Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.
| Use definition of derivation:
\begin{align}
\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}
&=
\frac{\frac{\sqrt{1+x^2}-1}{x} - \frac{\sqrt {1+x}-1}{x}}{\frac{\sqrt {1+x^3}-1}{x} - \frac{\sqrt {{1+x}}-1}{x}}\\
&=
\frac{(\sqrt{1+x^2})'|_{x=0} - (\sqrt {1+x})'|_{x=0} }{(\sqrt {1+x^3})'|_{x=0} - (\sqrt {{1+x}})'|_{x=0} }\\
&=
\dfrac{0-\frac12}{0-\frac12}\\
&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably I am mistaken in solving.
| You can assume $a+b+c+d=1$, so we need to prove $$E:=\sum {1\over 1-x} \geq 16/3$$
Now, since for $x\in (0,1)$ we have $${1\over 1-x}\geq {16x\over 9}+{8\over 9}$$ we get $$E\geq {16\over 9}(a+b+c+d)+4{8\over 9} = {48\over 9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Impulse train: why an indeterminate result? I have an impulse train given by
$$\frac{1}{R+1}+\frac{\sum_{k=1}^R \cos(\frac{2k \pi x}{R+1})}{R+1}$$
It seems obvious to me that, for $x=0$, the function returns $1$. This is because $\cos (0)=1$, and we therefore end up with $\frac{1}{R+1}+\frac{R}{R+1}=\frac{R+1}{R+1}=1$.
However, my math software (Mathematica) gives an indeterminate result at $x=0$. Usually this means there is a division by $0$ somewhere. But I can't see any reason for this function to produce an indeterminate result. (This is about the maths of the situation, not the way the software works.)
Can anyone explain?
| I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$\sum_{k=1}^R \cos \left(\frac{2 k\pi x}{R+1}\right)=\frac{1}{2} \left(\sin \left(\frac{\pi (2 R+1) x}{R+1}\right) \csc \left(\frac{\pi
x}{R+1}\right)-1\right)$$ making
$$\frac{1}{R+1}+\frac{\sum_{k=1}^R \cos(\frac{2k \pi x}{R+1})}{R+1}=\frac{\sin \left(\frac{\pi (2 R+1) x}{R+1}\right) \csc \left(\frac{\pi
x}{R+1}\right)+1}{2 (R+1)}$$ Probably, the indetermination comes from the limit of $\csc(t)$ when $t\to 0$ while, using Taylor expansion
$$\sin \left(\frac{\pi (2 R+1) x}{R+1}\right)=\frac{( \pi (2 R+1) ) }{R+1}x-\frac{( \pi (2R+1) )^3 }{6
(R+1)^3}x^3+O\left(x^5\right)$$
$$\csc \left(\frac{\pi x}{R+1}\right)=\frac{R+1}{\pi x}+\frac{\pi x}{6 (R+1)}+\frac{7 \pi ^3 x^3}{360
(R+1)^3}+O\left(x^5\right)$$
$$\sin \left(\frac{\pi (2 R+1) x}{R+1}\right)\,\csc \left(\frac{\pi x}{R+1}\right)=(2 R+1)-\frac{2 \pi ^2 R (2 R+1) }{3 (R+1)}x^2+O\left(x^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$\frac{bc}{x-a} + \frac{ab}{x-c} + \frac{ac}{x-b} \leq \frac{1}{2} \Bigl(x\Bigl)$$
This didn't help so instead I wrote it as $$2\Biggl(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}\Biggl ) \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
Not sure what to to do next.
Any help would be appreciated.
| Hint: Remember that $HM(a, b)$ can also be written as $\dfrac {2ab}{a+b}$, which looks a lot like the LHS...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$
For any acute angled triangle ABC , find the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$ .
Attempt:
As $A+B+C=\pi$
$C=\pi -(A+B)$
After differentiating it
$dA+dB+dC=0$
Now :
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin (A+B)}{\pi-(A+B)}$
$(\frac{A\cos A-\sin A}{A^2})dA + (\frac{B\cos B- \sin B}{B^2})dB + (\frac{C\cos C-\sin c}{C^2})dC =0$
But could not solve further .
| Here's one approach to doing this:
*
*Take $f(x)=\frac{\sin x}{x}$ and show that $f''(x)<0$ for $x\in(0,\pi/2)$.
*Use Jensen's inequality to conclude that for any $A,B,C\in(0,\pi/2)$ with $A+B+C$ fixed, $f(A)+f(B)+f(C)$ is maximised when they are all equal.
(Further hint for 1: write $f''(x)$ as a fraction and differentiate the numerator.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomial problem with unknown coefficients $a, b, c$ $p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.
Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-\frac{a^2}{a+1}$ and then getting $c=-\frac{a^4}{a+1}$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.
| Put $n=a+1$, then: $$b={-a^2\over a+1} = -{(n-1)^2\over n} = -{n^2-2n+1\over n} = -n+2-{1\over n} \in\mathbb{Z}$$
So $$n\mid 1\implies n =\pm 1 \implies a=0 \vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In how many ways a student can get $2m $ Marks An examination contains four Question papers each paper carrying maximum marks as $m$. Find number of ways a student appearing for all the four papers gets a total of $2m$ Marks.
I used generating Polynomial method that is to find coefficient of $x^{2m}$ in
$$(1+x+x^2+\cdots+x^m)^4$$ which is
$$(1-x^{m+1})^4(1-x)^{-4}$$ which gives the coefficient of $x^{2m}$ as
$$\binom{2m+3}{3}-4 \times \binom{m+2}{m-1}$$
But the answer is just $\binom{2m+3}{3}$. What went wrong?
| Let us consider the marks in all the $4$ papers as $a,b,c,d$; $0\le a,b,c,d\le m$
$$a+b+c+d=2m$$
Now we need to find the integer solution of the above equation in the given condition.
Number of solutions $=\dbinom{2m+4-1}{4-1}=\dbinom{2m+3}{3}$, but note that this also contains those solutions in which any variable is grater than $m$.
Hence we need to subtract those solutions for $a\ge m+1$ and let $t=a-(m=1),t\ge0$
$a+b+c+d=2m\implies t+b+c+d=m-1$.$$\dbinom{m-1+4-1}{4-1}=\dbinom{m+2}{3}$$and for $b\ge m+1,c\ge m+1, d\ge m+1$. Finally, $\dbinom{2m+3}{3}-4\times \dbinom{m+2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A problem related with limit and continuity.
What is the limit as $x$ tends to $1$ in the function $$\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}\quad?$$
I tried factoring out the $x$ term then again rewriting the factored term. But nothing came out at all. Could anyone help me out?
| Define
$$f(x)=x+x^2+x^3+\cdots+x^n.$$
Then $$f'(x)=1+2x+3x^2+\cdots+nx^{n-1}.$$
It follows that
\begin{align*}
\lim_{x \to 1}\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}&=\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}=f'(1)\\&=1+2+3+\cdots+n\\&=\frac{n(n+1)}{2}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n-1}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}\right)<\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}$$
Please check my solution for me and give me some idea.
| I think your way gets something wrong.
By your work:
For all $n\geq2$ we obtain:
$$\frac{1}{n\sqrt{n}}<\left(\frac{1}{\sqrt{n-1}}-\frac{1}
{\sqrt{n}}\right)\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}=$$
$$=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\sqrt{\frac{n}{n-1}}\right)\leq\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)(1+\sqrt2).$$
Thus, $$\sum_{k=1}^n\frac{1}{k\sqrt{k}}=1+\sum_{k=2}^n\frac{1}{k\sqrt{k}}\leq1+(1+\sqrt2)\left(1-\frac{1}{\sqrt{n}}\right).$$
Id est, it's enough to prove that
$$1+(1+\sqrt2)\left(1-\frac{1}{\sqrt{n}}\right)<2\sqrt2,$$ which is wrong for $n\rightarrow+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Incircle bisectors and related measures
This question was inspired by
in-triangle-abc-d-is-a-point-on-ac...,
show-that-am2-pp-a.
Cevians $|AD_a|=d_a$,
$|BD_b|=d_b$, $|CD_c|=d_c$
divide $\triangle ABC$ into three pairs of triangles,
($\triangle ABD_a$, $\triangle AD_aC$),
($\triangle BCD_b$, $\triangle BD_bA$),
and
($\triangle CAD_c$, $\triangle CD_cB$)
in such a way that
incircles for each pair have the same inradius,
$r_a,r_b$ and $r_c$, respectively
(in the image the circle centers are marked
with the corresponding radius).
Is there a known name for such cevians?
Something like "incircle bisectors"?
These three cevians, which lengths are defined as
\begin{align}
d_a&=\sqrt{\rho(\rho-a)}
,\quad
d_b=\sqrt{\rho(\rho-b)}
,\quad
d_c=\sqrt{\rho(\rho-c)}
\tag{1}\label{1}
\end{align}
uniquely define
the sides of the triangle:
\begin{align}
a&=\frac{d_b^2+d_c^2}{\sqrt{d_a^2+d_b^2+d_c^2}}
,\\
b&=\frac{d_c^2+d_a^2}{\sqrt{d_a^2+d_b^2+d_c^2}}
,\\
c&=\frac{d_a^2+d_b^2}{\sqrt{d_a^2+d_b^2+d_c^2}}
\tag{2}\label{2}
.
\end{align}
They provide a nice relations to semiperimeter
$\rho=\tfrac12(a+b+c)$
of $\triangle ABC$:
\begin{align}
\rho^2&=d_a^2+d_b^2+d_c^2
\tag{3}\label{3}
,
\end{align}
area
\begin{align}
S_{\triangle ABC}
&=
\frac{d_a d_b d_c}{\sqrt{d_a^2+d_b^2+d_c^2}}
\tag{4}\label{4}
\end{align}
and inradius $r$ of the triangle $ABC$:
\begin{align}
r&=
\frac{d_a d_b d_c}{d_a^2+d_b^2+d_c^2}
\tag{5}\label{5}
\end{align}
as well as the circumradius
\begin{align}
R&=\frac{(d_a^2+d_b^2)(d_b^2+d_c^2)(d_c^2+d_a^2)}
{4d_a d_b d_c(d_a^2+d_b^2+d_c^2)}
\tag{6}\label{6}
\end{align}
Unfortunately, in general, as the image illustrates,
these cevians are not concurrent.
Next, these "incircle bisectors"
introduce three inradii
\begin{align}
r_a&=\frac{r}{1+\sqrt{1-\frac{a}\rho}}
,\\
r_b&=\frac{r}{1+\sqrt{1-\frac{b}\rho}}
,\\
r_c&=\frac{r}{1+\sqrt{1-\frac{c}\rho}}
\tag{7}\label{7}
,
\end{align}
which also demonstrate plenty of nice relations, for example,
\begin{align}
\left(\frac{r}r_a-1\right)^2
+\left(\frac{r}r_b-1\right)^2
+\left(\frac{r}r_c-1\right)^2
&=1
\tag{8}\label{8}
,\\
\left(\frac{r}r_a-1\right)
\left(\frac{r}r_b-1\right)
\left(\frac{r}r_c-1\right)
&=\frac{r}{\rho}
\tag{9}\label{9}
,
\end{align}
\begin{align}
\rho&=
\frac{r r_a r_b r_c}{(r-r_a)(r-r_b)(r-r_c)}
\tag{10}\label{10}
,\\
d_a&=\rho\left(\frac{r}{r_a}-1\right)
\tag{11}\label{11}
,\\
d_a&=
\frac{r r_b r_c}{(r-r_b)(r-r_c)}
\tag{12}\label{12}
.
\end{align}
Angles at the feet of the "incircle bisectors"
$\delta_a=\angle AD_a C$,
$\delta_b=\angle BD_b A$,
$\delta_c=\angle CD_c B$,
defined as
\begin{align}
\cos\delta_a&=\frac{c-b}a
,\\
\cos\delta_b&=\frac{a-c}b
,\\
\cos\delta_c&=\frac{b-a}c
\tag{13}\label{13}
,
\end{align}
also provide some interesting identities, like
\begin{align}
\cos\delta_a+\cos\delta_b+\cos\delta_c&=
\frac{(a-b)(b-c)(c-a)}{abc}
\\
&=\frac{a}b+\frac{b}c+\frac{c}a-\frac{a}c-\frac{c}b-\frac{b}a
\tag{14}\label{14}
,\\
\cos\delta_a\cos\delta_b\cos\delta_c&=
-(\cos\delta_a+\cos\delta_b+\cos\delta_c)
\\
&=\frac{(a-c)(c-b)(b-a)}{abc}
\tag{15}\label{15}
.
\end{align}
\begin{align}
\cos\delta_a\cos\delta_b+\cos\delta_b\cos\delta_c
+\cos\delta_c\cos\delta_a
&=\frac{a}b+\frac{b}c+\frac{c}a+
\frac{a}c+\frac{c}b+\frac{b}a
-\left(\frac{a^3+b^3+c^3}{abc}\right)-3
\tag{16}\label{16}
\\
&=\frac{2r}R-1
\tag{17}\label{17}
,\\
\sin\delta_a\sin\delta_b\sin\delta_c
&=\frac{2r}R
\tag{18}\label{18}
,\\
S&=\tfrac12\,\rho\, R\sin\delta_a\sin\delta_b\sin\delta_c
\tag{19}\label{19}
.
\end{align}
Two of them define the third one, like the angles of triangle:
\begin{align}
\cos\delta_c&=
-\frac{\cos\delta_a+\cos\delta_b}{1+\cos\delta_a\cos\delta_b}
\tag{20}\label{20}
.
\end{align}
Are there any known references?
Famous
Baker's collection of formulae for the area of a plane triangle
does not mention these cevians
and neither do the [wiki entries on
wiki-Triangle,
wiki-Triangle_inequalities
and
Cevian.
Search on Google Scholar
was also futile (did I missed something trivial?).
Also I can not remember seen any reference
of these parameters used as a triplets,
only a single instance, without any special name,
like in already mentioned
in-triangle-abc-d-is-a-point-on-ac...,
show-that-am2-pp-a.
Summarizing the question:
1) Are there any known references,
where such cevians and identities are discussed/mentioned?
2) Is there a known name/notation for such cevians?
Something like "incircle bisectors"?
| Accidentally, I've found this open-access reference:
Yiu, Paul. The Congruent-Incircle Cevians of a Triangle. Missouri J.
Math. Sci. 15 (2003), no. 1, 21--32. doi:10.35834/2003/1501021.
https://projecteuclid.org/euclid.mjms/1567216820
For such cevians
they use a term "the congruent-incircle cevians of a triangle".
| {
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"url": "https://math.stackexchange.com/questions/2876870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to find $ f(x)$ if $f(1-f(x))=x$ for all $x$ $\in \mathbb{R}$ How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?
I have already recognized one problem caused from this: it
follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?
Best regards and thanks,
John
| This is a partial answer.
We know that $f(x)$ is invertible, because $f^{-1}(x)=1-f(x),$ from the original; from here we get the very interesting relationship of $f(x)+f^{-1}(x)=1.$ Suppose we try to find out what $f(0)$ is (set it equal to $a$). By repeated alternating applications of $f$ and the equation $f^{-1}(x)=1-f(x),$ we wind up with the following interesting table:
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
0 &a &1-a \\ \hline
1-a &0 &1 \\ \hline
1 &1-a &a \\ \hline
a &1 &0
\end{array}
$$
One more step gets you where you started. In studying this table, we see that if $f$ and $f^{-1}$ are to be well-defined, we cannot have $a=0, 1/2,$ or $1$. We get a similar table if we start off with $x=-1:$
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
-1 &b &1-b \\ \hline
1-b &-1 &2 \\ \hline
2 &1-b &b \\ \hline
b &2 &-1
\end{array}
$$
From here we find that $b\not=-1, -3, 2, 1/2.$ Yet another table generates when we start with $x=-2:$
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
-2 &c &1-c \\ \hline
1-c &-2 &3 \\ \hline
3 &1-c &c \\ \hline
c &3 &-2
\end{array}
$$
From this we get that $c\not=3, 1/2, -2.$ This generalizes to the following table:
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
1-n &m &1-m \\ \hline
1-m &1-n &n \\ \hline
n &1-m &m \\ \hline
m &n &1-n
\end{array}
$$
From here, we can see that $n=1/2$ forces $m=1/2,$ which would be consistent in this table. So $f(1/2)=1/2.$
Moving on, we can see that the following are true:
\begin{align*}
f(1-x)&=y \\
f(1-y)&=1-x \\
f(x)&=1-y \\
f(y)&=x.
\end{align*}
Combining two of these equations yields $f(1-x)=1-f(x)$. Differentiating yields $f'(1-x)=f'(x).$
These are mostly negative results, obviously. My hope is that perhaps these ideas might spur someone else on to a solution.
| {
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Which is the value of $t$ ? It is given $t\in \{0,1,2,3,4\}$, a group $G$ and an element $a$ of $G$ such that $a^8=a^t$ and $a^{88}=a^{74}$ in $G$. If $a^2\neq a^0$ which is $t$ ?
For $a\neq 0$ we have the following:
\begin{align*}a^{88}=a^{74}&\Rightarrow a^{14}=a^0\Rightarrow a^{2\cdot 8-2}=a^0\\ & \Rightarrow \left (a^8\right )^2=a^2\Rightarrow a^{2t}=a^2\\ & \Rightarrow \left (a^{2t}\right )^4=\left (a^2\right )^4\Rightarrow a^{8t}=a^8\\ & \Rightarrow \left (a^8\right )^t=a^8\Rightarrow \left (a^t\right )^t=a^t\\ & \Rightarrow a^{t^2}=a^t\Rightarrow a^{t^2-t}=a^0\end{align*}
Is everything correct so far?
How could we continue?
How could we use the fact that $a^2\neq a^0$ ?
| We know that $a^{14} = 1$, so the order of $a$ divides $14$.
We know that $a^2 \ne 1$, so the order of $a$ does not divide $2$.
We know that $a^{8-t} = 1$, so the order of $a$ is at most $8-t$, i.e. at most $8$.
Therefore, the order of $a$ is $7$, so $a^8 = a$, so $t = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Real value of equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$ Find the real value of x in the equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$
I tried to square the whole term and after expansion not getting the result.
| We may assume $x>0$. From $\sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}$, subtract $\sqrt{x-1}$ from both side and squaring gives us $$x^2-1=x^3+x-1-2x\sqrt{x^2-x}$$
Simplifying, $$(x^2-x)-2\sqrt{x^2-x}+1=0$$
Can you go on from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to proof the reason? I have this statement:
If $\frac{a}{b} = \frac{c}{d},$ prove that
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
I tried to add 1, multiply 1 and nothing.
My development was:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
$\frac{a-b}{b} = \frac{c-d}{d}$
$\frac{b}{a-b} = \frac{d}{c-d}$ (I raised to $^{-1}$)
So far I have arrived, without much success.
How can I prove it? Thanks in advance.
| Let, $\frac{a}{b}=\frac{c}{d}=k$ or, $a=bk,c=dk$. Now consider the LHS:$$\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{k+1}{k-1}$$
Also, $$\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{k+1}{k-1}~~~~~~~~~\boxed{}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Convergence of $\sum_{n=1}^{\infty}\left(\frac{\sin(2\alpha^n)}{\alpha^n}-2\right)$ Determine values of $\alpha \in \left[\frac{1}{2} ; \frac{3}{2}\right]=:I$ s.t. the following series converges:
$$\sum_{n=1}^{\infty}\left(\frac{\sin(2\alpha^n)}{\alpha^n}-2\right)$$
Am I doing it wrong or this series fails the limit test for all of the $\alpha\in I$? In case I got the wrong, how to handle this series?
Edit: I handled $2\alpha^n$ as $(2α)^n$ when evaluating the limit and so it goes to $0$ when $α<1$ and for the other values of α the result is $−2$.
| Let $a_n = \frac{\sin(2\alpha^n)}{\alpha^n}-2$ be the general term for your series.
If $\alpha = 1$, then the $a_n$ are constant and equal to $\sin(2) - 2 \neq 0$, so the series diverges.
If $\alpha > 1$, then $\lim_{n\to\infty} a_n = -2$.
Since it is not $0$, the series does not converge.
Now, suppose $\alpha<1$, and write $a_n = 2\left(\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right)$$.
We have have
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$
so that for $x \neq 0$
$$\frac{\sin(x)}x - 1 = - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots$$
In particular, for $0<x<1$, we have
$$- \frac{x^2}{3!} < \frac{\sin(x)}x - 1 < 0 \implies
\left|\frac{\sin(x)}x - 1\right| < \frac{x^2}{3!}.$$
It hence follows that
\begin{align}
\frac12 S
&=
\sum_{n=1}^{\infty}\,\left(\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right)
\\&\leqslant
\sum_{n=1}^{\infty}\,\left|\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right|
\\&\leqslant
\sum_{n=1}^{k}\,\left|\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right|
+
\sum_{n=k+1}^{\infty}\,\frac{{(2\alpha^n)}^2}{3!}
\\&=
\sum_{n=1}^{k}\,\left|\frac{\sin(2\alpha^n)}{2\alpha^n}-1\right|
+
\frac23\,\sum_{n=k+1}^{\infty}\,{\left(\alpha^2\right)}^n < \infty,
\end{align}
where $k$ is such that $n>k \implies 2\alpha^n < 1$, and the last step follows from the fact that $|\alpha^2|<1$ so that the geometric series converges.
In particular, $S$ converges absolutely and hence converges.
While the bound on $\sin(x)/x - 1$ is in fact always valid, I suppose it's easier to convince yourself purely from the power series expansion that it holds on the interval $[0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the oblique asymptote of $\sqrt{x^2+3x}$ for $x\rightarrow-\infty$ I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$
For $x\rightarrow+\infty$ we have:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x}{x}=1$$
which means that the function grows linearly.
$$f(x)-mx=\sqrt{x^2+3x}-x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}-1\right)\sim x\left(\frac {1}{2}\cdot\frac{3}{x}\right)=\frac{3}{2}$$
The oblique asymptote is $y=x+\frac 3 2$ which is correct. For $x\rightarrow-\infty$ we have:
$$\frac{f(x)}{x}=\frac{|x|}{x}=\frac{-x}{x}=-1$$
This means that
$$f(x)-mx=\sqrt{x^2+3x}+x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty$$
Which is not what my textbook reports ($-\frac{3}{2}$). Any hints on what I did wrong to find the $q$ for $x\rightarrow-\infty$?
| $$f(x)=\sqrt{x^2+3x}=|x|\left(1+{3\over x}\right)^{1\over2}$$ which, as $x\rightarrow -\infty$, is equal to $$-x\left(1-{3\over{2x}}\right)^{1\over2} \sim -x\left(1+{3\over{2x}}\right)= -x -{3\over2}$$ using Taylor or Bernoulli
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Is $f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right)$ uniformly convergent on $\left[0,\:2\pi\right]$ Is $f_{n}\left(x\right)$ uniformly convergent on $\left[0,\:2\pi\right]$?
\begin{equation}
f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right)
\end{equation}
We can see that $f_{n}\left(x\right)$ converges to $f\left(x\right)=\sin\left(x\right)$
point-wise.
Then how to proceed?
\begin{equation}
\lim_{n\rightarrow\infty}\sup_{x\in\left[0,\:2\pi\right]}\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin\left(x\right)\right|
\end{equation}
And hint? or there is another way without using $\lim\sup$ ? Thank you very much.
| Use that
$$\sin\left(x+\frac{x^{2}}{n}\right)=\sin x\cos \dfrac{x^2}{n}+\cos x\sin \dfrac{x^2}{n}.$$
Thus
\begin{align}\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|&= \left|sin x\cos \dfrac{x^2}{n}+\cos x\sin \dfrac{x^2}{n}-\sin x\right| \\ &\le |\sin x|\left|1-\cos\dfrac{x^2}{n}\right|+|\cos x|\left|\sin\frac{x^2}{n}\right|.\end{align}
Now using that
$$|\sin t|\le t, |1-\cos t|\le t, \forall t\in [0,\infty),$$
we get
$$\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|\le \dfrac{x^3}{n}+\dfrac{x^2}{n}=\dfrac{4\pi^2(2\pi+1)}{n}, \forall x\in[0,2\pi],\forall n\in \mathbb{N}.$$
So, given $\epsilon>0$ there exists $N\ge \dfrac{4\pi^2(2\pi+1)}{n}$ such that
$$n\ge N\implies\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|\le\epsilon, \forall x\in[0,2\pi],\forall n\in \mathbb{N}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$
| $\dfrac{1+\tan t}{1-\tan t} =$
$\dfrac{\cos t + \sin t}{\cos t -\sin t}=$
$\dfrac{(\cos t+\sin t)^2}{\cos^2 t - \sin^2t}=$
$\dfrac{1+2\sin t \cos t}{\cos^2 t-\sin^2 t}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find $z^3+bz^2+c=0$ Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5$
We have $z_1=(-\sqrt{2}-\sqrt{2}i)^5=32e^{i\frac{15\pi}{4}}$
and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=32e^{-i\frac{15\pi}{4}}$
Can we conclude straight away something about $b,c$?
| Using Vieta's formulas, we get that the product $z_1 z_2 z_3$ of the roots is equal to $-c$. Hence $z_3 = -2^{-10}c$. Plugging this value into the equation you get
$$-2^{-30} c^3 + 2^{-20}bc^2 +c=0$$ As $c \neq 0$ we get the equation $-c^2+2^{10}bc + 2^{30}=0 \tag{1}.$
Also $$z_1z_2+z_1z_3+z_2z_3=2^{10}-2^{10}c(z_1+z_2)=2^{10}-2^{15}\sqrt{2}c=0.$$
Hence $c=\dfrac{2^{-5}}{\sqrt{2}}$ and you get the value of $b$ by plugging in this value in equation $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$ Find the limit without using L'hopital
$$\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$$
I have tried multiply by the conjugate of the denominator and numerator but didn't work
Any hint would be appreciated
Thank you
| There are two conjugates you need to multiply:
\begin{align*}
\lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} &= \lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1} \\
&= \lim_{x\to0} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{(\sqrt[3]{1 + x} - 1)(\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1)} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\
&= \lim_{x\to0} \frac{\sqrt{1 + x}^2 - 1^2}{\sqrt[3]{1 + x}^3 - 1^3} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\
&= \lim_{x\to0} \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\
&= \frac{\sqrt[3]{1 + 0}^2 + \sqrt[3]{1 + 0} + 1}{\sqrt{1 + 0} + 1} = \frac{3}{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Trouble with argument in a complex number We have:
$z_1 = 2 + 2i$ and $z_2 = -1 -\sqrt{3} i$
I am asked for obtaining arg$\left(\frac{z_1}{z_2}\right)$ and arg($z_1z_2$)
As we know:
$$\theta = \tan^{-1} \left(\frac{y}{x}\right)$$
The division of complex numbers is:
$$\frac{z_1}{z_2} = z_1z_2^{-1} = \frac{z_1z_2^*}{z_2z_2^*}$$
Doing so I got $\frac{\sqrt{3}-1}{-1-\sqrt{3}}$ but I do not know how to get the angle without calculator using this result.
The same happened to me in the product:
$$arg(z_1z_2) = \tan^{-1}\left(\frac{-1-\sqrt{3}}{-1+\sqrt{3}}\right) = ...$$
I know it has to be a silly thing but I am stuck here.
| I do not really know if you wanted this. But, try using the following steps to find the value
$$tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{- 1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} \right)$$
$$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{\tan \left( \dfrac{\pi}{4} \right) + \tan \left( \dfrac{\pi}{3} \right)}{1 - \tan \left( \dfrac{\pi}{4} \right) \cdot \tan \left( \dfrac{\pi}{3} \right)} \right)$$
$$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \tan \left( \dfrac{7 \pi}{12} \right) \right) = \dfrac{7 \pi}{12}$$
You can try the same method for other as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate correlation coefficient for discrete random variable From a population consisting of the numbers: $\lbrace 1,2 \ldots 10 \rbrace$, two samples are chosen from it without replacement. If the random variable denoting the first choice is X and the second choice is $Y$, what is the correlation coefficient ($\rho$) between $X$ and $Y$
| Assuming $X$ is uniform on $\{1, 2, \dots, 10\}$ we have
$$
\mathbb E X = \frac{1 + 2 + \dots + 10}{10} = 5.5, \quad \mathbb{V}ar X = \frac{1^2 + \dots + 10^2}{10} - \mathbb E X^2 = 8.25.
$$
To compute the expected value of $Y$ write
$$
\mathbb E Y = \sum_{x \in [10]} \mathbb E[Y ~|~ X=x] \mathbb{P}(X = x),
$$
where $[n] := \{1, 2, \dots, n \}.$ Let us first compute $\mathbb E Y:$
$$
\mathbb E Y = \frac 1{10}\left[\frac {1 + \dots + 9}{9} + \dots + \frac{2 + \dots + 10}{9}\right] = \frac{55}{10} = 5.5.
$$
$$
\mathbb Var Y = \frac 1{10}\left[\frac {1^2 + \dots + 9^2}{9} + \dots + \frac{2^2 + \dots + 10^2}{9}\right] - \mathbb E Y^2 = 8.25
$$
For $\mathbb E XY$ we have
$$
\mathbb E XY = \frac{1}{10}\left[\frac{1 + 2 + \dots + 10}{9\cdot 10} \cdot \sum_{x, y \in [10] \setminus x} y \right] = 30.25.
$$
Hence,
$$
\rho(X, Y) = \frac{\mathbb Cov(X, Y)}{\sqrt{\mathbb Var X \mathbb Var Y}} = \frac{30.25 - 5.5^2}{8.25} = 0.
$$
As for me, it is counterintuitive that the answer is $0$ and I suspect that there must be much simpler solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x+8y+48
\end{align}$$
Is it okay? Thank you
| Here's another approach altogether:
$$\begin{align}
|z-4i|=2|z+4|
&\iff\left|z-4i\over z+4 \right|=2\\
&\iff{z-4i\over z+4 }=2e^{i\theta}\quad\text{for some }\theta\in\mathbb{R}\\
&\iff z={8e^{i\theta}+4i\over1-2e^{i\theta}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Find $\lim\limits_{t\to\infty}x(t)$ if $x'= (x-y)(1-x^2-y^2)$, $y' = (x+y)(1-x^2-y^2)$ Suppose $x_0,y_0$ are reals such that $x_0^2+y_0^2>0.$ Suppose $x(t)$ and $y(t)$ satisfy the following:
*
*$\frac{dx}{dt} = (x-y)(1-x^2-y^2),$
*$\frac{dy}{dt} = (x+y)(1-x^2-y^2),$
*$x(0) = x_0,$
*$y(0) = y_0.$
I am asked to find $\lim_{t\to\infty}x(t).$
Dividing the two differential equations, we have $$\frac{dy}{dx} = \frac{x+y}{x-y} = \frac{1 + \frac{y}{x}}{1-\frac{y}{x}}.$$
Let $v(x)=\frac{y}{x},$ so that $y = xv(x),$ and $$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$ This implies that $$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}.$$ Rearranging and integrating, we get $$\log(x) = \arctan(v)-\frac{1}{2}\log(1+v^2)+c,$$ so $$\log(x) = \arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log\left(1+\left(\frac{y}{x}\right)^2\right)+c.$$ By writing $1+\left(\frac{y}{x}\right)^2$ as $\frac{x^2+y^2}{x^2},$ we can simplify the above equation to $$\frac{1}{2}\log(x^2+y^2) = \arctan\left(\frac{y}{x}\right) + c.$$
However, I'm not sure how to proceed from this implicit equation. Any help would be much appreciated!
| We first pass to polar coordinates, viz:
$x = r\cos \theta; \; y = r \sin \theta; \tag 1$
$\dot x = \dot r \cos \theta - r\dot \theta \sin \theta; \tag 2$
$\dot y = \dot r \sin \theta + r \dot \theta \cos \theta; \tag 3$
$\begin{pmatrix} \dot x \\ \dot y \end{pmatrix} = \begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r\cos \theta \end{bmatrix} \begin{pmatrix} \dot r \\ \dot \theta \end{pmatrix}; \tag 4$
$\begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r\cos \theta \end{bmatrix}^{-1} = \dfrac{1}{r} \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} ; \tag 5$
$\begin{pmatrix} \dot r \\ \dot \theta \end{pmatrix} = \dfrac{1}{r} \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{pmatrix} \dot x \\ \dot y \end{pmatrix}; \tag 6$
$\dot x = (x - y)(1 - x^2 - y^2) = r(\cos \theta - \sin \theta)(1 - r^2); \tag 7$
$\dot y = (x + y)(1 - x^2 - y^2) = r(\cos \theta + \sin \theta)(1 - r^2); \tag 8$
$\begin{pmatrix} \dot r \\ \dot \theta \end{pmatrix} = \dfrac{1}{r} \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{pmatrix} r(\cos \theta - \sin \theta)(1 - r^2) \\ r(\cos \theta + \sin \theta)(1 - r^2)\end{pmatrix}$
$= (1 - r^2) \begin{bmatrix} r\cos \theta & r\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{pmatrix} \cos \theta - \sin \theta \\ \cos \theta + \sin \theta \end{pmatrix} = (1 - r^2) \begin{pmatrix} r \\ 1 \end{pmatrix}; \tag{10}$
$\dot r = r(1 - r^2); \tag{11}$
$\dot \theta = 1 - r^2. \tag{12}$
Since
$x_0^2 + y_0^2 > 0, \tag{13}$
we have
$r_0^2 = x_0^2 + y_0^2 > 0 \Longrightarrow r_0 > 0; \tag{14}$
by (11),
$0 < r < 1 \Longrightarrow \dot r > 0; \tag{15}$
$1 < r \Longrightarrow \dot r < 0; \tag{16}$
$r = 1 \Longrightarrow \dot r = 0; \tag{17}$
it is relatively easy to see that (14)-(17) in concert imply that
$\displaystyle \lim_{t \to \infty} r = 1; \tag{18}$
also, (11) and (12) together yield
$r \ne 0 \Longrightarrow \dot{(\ln r)} = \dfrac{\dot r}{r} = 1 - r^2 = \dot \theta, \tag{19}$
whence, integrating over $t$,
$\ln r - \ln r_0 = \theta - \theta_0, \tag{20}$
or
$\theta = \ln r - \ln r_0 + \theta_0, \tag{21}$
and so via (18)
$\displaystyle \lim_{t \to \infty} \theta = \lim_{t \to \infty} \ln r - \ln r_0 + \theta_0 = \theta_0 - \ln r_0; \tag{22}$
finally,
$\displaystyle \lim_{t \to \infty} x(t) = \lim_{t \to \infty} ( r \cos \theta) = (\lim_{t \to \infty} r)( \lim_{t \to \infty} \cos \theta) = \cos (\theta_0 - \ln r_0). \tag{23}$
| {
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Find all $n$ such that $x+y$ is a power of $2$ whenever $xy=n$.
Let $n$ be a positive integer. $n$ is called a golden integer if $n$
is composite, and if we write $n$ as $n=xy$, where $x$ and $y$ are
positive integers, then $x+y$ is a power of two ($x+y=2^{r}$). Find
all golden integers.
It is obvious by definition and the fact that $n=1\times n$ that there is an integer $a$, such that $n=2^{a}-1$. $n$ is composite, hence there are two integers, say $x$ and $y$, such that $1<x\leq y<n$ and $n=xy$. Thus we have the following system,
$$x+y=2^{b}$$
$$xy=2^{a}-1$$
Consequently,
$$(x+1)(y+1)=2^{b}(2^{a-b}+1)$$
$$(x-1)(y-1)=2^{b}(2^{a-b}-1)$$
Does anyone know how to continue this solution?
| If $n$ is a golden number and $n=xy$ is a decomposition with $x,y>1$, then indeed
$$x+y=2^b\qquad\text{ and }\qquad xy=2^a-1,$$
for positive integers $a$ and $b$. Note that $b<a$ because
$$2^b=x+y<xy=2^a-1.$$
Suppose $a$ is prime, say $a=p$. Then $x$ and $y$ divide $2^p-1$ and hence${}^1$ $x\equiv y\equiv1\pmod{p}$. Then because $b<a=p$ and
$$2^b=x+y\equiv2\pmod{p},$$
we see that $b-1$ strictly divides $p-1$, so $2(b-1)\leq p-1$. But also
$$2^p-1=xy=x(2^b-x)\leq 2^{2b-2},$$
which shows that $p\leq 2b-2$. This means $2b-2\leq p-1<p\leq 2b-2$, a contradiction.
Hence $a$ is composite, say $a=uv$ with $u,v>1$. Then
$$n=2^a-1=(2^u-1)\sum_{k=0}^{v-1}2^{ku},$$
and because $n$ is golden we have
$$(2^u-1)+\sum_{k=0}^{v-1}2^{ku}=2^c,$$
for some integer $c$. This implies $v=2$, and by symmetry also $u=2$ and so $n=15$.
*
*Note that $2^m-1$ and $2^n-1$ are coprime if and only if $m$ and $n$ are coprime. So if $q$ is a prime dividing $2^p-1$ then for all $m<p$ the prime $q$ does not divide $2^m-1$. So modulo $q$ the number $2$ has order $p$, which implies that $p\mid q-1$ and so $q\equiv1\pmod{p}$. It follows that all divisors of $2^p-1$ are congruent to $1$ modulo $p$.
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Reversing digits of power of 2 to yield power of 7 Are there positive integers $n$ and $m$ for which reversing the base-10 digits of $2^n$ yields $7^m$?
I've answered this question for powers of 2 and powers of 3 in the negative. Permutations of the digits of a number divisible by 3 yields another number divisible by 3 in base-10. This arises from the fact that the sum of base-10 digits of a number divisible by 3 is itself divisible by 3.
Thus since $2^n$ is not divisible by 3, reversing the digits can't yield another number divisible by 3, and hence no natural number power of 2 when reversed will be a natural power of 3.
I'm currently trying to put limitations on n and m by considering modular arithmetic. Suggestions for techniques in the comments would also be appreciated.
| Some basic observations:
$n$ must be even: one has $2^n=7^m=1 \pmod{3}$ and so $n$ is even.
$n-m$ must be divisible by 3: one has $2^n=7^m=(-2)^m \pmod{9}$ and so $2^{n-m}=(-1)^m \pmod{9}$ and so $n-m$ is divisible by 3.
$n-2m$ is divisible by 10 as the following two paragraph show.
If the number of digits is even, then $2^n+7^m$ is a multiple of 11. Therefore, $2^n+(-4)^m =0 \pmod{11}$. Since $n$ is even, we let $n=2l$, and so $4^l+(-4)^m=0 \pmod{11}$. It follows that $m$ is odd and $l-m$ is divisible by 5.
If the number of digits is odd, then $2^n=7^m \pmod{11}$. Again we have $4^l=(-4)^m \pmod{11}$ and so $4^{l-m}=(-1)^m \pmod{11}$, which implies that $m$ is even and $l-m$ is divisible by 5.
The number of digits of $2^n$ is given by the inequality $10^s<2^n<10^{s+1}$. So $s<n \log_{10} 2<s+1$. In particular $|n\log 2 - m \log 7|<1$.
| {
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Evaluate $\int \frac{\sqrt{1+x^8} dx}{x^{13}}$
Evaluate: $\displaystyle\int \frac{\sqrt{1+x^8}}{x^{13}}dx$
My attempt:
I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^{13}$ in the denominator as $x^{16}.x^{-3}$ hoping that it would bring the integrand into some form but that too did not work.
| The substitution $x=\tan^{1/4}t$ gives $$\int\frac{\sqrt{1+x^8}}{x^{13}}dx=\int\frac{1}{4}\sin^{-4}t\cos tdt=-\frac{1}{12}\sin^{-3}t+C=-\frac{1}{12}\bigg(\frac{x^8}{1+x^8}\bigg)^{-3/2}+C.$$
| {
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Can't understand how $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}(2k+4)!(k+2)!^2$ is correct I'm following my teacher's notes and I cannot understand how $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}(2k+4)!(k+2)!^2$ is correct. Shouldn't it be $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}\frac{(2k+4)!}{(k+2)!^2}$ ?
Not sure if relevant, but the exercise (with which I don't need help with!) is to prove by induction ${0 \choose 0} + {2 \choose 1} + ... + {2n \choose n} < \frac{1}{2} {2n + 2 \choose n + 1}$ where $n \geqslant 2$.
Thanks
| Indeed by definition of the binomial coefficient
$$\frac{1}{2}\binom{2k+4}{k+2}=\frac{1}{2}\frac{(2k+4)!}{(k+2)!^2}.$$
Plugging in any value, for example simply $k=0$, shows that
$$\frac{1}{2}\frac{(2k+4)!}{(k+2)!(k+2)!}\neq\frac{1}{2}(2k+4)!(k+2)!^2,$$
so this is indeed a misprint as suggested in the comments.
| {
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On Lame's Theorem I was trying to study Lame's theorem but I'm a little bit stuck on one specific step... I write here the theorem and the proof:
Lame's theorem:
using the Euclidean algorithm to find the greatest common divisor of two positive integers has number of divisions less than or equal five times the number of decimal digits in the minimum of the two integers.
Proof:
Let $a$ and $b$ be two positive integers where $a > b$.
Applying the Euclidean algorithm to find the greatest common divisor of two integers with $a = r_{0}$ and $b = r_{1}$, we get
$r_{0} = r_{1}q_{1}+r_{2}$, $0≤r_{2}<r_{1}, $
$r_{1} = r_{2}q_{2}+r_{3}$, $0≤r_{3}<r_{2}, $
. . .
$r_{n-2} = r_{n-1}q_{n-1}+r_{n}$, $0≤r_{n}<r_{n-1}, $
$r_{n-1} = r_{n}q_{n}$
Notice that each of the quotients $q_{1}, q_{2}, ..., q_{n-1} $ are all greater than $1$ and $q_{n} ≥ 2$ and this is because $r_{n} < r_{n-1}.$
Thus we have
$r_{n} ≥ 1=f_{2}$,
$r_{n-1} ≥ 2r_{n} ≥ 2f_{2} = f_{3}$,
$r_{n-2} ≥ r_{n-1} + r_{n} ≥ f_{3} + f_{2} = f_{4}$,
$r_{n-3} ≥ r_{n-2} + r_{n-1} ≥ f_{4} + f_{3} = f_{5}$,
...
$r_{2} ≥ r_{3} + r_{4} ≥ f_{n-1} + f_{n-2} = f_{n}$,
$b = r_{1} ≥ r_{2} + r_{3} ≥ f_{n} + f_{n-1} = f_{n+1}$.
Thus notice that $b≥f_{n+1}$.
By a Lemma I don't report here, we have $f_{n+1}>α^{n−1}$ for $n>2$.
As a result, we have $b > α^{n−1}$.
Now notice since $\log_{10} \alpha > \frac{1}{5}$
we see that
$\log_{10}b > (n − 1)/5$.
Thus we have
$(n - 1)< 5 log_{10}b$
Now let $b$ has $k$ decimal digits. As a result, we have $b < 10^k$ and thus $log_{10}b < k$. Hence we conclude that $(n − 1) < 5k$.
Since $k$ is an integer, we conclude that $n ≤ 5k$.
What I really don't understand is just this line:
Now notice since $\log_{10} \alpha > \frac{1}{5}$,
My question is: why $\frac{1}{5}$ has been chosen?
Where does it come from?
Has it been chosen because the theorem says:
[...]less than or equal five times [...]
?
Thank you
| You can easily expand out the polynomial
$({1+\sqrt{5}})^{5}$ and confirm it is greater than $10*2^{5}$
| {
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How to calculate the limit $\lim\limits_{n\to\infty} \int\limits_0^1 \frac{n(2nx^{n-1}-(1+x))}{2(1+x)}\,dx$? How to calculate the limit $\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$?
I have to calculate the limit when solving
Find $a,b$ for $\displaystyle\int_0^1 \dfrac{x^{n-1}}{x+1} dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}) (n\to\infty)$
First I calculated that
$\lim\limits_{n\to\infty} \displaystyle\int_0^1 \dfrac{nx^{n-1}}{x+1} dx=\dfrac{1}{2}$, thus $a=\dfrac{1}{2}$, then
$2b=\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$.
However, I cannot find a good way to calculate it.
| Your evaluation of $a$ is correct. As regards $b$, note that, by integration by parts applied twice, we have that
$$\begin{align}
\int_0^1 \dfrac{x^{n-1}}{x+1} dx&=\frac{1}{2n}+\frac{1}{n}\int_0^1 \frac{x^{n}}{(x+1)^2} dx
\\
&=\frac{1}{2n}+\frac{1}{4n(n+1)}+\frac{2}{n(n+1)}\int_0^1 \frac{x^{n+1}}{(x+1)^3} dx.\end{align}$$
Morever
$$0\leq \int_0^1 \frac{x^{n+1}}{(x+1)^3} dx\leq \int_0^1 x^{n+1}dx=\frac{1}{n+2}.$$
Hence
$$\int_0^1 \dfrac{x^{n-1}}{x+1} dx=\frac{1}{2n}+\frac{1}{4n^2}+O(1/n^3)$$
and it follows that $b=1/4$ (and $a=1/2$).
| {
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How do I show that $\cos^4x=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x )+\frac{3}{8}$ I know how to prove that
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
by substituting $\cos(2x)$ with $2\cos^2x-1$ according to the double angle identity
$$\cos(2x)=2\cos^2x-1$$
However, how do I do that for $\cos^4x$?
Do I do it by writing $\cos^4x$ as $$\cos^2(x)\cdot \cos^2(x)$$ and thus get it by squaring the LHS of $$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
Im not sure how to proceed. Any ideas?
| As an alternative we can use that
$$\cos x=\Re(e^{ix})$$$$\implies \cos 4x=\Re(e^{i4x})=\Re[(\cos x+i\sin x)^4]=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$
that is
$$\cos 4x=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$
$$\cos 4x=\cos^4 x-6\cos^2x (1-\cos^2 x)+(1-\cos^2 x)^2$$
$$\cos 4x=\cos^4 x-6\cos^2x+6\cos^4x+1-2\cos^2x+\cos^4 x$$
$$8\cos^4 x=\cos 4x+8\cos^2x-1$$
$$8\cos^4 x=\cos 4x+8\left(\frac{1}{2}+\frac{1}{2}\cos(2x)\right)-1$$
$$8\cos^4 x=\cos 4x+4\cos(2x)+3$$
$$\cos^4 x=\frac18\cos 4x+\frac12\cos(2x)+\frac38$$
| {
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Find value of $\prod(1+\alpha^2)$ if $\alpha,\beta,\gamma,\delta$ are the roots of equation $x^4+4x^3-6x^2+7x-9$ then find $\prod(1+\alpha^2)$
I know
*
*$\sum \alpha=-4$
*$\sum \alpha\beta=-6$
*$\sum \alpha\beta\gamma=-7$
*$\alpha\beta\gamma\delta=-9$
I know the sum and products and other things about the roots but I am not being to rearrange them such that I am able to get the required answer
| Transforming roots from $\alpha$ to $1+\alpha^2$,
$$y=1+\alpha^2$$
$$\alpha=\sqrt{y-1}$$
As $\alpha$ satisfies our given polynomial, Substitute $\alpha$ into that.
We get new equation with roots $1+\alpha^2,1+\beta^2,\ldots$
$$(x-1)^2+4(x-1)\sqrt{x-1}-6(x-1)+7\sqrt{x-1}-9=0$$
Rearranging and squaring,
$$(x-1)^2(x-7)^2+81-18(x-1)(x-7)=(x-1)(4x+3)^2$$
$$(x^2+1-2x)(x^2+49-14x)+81-18(x^2+7-8x)=(x-1)(16x^2+9+24x)$$
$$x^4-32x^3+34x^2+45x+13=0$$
$$\text{Product of roots}=\frac{13}{1}=13$$
| {
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The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by both $2$ and $3$
$$AABB \equiv 0 \pmod{2}\tag{1}$$
$$AABB \equiv 0 \pmod{3}\tag{2}$$
Reducing $AABB \equiv 0 \pmod{2}$,
$$AABB \equiv 0 \pmod{2} \implies 1000A + 100A + 10B + B \implies B \equiv 0 \pmod{2}$$
Now reducing $AABB \equiv 0 \pmod{3}$,
$$AABB \equiv 0 \pmod{3} \implies 1000A + 100A +10B + B \implies A + A + B + B \equiv 0 \pmod{3} \implies 2(A+B) \equiv 0 \pmod{3}$$
I think I've gone wrong so far.
| You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
| {
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How do you calculate this infinite series $\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}$?
Evaluate the infinite series $$\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $\frac{1}{12}$. For all values of $p$ I checked the result was a reciprocal of a natural number. How would you calculate this without a calculator for single values of $p$ and for any $p$ in general (natural numbers).
| Hint. Note that
$$\begin{align}\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}&=
2\sum_{n=0}^\infty \frac{\binom{2n+1}{n+p+1}}{(2n+1)(2.5)^{2n+1}}\\
&=\frac{4}{5}\sum_{n=0}^\infty \frac{\binom{2n}{n+p}x^n}{n+p+1}
\end{align}$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$\sum_{n=0}^{\infty} \frac{\binom{2n}{n}x^n}{n+1} = \frac{1-\sqrt{1-4x}}{2x}=\frac{2}{1+\sqrt{1-4x}}$$
(see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$,
$$\sum_{n=0}^{\infty} \frac{\binom{2n}{n+p}x^n}{n+p+1} = \left(\frac{2}{1+\sqrt{1-4x}}\right)^{2p+1}\cdot \frac{x^p}{2p+1}$$
| {
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Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$
Then, we have:
$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$.
Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.
Then, I used the following substitution:
$c + d = t$ and $a + b = 4 - t$.
We assume wlog that $a \leq b \leq c \leq d$.
Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.
Then, what we have to prove is:
$9t^2(4 - t)^2 \geq 256$.
We can rewrite this as:
$(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or
$(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,
at which point I got stuck.
| Alternatively, you could solve
$min_{(a, b, c, d) |a+b+c+d=4} f(a, b, c, d),$
$f$ being your term with the squares.
The convexity guarantees a global minimum. And the symmetry of the problem will yield the answer.
Edit
According to the comments, $f$ is not convex but it is subharmonic (thanks to Jack D'Aurizio), i.e. for the given $f$ we have $\Delta f \geq 0$ on $\mathbb{R}^4$. $f$ is obviously not constant. So by the maximum principle it cannot have a maximum on the interior of its domain. Hence, it has a unique (unconstrained) minimum in $\mathbb{R}^4$. The constraint defining function $g(a, b, c, d) =a+b+c+d$ is harmonic. I did not go further but maybe it then can be shown that the constrained min problem has a unique solution and that it must be symmetric because otherwise we could move towards the origin (0,0,0,0) and further minimize the function (using a closed ball argument based in the definition of an harmonic function according to Wikipedia).
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Solving an equality with 3 equations, and 3 variables Following is the question asked in a recent aptitude exam:
Given that :
$$ a+b+ab=10\\ b+c+bc=20 \\ c+a+ac=30$$
What is the value of $a+b+c+abc$ ?
I can solve it by finding the individual values by:
$$ a = \frac{10-b}{b+1} ,\\ c = \frac{20-b}{b+1}$$
Putting these in the third equation:
$$ \frac{10-b}{b+1}+\frac{20-b}{b+1} + \frac{(10-b)\times(20-b)}{(b+1)^2}=30\\
(30-2b)\times(b+1)+(b-20)\times(b-10)=30\times(b+1)\\{30\times(b+1)}-2b(b+1)+b^2-30b+200={30\times(b+1)}\\-2b^2-2b+b^2-30b+200=0\\-b^2-32b+200=0\\b^2+32b-200=0$$
Which can be solved to:
$$b= \frac{-32+-\sqrt{32^2-4\times(-200)}}{2} \approx \frac{-32+-42.71}{2} = 5.35 \text{ or} -37.35 $$
And we can get the values for $a$ and $c$ as well:
$$a \approx 9.157 \text{ or} -1.297\\ c\approx 2.30 \text{ or} -1.577$$
Here, the for the first two equations, $b=-37.35$, $a=-1.297$ and $c=-1.577$ works. Whereas, the third equation is satisfied by $a=9.157$ and $c=2.30$. This doesn't seem correct. Since only a single value of $b$ works for the first two equations, therefore the other value of $b$ is rejected. This also rejects the derived values of $a$ and $c$. So, the positive values of $a$ and $c$ should not be used. But the former values don't satisfy the third equation, but the latter values do. What is the problem here?
Moreover, is there any shorter way of solving this problem?
| Just add one
Add one to each equation and factorize to get :
$$
(a+1)(b+1) = 11 \\
(b+1)(c+1) = 21 \\
(a+1)(c+1) = 31 \\
$$
Now, set $x ,y,z = a+1,b+1,c+1$ respectively.This gives $xy = 11, yz = 21,zx = 31$, so $x^2y^2z^2 = 11 \times 21\times 31$ by multiplying these.
Now divide this equation by the other equations suitably to get the values of $x,y,z$, and use the fact that $a+b+c + abc = x+y+z - 3 + (x-1)(y-1)(z-1)$ to get the answer.
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Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$\text{Given } \sin{t} + \cos{t} = a, \text{ find an equivalent expression for } \sin^4{t} + \cos^4{t} \text{ in terms of } a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(\sin{t} + \cos{t})^4$ and using whatever identities I could remember;
$(\sin{t} + \cos{t})^4 = \sin^4{t} + \cos^4(t) + 4\sin{t}\cos^3{t} + 6\sin^2{t}\cos^2{t} + 4\sin^3{t}\cos{t}$
so then
$\begin{align}
\sin^4{t} + \cos^4{t} &= (\sin{t} + \cos{t})^4 - 4\sin{t}\cos^3{t} - 6\sin^2{t}\cos^2{t} - 4\sin^3{t}\cos{t}\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(2\cos^2{t} + 3\sin{t}\cos{t} + 2\sin^2{t})\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)\\
&= a^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)
\end{align}$
Couldn't get further than this, felt like I was overthinking it.
| $$(\sin t + \cos t )^2= a^2\rightarrow \sin t\cos t=\dfrac{a^2-1}{2}$$
then
$$\cos^4t + \sin^4t=(\cos^2t + \sin^2t)^2-2\cos^2t \sin^2t=1-2\left(\dfrac{a^2-1}{2}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\frac{1}{\sin(2x)} + \frac{1}{\sin(4x)} + \frac{1}{\sin(8x)} + \frac{1}{\sin(16x)}$
Evaluate
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(4x)} + \dfrac{1}{\sin(8x)} + \dfrac{1}{\sin(16x)}$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(2x) \cos (2x)} + \dfrac{1}{ 2\big [2\sin (2x)\cos (2x)\cos (4x)\big ]} + \dfrac{1}{\sin(16x)}$$
What am I missing?
Regards
| $$\sin(A-B)=\sin A \cos B-\cos A \sin B$$
$$\frac{1}{\sin 2x}=\dfrac{\sin(2x-x)}{\sin2x\sin x}=\cot x-\cot2x$$
$$\frac{1}{\sin 4x}=\dfrac{\sin(4x-2x)}{\sin4x\sin 2x}=\cot 2x-\cot4x$$
$$\frac{1}{\sin 8x}=\dfrac{\sin(8x-4x)}{\sin8x\sin 4x}=\cot 4x-\cot8x$$
$$\frac{1}{\sin 16x}=\dfrac{\sin(16x-8x)}{\sin16x\sin 8x}=\cot 8x-\cot16x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Geometric series: two ways to tackle same problem giving me issues I am trying to get a closed form expression for the following sum:
$$x + 2x^2 + 3x^3 + \cdots + nx^n$$
so by perturbation method:
$$S = x + 2x^2 + 3x^3 + \cdots + nx^n$$
$$xS= x^2 + 2x^3 + \cdots + (n-1)x^n + nx^{n+1}$$
$$(1-x)S = x + x^2 + x^3 + \cdots + x^n - nx^{n+1}$$
Now here is where it gets weird. The professor writes that this can be simplified to this well-known sum: $1+ x+ x^2 + \cdots + x^n = \frac{1-x^{n+1}}{1-x}$
Giving us:
$$(1-x)S = \frac{1-x^{n+1}}{1-x} - nx^{n+1} - 1$$
I see the $-1$ is because there is no $1$ in our sum but can't I do the following:
$$(1-x)S = x(1 + x + x^2+\cdots+x^{n-1}) - nx^{n+1}$$
$$(1-x)S = x \left( \frac{1-x^n}{1-x} \right) - nx^{n+1}$$
So I went to wolfram to see if this was possible and got a peculiar result when I went to test:
Now I might be using wolfram wrong but is it fair to assume that these two statements:
1.) $(1-x)S = (1-x^(n+1))/(1-x) - nx^(n+1) - 1$
2.) $(1-x)S = x((1-x^n)/(1-x)) - nx^(n+1)$
are equal? And if not: where did I go wrong to get the second statement?
| $$S = x + 2x^2 + \cdots + nx^n$$ implies $$(1-x)S = (x + x^2 + \cdots + x^n) - nx^{n+1}.$$ This much is clear for you. If you do not already know the sum for a finite geometric series, this too can be obtained by the same perturbation method you just applied to $S$. For instance, let $$G = 1 + x + x^2 + \cdots + x^n.$$ Then $$xG = x + x^2 + x^3 + \cdots + x^{n+1},$$ and $$(1-x)G = 1 - x^{n+1},$$ from which it follows that $$G = \frac{1 - x^{n+1}}{1-x}, \quad x \ne 1,$$ and in turn, $$S = \frac{(G-1) - nx^{n+1}}{1-x} = \frac{1 - x^{n+1} - 1 + x - n(1-x)x^{n+1}}{(1-x)^2} = \frac{x + (nx-n-1)x^{n+1}}{(1-x)^2}.$$
Second, judging from the output you obtained from Wolfram Alpha, your input was probably inappropriate. The triple equals sign === is the infix notation for the expression SameQ[] in the Wolfram language, and it gives True only if the LHS and RHS are exactly the same expression. So, if I were to guess, you might have used the word "same" in your input, rather than "equal to." Instead of dealing with these subtleties, it is much more meaningful to simply test the formula for a few carefully chosen values of $x$ and $n$. If you really want to, you could also input
Sum[x^k, {k,0,n}]
or you could even input
Sum[k x^k, {k,1,n}]
to get the answer to the original question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
| Hint: Write the function as $$\frac{\sin^2(x)-x^2}{x^4}\times \frac{x^2}{\sin^2(x)}$$ Otherwise use the Talor's expantion if you know it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
} |
If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$
But could not proceed further please help.
| Assuming $x\in \Bbb R.$ For brevity let $c=\cos x$ and $s=\sin x.$ Let $p=c^2s^2.$ We have $$ c^8+s^8=\frac {3}{8}\iff$$ $$ 1=(c^2+s^2)^4=(c^8+x^8)+c^2s^2(4c^4+6c^2s^2+4s^4)=$$ $$=\frac {3}{8}+c^2s^2(4(c^2+s^2)^2-2c^2s^2)=$$ $$=\frac {3}{8}+c^2s^2(4-2c^2s^2)\iff$$ $$\iff(0\leq p\leq 1\land \frac {5}{16}=2p-p^2)$$ $$\iff p=1- \sqrt {11}\;/4\iff$$ $$\iff |\sin 2x|=\sqrt {4p}=\sqrt {4-\sqrt {11}}.$$
Note: The 2nd, 3rd, and 4th displayed lines are a single sentence that is true iff $c^8+x^8=\frac {3}{8}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Find sum of all integer solution of $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ Given $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ , which $a,b,c ,d$ are an positive integer , all solution of $a,b,c,d$ are members of Set "S" ,then find sum all of members in set "S"
I don't know how to start to solve this problem but I've got 2 solution that is $\frac{2}{52} + \frac{3}{598} + \frac{4}{437} + \frac{5}{266} = \frac{1}{14}$ and $\frac{2}{224} + \frac{3}{304} + \frac{4}{437} + \frac{5}{115} = \frac{1}{14}$
question 2 If problem say that $a,b,c,d$ are an all integer , they will have negative integer to this solution ? please give me hint or theorem relevant. Thank you in advance .
| For Question 2, there are infinitely many integer solutions. Take $$(a,b,c,d):=(4,-7,5k,-4k)\,,\text{ where }k\in\mathbb{Z}_{\neq 0}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How can I prove this sequence convergent? I am trying to prove the sequence $(u_n)$ so that $u_1=1$ and $u_{n+1}=1+\dfrac{1}{u_n}$ is convergent. With some first items, I guess that the subsequence $(u_{2n})$ is decreasing and the subsequence $(u_{2n-1})$ is increasing, but I can't proof this. How can I proof?
| Hint: We can write
$$u_{3} = 1 + \frac{1}{1 + \frac{1}{u_1}} = 1 + \frac{u_1}{u_1+1} =
\frac{u_1+1}{u_1+1} + \frac{u_1}{u_1+1}
= \frac{2u_1+1}{u_1+1}$$
$$u_{4} = 1+\frac{u_1+1}{2u_1+1} =
\frac{2u_1+1}{2u_1+1} + \frac{u_1+1}{2u_1+1} =
\frac{3u_1+2}{2u_1+1}$$
$$u_{5} = 1 + \frac{2u_1+1}{3u_1+2} =
\frac{3u_1+2}{3u_1+2} + \frac{2u_1+1}{3u_1+2} =
\frac{5u_1+3}{3u_1+2}$$
Notice a pattern? You can prove it using induction!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find all non-negative integer $n$ that satisfy $f(x+1)+f(x-1)=\sqrt nf(x)$
Find all non-negative integer $n$ that there exists a non-periodic function $f:\Bbb R->\Bbb R\quad f(x+1)+f(x-1)=\sqrt nf(x)\forall x$.
My attempt:
$f(x+2)+f(x)=\sqrt n\cdot f(x+1)$
$\sqrt n \cdot (f(x+1)+f(x+3))=n\cdot f(x+2)$
$f(x+2)+f(x+4)=\sqrt n\cdot f(x+3)$
Therefore $$f(x)+f(x+4)=(n-2)\cdot f(x+2)\tag{*}$$
$n=0\implies f(x)=-f(x+2)\implies f(x)=f(x+4)$
$n=1\implies f(x+2)+f(x)=f(x+1)\,$ and $\,f(x+3)+f(x+1)=f(x+2)\implies f(x)=f(x+3)$
$n=2\quad(*)\implies f(x)+f(x+4)=0\,$ A similar argument hold for $n=0$.
$n=3\quad(*) \implies f(x)+f(x+4)=f(x+2).$ A similar argument hold for $n=1.$
$n=4\quad f(x)=x$ works.
$\forall n>4,f(x)$ can be defined recursively:
$i)\;$ $f(x)=f(\lfloor x\rfloor)\forall x$
$ii)\;$ $f(0)=0,\;f(1)=1$
$iii)\;$ $f(x)+f(x+2)=\sqrt n \cdot f(x+1)\;\forall x\in\Bbb Z$
It's easy to show that this satisfy the riginal functional equation, but we still need to prove it's non-periodic. It's sufficient to prove $iii)$ is monotonic.
$f(x+2)=\sqrt n\cdot f(x+1)-f(x)>2\cdot f(x+1)-f(x)\ge f(x+1)$
Is my proof correct?
Any help appreciated.
| Hint.
Assuming $f(x) = \lambda^x$ we have
$$
\lambda^{x+1}+\lambda^{x-1} = \sqrt{n}\lambda^x\Rightarrow \left(\lambda+\frac{1}{\lambda}=\sqrt n\right)\lambda^x
$$
hence
$$
\lambda = \frac 12\left(\sqrt n\pm\sqrt{n-4}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) $
Let $b>a>0$ and $x_1, x_2,\ldots,x_n,y_1, y_2,\ldots,y_n\in [a,b]$. If $$x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2\,,$$
then
$$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)\,.$$
My idea:
I proved that $\dfrac {a}{b}\leq \frac {x_k}{y_k}\leq \dfrac {b}{a} $.
$$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+...+\frac {x_n^3}{y_n}= (x_1y_1 \frac {x_1^2}{y_1^2}+...+x_ny_n\frac {x_n^2}{y_n^2}) $$
$x_1y_1+...+x_ny_n\leq x_1^2+...+x_n^2$ by Cauchy Schwartz.
Unfortunately $\frac {x_k}{y_k} $ is not smaller than $\frac {a^4+b^4}{ab (a^2+b^2)}$.
| Note that an equality condition is possible. Condider $n$ even. Let half of the $x_i$ equal to $a$ and let the corresponding $y_i$ equal to $b$. For the other half of the variables, exchange $a$ and $b$. Then the condition $x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2 = \frac{n}2 (a^2 + b^2)$ holds. Inserting in the inequality, indeed we have
$$
\frac{n}2 (\frac{a^3}{b} + \frac{b^3}{a}) = \frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n} = \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) = \frac {a^4+b^4}{ab (a^2+b^2)} \frac{n}2 (a^2 + b^2)
$$
with equality.
Now let us show that this constellation is extremal.
Let $q_i = \frac{x_i}{y_i}$. Choose a set of $x_i$. W.l.o.g., order the $x_i^2$ in ascending order. We have for $s$, the sum of squares: $n a^2 \leq s = x_1^2+x_2^2+\ldots+x_n^2 =y_1^2+y_2^2+\ldots+y_n^2 \leq n b^2$
The inequality is
$$
(x_1^2 q_1+x_2^2 q_2+\ldots+x_n^2 q_n) \leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)
$$
Now we argue for some extremal set of $y_i$.
By the rearrangement inequality, the maximum value of the LHS is obtained if the highest factors $q_i$ are chosen for the highest $x_i^2$. This can be achieved by letting as many $y_i = a$ for the highest indices $i$. The number $k$ of these settings $y_i = a$ is limited by $s =y_1^2+y_2^2+\ldots+y_{n-k}^2 + k a^2$. The number $k$ can be made as large as possible if all of the previous $y_1 = \ldots = y_{n-k} = b$. Then we have $s =(n-k) b^2 + k a^2$. This is $k = \frac{nb^2 -s}{b^2 - a^2}$ or the nearest possible smaller integer. So we have to show
$$
\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n} \leq \frac {x_1^3}{b}+\frac {x_2^3}{b}+\ldots+ \frac {x_{n-k}^3}{b} + \frac {x_{n-k+1}^3}{a} + \ldots + \frac {x_n^3}{a} \leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2)
$$
or
$$
\frac{\frac {x_1^3}{b}+\frac {x_2^3}{b}+\ldots+ \frac {x_{n-k}^3}{b} + \frac {x_{n-k+1}^3}{a} + \ldots + \frac {x_n^3}{a}}{x_1^2+x_2^2+\ldots+x_n^2} \leq \frac {a^4+b^4}{ab (a^2+b^2)}
$$
Now again, for the LHS the maximum value has to be found by considering settings of the $x_i$. This maximal setting of the $x_i$ is obtained if most of the $x_i$ with high indices $i$ are chosen as high as possible, i.e. $x_i=b$. With the same argument as already given, this is possible for $n-k$ many $b$. This gives that we have to show (for $k \leq n/2$):
$$
\frac{\frac {x_1^3}{b}+\frac {x_2^3}{b}+\ldots+ \frac {x_{n-k}^3}{b} + \frac {x_{n-k+1}^3}{a} + \ldots + \frac {x_n^3}{a}}{x_1^2+x_2^2+\ldots+x_n^2} \leq
\frac{k \frac {a^3}{b}+ (n - 2k)\frac {b^3}{b}+ k \frac {b^3}{a} }{(n-k) b^2 + k a^2} \leq
\frac {a^4+b^4}{ab (a^2+b^2)}
$$
Now in the LHS, we have that
$$
\frac{k \frac {a^3}{b}+ (n - 2k)\frac {b^3}{b}+ k \frac {b^3}{a} }{(n-k) b^2 + k a^2}
$$
is $1$ for $k=0$ and maximal for $k = n/2$ where we have
$$
\frac{k \frac {a^3}{b}+ (n - 2k)\frac {b^3}{b}+ k \frac {b^3}{a} }{(n-k) b^2 + k a^2} \quad {{(k = n/2)}\atop {=}} \quad
\frac {a^4+b^4}{ab (a^2+b^2)}
$$
The same method can be applied for $k \geq n/2$ where again the LHS is $1$ for $k=n$.
For odd $n$ we have to take extra care of the central term with $i = (n+1)/2$.
This proves the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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I was asked to evaluate the determinant of the I was asked to evaluate the determinant of the $n$x$n$ matrix
$$ A=
\begin{bmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{bmatrix}
$$
I tried starting from a 1x1 matrix to 5x5 matrix, and saw the pattern
$$\det{A}=(x-2)^{n-1}(x+2(n-1)).$$
Now I need to prove that this is the case for any nxn matrix of type A, so I tried using induction.The base case (n=1) holds.
$$A=
\begin{bmatrix}
x
\end{bmatrix}
$$
And clearly $\det{A}=x$ using both the definition of a determinant and the explicit formula above.
So now assume it holds for any $n$, and test for $n+1$.
$$\det{A}=(x-2)^n(x+2n).$$
What do I do now?
| We add all rows to the first one, then factor $x+2(n-1)$ from this first row, than use it (doubled) to eliminate in the other rows. With this strategy we are done quickly:
$$
\begin{aligned}
\det A
&=
\begin{vmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{vmatrix}
\\
&=
\begin{vmatrix}
x+2(n-1) & x+2(n-1) & x+2(n-1) & \cdots & x+2(n-1) \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{vmatrix}
\\
&=
(x+2(n-1))
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{vmatrix}
\\
&=
(x+2(n-1))
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
2-2 & x-2 & 2-2 & \cdots & 2-2 \\
2-2 & 2-2 & x-2 & \cdots & 2-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2-2 & 2-2 & 2-2 & \cdots & x-2 \\
\end{vmatrix}
\\
&=
(x+2(n-1))
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
0 & x-2 & 0 & \cdots & 0 \\
0 & 0 & x-2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & x-2 \\
\end{vmatrix}
\\
&=(x+2(n-1))(x-2)^{n-1}\ .
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the plane $P$ passing through the origin such that the three planes $P$, $P_1=(x+y+z=1)$ and $P_2= (x-y+z=2)$ meet along a line in R3. What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation and get $t$. Then I sub $t$ into the equation to obtain the equation of plane.
| Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B for all t. In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0. That is, z= x.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$ Problem
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
Attempt to solve
By solving $2x^2-5x-2=0$ we can get $\forall x \not\in X$.
$$ x= \frac{5\pm \sqrt{(-5)^2-4 \cdot 2\cdot(-2)}}{2 \cdot 2} $$
$$ x= \frac{5 \pm \sqrt{41}}{4} $$
Meaning :
$$ X=\mathbb{R}\setminus \{\frac{5-\sqrt{41}}{4},\frac{5+\sqrt{41}}{4}\} $$
In order to $f$ to be continuous $\forall x \
\not\in X$ have to have limits in these undefined points.
$$ \lim_{x \rightarrow \frac{5-\sqrt{41}}{4}} \frac{6x^3-11x^2-16x-4}{2x^2-5x-2}
=\frac{1}{4}(23-3\sqrt{41})$$
and
$$ \lim_{x \rightarrow \frac{5+\sqrt{41}}{4}}\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = \frac{1}{4}(23+3\sqrt{41})$$
Meaning if i set
$$
\begin{cases}
f(\frac{5-\sqrt{41}}{4})=\frac{1}{4}(23-3\sqrt{41}) \\
f(\frac{5+\sqrt{41}}{4}=\frac{1}{4}(23+3\sqrt{41})
\end{cases}
$$
is continous $\forall x \in \mathbb{R}$
Now only problem is that is this correct solution and if it is how do i solve it more simply than this ? I didn't compute these limits by hand, maybe there is more simpler solution ?
| You can simply perform polynomial division to get
$$6x^3 -11x^2 -16x -4 = 3x(2x^2-5x-2) + 4x^2 - 10x-4 =\\
= 3x(2x^2-5x-2) + 2(2x^2 - 5x-2) = (3x+2)(2x^2 - 5x-2)$$
This explains the fact that the function has finite limits at points where the denominator $2x^2-5x-2$ is zero.
Thus, the desired extension is
$$f(x)=3x+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the value of $r$ such that $\frac{1}{\binom{9}{r}} - \frac{1}{\binom{10}{r}} = \frac{11}{6\binom{11}{r}}$? Find the value of $r$:
$$\frac{1}{\dbinom{9}{r}} - \frac{1}{\dbinom{10}{r}} = \frac{11}{6\dbinom{11}{r}}$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.
| Since
$$\binom{n}{k} =
\begin{cases}
\dfrac{n!}{k!(n - k)!} & \text{if $0 \leq k \leq n$}\\[2 mm]
0 & \text{if $k > n$}
\end{cases}$$
we require that $k \leq 9$ since otherwise we would be dividing by $0$.
\begin{align*}
\frac{1}{\binom{9}{k}} - \frac{1}{\binom{10}{k}} & = \frac{11}{6\binom{11}{k}}\\
\frac{1}{\dfrac{9!}{k!(9 - k)!}} - \frac{1}{\dfrac{10!}{k!(10 - k)!}} & = \frac{11}{6 \cdot \dfrac{11!}{k!(11 - k)!}} && \text{by definition of $\binom{n}{k}$}\\
\frac{k!(9 - k)!}{9!} - \frac{k!(10 - k)!}{10!} & = \frac{11k!(11 - k)!}{6 \cdot 11!} && \text{division is multiplication by the reciprocal}\\
\frac{10k!(9 - k)!}{10!} - \frac{k!(10 - k)!}{10!} & = \frac{k!(11 - k)!}{6 \cdot 10!} && \text{form common denominator on LHS, cancel on RHS}\\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && \text{multiply by $6 \cdot 10!$}\\
6[10 - (10 - k)] & = (11 - k)(10 - k) && \text{divide by $k!(9 - k)!$}\\
6k & = 110 - 21k + k^2 && \text{simplify}\\
0 & = k^2 - 27k + 110 && \text{set quadratic equal to zero}
\end{align*}
When you solve the quadratic equation, keep in mind the restriction that $k \leq 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The probability of getting a 7 This problem is probably too easy for all you math geniuses out there, but I am having trouble understanding it, so here it is:
An athlete has 10 cards, numbered 1 to 10. Every day, he picks a card randomly and does that number of pushups. He doesn't replace the card, and picks until all the cards are gone. If after a certain number of tries, the athlete has done a total of 12 pushups, then the probability of him/her picking a 7 as the very next card is $\frac{a}{b}$. What is a + b?
I know that I have to count the possibilities of getting a 12:
*
*2 cards
The possibilities would be 2 and 10, 3 and 9, 4 and 8, 5 and 7, which would make 4 possibilities.
*
*3 cards
1, 2, 9 | 1, 3, 8 | 1, 4, 7 | 1, 5, 6 | 2, 3, 7 | 2, 4, 6 | 3, 4, 5 | which would be 7 possibilities.
*
*4 cards
1, 2, 3, 6 | 1, 2, 4, 5 | which is 2 possibilities.
There cannot be more than 4 cards: 1, 2, 3, 4, 5 would make 15.
The total number of possibilities for getting a 12 is 17, and out of those, there are 3 of them that will pick the card 7, which means 14 out of the 17 possibilities will be valid for the problem.
I don't know how to continue from here. Should I solve for the probability each of the three cases listed above separately, then add them, or is there a better method?
| First calculate the relative probabilities of getting a $2, 3$ or $4$ card sum of $12$.
These are:$$\frac{4}{\binom{10}{2}};\frac{7}{\binom{10}{3}}; \frac{2}{\binom{10}{4}} = \frac{4}{45}; \frac{7}{120}; \frac{2}{210} = \frac{448}{5040}; \frac{294}{5040}; \frac{48}{5040}$$
Given the sum is $12$, the actual probabilities of being a $2, 3$ card or $4$ card sum are:
$$\frac{448}{790}; \frac{294}{790}; \frac{48}{790}$$
The sum of all $3$ probabilities of drawing a $7$ from each of the $3$ card quantities is therefore:
$$P(7) = \frac{448}{790}\cdot \frac{3}{4}\cdot \frac{1}{8} + \frac{294}{790}\cdot \frac{5}{7}\cdot \frac{1}{7} + \frac{48}{790}\cdot \frac{1}{6} = 0.101266$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Recurrence relation for Pell's equation $x^2-2y^2=1$ I am wondering how to find the recurrence relation for solutions for $x$ in the Pell's equation $x^2-2y^2=1$.
I know the formula for the general term.
It is $$\frac{(3+2\sqrt2)^n+(3-2\sqrt2)^n}{2}$$ for $x_n$, the $n^{th}$ smallest solution for $x$.
Any help would be appreciated! Thanks!
I got a feeling that the recurrence formula is $x_n=6x_{n-1}-x_{n-2}$, but I wonder how to prove this relation true/false and how to derive/generate the recurrence relation.
Note that $x_{-1}=1$ and this recurrence formula applies to all nonnegative integral $n$.
| The fundamental solution is $9-8 = 1,$ meaning $3^2 - 2 \cdot 2^2 = 1.$ As a result, we have the matrix
$$
A =
\left(
\begin{array}{cc}
3 & 4 \\
2 &3
\end{array}
\right)
$$
which solves the automorphism relation, $A^T H A = H,$ where
$$
H =
\left(
\begin{array}{cc}
1 & 0 \\
0 & -2
\end{array}
\right)
$$
That is
$$ (3x+4y)^2 - 2 (2x+3y)^2 = x^2 - 2 y^2. $$
Next,
$$ A^2 - 6 A + I = 0. $$
Since
$$
A
\left(
\begin{array}{c}
x_n \\
y_n
\end{array}
\right) =
\left(
\begin{array}{c}
x_{n+1} \\
y_{n+1}
\end{array}
\right)
$$
and
$$
A^2
\left(
\begin{array}{c}
x_n \\
y_n
\end{array}
\right) =
\left(
\begin{array}{c}
x_{n+2} \\
y_{n+2}
\end{array}
\right)
$$
we find
$$ x_{n+2} - 6 x_{n+1} + x_n = 0 $$
$$ y_{n+2} - 6 y_{n+1} + y_n = 0 $$
This is just Cayley-Hamilton.
Caution: This is for $x^2 - 2 y^2 = 1.$ If we change the problem to $x^2 - 2 y^2 = 119 = 7 \cdot 17,$ the recurrence still holds, except that there are now four such families, each using the same recursion:
$$ (11,1) \; \; \; \; (37,25) \; \; \; \; (211,149) ... $$
$$ (13,5) \; \; \; \; (59,41) \; \; \; \; (341,241) ... $$
$$ (19,11) \; \; \; \; (101,71) \; \; \; \; (587,415) ... $$
$$ (29,19) \; \; \; \; (163,115) \; \; \; \; (949,671) ... $$
If you don't mind negative values for $x,y$ you can combine the above four into two families going both forth and back...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Eigenvalues of $aI + bJ$ Just as context, this problem arose when I was looking at the adjacency matrices of moore graphs of diameter 2.
Given that $I$ is the identity matrix and $J$ is the all 1 matrix. I have constructed some matrix $aI + bJ$. I am completely stuck in terms of how to find the eigenvalues of this matrix. I have considered trying to find the characteristic polynomial, but I have no idea how I would find the determinant of such a matrix.
| here is a matrix that shows a basis of the eigenvectors as columns, these being perpendicular to each other as well. Below is the 10 by 10 case. In smaller dimension, take the upper left square corner.
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n$ is divisible by $6$.
If the quadratic equations $x^2-mx+n=0$ and $x^2+mx-n=0$ both have integral roots, prove that $6|n$.
I've proved that $3|n$, and that $2|n$ for odd $m$, but I can't seem to prove it for even $m$.
Please help.
| Hint: Since both eqaution have intergal solution we have $$m^2-4n=a^2\;\;\;{\rm and}\;\;\;\;m^2+4n =b^2$$
for some integers $a,b$.
So $2m^2 =a^2+b^2$ implies
a) if $3\mid m$ then $3\mid a$ and $3\mid b$ implies $3\mid a^2-b^2 =4n$ so $3\mid n$
b) if $3\nmid m$ then $a^2\equiv_3 1$ and $b^2\equiv_3 1$ implies $$4n =a^2-b^2 \equiv _3 0 \implies 3\mid 4n\implies 3\mid n$$
Say $x_1$ and $x_2$ are solution to first equation. By Vieta formula $x_1+x_2 = m$ we see that if one is integer solution then so is the second. Now from $n=x_1x_2$ we get if $n$ is odd, then so are $x_1$ and $x_2$. Thus $m$ is even and so $a=2c$ and $b=2d$ from some integers $c,d$. Now we have $$2n =x^2-y^2=(x-y)(x+y)$$
which means that $x\equiv_2 y$. But then $2\mid n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any mistake in my approach for solving $ \int_0^{\pi/2} \frac{ \cos x}{3 \cos x + \sin x} \, dx $ ?? I had to evaluate this integral .
$$
\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx
$$
Here is how I proceeded
Dividing $N^r$ And $D^r$ by $\cos^3 x$
$$
\int_0^{\pi/2} \frac{ \sec^2 x}{3 \sec^2 x + \tan x \sec^2 x}\, dx \\
$$
Substituting $\tan x = t$
$$
\int_0^\infty \frac{ 1 }{(1+t^2)(t+3)} \, dt \\
$$
Then by using Partial Fractions , I got the answer as
$$
\frac{1}{10} \log (t+3) - \frac{1}{20} \log (t^2 + 1) + \frac{3}{10} \arctan (t) \biggr|_{0}^{\infty}
$$
But while substituting the limits , the answer comes out be be infinity which is wrong .
Is there any mistake in my approach ??
| $$\int_0^\infty \frac{ 1 }{(1+t^2)(t+3)} \, dt =\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{10}}}{\displaystyle\int_0^\infty}\dfrac{1}{t+3}\,\mathrm{d}t-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{10}}}{\displaystyle\int_0^\infty}\dfrac{t-3}{t^2+1}\,\mathrm{d}t\\=\dfrac{\ln\left(\left|t+3\right|\right)}{10}-\dfrac{\ln\left(t^2+1\right)}{20}+\dfrac{3\arctan\left(t\right)}{10}\biggr|_{0}^{\infty}\\$$
$$=\frac1{20}\ln\frac{(t+3)^2}{1+t^2}+\dfrac{3\arctan\left(t\right)}
{10}\biggr|_{0}^{\infty}$$
$$\lim_{t \to \infty}\frac1{20}\ln\frac{(t+3)^2}{1+t^2}=\frac 1 {20}\ln1=0$$
$$=0+\frac{3 \pi}{20}-\left(\frac{1}{20}\ln9+0\right)$$
$$=\frac{3\pi-\ln9}{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}\ \ (1)$$
By substituting (1) in the LHS of the original expression,
it is straightforward to show that it is equal to the RHS.
But my problem is on how to show without the un-nesting, by another approach. I've used a standard approach of
multiplying both terms by $\sqrt{2}+\sqrt{2+\sqrt{3}}$, leading to
$$\frac{4+\sqrt{3}+2\sqrt{4+2\sqrt{3}}}{-\sqrt{3}}\Leftrightarrow \frac{(4+\sqrt{3}+2\sqrt{4+2\sqrt{3}})\sqrt{3}}{-3}.$$
But I was not able to show that this last expression is equal to RHS by this approach.
Hints and answers not using my first un-nesting approach (if possible) will be appreciated. Sorry if this is a duplicate.
| The simple, "natural" solution is using direct radical denesting, as noted by the OP. But any algebraic solution will likely involve some form of "denesting" eventually, as noted by Misha Lavrov in a comment. For example, the following is a rather convoluted, very "artificial" solution using polynomial resultants.
Let $\,t_{1,2} = \sqrt{2} \pm \sqrt{2+\sqrt{3}}\,$, then $\,t_1+t_2=2 \sqrt{2}\,$ and $\,t_1t_2=-\sqrt{3}\,$, so $\,t_{1,2}\,$ are the roots of:
$$
t^2 - 2 \sqrt{2} \,t - \sqrt{3} = 0 \tag{1}
$$
Let $\,u = t_1 / t_2\,$, then $\,t_2\,$ and $\,u t_2 = t_1\,$ both satisfy $\,(1)\,$, so there is a common root between $\,(1)$ and:
$$
u^2 t^2 - 2 \sqrt{2} \,ut - \sqrt{3} = 0 \tag{2}
$$
Eliminating $\,t\,$ between $\,(1)\,$ and $\,(2)\,$ by means of resultants gives a quartic in $\,u\,$, namely $\,\operatorname{res}_{t}(\,t^2 - 2 \sqrt{2} \,t - \sqrt{3}\,,\; u^2 t^2 - 2 \sqrt{2} \,ut - \sqrt{3}\,) = 0\,$. Obviously $\,u=1\,$ must be a root, and in fact a double root, which helps with factoring it as:
$$
3 u^4 + 8 \sqrt{3} u^3 - 16 \sqrt{3} u^2 - 6 u^2 + 8 \sqrt{3} u + 3 = (u-1)^2\big(3 u^2 + 2\,(3 + 4 \sqrt{3})\, u + 3\big) \tag{3}
$$
Since we are looking for $\,u \ne 1$, our $\,u\,$ must be a root of the latter quadratic factor. It can be shown that it's the smaller of the two roots, so $\,u = \frac{1}{3}\left(-(3+4\sqrt{3}) - 2 \sqrt{6\,(2+\sqrt{3})}\right)\,$. However, simplifying it to $\,-3 - 2 \sqrt{3}\,$ requires the same radical denesting that would have easily solved the problem in the first place.
[ EDIT ] If, however, the end result is known in advance, then it is fairly easy to prove it directly without any explicit denesting. By rearranging the terms and isolating the nested radical:
$$
\begin{align}
\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3} \quad\iff\quad \sqrt{2+\sqrt{3}} &= \sqrt{2}\,\frac{4+2 \sqrt{3}}{2+ 2 \sqrt{3}}\\
&= \sqrt{2}\,\frac{2 + \sqrt{3}}{1 + \sqrt{3}} \cdot \color{red}{\frac{\sqrt{3}-1}{\sqrt{3}-1}} \\
&= \sqrt{2}\,\frac{1+\sqrt{3}}{2}
\end{align}
$$
Since both sides of the latter equality are positive, it is enough to show that their squares are equal:
$$
\left(\sqrt{2+\sqrt{3}}\right)^2 = \left(\sqrt{2}\,\frac{1+\sqrt{3}}{2}\right)^2 \quad\iff\quad 2 + \sqrt{3} = 2 \cdot \frac{4 + 2 \sqrt{3}}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that of a hyperbola.
| Here is a "non-calculus" way that plays the whole show back to AMGM. It is a bit cumbersome but works.
I prefer giving all stepwise substitutions to show how to bring the whole expression back to hyperbolic functions where AMGM suddenly gives all. The basic idea behind it is that $\cosh t = \sqrt{\sinh^2 t + 1}$:
$$\color{blue}{y=} x + 2\sqrt{x^2 - \sqrt{2}x + 1}$$
$$x^2 - \sqrt{2}x + 1 = (x - \frac{\sqrt{2}}{2})^2 + 1 - \frac{1}{2} = (x - \frac{\sqrt{2}}{2})^2 + \frac{1}{2}$$
$$\color{green}{u =x - \frac{\sqrt{2}}{2}} \Rightarrow \color{blue}{y=} u + \frac{\sqrt{2}}{2} +2\sqrt{u^2+\frac{1}{2}} = \color{blue}{\frac{\sqrt{2}}{2} + u + \sqrt{2}\sqrt{\left(\sqrt{2} u\right)^2 + 1}}$$
$$\color{green}{v = \sqrt{2} u} \Rightarrow \color{blue}{y =} \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} v + \sqrt{2}\sqrt{v^2 + 1} = \color{blue}{\frac{\sqrt{2}}{2} + \sqrt{2} \left(\boxed{ \frac{v}{2} + \sqrt{v^2 + 1}} \right)}$$
$$\color{green}{v = \sinh t} \Rightarrow \boxed{ \frac{\sinh t}{2} + \cosh t} = \frac{e^t - e^{-t}}{4} + \frac{e^t + e^{-t}}{2} = \frac{3}{4}e^t + \frac{1}{4} e^{-t} \boxed{\stackrel{AMGM}{\geq}} \sqrt{\frac{3}{4}} = \boxed{\frac{\sqrt{3}}{2}}$$
Setting $\color{green}{w= e^t}$, equality holds for
$$3w = \frac{1}{w} \stackrel{w>0}{\Leftrightarrow} w =\frac{1}{\sqrt{3}} \Rightarrow
\color{blue}{y \geq} \frac{\sqrt{2}}{2} + \sqrt{2}\frac{\sqrt{3}}{2} = \color{blue}{\frac{\sqrt{2}}{2}\left( 1 + \sqrt{3}\right)}$$
Note that
$$\left( \frac{\sqrt{2}}{2}\left( 1 + \sqrt{3}\right) \right)^2 =\frac{1}{2}\left( 1 + 2\sqrt{3} + 3\right) = 2 + \sqrt{3} $$
Backwards substitution yield $x$:
$$\color{green}{t =} \ln \frac{1}{\sqrt{3}} = \color{green}{-\ln \sqrt{3}} \Rightarrow \color{green}{v =} \sinh t = \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{2} = \color{green}{-\frac{\sqrt{3}}{3}}$$ $$ \Rightarrow \color{green}{x =} \frac{\sqrt{2}}{2}v+\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \cdot \left(- \frac{1}{\sqrt{3}} \right) + \frac{1}{\sqrt{2}} = \color{green}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$. Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.
$$\begin{align}
x^2 + y^2 &= r^2\sin^2(\theta)\\
z^2 &= r^2 \cos(\theta) \\
x^2 + y^2 - z^2&=r^2(\sin^2(\theta) - \cos^2(\theta)) = 4
\end{align}$$
Thus, $$-r^2(\cos(2\theta)) = 4$$
Is this right?
| The equation given is correct, but we can simplify further. Multiply by $-\sec 2\theta$ and take the square root to solve for $r$ which is $>0$. Thus $r=2\sqrt{-\sec(2\theta)}$.
Note that the result appears to be the square root of a negative number ... unless the secant function is negative. Recalling where the cosine function and thus the secant function is negative, conclude that $\pi/4<\theta<3\pi/4$. This goes along with the hyperboloid shape you might recognize from the rectangular equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The ways of covering a $4\times 4$ square by $1\times 2$ colored dominoes I'm stuck with this question
We have eight $1\times 2$ tiles that each one of them has one $1\times 1$ blue square and one $1\times 1$ red square. We want to cover a $4\times 4$ area with these tiles in a way that every row and every column of this area had exactly $2$ blue and $2$ red $1\times 1$ squares. In how many ways we can do this?
First of all, I tried to find out all possible ways which I can cover a $4\times 4$ square by $1\times 2$ dominoes. If I know all such possibilities, then for each such a covering, by obtaining the number of ways that some one can tiles this dominoes in blue-red squares in the manner that described above, we can get the answer. In the article How Many Ways Can We Tile a Rectangular Chessboard With Dominoes? the writer claims that the number of ways which we can tile a $4\times 4$ rectangle is $36$. But he did not described all this $36$ ways.
The above method is very prolix and also, I need to obtain all the possibilities of covering $4\times 4$ square by $1\times 2$ tiles and for each of such covering, all the ways which can satisfied by the conditions of question. Dose any one know a simpler method?
I should point out that the above question is designed for high school students and so I think it must have a simple solution.
| It actually should not be too bad (if a little tedious) to compute case-by-case with judicious use of symmetry.
I'll use a $4 \times 4$ matrix with entries $b$ (for blue), $r$ (for red) and $\cdot$, to denote the number of tilings of the $\cdot$ spaces given the colours of the $b$ and $r$ spaces. By symmetry,
the cases where the top left square is either $b$ or $r$ and either in a horizontal or vertical domino have equal tiling counts, thus
$$ \text{answer} = 4 \pmatrix{b & r & \cdot& \cdot\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr}$$
Now the other two elements in the top row could either be in one domino (which could be in either orientation, or two vertical dominos, of which the second must be the reverse of the first.
By symmetry, these cases with vertical dominos have the same counts.
Thus
$$ \pmatrix{b & r & \cdot& \cdot\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} = \pmatrix{b & r & b& r\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} + \pmatrix{b & r & r & b\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} + 2 \pmatrix{b & r & r & b\cr \cdot & \cdot & b& r\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} $$
Next, consider the first term on the right above. The (2,1)
entry could be $b$ in a horizontal domino (in which case there must be two $(r,b)$ horizontal dominos below it), or $r$ in a horizontal domino (related by symmetry to the last term above),
or $b$ or $r$ in a vertical domino (in either case with a $(r,b)$ horizontal domino below it). Thus
$$ \pmatrix{b & r & b& r\cr \cdot & \cdot& \cdot& \cdot\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} =
\pmatrix{b & r & b& r\cr b & r & \cdot& \cdot\cr r &b &\cdot &\cdot\cr r & b& \cdot&\cdot\cr} + \pmatrix{b & r & r & b\cr \cdot & \cdot & b& r\cr\cdot &\cdot &\cdot &\cdot\cr\cdot & \cdot& \cdot&\cdot\cr} + \pmatrix{b & r & b& r\cr b & \cdot& \cdot& \cdot\cr r &\cdot &\cdot &\cdot\cr r & b & \cdot&\cdot\cr}
+ \pmatrix{b & r & b& r\cr r & \cdot& \cdot& \cdot\cr b &\cdot &\cdot &\cdot\cr r & b & \cdot&\cdot\cr} $$
I'll leave the rest to you...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
| That the square roots are $\pm(\sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
\begin{align}
& \left|-1 + i4\sqrt 3\right| = \sqrt{(-1)^2 + (4\sqrt 3)^2 } = 7. \\[10pt]
\text{Therefore } & -1+i4\sqrt 3 = 7(\cos\varphi + i\sin\varphi). \\[10pt]
\text{Therefore } & \pm\sqrt{-1+i4\sqrt 3} = \pm\sqrt 7 \left( \cos \frac \varphi 2 + i \sin\frac\varphi 2 \right).
\end{align}
Notice that
$$
\sin \varphi = \frac{4\sqrt 3} 7 \quad \text{and} \quad \cos\varphi = \frac{-1} 7
$$
and recall that
\begin{align}
\tan\frac\varphi 2 & = \frac{\sin\varphi}{1+\cos\varphi} \\[12pt]
\text{so we have }\tan\frac\varphi 2 & = \frac{4\sqrt 3}{7-1} = \frac 2 {\sqrt 3}. \\[10pt]
\text{Therefore } \sin\frac\varphi2 & = \frac 2 {\sqrt 7} \quad \text{and} \quad \cos\frac\varphi2 = \frac{\sqrt3}{\sqrt7}.
\end{align}
Thus the desired square roots are
$$
\pm \left( \sqrt 3 + 2i \right).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
A hotel has 6 single rooms, 6 double rooms and 4 rooms in which 3 persons can stay. In how many ways can 30 persons be accommodated in this hotel? I wanted to know if my answer and the method I used is correct or not...
First, we pick $6$ out of $30$ people for the $6$ single rooms = $^{30}C_6$
They can be in any of rooms $\implies ^{30}C_6 \cdot 6!$
Then, we pick $12$ people out of remaining $24$ and pair them up for $6$ dual rooms $= \dfrac{^{24}C_{12} \cdot 12!}{(2^6 \cdot 6!)}$
Next, we take remaining people and form groups of $\implies \dfrac{12!}{(4^3 \cdot 3!)}$
Finally,we multiply all three terms...
Ans $ = \dfrac{^{30}C_6 \cdot 6! \cdot 12! \cdot {^{24}C_{12} \cdot 12!}}{(2^6 \cdot 6!) \cdot {(4^3 \cdot 3!)}} $
But there can be other ways like first choosing triplets then pairs then singles or pairs then triplets then singles and so on, so do we ignore them or add them?
| As stated in the comments, the flaw in your answer is that while you treated the rooms as distinguishable for the single rooms, you did not do so for the rooms that accommodate more than one person.
In how many ways can a hotel with six single rooms, six double rooms, and four rooms in which three people can stay assign $30$ guests to those rooms?
Let's assume that the rooms are numbered and that we assign people to rooms of the same size in order of increasing room number.
We can select six lucky people to receive single rooms in $\binom{30}{6}$ ways and assign them to those rooms in $6!$ ways. Thus, we can assign six people to the single rooms in
$$\binom{30}{6}6!$$
ways.
That leaves us with $24$ people. We can assign two of them to each of the six rooms that accommodate two people in
$$\binom{24}{2}\binom{22}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2}$$
ways.
That leaves us with $12$ people. We can assign three of them to each of the four rooms that accommodate three people in
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
ways.
Thus, the number of ways to assign $30$ people to the rooms in the hotel is
$$\binom{30}{6}6! \cdot \binom{24}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2} \cdot \binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
Simplifying this expression yields
$$\binom{30}{1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3} = \frac{30!}{1!1!1!1!1!1!2!2!2!2!2!2!3!3!3!3!}$$
To see why, imagine lining up $30$ people in some order, assigning the first six people to the singles in increasing room number order, assigning the next six (disjoint) pairs of people to the doubles in increasing room number order, and assigning the next four (disjoint) trios of people in increasing room number order. The numbers in the denominator represent the number of orders in which we assign the same set of people to the same room without creating a different set of room assignments.
Note that this number does not depend on the order in which the people are assigned to the rooms.
Where did you make a mistake?
You correctly counted the number of assignments of six people to single rooms.
The number of ways we can assign twelve of the remaining twenty-four people to the six doubles is
$$\binom{24}{2}\binom{22}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2} = \frac{24!}{2!2!2!2!2!2!12!}$$
which can be expressed in the form
$$\binom{24}{12} \cdot \frac{12!}{2!2!2!2!2!2!}$$
By dividing this number by $6!$, you treated the rooms as indistinguishable.
You made the analogous error in assigning the remaining twelve people to the four triples.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability kth outcome is not a repetition
A fair die is tossed $k$ times. What is the probability that the kth
outcome is ${\bf not}$ a repetition?
attempt.
First, the size of our sample space is $6^k$. We only have six possible outcomes, so for any outcome, say $O$, that occur at the kth try, we want the previous outcomes (k-1 tries) to be not $O$, so in how many ways can we do this?
Well, $5 \cdot ... \cdot 5 \cdot 1= 5^k$ so
$$ Prob = \frac{5^k}{6^k} $$
is this correct? How would we handle the case when we have $2$ dice instead?
| Your approach looks sensible, but would lead to $\dfrac{5^{k-1}}{6^{k-1}}$ for one die
Another way to get there would be consider the possible outcomes $1,2,3,4,5,6$ for the $k$th die and combine the probabilities of these with the conditional probabilities of the previous dies being different so $\frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1} + \frac16\left(\frac56\right)^{k-1}$ to give the same result
So for two dice following a similar approach you would then get $\frac1{36}\left(\frac{36-1}{36}\right)^{k-1} + \frac2{36}\left(\frac{36-2}{36}\right)^{k-1} + \frac3{36}\left(\frac{36-3}{36}\right)^{k-1} + \frac4{36}\left(\frac{36-4}{36}\right)^{k-1} +$ $\frac5{36}\left(\frac{36-5}{36}\right)^{k-1} + \frac6{36}\left(\frac{36-6}{36}\right)^{k-1} + \frac5{36}\left(\frac{36-5}{36}\right)^{k-1} + \frac4{36}\left(\frac{36-4}{36}\right)^{k-1} + $ $\frac3{36}\left(\frac{36-3}{36}\right)^{k-1} + \frac2{36}\left(\frac{36-2}{36}\right)^{k-1} + \frac1{36}\left(\frac{36-1}{36}\right)^{k-1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Triangle numbers that are squares of triangle numbers. What are the triangle numbers the are squares of other triangle numbers? I have found $1^2=1$ and $6^2=36$, but other than these examples I can't find any other triangle numbers that are squares of other triangle numbers, and I used a program to check this idea into the thousands.
Finding triangle numbers that are squares of other triangle numbers corresponds to finding integers $n$ and $k$ such that $n(n+1)/2=[k(k+1)/2]^2$, or such that
$$2n(n+1)=k^2(k+1)^2 .$$
I believe the only positive integer solutions to this equation are $(1,1)$ and $(3,8)$, but I don't know how to prove it. Are there any others?
| Note that $n$ and $n+1$ are coprime, as are $k$ and $k+1$. We can have both sides zero if $n=0,-1, k=0,-1$. Otherwise we must have either $n=k^2,2(n+1)=(k+1)^2$ or $n+1=k^2,2n=(k+1)^2$. The first gives
$$n=k^2\\2(n+1)=(k+1)^2\\2k^2+2=k^2+2k+1\\k^2-2k+1=0\\k=1\\n=1$$ while the second gives $$n+1=k^2\\2n=(k+1)^2\\2k^2-2=k^2+2k+1\\k^2-2k-3=0\\k=-1,3\\n=0,8$$
and you have found all the positive solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$
So
$$
\begin{split}
\left|z-\frac{1}{z}\right|
&=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\
&=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\
&=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}
\end{split}
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$
Now, comes the maximum value.
We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$
$$=\sqrt{6+\dfrac{1}{4}-a^2}$$
Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
| Hint: $z - \frac 1z = z - \frac {\overline z}{z\overline z} = z - \frac {\overline z}{|z|^2} = z - \frac {\overline z}4 = \frac 34 Re(z) -i\frac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $\sqrt {\frac {9}{16}a^2 + \frac {25}{16}b^2}$ given that $a^2 + b^2 = 4$.
Intuitively[2] I'd say that as $\frac {25}{16} > \frac {9}{16}$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.
As $a^2, b^2 \ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.
So max is $\sqrt {\frac {25}{16}*4} = \frac 52$ and min is $\sqrt{\frac {9}{16}*4} =\frac 32$.
====
[1]. There probably no need for this, but we can create an identity:
$z \pm \frac 1z = (1\pm \frac 1{|z|^2})Re(z) + i(1 \mp \frac 1{|z|^2})Im(z)$ for any $z\ne 0$.
[2] And formally I'd say $\sqrt {\frac {9}{16}a^2 + \frac {25}{16}b^2} =\sqrt {\frac {25}{16}b^2 + \frac {9}{16}(4-b^2)}=\sqrt{b^2 + \frac 94}$ which is max/min when $b$ is max/min.
Actually formally was easier than my intuition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Polar integral $\int_{r=0}^{\infty} \frac{b r dr}{(r^2+b^2)^{(3/2)}(r^2+a^2)^{(1/2)}}$ We need to solve above integral which I obtained after manipulation. Now I used $r^2 = u $ to get:
$$\int_0^\infty \frac{bdu}{2 (u+b^2)^{(3/2)}(u+a^2)^{(1/2)}}$$
which is converted to:
$$\int_0^\infty \frac{bdu}{2 (u+b^2) \sqrt{\left(u+\frac{a^2+b^2}{2}\right)^2 - \left(\frac{a^2-b^2}{2}\right)^2}}$$
Now I am having trouble reducing it. Please give only a hint! Thank you a lot!
| HINT:
Denote $I$ the given integral. Integration by parts leads to a function and an integral that is a constant multiple of $I.$
Solution (not accomplished, based on the above hint)
For
$$I(b,a)=\int_{r=0}^{\infty} \frac{b r\; dr}{(r^2+b^2)^{(3/2)}(r^2+a^2)^{(1/2)}}$$ apply integration by parts considering $u(r)=\frac{1}{(r^2+a^2)^{1/2}}$ and $v '(r)=\frac{br}{(r^2+b^2)^{3/2}}.$
We get $$I=\left[\frac{-b}{\sqrt{(r^2+a^2)(r^2+b^2)}}\right]_{0}^{\infty}-\frac{b}{a}\int_{r=0}^{\infty} \frac{a r\; dr}{(r^2+b^2)^{(1/2)}(r^2+a^2)^{(3/2)}}
$$
$$I(b,a)=\left[0-\frac{-b}{|ab|}\right]-\frac{b}{a}I(a,b).$$ From this we can compute $I(b,a)+\frac{b}{a}I(a,b),$ but just $I(b,a)$ is wanted.
SOLUTION with Euler's substitution
$$I=\int_{0}^{\infty} \frac{b r\; dr}{(r^2+b^2)^{(3/2)}(r^2+a^2)^{(1/2)}}=\int_{0}^{\infty} \frac{b r\; dr}{(r^2+b^2)^{2}\cdot {\sqrt{\frac{r^2+a^2}{r^2+b^2}}}}$$
Set $t=\sqrt{\frac{r^2+a^2}{r^2+b^2}},$ then $r^2=\frac{a^2-b^2t^2}{t^2-1},$ consequently $rdr=\frac{t(b^2-a^2)}{(t^2-1)^2}dt$ and $r^2+b^2=\frac{a^2-b^2}{t^2-1}.$ We obtain $$I=\int_{|\frac{a}{b}|}^1 \frac{b}{b^2-a^2}dt,$$ easy to finish.
| {
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"url": "https://math.stackexchange.com/questions/2955315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the remainder when $13^{13}$ is divided by $25$. Find the remainder when $13^{13}$ is divided by $25$.
Here is my attempt, which I think is too tedious:
Since $13^{2} \equiv 19 (\text{mod} \ 25),$ we have $13^{4} \equiv 19^{2} \equiv 11 (\text{mod} \ 25)$ and $13^{8} \equiv 121 \equiv 21 (\text{mod} \ 25).$
Finally, we have $13^{8+4} \equiv 13^{12} \equiv 21\times 11 \equiv 231 \equiv 6 (\text{mod} \ 25)$ and hence $13^{13} \equiv 3 (\text{mod} \ 25).$
Is there a less tedious way to find the remainder? Thank you.
| $\!\bmod 25\!:\ 13\equiv \overbrace{2^{\large -1}\!\equiv 3^{\large -3}}^{\Large 2\ \, \equiv\,\ 3^{\LARGE 3}}$ $\Rightarrow 13^{\large 13}\!\equiv 3^{\large -39}\!\equiv 3,\ $ by $\ 3^{\large 40}\!\equiv\!\!\! \overbrace{(3^{\large \color{#c00}{20}})^{\large 2}\!\equiv 1}^{\large\quad\ \ \color{#c00}{20}\ =\ \phi(25)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956834",
"timestamp": "2023-03-29T00:00:00",
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solve for $x$: $(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$ There is an equation that I think it is complicated ,a little!
$$(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$$
Actually we must solve for $x$ here.
I want you to hint me how can I simplify the equation and solve it.
| Hint:
Observe that for real $x$
each of the summands must be $=0$
Now $x^2+x-6=(x+3)(x-2)$
$$x^4-13x^2+36=(x^2-4)(x^2-9)$$
$$x^3-7x+6=x^3-2^3-7(x-2)$$ or try divide $x^3-7x+6$ by $x-2$
| {
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"url": "https://math.stackexchange.com/questions/2963146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need a hint on how to solve this inequality I want to show that, whenever $\frac{u_1^2}{p^2}+\frac{u_2^2}{q^2} \leq 1$ and $\frac{v_1^2}{p^2}+\frac{v_2^2}{q^2} \leq 1$, then
$$
\frac{(\lambda \, u_1 + (1 - \lambda) \, v_1)^2}{p^2} + \frac{(\lambda \, u_2 + (1 - \lambda) \, v_2)^2}{q^2} \leq 1
$$
for all $0 \leq \lambda \leq 1$.
My (failed) attempt: I tried to apply the Cauchy-Schwarz inequality and got
$$
\frac{(\lambda \, u_1 + (1 - \lambda) \, v_1)^2}{p^2} + \frac{(\lambda \, u_2 + (1 - \lambda) \, v_2)^2}{q^2} \leq
\left(\lambda^2 + (1-\lambda)^2\right)
\left(
\frac{u_1^2}{p^2} + \frac{u_2^2}{q^2} + \frac{v_1^2}{p^2} + \frac{v_2^2}{q^2}
\right) \leq 2
$$
What is the correct way to approach this?
| Using simpler variable names,
we are given
$\frac{a^2}{p^2}+\frac{c^2}{q^2}
\le 1$
and
$\frac{b^2}{p^2}+\frac{d^2}{q^2}
\le 1$.
$\begin{array}\\
\frac{(ra + (1 - r) b)^2}{p^2} + \frac{(rc + (1 - r)d)^2}{q^2}
&=\frac{r^2a^2+2r(1-r)ab+(1-r)^2b^2}{p^2} + \frac{r^2c^2+2r(1-r)cd+(1-r)^2d^2}{q^2}\\
&=r^2(\frac{a^2}{p^2}+\frac{c^2}{q^2})+2r(1-r)(\frac{ab}{p^2}+\frac{cd}{q^2})+(1-r)^2(\frac{b^2}{p^2}+\frac{d^2}{q^2})\\
&\le r^2+2r(1-r)(\frac{ab}{p^2}+\frac{cd}{q^2})+(1-r)^2\\
\end{array}
$
If we can show that
$\frac{ab}{p^2}+\frac{cd}{q^2}
\le 1$,
the upper bound is
$r^2+2r(1-r)+(1-r)^2
=(r+(1-r))^2
=1
$
and we are done.
But
$\begin{array}\\
(\frac{ab}{p^2}+\frac{cd}{q^2})^2
&=(\frac{a}{p}\frac{b}{p}+\frac{c}{q}\frac{d}{q})^2\\
&\le(\frac{a^2}{p^2}+\frac{c^2}{q^2})(\frac{b^2}{p^2}+\frac{d^2}{q^2})
\quad\text{by Cauchy-Schwarz}\\
&\le 1\\
\end{array}
$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
| Easy to see that $ab<0$.
Let $ab=x$.
Thus, we have $$(a^4+a^3)(b^4+b^3)=1$$ or
$$a^3b^3(ab+a+b+1)=1$$ or
$$x^3(x+a+b+1)=1$$ or
$$a+b=\frac{1-x^3-x^4}{x^3}.$$
Also, we have
$$\frac{a^4+a^3-1-(b^4+b^3-1)}{a-b}=0$$ or
$$a^3+a^2b+ab^2+b^3+a^2+ab+b^2=0$$ or
$$(a+b)^3-2ab(a+b)+(a+b)^2-ab=0,$$ which gives $$\left(\frac{1-x^3-x^4}{x^3}\right)^3-2x\cdot\frac{1-x^3-x^4}{x^3}+\left(\frac{1-x^3-x^4}{x^3}\right)^2-x=0$$ or
$$(x^6+x^4+x^3-x^2-1)(x^6-x^4-x^3-x^2+1)=0$$ and since
$x^6-x^4-x^3-x^2+1>0$ for $x<0$,
we obtain that one of roots it's $ab$ and the second it's $-\frac{1}{ab}.$
The proof that $ab<0$.
Indeed, let $f(x)=x^4+x^3-1$.
Thus, $f'(x)=x^2(4x+3),$ which says that $f$ increases on $\left[-\frac{3}{4},+\infty\right)$ and on $[0,1]$.
But $f(0)f(1)<0$, which says that $f$ has unique positive root.
Also, $f(-1)f(-2)<0$ and since $f$ decreases on $\left(-\infty,-\frac{3}{4}\right],$
we see that $f$ has unique negative root.
Id est, $ab<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$ The equation is
$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2\cdot\sqrt{-\frac{(x-4)^2}{x^2+x}}+4\cdot\sqrt{x^2-12}=x^2-8-\frac{4-x}{x^2+x}$$
| Left side exist only iff $${\frac{4-x}x}\geq 0\iff x\in (0,4]$$
and $${\frac{x-4}{x+1}}\geq 0\iff x\in (-\infty,-1)\cup [4,\infty)$$
So the only legitimate value for left side is $4$ which works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Glueing two affine curves along a map Let $C = V(y^2 - x^4 - 1) \subset \Bbb A^2$ and set
$C_0 = C \times \{0\}, C_1 = C \times \{1\} \subset \Bbb A^3$.
Consider the space $X$ obtained by quotienting $C_0 \sqcup C_1$ but the relation $((x, y) ; 0) \sim (1/x, y/x^2)$ for all $y$ and all $x \neq 0$.
According Miranda III.1.7, $X$ is a smooth projective curve (and hyperelliptic).
Thus $X = Proj(S)$ for some graded ring $S$.
My question: can you describe explicitly the ring $S$ (up to isomorphism of $k$-algebras)?
Typically, the comments in this question tells that if we replace $C$ by $\Bbb A^1$ and $\sim$ by $(x;0) \simeq (1/x, 1)$, we get $S = k[x_0,x_1]$, yielding $X = \Bbb P^1_k$. But what is $S$ in our case?
| Let $S = \frac{k[X,Y,Z]}{(Y^2 - (X^4 + Z^4))}$, where we give $Y$ weight $2$, and $X$ and $Z$ weight $1$. Then the polynomial $F = Y^2 - (X^4 + Z^4)$ is weighted homogeneous of degree $4$, so the quotient $S$ is a graded ring. Thus the completed curve $\overline{C}$ is given by the equation $Y^2 = X^4 + Z^4$ in the weighted projective space $\mathbb{P}(1,2,1)$. (In general, one usually considers a hyperelliptic curve of genus $g$ as living in $\mathbb{P}(1, g+1, 1)$.)
We can see that this is the ring obtained from the gluing construction Miranda gives by looking at affine open subsets of $\operatorname{Proj}(S)$. On the affine open $U_2$ where $Z \neq 0$, we have affine coordinates
$$
x = \frac{X}{Z} \quad \text{and} \quad y = \frac{Y}{Z^2}
$$
and the dehomogenization of $F$ with respect to $x$ and $y$ is
$$
y^2 = \frac{Y^2}{Z^4} = \frac{X^4}{Z^4} + 1 = x^4 + 1 \, .
$$
On the affine open $U_0$ where $X \neq 0$, we have affine coordinates
$$
z = \frac{Z}{X} \quad \text{and} \quad w = \frac{Y}{X^2}
$$
and the dehomogenization of $F$ with respect to $z$ and $w$ is
$$
w^2 = \frac{Y^2}{X^4} = 1 + \frac{Z^4}{X^4} = 1 + z^4 \, .
$$
The change of coordinates map on $U_0 \cap U_2$ is given by
$$
z = \frac{Z}{X} = \frac{1}{x} \qquad \qquad w = \frac{Y}{X^2} = \frac{Y/Z^2}{X^2/Z^2} = \frac{y}{x^2} \, ,
$$
which shows that we recover Miranda's gluing construction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Tips on additional methods to solve inequality involving multiple variables So I want some feedback about my proof of the following implication:
Let $a,b,c,d\in\mathbb{Z}$ where $a,b,c,d>0$. Prove that if $\frac{a}{b}<\frac{c}{d}$, then
$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
My attempt went as follows:
If $\frac{a}{b}<\frac{c}{d}$, then
$$ad<cb$$
Case 1: $\frac{a}{b}<\frac{a+c}{b+d}$
$$\frac{a}{b}=\frac{a}{b}\left(\frac{b+d}{b+d}\right)=\frac{ab+ad}{b(b+d)}$$
Since $ad<cb$
$$ab+ad<ab+cb\implies\frac{ab+ad}{b(b+d)}<\frac{ab+cb}{b(b+d)}\implies\frac{ab+ad}{b(b+d)}<\frac{b(a+c)}{b(b+d)}$$
$$\implies\frac{ab+ad}{b(b+d)}<\frac{a+c}{b+d}\implies\frac{a}{b}<\frac{a+c}{b+d}$$
Case 2: $\frac{a+c}{b+d}<\frac{c}{d}$
$$\frac{c}{d}=\frac{c}{d}\left(\frac{b+d}{b+d}\right)=\frac{cb+cd}{d(b+d)}$$
Since $ad<cb$
$$ad+cd<cb+cd\implies\frac{ad+cd}{d(b+d)}<\frac{cb+cd}{d(b+d)}\implies\frac{d(a+c)}{d(b+d)}<\frac{cb+cd}{d(b+d)}$$
$$\implies\frac{a+c}{b+d}<\frac{cb+cd}{d(b+d)}\implies\frac{a+c}{b+d}<\frac{c}{d}$$
So
$$\frac{a}{b}<\frac{a+c}{b+d}\land\frac{a+c}{b+d}<\frac{c}{d}\implies\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
So I'm quite sure this proof is a valid one, but I was wondering if there was a faster way of doing this. Maybe not requiring to break into two different cases but instead solving the compound inequality at once for both sides, or if there won't be much of a difference for both ways. Any tips would be appreciated.
| I think your proof is right, but I like the following way.
$$\frac{a+c}{b+d}-\frac{a}{b}=\frac{bc-ad}{b(b+d)}>0$$ and
$$\frac{c}{d}-\frac{a+c}{b+d}=\frac{bc-ad}{d(b+d)}>0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having trouble in figuring out the best approach to simplify $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$
so that I can find the value of $
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Hope somebody has an idea.
| Square both sides to get
$$b=-\sqrt{(a+b)(b+c)}$$
square again
$$0=ab+ac+bc$$
divide by $abc$
and finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $z_n$ is a perfect square if $z_0 = z_1 = 1$ and $z_{n+1} = 7z_n − z_{n−1} − 2$ Let $z_0 = z_1 = 1$ and $$z_{n+1} = 7z_n − z_{n−1} − 2$$ for all positive integers $n$. How is it possible to show that $z_n$ is a perfect square for all $n$?
| I answered the last one this way, let me make it easier reading
LEMMA if $$w_{n+3} - (M+1)w_{n+2} + (M+1)w_{n+1} - w_n = 0,$$ then $w_{n+2} - M w_{n+1} + w_n$ is CONSTANT
PROOF: $$ (w_{n+3} - M w_{n+2} + w_{n+1}) - w_{n+2} + M w_{n+1} - w_n = 0 \; , $$
$$w_{n+3} - M w_{n+2} + w_{n+1} = w_{n+2} - M w_{n+1} + w_{n} $$
This is constant for all $n$ by induction.
Let $(x_n,y_n)$ begin with $(1,0),$ $(1,1),$ $(2,3)$ be solutions to $x^2 + xy-y^2 = 1,$ with the rule for getting to the next solution is
$$
\left(
\begin{array}{cc}
1&1\\
1&2
\end{array}
\right)
\left(
\begin{array}{c}
x_n\\
y_n
\end{array}
\right) =
\left(
\begin{array}{c}
x_{n+1}\\
y_{n+1}
\end{array}
\right)
$$
It is easy to check that if $(a,b)$ satisfies $x^2 + xy-y^2 = 1,$ so does $(a+b,a+2b).$
It follows from Cayley-Hamilton that $x_n$ satisfies
$$ x_{n+2} - 3 x_{n+1} + x_n = 0. $$
Analogous for $y_n.$
Next, from Fricke and Klein (1897) or direct calculation,
$$
\left(
\begin{array}{ccc}
1&2&1\\
1&3&2 \\
1&4&4
\end{array}
\right)
\left(
\begin{array}{c}
x_n^2\\
x_n y_n \\
y_n^2
\end{array}
\right) =
\left(
\begin{array}{c}
x_{n+1}^2\\
x_{n+1 } y_{n+1} \\
y_{n+1}^2
\end{array}
\right)
$$
The characteristic polynomial of the three by three matrix is
$$ \lambda^3 - 8 \lambda^2 + 8 \lambda - 1 $$
Again by Cayley-Hamilton, we get
$$ x_{n+3}^2 - 8 x_{n+2}^2 + 8 x_{n+1}^2 - x_n^2 = 0 $$
From the lemma at the beginning,
$$ x_{n+2}^2 -7 x_{n+1}^2 + x_n^2 $$
is constant. Since $4 - 7 \cdot 1 + 1 = 5-7 = -2,$ we have
$$ x_{n+2}^2 -7 x_{n+1}^2 + x_n^2 = -2 $$
$$ x_{n+2}^2 =7 x_{n+1}^2 - x_n^2 -2 $$
| {
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"url": "https://math.stackexchange.com/questions/2970014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 0
} |
Continuous Bijection From $S^1$ to $[0,2\pi)$ I came up with the following (I believe) continuous bijection $\theta: S^1 \rightarrow [0,2\pi)$:
$\begin{align} \theta(x,y) & =
\begin{cases}
\arctan(\frac{y}{x}) & x > 0, 0 \leq y < 1 \\
\frac{\pi}{2} & x = 0, y = 1 \\
\arctan(\frac{-x}{y}) + \frac{\pi}{2} & x < 0, 0 < y < 1 \\
\pi & x = -1, y = 0 \\
\arctan(\frac{-y}{-x}) + \pi & x < 0, -1 < y < 0 \\
\frac{3\pi}{2} & x = 0, y = -1 \\
\arctan(\frac{x}{-y}) + \frac{3\pi}{2} & x > 0, -1 < y < 0 \\
\end{cases}
\end{align}$
This looks ugly though. Is there a more natural way to write a continuous bijection from $S^1$ to $[0,2\pi)$?
| you cannot find such a bijection since the image of a compact set by a continuous function is compact, $S^1$ is compact and $[0,2\pi)$ is not compact.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971086",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right)$ I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$
I have computed
\begin{align*}
a_{n+1}-a_n & = \frac{1}{\sqrt{(2n+1)(2n+2)}}-\frac{1}{\sqrt{n(n+1)}}\\
& = \frac{1}{\sqrt{n+1}}\left( \frac{1}{\sqrt{2(2n+1)}}-\frac{1}{\sqrt{n}} \right).
\end{align*}
Now I'm stuck.
| HINT
The hint provided by Chinnapparaj R is the key point, to evaluate the bounding sum we can use that
$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}=\sum_{k=1}^n \frac{1}{n+k}=\frac1n \sum_{k=1}^n \frac{1}{1+\frac kn}$$
which is a Riemann's sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Sum of the series $(i)$ and $(ii)$ Find the sum of the series: $(i)1+\frac{2}{9}+\frac{2.5}{9.18}+\frac{2.5.8}{9.18.27}+\cdots +\infty$$(ii)1+\frac{3}{4}+\frac{7}{16}+\frac{13}{64}+\cdots +\infty$The answer providing my book is :$(i)\frac{9}{4}^{\frac{1}{3}},(ii)\frac{8}{3}$.i couldn't think of how to start.Because i didn't find any helpful pattern to go further.Any hints or solution will be appreciated.
Thanks in advance.
| I still don't know what it's supposed to be .
1) I supposed that numerator always increased by next even number : $3 = 1 + \color{blue}{2}$, then $7 = 3 +\color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $\displaystyle \sum_{k=0}^{\infty} \frac{k(k+1)+1}{4^{k}} = \sum_{k=0}^{\infty} \frac{k(k+1)}{4^{k}} + \sum_{k=0}^{\infty} \frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $\displaystyle S(x) = \sum_{k=0}^{\infty} \frac{x^{k+1}}{4^{k}} = x^{1}\sum_{k = 0}^{\infty} \frac{x^{k}}{4^{k}} = \frac{4x^{1}}{4-x}$ , so you have $\displaystyle \sum_{k=0}^{\infty} \frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (\frac{4x}{4-x})''|_{x=1} = \frac{32}{27}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the laurent series with an added imaginary component Find the laurent series of
$$f(z)=\frac{i}{z^2-iz+2}$$ with the values of the region $a,b$ in which it is valid $$ a<|z-1|<b $$
My attempt at a solution:
I did a partial fraction expansion and found
$$ f(z)=\frac{1}{3(z-2i)} - \frac{1}{3(z+i)} $$
Expanding about the point $|z-1|$, for the upper and lower limits
$$ f(z)=\frac{1}{3(1-2i)(1-(-\frac{(z-1)}{1-2i})} - \frac{1}{3(1+i)(1-(-\frac{z-1}{1+i}))} $$
$$ f(z)=\frac{1}{3(z-1)(1-(-\frac{(1-2i)}{z-1})} - \frac{1}{3(z-1)(1-(-\frac{1+i}{z-1}))} $$
The next step through geometric expansion would be (for the first line above
$$\frac{1}{3(1-2i)} \sum (-\frac{(1-2i)}{z-1})^n -\frac{1}{3(1+1)} \sum (-\frac{(1+i)}{z-1})^n
$$
And so on...
But my final answer gives an
$$a=\sqrt{5}, b=\sqrt{2}$$ thus $$ \sqrt{5} < |z-1| < \sqrt{2} $$
This does not seem correct, but analysing a diagram of the system shows an annulus where the laurent series cannot have an analytical expansion (its either within a disc of radius less than $\sqrt{2}$ or everywhere in the complex plane excluding a disc of radius $\sqrt{5}$ about the point. Is anyone able to give me any insight? Have I made a mistake in my reasoning? Especially assuming I can take a geometric series with imaginary terms?
Much appreciated.
| You have;
$$f(z)=\frac{1}{3(1-2i)}\frac{1}{(1-(-\frac{(z-1)}{1-2i})}-\frac{1}{3(1+i)}\frac{1}{(1-(-\frac{z-1}{1+i}))}$$
Expanding in a geometric series gives;
$$f(z)=\frac{1}{3(1-2i)}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^{n} -\frac{1}{3(1+i)}\sum_{n=0}^{\infty}\left(-\frac{z-1}{1+i}\right)^{n}$$
A geometric series, $\sum_{n=0}^{\infty}r^{n}$ converges for $|r|<1$, so the the first term converges for $|z-1|<|2i-1|=\sqrt{5}$, and the second converges for $|z-1|<|1+i|=\sqrt{2}$, so as $\sqrt{2}<\sqrt{5}$, the expansion $f(z)$ converges for $|z-1|<\sqrt{2}$. Then as the modulus of a complex number is non negative, we have that this expansion converges for;
$$0\leq|z-1|<\sqrt{2}$$
Which you can confirm by drawing a diagram and looking at the positions of the poles. Looking at the poles also tells you that there should also be another expansion valid in the annulus $\sqrt{2}<|z-1|<\sqrt{5}$, and one in the region $\sqrt{5}<|z-1|$.
For the annular expansion write;
$$\frac{1}{1-(-\frac{z-1}{1+i})}=\frac{1+i}{z-1}\frac{1}{1+\frac{1+i}{z-1}}=\frac{1+i}{z-1}\sum_{n=0}^{\infty}\left(\frac{1+i}{z-1}\right)^{n}$$
Which converges for $|1+i|=\sqrt{2}<|z-1|$, so that the expansion;
$$f(z)=\frac{1}{3(1-2i)}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^{n} -\frac{1}{3(z-1)}\sum_{n=0}^{\infty}\left(\frac{1+i}{z-1}\right)^{n}$$
Is valid in the annulus $\sqrt{2}<|z-1|<\sqrt{5}$.
Finally, for the region $|z|>\sqrt{5}$, we expand the first term as;
$$
\frac{1}{3(1-2i)}\frac{1}{1-(-\frac{z-1}{1-2i})}
=
\frac{1}{3(z-1)}\frac{1}{1+\frac{1-2i}{z-1}}
=
\frac{1}{3(z-1)}
\sum_{n=0}^{\infty}\left(\frac{1-2i}{z-1}\right)^{n}
$$
Which converges for $|1-2i|=\sqrt{5}<|z-1|.$
So, as $\sqrt{2}<\sqrt{5}$, we have that the expansion;
$$
f(z)=
\frac{1}{3(z-1)}\sum_{n=0}^{\infty}\left(\frac{1+i}{z-1}\right)^{n}
-
\frac{1}{3(z-1)}
\sum_{n=0}^{\infty}\left(\frac{1-2i}{z-1}\right)^{n}
$$
converges for $\sqrt{5}<|z-1|<\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Functional equation - Cyclic Substitutions Please help solve the below functional equation for a function $f: \mathbb R \rightarrow \mathbb R$:
\begin{align}
&f(-x) = -f(x) , \text{ and } f(x+1) = f(x) + 1, \text{ and } f\left(\frac 1x\right) = \frac{f(x)}{x^2} \\
&\text{ for all } x \in \mathbb R \text{ and } x \ne 0 .
\end{align}
I know this will be solved by cyclic substitutions, but I'm unable to figure out the exact working. Can someone explain step wise?
| I don't know if this is what is meant with "cyclic substitutions", but it is a solution.
First, we observe that
$$
f(0)=0, f(1)=1, f(-1)=-1
$$
has to be true. Also, it suffices to determine $f(x)$ for $x>0$, because the rest follows from the condition $f(-x)=-f(x)$.
Let $x>0$.
Applying some of the conditions, we have
$$
\begin{aligned}
f(x)+1
&= f(x+1) \\
&= f(1(x+1)^{-1})(x+1)^2 \\
&= f(1-x(x+1)^{-1})(x+1)^2\\
&= (1+f(-x(x+1)^{-1}))(x+1)^2\\
&= (1-f(-x(x+1)^{-1}))(x+1)^2\\
&= (1-f(-x(x+1)^{-1}))(x+1)^2\\
&= (1-f((x+1)x^{-1})x^2(x+1)^{-2})(x+1)^2\\
&= (1-f(1+x^{-1})x^2(x+1)^{-2})(x+1)^2\\
&= (1-(1+f(x^{-1}))x^2(x+1)^{-2})(x+1)^2\\
&= (1-(1+f(x)x^{-2})x^2(x+1)^{-2})(x+1)^2\\
&= (x+1)^2-x^2-f(x)\\
\end{aligned}
$$
This yields $f(x)=x$.
| {
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Counting Credit Cards The credit cards (VISA CARDS & MASTER CARDS) numbers have the following properties;
Let $N$ be the card number
*
*$N$ has $16$ digits; $N = a_{1}a_{2}a_{3}...a_{16}$ where $a_{k}$ is the $k$-th digit of $N$
*$a_1$ $\neq 0$
*$2(a_1+a_3+a_5 + ... + a_{15}) + (a_{2}+a_{4}+a_{6}+...+a_{16})$ $+$ count$(\geq 5; a_1,a_3,a_5,...,a_{15})$ $\equiv0$ (mod $10$)
Now consider the following example to clarify the question;
The card number $N$ is $5529\,4203\,5061\,5465$
Clearly $N$ has $16$ digits,
$a_{1}\ne0$,
and $2(5+2+4+0+5+6+5+6) + (5+9+2+3+0+1+4+5)$ $+$ count$(\geq 5; 5,2,4,0,5,6,5,6)$
$=2(33)+29+5=100$ $\equiv 0$ (mod $10$)
How many $N$ are there?
| The answer is $9\cdot10^{14}$. That is, there are $9$ choices for $a_1$ followed by $10$ choices for each of $a_2$ to $a_{15}$, and ending with the unique value
$$a_{16}=-2(a_1+a_3+\cdots+a_{15})-(a_2+a_4+\cdots+a_{14})-\text{count}(\ge5;a_1,a_3\ldots,a_{15})\mod 10$$
Remark: The OP's formula looks different from the Luhn formula for the check digit, which says to double the values of $a_1,a_3,\ldots,a_{15}$ and then sum the digits of those numbers, along with the digits $a_2,a_4,\ldots,a_{14}$. What reconciles the two formulas is the observation that the sum of the digits of $2a$ is congruent mod $10$ to $2a+1$ if $a\ge5$.
| {
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"url": "https://math.stackexchange.com/questions/2976011",
"timestamp": "2023-03-29T00:00:00",
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find all integers $n$ such that $2^{n-1}*n+1$ is a perfect square. Clearly $n=0$
and i found that also $n=5$ gives a perfect square
And By representing the two functions , we found that there are only two solutions that are $n=0,5$
But I don't know how to prove that using elementary number theory.
| Robert Israel has already provided a nice answer, but it seems that you don't get it.
The idea in this answer is the same as the one in his answer. One can make his argument a bit simpler.
Let us prove that $2^{n-1}n+1$ is not a perfect square for $n\gt 5$.
Suppose that there exist positive integers $n,a$ such that $n\gt 5$ and
$$2^{n-1}n+1=a^2$$
Then, we have
$$2^{n-1}n=(a-1)(a+1)$$
Since LHS is even, RHS has to be even. Since $a-1$ and $a+1$ have the same parity, we see that both $a-1$ and $a+1$ have to be even. Therefore, there exist positive integers $p,q,r,s$ such that
$$a-1=2^pr,\qquad a+1=2^qs,\qquad p+q=n-1,\qquad rs=n$$
Suppose here that $p\ge 2$ and $q\ge 2$. Then, since both $a-1$ and $a+1$ are divisible by $4$, it follows that $2=(a+1)-(a-1)\ge 4$ which is impossible.
So, we have either $p=1$ or $q=1$.
If $p=1$, then $a+1=2^qs=2^{n-2}s\ge 2^{n-2}$ and $a-1\ge 2^{n-2}-2$ to have
$$2^{n-1}n\ge 2^{n-2}(2^{n-2}-2)\implies 2^{n-3}\le n+1$$
If $q=1$, then $a-1=2^pr=2^{n-2}r\ge 2^{n-2}$ and $a+1\ge 2^{n-2}+2$ to have
$$2^{n-1}n\ge 2^{n-2}(2^{n-2}+2)\implies 2^{n-3}\le n-1\implies 2^{n-3}\le n+1$$
So, in either case, it follows from the following lemma that $n\le 5$ which contradicts $n\gt 5$. $\quad\blacksquare$
Lemma : If $n$ is an integer such that $2^{n-3}\le n+1$, then $n\le 5$.
Proof for the lemma :
It is sufficient to prove that if $n\ge 6\in\mathbb Z$, then $2^{n-3}\gt n+1$.
For $n=6$, we get $2^{n-3}=8\gt 7=n+1$.
Supposing that it holds for $n$ gives
$$2^{(n+1)-3}=2\cdot 2^{n-3}\gt 2(n+1)=(n+1)+1+n\gt (n+1)+1\qquad\square$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$ Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$
My working out so far:
$|x-1|$ =
$x-1$ when $x≥1$
$1-x$ when $x<1$
Then we compute $x^2 + |x-1|$ as follows:
$x^2 + |x-1|$ =
$x^2 +x -1$ when $x≥1$
$x^2 - x + 1$ when $x<1$
So we have the following 2 cases:
i) When $x≥1$, we get $x^2 + |x-1| = x^2 +x -1$ and substituting this into the inequality $x^2 + |x-1| ≤ 1$ gives $x^2 + x-2≤0$. Hence we get $x≤1$ when $x≥1$.
ii) When $x<1$ we get $x^2 + |x-1| = x^2 -x +1$ and substituting into the inequality we get $x^2 - x +1 ≤ 1$ therfore $x^2-x ≤ 0$, factorising we get $x(x-1)≤0$, so we have $x≤0$ and $x≤1$
Not sure if this is correct, I feel there's a contradiction in the second part. I would really appreciate the help.
| You may rewrite the inequality first:
$$x^2 + |x-1| \leq 1 \Leftrightarrow |1-x| \leq 1-x^2 = (1-x)(1+x)$$
*
*$\color{blue}{x= 1}$ is an obvious solution.
For $x \neq 1$ you get
$$|1-x| \leq (1-x)(1+x) \stackrel{\color{blue}{x \neq 1}}{\Longleftrightarrow} \begin{cases} 1 \leq -(1+x) & \mbox{ for } x > 1 \\ 1\leq 1+x & \mbox{ for } x <1 \end{cases} \Leftrightarrow \begin{cases} x \leq -2 & \mbox{ for } x > 1 \color{red}{\mbox{ no solution}} \\ \color{blue}{0 \leq x} & \mbox{ for } \color{blue}{x <1} \end{cases}$$
All together
$$\color{blue}{\boxed{0 \leq x \leq 1}}$$
| {
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Applying rules of algebra when working with multiplication and exponents I'm taking an online course and help is hard to find. This specific problem has to do with recurrence relation. I apologize for being too general but I'm just looking for help in how to go about solving this problem.
The entire problem is:
Let $d_0,d_1,d_2,\ldots$ be defined by the formula $d_n=3^n-2^n$ for all integers $n\ge0$. Show that this sequence satisfies the recurrence relation: $d_k=5d_{k-1}-6d_{k-2}$.
The step I can do without any trouble is finding the statements that represent the values we're dealing with in the relation, $d_{k-1}$ and $d_{k-2}$:
$d_k=3^k-2^k$
$d_{k-1}=3^{k-1}-2^{k-1}$
$d_{k-2}=3^{k-2}-2^{k-2}$
However, when it comes time to plugging that into the relation and simplifying it down to the original definition of $d_k=3^k-2^k$ I fail miserably.
$d_k=5d_{k-1}-6d_{k-2}$
$=5\left(3^{k-1}-2^{k-1}\right)-6\left(3^{k-2}-2^{k-2}\right)$
$=5\cdot3^{k-1}-5\cdot2^{k-1}-6\cdot3^{k-2}+6\cdot2^{k-2}$
$=5\cdot\frac{3^k}{3}-6\cdot\frac{3^k}{3^2}-5\cdot\frac{2^k}{2}+6\cdot\frac{2^k}{2^2}$
Is that a decent start? Is there a better way to go, like breaking everything down into their simplest components, like so:
$d_k=5d_{k-1}-6d_{k-2}$
$=(2+3)\left(3^{k-1}-2^{k-1}\right)-(2\cdot3)\left(3^{k-2}-2^{k-2}\right)$
$=2\cdot\frac{3^k}{3}-2\cdot\frac{2^k}{2}+3\cdot\frac{3^k}{3}-3\cdot\frac{2^k}{2}-\left(\left(2\cdot\frac{3^k}{3^2}-2\cdot\frac{2^k}{2^2}\right)-\left(3\cdot\frac{3^k}{3^2}-3\cdot\frac{2^k}{2^2}\right)\right)$
But then what?
If I go either of these routes, I get stuck. I feel I have two issues: a.) identifying the way to go that seem the most logical; and, b.) working towards a solution. I don't know if I missed a big chunk in Algebra or if my brain just doesn't see what's going on.
What am I missing? Are either of these steps valid things to try? What are some general rules to follow to work these out?
Also, what specific discipline of Algebra is this? I don't think my course is introducing us to this stuff. I think it assumes we already know how to work these out.
| I start from what you wrote down:
$$5d_{k-1}-6d_{k-2}=5\cdot3^{k-1}-6\cdot3^{k-2}-5\cdot2^{k-1}+6\cdot2^{k-2}$$
and I want to get to $d_k=3^k-2^k$.
I rewrite $3^{k-1}=3\cdot3^{k-2}$, and similarly for $2^{k-1}$:
$$5\cdot3\cdot3^{k-2}-6\cdot3^{k-2}-5\cdot2\cdot2^{k-2}+6\cdot2^{k-2}$$
$$=15\cdot3^{k-2}-6\cdot3^{k-2}-10\cdot2^{k-2}+6\cdot2^{k-2}$$
I collect $3^{k-2}$ and $2^{k-2}$:
$$=(15-6)3^{k-2}-(10-6)2^{k-2}=9\cdot3^{k-2}-4\cdot2^{k-2}=3^2\cdot3^{k-2}-2^2\cdot2^{k-2}$$
From here I collapse the exponents to get the desired result.
$$=3^k-2^k$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^5-k)$ is true. Then there is $m \in \mathbb{Z}^+$ such that $$k^5 - k = 5m.$$ We must show that this statement is true for $n = k+1$, i.e. show that there is $\ell \in \mathbb{Z}^+$ such that $$(k+1)^5 - (k+1) = 5\ell.$$ Note that $(k+1)^5 - (k+1)$ expands as $k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$.
We can try to find a polynomial $P(x)$ such that $$(k^5-5) + P(x) =(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$$ so as to try to add $P(x)$ to both sides of the assumption. We find that
$$\begin{align}P(x) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k - (k^5-5) \\
&=5k^4 + 10k^3 + 10k^2 + 5k \end{align}$$ and we can also observe that since $k\in\mathbb{Z}^+$, we have that $\frac{1}{5}P(x) = k^4 + 2k^3 + 2k^2 + k$ is a positive integer. Thus we add this to both sides of our assumption $$ \begin{align}
k^5 - k &= 5m \\
(k^2 - k) + P(x) &= 5m + P(x) \\
(k+1)^5 - (k+1) &= 5\left(m + \tfrac{1}{5}P(x)\right)
\end{align}.$$ Since $\frac{1}{5}P(x),m \in \mathbb{Z}^+$, it follows that $m + \tfrac{1}{5}P(x) \in \mathbb{Z}^+$. Thus $5 \left| \big[ (k+1)^5 - (k+1) \big] \right.$
By PMI, $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$.
My questions
*
*Is this proof valid?
*What other ways can this be proved by induction? The polynomial expansions took a while to deal with, so I was wondering if there are any alternate methods. (Just FYI, I only have college first-year-level knowledge)
| In $\mathbb{Z}_5$ the integers mod $5$, which is a field, all elements obey $x^5=x$, or $0= x^5 -x$, from which the statement follows right away.
| {
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Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have no clue on how to approach the problem. Most surely not by direct substitution, but I'm not seeing how. Some hint will be appreciated.
| I am going to expand the hint provided in the comments: by letting $\omega=e^{2\pi i/3}$, one if free to conjecture (especially if he/she knows how the Galois group of a cubic polynomial is usually structured) that the algebraic conjugates of
$$ \alpha_1 = \frac{1}{\sqrt[3]{9}}\left[1+\sqrt[3]{-2}+\sqrt[3]{4}\right] $$
over $\mathbb{Q}(\sqrt[3]{6})$ are given by
$$ \alpha_2 = \frac{1}{\sqrt[3]{9}}\left[\omega+\sqrt[3]{-2}+\omega^2\sqrt[3]{4}\right] $$
$$ \alpha_3 = \frac{1}{\sqrt[3]{9}}\left[\omega^2+\sqrt[3]{-2}+\omega\sqrt[3]{4}\right]. $$
Let us see what the elementary symmetric functions of these $\alpha_1,\alpha_2,\alpha_3$ are.
$$ \alpha_1+\alpha_2+\alpha_3 = -\sqrt[3]{6} $$
$$ \alpha_1\alpha_2\alpha_3 = \frac{1}{9}\left[1^3+\sqrt[3]{-2}^3+\sqrt[3]{4}^3-3\sqrt[3]{1\cdot(-2)\cdot 4}\right]=1 $$
$$ \alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3 = \frac{3}{\sqrt[3]{9}^2}\left[\sqrt[3]{-2}^2-1\cdot\sqrt[3]{4}\right]=0$$
hence, yippie ya yeah, $\alpha_1,\alpha_2,\alpha_3$ are exactly the roots of $x^3+\sqrt[3]{6}x^2-1$.
As an alternative, it is possible to recall that $q(x)=x^3-6x^2+8$ is the minimal polynomial of $\sec\frac{2\pi}{9}$ over $\mathbb{Q}$. The algebraic conjugates of $\sec\frac{2\pi}{9}$ are $\sec\frac{4\pi}{9}$ and $\sec\frac{8\pi}{9}$, and the polynomials $x^3+\sqrt[3]{6}x^2-1$ and $q(x)$ are related via an affine map.
| {
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Solution to differential equation system and solution to its conversion into 2nd order differential equation Following is the differential equation system with IVP
$\vec{x}'=\small\begin{pmatrix}3&-9\\4&-3\end{pmatrix}\vec{x},
\vec{x}(0)=\small\begin{pmatrix}2\\-4\end{pmatrix}$ The particular solution to this differential equation system is given below.
$\vec{x}(t)=\frac23\small\begin{pmatrix}3\cos{(3\sqrt{3}t)}\\\cos{(3\sqrt{3}t)}+\sqrt{3}\sin{(3\sqrt{3}t)}\end{pmatrix}+\frac{14}{3\sqrt{3}}\small\begin{pmatrix}3\sin{(3\sqrt{3}t)}\\\sin{(3\sqrt{3}t)}-\sqrt{3}\cos{(3\sqrt{3}t)}\end{pmatrix}...(1)$
When i converted this differential equation system into 2nd order differential equation, I got $y"+13y'-7y=0, y(0)=2,y'(0)=-4$
Now, the particular solution to this 2nd order equation is $-\frac{2\sqrt{6}}{3}\sin{(\sqrt{6}t)}+2\cos{(\sqrt{6}t)}...(2)$
Now why there is a difference between these two solutions namely (1) and (2)
| We have the general case of
$$x' = a x + b y \\ y' = c x + d y$$
Taking the derivative of the first equation yields
$$x'' = a x' + b y' = a x' + b(c x + d y)$$
However, from the first equation, we also have that $y = \dfrac{1}{b}(x' - ax)$, for $b \neq 0$, and upon substituting
$$x'' = a x' + b c x + d (x' - a x) = 3x' - 36 x -3(x' - 3 x) = -27 x \implies x'' + 27 x = 0$$
We need two initial conditions and already have $x(0) = 2$, but need $x'(0)$. Using the first equation
$$x'(0) = a x (0) + b(y(0)) = 3(2) - 9(-4) = 42$$
We now have to solve the second order system
$$x'' + 27 x = 0, x(0) = 2, x'(0) = 42$$
We get
$$x(t) = 2 \cos (3 \sqrt{3} t) + \dfrac{14\sqrt{3}}{3} \sin \left(3 \sqrt{3} t\right)$$
Compare that to your current solution. Also, it is now just taking derivatives of this to find $y(t)$.
| {
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Trying to prove $\log\left(n!\right)$ is greater than or equal to $\left(\frac{n}{2}\log\left(n\right)\right)$ I'm trying to prove that $\log\left(n!\right)$ is greater than $\left(\frac{n}{2}\log\left(n\right)\right)$ and I'm kinda stuck.
I could only prove $\log\left(n!\right)\ge\ \left(\frac{n}{2}\log\left(\frac{n}{2}\right)\right)$ and following is my prove:
\begin{align}
\log(n!) &=\log(n\cdot(n-1)\cdot(n-2)\cdots(1))\\
&=\log(n)+\log(n-1)+\log(n-1)+\cdots+\log(1) \\
&\geq \log(n)+\cdots+\log(\frac{n}{2}) \\
&\geq \log(\frac{n}{2})+\log(\frac{n}{2})+\log(\frac{n}{2})+\cdots+\log(\frac{n}{2})\\
&= \frac{n}{2}\log(\frac{n}{2})
\end{align}
But this is not helping me either because $\frac{n}{2}\log(n) > \frac{n}{2}\log(\frac{n}{2})$
and $\frac{n}{2}\log(\frac{n}{2}) = \frac{n}{2}\log(n) - \frac{n}{2}\log(2)$
What am I missing?
| You're almost there -- just don't discard the second half of the sum (and deal with the "good" term $\log n$ separately):$$\begin{align}
\log(n!) &= \sum_{k=1}^n \log k = \log n + \sum_{k=\frac{n}{2}+1}^{n-1} \log k + \sum_{k=2}^{\frac{n}{2}}\log k \\
&\geq \log n + \left(\frac{n}{2}-1\right)\log \frac{n}{2} + \left(\frac{n}{2}-1\right) \log 2 \\
&= \log n + \left(\frac{n}{2}-1\right)\log n - \left(\frac{n}{2}-1\right)\log 2 + \left(\frac{n}{2}-1\right) \log 2 \\
&= \frac{n}{2}\log n
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$
$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$
$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\sec^2(x)-1}=\sin(\sqrt{5})\sec(x)\tag5$$
finally after all the hard work $(5)$ reveals
$$x+x=\sqrt{5}\tag6$$
$$x=\frac{\sqrt{5}}{2}\tag7$$
Is there another easy way solving this $(1)?$
| There is another way to solve this.
Using the fact that $\sqrt{x^2} = |x|$. Your equation turns out to be $$|x|+x=\sqrt{5}$$
First look for solutions where $x\geq 0$ in this case you get $x+x=\sqrt{5}$ and so $x=\frac{\sqrt{5}}{2}$.
Now if $x<0$ then $|x|=-x$ and so you get $-x+x=\sqrt{5}$ and clearly no $x$ satisfies this condition.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\epsilon-\delta$ proof that $f(x) = x \sin(1/x)$ on $(0,1)$ is uniformly continuous I have to prove
$$
f: (0,1) \to [-1,1], \ \ f(x)=x\sin(1/x)
$$
using $\epsilon-\delta$.
Try
Fix $\epsilon >0$. I have
$$
\left|x \sin(1/x) - y \sin (1/y)\right| \le |\sin (1/x)| |x-y| + |y| |\sin (1/x) - \sin(1/y)|
$$
I can control $|\sin (1/x)| |x-y|$ part, since I can control $|x-y|$ and $|\sin(1/x)| \le 1$. But I don't know how to deal with $|y| |\sin (1/x) - \sin(1/y)|$ part.
Any help will be appreciated.
| Using the prosthaphaereis formula
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| = 2|y|\left|\sin \frac{x^{-1}-y^{-1}}{2} \right| \left|\cos \frac{x^{-1}+y^{-1}}{2} \right| \leqslant 2|y|\left|\frac{x^{-1} - y^{-1}}{2} \right| \\\leqslant \frac{|x-y|}{|x|}$$
WLOG assume $x > y$ -- otherwise, work with the alternative estimate $|x| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right|$.
Note that we also have
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| \leqslant 2|y|$$
Suppose $|x - y| < \frac{\epsilon^2}{2}$. Either $|y| < \frac{\epsilon}{2}$ and we have
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| \leqslant 2\frac{\epsilon}{2} = \epsilon,$$
or
$$|x| > |y| \geqslant \frac{\epsilon}{2},$$
and
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right|\leqslant \frac{|x-y|}{|x|} < \frac{2}{\epsilon} \frac{\epsilon^2}{2} = \epsilon$$
Thus, $|x-y| < \frac{\epsilon^2}{2}$ implies $|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right|\leqslant \epsilon$ for all $x,y \in (0,1)$.
The estimate on the first line is only sharp enough to prove uniform continuity on an interval $[a,1)$ where $a > 0$. However, we know that $x \sin \frac{1}{x} \to 0 $ as $x \to 0$ and this implies continuous extendibility to $[0,1)$ and, hence uniform continuity on $(0,1)$. The argument above combines both of these elements into an $\epsilon - \delta$ proof.
| {
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Finding the side length of an equilateral triangle having three inscribed $120^\circ$ sectors in a certain arrangement
How do i even start this question? I thought of using length of tangent equal from exterior point but still of no use.
|
Let $G$ be the common centroid for the two triangles.
Pick an edge, say $AB$, let $H$ be the vertex on the small triangle which is the
apex for the circular sector touching edge $AB$. Let $G'$ and $H'$ be the foot of
$G$ and $H$ on edge $AB$.
It is clear the distance of $G$ to line $AB$ is
$$|GG'| = |HH'| - |GH|\cos(\theta + \frac{\pi}{6})$$
where $\theta$ is the angle illustrated in above diagram.
We know $|HH'| = 20$, the radius of the circular sectors. Since the small triangle has side
$9$, we also know $|GH| = 3\sqrt{3}$. In the diagram above, we find
$$\cos\theta = \frac{|HH'|}{|DH|} = \frac{20}{29}\quad\implies\quad\sin\theta = \frac{21}{29}$$
Combine these, we get
$$\begin{align}|GG'|
&= 20 - 3\sqrt{3}\left(\cos\theta\cos\frac{\pi}{6} - \sin\theta\sin\frac{\pi}{6}\right)\\
&= 20 - 3\sqrt{3}\left(\frac{20}{29}\cdot\frac{\sqrt{3}}{2} - \frac{21}{29}\cdot\frac{1}{2}\right)\\
&= \frac{980+63\sqrt{3}}{58}
\approx 18.77791725649723
\end{align}
$$
As a corollary,
$$|AB| = |BC| = |CA| = 2\sqrt{3}|GG'| = \frac{189+980\sqrt{3}}{29} \approx 65.04861349715516$$
| {
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"answer_id": 0
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Proving a sequence is bounded from below Let $$a_1=3\quad , \quad a_{n+1}=\frac{3a_n}{4}+\frac{1}{a_n}$$
Show that $a_n$ converges.
I know that I need to prove that $a_n$ is monotonic and bounded.
I've assumed that the limit exists, made the calculation and get that $L=2$, hence claiming two claims:
*
*$a_n\ge 2$ (I know that 0 is a simpler bound)
*$a_n$ is monotonically decreasing.
Now, if I can prove the first claim. then I can write $$a_{n+1}=\frac{3a_n}{4}+\frac{1}{a_n}=\frac{3a_n}{4}+\frac{a_n}{a_n^2}\le \frac{3a_n}{4}+\frac{a_n}{4}=a_n$$ which proves the second claim.
My only problem is proving the first one. If it is not the right direction, please hint me.
| To show boundedness from below, you may use AM-GM:
$$\frac{3a_n}{4}+\frac{1}{a_n}\geq 2 \sqrt{\frac{3a_n}{4}\cdot \frac{1}{a_n}}= \sqrt{3}$$
For convergence you may proceed as follows:
*
*$f(x) = \frac{3x}{4} + \frac{1}{x}$ has a fixpoint for $x^{\star} = 2$
*$f'(x) = \frac{3}{4}-\frac{1}{x^2}$
*A quick calculation shows that $|f'(x)| < 1$ for $x > \frac{2}{\sqrt{7}} \Rightarrow |f'(x)| \leq \color{blue}{q} < 1$ for $x \geq \sqrt{3}$
So, for any starting value $a_0 > 0$ you get
$$|2-a_{n+1}| = |f'(\xi_n)||2-a_n| < \color{blue}{q} |2-a_n| < \color{blue}{q^n}|2-a_1|\stackrel{n\to \infty}{\longrightarrow}0 $$
Edit after comment:
Specifically for your question concerning $a_n \geq 2$:
$$ \color{blue}{a_{n+1}-2} = \frac{3a_n}{4}+\frac{1}{a_n} - \left(\frac{3\cdot 2}{4}+\frac{1}{2} \right)= \frac{3}{4}\left( a_n - 2 \right) - \frac{a_n - 2}{2a_n}$$
$$ = \left(\frac{3}{4} - \frac{1}{2a_n} \right)(a_n -2) \stackrel{\color{blue}{a_n \geq 2}}{\geq} \frac{1}{2}(a_n -2) \color{blue}{\geq 0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Problems with proof by induction $\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$?
$$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Prove for $n=1$: $$\frac1{1\times2}=\frac1{1+1}=\frac12$$
Hip: $$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Demonstration: $$\frac1{n+1} + \frac1{(n+1)(n+2)}=\dots=\frac1{(n+1)+1}$$
My problem is that I can't find the correct algebra steps to get from the beginning of the demonstration to the end of the demonstration.
| $\frac {1}{1\times 2}=1-1/2$
$\frac{1}{2\times 3}=1/2 - 1/3$
........
$\frac {1}{n(n+1)} =1/n- \frac {1}{n+1}$
Add them up and cancel the middle terms to get $1-\frac {1}{n+1}=\frac {n}{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Given $3^x = 12$ find $(\sqrt{4/3})^\frac{1}{x-2}$ Given $3^x = 12$ find $$\left(\sqrt{\dfrac{4}{3}}\right)^\dfrac{1}{x-2}$$ in simple form.
I've faced this problem, in a high school book, but failed to solve.
I've tried to calculate
*
*$3^{x-2} = \dfrac{12}{9}$
*$x -2 = \log_3 \dfrac{12}{9}$
*$\dfrac{1}{(x -2)} = \dfrac{ \log 3} { \log (\frac43)}$
However, replacing $\dfrac{1}{(x -2)}$ on the power doesn't help for me.
$$\left(\dfrac{4}{3}\right)^\dfrac{1}{\dfrac{2 \log 3} { \log (\frac43)} } = \left(\dfrac{4}{3}\right)^\dfrac{\log \left(\frac{4}{3}\right)}{2 \log 3 } = \left(\dfrac{4}{3}\right)^\dfrac{2 \log 2 - \log 3}{2 \log 3 } $$
Any hint?
| You have found $3^{x-2}=\frac{12}{9}=\frac{4}{3}$. So replace the $\frac{4}{3}$ in your expression: $$\left(\sqrt{\frac43}\right)^{1/(x-2)}=\left(\sqrt{3^{x-2}}\right)^{1/(x-2)}=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating a Limit using Riemann integration We want to evaluate
$$ \lim_{n \to \infty} \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} $$
I have spent almost an hour in this problem. Here is my thought
First, I tried to evaluate by comparing with an integral of the form $\int_0^b f(x) dx $ using right or left-endpoints, but that doesnt work. Now, if we take the midpoint $\bar{x_i} = \dfrac{x_i + x_{i+1}}{2}$ and $\Delta x = \frac{b}{n} $, we can get somewhere, but first, we do $m=3n$ and simplify the expression to
$$ \lim_{m \to \infty} \sum_{i=1}^m \sqrt{ 9 - \left(\frac{6i-1}{m} \right)^2 } \frac{3}{2m} $$
With $x_i = 0 + b i\Delta x $, we have $\bar{x_i} = \frac{2 bi \Delta x + \Delta x }{2} = b \Delta x \left( \frac{2i + 1}{2} \right) = \frac{2}{b} \left( \frac{2i + 1}{2m} \right) $
We are getting closer here but still not what we desired. My feeling is that the integral is integrating over area of a circle of radius $6$, but im having trouble with the transformations that must be done to get to the right form.
Any suggestion?
| You may proceed as follows:
$$\sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i}{n} \right)^2 } \cdot \frac{1}{n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-1}{n} \right)^2 } \cdot \frac{1}{n}$$
$$3\sum_{i=1}^{3n} \sqrt{36 - 9\left(\dfrac{i}{3n} \right)^2 } \cdot \frac{1}{3n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} \leq 3\sum_{i=1}^{3n} \sqrt{36 - 9\left(
\dfrac{i-1}{3n} \right)^2 } \cdot \frac{1}{3n}$$
For the left and right sum you get in the limit
$$9\int_0^1 \sqrt{4-x^2}\; dx = \frac{3}{2}(3\sqrt{3} + 2\pi)\approx 17.219$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving a set of vectors is a basis for the quotient map between two vector spaces I want to see if my work is justifiable. I am tasked with the following:
I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).
$$\text{Implication is} \ \mathscr {B}_{\mathbb R^4 / \ ker \ T} =
\left\{
\begin{bmatrix}
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
\end{bmatrix},\begin{bmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}
\end{bmatrix}
\right\} \text{, where $\mathscr B$ is the basis for $\mathbb R^4 / \ ker \ T$}$$
$$
\implies \text{for some $[v] \in \mathbb R^4 / \ ker \ T$,}$$
$$[v] = a_1 \begin{bmatrix}
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
\end{bmatrix} + a_2 \begin{bmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}
\end{bmatrix}$$
$$\implies
\begin{bmatrix}
\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}
\end{bmatrix}
=
a_1 \begin{bmatrix}
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
\end{bmatrix} + a_2 \begin{bmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}
\end{bmatrix}$$
$$\implies
\begin{bmatrix}
\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}
\end{bmatrix}
=
\begin{bmatrix}
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix}
\end{bmatrix} $$
$$\implies
\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix} - \begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix} \in ker \ T$$
$$
\implies
\begin{pmatrix}
x \\
y-a_1 \\
z-a_2 \\
t \\
\end{pmatrix} = b_1 \begin{pmatrix}
1 \\
1 \\
0 \\
0 \\
\end{pmatrix} + b_2 \begin{pmatrix}
0 \\
0 \\
1 \\
1 \\
\end{pmatrix} \text{, where $b_1, b_2 \in ker \ T$}$$
After some algebra...
$$\implies a_1 = y - x$$
$$\implies a_2 = z - t$$
Which I think implies $\mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $\mathscr B$ is linearly independent. I may skip a step or two.
$$
\begin{bmatrix}
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix}
\end{bmatrix} = [0] \iff a_1 = a_2 = 0 $$
$$\begin{bmatrix}
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix}
\end{bmatrix} = [0] \implies
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix} - \begin{pmatrix}
0 \\
0 \\
0 \\
0 \\
\end{pmatrix} \in Ker \ T$$
$$\implies
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix} = \begin{pmatrix}
b1 \\
b1 \\
b2 \\
b2 \\
\end{pmatrix} \implies a_1 = a_2 = 0$$
Hence, basis. Has what I done made sense?
| It's all correct.
Note that the 4 given vectors form a basis of $\Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of $x$ when equality is given $$ x + y = \sqrt{x} + \sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.
| Try this method of completing the square.
Let $a=\sqrt x$ and $b=\sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
| Just a thought, that may be worth mentioning:
The expression:
$$p=(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = \frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$\frac{1.3.5.7}{2.4.6.8}=\frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=\frac{odd_numbers}{even_numbers}=\frac{p_o}{p_e}=$$
$$p=\frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<\frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<\frac{(n!)}{((n!)2^{n})^2}$$
$$p<\frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<\frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
| Let $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=\sum_{cyc}\frac{ab}{(b-c)(c-a)}=\frac{\sum\limits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$\sum_{cyc}\left|\frac{a}{b-c}\right|=\sqrt{\left(|x|+|y|+|z|\right)^2}=\sqrt{x^2+y^2+z^2+2\sum\limits_{cyc}|xy|}=$$
$$=\sqrt{(x+y+z)^2+2+2\sum\limits_{cyc}|xy|}\geq\sqrt{2+2}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Maximize $x_1^3+x_2^3+\cdots + x_n^3$ This is from a Brazilian math contest for college students (OBMU):
Given a positive integer $n$, find the maximum value of
$$x_1^3+x_2^3+ \cdots + x_n^3$$
where $x_j$ is a real number for all $j \in \{1,2,\cdots, n\}$ such that $x_1 + x_2 + \cdots + x_n = 0$ and $x_1^2 + x_2^2 + \cdots + x_n^2 = 1$ .
| I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain
$$ S =\{ (x_1,x_2,\cdots,x_n) \in {\mathbb{R}}^n | x_1 + x_2 + \cdots + x_n = 0 \text{ and } x_1^2 + x_2^2 + \cdots + x_n^2 = 1\} $$
is closed (since the functions $(x_1,x_2,\cdots,x_n ) \mapsto x_1 + x_2 + \cdots + x_n$ and $(x_1,x_2,\cdots,x_n ) \mapsto x_1^2 + x_2^2 + \cdots + x_n^2$ are continuous) and bounded (since $x_1^2 + x_2^2 + \cdots + x_n^2 = 1$). Then S is compact.
Define
$$f(x_1,x_2,\cdots,x_n ) = x_1^3 + x_2^3 + \cdots+ x_n^3 $$
$$g(x_1,x_2,\cdots,x_n ) = x_1^2 + x_2^2 + \cdots+ x_n^2 -1 $$
$$h(x_1,x_2,\cdots,x_n ) = x_1 + x_2 + \cdots+ x_n $$
Since $S$ is compact and $f$ is continuous, then $f$ reaches a maximum value in
$S$. We can find the maximum using Lagrange Multipliers:
$$\nabla f = \lambda_1 \nabla g + \lambda_2 \nabla h$$
Thus,
$$ 3(x_1^2,x_2^2,\cdots,x_n^2) = 2\lambda_1(x_1,x_2,\cdots,x_n ) + \lambda_2(1,1,\cdots,1)$$
Then,
$$3x_i^2 = 2\lambda_1 x_i+ \lambda_2 \text{ } \forall i \in \{1,2,\cdots,n\} \label{1}\tag{1}$$
If we add all the equations and use the constraints equations, we obtain $\lambda_2= \frac{3}{n}$. Then,
$$3x_i^2 = 2\lambda_1 x_i+ \frac{3}{n} $$
Solving for $x_i$, we obtain
$$ x_i = \frac{\lambda_1 \pm \sqrt{\lambda_1^2 + \frac{9}{n}}}{3} $$
Let $k$ be a natural number less or equal to $n$. Therefore we can suppose
$$ x_i = \frac{\lambda_1 + \sqrt{\lambda_1^2 + \frac{9}{n}}}{3} \forall i \in \{1,2,\cdots,k\}$$
$$ x_i = \frac{\lambda_1 - \sqrt{\lambda_1^2 + \frac{9}{n}}}{3} \forall i \in \{k+1,k+2,\cdots,n\}$$
Firtly, notice that $k$ is different of $0$ and $n$, because of the constraint $x_1+ x_2 + \cdots + x_n =0$. Using the constraint $x_1^2 + x_2^2 + \cdots + x_n^2 = 1$ and after some simplifications, we obtain
$$\lambda_1 \left(n\lambda_1 + (2k-n) \sqrt{\lambda_1^2 + \frac{9}{n}} \right) = 0\label{2}\tag{2}$$
And using $x_1+ x_2 + \cdots + x_n =0$, we get
$$ n\lambda_1 + (2k-n) \sqrt{\lambda_1^2 + \frac{9}{n}} = 0 \label{3}\tag{3}$$
So, we just need to satisfy \ref{3}, because with that \ref{2} is automatically satisfied. Then, we get
$$\lambda_1 = \frac{3(n-2k)}{2\sqrt{(n-k)nk}} $$
Notice that we should consider the other solution with the other sign, but in the end we obtain the same maximum value for both cases. Multiplying \ref{1} by $x_i$:
$$ 3x_i^3 = 2\lambda_1x_i^2 + \frac{3x_i}{n} $$
Adding in all i's:
$$3f = 2\lambda_1 $$
Thus,
$$ f(k) = \frac{n-2k}{\sqrt{(n-k)nk}} = \frac{1}{\sqrt{n}}\left(\sqrt{\frac{n-k}{k}}-\sqrt{\frac{k}{n-k}}\right)$$
Remember that $k \in {2,3,\cdots, n-1}$. Making $x = \sqrt{\frac{k}{n-k}}$ and analyzing the derivative of
$$y: (0,1) \rightarrow \mathbb{R}$$
$$ x \mapsto\frac{1}{x}-x $$
We see that the maximum value of $f$ is when $k=1$, that is,
$$ \frac{n-2}{\sqrt{(n-1)n}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Lyapunov stability of 4x4 matrix. Consider the following continuous-time state space representation of the form:
$\frac{d}{dx}x(t) = Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t\in \mathbb{R}^{+}$
$A=\begin{bmatrix}-1&3&0&0\\-3&-1&0&0\\0&0&0&3\\0&0&-3&0 \end{bmatrix} \quad B = \begin{bmatrix}0\\1\\0\\0 \end{bmatrix} \quad C=\begin{bmatrix}1&0&0&1 \end{bmatrix}$
The corresponding eigenvalues are: $-1+3i, \ -1-3i, \ 0+3i \ \text{and} \ 0-3i$.
The answer states that this system is Lyaponov stable.
But I'm wondering why.
Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?
| Write $I_n$ for a $n \times n$ identity matrix.
Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:
$$
\begin{align}
\dot{V}(x) &= \dot{x}^T P x + x^T P \dot{x} \\
&= x^T A^T P x + x^T P A x \\
&= x^T(A^T P + P A) x \\
&= x^T Q x,
\end{align}
$$
insert $A$ and $P$ and derive $Q$:
$$
\begin{align}
Q = A^T P + P A &=
\begin{bmatrix}
-1 & -3 & 0 & 0 \\
3 & -1 & 0 & 0 \\
0 & 0 & 0 & -3 \\
0 & 0 & 3 & 0
\end{bmatrix} \frac{1}{2} I_4 + \frac{1}{2} I_4
\begin{bmatrix}
-1 & 3 & 0 & 0 \\
-3 & -1 & 0 & 0 \\
0 & 0 & 0 & 3 \\
0 & 0 & -3 & 0
\end{bmatrix} \\ &=
\begin{bmatrix}
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{bmatrix} .
\end{align}
$$
So your directional derivative is $\dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$
Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$
I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $(n)$ into the $Sn \left(-\frac{1}{6}+\frac{1}{2}-\frac{1}{8}+\frac{1}{4}-\frac{1}{10}\right)$ , $Sn$ just keeps going on and on. So how do I solve this problem and what does this series converge to?
| More generally,
if
$s_m(n)
=\sum_{k=m+1}^n \frac{1}{k^2-m^2}
$
then,
if $n > 3m$,
$\begin{array}\\
s_m(n)
&=\sum_{k=m+1}^n \frac{1}{k^2-m^2}\\
&=\sum_{k=m+1}^n \frac1{2m}(\frac{1}{k-m}-\frac1{k+m})\\
&=\frac1{2m}\sum_{k=m+1}^n \frac{1}{k-m}-\frac1{2m}\sum_{k=m+1}^n\frac1{k+m}\\
&=\frac1{2m}(\sum_{k=1}^{2m} \frac{1}{k}+\sum_{k=2m+1}^{n-m} \frac{1}{k})-(\frac1{2m}\sum_{k=2m+1}^{n-m}\frac1{k}+\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k})\\
&=\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}-\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k}\\
\end{array}
$
so
$\begin{array}\\
s_m(n)-\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}
&=-\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k}\\
\text{so that}\\
|s_m(n)-\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}|
&=\frac1{2m}|\sum_{k=n-m+1}^{n+m}\frac1{k}|\\
&\le\frac1{2m}|\sum_{k=n-m+1}^{n+m}\frac1{n-m+1}|\\
&=\frac1{2m}|\frac{2m}{n-m+1}|\\
&=\frac{1}{n-m+1}\\
&\to 0
\quad\text{ as } n \to \infty\\
\end{array}
$
Therefore
$\lim_{n \to \infty} s_m(n)
=\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}
$.
For $m=1$
the sum is
$\frac1{2}(\frac1{1}+\frac1{2})
=\frac34
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.