Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$ How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$
I tried to factor and I got $2\cos^4(x)+(-2\sin^2(x)-3)(\cos^4(x)+\sin^4(x))$ but that doesn't lead me to my goal.
I also tried to write all the cosines in terms of sines to have: $-3\sin^6(x)-9\sin^2(x)-2-3\sin^4(x)$
But I don't see how to continue
Any hint is welcome! thnxx
| Apply $a^2+b^2=(a+b)^2-2ab$ in $$\sin^4x+\cos^4x=(\sin^2x)^2+(\cos^2x)^2$$
and $a^3+b^3=(a+b)^3-3ab(a+b)$ in $$\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^2x)^3$$
Finally the re-arrange the terms
| {
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"url": "https://math.stackexchange.com/questions/742358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Visually deceptive "proofs" which are mathematically wrong Related: Visually stunning math concepts which are easy to explain
Beside the wonderful examples above, there should also be counterexamples, where visually intuitive demonstrations are actually wrong. (e.g. missing square puzzle)
Do you know the other examples?
| This is my favorite.
\begin{align}-20 &= -20\\
16 - 16 - 20 &= 25 - 25 - 20\\
16 - 36 &= 25 - 45\\
16 - 36 + \frac{81}{4} &= 25 - 45 + \frac{81}{4}\\
\left(4 - \frac{9}{2}\right)^2 &= \left(5 - \frac{9}{2}\right)^2\\
4 - \frac{9}{2} &= 5 - \frac{9}{2}\\
4 &= 5
\end{align}
You can generalize it to get any $a=b$ that you'd like this way:
\begin{align}-ab&=-ab\\
a^2 - a^2 - ab &= b^2 - b^2 - ab\\
a^2 - a(a + b) &= b^2 -b(a+b)\\
a^2 - a(a + b) + \frac{a + b}{2} &= b^2 -b(a+b) + \frac{a + b}{2}\\
\left(a - \frac{a+b}{2}\right)^2 &= \left(b - \frac{a+b}{2}\right)^2\\
a - \frac{a+b}{2} &= b - \frac{a+b}{2}\\
a &= b\\
\end{align}
It's beautiful because visually the "error" is obvious in the line $\left(4 - \frac{9}{2}\right)^2 = \left(5 - \frac{9}{2}\right)^2$, leading the observer to investigate the reverse FOIL process from the step before, even though this line is valid. I think part of the problem also stems from the fact that grade school / high school math education for the average person teaches there's only one "right" way to work problems and you always simplify, so most people are already confused by the un-simplifying process leading up to this point.
I've found that the number of people who can find the error unaided is something less than 1 in 4. Disappointingly, I've had several people tell me the problem stems from the fact that I started with negative numbers. :-(
Solution
When working with variables, people often remember that $c^2 = d^2 \implies c = \pm d$, but forget that when working with concrete values because the tendency to simplify everything leads them to turn squares of negatives into squares of positives before applying the square root. The number of people that I've shown this to who can find the error is a small sample size, but I've found some people can carefully evaluate each line and find the error, and then can't explain it even after they've correctly evaluated $\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/743067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "137",
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Find the length of... Find the length of $\overline{AD}$ knowing that it is divided into three equal parts by to tangent circles with radius respectively $3\sqrt{3}$ and $\sqrt{3}$ . Here's the graph: so the segments $\overline{AB}$, $\overline{BC}$ and $\overline{BC}$ are equal.
Thank you very much!
|
Let $AB=BC=CD=x$.
$O1M$, $O2E$ $⊥AD$. (Perpendicular bisectors)
$O1MEO2$ - rectangular trapezoid.
$AM=MB$ and $CE=ED$. Hence $ME = 2x$.
From right triangles $O1AM$ and $O2ED$ we got $O1M = \sqrt{R^2 - \frac{x^2}{4}}$ and $O2E = \sqrt{r^2 - \frac{x^2}{4}}$
Use Pythagorean theorem for triangle $O1FO2$ ($O2F ⊥ O1M$):
$(\sqrt{R^2 - \frac{x^2}{4}} - \sqrt{r^2 - \frac{x^2}{4}})^{2} + (2x)^2 = (R+r)^2$
$\equiv$
$R^2 - \frac{x^2}{4} + r^2 - \frac{x^2}{4} - 2\sqrt{R^2 - \frac{x^2}{4}} \sqrt{r^2 - \frac{x^2}{4}} + 4x^2 = R^2 + 2Rr+r^2$
$\equiv$
$\frac{7}{2}x^2-2Rr=2\sqrt{R^2 - \frac{x^2}{4}} \sqrt{r^2 - \frac{x^2}{4}}$
$\equiv$
$\frac{49}{4}x^4-14x^2Rr+4R^2r^2=4R^2r^2+\frac{x^4}{4}-x^2(R^2+r^2)$
$\equiv$
$12x^4+x^2(R^2+r^2-14Rr)=0$
Where we get $x=\frac{1}{2\sqrt{3}} \sqrt{14Rr-R^2-r^2} = 2\sqrt{2}$
By condition $AD=3x$. So $AD=6\sqrt{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $Y_1, Y_2,\ldots,Y_n$ denote a random sample from the uniform distrib... Help find finding $ \text{Var}\left[\hat{\theta}_{2}\right]$ Let $Y_1, Y_2,\ldots,Y_n$ denote a random sample from the uniform distribution on the interval $(θ, θ + 1)$. Let $$ \hat{\theta}_2 = Y_{(n)} - \frac{n}{n+1}$$
Find the efficiency of $θ^1$ relative to $θ^2$
We have $Y_i\sim\mathcal{U}(\theta,\theta+1)$ and CDF of $Y_i$ based on Wikipedia
$$
G_{Y_i}(y)=\Pr[Y_i\le y]=\frac{y-\theta}{\theta+1-\theta}=y-\theta.
$$
Here, $Y_{(n)}$ is $n$-th order statistics. Therefore, $Y_{(n)}=\max[Y_1,\ldots, Y_n]$. Note that $Y_{(n)}\le y$ equivalence to $Y_i\le y$ for $i=1,2,\ldots,n$. Hence, for $\theta< y<\theta+1$, the fact that $Y_1,Y_2,\ldots, Y_n$ are i.i.d. implies
$$
G_{Y_{(n)}}(y)=\Pr[Y_{(n)}\le y]=\Pr[Y_1\le y,Y_2\le y,\ldots, Y_n\le y]=(\Pr[Y_i\le y])^n=\left(y-\theta\right)^{n}.
$$
The PDF of $Y_{(n)}$ is
$$
g_{Y_{(n)}}(y)=\frac{d}{dy}G_{Y_{(n)}}(y)=\frac{d}{dy}(y-\theta)^n=n(y-\theta)^{n-1}.
$$
The expected value of $Y_{(n)}$ is
$$
\begin{align}
\text{E}\left[Y_{(n)}\right]&=\int_{y=\theta}^{\theta+1}yg_{Y_{(n)}}(y)\ dy\\
&=\int_{y=\theta}^{\theta+1}yn(y-\theta)^{n-1}\ dy\\
&=n\int_{y=\theta}^{\theta+1}y(y-\theta)^{n-1}\ dy.
\end{align}
$$
I have trouble finding $$
\text{Var}\left[\hat{\theta}_{2}\right]=\text{Var}\left[Y_{(n)}-\frac{n}{n+1}\right]=\text{Var}\left[Y_{(n)}\right]=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2.
$$
I found that
$$
\begin{align}
\text{E}\left[Y_{(n)}\right]&=n\left[\frac{y(y-\theta)^n}{n+1}-\frac{\theta(y-\theta)^n}{n(n+1)}\right]_{y=\theta}^{\theta+1}\\
&=\frac{n(\theta+1)}{n+1}+\frac{\theta}{n+1}\\
&=\theta+\frac{n}{n+1}.
\end{align}
$$
The way I calculate $E(Y_{(n)}^2)$ is the following:
$$E(Y_{(n)}^2) = ny^2(y-\theta)^{n-1} = n\left[\left.y^2\frac{(y-\theta)^n}{n} \right|_\theta^{\theta+1} - \frac{2}{n} \int_\theta^{\theta+1} y(y-\theta)^n \,dy\right]$$
$$= \left.(\theta+1)^2 - 2\left(y\frac{(y-\theta)^{n+1}}{n+1} \right|_\theta^{\theta+1} - \int_\theta^{\theta+1} \frac{(y-\theta)^{n+1}}{n+1} dy\right) = (\theta+1)^2 -2 \left(\frac{\theta+1}{n+1} - \left.\frac{(y-\theta)^{n+2}}{(n+1)(n+2)}\right|_\theta^{\theta+1}\right)$$
$$= (\theta+1)^2- 2\frac{\theta +1}{n+1} - \frac{1}{(n+1)(n+2)}$$
Then I use $E(Y_{(n)}^2) - E(Y_{(n)})^2$, based on wolframalpha which gives me the following result ...please go here ...which is different from the correct answer $\text {Var} [\hat{\theta}_2]= V(Y(n))=\frac{n}{(n+2)(n+1)^2}$....Could anyone please check why?
| Variance can be evaluated as follows
$$
\text{Var}\left[\hat{\theta}_{2}\right]=\text{Var}\left[Y_{(n)}-\frac{n}{n+1}\right]=\text{Var}\left[Y_{(n)}\right]=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2.
$$
First, we calculate $\text{E}\left[Y_{(n)}^2\right]$. Using the result from here, we obtain
$$
\text{E}\left[Y_{(n)}^2\right]=n\int_{y=\theta}^{\theta+1}y^2(y-\theta)^{n-1}\ dy.
$$
Using IBP, let $u=y^2\Rightarrow du=2y\ dy$ and $dv=(y-\theta)^{n-1}\ dy\Rightarrow v=\dfrac{(y-\theta)^n}{n}$. Hence
$$
\begin{align}
n\int_{y=\theta}^{\theta+1}y^2(y-\theta)^{n-1}\ dy&=n\left[\left.\dfrac{y^2(y-\theta)^n}{n}\right|_{y=\theta}^{\theta+1}-\int_{y=\theta}^{\theta+1}\dfrac{(y-\theta)^n}{n} 2y\ dy\right]\\
&=(\theta+1)^2-2\int_{y=\theta}^{\theta+1}y(y-\theta)^{n}\ dy\;\;\;\Rightarrow\;\;\;\text{again we use IBP here.}\\
&=(\theta+1)^2-2\left[\left.\dfrac{y(y-\theta)^{n+1}}{n+1}\right|_{y=\theta}^{\theta+1}-\int_{y=\theta}^{\theta+1}\dfrac{(y-\theta)^{n+2}}{n+1} dy\right]\\
&=(\theta+1)^2-2\left[\dfrac{\theta+1}{n+1}-\left.\dfrac{(y-\theta)^{n+2}}{(n+1)(n+2)}\right|_{y=\theta}^{\theta+1}\right]\\
\text{E}\left[Y_{(n)}^2\right]&=(\theta+1)^2-\dfrac{2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}.
\end{align}
$$
You already have
$$
\text{E}\left[Y_{(n)}\right]=\theta+\frac{n}{n+1}.
$$
Thus
$$
\begin{align}
\text{Var}\left[\hat{\theta}_{2}\right]&=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2\\
&=(\theta+1)^2-\dfrac{2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}-\left(\theta+\frac{n}{n+1}\right)^2\\
&=\theta^2+2\theta+1-\dfrac{2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}-\theta^2-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{(2\theta+1)(n+1)-2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{2n\theta+2\theta+n+1-2\theta-2}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{2n\theta+n-1}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{n-1}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{(n-1)(n+1)+2}{(n+1)(n+2)}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{n(n+1)}{(n+1)(n+2)}-\frac{n^2}{(n+1)^2}\\
&=\dfrac{n}{n+2}-\frac{n^2}{(n+1)^2}\\
&=\frac{n(n+1)^2-n^2(n+2)}{(n+2)(n+1)^2}\\
&=\frac{n}{(n+2)(n+1)^2}.
\end{align}
$$
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
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"url": "https://math.stackexchange.com/questions/750703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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The number of primes in the factorization of $N!$ Is there an approximation to the number of primes in the factorization of $N!$?
For example:
*
*For $N=10$, this number is $15$.
*For $N=100$, this number is $239$.
*For $N=1000$, this number is $2877$.
*For $N=10000$, this number is $31985$.
*For $N=100000$, this number is $343614$.
*For $N=1000000$, this number is $3626619$.
There seem to be a gradual ascendant towards $4N$, but has that been proved as an upper limit?
| This does not completely answer your question but is helpful to know and the argument can be made completely rigorous.
The highest power of a prime $p$ dividing $N!$ is
$$\left\lfloor \dfrac{N}p \right\rfloor + \left\lfloor \dfrac{N}{p^2} \right\rfloor + \cdots \sim \dfrac{N}{p-1}$$ and the number of primes less than $N$ is $\sim \dfrac{N}{\log(N)}$ and $\displaystyle \sum_{p \leq N} \dfrac1p \sim \ln(\ln(N))$
and hence
$$\sum_{p \leq N} \dfrac{N}{p-1} \sim N \ln(\ln(N))$$
Hence, my possible guess is $\sim N \ln(\ln(N))$.
For instance, if you take $N$ to be $10^m$, we then have the highest power of $2$ dividing $10^m$ as $$\dfrac{N}2 + \dfrac{N}{2^2} + \cdots \dfrac{N}{2^m} = \dfrac{N}2 \dfrac{1-1/2^m}{1-1/2} = N - \dfrac{N}{2^m}$$
Similarly, the highest power of $5$ dividing $10^m$ as $$\dfrac{N}5 + \dfrac{N}{5^2} + \cdots \dfrac{N}{5^m} = \dfrac{N}5 \dfrac{1-1/5^m}{1-1/5} = \dfrac{N}4 - \dfrac{N}{4 \cdot 5^m}$$
Hence, the number of prime factors of $N = 10^m$ is $$\dfrac{5N}4 - 5^m - 2^{m-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove the inequality.Let a, b and c be nonnegative real numbers. Let $a$, $b$ and $c$ be nonnegative real numbers. Prove that $a^4+b^4+c^2\ge 8^{½}abc$
| Another way:
$$
a^4+b^4+\frac{c^2}{2}+\frac{c^2}{2}\geq 4\left(a^4b^4\frac{c^2}{2}\frac{c^2}{2}\right)^\frac 14=2^\frac 32 abc.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/753719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How many 6 letter words can be made with these conditions? The letters that can be used are A, I, L, S, T.
The word must start and end with a consonant. Exactly two vowels must be used. The vowels can't be adjacent.
| There are $3$ ways to choose the first consonant, and $3$ ways to choose the last one.
Of the four remaining letters, exactly two must be vowels. There are $$\binom{4}{2}-3=6-3=3$$ ways to choose which spots the vowels can be in, since they cannot be adjacent. Then, there are $2$ choices for the first vowel, and $2$ choices for the second.
Finally, there are $3$ ways to choose each of the remaining two consonants.
Multiplying all of these numbers together, we find the total number of $6$-letter words that fit all the constraints: $$3 \cdot 3 \cdot \left(\binom{4}{2}-3\right) \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 2^2 \cdot 3^5 = \boxed{972}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754888",
"timestamp": "2023-03-29T00:00:00",
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Lagrange multipliers from hell I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:
"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".
What I did:
the distance for any point $(x,y,z)$ from the origin is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$. so using lagrange multipliers we have:
$d(x,y,z)=\sqrt{x^2+y^2+z^2}$
$C_1(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2-9$
$C_2(x,y,z)=x-2z$
$L(x,y,z) = d-\lambda_1C_1-\lambda_2C_2 $ meaning:
$L(x,y,z)=\sqrt{x^2+y^2+z^2}-\lambda_1[(x-1)^2+(y-2)^2+(z-3)^2-9]-\lambda_2(x-2z)$
Let's derive and solve when derivatives are zero:
$\frac{\partial L}{\partial x}= \frac{x}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(x-1)-\lambda_2=0$
$\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(y-2)=0$
$\frac{\partial L}{\partial z} = \frac{z}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(z-3)+2\lambda_2=0$
$\frac{\partial L}{\partial \lambda_1} = -(x-1)^2-(y-2)^2-(z-3)^2+9=0$
$\frac{\partial L}{\partial \lambda_2} = 2z-x=0$
Solving this monstrous system seems very unlikely, and very difficult, and not how the question is meant to be solved. am I missing something?
| $\newcommand{\e}{\mathbf{e}}$Here's an algebraic approach:
Let $P$ denote the plane with equation $x - 2z = 0$ and $S$ the sphere
$$
(x - 1)^{2} + (y - 2)^{2} + (z - 3)^{3} = 9.
$$
The vectors $\e_{1} = (2, 0, 1)/\sqrt{5}$ and $\e_{2} = (0, 1, 0)$ are an orthonormal basis of $P$. The orthogonal projection to $P$ of $c = (1, 2, 3)$, the center of $S$, is
$$
\langle c, \e_{1}\rangle \e_{1} + \langle c, \e_{2}\rangle \e_{2}
= \sqrt{5}\e_{1} + 2\e_{2}
= (2, 2, 1)
= c_{0}.
$$
Since the distance from $c$ to $c_{0}$ is $\sqrt{5}$, the intersection of $S$ and $P$ is a circle of radius $\sqrt{3^{2} - 5} = 2$ centered at $c_{0}$.
In the Cartesian coordinates defined by $\{\e_{1}, \e_{2}\}$, the center $c_{0}$ has coordinates $(\sqrt{5}, 2)$, and a bit of algebra shows the nearest point to the origin is $(\sqrt{5}, 2)/3$, while the furthest point is $5(\sqrt{5}, 2)/3$.
Converting back to spatial coordinates, the nearest point to the origin is $(2/3, 2/3, 1/3)$ and the furthest point is $(10/3, 10/3, 5/3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Calculate $\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$ I am trying to calculate:
$$\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $\ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $\ln(1-x)$ then,
\begin{align*}
=&-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\\
=&-(x-x^2)-\frac{(x-x^2)^2}{2}-...\\
=&-x(1-x)+\frac{x^2}{2}(1-x)^2-\frac{x^3}{3}(1-x)^3\\
\text{thus the pattern is:}\\
=&\frac{x^n(1-x)^n}{n}
\end{align*}
Am I right?
Then our Integral would be: $$\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$\underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
\begin{align*}
\frac{n(n-1)...1}{(n+1)(n+2)...(2n)}\int_0^1 x^{2n} dx
\end{align*}
This shows a pattern:
\begin{align*}
=&\frac{(n!)^2}{(2n)!} (\frac{1}{2n+1})\\
=& \frac{(n!)^2}{(2n+1)!}
\end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
| Differentiation under the integral sign can be applied to this.
Consider
\begin{align}
I(a)&=\int_0^1 \, \frac{\ln{(1-a\,(x-x^2))}}{x-x^2}\, dx \tag 1\\
\frac{\partial}{\partial a} I(a)&=\int_0^1\, -\frac{1}{1-a\,(x-x^2)}\, dx \\
&= -\frac{4 \, \sqrt{-y^{2} + 4 \, y} \arctan\left(\frac{\sqrt{-y^{2} + 4 \, y}}{y - 4}\right)}{y^{2} - 4 \, y}\tag 2
\end{align}
Integrating $(2)$ gives us (I used Sage for that)
\begin{align}
I(a)=\sqrt{a} \sqrt{-a + 4} \arcsin\left(\frac{1}{2} \, a - 1\right) + 2 \, \arcsin\left(\frac{1}{2} \, a - 1\right)^{2} - 4 \, \arctan\left(\frac{{\left(a - 2\right)} \sqrt{a} \sqrt{-a + 4}}{a^{2} - 4 \, a}\right) \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right) + 4 \, \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right)^{2} + 4 \, \arcsin\left(\frac{1}{2} \, a - 1\right) \arctan\left(\frac{\sqrt{-a^{2} + 4 \, a}}{a - 4}\right) - \sqrt{-a^{2} + 4 \, a} \arcsin\left(\frac{1}{2} \, a - 1\right) + C \\
\tag 3
\end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-\frac{\pi^2}{2}$, hence:
\begin{align}
I(a)=\sqrt{a} \sqrt{-a + 4} \arcsin\left(\frac{1}{2} \, a - 1\right) + 2 \, \arcsin\left(\frac{1}{2} \, a - 1\right)^{2} - 4 \, \arctan\left(\frac{{\left(a - 2\right)} \sqrt{a} \sqrt{-a + 4}}{a^{2} - 4 \, a}\right) \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right) + 4 \, \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right)^{2} + 4 \, \arcsin\left(\frac{1}{2} \, a - 1\right) \arctan\left(\frac{\sqrt{-a^{2} + 4 \, a}}{a - 4}\right) - \sqrt{-a^{2} + 4 \, a} \arcsin\left(\frac{1}{2} \, a - 1\right)-\frac{\pi^2}{2}
\end{align}
Therefore, the required answer is
$$
I(1)=-\frac{\pi^2}{9}\approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
\begin{align}
I(a)&=\int_0^1 \, \frac{\ln{(1+a\,(x-x^2))}}{x-x^2}\, dx
\end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
\begin{align}
I(a)&=\ln\left( \frac{a+2 -\sqrt{a^{2} + 4 \, a}}{2}\right)^{2}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/756598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Prove determinant of $n \times n$ matrix is $(a+(n-1)b)(a-b)^{n-1}$? Prove $\det(A)$ is $(a+(n-1)b)(a-b)^{n-1}$ where $A$ is $n \times n$ matrix with $a$'s on diagonal and all other elements $b$, off diagonal.
| Let
$A=\begin{bmatrix}
a & b & b & b & \ldots & b \\
b & a & b & b & \ldots & b \\
b & b & a & b & \ldots & b \\
b & b & b & a & \ldots & b \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & b & \ldots & a
\end{bmatrix}$
a $n\times n$ matrix, cause adding/substracting a scalar multiple of a row to another row doesn't change the value of the determinant of a matrix we can add rows 2nd, 3rd, 4th, $\dots$ and $n$th to the first row, so
$$\det (A) = \det \begin{bmatrix}
a+(n-1)b & a+(n-1)b & a+(n-1)b & a+(n-1)b & \ldots & a+(n-1)b \\
b & a & b & b & \ldots & b \\
b & b & a & b & \ldots & b \\
b & b & b & a & \ldots & b \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & b & \ldots & a
\end{bmatrix}$$
Now, we can factor out $a+(n-1)b$ from the first row
$$\det (A) = \left[a+(n-1)b\right]\det \begin{bmatrix}
1 & 1 & 1 & 1 & \ldots & 1 \\
b & a & b & b & \ldots & b \\
b & b & a & b & \ldots & b \\
b & b & b & a & \ldots & b \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & b & \ldots & a
\end{bmatrix}$$
Also, if we substract $b$ times row 1 to rows 2nd, 3rd, 4th, $\ldots$ and $n$th we get
$$\det (A) = \left[a+(n-1)b\right]\det \begin{bmatrix}
1 & 1 & 1 & 1 & \ldots & 1 \\
0 & a-b & 0 & 0 & \ldots & 0 \\
0 & 0 & a-b & 0 & \ldots & 0 \\
0 & 0 & 0 & a-b & \ldots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \ldots & a-b
\end{bmatrix}$$
The last matrix is a lower triangular matrix and its determinant is the product of elements in the main diagonal. Therefore
$$\det (A) = \left[a+(n-1)b\right](a-b)^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How prove this $a^n-b^n$ always have prime factor $P$ and $P>n$
Let $p_{1},p_{2},p_{3}$ be different prime numbers, and let the positive integer $n$, be defined by
$$n=p_{1}p_{2}p_{3}.$$
Show that:
For any two positive integer $a,b$ ,then $a^n-b^n$ always has a prime factor $P$ satisfying $P>n$
This problem is from a maths exam a few days ago.
We only need to prove that $$a^{p_{1}p_{2}p_{3}}-b^{p_{1}p_{2}p_{3}}$$ has a prime factor $P>n$.
But I can't. Thank you very much!
| If $a=b$ then $a^n-b^n=0$ is divisible by any prime, so we are done since there are infinitely many primes.
Otherwise we may WLOG assume $a>b$. Let $d=\gcd(a, b)$. By Zsigmondy's theorem, $\left(\frac{a}{d}\right)^n-\left(\frac{b}{d}\right)^n$ has a prime factor $p$ that does not divide $\left(\frac{a}{d}\right)^k-\left(\frac{b}{d}\right)^k$ for any positive integer $k<n$. (Note $n \not=2, 6$)
We cannot have $p \mid \frac{a}{d}$, for otherwise $p \mid \left(\frac{a}{d}\right)^n-\left(\frac{b}{d}\right)^n$ implies $p \mid \frac{b}{d}$ as well, so $p \mid \frac{a}{d}-\frac{b}{d}$ and $1<n$, a contradiction. Thus $p \nmid \frac{a}{d}$, and simililarly $p \nmid \frac{b}{d}$.
The order of $\left(\frac{a}{d}\right)\left(\frac{b}{d}\right)^{-1} \pmod{p}$ is thus $n$. Since $p \nmid \left(\frac{a}{d}\right)\left(\frac{b}{d}\right)^{-1}$, we have by Fermat's little theorem that $\left(\left(\frac{a}{d}\right)\left(\frac{b}{d}\right)^{-1}\right)^{p-1} \equiv 1\pmod{p}$. Thus $n \mid p-1$. This gives (since clearly $p>1$) that $n \leq p-1$ so $p>n$.
Finally since $a^n-b^n=d^n\left(\left(\frac{a}{d}\right)^n-\left(\frac{b}{d}\right)^n\right)$ we have $p \mid a^n-b^n$ and $p>n$, so we are done.
A simpler proof which does not appeal to Zsigmondy's theorem might be possible, as we only require a special case. (Here $n=p_1p_2p_3$) and may not require the full machinery used to prove Zsigmondy's theorem (Cyclotomic polynomials)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Range of f(x) = $\frac{\sqrt3\,\sin x}{2 + \cos x}$ Can you give any idea about the range of the following function?
$$f(x) = \frac{\sqrt{3}\,\sin x}{2 + \cos x}$$
| Consider $f^2(x) = \dfrac{3 - 3cos^2x}{cos^2x + 4cosx + 4}$. Now let $t = cosx$, then look at $f(t) = \dfrac{3 - 3t^2}{t^2 + 4t + 4}$. We find $f_{max}$. $f'(t) = \dfrac{-6(2t + 1)}{t + 2} = 0$ when $t = \dfrac{-1}{2}$. So $f(-1) = 0$, $f(\frac{-1}{2}) = 1$, and $f(1) = 0$. So $f_{max}^2 = 1$. This means that: $f^2(x) \le 1$, and so: $-1 \le f(x) \le 1$. So the range of $f$ is: $[-1, 1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Successively longer sums of consecutive Fibonacci numbers: pattern? Consider the following:
$$\begin{align}
F_{n-1}+F_{n-2}&=F_n\\
F_{n-1}+F_{n-2}+F_{n-3}&=F_{n-1}+F_{n-1}\\
&=2F_{n-1}\\
F_{n-1}+F_{n-2}+F_{n-3}+F_{n-4}&=F_n+F_{n-2}\\
&=L_{n-1}\\
F_{n-1}+F_{n-2}+F_{n-3}+F_{n-4}+F_{n-5}&=F_{n-1}+L_{n-2}\\
&=F_{n-1}+F_{n-1}+F_{n-3}\\
&=2F_{n-1}+F_{n-3}\\
F_{n-1}+F_{n-2}+F_{n-3}+F_{n-4}+F_{n-5}+F_{n-6}&=2(F_{n-1}+F_{n-4})
\end{align}$$
Is there some logic to this pattern? Can it be predicted? Is there a formula that can be used to compute the value of
$$\sum_{i=1}^{r}{F_{n-i}}$$
in terms of only Fibonacci and Lucas numbers or their multiples, no constant terms, without resorting to a summation?
| $$F_n = \left(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n\right)_{2,1}$$
which let's us treat it like any other summation except using matrices instead of scalar numbers, just remember that matrix multiplication doesn't commute and everything else is the same.
Using $X = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, proceed:
$$\begin{align}
S &= \sum_{i=1}^r F_{n - i} \\
&= \left(\sum_{i=1}^r X^{n-i}\right)_{2,1}\\
&= \left(\sum_{i=n - r}^{n - 1} X^i\right)_{2,1} \\
&= \left(X^{n - r}
\underbrace{\sum_{i=0}^{r - 1} X^i}
_\text{Q = Geometric Series}
\right)_{2,1}\\
\end{align}$$
Geometric series,
$$\begin{align}
Q &= \sum_{i=0}^{r - 1} X^i\\
&= \left(X^{r} - I\right)
\left(X - I\right)^{-1}\\
&= \left(X^{r} - I\right)
X\\
&= X^{r+1} - X
\end{align}$$
$$\begin{align}
S &= \left(X^{n - r}\right)
\left(X^{r+1} - X\right)_{2,1}\\
&= \left(X^{n+1} - X^{n+1-r}\right)_{2,1}\\
&= F_{n + 1} - F_{n + 1 - r}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the monic generator of and ideal. Let $\mathbb{F}$ be a subfield of complex numbers, and let
$$
A = \begin{bmatrix}
1 & -2 \\[0.1em]
0 & 3 \\[0.1em]
\end{bmatrix}
$$
Find the monic generator of the ideal of all polynomials $f$ in $F[x]$ such that $f(A)=0.$
| We can solve this in a reasonably simple manner.
Consider some $f \in F[x]$ such that $f(A) = 0$. Suppose that $deg(f) = 1$. Then, $f$ is of the form $$ f(x) = cx + d. $$ However, we need $f(A)$ to be zero and all values are in $\mathbb R^{2x2}$, so
\begin{align*}
f(A) &= cA + d \\
&= c\begin{bmatrix}1&-2\\0&3\end{bmatrix} + \begin{bmatrix}d&0\\0&d\end{bmatrix} \\
&= \begin{bmatrix}c+d&-2c\\0&3c+d\end{bmatrix}
\end{align*}
So, unless we set $c=0$ and $d=0$, we cannot scale or add by any $c$ and $d$ such that $f(A)=0$. We can, however, try a function of degree 2, written as $$f(x)=x^2+cx+d.$$
Rewriting a bit, we see that
\begin{align*}
0 &= f(A) \\
&= A^2 + cA + d \\
&= \begin{bmatrix}1&-8\\0&9\end{bmatrix}+\begin{bmatrix}c+d&-2c\\0&3c+d\end{bmatrix}\\
&= \begin{bmatrix}1+c+d&-8-2c\\0&9+3c+d\end{bmatrix}
\end{align*}
Because each entry of $f(A)$ needs to be zero, we can find $c$ and $d$ by solving this system of linear equations:
$$
\begin{cases}
-1=c+d \\-8=2c \\-9=3c+d
\end{cases}
$$
This is solved by $c=-4$ and substituting into either equation to find that $d=3$. Thus, we have that $$ f(x)=x^2-4x+3 $$ and that $$ f(A)=A^2-4A+3 = 0, $$ as desired. Thus, $f$ is the monic polynomial that generates the ideal $fF[x]$; as such, $f$ is the gcd for all polynomials $g$ such that $g(A)=0$.
(Note that I am implicitly applying the isomorphism from the set of polynomials over $F$, denoted as $F[x]$, to the set of all polynomial functions over $F$, which I usually denote as $\mathbb P_\infty$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Formula for $\sum_{k=0}^n k^d {n \choose 2k}$ If $d \geq 1$ is an integer, is there a general formula for $$\sum_{k=0}^n k^d {n \choose 2k}\,?$$
We know that $\sum_{k=0}^n k {n \choose 2k} = \frac{n2^n}{8}$ and $\sum_{k=0}^n k^2 {n \choose 2k} = \frac{n(n+1)2^n}{32}$.
Note that ${n \choose 2k} = 0$ when $2k > n$.
| The answer that I presented in my other post is more complicated than
it needs to be.
Suppose we seek to evaluate
$$\sum_{k=0}^n k^d {n\choose 2k}.$$
We observe that
$$k^d = \frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{d+1}} \exp(kz) \; dz.$$
This yields for the sum
$$\frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{d+1}}
\sum_{k=0}^n {n\choose 2k} \exp(kz) \; dz$$
which is
$$\frac{1}{2}\frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{d+1}}
\left(\sum_{k=0}^n {n\choose k} \exp(kz/2)
+ \sum_{k=0}^n {n\choose k} (-1)^k \exp(kz/2)\right)
\; dz.$$
This yields two pieces, call them $A_1$ and $A_2.$ Piece $A_1$ is
$$\frac{1}{2}\frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{d+1}}
(1+\exp(z/2))^n \; dz$$
and
$$\frac{1}{2}\frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{d+1}}
(1-\exp(z/2))^n \; dz.$$
Now to evaluate $A_1$ put $z=2w$ to get
$$\frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(2w)^{d+1}}
(1+\exp(w))^n \; dw
\\ = \frac{d!}{2^{d+1} \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{d+1}}
(2+\exp(w)-1)^n \; dw
\\ = \frac{d!}{2^{d+1} \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{d+1}}
\sum_{q=0}^n {n\choose q} 2^{n-q} (\exp(z)-1)^q \;dz
\\ = \sum_{q=0}^n {n\choose q} 2^{n-q} \times q! \times
\frac{d!}{2^{d+1} \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{d+1}}
\frac{(\exp(z)-1)^q}{q!} \;dz.$$
Recall the species equation for labelled set partitions:
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields the bivariate generating function of the Stirling numbers
of the second kind
$$\exp(u(\exp(z)-1)).$$
Substitute this into the sum to get
$$\sum_{q=0}^n {n\choose q} 2^{n-q-d-1} \times q! \times
{d\brace q}
= 2^{n-d-1} \sum_{q=0}^n {n\choose q} 2^{-q} \times q! \times
{d\brace q}.$$
Now observe that when $n\gt d$ the Stirling number for $d\lt q\le n$
is zero, so we may replace $n$ by $d.$ Similarly, when $n\lt d$ the
binomial coefficient for $n\lt q\le d$ is zero so we may again replace
$n$ by $d.$ This gives the following result for $A_1:$
$$2^{n-d-1} \sum_{q=0}^d {n\choose q} 2^{-q} \times q! \times
{d\brace q}.$$
Moving on to $A_2$ we observe that when $d\lt n$ the contribution is
zero because the series for $1-\exp(z/2)$ starts at $z.$ We get
$$\frac{d!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(2w)^{d+1}}
(1-\exp(w))^n \; dw
\\ = \frac{(-1)^n d!}{2^{d+1}\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{d+1}}
(\exp(w)-1)^n \; dw
\\ = \frac{(-1)^n \times n!\times d!}{2^{d+1}\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{w^{d+1}}
\frac{(\exp(w)-1)^n}{n!} \; dw.$$
Recognizing the Stirling number we get
$$2^{-d-1} \times (-1)^n \times n! \times {d\brace n}.$$
which correctly represents the fact that we have a zero contribution
when $d\lt n.$
This finally yields the closed form formula
$$2^{n-d-1} \sum_{q=0}^d {n\choose q} 2^{-q} \times q! \times
{d\brace q}
+ 2^{-d-1} \times (-1)^n \times n! \times {d\brace n}$$
confirming the previous results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$ I am trying to evaluate the following limit
$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$$
If we use the binomial expansion of the numerator term, the answer is $\frac{1}{4}$. The same answer is obtained if we apply L'Hospital's rule.
Is it possible to solve this using something even more basic ? i.e. without recourse to binomial expansion or L'Hospital's rule or definition of derivative ? The reason I ask this is because this problem (from a textbook) comes up before the introduction of L'Hospital's rule.
user37238 wanted to know how the binomial expansion can be used.Basically, we have
$(1+x)^n= 1 + nx + \frac{n (n-1)}{2} + \frac{n (n-1)(n-2)}{2!} + \ldots$
Similarly,
$(1-x)^n= 1 - nx + \frac{n (n-1)}{2} - \frac{n (n-1)(n-2)}{2!} + \ldots$
Using these expansions with $n=\frac{1}{8}$ in the given problem gives the answer.
| $$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x} =\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1+1-(1-x)^\frac{1}{8} }{x}=\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1 }{x}+\lim_{x \rightarrow 0} \frac{(1-x)^\frac{1}{8}-1 }{-x}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} $$
We applied the known limit for $a\in R$:
$$\lim_{x \rightarrow 0} \frac{(1+x)^a-1 }{x}=a $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/768531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show the limit of the following is $\dfrac{1}{12}$
Show that $$\lim_{n\to \infty}n^2 \log \left(\dfrac{r_{n+1}}{r_{n}} \right)=\dfrac{1}{12}$$ where $r_{n}$ is defined as;
$$r_{n}=\dfrac{\sqrt{n}}{n!} \left(\dfrac{n}{e} \right)^n$$.
Now I simplified $\dfrac{r_{n+1}}{r_{n}}$ to $$\dfrac{r_{n+1}}{r_{n}}= \left(\dfrac{n+1}{n} \right)^{n+0.5} \dfrac{1}{e}$$
And so simplified $n^{2} \log \left(\dfrac{r_{n+1}}{r_{n}} \right)$ to $$n^{2}\left((n+0.5) \log \left(\dfrac{n+1}{n} \right)-1 \right).$$
Now I checked this on wolfram and as $n \rightarrow \infty$ this does converge t $\dfrac{1}{12}$ but I am not able to show it.
I have tried L'Hopital i.e $$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}}$$
Now as $n \rightarrow \infty$ $\log \left(\dfrac{n+1}{n} \right) \rightarrow 0$ but you end up with a $\dfrac{-1}{0}$ type limit.
Any help with this would be much appreciated.
EDIT: I have realised I have made a mistake and that $$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}}$$ gives you a $\dfrac{0}{0}$ type limit, will delve into the algebra and get back.
| Application of the rule of L'Hospital is a good idea:
$$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}} =\lim_{x\to 0} \dfrac{\dfrac{1}{x} \log(1+x) +0.5 \log (1+x)-1}{x^2} =(L'H) \lim_{x\to 0} \dfrac{\dfrac{-1}{x^2} \log(1+x)+\dfrac{1}{x(1+x)}+\dfrac{1}{2(1+x)} }{2x} =(L'H)\lim_{x\to 0} \dfrac{- \log(1+x)+\dfrac{x(x+2)}{2(1+x)} }{2x^3} =\lim_{x\to 0} \dfrac{- \dfrac{1}{1+x}+\dfrac{x^2+2x+2}{2(1+x)^2} }{6x^2} =\lim_{x\to 0} \dfrac{1}{12(1+x)^2}= \dfrac{1}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\int\frac1{1+2x}dx$? What approach would be ideal in finding the solution to the integration problem $\int\frac1{1+2x}dx$?
| In general
\begin{align}
\int\frac{a}{bx+c}dx=\frac{a}{b}\ln (bx+c)+C
\end{align}
Proof:
Let $u=bx+c$ then $x=\cfrac{u-c}{b}$ and $dx=\cfrac{du}{b}$. Hence
\begin{align}
\int\frac{a}{bx+c}dx=\int\frac{a}{u}\cdot\frac{du}{b}=\frac{a}{b}\int\frac{du}{u}=\frac{a}{b}\ln u+C=\frac{a}{b}\ln (bx+c)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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series convergence proof. Use Theorem 8.8 to show that if $0<x\leq 1$ then ln$(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}x^k$.
Theorem 8.8 states that for each natural number $n$ and each number $x>-1$ there is a number $c$ strictly between 0 and $x$ such that ln$(x+1)=x-\frac{x^2}{2}+..+\frac{(-1)^{n+1}x^n}{n}+\frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}$
Proof:
Assume that $0<x\leq 1$ and it follows from theorem 8.8 that ln$(x+1)=\sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}+\frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}\implies \lim_{n\to\infty} ln(1+x)=\lim_{n\to\infty} \sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}+\lim_{n\to\infty} \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}$ for some $c$ between $0$ and $x$. We see that $\lim_{n\to\infty} ln(1+x)=ln(1+x)$ and $\lim_{n\to\infty} \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}=0$. Also $\lim_{n\to\infty} \sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{k} $ by definition. Thus $ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{k}$. Am I on the right track here or no? Thanks
| I think you are on the right track. I reorganize your proof here, and add one or two details. For $x \in (0, 1]$, let $S_n(x) = \displaystyle \sum_{k=1}^n \dfrac{(-1)^{k+1}x^k}{k}$, then you need to prove: $\displaystyle \lim_{n \to \infty} \left(S_n(x) - ln(1+x)\right) = 0$, and using theorem 8.8, we have: $S_n(x) - ln(1+x) = \dfrac{1}{n+1}\cdot \left(\dfrac{-x}{1+c}\right)^{n+1}$. Next since $-1 < \dfrac{-x}{1+c} < 1$. So: $0 < \big|\dfrac{1}{n+1}\cdot \left(\dfrac{-x}{1+c}\right)^{n+1}\big| < \dfrac{1}{n+1}$, so by squeeze theorem, $\displaystyle \lim_{n \to \infty} \left(S_n(x) - ln(1+x)\right) = 0$,and the statement is proved.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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using substitution wrongly Solving integral, first way:
$$\int \frac{du}{u^2-9}=-\int \frac{du}{9-u^2}$$
$$u={3\sin v}$$
$$du=3\cos vdv$$
$$-\int \frac{3\cos vdv }{9-9\sin^{2}v}=-\frac 13\int\frac{dv}{\cos v}=-\frac 13\ln\left(\sec v+\tan v\right)=-\frac13 \ln \frac {\frac u3}{\sqrt{1-\frac{u^2}{9}}}$$
$$-\frac13 \ln \frac{u}{\sqrt{9-u^2}}=-\frac16 \ln \frac {u^2}{9-u^2}=\frac16 \ln \frac{9-u^2}{u^2}$$
second way:
$$\int \frac{du}{u^2-9}=\frac 16\int \frac{du}{u-3}-\frac16 \int \frac{du}{u+3}=\frac16 \ln \frac {u-3}{u+3}$$
what's wrong with first way?
Thanks all. $1$ from $\sec$ is missed, must be $1+\frac u3$
| As Grid commented, before any simplification
$$\ln\left(\sec v+\tan v\right)=\log \left(\frac{u}{3 \sqrt{1-\frac{u^2}{9}}}+\frac{1}{\sqrt{1-\frac{u^2}{9}}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Max. value $M$ of $c^2$ for which points lie opposite of $x(a+b)^2-(ab+bc+ca+1)y+2 = 0$ Let $M$ be the the maximum value of $c^2$ for which $O(0,0)$ and $A(1,1)$ does not lie on
opposite side of straight line $x(a+b)^2 -(ab+bc+ca+1)y +2=0.$Then value of $M$ is
$\bf{My\; Solution::}$ Given $O(0,0)$ and $A(1,1)$ does not lie on opposite side of line $L_{(x,y)}$ means
These two points lie on same side of line. So $L_{(0,0)}\times L_{(1,1)}>0$
So here $L_{(x,y)} = x(a+b)^2 -(ab+bc+ca+1)y +2.$
So $L_{(0,0)} = 2$ and $L_{(1,1)} = (a+b)^2-(ab+bc+ca+1)+2$
So $(a+b)^2-(ab+bc+ca+1)+2>0$
Now How can I solve after that
Help Required
Thanks
| Hint:
\begin{align}
(a+b)^2-ab -bc - ca +1 &\ge (a+b)^2 - \tfrac12(a^2+b^2) - \tfrac16(9b^2+c^2)-\tfrac16 (c^2+9a^2)+1 \\
&= -(a-b)^2+1 - \tfrac13 c^2 \\
&\ge 1- \tfrac13 c^2
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/772382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
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Squares constructed externally on the sides of a triangle and concurrent lines On the sides $BC, CA$ and $AB$ of the triangle $ABC$ we construct externally the squares $BCDE, ACFG $ and $ABHI$. Denote $A', B'$ and $C'$ the intersectiond points of the lines $BF$ and $CH$, $AD$ and $CI$, respectively $AE$ and $BG$. Prove, that $AA', BB'$ and $CC'$ are concurrent.
Any nice idea? Thank you!
| The lines $AA',BB',CC'$ are the altitudes of the triangle, so they intersect in the orthocenter.
To show that these are in fact altitudes, you can use a computation on coordinates. W.l.o.g. assume the following coordinates:
$$A=\begin{pmatrix}-1\\0\end{pmatrix}\qquad
B=\begin{pmatrix}1\\0\end{pmatrix}\qquad
C=\begin{pmatrix}x_C\\y_C\end{pmatrix}$$
From these you get
$$
G = A+\begin{pmatrix}0&-1\\1&0\end{pmatrix}(C-A) =
\begin{pmatrix}-1-y_C\\1+x_C\end{pmatrix} \\
E = B+\begin{pmatrix}0&1\\-1&0\end{pmatrix}(C-B) =
\begin{pmatrix}1+y_C\\1-x_C\end{pmatrix} \\
AE: (x_C-1)x+(y_C+2)y=1-x_C \\
BG: (x_C+1)x+(y_C+2)y=1+x_C \\
C' = \begin{pmatrix}x_C\\\tfrac{1-x_C^2}{2+y_C}\end{pmatrix}
$$
so you see that $C$ and $C'$ have the same $x$ coordinate, therefore $CC'\perp AB$. The other orthogonalities follow from the symmetry of the construction.
The lines joining the points as well as their intersection were actually computed using cross products of homogeneous coordinates:
$$
g_{AE} = A_h\times E_h =
\begin{pmatrix}-1\\0\\1\end{pmatrix}\times
\begin{pmatrix}1+y_C\\1-x_C\\1\end{pmatrix}=
\begin{pmatrix}x_C-1\\y_C+2\\x_C-1\end{pmatrix} \\
g_{BG} = G_h\times B_h =
\begin{pmatrix}-1-y_C\\1+x_C\\1\end{pmatrix}\times
\begin{pmatrix}1\\0\\1\end{pmatrix}=
\begin{pmatrix}x_C+1\\y_C+2\\-x_C-1\end{pmatrix} \\
C'_h = g_{AE}\times g_{BG} =
\begin{pmatrix}x_C-1\\y_C+2\\x_C-1\end{pmatrix}\times
\begin{pmatrix}x_C+1\\y_C+2\\-x_C-1\end{pmatrix}=
\begin{pmatrix}-2x_Cy_C-4x_C\\2x_C^2-2\\-2y_C-4\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/774381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the inequality $a,b,c$ are the sides of a triangle.Then show that
$$(a+b+c)^3 > 27(a+b-c)(b+c-a)(c+a-b)$$ Also give the case where equality holds i.e. $$(a+b+c)^3=27(a+b-c)(b+c-a)(c+a-b)$$
I tried triangle inequality, AM-GM inequality in many way but cannot help my cause.
| First apply AM-GM inequality: $(a+b+c)^3 \geq 27abc$, then you need to prove: $abc \geq (a+b-c)(b+c-a)(c+a-b)$, and this follows from:
$a^2 \geq a^2 - (b-c)^2$
$b^2 \geq b^2 - (c-a)^2$
$c^2 \geq c^2 - (a-b)^2$.
by multiply all three inequalities and taking square root.
| {
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} |
Binomial coefficients (Concrete mathematics 5.39)
Show that if $xy = ax+by$ then $$x^ny^n = \sum_{k=1}^n \binom{2n-1-k}{n-1} (a^nb^{n-k}x^k + a^{n-k}b^ny^k)$$ for all $n>0$. Find a similar formula for the more general product $x^my^n$. (There formulas give useful partial fraction expansions, for example when $x=1/(z-c)$ and $y=1/(z-d)$.)
I have no idea what I could do with this problem. I tried to substitute, but failed..
Hope someone can help
| (a) We prove it by induction. Initial case, $n=1$:
\begin{align}
\sum_{k=1}^n{\binom{2n-1-k}{n-1} (a^nb^{n-k}x^k + a^{n-k}b^ny^k)} & = \binom{0}{0}(a^1b^0x^1 + a^ob^1y^1) \\
& = ax+by \\
& = x^ny^n.\qquad\checkmark
\end{align}
Now assume the statement is true for some $n\gt 0$ and consider the case for $n+1$.
\begin{align}
(xy)^{n+1} & = (ax+by) x^ny^n \\
(xy)^{n+1} & = (ax+by) \sum_{k=1}^n{\binom{2n-1-k}{n-1} (a^nb^{n-k}x^k + a^{n-k}b^ny^k)} \qquad\text{by inductive assumption} \\
& = \sum_{k=1}^n{\binom{2n-1-k}{n-1} (a^{n+1}b^{n-k}x^{k+1} + a^{n-k}b^{n+1}y^{k+1} + a^nb^{n-k+1}x^ky + a^{n-k+1}b^ny^kx)}
\end{align}
We have a term in $x^ky$ that we need to convert to an expression with $x$ and $y$ separated.
\begin{align}
x^ky & = x^{k-1}(ax+by) \\
& = ax^k + bx^{k-1}y \\
& = ax^k + bx^{k-2}(ax+by) \\
& = ax^k + abx^{k-1} + b^2x^{k-2}y \\
& = \cdots \\
& = ax^k + abx^{k-1} + ab^2x^{k-2} + \cdots + ab^{k-1}x + b^ky.
\end{align}
The $y^kx$ term can be similarly expanded.
Then for any $j=1,\ldots ,n+1$ the $x^j$ term in the above summation has the following components:
\begin{align}
\text{For } k & =j-1: & \ \qquad \binom{2n-j}{n-1} a^{n+1}b^{n-j+1}x^j \\
k & =j: & \ \qquad \binom{2n-1-j}{n-1} a^{n+1}b^{n-j+1}x^j \\
k & =j+1: & \ \qquad \binom{2n-2-j}{n-1} a^{n+1}b^{n-j+1}x^j \\
& \ \ldots & \\
k & =n: & \ \qquad \binom{n-1}{n-1} a^{n+1}b^{n-j+1}x^j. \\
\end{align}
Summing these, we get $c_j$, the coefficient of the $x^j$ term:
\begin{align}
c_j & = \sum_{i=0}^{n-j+1}{\binom{2n-j-i}{n-1}} \\
& = \sum_{i=0}^{n-j+1}{\binom{n-1+i}{n-1}} \qquad\text{reversing the order of summation}\\
& = \binom{2n-j+1}{n} \qquad\text{by Identity 5.9 with $n:=n-j+1,\; r:=n-1$} \\
& = \binom{2(n+1) - 1 - j}{(n+1) - 1}.
\end{align}
Identity 5.9 from Concrete Mathematics can also be found here after Equation 8.
A similar argument works to determine the $y^j$ term and with that we have shown that the statement holds for $n+1$ and the proof is complete.
(b) Some trial and error indicates that the equivalent formula for $x^my^n$ is:
$$x^my^n = \sum_{k=1}^m{\binom{m+n-1-k}{n-1} a^nb^{m-k}x^k} + \sum_{k=1}^n{\binom{m+n-1-k}{m-1} a^{n-k}b^my^k}.$$
I imagine this formula could also be proved by induction.
| {
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Group theory, conjugation of permutations I have a past exam question that says...
Decompose the following permutations into a product of disjoint cycles. Are the two permutations conjugate?
$$\alpha= \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
7 & 4 & 5 & 3 & 8 & 6 & 9 & 1 & 2\\
\end{bmatrix}$$
and
$$\beta= \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5\\
\end{bmatrix}$$
I think that you have to find $x^{-1}\alpha x=\beta$, however I do not know how to find this $x$. Any help with this would be much appreciated. Thank you!
| Decomposition in cycle is fairly easy, and it yields
$$\alpha = (1\ 7\ 9\ 2\ 4\ 3\ 5\ 8)$$
$$\beta = (1\ 6\ 2\ 7\ 3\ 8\ 4\ 9\ 5)$$
Notice $6$ does not appear in the cycle decomposition of $\alpha$ (it would be a cycle of length 1, usually they are not written).
Two permutations of $S_n$ are conjugate iff they have the same cycle structure (see here): basically you only rename elements. Hence $\alpha$ and $\beta$ cannot be conjugate.
Where $\alpha$ and $\beta$ of same "cycle type", it would be easy to find a permutation to transform one into the other.
Say
$$\alpha=(1\ 7\ 4)(3\ 2\ 6\ 5)(8\ 9)$$
$$\beta=(3\ 4\ 7\ 6)(5\ 2\ 9)(1\ 8)$$
Then send each cycle to a cycle of same length: $1 \rightarrow 5$, $7 \rightarrow 2$...
So this one would do (it's not unique, just try to rotate any cycle)
$$x= \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
5 & 4 & 3 & 9 & 6 & 7 & 2 & 1 & 8\\
\end{bmatrix}$$
Or $x=(1\ 5\ 6\ 7\ 2\ 4\ 9\ 8)$
Then $x\alpha x^{-1}=\beta$ if you use the convention $(xy)(i) = x(y(i))$, or $x^{-1}\alpha x=\beta$ with the other convention $(xy)(i)=y(x(i))$. In France it's usually the former, and in english-speaking countries, the latter.
For example, $x^{-1}({\color{red}{5}}) = 1, \alpha(1)=7, x(7)={\color{red}{2}}$, and $\beta({\color{red}{5}})={\color{red}{2}}$.
To answer mayi's comment below, let's try with $\alpha=(1\ 3)(4\ 7\ 6)$ and $\beta=(1\ 5)(2\ 6\ 4)$.
They have the same cycle structure, hence they are conjugate. The trick is you have to work in $S_7$, the group of permutations of $\{1,2,3,4,5,6,7\}$, so that $\alpha$ and $\beta$ share the same range. You may then use the permutation
$$\pi= \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7\\
1 & - & 5 & 2 & - & 4 & 6\\
\end{bmatrix}$$
The images of $2$ and $5$ can be chosen freely, since they do not appear in $\alpha$, and the only remaining elements are $3$ and $7$. So, either $2\to3$ and $5\to7$, either $2\to7$ and $3\to5$. For instance:
$$\pi= \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7\\
1 & 3 & 5 & 2 & 7 & 4 & 6\\
\end{bmatrix}$$
Or in cycle notation, $\pi=(2\ 3\ 5\ 7\ 6\ 4)$. Then, with the "french" convention of permutation composition, $\beta=\pi\alpha\pi^{-1}$, and with the "english" convention, $\beta=\pi^{-1}\alpha\pi$.
But $\pi$ is of course not the only possible permutation that transforms $\alpha$ into $\beta$. If you write $\beta=(1\ 5)(4\ 2\ 6)$, it's easy to see that $\pi=(2\ 7)(3\ 5)$ is also valid. To find your $x=(1\ 5\ 3)(2\ 7)$, write instead $\beta=(5\ 1)(4\ 2\ 6)$ (still with $\alpha=(1\ 3)(4\ 7\ 6)$), and
$$\pi= \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7\\
5 & - & 1 & 4 & - & 6 & 2\\
\end{bmatrix}$$
Again, the images of $2$ and $5$ can be chosen freely. With $2\to7$ and $5\to3$, you get $\pi=(1\ 5\ 3)(2\ 7)$, your $x$. With the other choice, you would get $\pi=(1\ 5\ 7\ 2\ 3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this $ n = \frac {a^3 + 2b^3} {c^3 + 2d^3}. $ infinitely many special numbers Qustion
A special number is a positive integer $n$ for which there exists positive integers $a$, $b$, $c$, and $d$ with $$ n = \dfrac {a^3 + 2b^3} {c^3 + 2d^3}. $$ Prove that
i) there are infinitely many special numbers;
ii) $2014$ is not a special number.
I have solve (1):let $$b=c=d=1,a=3k+1$$ then
$$n=9k^3+9k^2+3k+1$$ is sepecial for any postive integer $k$,Clearly,these numbers are infinitely many.
But for (2),Assmue that
$$2014(c^3+2d^3)=a^3+2b^3\Longrightarrow a^3+2b^3=2\cdot 19\cdot 53(c^3+2d^3)$$
then I can't prove contradiction.I guess can use (1) to solve (2).But I can't. Thank you
| First note that $-2$ is not a cubic residue modulo $19$. To see this, suppose
$x^3\equiv -2\pmod{19}$. Then $x^{18}\equiv (-2)^6 \equiv 64 \equiv 7\pmod{19}$
Clearly $19$ does not divide $x$, or we would have $x^3\equiv 0\pmod{19}$, and so by Fermat's Little Theorem we obtain $x^{18}\equiv 1\pmod{19}$, which is a contradiction.
Now suppose that $19 \mid (x^3+2y^3)$. If $19$ divides either $x$ or $y$ then it must divide both. Suppose $19$ divides neither $x$ nor $y$ and let $z$ be the multiplicative inverse of $y$ modulo $19$. We then have
$x^3\equiv -2y^3\pmod{19}$ and so $(xz)^3\equiv -2\pmod{19}$ which is a contradiction.
We see that if $19\mid(x^3+2y^3)$ for some $x$ and $y$ then it follows that $19\mid x$ and $19\mid y$.
Now returning to the problem:
Suppose that $(a^3+2b^3)=2014(c^3+2d^3)$, and - furthermore - suppose that this quadruple $(a,b,c,d)$ is the smallest in the sense that $a$ has the minimum possible value for any quadruple $(a,b,c,d)$ satisfying this equation. We'll show that there is a solution will a smaller value for a, which is a contradiction.
Since $19\mid 2014$ we see that $19\mid(a^3+2b^3)$ and so $19\mid a$ and $19\mid b$ by our observations above.
Let $a=19a_1$ and $b=19b_1$
Then $19^2(a_1^3+b_1^3)=106(c^3+2d^3)$. We see that $19\mid(c^3+2d^3)$ and so we also have $19\mid c$ and $19\mid d$. Now let $c=19c_1$ and $d=19d_1$. We then obtain
$a_1^3+2b_1^3=2014(c_1^3+2d_1^3)$. This gives us a solution to the equation with a smaller value for $a$ (namely $a_1=\frac{a}{19}$), which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
| From $a^2-b^2=(a+b)(a-b)$ we have
$$\sqrt{n^2+n}-n=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=
\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
from which the result follows immediately.
| {
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"url": "https://math.stackexchange.com/questions/783536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Halving a tiny angle by doubling a side of a triangle. You have a triangle $ABC$ with a right angle $\angle CAB$. The line stretch between $A$ and $B$ shall be very long in comparison to the other sides. Now you are going to measure the angle $\angle ABC$ which is very tiny, e.g. $1'$.
Now you keep the line stretch between $A$ and $C$ constant. Next you double the length of $AB$ and $BC$ is adapted according to the Pythagorean theorem.
According to my physiology textbook: If $\angle ABC$ is tiny and I am going to double $AB$ as above, $\angle ABC$ should be halved.
How to prove that?
| The rough idea is as Vizzy says: for small angles $\theta$, $\tan \theta \approx \theta$.
We can make this rigorous.
Suppose the original angle is $\theta$, and that originally $AC = l$ and $AB = L$
($L \gg l$).
Then
$$
\tan \theta = \frac{l}{L}
$$
so after doubling the length of $AB$, the new angle $\theta_1$ will satisfy
$$
\tan \theta_1 = \frac{l}{2L} = \frac{\tan \theta}{2}
$$
Therefore,
$$
\theta_1 = \arctan \left( \frac{\tan \theta}{2} \right).
$$
Let $f(x) = \arctan \left( \frac{\tan x}{2} \right)$.
Then we can compute derivatives of $f$:
\begin{align*}
f(x) &= \arctan \left( \frac{\tan x}{2} \right)
& f(0) &= 0 \\
f'(x) &= \frac{4}{3 \cos(2x) + 5}
& f'(0) &= \frac{1}{2} \\
f''(x) &= \frac{24 \sin(2x)}{(3 \cos(2x) + 5)^2}
& f''(0) &= 0 \\
\end{align*}
If you continue in this way, you get the Taylor series for $f$:
$$
f(x) = \frac12 x + \frac18 x^3 + \frac{1}{32} x^5 + \frac{11}{1920} x^7 + \cdots
$$
In particular, for small values of $x$, the second taylor approximation
$\frac12 x$ is pretty accurate.
More precisely, if the third derivative is bounded by $M$, by the Lagrange remainder theorem, we have
$$
f(x) \approx \frac12 x
$$
with error no more than
$M \frac{x^3}{3!}$ -- an incredibly insignificant value when $x$ is small.
That is, we can approximate
$$
\theta_1 = \arctan \left( \frac{\tan \theta}{2} \right)
\approx \frac{\theta}{2}
$$
with error on the order of $\theta^3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $[E(Y^2)]$. When $Y = 3 * X - 5$ and $X$ is distributed in range $[0, 5]$ Need help with exercise.
Random variable $X$ is evenly distributed in range $[0, 5]$. Need to find $E[Y^2]$ when $Y = 3X - 5$
Every hint/tip will be appreciated.
Thank you
Alternative solution
$$E[Y^2] = E[(3X-5)^2] = E[9X^2-30X+25] = 9E[X^2]-30E[X]+25 =9(E[X]^2 + D[X]) - 30 *2.5 +25 =9*(2.5^2 + 25/12) - 75+25 = 25$$
| I assume that $X$ is uniformly distributed. The CDF of $X\sim\mathcal{U}(0,5)$ is
$$
F_X(x)=\Pr[X\le x]=\frac{x-0}{5-0}=\frac x5.
$$
Therefore
$$
\begin{align}
\Pr[Y\le y]&=\Pr[3X-5\le y]\\
F_Y(y)&=\Pr[3X\le y+5]\\
&=\Pr\left[X\le \frac{y+5}{3}\right]\\
&=F_X\left(\frac{y+5}{3}\right)\\
&=\frac{y+5}{15}.
\end{align}
$$
The region $0\le x<5$ is corresponding to $-5\le y<10$. Hence $Y\sim\mathcal{U}(-5,10)$. The pdf of $Y$ is
$$
f_Y(y)=\frac1{10-(-5)}=\frac1{15},
$$
then
$$
\begin{align}
\text{E}\left[Y^2\right]&=\int_{-5}^{10} y^2f_Y(y)\ dy\\
&=\frac1{15}\int_{-5}^{10} y^2\ dy\\
&=\frac1{15}\left[\frac13y^3\right]_{-5}^{10}\\
&=\frac1{45}(10^3-(-5)^3)\\
&=\Large\color{blue}{25}.
\end{align}
$$
| {
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value of $a+b$ of the following function $f(x)=x^3-3x^2+5x\;$ and
$\;f(a)=1,f(b)=5.\;$ Find $a+b$.
I know only one real root exist for each equation as derivative of the function is always positive .I do not intend to use the formula of roots of cubic equation. How should i go about this problem ????
| I assume that $a$ and $b$ are real numbers (this can be guaranteed since $f(a)=1$ and $f(b)=5$ have real solutions.
As you said that $f^\prime$ is always positive that means $f$ is a strictly increasing function. Since $f(0)=0$ then $0<a<b$.
Note that we have
$$f(\frac{a+b}2)-\frac{f(a)}2-\frac{f(b)}2=-\frac38(b-a)^2(a+b-2).$$
So, If $a+b\leq 2$ then $$f(\frac{a+b}2)-\frac{f(a)}2-\frac{f(b)}2\ge0\iff f(\frac{a+b}2)\ge 3\iff\frac{a+b}2\ge 1\iff a+b\ge 2 $$
which gives $a+b=2$.
If $a+b>2$, then have $$f(\frac{a+b}2)-\frac{f(a)}2-\frac{f(b)}2<0\iff f(\frac{a+b}2)< 3\iff\frac{a+b}2< 1\iff a+b< 2 $$ that leads to contradiction.
In conclusion, we must have $a+b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Parametric equations where sin(t) and cos(t) must be rational Suppose there are parametric equations
$$
x(t) = at - h\sin(t)
$$
$$
y(t) = a - h\cos(t)
$$
and it is required that both $\sin(t)$ and $\cos(t)$ should be rational.
What the values of $t$ should be in that case?
Thanks.
| Your question (at least how you wrote it) is really,
For what values of $t$ are both $\cos t$ and $\sin t$ rational?
Note that if $x$ and $y$ are real numbers with $x^2 + y^2 = 1$, then there exists a real number $t$ such that $x = \cos t$, $y = \sin t$.
Therefore, we want solutions to $x^2 + y^2 = 1$ for $x, y \in \mathbb{Q}$.
For this we let $x = \frac{a}{c}$, $y = \frac{b}{d}$, $a,b,c,d \in \mathbb{Z}$, with the fractions in reduced form. Then we get
$$
a^2 d^2 + b^2 c^2 = c^2 d^2 \tag{1}
$$
Let $c = c'k$, $d = d'k$ with $c', d'$ relatively prime. Then the above gives
$$
a^2 d'^2 + b^2 c'^2 = k^2 c'^2 d'^2
$$
Mod $d'$, we find $b^2 \equiv 0$.
Since $\frac{b}{d}$ was in reduced form, $d' = 1$.
Similarly $c' = 1$.
So we get $c = d = k$.
Then we have, coming back to (1),
$$
a^2 + b^2 = c^2
$$
so $(a,b,c)$ are a Pythagorean triple.
In particular they are a primitive Pythagorean triple,
because $(a,c) = (b,c) = 1$.
That means all possible $a, b, c$ (up to sign and switching $a$ and $b$) are given by
$$
(a,b,c) = (2mn, m^2 - n^2, m^2 + n^2)
$$
where $m,n$ are relatively prime and not both odd.
Therefore
for any $a,b,c$ the solution $t$
is given by
$$
t = \pm \cos^{-1}\left(\frac{a}{c}\right) + 2\pi n
= \pm \cos^{-1} \left(\pm \frac{2mn}{m^2 + n^2} \right)
\text{ or }
\pm \cos^{-1} \left(\pm \frac{m^2 - n^2}{m^2 + n^2} \right)
$$
for some $m,n$.
Unfortunately we cannot in general simplify the $\cos^{-1}$ out of the formula
as far as I know.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how can we get from $x \equiv 4 \pmod 9$ to $x \equiv 1 \pmod 3$ I want to solve the system:
$$ x \equiv 1 \pmod 3 , x \equiv 2 \pmod 5, x \equiv 3 \pmod 7, x \equiv 4 \pmod 9, x \equiv 5 \pmod {11}$$
The numbers $3,5,7,9,11$ are not pairwise coprime,especially the numbers $3,9$ are not coprime.
According to my notes, $x \equiv 4 \pmod 9 \Rightarrow x \equiv 4 \pmod 3 \Rightarrow x \equiv 1 \pmod 3$
But..how can we get from the relation $x \equiv 4 \pmod 9$ to this one: $x \equiv 1 \pmod 3$ ?
| $ x \equiv 4 \mod 9 $ means $x = 9k + 4$ for some integer $k$. that means $x = 3(3k + 1) + 1$ and $3k+1$ is an integer which means $x \equiv 1 \mod 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\lim_{x \to \infty} x - \sqrt{x^2 + 2x}$ without derivations How can I calculate $\displaystyle \lim_{x \to \infty} x - \sqrt{x^2 + 2x}$?
Here is what I've done so far:
Multiplying by $\displaystyle \frac{x + \sqrt{x^2 + 2x}}{x + \sqrt{x^2 + 2x}}$
I got $\displaystyle \frac {-2x}{x+\sqrt{x^2 + 2x}}$
Multiplying this by $\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$
I got $\displaystyle \frac{-2}{1+\frac{\sqrt{x^2 + 2x}}{x}}$
I know I'm very close to the answer wich is $-1$, but I have no idea what to do next. I can´t just say that $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1$, as far as I know...
| Try:
$$
\frac{\sqrt{x^2 + 2x}}{x} = \sqrt{\frac{x^2 + 2x}{x^2}} = \sqrt{1 + \frac{2}{x}}
$$
By continuity of the square root function this means
$$
\lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1.
$$
| {
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"url": "https://math.stackexchange.com/questions/789772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Understanding substitution to compute primitive To compute the primitive
$$\int x^2 \sqrt[3]{x^3+3}\ dx$$
I am trying this:
$t = x^3+3$
$dt = 3x^2dx$
but
$\int x^2 \sqrt[3]{x^3+3}\ dx=\ ?$
How to continue from here?
| Notice that $3x^2$ is the derivative of $x^3+2$.
So if you set $t=x^3+2$, you get $\dfrac{dx}{dt}=3x^2$ which gives $dx = 3x^2 dt$.
Now the whole thing becomes:
$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{3}\int 3x^2\sqrt[3]{x^3+3} \ dx = \frac13 \int t^{1/3} dt = \frac{1}{3}\frac{1}{1+\frac{1}{3}} t^{1+\frac{1}{3}} + C$$
$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{4}(x^2+3)^{4/3} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the area of the surface of revolution generated by revolving about the $x$-axis the hypocycloid $x=a\cos^3\theta$, $y=a\sin^3\theta$ Find the area of the surface of revolution generated by revolving about the $x$-axis the hypocycloid $x=a\cos^3\theta$, $y=a\sin^3\theta$ ($0 \leq \theta \leq \pi$)
I know you have to integrate $2\pi y ds$ for the limits $0$ to $\pi$ and I know that $ds$ is the square root of the sum of each derivative squared but I'm stuck on the integration, how do we do it? Please help!
| The standard formula for rotation about the $x$-axis, through an angle of $2\pi$-radians, when $x$ and $y$ are given as functions of $\theta$ is
$$A = 2\pi \int_{\theta_1}^{\theta_2} y\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^{\!\!2} + \left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^{\!\!2}}~\mathrm{d}\theta$$
In your case $x=a\cos^3\theta$ and $y=a\sin^3\theta$. It follows that
\begin{eqnarray*}
\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^{\!\!2} + \left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^{\!\!2} &=& \left(-3a\cos^2\theta\sin\theta\right)^{\!2}+\left(3a\sin^2\theta\cos\theta\right) \\ \\
&=&9a^2\cos^4\theta\sin^2\theta+9a^2\sin^4\theta\cos^2\theta \\ \\
&\equiv& 9a^2\cos^2\theta\sin^2\theta
\end{eqnarray*}
Since $0 \le \theta \le \pi$ we need to take care because $\cos\theta$ changes sign. We have
\begin{eqnarray*}
A &=& 2\pi\int_0^{\pi} a\sin^3\theta\sqrt{9a^2\cos^2\theta\sin^2\theta}~\mathrm{d}\theta \\ \\
&=& 6\pi a^2 \int_0^{\pi} \sin^3\theta \cdot |\cos\theta| \cdot |\sin\theta|~\mathrm{d}\theta
\end{eqnarray*}
Since $\sin \theta \ge 0$ for all $0 \le \theta \le \pi$ we have $|\sin\theta| \equiv \sin\theta$. However, $\cos\theta \ge 0$ for all $0 \le \theta \le \frac{\pi}{2}$ and $\cos\theta \le 0$ for all $\frac{\pi}{2} \le \theta \le \pi$. Hence $|\cos\theta| \equiv \cos\theta$ for all $0 \le \theta \le \frac{\pi}{2}$, while $|\cos\theta| \equiv -\cos\theta$ for all $\frac{\pi}{2} \le \theta \le \pi$. It follows that
\begin{eqnarray*}
A &=& 6\pi a^2 \int_0^{\pi/2} \sin^4\theta\cos\theta~\mathrm{d}\theta - 6\pi a^2\int_{\pi/2}^{\pi} \sin^4\theta\cos\theta~\mathrm{d}\theta \\ \\
&=& 6\pi a^2\left[\frac{1}{5}\sin^5\theta\right]_0^{\pi/2}-6 \pi a^2\left[\frac{1}{5}\sin^5\theta\right]_{\pi/2}^{\pi} \\ \\
&=& \frac{12}{5}\pi a^2
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area between the functions I have a homework where I have to find the area between the functions $y=x-2$ and $y=\sqrt{7-2x}$.
Computing the integral I have found $A=\frac {x^2}2-2x-\frac13 (7-2x)^{3/2}$; The upper limit is $\frac72$ and the downlimit is $2$.
After make the computing I have found $A= -0.607051$.
I think there is something wrong with the answer.
| First, you need to draw a picture:
The two curves cross, so you will need to divide this into two separate integrals. They cross when $x-2=\sqrt{7-2x}$, so that $(x-2)^2 = x^2-4x+4 = 7-2x$. Solving gives $x=3$ (and $x=-1$, which you don't care about here). The integrals of the two functions are
$$\int (x-2)\,dx = \frac{x^2}{2} - 2x + C,\qquad
\int \sqrt{7-2x}\,dx = -\frac{1}{3}(7-2x)^{3/2} + C$$
(note that the integral in your post above is incorrect). Therefore the enclosed area is
$$
\int_2^3 (\sqrt{7-2x}-(x-2))\,dx + \int_3^{7/2} ((x-2)-\sqrt{7-2x})\,dx,$$
which evaluates to $-\frac{13}{24} + \sqrt{3}\approx 1.19038$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$ Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$, where $n$ is a non-negative integer.
So, I went about proving this using Fermat's theorem.
So I had $11^{n}\cdot 11^2+(12^2)^n= 0\ (\mod 133)$
then $11^n\cdot 11^2+1728^n=0\ (\mod133)$
and finally $1^n+132^n=0\ (\mod 133)$.
Then I said $133^n=0\ (\mod133)$ because $133$ raised to any power will have a remainder of $0$. Did I apply Fermat's theorem correctly?
| since $133=7*19$, first show that $$7\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$
then $$19\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$
Since $gcd(7,19)=1$, the result follows.
$$ 11^{n+2}\equiv 11^{n}\cdot 11^2\equiv 2\cdot11^{n}\bmod 7$$
$$12^{2n+1}\equiv 5\cdot 144^{n}\equiv 5\cdot 11^{n}\bmod 7$$
So $$11^{n+1} + 12^{2n+1} \equiv 7\cdot11^n\equiv 0 \bmod 7$$
$$..........................................................................$$
$$11^{n+2}\equiv 121\cdot 11^n\equiv 7\cdot11^{n}\bmod 19$$
$$12^{2n+1}\equiv 12\cdot 144^{n}\equiv 12\cdot 11^{n}\bmod 19$$
So
$$11^{n+1} + 12^{2n+1} \equiv 19\cdot11^n\equiv 0 \bmod 19$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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if $\frac{1}{1+x+f(y)}+\frac{1}{1+y+f(z)}+\frac{1}{1+z+f(x)}=1$ find the function $f(x)$ Find all functions $f(x):(0,\infty)\to(0,\infty) $satisfying
$$\dfrac{1}{1+x+f(y)}+\dfrac{1}{1+y+f(z)}+\dfrac{1}{1+z+f(x)}=1$$
whenever $x,y,z$ are positive numbers and $xyz=1$
I know this if
$$xyz=1\Longrightarrow \dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}=1$$
because
\begin{align*}\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}&=\dfrac{1}{1+x+xy}+\dfrac{x}{x+xy+xyz}+\dfrac{xy}{xy+xyz+x^2yz}\\
&=\dfrac{1+x+xy}{1+x+xy}\\
&=1
\end{align*}
so I guess
$$f(x)=\dfrac{1}{x}$$
But I can't prove it.Thank you
| Consider a function $f$ that satisfies the proposed condition.
*
*Setting $(x,y,z)=(1,1,1)$ we see that $f(1)=1$.
*Now, consider $t>0$ and let $a=f(t)$, $b=f(1/t)$. Setting $(x,y,z)=(t,1/t,1)$ we get
$$
\frac{1}{1+t+b}+\frac{1}{2+1/t}+\frac{1}{2+a}=1\tag{1}
$$
and setting $(x,y,z)=(1/t,t,1)$ we get
$$
\frac{1}{1+1/t+a}+\frac{1}{2+t}+\frac{1}{2+b}=1\tag{2}
$$
Now it is easy to check that this system of two equations with unknowns $a$ and $b$ has a unique solution $(a,b)=(1/t,t)$. In particular, $f(t)=1/t$.
Indeed, from the $(1)$ we get
$$
b=\frac{1+3t-a t^2}{1+a+a t}
$$
and from the $(2)$ we get
$$
b=\frac{3t+(1-a)t^2}{1+at+a t^2}
$$
Equating these two expressions yields a simple equation equivalent to $(at-1)^2=0$, (for $t\ne1$.) So, $a=1/t$.
*Conversely, it is easy to check that $t\mapsto 1/t$ is a solution to the proposed problem. So, it is the only one.$\qquad\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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Constructing a polynomial with certain zeroes. I want to construct a polynomial $f(x)$ that has zeroes at $-9,\,-5,\,0,\, 5,\, 9$.
Can somebody provide a method (or perhaps some hints) for solving this?
| Zeroes at $-9$, $-5$, $0$, $5$, $9$... This means that the polynomial will be of the form:
$$f(x)=a(x)(x+9)(x+5)(x-5)(x-9)$$
Where $a$ is a constant. Any value of $a$ will work here. These are all correct answers:
$$f(x)=x(x+9)(x+5)(x-5)(x-9)$$
$$f(x)=\sqrt 2(x)(x+9)(x+5)(x-5)(x-9)$$
$$f(x)=\pi x(x+9)(x+5)(x-5)(x-9)$$
How did I construct this? The zeroes are $-9$, $-5$, $0$, $5$, $9$, which means that these $5$ values of $x$ will make $f(x)$ equal to $0$. Anything times $0$ will make the product $0$. These $5$ values will make $1$ factor equal to $0$, which makes $f(x)=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving implicit equation for x or y I want to solve the following equation for $x$ or $y$ (does not matter wich one) analytically.
$$\sqrt[3]{x+y} + \sqrt[3]{x-y} = 1$$
Wolframalpha returns following solution, but I could not think of a way how to get there:
$$ x+1 \neq 0 \qquad y = \frac{(x+1) \sqrt{8 x-1}}{3 \sqrt{3}}$$
Is there a nice 'tool' I do not know for solving this?
| $$(\sqrt[3]{x+y} + \sqrt[3]{x-y})^3 = 1$$
$$x+y+3(\sqrt[3]{x+y}\cdot\sqrt[3]{x-y})\underbrace{(\sqrt[3]{x+y} + \sqrt[3]{x-y})}_1+x-y=1$$
$$2x+3(\sqrt[3]{x+y}\cdot\sqrt[3]{x-y})=1$$
$$3(\sqrt[3]{x+y}\cdot\sqrt[3]{x-y})=1-2x$$
$$3(\sqrt[3]{x^2-y^2})=1-2x$$
$$27(x^2-y^2)=(1-2x)^3$$
$$27y^2 = 8 x^3+15 x^2+6 x-1$$
$$27y^2 = (x+1)^2 (8 x-1)$$
$$y = \frac{(x+1) \sqrt{8 x-1}}{3 \sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the average value of a function! over a region! Completely related to this question:
Finding surface area of part of a plane that lies inside a cylinder???
Find the average value of the function $f(x, y, z) = x^2yz$ over $S$
How does one do that?? I would think it would be just taking an average over the area, but the area is $S$ so I don't know how to be that specific!! Please help!!!
| For the two linked problems, here are two views of a graph of the cylinder $ \ x^2 \ + \ y^2 \ = \ 3 \ $ and the oblique plane $ x \ + \ 2y \ + \ 3z \ = \ 1 \ $ :
In order to find the average value of $ \ f(x,y,z) \ = \ x^2 \ y \ z \ $ over the region $ \ S \ $ of the plane lying within the cylinder, it will help to "evaluate" the function on the plane. We can re-write the equation of the plane to express $ \ z \ $ as the function $ \ z \ = \ \frac{1}{3} \ ( \ 1 \ - \ x \ - \ 2y \ ) \ $ . We then want to integrate $ \ f \ $ over the projected area of $ \ S \ $ onto the $ \ xy-$ plane:
$$ \iint_S \ x^2 \ y \ \cdot \frac{1}{3} ( \ 1 \ - \ x \ - \ 2y \ ) \ \ dA \ \ . $$
It will be convenient to use polar coordinates, since the projected area is simply a circle of radius $ \ \sqrt{3} \ $ centered on the origin; this gives us
$$ \frac{1}{3} \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^2 \ \cos^2 \theta) \ (r \ \sin \ \theta) \ ( \ 1 \ - \ r \ \cos \ \theta \ - \ 2r \ \sin \ \theta \ ) \ \ r \ dr \ d\theta $$
$$ = \ \ \frac{1}{3} \ \left[ \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^4 \ \cos^2 \theta \ \sin \ \theta) \ \ dr \ d\theta \ \ - \ \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^5 \ \cos^3 \theta \ \sin \ \theta) \ \ dr \ d\theta \quad - \ \ 2 \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^5 \ \cos^2 \theta \ \sin^2 \theta) \ \ dr \ d\theta \ \right] \ \ . $$
The first two integrals produce antiderivatives which are just powers of cosine, making the integral values zero over one full period. So we only need to calculate
$$ -\frac{2}{3} \ \int_0^{2 \pi} \cos^2 \theta \ \sin^2 \theta \ \ d\theta \ \int_0^{\sqrt{3}} \ r^5 \ \ dr \ = \ -\frac{2}{3} \ \cdot \ \frac{\pi}{4} \ \cdot \ \frac{27}{6} \ = \ -\frac{3 \pi}{4} \ \ . $$
As for the average value of the function, I am taking it that since the question asks for the average to be taken over the region $ \ S \ $ , we would divide this last result by the area found by john in the linked problem, thus,
$$ \langle \ f \ \ \rangle_S \ = \ \frac{- \ 3 \pi / 4}{\pi \ \sqrt{14}} \ = \ -\frac{3}{4 \ \sqrt{14}} \ \ \text{or} \ \ -\frac{3 \ \sqrt{14}}{56} \ \ . $$
Is this a plausible result? If we look at the function $ \ x^2 \ y \ z \ $ , its sign is determined by the factors $ \ y \ $ and $ \ z \ $ . The oblique plane crosses the $ \ xy-$ plane on the line $ \ x \ + \ 2y \ = \ 1 \ $ , so $ \ z \ $ is negative "below" that line in the $ \ xy-$ plane and positive "above" it. On the right below is a graph depicting the portions of the region $ \ S \ $ where $ \ x^2 \ y \ z \ $ is positive [green] and negative [red] ; as our region of integration is dominated by "negative" sections. So it is credible that the average value $ \langle \ f \ \ \rangle_S \ $ should be negative. The graph of $ \ \frac{1}{3} \ x^2 \ y \ ( \ 1 \ - \ x \ - \ 2y \ ) \ $ at left below supports this expectation.
| {
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"url": "https://math.stackexchange.com/questions/799914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating the limit of $\lim_{x\to\infty}(\sqrt{\frac{x^3}{x+2}}-x)$. How do I evaluate this limit:
$$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)$$
I tried to evaluate this using rationalizing the denominator, numerator and L'Hospital rule for nearly an hour with no success.
| Rationalizing, observe that:
\begin{align*}
\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)
&= \lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)\left(\dfrac{\sqrt{\dfrac{x^3}{x+2}} + x}{\sqrt{\dfrac{x^3}{x+2}} + x}\right) \\
&= \lim_{x\to\infty} \dfrac{\dfrac{x^3}{x+2} - x^2}{\sqrt{\dfrac{x^3}{x+2}} + x} \\
&= \lim_{x\to\infty} \dfrac{\dfrac{-2x^2}{x+2}}{\sqrt{\dfrac{x^3}{x+2}} + x} \cdot \frac{\dfrac{1}{x}}{\dfrac{1}{x}}\\
&= \dfrac{\lim\limits_{x\to\infty} \dfrac{-2x}{x+2}}{\lim\limits_{x\to\infty} \left(\sqrt{\dfrac{x}{x+2}} + 1\right)}\\
&= \frac{-2}{\sqrt{1}+1} \\
&= -1
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the integral $\int_0^{2\pi}\ln|a+b\sin x|dx$ where $0\lt a\lt b$ Find the value of the integral $$\int_0^{2\pi}\ln|a+b\sin x|dx$$ where $0\lt a\lt b$. What is the use of this inequality. I tried to integrate the integral by parts, but the integral of the 2nd term was quite messy.Please help.
| This is just for $0<b < a $
\begin{align*}
I &= \int_0^{2\pi}\log(a+\sqrt{a^2 - b^2}+be^{i\theta})d\theta + \int_0^{2\pi}\log(a+\sqrt{a^2 - b^2}+be^{-i\theta})d\theta\\
&= \int_0^{2\pi} \log \left(2a^2 +2a\sqrt{a^2 - b^2} + (a + \sqrt {a^2 - b^2}) b (e^{i\theta} + e^{-i\theta}) \right )d\theta\\
&= \int_0^{2\pi} \log (2 (a + \sqrt{a^2 - b^2}))d\theta + \int_0^{2\pi }\log(a + b \cos\theta)d\theta\\
\end{align*}
By Gauss MVT, the value of top two integral is
$$4\pi\log(a + \sqrt{a^2 -b^2})$$
Hence the for $0<b<a$ is
$$\int_0^{2\pi} \log(a+ b\cos\theta)d\theta = 2 \pi \log \left(a + \sqrt{a^2 - b^2}\over 2\right)$$
EDIT:: The integral for $\log (a + b \sin \theta )$ is equal to that of $\log (a + b \cos \theta )$ as $\theta$ goes around $2 \pi$. It can also be done by changing $be^{-i\theta}$ to $-b e^{-i\theta}$ is above derivation.
For $0<a < b$, the integral is equal to real part of the above solution as mentioned by @Lucian in comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/801136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int_0^{\pi/2} \ \frac{\cos{x}}{\sqrt{1+\cos{x}}} \, \mathrm{d}x.$ I need to somehow evaluate the following:
$$ \int_0^{\pi/2} \ \dfrac{\cos{x}}{\sqrt{1+\cos{x}}} \, \mathrm{d}x. $$
Can anyone give me any hints/pointers? I've tried to use parts, and some feeble substitutions, but to no avail :(
Thanks
| Hint: $$\frac{1+\cos x}{2} = \cos^2\left(\frac x2\right)\Rightarrow\\\int\frac{\cos x}{\sqrt{1+\cos x}} \, \mathrm{d}x=\\\int\frac{\cos x}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\\int \frac{2\cos^2\left(\frac x2\right)-1}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\\int \sqrt{2}\cos\left(\frac x2\right) \, \mathrm{d}x-\frac{1}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\\sqrt{2}\int \cos\left(\frac x2\right) \, \mathrm{d}x- \int \frac{1}{\sqrt{2}\cos\left(\frac x2\right)} \, \mathrm{d}x=\\2\sqrt{2}\sin\left(\frac x2\right)-\frac{1}{\sqrt2}\int \sec\left(\frac x2\right) \, \mathrm{d}x=\\2\sqrt{2}\sin\left(\frac x2\right)-\frac{1}{\sqrt2}\cdot 2 \left(-\ln\left(\cos\left(\frac x4\right)-\sin\left(\frac x4\right)\right)+\ln\left(\cos\left(\frac x4\right)+\sin\left(\frac x4\right)\right)\right)+C=\\\sqrt{2}\left(\sqrt2\sin\left(\frac x2\right)+\ln\left(\cos\left(\frac x4\right)-\sin\left(\frac x4\right)\right)-\ln\left(\cos\left(\frac x4\right)+\sin\left(\frac x4\right)\right) \right)+C$$
Thus $$\int_0^{\pi/2} \ \dfrac{\cos{x}}{\sqrt{1+\cos{x}}} \, \mathrm{d}x=\\\sqrt{2}\left(\sqrt2\sin\left(\frac \pi4\right)+\ln\left(\cos\left(\frac \pi8\right)-\sin\left(\frac \pi8\right)\right)-\ln\left(\cos\left(\frac \pi8\right)+\sin\left(\frac \pi8\right)\right) \right)-\sqrt{2}\left(\sin\left(0\right)+\ln\left(\cos\left(0\right)-\sin\left(0\right)\right)-\ln\left(\cos\left(0\right)+\sin\left(0\right)\right)\right)=\\2-2 \sqrt2 \cdot \frac{1}{\tanh\left(\tan(\frac \pi8)\right)}\approx 0,75$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/801483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove homogeneous inequality Prove that for all $a,b,c>0$ we have $$\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2}\leq\frac{1}{4}+\frac{4abc}{(a+b)(b+c)(c+a)}.$$
Please help me prove this homogeneous inequality.
| Let
$$
D=\frac{1}{4}+\frac{4abc}{(a+b)(b+c)(c+a)}-
\Bigg(\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2}\Bigg) \tag{1}
$$
We must show that $D$ is nonnegative. It will suffice to show
that
$$
D=\frac{1}{4}.\frac{(b-a)^2(c-a)^2(c-b)^2}{(a+b)^2(b+c)^2(c+a)^2} \tag{2}
$$
But this follows easily enough from the expanding the identity
$$
(b-a)^2(c-a)^2(c-b)^2=((b+a)^2-4ab)((c+a)^2-4ac)((c+b)^2-4bc)
$$
(I do not put in all the details as this is homework).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Partial fraction integral Question:
$\int \dfrac{5 }{(x+1) (x^2 + 4) } dx $
Thought process:
I'm treating it as a partial fraction since it certainly looks like one.
I cannot seem to solve it besides looking at it in the "partial fraction" way.
My work:
1) Focus on the fraction part first ignoring the $\int $ and $dx$ for the moment.
Multiply $(x+1) (x^2+4)$ on both sides of the equation and get:
$5$ = $\dfrac{A }{(x+1) } $ + $\dfrac{Bx+C }{(x^2+4) } $
Note: the x^2 + 4 is irreducible which explains the Bx+c as the numerator.
$ 5 = A(x^2+ 4) + Bx + C(x+1) $
I tried x = -1 which knocks out C:
$5 = A5 + - B $
I also tried x = 0 which knocks out B
$5 = A4 + C$
A is a lot harder to knock out since the squared changes the picked value to be positive.
I decided to add the two found equation together and get
$10 = A9 - B + C $
Now I am officially stuck at this step.
| Note as in the comments you will have
$$
\dfrac{5 }{(x+1) (x^2 + 4) }
= \frac{A}{x+1} + \frac{Bx + C}{x^2+4}
$$
Cross multiplying and solving the equations gives
$$
\dfrac{5 }{(x+1) (x^2 + 4) }
= \frac{1}{x+1} - \frac{x - 1}{x^2+4}
$$
Which is easier to integrate. A sneakier way than solving the equations is the following
$$
\frac{5}{(x+1)(x^2+4)}
=
\frac{(x^2 + 4) + (1 - x^2)}{(x+1)(x^2+4)}
= \frac{1}{x+1} + \frac{(1-x)(1+x)}{(x+1)(x^2+4)}
$$
As wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
log and poisson-like integral Here is a fun looking one some may enjoy.
Show that:
$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$
| Denote
$$
I(r)
=\int_{0}^{1}\log\left(\frac{x^{2}+2x r +1}{x^{2}-2x r+1}\right)\cdot \frac{1}{x}dx
$$
then
$$
\begin{align}
\frac{dI}{dr}
&=\int_0^1 \frac{4 \left(x^2+1\right)}{\left(2-4 r^2\right) x^2+x^4+1} dx\\
&=\int_0^1 \left(\frac{2}{x^2+2rx+1}+\frac{2}{x^2-2 r x+1} \right)dx\\
&=\frac{2 \tan ^{-1}\left(\frac{x+r}{\sqrt{1-r^2}}\right)}{\sqrt{1-r^2}}\Biggl|_0^1
+\frac{2 \tan ^{-1}\left(\frac{x-r}{\sqrt{1-r^2}}\right)}{\sqrt{1-r^2}}\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\left(\tan ^{-1}\left(\frac{x+r}{\sqrt{1-r^2}}\right)+\tan ^{-1}\left(\frac{x-r}{\sqrt{1-r^2}}\right)\right)\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\tan^{-1}\frac{\frac{x+r}{\sqrt{1-r^2}}+\frac{x-r}{\sqrt{1-r^2}}}{1-\frac{x+r}{\sqrt{1-r^2}}\frac{x-r}{\sqrt{1-r^2}}}\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\tan^{-1}\frac{2x\sqrt{1-r^2}}{1-x^2}\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\frac{\pi}{2}\\
&=\frac{\pi}{\sqrt{1-r^2}}\\
\end{align}
$$
Since $I(0)=0$, then
$$
I(\cos a)=I(0)+\int_0^{\cos a} \frac{dI}{dr}dr=\int_0^{\cos a}\frac{\pi}{\sqrt{1-r^2}}dr=\pi\sin^{-1}s|_0^{\cos a}=\frac{\pi^{2}}{2}-\pi a
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_0^\infty\frac{\ln x}{1+x^2}dx$ Evaluate $$\int_0^\infty\frac{\ln x}{1+x^2}\ dx$$
I don't know where to start with this so either the full evaluation or any hints or pushes in the right direction would be appreciated. Thanks.
| Another general approach :
Consider
$$
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx.\tag1
$$
Rewrite $(1)$ as
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac1{a^2}\int_0^\infty\frac{x^{b-1}}{1+\left(\frac{x}{a}\right)^2}\ dx.\tag2
\end{align}
Putting $x=ay\;\color{blue}{\Rightarrow}\;dx=a\ dy$ yields
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=a^{b-2}\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy,
\end{align}
where
\begin{align}
\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy=\frac{\pi}{2\sin\left(\frac{b\pi}{2}\right)}.
\end{align}
Hence
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac\pi2\cdot\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}.\tag3
\end{align}
Differentiating $(3)$ with respect to $b$ and setting $b=1$ yields
\begin{align}
\int_0^\infty\frac{\partial}{\partial b}\left[\frac{x^{b-1}}{a^2+x^2}\right]_{b=1}\ dx&=\frac\pi2\frac{\partial}{\partial b}\left[\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}\right]_{b=1}\\
\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx&=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
Thus
$$
\int_0^\infty\frac{\ln x}{x^2+1}\ dx=\large\color{blue}{0}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 7,
"answer_id": 0
} |
Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? For example:
$10 = 1 + 2 + 3 + 4$
$11 = 1 - 2 + 3 \times 4$
$12 = 3 \times 4$
$13 = -(1 - 2) + 3 \times 4$
$14 = 2 + 3 + 4 + 5$
$15 = 1 + 2 + 3 + 4 + 5$
$16 = (2/3)(4/5)(6 + 7 + 8 + 9)$
$17 = 1 + (2/3)(4/5)(6 + 7 + 8 + 9)$
$18 = (1 + 2)3!$
$19 = 4! - 5$
Obviously any triangular number has such an expression, as well as numbers one less than a triangular number. But how to prove it for other numbers? Or is there a counterexample?
| Every integer $n$ can be written as
$$n= (n+1)+(n+2)-(n+3)$$
where $(n+1)$ and $(n+2)$ are of course two consecutive integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of $\sin {\frac{31 \pi}3}$ The task was to find out the value of $$\sin\frac{31\pi}3$$
This is a example in my book in which following steps are shown:
$$\begin{align}
\sin\frac{31\pi}3& =\sin\left(10\pi +\frac\pi3\right)\\
&=\sin\frac\pi3\\
&=\frac{\sqrt3}2
\end{align}$$
I cannot understand step 3, $=\sin\frac\pi3$
| You can confirm this by using the angle-sum formula:
$$\begin{align} \sin(10\pi + \pi/3) & = \sin(10\pi)\cos(\pi/3) + \cos(10 \pi)\sin(\pi/3) \\ \\ & = 0\cdot\frac 12 + 1 \cdot \sin (\pi/3) \\ \\ & = \sin(\pi/3)\\ \\ & = \sqrt 3/2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/812611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Sudoku puzzle with exactly 3 solutions While published sudoku puzzles typically have a unique solution, one can easily conceive of a sudoku puzzle with two solutions. However, is it possible to construct a sudoku puzzle with exactly 3 different solutions?
Inspired by https://puzzling.stackexchange.com/a/6.
| You could use this as a model (this works for $4 \times 4$ but needs to be adapted for $9\times 9$):
$\begin{array}{cccc}
1 & 2 & & \\
3 & 4 & 1 & 2 \\
& 1 & & \\
& 3 & & 1 \end{array} $
With solutions
$\begin{array}{cccc}
1 & 2 & 3 & 4 \\
3 & 4 & 1 & 2 \\
2/4 & 1 & 4/2 & 3 \\
4/2 & 3 & 2/4 & 1 \end{array} $
and
$\begin{array}{cccc}
1 & 2 & 4 & 3 \\
3 & 4 & 1 & 2 \\
2 & 1 & 3 & 4 \\
4 & 3 & 2 & 1 \end{array} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Multiplication of odds vs. multiplication of probabilities I always believed that probabilities could be multiplied, until I encountered a statement in Machine Learning by Peter Flach about odds:
"Bayes’ rule tells us that we should simply multiply them: 1:6 times 4:1 is 4:6, corresponding to a spam probability of $0.4$."
But converting to probabilities, $1:6 = 1/7$ and $4:1 = 4/5$.
Multiplying the probabilities we get $4/35$ ~$= 0.11$
Does Bayes' Theorem work differently for odds and probabilities? What seems to be the problem?
| Suppose that the prior probability of a message being spam is $\dfrac{1}{7}$, and the conditional probability of a message triggering the spam filter is $k$ if it is spam but $\dfrac{k}{4}$ if it is not spam for some $k$.
Then, given the spam filter is triggered, the posterior probability the message is spam is $\dfrac{\frac{1}{7}\times k}{ \frac{1}{7}\times k+ \frac{6}{7}\times \frac{k}{4} } = \dfrac{4}{10}$.
This is equivalent to the same statement that if the prior odds for spam are $1:6$ and the likelihood ratio for the spam filter being triggered is $4:1$ then the posterior odds for spam given the spam filter is triggered is $4:6$.
Effectively you have said the earlier calculation is equivalent to $\dfrac{\dfrac{\frac{1}{7}\times k}{ \frac{1}{7}\times k+ \frac{6}{7}\times \frac{k}{4} }}{\dfrac{\frac{6}{7}\times \frac{k}{4}}{ \frac{1}{7}\times k+ \frac{6}{7}\times \frac{k}{4} }} = \dfrac {\dfrac{4}{10}}{\dfrac{6}{10}}$ or in a simplified form $\dfrac{1}{6} \times \dfrac{4}{1}=\dfrac{4}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
if A diagonalizable then show that $a=0$ Let $A= \begin{pmatrix}
2 & 0 & 0\\
a & 2 & 0 \\
b & c & -1\\
\end{pmatrix}$
if A diagonalizable then show $a=0$.
$P_A(x)=|xI-A|=(x-2)^2(x+1)=x^3-3x^2+4$
since A diagonalizable there is a P matrix such as $A=P^{-1}DP$
$D= \begin{pmatrix}
2 & 0 & 0\\
0 & 2 & 0 \\
0 & 0 & -1\\
\end{pmatrix}$
from cayley hamil. $A^3-3A^2+4I=0$ how do we continue?
| Check whether there are 2 different eigenvectors to $2$. An $2$-eigenvector $v$ satisfies
$$
\begin{pmatrix}
0 & 0 & 0\\
a & 0 & 0 \\
b & c & -3
\end{pmatrix}
\begin{pmatrix}
v_1 \\ v_2 \\ v_3
\end{pmatrix}=0.
$$
This has two independent solutions, iff the above matrix has rank 1, which is iff $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to find the limit without l'Hopital's Rule or series Is it possible to find $$\lim_{x\to0}\frac{\sin(1-\cos(x))}{x^2e^x}$$
without using L'Hopital's Rule or Series or anything complex but just basic knowledge (definition of a limit, limit laws, and algebraic expansion / cancelling?)
| \begin{align}
\lim_{x \to 0} \frac{\sin(1 - \cos x)}{\mathrm{e}^x \cdot x^2}
&= \lim_{x \to 0} \frac{\sin(1 - \cos x)}{1 - \cos x}
\cdot \frac{1 - \cos x}{x^2} \cdot \mathrm{e}^{- x} \\
&= \lim_{u \to 0} \frac{\sin u}{u}
\cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2}
\cdot \lim_{x \to 0} \mathrm{e}^{- x} \\
&= 1\cdot \lim_{x \to 0} \frac{2 \sin^2 \frac{x}{2}}{x^2} \cdot 1 \\
&= \lim_{u \to 0} \frac{2 \sin^2 u}{4 u^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Use Lagrange Multipliers to determine max and min Using Lagrange Multipliers, determine the maximum and minimum of the function
$f(x,y,z) = x + 2y$
subject to the constraints
$x + y + z = 1$ and $y^2 + z^2 = 4$:
Justify that the points you have found give the maximum and minimum of $f$.
So,
$$
\nabla f = (λ_1)\nabla g_1 + (λ_2)\nabla g_2
$$
I get to this point
$$
(1,2,0) = λ_1(1,1,1) + λ_2(0,2y,2z)
$$
Where do I go from here to find the critical points ect.
| We can find the extreme values using one constraint only: $z = 1 - x - y$, so $y^2 + (1-x-y)^2 = 4$.
Thus: $g(x,y) = y^2 + 1 + x^2 + y^2 - 2x - 2y + 2xy - 4 = 0$.
So: $\nabla f = \lambda\nabla g$ gives:
$(1,2) = \lambda(2x-2+2y,4y-2+2x)$. Thus:
$\dfrac{1}{2x-2+2y} = \dfrac{2}{4y-2+2x}$. Hence:
$4y-2+2x = 4x - 4 + 4y$. So $x = 1$, and $y^2 + y^2 = 4$. So: $y^2 = 2$, and $y = \pm \sqrt{2}$. Thus we easily see that extrema are:
$f_{min} = f(1,-\sqrt{2}) = 1 - 2\sqrt{2}$, and $f_{max} = f(1,\sqrt{2}) = 1 + 2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
formula for number triangles Hi, I have a triangle starting from $0$ and going up by one on the bottom row until there are $r$ items on the bottom row and there are $r$ rows a number is formed by adding the two numbers towards the fat end of the triangle together.
Example:
$$x \downarrow\ $$
$$
\begin{matrix}12\end{matrix}\\
\begin{matrix}4 & 8\end{matrix}\\
\begin{matrix}1 & 3 & 5\end{matrix}\\
\begin{matrix}0 & 1 & 2 & \boxed{3}\end{matrix}
$$
$$\begin{matrix}& & r - 1 \uparrow\end{matrix}$$
$\therefore$ when $r = 4$, $x = 12$.
When $r = 8$
$$
\begin{matrix}0\ \ & 1\ \ & 2\ \ & 3\ \ & 4\ \ & 5\ \ & 6\ \ & 7\ \ \end{matrix}\\
\begin{matrix}1\ \ & 3\ \ & 5\ \ & 7\ \ & 9\ \ & 11\ & 13\ \end{matrix}\\
\begin{matrix}4\ \ & 8\ \ & 12\ & 16\ & 20\ & 24\ \end{matrix}\\
\begin{matrix}12\ & 20\ & 28\ & 36\ & 44\ \end{matrix}\\
\begin{matrix}32\ & 48\ & 64\ & 80\ \end{matrix}\\
\begin{matrix}80\ & 112 & 144 \end{matrix}\\
\begin{matrix}192 & 256 \end{matrix}\\
\begin{matrix}448\end{matrix}\\
$$
$\therefore x = 448$
How do I work out $x$ from $r$? What is the formula? Any help by suggesting a new viewpoint or formula(e) would be greatly appreciated. Thank you.
| A recursive solution: number the rows $0$ to $r-1$ with $0$ being the fattest, notice that the difference between consecutive numbers in row $k$ is $2^k$.
Denote by $f(n)$ the number in the last row of a triangle with $n$ rows. Then $f(n)=2f(n-1)+2^{n-2}$
Notice $f(1)=0,f(2)=1,f(3)=4$
From this recurrence we pass to $f(n)=(n-1)2^{n-2}$
The proof is by induction. Suppose $f(n)=(n-1)2^{n-2}$
then $f(n+1)=(n-1)2^{n-1}+2^{n-1}=n(2^{n-1})$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to find the minimum/maximum distance of a point from elipse I have the point $(1,-1)$
and the ellipse $$x^2/9 + y^2/5 = 1 $$
How to find the minimum and maximum distance of the point from the ellipse ?
from exploring the ellipse I know that $$a = 3$$ , $$b =\sqrt{5}$$
$$ c = \sqrt{a^2-b^2} =\sqrt{9-5} = \sqrt{4}=2$$
the eccentricity of the ellipse is $$e=c/a = 2/3 $$
the center is $(0,0)$ and the guides are $$x=3,~~x=-3,~~y=\sqrt{5},~~y=-\sqrt{5}$$
the focus points are : $(2,0)$ and $(-2,0)$
how from all of that do I find the requested in the question?
| The lack of symmetry in the geometric arrangement doesn't work in our favor, but fortunately the coefficients in the equations are only unpleasant to work with near the very end. As indicated by Padmanabha P Simha, we can use the Lagrange-multiplier method to aid in starting our work. We can extremize the "distance-squared" function, as Kaj Hansen observes, since the distance from $ \ (1, \ 1) \ $ to a point $ \ (X, \ Y) \ $ on the ellipse can only be non-negative. We will derive the constraint function from the equation for the ellipse as
$$ \frac{x^2}{9} \ + \ \frac{y^2}{5} \ = \ 1 \ \ \rightarrow \ \ g(x,y) \ = \ 5x^2 \ + \ 9y^2 \ - \ 45 \ \ . $$
For the distance-squared function $ \ f(x,y) \ = \ (x - 1)^2 \ + \ (y + 1)^2 \ $ , the Lagrange equations are
$$ \frac{\partial f}{\partial x} \ = \ \lambda \ \cdot \ \frac{\partial g}{\partial x} \ \ \Rightarrow \ \ 2 \ (x - 1) \ = \ \lambda \ \cdot \ 10x \ \ , $$
$$ \frac{\partial f}{\partial y} \ = \ \lambda \ \cdot \ \frac{\partial g}{\partial y} \ \ \Rightarrow \ \ 2 \ (y + 1) \ = \ \lambda \ \cdot \ 18y \ \ . $$
The result for the multiplier is $ \ \lambda \ = \ \frac{1}{5} \ \frac{x - 1}{x} \ = \ \frac{1}{9} \ \frac{y + 1}{y} \ $ . (We will make use of this relation shortly.)
Alternatively, we can show (for instance, by applying Lagrange multipliers) that the segments from a point external to a curve to an extremal point of the curve are perpendicular (normal) to the curve. If we differentiate the equation for the ellipse implicitly with respect to $ \ x \ $ , we obtain
$$ \frac{d}{dx} \ [ \ 5x^2 \ + \ 9y^2 \ ] \ = \ \frac{d}{dx} \ [45] \ \ \Rightarrow \ \ \frac{dy}{dx} \ = \ - \frac{5}{9} \ \frac{x}{y} \ \ , $$
giving us the slope of a tangent line to a point on the ellipse. The slope of a normal line at that point is then $ \ \frac{9}{5} \ \frac{y}{x} \ $ . The line segment from $ \ ( 1, \ 1) \ $ to that point has a slope of $ \ \frac{y - (-1)}{x - 1} \ $ , so we find $ \ \frac{y + 1}{x - 1} \ = \ \frac{9}{5} \ \frac{y}{x} \ $ , which is equivalent to our Lagrange result (since this is what the method locates for us, as nicely illustrated by heropup's tangent circles).
We can re-arrange this equation as $ \ y \ = \ \frac{5x}{4x - 9} \ $ , which is a hyperbola. Upon graphing this along with our ellipse, we see that the two curves intersect where the local normals correspond to the lines also passing through $ \ ( 1, \ 1) \ $; the added rays indicate those normal lines.
We need to locate those intersections, so we may insert our expression for $ \ y \ $ into the ellipse equation to produce
$$ 5x^2 \ + \ 9 \ \left( \ \frac{5x}{4x - 9} \ \right)^2 \ - \ 45 \ = \ 0 \ \ \Rightarrow \ \ x^2 \ - \ 9 \ + \ \frac{9 \cdot 5x^2}{(4x - 9)^2} \ = \ 0 $$
$$ \Rightarrow \ \ \frac{(x^2 - 9)(4x - 9)^2 \ + \ 45x^2}{(4x - 9)^2} \ = \ 0 \ \ . $$
Since $ \ 4x - 9 \ = \ 0 \ $ is not in the domain of the rational function for $ \ y \ $ , we can simply solve for the values of $ \ x \ $ at which the numerator of the left-hand-side ratio is zero. Multiplying out the factors and simplifying leaves us to solve the quartic equation $ \ 16x^4 \ - \ 72x^3 \ - \ 18x^2 \ + \ 648x \ - \ 729 \ = \ 0 \ $ , which I left to WolframAlpha (the exact values for the two real and two complex conjugate solutions are rather horrifying). The closest and most distant points on the ellipse to $ \ (1, \ -1) \ $ and the minimal and maximal distances are then as given in heropup's answer and comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove this inequality: If $|x+3|< 0.5$, show that $|4x+13| < 3$ If $|x + 3| < 0.5$, show that $|4x + 13| < 3$
This is what I've got so far:
$|4x + 13| = |(x + 3) + (3x + 10)|$
by the Triangle Inequality:
$|(x + 3) + (3x + 10)| \le |x + 3| + |3x + 10|$
Now I continue to apply the Triangle Inequality to reach:
$|(x + 3) + (3x + 10)| \le |x + 3| + |x + 3| + |x + 3| + |x + 3| + |1|$
So I come to the conclusion: $|4x + 13| \le 4|x + 3| + 1$
Since $|x + 3| < 0.5$, then $4|x + 3| + 1 < 4\cdot 0.5 + 1 = 3$
Then $|4x + 13| < 3$
Please take a look and let me know if there are errors, if so, enlighten me.
Thank you :)
| Yes, your solution is correct. A quicker approach could go like this: $|4x+13|=|4(x+3)+1|\leq 4|x+3|+1<4\cdot\frac12+1=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/832063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Mass of ellipsoid
We are given the elipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1$$ with density function $$p(x,y,z)=x^2+y^2+z^2$$ Find the mass of the elipsoid.
I used the transformation $x=ar\sin\theta \cos\phi$, $y=br\sin \theta \sin \phi$ and $z=cr\cos\theta$. Spherical coordinates but I multiplied $x,y,z$ by $a,b,c$ respectively.
I calculate the jacobian of this transformation, it is $abcr^2\sin\theta$.
So the mass is $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi$$
But I'm having trouble calculating this integral. I thought maybe we can use that $$\sin^2x+\cos^2x=1$$ but because the coefficients are different it's difficult to use that here. How would I calculate this integral?
| $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi=$$
$$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}a^3bcr^4\sin^3\theta \cos^2\phi drd\theta d\phi+\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}ab^3cr^4\sin^3 \theta \sin ^2\phi dr d\phi+\\\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}abc^3r^4\cos^2 \theta \sin\theta drd\theta d\phi=$$
$$a^3bc\int_{0}^{2\pi} sin^3\theta d\theta\int_{0}^{\pi}cos^2\phi d\phi\int_{0}^{1}r^4dr +ab^3c\int_{0}^{2\pi} sin^3 \theta d\theta \int_{0}^{\pi}\sin ^2\phi d\phi\int_{0}^{1}r^4dr +\\abc^3\int_{0}^{2\pi}cos^2 \theta \sin\theta d\theta\int_{0}^{\pi}d\phi\int_{0}^{1}r^4dr $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/832985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
The Geodesics of a Sphere. I need to find the geodesics of a sphere. Then in polar coordinates
$$x=a \sin\theta \cos\phi \\y=a \sin\theta \sin\phi\\
z=a\cos\theta$$
Then $ds^2=dx^2+dy^2+dz^2$.
Can someone please tell me how is $ds^2=a^2(d\theta^2+\sin^2\theta \, d\phi ^2)$ obtained. I don't know much about spherical coordinates.
| Rather than discuss via comments, here's an explicit list of steps/hints. Perhaps you can look at them only when stuck:
Step 1. Admittedly, the integral you mentioned is not easy. Here's one approach:
\begin{align*}
\phi &= \int \frac{k \, d\theta}{\sin \theta \sqrt{ \sin^2 \theta-k^2}}
= \int\frac{k \, d \theta}{\sin^2\theta\sqrt{1-k^2\csc^2\theta}}
= \int\frac{k \csc^2 \theta \, d \theta}{\sqrt{1-k^2(1+\cot^2 \theta)}}.
\end{align*}
Now use a $u$-sub.
Step 2. Setting $u=\cot \theta$, we have $du=d(\cot \theta)=-\csc^2 \theta \, d\theta$, hence
\begin{align*}
\phi &= \int \frac{-k \, du}{\sqrt{1-k^2(1+u^2)}} = -\int \frac{ \, du}{\sqrt{\left(\frac{1-k^2}{k^2}\right) -u^2}}.
\end{align*}
Step 3. Using the formula $\int (a^2-x^2)^{-1/2} \, dx = \sin^{-1}(x/a)$, we have
\begin{align*}
\phi= - \sin^{-1}\left(\frac{u}{\sqrt{(1-k^2)/k^2}}\right)+C= -\sin^{-1}\left(\frac{k \cot \theta}{\sqrt{1-k^2}}\right)+C.
\end{align*}
Step 4. Compose with sine and use sum/difference formulas: $$k \cot \theta / \sqrt{1-k^2} = \sin(C-\phi)=\sin(C)\cos(\phi)-\cos(C)\sin(\phi).$$
Multiplying by $\sin \theta$, this gives
$$ k \cos \theta /\sqrt{1-k^2} = \sin (C) \cos (\phi) \sin (\theta)-\cos (C)\sin(\phi) \sin(\theta).$$
Step 5. Switch back to Cartesian coordinates:
$$\frac{k}{ \sqrt{1-k^2}}\cdot \frac{z}{a} = \sin(C) \cdot \frac{x}{a} - \cos(C) \cdot \frac{y}{a}.$$
Note that $k$ and $C$ are constants, so this is a linear relation of $x$, $y$, and $z$. What set of points satisfies such a relation? When you intersect the sphere with this set, you get a geodesic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
How to prove inequality $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ Question:
Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$
when I solved this problem, I have see following three variables inequality:
Assumming that $a,b,c>0,a+b+c=3$, show that :$$f(a,b,c)=\dfrac{a}{a+bc}+\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\ge\dfrac{3}{2}$$
solution can see:inequality
I found this three answer all is not true,
1、such as dear @Macavity, in fact
$$\sum_{cyc}\dfrac{a^2}{a^2+abc}\ge\dfrac{16}{\sum_{cyc}(a^2+abc)}$$
the Right not
$\dfrac{16}{4+\sum_{cyc}abc}$
2、and the @ante.ceperic is also not true.in fact
$$a^2+b^2+c^2+d^2+abc+bcd+cda+dab\le 8$$ is not true with $a+b+c+d=4$
such let $a=3$
| Take $1^{st}$ and $3^{rd}$ term apply the inequality, then take $2^{nd}$ and $4^{th}$ term and apply the inequality, you get
$$\ge 2\left(\frac1{1+\sqrt{bd}}+\frac1{1+\sqrt{ac}}\right)$$
again apply the inequality you get $\displaystyle \ge 2\left(\frac2{1+\sqrt{abcd}}\right)$
for $abcd =1$ you get your equality ie $a=b=c=d=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
} |
$\alpha$ and $\beta$ are solution of $a \cdot \tan\theta + b \cdot \sec\theta = c$ show $ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$ If $\alpha$ and $\beta$ are the solution of $$a \cdot \tan\theta + b \cdot \sec\theta = c$$, then show that $$ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$$
I did the following:
$$a \cdot \tan\theta + b \cdot \sec\theta = c$$
Squaring both sides, $$a^2 \cdot \tan^2\theta + b^2 \cdot \sec \theta + 2 \cdot a \cdot b \cdot \tan \theta \cdot \sec \theta = c^2$$
This implies $$(a^2 + b^2) \cdot \sin^2 \theta + 2 \cdot a \cdot b \cdot \sin \theta + (b^2 - c^2) = 0$$
Hence I get $$\sin \alpha + \sin \beta = \frac{-2ab}{a^2 + c^2}$$ and $$\sin \alpha \cdot \sin \beta = \frac{b^2 - c^2}{a^2 - c^2}$$
But I am not able to solve it further.
| Consider the following:
\begin{align*}
c-a\tan \theta & = b \sec \theta\\
(c-a\tan \theta)^2 & = (b\sec \theta)^2\\
(a^2-b^2)\tan^2 \theta -2ac \tan \theta+(c^2-b^2) & = 0.
\end{align*}
Think of this as a quadratic in $\tan \theta$. It has two solutions $\tan \alpha$ and $\tan \beta$. So
\begin{align*}
\tan \alpha + \tan \beta & = \frac{2ac}{a^2-b^2}\\
\tan \alpha \cdot \tan \beta & = \frac{c^2-b^2}{a^2-b^2}
\end{align*}
Now you can compute using
$$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/838082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A Sine-Cosine Integral $\int_{0}^{\frac{\pi}{2}} \frac{ \sin(kx) \ +\ \cos(kx)}{\sin x \ +\ \cos x} \ dx$ What is the value of the integral
\begin{align}
I(k) = \int_{0}^{\frac{\pi}{2}} \frac{ \sin(kx) + \cos(kx)}{\sin(x) + \cos(x)} \ dx
\end{align}
where $k$ is an integer ?
Is it possible to evaluate a related integral
\begin{align}
I(k) = \int_{0}^{\frac{\pi}{2}} \frac{ \sin(kx) - \cos(kx)}{\sin(x) - \cos(x)} \ dx
\end{align}
where $k$ is an integer ?
| Depending on even or odd $k$, the integrals
\begin{align}
I_k = \int_{0}^{\frac{\pi}{2}} \frac{ \sin(kx) + \cos(kx)}{\sin x + \cos x} \ dx
\end{align}
exhibit different behaviors. Their close-forms are derived separately below
\begin{align}
I_{2m+1}
=& \>2\cos\frac{m\pi}2\int_0^{\frac\pi4}\frac{\cos(2m+1)x}{\cos x}dx
=\cos\frac{m\pi}2\left(\frac\pi2-\sum_{k=1}^m \frac{2}k \sin\frac{k \pi}2 \right)
\\
\\
I_{2m}
=& \>2\sin\frac{(2m+1)\pi}4\int_0^{\frac\pi4}\frac{\cos(2mx)}{\cos x}dx\\
=&\>2\cos\frac{(2m+1)\pi}4\left(\coth^{-1}\sqrt2-\sum_{k=1}^m \frac{2}{2k-1} \sin\frac{(2k+1)\pi}4\right)
\end{align}
from which
\begin{align}
&I_1=\frac\pi2,\>\>\>I_5=2-\frac\pi2,\>\>\>I_9=\frac\pi2-\frac43,
\>\>\>I_{13}=\frac{26}{15}-\frac\pi2,\cdots\\
&I_3=I_7=I_{11}=I_{15}=\cdots = 0\\
&I_2=2-\sqrt2 \coth^{-1}\sqrt2,\>\>\>
I_4=\frac43-\sqrt2 \coth^{-1}\sqrt2,\>\>\>\\
&I_6=\sqrt2 \coth^{-1}\sqrt2-\frac{14}{15},\>\>\>
I_8=\sqrt2 \coth^{-1}\sqrt2- \frac{128}{105},\cdots\>\>\>
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate the integral $\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$? I have no idea how to start, it looks like integration by parts won't work.
$$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$
If someone could shed some light on this I'd be very thankful.
| Note
\begin{eqnarray}
I&=&\int_0^\infty \frac{x[(x^2+1)-1]^2\sin x}{(1+x^2)^3}dx\\
&=&\int_0^\infty \frac{x\sin x}{1+x^2}dx-2\int_0^\infty \frac{x\sin x}{(1+x^2)^2}dx+\int_0^\infty \frac{x\sin x}{(1+x^2)^3}dx.
\end{eqnarray}
From
$$ \int_0^\infty \frac{\cos(ax)}{b^2+x^2}dx=\frac{\pi}{2b}e^{-ab}, a>0, b>0, $$
we have
\begin{eqnarray}
\int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dx&=&=-\frac{d}{da}\int_0^\infty \frac{\cos(ax)}{b^2+x^2}dt=\frac{\pi}{2e^{-ab}},\\
\int_0^\infty \frac{x\sin x}{(b^2+x^2)^2}dx&=&-\frac{1}{2b}\frac{d}{db}\int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dt=\frac{a\pi}{4be^{-ab}},\\
\end{eqnarray}
and
\begin{eqnarray}
\frac{d^2}{db^2}\int_0^\infty \frac{x\sin (ax)}{b^2+x^2}dx&=&8b^2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx-2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^2}dx.
\end{eqnarray}
The latter one implies
\begin{eqnarray}
\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx=\frac{a(ab+1)\pi}{16b^3e^{-ab}}.
\end{eqnarray}
Thus
$$ I=\frac{\pi}{8e}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
How can I plot the set $M:=\left\{\left(\begin{array}{c}x\\y\end{array}\right)\in \mathbb R^2:9x^4-16x^2y^2+9y^4\leq9\right\}$ I need to plot the following set:
$$M:=\left\{\left(\begin{array}{c}x\\y\end{array}\right)\in \mathbb R^2:9x^4-16x^2y^2+9y^4\leq9\right\}$$
I have solved the equation $9x^4-16x^2y^2+9y^4-9=0$ for $y$, thinking this would make the task easier, but it turned out not to bring much help. However this is what I got:
$$y_{1,2,3,4}=\pm \sqrt{\frac89x^2\pm\frac19\sqrt{81-17x^4}}$$
Not so obvious how one can plot any of these functions. I might not see some trick that would help me to visualize the set (maybe switch to polar coordinates? Tried this, but didn't help me either). Of course after having struggled an hour or so, I asked Mathematica for help and got this:
The question is how it can be done without using the magic of Mathematica.
| Note that in polar co-ordinates $x=r\cos \theta, y=r\sin \theta$ the function becomes $$r^4(9\cos^4\theta-16\cos^2\theta\sin^2\theta+9\sin^4\theta)$$
Then $\cos^4\theta+\sin^4\theta =(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta$ so the trigonometric part becomes $$9-34\cos^2\theta\sin^2\theta=9-\frac {17}2\sin^2 2\theta$$
Then $\sin^2 2\theta=\frac 12-\frac 12\cos 4\theta$ so we get $$9-\frac {17}4+\frac {17}4\cos4\theta=\frac {19+17\cos4\theta}{4}$$
The boundary of the region is therefore $$r^4=\frac {36}{19+17\cos4\theta}$$
The right-hand side is always positive, and has a maximum value $18$ when $\cos 4\theta=-1$, minimum value $1$ when $\cos 4\theta=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/840608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the series $\sum_{n=1}^\infty\frac{\sin n}{\ln n+\cos n}$ converge? $$\sum_{n=1}^\infty\frac{\sin n}{\ln n+\cos n}$$
My guess is "yes", but I can't prove it.
| This method was inspired by the OP's heuristic argument in a comment to the question. We approximate the summand by Taylor polynomials, but with more and more terms as $n$ grows.
We need the fact that for every integer $k\ge1$,
$$
\bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| = O(k^7)
$$
uniformly for integers $B<C$. To see this, verify that $\sin n \cdot \cos^k n = q_k(e^{in})$, where $q_k(t)=\sum_{j=-k-1}^{k+1} c_{k,j}t^j$ is a Laurent polynomial with no constant term that satisfies $\sum_{j=-k-1}^{k+1} |c_{k,j}| = 2^{-k}\binom{k}{\lfloor k/2\rfloor} \le 1$. Then
\begin{align*}
\sum_{B\le n< C} \sin n \cdot \cos^k n &= \sum_{B\le n< C} q_k(e^{in}) \\
&= \sum_{j=-k-1}^{k+1} \sum_{B\le n< C} c_{k,j}e^{ijn} = \sum_{j=-k-1}^{k+1} c_{k,j} \frac{e^{ijC}-e^{ijB}}{e^{ij}-1},
\end{align*}
so that
$$
\bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| \le \sum_{j=-k-1}^{k+1} |c_{k,j}| \frac{2}{|e^{ij}-1|}.
$$
The fact that $\pi$ has an irrationality measure of less than $8$ (Salikhov, 2012) means that $|e^{ij}-1|\gg j^{-7}$. Therefore
$$
\bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| \ll k^7 \sum_{j=-k-1}^{k+1} |c_{k,j}| \ll k^7.
$$
Also note that for any positive integer $A$ and any real number $x\in[-\frac12,\frac12]$,
$$
\frac1{1+x} = \sum_{k=0}^{A-1} (-x)^k + r(x)
$$
where $|r(x)| \le |2x|^A$ (a consequence of Taylor's theorem).
Now we establish the convergence of the series in question by showing that it is Cauchy, i.e., that
$$
\sum_{n=M}^N \frac{\sin n}{\log n+\cos n} \to 0
$$
as $M\to\infty$ (and $N>M>e^{2e^2}$, say). For each $n$, we choose $x=\frac{\cos n}{\log n}$ and $A=\lceil{\log n}\rceil$, giving
$$
\sum_{n=M}^N \frac{\sin n}{\log n} \bigg( \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k + s(n) \bigg)
$$
where $|s(n)| \le |2\frac{\cos n}{\log n}|^A \le (\frac2{\log n})^{\log n} \le \frac1{n^2}$ for $n>e^{2e^2}$. Therefore the contribution from $\sum_{n=M}^N \frac{\sin n}{\log n}s(n)$ is $O(\frac1M)$. As for the other terms, we can write
\begin{multline*}
\sum_{n=M}^N \frac{\sin n}{\log n} \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k = \sum_{k<\log M} (-1)^k \sum_{n=M}^N \frac{\sin n \cdot \cos^k n}{\log^{k+1} n} \\
+ \sum_{\log M<k<\log N} (-1)^k \sum_{e^k < n \le N} \frac{\sin n \cdot \cos^k n}{\log^{k+1} n}.
\end{multline*}
Let $S_k(t) = \sum_{n\le t} \sin n \cdot \cos^k n$, which is $O(k^7)$ as noted above. Then
\begin{align*}
\sum_{B<n\le C} \frac{\sin n \cdot \cos^k n}{\log^{k+1} n} = \int_B^C \frac{dS_k(t)}{\log^{k+1} t} &= \frac{S_k(t)}{\log^{k+1} t} \bigg|_B^C + (k+1) \int_B^C \frac{S_k(t)}{t\log^{k+2} t}\,dt \\
&\ll k^7 \bigg(\frac1{\log^{k+1} B} + (k+1) \int_B^C \frac1{t\log^{k+2} t}\,dt \bigg) \\
&\ll k^7 \frac1{\log^{k+1} B}.
\end{align*}
In particular,
\begin{align*}
\sum_{n=M}^N \frac{\sin n}{\log n} \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k &\ll \sum_{k<\log M} \frac{k^7}{\log^{k+1} M} + \sum_{\log M<k<\log N} \frac{k^7}{\log^{k+1} e^k} \\
&\ll \sum_{k=1}^\infty \frac{k^7}{\log^{k+1} M} + \sum_{k=2}^\infty \frac1{k^{\log M-6}},
\end{align*}
and both these series tend to $0$ as $M\to\infty$ by the dominated convergence theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/841080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 4,
"answer_id": 1
} |
$\left(\sqrt{8}+\sqrt{2}\right)^2$ = 18 why?? I would like to now why
$$\left(\sqrt{8}+\sqrt{2}\right)^2$$
is equal to $18$? Please provide me
the proccess. Thank you.
| It does look counter-intuitive, doesn't it? But if you step through the multiplication, I think it will become crystal clear to you. First of all, $$(\sqrt{8} + \sqrt{2})^2 = (\sqrt{8} + \sqrt{2})(\sqrt{8} + \sqrt{2})$$, right? Applying FOIL (First, Outer, Inner, Last), we get $$\sqrt{8} \sqrt{8} = 8$$ $$\sqrt{8} \sqrt{2} = 4$$ $$\sqrt{2} \sqrt{8} = 4$$ $$\sqrt{2} \sqrt{2} = 2$$ And then $8 + 4 + 4 + 2 = 18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/844436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Find $zw, \frac{z}{w},\frac{1}{z}$ for $ z=2\sqrt{3}-2i, w=-1+i$ Find $zw, \frac{z}{w},\frac{1}{z}$ for $ z=2\sqrt{3}-2i, w=-1+i$
I went wrong somewhere, this is what I have so far (this is in polar):
$z=4\left(\cos\left(\frac{11\pi}{6}\right)+\sin\left(\frac{11\pi}{6}\right)\right) $
$w=\sqrt2\left(\cos\left(\frac{7\pi}{4}\right)+\sin\left(\frac{7\pi}{4}\right)\right) $
Now my setup should be:
$zw=4\sqrt2\left(\cos\left(\frac{11\pi}{6}+\frac{7\pi}{4}\right)+\sin\left(\frac{11\pi}{6}+\frac{7\pi}{4}\right)\right) $
The common denominator is $12$ so
$zw=4\sqrt2\left(\cos\left(\frac{22\pi}{12}+\frac{21\pi}{12}\right)+\sin\left(\frac{22\pi}{12}+\frac{21\pi}{12}\right)\right) $
which then should equal out to
$zw=4\sqrt2\left(\cos\left(\frac{43\pi}{12}\right)+\sin\left(\frac{43\pi}{12}\right)\right) $
The answer in the book says:
$zw=4\sqrt2\left(\cos\left(\frac{7\pi}{12}\right)+\sin\left(\frac{7\pi}{12}\right)\right) $
Where did I go wrong?
I haven't even tried the other problems yet.
| You have the angle wrong for $w$. It should be $3\pi/4$, right?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Integer sum as binomial coefficient What's the rule for expressing integer sums as binomial coefficients? That is, for $p=1$ it is $$\sum_{n=1}^N n^p = {{N+1}\choose 2} $$
What is it for higher powers?
| $\def\str#1#2{\left\{#1\atop#2\right\}}$
Let $\str{n}{m}$ be the Stirling number of the second kind, It can be defined by the formula
$$
X^p=\sum_{m=0}^p\str{p}{m}X(X-1)\cdots(X-m+1)\tag 1
$$
This implies that
$$\eqalign{
n^p&=\sum_{m=0}^p m!\str{p}{m}\binom{n}{m}\tag 2\cr
&=\sum_{m=0}^p m!\str{p}{m}\left(\binom{n+1}{m+1} -\binom{n}{m+1}\right)\cr
}
$$
So, taking the sum for $n=1,2,\ldots,N$ we get
$$
\sum_{n=1}^Nn^p=\sum_{m=0}^pm!\str{p}{m}\left(\binom{N+1}{m+1} -\binom{1}{m+1}\right)
$$
Noting that $\str{p}{0}=0$ for $p>0$ we conclude that
$$
\sum_{n=1}^Nn^p=\sum_{m=1}^pm!\str{p}{m} \binom{N+1}{m+1}
$$
For $p=1$, we have $\str{1}{1}=1$, and we get the OP's formula. For $p=2$ we have
$\str{2}{1}=1$ and $\str{2}{2}=1$,
$$
\sum_{n=1}^Nn^2= \binom{N+1}{2} +2\binom{N+1}{3}
$$
and for $p=3$ we have
$\str{3}{1}=1$, $\str{3}{2}=3$ and $\str{3}{3}=1$, hence
$$
\sum_{n=1}^Nn^3= \binom{N+1}{2} +
6 \binom{N+1}{3}
+6 \binom{N+1}{4} .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/846144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Algebraic manipulation of floors and ceilings I am trying to solve the summation
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=3}^{n} \left\lceil\frac{i-2}{2}\right\rceil
$$
I will list some of the simplifications that I've found so far, and then get around to asking my question. Please feel free to point out any logical errors that I have (in all likelihood) made.
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=1}^{n-2} \left\lceil\frac{i}{2}\right\rceil
$$
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \left(\sum_{i=1}^{\left\lceil\frac{n-2}{2}\right\rceil} i+ \sum_{i=1}^{\left\lfloor\frac{n-2}{2}\right\rfloor} i\right)
$$
Since the upper bound in both of the inner summations will be an integer, I have:
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \left(\frac{\left\lceil\frac{n-2}{2}\right\rceil\left(\left\lceil\frac{n-2}{2}\right\rceil + 1\right)}{2}+\frac{\left\lfloor\frac{n-2}{2}\right\rfloor\left(\left\lfloor\frac{n-2}{2}\right\rfloor + 1\right)}{2}\right)
$$
How do I simplify a floor times a floor and a ceiling times a ceiling algebraically?
I'm thinking that the identity
$$
n = \left\lceil\frac{n}{2}\right\rceil + \left\lfloor\frac{n}{2}\right\rfloor
$$
might come in handy, but I have very little experience manipulating floors and ceilings algebraically. Let me be clear: I'm not looking for an answer to the summation, just guidance in simplifying this current step.
| \begin{align*}S&=\sum_{n=3}^\infty\frac{1}{2^n}\sum_{i=3}^n\left\lceil\frac{i-2}{2}\right\rceil&=\sum_{n=3}^\infty\frac{1}{2^n}\sum_{i=3}^n\left\lfloor\frac{i-1}{2}\right\rfloor\\ &=\sum_{n=3}^\infty\frac{1}{2^n}\sum_{i=2}^{n-1}\left\lfloor\frac{i}{2}\right\rfloor &=\sum_{n=3}^\infty\frac{1}{2^n}\left\lfloor\frac{(n-1)^2}{4}\right\rfloor\end{align*}
with $n=2m\quad$ then $\displaystyle\quad S_{2m}=\sum_{m=2}^\infty\frac{1}{2^{2m}}\left\lfloor\frac{(2m-1)^2}{4}\right\rfloor=\sum_{m=2}^\infty\frac{m^2-m}{2^{2m}}=\sum_{m=1}^\infty\frac{m^2+m}{2^{2m+2}}$
with $n=2m+1\quad$ then $\displaystyle\quad S_{2m+1}=\sum_{m=1}^\infty\frac{1}{2^{2m+1}}\left\lfloor\frac{(2m)^2}{4}\right\rfloor=\sum_{m=1}^\infty\frac{m^2}{2^{2m+1}}$
So $\displaystyle\quad S=S_{2m}+S_{2m+1}=\sum_{m=1}^\infty\frac{3m^2+m}{4^{m+1}}=\boxed{\frac{2}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/847183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.
My gut feeling is telling me to multiply the above series by 8, then subtract it from the original, but it doesn't quite get me there because I'm going to have a whole lot of extra terms that I do not want.
| By factorizing $2^3 = 8$ from the even terms, we can re-express the sum of the even terms:
$$2^3 + 4^3 + \dots + (n-1)^3 = 8\left(1^3 + 2^3 + \dots + \left(\frac{n-1}{2}\right)^3\right)$$
Simply subtract this from $1^3 + 2^3 + \dots + n^3$ to give the sum of the remaining odd terms. You can compute the formula for the RHS of the above equation using the equation you mentioned.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
limits using $ \epsilon - \delta $ to prove two variable function I'm trying to use the $ \epsilon - \delta $ argument to prove $\lim_{(x,y) \rightarrow (1.1)} \frac{2xy}{x^2+y^2} =1$. I know that I need to show that $\forall \epsilon>0, \exists \delta>0$ s.t. for all (x,y) in the domain of f, $| \frac{2xy}{x^2+y^2} -1| < \epsilon$ whenever $0< \sqrt{(x-1)^2 + (y-1)^2} <\delta$.
I've made $| \frac{2xy}{x^2+y^2} -1|$ to $\frac{(x-y)^2}{x^2 + y^2} $, but I can't get it to look like $\sqrt{(x-1)^2 + (y-1)^2}$..... what should I do from now? :(
| So you have
$$
\frac{(x-y)^2}{x^2+y^2}.
$$
Since you can always assume that $x> 1/2$ and $y>1/2$,
$$
x^2+y^2 > \frac{1}{2},
$$
and
$$
\frac{(x-y)^2}{x^2+y^2} < 2(x-y)^2.
$$
Now, $(x-y)^2 = |(x-y)(x-y)|< (|x|+|y|)|x-y| < \left(\frac{3}{2}+\frac{3}{2} \right) |x-y|$ because you can also assume $x<3/2$, $y<3/2$. I leave the verification that $|x-y|$ can be made arbitrarily small if $x \to 1$, $y \to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
A simple conditional probability problem Assume that two fair dice are rolled one at a time. Given that the sum of the two numbers that occured was at least $7$, compute the probability that it was equal to $7$.
I tried computing the probability as follows:
$$P \left( X+Y=7| X+Y \geq 7 \right)=\frac{P \left(X+Y=7 \right)}{P \left(X+Y \geq 7 \right)}$$
And then I examined all different pairs whose sum is $7$ and then those whose sum is greater or equal to $7$. I am getting $3/12$ but the right answer is $6/21$. Could you please help me understand what I am doing wrong?
Thank you.
| $P(X+Y=7) = 1/6$
By Symmetry
$P(X+Y>7) = P(X+Y<7) = (1 - 1/6)/2$
$P(X+Y>=7) = P(X+Y>7)+P(X+Y=7) = 5/12 + 1/6$
$P( X+Y=7| X+Y >= 7) = \frac{1/6}{5/12+1/6} = 6/21$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Two diophantine equations with lots of unknowns Is it possible (tractable) to determine if the following system of equations has any nontrivial solutions (ie, none of the unknowns are zero) in the domain of integers?
$$A^2 + B^2=C^2 D^2$$
$$2 C^4 + 2 D^4 = E^2 + F^2$$
| To solve,
$$A^2+B^2=C^2 D^2\\
(2C)^4+(2D)^4=E^2+F^2$$
Choose,
$$\begin{aligned}
A&=2(ac-bd)(ad+bc)\\
B&=(ac-bd)^2-(ad+bc)^2\\
C&=a^2+b^2\\
D&=c^2+d^2\\
E&=(a^2+b^2 )^2-(c^2+d^2 )^2\\
F&=(a^2+b^2 )^2+(c^2+d^2 )^2\\
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
$f(x,y)=(x^2-y^2,2xy)$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$. What is the set $f(A)$ Let $f:\mathbb R^2\to\mathbb R^2$ be defined by the equation
$$f(x,y)=(x^2-y^2,2xy).$$
Show that $f$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$.
What is the set $f(A)?$
| The inverse function theorem does not seem to help since we are talking about global invertability. To show injective assume that
$$x^2-y^2=w^2-z^2$$ and
$$2xy=2wz$$
We then have on squaring and adding,
$$(x^2+y^2)^2=(w^2+z^2)^2$$ this implies that
$$x^2+y^2=w^2+z^2$$ from which we get $x^2=w^2$ and since we assume that these are positive we have $x=w>0$ and now the second equation gives $y=z$ proving injectivity.
To determine the image
set $$a=x^2-y^2$$
$$b=2xy$$ then we see that
$$x=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}$$
Now the condition $x=0$ gives
iff $a+\sqrt{a^2+b^2}=0$ that is $b=0$ and $a \leq 0$, so the negative $x$ axis is not in the image, all other points are.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x<2\\
\frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3
\end{cases}
$$
My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.
If someone could shed light on this topic, that would be much appreciated.
| Hint: integrate $ 3-x$. Say, integral is $f(x)$. Plug in $f(3)=1$ to get the value of constant
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of a line tangent to a circle: why it works? I've the problem following: "Determine the points on the circumference of the circle $x^2 + y^2 = 1$ whose tangent lines are
and the point $(3,0)$." I found the derivative $\dfrac{\mathrm{d} y}{\mathrm{d} x}=-\dfrac{x}{y}$ and I managed to solve the exercise by doing something like:
$y^2 = 1 - x^2\\
y - f(p) = f'(p)\cdot(x-p) \quad (1)\\
y - 0 = -\dfrac{x}{y}\cdot(x - 3)\\
y^2 = -x^2 + 3x\\
x = \dfrac{1}{3}\\$
From there I managed to get the correct equation of the line tangent to the circle as required by the exercise. But when I do $x=3$ in the line (1), I get:
$\sqrt{1-x^2}=\dfrac{-3}{\sqrt{1-3^2}}\cdot(x-3)$,
which results in some strange stuff. Then I don't know if the way I solved it is the correct one. If it is, why I can't obtain $x=\dfrac{1}{3}$ when $x=3$?
Thanks in advance.
| We do it the calculus way (there is a much simpler geometric way).
Let the point(s) of tangency be $(a,b)$. Then by your calculation, the slope of the tangent line is $-\frac{a}{b}$.
The tangent line, we are told, goes through the external point $(3,0)$. The line joining $(a,b)$ to $(3,0)$ has slope $\frac{b}{a-3}$. Thus we arrive at
the equation
$$\frac{b}{a-3}=-\frac{a}{b}.$$
Simplification gives $b^2=-a(a-3)$ and therefore $a^2+b^2=3a$. But $a^2+b^2=1$, and therefore $3a=1$. It follows that $a=\frac{1}{3}$. A picture, or algebra, shows there are two tangent lines, since $a^2+b^2=1$ yields two values of $b$.
Much closer to what you did is to find the equation of the tangent line. The tangent line has slope $-\frac{a}{b}$ and passes through $(a,b)$. It follows that an equation of the tangent line(s) is
$$y-b=-\frac{a}{b}(x-a).$$
Substitute the point $(3,0)$ for $(x,y)$. Some algebra, together with $a^2+b^2=1$, gives us the desired result.
Remark: It is very important not to let $(x,y)$ be the mystery point(s) of tangency, for that risks confusion with the variables used in describing the equation of the tangent line.
For an approach using analytic geometry, let the point(s) of tangency be $P=(a,b)$. Note that the line through $P$ and $(3,0)$ is perpendicular to the line $OP$. It follows by the Pythagorean Theorem that
$$(a-3)^2+b^2+1^2=3^2.$$
Some algebra, together with $a^2+b^2=1$, yields $a=\frac{2}{6}$.
Or else we can use the fact that the line $OP$ has slope $\frac{b}{a}$ and therefore the tangent line at $P$ has slope $-\frac{a}{b}$. Now use the calculation of the first solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the value of $\lfloor{100N}\rfloor$ What is the value of $\lfloor{100N}\rfloor$ where $\displaystyle N= \lim\limits_{a\,\rightarrow\,\infty}\sum\limits_{x=-a}^{a}\frac{\sin x}{x}$.
This is a part of a bigger problem that I was solving. I need the exact integer value of $\lfloor{100N}\rfloor$ .
I was only able to simplify $\displaystyle \lim\limits_{a\,\rightarrow\,\infty}\sum\limits_{x=-a}^{a}\frac{\sin x}{x}= 1+2\left(\lim\limits_{a\,\rightarrow\,\infty}\sum\limits_{x=1}^{a}\frac{\sin x}{x}\right)$. I don't know how to proceed further. Please help
| Note
\begin{eqnarray*}
\sum_{k=1}^n\cos(kx)=-\frac{1}{2}+\frac{\sin(\frac{n}{2}+1)x}{2\sin\frac{x}{2}}
\end{eqnarray*}
and hence
\begin{eqnarray*}
\sum_{k=1}^n\frac{\sin k}{k}&=&\int_0^1\sum_{k=1}^n\cos(kx)dx\\
&=&-\frac{1}{2}+\int_0^1\frac{\sin(\frac{n}{2}+1)x}{2\sin\frac{x}{2}}dx\\
\end{eqnarray*}
Using the following result:
Let $f(x)$ be monotonic in $[0,A]$. Then
$$ \lim_{n\to\infty}\int_0^Af(x)\frac{\sin(nx)}{x}dx=f(0)\frac{\pi}{2}. $$
we have
\begin{eqnarray*}
N&=&\lim\limits_{a\to\infty}\sum\limits_{k=-a}^{a}\frac{\sin k}{k}=1+2\lim_{a\to\infty}\sum\limits_{k=1}^{a}\frac{\sin k}{k}\\
&=&1+2\left(-\frac{1}{2}+\lim_{a\to\infty}\int_0^1\frac{\sin(\frac{a}{2}+1)x}{\sin\frac{x}{2}}dx\right)\\
&=&\lim\limits_{a\to\infty}\int_0^1\frac{x}{\sin\frac{x}{2}}\frac{\sin(\frac{a}{2}+1)x}{x}dx\\
&=&\pi
\end{eqnarray*}
and hence
$$ \lfloor{100N}\rfloor=314. $$
Here we use the fact: the function $\frac{x}{\sin\frac{x}{2}}$ is increasing in (0,1] and
$$ \lim_{x\to 0^+}\frac{x}{\sin\frac{x}{2}}=2. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Integrating $\int_1^2 \int_0^ \sqrt{2x-x^2} \frac{1}{((x^2+y^2)^2} dydx $ in polar coordinates I'm having a problem converting $\int\limits_1^2 \int\limits_0^ \sqrt{2x-x^2} \frac{1}{(x^2+y^2)^2} dy dx $ to polar coordinates.
I drew the graph using my calculator, which looked like half a circle on the x axis.
I know that $\frac{1}{(x^2+y^2)^2} dydx$ turns to $ \frac{r}{(r^2)^2}drd\theta$, which would be $ \frac{1}{r^3}$
The region of integration in $\theta $ is I guess $0 \leq \theta \leq \frac{1}{4} \pi $, but I'm stuck for r. The lower boundary is $x=1=rcos\theta \rightarrow r=1/cos\theta=sec\theta$, but what is the upper boundary?? Is it $\sqrt{2rcos\theta-r^2cos^2\theta}$? I tried this but I couldn't integrate my equation...
Could someone please help me out?
Also, how can I draw the graph of $y= \sqrt{2x-x^2} $? I could only know how it looked like by using my calculator...
| $\int_{1}^{2} \int_{0}^{\sqrt{2 x-x^{2}}} \frac{1}{\left(x^{2}+y^{2}\right)^{2}} d y d x$ is an integration over region $y=0$ to $y=\sqrt{2 x-x^{2}}$ and $x=1$ to $x=2$ as shaded region in below graph. Now we will change $xy$ to polar co-ordinate. Using $x=r \cos \theta$, $y=r \sin \theta$ and $dxdy=r dr d\theta$. After changing into polar co-ordinate shaded region in polar co-ordinate will be $r=\sec \theta$ to $r=2 \cos \theta$ and $\theta=0$ to $\theta=\pi/4$.
Graph for the region
Now our integration is
$$
\begin{aligned}
\int_0^{\pi/4}\int_{\sec \theta}^{2 cos \theta}\frac{rdrd\theta}{r^4} & =\int_0^{\pi/4}\int_{\sec \theta}^{2 cos \theta}\frac{drd\theta}{r^3},\\
&=\int_0^{\pi/4} \frac{-1}{2r^2}|_{\sec \theta}^{2 \cos \theta} d\theta,\\
&=\int_0^{\pi/4}\frac{-1}{8\cos^2\theta}+ \frac{\cos^2\theta}{2} d\theta,\\
&=\int_0^{\pi/4}\frac{\cos^2\theta}{2}-\frac{\sec^2\theta}{8},\\
&=\frac{\theta}{4}+\frac{\sin2\theta}{8}-\frac{\tan \theta}{8}|_0^{\pi/4},\\
&=\frac{\pi}{16}.
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Orthogonal Projection of $v$ on sub-space $U$
What is the orthogonal projection of $v = (1,5,-10)$ on the sub-space $U = Sp\{(5,-2,1),(1,2,-1)\}$?
Well, I managed to compute it on the way I've been taught by the book, but it doesn't seem to work. Any chance you guys can solve it? If needed, I can provide the way I tried to solve it.
| Another approach is to compute the normal to the subspace using the cross product:
$$
\begin{pmatrix}5,-2,1\end{pmatrix}\times\begin{pmatrix}1,2,-1\end{pmatrix}=\begin{pmatrix}0,6,12\end{pmatrix}
$$
Then subtract the component of $\begin{pmatrix}1,5,-10\end{pmatrix}$ in that direction:
$$
\begin{align}
&\begin{pmatrix}1,5,-10\end{pmatrix}-\overbrace{\frac{\begin{pmatrix}1,5,-10\end{pmatrix}\cdot\begin{pmatrix}0,6,12\end{pmatrix}}{\begin{pmatrix}0,6,12\end{pmatrix}\cdot\begin{pmatrix}0,6,12\end{pmatrix}}\begin{pmatrix}0,6,12\end{pmatrix}}^{\begin{array}{c}\text{orthogonal component of $\begin{pmatrix}1,5,-10\end{pmatrix}$}\\\text{ in the direction of $\begin{pmatrix}0,6,12\end{pmatrix}$}\end{array}}\\
&=\begin{pmatrix}1,5,-10\end{pmatrix}-\frac{-90}{180}\begin{pmatrix}0,6,12\end{pmatrix}\\
&=\begin{pmatrix}1,5,-10\end{pmatrix}+\begin{pmatrix}0,3,6\end{pmatrix}\\[3pt]
&=\begin{pmatrix}1,8,-4\end{pmatrix}
\end{align}
$$
Check
$$
\begin{pmatrix}1,8,-4\end{pmatrix}=\frac72\begin{pmatrix}1,2,-1\end{pmatrix}-\frac12\begin{pmatrix}5,-2,1\end{pmatrix}
$$
so the computed vector is in the given span.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
How to show that $\lim\limits_{n\to\infty}n^{2/3}a_{n}=\sqrt[3]{2}/\Gamma{(1/3})$ Let $$\left(\dfrac{1+x}{1-x}\right)^{1/3}=\sum_{n=0}^{\infty}a_{n}x^n,|x|<1$$
Show that
$$\lim_{n\to\infty}n^{2/3}a_{n}=\dfrac{\sqrt[3]{2}}{\Gamma{\left(\dfrac{1}{3}\right)}}$$
| Write
$$
\left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{(1+x)^{1/3} - 2^{1/3}}{(1-x)^{1/3}} + \frac{2^{1/3}}{(1-x)^{1/3}} =: f(x)+g(x).
$$
The function $f$ is continuous on the disk $|z| \leq 1$, so by Darboux's method (Thm. VI.14 in Analytic Combinatorics) we have
$$
[x^n] f(x) = o\left(\frac{1}{n}\right).
$$
The coefficients of the second series are given by
$$
[x^n] g(x) = 2^{1/3} (-1)^n \binom{-1/3}{n} = \frac{2^{1/3}}{\Gamma(1/3) n^{2/3}} + O\left(\frac{1}{n^{5/3}}\right)
$$
as shown by Jack D'Aurizio.
Combining these yields
$$
\begin{align}
[x^n]\left(\dfrac{1+x}{1-x}\right)^{1/3} &= [x^n] f(x) + [x^n] g(x) \\
&= \frac{2^{1/3}}{\Gamma(1/3) n^{2/3}} + o\left(\frac{1}{n}\right),
\end{align}
$$
so that
$$
n^{2/3} [x^n]\left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{2^{1/3}}{\Gamma(1/3)} + o\left(\frac{1}{n^{1/3}}\right).
$$
| {
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"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Infinite product: $\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}$ I am trying to find $$\lim_{n\rightarrow \infty}\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}.$$I have no idea how I can start. Please help me.
Thanks!
| The main idea is to use a telescope product of the form $ \prod \frac{b_j}{b_{j+2}} \frac{a_{j+1}}{a_j}$. We have$$ \lim_{n \to \infty} \prod_{k=2}^n \frac{k^3-1}{k^3+1}=\lim_{n \to \infty}\prod_{k=2}^n \frac{k-1}{k+1} \frac{k^2+k+1}{k^2-k+1}=\lim_{n \to \infty}\prod_{k=2}^n \frac{k-1}{(k+2)-1} \frac{(k+1)^2-(k+1)+1}{k^2-k+1}=\lim_{n \to \infty} \frac{1 \cdot 2}{n \cdot (n+1)}\frac{n^2+n+1}{4-2+1}=\lim_{n \to \infty} \frac{2}{3} \frac{n^2+n+1}{n^2+n}=\frac{2}{3}.$$
| {
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} |
How does $n+2\mid(n^2+n+1)(n^2+n+2)$ imply $n+2\mid12$? I'm trying to show that if $n+2\mid(n^2+n+1)(n^2+n+2)$, then $n+2 \mid 12$ for all $n\in \mathbb{N}$.
| Hint
$$(n^2+n+1)(n^2+n+2)=n^4+2n^3+4n^2+3n+2$$
and
$$ n^4+2n^3+4n^2+3n+2=(n+2)(n^3+4n-5)+12$$
| {
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} |
closed form for $\int_0^1x^{a+1}(1-x^2)^bJ_a(cx)dx$ $$\int_0^1x^{a+1}(1-x^2)^bJ_a(cx)dx$$
my friend posted the integral on the fb and I tried to solve it but I faild because I have little information about bessel function
so could some one help ?
| Using the series expansion for the Bessel function the integral is as follows:
\begin{align}
I &= \int_{0}^{1} x^{a+1} (1-x^{2})^{b} J_{a}(cx) \ dx \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a)!} \ \int_{0}^{1} x^{2n+2a+1} (1-x^{2})^{b} \ dx.
\end{align}
Making the substitution $x = \sqrt{t}$ the integral now becomes
\begin{align}
I &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a)!} \ \int_{0}^{1} t^{n+a} (1-t)^{b} \ dt \\
&= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a)!} \ B(n+a+1, b+1) \\
&= \frac{b!}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} (c/2)^{2n+a}}{n!(n+a+b+1)!} \\
&= \frac{b!}{2} \left(\frac{2}{c}\right)^{b+1} \ J_{a+b+1}(c).
\end{align}
Hence
\begin{align}
\int_{0}^{1} x^{a+1} (1-x^{2})^{b} J_{a}(cx) \ dx = \frac{b!}{2} \left(\frac{2}{c}\right)^{b+1} \ J_{a+b+1}(c).
\end{align}
| {
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} |
Compute integral: $\int_{0}^{\pi/2}\log(a^2\sin^2 x+b^2\cos^2 x )dx$ This is a integral for a calculus exam, and I have no idea how to solve it.
$$\int_0^{\frac{\pi}{2}} \log \big( a^2 \sin^{2}(x)+b^2 \cos^{2}(x) \big) \, \mathrm{d}x$$
| Consider
$$I(b)=\int_0^{\pi/2} \ln\left(a^2\sin^2x+b^2\cos^2x\right)\,dx $$
$\displaystyle
\begin{aligned}
\Rightarrow \frac{\partial I}{\partial b} &=2b\int_0^{\pi/2} \frac{\cos^2x}{a^2\sin^2x+b^2\cos^2x}\,dx\\
&=2b\int_0^{\pi/2} \frac{dx}{a^2\tan^2x+b^2}\\
&=2b\int_0^{\infty} \frac{dt}{(a^2t^2+b^2)(1+t^2)}\,\,\,\,\,\,\,(\tan x=t)\\
&=\frac{2b}{b^2-a^2}\left(\int_0^{\infty}\frac{dt}{1+t^2}-a^2\int_0^{\infty} \frac{dt}{a^2t^2+b^2}\right)\,\,\,\,\,\,(\text{partial fractions})
\\&=\frac{2b}{b^2-a^2}\left(\frac{\pi}{2}-\frac{\pi}{2}\frac{a}{b}\right)\\
&=\pi\frac{b-a}{b^2-a^2}=\frac{\pi}{a+b}
\end{aligned}$
Hence,
$$\Rightarrow I(b)=\pi \ln(a+b)+C$$
To evaluate $C$, I compute $I(0)$ i.e
$$I(0)=\int_0^{\pi/2} \ln(a^2\sin^2x)\,dx=\pi\ln\frac{a}{2}$$
$$\Rightarrow I(0)=\pi\ln a+C=\pi\ln\frac{a}{2} \Rightarrow C=-\pi\ln 2$$
Hence,
$$I(b)=\pi\ln\left(\frac{a+b}{2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Double Angle Trigonometry Question So there is this question which consists of 2 parts.
$$
a) \text{ Simplify } \frac{\sin2x}{1+\cos2x} \\
b) \text{ Hence, find the exact value of tan 15.}
$$
So far I've discovered that $ \text{a)} \tan x $
But I have no idea how to begin on part $b$, although I'm guessing the answer's correlated with a specific part of the working for part $a$. Can someone help me? Thanks in advance!
| We have: $\dfrac{\sin(2x)}{1+\cos(2x)}$
$=\hspace{12 mm}\dfrac{2\sin(x)\cos(x)}{1+2\cos^{2}(x)-1}$
$=\hspace{12 mm}\dfrac{2\sin(x)\cos(x)}{2\cos^{2}(x)}$
$=\hspace{12 mm}\dfrac{\sin(x)}{\cos(x)}$
$=\hspace{12 mm}\tan(x)$
Then, we want to evaluate $\tan(15)$.
We can do this using the original expression $\dfrac{\sin(2x)}{1+\cos(2x)}$:
$\Rightarrow \tan(15)=\dfrac{\sin(2\times15)}{1+\cos(2\times15)}$
$\hspace{19.5 mm}=\dfrac{\sin(30)}{1+\cos(30)}$
$\hspace{19.5 mm}=\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}}$
$\hspace{19.5 mm}=\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}}$
$\hspace{19.5 mm}=\dfrac{1}{2+\sqrt{3}}$
$\hspace{19.5 mm}=\dfrac{1}{2+\sqrt{3}}\times\dfrac{2-\sqrt{3}}{2-\sqrt{3}}$
$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{2^{2}-(\sqrt{3})^{2}}$
$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{4-3}$
$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{1}$
$\hspace{19.5 mm}=2-\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{align} x(x+1)=y+1 \\
y(y+1)=z+1\\
z(z+1)=x+1 \end{align}$$
Multiplying all we get $$xyz=1 \tag{1}$$ and adding all we get
$$x^2+y^2+z^2=3 \tag{2}$$
Now from Original Equations
$$\begin{align} x^2=y+1-x\\
y^2=z+1-y\\
z^2=x+1-z \end{align}$$ Multiplying all and Using $(1)$ we get
$$(y+1-x)(z+1-y)(x+1-z)=1 $$ $\implies$
$$xy+yz+zx-3=(x-y)(y-z)(z-x) \tag{3}$$ I am unable to proceed further..
| $\textbf{Hint:}$Note that $\displaystyle x^2-x-1=y$ is equal $\displaystyle (x-\frac{1}{2})^2-\frac{5}{2}=y-\frac{1}{2}$, so if you substitute $x_1=x-\frac{1}{2}$, $y_1=y-\frac{1}{2}$ and $z_1=z-\frac{1}{2}$ you get:
$$x_1^2-\frac{5}{2}=y_1$$
$$y_1^2-\frac{5}{2}=z_1$$
$$z_1^2-\frac{5}{2}=x_1$$
Substitute (1) to (2), next (2) to (3), you get polynomial equation to calculate $x_1$. The same with $y_1$ and $z_1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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} |
Calculus area between graph and x axis
How to solve this question ?
The integral would be 0.3x^3+xc^2 + c
0.3(64)^3 +(64)c^2 +c =0
c= 78643.2/( -64+c)
am I right ?
| The problem should be restated because the area "between" the function $y=x^2-c^2$ and $y=0$ is always equal to $\infty$.
I would restated the problem as follows
Find the exact positive value of $c$ if the area of the region bounded by $y=x^2-c^2$ and the x-axis is 64.
Next, make some sketches.
Clearly the region bounded by the curve and the x-axis is the region in blue. The area between the 2 curves is the |pink area|+|blue area|.
The area bounded would be
$$\bigg|\int_{-c}^c(x^2-c^2)dx\bigg|$$
$$= \bigg|\frac{x^3}{3}-c^2x\bigg|_{-c}^c\bigg| = \bigg|\bigg((\frac{c^3}{3}-c^3) - (-\frac{c^3}{3}-(-c^3))\bigg)\bigg|$$
$$= \bigg|-\frac{2}{3}c^3- \frac{2}{3}c^3\bigg|$$
$$= \bigg|-\frac{4}{3}c^3\bigg| = \frac{4}{3}c^3$$
But
$$\bigg|\int_{-c}^c(x^2-c^2)dx\bigg| = 64$$
so
$$\frac{4}{3}c^3 = 64$$
| {
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} |
How to calculate $\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx$? I want to calculate the following integral: $$I:=\displaystyle\int_{-\infty}^\infty\underbrace{\frac{x^2+2x}{x^4+x^2+1}}_{=:f(x)}dx$$ Of course, I could try to determine $\int f(x)\;dx$ in terms of integration by parts. However, I don't think that's the way one should do this. So, what's the trick to calculate $I$?
| As noted in the comments, the integral is:
$$I=\int_{-\infty}^{\infty} \frac{x^2}{x^4+x^2+1}\,dx=2\int_0^{\infty} \frac{x^2}{x^4+x^2+1}\,dx\,\,\,(*)$$
With the change of variables $x\mapsto 1/x$,
$$I=2\int_{0}^{\infty} \frac{1}{x^4+x^2+1}\,dx\,\,\,\,\,\,(**)$$
Add $(*)$ and $(**)$ i.e
$$2I=2\int_0^{\infty} \frac{1+x^2}{x^4+x^2+1}\,dx \Rightarrow I=\int_0^{\infty} \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}\,dx$$
Rewrite the denominator as $\left(x-\dfrac{1}{x}\right)^2+3$ and use the substitution $x-\dfrac{1}{x}=t$,
$$\Rightarrow I=\int_{-\infty}^{\infty} \frac{dt}{t^2+3}=\boxed{\dfrac{\pi}{\sqrt{3}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we apply the forward Euler method to $x''=x^2$? If we want to apply the forward Euler method to $x''=x$ with $x(0)=0, x'(0)=1$, we can introduce a new function $$u:=\begin{bmatrix}x'\\x \end{bmatrix}$$
then $$u'=\begin{bmatrix}0&1\\1&0\\ \end{bmatrix}u \ \ \text{and} \ \ u(0)=\begin{bmatrix}1\\0 \end{bmatrix}.$$
So then first step of the forward Euler would be
\begin{align*}
u_1 & = u(0)+hu'(0)
\\ & = \begin{bmatrix}1\\0 \end{bmatrix} + h \begin{bmatrix}0&1\\1&0\\ \end{bmatrix}\begin{bmatrix}1\\0 \end{bmatrix}
\\ & = \begin{bmatrix}1\\h \end{bmatrix}.
\end{align*}
And if we translate it back we get $x_1=h$.
How can we do a similar thing with the equation $x''=x^2$ (with $x(0)=0, x'(0)=1$)? It doesn't look like it can be written as a system, so there should be some other trick.
| Let $u = \begin{bmatrix} x \\ x'\end{bmatrix}$. Then, $u' = \begin{bmatrix} x' \\ x''\end{bmatrix} = \begin{bmatrix} x' \\ x^2\end{bmatrix}$ with $u(0) = \begin{bmatrix} 0 \\ 1\end{bmatrix}$.
So, taking forward Euler steps yields:
$u(h) \approx u(0) + hu'(0) = \begin{bmatrix} 0 \\ 1\end{bmatrix} + h\begin{bmatrix} 1 \\ 0^2\end{bmatrix} = \begin{bmatrix} h \\ 1\end{bmatrix}$
$u(2h) \approx u(h) + hu'(h) = \begin{bmatrix} h \\ 1\end{bmatrix} + h\begin{bmatrix} 1 \\ h^2\end{bmatrix} = \begin{bmatrix} 2h \\ 1+h^3\end{bmatrix}$
$u(3h) \approx u(2h) + hu'(2h) = \begin{bmatrix} 2h \\ 1+h^3\end{bmatrix} + h\begin{bmatrix} 1+h^3 \\ (2h)^2\end{bmatrix} = \begin{bmatrix} 3h+h^4 \\ 1+5h^3\end{bmatrix}$
and so on. The only difference is that $u'$ is a non-linear function of $u$.
| {
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Is $\int^x \cos \frac1t$ differentiable at zero? From Spivak's Calculus, 4th ed., exc 14-20:
Let $$f(x) = \begin{cases} \cos \frac1x, & x\neq 0\\ 0, &x=0.
\end{cases}$$ Is the function $\int_0^xf$ differentiable at zero?
I'm having trouble with this one; I'm sure I'm just missing something easy.
We seek
$$\lim_{h\to0} \frac1h \int_0^h \cos\frac1\xi d\xi.$$
A crude estimation tells us that $\frac1h \int_0^h \cos\frac1\xi d\xi$ is between $-1$ and $1$.
Intuitively, it seems that $\int_0^h\cos\frac1\xi d\xi$ is going to zero like $h$.
I considered changing variables (although Spivak has not yet introduced this technique, so it shouldn't be necessary) to
$$\frac1h \int_{1/h}^\infty \frac{1}{u^2} \cos u,$$
but that didn't really help.
| Without that substitution, we can note that
$$g(x) = \begin{cases} -x^2\sin \frac{1}{x} &, x \neq 0 \\ \quad 0 &, x = 0\end{cases}$$
is differentiable everywhere, with $g'(0) = 0$ and
$$g'(x) = \cos \frac{1}{x} -2x\sin\frac{1}{x}$$
for $x\neq 0$. So
$$\int_0^h \cos \frac{1}{\xi}\,d\xi = \int_0^h g'(\xi) + 2\xi\sin \frac{1}{\xi}\,d\xi = g(h) + 2\int_0^h \xi\sin \frac{1}{\xi}\,d\xi.$$
Now it is easy to see that
$$\frac{1}{h} \int_0^h \cos \frac{1}{\xi}\,d\xi = \frac{g(h)}{h} + \frac{2}{h}\int_0^h \xi\sin \frac{1}{\xi}\,d\xi$$
tends to $0$ for $h\to 0$ by the mean value theorem for integrals, since the integrand of the remaining integral is bounded by $\lvert h\rvert$ in absolute value.
The substitution $u = \frac{1}{\xi}$ leads to
$$\frac{1}{h} \int_{1/h}^\infty \frac{\cos u}{u^2}\,du,$$
and we can get further with an integration by parts:
$$\begin{align}
\int_a^b \frac{\cos u}{u^2}\,du &= \left[\frac{\sin u}{u^2}\right]_a^b + 2\int_a^b \frac{\sin u}{u^3}\,du\\
&= \frac{\sin b}{b^2} - \frac{\sin a}{a^2} + 2\int_a^b \frac{\sin u}{u^3}\,du.\tag{1}
\end{align}$$
We can estimate the remaining integral
$$\left\lvert 2\int_a^b \frac{\sin u}{u^3}\,du\right\rvert \leqslant 2\int_a^\infty \frac{du}{u^3} = \frac{1}{a^2}.$$
So we find
$$\left\lvert \frac{1}{h} \int_0^h \cos \frac{1}{\xi} \,d\xi\right\rvert \leqslant \frac{1}{h}\left(\frac{2}{1/h^2} + \frac{1}{1/h^2}\right) \leqslant 3h$$
for $h > 0$. The case $h < 0$ is symmetric, so we see that the integral $F$ of $f$ is differentiable in $0$ with $F'(0) = 0$.
| {
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"question_score": "6",
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} |
Evaluate expression in the form $a+bi$. So, I have to evaluate $\sqrt{-3}\sqrt{-12}$ into the form $a+bi$.
I know that $i^2 = -1$ so $i = \sqrt{-1}$
What I have done is:
$$\begin{align}\sqrt{-3}\sqrt{-12}
&= \sqrt{3(-1)}\sqrt{12(-1)}\\
&= \sqrt{3}i\sqrt{12}i\\
&= \sqrt{3}\cdot i\cdot 2\sqrt{3}\cdot i\\
&= 3\cdot 2\cdot (i^2)\\
&= 6\,i^2\\
&= 6\cdot (-1)\\
&=-6\end{align}$$
So, my question is if I have to have the expression in the form $a+bi$ is my answer of $-6$ still correct? Thanks.
| Yes it's correct. Just take $a = -6$ and $b = 0$.
| {
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How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$
Find the value:
$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$
I use computer have this reslut
$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{2}\dfrac{\pi\ln^2{2}}{\sqrt{2}}-\dfrac{3}{2}\dfrac{\pi^2\ln{2}}{\sqrt{2}}$$
where $C$ is Catalan constant.
My idea: let $$\tan{x}=t,\sin{2x}=\dfrac{2t}{1+t^2},dx=\dfrac{1}{1+x^2}$$
then
$$I=\int_{0}^{\infty}\ln^2{(1+t^4)}\dfrac{\dfrac{2}{1+t^2}}{2-\left(\dfrac{2t}{1+t^2}\right)^2}\cdot\dfrac{dt}{1+t^2}=\int_{0}^{\infty}\dfrac{\ln^2{(1+t^4)}}{(t^2-1)^2}dt$$
Then I can't.Thank you
| After performing the initial arctangent substitution, the resultant integral may be rewritten as the second derivative of a beta function and evaluated accordingly:
$$\begin{align}
I
&=\int_{0}^{\frac{\pi}{2}}\frac{2\cos^2{\theta}}{2-\sin^2{(2\theta)}}\log^2{\left(1+\tan^4{\theta}\right)}\,\mathrm{d}\theta\\
&=\int_{0}^{\infty}\frac{\log^2{(1+x^4)}}{1+x^4}\mathrm{d}x\\
&=\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\int_{0}^{\infty}\frac{\mathrm{d}x}{(1+x^4)^{\alpha}}\\
&=\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\int_{0}^{\infty}\frac{\frac14u^{-\frac34}}{(1+u)^{\alpha}}\mathrm{d}u\\
&=\frac14\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\operatorname{B}{\left(\frac14,\alpha-\frac14\right)}\\
&=\frac14\Gamma{\left(\frac14\right)}\Gamma{\left(\frac34\right)}\left[-\zeta{(2)}+\left(\gamma+\Psi{\left(\frac34\right)}\right)^2+\Psi^\prime{\left(\frac34\right)}\right]\\
&=\frac{\sqrt{2}\,\pi}{4}\left[-\frac{\pi^2}{6}+\left(\frac{\pi}{2}-\log{8}\right)^2+\pi^2-8C\right]\\
&=-2\sqrt{2}\,\pi\,C+\frac{13\pi^3}{24\sqrt{2}}+\frac{\pi\log^2{8}}{2\sqrt{2}}-\frac{\pi^2\log{8}}{2\sqrt{2}}.~~~\blacksquare
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers which are not quadratic squares
Prove that $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers $0<a<\frac{p}{2}$ which are not quadratic squares $\pmod p$ ($p\equiv3\bmod4$)
I don't know really from where to start (we know from wilson that $(p-1)!\equiv -1\pmod p$ but that won't help here as far as I see). The lecturer suggested lookigng at $p-n$ where $1\le n\le p-1$. I'd be glad for any hint.
| $$\left ( \left ( \frac{p-1}{2} \right )! \right )^2= 1^2 \cdot 2^2 \cdot 3^2 \cdots \left (\frac{p-1}{2} \right )^2=1 \cdot 2 \cdot 3 \cdots \frac{p-1}{2} \cdot 1 \cdot 2 \cdot 3 \cdots \frac{p-1}{2} \\\equiv 1 \cdot 2 \cdot 3 \cdots \frac{p-1}{2} \cdot (-1) \cdot (p-1) \cdot (-1) \cdot (p-2) \cdots (-1) \cdot \frac{p+1}{2} \overset{Wilson}{ \equiv} \ \ (-1) (-1)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
$$p \mid \left ( \left ( \frac{p-1}{2} \right )! \right )^2-(-1)^{\frac{p+1}{2}} \Rightarrow p \mid \left( \left ( \frac{p-1}{2} \right )!-(-1)^{\frac{p+1}{2}} \right ) \left( \left (\frac{p-1}{2} \right )!+(-1)^{\frac{p+1}{2}} \right )$$
As $p$ is a prime, $p \mid \left( \left ( \frac{p-1}{2} \right )!-(-1)^{\frac{p+1}{2}} \right )$ OR $p \mid \left( \left (\frac{p-1}{2} \right )!+(-1)^{\frac{p+1}{2}} \right )$
Therefore, $ \left(\frac{p-1}{2} \right )! \equiv (-1)^{\frac{p+1}{2}} \pmod p$ OR $ \left(\frac{p-1}{2} \right )! \equiv (-1)^{\frac{p+3}{2}} \pmod p$
Now,it remains to show that the number of integers $0<a<\frac{p}{2}$ which are not quadratic squares $\pmod p$ is either $\frac{p+1}{2}$ or $\frac{p+3}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $ Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $$
I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a multiple of $6$.
| If $1+\zeta+\zeta^2=0$, try to compute $(1+\zeta^0)^n + (1+\zeta)^n + (1+\zeta^2)^n$. Hint: $\zeta^3=1$, and, if $k$ is not divisible by $3$, $1+\zeta^k+\zeta^{2k}=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c^2(a^{n-2})=a^{n-2}(b^2+c^2)=a^{n-2}\times a^2=a^n \implies b^n+c^n<a^n$
But I stuck in part (b). Can I prove it in a way I proved part (a)?
| These may help you:
First prove for $n = 1$ (Easy, as $a = \sqrt{b^2+c^2} < \sqrt{b^2+c^2 + 2bc} = b +c$).
Then, separately try for $n<1$ and $1<n\leq2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving Fibonacci inequality I didn't see a question regarding this particular inequality, but I think that I have shown by induction that, for $n>1$. I am hoping someone can verify this proof.
$$\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}<F_{n+1}<\left(\frac{1+\sqrt{5}}{2}\right)^{n}$$
Here we state that $F_0=0, F_1=1$. The base case is established since
$$\left(\frac{1+\sqrt{5}}{2}\right)<F_{3}<\left(\frac{1+\sqrt{5}}{2}\right)^{2}$$
$$1.61803...<2<2.61803...$$
Now assume that $n=k$ and that the following holds:
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}<F_{k+1}<\left(\frac{1+\sqrt{5}}{2}\right)^{k}$$
Starting with the left side of the inequality, and looking at $F_{k+2}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k+2}$
$$F_{k+2}>\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}=\frac{2}{\sqrt{5}+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}>\frac{2}{3+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}=\left(\frac{2}{1+\sqrt{5}}\right)^2\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}=\left(\frac{1+\sqrt{5}}{2}\right)^{k}$$
So the left side didn't need the induction hypothesis, just some careful analysis. Now the right side of the inequality. By the induction hypothesis,
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k}>F_{k+1} \Rightarrow \left(\frac{1+\sqrt{5}}{2}\right)^{k+1}>\left(\frac{1+\sqrt{5}}{2}\right)F_{k+1} $$
But since $F_{k+1}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k+1}$
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}>\left(\frac{1+\sqrt{5}}{2}\right)F_{k+1}=F_{k+2} $$
Therefore,
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k}<F_{k+2}<\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}$$
I feel as though the analysis is good. Is there any holes or misstatements?
| Your proof is valid, but i like a little less tedious proof:
Note that $$\phi^2=\phi+1$$
$$\phi^3=\phi^2+\phi=2\phi+1$$
$$\phi^4=2\phi^2+\phi=3\phi+2$$
$$\phi^5=3\phi^2+2\phi=5\phi+3$$
see the Fibonacci numbers in the coeficcients, now generally:
$$\phi^n=F_n\phi+F_{n-1}$$, so we can rewrite your problem:
$$\phi^{n-1}=F_{n-1}\phi+F_{n-2}<F_{n+1}<F_n\phi+F_{n-1}=\phi^n$$
since $F_n+F_{n-1}=F_{n+1}$, and $1<\phi$ it follows that $F_n<F_n\phi$ and $$F_{n+1}=F_n+F_{n-1}<F_n\phi+F_{n-1}$$
since $2F_{n-1}+F_{n-2}=F_{n}+F_{n-1}=F_{n+1}$ and $\phi<2$, it follows that $F_{n-1}\phi<2F_{n-1}$ and
$$F_{n-1}\phi+F_{n-2}<2F_{n-1}+F_{n-2}=F_{n+1}$$
thus we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$? As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.
FIRST
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
&= \sqrt{16}\\
&=4
\end{align*}
SECOND
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\
&= 4i \times i\\
&=4i^2\\
&=-4
\end{align*}
Incidentally if the above is input in complex mode of Casio scientific calculator, the result comes out as $-4$.
Which of the above solutions is correct?
| $$\sqrt a\cdot\sqrt b=\sqrt{ab}$$ only work if $a,b\ge0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| Or: $1 + 2\sin{x }\cos{x} = \frac{1}{4},$ so $\sin{x}\cos{x} = \frac{-3}{8},$ leading to $\sin{x} - \cos{x} = \pm \sqrt{ \frac{1}{4} + \frac{3}{2}} = \pm \frac{\sqrt{7}}{2},$ allowing you to solve for $\sin{x}$ and $\cos{x}$ (making use of the fact that $0 < x < \pi$ (radians) so that $\sin{x} >0$).
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 7
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Finding common ratio from two sums I'm struggling with this very basic question on the binomial theorem:
The sum of the first and second terms of a geometric progression is 12, and the sum of the third and fourth term is 48. Find the two possible values of the common ratio and the corresponding values of the first term.
So I tried approaching this in two ways. The first:
$$
u_1 + u_2 = a + ar = 12 \\
u_3 + u_4 = ar^2 + ar^3 = 48 \\
4a(1 + r) = ar^2 + ar^3 \\
4 + 4r = r^2 + r^3 \\
r^3 + r^2 - 4r - 4 =0
$$
Okay, no solution there. So then I tried:
$$
S_2 = \frac{a(r^2 - 1)}{r-1} = 12 \\
S_4 = \frac{a(r^4 - 1)}{r-1} = 48 \\
\frac{4a(r^2 - 1)}{r-1} = \frac{a(r^4 - 1)}{r-1} \\
4r^2 - 4 = r^4 - 1 \\
r^4 - 4r^2 + 3 =0
$$
Am I missing something obvious here?
| From your first two equations we get
$$\frac{48}{12}=\frac{ar^2+ar^3}{a+ar}=r^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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