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Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\left( \frac{x+2}{3-x}-1\right)=x\left(\frac{2x-1}{3-x}\right)=x\left(\frac{2x}{-x}\right)=-2x\rightarrow-\infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
|
$\frac{x+2}{3-x} = \frac{5}{3-x} - 1 < 0$ when $x\to \infty$.
Hence $\underset{x\to\infty}{\text{lim}}f(x) = \underset{x\to\infty}{\text{lim}} x\log(1 + \frac{5}{x - 3}) = \underset{x\to\infty}{\text{lim}}x \frac{5}{x - 3} = 5$
where the second to last equality follows from $\log(x + 1) = x + o(x) \text{ for } x\to 0 $ thanks to @gimusi's suggesion.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$
Dividing numerator and denominator by x:
$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$
$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$
$$= \frac{7}{\sqrt{4+0}-2}$$
$$=\frac{7}{2-2}$$
$$=\infty$$
Conclusion: Limit does not exist.
Why is my solution wrong?
Correct answer: $\frac{-7}{4}$
|
It's a very common error: $\sqrt{x^2}=-x$, when $x<0$.
I usually suggest the substitution $t=-1/x$, so the limit becomes
$$
\lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}-\frac{7}{t}}-\frac{2}{t}\right)
=
\lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t}
$$
which is an easy derivative:
$$
f(t)=\sqrt{4-7t}
\qquad
f'(t)=\frac{-7}{2\sqrt{4-7t}}
\qquad
f'(0)=-\frac{7}{4}
$$
or, with a Taylor expansion,
$$
\lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t}
=
\lim_{t\to0^+}\frac{2(\sqrt{1-7t/4}-1)}{t}
=
\lim_{t\to0^+}\frac{2\bigl(1-\frac{1}{2}\frac{7t}{4}+o(t)\bigr)}{t}=-\frac{7}{4}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Interpolation and Approximation A quadratic polynomial $p(x)$ is constructed by interpolating the data points $(0,1)$, $(1,e)$, $(2,e^2)$. If $\sqrt{e}$ is approximated by using $p(x)$ then its approximated value is.
|
Assume $p(x)=ax^2+bx+c$ this quadratic polynomial satisfies the given points
$$p(0)=1=c$$
$$p(1)=e=a+b+c$$
$$p(2)=e^2=4a+2b+c$$
you got $a=\frac{(e-1)^2}{2}$, $b=\frac{(e-1)(3-e)}{2}$, $c=1$
thus your polynomial is $$p(x)=\frac{(e-1)^2}{2}x^2+\frac{(e-1)(3-e)}{2}x+1$$
$$\frac{(e-1)^2}{2}x^2+\frac{(e-1)(3-e)}{2}x+1=0$$
$$(e-1)^2x^2+(e-1)(3-e)x+2=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
An Euler type sum: $\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$, where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$ I've been trying to compute the following series for quite a while :
$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$$
where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$ are the generalized harmonic numbers of order 2.
I've already successfully computed the value of a lot of similar-looking series involving harmonic numbers and central binomial coefficients (by using Abel's transformation, finding the generating function...), and they all had closed forms in terms of the Riemann Zeta function, so I'm confident this one too has a nice closed form ; but it's a tough guy that resists my previous methods.
By numerical test, I suspect that the value is $\frac{3}{2}\zeta(3)$.
Any idea for derivation ?
|
Using the well-known identity
$$\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$
Divide both sides by $x$ then integrate , we get
$$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+C $$
set $x=0,\ $ we get $C=2\ln2$
Then
$$\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+2\ln2\tag1$$
Multiply both sides of (1) by $-\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $1$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
\begin{align}
\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}&=2\underbrace{\int_0^1\frac{\ln(1+\sqrt{1-x})\ln(1-x)}{x}dx}_{\sqrt{1-x}=y}-2\ln2\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{-\zeta(2)}\\
&=8\int_0^1\frac{y\ln(1+y)\ln y}{1-y^2}dy+2\ln2\zeta(2)\\
&=4\int_0^1\frac{\ln(1+y)\ln y}{1-y}-4\int_0^1\frac{\ln(1+y)\ln y}{1+y}+2\ln2\zeta(2)
\end{align}
where the first integral is
$$\int_0^1\frac{\ln y\ln(1+y)}{1-y}\ dy=\zeta(3)-\frac32\ln2\zeta(2)$$
and the second integral is
$$\int_0^1\frac{\ln y\ln(1+y)}{1+y}\ dy=-\frac12\int_0^1\frac{\ln^2(1+y)}{y}dy=-\frac18\zeta(3)$$
Combine the results of the two integrals we get
$$\boxed{\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)}$$
If we differentiate both sides of $\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
$$ \int_0^1x^{n-1}\ln x\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}-\zeta(2)}{n}\tag2$$
Now multiply both sides of $(2)$ by $ \frac{1}{4^n}{2n\choose n}$ the sum up from $n=1$ to $\infty$ we get
$$\sum_{n=1}^\infty \frac{H_n}{n^24^n}{2n\choose n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}-\zeta(2)\sum_{n=1}^\infty \frac{1}{n4^n}{2n\choose n}\\=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\=\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx}_{\text{Beta function:}7\zeta(3)-6\ln2\zeta(2)}-\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}$$
Substitute $\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)$ and $\sum_{n=1}^\infty\frac{1}{n4^n}{2n\choose n}=2\ln2$ we get
$$\boxed{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n4^n}{2n\choose n}=\frac32\zeta(3)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the general solution (i) $\cot \theta =-\dfrac {1} {\sqrt {3}}$ (ii)$4\cos ^{2}\theta =1$ my attempt for
(i)
$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$
$\cot ( \theta ) = - \frac { \sqrt { 3 } } { 3 }$
(ii)
$\left. \begin{array} { l } { \text { Let: } \cos ( \theta ) = u } \\ { 4 u ^ { 2 } = 1 } \end{array} \right.$
$\left. \begin{array} { l } { \text { Divide both sides by } 4 } \\ { \frac { 4 u ^ { 2 } } { 4 } = \frac { 1 } { 4 } } \end{array} \right.$
is it right way to find general solution for these equations?
|
$$1=4\cos^2t=2(1+\cos2t)$$
$$\iff\cos2t=?$$
$$2t=2n\pi\pm\dfrac{2\pi}3$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 0
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|
Making an expression stable for small values The following expression shows significant numerical differences in a program when I compile in x86 (32 bit) versus x64 (64 bit), when $a$ is small:
$$ \left( \dfrac{1}{a} - b \right) \left( 1- \exp(-a)\right)$$
Is there a way that I can refactor this expression so that it is more robust for small $a$? It is not completely clear to me that simply expanding the expression into four terms is the best solution.
|
You can approximate
$$
e^x \approx 1+x + \frac{x^2}{2!} + \frac{x^3}{3!}
\qquad
\Rightarrow
e^{-x} \approx 1-x + \frac{x^2}{2!} - \frac{x^3}{3!}
$$
Plugging this into the equation, we get
$$
\left(\frac{1}{x}-b\right) \left(1- \left[ 1-x + \frac{x^2}{2!} - \frac{x^3}{3!}\right]\right) = \left(\frac{1}{x}-b\right) \left(x - \frac{x^2}{2!} + \frac{x^3}{3!}\right) =
\left(\frac{1}{x}-b\right) x\left(1 - \frac{x}{2!} + \frac{x^2}{3!}\right)
$$
Then assume $x\neq 0$ ... Can you continue from here?
|
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|
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqrt x}+\frac{10}{x^2\sqrt x}\right)dx$$
$$=\int \left(x^2x^{-\frac{3}{2}}-4x^1x^{-\frac{3}{2}}+10x^{-\frac{3}{2}}\right)dx$$
$$=\int \left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}+10x^{-\frac{3}{2}}\right)dx$$
I then integrated the expression to get
$$\frac{2x^{\frac{3}{2}}}{3}-8x^{\frac{1}{2}}-20x^{-\frac{1}{2}}$$
This is, however, wrong. Any ideas as to why?
Thanks in advance $:)$
|
It is $x^{-5/2}$ and not $x^{-3/2}$.
|
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|
Probability of real roots for $x^2 + Bx + C = 0$ Question: The numbers $B$ and $C$ are chosen at random between $-1$ and $1$, independently of each other. What is the probability that the quadratic equation $$x^2 + Bx + C = 0$$ has real roots? Also, derive a general expression for this probability when B and C are chosen at random from the interval $(-q, q)$ for and $q>0$.
My approach: since we're trying to find the probability of real roots. We should first realize when it has imaginary roots. So, $$B^2 - 4aC < 0$$ $$a = 1$$ $$B^2 < 4C $$ I'm not sure where to go from here. How do I now find the probability of $B$ being greater than $4C$?
|
I assume you mean that $B,\,C\sim U(-1,\,1)$. We'll get the answer as a function of a fixed value for $C$, then average it out. For $C<0$ (which has probability $1/2$), the result is $0$; for $C> 1/4$ (which has probability $3/8$), the result is $1$; for $0\le C\le\frac{1}{4}$ (which has probability $\frac{1}{8}$), the condition $-2\sqrt{C}\le B\le 2\sqrt{C}$ has probability $4\sqrt{C}$. So the final result is $$\frac{3}{8}+\frac{1}{8}\int_0^{1/4} 4\sqrt{C}dC=\frac{3}{8}+\frac{1}{3}\bigg(\frac{1}{4}\bigg)^{3/2}=\frac{5}{12}.$$
Edit: the above is the probability of non-real complex roots; the probability of real roots is $1-\frac{5}{12}=\frac{7}{12}$.
|
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|
Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.
Do I have to use something else? Or am I missing some easy manipulation?
|
There still is the brutal way:
$$ \frac{r+2}{r(r+1)(r+3)}\stackrel{\text{PFD}}{=}\frac{2}{3}\cdot\frac{1}{r}-\frac{1}{2}\cdot\frac{1}{r+1}-\frac{1}{6}\cdot\frac{1}{r+3} \tag{1}$$
leads to:
$$ \frac{r+2}{r(r+1)(r+3)} = \frac{1}{6}\int_{0}^{1} x^{r-1}\left(4-3x-x^3\right)\,dx \tag{2}$$
then by summing both sides over $r\geq 1$:
$$ \sum_{r\geq 1}\frac{r+2}{r(r+1)(r+3)}=\frac{1}{6}\int_{0}^{1}\frac{4-3x-x^3}{1-x}\,dx=\frac{1}{6}\int_{0}^{1}\left(4+x+x^2\right)\,dx \tag{3} $$
and simplifying:
$$ \sum_{r\geq 1}\frac{r+2}{r(r+1)(r+3)}=\frac{1}{6}\left(4+\frac{1}{2}+\frac{1}{3}\right)=\color{red}{\frac{29}{36}}.\tag{4}$$
The partial sums are more directly computed via $(1)$:
$$\begin{eqnarray*} \sum_{r=1}^{n}\frac{r+2}{r(r+1)(r+3)}&=&\frac{2}{3}H_n-\frac{1}{2}\left(H_{n+1}-1\right)-\frac{1}{6}\left(H_{n+3}-\frac{11}{6}\right)\\&=&\frac{157 n+138 n^2+29 n^3}{36 (n+1)(n+2)(n+3)}.\tag{5} \end{eqnarray*}$$
|
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|
Evaluate $(\sqrt{3})^{11}$ (Evaluate Square Root $3$ to the power of $11$) I know the answer is $243 \sqrt 3$ but in my maths book they got $(\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3)$ but then they only took the first $5$ out of the $11$ $(\sqrt 3)$s and then got $(3) (3) (3) (3) (3) (3) \sqrt 3$ and then got $3^5 \sqrt 3$
Why are they only using to the first $5$ $(\sqrt 3)$s? What happened to the other $6$ and how did the answer $243 \sqrt3$ arise? Thank you
|
$$ \sqrt{3}^{11} = \sqrt{3} \cdot \sqrt{3}^{10} = \sqrt{3} \cdot \left(\sqrt{3}^2\right)^5 = \sqrt{3} \cdot 3^5 = 243\sqrt{3}. $$
|
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|
Least Common Multiple and the product of a sequence of consecutive integers Let $x>0, n>0$ be integers.
Let LCM$(x+1, x+2, \dots, x+n)$ be the least common multiple of $x+1, x+2, \dots, x+n$.
Let $v_p(u)$ be the highest power of $p$ that divides $u$.
It seems to me that:
$$\frac{(x+1)(x+2)\times\dots\times(x+n)}{\text{LCM}(x+1, x+2, \dots, x+n)} \le (n-1)!$$
Here is my thinking:
*
*Let $p$ be any prime that divides $(x+1)\times\dots\times(x+n)$.
*Let $x+i$ be the integer in $\{x+1,x+2. \dots, x+n\}$ that is divisible by the highest power of $p$.
*All integers in the sequence $\{x+1, x+2, x+3, \dots, x+n\}$ that are divisible by $p$ will have the form $x + i \pm ap$ where $0 < a \le \left\lfloor\dfrac{n-1}{p}\right\rfloor$.
*$v_p\left(\dfrac{(x+1)(x+2)\times\dots\times(x+n)}{\text{LCM}(x+1,x+2,\dots,x+n)}\right) \le v_p(p) + v_p(2p) + \dots + v_p\left(\left\lfloor\dfrac{n-1}{p}\right\rfloor p\right) = v_p((n-1)!)$
*So, it follows that the maximum power of $p$ that divides $\dfrac{(x+1)\times\dots\times(x+n)}{\text{LCM}(x+1,\dots,x+n)}$ is less than or equal to the maximum power of $p$ that divides $(n-1)!$
Edit: Made changes to third and fourth bullet point with an attempt to make them cleaner.
Based on feedback received.
Edit 2: I originally put the "same" power in the last bullet point. That is not correct. It should be less or equal. I have changed it.
Edit 3: Fixed a typo. n! should be (n-1)!.
|
Your claim is true according to a wiki page.
Changing $n,k$ to $x+n,n-1$ respectively in$$\binom nk\le\frac{\text{LCM}(n-k,n-k+1,\cdots, n)}{n-k}$$
we get
$$\binom{x+n}{n-1}\le\frac{\text{LCM}(x+1,x+2,\cdots, x+n)}{x+1}$$
which is your claim.
|
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|
Contour Integral of irrational polynomial from -1 to 1 I've been stuck at htis contour integral problem for a few hours now, and seem to be hitting brick walls.
$$
\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^4}dx\,,
$$
I tried a trig substitution $x=\cos{\theta}$ but noticed all the poles were the unit circle and I didn't know how to proceed.
$$
\frac{1}{2}\int_{0}^{2\pi} \frac{\sin^2{\theta}}{1+\cos^4{\theta}}d\theta\,,
$$
Next I thought maybe a rectangular contour but the contours that go vertically from $1$ to $1+i\infty$ and $-1+i\infty$ to $-1$ didn't seem to cancel, or I wasn't able to show that it does. I manipulated it for awhile.
$$
\int_{0}^\infty \frac{\sqrt{1-(1+iy)^2}}{1+(1+iy)^4}dy + \int_{0}^{-\infty} \frac{\sqrt{1-(1+iy)^2}}{1+(1+iy)^4}dy
$$
I naively wrote the trig integral in terms of z, but the denominator is an 8th order polynomial.
$$
i\oint \frac{2z^5-4z^3+2z}{z^8+4z^6+22z^4+4z^2+1} dz
$$
I think this might be getting closer to a solution, but how can I find the poles by hand?
|
As pointed out by @paul garrett, dog-bone contour (dumbbell contour) works perfectly. Indeed, consider the contour $\mathcal{C}$ given as follows:
$\hspace{12em}$
With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero,
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^4+1)} \, dz
&\longrightarrow
\int_{-1-0^+i}^{1-0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz
- \int_{-1+0^+i}^{1+0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz \\
&\quad = 2\int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^4+1} \, dx.
\end{align*}
On the other hand, Residue Theorem tells that
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz
&= - 2\pi i \sum_{a \ : \ a^4 + 1 = 0} \underset{z=a}{\mathrm{Res}} \, \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \\
&= \pi \sqrt{2(\sqrt{2}-1)}.
\end{align*}
(In order to use Residue Theorem, consider a large circle, apply Residue Theorem to the region enclosed by this circle and $\mathcal{C}$, and then let the radius of the circle go to $\infty$.) Therefore
$$ \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^4+1} \, dx = \frac{\pi}{2} \sqrt{2(\sqrt{2}-1)}. $$
|
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|
Help: $ |\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that
On other hand, a short calculation yields
$$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$$
Image of the page :-
Here,
$$\left(\frac{xz}{y^2}\right)^k= \frac{(a+1)(ab^2+1)}{(ab+1)^2}$$ and $ b \geq 2, a\geq 2^{49},k\geq 50 $ (see page $8, 9$).
So, how do we prove the following?
$$ \left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|\leq \frac{1}{b}$$
|
I have got $$\left|\frac{a+1}{a}-\frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|=\left|{\frac { \left( a+1 \right) \left( 2\,ab-a+1 \right) }{a \left( ab+1
\right) ^{2}}}
\right|$$
I have compute $$\frac{1}{4}-f(a,b)^2=\frac{\left(a^3 b^2-2 a^2 b+2 a^2-4 a
b+a-2\right) \left(a^3 b^2+6 a^2 b-2 a^2+4 a
b+a+2\right)}{4 a^2 (a b+1)^4}$$ and this is positive if $$b\geq 2,a\geq 2^{29}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Express $x$ in terms of $a$ and $b$: $\sin^{-1} {\frac{2a}{1+a^2}} + \sin^{-1}{\frac{2b}{1+b^2}} = 2\tan^{-1}x$
Find the value of $x$ from the following equation in terms of $a$ and $b$
$$\sin^{-1} {2a\over{1+a^2}} + \sin^{-1}{2b\over{1+b^2}} = 2\tan^{-1}x$$
I tried to expand the LHS using the formula $$\sin^{-1}c+\sin^{-1}d = \sin^{-1}\left(c\sqrt{1+d^2} + d\sqrt{1+c^2}\right)$$
But it didn't work out. Could anyone help me out?
|
Divide both sides by $2$, then using the $\arctan(x)=y \implies x=\tan(y)$ you will get that $$x=\tan\left(\frac{\arcsin\left(\frac{2a}{1+a^2}\right)+\arcsin\left(\frac{2b}{1+b^2}\right)}{2}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2909590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
continued fraction of $\sqrt{10k+3}$ Could you please help me to find the continued fraction of
$$\sqrt{10k+3}.$$
Where $k$ is a positive integer.
All the best,
|
\begin{equation*}
x+1=\sqrt{10k+3}
\end{equation*}
\begin{equation*}
x^{2}+2x+1=10k+3
\end{equation*}
\begin{equation*}
x^{2}+2x=10k+2
\end{equation*}
\begin{equation*}
x\left(x+2\right)=10k+2
\end{equation*}
\begin{equation*}
x=\dfrac{10k+2}{2+x}
\end{equation*}
\begin{equation*}
x=\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dots}}}
\end{equation*}
Then,
\begin{equation*}
\sqrt{10k+3}=1+x
\end{equation*}
\begin{equation*}
\sqrt{10k+3}=1+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dots}}}
\end{equation*}
|
{
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|
Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
So I have forgotten how to do this. Do I:
- reduce $A$ to reduced row-echelon form (why do I do this? Is it because it's easy to solve once you have pivot columns and free variables)?
- rewrite the system of equations
Is this right:
\begin{align}
A &= \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & -5 & -2 & 7 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 & \frac{1}{5} & \frac{-29}{5} \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix}
\end{align}
so: $v_4 = t, v_3 = s, v_2 = \frac{-2}{5}s - \frac{7}{5}t, v_1 = \frac{-1}{5}s + \frac{29}{5}t$
$$\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} \frac{-1}{5}s + \frac{29}{5}t \\ \frac{-2}{5}s - \frac{7}{5} t \\ s + 0t \\ 0 + t \end{bmatrix} = s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix} $$
Is this the set of all $v$?
EDIT
I messed up, the first $\frac{7}{5}$ should be a $\frac{-7}{5}$
|
Hint: Augment the matrix $A$ with the column $\begin{pmatrix} 1\\2\end{pmatrix}$ and row-reduce the resulting augmented matrix.
So, row reduce $\left (\begin{array}{rrrr|r}2&-1&0&1&1\\1&2&1&-3 &2\end{array}\right ) $.
|
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|
maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve, thanks in advance
|
With a bit of arrangements I did on paper, you could show that
\begin{equation}
f'(x)
=
-\dfrac{6\sqrt{-x^2+4x-3}+8x-16}{\sqrt{-x^2+4x-3}}
\end{equation}
which is zero when
\begin{equation}
6\sqrt{-x^2+4x-3}= -8x+16
\end{equation}
square both sides
\begin{equation}
-36x^2+ 144x - 108 = 64x^2 -256x + 256
\end{equation}
which is
\begin{equation}
100x^2 -400x +364 = 0
\end{equation}
Roots are
\begin{equation}
x_{1,2}
=
\frac{400 \pm 120}{200}
\end{equation}
i.e.
\begin{align}
x_1 &= 2.6 \\
x_2 &= 1.4
\end{align}
Either do a double derivative (which requires much time) to show that $f(x_2)$ is a maximum, or wiggle around $x_2$ if you're in a hurry.
|
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|
Show that $\sum_{\text{cyc}} \frac{1}{b^2+c^2+5bc-a^2} \leq \frac{\sqrt3}{8S}$ for a triangle with sides $a$, $b$, $c$ and area $S$ Let be $a$, $b$, $c$ sides of a triangle and $S$ his area. Prove that $$\sum_{\text{cyc}} \frac{1}{b^2+c^2+5bc-a^2} \leq \frac{\sqrt3}{8S}$$
My idea: $b^2+c^2-a^2 = 2bc \cos A$, so the inequality is equivalent to:
$$\sum_{\text{cyc}} \frac{\sin A}{2\cos A+5}\leq\frac{\sqrt{3}}{4}$$
|
Now, you can use Jensen because
$$\left(\frac{\sin\alpha}{5+2\cos\alpha}\right)''=\frac{\sin\alpha(10\cos\alpha-17)}{(5+2\cos\alpha)^3}<0$$ for all $0<\alpha<\pi.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
First derivative of $f(x)= \frac{2}{x+1} +3$ I'm struggling to find the first derivative of $f(x)= \frac{2}{x+1} +3$ using the limit definition of derivative. I keep coming up with $f'(x) = \frac{-5}
{(x+1)^2}$ but I should be getting $f'(x) = \frac{-2}{(x+1)^2}$
\begin{align}
\lim_{x\to 0} &= \frac{(\frac{2}{x+h+1}+3)-(\frac{2}{x+1}+3)}{h}\\
\lim_{x\to 0} &= \frac{\frac{2(x+1)+3(x+1)-(2(x+h+1)+3(x+h+1))}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= \frac{\frac{2x+2+3x+3-2x-2h-2-3x-3h-3}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= \frac{\frac{-5h}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= (\frac{-5h}{(x+h+1)(x+1)})*\frac{1}{h}\\
\lim_{x\to 0 }&= \frac{-5}{(x+1)^2}\\
\end{align}
|
The $3$ in the numerator should cancel out.
Even if you want to keep it for a while, it shouldn't be $3(x+1)$ and $3(x+h+1)$, it should be $3(x+1)(x+h+1)$ for both of them.
|
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|
How many even/odd numbers can be formed within a specified range?
Consider
$$A = \{ 1,2,3,4,5,6\}$$
a) How many $3$ digit even numbers between $200$ and $500$ can be formed?
b) How many $3$ digit odd numbers between $240$ and $600$ can be formed?
For the part a, we have that
$$P = (5)(6)(3) = 90$$
For the part b, we have that
$$P = (5)(2)(3) = 30$$
Am I wrong?
EDIT: Repetation is ALLOWED.
Regards.
|
For $(a)$, they have first digit $2,3$ or $4$, and last digit $2,4$ or $6$. So, $3\cdot 6\cdot 3=54$.
For $(b)$, first digit can be $2,3,4$ or $5$. Second digit can be $4,5$ or $6$ when the first digit is $2$, and the last digit is $1,3$ or $5$. So $3\cdot3=9$ for when the first digit is two. Then when the first digit is $3,4$ or $5$, we have any of the six for the second, and $1,3$ or $5$ for the third. So, $3\cdot 6\cdot 3=54$. So we get $54+9=63$.
|
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|
Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit Problem
If there exists $a \in \mathbb{R}$ such that:
$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$
has limit in $-2$. If such $a$ exists what is limit in $-2$ ?
Attempt to solve
My idea was first to try factorize denominator and then find factor of nominator in a way that these two cancel each other.
factorizing denominator gives:
$$ \lim_{x \rightarrow -2}\frac{3x^2+ax+a+3}{(x-1)(x+2)} $$
Now if it is possible to find solution to $3x^2+ax+a+3=0$ when $x=-2$
$$ 3(-2)^2+a(-2)+(-2)+3=0 $$
$$ 12-2a-2+3=0 $$
$$ 2a=13 \iff a = \frac{13}{2} $$
factorizing nominator gives:
$$ 3x^2+\frac{13x}{2}+\frac{19}{2}=0 $$
$$ x= \frac{-\frac{13}{2}\pm \sqrt{(\frac{13}{2})^2-4\cdot 3 \cdot (\frac{19}{2})}}{2\cdot 3} $$
Only problem is this equation is never zero with $x\in \mathbb{R}$ since there is negative value under square root. Now there is contradiction between $a=\frac{13}{2}$ and that equation $3x^2+ax+a+3=0$ when $a=\frac{13}{2}$ and $x=-2$
$$ 3(-2)^2+\frac{13}{2}\cdot (-2)+\frac{13}{2}+3 \neq 0 $$
There is clearly something wrong but i cannot see where this went wrong.
|
You have $3x^2+ax+a+3=0$ and you set $x = -2$ and got $3(-2)^2+a(-2)+(-2)+3=0$
But you should have gotten $3(-2)^2+a(-2)+a+3=0$ .
That is an "$a$" between the $ax$ and the $3$. It is not an "$x$".
So you should have gotten $12 -2a + a + 3 = 0$ or $15 -a = 0$ or $a = 15$
.......
so $3x^2 + 15x+18 = 0$ or in other words $3(x^2 + 5x + 6)=0$ so $3(x+2)(x+3)=0$ so you get $\lim \frac {3(x+2)(x+3)}{(x-1)(x+2)} = \lim \frac {3(x+3)}{x-1} = \frac {3(-2 +3)}{-2-1}=-1$.
|
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|
Proof verification of $\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$ using $\epsilon$-$\delta$ I am trying to prove that
$$\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$$
and this is my approach.
Let $\epsilon>0,\delta>0\,$ so that
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|<\epsilon\quad\text{whenever}\quad-2<x<-2+\delta$$
We have that
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}$$
Let $\delta<1$
$$-2<x<-2+\delta<-1\implies-3<x-1<-2\implies2<|x-1|<3$$
so
$$|x-1|\sqrt{x+2}\leq3\sqrt{x+2}<\epsilon$$
if we pick $\,\delta\leq\min\{1,\epsilon^2/9\}.$
Would this be correct?
|
Seems fine, just write it in the forward direction after working backward.
Given $\epsilon >0$, we let $\delta = \min\{ 1, \frac{\epsilon^2}{9}\}$,
then if $-2 < x < -2 + \delta$, we have $-3<x-1< -2+\delta \implies 2<|x-1|<3$.
hence,
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=\left|\frac{(x-1)(x+2)}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}<3\sqrt{x+2} \le 3\sqrt{\frac{\epsilon^2}{9}}=\epsilon$$
|
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|
Is it true that $\frac{1}{\pi^{2n+1}} \int_0^{\theta} \ln^{2n}\left(\frac{\sin x}{\sin\left(\theta-x\right)}\right)\,dx$ is a rational... I was trying to evaluate $\displaystyle \int_0^{\frac{\pi}{6}}\ln^2\left(2\sin x\right)\,dx$ in an elementary way (no complex variable) so i have considered:
$\displaystyle \int_0^{\frac{\pi}{6}} \ln^2\left(\frac{\sin x}{\sin\left(\frac{\pi}{6}-x\right)}\right)\,dx$.
Using lindep a function in PARI GP i have conjectured that this integral is equal to a rational times $\pi^3$*.
Then i have considered:
$\displaystyle \frac{1}{\pi^5}\int_0^{\frac{\pi}{6}} \ln^4\left(\frac{\sin x}{\sin\left(\frac{\pi}{6}-x\right)}\right)\,dx,\frac{1}{\pi^7} \int_0^{\frac{\pi}{6}} \ln^6\left(\frac{\sin x}{\sin\left(\frac{\pi}{6}-x\right)}\right)\,dx$ and it seems that these integrals are rational numbers.
then i have considered:
$\displaystyle \frac{1}{\pi^5}\int_0^{\frac{\pi}{7}} \ln^4\left(\frac{\sin x}{\sin\left(\frac{\pi}{7}-x\right)}\right)\,dx,\frac{1}{\pi^7} \int_0^{\frac{\pi}{7}} \ln^6\left(\frac{\sin x}{\sin\left(\frac{\pi}{7}-x\right)}\right)\,dx$
same things happen.
Then i have considered:
$\displaystyle \frac{1}{\pi^3}\int_0^{\sqrt{2}} \ln^2\left(\frac{\sin x}{\sin\left(\sqrt{2}-x\right)}\right)\,dx$.
and lindep doesn't show that this number is rational. (it's not a proof).
i have tested much more values ($\frac{\pi}{7}+\frac{1}{10000}$ for example)
My question:
is it true that:
$0< \theta <\pi$, a real
for all $n$, natural integer
$\displaystyle \frac{1}{\pi^{2n+1}} \int_0^{\theta} \ln^{2n}\left(\frac{\sin x}{\sin\left(\theta-x\right)}\right)\,dx$ is a rational
if only if $\theta=r\pi$, $0< r<1$ a rational.
*: i think i have a proof for this.
PS:
The idea of this came after reading: Evaluation of $\int_0^{\pi/3} \ln^2\left(\frac{\sin x }{\sin (x+\pi/3)}\right)\,\mathrm{d}x$
|
The integral can be modified as
\begin{align}
I&= \frac{1}{\pi^{2n+1}} \int_0^{\theta} \ln^{2n}\left(\frac{\sin x}{\sin\left(\theta-x\right)}\right)\,dx\\
&=\frac{1}{\pi^{2n+1}} \int_{-\theta/2}^{\theta/2} \ln^{2n}\left(\frac{\sin \left( \theta/2+y \right)}{\sin\left(\theta/2-y\right)}\right)\,dy\\
&=\frac{2}{\pi^{2n+1}} \int_{0}^{\theta/2} \ln^{2n}\left(\frac{\sin \left( \theta/2+y \right)}{\sin\left(\theta/2-y\right)}\right)\,dy\\
\end{align}
Denoting $\varphi=\theta/2$,
\begin{align}
I&=\frac{2}{\pi^{2n+1}} \int_{0}^{\varphi} \ln^{2n}\left(\frac{\sin\varphi\cos y+\sin y\cos\varphi}{\sin\varphi\cos y-\sin y\cos\varphi}\right)\,dy\\
&=\frac{2}{\pi^{2n+1}} \int_{0}^{\varphi} \ln^{2n}\left(\frac{1+\tan y\cot\varphi}{1-\tan y\cot\varphi}\right)\,dy\\
&=\frac{2^{2n+1}}{\pi^{2n+1}} \int_{0}^{\varphi} \operatorname{arctanh}^{2n}\left(\tan y\cot\varphi\right)\,dy
\end{align}
Now, changing $\tan y=\tan\varphi \tanh u$,
\begin{align}
I&=\frac{2^{2n+1}\tan\varphi}{\pi^{2n+1}} \int_{0}^{\infty}\frac{u^{2n}}{1+\tan^2\varphi \tanh^2 u}\frac{du}{\cosh^2u}\\
&=\frac{2^{2n+1}\sin\varphi\cos\varphi}{\pi^{2n+1}} \int_{0}^{\infty}\frac{u^{2n}\,du}{\cosh^2u-\sin^2\varphi}\\
&=\frac{2^{2n+1}\sin\theta}{\pi^{2n+1}} \int_{0}^{\infty}\frac{u^{2n}\,du}{\cosh 2u+\cos2\varphi}\\
&=\frac{\sin\theta}{\pi^{2n+1}} \int_{0}^{\infty}\frac{v^{2n}\,dv}{\cosh v-\cos(\pi-\theta)}
\end{align}
and using the integral representation of the Bernoulli polynomials DLMF:
\begin{equation}
B_{2n+1}\left(\frac{\pi-\theta}{2\pi}\right)=(-1)^{n+1}\frac{2n+1}{(2\pi)^{2n+1}}\sin\theta\int_{0}^{\infty}\frac{v^{2n}\mathrm{d}v}{\cosh v-\cos\theta}
\end{equation}
we obtain the closed form expression
\begin{equation}
I=\frac{(-1)^{n+1}2^{2n+1}}{2n+1}B_{2n+1}\left(\frac{\pi-\theta}{2\pi}\right)
\end{equation}
Thus, if $\theta=r\pi$ where $r$ is a rational, then the argument of the Bernoulli polynomial is a rational and the integral also. Reciprocally, however, other values of $0<\theta<\pi$ exist which make $ B_{2n+1}\left(\frac{\pi-\theta}{2\pi}\right)$ (and thus $I$) rational. For instance, one can check numerically that, $I_{n=1,\theta=\theta^*}=4/75$ for $\theta^*=\pi\left( 3^{-1/2}\cos\Phi+\sin\Phi \right)$, where $\Phi=3^{-1}\arctan\left( \sqrt{3}\sqrt{517}/18 \right)$(!) (This was obtained, by solving $B_3(x)=1/50$ using a CAS).
|
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|
Closed form of $\int_0^\infty \left(\frac{\arctan x}{x}\right)^ndx$ I know that for $n=1$ the integral is divergent and that for $n=2$ the integral has a closed form. However, I wonder if the general expression has a closed form.
My attempt: $$\int_0^\infty \left(\frac{\arctan(x)}{x}\right)^ndx=\frac{n}{1-n}\int_0^\infty\frac{\arctan^{n-1}(x)}{x^{n-1}(x^2+1)}dx=\frac{n}{1-n}\int_0^{(\frac{\pi}{2})^{n-1}} u^{n-1}\cot^{n-1}\left(u^{1/(n-1)}\right)du$$
I don't know if I'm on the right track here or not but I do not know through what methods to evaluate the last integral. Any help is appreciated.
|
As user90369 doesn't have the time apparently I thought I repesent my own solution explicitly for those interested.
Starting with
\begin{align}
\int_{-\pi/2}^{\pi/2} \frac{x^n \left(\cos x\right)^{n-2}}{\left(\sin x\right)^n} \, {\rm d}x &\stackrel{y=2x}{=} 2^{-n+1} \, i^n \int_{-\pi}^{\pi} y^n \, e^{-iy} \, \frac{\left(1+e^{-iy}\right)^{n-2}}{\left(1-e^{-iy}\right)^n} \, {\rm d}y \\
&\stackrel{z=e^{iy}}{=} -i \, 2^{-n+1} \int_\gamma \left(\log z\right)^n \, \frac{(z+1)^{n-2}}{(z-1)^n} \, {\rm d}z
\end{align}
where $\gamma$ is the unit circle in positive direction. The integrand is holomorph off the negative real line and the contour can be deformed to only enclose the cut $[-1,0]$.
Then
\begin{align}
&=i \, 2^{-n+1} (-1)^n \int_0^1 \frac{(1-z)^{n-2}}{(1+z)^n} \, \left\{ \left(\log z + i\pi\right)^n - \left(\log z - i\pi\right)^n \right\} {\rm d}z \\
&=2^{-n+2} (-1)^{n+1} \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{q=0}^\infty \begin{pmatrix} n \\ k \end{pmatrix} \begin{pmatrix} n-2 \\ p \end{pmatrix} \begin{pmatrix} -n \\ q \end{pmatrix} (-1)^p \pi^{n-k} \sin\left(\frac{\pi}{2}(n-k)\right) \\
&\qquad \times \int_0^1 z^{p+q} \, \left( \log z \right)^k \, {\rm d}z
\end{align}
and using the two formulas
\begin{align}
\int_0^1 x^n \, \left(\log x\right)^m \, {\rm d}x &= \frac{(-1)^m \, m!}{(n+1)^{m+1}} \tag{1} \\
\begin{pmatrix} -n \\ q \end{pmatrix} &= (-1)^q \begin{pmatrix} n+q-1 \\ q \end{pmatrix} = (-1)^q \, \frac{(q+1)\cdots(q+n-1)}{(n-1)!} \\ &= \frac{(-1)^q}{(n-1)!} \sum_{m=0}^{n-1} \left[ {n-1 \atop m} \right] (q+1)^m \tag{2}
\end{align}
with the Stirling numbers of the first kind we arrive at
\begin{align}
=2^{-n+2} (-1)^{n+1} \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{m=0}^{n-1} \sum_{q=0}^\infty \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \frac{(q+1)^m (-1)^{k+p+q} \pi^{n-k} n! \sin\left(\frac{\pi}{2}(n-k)\right)}{(p+q+1)^{k+1} (n-1)! (n-k)!} \, .
\end{align}
Next we reverse the summation $k \rightarrow n-k$ and decompose another time the factor $(q+1)^{m}=(p+q+1-p)^{m}$ in the nominator
\begin{align}
&=-2^{-n+2}\,n \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{m=0}^{n-1} \sum_{l=0}^m \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \begin{pmatrix} m \\ l \end{pmatrix} (-p)^{m-l} \\
&\qquad \frac{(-\pi)^{k} \sin\left(\frac{\pi}{2}k\right)}{k!} \sum_{q=0}^\infty \frac{(-1)^{q+p}}{(p+q+1)^{n-k-l+1}} \\
&=-2^{-n+2} \, n \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{l=0}^{n-1} \sum_{m=l}^{n-1} \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \begin{pmatrix} m \\ l \end{pmatrix} (-p)^{m-l}\\
&\qquad \frac{(-\pi)^{k} \sin\left(\frac{\pi}{2}k\right)}{k!} \left\{ \eta\left(n-k-l+1\right) -\sum_{q=1}^p \frac{(-1)^{q+p}}{(p-q+1)^{n-k-l+1}} \right\} \tag{3}
\end{align}
where $\eta(s)$ is the Dirichlet Eta-Function.
It seems as if
\begin{align}
&\quad \sum_{p=0}^{n-2} \sum_{q=1}^{p} \sum_{l=0}^{n-1} \sum_{m=l}^{n-1} \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \begin{pmatrix} m \\ l \end{pmatrix} \frac{(-p)^{m-l} (-1)^{q+p}}{(p-q+1)^{n-k-l+1}} \\
&=\sum_{p=0}^{n-2} \sum_{q=1}^{p} \begin{pmatrix} n-2 \\ p \end{pmatrix} \frac{(-1)^{q+p}}{(p-q+1)^{n-k+1}} \sum_{m=0}^{n-1} \left[ {n-1 \atop m} \right](-q+1)^m \\
&=(-1)^{n-1}(n-1)! \sum_{p=0}^{n-2} \sum_{q=1}^{p} \begin{pmatrix} n-2 \\ p \end{pmatrix} \begin{pmatrix} q-1 \\ n-1 \end{pmatrix} \frac{(-1)^{q+p}}{(p-q+1)^{n-k+1}} \\
&= 0 \, ,
\end{align}
since $q\leq n-2$ and $\begin{pmatrix} n-3 \\ n-1 \end{pmatrix}=0$, so the second term in the curly bracket does not contribute.
Here is a Maple Code of the result (3) for verification:
restart;
n := 5;
eta := proc (s) options operator, arrow; limit((1-2^(1-S))*Zeta(S), S = s) end proc;
r1 := simplify(-2^(-n+2)*n*add(add(add(add(binomial(n-2, p)*Stirling1(n-1, m)*(-1)^(n-1-m)*binomial(m, l)*(-p)^(m-l)*(-Pi)^k*sin((1/2)*Pi*k)*eta(n-k-l+1)/factorial(k), m = l .. n-1), l = 0 .. n-1), p = 0 .. n-2), k = 0 .. n));
r2 := simplify(int(x^n*cos(x)^(n-2)/sin(x)^n, x = -(1/2)*Pi .. (1/2)*Pi));
evalf([r1, r2])
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ we get $a^2-a \gt b-b^2$
Since $a \geq b$ we can get $a^3-a^2 \gt b^2-b^3$ $\Rightarrow$ $a^3+b^3 \gt a^2+b^2$
If this solution is incorrect, please explain why and attach the correct solution. Thank you.
|
It is correct, but I would explain the "since $a\ge b$ then $a^3 - a^2 > b^3 - b^2$" step a bit more.
\begin{align*}a^2 - a > b^2-b &\iff a(a^2-a) > a(b^2-b) \quad \text{(since $a>0$)} \\ &\iff a(a^2-a)>a(b^2-b)\ge b(b^2-b) \quad \text{(since $b\le a$)} \\ &\iff a^3-a^2 > b^3 -b^2 \\ &\iff a^3 + b^3 > a^2 + b^2\end{align*}
|
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|
How can I show a decreasing derivative? I have a derivative
$f'(x) = -1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x}$
And we have to show that it's decreasing in all intervals on the form $[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}$ for n>= 1
Our hint is to use the derivative such as $\frac {6}{(12n + 7)\pi} <= \frac{6}{19 \pi} < \frac{1}{8}$
I have no idea what to do other than
$f'(x) = - 1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x} < \frac {6}{(12 n + 11) \pi } $
Please help. And thank you.
|
Hint: $$x \in \left[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}\right] \iff \frac1x \in \left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$$
We have $x,y \in \left[\frac {6}{(12 n + 11) \pi },\frac {6}{(12n + 7)\pi}\right]$ and $x<y$ iff $\frac1x>\frac1y$ in $\left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$. So this question can be rewritten as proving that $g(x) = 1+8\cos x+4\sin x$ is increasing in $\left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$. Since $\sin$ and $\cos$ are periodic with period $2\pi$, the "$2n$"'s in the previous interval are redundant. You may continue from here. By either calculating $g'(x)$ OR rewriting the two trigonometric terms as one single term with a phase change.
|
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|
How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)$$
I have tried to factor out the $(1+\frac{1}{x})$ use the distributive property, and failed multiple times. Every online calculator I tried gives me different results.
|
Hint
$$-\frac{6}{x^2}\left(1+\frac{1}{x}\right)+\frac{6}{x^3}\left(1+\frac{1}{x}\right)^2=\frac{6}{x^2}\left(1+\frac 1x\right)\left(\frac 1x+\frac 1{x^2}-1\right).$$
|
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|
If $a_n=\frac{(-1)^n}{\sqrt{1+n}}$ and $c_n=\sum_{k=0}^n a_{n-k}a_k$, does $\sum_{n=0}^\infty c_n$ converge?
Suppose for $n \in \mathbb{N} \cup \{0\}$, $a_n=\frac{(-1)^n}{\sqrt{1+n}}$. Define $c_n=\sum_{k=0}^na_{n-k}a_k$. Does $\sum_{n=1}^\infty c_n$ converge?
What I attempted:-
\begin{equation}
\begin{aligned}
c_n&=\sum_{k=0}^na_{n-k}a_k \\
&=a_na_0+a_{n-1}a_1+\dots+a_0a_n\\
&=\frac{(-1)^n}{\sqrt{1+n}}\times 1+\frac{(-1)^{n-1}}{\sqrt{1+n-1}}\times \frac{(-1)}{\sqrt{1+1}}+\frac{(-1)^{n-2}}{\sqrt{1+n-2}}\times \frac{(-1)^2}{\sqrt{1+2}}\dots+1 \times \frac{(-1)^n}{\sqrt{1+n}} \\
&=(-1)^n\left(\frac{1}{\sqrt{1+n}}+\frac{1}{\sqrt{2n}}+\frac{1}{\sqrt{3(n-1)}}+\dots+\frac{1}{\sqrt{1+n}}\right)\\
&=(-1)^nd_n
\end{aligned}
\end{equation}
$(d_n)_{n=1}^\infty $ is a decreasing sequence and $\lim_{n \to \infty}d_n=0$. Hence $\sum_{n=0}^{\infty}(-1)^nd_n=\sum_{n=0}^{\infty}c_n$ converges. As a result $\sum_{n=1}^{\infty}c_n$ also converges.
Am I correct?
|
Hint. Note that
$$d_n=\sum_{k=0}^n\frac{1}{\sqrt{(1+k)(n+1-k)}}\geq \sum_{k=0}^n\frac{2}{(1+k)+(n+1-k)}=\frac{2(n+1)}{n+2},$$
and therefore $c_n=(-1)^n d_n$ does not tend to zero as $n$ goes to infinity.
|
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|
Proving $a\sqrt{7}>\frac{1}{c}$, where $a$ is an integer and $c = \lceil\frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$
Let $a \in \Bbb{N}$, and $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Prove that $$a\sqrt{7}>\frac{1}{c}$$
So the very original problem sounds like this:
It is given that $a,b \in \Bbb{N}$, $\sqrt{7}-\frac{a}{b}>0$. Prove that $\sqrt{7}-\frac{a}{b}>\frac{1}{ab}$.
Out of the first inequality, I expressed $b>\frac{a}{\sqrt{7}}$. So I thought that the least possible value of $b$ is $\lceil\frac{a}{\sqrt{7}}\rceil$. Also, I changed the inequality that I have to prove to $\frac{a^2+1}{ab}<\sqrt{7}$. I decided to change $b$ with $\lceil\frac{a}{\sqrt{7}}\rceil$, so the value of $\frac{a^2+1}{ab}$ would be as big as possible. I got $\frac{a^2+1}{a\lceil\frac{a}{\sqrt{7}}\rceil}<\sqrt{7}$. Then I made a $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Therefore, what I had to prove was $$\frac{a^2+1}{a\left(\frac{a}{\sqrt{7}}+c\right)}<\sqrt{7}$$ which is equivalent to $$a^2+1<a^2+ac\sqrt{7}$$ which is equivalent to $$a\sqrt{7}>\frac{1}{c}$$ Somehow, this inequality is not correct. If you could point me a mistake I made, I'd be insanely grateful.
|
For the original question the critical thing to realize is that if $7b^2-a^2\gt 0$, it is at least $3$. The squares $\bmod 7$ are $1,2,4$ which gives this, as $a^2$ cannot be $5$ or $6\ \bmod 7$
First we take care of a side case. If $b\sqrt 7 -a \gt 1, b\sqrt 7-a \gt \frac 1a$.
Given $1 \gt b\sqrt 7-a\gt 0$ we have $$b\sqrt 7 -a \gt 0\\
7b^2-a^2 \gt 0\\7b^2-a^2 \ge 3\\b\sqrt 7-a \ge \frac 1{b\sqrt 7+a}\\
b\sqrt 7-a \gt \frac 3{2a+1}\\ b\sqrt 7 - a \gt \frac 1a\\\sqrt 7 - \frac ab \gt \frac 1{ab}$$
|
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|
Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution!
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$
So, make the substitution
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x = \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$
Am I making the correct substitutions here? Substituting $d\theta$ a quantity of $(\sqrt{1}\sec\theta)$ will cancel from the denominator. Somewhere along the line I need to use the identity $\sin(2\theta)=2\sin(\theta)\cos(\theta).$
|
Setting
$x=\frac{\sqrt{\theta^{2}+1}}{2}$
$dx=\frac{\theta}{2\sqrt{\theta^{2}+1}}$,
and the funcion to integrate is:
$\frac{4\theta^{2}}{(\theta^{2}+1)^{2}}$,
whose integral is:
$2. atan(\theta)-\frac{2\theta}{\theta^{2}+1}$.
Substituting the value of
$\theta=\sqrt{4x^{2}-1}$,
we get:
$I=2. atan(\sqrt{4x^{2}-1})-\frac{\sqrt{4x^{2}-1}}{2x^{2}}$.
|
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|
$s_n=s_{n-1}+(n-1)s_{n-2}$ prove $s_n>\sqrt{n!n}$ for $n\ge4$ Define sequece as follows: $s_1=1,~s_2=2, s_n=s_{n-1}+(n-1)s_{n-2}$. I want to prove that $s_n>\sqrt{n!n}$ for $n\ge4$. I'd tried to use traditional induction on $n$, but it involves both two terms that are in front of the current term. Should I use strong induction? Or even should I solve such sequence first? (I'm not good at transform the recurrence relation to explicit formula)
|
Let's try an induction :
You have $s_1 = 1$ and $s_2 = 2$, so $s_3 = 4$ and $s_4 = 10$ and $s_5 = 26$.
Therefore, you have $s_4 = 10 = \sqrt{100} > \sqrt{96} = \sqrt{4! \times 4}$. And $s_5 = 26 = \sqrt{676} > \sqrt{600} = \sqrt{5! \times 5}$.
Now, let's suppose your result is true for two successive ranks $n$ and $n+1$, that is you have $s_n > \sqrt{n! n}$ and $s_{n+1} > \sqrt{(n+1)!(n+1)}$. You have then
$$s_{n+2} = s_{n+1} + (n+1)s_n$$
so
$$s_{n+2}^2 = s_{n+1}^2 + (n+1)^2s_n^2 + 2 (n+1)s_{n+1}s_n$$
so you get
$$s_{n+2}^2 > (n+1)!(n+1) + (n+1)^2n! n + 2(n+1) \sqrt{(n+1)!(n+1)n!n}$$
$$=(n+1)!(n+1)^2 + 2(n+1)!(n+1)\sqrt{n}$$
$$=(n+1)!((n+1)^2 + 2(n+1)\sqrt{n})$$
Now $n \geq 4$, so $\sqrt{n} \geq 2$, so you get
$$s_{n+2}^2 > (n+1)!((n+1)^2 + 4(n+1)) = (n+1)!((n+2)^2 + 2n +1)$$ $$> (n+1)!(n+2)^2 = (n+2)!(n+2)$$
Therefore you have
$$s_{n+2} > \sqrt{(n+2)!(n+2)}$$
which is your equality at the rank $n+2$. This completes your proof.
|
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|
Deriving polar graphs There is just some confusion I have with polar graphs, for instance there are some well known general forms of the polar graphs such as the circle, Limaçon, rose, Lemniscate.
The general equation for a circle with radius $\frac{a}{2}$ is given by either:
$$r = a \cos\theta \space \space \text{or}\space \space r = a \sin \theta$$
Then for Cardioids or Limacons are:
$$r = a \pm b\cos\theta \space \space \text{or}\space \space r = a \pm b \sin \theta$$
For Roses, we have:
$$r = a \cos n\theta \space \space \text{or}\space \space r = a\sin n\theta$$
Finally for Lemniscates we have:
$$r^2 = a^2 \cos n\theta \space \space \text{or}\space \space r^2 = a^2 \sin n\theta$$
So on and so forth. However what I am truly confused about is how does on come to derive that these shapes are indeed these algebraically? For instance I think I can derive the circle equation using $x$, $y$ for instance:
$$\bigg(x-\frac{a}{2}\bigg)^2+y^2 = \bigg(\frac{a}{2}\bigg)^2$$
$$x^2+y^2 = ax$$
$$r^2 = ar\cos\theta$$
$$r = a\cos\theta$$
Hence the above polar equation has the circle equation with center $(\frac{a}{2},0)$
So My QUESTION is how can we derive the other $3$ special cases for polar graphs starting for our $x$ and $y$ like we have for the circle case?
|
Degree 1 polynomials describe lines
Degree 2 polynomials describe lines the conic sections, parabola, circle, ellipse, hyperbola.
Degree 3 polynomials describe "elliptic curves"
You will need a $4^{th}$ degree$^+$ polynomial.
A cartiod
$r = a(1+\cos \theta)\\
r^2 = a(r+r\cos \theta)\\
x = r\cos \theta, r = \sqrt {x^2 + y^2}\\
x^2 + y^2 - ax = a\sqrt {x^2 + y^2}\\
(x^2 + y^2 - ax)^2 = a^2(x^2 + y^2)$
The rest can be found using similar approaches.
4-petaled rose.
$r = \cos 2\theta\\
r = \cos^2\theta - \sin^2\theta\\
r^3 = (r\cos\theta)^2 - (r\sin\theta)^2\\
(x^2 + y^2)^\frac 32 = x^2 - y^2\\
(x^2 + y^2)^3 = (x^2 - y^2)^2$
3 petaled rose
$r = \cos 3\theta\\
r = \cos^3 \theta - 3\cos\theta\sin^2\theta\\
r^4 = r^3\cos^3 \theta - 3r\cos\theta r^2\sin^2\theta\\
(x^2+y^2)^2 = x^3 - 3xy^2 $
|
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|
Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}}
\end{align}
|
Set
$$J=\int{\frac{dx}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}}$$
Use the following substitution:
$$u=\frac{2x-\left( a+b \right)}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}$$
once get:
$$du=\left( -\frac{{{\left( 2x-\left( a+b \right) \right)}^{2}}}{2{{\left( \sqrt{{{x}^{2}}-\left( a+b \right)x+ab} \right)}^{3}}}+\frac{2}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}} \right)dx$$
now use
$$\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}=\frac{2x-\left( a+b \right)}{u}$$
and after some algebraic simplifications you should have:
$$dx=\frac{2\left( 2x-\left( a+b \right) \right)}{4u-{{u}^{3}}}du$$
Finally,
$$J=2\int{\frac{du}{4-{{u}^{2}}}}$$
|
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|
convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x>0$ While checking the convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x>0$, we used ratio test to say that for $0< x < \frac1e$ $\sum u_n$ is convergent and for $\frac1e < x<\infty$ $\sum u_n$ is divergent.
We use Logarithimic Test for the case $x = \frac1e$, where we came across computing the limit $$\lim_{n \to \infty} n+n^2 \log \frac{n}{n+1}$$ but I am stuck in finding the limit.
|
$$
n+n^2\log\frac{n}{n+1} = n-n^2\log\frac{n+1}{n} = n-n^2\log\left(1+\frac{1}{n}\right)
$$
So do some asymptotics. As $n \to +\infty$, we have
\begin{align}
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)
\\
n^2\log\left(1+\frac{1}{n}\right) &= n-\frac{1}{2}+o(1)
\\
n-n^2\log\left(1+\frac{1}{n}\right) &= \frac{1}{2}+o(1)
\end{align}
|
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|
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
|
Yes, we can. If $f(n)=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$, then $f(n+1)-f(n)=n^2$.
|
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|
Solving the system $a^2-c^2=x^2-z^2$, $ab=xy$, $ac=xz$, $bc=yz$ I've stumbled upon these equations, and am struggling to find a manual way to solve this in $\Re$:
$$\begin{align}
a^2-c^2&=x^2-z^2 \\
ab&=xy \\
ac&=xz \\
bc&=yz\end{align}$$
I've used Wolfram Alpha to compute this and I found that this is only possible if:
$$a=x, \quad y=b, \quad z=c$$
Is there an easy manual way to compute this?
I've tried moving around variables with no luck.
|
This is an homogeneous system so making $y = \lambda x$ and $z = \mu x$ we have
$$
a^2-c^2=x^2(1-\mu^2)\\
ab = x^2\lambda\\
ac = x^2\mu\\
bc = x^2\lambda\mu
$$
and making now $A = \frac ax,B = \frac bx, C = \frac cx$
$$
A^2-C^2=1-\mu^2\\
AB = \lambda\\
AC = \mu\\
BC = \lambda\mu
$$
so we conclude easily
$$
A^2B^2C^2=\lambda^2\mu^2\\
A^2-C^2= 1-A^2C^2\Rightarrow A^2=1\Rightarrow a^2=x^2\Rightarrow b^2= y^2\Rightarrow c^2= z^2
$$
|
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"url": "https://math.stackexchange.com/questions/2946743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Convergence for series failed using "Ratio Test" $$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$
Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$
which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$
the result for latter is q=1. seemingly this is unclear if divergent or convergent.
The answer for this is convergent....
|
$$a_k=\prod_{k=1}^{n}\frac{2k-1}{3k-2}=\left(\frac{2}{3}\right)^n\cdot\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\cdot B\left(n+\tfrac{1}{2},\tfrac{1}{6}\right)\tag{1}$$
implies
$$\begin{eqnarray*} \sum_{k\geq 1}a_k &=& \frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n (1-x)^{-5/6} x^{n-1/2}\,dx \\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{\pi/2}\frac{4(1-\cos^2\theta)}{\cos^{2/3}\theta(1+2\cos^2\theta)}\,d\theta\\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}12\sqrt{1-z^6}\,\frac{dz}{1+2z^6}\tag{2}\end{eqnarray*}$$
hence trivially
$$ \sum_{k\geq 1}a_k = \tfrac{1}{2}\cdot\phantom{}_2 F_1\left(1,\tfrac{3}{2};\tfrac{5}{3};\tfrac{2}{3}\right) \leq \frac{3}{\pi}2^{4/3}\,\Gamma\left(\tfrac{2}{3}\right)^2.\tag{3}$$
In general, $\phantom{}_{p+1}F_p(\ldots; z)$ is convergent for any $|z|<1$.
|
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|
Find the order of the matrices $A$ and Matrice $B$?
Find the order of the matrices $A$ and $B$ in the Group $GL_2(\mathbb{F}_7)$: $$A=\begin{bmatrix} 1 &1 \\ 0& 1 \end{bmatrix}\;\text{and}\; B=\begin{bmatrix} 2 &0 \\ 0 & 1 \end{bmatrix}$$
For $A$ I take $
\pmatrix{1&x\\0&1}\pmatrix{1&y\\0&1}=\pmatrix{1&x+y\\0&1}\,,
$
and
for $ B$ we have
$
\pmatrix{a&0\\0&1}\pmatrix{b&0\\0&1}=\pmatrix{ab&0\\0&1}\,.
$
After that im not able to proceed further ,
any hints/solution
thanks in advance
|
For your matrix $A$, we have $$\begin{pmatrix}1 &1 \\0 &1 \end{pmatrix}^n=\begin{pmatrix}1 &n \\0 &1 \end{pmatrix}$$
so, we have to find a least $n$ for which $$\begin{pmatrix}1 &n \\0 &1 \end{pmatrix}=\begin{pmatrix}1 &0 \\0 &1 \end{pmatrix}$$ Now, clearly for $n=1,2,\cdots,6$ it does't happen but in the seventh power it becomes identity.
So $$\text{order of}\;\begin{pmatrix}1 &1 \\0 &1 \end{pmatrix}=7$$
In general order of $\begin{pmatrix}1 &1 \\0 &1 \end{pmatrix}$ in $GL(2,\Bbb{Z}_p)$ is $p$ where $p$ is prime
Similarly work for the second matrix
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rational fraction expression for triangular powers of 2 Why does the following pattern hold?
$$1=1=2^0$$
$$\frac{2(2^2-1)}{3}=2=2^1$$
$$\frac{3^2(3^2-1)(3^2-2^2)}{3^2 \times 5}=8=2^3$$
$$\frac{4^2(4^2-1)^2(4^2-2^2)(4^2-3^2)}{3^3 \times 5^2\times 7}=64=2^6$$
$$\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9}=1024=2^{10}$$
$$ n^{\lceil n/2\rceil}\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}=2^{n(n-1)/2}.$$
The motivation for the question comes from the following two questions:
Coefficients of binomial continued fractions
Determinants of products of binary matrices and binomial coefficients
|
a partial answer:
$$
\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9}
$$
may be re-written as
$$
\frac{\frac{9!}{0!}\frac{7!}{2!}\frac{5!}{4!}}{\frac{9!}{4!2^4}\frac{7!}{3!2^3}\frac{5!}{2!2^2}\frac{3!}{1!2^1}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $A+2B$ if $3\cos^2A+2\cos^2B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$
$A$ and $B$ are positive acute angles satisfying $3\cos^2A+2\cos^2B=4$ and $\dfrac{3\sin A}{\sin B}=\dfrac{2\cos B}{\cos A}$, then find the value of $A+2B$ ?
My Attempt
$\cos2B=2\cos^2B-1=3-3\cos^2A$ and $\sin2B=2\sin B\cos B=3\sin A\cos A$
\begin{align}
\sin(A+2B)&=\sin A\cos2B+\cos A\sin2B\\
&=\sin A.[3-3\cos^2A]+\cos A.[3\sin A\cos A]=3\sin A\\
\sin(A+2B)=3\sin A\implies \color{red}{?}
\end{align}
How do I prove that $A+2B=\dfrac{\pi}{2}$ from $\sin(A+2B)=3\sin A$ ?
What I know
\begin{align}
\cos(A+2B)&=\cos A\cos2B-\sin A\sin2B\\
&=\cos A.[3-3\cos^2A]+\sin A.[3\sin A\cos A]\\
&=3\cos A-3\cos^3A+3\sin^2A\cos A\\
&=3\cos A-3\cos A.[\cos^2A+\sin^2A]=0\implies \boxed{A+2B=\dfrac{\pi}{2}}
\end{align}
|
Hint:
Using double angle formula,
$$2\sin2B=6\sin A\cos A=3\sin2A$$
$$2\cos2B=6(1-\cos^2A)=6(2\sin^2A)=3(1-\cos2A)$$
Squaring and adding we get $\dfrac49=2-2\cos2A=2(2\sin^2A)$
$\implies\sin A=\dfrac13,\cos A=+\dfrac{2\sqrt2}3$ as $0<A<\dfrac\pi2$
$$\sin2B=3\sin A\cos A=\dfrac{2\sqrt2}3,$$
$$\cos2B=3\sin^2A=\dfrac13$$
Hope you can take it from here using $\cos(2B+A)$ or $\sin(2B+A)$!
|
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|
Is $\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ convergent?
$\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ : convergent?
My attempt
$$
(\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} = (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{n}}) + \cdots + (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n}})
$$
But I stuck here.
|
The limit is not convergent. You can use the integral test to see this.
\begin{align}
\sum_{m=1} ^n \frac{1}{\sqrt{m}} & \geq \int_{1}^{n} \frac{1}{\sqrt{x}} \ dx \\ & = 2\sqrt{n} - 2 .
\end{align}
So we can conclude that
$$\sum_{m=1} ^n \frac{1}{\sqrt{m}} - \sqrt{n} \ \geq \sqrt{n}- 2.$$
Letting $n \to \infty$ givs us what we want.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Minimum value of the given expression If I have $a>0$ and $ b >0 $ and $a+b=1$ then how can I find the minimum value of $(a+1/a)^2 + (b+1/b)^2$. I just tried to do it by expanding the expression just because I knew the graph of $x^2+1/x^2 $ and for that I had the minimum value as 2 but since there are two variables $a,b$ in the problem with the condition that both are less than 1 I'm not able to proceed from here. Any help will be appreciated .Thanks in advance.
|
This problem is as old as a volkswagen beetle classic ( that I have ,haha ). My answer is problably one of the $20$ different "proofs" that you can find here or google. I try to be a little different if possible...
Using Cauchy-Schwarz inequality twice: $(1^2+1^2)\left(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\right)\ge \left(1\cdot\left(a+\dfrac{1}{a}\right)+1\cdot\left(b+\dfrac{1}{b}\right)\right)^2=\left(a+b+\dfrac{1^2}{a}+\dfrac{1^2}{b}\right)^2\ge \left(a+b+\dfrac{(1+1)^2}{a+b}\right)^2=\left(1+\dfrac{2^2}{1}\right)^2= 5^2=25\implies \left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\ge \dfrac{25}{2}$, with $ =$ occurs when $a = b = \dfrac{1}{2}$ .
|
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|
If $\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2} = \frac{a} {(x + 2)^2}$, then what is $a$? $$\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}$$
The expression above is equivalent to $$\frac{a} {(x + 2)^2}$$
where $a$ is a positive constant and $x \neq -2$.
What's the value of $a$?
|
There is no need to do any algebra.
Consider
$$
\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}
=
\frac{a} {(x + 2)^2}
$$
Since this equality holds for all values of $x\ne -2$,
set $x=0$ and get
$$
\frac{6} {2^2}- \frac{2} {2}
=
\frac{a} {2^2}
$$
which gives $a=2$.
|
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|
How to obtain the lower bound of $\dfrac{n}{\sqrt[n]{n!}}$ by Taylor's series? We want to prove $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
I have some solutions for this, but I want to find another method applying the squeeze theorem. Thus, a natrual thought is to find the upper bound and the lower bound of $a_n$.
Notice that
$$e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
If we substitute $x$ for $n$, then
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.\tag1$$
Thus, we obtain
$$e>\frac{n}{\sqrt[n]{n!}},$$
which shows $e$ is a upper bound of $a_n$.
But how to obtain the lower bound by $(1)$? Say it again. I have other methods to deal with it. I just wonder whether there is some method depending on $(1)$ only or not.
|
\begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots\\
&<(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\
&< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\
&=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\
&=(2n+1)\cdot \frac{n^n}{n!}.
\end{align*}
Thus
$$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.$$
|
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|
How can I find $\gcd(n^a-1,m^a-1)$? From Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ , we have $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$
for every positive integers $a,n,m$.
I reversed $a$ with $n,m$, and I had this question:
Find $\gcd(n^a-1,m^a-1)$ for every positive integers $a,n,m$
My attempt is to find the greatest common divisor of $n^p-1$ and $m^p-1$ for every prime $p$ such that $p|a$, but I couldn't get any further.
How can I find the general formula of $\gcd(n^a-1,m^a-1)$ ?
(sorry for my grammar mistake, English is my second language)
|
Welcome. If n and m are as following forms:
$n=k b+1$ ⇒ $n≡1\mod b$⇒ $n^a≡1\mod b$
$m=t b+1$ ⇒ $m≡1 \mod b$⇒$m^a≡1\mod b$
Then the common divisor will be:
$gcd(n^a-1, m^a-1)=b$
In other words if $n-1$ and $m-1$ have a common divisor like b then $n^a-1$ and $m^a-1$ have a common divisor like b. This can also seen in following relations:
$n^a-1=(n-1)(1+n+n^2+ . . .+n^{a-1})$
$m^a-1=(m-1)(1+m+m^2+ . . .+m^{a-1})$
Finally the possible forms may be as follows:
$gcd(n^a-1, m^a-1)= gcd(n-1, m-1)$
Example: $n=4, m=7$ ⇒ $gcd(4^a-1, 7^a-1)=3$
$gcd(n^a-1, m^a-1)= gcd(n-1, 1+m+m^2+. . .+m^{a-1})$
Example: $n=32$, $m=2$, $a=5$ ⇒ gcd(32^5-1, 2^5-1)= 31
$gcd(n^a-1, m^a-1)= gcd(m-1, 1+n+n^2+. . . + n^{a-1})$
$gcd(n^a-1, m^a-1)= gcd[(1+m+m^2+. . .+m^{a-1}),(1+n+n^2+. . . + n^{a-1})]$
Example: $n=2$, $m=4$, $a=3$ ⇒ $gcd(2^3-1, 4^3-1)=7$
|
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|
Integration-by-part Here is the given question
$$\int{\frac{x^3}{\sqrt{1+x^2}}dx}$$
I solved using integration by part as follow:
$$
\int\frac{x}{\sqrt{1+x^2}}x^2\,dx = x^2\int\frac{x}{\sqrt{1+x^2}}\,dx - \int\biggl(\frac{dx^2}{dx}\int\frac{x}{\sqrt{1+x^2}}dx\biggr)dx\tag{i}\label{i}
$$
Solving for $\int \frac{x}{\sqrt{1+x^2}}$ and taking $1+x^2 = t \rightarrow 2xdx = dt \rightarrow xdx = \frac{dt}{2}$.
\begin{align}
&\int \frac{xdx}{\sqrt{1+x^2}} = \int\frac{dt}{2\sqrt{t}}\\
&\int\frac{dt}{2\sqrt{t}} = \frac{1}{2}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \rightarrow \sqrt{t} = \sqrt{1+x^2}
\end{align}
Pluggin in equation \eqref{i}
$$\int{\frac{x}{\sqrt{1+x^2}}x^2dx} = x^2\sqrt{1+x^2} - 2\int{x\sqrt{1+x^2}dx}\tag{ii}\label{ii}$$
Now solving for $\int{x\sqrt{1+x^2}dx}$.
Assuming $1+x^2 = t \rightarrow xdx = \frac{dt}{2}$
$$\frac{1}{2} \int\sqrt{t}dt \rightarrow \frac{\sqrt[3]{t}}{3} \text{ or }
\int{x\sqrt{1+x^2}dx} = \frac{\sqrt[3]{1+x^2}}{3}$$
Pluggin in equation
$$\int\frac{x}{\sqrt{1+x^2}}x^2dx = x^2\sqrt{1+x^2} - 2 \frac{\sqrt[3]{1+x^2}}{3} + C$$
But given answer is:
$$\frac{1}{3}(1+x^2)^{3/2}-(1+x^2)^{1/2}+C$$
What is wrong with my solution?
|
It should be
$$\int{\frac{x}{\sqrt{1+x^2}}*x^2dx} = x^2\sqrt{1+x^2} - 2\cdot \frac{(1+x^2)^{3/2}}{3} + C$$
Use
$$x^2\sqrt{1+x^2}=(x^2+1-1)\sqrt{1+x^2}=(1+x^2)^{3/2}-(1+x^2)^{1/2}$$ to match with the given answer
|
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|
Die is rolled until 1 appears. What is the probability of rolling it odd number of times? Problem: Die is rolled until 1 appears. What is the probability of rolling it odd number of times?
So, so far I have this:
$\frac{1}{6}$ - this is a probability of rolling "1" on first try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}$ - three tries (two times something else than "1" and "1" on the 3rd try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot
\frac{1}{6}$ - five tries
and so on...
By the looks of it, it seems I could come up with following formula
$$ p = \left( \frac{5}{6}\right)^{n-1} \cdot \frac{1}{6}$$
where p would be a probability for nth try.
However, I cannot think of any way to get probability of all odd number of tries and not sure if I am even on right track here.
|
Yet another wording for the same argument. See if this thinking/wording is convincing.
$P_m = $ prob of the first $1$ occurring on the $m$-th roll $= (\frac 56)^{m-1}\cdot \frac 16$.
Rolling the first one on the $m$th roll and rolling the first one on the $k$th roll are mutually exclusive events. So
$P_{m\lor k} =$ the probability of rolling the first $1$ occurring on the $m$th roll OR on the $k$th roll $= P_m + P_k$.
So the probability of rolling the first $1$ occurring on one of the odd rolls is:
$P_{odd} = \sum\limits_{m\text{ is odd}} P_m = $
$\sum\limits_{k = 0}^{\infty} P_{2k + 1} = $
$\sum\limits_{k = 0}^{\infty} (\frac 56)^{2k}\cdot \frac 16 =$
$\frac 16(\sum\limits_{k=1}^{\infty}[(\frac 56)^2]^k) = $
$\frac 16(\sum\limits_{k=1}^{\infty}(\frac {25}{36})^k) = $
(.... and as $|\frac {25}{36}|<1$ and as for all $|a|< 1$ we know $1 + a + a^2 + a^3 + .... = \frac 1{1-a}$ we have.....)
$=\frac 16( \frac 1{1- \frac {25}{36}})=\frac 16(\frac 1{\frac {11}{36}})=\frac 16\frac {36}{11}= \frac 6{11}$
|
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|
Bezout's Identity: Finding A Pair of Integers
There is only one integer $x$, between 100 and 200 such that the integer pair $(x, y)$ satisfies the equation $42x + 55y = 1$. What's the value of $x$ in this integer pair?
We know that
$$\begin{align} x &= x_0 + 5t \\ y &= y_0 - 4t \end{align}$$
But we need to know what $x_0, y_0$ are. By applying the GCD algorithm we can get the answer to be $x_0 = 17$ and $y_0 = 13$. So we need to find $100 \leq 17 + 5t \leq 200$. But treating this parametrically yields too many solutions. How do I discover the one solution?
|
First, you find a solution to $42x + 55y = 1$
\begin{array}{c|ccc|l|l}
& 55 & 1 & 0 & &55 = 55(1)+42(0) \\
-1 & 42 & 0 & 1 & (55,1,0)+(-1)(42, 0, 1)=(13,1,-1) & 42 = 55(0)+42(1) \\
\hline
-3 & 13 & 1 & -1 & (42, 0, 1) + (-3)(13,1,-1)=(3,-3,4)& 13 = 55(1)+42(-1) \\
-4 & 3 & -3 & 4 & (13,1,-1)+(-4)(3,-3,4)=(1, 13, -17)& 3=55(-3)+42(4) \\
& 1 & 13 & -17 & & 1=55(13)+42(-17)
\end{array}
So $42(-17)+55(13)=1$. We get $(x_0,y_0)=(-17, 13)$.
Next you characterize the general solutions.
$$(x,y)=(-17+55t, 13-42t)$$
Next, you search for the value of t that will put $x$ between $100$ and $200$.
\begin{equation}
100 < -17 + 55t < 200 \\
117 < 55t < 217 \\
\dfrac{117}{55} < t < \dfrac{217}{55} \\
2 \frac{7}{55} < t < 3 \frac{52}{55} \\
t = 3 \\
\end{equation}
Finally, you compute $x$, $x = -17+55(3) = 148$
|
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|
Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I inserted some $n$'s to get an idea on how the sequence is going to look like.
I got:
\begin{align}
a_0&=3\\
a_1&=1\\
a_2&=\frac{7}{9}\approx 0.\overline{7}\\
a_3&=\frac{3}{4}=0.75
\end{align}
Assumption: The sequence is monotonic for $\forall n\in \mathbb{N}$
Therefore, I show that
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I am having problems when trying to prove the inequalities above:
\begin{align}
& a_n \geq a_{n+1}\Longleftrightarrow \left|\frac{a_{n+1}}{a_n}\right |\leq 1\\
& = \left|\dfrac{\dfrac{(n+1)^2+3}{(n+2)^2}}{\dfrac{n^2+3}{(n+1)^2}}\right|\\
& = \frac{4 + 10 n + 9 n^2 + 4 n^3 + n^4}{12 + 12 n + 7 n^2 + 4 n^3 + n^4}\\
& = \cdots \text{ not sure what steps I could do now}
\end{align}
Boundedness:
The upper bound with $a_n<s_o;\; s_o \in \mathbb{N}$ is obviously the first number of $\mathbb{N}$:
\begin{align}
a_0=s_o&=\frac{0^2+3}{(0+1)^2}\\
&=3
\end{align}
The lower bound $a_n>s_u;\; s_u \in \mathbb{N}$
$s_u$ should be $1$, because ${n^2+3}$ will expand similar to ${n^2+2n+1}$ when approaching infinity. I don't know how to prove that formally.
Convergence
Assumption (s.a) $\lim_{ n \to \infty} a_n =1$
Let $\varepsilon$ contain some value, so that $\forall \varepsilon > 0\, \exists N\in\mathbb{N}\, \forall n\ge N: |a_n-a| < \varepsilon$:
\begin{align}
\mid a_n -a\mid&=\left|\frac{n^2+3}{(n+1)^2}-1\right|\\
&= \left|\frac{n^2+3}{(n+1)^2}-\left(\frac{n+1}{n+1}\right)^2\right|\\
&= \left|\frac{n^2+3-(n+1)^2}{(n+1)^2}\right|\\
&= \left|\frac{n^2+3-(n^2+2n+1)}{(n+1)^2}\right|\\
&= \left|\frac{2-2n}{(n+1)^2}\right|\\
&= \cdots \text{(how to go on?)}
\end{align}
|
In order to analyse the sequence, $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ We can look at the function, $$f(x) =\frac {x^2+3}{(x+1)^2}$$
$$f'(x) = \frac { x^2-2x-3}{(x+1)^4} >0 \text { for x>3 }$$ Therefore the sequence is increasing for $n>3$
Note that $$\lim _{x\to \infty } f(x) =1$$ Thus $$\lim _{n\to \infty } a_n=1$$
Since every convergent sequence is bounded, so is our sequence $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ is bounded.
For example we can see $|a_n|<4$ for all positive integers.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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|
If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_1=F_2=1$, we suppose that $n \geq 3$.
$$\begin{align}
F_{n+1}&=F_n+F_{n-1} \\
F_{n+1}^2&=(F_n+F_{n-1})^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2
\end{align}$$
$$\begin{align}
F_{n-1} &=F_{n-2}+F_{n-3} \\
F_{n-1}^2&=(F_{n-2}+F_{n-3})^2=F_{n-2}^2+2F_{n-2}F_{n-3}+F_{n-3}^2
\end{align}$$
so that
$$
F_{n+1}^2-F_{n-1}^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2-F_{n-2}^2-2 F_{n-2} F_{n-3}-F_{n-3}^2
$$
How can we deduce that the latter is divisible by $4$?
Or do we show it somehow else, for example by induction?
|
We have more general answer:
If $k\mid a-b$ then $k^2\mid a^k-b^k$
Proof: Write $a-b=km$ for some integer $m$. Then we have
$$a^k-b^k=(a-b)(\underbrace{a^{k-1}+a^{k-2}b+a^{k-3}b^2+...+b^{k-1}}_E)$$
\begin{eqnarray}E&\equiv_k& b^{k-1}+b^{k-2}b+b^{k-3}b^2+...+b^{k-1}\\
&\equiv_k&k\cdot b^{k-1} \\
&\equiv_k&0 \\
\end{eqnarray}
so $E = k\cdot p$ for some interger $p$, so we have $k^2\mid a^k-b^k$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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|
Diagonalizing a matrix with complex numbers hope everyone is doing well.
I'm a bit stuck on the following problem:
$T: C^2 \rightarrow C^2$, where $T$ is defined by $T(x,y)=(2x + (3-3i)y, (3+3i)x + 5y$
So I perform the usual calculations to diagonalize the matrix, right up until I find that the roots of the characteristic polynomial are $\lambda_1 = -1, \lambda_2 = 8$
From here I am having a hard time performing the row operations to get my eigenvectors. I realize that the calculations are fairly trivial but they are confusing me.
If someone could help me with, for example, the matrix associated to $\lambda_1$ I would be grateful:
$$
\begin{matrix}
3 & 3+3i \\
3-3i & 6 \\
\end{matrix}
$$
|
We may write
$T(x,y)=(2x + (3-3i)y, (3+3i)x + 5y \tag 1$
as
$T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix}, \tag 2$
and then he characterisic polynomial of $T$ is
$\det (T - \lambda I) = \det \left ( \begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} - \lambda I \right ) = \det \left ( \begin{bmatrix} 2 - \lambda & 3 - 3i \\ 3 + 3i & 5 - \lambda \end{bmatrix} \right) = \lambda^2 - 7 \lambda - 8; \tag 3$
it is easy to see that the roots of
$\lambda^2 - 7 \lambda - 8 = 0 \tag 4$
are
$\lambda = -1, \; 8; \tag 5$
since I'm not too swift at row operations myself, I'll solve for the eigenvectors directly from
$\begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}; \tag 6$
writing out the first row of this equation yields
$2x + (3 - 3i)y = \lambda x, \tag 7$
or
$(2 - \lambda)x + (3 - 3i) y = 0, \tag 8$
or
$y = -\dfrac{(2 - \lambda)x}{3 - 3i} = \dfrac{(\lambda - 2)x}{3 - 3i}; \tag 9$
since eigenvectors are are scalable, we may take $x = 1$; then the two eigenvectors are characteized by the two $y$-values
$y = \dfrac{(-1 - 2)x}{3 - 3i} = \dfrac{-3}{3 - 3i} = \dfrac{-1}{1 - i} = \dfrac{-1 - i}{2}, \tag{10}$
$y = \dfrac{(8 - 2)x}{3 - 3i} = \dfrac{6}{3 - 3i} = \dfrac{2}{1 - i} = \dfrac{2 + 2i}{2} = 1 + i; \tag{11}$
we may thus take the two eigenvectors to be
$v_{-1} = \begin{pmatrix} 1 \\ \dfrac{-1 - i}{2} \end{pmatrix}, \; v_8 = \begin{pmatrix} 1 \\ 1 + i \end{pmatrix}; \tag{12}$
we note that $v_{-1}$ and $v_8$ are in fact orthogonal with respect to the Hermitian inner product on $\Bbb C^2$ defined by
$\langle w, z \rangle = \bar w_1 z_1 + \bar w_2 z_2, \tag{13}$
where $w, z \in \Bbb C^2$:
$w = \begin{pmatrix} w_1 \\ w_2 \end{pmatrix}, \; z = \begin{pmatrix} z_1 \\ z_2 \end{pmatrix}; \tag{14}$
indeed,
$\langle v_{-1}, v_8 \rangle = 1 \cdot 1 + \dfrac{-1 + i}{2} \cdot (1 + i) = 1 - \dfrac{2}{2} = 0; \tag{15}$
the orthogonality of $v_{-1}$, $v_8$ should not surprise us since they belong to distinct eigenvalues of the matrix $T$, which is Hermitian: $T^\dagger = (T^\ast)^T = T$. $v_{-1}$ and $v_8$ form the columns of $V$, a diagonalizing matrix for $T$:
$V = \begin{bmatrix} v_{-1} & v_8 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ \dfrac{-1 - i}{2} & 1 + i \end{bmatrix}, \tag{16}$
the determinant of which is
$\det(V) = 1 + i + \dfrac{1 + i}{2} = \dfrac{3}{2}(1 + i); \tag{17}$
the inverse of $V$ is thus
$V^{-1} = \dfrac{2}{3(1 + i)}\begin{bmatrix} 1 + i & - 1 \\ \dfrac{1 + i}{2} & 1 \end{bmatrix} = \begin{bmatrix} \dfrac{2}{3} & -\dfrac{2}{3(1 + i)} \\ \dfrac{1}{3} & \dfrac{2}{3(1 + i)} \end{bmatrix}; \tag{18}$
finally, since
$TV = T\begin{bmatrix} v_{-1} & v_8 \end{bmatrix} = \begin{bmatrix} Tv_{-1} & Tv_8 \end{bmatrix} = \begin{bmatrix} -v_{-1} & 8v_8 \end{bmatrix}, \tag{19}$
we may write
$V^{-1}TV = V^{-1} \begin{bmatrix} -v_{-1} & 8v_8 \end{bmatrix} = \begin{bmatrix} -V^{-1} v_{-1} & 8V^{-1} v_8 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}. \tag{20}$
|
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"url": "https://math.stackexchange.com/questions/2974917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational I tried with both squaring and cubing the statement, it got messy, here's my latest attempt:
Assume for the sake of contradiction: $\sqrt[3]{5} + \sqrt{2}$ is rational
$\sqrt[3]{5} + \sqrt{2}$ = $\frac{a}{b}$ $a,b$ are odd integers $> 0$ and $ b\neq 0$
${(\sqrt[3]{5} + \sqrt{2})}^3$ = $\frac{a^3}{b^3}$
by multiplying by $b^3$:
${(\sqrt[3]{5} + \sqrt{2})}^3 \times b^3 $ = ${a^3}$
so: $a^3$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}^3$ which means $a$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}$
doing the same thing with $b$ i found :
$\frac{a^3}{{(\sqrt[3]{5} + \sqrt{2})}^3} $ = ${b^3}$
so: $b^3$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}^3$ which means $b$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}$ (wrong)
${(\sqrt[3]{5} + \sqrt{2})}$ is a common divisor for both $a$ & $b$ which is a contradiction, thus $\sqrt[3]{5} + \sqrt{2}$ is irrational. (wrong)
|
Assume that $$ \sqrt[3]{5} + \sqrt{2}=r$$ where r is a rational number.
We have $$ \sqrt[3]{5} =r-\sqrt{2}$$
Raise to the third power to get $$5=r^3-3r^2 \sqrt 2 +6r - 2\sqrt 2 $$
Solving for $\sqrt 2$ we get $$ \sqrt 2 = \frac {5-r^3-6r}{-3r^2-2}$$
The $RHS$ is a rational number which is impossible.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2975313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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|
Proof that $\sin^{2n + 1}(x) = \sum_{i = 0}^{n} a_{i} \sin\left(\left(2i + 1\right)x\right)$ In playing around on wolframalpha, I noticed that $\sin^{2n + 1}(x)$ where $n \in \mathbb{N} \cup \{0\}$ takes the form:
$$\sin^{2n + 1}(x) = \sum_{i = 0}^{n} a_{i} \sin\left(\left(2i + 1\right)x\right)$$
Where $a_{i} \in \mathbb{Q}$
Does anyone have any ideas on how one would go about (dis)proving this?
Any tips would be greatly appreciated.
|
Just a thought that came to mine on how this could be proven
Employ proof my induction
Step (1) : Prove true for $n = 0 $
$$\sin^{1}(x) = 1\cdot\sin(1\cdot x) = \sum_{i = 0}^{0} a_{i}^{(0)}\sin\left(\left(2i + 1\right)x\right)$$
Hence true.
Step (2) : Inductive Step - Assume true for $n = k$:
$$\sin^{2k + 1}(x) = \sum_{i = 1}^{k + 1}a_{i}^{(k)}\sin\left(\left(2i + 1\right)x\right)$$
Step (3) : Prove true for $n = k + 1$. Hence show,
$$\sin^{2(k + 1) + 1}(x) = \sum_{i = 1}^{k + 1}a_{i}^{(k + 1)}\sin\left(\left(2i + 1\right)x\right)$$
Now,
\begin{align}
\sin^{2k + 3}(x) &= \sin^{2}\left(x\right)\sin^{2k + 1}(x) \\
&= \left[\frac{1 - \cos(2x)}{2} \right]\sum_{i = 0}^{k}a_{i}^{(k)}\sin\left(\left(2i + 1\right)x\right) \\
&= \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{2}\sin\left(\left(2i + 1\right)x\right) - \sum_{i = 0}^{k }\frac{a_{i}^{(k)}}{2}\cos(2x)\sin\left(\left(2i + 1\right)x\right) \\
&= \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{2}\sin\left(\left(2i + 1\right)x\right) - \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{4}\left[\sin\left(\left(2i + 3\right)x\right) + \sin\left(\left(2i - 1\right)x\right) \right]
\\
&= \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{2}\sin\left(\left(2i + 1\right)x\right) - \sum_{i = 1}^{k + 1}\frac{a_{i-1}^{(k)}}{4}\sin\left(\left(2i + 1\right)x\right) -\sum_{i = -1}^{k - 1}\frac{a_{i + 1}^{(k)}}{4} \sin\left(\left(2i + 1\right)x\right) \\
&=\sum_{i = 1}^{k - 1}\left[\frac{2a_{i}^{(k)} - a_{i - 1}^{(k)} - a_{i+1}^{(k)}}{4}\right]\sin\left(\left(2i + 1\right)x\right) + \left[\frac{3a_{0}^{(k)} - a_{1}^{(k)}}{4} \right]\sin\left(\left(2\cdot0 + 1\right)x\right)\\
&\qquad - \left[\frac{2a_{k}^{(k)} - a_{k - 1}^{(k)}}{4}\right]\sin\left(\left(2k + 1\right)x\right)
- \frac{a_{k}^{(k)}}{4}\sin\left(\left(2\left(k + 1\right) + 1\right)x\right) \\
&= \sum_{i = 0}^{k + 1} a_{i + 1}^{(k + 1)} \sin\left(\left(2i + 1\right)x\right)
\end{align}
Where
$$a_{i + 1}^{(k + 1)} = \begin{cases}
\frac{3a_{0}^{(k)} - a_{1}^{(k)}}{4} & i = 0\\
\frac{2a_{i}^{(k)} - a_{i - 1}^{(k)} - a_{i+1}^{(k)}}{4} & 1\leq i\leq k - 1 \\
\frac{2a_{k}^{(k)} - a_{k - 1}^{(k)}}{4} & i = k \\
-\frac{a_{k}^{(k)}}{4} & i = k + 1
\end{cases}$$
As required.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2981137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Asymptotic behavior of solution to nonlinear ODE I am trying to determine the asymptotic behavior as $t \to \infty$ of the solution to the IVP:
$$y''(t) - y(t) + \frac{1}{[y(t)]^3} = 0\\y(0) = 1;\, y'(0) = 1$$
Without the nonlinearity, we have solutions of the form $e^{\pm t}$. I think the nonlinear solution has similar behavior, but I'm not sure how to show this or determine the precise asymptotic form.
|
Let's see if we can solve this exactly.
Inverting the function gives us:
$$-\frac{t''}{t'^3} - y + \frac{1}{y^3} = 0 \\ t''+\left(y-\frac{1}{y^3} \right) t'^3=0$$
$$t'(y)=f(y)$$
$$f'=\left(\frac{1}{y^3}-y \right) f^3$$
$$-\frac{1}{2 f^2}=-\frac{1}{2 y^2}-\frac{y^2}{2}+C$$
$$\frac{1}{f^2}=\frac{1}{y^2}+y^2+C$$
$$f^2=\frac{y^2}{1+Cy^2+y^4}$$
$$f=\frac{y}{\sqrt{1+2C_1y^2+y^4}}$$
$$t(y)= \int \frac{y ~dy}{\sqrt{1+2C_1y^2+y^4}}=\frac{1}{2} \int \frac{du}{\sqrt{1+2C_1 u+u^2}}=\frac{1}{2} \log \left(C_1+u+\sqrt{1+2C_1 u+u^2} \right)+C_2$$
So we have:
$$t(y)=\frac{1}{2} \log \left(C_1+y^2+\sqrt{1+2C_1 y^2+y^4} \right)+C_2$$
Substituting the first condition gives us:
$$C_2=-\frac{1}{2} \log \left(C_1+1+\sqrt{2(C_1+1)) } \right)$$
For the second condition we find:
$$t'(1)=\frac{1}{\sqrt{2(C_1+1)}}=1$$
$$C_1=-\frac{1}{2}$$
$$C_2=-\frac{1}{2} \log \frac{3}{2}$$
So we get finally the exact implicit solution:
$$t(y)=\frac{1}{2} \log \left(y^2-\frac12+\sqrt{y^4-y^2+1} \right)-\frac{1}{2} \log \frac{3}{2}$$
We can find $y(t)$ by solving the above equation, which would reduce to a quadratic one:
$$y^2-\frac12+\sqrt{y^4-y^2+1}=\frac{3}{2} e^{2t}$$
$$\sqrt{y^4-y^2+1}=\frac{3}{2} e^{2t}+\frac12 -y^2$$
$$y^4-y^2+1=y^4-\left(1+3 e^{2t} \right)y^2+\left(\frac{1}{2}+\frac{3}{2} e^{2t} \right)^2 $$
$$3 e^{2t} y^2=\left(\frac{1}{2}+\frac{3}{2} e^{2t} \right)^2 -1$$
$$y(t)= \frac{e^{-t}}{2\sqrt{3}} \sqrt{\left(1+3 e^{2t} \right)^2 -4}$$
$$y(t)= \frac{e^{-t}}{2} \sqrt{3 e^{4t}+2e^{2t} -1}$$
This is the exact solution, which can be checked by direct substitution. From this we can find the asymptotic.
For $t \to +\infty$ we have:
$$y(t) \asymp \frac{\sqrt{3} }{2}e^{t}$$
The asymptotic gives a very good approximation for $t>2$, see the plot (blue is the exact solution, orange the asymptotic):
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$ As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$
But unfortunately I do not have skills in techniques to handle double summation .
Help appreciated.
I've made some research in MSE and found several questions which could be helpful:
1) $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=-\frac{17\pi^4}{1440}$
2) $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{\pi^6}{12960}$
3) $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}= \frac{1}{3}\zeta(6)$
|
Partial Fractions gives
$$
\frac{m^2+n^2}{\left(m^2-n^2\right)^2}=\frac{1/2}{(m-n)^2}+\frac{1/2}{(m+n)^2}\tag1
$$
First, we will compute
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m-n)^2}
&=2\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{mn(m-n)^2}\tag{2a}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{(m+n)nm^2}\tag{2b}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac1{m^3}\tag{2c}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac1{nm^2}\tag{2d}\\
&=\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac{m+n}{nm^3}\tag{2e}\\
&=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{m^2n^2}\tag{2f}\\[3pt]
&=\frac{\pi^4}{36}\tag{2g}
\end{align}
$$
Explanation:
$\text{(2a)}$: symmetry between $m\lt n$ and $n\lt m$
$\text{(2b)}$: substitute $m\mapsto m+n$
$\text{(2c)}$: partial fractions
$\text{(2d)}$: swap $m$ and $n$ in $\text{(2b)}$ then partial fractions
$\text{(2e)}$: average $\text{(2c)}$ and $\text{(2d)}$
$\text{(2f)}$: simplify
$\text{(2g)}$: evaluate $\zeta(2)^2$
Next, we will compute
$$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn(m+n)^2}
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\color{#C00}{\frac1{nm^2(n+m)}}-\color{#090}{\frac1{m^2(n+m)^2}}\right)\tag{3a}\\
&=\color{#C00}{\frac{\pi^4}{72}}-\color{#090}{\frac{\pi^4}{120}}\tag{3b}\\[3pt]
&=\frac{\pi^4}{180}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: partial fractions
$\text{(3b)}$: the red sum is $\frac12$ of $\text{(2b)}$, the green sum is $\frac12\left(\zeta(2)^2-\zeta(4)\right)$
$\text{(3c)}$: simplify
Therefore,
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m+n)^2}
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn(m+n)^2}-\frac14\zeta(4)\tag{4a}\\
&=\frac{\pi^4}{180}-\frac{\pi^4}{360}\tag{4b}\\[9pt]
&=\frac{\pi^4}{360}\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: subtract the terms where $m=n$
$\text{(4b)}$: apply $\text{(3c)}$
$\text{(4c)}$: simplify
Thus.
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{m^2+n^2}{mn\left(m^2-n^2\right)^2}
&=\frac12\frac{\pi^4}{36}+\frac12\frac{\pi^4}{360}\tag{5a}\\
&=\frac{11\pi^4}{720}\tag{5b}
\end{align}
$$
Explanation:
$\text{(5a)}$: apply $(1)$, $(2)$, and $(4)$
$\text{(5b)}$: simplify
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inverse of tridiagonal Toeplitz matrix Consider the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse $A^{-1}$?
Clearly, $A^{-1}$ is symmetric.
I look for a proof of the following conjecture that $A^{-1}$ is given as follows:
If $A_{i, j}^{-1}$ such that $j$ is odd and $i =1+j + 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$. From which follows by symmetry:
If $A_{i, j}^{-1}$ such that $j$ is even and $i =-1+j - 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$.
All other $A_{i, j}^{-1} = 0$.
Here is an example, computed with Matlab, for $n=10$ which shows the structure:
$${A_{10 \times 10}^{-1}} = \left[ {\begin{array}{*{20}{r}}
0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 \\
1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 & 0
\end{array}} \right]$$
|
Firstly Matrix is Toeplitz. This means it represents multiplication by power series expansion. This means matrix inversions corresponds to multiplicative inversion
Therefore, consider $$x+x^{-1}=\frac{x^2+1}x$$ Now it's multiplicative inverse:
$$(x+x^{-1})^{-1}=\frac x{x^2+1}$$
Now you can expand with geometric series / Taylor expansion for $$\frac 1{x^2+1}=\frac 1{1-(-1\cdot x^2)}$$
And substitute with
$$\frac{1}{1-t}=1+t+t^2+\cdots, t=-x^2$$
and then finish. You will notice the flipping sign pattern and that the odd exponents disappear when you substitute and do the expansion.
|
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|
How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbers and the inequality $$n+1<2^n$$ In this case, i can use it because i suppose it for natural n such that n>1. I'd like to have something like $$....+(n+1)<2(2^{n+1})$$ because you can divide both sides by $2^{n+1}$, and so, you get $$...+\frac{n+1}{2^{n+1}}<2$$
I´ve noticed that $$2^n+n+1<2^n+2^n=2^{n+1}<2(2^{n+1})$$ and it leads me to think $$...+n+1<2^n+n+1$$ i´ve tried dividing both sides ,of the inequality i´m assuming as a fact, by $2^n/2$ in order to get $2^n$
in the right hand of the inequality and then, add $n+1$ to both sides: $$(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n})\frac{2^n}{2}+n+1<2(\frac{2^n}{2})+n+1=2^n+n+1$$ but at this part i get stuck because if i continue it doesn't lead me to what i´d like to show.
Please, help me. I hope someone here have already proved this before, because i don't find anything similar to this at any part of the web.
|
Hint:
$$\begin{aligned}
&\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}\\
=&\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}+\frac{1}{2^{n+1}}\right) \\
&+\left(\frac{1}{4}+\frac{2}{8}+...+\frac{n-1}{2^n}+\frac{n}{2^{n+1}}\right)\\
\end{aligned}$$
Now try to show each bracket is less than $1$, the second bracket looks like a good candidate for the induction hypothesis.
|
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|
Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$
What is
$$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$
for a triangle with sides $2$, $3$, and $4$?
One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc)$ to get the cosines. But that's lots of calculate.
Is there a better way to get the answer? Thanks!
|
There are well-known identities for $\triangle ABC$
with the angles $A,B,C$,
sides $a,b,c$,
semiperimeter $\rho=\tfrac12(a+b+c)$,
area $S$,
radius $r$ of inscribed and
radius $R$ of circumscribed circles,
\begin{align}
\sin A+\sin B+\sin C
&=\frac\rho{R}
\tag{1}\label{1}
,\\
\cos A+\cos B+\cos C
&=\frac{r+R}{R}
\tag{2}\label{2}
,
\end{align}
so
\begin{align}
x&=
\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}
=\frac{\rho}{r+R}
\tag{3}\label{3}
,
\end{align}
we also know that
\begin{align}
R&=\frac{abc}{4S}
,\\
r&=\frac{S}{\rho}
,\\
S&=\tfrac14\sqrt{4(ab)^2-(a^2+b^2-c^2)^2}
,\\
\end{align}
thus we can find that for $a=2,b=3,c=4$
\begin{align}
\rho&=\frac{9}{2}
,\\
S&=\frac{3\sqrt{15}}{4}
,\\
R&=\frac{8\sqrt{15}}{15}
,\\
r&=\frac{\sqrt{15}}{6}
,\\
x&=\frac{\rho}{r+R}
=\frac{3\sqrt{15}}{7}
\approx 1.6598500
.
\end{align}
|
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|
Minimize the expression This was a problem given in a no calculator math contest (HMMT):
What is the minimum value of $(xy)^2+(x+7)^2+(2y+7)^2$ where $x$ and $y$ are reals. My friends and I discussed about using calculus, but we reasoned that there must be a faster trick (calc would take too long). Any ideas?
|
Since another user has given the essential hint of this problem. Let's us look at a generalization of the problem and spell out some details.
For any $a,b,c > 0$, consider the function
$$G(x,y) = x^2y^2 + (ax+c)^2 + (by+c)^2$$
Expanding RHS out, we get
$$\begin{align}G(x,y)
&= x^2y^2 + a^2x^2 + b^2y^2 + 2c(ax+by) + 2c^2\\
&= (x^2+b^2)(y^2 + a^2) + 2c(ax+by) + 2c^2 - a^2b^2
\end{align}$$
Notice
$$(x^2+b^2)(y^2+a^2) = |(x+bi)(y+ai)|^2 = |(xy-ab) + (ax+by)i|$$
We find
$$\begin{align}G(x,y)
&= (xy-ab)^2 + (ax+by)^2 + 2c(ax+by) + 2c^2 - a^2b^2\\
&= (xy - ab)^2 + (ax+by+c)^2 + c^2 - a^2b^2
\end{align}
$$
This implies $G(x,y) \ge c^2 - a^2b^2$ for all $(x,y) \in \mathbb{R}^2$.
The locus of $xy-ab$ is a hyperbola. If one look at its lower-left branch,
it has negative $x$ and negative $y$ axis as asymptotes. It is clear
the two "ends" of this branch lies below the line $ax+by+c = 0$.
The point $(-b,-a)$ belong to this branch. Since $ax + by + c$ evaluates
to $c-2ab$ at this point. If $c > 2ab$, this point lies above the line $ax+by+c = 0$.
In this case, the hyperbola $xy-ab$ intersect the line $ax+by+c = 0$ at two points
$(t_1,\frac{ab}{t_1})$, $(t_2,\frac{ab}{t_2})$ for some $t_1 < -b < t_2 < 0$.
In short, when $c > 2ab$, $G(x,y)$ achieves the value $c^2-a^2b^2$ at some point.
As a result, we have proved that
Given $a,b,c > 0$. If $c > 2ab$, then
$$\min_{(x,y)\in\mathbb{R}^2} \left( x^2y^2 + (ax+c)^2 + (by+c)^2 \right) = c^2 - a^2b^2$$
For the function at hand, $(a,b,c) = (1,2,7)$ and $c > 2ab$. The minimum we seek is
$7^2 - 1^2 2^2 = 45$.
|
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|
Find the sum of the series $\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}$
Find the sum of the series $\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}$
My attempt:
I tried partial fractions decomposition and I get :
$$\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac {3}{2n}-\frac 1{n+1}-\frac 1{2(n+2)}=\frac{1}{2}\sum_{n=1}^{\infty}\frac 3{n}-\frac 2{n+1}-\frac 1{n+2}.$$
This should be the solution but the partial sum sequence I can't figure out the formula... what should I do in this case?
|
Good answers have already been given; here's some intuition:
$$\sum_{n=1}^{\infty}\frac{3}{n}-\frac{2}{n+1} - \frac{1}{n+2} = 3\left(1+\frac{1}2 +\frac{1}3 + \ldots\right) - 2\left(\frac{1}2 +\frac{1}3 + \ldots\right)-1\left(\frac{1}3 + \ldots\right)= 3+\frac{3}{2}+3\left(\frac{1}3 + \ldots\right) -1 -2\left(\frac{1}3 + \ldots\right) -1\left(\frac{1}3 + \ldots\right) = 3+\frac{1}{2}+3\left(\frac{1}3 + \ldots\right) -3 \left(\frac{1}3 + \ldots\right) = 3+\frac{1}{2}.$$
Now divide by two to get your answer.
|
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|
Solving $8x^3 - 6x + 1$ using Cardano's method Solve for the first root of $8x^3 - 6x + 1 = 0$
After solving I get $\sqrt[3]{\frac{-1 + \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$, which is not a solution to the cubic equation:
Here's how I come up with:
Using Cardano's method:
let $x = y - \frac{b}{3a}$
Then compress:
Since $b$ is $0$, then it turns out to be $8y^3 - 6y + 1 = 0$.
Divide $8$ to both sides.
$$ y^3 - \frac{3}{4}y + \frac{1}{8} = 0 $$
Let $3st = \frac{-3}{4}$ and $s^3 - t^3 = \frac{-1}{8}$
Now:
\begin{align}
\left(\frac{-1}{4t}\right)^{3} - t^{3} &= \frac{1}{8}\\
...\\
8t^6 - t^3 + \frac{1}{8} &= 0\\
\end{align}
I'll uncompress the equation above so it becomes quadratic:
\begin{align}
8t^2 - t + \frac{1}{8} &= 0\\
...\\
\left(\frac{1 + \sqrt{3}i}{16}\right)\left(\frac{1 - \sqrt{3}i}{16}\right)\\
\end{align}
Then take the cuberoot to get $t$ (only taking the positive root):
$$
\sqrt[3]{\frac{1 + \sqrt{3}i}{16}}
$$
Since $x = y - \frac{0}{24}$, which is similar to $x = y$
and $y = s - t$, then:
$t = \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$
$s = \sqrt[3]{\frac{1 - \sqrt{3}i}{16}}$
\begin{align}
y &= s - t\\
y &= \sqrt[3]{\frac{1 - \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}\\
x &= y + 0\\
x &= \sqrt[3]{\frac{1 - \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}\\
\end{align}
Checking if it is a solution, it turns out that it is not. Note: I'm new to this method so its unclear to me why it dont work and PLEASE dont mark this as a duplicate. Thanks.
|
You may have a sign error. I render $s^3$ as $-(1-i\sqrt{3})/16$ but you seem to have $+(1-i\sqrt{3})/16$. But even with the sign correction (which probably does get you a good answer), your approach is not best. See below.
When you have an expression with two different complex cube roots there are nine choices for this combination. Any of three roots or the first cube root could be paired with any of three cube roots for the second. Yet, of course, only three of the combinations can be correct for the cubic equation you started with. If your calculator/computer program does not choose the "principal values" of the cube roots in the right way for this particular equation, you go wrong (even without other errors).
The solution is easy. When you get a root for $t$, do not solve independently for $s$. Use the fact that $st=-(1/4), s=-(1/4t)$ to get an expression for $s$ that has the same cube root radical as the one for $t$. Now your intended difference $s-t$ contains only a single cube root radical, and its three possible values correspond properly to the three roots of the cubic equation.
|
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|
Sandwich Theorem not working?
This is the limit I need to solve:
$$\lim_{n \to \infty} \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4}$$
I simplified it to this:
$$\lim_{n \to \infty} \frac{2(4 \cos(n) - 3n^2)}{(6n^3 + 5n \sin(n))}.$$
At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit.
I use the fact that $\lim_{n \to \infty} \frac{a}{b} = \frac{\lim_{n \to \infty} a}{\lim_{n \to \infty} b}$ when $b\ne 0$.
By the Sandwich Theorem both the Numerator and Denominator is $\infty$.
Hence the answer is 1.
But if I calculate the limit whole without splitting it into two I get $\frac{3}{2}$. Which answer is correct? Please Help!
|
We have that
$$\frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$\frac{(\pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\sim \frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
|
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|
Complex root of equation 1 let $a,b$ & $c$ are the roots of cubic $$x^3-3x^2+1=0$$
Find a cubic whose roots are $\frac a{a-2},\frac b{b-2}$ and $\frac c{c-2} $
hence or otherwise find value of $(a-2)(b-2)(c-2)$
|
You know that: $$a+b+c=3$$ $$ab+bc+ac=0$$ $$abc=-1$$
Also, $$(a-2)(b-2)(c-2)=abc-2(ab+bc+ac)+4(a+b+c)-8$$
So $$(a-2)(b-2)(c-2)=-1-2(0)+4(3)-8=3$$
For cubic with roots as $$\frac a{a-2},\frac b{b-2},\frac c{c-2}$$
Find out $$\frac a {a-2}+\frac b{b-2}+\frac c{c-2}$$
and $$\frac a{a-2}\cdot \frac b{b-2} + \frac b{b-2}\cdot \frac c{c-2} + \frac a{a-2}\cdot \frac c{c-2}$$ by taking LCM and rigorous solving.
You already know that $$\frac a{a-2}\cdot \frac b{b-2} \cdot \frac c{c-2}=3$$
|
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|
Probability that 2 heads do not come consecutively.
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?
Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{HTHTHTHTH}$
Then we'll be left with 1 tail which can be placed at 6 places.
So number of cases = $6$
2) $4$ heads, $6$ tails
$\mathrm{HTHTHTH}$
Left with $3$ tails.
So all $3$ can be placed together at $5$ places.
Or $3 = 2+1$, in which case we have $5 \times 4 = 20$ cases.
Total number for #2 $= 20 + 5 = 25$
3) 3 heads, 7 tails.
$\mathrm{HTHTH}$
Left with $5$ tails.
All 5 can be placed together, so in that case we have $4$ cases.
Or $5 = 2+3= 1+4$
For $5 = 2+3 = 1+4$, we have $4\times 3$ cases each.
Therefore total number of cases for #3 $= 12+12+ 4 = 28$
4) 2 heads 8 tails.
$\mathrm{HTH}$
$7$ left
$7 = 7+ 0$ , $3$ cases.
$7 = 6+1= 5+2= 3+4$ , $6$ cases each
So number of cases for #4 = $6\times 3 + 3 = 21$
5) 1 head 9 tail, 10 cases
6) 0 heads, 1 case.
So total number of cases = $6+25+28+21+10 +1 = 91$
I tried thrice and got $91$ cases only.
So answer should be $P(E) = \dfrac{91}{2^{10}}$
But that's not the right answer.
Please tell me my mistake.
|
Another way: Let the number of solutions for $n$ tosses be $a_n$. For $n \ge 2$, the solutions are either T plus a solution for $n - 1$ tosses, or HT plus a solution for $n - 2$ tosses. So $a_n = a_{n-1} + a_{n-2}$. Since $a_1 = 2$ and $a_2 = 3$ we see that $a_n$ is the Fibonacci number $F_{n+2}$, which makes the probability of success $F_{n+2}/2^n$. For $n = 10$ this gives the answer found by Key Flex.
|
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|
How to calculate $\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx$? I get the answer that it equals $\frac{\pi(2\cos2+\sin2)}{2e^4}$ by mathematica.
But I don't know how it can equal this form.
My idea is to construct equations so that problems can be transformed.Like these:
$$\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx=\int_{-\infty}^{+\infty} \frac{(y+4) \sin (y+4)}{y^2+4y+20}\, dy$$
$$\int_{-\infty}^{+\infty} \frac{x \cos x}{x^2+4x+20}\, dx=-\int_{-\infty}^{+\infty} \frac{(y+4) \cos (y+4)}{y^2+4y+20}\, dy$$
So only by calculating $\int_{-\infty}^{+\infty} \frac{ \sin x}{x^2+4x+20}\, dx$ and $\int_{-\infty}^{+\infty} \frac{ \cos x}{x^2+4x+20}\, dx$ then we can get the values of $\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx$ and $\int_{-\infty}^{+\infty} \frac{x \cos x}{x^2+4x+20}\, dx$.
What's more I found that
$$\int_{-\infty}^{+\infty} \frac{ \sin x}{x^2+4x+20}\, dx=- \frac{\pi \sin 2}{4e^4}$$
$$\int_{-\infty}^{+\infty} \frac{ \cos x}{x^2+4x+20}\, dx= \frac{\pi \cos 2}{4e^4}$$
So I guess I can solve the problem only by calculating
$$\int_{-\infty}^{+\infty} \frac{ \sin x}{x^2+ c}\, dx$$ and
$$\int_{-\infty}^{+\infty} \frac{ \cos x}{x^2+ c}\, dx$$
Can you help me?
|
As you've already noted, the substitution $y=x-2$ gives$$\int_{\mathbb{R}}\frac{x\sin xdx}{x^{2}+4x+20} =2C_{0}\sin 2-2S_{0}\cos 2+S_{1}\cos 2-C_{1}\sin 2$$with $$C_{n}:=\int_{\mathbb{R}}\frac{y^{n}\cos ydy}{y^{2}+16},\,S_{n}:=\int_{\mathbb{R}}\frac{y^{n}\sin ydy}{y^{2}+16},$$i.e. $$C_{n}+iS_{n}=\int_{\mathbb{R}}\frac{\exp iydy}{y^{2}+16}.$$The characteristic function of the Cauchy distribution is$$\int_{\mathbb{R}}\frac{\gamma}{\pi}\frac{\exp itydy}{\left(y-y_{0}\right)^{2}+\gamma^{2}}=\exp\left(iy_{0}t-\gamma\left|t\right|\right),$$so in particular$$\int_{\mathbb{R}}\frac{\exp itydy}{y^{2}+16}=\frac{\pi}{4}\exp-4\left|t\right|.$$Differentiating with respect to t,$$\int_{\mathbb{R}}\frac{y\exp itydy}{y^{2}+16}=\pi i\operatorname{sgn}t\exp-4\left|t\right|.$$Hence$$C_{0}+iS_{0}=\frac{\pi}{4e^{4}},\,C_{1}+iS_{1}=\frac{\pi i}{e^{4}}\implies C_{1}=S_{0}=0,\,C_{0}=\frac{\pi}{4e^{4}},\,S_{1}=\frac{\pi}{e^{4}}.$$Finally,$$\int_{\mathbb{R}}\frac{x\sin xdx}{x^{2}+4x+20}=\frac{\pi}{2e^{4}}\left(2\cos2+\sin2\right).$$
|
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|
Show that these non-linear recursions produce integers only The recurrence is of third order:
Start with
\begin{align*}a_0(x)&=1\\ a_1(x)&=1\\ a_2(x)&=x \end{align*} and then
\begin{align*}a_{n+3}(x)&=\frac{a_{n+2}^2(x)-a_{n+1}^2(x)}{a_{n}(x)}.\end{align*}
The calculation and factorization of $a_n(x)$ for $3\le n \le 6$ gives the following results:
\begin{align*} a_3(x)&= (x+1)(x-1)\\
a_4(x)&=(x^2 + x -1)(x^2-x-1)\\
a_5(x)&=x(x^2 -2)(x^4-4x^2+2)\\
a_6(x)&=(x^6 +x^5-5x^4-4x^3+6x^2+3x-1)(x^6 -x^5-5x^4+4x^3+6x^2-3x-1)\\
\end{align*}
Questions:
*
*Any reference for this ?
*Show that $a_n(x)$ is an integer coefficient polynomial. It would be of
degree $F_{n+1} -1$, where $F_n$ is a Fibonacci number.
*Is it true that $a_n(x)$ divides $a_{2n+1}(x)$?
*For $k \in \mathbb N, k>1$, is there a general close form for $a_n(k)$?
For $k=2$, we obviously have $a_n(2)=F_{n+1}$.
For $k=3$, we have $a_n(3)=F_{2F_{n+1}}$.
What about $a_n(4)$?
|
Here is another proof that $a_n$ is integer, which is somewhat more elementary as it does not require the knowledge of Chebyshev Polynomials. It follows the same lines of reasoning as in Malouf, J.L., An integer sequence from a rational recursion, Discrete Mathematics 110 (1992)
257-261.
We suppose (induction hypothesis) that all the $a_j$ are integers up to $j=n$ with $n\ge5$.
We let $B_6=a_{n+1}$, $B_5=a_n$, $B_4=a_{n-1}$,, $B_3=a_{n-2}$, $B_2=a_{n-3}$, $B_1=a_{n-4}$ and $B_0=a_{n-5}$.
It is easy to verify that $a_0, a_1, a_2, a_3, a_4$ and $a_5$ are integers, then it suffices to show that $B_6$ is integer. This amounts at showing that $$\tag0 B_5^2-B_4^2\equiv 0 \pmod {B_3}.$$
If there exists a prime $p$ such that $p$ divides two consecutives members $a_j$ and $a_{j-1}$ of the sequence (up to the index $n$, all the members are integers by hypothesis), then since $a_{j-2}^2= a_{j-1}^2-a_ja_{j-3}$, then $p$ divides $a_{j-2}$, which by repetition leads to $p$ divides $a_1=1$, which is impossible, so we have shown that two consecutive members of the sequence are coprime, as long as they are integers.
We also show that $B_1$ is coprime to $B_3$ for suppose $p$ divides $B_3$, then $p$ does not divide $B_1$, otherwise since $B_1B_4 =B_3^2-B_2^2$, we would have $p$ divides $B_2$, which is impossible.
We have \begin{align*}
\tag1 B_0B_3 &=B_2^2-B_1^2 \equiv 0 \pmod {B_3}\\
\tag2 B_1B_4 &=B_3^2-B_2^2\equiv -B_2^2\pmod {B_3}\\
\tag3 B_2B_5 &=B_4^2-B_3^2 \equiv B_4^2\pmod {B_3} \end{align*}
From $(2)$ and $(1)$, we obtain $B_1B_4 \equiv -B_1^2\pmod {B_3}$, but since $B_1$ and $B_3$ are coprime that is $$\tag4 B_4 \equiv -B_1\pmod {B_3}.$$
From $(3)$, $(4)$ and $(1)$, we obtain $B_2B_5 \equiv B_2^2\pmod {B_3}$, but since $B_2$ and $B_3$ are coprime that is $$\tag5 B_5 \equiv B_2\pmod {B_3}.$$
Finally $(0)$ follows from $(4)$,$(5)$ and $(1)$. $\ \ \ \ \ \ \square$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I show that the system is hyperbolic if $u^2 + v^2 > c^2$ I know that for a system to be hyperbolic it must have 2 real distinct eigenvalues $\lambda$ where $\det(B-\lambda A)=0$. My system of equations are:
\begin{aligned}
(pu)_x + (pv)_y &= 0 \\
p(uu_x + vu_y) + c(p)^2p_x &= 0 \\
p(uv_x +vv_y)+c(p)^2p_y &= 0
\end{aligned}
I need to show that the system is hyperbolic if $u^2 + v^2 > c^2$. I am stuck on how to formulate the matrices $A,B$. Could someone please help?
|
In order for the system to be hyperbolic, it needs to have $n$ distinct real eigenvalues if you are working with $n$ dependent variables. In this case, $n=3$.
The first thing to do is to write it in the form
$$\mathbf A\frac{\partial \mathbf u}{\partial x} + \mathbf B\frac{\partial \mathbf u}{\partial y} = \mathbf c$$
where $\mathbf u = (p,u,v)^T$ is the vector of dependent variables.
All you need to do is expand out the derivatives in the given equations and write them out neatly:
\begin{matrix}
up_x & + & pu_x & + & & & vp_y & + & & & pv_y & = & 0 \\
c^2p_x & + & puu_x & + & & & & & pvu_y & & & = & 0 \\
& & & & puv_x & + & c^2p_y & + & & & pvv_y & = & 0
\end{matrix}
Notice how I have arranged each equation in the order $p_x$, $u_x$, $v_x$, $p_y$, $u_y$, $v_y$. We can now write it in the form as previously mentioned:
$$
\begin{pmatrix}
u & p & 0 \\
c^2 & pu & 0 \\
0 & 0 & pu
\end{pmatrix}\frac{\partial}{\partial x} \begin{pmatrix} p \\ u \\ v\end{pmatrix} + \begin{pmatrix}
v & 0 & p \\
0 & pv & 0 \\
c^2 & 0 & pv
\end{pmatrix}\frac{\partial}{\partial y} \begin{pmatrix} p \\ u \\ v\end{pmatrix} = 0
$$
There you have your $\mathbf A$ and $\mathbf B$, and the next step should be straightforward.
You then compute $\text{det}(\mathbf B- \lambda \mathbf A)$:
\begin{align}
\text{det}(\mathbf B- \lambda \mathbf A) = & \text{det} \begin{pmatrix} v - \lambda u & - \lambda p & p \\ -\lambda c^2 & pv-\lambda pu & 0 \\ c^2 & 0 & pv - \lambda pu \end{pmatrix} \\
= & (v-\lambda u)(pv-\lambda pu)(pv-\lambda pu)-p(pv-\lambda pu)c^2-(pv-\lambda pu)(-\lambda p)(-\lambda c^2) \\
= & p^2(v-\lambda u)\big[(v-\lambda u)^2-c^2-\lambda ^2 c^2\big] \\
= & p^2(v-\lambda u)\big[(u^2-c^2)\lambda^2-2uv\lambda+v^2-c^2\big]
\end{align}
Set this to $0$ and solve for $\lambda$:
\begin{align}
& \text{det}(\mathbf B- \lambda \mathbf A)=0 \\
\implies & p^2(v-\lambda u)\big[(u^2-c^2)\lambda^2-2uv\lambda+v^2-c^2\big]=0 \\
\implies & \lambda = \frac vu, \frac{uv\pm \sqrt{u^2v^2-(u^2-c^2)(v^2-c^2)}}{u^2-c^2} \\
\implies & \lambda = \frac vu, \frac{uv\pm c\sqrt{u^2+v^2-c^2}}{u^2-c^2}
\end{align}
In order for the system to by hyperbolic, these three roots must be real and distinct.
$\frac vu \neq \frac{uv\pm c\sqrt{u^2+v^2-c^2}}{u^2-c^2}$ because $\lambda = \frac vu$ does not satisfy $\big[(u^2-c^2)\lambda^2-2uv\lambda+v^2-c^2\big]=0$, whereas $\frac{uv + c\sqrt{u^2+v^2-c^2}}{u^2-c^2} \neq \frac{uv - c\sqrt{u^2+v^2-c^2}}{u^2-c^2}$ as long as $u^2+v^2-c^2 \neq 0$ (i.e. $u^2+v^2 \neq c^2$).
Finally, these roots are real if the expression in the square root is non-negative, i.e. if $u^2+v^2 \geq c^2$.
Hence, the system is hyperbolic if $u^2+v^2>c^2$.
|
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|
Real solution of $(\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.$
Real solution of equation $$(\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.$$
Try: Using Half angle formula
$\displaystyle \cos x=\frac{1-\tan^2x/2}{1+\tan^2 x/2}$ and $\displaystyle \sin x=\frac{2\tan^2 x/2}{1+\tan^2 x/2}$
Substuting These values in equation
we have an polynomial equation in terms of $t=\tan x/2$
So our equation $$3t^{4}+6t^{3}+8t^{2}-2t-3=0$$
Could Some Help me how to Factorise it.
OR is there is any easiest way How to solve it, Thanks
|
HINT:
Since $\sin x=\frac{2t}{1+t^2}$, the equation becomes $$\left(\frac{1-t^2-2t}{1+t^2}\right)\left(\frac{2t+1+t^2}{1-t^2}\right)+2=0\implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3015467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to show that matrix $A,B$ are not similar? Let $A, B$ be $4\times 4$ square matrix given by,
$$A=\begin{bmatrix}0 &0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\\\end{bmatrix}, B=\begin{bmatrix}0 &0&1&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\\\end{bmatrix}$$
\begin{array}{|c|c|c|}
\hline
\text{Property}& A & B\\\hline
\text{Determinant}& 0& 0\\ \hline
\text{Trace}& 0 &0 \\ \hline
\text{Rank}& 2 & 2\\ \hline
\text{Char. Polynomial}& x^4 &x^4\\ \hline
\end{array}
I want to show that $A$ and $B$ are not similar. All the problems I encountered so far were usually solved by showing that one of these above mentioned properties does not match.
How to approach this?
|
According to the comments, $A^2,B^2$ are not similar because
$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}\not= 0$.Yet, even if the previous product had been zero, $A,B$ would not have been similar. More generally
let $A=\begin{pmatrix}0&M\\0&0\end{pmatrix},B=\begin{pmatrix}0&U\\0&V\end{pmatrix}$ where each block has dimension $n\times n$, $M$ is invertible and $V\not=0$. Then $A,B$ are not similar (even if $V^2=0,UV=0$).
Proof. We show that if $Z=\begin{pmatrix}P&Q\\R&S\end{pmatrix}$ and $AZ=ZB$, then $\det(Z)=0$.
We obtain $MR=0$ and then $R=0$;
$SV=0$, then $\det(S)=0$;
$\det(Z)=\det(P)\det(S)=0$. $\square$
|
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"language": "en",
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"source": "stackexchange",
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|
Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.
But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?
|
Hint.
Calling $f(x) = \sqrt{x^2+(x+1)^2}$ we seek for
$$
\min_x f(x) + f(x-3)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\tag {ii}$$
For $(\mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$
|
(i)
\begin{align*}
(\sec^2x + \tan^2x)(\csc^2x + \cot^2x) & = (1 + \tan^2x + \tan^2x)(1 + \cot^2x + \cot^2x)\\
& = (1 + 2\tan^2x)(1 + 2\cot^2x)\\
& = 1 + 2\cot^2x + 2\tan^2x + 4\\
& = 5 + 2(\csc^2x - 1) + 2(\sec^2x - 1)\\
& = 5 + 2\csc^2x - 2 + 2\sec^2x - 2\\
& = 1 + 2\csc^2x + 2\sec^2x\\
& = 1 + 2\left(\frac{1}{\sin^2x} + \frac{1}{\cos^2x}\right)\\
& = 1 + 2\left(\frac{\cos^2x + \sin^2x}{\sin^2x\cos^2x}\right)\\
& = 1 + \frac{2}{\cos^2x\sin^2x}\\
& = 1 + 2\sec^2x\csc^2x
\end{align*}
(ii)
\begin{align*}
\frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x} & = \frac{\cos x}{1 - \tan x} \cdot \frac{\cos x}{\cos x} + \frac{\sin x}{1 - \cot x} \cdot \frac{\sin x}{\sin x}\\
& = \frac{\cos^2x}{\cos x - \sin x} + \frac{\sin^2x}{\sin x - \cos x}\\
& = \frac{\cos^2x}{\cos x - \sin x} - \frac{\sin^2x}{\cos x - \sin x}\\
& = \frac{\cos^2x - \sin^2x}{\cos x - \sin x}\\
& = \frac{(\cos x + \sin x)(\cos x - \sin x)}{\cos x - \sin x}\\
& = \cos x + \sin x
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of one-one function from sets $A$ to $B$. Let $A=\{1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5\}$. Number of one-one function from $A$ to $B$ such that $f(1) \neq 0$ and $f(i)\neq i$ for $i={1,2,3,4,5}$ is _______ .
So I know one one function means for each $x$ there will be only one $y$.
But here if I take for $1$ from $A$ then total number of functions $=480$ as ${{4_C}_1}×{{5_P}_4}$
But for every for every other $i$ from $A$ it will be ${{5_C}_1}×{{5_P}_4}$ so total cases will be $480+2400=2880$.
But I don't know how to further proceed.
|
Definition. A number $c$ is said to be a fixed point of a function if $f(c) = c$.
Let $A = \{1, 2, 3, 4, 5\}$; let $B = \{0, 1, 2, 3, 4, 5\}$.
If there were no restrictions, we would have $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 6!$ ways to map the elements in the domain to distinct elements in the codomain. Hence, there are $6!$ injective functions $f: A \to B$. From these, we must exclude those that violate the restrictions that $f(0) \neq 1$ and that $f(i) \neq i, 1 \leq i \leq 5$.
We will use the Inclusion-Exclusion Principle.
A number in the domain is mapped to a prohibited value of the codomain: We consider two cases, depending on whether or not that number is $1$.
The number $1$ is mapped to $0$ or $1$: There are two ways to choose the image of $1$. That leaves five elements in the codomain to which the remaining four numbers in the domain can be mapped. The remaining numbers in the domain can be mapped to distinct elements of that subset of the codomain in $5 \cdot 4 \cdot 3 \cdot 2 = 5!$ ways. Hence, there are
$$\binom{2}{1}5!$$
such injective functions.
A number other than $1$ is a fixed point: There are four ways to choose which of the four numbers larger than $1$ is a fixed point. That leaves five numbers in the codomain to which the remaining four elements in the domain can be mapped. The remaining numbers in the domain can be mapped to distinct elements of that subset of the codomain in $5!$ ways. Hence, there are
$$\binom{4}{1}5!$$
such injective functions.
Hence, there are
$$\binom{2}{1}5! + \binom{4}{1}5!$$
injective functions in which a number in the codomain is mapped to a prohibited element of the codomain.
Two numbers in the domain are mapped to probibited values of the codomain: We again have to consider two cases, depending on whether or not $1$ is one of the numbers that is mapped to a prohibited value.
The number $1$ is mapped to $0$ or $1$ and another number is a fixed point: There are $2$ ways to choose the image of $1$ and four ways to pick which of the other four numbers in the codomain is a fixed point. That leaves four numbers in the codomain to which the remaining three elements in the codomain can be mapped. The remaining numbers in the domain can be mapped to distinct elements of that subset of the codomain in $4 \cdot 3 \cdot 2 = 4!$ ways. Hence, there are
$$\binom{2}{1}\binom{4}{1}4!$$
such injective functions.
Two numbers larger than $1$ are fixed points: There are $\binom{4}{2}$ ways to choose the two fixed points. That leaves four numbers in the codomain to which the remaining three elements in the domain can be mapped. The remaining numbers in the domain can be mapped to distinct elements that subset of the codomain in $4!$ ways. Hence, there are
$$\binom{4}{2}4!$$
such injective functions.
Three numbers in the domain are mapped to prohibited values of the codomain: As above, we have two cases.
The number $1$ is mapped to $0$ or $1$ and two other numbers are fixed points: We choose the image of $1$ and which two of the other four numbers are fixed points. That leaves three elements in the codomain to which the remaining two elements in the domain can be mapped. We choose how to map those two elements to distinct elements of that subset of the codomain. There are
$$\binom{2}{1}\binom{4}{2}3!$$
such injective functions.
Three numbers larger than $1$ are fixed points: We choose which three of those four numbers are fixed points. That leaves three elements in the codomain to which the remaining two elements in the domain can be mapped. We choose how to map those two elements to distinct elements of that subset of the codomain. There are
$$\binom{4}{3}3!$$
such injective functions.
Hence, there are
$$\binom{2}{1}\binom{4}{2}3! + \binom{4}{3}3!$$
injective functions in which three numbers in the domain are mapped to prohibited elements in the codomain.
Four elements in the domain are mapped to prohibited values in the codomain: Again, there are two cases.
The number $1$ is mapped to $0$ or $1$ and three of the other four numbers are fixed points: Choose the image of $1$ and which three of the other four numbers are fixed points. Assign the remaining number in the domain to one of the two remaining numbers in the codomain. There are
$$\binom{2}{1}\binom{4}{3}2!$$
such injective functions.
All four numbers larger than $1$ are fixed points: That leaves two ways to assign $1$ to one of the remaining elements in the codomain (yes, we have already counted this case, but the Inclusion-Exclusion Principle compensates for that). There are
$$\binom{4}{4}2!$$
such injective functions.
Hence, there are
$$\binom{2}{1}\binom{4}{3}2! + \binom{4}{4}2!$$
injective functions in which four elements of the domain are mapped to probibited elements of the codomain.
All five elements of the domain are mapped to prohibited values in the codomain: There are two possibilities. Either $1$ is mapped to $0$ and the other four numbers are fixed points or all five numbers are fixed points. There are
$$\binom{2}{1}\binom{4}{4}$$
such functions.
By the Inclusion-Exclusion Principle, the number of injective functions $f: A \to B$ such that $f(1) \neq 0$ and $f(i) \neq i, 1 \leq i \leq 5$, is
$$6! - \binom{2}{1}5! - \binom{4}{1}5! + \binom{2}{1}\binom{4}{1}4! + \binom{4}{2}4! - \binom{2}{1}\binom{4}{2}3! - \binom{4}{3}3! + \binom{2}{1}\binom{4}{3}2! + \binom{4}{4}2! - \binom{2}{1}\binom{4}{4}$$
|
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|
What are the Legendre symbols $\left(\frac{10}{31}\right)$ and $\left(\frac{-15}{43}\right)$? I have the following two Legendre symbols that need calculated:
$\left(\frac{10}{31}\right)$ $=$ $-\left(\frac{31}{10}\right)$ $=$ $-\left(\frac{1}{10}\right)$ $=$ $-(-1)$ $=$ $-1$
$\left(\frac{-15}{43}\right)$ $=$ $\left(\frac{43}{15}\right)$ $=$ $\left(\frac{13}{15}\right)$ $=$ $-1$
is that correct?
I just want to make sure I am understanding this concept.
|
Here is an approach I would take. Note that
$$\left(\frac{10}{31}\right)=\left(\frac{-21}{31}\right)=\left(\frac{-1}{31}\right)\left(\frac{3}{31}\right)\left(\frac{7}{31}\right)=(-1)\Biggl(-\left(\frac{31}{3}\right)\Biggr)\Biggl(-\left(\frac{31}{7}\right)\Biggr)\,.$$
That is, $$\left(\frac{10}{31}\right)=-\left(\frac{1}{3}\right)\left(\frac{3}{7}\right)=-(+1)\Biggl(-\left(\frac{7}{3}\right)\Biggr)=\left(\frac{1}{3}\right)=+1\,.$$ This can be verified by noting that $6^2\equiv 5\pmod{31}$ and $8^2\equiv 2\pmod{31}$, so $$17^2\equiv (6\cdot 8)^2\equiv 5\cdot2=10\pmod{31}\,.$$
For the second part, note that
$$\left(\frac{-15}{43}\right)=\left(\frac{-1}{43}\right)\left(\frac{3}{43}\right)\left(\frac{5}{43}\right)=(-1)\Biggl(-\left(\frac{43}{3}\right)\Biggr)\left(\frac{43}{5}\right)\,.$$
Therefore, $$\left(\frac{-15}{43}\right)=\left(\frac{1}{3}\right)\left(\frac{3}{5}\right)=\left(\frac{3}{5}\right)\,.$$ It is easy to verify that $\left(\dfrac{3}{5}\right)=-1$, whence $$\left(\frac{-15}{43}\right)=-1\,.$$ You can check that $12^2\equiv 15\pmod{43}$, so $\left(\dfrac{15}{43}\right)=+1$, whereas $\left(\dfrac{-1}{43}\right)=-1$, confirming the calculations.
|
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|
Find the minimum value of $\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$ Given that $0\lt x\lt 2$ and $0\lt y\lt 2$ then find the minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$$
My try:
On factorisation we need minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {(y-2)^2+x^2} +\sqrt {(x-2)^2+(y-2)^2}$$
On seeing it for first time, the only thing that popped up was using the Minkowski inequality but I am not getting proper sequences for its application. I tried as much as I could to use this inequality but failed.
I tried substituting $x=2\cos \alpha$ and $y=2\cos \beta$ (where $\alpha, \beta \in \left(0,\frac {\pi}{2}\right)$ )on seeing the constraints on $x$ and $y$ but continuing it was very cumbersome so dropped the method.
Any help would be greatly appreciated.
P.S : It would be very great if someone hints at how I could use the Minkowski inequality efficiently. Thanks!!!
|
You can dress up your geometric argument as an inequality in $\mathbb{C}$.
Let $z=x+iy$, $a=2i$, $b=2-2i$. Then
\begin{align}
\sqrt{2x^2+2y^2} &= \lvert (1-i)z\rvert\\
\sqrt{y^2+x^2-4y+4} &= \lvert z-a\rvert\\
\sqrt{x^2+y^2-4x-4y+8} &= \lvert -iz-b\rvert
\end{align}
So the triangle inequality gives
$$
\lvert (1-i)z\rvert + \lvert z-a\rvert + \lvert -iz-b\rvert \ge \lvert a-b\rvert
$$
with equality if and only if the four points $a,z,-iz,b$ are all on the same line and in this order. Now argue why such a $z$ exists.
|
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|
Simplify third degree polynomial equations. Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3\over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3\over2})(6x^2 + 4x + 2) = 0$
|
It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $\deg r<\deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
\begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
\end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} \quad (*)$
Consider:
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 0, x_{3} \geq 4, x_{4} \geq 0 \quad (**)$
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 5, x_{3} \geq 4, x_{4} \geq 6 \quad (***)$
Let $f$ is the function that compute the number of non-negative solutions of an equation.
$\implies f(*) = f(**) - f(***)$
Thus, the number of non-negative solutions of (*) is $\sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 \choose 3 - 1} - {25-x_{1}+3 - 1 \choose 3 - 1}) = 3045$
I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!
|
It's the coefficient of $x^{40}$ of the product polynomial
$$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + \ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$
Or equivalently the coefficient of $x^{34}$ of
$$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + \ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$
which can be found using (generalised) binomials etc.
|
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|
Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers
If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but second is not an extreme value. Comment on particular cases when (i) ${a+b+c=0}$, (ii) ${a=b=c}$.
My Attempt:
$${F=f+\lambda \phi =x^3 + y^3 + z^3 - 3xyz + 3\lambda(ax + by + cz-1)}$$
$${\frac13F_x = x^2-yz+\lambda a = 0}{\text{ ...(1)}}$$
$${\frac13F_y = y^2-xz+\lambda b = 0}{\text{ ...(2)}}$$
$${\frac13F_z = z^2-xy+\lambda c = 0{\text{ ...(3)}}}$$
$${(1)x+(2)y+(3)z \implies f+\lambda (1) = 0 \implies \lambda = -f}$$
$${(1)+(2)+(3) \implies}{x^2+y^2+z^2-xy-yz-zx=(a+b+c)f}$$
$${\implies f/(x+y+z)=(a+b+c)f}\,,$$
then ${f=0}$ or ${x+y+z=1/(a+b+c)}$.
Also, ${(1)-(2) \implies x^2-y^2-z(x-y)=f(a-b) \implies \frac{x-y}{a-b}=f\frac{a+b+c}{x+y+z}}$
Similarly ${(2)-(3)}$ and ${(3)-(1)}$, then we get
${\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}=f\frac{a+b+c}{x+y+z}}$
I don't know how to proceed further. I couldn't get the other stationary value of ${f}$. I need help in proceeding further to calculate stationary points and/or stationary values, if at all, my method is correct.
|
Without Lagrange Multipliers.
Making the change of variables $y = \lambda x, z = \mu x$ and substituting we get
$$
\min\max x^3(1+\lambda^3+\mu^3-3\lambda\mu)\ \ \mbox{s. t. }\ \ x(a+\lambda b+\mu c) = 1
$$
or equivalently
$$
\min\max f(\lambda,\mu) = \frac{1+\lambda^3+\mu^3-3\lambda\mu}{(a+\lambda b+\mu c)^3}
$$
whose stationary points are solved by
$$
\left\{
\begin{array}{rcl}
\left(\lambda ^2-\mu \right) (a+c \mu )-b \left(\mu ^3-2 \lambda \mu +1\right)& = & 0 \\
c \left(\lambda ^3-2 \mu \lambda +1\right)+(a+b \lambda ) \left(\lambda -\mu ^2\right) & = & 0 \\
\end{array}
\right.
$$
giving the points
$$
\begin{array}{ccc}
\lambda & \mu & f(\lambda,\mu)\\
1 & 1 & 0 \\
\frac{b^2-a c}{a^2-b c} & \frac{c^2-a b}{a^2-b c} & \frac{1}{a^3+b^3+c^3-3 a b c} \\
\end{array}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Lagrange Error Value for $f(x)=\frac{1}{x}$ I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= \frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$
Step $1$ is to calculate the Taylor-Polynomial:
$$T_3(x,5)= \frac{1}{5}-\frac{(x-5)}{25}-\frac{(x-5)^2}{125}-\frac{(x-5)^3}{625}$$
I then used the Remainder formula $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!} \cdot (x-a)^{(n+1)}$. From this I get:
$$R_3(x)=\frac{f^{(4)}(\xi)}{4!} \cdot (x-5)^4 =\frac{-1}{\xi^4} \cdot (x-5)^4 = \frac{-(x-5)^4}{\xi^4}$$
Here's where I'm not sure how to proceed. I thought that I could make $2$ cases to try to find the Worst-Case scenario by checking which value of $\xi_1=4, \xi_2=6$ gives the Worst-Case scenario btu then I'm still left with either
$$R_{n_1}(x) =\frac{-(x-5)^4}{4^4} \text{ or } R_{n_2}(x) =\frac{-(x-5)^4}{6^4}$$
but in either case I'm still left without an exact value for the error, which makes my answer incorrect.
How can I obtain an exact value for the error of the Taylor-Polynomial in the interval $x \in [4,6]$?
|
Consider the function
$$f(x) = \dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$\begin{array}{|c|c|}
\hline \text{Derivatives} & \text{Values at x} =5 \\\hline
f(x) = \dfrac{1}{x} & \dfrac{1}{5} \\\hline
f'(x) = -\dfrac{1}{x^2} & -\dfrac{1}{25} \\\hline
f''(x)=\dfrac{2}{x^3} & \dfrac{2}{125} \\\hline
f^{(3)}(x) = -\dfrac{6}{x^4} & -\dfrac{6}{625} \\\hline
f^{(4)}(x) = \dfrac{24}{x^5} & \dfrac{24}{3125} \\\hline
\end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= \frac{1}{5}-\frac{(x-5)}{25}-\frac{(x-5)^2}{125}-\frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$\displaystyle \left|R_n(x)\right|=\left|\dfrac{f^{(n+1)}(\xi)}{(n+1)!}\right| \cdot \left|(x-a)^{(n+1)}\right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=\left|\dfrac{f^{(4)}(\xi)}{(4)!}\right| \cdot \left|(x-5)^{4}\right| = \left|\dfrac{1}{\xi^5}\right| \cdot \left|(x-5)^{4}\right|$$
On the interval $4 \le \xi \le 6$, the maximum of $\left|\dfrac{1}{\xi^5}\right|$ is at $\xi = 4$, hence, the maximum error is given by
$$|R_3(x)| = \left|\dfrac{f^{(4)}(\xi)}{(4)!}\right| \cdot \left|(x-5)^{4}\right| = \dfrac{1}{24576} \cdot \max \left|(x-5)^{4}\right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
|
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|
Finding the Laurent series (complex numbers) I have
$$
f(z)={\frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=\infty$.
I did
$$
{\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
|
We have
$$
\eqalign{
& {1 \over {z\left( {1 - z} \right)}} = \cr
& = \left\{ \matrix{
- \left( {{1 \over z} + {1 \over {\left( {1 - z} \right)}}} \right)\quad
\Rightarrow \quad - {1 \over z} - \sum\limits_{0\, \le \,n} {z^{\,n} } \quad \left| {\,z \to 0} \right. \hfill \cr
{1 \over {\left( {z - 1} \right)}} - {1 \over {\left( {1 + \left( {z - 1} \right)} \right)}}\quad \Rightarrow \quad {1 \over {\left( {z - 1} \right)}}
- \sum\limits_{0\, \le \,n} {\left( { - 1} \right)^{\,n} \left( {z - 1} \right)^{\,n} } \quad \left| {\,z \to 1} \right. \hfill \cr
- \left( {{1 \over z}} \right)\left( {1 - {1 \over {\left( {1 - {1 \over z}} \right)}}} \right)\quad
\Rightarrow \quad \sum\limits_{0\, \le \,n} {\left( {{1 \over z}} \right)^{\,n + 2} } \quad \left| {\,z \to \infty } \right. \hfill \cr} \right. \cr}
$$
|
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|
Integral$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$ $$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$$
I tried using the substitution $t^3=\tan x$. Which gives me
$$\int\frac{3t^3}{(t+1)(t^6+1)}$$
How should I proceed?
|
As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition
Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$\frac{3t^3}{(t+1)(t^6+1)}=-\frac{3}{2 (t+1)}-\frac{t+1}{2 \left(t^2+1\right)}+\frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.
|
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|
How to show $7^{th}$ degree polynomial is non-positive in $[0,1]$
Let $0\le x\le 1$, show that inequality
$$99x^7-381x^6+225x^5-415x^4+157x^3-3x^2-x-1\le 0$$
This problem comes from the fact that I solved a different inequality.I tried to solve it by factorizing it to see if I could get symbols.but I failed.
this inequality is hold by wolfapha Test it.
|
Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)\geq$$
$$\geq370x^4+3x^2+x+1-157x^3=6\cdot\frac{185}{3}x^4+3x^2+x+1-157x^3\ge$$
$$\geq9\sqrt[9]{\left(\frac{185}{3}x^4\right)^6\cdot3x^2\cdot x\cdot1}-157x^3=\left(9\sqrt[9]{\left(\frac{185}{3}\right)^6\cdot3}-157\right)x^3\geq0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong
|
The square of any prime $p\equiv1\pmod3$ appears as a factor of $x^2+x+1$ for some choice of $x$.
This is seen as follows.
The multiplicative group $\Bbb{Z}_{p^2}^*$ of coprime residue classes modulo $p^2$ is known to be cyclic of order $p(p-1)$. It follows that there is an element of order three in that group. Let the residue class of $x$ be one such. Because $p>3$ the order of $x$ modulo $p$ is also equal to three. Implying that $x-1$ is not a multiple of $p$. But $$x^3-1=(x-1)(x^2+x+1)\equiv1-1=0\pmod{p^2}$$
by construction,
so we can conclude that $$x^2+x+1\equiv0\pmod{p^2}.$$
A non-deterministic way of finding such an $x$ is to take a random integer $a$, and calculate the remainder $x$ of $a^{p(p-1)/3}$ modulo $p^2$. If the result is $x\neq1$, then we have found the required element of order three.
For example, with $p=31$, $a=3$ we find that
$$
3^{31(31-1)/3}=3^{310}\equiv521\pmod{31^2}.
$$
And with $x=521$ we get
$$
521^2+521+1=271963=31^2\cdot283
$$
as promised.
On the other hand no prime $p\equiv-1\pmod3$ will appear as a factor of $x^2+x+1$ for any integer $x>1$. This is because the factorization $(x^3-1)=(x-1)(x^2+x+1)\equiv0\pmod p$ implies that $x$ has order $1$ or $3$ in the group $\Bbb{Z}_p^*$. In the former case $x\equiv1\pmod3$ and therefore
$x^2+x+1\equiv1+1+1=3\not\equiv0\pmod p$. In the latter case Lagrange's theorem from elementary group theory tells us that $3$ must be a factor of the order of the group $G=\Bbb{Z}_p^*$. As $|G|=p-1$ we can conclude that $p\equiv1\pmod3$.
By more or less the same argument we can show that for all $p\equiv1\pmod3$ the number $x^2+x+1$ can be made divisible by any power $p^k$. This time the group has order $p^{k-1}(p-1)$. As an example consider $p=7,k=5$. The above method produces $$3^{7^4(7-1)/3}\equiv1353\pmod{7^5}.$$ And, predictably,
$$1353^2+1353+1=7^5\cdot109.$$
|
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|
How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $\frac{x}{x + y} = 5$ then what is $\frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 \land x\neq y$ then what is $\frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $\sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
|
Hint:
1: $x - \dfrac{1}{x} = 3 \implies x^2 + \dfrac{1}{x^2} - 2 = 9$
Now use, $\big(x + \dfrac{1}{x}\big)^2 = x^2 + \dfrac{1}{x^2} + 2$
2: $\dfrac{x}{x+y} = \dfrac{1}{1 + \dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 \implies \dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression
|
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|
Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers
\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}{z}\right)\\
\frac{13}{z}=xyz\left(\frac{1}{y}+\frac{1}{x}\right)\end{cases}
Let $p=xyz.$ Then we have:
\begin{cases}\frac{1}{x}-\frac{p}{y}-\frac{p}{z}=0\\
\frac{p}{x}-\frac{8}{y}+\frac{p}{z}=0\\
\frac{p}{x}+\frac{p}{y}-\frac{13}{z}=0\end{cases}
Then we get
$$\frac{p+1}{x}=\frac{p+8}{y}=\frac{p+13}{z}$$
Any clue here?
|
Alternatively, denote $y=ax, z=abx$. Then:
$$\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases} \Rightarrow \begin{cases} ax^3(1+b)=1 \\ a^2x^3(1+ab)=8 \\a^2b^2x^3(1+a)=13 \end{cases}.$$
Divide $(2)$ by $(1)$ and $(3)$ by $(2)$:
$$\begin{cases} a(ab+1)=8(1+b) \Rightarrow b=\frac{8-a}{a^2-8} \\ 8b^2(1+a)=13(1+ab) \Rightarrow 8\cdot \frac{(8-a)^2}{(a^2-8)^2}\cdot (1+a)=13(1+a\cdot \frac{8-a}{a^2-8}) \Rightarrow \end{cases}\\
8(8-a)^2(1+a)=13\cdot 8\cdot (a-1)(a^2-8) \Rightarrow \\
(3a-10)(2a^2+7a-2)=0.$$
You can find $a$, then $b$, then $x$, $y$ and $z$.
|
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|
Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$ Any idea on how to solve the following definite integral?
$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$
I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.
|
\begin{equation}
I = \int_0^1 \frac{\ln\left(x^2 + 1\right)}{x + 1}\:dx
\end{equation}
Here let:
\begin{equation}
I(t) = \int_0^1 \frac{\ln\left(tx^2 + 1\right)}{x + 1}\:dx
\end{equation}
We observe that $I = I(1)$ and $I(0) = 0$. Thus,
\begin{align}
I'(t) &= \int_0^1 \frac{x^2}{\left(tx^2 + 1\right)\left(x + 1\right)}\:dx = \frac{1}{t + 1}\int_0^1 \left[\frac{1}{x + 1} + \frac{x}{tx^2 + 1} - \frac{1}{tx^2 + 1} \right]\:dx \\
&= \frac{1}{t + 1}\bigg[\ln(x + 1) + \frac{1}{2t}\ln\left(tx^2 + 1 \right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}x\right) \bigg]_0^1 \\
&=\frac{1}{t + 1} \bigg[\ln(2) + \frac{1}{2t}\ln\left(t + 1\right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}\right)\bigg]
\end{align}
Thus,
\begin{align}
I(t) &= \int\frac{1}{t + 1} \bigg[\ln(2) + \frac{1}{2t}\ln\left(t + 1\right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}\right)\bigg]\:dt \\
&= \int\frac{\ln(2)}{t + 1}\:dt + \int \frac{1}{2t\left(t + 1\right)}\ln\left(t + 1\right)\:dt + \int \frac{1}{\sqrt{t}\left(t + 1\right)}\arctan\left(\sqrt{t}\right)\:dt \\
&= I_1 + I_2 + I_3
\end{align}
Solve each individually:
For $I_1$:
\begin{equation}
I_1 = \int\frac{\ln(2)}{t + 1}\:dt = \ln(2)\ln(t + 1)
\end{equation}
For $I_2$:
\begin{align}
I_2 &= \int\frac{1}{2t\left(t + 1\right)}\ln\left(t + 1\right)\:dt = \frac{1}{2}\left[ \int \left(\frac{1}{t} - \frac{1}{t + 1} \right)\ln(t + 1)\:dt\right]\\
& =\frac{1}{2} \left[\int \frac{\ln(t + 1)}{t}\:dt - \int \frac{\ln(t + 1)}{t + 1}\:dt \right] \\
&= \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right]
\end{align}
Where $\operatorname{Li}_{2}(t)$ is Dilogarithm function.
For $I_3$:
\begin{equation}
\int \frac{1}{\sqrt{t}\left(t + 1\right)}\arctan\left(\sqrt{t}\right)\:dt = \arctan^2\left(\sqrt{t}\right)
\end{equation}
Combining we arrive at:
\begin{equation}
I(t) = \ln(2)\ln(t + 1) + \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] + \arctan^2\left(\sqrt{t}\right) + C
\end{equation}
Where $C$ is the constant of integration.
Using $I(0) = 0$ we see that $C = 0$. Thus
\begin{equation}
I(t) = \ln(2)\ln(t + 1) + \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] + \arctan^2\left(\sqrt{t}\right)
\end{equation}
And finally:
\begin{align}
I = I(1) &= \ln(2)\ln(1 + 1) + \frac{1}{2} \left[-\operatorname{Li}_{2}(-1) - \frac{1}{2}\ln^2(1 + 1) \right] + \arctan^2\left(\sqrt{1}\right) \\
&=\ln^2(2) + \frac{1}{2} \left[-\frac{\pi^2}{12} - \frac{1}{2}\ln^2(2)\right] + \frac{\pi^2}{16} \\
&= \frac{3}{4}\ln^2(2) - \frac{\pi^2}{48}
\end{align}
|
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"url": "https://math.stackexchange.com/questions/3054362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
minimum value of $I(l)$ in definite integration If $\displaystyle I(l)=\int^{\infty}_{0}\frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$
Then value of $l$ for which $I(l)$ is minimum
What i tried
Put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$\displaystyle I(l)=\int^{\infty}_{0}\frac{t^{4-l}}{2t^6+4t^5+3t^4+5t^3+3t^2+4t+2}dt=\int^{\infty}_{0}\frac{x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$
$\displaystyle 2I(l)=\int^{\infty}_{0}\frac{x^l+x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$
How do i solve it.Help me
Thanks in advance
|
Denote $P(x)$ the polynomial of denominator, $$I'(l)=\int_0^\infty\frac{x^l\ln x}{P(x)}dx\\
=\int_0^1\frac{x^l\ln x}{P(x)}dx+\int_1^\infty\frac{x^l\ln x}{P(x)}dx\\
=\int_0^1\frac{x^l\ln x}{P(x)}dx-\int_0^1\frac{x^{4-l}\ln x}{P(x)}dx\text{ (Sub $x\mapsto 1/x$)}\\
=\int_0^1\frac{(x^l-x^{4-l})\ln x}{P(x)}dx$$
$x^l-x^{4-l}>0$ if $0<l<2$,
$x^l-x^{4-l}<0$ if $2<l<5$,
$x^l-x^{4-l}=0$ if $l=2$.
Notice $\frac{\ln x}{P(x)}<0$ if $0<x<1$, can you continue?
Can you prove it is valid to differentiate under the integral sign?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $|2a + 3b + 2 \sqrt{3}(a \times b)|$ where $a,b$ are perpendicular unit vectors. Find $|2a + 3b + 2 \sqrt{3}(a \times b)|$ where &a,b$ are perpendicular unit vectors.
My attempt$:$
$(|2a+3b+2\sqrt(3)(a \times b)|)^{2}$ = $ 4a^{2} + 9b^{2} + 12 + 8\sqrt{3} a.\hat{n} +12\sqrt{3}b.\hat{n}$
$( a.b = 0, |a| = |b| = 1), \hat{n}$ is normal vector to $a$ and $b$.
$4+9+12+8\sqrt{3} (±1) + 12\sqrt{3}(±1)$
$25 ± 8\sqrt{3} ± 12\sqrt{3}|$
I have to eliminate one case $ 25 - 20\sqrt{3}$ and then my answer should be $(25+20(3)^{1/2})^{1/2}$ or $(25+ 4(3)^{1/2})^{1/2}$ or $(25-4(3)^{1/2})^{1/2}$
Is this right way to solve the problem. I am given only one place to anwer and I have found three answers.
|
$\{a,b,a\times b\}$ is an orthogonal set in $\mathbb{R}^3$ so the Pythagorean theorem gives:
$$|2a + 3b + 2\sqrt{3}(a\times b)|^2 = |2a|^2 + |3b|^2 + |2\sqrt{3}(a\times b)|^2 = 4 + 9 + 12 = 25$$
Hence $|2a + 3b + 2\sqrt{3}(a\times b)| = 5$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $a_n \in [0,2)$
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$
Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0) < 2$.
Inductive step:
Suppose that $0 \leq a_n < 2$
So $$\begin {split} 0 \leq a_n < 2 &\iff 0 \geq -a_n > -2 \\ &\iff 6 \geq 6-a_n > 6-2 \\ &\iff \frac{1}{6} \geq \frac{1}{6-a_n} > \frac{1}{4} \\ &\iff \frac{a_n}{6}+1 \geq \frac{6+a_n}{6-a_n} > \frac{3}{2} + \frac{a_n}{4} \end{split}$$
To be fair I have no idea if this is going somewhere.
|
The function $f(x) = \frac{6+x}{6-x}$ is strictly increasing, which you can check by taking the derivative:
$$f'(x) = \frac{(6-x) + (6+x) }{(6-x)^2} = \frac{12}{(6-x)^2} > 0$$
Hence if you assume $0 \le a_n < 2$, you get
$$a_{n+1} = \frac{6+a_n}{6-a_n} \ge \frac{6+0}{6-0} = 1 \ge 0$$
$$a_{n+1} = \frac{6+a_n}{6-a_n} < \frac{6+2}{6-2} = 2$$
Therefore $0 \le a_{n+1} < 2$ as well.
|
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|
Evaluate $ \lim\limits_{n \to \infty}\sum\limits_{k=2}^{n} \frac{1}{\sqrt[k]{n^k+n+1}+1} $ $$ \lim_{n \to \infty}\sum_{k=2}^{n} \frac{1}{\sqrt[k]{n^k+n+1}+1} $$
I expect the squeeze theorem helps us solving this but I can't find the inequality.
The result should be $1$.
|
Consider
$s(n)
=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}
$
where
$f(n) \ge 0$ and
$f(n)/n^{c} \to 0$
for some $c > 0$.
$\begin{array}\\
s(n)
&=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}\\
&\lt\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k}}\\
&=\sum_{k=2}^{n} \dfrac{1}{n}\\
&\to \ln(n)-1+\gamma\\
\end{array}
$
Similarly,
since
$(1+x)^k \ge 1+kx,
(1+x/k)^k \ge 1+x$
so
$(1+x)^{1/k} \le 1+x/k$,
$\begin{array}\\
s(n)
&=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}\\
&=\sum_{k=2}^{n} \dfrac{1}{n\sqrt[k]{1+f(n)/n^k}+1}\\
&=\sum_{k=2}^{n} \dfrac1{n}\dfrac{1}{\sqrt[k]{1+f(n)/n^k}+1/n}\\
&\ge\sum_{k=2}^{n} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}\\
\end{array}
$
Since $f(n)/n^c \to 0$
for some $c > 0$,
$f(n) < an^c$
for some $a > 0$
so
$\begin{array}\\
f(n)/(kn^k)
&\lt an^c/(kn^k)\\
&=a/(kn^{k-c})\\
&\lt a/(kn)
\qquad\text{for } k \ge c+1\\
&\lt 1/n
\qquad\text{for } k > a\\
\end{array}
$
Therefore,
letting
$p(n) = \max(c+1, a)$,
$f(n)/(kn^k) < 1/n$
for $k > p(n)$.
Therefore
$\begin{array}\\
s(n)
&\ge\sum_{k=2}^{n} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}\\
&\ge\sum_{k=2}^{p(n)} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}+\sum_{k=p(n)}^{n} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}\\
&\gt\sum_{k=p(n)}^{n} \dfrac1{n}\dfrac{1}{1+2/n}\\
&\gt\sum_{k=p(n)}^{n} \dfrac1{n}\dfrac{1}{2}\\
&\gt \frac12\ln(n/p(n))\\
&\to \infty
\qquad\text{since }n/p(n) \to \infty\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $B^*$, the dual basis Find a basis $B$ for
$$V = \left\{ \left[
\begin{array}{cc}
x\\
y\\
z
\end{array}
\right] \in \mathbb{R}^3 \vert x+y+z = 0\right\}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.
For example, if $V=\mathbb{R}^2$ and $ B = $$\left\{ \left[
\begin{array}{cc}
3\\
4
\end{array}
\right], \left[
\begin{array}{c}
5\\
7
\end{array}
\right]\right\} $, then $ A = $$ \left[
\begin{array}{cc}
3&5\\
4&7
\end{array}
\right]$, then $A^{-1} = $$ \left[
\begin{array}{cc}
7&-5\\
-4&3
\end{array}
\right]$, and the dual basis $B^* = (l_1, l_2)$ when :
$ l_1= \left( \left[
\begin{array}{cc}
x\\
y\\
\end{array}
\right]\right) = 7x_1 - 5x_2 $
$ l_2= \left(\left[
\begin{array}{cc}
x\\
y\\
\end{array}
\right]\right) = -4x_1 + 3x_2 $
However, when finding a basis for $V$, I get to the following solution vector
$$\left[\begin{array}{cc}
-y-z\\
y\\
z
\end{array}
\right] = y \left[\begin{array}{cc}
-1\\
1\\
0
\end{array}
\right] + z \left[\begin{array}{cc}
-1\\
0\\
1
\end{array}
\right]$$
If I build a matrix out of the basis$ \left\{ \left[\begin{array}{cc}
-1\\
1\\
0
\end{array}
\right] , \left[\begin{array}{cc}
-1\\
0\\
1
\end{array}
\right] \right\}$ I will have a matrix that is $3\times2$ and I cannot inverse this.
How to proceed from here?
|
Be
\begin{equation}
V = \left\lbrace \
\left[\begin{array}{c}
x \\
y \\
z \\
\end{array}\right] \in \mathbb{R}^3 : x+y+z = 0 \right\rbrace
\end{equation}
Then $z=-x-y$, thus basis $B$ is:
\begin{equation}
B = \left\lbrace\
\left[\begin{array}{c}
1 \\
0 \\
-1 \\
\end{array}\right]
,
\left[\begin{array}{c}
0 \\
1 \\
-1 \\
\end{array}\right] \ \right\rbrace
\end{equation}
Where $B=\{v_1,v_2\}$. The dual basis is given by $f_i(v_j)=\delta_{ij}$ where $f_i\in B^{*}$, $v_j\in B$ and $\delta_{ij}$ is the Kronecker delta since:
\begin{equation}
\delta_{ij}=\begin{cases}
1 \quad\textrm{ if } i=j \\
0 \quad\textrm{ if } i\neq j \\
\end{cases}
\end{equation}
Then for $f_1$:
\begin{eqnarray}
f_1\left(\
\left[\begin{array}{c}
x \\
y \\
z \\
\end{array}\right]\
\right)
=
f_1\left(\
\left[\begin{array}{c}
x \\
y \\
-x-y \\
\end{array}\right]\
\right)
&=&
xf_1\left(\
\left[\begin{array}{c}
1 \\
0 \\
-1 \\
\end{array}\right]\
\right)
+
yf_1\left(\
\left[\begin{array}{c}
0 \\
1 \\
-1 \\
\end{array}\right]\
\right) \\
f_1\left(\
\left[\begin{array}{c}
x \\
y \\
-x-y \\
\end{array}\right]\
\right)
&=&
x
\end{eqnarray}
And $f_2$:
\begin{eqnarray}
f_2\left(\
\left[\begin{array}{c}
x \\
y \\
z \\
\end{array}\right]\
\right)
=
f_2\left(\
\left[\begin{array}{c}
x \\
y \\
-x-y \\
\end{array}\right]\
\right)
&=&
xf_2\left(\
\left[\begin{array}{c}
1 \\
0 \\
-1 \\
\end{array}\right]\
\right)
+
yf_2\left(\
\left[\begin{array}{c}
0 \\
1 \\
-1 \\
\end{array}\right]\
\right) \\
f_2\left(\
\left[\begin{array}{c}
x \\
y \\
-x-y \\
\end{array}\right]\
\right)
&=&
y
\end{eqnarray}
Then $B^{*}=\{f_1,f_2\}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $x^4 + 2x^2 - x + 1$ irreducible in $\mathbb Z_7[x]$? How would one do this?
I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2.
How would I prove that it is or it isn't?
There are a lot of polynomials with this conditions.
|
The product of all the monic, irreducible polynomials over $\mathbb{F}_7$ wih degree $\leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $\mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1\pmod{f(x)}$, for instance through the Brauer chain $x^4\to x^8\to x^{16}\to x^{32}\to x^{48}$.
$$ x^4 \equiv -2x^2+x-1\pmod{f(x)} $$
$$ x^8 \equiv 3x^3-3x^2+2x-3\pmod{f(x)} $$
$$ x^{16} \equiv x^3+3x^2+2x-1\pmod{f(x)} $$
$$ x^{32} \equiv -x^3+2x^2+x-3\pmod{f(x)} $$
$$ x^{48}-1 \equiv -x^3-2x^2+x+2 \pmod{f(x)}$$
give
$$\gcd(x^{48}-1,x^4+2x^2-x+1)=\gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$
$$ = \gcd(x^3+2x^2-x-2,-x-3) $$
but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7\nmid 90$ and we have finished.
|
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|
$\ X_i $ is discrete random variable, Compute $\ \sum X_i = 97 $ Let $\ X_1, X_2, , \dots , X_{10} $ be a discrete random variable with uniform distribution between $\ 0 $ to $\ 10 $. Compute $\ P\{ \sum_{i=1}^{10} \ X_i = 97 \} $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?
$$\ P\{ \sum_{i=1}^{10} X_i = 97 \} = {10 \choose 1 } \cdot \frac{1}{11}^{10} + {10 \choose 2} \cdot \frac{1}{11}^{10} + {10 \choose 3} \cdot \frac{1}{11}^{10} $$
|
Your solution is almost correct.
The correct solution is:$$11^{-10}\left(\frac{10!}{9!1!}+\frac{10!}{8!1!1!}+\frac{10!}{7!3!}\right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $\binom{10}{2}=\frac{10!}{8!2!}$ must be replaced by $\frac{10!}{8!1!1!}$.
|
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|
Sum of series $\sqrt{1+\frac1{n^2}+\frac1{(n+1){}^2}}$ How to solve this sum?$$\sqrt{1+\frac1{1^2}+\frac1{2^2}}+\sqrt{1+\frac1{2^2}+\frac1{3^2}}+\cdots+\sqrt{1+\frac1{19^2}+\frac1{20^2}}$$
I assumed it to be a sum of $\sqrt{1+\dfrac{1}{n^2} + \dfrac{1}{(n+1)^2}}$ but It complicates the powers and I am unable to factorise further.
|
$$1+\dfrac1{n^2}+\dfrac1{(n+1)^2}=\dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=\dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$
$n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2\cdot1+2n\cdot1+(n)^2+1^2+2n^2\cdot n=(n^2+n+1)^2$
Now $$\dfrac{n^2+n+1}{n(n+1)}=1+\dfrac1{n(n+1)}=1+\dfrac{n+1-n}{n(n+1)}=?$$
See also: Telescoping series
|
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|
Find all $p \in \mathbb{Z}$ such that $ p^2+ 4p + 16 $ is a perfect square I've tried some things involving modular residues but they don't seem to work. Anyone know how to do this?
|
So $p^2 + 4p + 16 = m^2$
$p^2 + 4p + 4 =m^2 - 12$
$(p+2)^2 = m^2 - 12$
$m^2 - (p+2)^2 = 12$
$(m + p + 2)(m - p - 2) = 12$
So $m+p+2 = k$ and $m-p-2 = j$ for two integers so that $kj = 12$.
Also note, $k + j = 2m$.
So $k + j$ is even. So $\{k,j\} = \{\pm 2, \pm 6\}$ and .... thats it. All other factors involve an odd number and an even number and will have an odd sum.
So $2m = \pm2 \pm 6 =8$ and $m =\pm 4$ and $p^2 +4p + 16 = 16$ so $p(p+4) = 0$ so $p= 0$ or $p = -4$.
===
Another alternative. Note that $(n+1)^2 - n^2 = n^2 +2n +1 - n^2 =2n+1$ so that means $n^2 = (n-1)^2 + (2n-1)= (n-2)^2 + (2n-1) + (2n-3) = ...... = (n-n)^2 + (2n-1) + (2n-3) + ..... + 3 + 1= 1 + 3 + 5 + ..... + 2n-1$.
In other words $n^2 = $ the sum of the first $n$ odd numbers.
So if $p^2 + 4p + 16 = m^2$ then $(p-2)^2 - m^2 = 12$. But $(p-2)^2$ is the sum of the first $p-2$ odd numbers. And $m^2$ is the sum of the first $m$ odd numbers. So $(p-2)^2 - m^2$ is the sum of the $m + 1$st odd number to the $p-2$ odd number.
In other words: $12$ is a sum of a sequence of consecutive odd numbers.
The first such way of doing that is $5 + 7 = 12$. The next such way would need $4$ terms averaging $3$ but the lowest possible value is $1$ and $1 + 3+5 + 7$ is too big.
So $5 +7 = 12$ is the only option. So $(p-2)^2 = 1+3=4$ and $m^2 = 1+3 + 5+7 = 16$ and $(p-2)^2 - m^2 = 12$ and so $p-2 = \pm 2$ and $m =\pm 4$.
So $p =0$ or $4$ are the only solutions.
|
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|
Prove that $ \int\limits_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $ Is this conjecture true?
Conjecture:
$$
\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8}
$$
I found it myself based on numerical evidence. Need help in analytical proof. Thanks.
|
It is easier than I exspected in the first place. However, first of all lets denote your integral as
$$\mathfrak I~=~\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x$$
We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}\frac{1-x}{1+x}$ thus write $\mathfrak I$ as
$$\begin{align*}
\mathfrak I=\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x&=\int_0^\infty \frac1{1+\left(e^{2x}\frac{1-x}{1+x}\right)^2}\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}
\end{align*}$$
Now we can try the straightforward substitution $z=e^{2x}\frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to
$$\begin{align*}
\mathfrak I &=\int_0^\infty \frac1{1+\left(e^{2x}\frac{1-x}{1+x}\right)^2}\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}\\
&=\int_1^{-\infty} \frac1{1+z^2}\frac{-\mathrm dz}{2}\\
&=\frac12\int_{-\infty}^1 \frac{\mathrm d z}{1+z^2}\\
&=\frac12\left[\arctan{z}\right]_{-\infty}^1
\end{align*}$$
$$\therefore~\mathfrak I ~=~\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x~=~\frac{3\pi}8$$
Where we used the symmetry aswell as well-known values of the tangent function.
|
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|
Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
\begin{align*}
360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\
180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\
90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\
45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\
22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\
11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\
5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9
\end{align*}
As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!
|
Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
$$
n=17254,\quad
d(n)=1+7+2+5+4=19,
\quad
d(d(n))=1+9=10,
\quad
d(d(d(n)))=1+0=1=d^*(n)
$$
Note that $d(n)\le n$, equality holding if and only if $1\le n\le 9$. Also, $1\le d^*(n)\le 9$, because the process stops only when the digit sum obtained is a one-digit number.
The main point is that $n-d(n)$ is divisible by $9$: indeed, if
$$
n=a_0+a_1\cdot10+a_2\cdot10^2+\dots+a_n\cdot10^n,
$$
then
$$
n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+\dots+a_n(10^n-1)
$$
and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above)
$$
n-d^*(n)=\bigl(n-d(n)\bigr)+\bigl(d(n)-d(d(n))\bigr)+\bigl(d(d(n))-d(d(d(n)))\bigr)
$$
and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.
Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that
$d^*(n)=9$ if and only if $n$ is divisible by $9$.
Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.
When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.
What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually
\begin{align}
&45 \xrightarrow{\cdot10} 450 \xrightarrow{/2} 225 && d^*(225)=9 \\
&225 \xrightarrow{\cdot10} 2250 \xrightarrow{/2} 1225 && d^*(1225)=9
\end{align}
and so on. Note that the first step can also be stated as
$$
45 \xrightarrow{\cdot5} 225
$$
Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.
|
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"url": "https://math.stackexchange.com/questions/3064917",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "21",
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|
Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is: $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to make some changes but I don't know if they are on the right step or not:
First, I am able to simplify the left fractions numerator and the right fractions denominator:
$\sqrt{mn^3}=\sqrt{mn^2n^1}=n\sqrt{mn}$
$\sqrt{m^2c^4}=m\sqrt{c^2c^2} = mcc$
So my new expression looks like:
$$\frac{n\sqrt{mn}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{mcc}$$
From this point I'm really at a loss to my next steps. If I multiply them both together I get:
$$\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mcc)}.$$
Next, I was thinking I could multiply out the radical in the denominator but I feel like I need to simplify what I have before going forwards.
Am I on the right track? How can I simplify my fraction above in baby steps? I'm particularly confused by the negative exponents.
How can I arrive at the solution $\dfrac{\sqrt{mnc}}{a^9cmn}$?
|
Recall that $a^{-x}=\frac{1}{a^x}$, $a^{1/x}=\sqrt[x]{a}$, and $a^na^m=a^{m+n}$
Always convert everything to exponents first, then use arithmetic
$$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}}\frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}=m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4/2}$$
Rearrange
$$m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4}=(m^{1/2}\ m^{-2/2})\ (n^{3/2}\ n^{-2})\ (a^{-2}\ a^{-7})\ (c^{3/2}\ c^{-4/2})$$
Simplify
$$(m^{1/2-1})\ (n^{3/2-2})\ (a^{-2-7})\ (c^{3/2-2})=(m^{-1/2})\ (n^{-1/2})(a^{-9})(c^{-1/2})=\frac{1}{a^9\sqrt{mnc}}$$
Multiply by $\frac{\sqrt{mnc}}{\sqrt{mnc}}$
$$\frac{1}{a^9\sqrt{mnc}}\frac{\sqrt{mnc}}{\sqrt{mnc}}=\frac{\sqrt{mnc}}{a^9mnc}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Prove that $3^{2n} +7$ is divisible by 8
Prove by induction that $3^{2n} +7$ is divisible by 8 for $n \in \Bbb N$
So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?
If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$
Assume true for $n=k$ so $$ 8\vert 3^{2k} +7$$
If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A \in \Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B \in \Bbb Z$
$$3^2 \times 3^{2k}+7$$
$$3^2 \times (8A)=72A$$
$$72A =8(9A)=9B$$
So by induction $3^{2n} +7$ is divisibe by 8 $\forall n \in \Bbb N$
|
An option:
Step $n+1$:
$3^{2n+2}+7 =9 \cdot 3^{2n} +7=$
$(8+1)\cdot 3^{2n} +7=$
$(3^{2n}+7) +8;$
The first term is divisible by $8$ by hypothesis, so is the second term.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3067826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What is the probability that no cup is on a saucer of the same colour? A tea set comprises four cups and saucers in four distinct colours.
If the cups are placed at random on the saucers, what is the probability that no cup is on a saucer of the same colour?
MY ATTEMPT
As far as I understand, the given event's probability is given by
\begin{align*}
\textbf{P}(C^{c}_{1}\cap C^{c}_{2}\cap C^{c}_{3}\cap C^{c}_{4}) = 1 - \textbf{P}(C_{1}\cup C_{2}\cup C_{3}\cup C_{4})
\end{align*}
where the last expression can be calculated using the inclusion-exclusion principle based on the knowledge of the probabilities
\begin{cases}
\textbf{P}(C_{i})\quad\text{for}\quad1\leq i \leq 4\\
\textbf{P}(C_{i}\cap C_{j})\quad\text{for}\quad 1\leq i < j \leq 4\\
\textbf{P}(C_{i}\cap C_{j}\cap C_{k})\quad\text{for}\quad 1 \leq i < j < k \leq 4\\
\textbf{P}(C_{1}\cap C_{2}\cap C_{3}\cap C_{4})\\
\end{cases}
which can be easily obtained.
|
There are $4!$ ways to arrange the cups on the saucers. From these, we wish to exclude those arrangements in which a cup is placed on a saucer of its own color.
If cup $C_i$ sits on saucer $S_i$, there are $3!$ ways to arrange the remaining three cups on the remaining three saucers, so $|C_i| = 3!$, $1 \leq i \leq 4$.
If cups $C_i$ sits on saucer $S_i$ and cup $C_j$ sits on saucer $S_j$, then there are $2!$ ways to arrange the remaining two cups on the remaining two saucers, so $|C_i \cap C_j| = 2!$, $1 \leq i < j \leq 4$.
If cup $C_i$ sits on saucer $S_i$, cup $C_j$ sits on saucer $S_j$, and cup $C_k$ sits on saucer $S_k$, then there is one way to place the remaining cup on the remaining saucer, so $|C_i \cap C_j \cap C_k| = 1!$, $1 \leq i < j < k \leq 4$.
There is only one way to place each cup on its own saucer, so $|C_1 \cap C_2 \cap C_3 \cap C_4| = 0!$.
Hence, by the Inclusion-Exclusion Principle,
\begin{align*}
\Pr&(C_1^C \cap C_2^C \cap C_3^C \cap C_4^C) = 1 - \Pr(C_1 \cup C_2 \cup C_3 \cup C_4)\\
& = 1 - \Pr(C_1) - \Pr(C_2) - \Pr(C_3) - \Pr(C_4) + \Pr(C_1 \cap C_2) + \Pr(C_1 \cap C_3) + \Pr(C_1 \cap C_4) + \Pr(C_2 \cap C_3) + \Pr(C_2 \cap C_4) + \Pr(C_3 \cap C_4) - \Pr(C_1 \cap C_2 \cap C_3) - \Pr(C_1 \cap C_2 \cap C_4) - \Pr(C_1 \cap C_3 \cap C_4) - \Pr(C_2 \cap C_3 \cap C_4) + \Pr(C_1 \cap C_2 \cap C_3 \cap C_4)\\
& = 1 - \binom{4}{1}\Pr(C_1) + \binom{4}{2}\Pr(C_1 \cap C_2) - \binom{4}{3}\Pr(C_1 \cap C_2 \cap C_3) + \binom{4}{4}\Pr(C_1 \cap C_2 \cap C_3 \cap C_4)\\
& = 1 - 4 \cdot \frac{3!}{4!} + 6 \cdot \frac{2!}{4!} - 4 \cdot \frac{1!}{4!} + 1 \cdot \frac{0!}{4!}\\
& = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\\
& = \frac{24 - 24 + 12 - 4 + 1}{24}\\
& = \frac{9}{24}\\
& = \frac{3}{8}
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/3071423",
"timestamp": "2023-03-29T00:00:00",
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|
Generalization of $(a+b)^2\leq 2(a^2+b^2)$ We know that, $(a+b)^2\leq 2(a^2+b^2)$. Do we have anything similar for $$\left(\sum_{i=1}^N a_i\right)^2.$$
where $a_i\in \mathbb{R}\ \ \ \ \forall\ i\in \{1,\cdots,N\}$.
For $n=3$, we get
\begin{equation}
\begin{aligned}
(a_1+a_2+a_3)^2&\leq 2\left((a_1+a_2)^2+a_3^2\right)
\\&\leq 2\left(2(a_1^2+a_2^2)+a_3^2 \right).
\end{aligned}
\end{equation}
Do we have some sort of generalization?
|
It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)\geq(a_1+a_2+...+a_n)^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3072599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.