Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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 ;
$f(x) = \begin{cases}\frac{x^2-x-12}{x-4}&x\ne4\\7&x= 4\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
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:
$$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+... | $$\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$$
Edit:
\begin{align}
0 \le \lim_{x \to \infty}\frac{\sqrt{x^3}}{6^x} &\le \lim_{x \to \infty}\frac{x^2}{e^{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
replacing Inequalities I encountered a problem today:
Prove that:
$$\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}$$
I can prove ... | Note that
\begin{align*}
a^3+b^3+c^3 &= \frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\\
&\ge \frac{(a^2+b^2+c^2)^2}{a+b+c}.
\end{align*}
Then
\begin{align*}
\frac{a^3+b^3+c^3}{a^2+b^2+c^2} &\ge \frac{a^2+b^2+c^2}{a+b+c}\\
&\ge \frac{a+b+c}{3},
\end{align*}
given that
\begin{align*}
(a+b+c)^2 \le 3(a^2+b^2+c^2).
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have th... | Another way of stating what everyone has said:
$2x^2 - x= 2(x^2-x+1) + x - 2$
$\Rightarrow 2x^2-x = x-2 \mod x^2-x+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 16,
"answer_id": 9
} |
How to prove that $\sum_{m\leqslant x}\sum_{n\leqslant x}\Big\{\frac{x}{m+n}\Big\}=\Big(2\log2-\frac{\pi^2}{12}\Big)x^2+O(x\log x)$? Prove that $$\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 $x$... | 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}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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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} = \fr... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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 ... | $$\begin{align}\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}&=\log_3{24} - \frac{3\log_3{5}\times \log_3{2}}{\log_3{5}}\\
&=\log_3{24} - 3\log_3 2\\
\end{align}$$
Now you can use the exponent law again on the second term, and then the additive law. Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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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 ... | 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 sco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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 $$\sum^{\infty}_{r=0}\frac{(4r)!}{(4r+4)!}=\sum^{\infty}_{r=0}\frac{1}{(4r+1)... | Another pssible way is to consider partial sums
$$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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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 ... | $A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots+\dfrac{1}{2^{n+1}}$
$A=\dfrac{2^n}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\dfrac{2^{n-2}}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}$
$2A=\dfrac{2^{n+1}}{2^{n+1}}+\dfrac{2^{n}}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\cdots+\dfrac{2}{2^{n+1}}$
Then
$A=2A-A$
$=\left(\dfrac{2^{n+1}}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 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 ... | 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
$$|a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Inverse element in field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ Determine $(2\sqrt{2}-3\sqrt{3}+5)^{-1}$ in $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, which is of degree $4$ over $\mathbb{Q}$ and for $a= \sqrt{2} + \sqrt{3}$ follows $a^4-10a^2+1=0$.
I did calculate inverse elements in field e... | Let $\beta = 2\sqrt{2}-3\sqrt{3}+5$. Then
$(\beta-5)^2 = 35-12\sqrt{6}$ and $((\beta-5)^2-35)^2 = 864$, and so
$$
\beta^4 - 20 \beta^3 + 80 \beta^2 + 200 \beta = 764
$$
Writing this as
$$
\beta(\beta^3 - 20 \beta^2 + 80 \beta + 200) = 764
$$
gives $\beta^{-1} = \frac{1}{764}(\beta^3 - 20 \beta^2 + 80 \beta + 200)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Properties of modular arithmetic clarification The sum of cubes of any three consecutive integers is divisible by $9$. i.e $(n^3 + (n+1)^3 + (n+2)^3)\mod9=0$ for every $n\in\mathbb{Z}$.
They say it is sufficient to check out for values of $n=0,1,2,3,...8$ due to properties of modular arithmetic. Why this is so ?
Thank... | Alternatively:
$n^3+(n+1)^3+(n+2)^3$
$=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8$
$=3n^3+9n^2+15n+9$
$=9(n^2+1)+3n(n^2+5)$
In fact, you only need to check for $n=1;n=2;n=3$.
*
*If $n=1$ or $n \equiv 1 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$
*If $n=2$ or $n \equiv 2 \pmod{3}$ then $n^2\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\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)$$
From here I just ... | Note:
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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\... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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{... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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 tr... | This is the route I usually take but....
$$\begin{align}
x_{n+1}&=5^{n-1} x_n+3^n \\
&=5^{n-1}(5^{n-2} x_{n-1}+3^{n-1})+3^n \\
&\quad = 5^{2n-(1+2)}x_{n-1}+(3 \cdot 5)^{n-1} +3^n \\
&=5^{2n-(1+2)}(5^{n-3}x_{n-2}+3^{n-2})+(3\cdot5)^{n-1}+3^n \\
&\quad =5^{3n-(1+2+3)}x_{n-2}+5^{2n-(1+2)}3^{n-2}+5^{n-1}3^{n-1}+3^n \\
&\vd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to show that this integral is correct? How can one show that
$$\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}}$$
assume $\alpha\ge1$.
I can't see how to simplify $... | Using
$$\tan(x) \, \tan(x/2) = \frac{2 \, \sin^{2}(x/2)}{\cos(x)}$$
and
$$\sin(x) \, \tan(x/2) = 2 \, \sin^{2}(x/2)$$
then
\begin{align}
I &= \int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\tan(x)\tan\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx \\
&= 2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
How to solve the following trigonometrical equation? I have the following equations
\begin{align*}
R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\
R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}).
\end{align*}
From these equations how can I obtain the following solution
$$\omega T_2=\p... | 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
$$\sin \theta = \sin(\varphi+\psi) = \cos\varphi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find generating function of series $a_{n} = 2^{n} + 3^{n} $ Find generating function of series $a_{n} = 2^{n} + 3^{n} $
I'm having a problem with this because at first i have to find recursive equation for $a_{n}$ . I find that $a_{0} = 2$ , $a_{1} = 5$ but how do I find the rest? Also I can't simply add generating fun... | Hint:
$$\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... | {
"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
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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 t... | $$I=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}}, dx$$
$1)sin(x)+cos(x)=Rcos(x-a)\therefore R=\sqrt{2}cos(x-\frac{\pi}{4})$
$$sin(x)+cos(x)=\sqrt{2}cos(x-\frac{\pi}{4})$$
do the same for $cos(x)-sin(x)$, so
$$cos(x)-sin(x)=\sqrt{2}cos(x+\frac{\pi}{4})$$
$$\therefore\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Given matrix $A$, find a matrix $S$ such that result of $S^{-1} \circ A \circ S$ is a diagonal matrix
Given is matrix $A=\begin{pmatrix} 3 & 0 & 7\\ 0 & 1 & 0\\ 7 & 0 & 3
\end{pmatrix}$, find a matrix $S$ such that $D=S^{-1} \circ A \circ S$
where $D$ is a diagonal matrix.
So I'm not sure how this is supposed to ... | Find first the eigenvalues of $A$, these are the roots of the characteristic polynomial of $A$,
$$
(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 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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 valu... | 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 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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 ... | 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:
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$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... | Note that $$\bigg(x+\frac 1x\bigg)^2=x^2+\frac{1}{x^2}+2\to x+\frac 1x=\sqrt{a+2}$$
$$\bigg(x+\frac{1}{x}\bigg)\bigg(x^2+\frac{1}{x^2}\bigg)=x^3+\frac{1}{x^3}+x+\frac 1x\to x^3+\frac{1}{x^3}=(a-1)\sqrt{a+2}$$
$$\bigg(x^2+\frac{1}{x^2}\bigg)^2=x^4+\frac{1}{x^4}+2\to x^4+\frac{1}{x^4}=a^2-2$$
$$\bigg(x^4+\frac{1}{x^4}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Evaluate: $\lim_{x\to 0}\frac{\sin^3x-x^3\operatorname{sgn}\left(1-\left[\frac{x}{\sin^{-1}x}\right]\right)}{x\tan^2 x\sin(\pi\cos x)}$ $$\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)}$$
(Note that here $\operatorname{sgn}$ denotes the Signum ... | We have that
$$\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-\... | {
"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
} |
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_... | 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... | {
"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 $c(n)$ so... | 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 g... | {
"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 ... | 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
$$-3 - a^3 b + 2 a b^2=0 \qq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $S$ in term of $k$ (telescoping sums) Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.
Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.
I've tried to write the terms of $S$ in the follo... | Note that your sum is $$\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! $$
An equivalent statement is this:
Take a Sieve of Eratosthenes-like list of any fixed width $n$ and sieve out multiples of th... | Your conjecture does not hold for $n=8, k=25$ as all $8$ numbers
$$(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$$have factors less than ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FO... | Alternative method:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Express the integrand in the form $\lfrac{(2ax+b)·(x^2-3x+2)+(2cx+d)}{\sqrt{x^2-3x+2}}$ for some constants $a,b,c,d$.
Then split it into $( a(2x-3) + (3a+b) ) · \sqrt{x^2-3x+2} + \lfrac{c(2x-3)+(3c+d)}{\sqrt{x^2-3x+2}}$, so that as a sum of four terms the fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be t... | dont know of an elegant method:
on LHS substitute $c = a - b$
$LHS = \sin^2 a + \sin^2 b + (\sin a\cos b - \cos a \sin b)^2 $
$= \sin^2 a + \sin^2 b + \sin^2a \cos^2b + \cos^2a \sin^2b - 2\cos a\cos b \sin a \sin b$
$ = \sin^2 a + \sin^2 b + (1 - \cos^2a)\cos^2b + \cos^2a(1-\cos^2b)- 2\cos a\cos b \sin a \sin b $
$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A problem in trigonometry If $\tan(x+y) =a+b$
and, $\tan(x-y)=a-b$
Then prove that:$a \tan(x) -b \tan(y)=a^{2}-b^{2}$
I have used the formulae for $\tan(x+y)$ and $\tan(x-y)$ and then cross multiplied and then found $a$ and $b$ individually and then tried to put the values of $a$ and $b$ in "$a \tan(x) -b \tan(y)$"(l... | We are given
$$
\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{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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
$$
\begin{align}
\frac 81(\f... | We have that
$$\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... | 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
$$V=2\cdot \frac{b}{a}\cdot \frac{-1}2\cdot \int_{t=0}^{\pi}\int_{m=a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify the polynomial $xy(x+y)+(x+y)+(x+y)^2=13xy$ to the form $y^2=4x^3-g_{2}x-g_{3}$
I try to simplify a polynomial to the form: $y^2=4x^3-g_{2}x-g_{3}$, which is the elliptic curves. And the polynomial is $xy(x+y)+(x+y)+(x+y)^2=13xy$.
I try to let the $u=x+y$ and $v=x-y$, then I get the $u^3-uv^2+4u-9u^2+13v^2=... | Are you certain that you have stated the problem correctly? As currently stated
$$ 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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} = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
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... |
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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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}}$$ which seems to give a homogenous equation. (There is no answer given for th... | I made a summary with the most important steps:
STEP $1$
$$\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}$$
STEP... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hint Taylor Series $x^{2}\ln(x)$ about $a=1$ Find the taylor series of $g(x):= x^{2}\ln(x)$ about $a=1$
Idea: I would have thought we should find the Taylor series of $\ln(x)$ about $a=1$, which is
$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k!}(x-1)^{k}$ and then I would just multiply that with $x^{2}$. So, we'd get:
$\sum... | Your answer is no completely correct. As you would like the expansion around $a=1$, you should also expand $x^2$ around this point. In fact, you can easily show that
$$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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 integ... | 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 equ... | {
"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
$$
\begin{pmatrix}
x(t) \\
y(t... | I think the key here is to rewrite the linear system as the following matrix equation
$$\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}}}.$$
The characteristic equation is $\lambda^2+4=0,$ so you're correc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the prime factorization of $2^{22}+1$? The problem is
Find the prime factorization of $2^{22}+1$
I have a solution but I think there must be some better ways:
My Solution:
$2^{22}+1 = (2^{22} + 2 \cdot 2^{11} +1) - 2 \cdot 2^{11} = (2^{11}+1)^2 - 2^{12}$
and we can factor as
$(2^{11}+1-2^6)(2^11+1+2^6) = (204... | Suppose $p$ is a prime factor of $2^{22}+1$ greater than $5$. Then $2^{22}\equiv -1 \pmod{p}$. So we know the order of $2$ modulo $p$ is $44$, and thus $44 \mid p-1$. We conclude that $p = 44n+1$ for some $n$.
Using your factorization, we have that $\sqrt{2113} = 45.96\ldots$. There are no primes of the form $44n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Rationalize this I'm having trouble rationalizing the denominator of this fraction. Would you kindly explain this to a fellow self-learning math student?
$$\frac{10}{\sqrt[4]{3}-1}$$ knowing that $a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}\cdot b+a^{n-3}\cdot b^2+...+a\cdot b^{n-2}+b^{n-1}\right)$
$a,b\in \mathbb{R... | Couldn't see how the tip helps.
$$
\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}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding limit of multivariable function. I have a question that might be silly but I need to understand these kind of problems. So i have this limit:$$\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\thet... | 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) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
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 s... | 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 nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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}{ccc... | 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:
$$
\pmatrix{
1 & \alpha & -1 & 2 \\
2 & -1 & \alpha & 5 \\
1 & 10 & -6 & 1 \\
} \to
\pmatrix{
1 & \alpha & -1 & 2 \\
0 & -1 - 2\alpha & 2+\alpha & 1 \\
0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem with Complex Variables
Solve the equation $z^2 + z + 1 = 0$ for $z = (x,y)$ by writing $(x,y) (x,y) + (x,y) + (1,0) = (0,0)$ and then solving a pair of simultaneous equations in $x$ and $y$.
My main difficulty is finding the answer is using the suggestion in that book that there isn't a real solution implies ... | Note that, for a complex $z = (x,y)$, $x$ and $y$ are both reals. The suggestion is not to require $z$ to be real. We need to find $z$, which is a complex number, and, being a complex number, it has both its real and imaginary components real numbers. Lack of a real solution of $x^2+x+1=0$ suggests that $y \neq 0$.
An ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to derive this factorization $ax^2 + bx + c = a(x − x_1)(x − x_2) $? I faced this factorization formula $$ax^2 + bx + c = a(x − x_1)(x − x_2)$$ where $x_1$ is the first solution and $x_2$ is the second one. But I don't understand how the formula gets derived... Could anybody explain it to me? Thanks!
I know I can s... | You can also complete the square. Assuming $a \neq 0$ (necessary to be a quadratic):
\begin{align*}
ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) \\
&= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{c}{a} - \frac{b^2}{4a^2}\right) \\
&= a\left(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Finding Lyapunov function for a stable equilibruim of a non linear system Given the following system:
$$
\left\{
\begin{array}{c}
\dot x=y-x^2-x \\
\dot y=3x-x^2-y
\end{array}
\right.
$$
I need to find the equilibrium points, and if stable, to find a Lyapunov function.
I have found two equilibrium points: $(0,0), (... | Considering the surroundings for $(1,2)$ we have
$$
\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{arr... | {
"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 $... | 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 $43$:
$$(x - (2^{-1})y)^2 + 3(4)^{-1}y^2 \equiv 0 \pmod{43} \quad \iff \quad (x+21y)^2 + 33y^2 \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the parity-check matrix of a generator matrix in $\Bbb F_3$. I am currently looking into linear codes and thought I had understood building a parity-check matrix for a given generator matrix - but the presented solution is anything but what I have.
Given a generator matrix $G= \begin{pmatrix}I_3 \\ A\end{pmatr... | $$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\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 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 $$\cot x =t $$ $$I=\int_0^{\infty} \operatorname{arccot}^2 ... | The substitution $\sin(x) = \sqrt{t}$ leads to the expression
$$I = \frac{1}{2} \int \limits_0^1 \frac{t^{-1/4} \arcsin^2 (\sqrt{t})}{\sqrt{1-t}} \, \mathrm{d} t \, . $$
Now you can use the power series for $\arcsin^2$ (see for example this question) and integrate term by term (monotone convergence). Using the beta fun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
My thought on this are as follows:
I've solved a similar problem where I had to find the inverse of:
$$
g(x) = \sqrt[3]{x+\sqrt{x^2-1}} + \sqrt[3]{x-\sqrt{x^2-1}}
$$
So let $a = \sqrt[3]{x+\sqrt{x^2-1}}$ and ... | You want to solve $x^3 + 3x - 2y = 0$ as $x = f(y)$. The canonical way to do this is to write $x = a + b$, obtaining
$$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$$
Multiplyin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate
$\lim_{n \rightarrow \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know that $1 - \frac {2}{k(k+1)} = \fr... | We have that
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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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 h... | You can directly use the formula $$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$
$$x=\frac{2}{\sqrt{5}},y=\frac{3}{\sqrt{10}}$$
$$=\cos^{-1}\left(\left(\frac{2}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)-\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
integrate $2\sqrt{1-x^2/4-y^2/9}$ I am trying to find the volume of the ellipsoid $E$ given by $x^2/4 + y^2/9 + z^2 \leq 1$ by computing $\iiint_E dV$
I ended up with $$\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$$ is the same as $$\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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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 u... | 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 app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
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 i... | 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
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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$$
Could some help me how... | Alternatively, write it as:
$$\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$$
Note that the graph of $f(a)=\left(a+\frac12\cos x\right)^2-(\cos ^2x +\cos x+\frac 14\cos ^2x)$ is an opening-up parabola with the ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Show that $f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$ is strictly increasing I want to show that the function $f:\mathbb N \to \mathbb Q$ defined by$$f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$$
is strictly increasing. Could I have a hint on how to go about this? I have tried playing around with $f(n+1)-f(n)$ but it's all so messy ... | I will supplement @Markus Scheuer's answer by providing a proof of the claim (Corollary 3 of Monotonicity of Riemann Sums by D. Borwein, J. M. Borwein and B. Sims) in a slightly generalized setting.
Lemma. Let $f : [0, 1] \to \mathbb{R}$ and $\lambda \in \mathbb{R}$. Suppose that
$$F(x) = \lambda f(x) + (1-\lambda) ... | {
"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
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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 ... | 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
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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 can... | 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",
"answer_count": 2,
"answer_id": 0
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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 ... | $$\dfrac{x^{10}+1}{(x^2-1)(x^2+x+1)}=(x-1)(x^5-x^3+x^2+x-1)+\dfrac{{x^3+x}}{(x^2-1)(x^2+x+1)}$$
| {
"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}\ove... | $B=\ln(I+A)$, 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858233",
"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
$$
\begin{pmatrix}
0 & -1 & 1 \\
3 & 1 & 2 \\
... | Let's rewrite it as following $$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}$$since for any $v\in\Bbb R^3$ we have $v=a_1$ , $v=a_1\begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Find the second degree Taylor polynomial of $f(x,y)=e^{-x^2-y^2}cos(xy)$ at $x_0=0$, $y_0=0$. I would like to either confirm my solution is correct, or find the error in it. I used the following MATLAB code to try to check my answer, but the solution it gave differs from mine.
f = exp(-(x^2+y^2))cos(xy);
taylor(f,[x,y... | I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^{-x^2-y^2}\cos(xy) = (1-x^2-y^2+\cdots)(1-\frac{1}{2}(xy)^2+ \cdots )$$
But, $(xy)^2$ is order $4$ so the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the indefinite integral of $\frac{1}{x^{2/3}(1+x^{2/3})}$?
$$\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... | 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 $... | 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
$$ \Big(1 + \frac1t\Big) \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m}t) \Big(1 + \frac{x^{2m}}t\Big) =
\sum_{n=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
T... |
\begin{align}
p_1&=(\tfrac{\sqrt2}2,\tfrac{\sqrt2}2)
,\\
p_2&=(0,1)
,\\
p_3&=(-\tfrac{\sqrt2}2,\tfrac{\sqrt2}2)
,\\\ \dots
\end{align}
\begin{align}
S&=
(p_{1x}-p_{3x})^2+\tfrac32(p_{1x}-p_{3x})(p_{2y}-p_{1y})
=\frac{3\sqrt{2}+1}{2}
,\\
&=
\frac{3k\sqrt{2}+k}{2k}
,\\
a&=3k,\quad b=2,\quad c=k,\quad d=2k
,\\
a+b+c+d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| We have
$$\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}\... | {
"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^... | 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^... | {
"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
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $... | Hint: Use the factorisation formula $$\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}{\si... | {
"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... | 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 a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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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( \fra... | 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\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870680",
"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 clea... | 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
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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... | 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
$$t-1\geq\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Evaluate $\int_m(\nabla \times F)\cdot NdS$ Evaluate $\int_m(\nabla \times F)\cdot N\,dS$ where $F=(e^x+z^2,\sin(yz),x^3)$ and $M=\{(x,y,z):y=\frac{x^2}{2}+z^2-4,y\leq 0\}$ and $N$ points outwards
So my plan is:
*
*find the unit normal of $M$
*find the area element ($dS$) of $M$ by projecting $M$ onto $XZ$ plane
... | HINT
A possible way, since $div(\nabla \times F)=0$, is note that
$$\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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 n... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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:
$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac... | Write
$$
I(t)=\int_0^{\frac{1}{\sqrt{2}}} \frac{2\arcsin(tx)}{\sqrt{\frac{1}{2}-x^2}} \, {\rm d}x
$$
and calculate
\begin{align}
I'(t) &= \int_0^{\frac{1}{\sqrt{2}}} \frac{2x}{\sqrt{\left(\frac{1}{2}-x^2\right)\left(1-(tx)^2\right)}} \, {\rm d}x \\
&= \frac{\log\left(\sqrt{2}+t\right)-\log\left(\sqrt{2}-t\right)}{t} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
If $a_1, \ldots ,a_9$ are in harmonic progression, then find the value of the determinant If $a_1, \ldots ,a_9$ are in harmonic progression ,then find the value of the determinant
$$
\begin{vmatrix}
a_1 &a_2&a_3\\
5&4&a_6\\
a_7&a_8&a_9\\
\end{vmatrix}
$$
I calculated the terms as $a_1=\frac{20}{1},a_2=\frac{20}{... | after taking $20^3$ common and performing the following row operations
replace $R_2$ with $R_2-\frac{R1}{4}$
replace $R_3$ with $R_3-\frac{R1}{7}$
I got this
\begin{vmatrix}
1 & 1/2& 1/3\\
0 & 3/40& 1/12\\
0 & 3/56& 4/63\\
\end{vmatrix}
which didn't take much time now calculate the determinant about column 1
I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $a_n = \frac{n}{2n+1}+\frac{1}{n^3}$ is a Cauchy Sequence I want to show that
$$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}... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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.
| $$\begin{array}{rcl}
4x^2 + 12x + \dfrac{12}{x} + \dfrac{4}{x^2} &=& 47 \\
4\left(x^2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 47 \\
4\left(x^2 + 2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 55 \\
4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) &=& 55 \\
4\left(x + \dfrac1x\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ 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}$$
Solve for $x$
\begin{align*}
x&=\frac{(3a^2+2)\pm\sqrt{(3a^2+2)^2-12a(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 6
} |
Integrate $\int \frac{x^3 e^x \:dx}{\sqrt{x^2-1}}$ Integrate $$\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} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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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... | $$\begin{align*}xy-\sqrt{(1-x^2)(1-y^2)}&\le 1\\
\Leftrightarrow
(xy)^2&\le 1+(1-x^2-y^2+(xy)^2 )+2\sqrt{1-x^2)(1-y^2)}
\end{align*}$$
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 ... | $$−64t^6+80t^4−20t^2+1=(1-4t^2)(1-16t^2+16t^4)$$
| {
"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 multipl... | 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, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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 onl... | 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 p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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
$$(x-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^{4-k}x^... | I believe you are over-complicating it. The given system tells you that the quartic polynomial $p(n)=an+bn^2+cn^3+dn^4$ attains the values $2,5,6,1$ at $n=1,2,3,4$. By Lagrange interpolation such polynomial is
$$ p(n) = \frac{1}{24}\left(18n+37n^2-6n^3-n^4\right) $$
hence
$$ C_6 = \frac{p(6)}{6} = \color{red}{-8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$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\c... | 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
} |
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 t... | $$I=\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{x\sqrt{\left(2x+1\right)^2+3}}\,\mathrm{d}x$$
Substitute $u=2x+1$
$$I=2{\displaystyle\int}\dfrac{1}{\left(u-1\right)\sqrt{u^2+3}}\,\mathrm{d}u$$
Substitute $u=\sqrt{3}\tan\left(v\right)$
$$I=2{\displaystyle\int}\dfrac{\sqrt{3}\sec^2\left(v\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Maclaurin Series Expansion nth term expression My task is to show that the Maclaurin series of $\ln\frac{x+1}{x-1}$ is given as $\sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$
My approach is using the standard expansion of $\ln(x+1)$ to get $x-\frac{x^2}{2}+\frac{x^3}{3}$
I got $$\ln(x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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$
obvious... |
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 de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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... | 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,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
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... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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 } ... | $$\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\left(\frac{n+1}{n}\right)^n$$
Divide both denominator and numerator by $n^2 $
$$\frac{\frac{(n+1)}{n}\frac{(n+1)}{n}}{\frac{(2n+2)}{n}\frac{(2n+1)}{n}}\cdot\left(\frac{n+1}{n}\right)^n$$
$$\frac{(1+\frac{1}{n})(1+\frac{1}{n})}{(2+\frac{2}{n})(2+\frac{1}{n})}\cdot\left(1+\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Does $x^2+3x^4+o(x^4)=x^2+o(x^3)$ hold true? My textbook asks if the following holds true or false for $x\rightarrow0$:
$$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}$$
and we say $f(x)=o(g(x))$.
So to ... | Note that:
$$\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",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.