Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Removing a discontinuity? How would you "remove the discontinuity" of $f$ ? In other words, how would you define $f(4)$ in order to make $f$ continuous at $x=4$?
$$f(x) = \dfrac{x^2-x-12}{x-4}$$
| $f(x) = \frac{x^2-x-12}{x-4}$
$f(x) = \frac{(x-4)(x+3)}{(x-4)}$
$\lim_{x\to4}f(x) = \frac{x^2-x-12}{x-4} = \lim_{x\to4}\frac{(x-4)(x+3)}{(x-4)} = 7$
So redefine $f(x)$ as ;
$f(x) = \begin{cases}\frac{x^2-x-12}{x-4}&x\ne4\\7&x= 4\end{cases}$
| {
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"url": "https://math.stackexchange.com/questions/2763300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{x\to\infty} [\log_{10}{(5\cdot6^x+2\sqrt{x^3}})-\log_{10}{(\sin{x}+4\cdot6^x+\cos{x})]}$ Given this limit:
$$\lim_{x\to\infty} [\log_{10}{(5\cdot6^x+2\sqrt{x^3}})-\log_{10}{(\sin{x}+4\cdot6^x+\cos{x})]}$$
I simplified:
$$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+\cos{x}}}$$
Now, I can't understand why I can cancel $6^x$, $2\sqrt{x^3}$, $\sin{x}$, $6^x$ and $\cos{x}$ so that the solution is $\log_{10}{\frac{5}{4}}$ as my professor told me.
| $$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+\cos{x}}}= \lim_{x \to \infty}\log_{10}{\frac{5+2\frac{\sqrt{x^3}}{6^x}}{\frac{\sin{x}}{6^x}+4+\frac{\cos{x}}{6^x}}}=\log_{10}\frac54$$
Edit:
\begin{align}
0 \le \lim_{x \to \infty}\frac{\sqrt{x^3}}{6^x} &\le \lim_{x \to \infty}\frac{x^2}{e^{(\log 6) x}} \le \lim_{x \to \infty}\frac{x^2}{e^x} \le \lim_{x \to \infty}\frac{x^2}{1+x+\frac{x^2}{2}+\frac{x^3}{6}} = 0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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replacing Inequalities I encountered a problem today:
Prove that:
$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2} \ge \frac{a+b+c}{3}$$
for all $a,b,c>0$
I used the RMS-AM inequality to replace the LHS with
$$\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{3}}$$
and replaced the RHS using AM-GM inequality
$$\frac{3abc}{a^2+b^2+c^2}$$
I can prove the new inequality, but does that mean I have proved the original inequality? I couldn't find another way to prove the original inequality except for expanding the terms and using scalar products. Thanks in advance!
| Note that
\begin{align*}
a^3+b^3+c^3 &= \frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\\
&\ge \frac{(a^2+b^2+c^2)^2}{a+b+c}.
\end{align*}
Then
\begin{align*}
\frac{a^3+b^3+c^3}{a^2+b^2+c^2} &\ge \frac{a^2+b^2+c^2}{a+b+c}\\
&\ge \frac{a+b+c}{3},
\end{align*}
given that
\begin{align*}
(a+b+c)^2 \le 3(a^2+b^2+c^2).
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!
BACKGROUND: The background of this question is that I tried to find this integral:
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
| Another way of stating what everyone has said:
$2x^2 - x= 2(x^2-x+1) + x - 2$
$\Rightarrow 2x^2-x = x-2 \mod x^2-x+1$
| {
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How to prove that $\sum_{m\leqslant x}\sum_{n\leqslant x}\Big\{\frac{x}{m+n}\Big\}=\Big(2\log2-\frac{\pi^2}{12}\Big)x^2+O(x\log x)$? Prove that $$\sum_{m\leqslant x}\sum_{n\leqslant x}\Big\{\frac{x}{m+n}\Big\}=\Big(2\log2-\frac{\pi^2}{12}\Big)x^2+O(x\log x),$$
where $\{x\}$ is the fractional part of the real number $x$.
I know
$$\sum_{n\leqslant x} \Big\{\frac{x}{n} \Big\}=(1-\gamma)x+O\big(x^{1/2}\big),$$
where $\gamma$ is Euler's constant. But I don't know whether it is useful. Can you help me?
| Write $\{x\} = x - \lfloor x\rfloor$,s so that
\[\sum_{m \leq x} \sum_{n \leq x} \left\{ \frac{x}{m + n}\right\} = \sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} - \sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor.\]
Then note that
\[\sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor = \sum_{m \leq x} \sum_{n \leq x} \sum_{\ell \leq \frac{x}{m + n}} 1 = \sum_{\ell \leq \frac{x}{2}} \sum_{n \leq \frac{x}{\ell} - 1} \sum_{m \leq \frac{x}{\ell} - n} 1.\]
The sum over $m$ is $\lfloor x/\ell\rfloor - n$. The ensuing sum over $n$ is
\[\left\lfloor \frac{x}{\ell}\right\rfloor \left(\left\lfloor \frac{x}{\ell}\right\rfloor - 1\right) - \sum_{n \leq \frac{x}{\ell} - 1} n.\]
Via partial summation,
\[\sum_{n \leq \frac{x}{\ell} - 1} n = \left(\frac{x}{\ell} - 1\right) \left(\left\lfloor \frac{x}{\ell} \right\rfloor - 1\right) - \int_{1}^{\frac{x}{\ell} - 1} \lfloor t\rfloor \, dt,\]
and this integral is equal to $\frac{1}{2} \left(\frac{x}{\ell} - 1\right)^2 + O(\frac{x}{\ell})$. So the sum over $n$ and $m$ simplifies to
\[\frac{x^2}{2 \ell^2} + O\left(\frac{x}{\ell}\right).\]
The ensuing sum over $\ell$ is
\[\frac{x^2}{2} \sum_{\ell = 1}^{\infty} \frac{1}{\ell^2} - \frac{x^2}{2} \sum_{\ell > \frac{x}{2}} \frac{1}{\ell^2} + O(x \log x).\]
The first sum over $\ell$ is $\zeta(2) = \pi^2/6$. The second is $O(1/x)$. So this simplifies to
\[\frac{\pi^2 x^2}{12} + O(x \log x).\]
Now we deal with
\[\sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} = x \sum_{m \leq x} \sum_{n \leq x} \frac{1}{m + n}.\]
We deal with the sum over $n$ via partial summation: it is equal to
\[\frac{\lfloor x\rfloor}{m + x} + \int_{1}^{x} \frac{\lfloor t \rfloor}{(m + t)^2} \, dt = \log \frac{m + x}{m + 1} + O\left(\frac{1}{m}\right),\]
where we have used the fact that $\lfloor x \rfloor = x - \{x\} = x + O(1)$, the fact that the antiderivative of $t/(m + t)^2$ is $m/(m + t) + \log(m + t)$, and the fact that the antiderivative of $1/(m + t)^2$ is $-1/(m + t)$.
So it remains to evaluate
\[\sum_{m \leq x} \log \frac{m + x}{m + 1} = \lfloor x\rfloor \log \frac{2x}{x + 1} + (x + 1) \int_{1}^{x} \frac{\lfloor t\rfloor}{(t + x)(t + 1)} \, dt.\]
The antiderivative of $\frac{t}{(t + x)(t + 1)}$ is $\frac{x}{x - 1} \log \frac{t + x}{t + 1}$, and so after some simplification, we arrive at $(2\log 2) x + O(\log x)$.
| {
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If $n \ge 1$, does it follow that $\left(\frac{3}{2}\right)^n > n$ It seems to me the answer is yes. Here's my reasoning:
For $n=1$, $\frac{3}{2} > 1$
and, $\frac{d}{dn}\left(\frac{\left(\frac{3}{2}\right)^n}{n}\right) > 0 $
Using the quotient rule:
$\dfrac{d}{dn}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} = \frac{\left(\frac{3}{2}\right)^n\left(\ln \frac{3}{2}\right)n - \left(\frac{3}{2}\right)^n }{n^2} > 0$ for $n \ge 1$
Am I correct? Is there a more standard way to show this?
| Just check $f(x) = (\frac{3}{2})^x - x > 0$ for $x=1,2,3$.
Then for $x> 3$ you have $f'(x)= (\frac{3}{2})^x\ln \frac{3}{2} -1> 0$ anyway.
| {
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"timestamp": "2023-03-29T00:00:00",
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Logarithmic expression how to simplify $$ \log_{3}24 - 3\log_{3}5\times \log_{5}2$$
What I can get is:
$$ \log_3{24} - \log_3{5^3} \times \log_{5}2$$
Change of base rule to get it all in base 3:
$$ \log_5{2} = \frac{\log_3{2}}{\log_3{5}} $$
Now I have:
$$\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}$$
How to continue from here?
| $$\begin{align}\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}&=\log_3{24} - \frac{3\log_3{5}\times \log_3{2}}{\log_3{5}}\\
&=\log_3{24} - 3\log_3 2\\
\end{align}$$
Now you can use the exponent law again on the second term, and then the additive law. Can you finish from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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An ice cream parlor has 28 different ice cream flavors. How many different ways are there to choose 6 scoops of ice cream if at least An ice cream parlor has $28$ different ice cream flavors. How many
different ways are there to choose $6$ scoops of ice cream if at least $2$ scoops
must be chocolate?
My attempt
I used the formula $\binom {n+r-1}{r}$ and here $n=28, r=6$
$\binom {33}{6}-\binom{28}{1}$
Can anyone please explain me this
| No of ways of selecting scoops such that atleast 2 scoops must be chocolate
= total no of ways(S) - containing no chocolate(A) - containing exactly 1 chocolate(B)
There are 2 different answers depending on whether the arranging order from bottom to top matter or not
1) Order matters
S = $28^6$ since for every scoop you have 28 options
A = $27^6$ since we are excluding chocolate
B = There is exactly one chocolate which can be arranged in 6 places the rest of the places can be filled by $26^5$ ways = $6*26^5$
Answer = $S - A - B = 28^6 - 27^6 - (6*26^5)$
2) Order doesn't matter i.e. we are only interested in different number of combinations(example ABBB,BABB,BBAB,BBBA counts as 1 case instead of 4)
The simpler way to solve this using stars and bars method
$X_1+X_2+X_3+.......X_{28} = 6$ where $X_i$ represents the flavour of icecream
Non negative solution of above equation is given by $\binom{28+6-1}{6-1}$ =$\binom{33}{5}$
S = $\binom{33}{5}$
A = containing no chocolate = $X_1+X_2+X_3+.......X_{27} = 6$ == $\binom{27+6-1}{6-1}$ =$\binom{32}{5}$
B = containing exactly one chocolate =
$X_1+X_2+X_3+.......X_{27} + 1 = 6$ ($X_{28} = 1$)
=$X_1+X_2+X_3+.......X_{27} = 5$
Solutions = $\binom{27+5-1}{5-1}$ =$\binom{31}{4}$
Answer = $S-A-B$ = $\binom{33}{5}$ - $\binom{32}{5}$ - $\binom{31}{4}$ = 4495
Alternative way:
atleast 2 chocolate
$X_1+X_2+X_3+.......X_{28} = 6$
let $X_{28}$ be chocolate such that $X_{28}$ >=2
Let $X_{28}$ = $Y_{28}$ +2
$X_1+X_2+X_3+.......X_{27} +Y_{28} + 2 = 6$
=$X_1+X_2+X_3+.......X_{27} +Y_{28} = 4$
Solution of above equation = $\binom{28+4-1}{4-1}$ =$\binom{31}{3}$ = 4495
| {
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Finding Sum of $\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty$ Finding Sum of $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty\; \bf{terms}$$
Try: Writting it as $$\sum^{\infty}_{r=0}\frac{(4r)!}{(4r+4)!}=\sum^{\infty}_{r=0}\frac{1}{(4r+1)(4r+2)(4r+3)(4r+4)}$$
$$\frac{1}{3}\sum^{\infty}_{r=0}\bigg[\frac{1}{(4r+1)(4r+2)(4r+3)}-\frac{1}{(4r+2)(4r+3)(4r+4)}\bigg]$$
$$\frac{1}{6}\bigg[\bigg(\frac{1}{(4r+1)(4r+2)}-\frac{1}{(4r+2)(4r+3)}\bigg)-\bigg(\frac{1}{(4r+2)(4r+3)}-\frac{1}{(4r+3)(4r+4)}\bigg)\bigg]$$
I did not understand how can i solve further, thanks
| Another pssible way is to consider partial sums
$$S_p=\frac16\sum_{r=0}^p\left(
\frac{1}{4r+1}-\frac{3}{4r+2}+\frac{3}{4r+3}-\frac{1}{4r+4}\right)$$ and to use generalized harmonic numbers
$$\sum_{r=0}^p\frac{1}{4r+1}=\frac{1}{8} \left(2 H_{p+\frac{1}{4}}+\pi +\log (64)\right)$$
$$\sum_{r=0}^p\frac{1}{4r+2}=\frac{1}{4} \left(H_{p+\frac{1}{2}}+\log (4)\right)$$
$$\sum_{r=0}^p\frac{1}{4r+3}=\frac{1}{4} \left(H_{p+\frac{3}{4}}-\frac{\pi }{2}+\log (8)\right)$$
$$\sum_{r=0}^p\frac{1}{4r+4}=\frac{H_{p+1}}{4}$$ which make
$$S_p=\frac{1}{24} \left(H_{p+\frac{1}{4}}-3 H_{p+\frac{1}{2}}+3
H_{p+\frac{3}{4}}-H_{p+1}-\pi +\log (64)\right)$$ Now, using the asymptotics,
$$S_p=\frac{1}{24} (\log (64)-\pi )-\frac{1}{768
p^3}+O\left(\frac{1}{p^4}\right)$$ wcih shows the limit and how it is approached.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that 1/2 + 1/4 + 1/8 ....... = 1 I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact? I need it without Sigma so I can explain it to my little sister.
It is not a duplicate because this one does not use Sigma, and the one marked as duplicate does. I want it to use variables and equations.
| $A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots+\dfrac{1}{2^{n+1}}$
$A=\dfrac{2^n}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\dfrac{2^{n-2}}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}$
$2A=\dfrac{2^{n+1}}{2^{n+1}}+\dfrac{2^{n}}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\cdots+\dfrac{2}{2^{n+1}}$
Then
$A=2A-A$
$=\left(\dfrac{2^{n+1}}{2^{n+1}}+\dfrac{2^{n}}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\cdots+\dfrac{2}{2^{n+1}}\right)-\left(\dfrac{2^n}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\dfrac{2^{n-2}}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}\right)$
$=\dfrac{2^{n+1}}{2^{n+1}}-\dfrac{1}{2^{n+1}}=1-\dfrac{1}{2^{n+1}}$
If the expression $A$ has more fractions, $n+1$ will be larger, which means $\dfrac{1}{2^{n+1}}$ should be closer to zero, which means $A$ converges to $1$.
| {
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Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then for $a\geq 1$:
so then $$2-a=0\implies a=2$$ and $$\frac {a+2}3=2\implies a=4.$$
and for $a\leq1:$
so then $$2-a=3\implies a=-1$$ and $$\frac {a+2}3=0\implies a=-2.$$
Now my answers are in the type of interval. How do I know which interval to choose?
| Rewrite
$$f(x) = 1 + \frac{(a-1)x}{x^2+x+1}$$
and note that $f(x) \in [0,2]$ is equivalent to $(a-1)\frac{x}{x^2+x+1} \in [-1,1]$.
Notice that $\frac{x}{x^2+x+1} \in [-1,1]$:
$$\frac{x}{x^2+x+1} \le 1 \iff 0 \le x^2+1$$
$$\frac{x}{x^2+x+1} \ge -1 \iff 0 \le x^2+2x + 1$$
Now, if $a \in [0,2]$ then $|a-1| \le 1$ so
$$|a-1|\underbrace{\left|\frac{x}{x^2+x+1}\right|}_{\le 1} \le 1, \forall x \in \mathbb{R}$$
Conversely, if $(a-1)\frac{x}{x^2+x+1} \in [-1,1], \forall x \in \mathbb{R}$ then in particular for $x = -1$ we get $1-a \in [-1,1]$ which implies $a \in [0,2]$.
We conclude $f(x) \in [0,2], \forall x \in \mathbb{R}$ if and only if $a \in [0,2]$.
| {
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Inverse element in field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ Determine $(2\sqrt{2}-3\sqrt{3}+5)^{-1}$ in $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, which is of degree $4$ over $\mathbb{Q}$ and for $a= \sqrt{2} + \sqrt{3}$ follows $a^4-10a^2+1=0$.
I did calculate inverse elements in field extensions of degree $2$ over $\mathbb{Q}$, but am somehow stuck with this problem, any help is very much appreciated!
| Let $\beta = 2\sqrt{2}-3\sqrt{3}+5$. Then
$(\beta-5)^2 = 35-12\sqrt{6}$ and $((\beta-5)^2-35)^2 = 864$, and so
$$
\beta^4 - 20 \beta^3 + 80 \beta^2 + 200 \beta = 764
$$
Writing this as
$$
\beta(\beta^3 - 20 \beta^2 + 80 \beta + 200) = 764
$$
gives $\beta^{-1} = \frac{1}{764}(\beta^3 - 20 \beta^2 + 80 \beta + 200)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Properties of modular arithmetic clarification The sum of cubes of any three consecutive integers is divisible by $9$. i.e $(n^3 + (n+1)^3 + (n+2)^3)\mod9=0$ for every $n\in\mathbb{Z}$.
They say it is sufficient to check out for values of $n=0,1,2,3,...8$ due to properties of modular arithmetic. Why this is so ?
Thanks
| Alternatively:
$n^3+(n+1)^3+(n+2)^3$
$=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8$
$=3n^3+9n^2+15n+9$
$=9(n^2+1)+3n(n^2+5)$
In fact, you only need to check for $n=1;n=2;n=3$.
*
*If $n=1$ or $n \equiv 1 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$
*If $n=2$ or $n \equiv 2 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$
*If $n=3$ or $3|n$ then $3|n^2+5\Rightarrow9|n(n^2+5)$
Note that $9|9(n^2+1)$, this finishes the proof.
| {
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Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just tried to work out the bottom of the first fraction which I found to be $\frac{(3-x)(4-x)-x}{(3-x)(4-x)}$ now taking the reciprocal gives me $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$.
I did try factorising the bottom to get to $\frac{(3-x)(4-x)}{(x-6)(x-2)}$ and then using partial fractions gets me to $\frac{-15}{2(x-6)}+\frac{1}{2(x-2)}$ which is definitely not where I want to be.
I feel like I need to work from $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$ and somehow pull out a $1$ here but not entirely sure how.
Would really appreciate if anyone could help me.
Thank you.
| Note:
$$\frac{x^2-7x+12}{x^2-8x-12}=\frac{x^2-8x+12+x}{x^2-8x-12}=\frac{x^2-8x+12}{x^2-8x-12}+\frac{x}{x^2-8x-12}=1+\frac{x}{x^2-8x-12}=1+\frac{x}{(x-2)(x-6)}$$
Then Partial Fractions:
$$A(x-2)+B(x-6)=x$$
$$x=2\to-4B=2, B=-\frac 12$$
$$x=6\to 4A=6,A=\frac 32$$
Hence $$\frac{x}{(x-2)(x-6)}=-\frac {1}{2(x-2)}+\frac {3}{2(x-6)}$$
Signs are the wrong way around here, but this is simply because $x-2=-(2-x)$. So simply flip them over and flip signs.
$$\frac{x}{(x-2)(x-6)}=\frac 12\bigg(\frac{1}{(2-x)}-\frac {3}{(6-x)}\bigg)$$
as required.
| {
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} |
Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 3\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot 2^n\cdot n!}{(n!)^2}=2^n\cdot\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}.$
For $n=0$ we have:
$\frac{(2*0)!}{(0!)^2}=1=2^0.$
Now to our inductive step $n+1$:
\begin{align*}
& \frac{(2(n+1))!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(2n+2)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{1\cdot 2\cdot 3\cdot...\cdot 2n+1 \cdot 2n+2}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n \cdot 2n+2)}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot 2^{n+1} \cdot (n+1)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & 2^{n+1}\cdot \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)}{((n+1)!)}\geq2^{n+1}\\
\end{align*}
Is this proof correct/sufficient? Was induction even needed here or should I have argued instead that $\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}>0$? Is this step necessary even for the induction?
| This proof does not look complete. You reduced the question to a more manageable one, but you didn't actually prove that the more manageable one is true. It also never makes use of the inductive hypothesis - as it stands, it's basically a direct proof, not a inductive proof.
I'd also recommend rewriting it because it's somewhat unclear as it is - it moves backwards, starting at its conclusion and simplifying from there. This is logically fine - as you have the statements are equivalent - but it can be clarified by writing prose to indicate that you are reducing the problem to a simpler one. You could also be more precise by saying that you are actually just rewriting the left hand side - that is, we are dealing with a series of equal expressions more than with a series of equivalent statements.
Letting my $n$ be your $n+1$ to keep the expressions clean, you've shown that
$${2n\choose n}=\frac{(2n)!}{(n!)^2}=2^n\cdot \frac{1\cdot 3 \cdot \ldots \cdot (2n-1)}{1\cdot 2 \cdot \ldots \cdot n}.$$
You are trying to claim that this is at least $2^n$. So, what you need to show is that $\frac{1\cdot 3 \cdot \ldots (2n-1)}{1\cdot 2\cdot \ldots \cdot n}$ is at least $1$. An easy way to see that is to break it apart as $\frac{1}{1}\cdot \frac{3}1\cdot \ldots \cdot \frac{2n-1}n$, where every factor is at least $1$, so so is their product. (Or, more directly, you can use the method in @zhw's answer)
You can also argue this inductively by noting that if $\frac{1\cdot \ldots \cdot (2n-1)}{1\cdot 2 \cdot\ldots \cdot n}\geq 1$, then multiplying both sides by $\frac{2n+1}{n+1}\geq 1$ we get $\frac{1\cdot 3 \cdot \ldots\cdot (2n+1)}{1\cdot 2 \cdot\ldots\cdot (n+1)}\geq \frac{2n}{n+1}\geq 1$. Or, going that route, you can also note that
$$\frac{(2n+2)!}{((n+1)!)^2}=\frac{(2n+1)\cdot (2n+2)}{(n+1)^2}\cdot \frac{(2n)!}{(n!)^2} =2\cdot \frac{2n+1}{n+1}\cdot \frac{(2n)!}{(n!)^2}\geq 2\cdot {2n \choose n}$$
from which point the inductive proof is easy enough construct, since your expression is at least doubling at each step
| {
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Cyclotomic polynomials and products of cosines problem I've run into an inconsistency I can't figure out while trying to find products of the cosine of various roots of unity.
For example: $\cos(\frac{2\pi}{5})\cdot cos(\frac{4\pi}{5}) \cdot cos(\frac{6\pi}{5}) \cdot cos(\frac{10\pi}{5})$ Multiply everything by $\frac{\sin(\frac{2\pi}{5})}{sin(\frac{2\pi}{5})}$ and apply the trig rules and you get $\frac{1}{16}$
(which is correct)
These are all part of the 5th roots of unity which are solutions to $z^5 - 1 = 0$ and all of the form $\cos(x) + i\sin(x)$ So you can substitute that in expand and replace $\sin(x)^2 = 1 - cos(x)^2$ and split by the real and imaginary parts. For the reals you get $16\cos(x)^5 -10\cos(x))^3 + 5\cos(x) - 1 = 0$ and by the vieta formula this also implies that product of the cosine of the 5 roots is $\frac{1}{16}$ One root is 1 so this is consistent with first method.
But then I was looking at $\Phi_5 = z^4 + z^3 + z^2 + z + 1$ The four root besides one are roots of this polynomial and I should be able to do the same substitution from above.
let c = cos(x) and s =sin(x)
$(c + is)^4 + (c + is)^3 + (c + is)^2 + c + is + 1 = 0$
$(c^4 + 4ic^3s -6c^2s^2 -4ics^3 + s^4) + (c^3 + 3ic^2s - 3cs^2 - is^3) + (c^2 + 2ics - s^2) + (c + is) + 1 = 0$
For the real part:
$c^4 -6c^2s^2 + s^4 + c^3 - 3cs^2 + c^2 - s^2 + c + 1 = 0$
Then substituting $s^2 = 1 - c^2$ I get
$8\cos(x)^4 + 4\cos(x)^3 - 6\cos(x)^2 - 2\cos(x) + 1 = 0$ which implies that the product of the cosines is $\frac{1}{8}$
What's going on?
| Thanks that gave me the idea to figure out what went wrong. The key is its necessary to solve both the imaginary and real parts of the derived equation and throw out solutions that don't handle both. Its interesting what extra solutions did show up and how this method "eliminated" the duplicate roots.
| {
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Please help us to find the general solution of this recurrence: $x_{n+1}=5^{n-1} x_n+3^n$ Please help us to find the general solution of this recurrence:
$$x_{n+1}=5^{n-1} x_n+3^n.$$
We found the solution for the associated homogeneous recurrence
$x_{n+1}=5^{n-1} x_n$
which is
$$x_n = 5^{(n-1)(n-2)/2}x_1,$$
and we tried to use it to find a particular solution of the nonhomogeneous recurrence, but unfortunately we were not able to find it.
| This is the route I usually take but....
$$\begin{align}
x_{n+1}&=5^{n-1} x_n+3^n \\
&=5^{n-1}(5^{n-2} x_{n-1}+3^{n-1})+3^n \\
&\quad = 5^{2n-(1+2)}x_{n-1}+(3 \cdot 5)^{n-1} +3^n \\
&=5^{2n-(1+2)}(5^{n-3}x_{n-2}+3^{n-2})+(3\cdot5)^{n-1}+3^n \\
&\quad =5^{3n-(1+2+3)}x_{n-2}+5^{2n-(1+2)}3^{n-2}+5^{n-1}3^{n-1}+3^n \\
&\vdots \\
&\vdots \quad \text{seeing a pattern...after i steps} \\
&\vdots \\
&=5^{(i+1)n+\frac{(i+1)(i+2)}{2}}x_{n-i}+\sum_{k=0} ^{i} 5^{kn-\frac{k(k+1)}{2}}3^{n-k} \\
&\vdots \\
&\vdots \quad \text{supposing that you do it n times, and set i=n...} \\
\\
&\vdots \\
&=5^{\frac{(n+1)(3n+2)}{2}}x_0+\sum_{k=0} ^{n} 5^{kn-\frac{k(k+1)}{2}}3^{n-k} \\
\end{align}$$
I'm not sure if this is any help to you but I couldn't resist taking a stab.
| {
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show that this integral is correct? How can one show that
$$\int_0^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha} \tan(x) \tan\left(\frac{x}{2}\right)\right] \sqrt{\sin(x) \tan \left(\frac{x}{2}\right)} \, \mathrm dx=-\frac{\ln(\alpha)}{\sqrt{2}}$$
assume $\alpha\ge1$.
I can't see how to simplify $\tan(x)\tan(x/2)$ and $\sin(x)\tan(x/2)$.
| Using
$$\tan(x) \, \tan(x/2) = \frac{2 \, \sin^{2}(x/2)}{\cos(x)}$$
and
$$\sin(x) \, \tan(x/2) = 2 \, \sin^{2}(x/2)$$
then
\begin{align}
I &= \int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\tan(x)\tan\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx \\
&= 2\sqrt{2} \, \int_{0}^{\pi/2} \sin(x/2) \, \cos(x/2) \, \ln\left(\frac{2}{\alpha} \, \frac{\sin^{2}(x/2)}{\cos(x)} \right) \, dx
\end{align}
Let $u = \sin(x/2)$ to obtain
\begin{align}
I &= 4 \sqrt{2} \, \int_{0}^{1/\sqrt{2}} u \, \ln u \, du - 2\sqrt{2} \, \ln\left(\frac{\alpha}{2}\right) \, \int_{0}^{1/\sqrt{2}} u \, du - \sqrt{2} \,\int_{0}^{\pi/2} \sin(x) \, \ln(\cos(x)) \, dx \\
&= - \frac{1 + \ln 2}{\sqrt{2}} - \frac{1}{\sqrt{2}} \, \ln\left(\frac{\alpha}{2}\right) + \frac{1}{\sqrt{2}} \\
&= - \frac{\ln \alpha}{\sqrt{2}}.
\end{align}
| {
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"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
How to solve the following trigonometrical equation? I have the following equations
\begin{align*}
R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\
R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}).
\end{align*}
From these equations how can I obtain the following solution
$$\omega T_2=\pm \arccos\left(\frac{Q_1^2+Q_2^2-R_1^2+R_2^2}{2R_2\sqrt{Q_1^2+Q_1^2}}\right)-\arctan\left(\frac{Q_2}{Q_1}\right)+\phi_2+2k\pi .$$
| As mentioned in the comments by @Blue the problem can be reduced into $$ a \sin \theta + b \cos \theta +c = 0 $$
to be solved for $\theta$.
The way I have dealt with this equation in the past was to split the angle into two angles $\theta = \varphi + \psi$, and expand
$$\sin \theta = \sin(\varphi+\psi) = \cos\varphi \sin\psi + \sin\varphi \cos\psi$$
$$\cos \theta = \cos(\varphi+\psi) = \cos\varphi \cos \psi - \sin\varphi \sin\psi$$
Then my equation to be solved is
$$ \Bigl( \cos\varphi ( b \cos\psi + a \sin \psi) \Bigr) + \Bigl( \sin\varphi (a \cos\psi - b \sin\psi)+c \Bigr) =0 $$
and you can set each part equal to zero individually
$$\begin{aligned}
\cos\varphi ( b \cos\psi + a \sin \psi) & = 0 \\
\sin\varphi (a \cos\psi - b \sin\psi)+c & = 0
\end{aligned} $$
The first one is solved by $\boxed{\psi = -\tan^{-1} \left( \frac{b}{a} \right) + k \pi }$ which transforms the second one into
$$ \sin\varphi \sqrt{a^2+b^2}+c = 0 $$
with solution $\boxed{ \varphi = \sin^{-1}\left( \frac{c}{\sqrt{a^2+b^2}} \right) +2 k \pi}$
So all together you have
$$ \boxed{ \color{blue}{ \theta = \sin^{-1}\left( \frac{c}{\sqrt{a^2+b^2}} \right)-\tan^{-1} \left( \frac{b}{a} \right) +2 k \pi} } $$
| {
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"source": "stackexchange",
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"answer_count": 3,
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} |
Find generating function of series $a_{n} = 2^{n} + 3^{n} $ Find generating function of series $a_{n} = 2^{n} + 3^{n} $
I'm having a problem with this because at first i have to find recursive equation for $a_{n}$ . I find that $a_{0} = 2$ , $a_{1} = 5$ but how do I find the rest? Also I can't simply add generating functions, I have to solve recursive equation using generating functions. First step is to find recursive equation for $a_{n}$ then some other stuff that I can handle.
| Hint:
$$\sum_{n=0}^\infty a_nx^n = \sum_{n=0}^\infty (2^n+5^n)x^n = \sum_{n=0}^\infty (2x)^n + \sum_{n=0}^\infty (5x)^n = \frac1{1-2x} + \frac1{1-5x}$$
Now you can proceed to find the recursive relation:
$$\sum_{n=0}^\infty a_nx^n = \frac1{1-2x} + \frac1{1-5x} = \frac{2-7x}{1-7x+10x^2}$$
$$\sum_{n=0}^\infty a_nx^n - 7\sum_{n=1}^\infty a_{n-1}x^n + 10 \sum_{n=2}^\infty a_{n-2}x^n = (1-7x+10x^2)\sum_{n=0}^\infty a_nx^n = 2-7x$$
Comparing powers gives:
\begin{align}
x^0:& \quad a_0 =2\\
x^1:& \quad a_1 - 7a_0 = -7 \implies a_1 = 7\\
x^k \text{ for } k \ge 2:& \quad a_k - 7a_{k-1} + 10a_{k-2} = 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Convergence of integral with trigonometry
*
*$\displaystyle \int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}} dx$
Obviously, problem in $ -\frac{\pi}{4} $ as $ \lim_{x\to -\frac{\pi}{4}^+} = \infty $, integral is positive on whole segment, but I can't use any usual rules to prove it's convergence or simly divide it inside the root. I guess, i have to bound it with converging function, but can't find a good one aswell. Any thoughts?
| $$I=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}}, dx$$
$1)sin(x)+cos(x)=Rcos(x-a)\therefore R=\sqrt{2}cos(x-\frac{\pi}{4})$
$$sin(x)+cos(x)=\sqrt{2}cos(x-\frac{\pi}{4})$$
do the same for $cos(x)-sin(x)$, so
$$cos(x)-sin(x)=\sqrt{2}cos(x+\frac{\pi}{4})$$
$$\therefore\frac{\cos x - \sin x}{\cos x + \sin x}=\frac{\sqrt{2}cos(x+\frac{\pi}{4})}{\sqrt{2}cos(x-\frac{\pi}{4})}=\frac{cos(x+\frac{\pi}{4})}{cos(x-\frac{\pi}{4})}$$
Can you do anything with this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Given matrix $A$, find a matrix $S$ such that result of $S^{-1} \circ A \circ S$ is a diagonal matrix
Given is matrix $A=\begin{pmatrix} 3 & 0 & 7\\ 0 & 1 & 0\\ 7 & 0 & 3
\end{pmatrix}$, find a matrix $S$ such that $D=S^{-1} \circ A \circ S$
where $D$ is a diagonal matrix.
So I'm not sure how this is supposed to work, I tried to create some linear system but soon realized there are too many unknowns! Then I thought about the zero matrix, if it's actually a diagonal matrix. This should be the case because a diagonal matrix is a matrix where all non-diagonal entries are zero and this is satisfied in a zero matrix.
But the bad thing is when we want the inverse of that zero matrix, we have a problem because division by zero doesn't work : /
So I can't use that matrix. But how else do you solve this problem?
| Find first the eigenvalues of $A$, these are the roots of the characteristic polynomial of $A$,
$$
(x-10)(x-1)(x+4)\ .
$$
So the eigenvalues are $10$, $1$, $-4$. We subtract them from the diagonal and search for eigenvectors. Here are my choices:
$$
\begin{aligned}
\begin{bmatrix}
3-10 & 0 & 7\\
0& 1-10 & 0\\
7 & 0 & 3-10
\end{bmatrix}
\begin{bmatrix}
1\\0\\1
\end{bmatrix}
&=
\begin{bmatrix}
0\\ 0\\ 0
\end{bmatrix}
\ ,\\
\begin{bmatrix}
3-1 & 0 & 7\\
0& 1-1 & 0\\
7 & 0 & 3-1
\end{bmatrix}
\begin{bmatrix}
0\\1\\0
\end{bmatrix}
&=
\begin{bmatrix}
0\\ 0\\ 0
\end{bmatrix}
\ ,
\\
\begin{bmatrix}
3-(-4) & 0 & 7\\
0& 1-(-4) & 0\\
7 & 0 & 3-(-4)
\end{bmatrix}
\begin{bmatrix}
1\\ 0\\ -1
\end{bmatrix}
&=
\begin{bmatrix}
0\\ 0\\ 0
\end{bmatrix}
\ ,\\
\end{aligned}
$$
Equivalently,
$$
\begin{aligned}
\begin{bmatrix}
3 & 0 & 7\\
0& 1 & 0\\
7 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
1\\0\\1
\end{bmatrix}
&=
\begin{bmatrix}
1\\0\\1
\end{bmatrix}
\cdot[10]
\ ,\\
\begin{bmatrix}
3 & 0 & 7\\
0& 1 & 0\\
7 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
0\\1\\0
\end{bmatrix}
&=
\begin{bmatrix}
0\\1\\0
\end{bmatrix}
\cdot
[1]
\ ,
\\
\begin{bmatrix}
3 & 0 & 7\\
0& 1 & 0\\
7 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
1\\ 0\\ -1
\end{bmatrix}
&=
\begin{bmatrix}
1\\ 0\\ -1
\end{bmatrix}
[-4]
\ ,\\
\end{aligned}
$$
equalities of products of matrices of the shape
$(3\times 3)\cdot(3\times 1)$, and
respectively $(3\times 1)\cdot(1\times 1)$.
Now consider the matrix built from the eigenvectors, taken as columns.
We have the block matrix computation, subsumming the above three equalities:
$$
\underbrace{
\begin{bmatrix}
3 & 0 & 7\\
0& 1 & 0\\
7 & 0 & 3
\end{bmatrix}}_A
\underbrace{
\begin{bmatrix}
1 & 0 & 1\\
0 & 1 & 0\\
1& 0& -1
\end{bmatrix}}_S
=
\underbrace{
\begin{bmatrix}
1 & 0 & 1\\
0 & 1 & 0\\
1& 0& -1
\end{bmatrix}}_S
\underbrace{
\begin{bmatrix}
10 & & \\
&1 & \\
& & -4
\end{bmatrix}}_D
\ .
$$
We have $AS=SD$, so the above choices answer the request.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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find range for dependent
There are three vector $a$, $b$, $c$ in three-dimensional real vector space, and the inner product between them $a\cdot a=b\cdot b = a\cdot c= 1, a\cdot b= 0, c\cdot c= 4$. When setting $x = b\cdot c$, answer the following question: when $a, b, c$ are linearly dependent, find all possible values of $x$.
(dot here means dot product)
For dependent condition
$$\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}$$
So $\theta = 0, \pi$.
Then
$$\begin{align}
x&=|b||c| \cos \theta \\
x&=2 \cos \theta \\
\implies x &= 2 \cos 0, x = 2 \cos \pi \\
x &= \mp 2
\end{align}$$
Am I right using triple cross product? or should i find $\theta$?
| For the first identity, $a \times b$ is orthogonal to both $a,b$, hence $c$ must be in the plane spanned by the orthonormal $a,b$.
So we can write $c$ as
$$
c = \alpha a + \beta b
$$
plugging this in we see that
$$
1 = a \cdot c = \alpha a \cdot a = \alpha
$$
and
$$
x = b \cdot c = \beta b \cdot b = \beta
$$
hence the norm squared of $c$ is
$$
c\cdot c = \alpha^2 + \beta^2 = 1 + x^2 = 4
$$
this gives
$$
x = \pm \sqrt{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For $0 < x < \pi/2 $, prove that $x + \frac{x^{3}}{3} <\tan x $ I proved as follows:
Let $f(x) = \tan x - x - \frac{x^{3}}{3}.$
Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x > \tan x - x,$ since $\\tan x > x,$ $f''(x) > 0$ so $f'(x)$ is increasing.
Since $f '(0) = 0,$ $f '(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$
So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$
But my proof seems complex. Is there any simpler way to prove this?
| For $0 < x < \pi/2$ we have $\tan'(x) = 1 + \tan^2(x)$ and therefore
$$
\tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x 1 \, du = x \, .
$$
It follows that
$$
\tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x (1 +u^2) \, du = x + \frac 13 x^3 \, .
$$
This process can be repeated to get better lower bounds:
$$
\tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x (1 + (u + \frac {u^3}3)^2) \, du = x + \frac 13 x^3 + \frac{2}{15} x^5 + \frac{1}{54} x^7 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| Note that $$\bigg(x+\frac 1x\bigg)^2=x^2+\frac{1}{x^2}+2\to x+\frac 1x=\sqrt{a+2}$$
$$\bigg(x+\frac{1}{x}\bigg)\bigg(x^2+\frac{1}{x^2}\bigg)=x^3+\frac{1}{x^3}+x+\frac 1x\to x^3+\frac{1}{x^3}=(a-1)\sqrt{a+2}$$
$$\bigg(x^2+\frac{1}{x^2}\bigg)^2=x^4+\frac{1}{x^4}+2\to x^4+\frac{1}{x^4}=a^2-2$$
$$\bigg(x^4+\frac{1}{x^4}\bigg)\bigg(x+\frac 1x\bigg)=x^5+\frac{1}{x^5}+x^3+\frac{1}{x^3}$$
$$\to (a^2-2)(\sqrt{a+2})=x^5+\frac{1}{x^5}+(a-1)(\sqrt{a+2})$$
$$\to x^5+\frac{1}{x^5}=(a^2-a-1)(\sqrt{a+2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Evaluate: $\lim_{x\to 0}\frac{\sin^3x-x^3\operatorname{sgn}\left(1-\left[\frac{x}{\sin^{-1}x}\right]\right)}{x\tan^2 x\sin(\pi\cos x)}$ $$\lim_{x\to 0}\frac{\sin^3x-x^3\operatorname{sgn}\left(1-\left[\frac{x}{\sin^{-1}x}\right]\right)}{x\tan^2 x\sin(\pi\cos x)}$$
(Note that here $\operatorname{sgn}$ denotes the Signum function and $[x]$ denotes the greatest integer less than or equal to $x$.)
The above simplifies to:
$$\lim_{x→0} \frac{\sin^{3}(x)-x^{3}}{x\tan^{2}(x)\sin(\pi(1-\cos x))}$$
then after further simplification i get:
$$=\lim_{x→0} \frac{6}{π}\frac{\sin(x)-x}{x^{3}}$$
$$=\lim_{x→0} \frac{6}{\pi}\frac{-1}{6}=\frac{-1}{π},$$
but in this method I have to calculate the limit of
$$\lim_{x→0} \frac{\sin(x)-x}{x^{3}}=\frac{-1}{6},$$
which is lengthy. Is there any other way to do this without calculating this limit.
| We have that
$$\frac{\sin^{3}x-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))}
=\frac{x^3-\frac12x^5+o(x^5)-x^3}{x^3(x^2+o(x^2))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}
=\frac{-\frac12x^5+o(x^5)}{(x^5+o(x^5))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}
=\frac{-\frac12+o(1)}{(1+o(1))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}\to \frac{-\frac12}{1\cdot\frac \pi 2\cdot 1}=-\frac 1 \pi$$
Without Taylor's expansion note that we have
$$\frac{\sin^{3}x-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))} =\frac{\sin^{3}x-x^{3}}{x^5}\frac{1}{\frac{\tan^2x}{x^2} \frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}$$
then since by standard limits the second part $\to \frac 2 \pi $ we can analize
$$\frac{\sin^{3}x-x^{3}}{x^5}=\frac{(\sin x-x)(\sin^2 x+\sin x \cdot x+x^2)}{x^5}=\frac{\sin x-x}{x^3}\frac{\sin^2 x+\sin x \cdot x+x^2}{x^2}$$
and since by standard limits $\frac{\sin^2 x+\sin x \cdot x+x^2}{x^2}\to 3$ we need to consider
$$\frac{\sin x-x}{x^3}\to -\frac 16$$
which can be solved as indicated here Are all limits solvable without L'Hôpital Rule or Series Expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the limiting distribution of the sequence The following is a qualifying exam problem.
Let $\{X_n\}$ be an i.i.d. sequence of Binomial($n,1/2$) random variables. Define the sequence,
$$ Y_n = \left(1+\frac{1}{\sqrt{n}}\right)^{X_n}\left(1-\frac{1}{\sqrt{n}}\right)^{n-X_n} $$
Find the limiting distribution of $Z_n = \ln(Y_n)$.
I tried using a characteristic function approach. The chf of $Z_n$ is given by,
$$ \begin{eqnarray*}\phi_{Z_n}(t) &=& E[e^{itZ_n}] \\ &=& E[e^{it\ln(Y_n)}] \\
&=& E[e^{itX_n\ln\left(1+\frac{1}{\sqrt{n}}\right)}e^{it(n-X_n)\ln\left(1-\frac{1}{\sqrt{n}}\right)}]\\
&=& \sum\limits_{k=0}^ne^{it\ln\left(1+\frac{1}{\sqrt{n}}\right)k}e^{it\ln\left(1-\frac{1}{\sqrt{n}}\right)(n-k)}\binom{n}{k}\frac{1}{2^n}\\
&=& \left(\frac{(1+\frac{1}{\sqrt{n}})^{it}+(1-\frac{1}{\sqrt{n}})^{it}}{2}\right)^n \end{eqnarray*} $$
A computation in Mathematica reveals that,
$$ \lim\limits_{n\rightarrow\infty}\phi_{Z_n}(t) = e^{-\frac{1}{2}it - \frac{1}{2}t^2} $$
which we recognize as the chf of a Gaussian distribution with mean $-\frac{1}{2}$ and variance $1$.
A couple questions:
1) Is there a straightforward way of computing the limit above (without the use of CAS software)?
2) Is there a more time efficient approach here? Keep in mind that this was a qualifying exam problem.
| You can use the central limit theorem:
\begin{align}
&\ln Y_n \\
& = X_n \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}} + n \ln \left(1-\frac{1}{\sqrt{n}}\right) \\
%
% & = \underbrace{\sqrt{n} \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}}_{\xrightarrow[n\to\infty]{} \, 2} \times \sqrt{n}\left( \frac{X_n}{n} - \frac{1}{2} \right)
% + \frac{1}{2}n \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}
% + n \ln \left(1-\frac{1}{\sqrt{n}}\right) \\
%
%
%
& = \sqrt{\mbox{Var}(X_1)} \underbrace{\sqrt{n} \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}}_{\xrightarrow[n\to\infty]{} \, 2} \times \frac{\sqrt{n}}{\sqrt{\mbox{Var}(X_1)}}\left( \frac{X_n}{n} - \frac{1}{2} \right) +
\frac{n}{2} \ln \left( 1+\frac{1}{\sqrt{n}}\right) + \frac{n}{2} \ln \left( 1-\frac{1}{\sqrt{n}}\right) \\
%
%
%
&\xrightarrow[n\to\infty]{D} Z-\frac{1}{2}
\end{align}
where Var$(X_1)=1/4$, $Z\sim N(0,1)$, and in the last I used that
\begin{align}
\lim_{n\to\infty} \left( \left( 1+\frac{1}{\sqrt{n}}\right)\left( 1-\frac{1}{\sqrt{n}}\right)\right)^n
= \lim_{n\to\infty} \left( 1-\frac{1}{n}\right)^n = \frac{1}{e}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Upper Bounding with an Exponential Consider the following, where $n\geq 7$ is a natural number:
$$\frac{1}{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}.$$
I am convinced that there should be an upper bound of the form
$$\frac{1}{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}\leq e^{-c(n)},$$
for $c(n)$ some positive function of $n$ but I am failing dismally to derive one.
Any help would be gratefully appreciated.
Edit: I think $c(n)\approx 4.689$ works and suffices for my needs. This is independent of $n$.
| Note that we can write
$$\begin{align}
\frac1{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}&=e^{-\pi^2}\left(1+\frac{e^{\pi^2/n^2}-1}{2}\right)^{n^2}\\\\
&\le e^{-c(n)}
\end{align}$$
where $c(n)$ is given by
$$c(n)\le\pi^2-n^2\log\left(1+\frac{e^{\pi^2/n^2}-1}{2}\right)$$
The asymptotic expansion of $c(n)$ is given by
$$\begin{align}
c(n)&=\pi^2-n^2 \log\left(1+\frac{(\pi^2/n^2+\frac12(\pi^2/n^2)^2+\frac16(\pi^2/n^2)^3+O\left(1/n^8\right)}{2}\right)\\\\
&=\pi^2-n^2\left(\frac{\pi^2}{n^2}+\frac{\pi^4}{8n^4}+O(1/n^8)\right)\\\\
&=\frac{\pi^2}{2}-\frac{\pi^4}{8n^2}+O\left(\frac1{n^6}\right)\\\\
&=c_{\text{approx}}+O\left(\frac1{n^6}\right)
\end{align}$$
where $c_{\text{approx}}=\frac{\pi^2}{2}-\frac{\pi^4}{8n^2}$.
For $n=7$, we have
$$\begin{align}c(7) &\approx 4.68672854698357\\
c_{\text{approx}}(7) &=4.68630962137631
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the real roots and complex roots to a polynomial I have tried solving and finding the real roots to the polynomial
$ {x^5 - 5x + 3} $
by saying that one of the solutions has to be a factor of the $ +3 $, so the factors might be
$ \pm 1, or \pm 3 $
and when I place the supposed factors into the polynomial, I get that none of them zero out, so I guess it means no real numbers have the zeroes for this function. So how do I go about finding the complex roots if I don't have the real roots?
| Suppose that
$$x^5-5x+3=(x^2+a x+b)(x^3+c x^2+d x+e)$$ Expand and group to get
$$(b e-3)+x (a e+b d+5)+x^2 (a d+b c+e)+x^3 (a c+b+d)+x^4
(a+c)=0$$ n which all coefficients must be equal to $0$.
Then, successively, $c=-a$, $b+d=a^2\implies d=a^2-b$, $e=-a^3+2ab$ make that we are left with
$$-3 - a^3 b + 2 a b^2=0 \qquad \text{and} \qquad 5 - a^4 + 3 a^2 b - b^2=0$$ where $a=1$, $b=-1$ seem to be "obvious" solutions.
So, back to $c,d,e$
$$x^5-5x+3=(x^2+x-1)(x^3-x^2+2x-3)$$ The first term has two real roots and the second has then the equation has one real root and two non-real complex conjugate roots since $\Delta=-175$.
Now, have look here for the solution of the cubic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $S$ in term of $k$ (telescoping sums) Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.
Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.
I've tried to write the terms of $S$ in the following way:
\begin{align}
\frac{1275}{1276} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1275\times1276} \\
\frac{1276}{1277} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1276\times1277} \\
& \,\,\,\vdots \\
\frac{2548}{2549} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2548\times2549}
\end{align}
But it's just not enough, because in $k$ we don't have $\frac{1}{2\times3}$, $\frac{1}{4\times5}$, etc.
Can you please help me solve the problem? Thanks!
| Note that your sum is $$\sum_{n=1}^{1275}{\frac{1}{2n(2n-1)}}$$
We can rewrite this using partial fractions as:
$$\sum_{n=1}^{1275}{\bigg[\frac{1}{2n-1}-\frac{1}{2n}\bigg]}$$
and use telescoping from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Generalization of Opperman's Conjecture Does this conjecture have a name? What about a counterexample?:
$$ \forall n,k \in \mathbb{N}, k \gt 1, \exists d \in (kn-n,kn] \text{ s.t. } d \perp n! $$
An equivalent statement is this:
Take a Sieve of Eratosthenes-like list of any fixed width $n$ and sieve out multiples of the primes $\leq n$. While keeping the width fixed, let the length of the list extend to infinity. In every row there is an unsieved number (a number that is coprime to all the primes $\leq n$).
Note that for $k \leq n+2$, $d\perp n!$ is necessarily prime. Bertrand's Postulate sets $k=2$. Legendre's conjecture is about the union of rows $n+1$ and $n+2$. Opperman's conjecture is each of these rows individually. This generalization would extend Opperman to any row whatsoever (except the first).
Results to date: Erdos et. al. proved Bertrand. Legendre and Opperman are open. Prime sieves like this one are cyclic in nature, with the distribution of primes repeating every $\prod^{n}_{p \perp n} p$ rows. Ive checked by hand all possible rows in all the possible widths $\leq$ 7.
Examples of sieves:
$n = 5$:
\begin{array}{|c|c|c|c|c|}
\hline
1 & \color{grey}{2} & \color{grey}{3} & & \color{grey}{5} \\ \hline
& 7 & & & \\ \hline
11 & & 13 & & \\ \hline
& 17 & & 19 & \\ \hline
& & 23 & & \\ \hline
& & & 29 & \\ \hline
31 & & & & \\ \hline
& 37 & & & \\ \hline
41 & & 43 & & \\ \hline
& 47 & & 49 & \\ \hline
& & 53 & & \\ \hline
& & & 59 & \\ \hline
61 & & & & \color{white}{65} \\ \hline
\end{array}
$n = 6$:
\begin{array}{|c|c|c|c|c|c|}
\hline
1 & \color{grey}{2} & \color{grey}{3} & & \color{grey}{5} & \\ \hline
7 & & & & 11 &\\ \hline
13 & & & & 17 &\\ \hline
19 & & & & 23 &\\ \hline
& & & & 29 &\\ \hline
31 & & & & &\\ \hline
37 & & & & 41 &\\ \hline
43 & & & & 47 &\\ \hline
49 & & & & 53 &\\ \hline
& & & & 59 &\\ \hline
61 & \color{white}{62} & \color{white}{63} & \color{white}{64} & \color{white}{65} & \color{white}{66}\\ \hline
\end{array}
| Your conjecture does not hold for $n=8, k=25$ as all $8$ numbers
$$(201, 202, 203, 204, 205, 206, 207, 208)$$ have factors less than $8$, which are $(3, 2, 7, 2, 5, 2, 3, 2)$ respectively.
For prime $n$, smallest counterexample is $n=13, k=168$ because all $13$ numbers$$2185, 2186, \cdots, 2197$$have factors less than or equal to $13$, which are $(5, 2, 3, 2, 11, 2, 7, 2, 3, 2, 5, 2, 13)$.
I thought your conjecture is true before running a program because intuitively $\prod_{p_i\le n} (p_i-1)/p_i$, which is the ratio of numbers left on each stage of Sieve of Eratosthenes decreases much slower than the growth of $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.
PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
I LOVE YOU HONEY.
Anyone knows? Are we gonna get an integer number?
My attempt:
Does this help?
$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$
(long division)
\begin{align*}
I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\
&=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\
&=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2
\end{align*}
after the substitution
\begin{gather*}
(x-2)^{1/2}=u\\
du=\frac1{2(x-2)^{1/2}}dx\\
u^2=x-2\\
(x-1)^{1/2}=(u^2+1)^{1/2}
\end{gather*}
Update: This may help us proceed.
I tried to proceed:
$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$
after $u^2+1=t$ and $dt=2udu$
\begin{align*}
u^4-4u^2-4
&= (u^2+1)^2-(6u^2+5) \\
&= (u^2+1)^2-6(u^2+1)+1 \\
&= ((u^2+1)-3)^2-8
\end{align*}
I wonder whether this question can be solved from here?
Update:
This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.
Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.
| Alternative method:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Express the integrand in the form $\lfrac{(2ax+b)·(x^2-3x+2)+(2cx+d)}{\sqrt{x^2-3x+2}}$ for some constants $a,b,c,d$.
Then split it into $( a(2x-3) + (3a+b) ) · \sqrt{x^2-3x+2} + \lfrac{c(2x-3)+(3c+d)}{\sqrt{x^2-3x+2}}$, so that as a sum of four terms the first and third have obvious antiderivatives. The other two terms can be solved by standard techniques.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be the right.
| dont know of an elegant method:
on LHS substitute $c = a - b$
$LHS = \sin^2 a + \sin^2 b + (\sin a\cos b - \cos a \sin b)^2 $
$= \sin^2 a + \sin^2 b + \sin^2a \cos^2b + \cos^2a \sin^2b - 2\cos a\cos b \sin a \sin b$
$ = \sin^2 a + \sin^2 b + (1 - \cos^2a)\cos^2b + \cos^2a(1-\cos^2b)- 2\cos a\cos b \sin a \sin b $
$ = 2 - 2\cos^2a\cos^2b - 2\cos a\cos b \sin a \sin b$
$= 2 - 2 \cos a \cos b(\cos a\cos b + \sin a \sin b)$
$= 2 (1 - \cos a \cos b\cos(a-b)) = 2(1 - \cos a \cos b\cos c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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A problem in trigonometry If $\tan(x+y) =a+b$
and, $\tan(x-y)=a-b$
Then prove that:$a \tan(x) -b \tan(y)=a^{2}-b^{2}$
I have used the formulae for $\tan(x+y)$ and $\tan(x-y)$ and then cross multiplied and then found $a$ and $b$ individually and then tried to put the values of $a$ and $b$ in "$a \tan(x) -b \tan(y)$"(l.h.s) but it did not lead to the r.h.s.
| We are given
$$
\tan(x+y)=a+b\tag1
$$
and
$$
\tan(x-y)=a-b\tag2
$$
Solving for $a$ gives
$$
\begin{align}
a
&=\frac12(\tan(x+y)+\tan(x-y))\\[3pt]
&=\frac12\left(\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}+\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\right)\\
&=\frac{\tan(x)+\tan(x)\tan^2(y)}{1-\tan^2(x)\tan^2(y)}\tag3
\end{align}
$$
and solving for $b$ gives
$$
\begin{align}
b
&=\frac12(\tan(x+y)-\tan(x-y))\\[3pt]
&=\frac12\left(\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}-\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\right)\\
&=\frac{\tan(y)+\tan^2(x)\tan(y)}{1-\tan^2(x)\tan^2(y)}\tag4
\end{align}
$$
Therefore,
$$
\begin{align}
a\tan(x)-b\tan(y)
&=\frac{\tan^2(x)-\tan^2(y)}{1-\tan^2(x)\tan^2(y)}\\
%&=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\\[6pt]
%&=\tan(x+y)\tan(x-y)\\[9pt]
%&=a^2-b^2
\tag5
\end{align}
$$
Factoring the right hand side of $(5)$ leads quickly to the answer.
| {
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"answer_id": 2
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Linear equation help $$\frac{1-x}{4} + \frac{5x+1}{2} = 3 - \frac{2(x+1)}{8}$$
I got x=1 but the book says x=4/5 and I don't understand how to get to that, I tried working backwards too but I just can't figure it out. Any help would be appreciated, thanks.
Here's my work
using 8 as the lcm
$$
\begin{align}
\frac 81(\frac {(1-x)}4) + \frac 81(\frac {(5x+1)}2) &= \frac 81(3) - \frac 81\frac {2(x+1)}8 \\
2(1-x) + 4(5x+1) &= 24 - 2x + 2 \\
2 - 2x + 20x + 4 &= 24 - 2x + 2 \\
6 + 18x &= 26 -2x \\
20x &= 20 \\
x &= 1 \\
\end{align}
$$
| We have that
$$\frac{1-x}{4} + \frac{5x+1}{2} = 3 - \frac{2(x+1)}{8}\iff 2(1-x)+4(5x+1)=24-2(x+1)\\\iff2-2x+20x+4=24-2x-2\iff20x=16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What am I doing wrong in evaluating this triple integral? I am required to find the volume common to ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2} = 1$ and cylinder $x^2 + y^2 = ay$
I set up the following triple integral:
$$V=2\cdot \int_{}^{}\int_{R}^{}\int_{z=0}^{\frac{b}{a}\cdot \sqrt{a^2 - x^2 - y^2}}dz\cdot dR= 2\frac{b}a \cdot \int_{}^{}\int_{R}^{} \sqrt{a^2 - x^2 - y^2} dy dx$$ where R: $x^2+y^2=ay$
Using $x=r\cos(t), y=r\sin(t), dydx = rdrdt$
$$V=2\cdot \frac{b}a \int_{t=0}^{\pi}\int_{r=0}^{asin(t)} \sqrt{a^2 -r^2}\cdot r \;\; dr dt \; \; \; \; equation -(1)$$
Let $a^2-r^2=m \Rightarrow r\cdot dr=\frac{-1}2dm$
$$V=2\cdot \frac{b}{a}\cdot \frac{-1}2\cdot \int_{t=0}^{\pi}\int_{m=a^2}^{a^2cos^2(t)} \sqrt{m}\;\; dm dt = \frac23\cdot \frac{b}{a}\cdot a^3\cdot \int_{t=0}^{\pi}1-\cos^3(t) \; dt $$
Now $\cos^3(\pi-t)=-\cos^3(t) $ Hence its integration is $0$ over $0 $ to $\pi$
$$V=\frac23\cdot ba^2 \cdot(\pi) $$
However the answer given to be is $\frac43 ba^2 (\frac\pi2-\frac23)$
The thing confusing me is if I had used symmetry in equation $(1)$ and multiplied the integral by $2$ in order to reduce the limits of $t$ from $0$ to $\pi/2 $, I am getting the desired result as below :
From equation -$(1)$
$$V=2\cdot \frac{b}a \int_{t=0}^{\pi}\int_{r=0}^{asin(t)} \sqrt{a^2 -r^2}\cdot r \;\; dr dt $$
$$= 4\frac{b}a \int_{t=0}^{\pi/2}\int_{r=0}^{asin(t)} \sqrt{a^2 -r^2}\cdot r \;\; dr dt = 4\frac{b}a \cdot \frac{-1}2\int_{t=0}^{\pi/2}\int_{m=a^2}^{a^2cos^2(t)} \sqrt{m}\; dmdt$$
$$ = 2\frac{b}a \cdot a^3\cdot \frac23 \int_{0}^{\pi/2}1-cos^3(t)\; dt = \frac43\cdot ba^2 (\frac\pi2-\frac23)$$
Where am I going wrong in the first method?
| In the first integral the set up should be
$$V=2\cdot \int_{}^{}\int_{R}^{}\int_{z=0}^{\sqrt{b^2 - \frac{b^2}{a^2}x^2 - \frac{b^2}{a^2}y^2}}dz\cdot dR$$
It seems that the problem is in this step for $(a^2\cos^2 t)^\frac32=a^3|\cos t|^3$ and thus
$$V=2\cdot \frac{b}{a}\cdot \frac{-1}2\cdot \int_{t=0}^{\pi}\int_{m=a^2}^{a^2cos^2(t)} \sqrt{m}\;\; dm dt = \frac23\cdot \frac{b}{a}\cdot a^3\cdot \int_{t=0}^{\pi}1-\color{red}{|\cos(t)|^3} \; dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify the polynomial $xy(x+y)+(x+y)+(x+y)^2=13xy$ to the form $y^2=4x^3-g_{2}x-g_{3}$
I try to simplify a polynomial to the form: $y^2=4x^3-g_{2}x-g_{3}$, which is the elliptic curves. And the polynomial is $xy(x+y)+(x+y)+(x+y)^2=13xy$.
I try to let the $u=x+y$ and $v=x-y$, then I get the $u^3-uv^2+4u-9u^2+13v^2=0$. But how to get ahead?
| Are you certain that you have stated the problem correctly? As currently stated
$$ xy(x+y)+(x+y)+(x+y)^2=13xy $$
can be rewritten
\begin{eqnarray}
x^2y+xy^2+x+y+x^2+2xy+y^2&=&13xy\\
y^2(x+1)+y(x^2-11x+1)+x&=&0\\
y(x^2-11x+1)&=&-[y^2(x+1)+x]\\
y^2(x^2-11x+1)^2&=&[y^2(x+1)+x]^2
\end{eqnarray}
So if there exist constants $g_3,\,g_2$ such that
$$y^2=4x^3-g_{2}x-g_{3}$$
then
$$ (4x^3-g_{2}x-g_{3})(x^2-11x+1)^2=[(4x^3-g_{2}x-g_{3})(x+1)+x]^2 $$
This will resolve in an eighth degree polynomial equation of the form
$$ 16x^8-4x^7+P_6(x,g_2,g_3)=0 $$
Where $P_6$ is a degree six polynomial whose cofficients are function of $g_3,\,g_2$. But the coefficients of $x^8$ and $x^7$ are independent of the values of $g_3,\,g_2$.
So no constant values of $g_3,\,g_2$ can result in a solution for $x\ne0$.
Thus there cannot be values of $g_3,\,g_2$ for which $y^2=4x^3-g_{2}x-g_{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove that $\frac{x^2y}{1+x^4+y^2}$ has no global minimum I am not sure how to approach this problem. The usual methods do not work to find a minimum.
I can see that, but how to show that there must not exist a minimum?
| Since $(x^2-|y|)^2 = x^4+y^2 - 2 x^2|y| \ge 0$, we have $x^4+y^2 \ge 2 x^2 |y|$.
Hence ${x^2|y| \over 1+x^4+y^2} \le {x^4+y^2 \over 1+x^2 + y}<{1 \over 2}$.
If we let $x(t) = \sqrt{|t|}, y(t) = -t$, we have $\lim_{t \to\infty } {-t^2 \over 1+ 2 t^2} = -{1\over 2}$.
In particular, $\inf_{x,y} {x^2 y \over 1+x^4+y^2} = - {1 \over 2}$, but
the value of $-{1 \over 2}$ is never attained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How I show that $(b-a)^p \equiv b^p-a^p \mod{p}$?
If $b \ge a$ and p is prime, then $(b-a)^p \equiv b^p-a^p \mod{p}$.
I know this:
$b^p = (b - a+a)^p \equiv (b-a)^p +a^p \mod{p} $.
After all help in this comments I wrote this, I want to know if is correct my answer like this?
\begin{align}
b^p = (b -a +a)^p \equiv (b-a)^p +a^p \mod{p}
\end{align}
\begin{align}
b^p \equiv (b-a)^p +a^p \mod{p}
\end{align}
So, we have that:
\begin{align}
b^p\equiv a^p \mod{p} \leftrightarrow p | b^p -a^p
\end{align}
and,
\begin{align}
b-a^p \equiv b -a \equiv b^p -a^p\mod{p} \to\\
(b-a)^p\equiv b^p-a^p \mod{p} \to \\
p| (b-a)^p - (b^p-a^p) \to \\
\end{align}
$p|(b-a)^p \to p|(b-a)$ and $p|(b^p-a^p)$
So, we have that
\begin{align}
p | (b-a)^p - (b^p -a^p) \to (b-a)^p \equiv b^p -a^p.
\end{align}
|
I know this: $b^p = (b - a+a)^p \equiv (b-a)^p +a^p \mod{p}$
Well if you know that then just subtract $a^p$ from both sides:
$b^p -a^p\equiv (b-a)^p +a^p -a^p \equiv (b-a)^p\mod{p}$
......
But perhaps more simply:
By binomial expansion $(n + m)^k \equiv n^k + m^k \mod c$ so
$(b -a)^p \equiv b^p + (-a)^p\equiv b^p\pm a^p\mod p$. (With $\pm$ being $+$ if $p$ is even and $\pm$ being $-$ if $p$ is odd. And if $p = 2$ then $\pm$ could be either as $1 \equiv -1\mod 2$.)
As $p$ is prime, either $p$ is odd or $p$ is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Show that the substitution $x=X-1$ and $y=Y+3$ turns $\frac{dy}{dx}=...$ into a homogenous equation. $$\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$$
After substitution of the given variable I get $$\frac{dY}{dX} = \frac{4-\frac{Y}{X}}{2+\frac{Y}{X}}$$ which seems to give a homogenous equation. (There is no answer given for this part of the question)
But then the next question says find the particular solution with $x=0$ and $y=3$ giving your answer in the form $f(x,y)=c$. Then after this point I get an answer that is completely different to the answer given in the book.
Let $Y=vX$. Then $\frac{dY}{dX} = X\frac{dv}{dX}+v$
Replace, then $$X\frac{dv}{dX}+v=\frac{4-v}{2+v}$$
Making the Integrals
$$\int \frac{1}{\frac{4-v}{2+v}-v}dv=\int\frac{1}{X}dX$$
$$\frac{3\ln(1-v)}{5}-\frac{2\ln(4+v)}{5}-\ln(X)=c$$
Replacing everything back to $y$ and $x$.
$$\frac{3\ln(1-\frac{y-3}{x+1})}{5}-\frac{2\ln(4+\frac{y-3}{x+1})}{5}-\ln(x+1)=c$$
According to this $c=-\frac{2\ln(4)}{5}$
Even before the replacement stages the answer is already completely different to that of the answer given.
What did I do wrong?
Just for reference, answer given is $(x-y+4)^3(4x+y+1)^2 = 16$.
| I made a summary with the most important steps:
STEP $1$
$$\frac{dY}{dX}=\frac{4(X-1)-Y-3+7}{2(X-1)+Y+3-1}\ \rightarrow\ Y'=\frac{4X-Y}{2X+Y}\rightarrow\ Y'=\frac{4-\frac YX}{2+\frac YX}$$
$$Z=\frac YX\qquad \frac{dY}{dX}=Z+X\frac{dZ}{dX}$$
$$Z+XZ'=\frac{4-Z}{2+Z}\qquad\rightarrow\qquad XZ'=-\frac{Z^2+3Z-4}{Z+2}$$
STEP $2$
$$\ln|CX|=-\int\frac{Z+2}{(Z+4)(Z-1)}dZ$$
Solve the integral by Partial Fractions then
$$\ln|CX|=-\frac25\ln|Z+4|-\frac35\ln|Z-1|\\\ln|CX(Z+4)^{\frac25}(Z-1)^{\frac35}|=0\\CX(Z+4)^{\frac25}(Z-1)^{\frac35}=1\\(Y+4X)^2(Y-X)^3=C$$
STEP $3$
Substitute Y and X
GENERAL SOLUTION
$$(4x+y+1)^2(y-x-4)^3=C\\(4x+y+1)^2(x-y+4)^3=C$$
Check the answer
STEP 4
Particular Solution with $y(0)=3$
$$C=-16$$
$$(4x+y+1)^2(y-x-4)^3=-16\\(4x+y+1)^2(-(x-y+4))^3=-16$$
$$(4x+y+1)^2(-(x-y+4))^3=-16\qquad \rightarrow \qquad (4x+y+1)^2(x-y+4)^3=16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hint Taylor Series $x^{2}\ln(x)$ about $a=1$ Find the taylor series of $g(x):= x^{2}\ln(x)$ about $a=1$
Idea: I would have thought we should find the Taylor series of $\ln(x)$ about $a=1$, which is
$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k!}(x-1)^{k}$ and then I would just multiply that with $x^{2}$. So, we'd get:
$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k!}(x-1)^{k}x^2$
This cannot be simplified can it? Surely there must be a better without having to calculate the $n$-th derivative of $x^2\ln(x)$
| Your answer is no completely correct. As you would like the expansion around $a=1$, you should also expand $x^2$ around this point. In fact, you can easily show that
$$x^2 = 1 +2 (x-1) + (x-1)^2\;.$$
As a result, we have that
$$x^2 \ln x = \Bigl[1 +2 (x-1) + (x-1)^2 \Bigr] \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^k\;.$$
Or, after expanding
$$x^2 \ln x = \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^k + 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^{k+1} + \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^{k+2}\\
= \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^k + 2 \sum_{k=2}^\infty \frac{(-1)^{k}}{k-1} (x-1)^{k} + \sum_{k=3}^\infty \frac{(-1)^{k+1}}{k-2} (x-1)^{k}\\
= (x-1) + \frac32 (x-1)^2 +\sum_{k=3}^\infty (-1)^{k+1}\left[\frac{1}{k} - \frac{2}{k-1} + \frac{1}{k-2} \right] (x-1)^k.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Floor function of a real number The floor of a real number $x$, denoted by $⌊x⌋$, is the largest integer that is less than or equal to $x$. The ceiling of a real number x, denoted by $⌈x⌉$, is the smallest integer that is greater than or equal to $x$. Let $x$ and $y$ be any real numbers. Let $n, m$ and $k$ be any integers.
*
*$⌊x⌋ \le x < ⌊x⌋+1$, equivalently $x-1 < ⌊x⌋ \le x$.
*$⌈x⌉-1 < x \le ⌈x⌉$, equivalently $x \le ⌈x⌉ < x+1$.
*If $n \le x < n+1$ (or $x-1 < n \le x$), then $n=⌊x⌋$; similarly, if $n-1 < x \le n$ (or $x \le n < x+1$), then $n=⌈x⌉$.
*If $x \le y$, then $⌊x⌋ \le ⌊y⌋$ and $⌈x⌉ \le ⌈y⌉$. Similarly for $x \ge y$.
*If $n \le x$, then $n \le ⌊x⌋$. If $n \ge x$, then $n \ge ⌈x⌉$.
*If $x$ has an integer value, then $⌊x⌋=x=⌈x⌉$. If x has a non-integer value, then $⌈x⌉=⌊x⌋+1$.
Furthermore, we have the well-known inequality: $ \frac {x-1}{x} \le \ln x \le \frac {x^2 -1}{2x} \le x-1$
I am trying to find all pairs of positive real numbers $(x,y)$ such that $ \ln(⌊x+y⌋ )=⌊y⌋+⌊x⌋,$ but the above identities are hard to follow. Thank you
| Note that the right side is an integer so the left side must be, too. That means $\lfloor x+y\rfloor$ is an integer power of $e$ but it is also an integer, so it must be $1$ and the $\log$ is $0$. This gives
$$\lfloor x+y\rfloor=1\\ \lfloor x \rfloor + \lfloor y \rfloor=0$$
If either $x$ or $y$ is greater than or equal to $1$ the second will fail, so our final solution is $$0 \lt x \lt 1\\ 1-x \lt y \lt 1$$
or the part of the unit square above $x+y=1$ It is the region strictly inside the triangle below
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Exponential form of general integral of a linear system of ODE Consider the following linear system of differential equations:
$$
\begin{cases}
\dot{x}=-4y \\
\dot{y}=x
\end{cases}
$$
where $x(t)$ and $y(t)$ are unknown real functions.
One can simply verify that the general solution is
$$
\begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix}
= c_1
\begin{pmatrix}
\cos(2t) \\
\frac{1}{2}\sin(2t)
\end{pmatrix}
+c_2
\begin{pmatrix}
-2\sin(2t) \\
\cos(2t)
\end{pmatrix}
$$
where $c_1$ and $c_2$ are real parameters.
Question: Which is the exponential form of this expression?
It should by something like
$$
\begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix}
= k_1 e^{i2t}
\begin{pmatrix}
P_{11}\\
P_{21}
\end{pmatrix}
+k_2 e^{-i2t}
\begin{pmatrix}
P_{12} \\
P_{22}
\end{pmatrix}
$$
where $k_1$ and $k_2$ should (?) be complex parameters.
| I think the key here is to rewrite the linear system as the following matrix equation
$$\underbrace{\begin{pmatrix}
0& -4\\
1 & 0
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
\dot{x}\\
\dot{y}
\end{pmatrix}}_{A\vec{x}=\dot{\vec{x}}}.$$
The characteristic equation is $\lambda^2+4=0,$ so you're correct in using $\cos(2t)$ and $\sin(2t).$ Now, to get at a complex exponential solution, you need to find eigenvectors of $A$ with eigenvalues $-2i,2i.$ These are $$\begin{pmatrix}
2i\\
1
\end{pmatrix}\quad \text{and}\quad \begin{pmatrix}
-2i\\
1
\end{pmatrix}$$
respectively.
Consequently, you can write
$$\begin{pmatrix}
x(t)\\
y(t)
\end{pmatrix}=c
_1\begin{pmatrix}
2i\\
1
\end{pmatrix}e^{2it}+c_2\begin{pmatrix}
-2i\\
1
\end{pmatrix}e^{-2it}.$$
It is worth noticing that the eigenvectors, are attached to their associated eigenfunctions, and that the eigenvalue of the derivative operator is identical to that of the eigenvalue under the action of $A.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding the prime factorization of $2^{22}+1$? The problem is
Find the prime factorization of $2^{22}+1$
I have a solution but I think there must be some better ways:
My Solution:
$2^{22}+1 = (2^{22} + 2 \cdot 2^{11} +1) - 2 \cdot 2^{11} = (2^{11}+1)^2 - 2^{12}$
and we can factor as
$(2^{11}+1-2^6)(2^11+1+2^6) = (2048+1-64)(2048+1+64)= 1985 \cdot 2113$
From here it's not infeasible by hand, but I think there must be a better method.
| Suppose $p$ is a prime factor of $2^{22}+1$ greater than $5$. Then $2^{22}\equiv -1 \pmod{p}$. So we know the order of $2$ modulo $p$ is $44$, and thus $44 \mid p-1$. We conclude that $p = 44n+1$ for some $n$.
Using your factorization, we have that $\sqrt{2113} = 45.96\ldots$. There are no primes of the form $44n+1$ in the range between $5$ and $45$, so $2113$ is prime. To factor $1985$, the $5$ is easy, and we're left with $397$. Again, there are no primes of the form $44n+1$ in the range $5$ to $\sqrt{397}$, so it's prime too.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Rationalize this I'm having trouble rationalizing the denominator of this fraction. Would you kindly explain this to a fellow self-learning math student?
$$\frac{10}{\sqrt[4]{3}-1}$$ knowing that $a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}\cdot b+a^{n-3}\cdot b^2+...+a\cdot b^{n-2}+b^{n-1}\right)$
$a,b\in \mathbb{R}$ and $n\in \mathbb{N}$
| Couldn't see how the tip helps.
$$
\frac{10}{\sqrt[4]{3}-1} = \frac{10}{\sqrt[4]{3}-1}\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1} = \frac{10(\sqrt[4]{3}+1)}{\sqrt{3}-1}=\frac{10(\sqrt[4]{3}+1)}{\sqrt{3}-1}\frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{10(\sqrt[4]{3}+1)(\sqrt{3}+1)}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding limit of multivariable function. I have a question that might be silly but I need to understand these kind of problems. So i have this limit:$$\lim_{(x,y)\to(0,0)} \frac{x^3 + y^3}{x^2 + y^2}$$
To solve it I am going to use poolar coordinates, so the limit would be like this:
$$\lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2}=$$
$$\lim_{r\to0} {r(\cos^3\theta + \sin^3\theta)}$$
Now it is clear to me that r tends to $r\to0$ but, can I actually say something about $\cos^3\theta + \sin^3\theta?$
In case I could, then I would say that it is bounded, thus the limit would be 0. But it is not clear to me, could anyone help me with this?
| It works because
$$\left|\cos^3\theta + \sin^3\theta\right| \le \left|\cos\theta\right|^3 + \left|\sin\theta\right|^3 \le 2$$
Also without polar coordinates:
\begin{align}
\left|\frac{x^3+y^3}{x^2+y^2}\right| &= \frac{|x+y||x^2-xy+y^2|}{x^2+y^2} \\
&\le|x+y| \cdot \frac{x^2+y^2+|xy|}{x^2+y^2} \\
&\le \sqrt{2}(x^2+y^2) \cdot \frac{\frac32(x^2+y^2)}{x^2+y^2}\\
&= \frac{3\sqrt{2}}2\|(x,y)\|^2 \xrightarrow{(x,y) \to (0,0)} 0
\end{align}
| {
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Cube root of numbers such as $2+11i$ How can I find powers and roots of complex numbers with ugly argument such as cube root of 2+11i? I saw Find the solutions to $z^3 = 2 + 11i$., but the answer includes some guessing and I would like to have some algorithm for such a tasks, which I can program. I would also like to solve it symbolically and thus precisely. Of course I know the formula: $a^b=e^{b \ln(a)}$ But when I try to use it:
$\sqrt[3]{2+11i}=(2+11i)^\frac{1}{3}=e^{\frac{1}{3} \ln(2+11i)}=e^{\frac{1}{3} \big(\ln(\sqrt{125})+i \arctan(\frac{11}{2}) \big)}=$
$=\sqrt{5} \big(\cos(\frac{1}{3} \arctan(\frac{11}{2}))+i \sin (\frac{1}{3} \arctan(\frac{11}{2})) \big)$
And I cannot solve this further symbolically, of course when I solve it in floating point numbers I get the right solution $2+i$.
| Note that\begin{align}(a+bi)^3=2+11i&\iff\left\{\begin{array}{l}a^3-3ab^2=2\\3a^2b-b^3=11.\end{array}\right.\\&\iff\left\{\begin{array}{l}a(a^2-3b^2)=2\\b(3a^2-b^2)=11\end{array}\right.\end{align}Perhaps that there are no integer solutions, but if there are, then you can use the fact that both $2$ and $11$ are prime numbers. The only way of expressing $2$ as a product of two integers is when one of the factors is $\pm2$ and the other one is $\pm1$. So, can we take $a=2$ and $a^2-3b^2=1$. Yes, we can! Just take $b=\pm1$. And it turns out that $a=2$ and $b=1$ are also a solution of the second equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the rank of the Matrix $M$ for the different values of $\alpha$ Let's have $\
M=
\left[ {\begin{array}{ccccc}
1 & \alpha & -1 & 2 \\
2 & -1 & \alpha & 5 \\
1 & 10 & -6 & 1 \\
\end{array} } \right]$.
Using Gaussian elimination I've come to have the next matrix
$\
M'=
\left[ {\begin{array}{ccccc}
1 & \alpha & -1 & 2 \\
0 & 1 & \frac{-5}{10- \alpha} & \frac{-1}{10 - \alpha} \\
0 & 0 & \frac{-5(1+2\alpha)}{10- \alpha}+\alpha +2 & 1 \\
\end{array} } \right]$
So:
*
*$\alpha \neq 10$
*$\frac{-5(1+2\alpha)}{10- \alpha}+\alpha +2 = 1$ and then i calculate the values of $\alpha$
Is it ok what I did? Is the rank always going to be 3 no matter the value of $\alpha \neq 10$ ? What if $\alpha = 10$ ?
Since it is not a square matrix this is the only way to find out the rank, or is there another way?
| A good strategy for these problems is to do as many operations as you can without dividing by an expression involving a variable.
In particular, I find:
$$
\pmatrix{
1 & \alpha & -1 & 2 \\
2 & -1 & \alpha & 5 \\
1 & 10 & -6 & 1 \\
} \to
\pmatrix{
1 & \alpha & -1 & 2 \\
0 & -1 - 2\alpha & 2+\alpha & 1 \\
0 & 10-\alpha & -5 & -1 \\
}
$$
The rank of this matrix is always at least $2$, since the first and last rows are linearly independent. We can only have rank exactly two if the matrix
$$
\pmatrix{-1-2\alpha & 2+\alpha & 1\\
10-\alpha & -5 & -1}
$$
has rank $1$. That is, we can only achieve rank $2$ if the first and second columns are multiples of the third. In other words, we must have
$$
-(-1-2\alpha) = 10 - \alpha \implies \alpha = 3\\
(-(-5)) = 2 + \alpha \implies \alpha = 3
$$
So indeed, when $\alpha = 3$, the matrix has rank $2$. In all other cases, the matrix has rank $3$.
A more computational approach is to note that permuting the columns (or more generally "applying column operations") doesn't change the rank. So, we can have
$$
\pmatrix{
1 & \alpha & -1 & 2 \\
2 & -1 & \alpha & 5 \\
1 & 10 & -6 & 1 \\
} \to
\pmatrix{
1 & \alpha & -1 & 2 \\
0 & -1 - 2\alpha & 2+\alpha & 1 \\
0 & 10-\alpha & -5 & -1 \\
}
\\ \to
\pmatrix{
1 &2 & \alpha & -1 \\
0 &1 & -1 - 2\alpha & 2+\alpha \\
0 & -1 & 10-\alpha & -5 \\
}
\\ \to
\pmatrix{
1 &2 & \alpha & -1 \\
0 &1 & -1 - 2\alpha & 2+\alpha \\
0 & 0 & 9-3\alpha & -3 + \alpha \\
}
$$
It is clear that this matrix will have full rank whenever $\alpha \neq 3$, and rank $2$ when $\alpha = 3$.
| {
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"url": "https://math.stackexchange.com/questions/2837182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with Complex Variables
Solve the equation $z^2 + z + 1 = 0$ for $z = (x,y)$ by writing $(x,y) (x,y) + (x,y) + (1,0) = (0,0)$ and then solving a pair of simultaneous equations in $x$ and $y$.
My main difficulty is finding the answer is using the suggestion in that book that there isn't a real solution implies that $y \neq 0$, but I will work out the rest of this solution just to be sure that I grasp the logic.
First, using the fact that we have some complex variable $z = (x,y)$ and that we can represent a purely real number as $x = (x,0)$, meaning that $1 = (1,0)$, we can reason, as the problem suggests, that
\begin{equation}
z^2 + z + 1 = 0 \implies (x,y)(x,y) + (x,y) + (1,0) = (0,0)
\end{equation}
Using multiplication and addition of complex variables, we get
\begin{equation}
\left(x^2 - y^2, 2xy \right) + \left(x + 1, y\right) = (0,0) \implies \left(x^2 - y^2 + x + 1, 2xy + y\right) = (0,0)
\end{equation}
which then implies that
$x^2 - y^2 + x + 1 = 0$ and $2xy + y = 0$. From this second equation, we get $2xy + y = y(2x + 1) = 0$, which implies $y = 0$ or $2x + 1 = 0 \implies x = - \frac{1}{2}$. Plugging $y = 0$ into the first equation gives us $x^2 - 0^2 + x + 1 = x^2 + x + 1 = 0$, which doesn't have any real solutions.
Now, the suggestion states that the lack of a real solution solution, $x$, when $y = 0$ is justification for throwing out this value of $y$. If I'm not mistaken, and this may well be completely obvious, the logic for this is that we're trying to solve for some value $z = (x,y)$, so we obviously require a real solution even if our result is purely imaginary, though no such $x$ exists. I still I think am not completely convinced of this, as I'm not totally sure why we couldn't reason that we have a purely imaginary solution, i.e., $x = 0$.
But, I think my confusion is in the above step. After this, the algebra follows and gives me the same answer as the textbook, $z = \left(-\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)$.
| Note that, for a complex $z = (x,y)$, $x$ and $y$ are both reals. The suggestion is not to require $z$ to be real. We need to find $z$, which is a complex number, and, being a complex number, it has both its real and imaginary components real numbers. Lack of a real solution of $x^2+x+1=0$ suggests that $y \neq 0$.
An alternative way to solve $z^2 + z +1=0$ is to use the famous completing square method. That is,
$$z^2+z+1=\left(z+\frac{1}{2}\right)^2+\frac{3}{4}=0 \implies z+\frac{1}{2}=\pm \frac{\sqrt{3}}{2} i.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to derive this factorization $ax^2 + bx + c = a(x − x_1)(x − x_2) $? I faced this factorization formula $$ax^2 + bx + c = a(x − x_1)(x − x_2)$$ where $x_1$ is the first solution and $x_2$ is the second one. But I don't understand how the formula gets derived... Could anybody explain it to me? Thanks!
I know I can solve a quadratic equation and substitute the solutions into this formula $a(x − x_1)(x − x_2)$, but I want to understand how it works.
| You can also complete the square. Assuming $a \neq 0$ (necessary to be a quadratic):
\begin{align*}
ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) \\
&= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{c}{a} - \frac{b^2}{4a^2}\right) \\
&= a\left(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right).
\end{align*}
Then, using a difference of two squares factorisation,
$$ax^2 + bx + c = a\left(x + \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\right)\left(x + \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\right),$$
as requested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Finding Lyapunov function for a stable equilibruim of a non linear system Given the following system:
$$
\left\{
\begin{array}{c}
\dot x=y-x^2-x \\
\dot y=3x-x^2-y
\end{array}
\right.
$$
I need to find the equilibrium points, and if stable, to find a Lyapunov function.
I have found two equilibrium points: $(0,0), (1,2)$. By linearization I'v found $(0,0)$ to be unstable, and $(1,2)$ to be stable. I tried to find a Lyapunov function for $(1,2)$ with no success. How can I find such a function?
| Considering the surroundings for $(1,2)$ we have
$$
\left\{
\begin{array}{c}
\dot x=y-x^2-x \\
\dot y=3x-x^2-y
\end{array}
\right. \approx \left\{
\begin{array}{c}
\dot x=y-3x+1 \\
\dot y=x-y+1
\end{array}
\right.
$$
because linearizing about $(1,2)$ we have
$$
\left(\begin{array}{c}
y-x^2-x \\
3x-x^2-y
\end{array}
\right) = \left(\begin{array}{c}
0 \\
0
\end{array}
\right)+\left(
\begin{array}{cc}
-3 & 1 \\
1 & -1 \\
\end{array}
\right)\left(\begin{array}{c}
x-1 \\
y-2
\end{array}
\right)+O(x,y)
$$
now considering a local Lyapunov function such as
$$
V = \alpha (x-1)^2+\beta(y-2)^2
$$
we have
$$
\dot V =-2 (\alpha (x-1) (3 x-y-1)-\beta (y-2) (x-y+1))
$$
or
$$
\dot V = (x-1,y-2)^{\dagger}\cdot H \cdot(x-1,y-2)
$$
with
$$
H = \left(
\begin{array}{cc}
-6 \alpha & \alpha +\beta \\
\alpha +\beta & -2 \beta \\
\end{array}
\right)
$$
with eigenvalues
$$
\lambda = \pm\sqrt{2} \sqrt{5 \alpha ^2-2 \alpha \beta +\beta ^2}-3 \alpha -\beta
$$
now it is easy to show that there exists $\alpha > 0, \beta > 0$ such that
$$
\lambda < 0
$$
hence this affirm the existence of a local Lyapunov function to assure the stability of $(1,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$
$x^2 -xy + y^2 \equiv 0 \pmod{43}$
We could transform this into
$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$
and thus
$x^3 + y^3 \equiv 0 \pmod{43}$
From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $43$ itself, $84$ and $127$.
How should I proceed?
| If we were working in $\mathbb{Q}$ we could transform your equation into:
$$x^2 - xy + y^2 = 0 \qquad \iff \qquad \left(x- \frac{1}{2}y\right)^2 + \frac{3}{4}y^2 = 0.$$
Fortunately, both $2$ and $4$ are invertible modulo $43$:
$$(x - (2^{-1})y)^2 + 3(4)^{-1}y^2 \equiv 0 \pmod{43} \quad \iff \quad (x+21y)^2 + 33y^2 \equiv 0 \pmod{43}.$$
Let $z = x + 21y$. We have the equation $z^2 + 33y^2 \equiv 0 \pmod {43}$, or better yet,
$$z^2 \equiv 10y^2 \pmod{43}.$$
Now $10$ is a quadratic residue modulo $43$, since $15^2 \equiv 10 \pmod{43}$, so
$$z^2 \equiv (15y)^2 \pmod{43}$$
which means
$$z \equiv \pm 15y\pmod{43} \qquad \iff \qquad x + 21y \equiv \pm 15y \pmod{43}.$$
This gives us two diferent families of solutions:
$$x \equiv 37y \pmod{43} \qquad \text{and} \qquad x \equiv 7y\pmod{43}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the parity-check matrix of a generator matrix in $\Bbb F_3$. I am currently looking into linear codes and thought I had understood building a parity-check matrix for a given generator matrix - but the presented solution is anything but what I have.
Given a generator matrix $G= \begin{pmatrix}I_3 \\ A\end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \\1 & 0 & 1\\0&1&1\end{pmatrix}$
for a linear code in $\mathbb{F_3}$,the check matrix is $$H = (-A^{T(?)} | I_{n-m}) = (-A | I_{6-3}) =\begin{pmatrix}2&2&0&1&0&0 \\2&0&2&0&1&0\\0&2&2&0&0&1\end{pmatrix}.$$
I calculated the check matrix and only got this:
$$H_{wrong?} = \begin{pmatrix}1&1&0&1&0&0 \\1&0&1&0&1&0\\0&1&1&0&0&1\end{pmatrix}.
$$
The right-hand $3\times 3$ square is obviously the identity matrix, the left-hand $3\times 3$ square is my transposed matrix $A$ given from $G$.
Do I have to do further calculations on $A^T$ because of the linear code being in $\mathbb{F_3}$? Is the actual transposed matrix where I transposed 1's have to be -1's for the check matrix, to which I then add the 3? I am completely lost and just spitballing.
| $$A=\begin{pmatrix} 1 & 1 & 0 \\1 & 0 & 1\\0&1&1\end{pmatrix}\Rightarrow A^T=\begin{pmatrix} 1 & 1 & 0 \\1 & 0 & 1\\0&1&1\end{pmatrix}\Rightarrow-A^T=\begin{pmatrix} -1 & -1 & 0 \\-1 & 0 & -1\\0&-1&-1\end{pmatrix}.$$ Then, since $-1\equiv 2\bmod 3$; we conclude $$-A^T=\begin{pmatrix} 2 & 2 & 0 \\2 & 0 & 2\\0&2&2\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x} \, dx$ Greetings I am trying to find a closed form for: $$I=\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x}\,dx$$ If we rewrite the integral as $$I=\int_0^\infty x^2 \sqrt{\frac{1}{\sqrt{1+\cot^2 x}}}\,dx$$ now with $$\cot x =t $$ $$I=\int_0^{\infty} \operatorname{arccot}^2 (x)(1+x^2)^{-\frac{5}{4}}dx$$ and with https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms $$I=\frac{1}{4i}\int_0^{\infty}\log^2\left(\frac{z-i}{z+i}\right)(1+x^2)^{-\frac{5}{4}} \, dx$$ Now for the $\log$ I thought to expand into power series but since the radius of converge is abit smaller, this fails. Also integrating by parts or combining the initial integral with $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\cos x}\,dx$ wasn't much of a help, could you help me evaluate this integral ?
| The substitution $\sin(x) = \sqrt{t}$ leads to the expression
$$I = \frac{1}{2} \int \limits_0^1 \frac{t^{-1/4} \arcsin^2 (\sqrt{t})}{\sqrt{1-t}} \, \mathrm{d} t \, . $$
Now you can use the power series for $\arcsin^2$ (see for example this question) and integrate term by term (monotone convergence). Using the beta function you will find
\begin{align}
I &= \frac{1}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \int \limits_0^1 t^{n-\frac{1}{4}} (1-t)^{-\frac{1}{2}} \, \mathrm{d} t \\
&= \frac{1}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \operatorname{B}\left(n+\frac{3}{4},\frac{1}{2}\right) \\
&= \frac{\sqrt{\pi}}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \frac{\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)} \\
&= \frac{\sqrt{\pi} \, \Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \frac{\prod_{k=1}^n (4k-1)}{\prod_{l=1}^{n+1} (4l-3)} \\
&= \frac{\pi \sqrt{2 \pi}}{\Gamma\left(\frac{1}{4}\right)^2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (4n+1) (2n-1)!!} \prod \limits_{k=1}^n \frac{4k-1}{4k-3} \, .
\end{align}
Mathematica gives the following expression in terms of a hypergeometric function:
$$ I = \frac{6 \pi \sqrt{2 \pi}}{5 \Gamma\left(\frac{1}{4}\right)^2} \, {}_4 \! \operatorname{F}_3 \left(1,1,1,\frac{7}{4};\frac{3}{2},2,\frac{9}{4};1\right) \approx 1.208656578687 \, .$$
Inverse symbolic calculators do not seem to give any expression for this number, so this might be as good as it gets.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
My thought on this are as follows:
I've solved a similar problem where I had to find the inverse of:
$$
g(x) = \sqrt[3]{x+\sqrt{x^2-1}} + \sqrt[3]{x-\sqrt{x^2-1}}
$$
So let $a = \sqrt[3]{x+\sqrt{x^2-1}}$ and $b = \sqrt[3]{x-\sqrt{x^2-1}}$, then
$y = a + b$ and hence:
$$
y^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = \\
= 2x+3ab(a+b)=2x+3(a+b)
$$
But $y = a + b$ and then
$$
y^3 = 2x + 3y \iff x = \frac{y^3-3y}{2}
$$
This function is very similar to the one in the title.
The answer says the inverse $f^{-1}(x) = \sqrt[3]{x+\sqrt{x^2+1}} + \sqrt[3]{x-\sqrt{x^2+1}}, \; x\in \mathbb R$. So knowing the answer it's easy to show that it indeed appears to be the inverse of $f(x)$ using the approach above. But how could I achieve the same result given the inverse function is unknown?
| You want to solve $x^3 + 3x - 2y = 0$ as $x = f(y)$. The canonical way to do this is to write $x = a + b$, obtaining
$$a^3 + 3a^2b + 3ab^2 + b^3 + 3(a + b) - 2y = 0$$
The trick is specify that $3ab = -3$. The above equation then reduces to
$$a^3 - b^3 - 2y = 0$$
Which is the same as
$$a^3 - 1/a^3 - 2y = 0$$
Multiplying by $a^3$ gives
$$a^6 - 2ya^3 - 1 = 0$$
You can now use the quadratic formula to find an expression for $a^3$ in terms of $y$, and then you get three possible complex cube roots for $a$. You then can find $b$ through $3ab = -3$ and then you have three possible $x = a + b$. These are the solutions to your original equation. The simplest of these three will be given by the formula you have there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate
$\lim_{n \rightarrow \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know that $1 - \frac {2}{k(k+1)} = \frac{(k+2)(k-1)}{k(k+1)}$
After that I'm not able to proceed further.
Any hints/solution will be apprecaited.
Thanks you and thanks in advance.
| We have that
$$\left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)=$$
$$=\frac{1\cdot \color{green}4}{\color{red}2\cdot 3}\,\frac{\color{red}2\cdot \color{green}5}{\color{red}3\cdot \color{green}4}\,\frac{\color{red}3\cdot \color{green}6}{\color{red}4\cdot \color{green}5}\ldots\frac{\color{red}{(n-1)}\cdot \color{green}{(n+2)}}{\color{red}n\cdot \color{green}{(n+1)}}\,\frac{\color{red}n\cdot (n+3)}{(n+1)\cdot \color{green}{(n+2)}}=\frac{n+3}{3(n+1)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$ Evaluate:
$$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$$
We let
$$\alpha = \operatorname{arccos} \frac{2}{\sqrt5} \qquad \beta = \operatorname{arccos}\frac{3}{\sqrt{10}}$$
Then we have:
$$\begin{align}
\cos(\alpha) = \frac{2}{\sqrt5} &\qquad \cos(\beta) = \frac{3}{\sqrt{10}} \\[4pt]
\sin(\alpha) = \frac{1}{\sqrt5} &\qquad \sin(\beta) = \frac{1}{\sqrt{10}}
\end{align}$$
In order to evaluate, we are told, we first determine $\sin(\alpha + \beta)$; we wind up with $1/\sqrt2$, thus we have $\pi/4$.
What I am confused about is why we have to use sin($\alpha + \beta$). For example, if I were to use $\cos(\alpha + \beta)$, I would get the answer $7/(\sqrt{10}\sqrt5)$, which I do not know what to do with. I am having trouble finding out whether there is some kind of pattern to this kind of thing, or did the author just know to use $\sin(\alpha + \beta)$ since he/she checked cos and saw nothing comes out of this?
Any help is much appreciated, thank you
| You can directly use the formula $$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$
$$x=\frac{2}{\sqrt{5}},y=\frac{3}{\sqrt{10}}$$
$$=\cos^{-1}\left(\left(\frac{2}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)-\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}\right)$$
$$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{1-\frac45}\sqrt{1-\frac{9}{10}}\right)$$
$$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{\frac15}\sqrt{\frac{1}{10}}\right)$$
$$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{\sqrt{50}}\right)$$
$$=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$
$$\cos^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}\frac{3}{\sqrt{10}}=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
integrate $2\sqrt{1-x^2/4-y^2/9}$ I am trying to find the volume of the ellipsoid $E$ given by $x^2/4 + y^2/9 + z^2 \leq 1$ by computing $\iiint_E dV$
I ended up with $$\int_{-2}^{2}\int_{-3\sqrt{1-x^2/4}}^{3\sqrt{1-x^2/4}}\int_{-\sqrt{1-x^2/4-y^2/9}}^{\sqrt{1-x^2/4-y^2/9}}1dzdydx$$
After doing the inner integral I get $$\int_{-2}^{2}\int_{-3\sqrt{1-x^2/4}}^{3\sqrt{1-x^2/4}}2\sqrt{1-x^2/4-y^2/9}\space dydx$$
I am not sure about how to figure out the next one
The hint is to use trig substitution, though I am not sure if it is talking about this step or the next one
| $$\int_{-2}^{2}\int_{-3\sqrt{1-x^2/4}}^{3\sqrt{1-x^2/4}}2\sqrt{1-x^2/4-y^2/9}\space dydx$$ is the same as $$\int_{-2}^{2}\int_{-\sqrt{9-9x^2/4}}^{\sqrt{9-9x^2/4}}2\sqrt{1-x^2/4-y^2/9}\space dydx$$ then do trig sub so $\frac{y^2}{9}=1-\frac{x^2}{4}\sin^2{\theta}$
then $y=\sqrt{9-\frac{9x^2}{4}}\sin{\theta}$, $dy=\sqrt{9-\frac{9x^2}{4}}\cos{\theta}d\theta$
then you get $$\int_{-2}^{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}6(1-\frac{x^2}{4})\cos^2{\theta}d\theta dx=\int_{-2}^{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}3(1-\frac{x^2}{4})(1+\cos{2\theta})d\theta dx$$
then you can keep going to end up with $8\pi$
(this was the given solution)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Formulas for solving problems such as “how many positive factors of x are also multiples of y?” I recently came upon the question that follows.
“How many positive factors of 60 are also multiples of 4?”
Since 60 is relatively small I used a brute force method.
But I was wondering if there was a way you could do this using the prime factorization of 60.
Thanks!
| Yes, indeed you can. Suppose that $a\mid b$ and you ask how many factors of $b$ are also multiples of $a$. Suppose $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ and that $b=p_1^{b_1}p_2^{b_2}\cdots p_k^{b_k}p_{k+1}^{b_{k+1}}\cdots p_n^{b_n}$ (where the primes are potentially written out of order to make it so all primes appearing in $a$ occur before those which don't appear in $a$).
There will be $(b_1-a_1+1)(b_2-a_2+1)\cdots(b_k-a_k+1)(b_{k+1}+1)\cdots (b_n+1)$ number of such factors of $b$ which are also multiples of $a$.
This can be seen directly by multiplication principle.
For the specific case of $4$ and $60$, note $4=2^2$ and $60=2^2\cdot 3^1\cdot 5^1$ so there are $(2-2+1)(1+1)(1+1)=1\cdot 2\cdot 2 = 4$ such factors of $60$ which are also multiples of $4$. Namely $4,12,20,60$, or rewritten: $2^2\cdot 3^0\cdot 5^0, 2^2\cdot 3^1\cdot 5^0, 2^2\cdot 3^0\cdot 5^1, 2^2\cdot 3^1\cdot 5^1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Implement genarating function for a combinatorial question I've tried to solve the following question by using generating functions.I know It's pretty simple to do it with Inclusion–exclusion principle but I insist to solve it with other technique as well.
Q: how many combination to spread 10 balls into 5 boxes when in each box there is at least one ball.
A:
$
(x+x^2+x^3+x^4...)^5 = x^5(1+x+x^2+x^3...)^5 = x^5 \frac{1}{(1-x)^5} = x^5 \sum_{i=0}^\infty \binom{n+5-1}{5-1}x^n$
I want the coefficient of $x^{10}$ so $n=5$ which is $\binom{9}{4} = 126$
sadly it isn't the same answer like the Inclusion–exclusion principle, i guess my series isn't right but I can't see why.
| This is small enough to check by hand which of your results (if either) is correct.
$6$: $5$ arrangements
$5$, $2$: $5\cdot4=20$ arrangements
$4$, $3$: $5\cdot4=20$ arrangements
$4$, $2$, $2$: $5\cdot\binom42=30$ arrangements
$3$, $3$, $2$: $5\cdot\binom42=30$ arrangements
$3$, $2$, $2$, $2$: $5\cdot4=20$ arrangements
$2$, $2$, $2$, $2$, $2$: $1$ arrangement
The total is $126$, so the error seems to be in your application of the inclusion–exclusion principle.
| {
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"url": "https://math.stackexchange.com/questions/2849881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Range of $a$ in Trigonometric equation
If the inequality $\sin^2 x+a\cos x+a^2>1+\cos x$ hold for any $x\in \mathbb{R}.$ Then range of $a$ is
Try: $1-\cos^2 x+a\cos x+a^2-1-\cos x>0$
$$\cos^2 x+(1-a)\cos x-a^2<0$$
$$4\cos^2 x+4(1-a)\cos x-4a^2<0$$
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
Could some help me how to solve further, Thanks
| Alternatively, write it as:
$$\sin^2 x+a\cos x+a^2>1+\cos x \iff \\
a^2+\cos x\cdot a-(\cos ^2x+\cos x)>0 \iff \\
\left(a+\frac12\cos x\right)^2-(\cos ^2x +\cos x+\frac 14\cos ^2x)>0$$
Note that the graph of $f(a)=\left(a+\frac12\cos x\right)^2-(\cos ^2x +\cos x+\frac 14\cos ^2x)$ is an opening-up parabola with the vertex:
$$V\left(-\frac12\cos x,-\cos ^2x-\cos x-\frac14\cos^2x\right)$$
The extreme $y$-coordinates of the vertex are:
$$f'(x)=\left(-\cos^2x -\cos x-\frac14\cos^2x\right)'=\frac52\cos x\sin x+\sin x=0 \Rightarrow \\
1) \ \sin x\left(\frac52\cos x+1\right)=0 \Rightarrow
\sin x=0 \ \text{(which implies $\cos x=\pm 1)$}\ \text{or} \\
2) \ \cos x=-\frac25 \ \text{(which imples $\sin x=\pm \frac{\sqrt{21}}{5}$)}.$$
Hence, there are three critical points:
$$\cos x=-1 \ \left(\text{$V\left(\frac12,-\frac14\right)$ is local min}\right);\\
\cos x=-\frac25 \ \left(\text{$V\left(\frac15,\frac15\right)$ is local max}\right);\\
\cos x=1 \ \left(\text{$V\left(-\frac12,-\frac94\right)$ is local min}\right)$$
Note: 1) SOC are left as an exercise. 2) See the Desmos graph.
Hence, it is sufficient to consider the extreme cases:
$$\begin{align}f(a)&=\left(a-\frac12\right)^2-\frac14>0 \Rightarrow a\in (-\infty,0)\cup (1,+\infty)\\
f(a)&=\left(a-\frac15\right)^2+\frac15>0 \Rightarrow a\in (-\infty,+\infty)\\
f(a)&=\left(a+\frac12\right)^2-\frac94>0 \Rightarrow a\in (-\infty,-2)\cup (1,+\infty)\end{align}$$
which can be observed from the Desmos graph too.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$ is strictly increasing I want to show that the function $f:\mathbb N \to \mathbb Q$ defined by$$f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$$
is strictly increasing. Could I have a hint on how to go about this? I have tried playing around with $f(n+1)-f(n)$ but it's all so messy with the different denominators that I'm not sure what to do.
| I will supplement @Markus Scheuer's answer by providing a proof of the claim (Corollary 3 of Monotonicity of Riemann Sums by D. Borwein, J. M. Borwein and B. Sims) in a slightly generalized setting.
Lemma. Let $f : [0, 1] \to \mathbb{R}$ and $\lambda \in \mathbb{R}$. Suppose that
$$F(x) = \lambda f(x) + (1-\lambda) f(1-x) $$
is concave. If we write $\tau_n = \sum_{k=1}^{n} f\left(\frac{k}{n}\right)\frac{1}{n}$, then we have the following inquality
$$ \tau_{n+1} \geq \frac{1}{(n+1)^2}f(0) + \frac{n(n+2)}{(n+1)^2} \tau_n. \tag{*} $$
As a corollary, suppose that $f$ and $\lambda$ satisfy the assumption of Lemma. Then
*
*If $f(0)$ is the global maximum of $f$, then $\tau_{n+1} \geq \tau_n$ since $f(0) \geq \tau_n$.
*If $0$ is the unique maximum point of $f$, then we have strict inequality $\tau_{n+1} > \tau_n$.
Proof of Lemma. We may write
$$ \tau_{n+1} = \frac{f(1)}{n+1} + \sum_{k=1}^{n} F\left(\frac{k}{n+1}\right)\frac{1}{n+1}. $$
Now by using the identity $ \frac{k}{n+1} = \frac{k}{n+1}\left(\frac{k-1}{n}\right) + \frac{n+1-k}{n+1} \left( \frac{k}{n} \right) $ together with the concavity of $F$,
$$ F\left(\frac{k}{n+1}\right)
\geq \frac{k}{n+1} F\left(\frac{k-1}{n}\right) + \frac{n+1-k}{n+1} F\left( \frac{k}{n} \right). $$
Plugging this back, we find that
\begin{align*}
\tau_{n+1}
&\geq \frac{f(1)}{n+1} + \sum_{k=1}^{n} \left[ \frac{k}{(n+1)^2} F\left(\frac{k-1}{n}\right) + \frac{n+1-k}{(n+1)^2} F\left( \frac{k}{n} \right) \right] \\
\\
%
%
&= \frac{f(1)}{n+1} + \color{red}{\lambda \sum_{k=1}^{n} \frac{k}{(n+1)^2} f\left(\frac{k-1}{n}\right)} + \color{blue}{(1-\lambda) \sum_{k=1}^{n} \frac{k}{(n+1)^2} f\left(\frac{n-k+1}{n}\right)} \\
&\hspace{4.1em} + \color{blue}{\lambda \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left( \frac{k}{n} \right)} + \color{red}{(1-\lambda) \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left( \frac{n-k}{n} \right)} \\
\\
%
%
&= \frac{f(1)}{n+1} + \color{red}{\lambda \sum_{k=0}^{n-1} \frac{k+1}{(n+1)^2} f\left(\frac{k}{n}\right)} + \color{blue}{(1-\lambda) \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left(\frac{k}{n}\right)} \\
&\hspace{4.1em} + \color{blue}{\lambda \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left( \frac{k}{n} \right)} + \color{red}{(1-\lambda) \sum_{k=0}^{n-1} \frac{k+1}{(n+1)^2} f\left( \frac{k}{n} \right)} \\
\\
%
%
&= \frac{f(0)}{(n+1)^2} + \color{red}{\sum_{k=0}^{n-1} \frac{k+1}{(n+1)^2} f\left(\frac{k}{n}\right)} + \color{blue}{\sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left(\frac{k}{n}\right)} \\
\\
%
%
&= \frac{1}{(n+1)^2}f(0) + \frac{n(n+2)}{(n+1)^2} \tau_n
\end{align*}
as desired. ////
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
$$\frac{1}{x+1}-\frac{1}{x+5}=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
| Make the equations more symmetric by shifting $z:=x+3$ and simplify
$$\frac{1}{(z-2)(z-1)}+\frac{1}{(z-1)z}+\frac{1}{z(z+1)}+\frac{1}{(z+1)(z+2)}$$
$$=\frac{2z^2+4}{(z^2-4)(z^2-1)}+\frac2{z^2-1}$$
$$=\frac{4z^2-4}{(z^2-4)(z^2-1)}$$
$$=\frac4{z^2-4}$$
giving $z=\pm 3$ and $x=0,x=-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How do I prove the following inequality? How do I proceed to solve the inequality $$\frac{(a^2+b^2)}{(a+b)} + \frac {(b^2+c^2)}{(b+c)} + \frac{(a^2+c^2)}{(a+c)} \geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $\frac {(a^2+b^2)}{(a+b)}$ to $AM$.I have tried finding $a_1$ and $a_2$ by doing
$ (\frac{(a_1+a_2)}{2}\bigr)^2 \geq$ $\frac{(a^2+b^2)}{(a+b)}$
| If $a$, $b$, $c>0$ then
$$\frac{a^2+b^2}{a+b}\ge\frac{a+b}2$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$ Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$
I have done it until the the divisor is of second degree.
But here the degree of the remainder is $4$
This means the remainder will be of the form. : $ax^3+bx^2+cx+d$ which makes it difficult to solve. Is it possible to solve $4$ variants with $4$ equations?
| $$\dfrac{x^{10}+1}{(x^2-1)(x^2+x+1)}=(x-1)(x^5-x^3+x^2+x-1)+\dfrac{{x^3+x}}{(x^2-1)(x^2+x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $B$ if $B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
Let $$
\ A=\begin{bmatrix} 0 & a & a^2 & a^3 \\ 0 & 0 & a & a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$$ and
$B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ {{1}\over{2!}} B^2+ {{1}\over{3!}} B^3+...$
My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $n\geq 4$
$A^2= A.A=\begin{bmatrix} 0 & 0 & a^2 & 2a^3 \\ 0 & 0 & 0 & a^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^3= A^2.A=\begin{bmatrix} 0 & 0 & 0 & a^3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^4=A^3.A =\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^n=0$ for every $n\geq 4$, so:
$B= \begin{bmatrix} 0 & a & {{1}\over{2}} a^2 & {{1}\over {3}}a^3 \\ 0 & 0 & a & {{1}\over{2}} a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
But is there any way easier than my way ?
And what about $(ii)$ ?
| $B=\ln(I+A)$, and so $e^B=I+A\implies ii)$...
You asked if there was a better way of doing it... I believe the answer is to notice that you have the power series for $\ln$ and $\exp$ respectively; and then use the fact (probably covered in your course) that those series can be applied to matrices. See for instance .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Operator's matrix in canonical form A stupid question
Let $F:\Bbb R^3\rightarrow\Bbb R^3$ the operator such that
$F((1,0,0)=(0,-1,1))\\ F((0,1,0)=(3,1,2))\\ F((0,0,1))=(2,1,1)) $
Determine the matrix $A$ of $F$ in canonical basis.
I think that tha matrix $A$ is
$$
\begin{pmatrix}
0 & -1 & 1 \\
3 & 1 & 2 \\
2 & 1 & 1 \\
\end{pmatrix}
$$
But in the correction of test my professor said "no"! So, those vector should be used as columns of $A$? I don't understand why
| Let's rewrite it as following $$F\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}0\\-1\\1\end{bmatrix}\\ F\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}\\F\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}2\\1\\1\end{bmatrix}$$since for any $v\in\Bbb R^3$ we have $v=a_1$ , $v=a_1\begin{bmatrix}1\\0\\0\end{bmatrix}+a_2\begin{bmatrix}0\\1\\0\end{bmatrix}+a_3\begin{bmatrix}0\\0\\1\end{bmatrix}$ we obtain$$Fv=a_1F\begin{bmatrix}1\\0\\0\end{bmatrix}+a_2F\begin{bmatrix}0\\1\\0\end{bmatrix}+a_3F\begin{bmatrix}0\\0\\1\end{bmatrix}=a_1\begin{bmatrix}0\\-1\\1\end{bmatrix}+a_2\begin{bmatrix}3\\1\\2\end{bmatrix}+a_3\begin{bmatrix}2\\1\\1\end{bmatrix}=\begin{bmatrix}0&3&2\\-1&1&1\\1&2&1\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=Av$$where $$A=\begin{bmatrix}0&3&2\\-1&1&1\\1&2&1\end{bmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the second degree Taylor polynomial of $f(x,y)=e^{-x^2-y^2}cos(xy)$ at $x_0=0$, $y_0=0$. I would like to either confirm my solution is correct, or find the error in it. I used the following MATLAB code to try to check my answer, but the solution it gave differs from mine.
f = exp(-(x^2+y^2))cos(xy);
taylor(f,[x,y],[0,0])
ans =
x^4/2 + (x^2*y^2)/2 - x^2 + y^4/2 - y^2 + 1
My solution:
The second degree Taylor polynomial at the point $(a,b)$ is given by
$$p_2(x,y)=f(a,b)+Df(a,b)\left[\begin{array}{c}
x-a\\y-b
\end{array}\right]+\frac{1}{2}[x-a\mbox{ }y-b]Hf(a,b)\left[\begin{array}{c}
x-a\\y-b
\end{array}\right]$$
where $Df(a,b)$ is the Jacobian matrix and $Hf(a,b)$ is the Hessian matrix (i.e. first and second order partial derivatives, respectively). Thus, we begin by computing the partial derivatives:
\begin{equation*}
\begin{split}
\frac{\partial f}{\partial x}(x,y)=-e^{-x^2-y^2}(y\mbox{ sin}(xy)+2x\mbox{ cos}(xy))\\
\frac{\partial f}{\partial y}(x,y)=-e^{-x^2-y^2}(x\mbox{ sin}(xy)+2y\mbox{ cos}(xy))\\
\frac{\partial^2 f}{\partial x^2}(x,y)=e^{-x^2-y^2}((4x^2-y^2-2)\mbox{cos}(xy)+4xy\mbox{ sin}(xy))\\
\frac{\partial^2 f}{\partial y^2}(x,y)=e^{-x^2-y^2}(4xy\mbox{ sin}(xy)-(x^2-4y^2+2)\mbox{cos}(xy))\\
\frac{\partial^2 f}{\partial x\partial y}(x,y)=\frac{\partial^2 f}{\partial y\partial x}(x,y)=e^{-x^2-y^2}((2x^2+2y^2-1)\mbox{sin}(xy)+3xy\mbox{ cos}(xy))
\end{split}
\end{equation*}
Then, at the point $(x_0,y_0)=(0,0)$,
\begin{equation*}
\begin{split}
f(0,0)=1\\
\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0\\
\frac{\partial^2 f}{\partial x^2}(0,0)=\frac{\partial^2 f}{\partial y^2}(0,0)=-2\\
\frac{\partial^2 f}{\partial x\partial y}(0,0)=\frac{\partial^2 f}{\partial y\partial x}(0,0)=0
\end{split}
\end{equation*}
Therefore, the second degree Taylor polynomial at the point $(0,0)$ is
\begin{equation*}
\begin{split}
p_2(x,y)=f(0,0)+Df(0,0)\left[\begin{array}{c}
x-0\\y-0
\end{array}\right]+\frac{1}{2}[x-0\mbox{ }y-0]Hf(0,0)\left[
\begin{array}{c}
x-0\\y-0
\end{array}\right]\\
=1+[0\mbox{ }0]\left[
\begin{array}{c}
x\\y
\end{array}\right]+\frac{1}{2}[x\mbox{ }y]\begin{bmatrix}
-2&0\\0&-2
\end{bmatrix}\left[
\begin{array}{c}
x\\y
\end{array}\right]=1-x^2-y^2
\end{split}
\end{equation*}
| I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^{-x^2-y^2}\cos(xy) = (1-x^2-y^2+\cdots)(1-\frac{1}{2}(xy)^2+ \cdots )$$
But, $(xy)^2$ is order $4$ so the answer is merely,
$$f(x,y) = e^{-x^2-y^2}\cos(xy) = 1-x^2-y^2+\cdots $$
The above is the Taylor expansion of $f$ at $(0,0)$ to order $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the indefinite integral of $\frac{1}{x^{2/3}(1+x^{2/3})}$?
$$\int \frac{dx}{x^{2/3}(1+x^{2/3})}.$$
I substituted,
$$t=\frac{1}{x^{1/3}}$$
$$\frac{dt}{dx} = -\frac{1}{3x^{4/3}}$$
$$\frac{dt}{dx} = -\frac{t^4}{3}$$
Rewriting the question,
$$\int \frac{dx}{x^{2/3}+x^{4/3}}$$
$$-\frac{1}{3} \int \frac{dt}{t^4\Bigl(\frac{1}{t^2}+\frac{1}{t^4}\Bigr)}$$
We have,
$$-\frac{1}{3} \int \frac{dt}{t^2 + 1}$$
$$-\frac{1}{3}\tan^{-1}t+C$$
$$-\frac{1}{3}\tan^{-1}\Biggl(\frac{1}{x^{1/3}}\Biggr)+C$$
But the answer given is $$3\tan^{-1}x^{1/3}+C$$
Where am I wrong?
Any help would be appreciated.
| If $$t=x^{-2/3}$$ then $$dt=\frac{-2}{3}x^{-5/3}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
congruence of Ramanujan tau function mod 2 The Ramanujan tau function, studied by Ramanujan (1916), is the function ${\displaystyle \tau :\mathbb {N} \to \mathbb {Z} }$ defined by the following identity:
${\displaystyle \sum _{n\geq 1}\tau (n)q^{n}=q\prod _{n\geq 1}(1-q^{n})^{24}=\eta (z)^{24}=\Delta (z)}$.
Wiki says $\tau(n)=\sigma_{11}(n) \operatorname{mod} 2$ for $n$ odd, is there a simple way to see this (without using modular form)?
| Start with the Jacobi triple product
in the form
$$ \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m-1}y^2) (1 + x^{2m-1}/y^2) =
\sum_{n=-\infty}^{\infty} x^{n^2} y^{2n}. \tag{1}$$
Let $\ y^2 = x t \ $ to get
$$ \Big(1 + \frac1t\Big) \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m}t) \Big(1 + \frac{x^{2m}}t\Big) =
\sum_{n=-\infty}^\infty x^{n^2+n} t^n . \tag{2}$$
Divide both sides by $\ 1 + 1/t \ $ to get
$$ \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m}t)
\Big(1 + \frac{x^{2m}}t\Big) =
\sum_{n=1}^\infty x^{n^2-n}\frac{ (1/t^n + t^{n-1})}{(1/t + 1)}. \tag{3}$$
Take the limit as $\ t \to -1 \ $ to get
$$ \prod_{m=1}^\infty (1 - x^{2m}) (1 - x^{2m}) (1 - x^{2m}) =
\sum_{n=1}^\infty x^{n^2-n} (2n-1). \tag{4}$$
Define $\ E(q) := \prod_{n=1}^\infty (1 - q^n) \ $ and replace $\ x^2 \ $ with $\ q \ $ to get
$$ E(q)^3 = \sum_{n=1}^\infty (2n-1) q^{ n(n-1)/2}. \tag{5}$$
Taking modulo $2$ to get
$$ E(q)^3 \equiv \sum_{n=1}^\infty q^{ n(n-1)/2} \pmod{2}. \tag{6}$$
Raise both sides to the eighth power and using
$\ (a + b)^{2^k} \equiv a^{2^k} + b^{2^k} \pmod{2} \ $ implies
$$ E(q)^{24} \equiv \sum_{n=1}^\infty q^{4n(n-1)} \pmod{2}. \tag{6}$$
Multiply both sides by $\ q \ $ and rewrite to get
$$ \sum_{n=1}^\infty \tau(n) q^n \equiv \sum_{n=1}^\infty q^{(2n-1)^2} \pmod{2}. \tag{7}$$
If $\ n \equiv 1 \pmod{2},\ $ then $\ \sigma_k(n) \equiv \sigma_0(n) \pmod{2}. \ $ Also
$\ \sigma_0(n) \equiv 1 \pmod{2} \ $ iff $\ n \ $ is a square integer. This proves
that $\ \tau(n) \equiv \sigma_k(n) \ $ for $\ n \ $ odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
|
\begin{align}
p_1&=(\tfrac{\sqrt2}2,\tfrac{\sqrt2}2)
,\\
p_2&=(0,1)
,\\
p_3&=(-\tfrac{\sqrt2}2,\tfrac{\sqrt2}2)
,\\\ \dots
\end{align}
\begin{align}
S&=
(p_{1x}-p_{3x})^2+\tfrac32(p_{1x}-p_{3x})(p_{2y}-p_{1y})
=\frac{3\sqrt{2}+1}{2}
,\\
&=
\frac{3k\sqrt{2}+k}{2k}
,\\
a&=3k,\quad b=2,\quad c=k,\quad d=2k
,\\
a+b+c+d&=6k+2,
\quad k\in\mathbb{R}
.
\end{align}
In order to get $a+b+c+d=10$,
we need to have $k=\tfrac43$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| We have
$$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$ and
$$\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)\left(
\sqrt{49-x^2}+\sqrt{25-x^2}\right)=3\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right),$$
which gives
$$\sqrt{49-x^2}+\sqrt{25-x^2}=8$$ and from here
$$\sqrt{49-x^2}=5.5,$$ which gives the answer:
$$\left\{-\sqrt{18.75},\sqrt{18.75}\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Making the sum of 5th power of integers, a perfect square. Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.
But this question is something else. I haven't been able to solve this after $4$ hours.
Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.
I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.
I hope anyone can help me here.
| This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:
$$n^2 = (a+b)m + c^5 + d^5\quad\quad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
| Hint: Use the factorisation formula $$\cos a -\cos b =-2\sin{a+b\over 2}\sin{a-b\over 2}$$
and
$$\sin a -\sin b =2\sin{a-b\over 2}\cos{a+b\over 2}$$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} = \lim_{x\to0}\frac{2\color{red}{\sin 2x}\cdot \color{green}{\sin x}\cdot\color{blue}{x^2}}{\color{blue}{\sin (x^2)}\cos (2x^2)\cdot\color{red}{2x}\cdot \color{green}{x}} =2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 1
} |
Number of inflection points of $(x-2)^6(x-3)^9$
Find the number of inflection points of $(x-2)^6(x-3)^9$
Attempt:
If $f(x)$ has $n$ critical points then even $f(x+a)$ will also have $n$ critical points.
So we can simplify it to finding the number of inflection points of $(x)^6(x-1)^9$.
I have evaluated the double derivative to be:
$$6(x-1)^7x^4(35x^2-28x +5)$$
Clearly, the double derivative is $0$ at $4$ points, $0$, $1$ and the roots of the quadratic. But curvature is not changing around $x=0$ so it's not an inflection point.
Thus, there should be $3$ inflection points.
But answer given is $1$ inflection point only.
Please let me know what concept am I missing on.
I tried my level best to zoom into the graph and catch three critical points but there appears to be only one.
| It should be
$$f'(x)=3(x-3)^8(x-2)^5(5x-12)=0 \implies x=3,2,\frac{12}5$$
and
$$f''(x)=6(x-3)^7(x-2)^4(35x^2-168x+201)=0 \\\implies x_1=2,\,x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$
and since $p(x)=35x^2-168x+201\implies p(12/5)\neq 0$ the unique stationary inflection point is at $x=3$ otherwise, if we look at general inflection points, by the sign of $f''(x)$ the inflection points are three at
$$x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Binomial Coefficients inequality simplifying to answer involving "e" In doing some analysis, my professor wrote \begin{align*} \begin{pmatrix} n+ m \\ m \end{pmatrix} &= \frac{(n+m)!}{m!n!} \\ &= \frac{(n+m)(n+m-1)...(n+1)n!} {m!n!} \\ &= \frac{(n+m)}{m} \frac{(n+m-1)}{m-1} ... \frac{(n+1)}{1} \\ & \leq \bigg( \frac{e(n+m)}{m} \bigg)^m \end{align*}
I understand the first equality is the definition of binomial coefficients, the third equality cancels the $n!$. I am confused about how the last inequality. I understand that $$ e^m = \sum_{k=0}^{\infty} \frac{m^k}{k!} $$ but I don't quite see how I get the last inequality from this. If someone could tell me the explicit step I need to take I would greatly appreciate it.
| The following inequality chain is valid for positive integers $m$
\begin{align*}
\sqrt{2\pi m}\left(\frac{m}{e}\right)^m\leq m! \leq e\sqrt{m}\left(\frac{m}{e}\right)^m\tag{1}
\end{align*}
The reciprocal of the left part of (1) gives
\begin{align*}
\frac{1}{m!}\leq\frac{1}{\sqrt{2\pi m}}\left(\frac{e}{m}\right)^m\leq\left(\frac{e}{m}\right)^m\tag{2}
\end{align*}
We also have
\begin{align*}
(n+m)(n+m-1)\cdots(n+1)\leq (n+m)^{m}\tag{3}
\end{align*}
From(2) and (3) we obtain
\begin{align*}
\color{blue}{\frac{(n+m)(n+m-1)\cdots(n+1)}{m!}}&\leq \left(\frac{e}{m}\right)^m(n+m)^m\\
&\,\,\color{blue}{=\left(\frac{e(n+m)}{m}\right)^m}
\end{align*}
and the claim follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sum of a Sequence of Odd Numbers that are Squared What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + \ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.
This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):
I am a little stuck on what to do next and how to obtain $\frac{n (4n^2 - 1)}{3}$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)
| Your approach is almost correct. Check again your steps. At the end you should have
$$\begin{align}(k+\mathbf{2})^3-1
&=6(1+3^2+\dots+k^2)+12(1+3+\dots+k)+\underbrace{(8+8+\dots+8)}_{\text{$(k+1)/2$ times}}\\
&=6(1+3^2+\dots+k^2)+12\left(\frac{k+1}{2}\right)^2+8\mathbf{\left(\frac{k+1}{2}\right)}.\end{align}$$
Hence
$$6(1+3^2+\dots+k^2)=(k+\mathbf{2})^3-1-12\left(\frac{k+1}{2}\right)^2-8\mathbf{\left(\frac{k+1}{2}\right)}$$
and it follows that
$$\sum_{j=1}^n(2j-1)^2=1+3^2+\dots+k^2=\frac{k(k+2)(k+1)}{6}=\frac{n (4n^2 - 1)}{3}$$
where $n=(k+1)/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$
Or $\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$
Then i can't prove $\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$
I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QM\ge AM+GM$ with $n=2$ so is that true for $n=i$?
| Let $a^2+b^2=2t^2ab$, where $t>0$.
Thus, by AM-GM $t\geq1$ and we need to prove that
$$\frac{2ab}{\sqrt{a^2+b^2+2ab}}+\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}+\frac{\sqrt{a^2+b^2+2ab}}{2}$$ or
$$\sqrt{\frac{2}{t^2+1}}+t\geq1+\sqrt{\frac{t^2+1}{2}}$$ or
$$t-1\geq\sqrt{\frac{t^2+1}{2}}-\sqrt{\frac{2}{t^2+1}}$$ or
$$t-1\geq\frac{t^2-1}{\sqrt{2(t^2+1)}}$$ or
$$\sqrt{2(t^2+1)}\geq t+1,$$ which is true by C-S:
$$\sqrt{2(t^2+1)}=\sqrt{(1^2+1^2)(t^2+1)}\geq t+1.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_m(\nabla \times F)\cdot NdS$ Evaluate $\int_m(\nabla \times F)\cdot N\,dS$ where $F=(e^x+z^2,\sin(yz),x^3)$ and $M=\{(x,y,z):y=\frac{x^2}{2}+z^2-4,y\leq 0\}$ and $N$ points outwards
So my plan is:
*
*find the unit normal of $M$
*find the area element ($dS$) of $M$ by projecting $M$ onto $XZ$ plane
*find the limits of integration and calculate the integral (will got to polar coordinates)
How do I find the unit normal of such a surface?
I thought about look at a parameterisation $\phi(x,z)=(x,\frac{x^2}{2}+z^2-4,z)$
Then $\hat{n}=\frac{\phi_x\times\phi_z}{\|\phi_x\times\phi_z\|}=\frac{(x,-1,2z)}{\sqrt{x^2+4z^2+1}}$
now looking the projection we have $dS=\sqrt{1+x^2+4z^2}$
Now $\nabla \times F=(y\sin (yz),-3x^2+2z,0)$ so it will be $$\left(\left( \frac{x^2}{2} + z^2-4\right)\sin\left(\left(\frac{x^2}{2}+z^2-4\right) z\right), -3x^2+2z,0 \right)$$
And last we will have to calculate:
$$\int_0^{2\sqrt{2}}\int_o^2\left(\left(\frac{x^2}{2}+z^2-4\right) \sin\left(\left(\frac{x^2}{2}+z^2-4\right)z\right), -3x^2+2z,0\right) \cdot (x,-1,2z) \,dx\,dx?$$
| HINT
A possible way, since $div(\nabla \times F)=0$, is note that
$$\int_M(\nabla \times F)\cdot NdS=-\int_E(\nabla \times F)\cdot NdS$$
where $E$ is the surface enclosed by the boundary of $M$ at $y=0$ that is
$$E=\{(x,y,z):\frac{x^2}{2}+z^2-4\le 0,y= 0\}$$
with
$$(\nabla \times F)\cdot N=(y\sin (yz),-3x^2+2z,0)\cdot (0,1,0)=-3x^2+2z$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $I_n=\int_0^{\pi/2} \frac{1}{\left( a\cos^2x+b\sin^2x\right)^n} \, dx$ I would like to evaluate (using elementary methods if possible) : (for $a>0,\ b>0$)
$$
I_n=\int_0^{\pi/2} \frac{1}{( a\cos^2x+b\sin^2x)^n} \, dx,\quad \ n=1,2,3,\ldots
$$
I thought about using $u=\tan(x)$ or $u=\frac{\pi}{2}-x$ but did not work. wolfram alpha evaluates the indefinite integral but not definite integral???
| Per geometric summation
\begin{align}
\sum_{n\ge 1}I_n t^n=&
\int_0^{\frac{\pi}{2}} \sum_{n\ge 1}\frac{t^n}{(a \cos^2x+ b \sin^2x)^n}dx\\
=&\int_0^{\frac{\pi}{2}}\frac{t}{a\cos^2x+ b\sin^2x-t}dx
=\frac{\pi}{2}\frac{t}{\sqrt{(a-t )(b-t )}}\\
=&\sum_{n\ge 1}\frac{\pi}{2^{2n-1}}\sum_{i+j+1=n} \frac{1}{a^{i+1/2} b^{j+1/2}} {2i\choose i}{2j\choose j} t^n
\end{align}
where the Taylor series $ \frac{1}{\sqrt{a-t}}=\sum_{i\ge 0}\frac{2i\choose i }{2^{2i} a^{i+1/2}}t^i$ is applied in the last step. Thus
$$I_n= \frac{\pi}{2^{2n-1}\sqrt{ab}}\sum_{i+j+1=n} \frac{{2i\choose i}{2j\choose j}}{a^{i} b^{j}}
$$
which, in particular, results in
\begin{align}
&I_1 =\frac\pi{2\sqrt{ab}}\\
& I_2 =\frac\pi{4\sqrt{ab}}\left(\frac1a+\frac1b\right)\\
& I_3 =\frac\pi{16\sqrt{ab}}\left(\frac3{a^2}+\frac2{ab}+\frac3{b^2}\right)\\
&I_4=\cdots
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral $\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$ I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then:
$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac{\pi^2}{8}-\int_0^1 \frac{\arcsin^2 x}{(1+x^2)^{3/2}}dx$$ We can expand into power series the integral, we have: $\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$ and using the binomial series for $(1+x^2)^{-3/2}$ will result in: $$\sum_{n\geq1}\frac{2^{2n-1}x^{2n}}{n^2\binom {2n}n}\sum_{k\ge 0}\binom{-3/2}{k}x^{2k}$$ But I dont know how to simplify this. I tried one more thing, letting $\sin x= \sin^2 t$ gives:
$$I=2\int_0^\frac{\pi}{2}\frac{x\sin x}{\sqrt{1+\sin^2 x}}dx$$ Since $\int \frac{\sin x}{\sqrt{1+\sin^2x}}dx=-\arcsin\left(\frac{\cos x}{\sqrt 2} \right)+C$ we can integrate by parts to obtain: $$I=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\cos x}{\sqrt 2}\right)dx=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ But I am stuck, so I would appreciate some help.
Edit: By letting $\frac{\sin x}{\sqrt 2} =t $ We get: $$I=2\int_0^\frac1{\sqrt{2}} \frac{\arcsin x}{\sqrt{\frac12-x^2}}dx=2\text{Li}_2\left(\frac1{\sqrt 2}\right)-\frac{\pi^2}{24}+\frac{\ln^2 2}{4}$$ Where the latter integral was evaluated with wolfram. I would love to see a proof for that.
| Write
$$
I(t)=\int_0^{\frac{1}{\sqrt{2}}} \frac{2\arcsin(tx)}{\sqrt{\frac{1}{2}-x^2}} \, {\rm d}x
$$
and calculate
\begin{align}
I'(t) &= \int_0^{\frac{1}{\sqrt{2}}} \frac{2x}{\sqrt{\left(\frac{1}{2}-x^2\right)\left(1-(tx)^2\right)}} \, {\rm d}x \\
&= \frac{\log\left(\sqrt{2}+t\right)-\log\left(\sqrt{2}-t\right)}{t} \\
&= \frac{{\rm Li}_1 \left(\frac{t}{\sqrt{2}}\right) - {\rm Li}_1 \left(-\frac{t}{\sqrt{2}}\right)}{t}\, .
\end{align}
Then
\begin{align}
I(1) &= \int_0^1 I'(t) \, {\rm d}t \\
&={\rm Li}_2 \left(\frac{1}{\sqrt{2}}\right) - {\rm Li}_2 \left(-\frac{1}{\sqrt{2}}\right) \, .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
If $a_1, \ldots ,a_9$ are in harmonic progression, then find the value of the determinant If $a_1, \ldots ,a_9$ are in harmonic progression ,then find the value of the determinant
$$
\begin{vmatrix}
a_1 &a_2&a_3\\
5&4&a_6\\
a_7&a_8&a_9\\
\end{vmatrix}
$$
I calculated the terms as $a_1=\frac{20}{1},a_2=\frac{20}{2},a_3=\frac{20}{3},a_4=\frac{20}{4},a_5=\frac{20}{5},a_6=\frac{20}{6},a_7=\frac{20}{7},a_8=\frac{20}{8},a_9=\frac{20}{9}$ but now calculating the determinant is tedious.Is there some other elegant method possible?
| after taking $20^3$ common and performing the following row operations
replace $R_2$ with $R_2-\frac{R1}{4}$
replace $R_3$ with $R_3-\frac{R1}{7}$
I got this
\begin{vmatrix}
1 & 1/2& 1/3\\
0 & 3/40& 1/12\\
0 & 3/56& 4/63\\
\end{vmatrix}
which didn't take much time now calculate the determinant about column 1
I got $det(A)=\frac{20^3}{3360}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $a_n = \frac{n}{2n+1}+\frac{1}{n^3}$ is a Cauchy Sequence I want to show that
$$a_n = \frac{n}{2n+1} + \frac{1}{n^3}$$
is a Cauchy sequence.
My attempt:
$$|a_m-a_n|=|(\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3})|$$
$$\leq\frac{m}{2m+1}+\frac{n}{2n+1}+\frac{1}{m^3}+\frac{1}{n^3}$$
$$<\frac{1}{2}+\frac{1}{2}+\frac{1}{N^3}+\frac{1}{N^3}$$
$$=1+\frac{2}{N^3}$$
$$\leq \epsilon\Rightarrow N(\epsilon)\geq(\frac{2}{\epsilon-1})^{\frac{1}{3}}$$
But this seems a bit off?
Any thoughts?
| You got it wrong here:
$$\begin{align}|a_m - a_n| &= \left|\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3}\right| \\
&\leqslant
\frac{m}{2m+1}+\frac{n}{2n+1}+\frac{1}{m^3}+\frac{1}{n^3}
\end{align}$$
I mean, it is correct, but it won't help. The number inside the absolute value bars is small precisely because of those subtractions, and expanding it the way you did makes those subtractions vanish.
You should try to keep the subtractions around:
$$\begin{align}|a_m - a_n| &= \left|\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3}\right| \\
&\leqslant
\left|\frac{m}{2m+1}-\frac{n}{2n+1}\right|+\left|\frac{1}{m^3}-\frac{1}{n^3}\right|
\end{align}$$
Can you go on from here? [Hint: fix $m\leqslant n$. The other case, $n<m$, is similar.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A direct way to an inequality : Ferrari's identities I would like to submit a recent answer that I gave (and I have deleted) where someone tolds me that I was "total wrong" this is the following :
Begin
Prove that $$\frac{x^2+y^2+z^2}{2}\geq (\alpha\frac{x^4+y^4+z^4}{\beta})^{0.25}$$
With the condition $x+y+z=0$ and $\sqrt{\frac{x^2+y^2+z^2}{2}}\geq \frac{(2\alpha)^{0.25}}{\beta^{0.25}} $
Now the key of this problem is the following Ferrari's identities :
$$(a^2 + 2ac - 2bc - b^2)^4 + (b^2 - 2ab - 2ac - c^2)^4 + (c^2 + 2ab + 2bc - a^2)^4
= 2(a^2 + b^2 + c^2 - ab + ac + bc)^4 . $$
And
$$(a^2 + 2ac - 2bc - b^2)^2 + (b^2 - 2ab - 2ac - c^2)^2 + (c^2 + 2ab + 2bc - a^2)^2
= 2(a^2 + b^2 + c^2 - ab + ac + bc)^2 . $$
It remains to put :
$x=a^2 + 2ac - 2bc - b^2$
$y=b^2 - 2ab - 2ac - c^2$
$z=c^2 + 2ab + 2bc - a^2$
And finally get :
$$(a^2 + b^2 + c^2 - ab + ac + bc)^2\geq \frac{(2\alpha)^{0.25}}{\beta^{0.25}}(a^2 + b^2 + c^2 - ab + ac + bc)$$
Or $$(p)^2\geq \frac{(2\alpha)^{0.25}}{\beta^{0.25}}p$$
Wich is true with the conditions of the beginning !
End
Could someone tells me If I'm in a wrong way or not ?
| That's an interesting angle, but honestly a bit hard to follow as posted.
With your notations, as well as $\,p=a^2 + b^2 + c^2 - ab + ac + bc\,$, Ferrari's identities can be written as:
$$
\begin{align}
x^2 + y^2+z^2 &= 2p^2 \\
x^4+y^4+z^4 &= 2p^4
\end{align}
$$
Then the problem reduces to proving that $\,\sqrt{\dfrac{2p^2}{2}} \ge \sqrt[4]{\dfrac{2\alpha}{\beta}} \implies \dfrac{2p^2}{2} \ge \sqrt[4]{\dfrac{\alpha\cdot 2p^4}{\beta}}\,$ which is of course trivial since both inequalities are equivalent to $\,|p| \ge \sqrt[4]{\dfrac{2\alpha}{\beta}}\,$.
What's missing, however, is that the substitution you used has yet to be justified:
It remains to put :
$x=a^2 + 2ac - 2bc - b^2 \\
y=b^2 - 2ab - 2ac - c^2 \\
z=c^2 + 2ab + 2bc - a^2$
It is not a priori obvious that the system is solvable in $a,b,c$ for all $x,y,z$ such that $x+y+z=0$.
A secondary point is that you must show why the argument works if $\,a,b,c\,$ turn out to be complex.
| {
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"url": "https://math.stackexchange.com/questions/2878345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ given $4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$
$$4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$$ Solve for $x$.
I tried to put every single monomial under $x^2$ denominator, but that did not get me to anything I could solve for.
I appreciate any help.
| $$\begin{array}{rcl}
4x^2 + 12x + \dfrac{12}{x} + \dfrac{4}{x^2} &=& 47 \\
4\left(x^2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 47 \\
4\left(x^2 + 2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 55 \\
4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) &=& 55 \\
4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) - 55 &=& 0 \\
\left(x + \dfrac1x\right) &=& -5.5 ~\text{or}~ 2.5 \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$
My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
| $$x^3=y^3+2y+1\tag{1}$$
We are after integer solutions of $(1)$. On setting $y=x-a$, where $a\in\mathbb{Z}$, we have from $(1)$
$$x^3=(x-a)^3+2(x-a)+1=x^3-3a x^2+(3a^2+2)x-a^3-2a+1\tag{2}$$
which rearranges to
$$3a x^2-(3a^2+2)x+a^3+2a-1=0\tag{3}$$
Solve for $x$
\begin{align*}
x&=\frac{(3a^2+2)\pm\sqrt{(3a^2+2)^2-12a(a^3+2a-1)}}{6a}\\
&=\frac{(3a^2+2)\pm\sqrt{-3a^4-12a^2+12a+4}}{6a}\tag{4}
\end{align*}
The discriminant is
\begin{align*}
-3a^4-12a^2+12a+4&=-3a^4-4(3a^2-3a-1)\\
&=-3a^4-4[3\left((a-\tfrac12)^2-\tfrac7{12}\right)]\\
&=-3a^4-12(a-\tfrac12)^2+7\\
&=7-(3a^4+3(2a-1)^2)
\end{align*}
Now check various integral $a$ for solutions: $a=0$ gives division by zero in $(4)$; $a=1$ gives $x_+=1$ or $x_-=\frac23$; $a=-1$ gives discriminant of $-23$; and for $|a|\ge2$ we get the discriminant $7-(3a^4+3(2a-1)^2)<0$ (since $3a^4+3(2a-1)^2$ is nonnegative).
Hence $x=1$, $y=0$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 6
} |
Integrate $\int \frac{x^3 e^x \:dx}{\sqrt{x^2-1}}$ Integrate $$\int \frac{x^3 e^x \:dx}{\sqrt{x^2-1}}$$
My try:
I used Parts by writing integral as:
$$I=\int x^2 e^x {\frac{x\:dx}{\sqrt{x^2-1}}}$$ we get
$$I=x^2 e^x\sqrt{x^2-1}-\int (x^2-2x+2)e^x \sqrt{x^2-1}dx$$
But the other integral has become more complicated
| This is a modification which I hope will be helpful.
$$ \int (x^2 -2x+2)e^x \sqrt{ x^2-1}$$
$$\cot(cosec^{-1} x)=\sqrt{ x^2-1}$$
Let $cosec^{-1} x=u$ then:
$ cosec (u)=\frac{1}{\sin u} =x$
$dx=\frac{-\cos u}{\sin^2 u}du$
substituting we get:
$$ \int (x^2 -2x+2)e^x \sqrt{ x^2-1}=\int(\frac{1}{\sin^2 u}-\frac{2}{\sin u} +2)cotan (u) .\frac{-\cos u}{\sin^2 u} e^{1/\sin u}du$$
$\frac{-\cos u}{\sin^2 u} e^{1/\sin u}=(e^{1/\sin x})'$
$\frac{-\cos u}{\sin^2 u} e^{1/\sin u}du= dv$
$v=\frac{1}{\sin u}$
$(\frac{1}{\sin^2 u}-\frac{2}{\sin u} +2)cotan (u)=t$
I think you could now do it .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$ Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$
I tried:
$x=\sin\alpha$ and $y=\cos\beta$
$\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}$ but if I write $\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta$, it's not true because
$\cos\alpha \sin\beta$ can be negative.
Can someone give me an idea?
| $$\begin{align*}xy-\sqrt{(1-x^2)(1-y^2)}&\le 1\\
\Leftrightarrow
(xy)^2&\le 1+(1-x^2-y^2+(xy)^2 )+2\sqrt{1-x^2)(1-y^2)}
\end{align*}$$
which is true since $|x|\le 1$ then $1-x^2 > 0$ and similarly for $y$.
And do the same for the other side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 x$
Solving and re-arranging we get
$-64\sin^6 x+80\sin^4 x-20\sin^2 x+1=0$
Using the substitution $\sin^2 x=t$
$-64t^3+80t^2-20t+1=0$
I am not able to solve it from hence forth
| $$−64t^6+80t^4−20t^2+1=(1-4t^2)(1-16t^2+16t^4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Solve the system $x^3+y=3x+4$, $2y^3+z=6y+6$, $3z^3+x=9z+8$
Find the real solutions of the system below:
$$\begin{aligned} x^{3}+y &= 3x+4\\ 2y^{3}+z &= 6y+6\\ 3z^{3}+x &= 9z+8\end{aligned}$$
I wrote the system as:
$$x(x^{2}-3)=4-y$$
$$2y(y^{2}-3)=6-z$$
$$3z(z^{2}-3)=8-x$$
Then I tried to use the addition and multiplication of the equations above, but it didn't seem to be very useful.
| We have
$$(y-2)=-(x-2)(x+1)^2\,,$$
$$(z-2)=-2(y-2)(y+1)^2\,,$$
and
$$(x-2)=-3(z-2)(z+1)^2\,.$$
Multiply the three equation above to get
$$(x-2)(y-2)(z-2)\big(1+6(x+1)^2(y+1)^2(z+1)^2\big)=0\,.$$
Since $1+6(x+1)^2(y+1)^2(z+1)^2\geq 1>0$, we conclude that
$$(x-2)(y-2)(z-2)=0\,,$$
whence $x=2$, $y=2$, or $z=2$. However, this means $x=y=z=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Even numbers of the form $\frac {n(n+1)}{2}$ and twin primes If we take some even number of the form $\frac{n(n+1)}{2}$ and add $1$ to it and also subtract $1$ from it then we have a mapping $\frac{n(n+1)}{2}\to\left\{\frac{n(n+1)}{2}-1,\frac{n(n+1)}{2}+1\right\}$.
From some even numbers of that form that I checked only number $6=\frac{4\cdot 3}{2}$ maps to a twin prime pair $6 \to \{5,7\}$.
Is there any simple explanation why $\left\{\frac{n(n+1)}{2}-1, \frac{n(n+1)}{2}+1\right\}$ is so rarely a twin prime pair?
| Note that: $$\frac{n(n+1)}{2} - 1 = \frac{n^2 + n -2}{2} = \frac{(n+2)(n-1)}{2}$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How does recurrence relation works with system of equations? There are $$ a+b+c+d = 2\\2a+2^2b+2^3c+2^4d = 5\\ 3a+3^2b+3^3c+3^4d = 6\\4a+4^2b+4^3c+4^4d = 1$$
then I'm given
$$C_{n}= a+bn+cn^2+dn^3$$
from linear recurrence relation with repeated roots said that
$$(x-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^{4-k}x^{4-k}$$
and then
$$C_{n}=\sum_{k=1}^{4}\binom{4}{k}(-1)^{5-k}C_{n-k} = 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4}$$ $\forall n \geq 5$
I want to find $C_{6}$. Then i've plugged $C_{5}=-3$ into that relation which finally gave $C_{6}=-8$ but wolfram alpha told me that $C_{6} =-48
Question: How do I use recurrence relations to find out the correct $C_6$
| I believe you are over-complicating it. The given system tells you that the quartic polynomial $p(n)=an+bn^2+cn^3+dn^4$ attains the values $2,5,6,1$ at $n=1,2,3,4$. By Lagrange interpolation such polynomial is
$$ p(n) = \frac{1}{24}\left(18n+37n^2-6n^3-n^4\right) $$
hence
$$ C_6 = \frac{p(6)}{6} = \color{red}{-8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now
| HINT:
I'm sure you can solve the equation $x^2-1=0$.
For the second equation, you have that either $$x^2=0\implies x=0$$ or $$\cos\frac{\pi x}2=0\implies x=\frac2\pi\cdot\left(\frac\pi2+\pi k\right)$$ for some integer $k$.
But can $x=0$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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Evaluate: $\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}$
Evaluate
$$
\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} \cdotp
$$
My attempt:
$$
I
= \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}
= \int \frac{\mathrm{d}x}{x\sqrt{\left(x + \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}
$$
I thought completing the square would bring the integrand into some form, but it did not. Please help.
| $$I=\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{x\sqrt{\left(2x+1\right)^2+3}}\,\mathrm{d}x$$
Substitute $u=2x+1$
$$I=2{\displaystyle\int}\dfrac{1}{\left(u-1\right)\sqrt{u^2+3}}\,\mathrm{d}u$$
Substitute $u=\sqrt{3}\tan\left(v\right)$
$$I=2{\displaystyle\int}\dfrac{\sqrt{3}\sec^2\left(v\right)}{\left(\sqrt{3}\tan\left(v\right)-1\right)\sqrt{3\tan^2\left(v\right)+3}}\,\mathrm{d}v$$
$$I=2{\displaystyle\int}\dfrac{\tan^2\left(\frac{v}{2}\right)+1}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)\left(\frac{2\cdot\sqrt{3}\tan\left(\frac{v}{2}\right)}{1-\tan^2\left(\frac{v}{2}\right)}-1\right)}\,\mathrm{d}v$$
Substitute $w=\tan\left(\dfrac{v}{2}\right)$
$$I=\class{steps-node}{\cssId{steps-node-2}{4}}{\displaystyle\int}\dfrac{1}{w^2+2\cdot\sqrt{3}w-1}\,\mathrm{d}w$$
$$I=4{\displaystyle\int}\dfrac{1}{\left(w+\sqrt{3}-2\right)\left(w+\sqrt{3}+2\right)}\,\mathrm{d}w$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Maclaurin Series Expansion nth term expression My task is to show that the Maclaurin series of $\ln\frac{x+1}{x-1}$ is given as $\sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$
My approach is using the standard expansion of $\ln(x+1)$ to get $x-\frac{x^2}{2}+\frac{x^3}{3}$
I got $$\ln(x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3}$$
After this point I'm a little confused as to how to proceed. My guess would be to subtract $\ln(x-1)$ from $\ln(x+1)$ but I still don't understand how to get this form or what exactly it's asking.
Any help would be greatly appreciated.
| You have $$ \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...$$
and $$\ln (x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3}+...$$
Thus $$\ln\frac{x+1}{x-1}= \ln (1+x)-\ln (x-1)$$
$$=x-\frac{x^2}{2}+\frac{x^3}{3}...+x+\frac{x^2}{2}+\frac{x^3}{3}+....$$
$$= 2x+(2/3) x^3 + 2/5 (x^5) +(2/7)x^7 +..... $$
$$= \sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Partial fraction decomposition and trig substitution I am reviewing material for the CAA module 0 exam and part of it is integration. I had trouble doing a partial fractions problem and the work that they having showing the answer is convoluted and unclear.
I am asked to integrate:
$\int \frac{x+7}{x^2(x+2)}dx$
obviously when you set it up you get:
$\frac{A}{x}+\frac{B}{x^2}+\frac{c}{x+2}=\frac{x+7}{x^2(x+2)}\tag{$*$}$
I had no problems here I had problems in the next step.
multiplying through I got:
$A(x^2)(x+2)+B(x)(x+2)+C(x^2)(x)=x+7$
Somehow they got $Ax(x+2)+B(x+2)+C(x^2)=x+7$
That is not even possible unless you factor out an x on the left hand side and divide both sides by x which would give you $\frac{x+7}{x}$ on the right hand side. So either their work is incorrect or I am missing something.
I also tried solving a trig substitution problem: (ignoring the constants they get pulled out anyway)
$\int\frac{1}{x^4\sqrt{x^2-7}}dx$
obviously $x=\sqrt{7}\sec(\theta)$
and eventually you get to: $\int \cos^3(\theta) dx$
peel off a cos: $\int \cos^2(\theta)\cos(\theta)\,d(\theta)$
use an identity:
$\int 1-\sin^2(\theta) \cos(\theta)$
let $u=\sin(\theta),\enspace du=\cos(\theta)\,d(\theta) \to d(\theta)=\frac{1}{\cos(\theta)}du$
which gives an integral of:
$\int 1-u^2 du$ because cos gets canceled out with substituting $d(\theta)$
which comes out to be $u-\frac{1}{3}u^3+c$
now in the reference triangle the Adjacent side $= \sqrt{7}$ Hypotenuse $=x$
Now is this right? This has to be right integration wise because even the answer showed an integral of cos cubed but then it went off and used some integration technique other than what I used which is the only sensible technique to peel off the cos. Their answer gave something retarded like $\frac{(2x^2+7)\sqrt{x^2-7}}{x^2 \times x}$ from the integral of $\cos^3(\theta)$ which came out to be $2+\cos^2(\theta)\sin(\theta)$. Am I going crazy? My method gives a valid solution, correct? and they used an unnecessary method right? I mean I was taught this method way back in calc II and I am going to be very frustrated if for some magical reason its something different.
|
Somehow they got $Ax(x+2)+B(x+2)+C(x^2)=x+7$
That is not even possible unless you factor out an x on the left hand side and divide both sides by x which would give you $\frac{x+7}{x}$ on the right hand side. So either their work is incorrect or I am missing something.
Yes, you're missing the notion of least common denominator, i.e the l.c.m. of the denominators:
$$\frac{A}{x}+\frac{B}{x^2}+\frac{c}{x+2}=\frac{x+7}{x^2(x+2)}$$
Here it is $x^2(x+2)$, and
$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2}= \frac{Ax(x+2)}{x^2(x+2)}+\frac{B(x+2)}{x^2(x+2)}+\frac{Cx^2}{x^2(x+2)},$$
so that multiplying by the denominator, you obtain indeed
$$Ax(x+2)+B(x+2)+Cx^2=x+7.$$
*
*Now setting $x=0$, you get $2B=7$, so $B=\frac72$.
*Next $x=-2$ yields $4C=5$, or $C=\frac54$.
*Finally the coefficient of $x^2$ in the l.h.s. must be $0$, so $A+C=0$, and $A=-\frac 54$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all triples $(p,x,y)$ such that $ p^x=y^4+4$ Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x = (y^2+2y+2)(y^2-2y+2)$
Then: $p^k = y^2+2y+2$ and $p^j= y^2-2y+2$
The solutions I saw were different from here and they just wanted to prove that $(5,1,1)$ was the only solution. I´ll just ask what I didn't understand:
1)
Therefore, $p^k = (y+1)^2+1$ and $p^j= (y-1)^2+1$
And then: $(y+1)^2 \equiv (y-1)^2 \equiv -1 \pmod p$
So here $-1$ is a quadratic residue, and therefore $p= 4n+1$. I understood his solution until this part.
*
*I know that $p= 4n+1$ because I saw that was one of the properties of quadratic residues, but would like to see a proof of it if possible.
*He then says that by$\mod 8$ , $ k$ and $j$ had different parity, so $x$ was odd.
Then he assumes that $p, k >1$ and says:
*
*If $k=2m$, then: $(y+1)^2= (p^m+1)(p^m -1)$ or if $j= 2m$: then $(y-1)^2= (p^m+1)(p^m -1)$ and affirms those equations have no solution, that might be a property but want to know about it anyway.
Then he says there are no solution for $x>1$, then $p = (y^2+2y+2)(y^2-2y+2)$, so $1=y^2-2y+2$ and that was the triple $(5,1,1)$, therefore that is the only solution.
2)
He started off by saying: if $y$ is even then $ p^k \equiv p^j \equiv 2 \pmod 4$ then $a = b =1$ and that is not possible, therefore $y$ is odd.
*
*Since $ 4y = p^{b} - p^{a}$, therefore $ p^{b-a}=5$, $ p^{a}=y$
*$ 2y^{2}+4=p^{b}+p^{a}$, if $ a\neq 0$, then $ y^2 \equiv 3\mod 5$, which is impossible.
I don´t know how you can deduce those things. After that he affirmed that $(5,1,1)$ was the only solution.
3) He showed that $y$ was odd and said:
*
*$gcd (y^2+2y+2, y^2-2y+2) = gcd (y^2+2y+2, 4y) = 1$ (I think this is also a property but I don´t even know it).
And since $y^2+2y+2 > y^2-2y+2$, then $y^2+2y+2=p^x$ and $y^2-2y+2 =1$, so $y=1$ and $(5,1,1)$ was the only solution.
Sorry if I asked many things, I just thought those things may be useful for other problems. I pointed out the things I didn´t understood so that it was easy to see them. Thanks in advance.
| There is a MUCH easier way to finish.
Starting from $p^k=y^2+2y+2$; $p^j=y^2-2y+2$ we note on the one hand, the following inequality $y^2+2y+2 > y^2-2y+2$ (for natural numbers $y$) and so $j<k$, and therefore
$y^2+2y+2$ must be of the form $p^i(y^2-2y+2)$ for some prime and some positive integer $i$.
On the other hand, we note the following: $y^2+2y+2 < 2(y^2-2y+2)$ for all $y \geq 6$.
So we conclude e.g., $y \le 5$ and $p^x \le 5^4+4$. Furthermore using the above reasoning you can conclude $p \le 5$ if $y \in \{2,3,4,5\}$. This leaves us only a very small set of triples to check via brute force.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that $\left|\operatorname{Re}(3+i+2\bar z^2 -iz)\right| \le 6$, when $|z| \le 1$ So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y^2-ix+y)\right|$
$\left|3+2x^2-2y^2+y\right|$
$3+2x^2-2y^2+y$
And then I'm not really sure where to go. Pretty sure I took the wrong approach because not using the initial information that $|z| \le 1$
| You need to prove $\left|3+2x^2-2y^2+y\right|\le 6$ subject to $|z|\le 1 \iff x^2+y^2\le 1$.
Note:
$$\left|3+2x^2-2y^2+y\right|=\left|3+2(x^2+y^2)-4y^2+y\right|\le |5-4y^2+y|=|4y^2-y-5|\le6 \Rightarrow \\
\begin{cases}4y^2-y-5\ge -6\\ 4y^2-y-5\le 6\end{cases} \Rightarrow \begin{cases}4y^2-y+1\ge 0\\ 4y^2-y-11\le0\end{cases} \Rightarrow \begin{cases}y\in(-\infty,+\infty)\\ y\in\left[\frac{1-\sqrt{177}}{8}\approx -1.538,\frac{1+\sqrt{177}}{8}\approx 1.788\right]\end{cases}.$$
Also note:
$$x^2+y^2\le 1 \Rightarrow -1\le y\le 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $
I understand every step until here
$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$
How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
| $$\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\left(\frac{n+1}{n}\right)^n$$
Divide both denominator and numerator by $n^2 $
$$\frac{\frac{(n+1)}{n}\frac{(n+1)}{n}}{\frac{(2n+2)}{n}\frac{(2n+1)}{n}}\cdot\left(\frac{n+1}{n}\right)^n$$
$$\frac{(1+\frac{1}{n})(1+\frac{1}{n})}{(2+\frac{2}{n})(2+\frac{1}{n})}\cdot\left(1+\frac{1}{n}\right)^n$$
$$\lim_{n \to \infty}\frac{(1+\frac{1}{n})(1+\frac{1}{n})}{(2+\frac{2}{n})(2+\frac{1}{n})}=\frac{1\cdot 1}{2 \cdot 2}=\frac{1}{4}$$
and$$\lim_{n \to \infty} (1+\frac{1}{n})^n=e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
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Does $x^2+3x^4+o(x^4)=x^2+o(x^3)$ hold true? My textbook asks if the following holds true or false for $x\rightarrow0$:
$$x^2+3x^4+o(x^4)=x^2+o(x^3)$$
I think I understand what the little-o notation means. For $x\rightarrow x_o$ we have:
$$\lim_{x\rightarrow x_0}{\frac{f(x)}{g(x)}=0}$$
and we say $f(x)=o(g(x))$.
So to find an answer to the problem I evaluated the following limit:
$$\lim_{x\rightarrow0}{\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}}$$
For $x\rightarrow0$, I think we have
$$\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}\sim \frac{x^2+o(x^4)}{x^2+o(x^3)}=\dots$$
Any hints on how to continue? My textbook says the answer should be true. Sorry I couldn't find the appropriate tag for this little-o notation question.
| Note that:
$$\lim_{x\rightarrow0}{\frac{x^2+o(x^4)}{x^2+o(x^3)}}=\lim_{x\rightarrow0}{\frac{x^2[1+o(x^2)]}{x^2[1+o(x)]}}=\lim_{x\rightarrow0}{\frac{1+o(x^2)}{1+o(x)}}=\frac{1+0}{1+0}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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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.