Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the smallest number such that when you divide with $4,6,8,10,12$ the remainder is $2,4,6,8,10$
Find the smallest number such that when you divide with $4,6,8,10,12$ the remainder is $2,4,6,8,10$.
This is the solution:
Let $n \in \mathbb N$ such that $n = 4q_1 + 2$, for some $q_1$, $n = 6q_2 + 4$, for some $q_2$, $n = 8q_3 + 6$, for some $q_3$, $n = 10q_4 + 8$, for some $q_4$, $n = 12q_5 + 10$, for some $q_5$.
Then $4 \mid n + 2$, $6 \mid n + 2$, $8 \mid n + 2$, $10 \mid n + 2$, $12 \mid n + 2$, so $[4, 6, 8, 10, 12] \mid n + 2$ so $120 \mid n + 2$ as $[4, 6, 8, 10, 12] = 120$ so the smallest number is $118$.
I do not understand how he get that $4 \mid n + 2$, $6 \mid n + 2$, $8 \mid n + 2$, $10 \mid n + 2$, $12 \mid n + 2$, can you explain it to me?
| $$n+2=4q_1+2+2=4(q_1+1)\\n+2=6q_2+4+2=6(q_2+1)\\n+2=8q_3+6+2=8(q_3+1)\\\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the radius of convergence for $\frac{1}{(3 - x)(1 + x)^{2}}$ at $x_{0} = 0$ and $x_{0} = 1$ I am having trouble with the problem
Find the radius of convergence for $f(x) = \frac{1}{(3 - x)(1 +
x)^{2}}$ at $x_{0} = 0$ and $x_{1} = 1$.
Usually, when I'm trying to find radius of convergence, I'm given some sort of series (like, a summation). This isn't the case here, and I'm confused as to what I'm supposed to do. This is a problem in the exercise. It is in the section for power series. Also, I have this formula for the radius of convergence $R$:
$$R = \frac{1}{\limsup_{n\to\infty} |a_{n}|^{1/n}},$$
but I'm not sure how I'm supposed to use that here either.
Any help is appreciated.
EDIT: Maybe it has something to do with the standard Taylor series form? Like,
$$f(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + f''(x_{0})(x - x_{0})^{2}/2! \ldots $$
Also, in case it matters, by Partial Fractions, you can write
$$\frac{1}{(3-x)(1 + x)^{2}} = \frac{1}{16(x + 1)} + \frac{1}{4(x+1)^{2}} - \frac{1}{16(x - 3)}. $$
The first few terms of the Taylor series expansion about $x_{0} = 0$ is given by
$$f(x) = \frac{1}{3} - \frac{5}{9}x + \frac{22}{27}x^{2} - \frac{86}{81}x^{3} + \frac{319}{243}x^{4} + \cdots $$
(I don't have a closed term for the sum, but these are the first terms explicitly computed. Looks like the denominator is just powers of $3$. Not sure about the numerator though...)
| Here are my thoughts. This might not be the original purpose of the exercise, but this is my guess.
According to your partial fraction decomposition, and using the well-know power series $(1-x)^{-1} = \sum_0^\infty x^n$ for $x \in (-1,1)$, at $x=0$, we have
\begin{align*}
f(x)&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac {\mathrm d}{\mathrm dx} \left( \frac 1{4(x+1)} \right) + \frac 1{48} \cdot \frac 1{1 - x/3} \\
&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac 14 \cdot\frac {\mathrm d }{\mathrm dx}\left( \sum_0^\infty (-1)^n x^n \right) + \frac 1{48} \cdot \sum_0^\infty \frac {x^n}{3^n}\\
&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac 14 \sum_0^\infty (n+1)(-1)^{n+1} x^{n} + \frac 1{48} \cdot \sum_0^\infty \frac {x^n}{3^n}\\
&= \sum_0^\infty \left(\frac {(-1)^n}{16} + \frac 14 (n+1)(-1)^{n+1} +\frac 1{16 \cdot 3^{n+1}} \right) x^n,
\end{align*}
and all of these computations are valid iff $|x|<1$ and $|x/3|<1$. Conclusively the radius is $1$. You might try to compute again by the formula.
At $x=1$, just let $y = x-1$, then $x = y+1$ and
$$
f(x) =g(y) = \frac 1{(2-y)(2+y)^2},
$$
you could do the similar thing like above to give the power series at $y=0$. You could see that the operations are valid iff $|y/2|<1$, hence $|y|<2$ and the radius is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What are the possible values of these letters? Out of all the questions I answered in a math reviewer, this one killed me (and 7 more).
Let $J, K, L, M, N$ be five distinct positive integers such that
$$
\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{JKLMN} = 1.
$$
Then, what is $J + K + L + M + N$?
I have been thinking about this for nearly 6 days.
| Induction could lead you to the answer.
The equation is :
$$
\frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_{n}} + \frac 1 { x_1 x_2 \dots x_{n}} = 1
$$
Case $ n = 0 $: the empty set solves the equation as an empty product is 1
Case $ n = 1 $: the obvious solution is $ x_1 = 2 $.
Case $ n = 2 $: a bit more difficult, but you can find $ x_1 = 2, x_2 = 3 $.
Doing this, I noticed one thing: assuming that you solved the $ (n-1) $-th equation, you can pick $ x_n $ so that $ + \frac 1 {x_{n}} $ in the first part of the equation compensates the factor $ \frac 1 {x_n} $. Let’s check.
Case any $ n $: assuming that $ x_1, \dots x_{n} $ solves the equation, we require $ x_{n+1} $ so that
$$
\frac 1 {x_n} + \frac 1 {x_2} + \dots + \frac 1 {x_{n+1}} + \frac 1 { x_1 x_2 \dots x_{n+1}} = \frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_n} + \frac 1 { x_1 x_2 \dots x_n}
$$
Removing identical summands:
$$
\frac 1 {x_{n+1}} + \frac 1 { x_1 x_2 \dots x_{n+1}} = \frac 1 { x_1 x_2 \dots x_n }
$$
Multiplying tops by $ x_1 x_2 \dots x_{n+1} $ :
$$
x_1 x_2 \dots x_{n} + 1 = x_{n+1}
$$
Solved!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 2
} |
Verification of Laurent series for the function $f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}$ Suppose that $$f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
\begin{align}
f(z)&=-2\ \frac{d}{dz}\left(\frac{1}{z+2}\right)-\frac{5}{(z-2)-2} \\
&=-2\ \frac{d}{dz}\left(\frac{1}{4}\left(\frac{1}{1+\frac{(z-2)}{4}}\right)\right)+\frac{5}{2}\left(\frac{1}{1-\frac{(z-2)}{2}}\right) \\
&=-2\ \frac{d}{dz}\left(\frac{1}{4}\sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{4}\right)^n\right)+\frac{5}{2}\left(\sum_{n=0}^{\infty} \left(\frac{z-2}{2}\right)^n\right) \\
&=-2\sum_{n=1}^{\infty} (-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}+5\sum_{n=0}^{\infty} \frac{(z-2)}{2^{n+1}}^n.
\end{align}
I believe this is the right answer, yet the answer provided (using a different method) states the answer is $$f(z)=\sum_{n=0}^{\infty}c_n(z-2)^n, \ \ c_n=(-1)^n\frac{2(n+1)}{4^{n+3}}+\frac{5}{2^{n+1}}.$$
| After the last $=$ sign, what you got was:$$-2\sum_{n=1}^\infty(-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$\sum_{n=1}^\infty(-1)^{n-1}\frac{2n(z-2)^{n-1}}{4^{n+1}}=\sum_{n=0}^\infty(-1)^n\frac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theorem,
$$2^{4}\equiv 1\pmod5$$
Raising to the power k we get,
$$2^{4k}\equiv 1\pmod5$$
$$2^{4k+3}\equiv 8\pmod5$$
$$2^{4k+3}\equiv 3\pmod5$$
So, $5\mid 2^{n}-3\quad\forall \quad n=4k+3$ where $k$ is any non-negative integer.
Similarly, by Fermat's theorem
$$2^{12}\equiv 1\pmod{13}$$
$$2^{12k}\equiv 1\pmod{13}$$
$$2^{12k+4}\equiv 16\pmod{13}$$
$$2^{12k+4}\equiv 3\pmod{13}$$
Therefore, $13\mid 2^{n}-3\quad\forall \quad n=12k+4$
How do I show that it contains no term which is divisible by 65?
Thank you!
| You're on the right track. By Fermat's little theorem, $2^n\pmod{5}$ and $2^n\pmod{13}$ depend only on $n\pmod{4}$ and $n\pmod{12}$, respectively. You've shown that $2^n\equiv 3 \pmod{5}$ if $n\equiv 3\pmod{5}$, but that's the opposite of what you want to prove; you want to show that $2^n\equiv 3\pmod{5}$ only if $n\equiv 3\pmod{5}$, and similarly for $13$. The result you'll get is that $2^n\equiv 3\pmod{5}$ only if $n\equiv 3\pmod{4}$ and $2^n\equiv 3\pmod{13}$ only if $n\equiv 4\pmod{12}$. Those two conditions can clearly never be simultaneously satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit of a sequence of integrals involving continued fractions The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.
Find $\lim_{n \to \infty} A_n $ if
$$ A_1 = \int\limits_0^1 \frac{dx}{1 + \sqrt{x} }, \; \; \; A_2 =
\int\limits_0^1 \frac{dx}{1 + \frac{1}{1+\sqrt{x}} }, \; \; \; A_3 =
\int\limits_0^1 \frac{dx}{1 + \frac{1}{1+\frac{1}{1+\sqrt{x}}} },
...$$
First of all, I think this is a hard question for a midterm exam, but anyway, notice that we can calculate $A_1$ by making $t=\sqrt{x}$
$$ A_1 = \int\limits_0^1 \frac{2 t dt }{1+t} = 2 \int\limits_0^1 dt - 2 \int\limits_0^1 \frac{dt}{1+t}=2-2(\ln2)=2-\ln2^2 $$
Now, as for $A_2$ I would do $t = \frac{1}{1+\sqrt{x}}$ which gives $d t = \frac{ dx}{2 \sqrt{x} (1+\sqrt{x})^2} = \frac{t^2 dx}{2 (t-1)}$ thus upon sustituticion we get
$$ A_2 = - \int\limits_1^{1/2} \frac{2 (t-1) }{t^2(1+t) } dt $$
which can easily solved by partial fractions or so. But, apparently this is not the way this problem is meant to be solved as this exam contained 4 questions to be solved in an hour. What is the trick, if any, that can be used to solve this problem without doing the unellegant partial fractions?
| $\sqrt{x}$
is a McGuffin.
More generally,
let
$f_1 = \frac{1}{1 + g(x) }
$
where
$g'(x) > 0,
g(0) = 0
$,
$f_n(x)
=\frac{1}{1+f_{n-1}(x)}
$,
and
$A_n = \int_0^1 f_n(x) dx
$.
Then
$f_n(x)
\to \dfrac{\sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$\begin{array}\\
f_n(x)
&=\frac{1}{1+\frac{1}{1+f_{n-2}(x)}}\\
&=\frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\\
&=\frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\\
\end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$\frac12 < f_n(x)
\lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
\lt 1$.
$\begin{array}\\
f_n(x)-f_{n-2}(x)
&=\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\\
&=\dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\\
&=\dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\\
\end{array}
$
$\begin{array}\\
f_n(x)+f_n^2(x)
&=\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\\
&=\dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+\dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
&=\dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
&=\dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
\text{so}\\
1-f_n(x)-f_n^2(x)
&=\dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\\
&=\dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
\end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
\lt \frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
\to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = \frac{\sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x \ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
\to x_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
Help with a system of inequalities with absolute values I'm trying to solve this system on inequalities
$$
\left\{
\begin{array}{c}
|x-3|<2x \\
|2x+5|>3
\end{array}
\right.
$$
The steps I'm taking are:
Finding the absolute values sings, so for
$x-3 \geq 0$ we have $x \geq 3$ therefore
$$|x-3| =
\left\{
\begin{array}{c}
x-3 & \text{for $x \geq 3$} \\
-x+3 & \text{for $x < 3$}
\end{array}
\right. $$
and
$2x+5 \geq 0$ we have $x \geq \frac{-2}{5}$ therefore
$$|2x+5| =
\left\{
\begin{array}{c}
2x+5 & \text{for $x \geq \frac{-2}{5}$} \\
-2x-5 & \text{for $x<\frac{-2}{5}$} \\
\end{array}
\right. $$
So I build a few systems with the complete inequalities, for the first one we have:
$$
\left\{
\begin{array}{c}
x \geq 3 \\
x-3<2x = x>-3
\end{array}
\right. $$
So the solution here would be $x>3$, then:
$$
\left\{
\begin{array}{c}
x<3 \\
-x+3<2x = x>1
\end{array}
\right. $$
The solution would be $1<x<3$. Then
$$
\left\{
\begin{array}{c}
x \geq \frac{-2}{5} \\
2x+5>3 = x>-1
\end{array}
\right. $$
So the solution of the system is $x>-1$, then
$$
\left\{
\begin{array}{c}
x< \frac{-2}{5} \\
-2x-5>3 = x<-4
\end{array}
\right. $$
And the solution is $x<-4$
Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?
| There is actually only the first inequality to be solved:
$$ |2x+5| \stackrel{\color{blue}{0}\leq|x-3| \color{blue}{< 2x}}{>} |x-3|+5 > 3$$
So, you only need to solve $|x-3| < 2x$ while $x>0$:
$$
-2x < x-3 < 2x \Leftrightarrow
\begin{cases}
3x > 3 \Leftrightarrow \color{blue}{x> 1} \\
x > -3 \mbox{ does not extend the solution}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why does $\int_1^\sqrt2 \frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$ equal to $0$? In this question, the OP poses the following definite integral, which just happens to vanish:
$$\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=\sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=\sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=\sqrt[3]{2}$ to $x=\sqrt{2}$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes:
$$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{tx-sx^2+x^3}\bigg)dx=0$$
| Put in other words we desire to show that:
$$I=\int_1^\sqrt2 \ln\left(\frac{2-2x^2+x^4}{2-2x+x^2}\right)\frac{dx}{x}=\int_1^\sqrt 2\frac{1}{x}\ln x dx=\frac18\ln^2 2$$
Let's take the LHS integral and split it in two parts.
$$I=\underbrace{\int_1^\sqrt2 \frac{\ln(2-2x^2+x^4)}{x}dx}_{\mathcal K}-\underbrace{\int_1^\sqrt2 \frac{\ln(2-2x+x^2)}{x}dx}_{\mathcal J}$$
$$\mathcal J=\int_1^\sqrt2 \frac{\ln(2-2x+x^2)}{x}dx\overset{\large x\to \frac{2}{x}}=\int_{\sqrt 2}^2 \frac{\ln(2-2x+x^2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$
$$\Rightarrow 2\mathcal J=\int_1^2\frac{\ln(2-2x+x^2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$
$$\mathcal J\overset{x\to x^2}=\int_1^\sqrt 2\frac{\ln(2-2x^2+x^4)}{x}dx-\frac{1}{8}\ln^22=\mathcal K-\frac{1}{8}\ln^2 2$$
$$\Rightarrow I=\frac{1}{8}\ln^2 2$$
Your conjecture is also correct, since by the same method we can show that:$$\int_1^{\sqrt{t}}\ln\left(\frac{t+sx^2+x^4}{t+sx+x^2}\right)\frac{dx}{x}=\int_1^\sqrt{t} \frac{\ln x}{x}dx=\frac{1}{8}\ln^2t$$
This time after we split the LHS integral into two parts, we will substitute in the second integral $\displaystyle{x\to\frac{t}{x}}$,
followed by an addition with the original $\mathcal J$ from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 1,
"answer_id": 0
} |
What is the limit of $\lim_{n\to \infty} (1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16}) \cdots (1 - \frac{1}{(n+1)^2})$? What is the evaluation of the following infinite series?
$$\lim_{n\to \infty} \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) \cdots \left(1- \frac{1}{(n+1)^2}\right)$$
I've tried to simplify each expression which left me with:
$$\lim_{n\to \infty} \frac{3\times8\times15\times24\times\cdots\times((n+1)^2-1)}{4\times9\times16\times25\times\cdots\times(n+1)^2}$$
Is this a good way to approach this problem?
| hint
Take logarithm and use
$$
\sum_{k=2}^{n+1}\ln\left(1-\frac{1}{k^2}\right)
= \sum_{k=2}^{n+1}\left(\ln\left(\frac{k-1}{k}\right)+\ln\left(\frac{k+1}{k}\right)\right)
=\ln\left(\frac{1}{n+1}\right)+\ln\left(\frac{n+2}{2}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Sequence in which adding 2 produces a square Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 \cdot 7 - 2 = 47, x_5=47 \cdot 7 - 7=322, x_6=322 \cdot 47 - 7 =15127, x_7=15127 \cdot 322 -47=4870847,x_8=4870847 \cdot 15127-322=73681302247.$
My spreadsheet cannot calculate any more, but notice that $x_n+2$ is always square:
$\sqrt{x_4+2}=7$, $\sqrt{x_5+2}=18$, $\sqrt{x_6+2}=123$, $\sqrt{x_7+2}=2207$,$\sqrt{x_8+2}=271443.$ Is this a coincidence?
| We have
\begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\\
\end{align*}
then
\begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\\
\end{align*}
then by telescoping
\begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\\
\end{align*}
Now we let $x_n=z_n-2$. We then have
\begin{align*}
z_{n+1}z_{n-2}-2z_{n+1}-2z_{n-2}+4&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4-4\\
z_{n+1}z_{n-2}&= z_n^2-4z_n+z_{n-1}^2-4z_{n-1}+2z_{n+1}+2z_{n-2}\\
z_{n+1}z_{n-2}&= z_n^2-4z_n+z_{n-1}^2-4z_{n-1}+2x_{n+1}+4+2z_{n-2}\\
\end{align*}
But \begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
\end{align*}
then by substitution and reduction, we find
\begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
\end{align*}
from which a proof by induction easily follows.
Addition
We show that $v_n=\sqrt{\frac{x_{n}-2}{5}}$ is A101361 (shifted).
We have $z_n = 5v_n^2+4$ and $x_n= 5v_n^2+2$. Then
\begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}+4)^2\\
(5v_{n+1}^2+4)(5v_{n-2}^2+4)&=(5v_{n}^2+5v_{n-1}^2+4)^2
\end{align*}
which, after algebric manipulations, can be rearranged as
$$5v_{n+1}^2v_{n-2}^2-5(v_{n}^2-v_{n-1}^2)=4\left(5v_{n}^2v_{n-1}^2 +2v_n^2 +2v_{n-1}^2 -(v_{n+1}^2+v_{n-2}^2)\right) \tag1 $$
On the other hand, we have
\begin{align*}
x_{n+1}+x_{n-2}&=x_nx_{n-1}\\
5v_{n+1}^2+2+5v_{n-2}^2+2&=(5v_{n}^2+2)(5v_{n-1}^2+2)
\end{align*}
which can be simplified so that
$$v_{n+1}^2+v_{n-2}^2 = 5v_{n}^2v_{n-1}^2 +2v_n^2 +2v_{n-1}^2 \tag2$$
Comparing (1) and (2), we see that it must be that $v_{n+1}v_{n-2}=v_n^2 - v_{n-1}^2$ which is the third-order nonlinear recursion that is satisfied by A101361. Moreover, it is easy to verify that the initial terms are equal (shifted). Also it is said in that OEIS page that $v_{n+1}= F_{2F_{n}}$, where $F_{n}$ is the Fibonacci number. Then we finally obtain a nice closed form for the OP sequence, in terms of Fibonacci and Lucas numbers:
$$ x_n = L_{2F_{n-1}}^2 -2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $2n+1$ and $3n+1$ are perfect squares, then prove that $8|n$.
If for some number $n\in \mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.
if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ with $8$ then remainder is $0,1,4$, so I need to show that $m^2$ and $k^2$ have the same remainder. But I do not know is this good way because I do not know how to prove that they must have the same remainder
| It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4\not \mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4\not \mid b-a$ and $4\mid b+a$.
As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.
So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.
AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.
And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.
So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$ I need to find
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$
Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.
Here are the two ways I approached this:
WAY I:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$
WAY II:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$
What am I doing wrong here?
Thanks!
| As an alternative by $y=x+5 \to 0$
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}=\lim_{y\rightarrow 0} \frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=\lim_{y\rightarrow 0} \frac{2y^2-20y}{2y^2-17y}=\lim_{y\rightarrow 0} \frac{2y-20}{2y-17}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$
Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?
| Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = \frac{(k+1)(2k+3)(7k+8)}{6}$$.
We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = \frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
$$ = \frac{(k+1)(2k+3)(7k+8)}{6}$$
Hope it helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the general solution in terms of Bessel functions: $t^2x'' + x' + x = 0, \quad t < 0, \text{ Hint: } s = 2\sqrt{t}$ I was asked the following question:
Find the general solution in terms of Bessel functions:
$$t^2x'' + x' + x = 0, \quad t < 0, \text{ Hint: } s = 2\sqrt{t}$$
My approuch
I think that what I have to do is transform the given equation in one that has a form of a Bessel equation, and for that it must be used the hint.
*
*$\frac{ds}{dt} = t^{-1/2}$
*$\frac{dx}{dt} = \frac{dx}{ds}\frac{ds}{dt} = t^{-1/2} \frac{dx}{ds}$
*$\frac{d^2x}{dt^2} = \frac{d}{dt} \Big( \frac{dx}{dt} \Big) = \frac{d}{dt} \Big( t^{-1/2} \frac{dx}{ds} \Big) = \frac{d}{dt}t^{-1/2} \frac{dx}{ds} + t^{-1/2}\frac{d}{dt} \Big( \frac{dx}{ds} \Big) = \frac{-1}{2} t^{-3/2} \frac{dx}{ds} + t^{-1/2} \frac{d}{ds} \Big( \frac{dx}{ds} \Big) \frac{ds}{dt} = \frac{-1}{2} t^{-3/2} \frac{dx}{ds} + t^{-1} \frac{d^2x}{ds^2}$
So, in the given equation, now we have:
$$
\begin{align}
t^2\Big( \frac{-1}{2} t^{-3/2} \frac{dx}{ds} + t^{-1} \frac{d^2x}{ds^2} \Big) + t^{-1/2}\frac{dx}{ds} + x &= 0 \\
\frac{-1}{2}t^{1/2}\frac{dx}{ds} + t\frac{d^2x}{ds^2} + t^{-1/2}\frac{dx}{ds} + x &= 0 \\
\frac{-1}{2} \frac{s}{2} \frac{dx}{ds} + \frac{s^2}{4} \frac{d^2x}{ds^2} + \frac{2}{s} \frac{dx}{ds} + x &= 0 \\
s^2 \frac{d^2x}{ds^2} + \Big( \frac{8}{s} - s \Big)\frac{dx}{ds} + 4x &= 0
\end{align}
$$
Which doesn't have a Bessel-equation form. So please, can anybody enlighten me? Thanks in advanced!
| We see that $t=0$ is not a regular singularity of the equation
$$x''+\dfrac{1}{t^2}x'+\dfrac{1}{t^2}x=0$$
then this is not a Bessel differential equation.
| {
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Use $\cos 5\theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$ I used $\cos(3\theta + 2\theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta,$ and hence show that
$$\text{the roots of }x(16x^4 - 20x^2 + 5) \text{ are: } 0,\cos\frac{\pi}{10}, \cos\frac{3\pi}{10},\cos\frac{7\pi}{10}, \cos\frac{9\pi}{10}$$
| *
*The equation $$\;x(16x^4 - 20x^2 + 5) = 0\; \tag 1$$ can have $5$ real solutions (incl. multiplicity).
*From $\;\cos 5\theta = 16\cos^5 \theta - 20\cos^3 \theta + 5\;$ we deduce that we will look for solutions of $(1)$ in the form of $\cos \theta\;$ where $\;\cos 5\theta = 0.$
*$\cos 5\theta = 0$ holds for $10$ different values of $\;\theta \in (-\pi,\pi]\;$ obtained when solving $$\cos 5\theta =\frac \pi2 +2k\pi\quad \text{or} \quad \cos 5\theta = -\frac \pi2 + 2k\pi,\; k=0,1,\dots,4.$$
These are $$\frac{\pi}{10},\; \frac{5\pi}{10},\; \frac{9\pi}{10},\;\frac{13\pi}{10},\;\frac{18\pi}{10}\;\text{and}\;-\frac{\pi}{10},\;\frac{3\pi}{10},\;\frac{7\pi}{10},\;\frac{11\pi}{10},\;\frac{15\pi}{10},\;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$? How would I get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$?
It make sense to me logically. When one multiplies it out, it would condense to $x^{p-1}-1$. But it's just not clicking. What is the arithmetic between these steps?
| First of all, we might as well set
$n = p - 1, \tag 1$
and then investigate the equation
$x^n - 1 \overset{?}{=} (x - 1)(x^{n - 1} + x^{n - 2} + \ldots + x + 1) = (x - 1)\displaystyle \sum_0^{n - 1} x^i. \tag 2$
One may in fact use a simple induction to validate (2); we check
$n = 1: \; x^1 - 1 = x^1 - 1 = (x - 1) \displaystyle \sum_0^0 x^i, \tag 3$
$n = 2: \; x^2 - 1 = (x - 1)(x + 1) = (x - 1)\displaystyle \sum_0^1 x^i, \tag 4$
$n = 3: \; x^3 - 1 = (x - 1)(x^2 + x + 1) = (x - 1)\displaystyle \sum_0^2 x^i; \tag 5$
now if
$\exists k \in \Bbb N, \; x^k - 1 = (x - 1)\displaystyle \sum_0^{k - 1} x ^i, \tag 6$
we have
$x^{k + 1} - 1 = x^{k + 1} - x^k + x^k - 1 = (x - 1)x ^k + x^k - 1, \tag 7$
and then using (6),
$x^{k + 1} - 1 = (x - 1)x ^k + (x - 1)\displaystyle \sum_0^{k - 1} x ^i, \tag 8$
whence
$x^{k + 1} - 1 = (x - 1) \left (x ^k + \displaystyle \sum_0^{k - 1} x ^i \right ) = (x - 1) \displaystyle \sum_0^k x ^i, \tag 9$
which shows inductively that (2) must bind.
As far as performing the actual aritmetical/algebraic operations required above is concerned, it is easy to write them out explicitly for equations (3)-(5), viz
$(x - 1)(x + 1) = x(x + 1) - 1(x + 1) = x^2 + x - x - 1 = x^2 - 1; \tag{10}$
it is apparent that the distributive law is used critically here, allowing us as it does to write both the first equality in (10) as well as
$x(x + 1) = x^2 + x \; \text{etc}; \tag{11}$
in fact, the distributive law is tacitly invoked in (7)-(9); indeed, it plays a central role in performing the arithmetic necessary to establish (2).
| {
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"timestamp": "2023-03-29T00:00:00",
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find all $k$ such that $k^5+3$ is divisible by $k^2+1$ That's it. I have a solution with substitutions, but it's tedious and not too generic. Is there a solution using some cool theorem for polynomials divisibility?
| Solution with substitutions.
If $k^5+3$ is divisible by $k^2+1$, it could be expressed as $(k^2+1)(k^3+n)$, which if expanded, leads us to $k^3+nk^2+n-3$.
Now let's make another substitution by introducing another variable $a: k=-(n+a)$, which will get us $-an^2+(1-2a^2)n-(a^3+3)$. It is solvable for $n$ only when $D\ge0$ or $-4a^2-12a+1\ge0$.
Roots of the last [expression turned into an] equation are $-3\pm\frac{\sqrt{10}}{2}$, and since we are looking only for integers, we need to check $a\in\{-3,-2,-1,0\}$.
Now we can start unrolling our substitutions. For all valid values of $a$, solve $-an^2+(1-2a^2)n-(a^3+3)$ for $n$, and get values of $k$ directly as $-(n+a)$
$$\begin {array} {r|r} \ equation & a & n & k\\
\hline
3n^2-17n+24=0 & -3 & 3 & 0 \\
3n^2-17n+24=0 & -3 & \frac83 & - \\
2n^2-7n+5=0 & -2 & 1 & 1 \\
2n^2-7n+5=0 & -2 & 2.5 & - \\
n^2-n-2=0 & -1 & -1 & 2 \\
n^2-n-2=0 & -1 & 2 & -1 \\
n-3=0 & 0 & 3 & -3 \\
\end {array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt{2}\sec x+\tan x=1$
Solve $\sqrt{2}\sec x+\tan x=1$
I understand it can be very easily solved by expanding in terms of $\sin x$ and $\cos x$, gives $x=2n\pi-\frac{\pi}{4}$. But, what if I do the following:
$$
\sqrt{2}\sec x+\tan x=1\\
\text{Differentiating}\implies\sqrt{2}\sec x\tan x+\sec^2 x=0\implies\sqrt{2}\tan x+\sec x=0
$$
Step 1
$$
\sec x=-\sqrt{2}\tan x=\frac{1-\tan x}{\sqrt{2}}\implies2\tan x=\tan x-1\implies\tan x=-1\\
\boxed{x=n\pi-\frac{\pi}{4}}
$$
Step 2
$$
\tan x=1-\sqrt{2}\sec x=\frac{-\sec x}{\sqrt{2}}\implies2\sec x-\sqrt{2}=\sec x\\
\implies\sec x=\sqrt{2}\implies\boxed{x=2n\pi\pm\frac{\pi}{4}}
$$
$$
x=n\pi-\frac{\pi}{4}\quad\&\quad x=2n\pi\pm\frac{\pi}{4}\\\implies \bigg[x=2n\pi-\frac{\pi}{4}\text{ or }x=2n\pi+\frac{3\pi}{4}\bigg]\quad\&\quad x=2n\pi\pm\frac{\pi}{4}\\
\implies \boxed{x=2n\pi-\frac{\pi}{4}}
$$
In my attempt why do I need Step 2 to get the complete solution ?
Can someone give a proper explanation to my attempt ?
| The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are.
It turns out that, by coincidence, $\sqrt{2} \tan\left(2n\pi - \frac{\pi}{4} \right) + \sec\left(2n\pi - \frac{\pi}{4} \right) = 0$, i.e. $\sqrt{2} \tan x + \sec x = 0$ does hold at the particular value of $x$ you want to find. As such, your manipulations are technically "valid" in that everything you say is true, but only because the answer that you are trying to find makes these expressions true. The argument, however, is circular, so the proof fails (you are starting by presuming what you want to show, i.e. $x = 2 \pi n - \frac{\pi}{4}$). If, for example, the original question were $\sqrt{2} \sec x + \tan x = 0$, it should be apparent that this method fails.
| {
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Congruence proof. Show $a$ is odd, $b$ is even, and $ a \equiv 1$ mod $4 $ under certain conditions. If $ p \equiv$ $1$ mod $4$, $ p = a^2 + b^2$, and $a + bi \equiv 1$ mod $2+2i$, then $a$ is odd and $b$ is even. Moreover, if $4|b$, then $ a \equiv 1$ mod $4$, and if $ 4 \nmid b$, then $ a \equiv -1$ mod $4$.
Proof:
$ a + bi \equiv 1(2+2i) $ implies that $ a + bi \equiv 1$ mod $2 $. and so $a$ is odd and $b$ is even. Since $ 4 = -2(i-1)(i+1) $ it follows that if $4 \mid b$ then $a+bi \equiv a \equiv 1$ mod $2+2i$. Taking conjugates $ a \equiv 1$ mod $2-2i $. Thus $(2+2i)(2-2i) = 8 \mid (a+1)^2 $ and $ a \equiv 1$ mod $4$. If $ 4 \nmid b $ then $ b = 4k + 2 $ for some $ k $. Thus $ a + bi \equiv a+2i \equiv 1$ mod $2+2i $. Since $ 2i \equiv -2$ mod $2+2i $ we have $ a \equiv 3 \equiv -1$ mod $2+2i $. As before $ 8 \mid (a+1)^2 $ and so $ a \equiv -1$ mod $4 $.
So I'm trying to understand this proof, but unfortunately I'm a beginner.So I don't understand this proof. Here are my questions:
1) Why implies $ a + bi \equiv 1(2+2i) $ that $ a + bi \equiv 1$ mod $2 $? And why is this enough to say that $a$ is odd and $b$ is even?
2) Why do we have $a+bi \equiv a \equiv 1$ mod $2+2i$, if $4|b$? ( second line of the proof )
3) Why do we have $(2+2i)(2-2i) = 8 \mid (a+1)^2 $ and $ a \equiv 1$ mod $4$ ?
I can imagine that these questions are very simple for you, but I really want to understand this proof. So please try to be accurate. Thank you.
| First, $a+bi \equiv 1\pmod{2+2i}$ means that $a+bi = 1 + k(2+2i) = 1+2k(1+i)$ for some integer $k$, so that $a+bi \equiv 1\pmod{2}$. As for your second question, $4 = -2(i-1)(i+1) = (1-i)(2+2i)$, so $4\mid b$ implies that $2+2i\mid b$ and thus $bi\equiv 0\pmod{2+2i}$. For your final question, I think the correct statement is $8\mid (a-1)^2$, not $(a+1)^2$: since $2+2i\mid a-1$ and $2-2i\mid a-1$, this is clear.
Here's another proof that you may find easier to understand: $a+bi\equiv 1\pmod{2+2i}$ means that $a+bi = 1+k(2+2i) = 1+2k(1+i)$ for some integer $k$. Rewriting as $a+bi = (2k+1)+2ki$ shows that $a$ is odd and $b$ is even. Now, if $4\mid b$, then $k$ must be even, so that $a=2k+1\equiv 1\pmod{4}$, while if $4\nmid b$, then $k$ must be odd, so that $a=2k+1\equiv -1\pmod{4}$.
| {
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$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$?
$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $\tan^{-1}x+\tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z$ ?
\begin{align}
\tan^{-1}x+\tan^{-1}y+\tan^{-1}z&=
\begin{cases}\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy < 1 \\[1.5ex]
\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y>0 \\[1.5ex]
-\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y<0 \end{cases}\\
&=
\end{align}
| Write $a:=\arctan x$ etc. so $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\frac{\tan a+\tan b + (1-\tan a\tan b)\tan c}{1-\tan a\tan b - (\tan a+\tan b)\tan c}.$$If $xy\ne 1$, we can cancel $1-\tan a\tan b$ to get $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\frac{\tan (a+b)+\tan c}{1-\tan (a+b)\tan c}.$$If $\tan (a+b)\tan c\ne 1$ i.e. $(x+y)z\ne 1-xy$ i.e. $xy+yz+zx=1$, we have $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Using the residue theorem to evaluate an integral $$I = \int_0^{\infty} \frac {dx}{x^6+1}$$
My thinking is that I can use the property of even functions to integrate across the whole domain from negative infinity to positive infinity.
Can the residue theorem be used to evaluate this integral?
Can you use a pie-shaped contour subtending an angle of: $$\frac {\pi}3$$?
| First approach:
$$\begin{eqnarray*}\int_{\mathbb{R}^+}\frac{dx}{x^6+1}&=&\frac{1}{2}\int_{\mathbb{R}}\frac{dx}{x^6+1}=\pi i\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\operatorname*{Res}_{z=\xi}\frac{1}{z^6+1}\\&=&\pi i\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\frac{1}{6\xi^5}=-\frac{\pi i}{6}\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\xi\\&=&\frac{\pi}{6}\left(1+2\sin\frac{\pi}{6}\right)=\color{red}{\frac{\pi}{3}}.\end{eqnarray*}$$
The contour integration is performed along a semicircle contour in the upper half plane: the ML lemma clearly applies since $\left|\frac{1}{x^6+1}\right|=o\big(|x|^{-2}\big)$ as $|x|\to +\infty$.
Second approach: by splitting $\mathbb{R}^+$ as $(0,1]\cup[1,+\infty)$ and performing the substitution $x\mapsto\frac{1}{x}$ on the second "half" we get
$$\begin{eqnarray*}\int_{\mathbb{R}^+}\frac{dx}{x^6+1}&=&\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx=\int_{0}^{1}\frac{1+x^4-x^6-x^{10}}{1-x^{12}}\,dx\\&=&\sum_{n\geq 0}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)\end{eqnarray*}$$
where the reflection formula for the $\psi$ (digamma) function gives
$$ \sum_{n\geq 0}\left(\frac{1}{an+b}-\frac{1}{an+(a-b)}\right)=\frac{\pi}{a}\cot\frac{\pi b}{a} $$
then
$$ \int_{\mathbb{R}^+}\frac{dx}{x^6+1} = \frac{\pi}{12}\left(\cot\frac{\pi}{12}+\cot\frac{5\pi}{12}\right)=\frac{\pi}{12}\left[(2+\sqrt{3})+(2-\sqrt{3})\right]=\color{red}{\frac{\pi}{3}}. $$
Third approach: by setting $\frac{1}{1+x^6}=u$ we have
$$ \int_{0}^{+\infty}\frac{dx}{x^6+1}=\frac{1}{6}\int_{0}^{1}u^{-1/6}(1-u)^{-5/6}\,du =\frac{\Gamma(5/6)\Gamma(1/6)}{6}=\frac{\pi}{6\sin\frac{\pi}{6}}=\color{red}{\frac{\pi}{3}}$$
by the Beta function and the reflection formula for the $\Gamma$ function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A functional equation (another) I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $x\in [1/2,1]$
$$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right).$$
I am actually not sure about existence of a solution.
| $$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right) \tag 1$$
As already pointed out this implies $f(1)=\frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=\frac23+c\left(\frac{x}{x-1}\right)^2 \tag 2$$
$c=$any constant.
$$f\left( \frac{2x}{1+x} \right)=\frac23+c\left(\frac{ \frac{2x}{1+x}}{ \frac{2x}{1+x}-1}\right)^2 \tag 3$$
After simplification :
$$f\left( \frac{2x}{1+x} \right)= \frac23+c\left(\frac{2x}{x-1}\right)^2$$
We put $(3)$ in $(1)$ :
$$\frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right)=\frac{1}{2} + \frac{1}{4}\left( \frac23+c\left(\frac{2x}{x-1}\right)^2\right)$$
After simplification :
$$\frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right)=\frac23+c\left(\frac{x}{x-1}\right)^2$$
Comparing to Eq.$(2)$ we see that :
$$\frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=\frac23+c\left(\frac{x}{x-1}\right)^2 \qquad (x\neq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=\frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The distributive law $4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
| Do it in steps.
You accept that $M(x+y) = Mx + My$
So replace $M$ with $(x+y)$ and you get:
$(x+y)(x+y) = M(x+y) =$
$Mx + My = $
$(x + y)x + (x+y)y$
Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:
$(x+y)(x+y) = M(x+y) =$
$Mx + My = $
$(x + y)x + (x+y)y=$
$(x+y)A + (x+y)B=$
$xA + yA + xB + yB =$
$xx + yx + xy + yy =$
$x^2 + 2xy + y^2$.
Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.
$(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.
$(x+y)(x+y) = (x+y)x + (x+y)y=$.
Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:
$(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.
And then some clean up:
$=x^2 + 2xy + y^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$
Finding sum of series $$\displaystyle \sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$
Try: Let $$S = \displaystyle \sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$
So, $$S =\sum^{\infty}_{k=1}\frac{k^2\cdot (2k-2)!}{(2k+2)!}=\frac1{3!}\sum^{\infty}_{k=0}\frac{(k+1)^2\cdot(2k)!\cdot 3!}{(2k+3+1)!}$$
with the help of identity $$B(m,n) = \int^{1}_{0}x^m(1-x)^ndx = \frac{\Gamma (m+1)\Gamma(n+1)}{\Gamma(m+n+2)}$$
$$B(m,n) = \frac{\Gamma (m+1)\Gamma(n+1)}{\Gamma(m+n+2)}=\frac{m!\cdot n!}{(m+n+1)!}$$
So $$S=\sum^{\infty}_{k=0}(k+1)^2\int^{1}_{0}(x)^{2k}(1-x)^3dx$$
$$S=\int^{1}_{0}x^{-2}(1-x)^3\sum^{\infty}_{k=1}(kx^k)^2dx$$
Can someone explain me how to calculate $\displaystyle \sum^{\infty}_{k=1}k^2x^{2k}$ in some short way . although I am trying to solve it but it is too lengthy.
Please explain to me ,thanks.
| Applying partial fractions along with the digamma function, we get that:
\begin{align*}
\sum_{k=1}^{\infty} \frac{k}{2\left ( 2k-1 \right )\left ( 2k+1 \right )\left ( 2k+2 \right )} &= \sum_{k=1}^{\infty} \left ( \frac{1}{24\left ( 2k-1 \right )}+ \frac{1}{8\left ( 2k+1 \right )} -\frac{1}{12\left ( k+1 \right )} \right ) \\
&=\sum_{k=0}^{\infty} \left ( \frac{1}{24\left ( 2k-1 \right )}+ \frac{1}{8\left ( 2k+1 \right )} -\frac{1}{12\left ( k+1 \right )} \right ) \\
&= \sum_{k=0}^{\infty} \left ( \frac{1}{24\cdot 2\left ( k-\frac{1}{2} \right )}+ \frac{1}{8 \cdot 2 \left ( k + \frac{1}{2} \right )} - \frac{1}{12\left ( k+1 \right )} \right )\\
&= -\frac{1}{48} \psi^{(0)} \left ( -\frac{1}{2} \right ) - \frac{1}{16} \psi^{(0)} \left ( \frac{1}{2} \right ) + \frac{1}{12} \psi^{(0)} (1) \\
&=-\frac{1}{48} \left ( 2-\gamma -2\log 2 \right ) - \frac{1}{16} \left ( -\gamma -2 \log 2 \right ) - \frac{\gamma}{12} \\
&= \frac{1}{24} \left ( \log 16 -1 \right ) \\
& =\frac{4 \log 2 -1}{24}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Method to factor an expression As the title says, i want to factorize an expression, but i don't have any clue how to proceed.
Here is the expression :
$2x² -7x +3$
And here is the factorized form :
$(x-3)(2x - 1)$
My question is, which method or rule to use to go from first to second ?
please note that I am a beginner, and the only question that i found which is closer to mine is this post.
Thank you for your help !
| In general ,if $x_1$ and $x_2$ are roots of $$\underbrace{a}_{\neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\text{sum of the roots}=-\frac{b}{a}\;\;\&\;\;\text{product of the roots}=\frac{c}{a}$$
So your case, $x_1+x_2= \frac{7}{2}$ and $x_1x_2=\frac{3}{2}$
Now solve these to get $x_1$ and $x_2$ to finish your conclusion
To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-\frac{7}{2}x+\frac{3}{2}=0$$ which means $$x^2-2\left(\frac{7}{4}\right)x=-\frac{3}{2}$$ which is same as $$x^2-2\left(\frac{7}{4}\right)x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}=\frac{25}{16}$$ so $$\left(x-\frac{7}{4}\right)^2=\frac{25}{16}$$ and so $$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}$$ so $$x=\frac{7}{4} \pm \frac{5}{4}=\frac{7\pm 5}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the area of trapezoid. In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $\angle BAD=\theta,$
$BD=h$,
$\angle ABD=90^\circ$
$\angle CBD=90^\circ-\theta$
$CD=2$because trapezoid is isosceles
Apply cosine law in triangle BDC,
$\cos(90-\theta)=\frac{h^2+2^2-2^2}{2\times 2\times h}=\frac{h}{4}$
$\sin\theta=\frac{h}{4}..(1)$
In right triangle $ABD,\sin \theta=\frac{h}{\sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2\sqrt3$
Area of $ABCD=\frac{1}{2}\times 2\times h+\frac{1}{2}\times 2\times 2\times \sin2\theta=3\sqrt3$
But the answer given is $2\sqrt2(\sqrt{5}+1)$
| I've got $3\sqrt 3$ using a slightly different method. $\angle BAD=\theta$, $\angle ABD =90^\circ$ means $\angle CBD=\angle BDA=90^\circ-\theta$. Since the trapezoid is isosceles, $\angle CDA=\theta$, and you can get $\angle CDB=2\theta -90^\circ$. Since $BC=AB=CD$ you get $\angle CDB=\angle CDB$ or $$2\theta-90^\circ=90^\circ-\theta$$ so $\theta=60^\circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2AB\cos 60^\circ=4$, and the height $h=AB\sin60^\circ=\sqrt 3$. Therefore the area is $$\frac12 (BC+AD)\cdot h=\frac12 6\sqrt3=3\sqrt 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the integral $\int_{-1}^1 \frac{1}{(2-x)\sqrt{1-x^2}}\,dx$ I started with the substitution $x=\sin(\theta)$ with $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$
And I got the integral $$\int_{-\pi/2}^{\pi/2} \frac{1}{2-\sin(\theta)}\,d\theta$$
which I am again not able to evaluate.
Even integration by parts is not working.
| Here is a slightly different approach.
Let
$$I = \int_{-1}^1 \frac{dx}{(2 - x) \sqrt{1 - x^2}} \qquad (1)$$
Enforcing a substitution of $x \mapsto - x$ gives
$$I = \int_{-1}^1 \frac{dx}{(2 + x) \sqrt{1 - x^2}} \qquad (2)$$
Adding (1) to (2) one obtains
$$I = 2 \int_{-1}^1 \frac{dx}{(4 - x^2) \sqrt{1 - x^2}} = 4 \int_0^1 \frac{dx}{(4 - x^2) \sqrt{1 - x^2}},$$
since the integrand is even between symmetric limits.
Now set $x = \sin \theta$. As $dx = \cos \theta \, d\theta$, doing so yields
$$I = 4 \int_0^{\pi/2} \frac{d\theta}{4 - \sin^2 \theta}.$$
Taking advantage of the identity $\sin^2 \theta + \cos^2 \theta = 1$, the dominator in the above integral can be rewritten as
$$I = 4 \int_0^{\pi/2} \frac{d\theta}{3 \sin^2 \theta + 4 \cos^2 \theta} = 4 \int_0^{\pi/2} \frac{\sec^2 \theta}{3 \tan^2 \theta + 4} \, d\theta.$$
Setting $u = \tan \theta, du = \sec^2 \theta \, d\theta$ one has
\begin{align}
I &= 4 \int_0^\infty \frac{du}{3u^2 + 4}\\
&= \frac{4}{3} \int_0^\infty \frac{du}{u^2 + \left (\frac{2}{\sqrt{3}} \right )^2}\\
&= \frac{4}{3} \left [\frac{\sqrt{3}}{2} \tan^{-1} \left (\frac{u \sqrt{3}}{2} \right ) \right ]_0^\infty\\
&= \frac{\pi}{\sqrt{3}}.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3046228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\int_0^\infty \ln(1+\frac{z}{\cosh(x)})dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$ and a closed form for $-1I observed graphically that $$f(z)=\int_0^\infty \ln\left(1+\frac{z}{\cosh(x)}\right)dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$$
Can anyone explain why this holds? I tried differentiating with respect to $z$ but I didn't really know how to continue further.
Also, if anyone knows how to obtain a closed form of $f(z)$ for $-1<z<1$ it would be greatly appreciated as I'm curious about specific values such as $f(-\frac{1}{2})=-\frac{7\pi^2}{72}$ and $f(\frac{1}{2})=\frac{5\pi^2}{72}.$
| Here is a more direct way.
The first thing to note is the function $f(z)$ is continuous for all $z > - 1$, including at the point $z = 1$.
From
$$f(z) = \int_0^\infty \ln (1 + z \operatorname{sech} x) \, dx,$$
on differentiating with respect to $z$ we have
\begin{align}
f'(z) &= \int_0^\infty \frac{\operatorname{sech} x}{1 + z \operatorname{sech}} \, dx\\
&= \int_0^\infty \frac{dx}{\cosh x + z}\\
&= 2 \int_0^\infty \frac{e^x}{e^{2x} + 2z e^x + 1} \, dx.
\end{align}
In the last line we have used the fact that $\cosh x = (e^{x} + e^{-x})/2$.
Making a substitution of $u = e^x, du = e^x \, dx$ leads to
$$f'(z) = 2 \int_1^\infty \frac{du}{(u + z)^2 + 1 - z^2}. \qquad (*)$$
Case 1: $-1 < z < 1$
For this case, as the term $(1 - z^2)$ will be positive we can write ($*$) as:
\begin{align}
f'(z) &= 2 \int_1^\infty \frac{du}{(u + z)^2 + (\sqrt{1 - z^2})^2}\\
&= \frac{2}{\sqrt{1 - z^2}} \left [\tan^{-1} \left (\frac{u + z}{\sqrt{1 - z^2}} \right ) \right ]_1^\infty\\
&= \frac{\pi}{\sqrt{1 - z^2}} - \frac{2}{\sqrt{1 - z^2}} \tan^{-1} \left (\frac{1 + z}{\sqrt{1 - z^2}} \right )\\
&= \frac{\pi}{\sqrt{1 - z^2}} - \frac{2}{\sqrt{1 - z^2}} \left [\frac{\pi}{2} - \tan^{-1} \left (\frac{\sqrt{1 - z^2}}{1 + z} \right ) \right ]\\
&= \frac{2}{\sqrt{1 - z^2}} \tan^{-1} \left (\frac{\sqrt{1 - z^2}}{1 + z} \right ), \qquad (**)
\end{align}
where we have made use of the result $\tan^{-1} (x) = \pi/2 - \tan^{-1} (1/x)$ for $x > 0$.
Noting that
$$\tan \left (\frac{1}{2} \cos^{-1} z \right ) = \frac{\sqrt{1 - z^2}}{1 + z},$$
the result in ($**$) can be rewritten as
$$f'(z) = \frac{1}{\sqrt{1 - z^2}} \cos^{-1} z.$$
Integrating up with respect to $z$ it is readily seen that
$$f(z) = \int \frac{\cos^{-1} z}{\sqrt{1 - z^2}} \, dz = -\frac{1}{2} (\cos^{-1} z )^2 + C.$$
To find the constant $C$, choosing $z= 0$ we see that $f(0) = 0$. Thus $C = \pi^2/8$, leading to
$$f (z) = \frac{\pi^2}{8} - \frac{1}{2} (\cos^{-1} z)^2, \, -1 < z < 1. \qquad (\dagger)$$
Case 2: $z > 1$
For this case, as the term $(1 - z^2)$ will be negative we can write ($*$) as:
\begin{align}
f'(z) &= 2 \int_1^\infty \frac{du}{(u + z)^2 - (\sqrt{z^2 - 1})^2}\\
&= -\frac{2}{\sqrt{z^2 - 1}} \left [\coth^{-1} \left (\frac{u + z}{\sqrt{z^2 - 1}} \right ) \right ]_1^\infty\\
&= \frac{2}{\sqrt{z^2 - 1}} \coth^{-1} \left (\frac{1 + z}{\sqrt{z^2 - 1}} \right )\\
&= \frac{2}{\sqrt{z^2 - 1}} \tanh^{-1} \left (\frac{\sqrt{z^2 - 1}}{z+ 1} \right ), \qquad (***)
\end{align}
where we have made use of the result $\coth^{-1} (x) = \tanh^{-1} (1/x), x \neq 0$.
Noting that
$$\tanh \left (\frac{1}{2} \cosh^{-1} z \right ) = \frac{\sqrt{z^2 - 1}}{z + 1},$$
the result in ($***$) can be written as
$$f'(z) = \frac{1}{\sqrt{z^2 - 1}} \cosh^{-1} z.$$
Integrating up with respect to $z$ it is readily seen that
$$f(z) = \int \frac{\cosh^{-1} z}{\sqrt{z^2 - 1}} \, dz = \frac{1}{2} (\cosh^{-1} z)^2 + C.$$
To find the unknown constant $C$, as the function $f(z)$ is continuous for all $z > - 1$, for $f$ to be continuous at $z = 1$, from ($\dagger$) we require $f(1) = \pi^2/8$. Thus $C = \pi^2/8$, leading to
$$f(z) = \frac{\pi^2}{8} + \frac{1}{2} (\cosh^{-1} z)^2, \qquad z > 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Extract the Square Root of $ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $
Extract the Square Root of $$ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $$
I found a method to do this in a book but couldn't understand how to do it. Can anyone explain it?
| $$a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)+\frac{41a}{16}-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+1$$
For convenience, let $x=\sqrt a$. Then we get
$$x^4-\dfrac 32x^3 +\frac{41}{16}x^2-\dfrac 32x+1$$
Use of the square root algorithm would produce this.
\begin{array}{r}
&& x^2 & -\dfrac 34x & +1\\
&&--- &--- &--- &--- &---\\
&\mid&x^4 &-\dfrac 32x^3 &+ \dfrac{41}{16}x^2 &-\dfrac 32x &+1 \\
&x^2 \mid & x^4 \\
&\mid&--- &--- &---\\
&\mid &&-\dfrac 32x^3 &+ \dfrac{41}{16}x^2 \\
&2x^2-\dfrac 34x \mid &&-\dfrac 32x^3 &+\dfrac{9}{16}x^2 \\
&\mid &&--- &--- &--- &--- \\
&\mid && &2x^2 &-\dfrac 32x &+1 \\
&2x^2-\dfrac 32x+1 \mid && &2x^2 &-\dfrac 32x &+1 \\
&&& &--- &--- &--- \\
&&&&&&0
\end{array}
Hence
$$a^2-\frac 32a\sqrt a+\frac{41}{16}a-\frac 32\sqrt a+1
= \left(a - \dfrac 34\sqrt a + 1\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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power series for $\frac1{x^2+x+1}$ So I am trying to find the sequence $(a_n)_{n\geq0}$ such that
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$a_0(x^2+x+1)+\sum_{n\geq1}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$x=0$:
$$a_0=1$$
So
$$x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$\frac{x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)}x=\frac1x$$
$$x+1+a_1(x^2+x+1)+\sum_{n\geq2}a_n(x^{n+1}+x^{n}+x^{n-1})=0$$
$x=0$:
$$a_1=-1$$
One more time:
$$x^2+x+1-x^3-x^2-x+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$-x^3+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=0$$
Divide both sides by $x^2$:
$$-x+a_2(x^2+x+1)+\sum_{n\geq3}a_n(x^{n}+x^{n-1}+x^{n-2})=0$$
$x=0$:
$$a_2=0$$
How do I find $a_n$? Am I doing things right so far? Thanks.
| You could have continue your way.
Starting from
$$\sum_{n=0}^\infty a_nx^{n+2}+\sum_{n=0}^\infty a_nx^{n+1}+\sum_{n=0}^\infty a_nx^n=1$$ you already found $a_0=1$ and $a_1=-1$.
Now, consider a given degree $m >1$. You then have the recurrence relation
$$a_{m-2}+a_{m-1}+a_m=0$$ and the roots of the characteristic equation
$$r^2+r+1=0 \implies r_\pm=\frac{-1 \pm i \sqrt 3} 2$$ making by the end (I let to you the intermediate steps)
$$a_m=\cos \left(\frac{2 \pi m}{3}\right)-\frac{1}{\sqrt{3}}\sin \left(\frac{2 \pi
m}{3}\right)$$ making $a_{3m}=1$, $a_{3m-1}=0$, $a_{3m+1}=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove Nice Determinant Equations I often come across these kinds of problems in A-level exam papers:
Prove that
$$
\begin{vmatrix}
(a+b)^2 & 1 & 1 \\
a^2 & (1+b)^2 & a^2 \\
b^2 & b^2 & (1+a)^2
\end{vmatrix}
= 2ab(1+a+b)^3
$$
or
Prove that
$$
\begin{vmatrix}
1 & 1 & 1 \\
x & y & z \\
yz & xz & xy
\end{vmatrix}
= (x-y)(y-z)(z-x)
$$
Questions which involve nice determinant results, but which are quite a pain to prove by expanding.
Are there any tricks that one can use to prove such results? I'm familiar with Vandermonde matrices for example, but I haven't come across anything that might help me with these, especially the first one.
| By subtracting the first column from the second and the third we get that :
$$
\begin{split}
\begin{vmatrix}
1 & 1 & 1 \\
x & y & z \\
yz & xz & xy
\end{vmatrix}
&=\begin{vmatrix}
1 & 0 & 0 \\
x & y-x & z-x \\
yz & z(x-y) & y(x-z)
\end{vmatrix}\\
\\
&=(x-y)(z-x)\begin{vmatrix}
1 & 0 & 0 \\
x & -1 & 1 \\
yz & z & -y
\end{vmatrix}\\
\\
&=(x-y)(z-x)\begin{vmatrix}
-1 & 1 \\
z & -y
\end{vmatrix}\\
\\
& =(x-y)(y-z)(z-x)
\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$??? I saw in a paper by @Jack D'aurizio the following integral
$$I=\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$$
Below is my attempt.
$$I=4\int_0^{\pi/2}\log^2(\cos x)\mathrm{d}x$$
Then we define
$$F(a)=\int_0^{\pi/2}\log^2(a\cos x)\mathrm{d}x$$
So we have
$$F'(a)=\frac2a\int_0^{\pi/2}\log(a\cos x)\mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.
| \begin{align}J=\int_0^{\frac{\pi}{2}} \ln^2\left(\cos x\right)\,dx\end{align}
Observe that,
\begin{align}I&=4J\\
J&=\int_0^{\frac{\pi}{2}} \ln^2\left(\sin x\right)\,dx\\
\int_0^{\frac{\pi}{2}} \ln\left(\sin x\right)\,dx&=\int_0^{\frac{\pi}{2}} \ln\left(\cos x\right)\,dx
\end{align}
(change of variable $y=\dfrac{\pi}{2}-x$ )
\begin{align}
K&=\int_0^{\frac{\pi}{2}} \ln^2 \left(2\sin x\cos x\right)\,dx\\
&=\int_0^{\frac{\pi}{2}} \ln^2 \left(\sin\left(2x\right)\right)\,dx\\
\end{align}
Perform the change of variable $y=2x$,
\begin{align}
K&=\frac{1}{2}\int_0^{\pi} \ln^2 \left(\sin x\right)\,dx\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}} \ln^2 \left(\sin x\right)\,dx+\frac{1}{2}\int_{\frac{\pi}{2}}^\pi \ln^2 \left(\sin x\right)\,dx\\
\end{align}
In the latter integral perform the change of variable $y=\dfrac{\pi}{2}-x$ and recall $\sin\left(\pi-x\right)=\sin x$ for $x$ real,
\begin{align}
K&=\int_0^{\frac{\pi}{2}} \ln^2 \left(\sin x\right)\,dx\\
&=\int_0^{\frac{\pi}{2}} \ln^2 \left(\cos x\right)\,dx\\
&=J
\end{align}
On the other hand,
\begin{align}
K&=\int_0^{\frac{\pi}{2}}\left(\ln 2+\ln(\sin x)+\ln(\cos x)\right)^2 \,dx\\
&=\frac{\pi}{2}\ln^2 2+\int_0^{\frac{\pi}{2}}\ln^2(\sin x)\,dx+\int_0^{\frac{\pi}{2}}\ln^2(\cos x)\,dx+2\ln 2\int_0^{\frac{\pi}{2}}\ln(\cos x)\,dx+\\
&2\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin x)\,dx+2\int_0^{\frac{\pi}{2}}\ln(\sin x)\ln(\cos x)\,dx\\
&=\frac{\pi}{2}\ln^2 2+2J+4\ln 2\int_0^{\frac{\pi}{2}}\ln(\cos x)\,dx+2\int_0^{\frac{\pi}{2}}\ln(\sin x)\ln(\cos x)\,dx\\
\end{align}
\begin{align}L&=\int_0^\infty \frac{\ln^2 x}{1+x^2}\,dx\end{align}
Perform the change of variable $x=\tan y$,
\begin{align}L&=\int_0^{\frac{\pi}{2}} \ln^2\left(\tan x\right)\,dx\\
&=\int_0^{\frac{\pi}{2}}\left(\ln\left(\sin x\right)-\ln\left(\cos x\right)\right)^2\,dx\\
&=2J-2\int_0^{\frac{\pi}{2}}\ln(\sin x)\ln(\cos x)\,dx\\
\end{align}
Therefore,
\begin{align}K+L&=\frac{\pi}{2}\ln^2 2+4J+4\ln 2\int_0^{\frac{\pi}{2}}\ln(\cos x)\,dx
\end{align}
Therefore (recall $K=J$),
\begin{align}J&=\frac{1}{3}L-\frac{\pi}{6}\ln^2 2-\frac{4}{3}\ln 2\int_0^{\frac{\pi}{2}}\ln(\cos x)\,dx\end{align}
On the other hand,
\begin{align}L&=\int_0^1 \frac{\ln^2 x}{1+x}\,dx+\int_1^\infty \frac{\ln^2 x}{1+x}\,dx\end{align}
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}L&=2\int_0^1 \frac{\ln^2 x}{1+x}\,dx\end{align}
But it is well known that,
\begin{align}\int_0^1 \frac{\ln^2 x}{1+x^2}\,dx&=\frac{\pi^3}{16}\\
\int_0^{\frac{\pi}{2}}\ln(\cos x)\,dx&=-\frac{1}{2}\pi\ln 2
\end{align}
Therefore,
\begin{align}J&=\frac{\pi^3}{24}-\frac{\pi}{6}\ln^2 2-\frac{4}{3}\ln 2\times -\frac{1}{2}\pi\ln 2\\
&=\boxed{\frac{\pi^3}{24}+\frac{1}{2}\pi\ln^2 2}\\
\end{align}
PS:
See: https://math.stackexchange.com/a/2942594/186817
(in this post i assume only the value of $\zeta(4)$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right)\rightarrow \sin0 $. However I have also $x$ before $\sin x$ and I don't know how to calculate it.
| The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.
$\sqrt{x^2+3}- \sqrt{x^2+2} = x(\sqrt{1+3/x^2}- \sqrt{1+2/x^2})=$
$= x(1+\frac{1}{2}\cdot(3/x^2)+O(1/x^4)- 1-\frac{1}{2}\cdot(2/x^2)+O(1/x^4))= \frac{1}{2x}+O(1/x^3)$.
Now $\sin(\frac{1}{2x}+O(1/x^3))= \frac{1}{2x}+O(1/x^2)$ as $x\rightarrow \infty$, thus the expression is $\frac{1}{2}+ O(\frac{1}{x})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my question is then why does $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)} = -\frac{1}{6}$$
As I though that it would be $\frac{1}{2}$.
| Let $\dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=\dfrac{am+b}{m(m+1)}$
$\implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$
Set $2b=2\iff b=1,ar=-r\iff a=-1$
$$\implies\sum_{r=1}^n\dfrac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(f(r)-f(r+1)\right)=f(1)-f(n+1)$$
Set $n\to\infty$ and show that $n\to\infty f(n)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! \approx \sqrt{2\pi\cdot 15}
\cdot \left(\frac{15}{e}\right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
| Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$ or $\frac{\pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
| Note that we have
$$\psi'(z)=\sum_{n=0}^\infty \frac{1}{(n+z)^2}$$
where $\psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^2}=\frac19 \psi'(1/3)$$
and
$$\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac19 \psi'(2/3)$$
Interestingly, since we have
$$\sum_{n=0}^\infty \left(\frac1{(3n+3)^2}+\frac1{(3n+2)^2}+\frac1{(3n+1)^2}\right)=\frac{\pi^2}{6}$$
we find that
$$\psi'(1/3)+\psi'(2/3) = 4\pi^2/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$AC^T = \det(A)I$ Let A = $\begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} &
\dots & a_{nn} \end{bmatrix}$
and the matrix of cofactors of $A$ is
$$C=\begin{bmatrix} C_{11} & C_{12} & \dots & C_{1n} \\ C_{21} & C_{22} & \dots & C_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ C_{n1} & C_{n2} &
\dots & C_{nn} \end{bmatrix}.
$$
I try to understand why $AC^T = \det(A)I$ necessarily.
Why is it that $a_{11}C_{21} + a_{12}C_{22} + \dots + a_{1n}C_{2n} = 0$?
| The expression
$$
\color{red}{a_{11}}C_{21} + \color{red}{a_{12}}C_{22} + \dots + \color{red}{a_{1n}}C_{2n}
$$
is the Laplace expansion along the second row of
$$
\begin{vmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\ \color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}\dots & \color{red}{a_{1n}} \\ a_{31} & a_{32} & \dots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} &
\dots & a_{nn}
\end{vmatrix}=\color{red}{a_{11}}C_{21} + \color{red}{a_{12}}C_{22} + \dots + \color{red}{a_{1n}}C_{2n}=0.
$$
Edit: take the matrix $A$ and do the determinant expansion along the second row
$$
\begin{vmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\ \color{blue}{a_{21}} & \color{blue}{a_{22}} & \color{blue}\dots & \color{blue}{a_{2n}} \\ a_{31} & a_{32} & \dots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn}
\end{vmatrix}=\color{blue}{a_{21}}C_{21} + \color{blue}{a_{22}}C_{22} + \dots + \color{blue}{a_{2n}}C_{2n}.
$$
Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate: $\lim\limits_{x\to0^-} \frac1{\ln(1-x)}+\frac1x$ without LHR/Expansions
How to calculate $$\lim\limits_{x\to0^-} \left(\frac1{\ln(1-x)}+\frac1x \right)$$ without using L'Hopital, expansions nor integration?
I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-\frac{x^2}2-x-1$
We get:
$0\le\frac{e^x-x-1}{x^2}-\frac1 2\le \frac{e^x-x-1}{x}$
Thus:
$\lim\limits_{x\to0^-} \frac{e^x-x-1}{x^2} = \frac12$
By substituting: $t=\ln(1-x)$ in the original limit we get:
$\lim\limits_{t\to0^+} \frac{1-e^t+t}{t(1-e^t)} = \lim\limits_{t\to0^+} \frac{e^t-t-1}{t^2}.\frac{t}{e^t-1} = \frac12$
| Note that
$$\begin{eqnarray}
\lim_{x\to0^-} \left(\frac1{\ln(1-x)}+\frac1x \right)&=&\lim_{x\to0^+}\frac{x-\ln(1+x)}{x\ln(1+x)} \\&= &\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}\cdot \lim_{x\to0^+}\frac{x}{\ln(1+x)}\\&=&\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}.
\end{eqnarray}$$Now, let $$g(x) =\begin{cases}\frac{x-\ln(1+x)}{x},\quad x\neq 0\\ 0,\quad x=0\end{cases}.
$$ We can see that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Hence by MVT, we have
$$
L=\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}=\lim_{x\to0^+}\frac{g(x)}{x}=\lim_{c\to0^+} g'(c) =\lim_{c\to 0^+} \frac{\frac{c^2}{1+c}-(c-\ln(1+c))}{c^2}=1-L.
$$ This gives $L=\frac{1}{2}$.
(Justification of taking the limit) Let $h(x) = \ln(1+x) - x +\frac{x^2}{2}$. We have $h(x) \ge 0$ since $h(0) = 0$ and $h'(x) = \frac{1}{1+x}-1+x\ge 0$. This shows $\frac{g(x)}{x}\le\frac{1}{2}$. By MVT, we know that for some $c\in (0,x)$,
$$
\frac{1}{1+x}\le \frac{1}{1+c}= \frac{g(c)}{c}+\frac{g(x)}{x} .
$$Thus we have
$$
\frac{1}{1+x}-\frac{1}{2}\le g(x) \le \frac{1}{2},
$$ and
$$
\lim_{x\to 0^+}g(x) =\frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Let $A= \begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}$. Find the entry in the first row and second column of $A^{2014}$ I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=\begin{pmatrix} 1&1 \\ -4&-1 \end{pmatrix}$
$D=\begin{pmatrix} 0&0 \\ 0&6 \end{pmatrix}$
$P^{-1}=\frac{1}{3}\begin{pmatrix} -1&-1 \\ 4&1 \end{pmatrix}$.
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.
| Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=\begin{bmatrix}1\\-1\end{bmatrix}\begin{bmatrix}8&2\end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=\left(\begin{bmatrix}1\\-1\end{bmatrix}\begin{bmatrix}8&2\end{bmatrix}\right)^{2014}=\begin{bmatrix}1\\-1\end{bmatrix}\left(\begin{bmatrix}8&2\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}\right)^{2013}\begin{bmatrix}8&2\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}6^{2013}\begin{bmatrix}8&2\end{bmatrix}=6^{2013}A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
For integers $a,b$ and prime $p$, are there infinitely many solutions to $p = (a-b)(a+b)$? I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p \mid (a-b)$ or $p \mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:
For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
$$p = (a-b)(a+b)$$
| If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=\frac{p+1}{2}$ for every prime p besides $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Minimising the Sum of distance of two line formed by joining some point on a line to two fixed points Consider the xy coordinate system.
Let $l$ be the line $y=-x$.
Consider two points $A=(0,2)$ and $B=(2,3)$
The Question asks to find a point $C$ on the line $l$ such that $|AC|+|BC|$ is minimum.
The only method I got to tackle this problem is to use parametric form of a point on the line $l$, that is $(t,-t)$ and use calculus to minimize the distance as a function of variable $t$.
I tried to find the critical points using differentiation.
Unfortunately it gives me a $4$ degree equation with leading coefficient being $24$. Also since it has $4$ roots I an also getting duplicate roots
Any other better approach?
| Let's see. We have $A = (0, 2)$, $B = (2, 3)$, and the point $C$ (as a function of $t$) $C(t) = ( t, -t)$. The Euclidean distance between two points is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, so the sum of distance from $A$ to $C$, and the distance from $C$ to $B$, as a function of $t$ is
$$\bbox{
L(t) = \sqrt{t^2 + (-t-2)^2} + \sqrt{(t - 2)^2 + (-t - 3)^2}
}$$
which simplifies to
$$\bbox{ L(t) = \sqrt{2 t^2 + 4 t + 4} + \sqrt{2 t^2 + 2 t + 13} }$$
Its extrema (minimum or maximum points) occur when its derivative is zero. Using the chain rule, $\frac{d \sqrt{f(x)}}{d x} = \frac{\frac{d f(x)}{d x}}{2 \sqrt{f(x)}}$, so we need to solve
$$\bbox{ \frac{d L(t)}{d t} = \frac{2 t + 2}{\sqrt{2 t^2 + 4 t + 4}} + \frac{2 t + 1}{\sqrt{2 t^2 + 2 t + 13}} = 0 }$$
for $t$. Note that neither polynomial within the square roots is always positive ($\gt 0$) for all real $t$, as they're parabolas; left one has a minimum $2$ at $t = -1$, and the right one has a minimum $12.5$ at $t = -0.5$. Or, if you prefer, because neither $A$ nor $B$ is on the line, and the square roots correspond to the distance from a point on the line to $A$ and $B$, respectively, the square roots must be greater than zero for all real $t$. This means both divisors are always greater than zero, and we can multiply the equation by their product. This gives us
$$\bbox{ \frac{d L(t)}{d t} = (2 t + 2) \sqrt{2 t^2 + 2 t + 13} + (2 t + 1)\sqrt{2 t^2 + 4 t + 4} = 0 }$$
and, moving the second summand to the right side,
$$\bbox{ (2 t + 2) \sqrt{2 t^2 + 2 t + 13} = (- 2 t - 1) \sqrt{2 t^2 + 4 t + 4} }$$
If we restrict our solutions to those where both sides have the same sign, i.e. $(2 t + 2)$ and $(- 2 t - 1)$ have the same sign, we can square both sides. This gives us
$$\bbox{ (2 t + 2)^2 ( 2 t^2 + 2 t + 13 ) = (- 2 t - 1)^2 ( 2 t^2 + 4 t + 4 ) }$$
Expanding and moving the right side back to the left side, we get
$$\bbox{ 42 t^2 + 92 t + 48 = 0 }$$
This second-degree function has two roots: $t = -\frac{6}{7}$ and $t = -\frac{4}{3}$. However, remeber the restriction earlier. For $t = -\frac{4}{3}$, $2 t + 2 \lt 0$ but $-2 t - 1 \gt 0$, so $t = -\frac{4}{3}$ is not a valid solution. For $t = -\frac{6}{7}$, $2 t + 2 \gt 0$ and $-2 t - 1 \gt 0$, so $t = -\frac{6}{7}$ is a valid answer.
Therefore, the point $C$ we are looking for is $C\left(-\frac{6}{7}\right)$, i.e. $C = \left(-\frac{6}{7}, \frac{6}{7}\right)$.
As Aretino already pointed out, that is the exact same point that you get if $A^\prime$ is the reflection of $A$ about the line $y = -x$ (i.e. $A^\prime = (-2, 0)$), and you find $C$ at the intersection between lines $y = -x$ and $A^\prime B$.
| {
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How many pairs $(x,y)$ such that $x+2=y$ have an $x$ or $y$ divisible by 5 and not 3? Let the set $S_{n}$ = {$(x,y):x,y \in \mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $\mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{9} =\{(3,5),(5,7),(7,9)\}$ and $f(9) = 3$.
$S_{13} = \{(3,5), (5,7), (7,9), (9,11),(11,13)\}$ and $f(13) = 5$
$S_{37} = \{(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35),(35,37)\}$
and $f(37) = 17$
It turns out that $f(n) = ((n-3)/2)$ for odd integers $n$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 5 and $x$ or $y$ is not divisible by 3, and $x \neq 5$ and $y\neq 5$.
Example: $g(37) = 2$ because there are only 2 pairs where $x$ or $y$ is divisible by 5 and not 3, and $x$ or $y$ is not equal to 5.
They are (23,25) and (35,37).
Note, the pairs (3,5) and (5,7) are not included since they contain 5, the pairs (13,15) and (15,17) are not included because 15 is divisible by 3, the pair (25,27) is not included because 27 is divisible by 3, and the pair (33,35) is not included because 33 is divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ ?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n \to\infty$?
Edit:
I believe $g(n)$ approaches $(1/3)(2/5)f(n)$ as $n \to\infty$.
Since $f(n)$ approaches $(n/2)$ as $n$ gets large, $g(n)$ approaches $n/15$ as $n \to\infty$.
I need somebody to confirm this.
The table below compares the actual value for $g(n)$ and $n/15$ for a few values of $n$ and they look comparable.
$g(n)$ approaches $n/15$ as $n \to\infty$">
| Integers to be avoided are multiples of $15$ so you have the following sequence for $g(\infty$): $$\{\color{red}{(23,25),(25,27)},\color{blue}{(33,35),(35,37)},\color{green}{(53,55),(55,57)},\cdots\}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
\begin{array}{c|c}n\pmod{30}&1&3&5&7&9\\\hline g(n)&\frac2{15}(n-1)-2&\frac2{15}(n-3)-2&\frac1{10}(n-5)&\frac2{15}(n-7)&\frac2{15}(n-9)\end{array}
\begin{array}{c|c}n\pmod{30}&11&13&15&17&19\\\hline g(n)&\frac2{15}(n-11)&\frac2{15}(n-13)&\frac2{15}(n-15)&\frac2{15}(n-17)&\frac2{15}(n-19)\end{array}
\begin{array}{c|c}n\pmod{30}&21&23&25&27&29\\\hline g(n)&\frac2{15}(n-21)&\frac2{15}(n-23)&\frac2{15}(n+5)-3&\frac2{15}(n+3)-2&\frac2{15}(n+1)-2\end{array}
Therefore, since $f(n)=\frac{n-3}2$ as you have calculated, for odd $n>23$, $$\lim_{n\to\infty}\frac{g(n)}{f(n)}=\begin{cases}\frac15\,\text{for}\,n\equiv5\pmod{30}\\\frac4{15}\,\text{otherwise}\end{cases}$$
| {
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Find minimum value of $a^2+b^2$ Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+\alpha x + 1)(x^2+\beta x +1)$. Then expanding and comparing coefficients we get $\alpha\beta=0$, meaning either $\alpha=0$ or $\beta=0$. Suppose $\alpha=0$. Then we see that $(x^2+\beta x+1)$ should have real roots, from which we get $\beta^2 \geq 4$. But we get $a=b=\beta$ from the comparison above. So $a^2+b^2 = 2\beta^2 \geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.
| $$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2-\frac{b^2}{4}x^2+(\frac{b}{2}x+1)^2=$$
$$=(x^2+\frac{a}{2}x)^2+x^2(2-\frac{a^2+b^2}{4})+(\frac{b}{2}x+1)^2$$
If $2-\frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(\frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2\ge 8$.
Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-\frac{a}{2}$ and $x=-\frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.
Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.
| {
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Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 \implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 \implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 \implies yxz =1$. If we take $G$ to be the group of $2\times 2$ matrices and let $x = \left( \begin{array} { c c } { 1 } & { 2 } \\ { 0 } & { 2 } \end{array} \right)$, $y = \left( \begin{array} { l l } { 0 } & { 1 } \\ { 2 } & { 1 } \end{array} \right)$ and $z = \left( \begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \\ { 1 } & { - 1 } \end{array} \right)$. Then $x y z = \left( \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) = 1$ but $y x z = \left( \begin{array} { c c } { 2 } & { - 2 } \\ { 5 } & { - 9 / 2 } \end{array} \right) \neq 1$.
I don't understand where my proof went wrong.
| For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
| {
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Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$
My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is not an easy inequality.
| Let $a=x+1$, $b=y+1$ and $c=z+1$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$3((x+1)^2+(y+1)^2+(z+1)^2+2(x-y+1))-(x+y+z+3)^2\geq0$$ or
$$x^2+y^2+z^2-xy-xz-yz+3x-3y+3\geq0$$ or
$$z^2-(x+y)z+x^2+y^2-xy+3x-3y+3=0,$$ for which it's enough to prove that
$$(x+y)^2-4(x^2+y^2-xy+3x-3y+3)\leq0$$ or
$$(x-y+2)^2\geq0.$$
We see that your inequality is true for all real variables.
| {
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Distinct prime number solution of an equation I am trying to solve the following question:
$$\textrm{Find all distinct prime numbers } p, q \textrm{ and } r \textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 \equiv p^2 \equiv 1(p \geq 5), 3(p = 3), 4(p = 2) \mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct.
$$3\times(1, 3)-5\times(1, 3)-4\times(1, 3, 4) \equiv 2 \mod{6}$$
Possible solutions would be
$$\begin{align}
3\times1-5\times3-4\times1&\equiv\\
3\times1-5\times3-4\times4&\equiv\\
3\times3-5\times3-4\times1&\equiv\\
3\times3-5\times3-4\times4&\equiv2\mod{6}
\end{align}$$
So $q \equiv 3 \mod{6} \Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
| If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 \equiv 1 \pmod{5}$$
or $$r^2 \equiv -2 \pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solve
$$3\cdot 5^4-5\cdot 3^4 -4r^2 = 26$$ to get $r=19.$
| {
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $
I made this far:
$$
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
| $$a_n=\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$
$$=\frac{n^3(\sqrt{n^4+1}-n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$
$$=\frac{n^3}{(\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n)(\sqrt{n^4+1}+n^2)}$$
$$=\frac{1}{(\sqrt{1+\sqrt{1+\frac{1}{n^4}}}+\sqrt{2})(\sqrt{1+\frac{1}{n^4}}+1)}$$
Now, apply $\lim_n\to \infty$ on both sides and get the result.
Hope it helps:)
| {
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Prove $n=m^3-3m^2+2m$, for any integer $m$, then $n$ is a multiple of $6$. Prove $n=m^3-3m^2+2m$, for any integer $m$, then $n$ is a multiple of $6$.
So far I have that $n=m(m-3)(m+3)$, which are $3$ consecutive integers so at least one must be a multiple of 2. I am not sure how to get any further? Maybe use 2 cases?
Thank you!
| This can be proved by induction. If $m=1$, then your number is $0$, which is a multiple of $6$. And if $m^3-3m^2+2m$ is a multiple of $6$; then, since\begin{align}(m+1)^3-(m+1)^2+2(m+1)-(m^3-3m^2+2m)&=3m^2-3m\\&=3m(m-1),\end{align}which is clearly a multiple of $6$, $(m+1)^3-(m+1)^2+2(m+1)$ is a multiple of $6$ too.
| {
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Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k}\text{ } \text{ } \text{} \text{ } \text{ }\text{the } n^{th} \text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
| Using the fact that $$\int_0^1x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$$
multiply both sides by $\ \displaystyle-\frac{2}{n4^n}{2n \choose n}$ then take the sum, we get,
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n}}{n^24^n}{2n \choose n}=-2\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(x^2)^n}{n4^n}{2n \choose n}\right)\ dx
\end{align}
I managed here to prove: $$\quad\displaystyle\sum_{n=1}^\infty \frac{x^n}{n4^n}{2n \choose n}=-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2$$
which follows:
\begin{align}
S&=4\underbrace{\int_0^1\frac{\ln(1-x)\tanh^{-1}{\sqrt{1-x^2}}}{x}\ dx}_{\text{IBP}}+4\int_0^1\frac{\ln(1-x)\ln x}{x}\ dx-4\ln2\int_0^1\frac{\ln(1-x)}{x}\ dx\\
&=-4\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx+4(\zeta(3))-4\ln2(-\zeta(2))\\
&=-4\left(\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)\right)+4\zeta(3)+\frac{2}{3}\pi^2\ln(2)\\
&\boxed{=\frac{23}4\zeta(3)-\frac{5}{6}\pi^2\ln2}
\end{align}
Credit goes to ComplexYetTrivial for nicely evaluating $\ \displaystyle\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx=\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)$
| {
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Infinite series problem
The sum of $$\frac{2}{4-1}+\frac{2^2}{4^2-1}+\frac{2^4}{4^4-1}+\cdots \cdots $$
Try: write it as $$S = \sum^{\infty}_{r=0}\frac{2^{2^{r}}}{2^{2^{r+1}}-1}=\sum^{\infty}_{r=0}\frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
| You may continue as:
$$\frac{2^n}{(2^n-1)(2^n+1)} =\frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =\frac 1{2^n-1} - \frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$\left(\frac 11 - \frac 13\right) + \left(\frac 13 - \frac 1{15}\right) + ... + \left(\frac 1{2^n-1} - \frac 1{(2^n-1)(2^n+1)}\right)$$
After cancelling the terms you are left with:
$$1 - \frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
| {
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Compute $\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$ Compute $$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}\implies I = 3\int_0^{\infty}\frac {u}{u^6+1}du=\frac 32\int_0^\infty \frac {1}{u^3+1}du=\frac 32 \int_0^\infty \frac {1}{(x+1)(x^2+x+1)}dx\\\implies I=\frac 32 \int_0^{\infty} \frac 1{x+1}-\frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
| Yet another approach.
\begin{align}
\int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &\overset{\large x \mapsto x^{3/2}}=\frac{3}{2} \int_0^\infty \frac{dx}{1 + x^3}\\
&= \frac{3}{2} \int_0^1 \frac{dx}{1 + x^3} + \frac{3}{2} \int_1^\infty \frac{dx}{1 + x^3}.
\end{align}
Enforcing a substitution of $x \mapsto 1/x$ in the second of the integrals leads to
\begin{align}
\int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &= \frac{3}{2} \int_0^1 \frac{1 + x}{1 + x^3} \, dx\\
&= \frac{3}{2} \int_0^1 \frac{dx}{x^2 - x + 1}\\
&= \frac{3}{2} \int_0^1 \frac{dx}{\left (x - \frac{1}{2} \right )^2 + \frac{3}{4}}\\
&= \sqrt{3} \left [\tan^{-1} \left (\frac{2x - 1}{\sqrt{3}} \right ) \right ]_0^1\\
&= \frac{\pi}{\sqrt{3}}.
\end{align}
| {
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On the evalution of an infinite sum
I wish to show that
$$\sum_{n = 0}^\infty (-1)^n \left [\frac{2n + 1/2}{(2n + 1/2)^2 + 1} + \frac{2n + 3/2}{(2n + 3/2)^2 + 1} \right] = \frac{\pi}{\sqrt{2}} \frac{\cosh \left (\frac{\pi}{2} \right )}{\cosh (\pi)}.$$
The reason I wish to find such a sum is as follows.
The question here called for the evaluation (I have added its value) of
$$\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx = \frac{\pi}{\sqrt{2}} \frac{\cosh \left (\frac{\pi}{2} \right )}{\cosh (\pi)}.$$
As one of the comments, the OP remarked that they would like to see different approaches to the evaluation of the integral so I thought I would try my hand at one that does not rely on contour integration and the residue theorem. My approach was as follows:
\begin{align}
\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx &= \int_0^1 \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx + \int_1^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx\\
&= \int_0^1 \frac{\cos (\ln x) (x + 1)}{\sqrt{x} (1 + x^2)} \, dx,
\end{align}
after a substitution of $x \mapsto 1/x$ has been enforced in the second of the integrals. Now if we enforce a substitution of $x \mapsto e^{-x}$ one arrives at
$$\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx = \int_0^\infty \frac{\cos x \cosh (x/2)}{\cosh x} \, dx.$$
Writing the hyperbolic functions in terms of exponentials we have
\begin{align}
\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx &= \int_0^\infty \frac{\cos x (e^{-x/2} + e^{-3x/2})}{1 + e^{-2x}} \, dx\\
&= \text{Re} \sum_{n = 0}^\infty (-1)^n \int_0^\infty \left [e^{-(2n + 1/2 - i) x} + e^{-(2n + 3/2 - i)x} \right ] \, dx\\
&= \text{Re} \sum_{n = 0}^\infty (-1)^n \left [\frac{1}{2n + 1/2 - i} + \frac{1}{2n + 3/2 - i} \right ] \tag1\\
&= \sum_{n = 0}^\infty (-1)^n \left [\frac{2n + 1/2}{(2n + 1/2)^2 + 1} + \frac{2n + 3/2}{(2n + 3/2)^2 + 1} \right],
\end{align}
which brings me to my sum.
Some thoughts on finding this sum
Rewriting the sum $S$ in (1) as follows:
\begin{align}
S &= \text{Re} \cdot \frac{1}{4} \sum_{n = 0}^\infty \left [\frac{1}{n + 1/8 - i/4} + \frac{1}{n + 3/8 - i/4} - \frac{1}{n + 5/8 - i/4} - \frac{1}{n + 7/8 - i/4} \right ]\\
&= \text{Re} \cdot \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 7/8 - i/4} \right ) + \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 5/8 - i/4} \right )\\
& \qquad - \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 3/8 - i/4} \right ) - \frac{1}{4} \sum_{n = 0}^\infty \left (\frac{1}{n + 1} - \frac{1}{n + 1/8 - i/4} \right )\\
&= \frac{1}{4} \text{Re} \left [\psi \left (\frac{7}{8} - \frac{i}{4} \right ) + \psi \left (\frac{5}{8} - \frac{i}{4} \right ) - \psi \left (\frac{3}{8} - \frac{i}{4} \right ) - \psi \left (\frac{1}{8} - \frac{i}{4} \right ) \right ].
\end{align}
Here $\psi (z)$ is the digamma function. I was rather hoping to use the reflexion formula for the digamma function, but alas it does not seem to take me any closer to a final real solution.
Final thought
While it would be nice to see how to evaluate this sum, perhaps my approach was not the best so alternative methods to evaluate the integral that avoid this sum and does not rely on contour integration would also be welcome.
| $$I=\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx$$
Here is an easy way to compute this integral.
Let's start with a classic integral:
$$\int_0^\infty \frac{y^{a-1}}{1+y}dy=\frac{\pi}{\sin(\pi a)};0<a<1$$
(First calculated by Euler)
The change of variable $y=x^2$ gives
$$I(a)=\int_0^\infty \frac{x^a}{1+x^2}dx=\frac{\pi}{2\cos(\frac{\pi}{2} a)};-1<a<1$$
Now let's note that $\frac{\pi}{2\cos(\frac{\pi}{2} a)}$ as a function of $a$ is regular on the whole complex plane excluding points
$$a=2n+1 ;n\in Z$$
That means it is an analytic continuation of integral $I(a)$ from $(-1,1)$ to the complex plane
So we can safely take $a$ as a complex variable, say $a+ib$
Thus
$$\int_0^\infty \frac{x^{a+ib}}{1+x^2}dx=\frac{\pi}{2\cos(\frac{\pi}{2} (a+ib))}$$
Now separate here real and imaginary parts taking into account that
$$x^{a+ib}=x^a\cos(b\ln x)+ix^a\sin(b\ln x)$$
Choosing real part we finally get
$$I=\int_0^\infty \frac{x^a \cos (b\ln x)}{x^2 + 1} \, dx$$
$$=\frac{\pi}{2}\frac{\cos(\frac{\pi}{2} a)\cosh(\frac{\pi}{2} b)}{\cos^2(\frac{\pi}{2} a)\cosh^2(\frac{\pi}{2} b)+\sin^2(\frac{\pi}{2} a)\sinh^2(\frac{\pi}{2} b)}$$
Choose here $a=\frac{1}{2}$ and $b=1$ to get desired integral
| {
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} |
Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
| Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+\lambda(a^2+b^2+c^2-1) \tag 1$$ Computing the partial derivatives
$$\frac{\partial F}{\partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a \lambda \tag 2$$
$$\frac{\partial F}{\partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b \lambda\tag 3$$
$$\frac{\partial F}{\partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c \lambda\tag 4$$
$$\frac{\partial F}{\partial \lambda}=a^2+b^2+c^2-1\tag 5$$
From $(2)$, $b=\frac{1}{12} (a \lambda +41 a-15 c)$; plug in $(3)$ to get $c=\frac{1}{15} (a \lambda +25 a)$ which makes $b=\frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$\frac{2}{15} a \lambda (\lambda +50)=0 \tag 6$$ So, we have three cases : $a=0$ , $\lambda=0$, $\lambda=-50$.
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $\lambda=0$ would make $c=\frac{5 }{3}a$ which, in turn, would make $\frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $\lambda=-50$ which makes $c=-\frac{5 }{3}a$ which gives again $\frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $\frac{a^2 \lambda ^2}{9}$ with $\lambda=-50$ and $\frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
| {
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Where to start with: $\lim_{n\to\infty} (\sqrt[3]{n^{48}+n} - \sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$ I have limit:
$\lim_{n\to\infty} (\sqrt[3]{n^{48}+n} - \sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube roots by multiply them with
$\dfrac{(\sqrt[3]{n^{48}+n})^2 + (\sqrt[3]{n^{48}+n})(\sqrt[3]{n^{48}+n^2})+ (\sqrt[3]{n^{48}+n^2})^2)}{(\sqrt[3]{n^{48}+n})^2 + (\sqrt[3]{n^{48}+n})(\sqrt[3]{n^{48}+n^2})+ (\sqrt[3]{n^{48}+n^2})^2)}$
which gives me:
$\dfrac{((n^{48}+n) - (n^{48}+n^2))((n^3 +3)^{12} - (n^4+4n)^9 )}{(\sqrt[3]{1n^{48}+n})^2 + (\sqrt[3]{1n^{48}+n})(\sqrt[3]{1n^{48}+n^2})+ (\sqrt[3]{1n^{48}+n^2})^2)}$
but I can not move from here other than just do
$\dfrac{(n - n^2)((n^3 +3)^{12} - (n^4+4n)^9) }{(\sqrt[3]{1n^{48}+n})^2 + (\sqrt[3]{1n^{48}+n})(\sqrt[3]{1n^{48}+n^2})+ (\sqrt[3]{1n^{48}+n^2})^2)}$
I am trying to find some known limits there or something I can grasp on, but can not find anything.
| Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\\
=((n^{36}+36n^{33}+594n^{30}+\cdots)-(n^{36}+36n^{33}+576n^{30}+\cdots))\\
=18n^{30}+\cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(\sqrt[3]{n^{48}+n}-\sqrt[3]{n^{48}+n^{2}})\\
=n^{16}(\sqrt[3]{1+n^{-47}}-\sqrt[3]{1+n^{-46}})\\
=n^{16}((1+\frac{1}{3}n^{-47}+\cdots)-(1+\frac{1}{3}n^{-46}+\cdots))\\
=n^{16}(-\frac{1}{3}n^{-46}+\frac{1}{3}n^{-47}+\cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+\cdots)$$
which equals $-6$ in the limit.
Hope that helps
| {
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Rewriting $2x^2-12x+11$ by completing the square I want to rewrite $2x^2-12x+11$ in square form. According to my book, the answer is $2(x-3)^2-7$. However, I get $2(x-3)^2+2$ for some reason.
My Steps:
$2x^2-12x+11$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+9$
$2(x - 3)^2=-2$
$2(x-3)^2+2$
What am I doing wrong?
| An alternative way to see how this all works is to note that completing the square doesn't change the value of the expression. What you are essentially doing is adding a cleverly-written $0$ to it so that it becomes easier to work with.
Observe, $$\begin{align}2x^2-12x+11 & = 2x^2 - 12x \qquad\qquad\qquad\qquad\qquad\quad +11 \tag{Leave space}\\ &= 2[x^2 - 6x]\qquad\qquad\qquad\qquad\qquad\;\; +11 \tag{Factor}\\&= 2[x^2 - 6x +\underbrace{\left({-6\over2}\right)^2 - \left({-6\over2}\right)^2}_\text{cleverly-written zero}] + 11 \\ &= 2[\underbrace{x^2 - 6x + 9}_{(x-3)^2} - 9] + 11\\ &= 2[(x-3)^2 - 9] + 11\\ &= 2(x-3)^2 - 18 + 11\\ &= 2(x-3)^2 - 7.\end{align}$$
| {
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$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$ $$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$$
My working:
$$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\int_{0}^{1}\frac{\ln x}{x^2-1}dx=-\sum_{r\ge 1}\int_0^{1}x^{2r}\ln x\,dx =\sum_{r\ge 1}\frac {1}{(2r-1)^2}=
\frac{\pi^2}{8}$$
Is there any other approach?
| \begin{align}J&=\int_{0}^{1}\frac{\ln x}{x^2-1}dx\\
&=\frac{1}{4}\int_{0}^{1}\frac{2x\ln(x^2)}{1-x^2}dx-\int_{0}^{1}\frac{\ln x}{1-x}dx\\
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx-\int_{0}^{1}\frac{\ln x}{1-x}dx\\
&=-\frac{3}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx\\
&=-\frac{3}{4}\zeta(2)\\
&=-\frac{3}{4}\times -\frac{\pi^2}{6}\\
&=\frac{\pi^2}{8}\\
\end{align}
NB:
I assume that,
\begin{align} \zeta(2)=\frac{\pi^2}{6}\\
\end{align}
And,
\begin{align}\int_{0}^{1}\frac{\ln x}{1-x}dx&=\int_0^1\left(\sum_{n=0}^\infty x^n\right)\ln x\,dx\\
&=\sum_{n=0}^\infty \int_{0}^{1}x^n\ln x\,dx\\
&-=\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\
&=-\zeta(2)
\end{align}
For $0\leq x<1$,
\begin{align}
\frac{1}{x^2-1}=\frac{x}{1-x^2}-\frac{1}{1-x}
\end{align}
| {
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Is there any other way to establish this trig identity? $\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $ I needed to verify this trig identity: $$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $$
what I did is I worked on both sides individually:
LHS: $$\frac{\frac1{\cos(x)} + 1}{\frac {\sin(x)}{\cos (x)}} =\frac{1 + \cos(x)}{\sin(x)}$$
RHS: $$\frac{\sin(x)}{1 - \cos(x)}=\frac {\sin(x)(1-\cos^2(x))}{1 -\cos^2(x)}=\frac {\sin(x)(1-\cos^2(x))}{\sin^2(x)}= \frac{1 +\cos(x)}{\sin(x)} $$
QED.
I can't think of other way to verify this, like just working on one side.
Is there any?
| Working both sides of
$$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)}$$
First multiply both sides by $\sin(x)$ giving
$$\frac{\tan(x) + \sin(x)}{\tan(x)} = \frac{\sin^2(x)}{1 - \cos(x)}$$
Since $\sin^2(x) = 1-\cos^2(x)$ we immediately have
$$1 + \cos(x) = 1 + \cos(x)$$
| {
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How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mess with the constant edits, I will stop editing and not touch this so people can review the question. Sorry
a) Prove that P(1) is true
$~$
$~$
$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$
Show that P(k+1) is true as well
$$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$
$~~$
$$ = \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$
| Hint:
It is a telescoping sum since, for any $k>0$,
$$\frac 1{k(k+1)}=\frac 1k-\frac 1{k+1}.$$
| {
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$.
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
| Nice observation!
It is indeed true that $ \sum_{k=0}^{n - 1} (2 k + 1) (n - k) = \sum_{k=1}^{n} k^2$.
This can be proved by using the formulas for $\sum 1, \sum k, \sum k^2$.
| {
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If $y=\frac {a+bz}{c+dz}$, $z=\frac{a+bx}{c+dx}$, $x=\frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$ I need to solve this problem and I don’t know how.
If $x, y, z$ are unequal and $y = \frac {a + bz}{c + dz}, z = \frac {a + bx}{c + dx}, x = \frac {a + by}{c + dy},$ then $ad + bc + b^2 + c^2 = 0$.
I see that I need to eliminate $x$, $y$, $z$ but I don’t know how to start.
I appreciate your help.
| Substitute formula for x into formula for z we have
$$z = \frac{ac + ady + ab + b^2y}{c^2 + cdy + ad + bdy}$$
Substitute this into formula for y we have
$$(c^3 + acd)y + (c^2d + bcd + ad^2 + b^2d)y^2 = (ac^2 + a^2d + abc + ab^2) + (bad + b^3)y$$
$$(c^3 + acd - abd - b^3)y + d(c^2 + bc + ad + b^2)y^2 = a(ad + bc + b^2 + c^2)$$
$$y(ad(c - b) + (c^3 - b^3)) + d(ad + bc + b^2 + c^2)y^2 = a(ad + bc + b^2 + c^2)$$
$$(ad + bc + b^2 + c^2)(dy^2 + (c - b)y - a) =0$$
Hence $ad + bc + b^2 + c^2 =0$
| {
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Find the power series expansion of $f(x)= \frac{e^x - 1}{x}$
Prove that the following function is analytical in 0 and find its power series centered in 0
$$f(x)= \frac{e^x - 1}{x}, f(0)=1$$
I'm trying to write $f$ as some kind of combination of function with known Taylor series expansion in an open interval around zero, which would prove the analicity:
$$f(x)= \frac{e^x}{x}- \frac{1}{x}=\frac{1}{x}\sum_{n=0}^{\infty} \frac{x^n}{n! } - \frac{1}{x} =\sum_{n=0}^{\infty} \frac{x^{n-1}}{n! } - \frac{1}{x}$$
but I don't know about any power series expansion of $\frac{1}{x}$
| You know that $e^x= \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot$, so
$$
e^x- 1= \sum_{n=1}^\infty \frac{x^n}{n!}= x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot.
$$
thus
$$
\frac{e^{x}- 1}{x}= \sum_{n= 1}^\infty \frac{x^{n-1}}{n!}=\sum_{n= 0}^\infty \frac{x^n}{(n+1)!}= 1+ \frac{x}{2!}+ \frac{x^2}{3!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot
$$
| {
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This sum $\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}g(n)$ and the golden ratio We got this strange sum?:
$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\left[8\cdot\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}-\frac{(n-\alpha)^{1/5}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(\alpha+1)^{1/5}\tag1$$
Where $\alpha\ge0$ and $\phi$ is the golden ratio
e.g
Let $\alpha=4$
$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\left[8\cdot\frac{(n-5)^{1/5}}{(2n-1)^2}-\frac{(n-4)^{1/5}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(5)^{1/5}$$
Using binomial series,
$$(n-5)^{1/5}=n^{1/5}-n^{-4/5}-2n^{-9/5}-6n^{-14/5}-21n^{-19/5}\cdots$$
$$(n-4)^{1/5}=n^{1/5}-\frac{4}{5}n^{-4/5}-\frac{32}{25}n^{-9/5}-\frac{384}{125}n^{-14/5}-\frac{5376}{625}n^{-19/5}\cdots$$
$$8\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{1}{(2n-1)^2}\left[n^{1/5}-n^{-4/5}-2n^{-9/5}-6n^{-14/5}-21n^{-19/5}\cdots\right]=A$$
$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{1}{(n+1)^2}\left[n^{1/5}-\frac{4}{5}n^{-4/5}-\frac{32}{25}n^{-9/5}-\frac{384}{125}n^{-14/5}-\frac{5376}{625}n^{-19/5}\cdots\right]=B$$
In general we have to evaluate
$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{n^a}{(2n-1)^2}=f(a)$$
$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{n^a}{(n+1)^2}=g(a)$$
Can we easily evaluate $f(a)$ and $g(a)$ and uses it to show that $(1)$ is true?
| This solution is only the prooving of the statement, I did not look for $f(a), f(b)$.
Let's split the original sum into two pieces:
$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}-\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n}}\frac{(n-\alpha)^{1/5}}{(n+1)^2}\tag1$
The first sum equals to:
$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}=\sum\limits_{n=1}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}+8(-\alpha-1)^{1/5}\tag2$
Reindexing the second sum in (2):
$\sum\limits_{n=0}^\infty\binom{2n+2}{n+1}^2\frac{1}{2^{5n+2}}\frac{(n-\alpha)^{1/5}}{(2n+1)^2}+8(-\alpha-1)^{1/5}\tag3$
Let's form the $\binom{2n+2}{n+1}$ in the following way:
$\binom{2n+2}{n+1}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)^2 n!^2}=\binom{2n}{n}\frac{2(2n+1)}{n+1}\tag4$
Put it back to (3) we get:
$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n}}\frac{(n-\alpha)^{1/5}}{4(2n+1)^2}(\frac{2(2n+1)}{n+1})^2+8(-\alpha-1)^{1/5}\tag5$
We can see that the difference of second sum in (1) and the sum in (5) is only the sign.
So the original sum is equal to: $8(-\alpha-1)^{1/5}=8e^{i\pi/5}(\alpha+1)^{1/5}\tag5$
Easy to see that $e^{i\pi/5}=\frac{1+\sqrt{5}}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}}$
Introducing $\phi$, we get: $\big(4\phi+4i\sqrt{\frac{\sqrt{5}}{\phi}}\big)(\alpha+1)^{1/5}=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(\alpha+1)^{1/5}\tag6$
And ready.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $\sum_{r=1}^n r^{\overline{a+1}}(r+b)=\big(\sum_{r=1}^n r^{\overline{a+1}}\big)\big(An+B\big)$?
Why is
$$\sum_{r=1}^n r^{\overline{a+1}}(r+b)=\bigg(\sum_{r=1}^n r^{\overline{a+1}}\bigg)\bigg(An+B\bigg)=\frac{n^{\overline{a+2}}}{a+2}\cdot (An+B)$$
where $A, B$ are rational numbers?
?
It is clear that LHS is a polynomial in $n$ of order $a+3$, but that does not necesarily imply that $\sum_{r=1}^n r^{\overline{a+1}}$ is a factor.
By extension, is it true that
$$\sum_{r=1}^n r^{\overline{a+1}}P(r)=\bigg(\sum_{r=1}^n r^{\overline{a+1}}\bigg)Q(n)=\frac{n^{\overline{a+2}}}{a+2}\cdot Q(n)$$
where $P(\cdot), Q(\cdot)$ are polynomials of the same degree?
| $$\begin{align}
\sum_{r=1}^n r^{\overline{a+1}}(r+b)
&=\sum_{r=1}^n r^{\overline{a+1}}(r+a+1)+ r^{\overline{a+1}}(b-a-1)\\
&=\sum_{r=1}^n r^{\overline{a+2}}+(b-a-1)\sum_{r=1}^n r^{\overline{a+1}}\\
&=\frac {n^{\overline{a+3}}}{a+3}+(b-a-1)\cdot \frac {n^{\overline{a+2}}}{a+2}\\
&=(n+a+2)\cdot \frac {a+2}{a+3}\cdot \frac {n^{\overline{a+2}}}{a+2}+(b-a-1)\cdot \frac{n^{\overline{a+2}}}{a+2}\\
&=\frac {n^{\overline{a+2}}}{a+2}\;\;\left[(n+a+2)\cdot \frac {a+2}{a+3}+(b-a-1)\right]\\
&=\frac {n^{\overline{a+2}}}{a+2}\;\;\left[\frac {a+2}{a+3}n+\frac {(a+2)^2-(a+1)(a+3)+b(a+3)}{a+3}\right]\\
&=\frac {n^{\overline{a+2}}}{a+2}\;\;\bigg[\underbrace{\frac {a+2}{a+3}}_A n+\underbrace{b+\frac 1{a+3}}_B\bigg]\\
&\equiv\left(\sum_{r=1}^n r^{\overline{a+1}}\right)\ \bigg(An+B\bigg)
\end{align}$$
For the general case, using consider the case where $P(\cdot)$ is of degree $m$, i.e.
$$P(r)=r^m+\lambda_1 r^{m-1}\lambda_2 r^{m-2}+\cdots +\lambda_m$$
This can be converted to a factorial polynomial of the same degree, i.e.
$$P(r)=r^\overline{m}+\mu_1 r^\overline{m-1}+\mu_2 r^\overline{m-2}+\cdots+\mu_m$$
Hence
$$\begin{align}\sum_{r=1}^n r^\overline{a+1}P(r)
&= \sum_{r=1}^n r^\overline{a+m+1}
+\mu_1 r^\overline{a+m}
+\mu_2 r^\overline{a+m-1}
+\cdots
+\mu_m r^\overline{a+1}\\
&=\frac {n^\overline{a+m+2}}{a+m+2}
+\mu_1 \frac {n^\overline{a+m+1}}{a+m+1}
+\mu_2 \frac {n^\overline{a+m}}{a+m+1}+\cdots
+\mu_m \frac {n^\overline{a+2}}{a+2}\\
&=\frac {n^\overline{a+2}}{a+2}\cdot Q(n)
\end{align}$$
since $\frac {n^\overline{a+2}}{a+2}$ is a factor of $\frac {n^\overline{a+i}}{a+i}$ for $i=3,4,5,\cdots, m+1$.
| {
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What is the value of $x+y$ if x and y are co-primes and $PR=\dfrac{x}{y}$ in the diagram?
$ST$ is the perpendicular bisector of $PR$ and $SP$ is the angle
bisector of $\angle QPR$. If $QS=9cm$ and $SR=7cm$ then
$PR=\dfrac{x}{y}$ where x, y are co-primes. $x+y$=?
Source: Bangladesh Math Olympiad 2015 junior category.
I found that $PS=7$, but I can't get the value of $PR$.
| Here's a solution with less algebra:
Note first that $R$ is the mirror image of $P$ across $ST$. Consider now the point $Q'$ obtained by reflecting $Q$ across $ST$. We obtain an isosceles trapezoid $PQQ'R$, which is a cyclic quadrilateral. We know that $|PS| = |RS| = 7$ and $|QS| = |Q'S| = 9$, and that $\angle QPS = \angle SPR$.
Let $a := |PR|, b := |PQ| = |Q'R|, c := |QQ'|$.
Since $\angle QPQ' = \angle Q'PR$, we know by the inscribed angle theorem that $b = c$, and by the intercept theorem through $S$ we find that $c : a = 9 : 7$, so $b = c = \frac{9}{7} a$.
Now we can use Ptolemy's theorem, which yields $ac+b^2 = (9+7)^2$. Replacing $b$ and $c$ we get $\frac{9}{7}a^2 + \frac{9}{7}\frac{9}{7}a^2 = 256$, so $a^2 = \frac{256}{\frac{9}{7} + \frac{9}{7}\frac{9}{7}} = \frac{12544}{144}$ and $|PR| = a = \frac{112}{12} = \frac{28}{3}$.
The solution is therefore $28 + 3 = 31$.
| {
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Find upper and lower bounds for the finite sum Find upper and lower bound for the following finite sum:
$\frac{1}{1} + \frac{1}{2^3} + \frac{1}{3^3} + ··· + \frac{1}{n^3}$
My attempt is:
Using the integral test:
we know that $\frac{1}{1} + \frac{1}{2^3} + \frac{1}{3^3} + ··· + \frac{1}{n^3}$ = $\sum_{i=1}^n 1/i^3$ = $\int_1^n$1/$i^3$di = $\int_1^n1/x^3$dx = $-1/2n^2$ + $1/2$
But now I'm stuck. How can this test give the lower and upper bouunds? Any help please?
| You do not know that $\sum_{i=1}^n 1/i^3= \int_1^n1/x^3\, dx$ as the equality is not exact. In fact $$\int\limits _1^{n+1}\frac{1}{x^3}\, dx \le \sum_{i=1}^n \frac{1}{i^3}\le 1+ \int\limits_1^n \frac{1}{x^3}\, dx $$
and using your earlier analysis, this becomes $$\frac12 - \frac{1}{2(n+1)^2} \le \sum\limits _{i=1}^n \frac{1}{i^3}\le \frac32 - \frac{1}{2n^2}.$$
This lower bound is not good, as clearly $\sum\limits_{i=1}^n \frac{1}{i^3}\ge 1$ when $n\ge 1$. As Will Jagy suggested, it might therefore be better to use as the lower bound $1+\int\limits _2^{n+1}\frac{1}{x^3}\, dx =$ so $$ \frac{9}{8}-\frac{1}{2(n+1)^2}\le \sum\limits _{i=1}^n \frac{1}{i^3}\le \frac32 - \frac{1}{2n^2}.$$
These bounds are not good for $n\ge 2$ and you could continue the process. An alternative approach might be to start at the other end and to note that the infinite sum is $\zeta(3)= \sum\limits_{i=1}^\infty \frac{1}{i^3} \approx 1.2020569$, known as Apéry's constant, and then subtract $\int\limits _n^{\infty}\frac{1}{x^3}\, dx$ or $\int\limits _{n+1}^{\infty}\frac{1}{x^3}\, dx$ so $$\zeta(3) -\frac{1}{2n^2} \le \sum\limits _{i=1}^n \frac{1}{i^3}\le \zeta(3) -\frac{1}{2(n+1)^2} $$ and these are tighter than the original suggestion for $n\ge 2$ and much tighter as $n$ increases.
| {
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Given that A, B and C do not lie on the same line...find the area of triangle ABC.
\begin{array} { c } { \text { Given that } A , B \text { and } C \text { do not lie on the same line.If } \vec { O A } + \vec { O B } + \vec { O C } = 0 , | \vec { O A } | = \sqrt { 6 } \text { , } } \\ { | \vec { O B } | = 2 \text { and } | \vec { O C } | = \sqrt { 14 } , \text { find the area of } \triangle A B C . } \end{array}
I found this problem online and I am not sure what should I do. I have tried some geometric transformations but couldn't find the answer. I guess it isn't as easy as I thought at first.
| Wlog $B=(2,0)$, $A=(x,y)$ with $x^2+y^2=6$, $C=(u,v)$ with $u^2+v^2=14$. Moreover, $x+u+2=0$ and $y+v=0$. This allows us to eliminate $u,v$:
$$ \begin{align}x^2+y^2&=6\\(x+2)^2+y^2&=14\\\implies\qquad 4x+4=14-6\end{align}$$
and thereby
$$ x=1,\;u=-3,\;y=\pm\sqrt 5,v=\mp\sqrt 5.$$
From $y+v=0$, we conclude that $BC$ intersects the $x$-axis at $\frac{x+u}{2}=-1$. Thus the $x$-axis splits $ABC$ into two triangles of base length $3$ and height $\sqrt 5$. The area is therefore $3\sqrt5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| You can always pass to polar coordinates to compute 2d limits:
Set $x=r \cos\theta , y=r\sin \theta$, and now we need to compute
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)}$$
Since this is a limit for the variable $r$, you can use L'Hôpital taking derivatives w.r.t. $r$. Note that if we end with something depending on $\theta$, the limit does not exist, becasuse the limit (if exists) is unique.
After L'Hôpital and some basic manipulation, we end with
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)} = \lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{\sin (2r\sin\theta) \sin \theta(1+r^2\cos^2\theta) + 2r\cos^2\theta}$$
You use the always useful $\sin (ar) \approx ar$ when $r \approx 0$ to rewrite the denominator of the RHS:
$$\lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{(2r\sin\theta)\sin\theta(1+r^2\cos^2\theta) + 2r\cos^2\theta} = \lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{2r\sin^2 \theta(1+r^2\cos^2\theta) + 2r\cos^2\theta}= \dfrac{1}{\sin^2 \theta + \cos^2\theta} = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$ Let $a,b,c$ be reals with $a+b+c=1$. Show that :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$
I have tried to something like this:
$$18a^4-24a^3+6a^2-12a+12\geq 0$$
$$18b^4-24b^3+6b^2-12b+12\geq 0$$
$$18c^4-24c^3+6c^2-12c+12\geq 0$$
After summing I get :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+24\geq24(a^3+b^3+c^3).$$
If someone has an ideea I would gratefully appreciate.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$
Thus, since our inequality is fourth degree, we see that it's equivalent to $f(w^3)\geq0,$
where $f$ is a linear function.
But the linear function gets a minimal value for an extreme value of $w^3$,
which happens for equality case of two variables.
Id est, it's enough to prove our inequality for $b=a$ and $c=1-2a,$ which gives
$$18(2a^4+(1-2a)^4)+6(2a^2+(1-2a)^2)+1\geq24(2a^3+(1-3a)^3)$$ or
$$324a^4-432a^3+180a^2-24a+1\geq0$$ or
$$(18a^2-12a+1)^2\geq0.$$
Done!
| {
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Maximum value of expression $a+b+c$
If $a,b,c$ are non negative integers such that $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?
My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^3+b^3+c^3}{3}\geq \bigg(\frac{a+b+c}{3}\bigg)^3\cdots (1)$$
From given condition $$\frac{a^3+b^3+c^3}{3}=\frac{(a+b+c)^2}{2}\cdots (2)$$
So we have $$\frac{(a+b+c)^2}{2}\geq \frac{(a+b+c)^3}{27}$$
$$a+b+c\leq \frac{27}{2}=13.5$$
equality hold when $a=b=c=4.5$
but $a,b,c$ are non negative integers
Could some help me to solve it, Thanks
| By your work $$a+b+c\leq13,$$ but since
$$3(a+b+c)^2=2(a^3+b^3+c^3)=2(a^3+b^3+x^3-3abc+3abc)=$$
$$=2(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+6abc,$$ we obtain that $a+b+c$ is divisible by $3$.
Thus, we see that $$a+b+c\leq12.$$ But $$(a,b,c)=(3,4,5)$$ is valid, which says that $12$ is a maximal value.
Actually, your inequality we can get also by Holder:
$$a^3+b^3+c^3=\frac{1}{9}(1+1+1)^2(a^3+b^3+c^3)\geq\frac{1}{9}(a+b+c)^3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$
since $x+y+z = 6$, $z = 6-x-y$, so
\begin{align*}
d &= \sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \\
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
\end{align*}
Find partial derivative $f_x$ and $f_y$ and critical points
\begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \\
&= 24-2y \quad (\text{set }= 0) \\
&= \text{critical point }y = 4 \\
f_y &= 2y + 2(-x-y+12) \\
&= 24 - 2x \quad (\text{set }= 0) \\
&= \text{critical point }x = 12 \\
\end{align*}
Plug in $x = 12$ and $y = 4$ to original equation
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
| You can draw a line from your point to the plane and then find the length of its projection onto the unit normal vector to the plane.
| {
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Solving the equation $2^x+2^{-x} = 10$ I'm trying to figure out how to solve this equation in terms of $x$.
Mistake I made so far is:
$$\log(2^x) + \log(2^{-x}) = \log(10)$$
$$x \cdot \log(2) - x\cdot \log(2) = 1$$
$$0 = 1$$
| Let $2^x=y$
The equation becomes: $y^2-10y+1=0$. This quadratic equation has two solutions:
$$y = \frac{10 \pm \sqrt{100-4}}{2} = 5 \pm 2\sqrt{6}.$$
So you get $x = \log_2(5+2\sqrt{6})$ and $x = \log_2(5-2\sqrt{6})$ as solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find some digits of $17!$
$17!$ is equal to $$35568x428096y00$$
Both $x$ and $y$, are digits. Find $x$ and $y$.
So, $$17!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13\times 17=(2^3\times 5^3)\times 2^{12}\times 3^6\times 7^2\times 11\times 13\times 17$$
If there`s a product of $(2\times 5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
| The alternating sum of digits must be divisible by $11$, i.e., $11\mid 18-x$. It follows that $x=7$.
| {
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Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$. I am new to multivariable limits and I am having a hard time arguing this limit:
Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$.
I can work out the overarching context of the proof as follows: Let $\epsilon>0$ be given. We need a $\delta$ so that whenever $\|(x,y)-(2,1)\|<\delta$ we have $\|f(x,y)-f(2,1)\|<\varepsilon.$ The difficulty I always have with these problems is that I know I need to rewrite $\|\sqrt{x+y-2}-1\|$ in terms of the given $\delta$ estimate but for some reason when it comes to these limit problems algebra turns into the most difficult thing in the world for me.
| Let $\varepsilon > 0$ and set $\delta = \min\left\{\frac{\varepsilon^2}{6}, \frac12\right\}$.
Let $\|(x,y) - (2,1)\| < \delta$. This implies $|x-2| < \delta$ and $|y-1| < \delta$. Also
$$x+y-2 = 1 + (x-2) + (y-1) \ge 1-|x-2|-|y-1| > 1-2\delta \ge 0$$
so $\sqrt{x+y-2}$ is well-defined.
We have
$$|x+y-2| \le |x-2| + |y-1| + 1 < 2\delta + 1$$
$$|x+y-2| \ge x+y-2> 1-2\delta$$
so
\begin{align}
\left(\sqrt{x+y-2}-1\right)^2 &= (x+y-2) - 2\sqrt{x+y-2}+1 \\
&< (2\delta+1) - 2\sqrt{1-2\delta} + 1 \\
&\le (2\delta+1) - 2(1-2\delta) + 1 \\
&\le 6\delta \\
&< \varepsilon^2
\end{align}
and hence $$\left|\sqrt{x+y-2}-1\right| < \varepsilon$$
| {
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maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please.
| For $a=b=c=\frac{1}{3}$ we get the value $\frac{7}{27}.$
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(a+b+c)(ab+ac+bc)-2abc\leq\frac{7(a+b+c)^3}{27}$$ or
$$\sum_{cyc}(7a^3-6a^2b-6a^2c+5abc)\geq0$$ or
$$7\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(a^2b+a^2c-2abc)\geq0$$ or
$$7\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}c(a-b)^2\geq0,$$
which is true by Schur.
| {
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Triangle Geometry question find minumum value of $n-m$ In the triangle shown, for $\angle A$ to be the largest angle of the triangle, it must be that $m<x<n$. What is the least possible value of $n-m$, expressed as a common fraction?
I found that $x<4.5$ and $x>1.6$ so I thought the answer was $4.5-1.6=\frac{29}{10}$ but this answer is wrong. Could someone help me see what I did wrong?
| By triangle inequality we know that
1) $3x + (x+4) > x+9$
2) $(x+4)+(x+9) > 3x$
3) $(x+9) + 3x > x+4$.
1) gives us $3x > 5$ or $x > \frac 53$.
2) gives us $x < 13$
And 3) gives us $3x > -5$ which is a no brainer as $x > 0$.
The fact that A is the largest angle gives us that $x+9$ is the largest side. Obviously $x + 9 > x+4$ so that doesn't help us but $x + 9 > 3x$ tells us that $x < \frac 92$.
Putting that together we get $\frac 53 < x < \frac 92$
Since we must have $x > m$ and we can have $x $ as close to $\frac 53$ as we like, the very most that $m$ can be is $\frac 53$. Likewise the very least $n$ can be as we must have $m > $ and $x$ may be as close to $\frac 92$ as we want.
So the very least $n-m$ can be will occur when $n$ is its least possible value $\frac 92$ and $m$ is its greatest possible value of $\frac 53$ so
$m - n = \frac 92 - \frac 53 = \frac {17}{6}$
You problem was rounding $\frac 53 = 1.6$..... Why in the HECK did you do that????????
| {
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Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c- b)$$
If $c\neq {\rm mid}\{a, b, c\}$, the inequality is obviously true!
If $c={\rm mid}\{a, b, c\}$, we have $(a- c)(c- b)= 0\Leftrightarrow c= \dfrac{c^2+ ab}{a+ b}$. I tried to prove that:
$$f(c)- f(\frac{c^2+ ab}{a+ b})= (a- c)(c- b)F\geqq 0$$
where $f(c)= 3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$ but without success! I found this inequality by using discriminant and some coefficient skills. Thank you so much
| Consider three cases.
*
*$a=\max\{a,b,c\}$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$
Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)\geq0;$$
*$b=\max\{a,b,c\}$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$
Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)\geq0$$ and
*$c=\max\{a,b,c\}$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$
Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3v\geq0$$ and we are done!
Actually, the following stronger inequality is also true.
Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq b(a+b-c)(a-c)(c-b).$$
| {
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If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite
If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite.
My attempt so far:
$$a^2+b^2+c^2+d^2$$
$$=a^2+d^2+2ad+b^2+c^2+2bc-2ad-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+bc+c^2)-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+2bc+c^2)$$
$$=(a+d)^2+(b+c)^2-2(b+c)^2$$
$$=(a+d)^2-(b+c)^2$$
$$=(a+b+c+d)(a-b-c+d)$$
Now I am trying to prove that $a-b-c+d\not=1$ so I tried to assume the contradiction but I am unable to finish. Any help would be appreciated.
| Firstly, if at least one of $a,b,c,d$ is greater than one than
$$a^2+b^2+c^2+d^2\gt a+b+c+d$$
$$\therefore a-b-c+d\gt 1$$
If they are all equal to $1$, then $1^2+1^2+1^2+1^2=4$ which is composite.
| {
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Find $ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $ with Taylor I have to calculate some limits and try to solve them in use of taylor.
$$
\lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right)
$$
In taylor pattern I have $x_0$ to put, but there $x_0$ is $\infty$ so I want to replace it with something other
$$
y = \frac{1}{x} \\
\lim_{y\to 0^+} \left(\frac{1}{y}-\frac{1}{y^2} \ln (1+y)\right) $$
Let $$ f(y) = \frac{1}{y}-\frac{1}{y^2} \ln (1+y) $$
$$f'(y) = -\frac{1}{y^2} + \left(-\frac{2}{y^3}\ln (1+y) - \frac{y^2}{1+y}\right) $$
but $f'(0)$ does not exists because I have $0$ in denominator.
| Proceeding with your substitution, since $\log \left( 1+y\right) =y-\frac{1}{
2}y^{2}+O\left( y^{3}\right) $, we have:
\begin{eqnarray*}
\frac{1}{y}-\frac{1}{y^{2}}\log \left( 1+y\right) &=&\frac{1}{y}-\frac{1}{
y^{2}}\left( y-\frac{1}{2}y^{2}+O\left( y^{3}\right) \right)
&=&\frac{1}{2}+O\left( y\right) \overset{y\rightarrow 0}{\longrightarrow }
\frac{1}{2}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $PA+PB$ is If $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is
what i try
let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$
Then $P(2\lambda-2,2\lambda-7,\lambda+2)$
$PA+PB=\sqrt{(2\lambda-7)^2+(2\lambda-10)^2+(\lambda-2)^2}+\sqrt{(2\lambda-3)^2+(2\lambda-8)^2+\lambda^2)}$
$PA+PB=\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}$
how do i minimize it b3cause derivative method is very tedious help me please
| Note that
$$
|PA|+|PB|=3\left(\sqrt{(\lambda-4)^2+1}+\sqrt{(\lambda-2)^2+1}\right).
$$ The expression in the parenthesis is equal to the sum of distance between $X=(\lambda,0)$ and $Y=(4,1)$ and distance between $X=(\lambda,0)$ and $Z=(2,-1)$. Then, by the triangle inequality, $|XY|+|XZ|
$ is minimized when $X$ is on the line segment between $Y$ and $Z$. This gives $$|XY|+|XZ|\ge |YZ|=\sqrt{8}=2\sqrt 2$$ hence giving
$$
|PA|+|PB|\ge 6\sqrt{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the following problem by using the Binomial formula If $n=10$ and $p=0.60$ find $P(X\geq 3)$.
The formula we've been given: $P(X=x) = \binom{n}{x} \cdot p^x \cdot q^{n-x}$, where $n$ is the number of trials, $p$ is the probability of success and $q = 1-p$.
I know that I could do $P(X\geq 3) = \binom{10}3 \cdot 0.6^3 \cdot 0.4^7$ adding until I reach $\binom{10}{3} \cdot 0.60^{10} \cdot 0.40^0$.
I know there is a way I could do this without having to individually calculate from 3 until 10. I'm thinking along the lines of
$P(X<3) = \binom{10}{2} \cdot 0.4^2 \cdot 0.6^8 + \binom{10}{1} \cdot 0.4^1 \cdot 0.6^9 + \binom{10}{0} \cdot 0.4^0 \cdot 0.6^{10}$
But I'm having doubts about this because I tried the long way and got $0.98771$ but with the other method I get $0.1673$.
| You have exchanged $p$ and $1-p$. As you said $p=0.6$. Therefore
$P(X\leq 2)={10 \choose 2} \cdot 0.6^2 \cdot 0.4^8 + {10 \choose 1} \cdot 0.6^1 \cdot 0.4^9 + {10 \choose 2} \cdot 0.6^0 \cdot 0.4^{10}=0.01229$
Thus $P(X\geq 3)=1-P(X\leq 2)=1-0.01229=0.98771$
| {
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Geometry high school math competition question
Let $ABC$ be an equilateral triangle with side length $2$.
Let the circle with diameter AB be $\gamma$.
Consider the two tangents from $C$ to $\gamma$, and let the tangency point closer to $A$ be $D$.
Find the area of triangle $CAD$.
I was able to figure out that $CD$ has to be $\sqrt2$. I can not figure out the height of triangle $CAD$.
I am trying to calculate the height from $D$ to $AC.$ Alternatively, if I could find the length of $AD$, then I should also be able to solve problem.
| Let $O$ be a center of the circle $\Gamma$.
Thus, by the Pythagoras's theorem
$$CD^2=CO^2-DO^2=\left(\sqrt3\right)^2-1^2=2$$ and $$CD=\sqrt2.$$
Id est, $$S_{\Delta ACD}=\frac{1}{2}CD\cdot AC\sin\measuredangle ACD=$$
$$=\frac{1}{2}\cdot\sqrt2\cdot2\sin\left(\arctan\frac{1}{\sqrt2}-30^{\circ}\right)=$$
$$=\sqrt2\left(\sin\arctan\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}-\cos\arctan\frac{1}{\sqrt2}\cdot\frac{1}{2}\right)=$$
$$=\sqrt2\left(\frac{\frac{1}{\sqrt2}}{\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}\cdot\frac{\sqrt3}{2}-\frac{1}{\sqrt{1+\left(\frac{1}{\sqrt2}\right)^2}}\cdot\frac{1}{2}\right)=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$
There is a solution without trigonometry.
Let $DK$ be an altitude of $\Delta ADO$.
Thus, since $\Delta OKD\sim\Delta CDO,$ we obtain:
$$\frac{DK}{DO}=\frac{DO}{CO}$$ or
$$\frac{DK}{1}=\frac{1}{\sqrt3},$$ which gives $$DK=\frac{1}{\sqrt3}$$ and
$$S_{\Delta ACD}=S_{\Delta ADO}+S_{\Delta OCD}-S_{\Delta ACO}=$$
$$=\frac{1}{2}\cdot\frac{1}{\sqrt3}\cdot1+\frac{1}{2}\cdot\sqrt2\cdot1-\frac{1}{2}\cdot\sqrt3\cdot1=\frac{1}{\sqrt2}-\frac{1}{\sqrt3}.$$
| {
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Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 + 1$$
$$1\cdot 1000 + (a-11)\cdot 100+(2-a-b)+1-a-b=0$$
So, our number will be something like this: $1(a-11)(2-a-b)(1-a-b)=0$. This is contradiction since the first digit is $1$, and on the right we have $0$.
Does that mean there are no $a$ and $b$ to satisfy this equality?
Alternatively, I did the following: since the last digit in $1a21$ is $1$, it has to be $a+b=L1$. $a$ and $b$ are digits, so there are couple of cases: $a=0$ and $b=1$, $a=1$ and $b=0$, $a=5$ and $b=6$, $a=6$ and $b=5$... None of those cases satisfy what we want, so there are no $a$ and $b$.
Are those two ways to do it the right ones? I don't have any solutions therefore I'm not sure.
Thank you.
| One example is 729+492=1221
a=2,b=9
7 a b
+4 b a
1 a 2 1
Hint: b+a is not equal to 1. So b+a=11 hence the solution
| {
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If $85_b = 58_c$, what is the smallest possible value of $b$? I have searched and watched videos online and can't find a method to solve this problem:
If $85_b=58_c$ for some positive integer bases $b$ and $c$, what is the smallest possible value for $b$?
| $85_b = 58_c$ means $8b + 5 = 5 c + 8$. Thus, $b = (3+5c)/8$. This is an integer whenever $c = 8n + 1$ for some integer $n$. In this case, $b = 5n+1$. Now, since $8$ appears as a digit, we must have $b > 8$. The smallest integer of the form $5n+1$ that is larger than $8$ is $11$.
So the smallest solution is $b = 11$, $c = 17$, as $85_{11} = 58_{17} = 93_{10}$
| {
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Easy exponential equation but… $$\left(\sqrt{5+2\sqrt6}\right)^x-\left(\sqrt{5-2\sqrt6}\right)^x=96$$
I was trying to use that $$(5+2\sqrt6)(5-2\sqrt6)=1$$ but I end up with $$t^2-98t+1=0$$ which has some ugly solutions and I think I should have some nice ones there. Thx
| $$\left(\sqrt{5+2\sqrt{6}}\right)^x - \left(\sqrt{5-2\sqrt{6}}\right)^x = 96$$
Take natural log of both sides
$$\ln \left[ \left(\sqrt{5+2\sqrt{6}}\right)^x - \left(\sqrt{5-2\sqrt{6}}\right)^x \right]= \ln96$$
Use the $\ln(a+b) = \ln(a) + \ln(1+b/a)$ property
$$\ln \left(\sqrt{5+2\sqrt{6}}\right)^x + \ln \left(1-\dfrac{\left(\sqrt{5-2\sqrt{6}}\right)^x}{\left(\sqrt{5+2\sqrt{6}}\right)^x}\right)=\ln96$$
Take out the $x$ from the log
$$x \left[\ln\left(\sqrt{5+2\sqrt{6}}\right)\right] + x\left[\ln\left(1-\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}\right)\right] =\ln96$$
Factor the $x$ and isolate it
$$x=\dfrac {\ln96} {\ln\left(\sqrt{5+2\sqrt{6}}\right) + \ln\left(1-\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}\right)}$$
The rest is just computation
| {
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Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of the variables scares me.
Here what I've done.
Using the AM-GM inequality and the Schur's inequality, we have that
$$a^3 + b^3 + c^3 \ge 3abc \implies P \ge \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$$
$$\ge \dfrac{9}{2}[ab(a + b) + bc(b + c) + ca(c + a)] = \dfrac{9}{2}[ab(2 - c) + bc(2 - a) + ca(2 - b)]$$
$$\ge \dfrac{9}{2}[2(ab + bc + ca) - 3abc] \ge \dfrac{27}{2}[2\sqrt[\frac{3}{2}]{abc} - abc]$$
Let $abc = m \implies m \le \left(\dfrac{a + b + c}{3}\right)^3 = \dfrac{8}{27}$
The problem becomes
Find the minimum value of $P' = 2\sqrt[\frac{3}{2}]{m} - m$ when $ 0 < m \le \dfrac{8}{27}$.
which is invalid because there isn't a minimum with the given condition.
| Let $a=b=c=\frac{2}{3}$. Thus, $P=8.$
We'll prove that it's a minimal value of $P$.
Indeed, we need to prove that
$$\sum_{cyc}(4a^3+5abc)\geq(a+b+c)^3$$ or
$$3\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is true by Schur.
| {
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Product of pairwise sum is perfect square For which $n$ can we divide $1,2,\ldots,2n$ into $n$ pairs so that the product of the sum of the $n$ pairs is a perfect square?
If $n$ is even, this is possible: match the first number with the last, second number with the second-to-last, etc., so every pair has sum $2n+1$ and the product is $(2n+1)^n$. This is a perfect square because $n$ is even.
For $n=1$ it is clearly impossible, but for $n=3$ it is possible: $(1+5)(2+4)(3+6) = 18^2$.
| You're almost done just take for $n$ odd $(1+5)(2+4)(3+6)(7+2n)(8+2n-1)\cdots=6^2\cdot 3^2\cdot (2n+7)^{n-3}$
| {
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Proving $\int_0^{n^2} \left\lfloor\sqrt t\right\rfloor \,dt=\frac{n(n-1)(4n+1)}{6}$ I want to prove
$$\int_{0}^{n^{2}} \left\lfloor \sqrt{t} \right\rfloor \,dt=\frac{n(n-1)(4n+1)}{6}$$
Is it correct to say that $\left\lfloor \sqrt{t} \right\rfloor=\sqrt{(k-1)}$ and $(k-1)^{2} < t < k^{2}$, where
$$\int_{0}^{n^{2}} \left\lfloor \sqrt{t} \right\rfloor \,dt =\sum_{k=1}^{n} \sqrt{(k-1)} \cdot (k^2 - (k-1)^2)$$
I would know how to do the rest, but how did we know that $\left\lfloor \sqrt{t} \right\rfloor=\sqrt{(k-1)}$, is it something you assume depending on the function? Because for a previous function, there was
$$\int_{0}^{n} \left\lfloor t^{2} \right\rfloor \,dt$$
and it was set $\left\lfloor t^{2} \right\rfloor=(k-1)^2$, I would really appreciate if someone would kindly explain this to me.
| You have
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \sum_{k=0}^{n-1} \int_{k^2}^{(k+1)^2} \lfloor \sqrt{t} \rfloor \mathrm{dt}$$
But if $k^2 < t < (k+1)^2$, then $\lfloor \sqrt{t} \rfloor = k$. So
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \sum_{k=0}^{n-1} k((k+1)^2-k^2) = \sum_{k=0}^{n-1} k(2k+1) = 2 \frac{n(n-1)(2n-1)}{6}+\frac{(n-1)n}{2} $$
i.e.
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \frac{n(n-1)(4n+1)}{6}$$
| {
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Determine all positive powers. Determine all positive powers of the $5 \times 5$ matrix...
$$C=\begin{bmatrix}
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0 \\
0&0&1&0&0 \\
0&0&0&1&0
\end{bmatrix}$$
I know that taking powers of $C$ just shifts each $1$ down a row, replacing each $1$ with a $0$. The right most $1$ is shifted out of the matrix and replaced with a $0$. This happens for $C^2$, $C^3$, $C^4$, and $C^5$, where the sub-diagonal of $1's$ is shifted down until $C^5$ at which time the matrix becomes the $5 \times 5$ zero matrix.
This is the solution I came up with...
\begin{align}
C^2 &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0 \\
0&0&1&0&0
\end{bmatrix}, \\
C^3 &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0
\end{bmatrix}, \\
C^4 &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
1&0&0&0&0
\end{bmatrix}, \\
C^k &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0
\end{bmatrix}, k\geq5
\end{align}
Is there a more general way to express $C^2$, $C^3$, and $C^4$ without explicitly writing them out? Or asked another way, is it possible to express $C^k$ for all positive powers without having to explicitly write out $C^2$, $C^3$, and $C^4$?
| $C$ is the adjacency matrix of a directed graph $5 \to 4 \to 3 \to 2 \to 1$. $(C^k)_{ij}$ is then the number of paths of length $k$ starting at $i$ and ending at $j$. This is $1$ if $i-j = k$, otherwise $0$. Of course if $k \ge 5$ all entries are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
| With
$y = Ax^2 + Bx + C, \tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, \tag 2$
viz,
$y' = 2Ax + B, \tag 3$
$y'' = 2A, \tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; \tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, \tag 5$
from which we infer
$-2A = 1, \tag 6$
$2(A - B) = 0, \tag 7$
$2A + B - 2C = 0; \tag 8$
thus
$A = B = -\dfrac{1}{2}, \tag 9$
$C = -\dfrac{3}{4}, \tag{10}$
and of course
$y = -\dfrac{1}{2}x^2 - \dfrac{1}{2}x - \dfrac{3}{4}. \tag{11}$
We Check:
From (11),
$y' = -x - \dfrac{1}{2}, \tag{12}$
$y'' = -1, \tag{13}$
$y'' + y' - 2y$
$= -1 - x - \dfrac{1}{2} - 2(-\dfrac{1}{2}x^2 - \dfrac{1}{2}x - \dfrac{3}{4}) = -1 - x - \dfrac{1}{2} + x^2 + x + \dfrac{3}{2} = x^2. \tag{14}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to justify the convergence of series $\left(\frac{1}{2}\right)^{2(2k-1)}$? I see this formula somewhere in a book, though the book doesn't provide the justification.
$$
\sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{2(2k-1)}
= \frac{4}{15}
$$
Any clue would be appreciated.
| Here is a way you may understand:
For any $x\in \Bbb R$ s.t. $|x|<1$ we have $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$
Proof is here.
Now in your case, let $\displaystyle S=\sum_{k=1}^\infty \frac{1}{4^{2k-1}}=\frac{1}{4}+\frac{1}{4^3}+\frac{1}{4^5}+\dots=\frac{1}{4}(1+\frac{1}{4^2}+\frac{1}{4^4}+\dots)=\frac{1}{4}\sum_{k=0}^\infty (\frac{1}{4^2})^k$
Now by above formula, [by putting $x=\frac{1}{4^2}$] $$S=\frac{1}{4}\frac{1}{1-\frac{1}{4^2}}=\frac{4}{15}$$
Hope this works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x - 3)^2 \implies a + b = \dfrac{a^2 - b^2}{a + b} = \dfrac{2x - 1}{(x - 3)^2}$.
$\left\{
\begin{align}
a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 106x + 80}{2(x - 3)^2}\\
b = \dfrac{(a + b) - (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 110x + 82}{2(x - 3)^2}
\end{align}
\right.$
| $ t = \sqrt{3-x}\geq 0$, then we get $$\sqrt{5-t^2}-t=t^4$$
so $$5-t^2 = t^2+2t^5+t^8$$
so we have $$t^8+2t^5+2t^2-5=0$$
Since $f(t)=t^8+2t^5+2t^2$ is strictly increasing for positive $t$, given equation has at most one real (positive) solution and that is $t=1$, or $x= 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$
If $a,b,c\in\mathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is
what i try
Cauchy schwarz inequality
$$(a^2+(2b)^2+c^2)(2^2+\frac{1}{2^2}+3^2)\geq (2a+b+3c)^2$$
How do i solve it without Cauchy schwarz inequality Help me please
| Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)\in P$ for $t\in\Bbb R$, then we get $53t=40$ and $x=\frac{40}{53}(4,1,6)$ follows. Hence we obtain
$$
a^2+(b')^2+c^2 \ge |x|^2=\frac{1600}{53}
$$ is the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Prove that there exists an integer $ x$ such that $ \frac{x^2-2}{p}$ is the square of an integer
Let $ p\equiv -1\pmod 8$ be a prime number. Prove that there
exists an integer $ x$ such that $ \frac{x^2-2}{p}$ is the square of
an integer
This problem is from :
AOPS,there are some In the discussion, there was one person who did something worth doing, but it wasn't right.
Here's his answer.
Let $(a,b)$ be a nontrivial solution to $x^2-py^2=1$. Then for $(c,d)=(2a,2b)$, $c^2-pd^2=4 \Longrightarrow c^2+d^2 \equiv 4 \pmod{8}$ $\Longrightarrow 4|c,d \Longrightarrow gcd(c-2,c+2)=2$. But we have $(c-2)(c+2)=pd^2$, so either:
i) $c-2=2pu^2$, $c+2=2v^2$, or
ii) $c-2=2u^2$, $c+2=2pv^2$, for some positive integers $u$, $v$.
ii) implies $ u^2-pv^2=-2 \Longrightarrow u^2+v^2+2 \equiv 0 \pmod{8}$, which is impossible, so we must have i).
But i) implies $v^2-pu^2=2 \Longleftrightarrow \frac{v^2-2}{p} = u^2$, which is a solution to the desired equation
As the last person pointed out, there was an important error in the answer：
$c^2-pd^2=4 \Longrightarrow c^2+d^2 \equiv 4 \pmod{8}$ $\Longrightarrow 4|c$ or $4|d$. So maybe $gcd(c-2,c+2)=4$.
so this case How to prove it
| Let $p \equiv 7 \pmod 8$ be prime. Let $u,v > 0$ give the smallest positive solution to $u^2 - p v^2 = 1.$ Indeed $u \geq 3.$ Then $u^2 - 1 = p v^2,$ so $(u+1)(u-1) = p v^2.$ As $\gcd(u+1, u-1)$ is either $1$ or $2,$ we have four possibilities with some $\gcd(g,h) = 1:$
(I)
$$ u+1 = p g^2 \; , \; \; u-1 = h^2 $$
Subtracting gives $2 = p g^2 - h^2,$ or $h^2 - p g^2 = -2,$ which is impossible as $-2$ is not a quadratic residue $\pmod p.$
(II)
$$ u+1 = g^2 \; , \; \; u-1 = ph^2 $$
Subtracting gives $2 = g^2 - ph^2.$
(III)
$$ u+1 = 2 p g^2 \; , \; \; u-1 = 2h^2 $$
Subtracting gives $2 = 2p g^2 - 2h^2,$ or $h^2 - p g^2 = -1,$ which is impossible as $-1$ is not a quadratic residue $\pmod p.$
(IV)
$$ u+1 = 2 g^2 \; , \; \; u-1 = 2ph^2 $$
Subtracting gives $2 = 2 g^2 - 2ph^2,$ or $g^2 - p h^2 = 1.$
The hypothesis was that $u^2 - p v^2 = 1$ was the solution with smallest positive entries, so we have a contradiction. Note $2 g^2 = u+1,$ so $g < u,$ as $u \geq 3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Series $\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$ Alright, this is another problem that I have been stuck on. The goal is to determine whether it is convergent or divergent.
$$\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$$
So to start off, Integral Test seems rough as the denominator is not factorable for partial fraction decomposition.
So then, I tried Direct Comparison Theorem, but...
$$\frac{1}{n^3} < \frac{n^2 - 5n}{n^3 + n + 1}[n > 5]$$
Although not for the intervals [0, 5].
$$\frac{1}{n^3} > \frac{n^2 - 5n}{n^3 + n + 1}[0<n<5]$$
So yeah, that's kind of confusing. Especially since the problem starts at n = 1 instead of n = 5.
However, I know that, by p-series
$$\sum_{n=1}^{\infty} \frac{1}{n^3} --> converges $$
And if the smaller value converges, then Direct Comparison Theorem tells us nothing.
So I decided to try the Limit Comparison Theorem:
$$b_n = \frac{1}{n^3} $$
$$\lim_{n\to0} \frac{n^2 - 5n}{n^3+n+1}*\frac{n^3}{1} = \lim_{n\to0} \frac{n^6 - 5n^4}{n^3 + n + 1} = {\infty}$$
So if bn is convergent by p series, but the limit is divergent, then LCT is useless.
So, now my question is where did I go wrong in attempting to prove convergence/divergence?
| For $n \ge 5$, $$\frac{n^2 - 5n}{n^3 + n + 1} > \frac{(n^2 + 2n + 1) - (7n + 7) + 6}{n^3 + 3n^2 + 3n + 1} > \frac{(n+1)^2 - 7(n+1)}{(n+1)^3} = \frac{1}{n+1} - \frac{7}{(n+1)^2}.$$ Since $$\sum_{n=5}^\infty \frac{1}{n+1}$$ does not converge but $$\sum_{n=5}^\infty \frac{7}{(n+1)^2}$$ is convergent, the original sum does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.
Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.
Approach:
*
*I defined a function $f: \mathbb{R} \to \mathbb{R}$ with $$ f(x) = \cosh(x \log{2}) - \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b}.
$$ In order to show the inequality, it suffices to show $f(x) \geq 0$ for all $x \geq 0$.
*I tried to plot the function for some chosen parameters like $a=1$ and $b=2$. In all those cases the function was non-negative, so I suppose that this inequality is true.
*It is $f(0) = 1-1 = 0 \geq 0$.
*Now $f$ is differentiable, so I computed
$$ f'(x) = \frac{\log{2}}{2(a+b)} \big(a (2^x - 2^{-x} - 2^{x+1}) + b(2^x-2^{-x}+2^{-x+1} ) \big). $$
*I would be done if I could show that $f'(x) \geq 0$ for all $x \geq 0$, so $f$ is monotonically increasing and we get our desired result. However, I can not see how this can be shown.
Could you please help me with this problem? That would be nice, thank you in advance!
| Big wall of algebra incoming:
\begin{align*}
\frac{a\cdot2^{x}+b\cdot2^{-x}}{a+b}\le\cosh\left(x\ln2\right)&\implies\frac{a\cdot2^{x}+\frac{b}{2^{x}}}{a+b}\le\frac{2^{x}+2^{-x}}{2}\\
&\implies\frac{a\cdot2^{x}+\frac{b}{2^{x}}}{a+b}\le\frac{2^{x}+\frac{1}{2^{x}}}{2}\\
&\implies\frac{a\cdot2^{2x}+b}{2^{x}\left(a+b\right)}\le\frac{2^{2x}+1}{2\cdot 2^{x}}\\
&\implies\frac{2^{2x}+b}{a+b}\le\frac{2^{2x}+1}{2}\\
&\implies\frac{a\cdot4^{x}+b}{a+b}\le\frac{4^{x}+1}{2}\\
&\implies\frac{a\cdot4^{x}+b}{a+b}-\frac{4^{x}+1}{2}\le0\\
&\implies\frac{2\left(a\cdot4^{x}+b\right)-\left(a+b\right)\left(4^{x}+1\right)}{2\left(a+b\right)}\le0\\
&\implies\frac{2a\cdot4^{x}+2b-a\cdot4^{x}-b\cdot4^{x}-a-b}{2\left(a+b\right)}\le0\\
&\implies\frac{4^{x}\left(2a-a-b\right)+b-a}{2\left(a+b\right)}\le0\\
&\implies\frac{4^{x}\left(a-b\right)+b-a}{2\left(a+b\right)}\le0\\
&\implies\frac{4^{x}\left(a-b\right)-\left(a-b\right)}{2\left(a+b\right)}\le0\\
&\implies\frac{\left(a-b\right)\left(4^{x}-1\right)}{2\left(a+b\right)}\le0\\
&\implies\left(a-b\right)\left(4^{x}-1\right)\le0\\
&\implies4^{x}-1\ge0\text{ since }a-b<0\\
&\implies4^{x}\ge1,
\end{align*}
which is true for all $x\ge0$.
If someone can find a pretty way to format this, then feel free to edit it.
| {
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"url": "https://math.stackexchange.com/questions/3167725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partial fraction decomposition of even function I need to do partial fraction decomposition of this function (to solve its integral):
$\frac{t^2}{t^4+4}$
Since $t^4+4=(t^2+2t+2)(t^2-2t+2)$ I would do:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}+C\frac{2t-2}{t^2-2t+2}+D\frac{1}{t^2-2t+2}$
And then continue with calculations.
According to the teacher's notes though, I can instead write it like this because the function is even:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}-A\frac{2t-2}{t^2-2t+2}+B\frac{1}{t^2-2t+2}$
But I can't understand how to get it, knowing that the function is even.
Can you please help me?
| Let $t=2u$
$$\dfrac{2u^2}{u^4+1}=\dfrac{u^2-1+u^2+1}{u^4+1}=\dfrac{1-1/u^2}{(u+1/u)^2-2}+\dfrac{1+/u^2}{(u-1/u)^2+2}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bounded recursive sequence - Proof by induction
Given the sequence $(x_n)$ defined by
\begin{cases}
x_1 &= 1\\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right),
\end{cases}
prove that $1 \leq x_n \leq \frac{3}{2}, \forall n \in N$.
I verified the base case for $n=1$ and $n=2$.
Assumed that the boundaries hold for all $k \leq n$. Used the induction hypothesis to show that $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{1}{2} \leq \frac{x_n}{2} \leq \frac{3}{4} $ and $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{2}{3} \leq \frac{1}{x_n} \leq 1$.
Adding term by term I got $1 \leq \frac{7}{6} \leq \frac{1}{2}(x_n + \frac{2}{x_n}) \leq \frac{7}{4} $.
Is my reasoning correct? How could I show that $\frac{3}{2}$ is also an upper boundary, since it is smaller than $\frac{7}{4}$?
PS: I found similar questions for which $x_n \leq \sqrt{2}$.
| Let $a \gt 0$.
Exercise 1: $\left(\frac{a}{2} + \frac{1}{a}\right)^2 = 2 \Leftarrow\Rightarrow a = \sqrt 2$.
Exercise 2: $\left(\frac{a}{2} + \frac{1}{a}\right)^2 \ge 2$.
Hint: Apply the quadratic formula to a quartic polynomial.
Using the exercises, if $n \gt 1$ then $x_n \gt \sqrt 2 \gt 1$, and we get 'one half' (and something stronger) of what the OP was looking for.
Since $x_2 = \frac{3}{2}$, $x_2 \le \frac{3}{2}$.
Assume for some fixed $k \gt 2$ that $x_k \leq \frac{3}{2}$.
We will show that $x_{k+1} \leq \frac{3}{2}$ is also true, allowing us to employ induction and find the bounds for the OP's sequence.
In general, if $\sqrt 2 \leq a \leq \frac{3}{2}$ then
$$\tag 1 \frac{a}{2} \leq \frac{3}{4}$$
and
$$\tag 2 \frac{1}{a} \leq \frac{1}{\sqrt 2}$$
So
$$\tag 3 \frac{1}{2}\left(a + \frac{2}{a}\right) = \frac{a}{2} + \frac{1}{a} \le \frac{3}{4} + \frac{1}{\sqrt 2} \lt \frac{3}{2}$$
Applying induction we conclude that for $n \gt 1$,
$$ \sqrt 2 \le x_n \le \frac{3}{2}$$
and we can also state that for $n \ge 1$ that
$$ 1 \le x_n \le \frac{3}{2}$$
With the above work completed and working out some terms of the sequence $x_n$, an observant analyst would be naturally drawn to the question
If $\delta \gt 0$ and $a = \sqrt 2 + \delta$, can we assert that $\frac{a}{2} + \frac{1}{a}$ is less than $a$?
The answer is yes and is demonstrated using simple algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $f(p) = (\frac{3p-3}{p^2-1}) + (\frac{1}{p-1}) \times \sum_{q=3}^{l(p), with \ q\ prime} (\frac{q+1}{q-1})$ is always <= 1? Let function $l(p)$ be defined as the largest prime number less than $p$.
For example: $l(7)=5$, $l(11)=7$, $l(17)=13$
Let the function $f(p)$ be defined as follows:
$$f(p) = \left(\frac{3p-3}{p^2-1}\right) + \left(\frac{1}{p-1}\right) \times \sum_{q=3}^{l(p), with \ q\ prime} \left(\frac{q+1}{q-1}\right)$$
where the summation is over prime numbers only. I.e. $q$ is always a prime.
Below are some examples of the values for $f(p)$:
$$f(3) = \frac{9-3}{9-1} + \frac{1}{3-1} \times 0$$
$$f(3) = \frac{6}{8}$$
$$f(5) = \frac{15-3}{25-1} + \frac{1}{5-1} \times \frac{3+1}{3-1}$$
$$f(5) = \frac{12}{24} + \frac{1}{4} \times 2$$
$$f(5) = 1$$
$$f(7) = \frac{21-3}{49-1} + \frac{1}{7-1} \times \left(\frac{3+1}{3-1} + \frac{5+1}{5-1}\right)$$
$$f(7) = \frac{18}{48} + \frac{1}{6} \times \left(2 + \frac {3}{2}\right)$$
$f(7) = 0.958333$
$f(11) = 0.733333$
$f(13) = 0.717063$
How to prove that $f(p)$ is always less than or equal to 1 for all prime numbers p?
Below is a graph of the values up to $p=101$. The curve appears to decline but there are exceptions, $f(31) > f(29)$ and $f(43) > f(41)$.
| The essence of this problem is finding a good upper bound for the sum over the primes. Let $\pi(n)$ denote the prime-counting function, i.e. $\pi(n)$ is the number of primes less than or equal to $n$, and let $H_n:=\sum_{k=1}^n\tfrac1k$ denote the $n$-th harmonic sum. Then
\begin{eqnarray*}
\sum_{\substack{q=3\\q\ \text{prime}}}^{l(p)}
\frac{q+1}{q-1}
&=&\sum_{\substack{q=3\\q\ \text{prime}}}^{p-1}\left(1+\frac{2}{q-1}\right)\\
&=&\left(\pi(p-1)-1\right)
+2\sum_{\substack{q=3\\q\ \text{prime}}}^{p-1}\frac{1}{q-1}\\
&\leq&\pi(p-1)-1+2\sum_{k=2}^{p-2}\frac1k\\
&=&\pi(p-1)-1+2(H_{p-2}-1)\\
&=&\pi(p-1)+2H_{p-1}-3.
\end{eqnarray*}
For all $n>1$ we have the well-known upper bounds
$$\pi(n)\leq\frac{n}{\ln(n)}\left(1+\frac{3}{2\log(n)}\right)
\qquad\text{ and }\qquad
H_n\leq\ln(n+1).$$
It follows that
\begin{eqnarray*}
f(p)
&=&\frac{3(p-1)}{p^2-1}+\frac{1}{p-1}\sum_{\substack{q=3\\q\ \text{prime}}}^{p-1}
\frac{q+1}{q-1}\\
&\leq&\frac{3(p-1)}{p^2-1}+\frac{1}{p-1}(\pi(p-1)+2H_{p-2}-3)\\
&\leq&\frac{3}{p+1}+\frac{1}{p-1}
\left(\frac{p-1}{\ln(p-1)}\left(1+\frac{3}{2\ln(p-1)}\right)
+\ln(p-1)-3\right)\\
&=&\frac{3}{p+1}+\frac{1}{\ln(p-1)}\left(1+\frac{3}{2\ln(p-1)}\right)
+\frac{\ln(p-1)}{p-1}-\frac{3}{p-1}.
\end{eqnarray*}
It is an exercise in basic calculus to show that the latter is strictly decreasing for $p\geq5$, and it is an exercise in basic algebra to show that the latter is less than $1$ for $p=5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Remainder of polynomial product, CRT solution via Bezout Given:
$$f(x) \pmod{x^2 + 4} = 2x + 1$$
$$f(x) \pmod{x^2 + 6} = 6x - 1$$
Define r(x) as:
$$f(x) \pmod{(x^2 + 4)(x^2+6)} = r(x)$$
What is $r(4)$?
The 3 equations can be restated as quotient · divisor + remainder:
$$f(x) = a(x)(x^2 + 4) + 2x + 1 $$
$$f(x) = b(x)(x^2 + 6) + 6x - 1 $$
$$f(x) = c(x)(x^2 + 4)(x^2 + 6) + r(x) = c(x)(x^4 + 10x^2 + 24) + r(x) $$
Note this isn't homework, and there are several different methods that can be used to solve this, one of which produces an f(x) based on the 2 given remainders, two of which produce r(x) without having to determine f(x), and a slight variation that produces r(4). I've looked at other polynomial remainder questions here at SE, but those did not involve all of the methods that I'm aware of that can be used to solve this particular problem, so I thought it might be interesting for others here at SE. Some, but not all of the methods are related to Chinese remainder theorem, so I wasn't sure if should also tag this question with Chinese remainder theorem. I found this problem at another forum site, so I'm not sure of the origins of this particular problem.
| Hint
Observe that $\gcd(x^2+4, x^2+6)=1$ and
$$\frac{1}{2}(x^2+6)-\frac{1}{2}(x^2+4)=1.$$
Now apply Chinese remainder theorem to the system
\begin{align*}
f(x) & \equiv 2x+1 \pmod{x^2+4}\\
f(x) & \equiv 6x-1 \pmod{x^2+6}
\end{align*}
To get something like:
$$f(x) \equiv \underbrace{(2x+1)(\ldots) + (6x-1)(\ldots)}_{r(x)} \pmod{(x^2+4)(x^2+6)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$ sin^2 x = \frac{1}{2}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sin x= \lvert\frac{1}{\sqrt2}\rvert$$
$$ \sin x= \frac{\sqrt2}{2} ; \sin x= -\frac{\sqrt2}{2}$$
So if I look for all the values of $x$ that solve the above I should get not only $ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°) but also $ x= \frac{3\pi}{4}$ (135°) ; $ x= \frac{7 \pi}{4}$ (315°).
What am I doing wrong?
| The issue :
when you write out $\sin x = \cos x$ implies $\sin^2 x = \cos^2 x$, there is no problem with the implication. However, an implication leads to a weaker statement, which may admit "extraneous" solutions. For example, the above implication is got by multiplying by squaring both sides, but then that ends up including the case when $\sin x = -\cos x$.
Therefore you need to ensure that your implications are both ways, so that the solutions of both equations are retained, or you have to eliminate extraneous solutions once you reach the end.
The two extraneous solutions $\frac{3\pi}4 , \frac{7\pi}4$ obtained are of the equation $\sin x = -\cos x$, which got included as solutions of the equation $\sin^2 x = \cos^2 x$ when you moved to this statement from the one you were given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
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