (3k+1)^2$.
So there is no such $1 + 3k$ (other than $1$ that will work).
So $a_0 = 1$ and $a_0 = 3k$ are all such starting points.
| {
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1)$
The Book's Answer: $4x^3(3x+1)(x+1)^7$
| Since $(\forall x\in\mathbb{R}):4(x^3+2x^2+x)^3(3x^2+4x+1)=4x^3(3x+1)(x+1)^7$, both answers are correct. But your answer is more natural.
| {
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Finding the inverse of $x^3+x^2+1$ in $\Bbb F_2[x]/(x^4+x^2)$ with the Euclidean algorithm My first thought was successful: $x^4+x^2=x^2(x^2+1)$ and $x^3+x^2+1=x^2(x+1)+1$ so it is its own inverse because $(x^2(x+1)+1)^2\equiv x^4(x+1)^2+1\equiv x^4(x^2+1)+1\equiv1.$
The given solution claims to use the Euclidean algorithm, succintly, but I don't quite get it:
$x^4+x^2=(x^3+x^2+1)(x+1)+(x+1); \\ x^3+x^2+1=x^2(x+1)+1$
hence $1=(x^3+x^2+1)(x^3+x^2+1)+x^2(x^4+x^2)$
What is the logic in the above lines? How does it work in general?
| For integers, if you're trying to find an inverse of $a$ mod $n$ using Euclidean algorithm, you would divide $n$ by $a$, find the remainder $r_1$, then divide $a$ by $r_1$, find the remainder $r_2$; divide $r_1$ by $r_2$, etc., until the remainder becomes 1. Here you're looking for the inverse of $x^3+x^2+1$ mod $x^4+x^2$, so you divide $x^4+x^2$ by $x^3+x^2+1$, find the remainder $x+1$, then divide $x^3+x^2+1$ by this remainder $x+1$, this time the remainder is already 1.
To find the inverse, you work backwards. The second equation tells you
$$1 = (x^3+x^2+1) - (x^2(x+1))$$
By the first equation,
$$x+1 = [x^4+x^2] - [(x^3+x^2+1)(x+1)]$$
Plug this into the above equation:
$$1=(x^3+x^2+1)- x^2(x^4+x^2-(x^3+x^2+1)(x+1)) \equiv (x^3+x^2+1)(1+x^3+x^2)$$
In short, you're doing exactly the same thing as when you solved congruence relations of integers using Euclidean algotithm.
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Show that the Huber-loss based optimization is equivalent to $\ell_1$ norm based. Dear optimization experts,
My apologies for asking probably the well-known relation between the Huber-loss based optimization and $\ell_1$ based optimization. However, I am stuck with a 'first-principles' based proof (without using Moreau-envelope, e.g., here) to show that they are equivalent.
Problem formulation
The observation vector is
\begin{align*}
\mathbf{y}
&= \mathbf{A}\mathbf{x} + \mathbf{z} + \mathbf{\epsilon} \\
\begin{bmatrix} y_1 \\ \vdots \\ y_N \end{bmatrix} &=
\begin{bmatrix}
\mathbf{a}_1^T\mathbf{x} + z_1 + \epsilon_1 \\
\vdots \\
\mathbf{a}_N^T\mathbf{x} + z_N + \epsilon_N
\end{bmatrix}
\end{align*}
where
*
*$\mathbf{A} = \begin{bmatrix} \mathbf{a}_1^T \\ \vdots \\ \mathbf{a}_N^T \end{bmatrix} \in \mathbb{R}^{N \times M}$ is a known matrix
*$\mathbf{x} \in \mathbb{R}^{M \times 1}$ is an unknown vector
*$\mathbf{z} = \begin{bmatrix} z_1 \\ \vdots \\ z_N \end{bmatrix} \in \mathbb{R}^{N \times 1}$ is also unknown but sparse in nature, e.g., it can be seen as an outlier
*$\mathbf{\epsilon} \in \mathbb{R}^{N \times 1}$ is a measurement noise say with standard Gaussian distribution having zero mean and unit variance normal, i.e. $\mathcal{N}(0,1)$.
We need to prove that the following two optimization problems P$1$ and P$2$ are equivalent.
P$1$:
\begin{align*}
\text{minimize}_{\mathbf{x},\mathbf{z}} \quad & \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1
\end{align*}
and
P$2$:
\begin{align*}
\text{minimize}_{\mathbf{x}} \quad & \sum_{i=1}^{N} \mathcal{H} \left( y_i - \mathbf{a}_i^T\mathbf{x} \right),
\end{align*}
where the Huber-function $\mathcal{H}(u)$ is given as
$$\mathcal{H}(u) =
\begin{cases}
|u|^2 & |u| \leq \frac{\lambda}{2} \\
\lambda |u| - \frac{\lambda^2}{4} & |u| > \frac{\lambda}{2}
\end{cases} .
$$
My partial attempt following the suggestion in the answer below
We attempt to convert the problem P$1$ into an equivalent form by plugging the optimal solution of $\mathbf{z}$, i.e.,
\begin{align*}
\text{minimize}_{\mathbf{x},\mathbf{z}} \quad & \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \\
\equiv
\\
\text{minimize}_{\mathbf{x}} \left\{ \text{minimize}_{\mathbf{z}} \right. \quad & \left. \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \right\}
\end{align*}
Taking derivative with respect to $\mathbf{z}$,
\begin{align}
0 & \in \frac{\partial}{\partial \mathbf{z}} \left( \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \right) \\
\Leftrightarrow & -2 \left( \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \right) + \lambda \partial \lVert \mathbf{z} \rVert_1 = 0 \\
\Leftrightarrow & \quad \left( \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \right) = \lambda \mathbf{v} \ .
\end{align}
for some $ \mathbf{v} \in \partial \lVert \mathbf{z} \rVert_1 $ following Ryan Tibshirani's lecture notes (slide#18-20), i.e.,
\begin{align}
v_i \in
\begin{cases}
1 & \text{if } z_i > 0 \\
-1 & \text{if } z_i < 0 \\
[-1,1] & \text{if } z_i = 0 \\
\end{cases}.
\end{align}
Then, the subgradient optimality reads:
\begin{align}
\begin{cases}
\left( y_i - \mathbf{a}_i^T\mathbf{x} - z_i \right) = \lambda \ {\rm sign}\left(z_i\right) & \text{if } z_i \neq 0 \\
\left| y_i - \mathbf{a}_i^T\mathbf{x} - z_i\right| \leq \lambda & \text{if } z_i = 0
\end{cases}
\end{align}
Also, following, Ryan Tibsharani's notes the solution should be 'soft thresholding' $$\mathbf{z} = S_{\lambda}\left( \mathbf{y} - \mathbf{A}\mathbf{x} \right),$$
where
\begin{align}
S_{\lambda}\left( y_i - \mathbf{a}_i^T\mathbf{x} \right) =
\begin{cases}
\left( y_i - \mathbf{a}_i^T\mathbf{x} - \lambda \right) & \text{if } \left(y_i - \mathbf{a}_i^T\mathbf{x}\right) > \lambda \\
0 & \text{if } -\lambda \leq \left(y_i - \mathbf{a}_i^T\mathbf{x}\right) \leq \lambda \\
\left( y_i - \mathbf{a}_i^T\mathbf{x} + \lambda \right) & \text{if } \left( y_i - \mathbf{a}_i^T\mathbf{x}\right) < -\lambda \\
\end{cases} .
\end{align}
Now, we turn to the optimization problem P$1$ such that
\begin{align*}
\text{minimize}_{\mathbf{x}} \left\{ \text{minimize}_{\mathbf{z}} \right. \quad & \left. \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \right\} \\
\equiv
\end{align*}
\begin{align*}
\text{minimize}_{\mathbf{x}} \quad & \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - S_{\lambda}\left( \mathbf{y} - \mathbf{A}\mathbf{x} \right) \rVert_2^2 + \lambda\lVert S_{\lambda}\left( \mathbf{y} - \mathbf{A}\mathbf{x} \right) \rVert_1
\end{align*}
*
*if $\lvert\left(y_i - \mathbf{a}_i^T\mathbf{x}\right)\rvert \leq \lambda$, then So, $\left[S_{\lambda}\left( y_i - \mathbf{a}_i^T\mathbf{x} \right)\right] = 0$.
the objective would read as $$\text{minimize}_{\mathbf{x}} \sum_i \lvert y_i - \mathbf{a}_i^T\mathbf{x} \rvert^2, $$ which is easy to see that this matches with the Huber penalty function for this condition. Agree?
*
*if $\lvert\left(y_i - \mathbf{a}_i^T\mathbf{x}\right)\rvert \geq \lambda$, then $\left( y_i - \mathbf{a}_i^T\mathbf{x} \mp \lambda \right)$.
the objective would read as $$\text{minimize}_{\mathbf{x}} \sum_i \lambda^2 + \lambda \lvert \left( y_i - \mathbf{a}_i^T\mathbf{x} \mp \lambda \right) \rvert, $$ which almost matches with the Huber function, but I am not sure how to interpret the last part, i.e., $\lvert \left( y_i - \mathbf{a}_i^T\mathbf{x} \mp \lambda \right) \rvert$. Please suggest...
| The idea is much simpler. Use the fact that
$$\min_{\mathbf{x}, \mathbf{z}} f(\mathbf{x}, \mathbf{z}) = \min_{\mathbf{x}} \left\{ \min_{\mathbf{z}} f(\mathbf{x}, \mathbf{z}) \right\}.$$
In your case, the solution of the inner minimization problem is exactly the Huber function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
What's the u-v equation for u and v defined as follows? I can not solve the u-v equation for $u=\frac{1-r^2}{1-2r \cos \theta + r^2}$ for $0 < r < 1$ and $v=\frac{2r\sin\theta}{1-2r \cos \theta + r^2}$ for $0 < r < 1.$ I assume that it is a circle because it is the polar form of a bilinear function $ w=\frac{i(1-z)}{1+z}.$ I have tried to equate $u^2+(v-t)^2=k$, for some t and k. And figure out what is $t$ and $k$. But I failed. Is there any suggestion ? Thank you!
| Your expressions should read
\begin{align}
w &= \frac{i(1-z)}{1+z} \\
u+vi &= \frac{2r\sin \theta+i(1-r^2)}{1+2r\cos \theta+r^2} \\
\color{red}{u} &= \frac{2r\sin \theta}{1\color{red}{+}2r\cos \theta+r^2} \\
\color{red}{v} &= \frac{1-r^2}{1\color{red}{+}2r\cos \theta+r^2}
\end{align}
*
*$\theta=0 \implies
(u,v)=\left( 0, \dfrac{1-r}{1+r} \right)$
*$\theta=\pi \implies
(u,v)=\left( 0, \dfrac{1+r}{1-r} \right)$
*$2t=\dfrac{1-r}{1+r}+\dfrac{1+r}{1-r} \implies t=\dfrac{1+r^2}{1-r^2}$
*$2\sqrt{k}=\dfrac{1+r}{1-r}-\dfrac{1-r}{1+r} \implies \sqrt{k}=\dfrac{2r}{1-r^2}$
*Try to verify $$u^2+\left(v-\frac{1+r^2}{1-r^2} \right)^2=\frac{4r^2}{(1-r^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the value of $\int^{1}_{0}(1+x)^{m}(1-x)^{n} \,\mathrm d x$
Find the value of $$\displaystyle \int^{1}_{0} (1+x)^{m}(1-x)^{n} \,\mathrm d x$$ where $m, n \geq 1$ and $m, n \in \mathbb{N}$.
Let
$$\displaystyle I_{m,n} = \int^{1}_{0}(1+x)^m(1-x)^n \,\mathrm d x = \int^{1}_{0}(2-x)^m\cdot x^n \,\mathrm d x$$
Put $x=2\sin^2 \theta$ and $dx = 4\sin \theta \cos \theta \,\mathrm d \theta$ and changing limits, We get
$$\displaystyle I_{m,n} = 2^{m+n+2}\int^{\frac{\pi}{2}}_{0}\cos^{2m+1}\theta\cdot \sin^{2n+1}\theta \,\mathrm d \theta$$
How can I form a recursive relation? Could someone help me? Thanks.
| By parts,
$$I_{m,n}:=\int_0^1(1+x)^m(1-x)^ndx\\
=\left.\frac{(1+x)^{m+1}(1-x)^n}{m+1}\right|_0^1+\frac n{m+1}\int_0^1(1+x)^{m+1}(1-x)^{n-1}dx.$$
This gives you the recurrence relation
$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}I_{m+1,n-1}.$$
From this,
$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}\left(-\frac1{m+2}+\frac{n-1}{m+2}I_{m+2,n-2}\right)\\
=-\frac1{m+1}-\frac n{(m+1)(m+2)}+\frac{n(n-1)}{(m+1)(m+2)}I_{m+2,n-2}\\
=-\frac1{m+1}-\frac n{(m+1)(m+2)}-\frac{n(n-1)}{(m+1)(m+2)(m+3)}+\frac{n(n-1)(n-2)}{(m+1)(m+2)(m+3)}I_{m+3,n-3}\\
=\cdots\\
=-\sum_{k=0}^n\frac{m!n!}{(m+k+1)!(n-k)!}+\frac{n!}{(m+n+1)!}I_{m+n,0}.$$
The final integral is elementary, $\dfrac1{m+n+1}$. I don't know how to simplify the summation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le \int_5^8 \left(1-\frac{12}{2x+5}\right)dx\le\frac{9}{7}$$ Whilee the first one could be reduced to what is needed, how can I deal with the right bound?
| $f(x)=\frac{2x-7}{2x+5}$ is a concave function on $[5,8]$, hence by the Hermite-Hadamard inequality the wanted integral is bounded between
$$ \frac{1}{2}f(5)+f(6)+f(7)+\frac{1}{2}f(8) = \frac{11043}{11305} > \frac{42}{43}$$
and
$$ f\left(\tfrac{11}{2}\right)+f\left(\tfrac{13}{2}\right)+f\left(\tfrac{15}{2}\right) = \frac{59}{60}<1.$$
As an alternative, one may notice that the exact value is given by $3-6\log\left(1+\frac{2}{5}\right)$, and by Padé approximants $\log(x+1)$ is bounded between $\frac{x}{1+\frac{x}{2}}$ and $\frac{x+\frac{x^2}{6}}{1+\frac{2x}{3}}$ over $[0,1]$. This gives that the wanted integral is between $\frac{93}{95}$ and $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
example of an inhomogenenous markov chain I have found a very interesting exercice about inhomogeneous markov chains, which I think you are also interested in and hope someone can solve it.
Let $X_{1},X_{2},...$ be a inhomogeneous markov chain with state space $\{0,1\}$ with transition probabilities
$$
\mathbb{P}(X_{n+1} = 1 | X_{n} = 0) = \frac{1}{n^{2}}\\
\mathbb{P}(X_{n+1} = 0 | X_{n} = 1) = \frac{1}{n^{2}}.
$$
*
*$X := \lim_\limits{n \to \infty} X_{n}$ almost surely exists.
*If $\mathbb{P}(X_{1} = 1) = \mathbb{P}(X_{1} = 0) = \frac{1}{2}$, then $\mathbb{P}(X = 1) = \frac{1}{2}$.
*If $X_{1} \equiv 1$ then $\mathbb{P}(X = 1) > \frac{1}{2}$.
Im very interested how you would solve it. The solution was discussed in the lecture but I hope that I see a different approach. Maybe with matrix calculation
| So the markov matrices is,
$$
P_n = \begin{pmatrix}
1-\frac{1}{n^2} & \frac{1}{n^2} \\
\frac{1}{n^2} & 1-\frac{1}{n^2}
\end{pmatrix} =
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\frac{1}{n^2}
$$ Hence
$$
P_n = I + A\frac{1}{n^2}
$$ Now this means that all matrices $P_n$ can be simultaniously diagonalize. Note that for $v_1 = (1,1)^t$ $Av_1 = 0$ and for $v_2 = (1,-1)^t$ we have
$$
Av_2 = -2 v_2
$$ Hence to find the final probability matrix we need to calculate
$$
\prod_{i=1}^nP_i \to c = \prod_{i=1}^n\left( 1 - \frac{2}{n^2} \right )
$$ Note https://www.wolframalpha.com/input/?i=prod+(1-2%2Fi%5E2)
Then we get probability matrix as
$$
\prod_{i=1}^nP_i = \frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&c\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = \begin{pmatrix}\frac{1+c}{2}&\frac{1-c}{2}\\\frac{1-c}{2}&\frac{1+c}{2}\end{pmatrix},
$$ with $c=-0.22$. Now the convergence of the product proves the limit and the evaluation of the matric putting numbers of c in it lead to the answers of 2,3, 3 looks wrong, note that I included the case $n=1$ this case just swaps the states and from this we note that 3 is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Wave problem $u_{tt}=u_{xx}-u_t, 0\le x\le 2\pi$ with $u(x,0)=\phi(x)$, $u_t(x,0)=\psi(x)$ and $u(0,t)=u(2\pi,t)=0$ Solve the wave problem $$\begin{cases}u_{tt}=u_{xx}-u_t, 0\le x\le 2\pi\\ u(0,t)=u(2\pi,t)=0\\ u(x,0)=\phi(x)\\ u_t(x,0)=\psi(x)\end{cases}$$ Write the eigenvalues and eigenfunctions explicitly.
Are they orthogonal?
We propose the separated solution $u(x,t)=X(x)T(t)$.
Then $\frac{(T''+T')}{T}=\frac{X''}{X}=\lambda$.
If we define $\lambda=\beta^2,$ we have $X(x)=C\cos\beta x+D\sin\beta x$ and $T(t)=Ae^{r_1t}+Be^{r_2t}, r=-1/2\pm\frac{\sqrt{1+4\beta^2}}{2}$.
Applying boundary condition, we find that $X_n(x)=D_n\sin(\lambda_nx),\lambda=\sqrt{n/2}$
Applying initial conditions we find that $T(0)=A+B=\phi(x)$ and $T'(0)=Ar_1+Br_2=\psi(x)$
My question here is how will I find $A$ and $B$ if we don't know $\phi$ nor $\psi$ ?
How will I find the eigenvalues ?
Help me please
| The boundary conditions give
$$ X(x) = \sin\left(\frac{n}{2}x\right) $$
As usual, we drop the remaining constant for now, since it'll absorbed into the constants of $T(t)$
This means $\lambda_n = X''/X = -n^2/4 = -\beta^2$ (you have a sign error). The other equation becomes
$$ T'' + T' + \frac{n^2}{4}T = 0 $$
The two roots of the characteristic are
$$ r = \frac{-1 \pm \sqrt{1-n^2}}{2} $$
Since $n \ge 1$, we rewrite
$$ r = \frac{-1 \pm \sqrt{-(n^2-1)}}{2} = \frac{-1 \pm i\sqrt{n^2-1}}{2} $$
This gives the solution
$$ T(t) = e^{-t/2}\left[A\cos\left(\frac{\sqrt{n^2-1}}{2}t\right) + B\sin\left(\frac{\sqrt{n^2-1}}{2}t\right) \right] $$
For ease of notation, let $\mu_n = \sqrt{n^2-1}$ to make things shorter, so we have
\begin{align} T(t) &= e^{-t/2}\left[A\cos\left(\frac{\mu}{2}t\right) + B\sin\left(\frac{\mu}{2}t\right) \right] \\
T'(t) &= e^{-t/2}\left[\left(-\frac{A}{2} + \frac{\mu B}{2}\right)\cos\left(\frac{\mu}{2}t\right) + \left(-\frac{B}{2} - \frac{\mu A}{2} \right)\sin \left(\frac{\mu}{2}t\right) \right] \end{align}
As usual, the full solution in series form is
$$ u(x,t) = \sum_n X_n(x)T_n(t) = \sum_{n=1}^\infty \sin\left(\frac{n}{2}x\right) e^{-t/2}\left[A\cos\left(\frac{\mu}{2}t\right) + B\sin\left(\frac{\mu}{2}t\right) \right] $$
We can now plug in the given initial conditions
\begin{align} u(x,0) &= \sum_n X_n(x)T_n(0) = \sum_{n=1}^\infty B_n \sin\left(\frac{n}{2}x\right) = \phi(x) \\
u_t(x,0) &= \sum_n X_n(x){T_n}'(0) = \sum_{n=1}^\infty \left(-\frac{A_n}{2} + \frac{\mu B_n}{2}\right) \sin\left(\frac{n}{2}x\right) = \psi(x)
\end{align}
You'll need to find the Fourier series of these to functions to obtain the coefficients
\begin{align}
B_n &= \frac{1}{\pi}\int_0^{2\pi} \phi(x) \sin\left(\frac{n}{2}x\right) dx \\
\frac{-A_n + \mu B_n}{2} &= \frac{1}{\pi} \int_0^{2\pi} \psi(x) \sin\left(\frac{n}{2}x\right) dx
\end{align}
and solve the resulting system of equations. There's no specific form for the solution here, but you'll know what to do when you are given specific $\phi(x)$ and $\psi(x)$, i.e. just do the integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829941",
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"source": "stackexchange",
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Trying to find the mean of a Random Variable but I cannot do the integration. Please consider the problem below and my partial solution to it. Is it right so far? I do not know how to perform the integration and I am hoping somebody can point me in the right direction.
Thanks,
Bob
Problem:
The joint pdf of a bivariate r.v. $(X,Y)$ is given by:
\begin{eqnarray*}
f_{XY}(x,y) &=& \frac{1}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) } \\
\end{eqnarray*}
Find the mean of $X$.
Answer:
\begin{eqnarray*}
u_x &=&
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x \frac{1}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) } \, dy \, dx \\
\end{eqnarray*}
Now to perform this double integral, I am going to do it in polar coordinates.
\begin{eqnarray*}
u_x &=&
\int_{0}^{2 \pi} \int_{0}^{\infty}
r^2 \cos{\theta} \frac{1}{\sqrt{3 \pi } }
e^{ -\frac{2}{3} ( r^2 - r^2 \cos{\theta} \sin{\theta} ) }
\, dr \, d\theta \\
u_x &=&
\int_{0}^{2 \pi} \int_{0}^{\infty}
r^2 \cos{\theta} \frac{1}{\sqrt{3 \pi } }
e^{ \frac{2}{3}r^2 ( \cos{\theta} \sin{\theta} - 1 ) }
\, dr \, d\theta \\
\end{eqnarray*}
How do I complete the integration?
| Don't go polar. Complete the square instead: $x^2 - xy + y^2 = \frac{3}{4}x^2 + (y - \frac{x}{2})^2$. The integrand $$\frac{x}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) }$$ becomes
$$\frac{x}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} e^{-\frac{2}{3}(y-\frac{x}{2})^2}$$
Integrate it first over $dy$. Can you continue?
| {
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If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ Where $M$ is the minimal polynomial
$x^3-x=x(x^2-1)=x(x-i)(x+i)$ so $A$ is diagonalizable
$A=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}P$
$A^2=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}^2P=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}^2P$
So $P_{A^2}(x)=x(x-1)^2$ but $A^2(A^2-1)=0$ so $M_{A^2}(x)=x(x-1)=x^2-x$
Is there another way to conclude it?
| From $M_A(x) = x(x-1)(x+1)$ we conclude that $A$ is diagonalizable and $\sigma(A) = \{0, -1, 1\}$.
Hence $A^2$ is also diagonalizable and $\sigma(A)^2 = \sigma(A)^2 = \{0,1\}$. Therefore, the minimal polynomial $M_{A^2}$ has only linear factors and has zeroes $0, 1$.
The only option is $M_{A^2}(x) = x(x-1)$.
| {
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"source": "stackexchange",
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Then general values of $\theta$ in inverse Trigo sum
If $\displaystyle \theta = \tan^{-1}\bigg(2\tan^2 \theta\bigg)-\frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg).$ Then general values of $\theta$ is
Try: Let $\alpha =\tan^{-1}\bigg(2\tan^2 \theta\bigg)\Rightarrow \tan \alpha =2 \tan^2 \theta$
and $\displaystyle \beta = \frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg)\Rightarrow \sin (2\beta) = \frac{3\sin 2 \theta}{5+4\cos 2 \theta} = 3\frac{2\tan \theta}{1+\tan^2 \theta}\cdot \frac{1}{5+4\bigg(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\bigg)}$
So $\displaystyle \sin (2\beta) = \frac{6\tan \theta}{9+\tan^2 \theta}.$
could some help me how to find range of $\theta$. Thanks
| Hint:
multiply the original equation by $2$, take the sine of both sides and remember that $$\cot = \frac 1\tan\\1+\tan^2 = \sec^2\\1 + \cot^2 = \csc^2$$
Added:
$$\theta = \tan^{-1}u - \frac 12 \sin^{-1}v\\2\theta = 2\tan^{-1}u -\sin^{-1}v\\\sin 2\theta = \sin\left(2\tan^{-1}u -\sin^{-1}v\right)\\\sin 2\theta = \sin(2\tan^{-1}u)\cos(\sin^{-1}v) - \cos(2\tan^{-1}u)\sin(\sin^{-1}v) $$
Now you just need to apply double angle formulas, then reduce the various "function of inverse function" combinations, which all have strictly algebraic expressions (hinted at by the formulas above).
In the end, you can reduce it down to an algebraic expression involving only $\sin \theta$ and $\cos \theta$, and since cosine is positive on the range of $\sin^{-1}$ and $\tan^{-1}$, you can convert that to strictly $\sin \theta$. Solve the resulting eequation for $\sin \theta$ (I haven't worked it that far, so I don't know how nasty that will be to do), and then back out the values of $\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Equation for product of sequence involving 2 Supposing for all even numbers less that or equal to n say 16,
for example the set:
2, 4, 6, 8, 10, 12, 14, 16
2.4.6.8.10.12.14.16 = 10,321,920 = $2^{15}. 3^{2}. 5^{1}. 7^1$
I would like to find an equation that gives me the products of 2's only not the other factors in the case of $n=16$ this would be:
$2\,.\,2\,.\,2^2\,.\,2\,.\,2^3\,.\,2\,.\,2\,.\,2\,.\,2^4\,=\,2^{15}$
Another example is n = 14
2.4.6.8.10.12.14 = 645,120 = $2^{11}. 3^2. 5^1. 7^1$
this would be $2\,.\,2\,.\,2^2\,.\,2\,.\,2^3\,.\,2\,.\,2\,.\,2\,=\,2^{11}$
Can it be done?
| Hint
$$2 \cdot 4 \cdot \cdots \cdot 2n = (2 \cdot 1) \cdot (2 \cdot 2) \cdot \cdots \cdot (2 \cdot n) = (2 \cdot 2 \cdot \cdots \cdot 2) \cdot (1 \cdot 2 \cdot \cdots \cdot n) .$$
Now, see this answer, which gives a formula for the highest power of a prime dividing $n!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this question by Ramanujan? click here for photo
$$1+2\sum_{k=1}^\infty \frac{1}{(4k)^3-(4k)}= \frac{3}{2}\ln(2)\,.$$
well i have attatched a photo which has been asked to prove without using calculus,but how to solve this using calculus ?
| I have no clue how to solve this without calculus. At least, I do not know how to define the natural logarithm without calculus. I am borrowing some part of this solution from mechanodroid's deleted solution.
First, write
$$\frac{1}{(4k)^3-(4k)}=-\frac{1}{4k}+\frac{1}{2(4k-1)}+\frac{1}{2(4k+1)}$$
for every positive integer $k$. Define
$$S_n:=1+2\,\sum_{k=1}^n\,\frac{1}{(4k)^3-(4k)}$$
for all $n=1,2,3,\ldots$. Hence, the $n$-th partial sum is given by $$
\begin{align}
S_{n}&=1+\sum_{k=1}^{n}\,\left(-\frac{1}{2k}+\frac{1}{4k-1}+\frac{1}{4k+1}\right)
\\&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2n}\right)+\left(\frac{1}{2n+1}+\frac{1}{2n+3}+\ldots+\frac{1}{4n+1}\right)
\\&=T_n+U_n\,,
\end{align}$$
where $T_n:=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2n}$ and $U_n:=\frac{1}{2n+1}+\frac{1}{2n+3}+\ldots+\frac{1}{4n+1}$. It remains to show that $$\lim_{n\to\infty}\,T_n=\ln(2)\text{ and }\lim_{n\to\infty}\,U_n=\frac{1}{2}\,\ln(2)\,.$$
The former is well known, whilst the latter follows from the fact that $2U_n$ is a Riemann sum for $\int_{2}^{4}\,\frac{1}{x}\,\text{d}x$. Alternatively, observe that $T_n=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$, and
$$\frac{1}{2} \,T_{n+1}\leq U_n\leq \frac{1}{2} T_{n+1}+\frac{1}{2n+1}-\frac{1}{4n+2}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Meaning of the sign of a complex number What is the meaning and the practical uses of the sign (signum function) of a complex number $z$ defined as $\frac{z}{|z|}$? Does it also extend to quaternions?
| You can think of a complex number $z=a+ib$ as a vector in the complex plane, and as such it has a direction and a length. The sign of $z$ gives the normalized direction of $z$ as a unit vector lying on the unit circle. We normalise $z$ by dividing by its modulus:
$$\operatorname{sgn}(z)=\frac{z}{|z|}=\frac{a+bi}{\sqrt{a^2+b^2}}$$
This gives a unit complex number. For example take $z=5+3i$. Then
$$\operatorname{sgn}(5+3i)=\frac{5+3i}{|5+3i|}=\frac{5}{\sqrt{34}}+\frac{3}{\sqrt{34}}i$$
where
$$\left(\frac{5}{\sqrt{34}}\right)^2+\left(\frac{3}{\sqrt{34}}\right)=1$$
For a quaternion $q=a+bi+cj+dk$, the case is similar:
$$\operatorname{sgn}(q)=\frac{a+bi+cj+dk}{|a+bi+cj+dk|}=\frac{a+bi+cj+dk}{\sqrt{a^2+b^2+c^2+d^2}}$$
For example take $q=1+2i+3j+5k$. Then
$$\operatorname{sgn}(1+2i+3j+5k)=\frac{1+2i+3j+5k}{|1+2i+3j+5k|}=\frac{1}{\sqrt{39}}+\frac{2}{\sqrt{39}}i+\frac{3}{\sqrt{39}}j+\frac{5}{\sqrt{39}}k$$
where
$$\left(\frac{1}{\sqrt{39}}\right)^2+\left(\frac{2}{\sqrt{39}}\right)+\left(\frac{3}{\sqrt{39}}\right)^2+\left(\frac{5}{\sqrt{39}}\right)^2=1$$
Note in the complex case we have the modulus $|z|=\sqrt{z\cdot\bar{z}}=\sqrt{(a+bi)\cdot(a-bi)}=\sqrt{a^2+b^2}$, where $\bar{z}$ is the complex conjugate. In the quaternion case we have $|q|=\sqrt{q\cdot\bar{q}}=\sqrt{(a+bi+cj+dk)\cdot(a-bi-cj-dk)}=\sqrt{a^2+b^2+c^2+d^2}$, where $\bar{q}$ is the quaternion conjugate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integral of product of error function difference In the course of my research I came across the following integral:
$$\int_{-\infty}^{\infty}\,\left(\operatorname{erf}\left(ax-b\right) -\operatorname{erf}\left(\frac{a}{\gamma} x-b-
\dfrac{ar}{\gamma}\right) \right)
*
\left(\operatorname{erf}\left(cx-d\right)-\operatorname{erf}\left(\frac{c}{\gamma}x-d-\frac{cr}{\gamma}\right)\right) \, dx$$
, where $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$ is the error function
Does anyone know if the integration given above can be calculated into closed-form?
Thank you in advance.
| Let us take $a \ge 0$, $b \ge 0$, $c \ge 0$, $d \ge 0$, $\gamma \ge 0$ and $r \ge 0$ and let us define:
\begin{eqnarray}
{\mathfrak I}^{(a,b)}_{c,d}(\gamma,r):=
\int_{-\infty}^{\infty}\,\left(\operatorname{erf}\left(ax-b\right) -\operatorname{erf}\left(\frac{a}{\gamma} x-b-
\dfrac{ar}{\gamma}\right) \right)
*
\left(\operatorname{erf}\left(cx-d\right)-\operatorname{erf}\left(\frac{c}{\gamma}x-d-\frac{cr}{\gamma}\right)\right) \, dx
\end{eqnarray}
Then by differentiating with respect to the parameter $b$ and using the identity:
\begin{equation}
\int\limits_{-\infty}^\infty e^{-u^2} \text{erf}(A u + B) du=\sqrt{\pi} \text{erf} \left( \frac{B}{\sqrt{1+A^2}}\right)
\end{equation}
we get:
\begin{eqnarray}
&&\frac{\partial }{\partial b} {\mathfrak I}^{(a,b)}_{c,d}(\gamma,r) =
\frac{2}{a} \left( \right. \\
&&
-\text{erf}\left(\frac{\frac{c (a r-b)}{a g}+d}{\sqrt{\frac{c^2}{a^2 g^2}+1}}\right)+\text{erf}\left(\frac{d-\frac{b c}{a}}{\sqrt{\frac{c^2}{a^2}+1}}\right)+g \left(\text{erf}\left(\frac{d-\frac{b c}{a}}{\sqrt{\frac{c^2}{a^2}+1}}\right)-\text{erf}\left(\frac{d-\frac{c (a r+b
g)}{a}}{\sqrt{\frac{c^2 g^2}{a^2}+1}}\right)\right) \\
&&\left.\right)
\end{eqnarray}
Now all we need to do is to integrate with respect to $b$ from minus infinity to $b$.
Since $\int \text{erf}(x) dx = \exp(-x^2)/\sqrt{\pi} + x \text{erf}(x)$ and since in the expression for the partial derivative the parameter $b$ enters the error function in a linear manner only it is clear that it is possible to construct the anti-derivative. Having done this we just take the values at $b$ and at minus infinity . It turns out that the later value is equal to zero and therefore we are left with the value at $b$ only which we simplified by hand. Now the final result reads:
\begin{eqnarray}
&&{\mathfrak I}^{(a,b)}_{c,d}(\gamma,r) = \frac{2}{a c} \left( \right. \\
&& -\frac{(g+1) \sqrt{a^2+c^2} e^{-\frac{(b c-a d)^2}{a^2+c^2}}}{\sqrt{\pi } } + \frac{\sqrt{a^2 g^2+c^2} e^{-\frac{(a c r+a d g-b c)^2}{a^2 g^2+c^2}}+\sqrt{a^2+c^2 g^2} e^{-\frac{(a c r-a d+b c g)^2}{a^2+c^2 g^2}}}{\sqrt{\pi } }+\\
&& (g+1) (a d-b c) \text{erf}\left(\frac{b c-a d}{\sqrt{a^2+c^2}}\right)+\\
&& (a c r-a d+b c g) \text{erf}\left(\frac{a c r-a d+b c g}{\sqrt{a^2+c^2 g^2}}\right)+(a c r+a d g-b c) \text{erf}\left(\frac{a c r+a d g-b c}{\sqrt{a^2 g^2+c^2}}\right)\\
\left. \right)
\end{eqnarray}
For[count = 1, count <= 100, count++,
{a, b, c, d, g, r} = RandomReal[{0, 1}, 6, WorkingPrecision -> 50];
I1 = NIntegrate[(Erf[(a x - b)] -
Erf[(a/g x - b - a r/g)]) (Erf[c x - d] -
Erf[c/g x - d - c r/g]), {x, -Infinity, Infinity},
WorkingPrecision -> 15];
2/Abs[a] NIntegrate[ (-Erf[(xi c - d a)/Sqrt[a^2 + c^2]] +
Erf[(-d a g + c (xi - a r))/Sqrt[a^2 g^2 + c^2]] +
Abs[g] (-Erf[(xi c - d a)/Sqrt[a^2 + c^2]] +
Erf[(-d a + c (xi g + a r))/Sqrt[
a^2 + c^2 g^2]])), {xi, -Infinity, b}];
I2 = 2/(
Abs[a] c) (-E^(-((b c - a d)^2/(a^2 + c^2))) Sqrt[a^2 + c^2] /
Sqrt[\[Pi]] (1 + g) +
1/ Sqrt[\[Pi]] (E^(-((-b c + a d g + a c r)^2/(c^2 + a^2 g^2)))
Sqrt[c^2 + a^2 g^2] +
E^(-((-a d + b c g + a c r)^2/(a^2 + c^2 g^2))) Sqrt[
a^2 + c^2 g^2]) + (a d - b c) (1 + g) Erf[(b c - a d)/Sqrt[
a^2 + c^2]] + (-a d + b c g + a c r) Erf[(-a d + b c g +
a c r)/Sqrt[
a^2 + c^2 g^2]] + (-b c + a d g + a c r) Erf[(-b c + a d g +
a c r)/Sqrt[c^2 + a^2 g^2]]);
If[Abs[I2/I1 - 1] > 10^(-3),
Print["results do not match", {a, b, c, d, g, r, {I1, I2}}];
Break[]];
If[Mod[count, 10] == 0, PrintTemporary[count]];
];
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate a supremum over the unit sphere of $\mathbb{C}^2$
I want to calculate
$$K=\sup\left\{\left|y\overline{x}+|y|^2\right|^2+1;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\right\}.$$
I try to solve the problem as follows:
let $x=r_1e^{i\theta_1}$ and $y=r_2e^{i\theta_2}$ then
$$ |x|^2+ |y|^2=1 \quad \Rightarrow \quad r^2_1+r_2^2 =1.$$
and
$$y\overline{x}+|y|^2= r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}.$$
so
$$|r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}|^2\leq r_2^4+r_1^2r_2^2.$$
Hence
$$K\leq\sup\{r_2^4+r_1^2r_2^2;\;\;r_1,r_2\geq0,\;r^2_1+r_2^2 =1\}.$$
| We have
$$
f(x,y) = (y\bar x+y\bar y)(\bar y x+\bar y y)+1 = |y|^2\left(|x|^2+|y|^2+y\bar x+\bar y x\right)+1 = |y|^2\left(y\bar x+\bar y x+1\right)+1
$$
now calling
$$
x = \rho_x e^{i\phi}\\
y = \rho_y e^{i\psi}
$$
$$
f(\rho_x e^{i\phi},\rho_y e^{i\psi}) = \rho_y^2\left(2\rho_2\rho_y\cos(\phi-\psi)+1\right)+1
$$
so the problem now reads
$$
\sup f(\rho_x e^{i\phi},\rho_y e^{i\psi}) \;\;\;\mbox{subjected to}\;\;\; \rho_x^2+\rho_y^2 = 1
$$
Here $\rho_x \ge 0, \rho_y \ge 0$ so the problem can be simplified to
$$
\sup \rho_y^2\left(2\rho_x\rho_y+1\right)+1 \;\;\;\mbox{subjected to}\;\;\; \rho_x^2+\rho_y^2 = 1
$$
This problem can be solved easily with the Lagrange multipliers technique giving the results
$$
\rho_x = \frac{84-28 \sqrt{2}-7 \sqrt{7} \left(1+2 \sqrt{2}\right)^{3/2}+3 \sqrt{7 \left(1+2 \sqrt{2}\right)}+6 \sqrt{14 \left(1+2
\sqrt{2}\right)}}{28 \left(\sqrt{2}-\sqrt{7 \left(1+2 \sqrt{2}-3\right)}\right)}\\
\rho_y = \frac{1}{14} \left(6-2 \sqrt{2}+7 \sqrt{\frac{4}{7}+\frac{8
\sqrt{2}}{7}}\right)
$$
and the maximum is
$$
\frac{3283+392 \sqrt{2}+10 \sqrt{7 \left(835+872 \sqrt{2}\right)}}{2401}\approx 2.09937
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$
$$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$
$$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$
and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$
Could some help me how to find its upper bound, Thanks
| Let $\boldsymbol{x}=(a_1,a_2,a_3,a_4) \in \mathbb{R}^4$,
$$E=\boldsymbol{x}
\begin{pmatrix}
2 & -1 & 0 & -1 \\
-1 & 2 & -1 & 0 \\
0 & -1 & 2 & -1 \\
-1 & 0 &-1 & 2
\end{pmatrix}
\boldsymbol{x}^T$$
*
*Eigenvalues: $$\lambda_1=4, \, \lambda_2=\lambda_3=2, \, \lambda_4=0$$
*Unit eigenvectors:
\begin{align}
\boldsymbol{v}_1 &= \frac{(-1,1,-1,1)}{2} \\
\boldsymbol{v}_2 &= \frac{(0,-1,0,1)}{\sqrt{2}} \\
\boldsymbol{v}_3 &= \frac{(-1,0,1,0)}{\sqrt{2}} \\
\boldsymbol{v}_4 &= \frac{(1,1,1,1)}{2} \\
\boldsymbol{v}_i \cdot \boldsymbol{v}_j &= \delta_{ij}
\end{align}
*
*Note that $\boldsymbol{v}_k$ satisfies $a_1+a_2+a_3+a_4=0$ for $k=1$, $2$ or $3$ whereas $\boldsymbol{v}_4$ is normal to the hyperplane $a_1+a_2+a_3+a_4=0$. Therefore,
$$\boldsymbol{x}=
\alpha \, \boldsymbol{v}_1+
\beta \, \boldsymbol{v}_2+
\gamma \, \boldsymbol{v}_3$$
$$\lambda_{2,3} \Vert \boldsymbol{x} \Vert^2
\le E \le
\lambda_{1} \Vert \boldsymbol{x} \Vert^2$$
$$2\le E \le 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but it is hard to guess here.
So I set $z=x+iy$ and got
$$(x^3-y^2x+2xy^2-2x^2+2y^2+\frac{1}{4})+i(-2x^2y+yx^2-y^3+4xy)=0$$
Looking at the imaginary part we get
$y(-x^2+4x-y^2)=0$ so or $y=0$ or $-x^2+4x-y^2=0\iff (x-2)^2+y^2=4$
But how I continue from here? and what does
I read and now I am given it a try:
for a. we look at $|z|<1$ and $\frac{1}{4}<|z|$ we look at the boundary
so fo $|z|=1$ $|z^3+\frac{1}{4}|<|-2z^2|$ so we choose $g(z)=z^3+\frac{1}{4}$ and $f(z)=-2z^2$
So $|z^3+\frac{1}{4}|\leq 1+\frac{1}{4}\leq 2=|-2z^2|$
So we can conclude that there are $2$ zeros in $|z|<1$?
| It is not too difficult to see that all three roots are real.
The roots are the same as roots of $$ x^3 -2x^2 +1/4 =0$$
Zeros could be approximated by Newton's method as $$ -0.32772....\\ 0.39462....\\ 1.9331...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$ I would like to work this out:
$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$
Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$
$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$
Making a sub: $u=\tan^2(y)$, $du=\sec^2(y)dy$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\ln\sec(y)}{\sec^2(y)\sqrt[3]{\tan^4(y)}}\mathrm dy$$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\cot(y)\cos^2(y)\ln\sec(y)}{\sqrt[3]{\tan(y)}}\mathrm dy$$
I can't continue.
Maybe there is another alternative way to simplify $I$
| In this solution we will not use the beta function or the digamma function.
Put
\begin{equation*}
f(s)=\int_{0}^{\infty}\dfrac{\ln(1+s^3x^3)}{(1+x^3)^2}\, \mathrm{d}x.
\end{equation*}
We want to calculate $f(1)$. However,
\begin{equation*}
f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s\tag{1}
\end{equation*}
and
\begin{equation*}
f'(s)=\int_{0}^{\infty}\dfrac{3s^2x^3}{(1+s^3x^3)(1+x^3)^2}\, \mathrm{d}x =[x^3=z] = \int_{0}^{\infty}\dfrac{s^2z^{1/3}}{(1+s^3z)(1+z)^2}\, \mathrm{d}z .
\end{equation*}
For $0<s<1$ we use keyhole contour integration and get
\begin{equation*}
f'(s) = \dfrac{2\pi\sqrt{3}}{9}\dfrac{2s+1}{(s^2+s+1)^2}\cdot s^2 .
\end{equation*}
Integration by parts in (1) will finally give us
\begin{equation*}
f(1) = \dfrac{2\pi}{27}\left(\sqrt{3}(3\ln 3 -1)-\pi\right).
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Factorization Manipulation Let $x$ and $y$ be real numbers. Consider $t=x^2+10y^2-6xy-4y+13$. So what is the $t$ as smallest number? The solution is $9$.
My trying:
$$t=x^2+10y^2-6xy-4y+13$$
$$=x^2+9+10y^2-6xy-4y+4$$
$$=(x+3)^2-3x+2y(5y-2)-6xy+4$$.
So let $x=-3$ and $y=0$. From this I got $t=13$. My answer is false. Can you help?
| Notice that
$$t=x^2+10y^2-6xy-4y+13=x^2-6y \cdot x+10y^2-4y+13,$$which could be regarded as a quadratic function with respect to the variable $x$.
Hence, $$t \geq \dfrac{4\cdot 1 \cdot (10y^2-4y+13)-(-6y)^2}{4 \cdot 1}=y^2-4y+13=(y-2)^2+9\geq 9,$$ with the equality holding if and only if $x=-\dfrac{-6y}{2 \cdot 1}$ and $y=2$,namely, $x=6,y=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equivalence in positive definiteness of $AB$ and $BA$ Let $A,B$ any two real matrices, with $AB$ and $BA$ not necessarily symmetric. I know that $AB$ and $BA$ have the same eigenvalues. Is it true that $AB$ is positive definite iff $BA$ is positive definite? The definition I use for positive definiteness of a general real matrix is that
$$ u^T A u > 0 $$
for all non-zero real vectors $u$. This is equivalent to the symmetric part of $A$ being positive definite.
| The answer is no.
Consider $A = \begin{bmatrix}2 & 1 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix}$.
Then $AB = \begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$ is positive definite:
$$\begin{bmatrix}x & y\end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}=\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}x+y \\ -x+y \end{bmatrix}=(x^2+xy)+(-xy+y^2)=x^2+y^2$$
But $BA = \begin{bmatrix} 2 & 1 \\ -2 & 0 \end{bmatrix}$ is not:
$$\begin{bmatrix}1 & 3\end{bmatrix}\begin{bmatrix} 2 & 1 \\ -2 & 0 \end{bmatrix}\begin{bmatrix}1 \\ 3 \end{bmatrix} = \begin{bmatrix}1 & 3\end{bmatrix}\begin{bmatrix}5 \\ -2 \end{bmatrix} = -1$$
| {
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| My answer is a little roundabout but without calculus and without pictures or symmetry:
Arithmetic-geometric inequality:
$$ a+b \geq 2\sqrt{ab} $$
Harmonic-geometric inequality and some rearrangement:
$$ \sqrt{ab} \geq \frac{2}{\frac{1}{a}+\frac{1}{b}}$$
$$ \frac{1}{a} + \frac{1}{b}\geq \frac{2}{\sqrt{ab}}$$
Add both results to get
$$ a+b+\frac{1}{a}+\frac{1}{b} \geq2\left(\sqrt{ab}+\frac{1}{\sqrt{ab}}\right)~~~~~~~~~~(1)$$
Also note that by the given constraint
$$ 0\leq(a-b)^2 = a^2+b^2-2ab = 1 - 2ab$$
and therefore
$$ \sqrt{ab} \leq \frac{1}{2}\sqrt{2}$$
Now let $x:=\sqrt{ab}$.
So $0\leq x \leq \frac{1}{2}\sqrt{2} < 1$
To finish off, all we need to show is that the right-hand side $2(x+\frac{1}{x})$ of inequality $(1)$ is minimal if $x$ is maximal and therefore $\frac{1}{2}\sqrt{2}$ because then
$$ a+b+\frac{1}{a}+\frac{1}{b} \geq2\left(\frac{1}{2}\sqrt{2}+\sqrt{2}\right) = 3 \sqrt{2}$$
is always true.
To show that, let's show that the function $f:x\mapsto x+\frac{1}{x}$ is decreasing on the interval $(0,1)$. That's easy with calculus. Without:
Let's choose $h,x$ arbitrarily such that $0<h<1$ and $0<x<x+h<1$.
Then rearrange equivalently or backwards-implicatively to get to our monotonicity claim from a true statement
$$x+h+\frac{1}{x+h} < x+\frac{1}{x}$$
$$ h+\frac{1}{x+h} < \frac{1}{x}$$
Multiply through
$$ hx(x+h) + x < x + h$$
And since $h+x < 1$:
$$\Leftarrow hx + x \leq x +h $$
Since also $x<1$:
$$\Leftarrow h + x \leq h+x $$
which is true. Since $x,h$ were arbitrary from $(0,1)$, this proves monotonicity and hence the claim.
| {
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} |
How to explain powers of $(x+1)^{2^n}$ appearing in the Babylonian approximation of $\sqrt x$? I'm working with this iteration used for approximating square roots and trying to see what I can draw out from it, and in doing so I found something very strange that I can't logically explain. I'm looking for any insight into why this is the case or perhaps a proof of it. The iteration is as follows:
$$\rho_{n+1}=\frac{(\rho_n)^2+x}{2\rho_n},\rho_0=1$$
which approximates $\sqrt x$
I'll list the first four general results here:
$$\rho_1=\frac{x+1}{2}$$
$$\rho_2=\frac{x^2+6x+1}{4x+4}$$
$$\rho_3=\frac{x^4+28x^3+70x^2+28x+1}{8x^3+56x^2+56x+8}$$
$$\rho_4=\frac{x^8+120x^7+1820x^6+8008x^5+12870x^4+8008x^3+1820x^2+120x+1}{16x^7+560x^6+4368x^5+11440x^4+11440x^3+4368x^2+560x+16}$$
The property I spotted was that in all cases, the co-efficients of $(x+1)^{2^n}$ appear in $\rho_n$, with the co-efficients of even powers on the numerator and of odd powers on the denominator of $\rho_n$, in such a way that they snake through the polynomial fractions. For example
$$(x+1)^8=x^8+8x^7+28x^6+56x^5+70x^4+56x^3+28x^2+8x+1$$ and a comparison with $\rho_3$ shows my point nicely.
I can show the general results as sums of multiples of powers of $(x+1)$ and $x$, e.g.
$$\rho_2=\frac{(x+1)^2+4x}{4(x+1)}$$
$$\rho_3=\frac{(x+1)^4+24x(x+1)^2+16x}{8(x+1)^3+32x(x+1)}$$
however I don't know how much use this is in explaining the property I found.
Any ideas will be appreciated.
| Possible fixed points of the iteration are $\sqrt x$ and $-\sqrt x$,
this suggests to look at
$$
\frac{\rho_{n+1} - \sqrt x}{\rho_{n+1} + \sqrt x} =
\frac{\rho_n^2 - 2 \sqrt x \rho_n + x}{\rho_n^2 + 2 \sqrt x \rho_n + x} =
\left( \frac{\rho_n - \sqrt x}{\rho_n + \sqrt x} \right)^2 \, .
$$
It follows that
$$
\frac{\rho_n - \sqrt x}{\rho_n + \sqrt x} = \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)^{2^n} \, .
$$
This implies (quadratic) convergence of $\rho_n$ to $\sqrt x$ because
the fraction on the right-hand side is of absolute value less than one
if $x > 0$.
It also gives the explicit formula
$$
\rho_n=\frac{(\sqrt x+1)^{2^n}+(\sqrt x-1)^{2^n}}
{(\sqrt x+1)^{2^n}-(\sqrt x-1)^{2^n}}\sqrt x
$$
which Lord Shark the Unknown found.
If we expand all expressions using the binomial formula then
all odd powers of $\sqrt x$ cancel in the numerator, and all even
powers of $\sqrt x$ cancel in the denominator. This gives the
expression (first found by achille hui in a now deleted answer):
$$
\rho_n = \frac{\sum_{k=0}^{2^{n-1}} \binom{2^n}{2k} x^k}{\sum_{k=0}^{2^{n-1}}\binom{2^n}{2k+1} x^k}
$$
where the coefficients of the binomial expansion of $(x+1)^{2^n}$
appear alternatingly in the numerator and denominator.
Note also that this is exactly Newton's method to find a zero
of $f(\rho) = \rho^2 - x$:
$$
\rho_{n+1} = \rho_n - \frac{f(\rho_n)}{f'(\rho_n)} =
\rho_n - \frac{\rho_n^2 - x}{2 \rho_n} = \frac{\rho_n^2 + x}{2 \rho_n} \, .
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
| Pulling a small rabbit from a hat, consider
$$f(a,b)=3+{b+1\over a+b-2}={3a+4b-5\over a+b-2}$$
It's clear that $a^2+b^2=1$ implies $a+b-2\lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2\le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5\le0$ if $a^2+b^2=1$. Thus $f(a,b)\ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f({3\over5},{4\over5})=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$
But I can't prove this inequality to prove $(1)$
| $$\color{brown}{\textbf{Final edition (13.08.18)}}$$
$\color{green}{\textbf{Task transformations.}}$
The problem is homogeneous with respect to unknowns $x,y,z.$
Let WLOG
$$x+y+z=6,\tag1$$
Then the equivalent inequality is
$$x^2y^2(6-x-y)^2\ge64(3-x)(3-y)(x+y-3),\tag2$$
where
$$(x,y,6-x-y)\in(0,6),$$
or
$$((3-y)+(x+y-3),(3-x)+(x+y-3),(3-x)+(3-y))\in(0,6).\tag3$$
Inequality $(2)$ is satisfied if one or three of the factors $(3-x),(3-x),(x+y-3)$ are non-positive.
On the other hand, conditions $(3)$ is not satisfied if two of this tactors are non-positive.
Thus, it remains to prove the inequality $(2)$ in the case
$$(x,y,x+y-3)\in(0,3)^3.\tag4$$
Denote
$$t= xy,\tag5$$
then
$$(3-x)(3-y)= xy - 3(x+y) +9 = t-3(6-z)+9 = t+3z-9,$$
$$t=xy\le\left(\frac{x+y}2\right)^2=\frac{(6-z)^2}4,$$
and the stronger problem is to prove the inequality
$$z^2t^2\ge 64(t+3z-9)(3-z),$$
or
$$f(z,t)=z^2t^2+64(z-3)t + 192(z-3)^2\ge0,\tag6$$
under the conditions
$$z\in(0,3),\quad t\in\left(0,\frac{(6-z)^2}4\right].\tag7$$
$\color{green}{\textbf{Proof.}}$
The least value of $f(z,p)$ can be achieved only in the stationary points or in the bounds.
The stationary points of $f(z,p)$ can be found from the system $f'z=f'_t = 0,$ or
\begin{cases}
zt^2+32t + 192(z-3) = 0\\
z^2t+32(z-3) = 0.
\end{cases}
Taking in account $(7),$ one can obtain
$$\begin{cases}
(zt+32-6z^2)t = 0\\
z^2t+32(z-3) = 0,
\end{cases}\rightarrow
\begin{cases}
zt=6z^2-32\\
z(6z^2-32)+32(z-3) = 0,
\end{cases}\rightarrow
\begin{cases}
z_s=2\sqrt[3]2\\
t_s=12\sqrt[3]2-8\sqrt[3]4
\end{cases}\\[8pt]
f(z_s,t_s) = 64(11-54\sqrt[3]2+36\sqrt[3]4)\approx7.085>0.
$$
Let us check the bounds.
Case $\mathbf{z\to 0,\quad t\in(0,9).}$
$$f(0, t) = 192(9 - t)>0.$$
Case $\mathbf{z\to 3, quad t > 0.}$
$$f(3, t) = 3t^2 >0.$$
Case $\mathbf{t=\frac{(6-z)^2}4,\quad z\in(0,3).}$
$$f\left(z, \frac{(6-z)^2}4\right) = \frac1{16}z^2(6-z)^4+16(z-3)(6-z)^2 + 192(z-3)^2 = \frac1{16}z^2(z-2)^2(z^2-20z+132)\ge0$$
(see also Wolfram Alpha).
$\textbf{Proved.}$
| {
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} |
Proving the equality case in triangle inequality Background
When plotted on a real number line, it may be deduced that if
$$a,b,c \in \mathbb{R} $$
$$a < b < c$$
then
$$\left| {a - c} \right| = \left| {a - b} \right| + \left| {b - c} \right|$$
Problem
But the problem is with the proof. How can the above statement be proven true from the properties of order structure and the definition of absolute value?
| Since both sides of the equality are nonnegative, we can square it:
\begin{align}
|a-c| = |a-b| + |b-c| &\iff |a-c|^2 = (|a-b| + |b-c|)^2\\
&\iff a^2-2ac+c^2 = a^2-2ab+b^2+b^2-2bc+c^2 + 2|a-b||b-c|\\
&\iff 0 = b^2 - ab - bc + ac + |a-b||b-c|\\
&\iff 0 = -(a-b)(b-c) + |a-b||b-c|\\
&\iff (a-b)(b-c) = |a-b||b-c|\\
&\iff a-b \,\text{ and } \,b-c \text{ have the same sign}\\
&\iff a \ge b \ge c \,\text{ or }\,a \le b \le c
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
prove this inequality $\sum\frac{a}{b}\ge \sum a^2$ Let $a,b,c>0$ such $a+b+c=3$, show that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge a^2+b^2+c^2$$
I have show this not stronger inequality
$$\sum\dfrac{a}{b}\ge \sum a$$
| Let $\sum $ denote cyclic sum, then observe:
$$\sum a(a-b)^2(b-2c)^2 \geqslant 0$$
$$\implies \sum ab^4 + \sum a^3b^2 + 2\sum a^2b^3+4abc\sum ab -8abc\sum a^2 \geqslant 0 $$
$$\implies 2\left( \sum a \right)^2\sum ab^2 + abc\left(\sum a\right)^2 \geqslant 21abc\sum a^2$$
$$\implies 6 \sum \frac{a}b + 3 \geqslant 7\sum a^2$$
Add the obvious $\sum a^2 \geqslant 3$ to the above to conclude. Equality is when $a=b=c=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof involving polynomial roots From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.
| Considering
\begin{align}
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \\
S_1 &= a+b+c+d \\
&= -1 \\
S_2 &= ab+ac+ad+bc+bd+cd \\
&= 0 \\
S_3 &= bcd+acd+abd+abc \\
&= 0 \\
S_4 &= abcd \\
&= -1 \end{align}
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),{a,b,c,d}] in Wolfram Alpha or so, we have:
\begin{align}
f(x) &= (x-\color{red}{ab})(x-\color{red}{ac})(x-\color{red}{ad})
(x-\color{red}{bc})(x-\color{red}{bd})(x-\color{red}{cd}) \\
&= x^6-S_2 x^{5}+(S_1 S_3-S_4)x^{4}+(2S_2 S_4-S_3^2-S_1^2 S_4)x^{3} \\
& \quad +(S_1 S_3 S_4-S_4^2)x^{2}-S_2 S_4^2 x+S_4^3 \\
&= x^6+x^4+x^3-x^2-1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857380",
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"source": "stackexchange",
"question_score": "7",
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inequality proving for intersection AB and CD are line segments, $AB=CD=1$, intersecting in point O, $\enspace$ $AB\cap CD =O$, $\angle AOC=60^{\circ}$. Prove that $AC+BD\geq1$. $\enspace$
What I tried:
$AO+BO=1$, $\enspace$ $CO+DO=1$
$AO=x$, $\enspace$ $CO=y$
$BO=1-x$, $\enspace$ $DO=1-y$
Law of cosines: $AC^2=x^2+y^2-xy$ $\enspace$ (1)$\enspace$ and $\enspace$ $BD^2=x^2+y^2-xy-x-y+1$ $\enspace$ (2)$\enspace$
But expressing AC and BD through (1) and (2) and replacing them in $AC+BD\geq1$, I get nowhere.
Could someone give me an idea to prove it?
| Let $ABDK$ be a parallelogram.
Thus, $AK=DB$, $AD=AB=DC=1$ and $\measuredangle ADC=\measuredangle AOC=60^{\circ}.$
Hence, $\Delta KDC$ is an equilateral triangle, which says $KC=1$ and by the triangle inequality
$$AC+BD=KA+AC\geq KC=1.$$
Also, you can end your work by the triangle inequality again:
$$AC+BD=\sqrt{x^2+y^2-xy}+\sqrt{(1-x)^2+(1-y)^2-(1-x)(1-y)}=$$
$$=\sqrt{\left(x-\frac{1}{2}y\right)^2+\frac{3}{4}y^2}+\sqrt{\left(1-x-\frac{1}{2}(1-y)\right)^2+\frac{3}{4}(1-y)^2}\geq$$
$$\geq\sqrt{\left(x-\frac{1}{2}y+1-x-\frac{1}{2}(1-y)\right)^2+\frac{3}{4}(y+1-y)^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857811",
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Calculate max/min of a 3 variable function, restricted to g(x,y,z)=0 Calculate extrema of $f(x,y,z)=xe^{yz}$ on boundary $3x^2 +y^2 +z^2 =27$
*
*I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + \lambda g_x$,$f_y + \lambda g_y,f_z + \lambda g_z,g(x,y,z)=0$
$e^{yz}+\lambda (6x)=0$
$xze^{yz}+\lambda (2y)=0$
$xye^{yz}+\lambda (2z)=0$
$3x^2 +y^2 +z^2 -27 = 0$
I tried a lot of combinations but I can't solve this. Can you help me ?
| Hint.
Make $\mu = \frac{\lambda}{e^{yz}}$ and then solve
$$
1+\mu 6 x = 0\\
x z + \mu 2 y = 0\\
xy + \mu 2z = 0\\
3x^2+y^2+z^2-27=0
$$
giving
$$
\left[
\begin{array}{ccccc}
x & y & z & \mu & f \\\
-3 & 0 & 0 & \frac{1}{18} & 0 \\
3 & 0 & 0 & -\frac{1}{18} & 0 \\
-\frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & \frac{1}{2 \sqrt{3}} & -\frac{13 e}{\sqrt{3}} \\
-\frac{1}{\sqrt{3}} & \sqrt{13} & \sqrt{13} & \frac{1}{2 \sqrt{3}} & -\frac{13 e}{\sqrt{3}} \\
\frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & -\frac{1}{2 \sqrt{3}} & \frac{13 e}{\sqrt{3}} \\
\frac{1}{\sqrt{3}} & \sqrt{13} & \sqrt{13} & -\frac{1}{2 \sqrt{3}} & \frac{13 e}{\sqrt{3}} \\
\end{array}
\right]
$$
| {
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"source": "stackexchange",
"question_score": "1",
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Solve $\frac{1}{x}+\frac{1}{y}= \frac{1}{2007}$ The number of positive integral pairs $(x<y)$ such that $\frac{1}{x}+\frac{1}{y}= \frac{1}{2007}$
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
| $$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4\cdot 223^2=(x-2007)(y-2007)$$
\begin{align}3^4\cdot 223^2 &=(3^0) \cdot (3^4\cdot 223^2)\\
&=(3^1) \cdot (3^3\cdot 223^2)
\\
&=(3^2) \cdot (3^2\cdot 223^2)
\\
&=(3^3) \cdot (3^1\cdot 223^2)\\
&=(3^4) \cdot (3^0\cdot 223^2)\\
&=(3^0\cdot 223) \cdot (3^4\cdot 223)\\
&=(3^1\cdot 223) \cdot (3^3\cdot 223)
\end{align}
I hope you can recover $x$ and $y$ from here.
| {
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Limits for sequences: Prove that $\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$
Prove that $$\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$$
Note, this is a sequence.
I know that a sequence $(c_n)_{n=1}^{\infty}$ converges to a finite value L if for every $\epsilon > 0$, there exists some $N>0$ so that for all $n$ it holds that:
$$n > N \implies |c_n - L| < \epsilon$$.
I've learnt this in school: $|c_n - L| < \epsilon$ is the distance between each sequence term and the epsilon-value. If we can write $n$ in terms of epsilon, then for any epsilon, we can easily 'choose' an N-value, so that the implication above holds.
I'm afraid I haven't gotten further than this... :
$$\left|\frac{2n^2 - 3n + 5}{9 - n^2} - (-2)\right| < \epsilon \iff \left|\frac{23 - 3n}{9 - n^2}\right| < \epsilon$$
I just can't seem to reduce $n$ in a way that would help me solve this inequality...
I don't know if there is some algebraic trick that is going over my head or if there is some "theoretical trick" which I could use to simplify. Please, a hint would be greatly appreciated.
| If we take $n > 3+\frac6\varepsilon$ we have $ \frac{3}{n-3} < \frac{\varepsilon}2$, and if we take $n > \sqrt{\frac{28}\varepsilon + 9}$ we have $ \frac{14}{n^2-9} < \frac{\varepsilon}2$.
Therefore for $n \ge \max\left\{4, 3+\frac6\varepsilon, \sqrt{\frac{28}\varepsilon + 9}\right\}$ we have
\begin{align}
\left|\frac{2n^2-3n+5}{9-n^2}+2\right| &= \frac{|23-3n|}{|9-n^2|}\\
&\le \frac{3n+23}{(n+3)(n-3)}\\
&= \frac{3(n+3)+14}{(n+3)(n-3)}\\
&= \frac{3}{n-3} + \frac{14}{n^2-9}\\
&< \frac{\varepsilon}2 + \frac{\varepsilon}2\\
&= \varepsilon
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to find the partial sum of $n/2^n$? I'm trying to find a formula for the partial sum of $n/2^n$.
I've tried this so far...
$$ S_n = \frac12 + \frac24 + \frac38 + \cdots + \frac{n}{2^n} $$
Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $\frac{2^n}n\cdot\frac{n+1}{2^{n+1}}$ (a terms that should take each term in the sequence to the next term.)
$$ \frac{n+1}{2n} S_n = \frac24 + \frac38 + \cdots + \frac{n+1}{2^{n+1}} $$
Then
$$ S_n- \frac{n+1}{2n} S_n = \frac12 - \frac{n+1}{2^{n+1}}$$
$$ S_n = \frac{\frac12 - \frac{n+1}{2^{n+1}}}{1-\frac{n+1}{2n}} $$
But, alas, this does not give the correct answers.
Can someone point out where I went wrong? Thanks
| Observe
$$2S_n = 1+ \frac{1+1}2 + \frac{2+1}4 + \cdots + \frac{(n-1)+1}{2^{n-1}}
\\=S_{n-1}+1+\frac12+\frac14\cdots+\frac1{2^{n-1}}
\\=S_n-\frac n{2^n}+2-\frac1{2^{n-1}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
| HINT
We can try with
$${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
$$\sqrt {x+1}{\sqrt {x- 1\over x}} + {\sqrt{x-1\over x}} = x$$
$$(\sqrt {x+1}+1){\sqrt {x- 1\over x}} = x$$
$${\sqrt {x- 1\over x}} = \frac{x}{\sqrt {x+1}+1}\frac{\sqrt {x+1}-1}{\sqrt {x+1}-1}=\sqrt {x+1}-1$$
$${\sqrt {x- 1}} =\sqrt {x^2+x}-\sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation
*
*${{x^2- 1\over x}}\ge 0$
*${{x- 1\over x}}\ge 0$
*$x\neq 0$
| {
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"source": "stackexchange",
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Maximum $n$ such that ${n \choose k} \,2^{1 - {k \choose 2}} < 1$ (where $k$ is a constant) Maximum value of $n$ such that the expression given below does not exceed 1. ($k$ is a constant)
$${n \choose k} 2^{1 - {k \choose 2}} < 1$$
Any hints on how to approach this problem.
Thanks.
Edit:
I tried doing the problem this way.
$${n \choose k} < \frac{n^k}{k!}$$
$${n \choose k} 2^{1 - {k \choose 2}} < \frac{n^k}{k!}2^{1 - {k \choose 2}}$$
$$\frac{n^k}{k!}2^{1 - {k \choose 2}} < 1 \Rightarrow n < \sqrt[k]{\frac{k!}{2^{1 - {k \choose 2}}}}$$
But, still $n_{max}$ can be greater than $\sqrt[k]{\frac{k!}{2^{1 - {k \choose 2}}}}$. How can i get the least upper bound for $n$.
| I shall prove that, for a positive integer $k$, the maximum natural number $n$, denoted by $n_k$, such that
$$\binom{n}{k}\leq 2^{\binom{k}{2}-1}$$
satisfies
$$\lim_{k\to\infty}\,\frac{n_k}{k\,2^{\frac{k}{2}}}=\frac{1}{\text{e}\sqrt{2}}\,.\tag{*}$$
I also provide some bounds for the value of $n_k$.
First, we note that
$$\frac{\left(n_k-\frac{k-3}{2}\right)^k}{k!}>\binom{n_k+1}{k}>2^{\binom{k}{2}-1}\,,$$
where we have applied the AM-GM Inequality
$$\begin{align}(n_k+2-k)&(n_k+3-k)\cdots (n_k)(n_k+1)\\
&\leq \left(\frac{(n_k+2-k)+(n_k+3-k)+\ldots +(n_k)+(n_k+1)}{k}\right)^k
\\&=\left(n_k-\frac{k-3}{2}\right)^k\,.\end{align}$$
That is,
$$n_k-\frac{k-3}{2}>\sqrt[k]{2^{\binom{k}{2}-1}\,k!}=2^{\frac{k-1}{2}}\,\sqrt[k]{\frac{k!}{2}}\geq 2^{\frac{k-1}{2}}\left(\frac{\pi k}{2}\right)^{\frac{1}{2k}}\,\left(\frac{k}{\text{e}}\right)>2^{\frac{k-1}{2}}\,\left(\frac{k}{\text{e}}\right)\,,$$
where we have used the Stirling Approximation $$k!\geq \sqrt{2\pi k}\left(\frac{k}{\text{e}}\right)^k\,.$$
This shows that $$n_k>2^{\frac{k-1}{2}}\,\left(\frac{k}{\text{e}}\right)+\frac{k-3}{2}\,.$$
Now,
$$\frac{(n_k+1-k)^k}{k!}<\binom{n_k}{k}\leq 2^{\binom{k}{2}-1}\,.$$
This gives
$$n_k-(k-1)<\sqrt[k]{2^{\binom{k}{2}-1}\,k!}\leq 2^{\frac{k-1}{2}}\,\left(\frac{\text{e}^2k}{4}\right)^{\frac{1}{2k}}\,\left(\frac{k}{\text{e}}\right)\,,$$
in which we have used the Stirling approximation
$$k!\leq \sqrt{\text{e}^2k}\,\left(\frac{k}{\text{e}}\right)^k\,.$$
That is,
$$n_k<2^{\frac{k-1}{2}}\,\left(\frac{k}{\text{e}}\right)\,\Biggl(1+O\left(\frac{\ln(k)}{k}\right)\Biggr)+k-1\,,$$
noting that
$$\left(\frac{\text{e}^2k}{4}\right)^{\frac{1}{2k}}=1+\frac{\ln(k)}{2k}+\frac{1-\ln(2)}{k}+O\left(\left(\frac{\ln(k)}{k}\right)^2\right)=1+O\left(\frac{\ln(k)}{k}\right)\,.$$
In conclusion, we have
$$2^{\frac{k-1}{2}}\,\left(\frac{k}{\text{e}}\right)+\frac{k-3}{2}<n_k<2^{\frac{k-1}{2}}\,\left(\frac{k}{\text{e}}\right)\,\Biggl(1+O\left(\frac{\ln(k)}{k}\right)\Biggr)+k-1\,.$$
This proves (*). In fact, we can also show that, for all $k\geq 22$,
$$2^{\frac{k-1}{2}}\,\left(\frac{k+\frac{1}{2}\,\ln(k)+\frac{1}{5}}{\text{e}}\right)<n_k<2^{\frac{k-1}{2}}\,\left(\frac{k+\frac{1}{2}\,\ln(k)+\frac{1}{3}}{\text{e}}\right)\,,$$
using
$$\left(\frac{\pi k}{2}\right)^{\frac{1}{2k}}=1+\frac{\ln(k)}{2k}+\frac{\ln(\pi)-\ln(2)}{2k}+O\left(\left(\frac{\ln(k)}{k}\right)^2\right)$$
and
$$\frac{1}{5}<\frac{\ln(\pi)-\ln(2)}{2}<1-\ln(2)<\frac{1}{3}\,.$$
Anyhow, the best approximator of $n_k$ is still $m_k:=2^{\frac{k-1}{2}}\,\sqrt[k]{\dfrac{k!}{2}}$, albeit being trickier to calculate, since we have
$$m_k+\frac{k-3}{2}< n_k < m_k+k-1\,.$$
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int_{-1}^{1} \cot^{-1} \left(\frac{1}{\sqrt{1-x^2}}\right) \cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-(x^2)^{|x|}}}\right)$
$$\int_{-1}^{1} \left(\cot^{-1} \dfrac{1}{\sqrt{1-x^2}}\right) \left(\cot^{-1}\dfrac{x}{\sqrt{1-(x^2)^{|x|}}}\right)= \dfrac{\pi^2(\sqrt a-\sqrt b )}{\sqrt c}$$
, where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.
Using $\int_a^b f(x)dx= \int_a^bf(a+b-x{) dx}$, I got:
$$2I =2\pi\int_{0}^1 \cot^{-1}\left(\dfrac{1}{\sqrt{1-x^2}}\right) dx$$
Then, letting $x = \sin \theta$
$I = \pi\displaystyle\int_{0}^{\pi/2}\arctan (\cos\theta) \cos \theta d\theta$
After this I tried integration by parts but it gets really complicated with that? How do I continue?
EDIT:
Please note that arccot(x) + arccot(-x)= $\pi$ $\ne 0$
Principal range of $\cot^{-1}x$ considered in the question is $(0,\pi)$
| Using the property $$\int_a^b f(x)=\int_a^b f(a+b-x)$$
The integral changes to $$I=\pi\int_0^1 \arctan \sqrt{1-x^2} dx$$
You might know the property that $$\int_a^b f(x)+\int_{f(a)}^{f(b)} f^{-1}(x) dx=-af(a)+bf(b)$$
Let $$J=\int_0^1 \arctan \sqrt{1-x^2} dx+\int_{\frac {\pi}{4}}^0 \sqrt {1-\tan ^2x } dx$$ Using above property $J=0$ Hence $$\pi\int_0^1 \arctan \sqrt{1-x^2} dx=\pi\int_0^{\frac {\pi}{4}} \sqrt {1-\tan ^2x } dx=I$$
Using the substitution $\tan x=\sin \theta$ in the right integral we get $$I=\pi\int_0^{\frac {\pi}{2}} \frac {\cos ^2\theta}{1+\sin^2\theta} d\theta=\pi\int_0^{\frac {\pi}{2}}\left( -1+\frac {2\sec^2\theta}{1+2\tan^2\theta}\right) d\theta$$
Hope you can continue further
| {
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"timestamp": "2023-03-29T00:00:00",
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Laurent Series $e^{\frac1{1-z}}$, $|z|>1$. I tried to find Laurent expansion for:
$e^{\frac1{1-z}}$, $|z|>1$.
I tried next: $\frac1{1-z}=-\sum_{n=1}^{\infty}\frac1{z^n}$,
then using $e^{\frac1{1-z}}=1+\frac1{1-z}+\frac{1}{{2!(1-z)}^2}+\frac{1}{{3!(1-z)}^3}+...$
Then I have $e^{\frac1{1-z}}$ = $1$ + $(-\frac1{z}-\frac1{z^2}-\frac1{z^3}-...) + (\frac1{2!})(\frac1{z^2}+\frac1{z^4}+…)+...$ = $1 - \frac1z-\frac1{2z^2}-...$.
The result is wrong, since in the book the answer is: $1 - \frac1z+ \frac1{2z^2}-\frac1{6z^3}+\frac1{24z^4}-\frac{19}{120z^5}...$
What I did wrong?
| With $z=\dfrac{1}{w}$ we have $|w|<1$ then by $e^z$ expansion
\begin{align}
e^{\frac{1}{1-z}}
&= e^{\frac{-w}{1-w}} \\
&= \sum_{n\geq0}\dfrac{1}{n!}\left(\frac{-w}{1-w}\right)^n \\
&= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1-w\right)^{-n} \\
&= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1+nw+\dfrac{n(n+1)}{2}w^2+\cdots\right) \\
&= \sum_{n\geq0}\dfrac{(-1)^n}{n!}\left(w^n+nw^{n+1}+\dfrac{n(n+1)}{2}w^{n+2}+\cdots\right)
\end{align}
after calculation some terms we let $w=\dfrac1z$.
Edit: One may find
\begin{align}
&1 \\
&-(w+w^2+w^3+w^4+w^5+\cdots) \\
&+\frac12(w^2+2w^3+3w^4+4w^5+\cdots) \\
&-\frac16(w^3+3w^4+6w^5+\cdots) \\
&+\cdots\\
&=1-w-\frac{w^2}{2}-\frac{w^3}{6}+\frac{w^4}{24}+\frac{19 w^5}{120}+\cdots\\
&=\color{blue}{1-\dfrac1z-\frac{1}{2z^2}-\frac{1}{6z^3}+\frac{1}{24z^4}+\frac{19}{120z^5}+\cdots}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{(1+(xi)^2)}}\\
&=-x^{-1}+\arctan{xi}+C
\end{split}
Is this correct?
Thanks in advance!
| $x = \tanh(u)$ then $dx = \operatorname{sech}^2(u) du$
$$
\begin{align}
\int \frac{dx}{x^2(1-x^2)}
&= \int \frac{\operatorname{sech}^2(u) du}{\tanh^2(u)(1-\tanh^2(u))}\\
&= \int \frac{\operatorname{sech}^2(u) du}{\tanh^2(u)(\operatorname{sech}^2(u))}\\
&= \int \frac{du}{\tanh^2(u)}\\
&= \int \frac{\cosh^2(u)}{\sinh^2(u)} du\\
&= \int (1 + \operatorname{csch}^2(u))du\\
&= u - \coth(u)\\
&= \operatorname{arctanh}(x) - \frac{1}{x} + C
\end{align}
$$
| {
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Finding the area enclosed by the graphs of: $y=x\left(x-4\right)^2$ ,$y=4x-x^2$.
Find the area enclosed by the graphs of:
$y=x\left(x-4\right)^2$
, $y=4x-x^2$.
My answer was $\frac{7}{12}+\frac{45}{4}$, but apparently this is wrong: the right answer is $\frac{37}{2}$ according to the textbook.
Intersections occur where $x(x - 4)^2 = x(4 - x)$.
So $x(x - 4)(x - 4 + 1) = 0$ which has solutions at $x = 0$, $x = 3$ and $x = 4$.
The difference equation is $y = x(x - 4)^2 - (4x + x^2) = x^3 - 9x^2 + 12x$.
\begin{align} \text{Enclosed area} & = \int_0^4 \lvert x^3 - 9x^2 + 12x \rvert \, dx \\
& = \int_0^3 x^3 - 9x^2 + 12x \, dx + \int_3^4 -x^3 + 9x^2 - 12x \, dx \\
& = \left[ \frac14 x^4 - 3x^3 + 6x^2 \right]_0^3 + \left[ -\frac14 x^4 + 3x^3 - 6x^2 \right]_3^4 \\
& = \left( \frac{81}4 - 81 + 54 \right) - \left( 0 \right) + \left( -64 + 192 - 96 \right) - \left( -\frac{81}4 + 81 - 54 \right) \\
& = -\frac{27}4 - 0 + 32 - \frac{27}4 \\
& = \frac{37}2 \end{align}
It appears that they changed the $-x^2$ to $x^2$ in the latter equation when finding the difference equation. Why did they do that?
| You are correct. That is a typo. Furthermore it propagates all the way to the end. Your answer is correct.
| {
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Finding integer solutions to $ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $ I was browsing through facebook and came across this image:
I was wondering if we can find more examples where this happens?
I guess this reduces to finding integer solutions for the equation
$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for integers a,b,c
Or can we even further extend to when they are all distinct that is finding solutions to
$$ \frac{a^3+b^3}{c^3+d^3} = \frac{a+b}{c+d} $$ for integers a,b,c
I don't really have that much knowledge in the number theory area so I have come here
| $$\frac{a^3+b^3}{a^3+c^3}=\frac{(a+b)(a^2-ab+b^2)}{(a+c)(a^2-ac+c^2)}=\frac{a+b}{a+c}$$
If $a+c \neq 0$ and $a+b \neq 0,$ then $$a^2-ab+b^2=a^2-ac+c^2,$$namely $$(b+c-a)(b-c)=0.$$
If $b=c$, the case is trivial. If $b \neq c$, then $$b+c=a.$$
| {
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Prove that: $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$ Given three positive numbers a,b,c satisfying $a+b+c=3$. Show that $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$
Things I have done so far:
$$a+b+c=3\Rightarrow b+c=3-a;0<a<3$$
$$\Rightarrow \frac{1}{(b+c)^2+a^2}=\frac{1}{(3-a)^2+a^2}=\frac{1}{2(a-1)^2+7-2a}\leq \frac{1}{7-2a}$$
Then, I tried to use the UCT to solve this problem.
I created the new inequality:
$$\frac{1}{7-2a}\leq \frac{1}{5}+m.(a-1)(*)$$
with $0<a<3$. I needed to find "m" which make (*) always true.After that, I found $$m=\frac{2}{25}$$
However,
$$(*)\Leftrightarrow \frac{1}{7-2a}\leq \frac{1}{5}+\frac{2}{25}.(a-1)\Leftrightarrow 4(a-1)^{2}\leq 0$$
which is wrong with any $$a\in \mathbb{R} $$
Can you show me what my mistake is?
I hope you can have "smart" way to solve this problem.
Sorry, I am not good at English.
| Hint: Use the Tangent Line method.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{(3-a)^2+a^2}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or
$$\sum_{cyc}\left(\frac{a^2-3a+2}{2a^2-6a+9}+\frac{1}{5}(a-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2(2a+1)}{2a^2-6a+9}\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| $\bmod n^2\!+1\!:\,\ \color{#c00}{n^2\equiv -1}\,\Rightarrow\, n^6\equiv (\color{#c00}{n^2})^3\equiv (\color{#c00}{-1})^3\equiv -1$ so $\,n^6-7\equiv -8$ is the remainder,
hence the quotient is $\ \dfrac{(n^6-7)-8}{n^2+1} = \dfrac{n^6+1}{n^2+1} = n^4 - n^2 + 1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get:
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong?
| A slightly different approach: you have $a^2+b^2=7$ and $a^2 b^2=1$ with $a,b\in\mathbb{R}^+$, such that
$$ \operatorname*{Res}_{x=ia}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ia}\frac{1}{(x+ia)(x^2+b^2)} = \frac{1}{2i}\cdot \frac{1}{a(b^2-a^2)}$$
and similarly
$$ \operatorname*{Res}_{x=ib}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ib}\frac{1}{(x+ib)(x^2+a^2)} = \frac{1}{2i}\cdot \frac{1}{b(a^2-b^2)}$$
so $\pi i$ times the sum of the residues at $ia$ and $ib$ equals $\frac{\pi}{2ab(a+b)}$. On the other hand the integral is blatantly positive, such that $ab(a+b)$ is the square root of $a^2 b^2(a^2+b^2+2\sqrt{a^2 b^2}) = 9$ and voilà the wanted outcome $\frac{\pi}{6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$
so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$
Can anyone please correct me If I have gone wrong anywhere?
| Yes, the solution is correct.
You are using:
$$\arctan a\pm \arctan b=\arctan\frac{a\pm b}{1\mp ab}$$
to simplify:
$$x\tan^{-1}(x+2)-x\tan^{-1}x=x\tan^{-1}\frac{(x+2)-x}{1+(x+2)x}=x\tan^{-1}\frac{2}{1+2x+x^2}.$$
However, to evaluate the limit you can avoid using the L'Hospital's rule:
$$\lim_{x \to \infty} x \tan^{-1} \frac {2}{1+2x+x^2}=
\lim_{x\to\infty} \frac{\tan^{-1}\frac{2}{1+2x+x^2}}{\frac{2}{1+2x+x^2}}\cdot \frac{\frac{2}{1+2x+x^2}}{\frac{1}{x}}=1\cdot 0=0,$$
where $\lim_\limits{x\to 0} \frac{\tan^{-1}x}{x}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method
By integrating factor method -
I put it into the form of $ \frac{dy}{dx} + \frac{4x}{x^2 + 1} \cdot y = \frac{x}{x^2 + 1} $
My integrating factor is $I = (x^2 + 1)^2 $
By formula, $ y (x^2 +1)^2 = \int{ (x^3 + x)} dx $
And thus my general solution is $ y = \frac{x^4 + 2x^2 + C}{4(x^2 + 1)^2} $
By separable method is where I got problems with...
I put it into the form of $\int{ (\frac{x}{x^2 + 1} )} dx = \int{ (\frac{1}{(1-4y)}} dy $
Getting $\frac{1}{2} \ln (x^2 + 1) + C = \frac{1}{4} \ln (1-4y) $
$2 \ln (x^2 +1) + C = \ln (1-4y) $
$1-4y = e^{\ln (x^2 + 1)^2} + e^{C} = (x^2 +1)^2 + C $
$ y = \frac{ (x^2+1)^2 + C - 1}{4} $
Why is my general solution from separable method very different from the integrating factor method ? I suspect that I have went wrong in the separable method but cannot identify it ...
| Yes, you have.
In separable form ,
$$\displaystyle \int \dfrac{1}{1-4y}~dy=\int\dfrac{x}{x^{2}+1}~dx$$
$\dfrac{-1}{4}\log(1-4y)=\dfrac{1}{2}\log(x^{2}+1)$ You have missed minus sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Non-PSD matrix with two different PSD submatrixes Let $A$ be an $n \times n$. Define $A_{-i}$ to be the matrix $A$ without the $i$-th column and row. For instance
$$
A=
\begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\implies
A_{-2}=
\begin{pmatrix}
1 & 3\\
7 & 9
\end{pmatrix}
$$
Is it possible to find $S$ symmetric and not positive semidefinite such that there exist $i_1\not= i_2$ with $A_{-i_1}, A_{-i_2}$ which are positive semidefinite?
| $$
\left(
\begin{array}{ccc}
5 & -3 & -3 \\
-3 & 5 & -3 \\
-3 & -3 & 5 \\
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
7 & -4 & -5 \\
-4 & 8 & -6 \\
-5 & -6 & 9 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
First four numbers of a sequence are $2,0,1,8,$. Each next is the last digit of the sum of the preceding four numbers. Will $2,0,1,8$ show up again? Let $a_n$ is a sequence of numbers. The firs four numbers are $2,0,1,8$ and each following number is the last digit of the sum of the preceding four numbers. The first ten numbers are $2,0,1,8,1,0,0,9,0,9$.
Will the succession $2,0,1,8$ show up again? Will the succession $2,0,1,9$ show in this sequence of numbers?
I couldn't do more.
$2+0+1+8=11$, $\enspace$ $11 \enspace mod \enspace 10=1$,$\enspace$ $a_5=1$
$a_n=(a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4})\enspace mod\enspace10$
Could someone give me an idea?
| Notice that given any four sequential numbers in the sequence, say $a_n, a_{n+1}, a_{n+2}, a_{n+3}$ with $n>1$, $a_{n-1}$ is determined uniquely; in fact, $a_{n-1} = a_{n+3} - a_{n+2} - a_{n+1} - a_{n} \bmod 10$.
Since there are only finitely many four-tuples of the digits $0-9$, the sequence of four-tuples must eventually repeat. That is, there are $m$ and $n$ with $m<n$ and $a_{m+i} = a_{n+i}$ for $i = 0,1,2,3$. Let $k$ be the least such $m$. We claim $k=1$. If not, we must have $a_{k+i} = a_{n+i}$ for $i = -1, 0, 1, 2$, contradicting the minimality of $k$. This shows that the sequence eventually produces $2,0,1,8$ again.
On the other hand, the sequence will never produce $2,0,1,9$ because modulo $2$ the sequence $0,0,1,0,1$ repeats indefinitely, so two odd digits can never appear in succession.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Taking square roots modulo $2^N$ I was trying to solve $y^2 - y \equiv 16 \pmod{512}$ by completing the square.
Here is my solution.
\begin{align}
y^2 - y &\equiv 16 \pmod{512} \\
4y^2 - 4y + 1 &\equiv 65 \pmod{512} \\
(2y-1)^2 &\equiv 65 \pmod{512} \\
2y - 1 &\equiv \pm 33 \pmod{512} &\text{Found by pointwise search.}\\
2y &\equiv 34, -32 \pmod{512} \\
y &\equiv 17, -16 \pmod{256} \\
y &\in \{17, 273, 240, 496\} \pmod{512} &\text{These values need to be verified.}\\
y &\in \{240, 273\}\pmod{512}
\end{align}
I had to solve $x^2 \equiv 65 \pmod{2^9}$ by a pointwise search. Is there any systematic method for solving equivalences of the form $x^2 \equiv a \pmod{2^N}$, or more generally $x^2 \equiv a \pmod{p^N}$ for a prime number $p$.
| This isn't a general method, but here's a neat trick if $2^k|a-1$ for some large enough $k$.
Claim. If $x^2\equiv a\bmod 2^n$ and $a\equiv 1\bmod 2^k$, with $n\leq 2k$, then
$$x\equiv \pm \frac{a+1}{2}\bmod 2^{n-1}.$$
Proof. As there are only two residue classes $\bmod 2^{n-1}$ that will satisfy this, it suffices to show that these both do. Setting $b=(a+1)/2$ we need to show that
$$b^2\equiv a=2b-1\bmod 2^n.$$
Indeed, this is equivalent to
$$2^n|(b-1)^2 \Leftrightarrow b\equiv 1\bmod 2^{\left\lceil\frac{n}{2}\right\rceil},$$
which is itself equivalent to
$$a\equiv 1\bmod 2^{\left\lceil\frac{n}{2}\right\rceil},$$
which is true as $\left\lceil\frac{n}{2}\right\rceil \leq k$.
So, since we have $x^2\equiv a\bmod 2^9$ and $a\equiv 1\bmod 2^6$, we have that $x\equiv \pm \frac{a+1}{2} = \pm 33\bmod 2^8.$
A similar result holds in general:
If $p$ is an odd prime with $x^2\equiv a\bmod p^n$ and $a\equiv 1\bmod 2^k$ with $n\leq 2k$, then
$$x\equiv \pm \frac{a+1}{2}\bmod p^n.$$
This can be proven in nearly identical fashion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple limit with asymptotic approach. Where's the error? Simply calculus question about a limit.
I don't understand why I'm wrong, I have to calculate
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\sin x}{1 - \cos\sqrt{x^3}}
$$
Using asymptotics, limits and De l'Hospital rule I would write these passages...
$$ = \lim_{x \rightarrow 0} \frac{x \, (2 - \sqrt[3]{8 - x^2})}{x^3/2}
= \lim_{x \rightarrow 0} \frac{\frac{2}{3}\frac{\sqrt[3]{8 - x^2}}{8-x^2}x}{x}
= \frac{1}{6}
$$
But the answers should be $\frac{5}{6}$. Thank you for your help.
| The mistake lies at the beginning :
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}(x-\frac{x^3}{6})}{1 - \cos\sqrt{x^3}} =\frac56$$
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\:x}{1 - \cos\sqrt{x^3}} =\frac16$$
At denominator $1-\cos(x^{3/2})$ is equivalent to $\frac12 x^3$. Thus one cannot neglect the $x^3$ terms in the numerator. So, the equivalent of $\sin(x)$ must not be $x$ but $x-\frac{x^3}{6}$ . This was the trap of the exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Product of binary matrices with binary eigenvalues Consider two binary matrices with obvious patterns:
$$
C=
\begin{bmatrix}
1 &0 &0 &0 &0 &0 &0\\
1 &0 &0 &0 &0 &0 &0\\
0 &1 &0 &0 &0 &0 &0\\
0 &1 &0 &0 &0 &0 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &0 &1 &0 &0 &0
\end{bmatrix}
$$
and
$$
T=
\begin{bmatrix}
1 &1 &0 &0 &0 &0 &0\\
0 &1 &1 &0 &0 &0 &0\\
0 &0 &1 &1 &0 &0 &0\\
0 &0 &0 &1 &1 &0 &0\\
0 &0 &0 &0 &1 &1 &0\\
0 &0 &0 &0 &0 &1 &1\\
0 &0 &0 &0 &0 &0 &1
\end{bmatrix}
$$
The eigenvalues of the matrices $T^n C,n=0,1,2,3$ are zeros and consecutive powers of $2$ equal to $0,1,2,4$. I'd like to have a proof of the generalization of this fact for matrices of larger size with the same patterns.
Note, the entries of $T^n C$ in the left upper corner are zeros and binomial coefficients for the power $n+1$.
A motivation for this question is in
Binary eigenvalues matrices and continued fractions
| Some thoughts:
Note that $T = I + N$, where $I$ is the identity matrix and
$$
N = \pmatrix{0&1\\&0&1\\&&0&1\\&&&0&1\\&&&&0&1\\&&&&&0&1\\&&&&&&0}
$$
Notably, $N^7 = 0$. Because $NI = IN$, we can compute $T^n = (I + N)^n$ by binomial expansion. That is, we have
$$
T^n = \binom n0 I + \binom n1 N + \cdots + \binom n6 N^6
$$
We can verify that $T^n$ is therefore the upper triangular Toeplitz matrix for which, in the notation of the linked wiki page, we have $a_{-k} = \binom nk$ whenever $0 \leq k \leq n$ and all other entries are $0$.
With that, we may compute
$$
T^n C = \pmatrix{\binom n0 + \binom n1 & \binom n2 + \binom n3 & \binom n4 + \binom n5 & \binom n6 &0&0&0\\
\binom n0 & \binom n1 + \binom n2 & \binom n3 + \binom n4 & \binom n5 & 0&0&0\\
0 & \binom n0 + \binom n1 & \binom n2 + \binom n3 & \binom n4 & 0&0&0\\
0 & \binom n0 & \binom n1 + \binom n2 & \binom n3 & 0&0&0\\
0 & 0 & \binom n0 + \binom n1 & \binom n2 & 0&0&0\\
0 & 0 & \binom n0 & \binom n1 & 0&0&0\\
0 & 0 & 0 & \binom n0 & 0&0&0\\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Parametrizing the intersection of a cylinder and a sphere I need to parametrize the intersection between the cylinder $
x^2 + y^2= \frac{1}{4}$ and the sphere $(x+ \frac{1}{2})^2 + y^2 +z^2 = 1$.
I tried parametrizing the first equation which gives $r(t) = (\frac{cos(t)}{2}, \frac{sin(t)}{2})$ since the radius is 1/2.
Then, I plugged in the values in the second equation which yields $(\frac{cos(t) +1}{2})^2 + \frac{sin^2(t)}{4} + z^2 = 1$. We isolate $z$ and get that $z = \sqrt{1-\frac{cos(t)}{2}}$. I'm stuck here because the parametrization would be incomplete if we choose the positive or negative root, am I doing something wrong? If so, what would a correct parametrization be?
EDIT: I've just recalled that $\sqrt{\frac{1-cos(t)}{2}} = sin(\frac{x}{2})$ or $cos(\frac{x}{2})$. Is this helpful?
| $x^2+y^2=\frac{1}{4}$
$\left( x+\frac{1}{2} \right)^2 + y^2 + z^2 = 1$
Expand
$x^2 + x + \frac{1}{4} + y^2 + z^2 = 1$
Collect
$\left(x^2+y^2\right)+x+\frac{1}{4}+z^2=1$
Substitute first equation
$\frac{1}{4}+x+\frac{1}{4}+z^2=1$
The key equations now being$\ldots$
$x+z^2=\frac{1}{2}$
$x^2+y^2=\frac{1}{4}$
We pick $z=t$ hoping for a cleaner solution...
$x=\frac{1}{2}-t^2$
$x^2+y^2=\left( \frac{1}{2}-t^2 \right)^2+y^2=t^4-t^2+\frac{1}{4}+y^2=\frac{1}{4}$
$y^2=t^2-t^4$
$y^2=t^2\left(1-t^2\right)$
$y=\pm t\sqrt{1-t^2}$
And we write...
$\left(\frac{1}{2}-t^2,\pm t\sqrt{1-t^2},t\right)$
for $t \in \left[-1,1\right]$
and I expect someone else to point out the algebra mistake if there is one because there's alot of expansion in my approach, but I checked it and I'm pretty sure it's consistent.
Because the cylinder is of lesser radius, the intersection is a single point at the xy plane. The vertical (xy) projection of the curve is a circle. The lateral (xz) cross section of the curve is a parabola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding closed form for $\sum_{k=1}^n k2^{k-1}$ I am trying to use the perturbation method to find a closed form for:
$$
S_n = \sum_{k=1}^n k2^{k-1}
$$
This is what I’ve tried so far:
$$
S_n + (n+1)2^n = 1 + \sum_{k=2}^{n+1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=1}^{n} (k+1)2^{k}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=0}^{n-1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + 0 + n2^{n-1} + \sum_{k=1}^{n} k2^{k-1}
$$
But I don’t think it is right since it would result in a $S_n - S_n = ...$
Could you please help me?
Thank you very much.
| You have
$$S_n=\sum_{k=1}^n k 2^{k-1}=1\cdot 2^0 +2\cdot 2^1+3\cdot 2^2+\cdots+n2^{n-1}=$$
$$=\big(2^0+2^1+2^2+\cdots+2^{n-1}\big)+\big(1\cdot 2^1+2\cdot 2^2+\cdots+(n-1)2^{n-1}\big)=\quad (*)$$
$$=\sum_{k=0}^{n-1}2^k+2\big(1\cdot 2^0+2\cdot 2^1+\cdots+(n-1)2^{n-2}\big)=\quad (**)$$
$$=(2^n-1)+2 S_{n-1}.$$
So,
$$S_n=(2^n-1)+2(S_n-n2^{n-1}).$$
And you can get $S_n$ from there.
(*) I'm using the fact that if I have
$$1\cdot 2^0+2\cdot 2^1+3\cdot 2^2+4\cdot 2^3+\cdots,$$
that means that I have $2^0$ once, $2^1$ added twice, $2^2$ added three times, $2^3$ four times, and so on. So next I took just one of each to form the sum
$$2^0+2^1+2^2+2^3+\cdots$$
and the original sum ends up whith one less $2^0$ (that is, with none), one less $2^1$ (so just only one), one less $2^2$ (just two), one less $2^3$ (just three, instead of the original four), etc.
Let's put it backwards: if you sum
$$2^0+2^1+2^2+2^3+\cdots$$
and
$$[0\cdot 2^0]+[1]\cdot 2^1 +2\cdot 2^2 +3\cdot 2^3+\cdots$$
(I put between squared brackets what you wouldn't usually type explicitly),
you would get the original
$$1\cdot 2^0+2\cdot 2^1+3\cdot 2^2+4\cdot 2^3+\cdots.$$
(**) Now, going back to the decomposition in two sums we've got: I finally took out a common factor $2$ from the second sum to get back all terms in which the exponent of the $2$ is one less than the coefficient by which it is multiplied, which caracterizes the terms of $S_n$.
Notes:
The answer is, naturally,
$$S_n=1-2^n+n2^n=1+(n-1)2^n.$$
This is a perturbation process, since I wrote $S_n$ in terms of $S_{n-1}$, from which I made $S_n$ reappear. I could have started from $S_{n+1}$, too, and reduce it to an expression on $S_n$, like in
$$S_{n+1}=2^{n+1}-1+2S_n.$$
Then I could have gone on like
$$S_n+(n+1)2^n=2^{n+1}-1+2S_n,$$
and the same result would arise from this equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the integers that double if the first and last digits are swapped Find the integers that double if their first and last digits are swapped.
In other words find the digits: $a$, $b$ and $n_1$ ... $n_m$ in such a way that:
[$a$ $n_1$ $n_2$ ... $n_m$ $b$] $\times$ $2$ = [$b$ $n_1$ $n_2$ ... $n_m$ $a$]
For example, using the rule above: $25$ turns into $52$ but unfortunately $52 \gt 25 \times 2$. The number $102$ transforms into $201$ but again $102 \times 2 \gt 201$.
UPDATE
Possible solution:
Case 1: the number ends in $1$ and in consequence its double ends in $2$
$\ldots 1 \times 2$
$\ldots2$
In conclusion the pair must be:
$2 \ldots 1 \times 2$
$1 \ldots 2$ (impossible)
Case 2: the number ends in $2$
$4 \ldots 2 \times 2$
$2 \ldots 4$ (impossible)
Case 3:
$6 \ldots 3 \times 2$
$3 \ldots 6$ (impossible)
Case 4:
$8 \ldots 4\times 2$
$4 \ldots 8$ (impossible)
Case 5:
$0 \ldots 5 \times 2$ (the first digit can not be zero)
$5 \ldots 0$
Case 6:
$2 \ldots 6 \times 2$
$6 \ldots 2$ (impossible)
Case 7:
$4 \ldots 7 \times 2$
$7 \ldots 4$ (impossible)
Case 8:
$6 \ldots 8 \times 2$
$8 \ldots 6$ (impossible)
Case 9: the number ends in $9$
$8 \ldots 9 \times 2$
$9 \ldots 8$ (impossible)
Case 10: the number ends in $0$
$0 \ldots 0 \times 2$ (the number can not start with $0$)
$0 \ldots 0$
| I'll give an algebraic proof. Your answer is good, though.
Given a number $l$, greater than $10$ since this is obviously not true for single digit numbers, write it as $l = 10^nx + 10y + z$, where $x$ and $z$ are single digit numbers i.e. $10^n \leq l < 10^{n+1}$. Essentially, we are isolating the first and last digits separately (for example, $166465 = 1 \times 10^5 + 6646 \times 10 + 5$)
Now, from the given condition, $2l = 10y + 10^n z + x$, since we have only swapped the last two digits. Of course, by ordinary multiplication, this is also equal to $2 \times 10^n x + 20y + 2z$, by multiplying the expression for $l$ by $2$.
Consequently, $10y + 10^n z + x = 2\times 10^n x + 20y + 2z$. A few transpositions give $10y = (10^n - 2) z - (2 \times 10^n-1)x$.
The last digit of the left hand side is $0$. Therefore, so is the last digit of the number on the right hand side. Note that $n \geq 1$, since we know $l$ has two digits. This gives the last digit of the right hand side as the last digit of $x - 2z$.
Consequently, $x - 2z$ must have last digit zero. That is, either $x = 2z$, or $x = 2z - 10$ (note that $-20 < x - 2z < 10$).
But $y \geq 0$, so $z \geq \frac{(2 \times 10^n - 1) x}{10^n - 2} \geq 2x \geq x$. So $x = 2z - 10$ is the only possibility : the other forces $x \geq z$.
But if $x = 2z - 10$ then $10 y = 10^n z - 2z - 2(10^n)(2z - 10) + 2z - 10 = - 3 \times 10 ^n z + 2 \times 10^{n+1} - 10$, and dividing by $10$ gives $y = 2 \times 10^n - 3 \times 10^{n-1}z - 1 = 10^{n-1}(20- 3z) - 1$.
This forces $z \leq 6$ from $y \geq 0$. Also, note that $2z - 10 \geq 0$ so $z \geq 5$.
Note that $z = 6$ is ruled out since this gives $x = 1$ but $x$ is the last digit of an even number, hence is even.
But if $z = 5 $ then $x = 0$, but $x \neq 0$ by assumption that $10^n \leq l$.
Thus, no such $l$ exists.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many solutions to each of the equations $2^x+3^y=5^z$, $2^x+5^y=3^z$, $3^x+5^y=2^z?$ Let $x,y,z\in\Bbb{N}$ How many total solutions are there to each of the three distinct equations below?
$$2^x+3^y=5^z \tag 1$$
$$2^x+5^y=3^z \tag 2$$
$$3^x+5^y=2^z \tag 3$$
I found 3 solutions to equation (1), 4 solutions to equation (2), 5 solutions to equation (3),(12 altogether) which might be all of them.
$$3^0+5^0=2^1$$
$$5^0+2^1=3^1$$
$$5^0+3^1=2^2$$
$$3^0+2^2=5^1$$
$$2^1+3^1=5^1$$
$$3^1+5^1=2^3$$
$$5^0+2^3=3^2$$
$$2^2+5^1=3^2$$
$$3^2+2^4=5^2$$
$$2^1+5^2=3^3$$
$$5^1+3^3=2^5$$
$$3^1+5^3=2^7$$
| Mahler proved this equation has only finite solutions.$a^x+b^y=c^z$
| {
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A divisibility problem without using induction or exhaustion
Problem: Determine the values of $X$ such that $24\mid(18+37X-6X^2-X^3)$ without using induction or exhaustion.
This question was motivated by the solution set $$[a\quad b\quad c\quad d]=\left[\frac34\quad\frac{37}{24}\quad-\frac14\quad-\frac1{24}\right]$$ in this question; that is, when is the value of $a+bX+cX^2+dX^3$ an integer?
It is easy to show that it holds for all odd integers $X$, after letting $X=2K+1$ and arriving at the expression $K(K-1)(2K+11)\equiv0\pmod6$.
For even integers, it is harder. Letting $X=2K$, we get $$18+148K-96K^2-64K^3\equiv0\pmod{24}$$ or that $$4K^3-K-9\equiv0\pmod{12}$$ and we can further simplify this to $$K(1-2K)(1+2K)\equiv3\pmod{12}.$$ How can we find for which $K$ this congruence holds without mathematical induction or proof by exhaustion?
| It is necessary and sufficient that $18+37X-6X^2-X^3$ is divisble by $8$ and by $3$. Reducing mod $3$ yields
$$18+37X-6X^2-X^3\equiv X+2X^3\pmod{3},$$
where $X+2X^3\equiv-X(X+1)(X+2)\pmod{3}$, so this is satisfied for all $X$.
Reducing mod $8$ yields
$$18+37X-6X^2-X^3\equiv2+5X+2X^2+7X^3\pmod{8},$$
which is zero precisely when $X\equiv1,3,5,6,7\pmod{8}$. So the expression is divisible by $24$ if and only if $X$ is odd or $x\equiv6\pmod{8}$.
Another way to see this, of course after the fact, is that
$$18+37X-6X^2-X^3=24(2X+1)-(X+1)(X+2)(X+3),$$
so the left hand side is divisible by $24$ precisely when $(X+1)(X+2)(X+3)$ is. Divisibility by $3$ is guaranteed as it is a product of three consecutive integers. The product is divisible by $8$ whenever
*
*either $X$ is odd, as then $X+1$ and $X+3$ are congruent to $0$ and $2$ modulo $4$, in some order,
*or $X+2$ is divisible by $8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$.
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} \\[10pt]
= {} & \frac{\frac{x^2+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}}}{x+\sqrt x^2 \sqrt \frac{x+\frac{1}{x}}{x+1}} \\
= {} & \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt x^2 \sqrt \frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}} = \frac{x+1}{x+ x \sqrt 1} = \frac{x+1}{2x} \to \frac {1}{2}
\end{align}
I used this for symplifying:
*
*${(x-\sqrt \frac{x^3+x}{x+1})} \cdot {{(x+\sqrt \frac{x^3+x}{x+1})}} = x^2 -\frac{x^3+x}{x+1}=> (a-b)(a+b) = a^2-b^2$.
*$\sqrt \frac{x^3+x}{x+1} = \sqrt {x^2\cdot\frac{({x+\frac{1}{x}})}{x+1}} = \sqrt x^2 \sqrt\frac{x+\frac{1}{x}}{x+1}$.
I didn't write everything formally correct but hopefully you'll get the idea. The problem I see is that sometimes I'm applying $\lim\limits_{x \rightarrow \infty}$ to just the lower part of a fraction, when I probably have to apply it to both parts of the fraction? I'm talking about this part especially: $\lim\limits_{x \rightarrow \infty}\frac{x+1}{1-\frac{1}{x}} \sim x+1$.
As always very grateful for any comments/help. Cheers.
| I think, there is a little mistake in your computations.
It should be
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3-x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} \\[10pt]
= {} & \frac{\frac{x^2-x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x-1}{1+\frac{1}{x}} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x-1}{1+\frac{1}{x}}}{x+x \sqrt \frac{x+\frac{1}{x}}{x+1}} \\
= {} & \frac{\frac{1-\frac{1}{x}}{1+\frac{1}{x}} }{1+\sqrt\frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}} \to \frac {1}{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find lines of invariant points? Every time I try a question on this topic I get it wrong.
My textbook says:
Invariant points satisfy $B\begin{pmatrix}u\\ v\end{pmatrix}=\begin{pmatrix}u\\ v\end{pmatrix}$
Re-write this as a system of equations.
Check whether both equations are in fact the same.
If so, they give a line of invariant points.
So I tried this question using that method (and the method needed for finding invariant lines):
For [this] matrix, find any lines of invariant points and any other invariant lines through the origin.
$\begin{pmatrix}3&-2\\ 4&-3\end{pmatrix}$
What I did was:
$\begin{pmatrix}3&-2\\ 4&-3\end{pmatrix}\begin{pmatrix}u\\ mu\end{pmatrix}=\begin{pmatrix}u\\ v\end{pmatrix}=\begin{pmatrix}3u-2mu\\ 4u-3mu\end{pmatrix}$
These equations are not the same, so no lines of invariant points.
Then, to find invariant lines:
$\begin{pmatrix}3&-2\\ 4&-3\end{pmatrix}\begin{pmatrix}u\\ mu\end{pmatrix}=\begin{pmatrix}u'\\ v'\end{pmatrix}=\begin{pmatrix}3u-2mu\\ 4u-3mu\end{pmatrix}$
$4u-3mu=m\left(3u-2mu\right)$
$2m^2-6m+4\:=0\:\Rightarrow \:m\:=1\: $or$ \:m=2$
So the invariant lines should be $y=2x$ or $y=x$. As far as I know this is sort of right, except $y=x$ is apparently also a line of invariant points. What did I do wrong?
| Let
$$\mathbf{B} = \left [ \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} \right ], \quad \mathbf{p} = \left [ \begin{matrix} u \\ v \end{matrix} \right ]$$
If $\mathbf{p}$ is an invariant point with respect to $\mathbf{B}$, then
$$\mathbf{B} \mathbf{p} = \mathbf{p} \tag{1}\label{NA1}$$
which is equivalent to
$$\left\lbrace \begin{aligned}
b_{11} u + b_{12} v &= u \\
b_{21} u + b_{22} v &= v \end{aligned} \right .$$
and
$$\left\lbrace \begin{aligned}
(b_{11} - 1) u + b_{12} v &= 0 \\
b_{21} u + (b_{22} - 1) v &= 0 \end{aligned} \right . \tag{2}\label{NA2}$$
To find invariant points, you solve $\eqref{NA2}$ for $u$ and $v$. In some cases, the solution is not a single point, but a line. If the matrix is all zeros, then all points are invariant.
Let's examine $\mathbf{B} = \left [ \begin{matrix} 1 & 4 \\ 2 & -1 \end{matrix} \right ]$. Substituting to $\eqref{NA2}$ we get
$$\left\lbrace\begin{aligned}
(1 - 1)u + 4 v &= 0 \\
2 u + (-1 - 1) v &= 0 \\
\end{aligned} \right . \quad \iff \quad \left\lbrace\begin{aligned}
4 v &= 0 \\
2 u - 2 v &= 0 \\
\end{aligned}\right .$$
The only solution to this is $u = v = 0$. So, this matrix has an invariant point at origin.
Let's examine $\mathbf{B} = \left [ \begin{matrix} 3 & -2 \\ 4 & -3 \end{matrix} \right ]$. Substituting to $\eqref{NA2}$ we get
$$\left\lbrace\begin{aligned}
2 u - 2 v &= 0 \\
4 u - 4 v &= 0 \\
\end{aligned}\right .$$
If we divide the upper one by 2, or the lower one by 4, we get the same equation: $u - v = 0$. Thus, this matrix has an invariant line $u = v$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+2n+1)+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+12n+6+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+13n+7$
$\Rightarrow$ $10^n \cdot 10^1 \gt 6n^2+13n+7$
$\Rightarrow$ $(6n^2+n)\cdot10 \gt 6n^2+13n+7$
$\Rightarrow$ $60n^2+10n \gt 6n^2+13n+7$
I am stuck at this point. What techniques or tricks are there to solve the rest?
It is of interest to me, because I am currently practicing a lot of exercises related to convergences, inequalities and mathematical induction.
Any hints guiding me to the right direction I much appreciate.
| You almost finished. Now you can transfer all to the left and group:
$$60n^2+10n \gt 6n^2+13n+7 \iff \\
54n^2-3n-7>0 \iff \\
44n^2+3n^2-3n+7n^2-7>0 \iff \\
44n^2+3n(n-1)+7(n^2-1)>0,$$
which is true because $n>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $ABCD$ be a parallelogram. Show that $\angle BQD = 90^ \circ$.
Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $\angle BQD = 90^ \circ$.
| Alternatively, draw the second diagonal and denote the center with $O$.
$\hspace{4cm}$
Using the formulas of medians:
$$\begin{align}DQ^2&=\frac{2CD^2+2PD^2-PC^2}{4};\\
BQ^2&=\frac{2BC^2+2BP^2-PC^2}{4};\\
OP^2&=\frac{2AP^2+2PC^2-AC^2}{4} \Rightarrow \color{red}{AC^2}=2AP^2+2PC^2-4OP^2=\color{red}{2BD^2+2PC^2-4OP^2}.\end{align}$$
Adding:
$$\begin{align}DQ^2+BQ^2&=\frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\\
&=\frac{(\color{red}{AC^2}+BD^2)+2\left(\left(\frac{BD}{2}+OP\right)^2+\left(\frac{BD}{2}-OP\right)^2\right)-2PC^2}{4}=\\
&=\frac{(\color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2\left(\frac{BD^2}{2}+2OP^2\right)-2PC^2}{4}=\\
&=BD^2.\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n+1}(2n+1)+1)}4=(n+1)(3^{n+1})+\frac{3(3^n(2n-1)+1))}4 $$
I am stuck afterward, please help, thanks.
| We have
$$\sum_{k=1}^{n+1} k\cdot 3^k=(n+1)\cdot 3^{n+1}+\sum_{k=1}^n k\cdot 3^k\stackrel{Ind. Hyp.}=(n+1)\cdot 3^{n+1}+\frac {3(3^n(2n-1)+1)} 4=$$
$$=\frac {4(n+1)\cdot 3^{n+1}+3^{n+1}(2n-1)+3} 4=\frac {3^{n+1}(6n+3)+3} 4=$$
$$=\frac {3(3^{n+1}(2(n+1)-1)+1)} 4=f(n+1)$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Evaluating an Integral with a $\sqrt{1-x^2}$ in it Below is a problem I did but my answer does not match the book's table of
integrals. I would like to know where I went wrong.
Thanks
Bob
Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int x^2 \sqrt{ 1 - x^2 } \,\, dx \\
\end{eqnarray*}
Answer
\begin{eqnarray*}
\text{Let } I &=& \int x^2 \sqrt{ 1 - x^2 } \,\, dx \\ \\
\text{Let }\sin u &=& x \\
dx &=& \cos{u} \, du \\
I &=& \int \sin^2{u} \sqrt{1 - \sin^2u} \,\, \cos{u} \, du \\
I &=& \int \sin^2{u} \cos^2{u} \,\, du \\
\end{eqnarray*}
Now recall the following two standard identities:
\begin{eqnarray*}
\sin^2{\theta} &=& \frac{1 - \cos{2 \theta}}{2} \\
\cos^2{\theta} &=& \frac{1 + \cos{2 \theta}}{2} \\
\end{eqnarray*}
Now applying the above two identities we have:
\begin{eqnarray*}
I &=& \int \frac{(1 - \cos{2 u})(1 + \cos{2 u})}{4} \,\, du \\
4I &=& \int 1 - \cos^2{2u} \,\, du = \int \sin^2{2u} \,\, du \\
4I &=& \int \frac{1 - \cos{4u}}{2} \,\, d\theta \\
8I &=& \int 1 - \cos{4u} \,\, du \\
\int \cos{4u} \,\, du &=& \frac{\sin{4 u}}{4} + C_1 \\
\end{eqnarray*}
Now recall the following two standard identities:
\begin{eqnarray*}
\sin{2 \theta} &=& 2 \sin{\theta} \cos{\theta} \\
\cos{2 \theta} &=& 1 - 2 \sin^2{\theta} \\
\end{eqnarray*}
\begin{eqnarray*}
\sin{4 u} &=& 2 \sin{2 u} \cos{2 u} = 4 \sin u \cos u ( 1 - 2 \sin^2 u) \\
\sin{4 u} &=& 4 \sin u \cos u - 8 \sin^3 u \cos u \\
\sin{4 u} &=& (4 \sin u - 8 \sin^3 u ) ( 1 - \sin^2 u)^\frac{1}{2} \\
\int \cos{4u} \,\, du &=&
( \sin u - 2 \sin^3 u ) ( 1 - \sin^2 u)^\frac{1}{2} + C_2 \\
8I &=& \sin^{-1} x - (x - 2x^3)\sqrt{1 - x^2} + C_2 \\
I &=& \frac{1}{8}\sin^{-1} x - \frac{(x - 2x^3)\sqrt{1 - x^2}}{8} + C \\
\end{eqnarray*}
However, the book's answer is:
\begin{eqnarray*}
\int x^2 \sqrt{1 - x^2} \,\, dx &=&
\frac{1}{8} \sin^{-1}{x} - \frac{x\sqrt{1-x^2}(1-2x^2)}{8} + C \\
\end{eqnarray*}
| Let's try a different method:
$$
x^2\sqrt{1-x^2}=\frac{x^2-x^4}{\sqrt{1-x^2}}=(x^3-x)\frac{-x}{\sqrt{1-x^2}}
$$
Let's do integration by parts:
\begin{align}
I&=\int x^2\sqrt{1-x^2}\,dx\\[6px]
&=\int(x^3-x)\frac{-x}{\sqrt{1-x^2}}\,dx\\[6px]
&=(x^3-x)\sqrt{1-x^2}-\int(3x^2-1)\sqrt{1-x^2}\,dx\\[6px]
&=(x^3-x)\sqrt{1-x^2}-3I+\int\sqrt{1-x^2}\,dx
\end{align}
The last integral is well known:
$$
\int\sqrt{1-x^2}\,dx=\frac{1}{2}\arcsin x+\frac{1}{2}x\sqrt{1-x^2}
$$
so we obtain
$$
4I=\frac{1}{2}\arcsin x+\frac{1}{2}(2x^3-x)\sqrt{1-x^2}+c
$$
which is the same as your solution, but also the same as the book's, because we can collect $x$:
$$
I=\frac{1}{8}\arcsin x+\frac{1}{8}x(2x^2-1)\sqrt{1-x^2}+c
$$
| {
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How can I determine general formula of this sequence? I am trying to find general formula of the sequence $(x_n)$ defined by
$$x_1=1, \quad x_{n+1}=\dfrac{7x_n + 5}{x_n + 3}, \quad \forall n>1.$$
I tried
put $y_n = x_n + 3$, then $y_1=4$ and
$$\quad y_{n+1}=\dfrac{7(y_n-3) + 5}{y_n }=7 - \dfrac{16}{y_n}, \quad \forall n>1.$$
From here, I can't solve it. How can I determine general formula of above sequence?
With Mathematica, I found $x_n = \dfrac{5\cdot 4^n-8}{4^n+8}$. I want to know a method to solve problem, than have a given formula.
| Here is a method that I read from a book. Yet I did not think deeply why it works in general.
If there exists real numbers $\alpha$, $\beta$ and $r$ such that
$$\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=r\cdot\frac{a_n-\beta}{a_n-\alpha}$$
for all $n\in\mathbb{N}$, then the sequence $\{b_n\}$, where $b_n=\frac{a_n-\beta}{a_n-\alpha}$, would be geometric and can be solved easily.
So our job is to find such $\alpha$, $\beta$ and $r$.
Substituting the recurring equation,
\begin{align}
\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}&=\frac{\frac{7a_n+5}{a_n+3}-\beta}{\frac{7a_n+5}{a_n+3}-\alpha} \\
&=\frac{7a_n+5-\beta(a_n+3)}{7a_n+5-\alpha(a_n+3)} \\
&=\frac{(7-\beta)a_n+(5-3\beta)}{(7-\alpha)a_n+(5-3\alpha)} \\
&=\frac{7-\beta}{7-\alpha}\cdot\frac{a_n-\left(-\frac{5-3\beta}{7-\beta}\right)}{a_n-\left(-\frac{5-3\alpha}{7-\alpha}\right)}
\end{align}
Hence the trick should work if there is a solution for $\alpha=-\frac{5-3\alpha}{7-\alpha}$ and $\beta=-\frac{5-3\beta}{7-\beta}$ and $r=\frac{7-\beta}{7-\alpha}$.
Noting that $\alpha$ and $\beta$ are roots of $u=-\frac{5-3u}{7-u}$.
\begin{align}
u&=-\frac{5-3u}{7-u} \\
u(7-u)&=-(5-3u) \\
u^2-4u-5&=0 \\
u&=-1\enspace\text{or}\enspace 5 \\
\end{align}
Take $(\alpha,\,\beta)=(-1,\,5)$. $r=\frac{7-5}{7-(-1)}=\frac{1}{4}$ follows.
$$b_1=\frac{a_1-\beta}{a_1-\alpha}=\frac{1-5}{1-(-1)}=-2$$
For all $n\in\mathbb{N}$,
$$b_n=r^{n-1}\cdot b_1=\left(\frac{1}{4}\right)^{n-1}(-2)=\frac{-8}{4^n}$$
Back substitute into $b_n=\frac{a_n-\beta}{a_n-\alpha}$.
\begin{align}
\frac{-8}{4^n}&=\frac{a_n-5}{a_n-(-1)} \\
-8(a_n+1)&=4^n(a_n-5) \\
a_n&=\frac{5\cdot 4^n-8}{4^n+8}
\end{align}
The same result as given by Mathematica.
| {
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Find the probability that the white ball labelled $1$ is drawn before all the black balls.
Suppose in an urn there are $20$ black balls labelled $1,2, \ldots , 20$ and $10$ white balls labelled $1,2, \ldots ,10$. Balls are drawn one by one without replacement. Find the probability that the white ball labelled $1$ is drawn before all the black balls.
My attempt $:$
If we want to draw the first white ball before all the black balls then I have to draw the first white ball in one of first $10$ steps.
Suppose I draw the first white ball in $k$-th step. Then in order to fulfil my requirement I have to draw white balls in first $k-1$ steps. That can be done in $\binom 9 {k-1} (k-1)!$ ways. For each of these ways remaining $30-k$ balls can be drawn in $(30-k)!$ ways. This $k$ can run from $1$ to $10$.
So the total number of ways to draw the first white ball before all the black balls is $$\sum\limits_{k=1}^{10} \binom 9 {k-1}(k-1)! (30-k)!$$ So the required probability is $$\frac1{30!}{\sum\limits_{k=1}^{10} \binom 9 {k-1}(k-1)! (30-k)!} =\sum\limits_{k=1}^{10} {\frac {9!(30-k)!} {30!(10-k)!}}$$
Now my instructor has given it's answer which is $\frac {1} {21}$. Does the above sum evaluate to $\frac {1} {21}$? Is there any other simpler way to do this? Please help me in this regard. Thank you very much.
| To evaluate $\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}$:
\begin{align}
\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\frac{(30-k)!}{20!(10-k)!} \\
&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\binom{30-k}{20} \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{20}{20}+\binom{21}{20}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{21}{21}+\binom{21}{20}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{22}{21}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{23}{21}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\
&=\cdots \\
&=\frac{9!\cdot 20!}{30!}\binom{30}{21} \\
&=\frac{9!\cdot 20!}{30!}\frac{30!}{21!\cdot 9!} \\
&=\frac{1}{21}
\end{align}
When handling the sum, the relations $\binom{n}{n}=\binom{n+1}{n+1}$ and $\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$ are used.
| {
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Is this proof correct (Rationality of a number)? Is $\sqrt[3] {3}+\sqrt[3]{9} $ a rational number? My answer is no, and there is my proof. I would like to know if this is correct:
Suppose this is rational. So there are positive integers $m,n$ such that $$\sqrt[3]{3}+\sqrt[3]{9}=\sqrt[3]{3}(1+\sqrt[3]{3})=\frac{m}{n}$$
Let $x=\sqrt[3]{3}$. We get $x^2+x-\frac{m}{n}=0 \rightarrow x=\frac{-1+\sqrt{1+\frac{4m}{n}}}{2}$.
We know that $x$ is irrational and that implies $\sqrt{1+\frac{4m}{n}}$ is irrational as well (Otherwise $x$ is rational). Write $x=\sqrt[3]{3}$, multiply both sides by $2$ and then raise both sides to the power of $6$ to get:
$$24^2=\left(\left(\sqrt{1+\frac{4m}{n}}-1\right)^3\right)^2\rightarrow 24=\left(\sqrt{1+\frac{4m}{n}}-1\right)^3=\left(1+\frac{4m}{n}\right)\cdot \sqrt{1+\frac{4m}{n}}-3\left(1+\frac{4m}{n}\right)+3\sqrt{1+\frac{4m}{n}}-1$$
Let $\sqrt{1+\frac{4m}{n}}=y,1+\frac{4m}{n}=k $.
We get: $25=ky-3k+3y\rightarrow y=\frac{25+3k}{k+3}$, So $y$ is rational. But we know $y$ is irrational (again, otherwise $\sqrt[3]{3}$ is rational) which leads to a contradiction. So the answer is No, $\sqrt[3]{3}+\sqrt[3]{9}$ is irrational.
Is this proof correct? Is there another way to prove this? Thanks!
| Your proof is correct.
Let $\alpha = \sqrt[3]{3} + \sqrt[3]{9} = \sqrt[3]{3}(1+\sqrt[3]{3})$. We have
$$\alpha^3 = 3(1+\sqrt[3]{3})^3 = 3(3 + 3\sqrt[3]{3} + 3\sqrt[3]{9} + 3) = 12 + 9(\sqrt[3]{3} + \sqrt[3]{9}) = 12 + 9\alpha$$
Hence $\alpha^3 - 9\alpha -12 = 0$.
However, the polynomial $x^3-9x-12$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion for $p = 3$ so it cannot have any rational roots. Therefore $\alpha$ is not rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of the power series $\sum_{n=1}^\infty n*(n+1)*x^n$ $\sum_{n=1}^\infty n*(n+1)*x^n$
Hello everyone, I need help in solving the question above.
I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue.
Thanks.
| Just to make the problem more general.
Consider
$$\sum_{n=1}^\infty (an^2+bn+c)\,x^n$$ rewrite $n^2=n(n-1)+n$ which makes
$$an^2+bn+c=a n(n-1)+a n+bn+c=an(n-1)+(a+b)n+c$$
$$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a \sum_{n=1}^\infty n(n-1)x^n+(a+b)\sum_{n=1}^\infty nx^n+c\sum_{n=1}^\infty x^n$$
$$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a x^2 \sum_{n=1}^\infty n(n-1)x^{n-2}+(a+b)x\sum_{n=1}^\infty nx^{n-1}+c\sum_{n=1}^\infty x^n$$
$$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a x^2\left(\sum_{n=1}^\infty x^n \right)''+(a+b)x\left(\sum_{n=1}^\infty x^n \right)'+c\left(\sum_{n=1}^\infty x^n \right)$$
If we had an $n^3$ term, the same idea
$$n^3=n(n-1)(n-2)+3n(n-1)+n$$ and so on for $n^k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Aperiodicity of $\sin[2x+\cos(\sqrt{2}x)]$
Is the function $f(x)=\sin[2x+\cos(\sqrt{2} \cdot x)]$ periodic?
I don't know how to prove that this function is not periodic. Can you help me out?
| First, some intuition. Here's a graph of $f(x)=\sin(2x+\cos(\sqrt{2} \cdot x))$:
We have to prove that the function is aperiodic. From the graph it appears that something stronger is true: even if we focus on any specific $y$-value $y_0$, the values of $x$ where $f(x) = y_0$ will not be evenly spaced. So let's pick $y_0 = 0$ for convenience.
Towards a contradiction, let's also assume the function is periodic, and it has period $a$. Then let $r$ be any root, where $f(r) = 0$. We have
$$
\cdots = f(r-2a) = f(r-a) = f(r) = f(r+a) = f(r+2a) = \cdots = 0.
$$
In order for all these to be zero, we need for all $k \ge 1$:
\begin{align*}
2r + \cos(\sqrt{2} \cdot r) &\in \pi \mathbb{Z} \tag{1} \\
2(r+ka) + \cos(\sqrt{2} \cdot (r+ka)) &\in \pi \mathbb{Z} \tag{2} \\
2(r-ka) + \cos(\sqrt{2} \cdot (r-ka)) &\in \pi \mathbb{Z} \tag{3} \\
\end{align*}
where $\pi \mathbb{Z}$ is the set of real numbers that are $\pi$ times an integer. Adding (2) and (3) we get
\begin{align*}
4r + \cos(\sqrt{2} \cdot (r+ka)) + \cos(\sqrt{2} \cdot (r-ka)) &\in \pi \mathbb{Z},
\end{align*}
and this can be simplified: note $\cos(x + y) + \cos(x - y) = 2 \cos x \cos y$. So
$$
4r + 2\cos(\sqrt{2} \cdot r) \cos(\sqrt{2} \cdot ka) \in \pi \mathbb{Z}. \tag{4}
$$
Now we can subtract two times (1) to get
$$
2 \cos(\sqrt{2} \cdot r) [\cos(\sqrt{2} \cdot ka) - 1] \in \pi \mathbb{Z}.
$$
But $\cos$ is always between $-1$ and $1$, so this expression is between $-4$ and $4$ (it's the product of $2$, something between $-1$ and $1$, and something between $-2$ and $2$). Therefore this value is $-\pi, 0,$ or $\pi$.
And it must be for all $k$. In particular it is either $0$, $\pi$, or $-\pi$ for infinitely many $k$.
This gives us a couple of cases. Either (i) $\cos(\sqrt{2} \cdot r) = 0$; or (ii) $\cos(\sqrt{2} \cdot ka)$ attains the same value for infinitely many $k$.
*
*Case (i). We know $\sqrt{2} \cdot r - \frac{\pi}{2} \in \pi \mathbb{Z}$.
So $(2 \sqrt{2})r \in \pi \mathbb{Z}$.
At the same time, we go back to equation (1), and we get that $2r \in \pi \mathbb{Z}$. If $r \ne 0$ then this implies that the ratio $\frac{(2 \sqrt{2})r}{r} = 2 \sqrt{2}$ is rational, contradiction (because $\sqrt{2}$ is irrational). So then $r = 0$. But then $\cos(\sqrt{2} \cdot r) = 1$, not $0$.
*Case (ii). If $\cos x = \cos y$, what can we say about $x$ and $y$? Either $x$ and $y$ differ by a multiple of $2 \pi$, or $x$ and $-y$ differ by a multiple of $2 \pi$ (prove this). But if we have infinitely many such values, we must find two that differ by a multiple of $2 \pi$. So we have $k_1$ and $k_2$ such that $\sqrt{2} \cdot k_1 a - \sqrt{2} \cdot k_2 a \in 2 \pi \mathbb{Z}$.
So $a (k_1 - k_2) \cdot \frac{\sqrt{2}}{\pi} \in \mathbb{Q}$, so $a$ is some rational number times $\frac{\pi}{\sqrt{2}}$.
This gets us very close to our contradiction. Go back to equations (2) and (3), and plug in values of $k$ such that $\sqrt{2} k a$ is a multiple of $2 \pi$, so it can be ignored. We then have that
$$
2r + 2ka + \cos(\sqrt{2} \cdot r) \in \pi \mathbb{Z}
$$
for all such values of $k$.
Subtracting two such values $k_3$ and $k_4$, $2(k_3 - k_4) a \in \pi \mathbb{Z}$, so $a$ is $\pi$ times a rational number.
But now $a$ is a rational number times $\pi$ and a rational number times $\frac{\pi}{\sqrt{2}}$. Since $a$ is nonzero (it's a period), this means that $\sqrt{2}$ is rational, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and we are back to where we started. So we assume $x$ and $y$ are coprime.
Testing the parity, we have sum of 2 cubes to be odd. WLOG, we can assume $x$ is even and $y$ is odd.
Trying mod $3$, we have $x+y=0 \pmod 3$. Since $x$ and $y$ are coprime, $x$ and $y$ must be congruent to $1$ and $-1$ or vice-versa.
If I assume $x=3m+1$ and $y=3n-1$, expand out and simplify, I get $27(m^3+n^3)+27(m^2-n^2)+9(m+n)=3^z$. If I assume $z \geq 3$, this gives $(m^3+n^3)+(m^2-n^2)+\frac{m+n}{3}=3^{z-3}$. But I don't see how to proceed.
I also tried mod $9$ but didn't get anywhere, it didn't cut down the possibilities by much.
I also tried letting $y=x+r$. Then
\begin{align*}
x^3+y^3 &= x^3+(x+r)^3 \\
&= x^3 + (x^3+3x^2r+3xr^2+r^3) \\
&= 2x^3+3x^2r+3xr^2+r^3 \\
&= 3^z
\end{align*}
Then $3\mid 2x^3+3x^2r+3xr^2+r^3$, and $3\mid 3x^2r+3xr^2$, so this implies $3 \mid 2x^3+r^3$. But this doesn't yield any contradiction.
Can anyone supply a proof? Or if my hypothesis is wrong, how to derive all the integer solutions?
Thank you.
| Solution by LTE, let $d=(x,y)>1$, $x=dx_1$, $y=dy_1$, $(x_1,y_q)=1$ then $$d^3\mid x^3+y^3=3^z$$ then $d\mid 3^a$ for a positive integer $a<z/3$. Dividing by $d^3$ we obtain $x^3_1+y^3_1=3^b$, for a positive integer $b=z-a$. This is a equation similar to the original, then we can assume $(x,y)=1$. If $3\nmid x+y$ then $x+y=1$ a contradiction. Then we have all the conditions to use LTE: $$v_3(x^3+y^3)=v_3(x+y)+v_3(3)=v_3(x+y)
+1=v_3(3^z)=z$$ From here we conclude that $x+y=3^{z-1}$ this implies that $x^2-xy+y^2=3$, $(x-y)^2+xy=3$, if $|x-y|\ge 2$ we have a contradiction, then there are two cases:
Case 1. $|x-y|=0$ this is not possible due $x$ and $y$ are coprimes and $x+y>2$.
Case 2. $|x-y|=1$ implying $xy=2$, with solutions $(x,y)=(2,1)$ or $(1,2)$, implying $3^{z-1}=2+1\Rightarrow z=2$.
Then the solutions to the original equation are: $(x,y,z)=(3^k,2\cdot 3^k,3k+2)$ or $(2\cdot 3^k, 3^k,3k+2)\forall k\in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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