Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$\int_0^\pi e^{\cos x}\cos(x-\sin x) dx$ I’m not quite sure how to approach this problem using complex analysis. Mainly I’m confused int the contour to use after changing the integral by changing
$\cos(x-\sin x)$
to become
$Re(e^{i(x-\sin x})$, so that the integral became
$$Re\int_0^\pi e^{\cos x}e^{i(x-\sin x)} dx$$
$$Re\int_0^\pi e^{(\cos x-i*\sin x)} e^{ix} dx$$
And then I subbed in
$z=e^{ix}$
So then
$\frac{dz}{iz}=dx$
$$Re\int_C \frac{1}{iz} dz$$
But I’m not sure what contour to use to solve this
| \begin{align}\int_0^\pi \text{e}^{\cos x}\cos(x-\sin x) dx&=\int_0^\pi\text{e}^{\cos x}\cos x\cos(\sin x)dx+\int_0^\pi\text{e}^{\cos x}\sin x\sin(\sin x)dx\\
&=\Big[\sin(\sin x)\text{e}^{\cos x}\Big]_0^\pi+2\int_0^\pi\text{e}^{\cos x}\sin x\sin(\sin x)dx\\
&=2\int_0^\pi\text{e}^{\cos x}\sin x\sin(\sin x)dx\\
&=2\int_0^\pi\cosh(\cos x)\sin x\sin(\sin x)dx+2\int_0^\pi\sinh(\cos x)\sin x\sin(\sin x)dx\\
&=2\int_0^\pi\cosh(\cos x)\sin x\sin(\sin x)dx\\
&=4\int_0^{\frac{\pi}{2}}\cosh(\cos x)\sin x\sin(\sin x)dx\\
&=4\int_0^{\frac{\pi}{2}}\left(\left(\sum_{n=0}^\infty\frac{(-1)^n\sin^{2n+2}x}{(2n+1)!}\right)\left(\sum_{m=0}^\infty\frac{\cos^{2m} x}{(2m)!}\right)\right)\,dx\\
&=4\sum_{n=0}^\infty \left(\sum_{j=0}^n\frac{(-1) ^j}{(2j+1)!(2(n-j))!}\int_0^{\frac{\pi}{2}}\sin^{2j+2}x\cos^{2(n-j)}x\,dx\right)
\end{align}
For $p,q$ integers,
\begin{align}\int_0^{\frac{\pi}{2}}\sin^{2p}x\cos^{2q}x\,dx&=\frac{1}{2}\text{B}\left(p+\frac{1}{2},p+\frac{1}{2}\right)\\
&=\frac{1}{2}\frac{\Gamma\left(p+\frac{1}{2}\right)\Gamma\left(q+\frac{1}{2}\right)}{\Gamma\left(p+q+1\right)}\\
&=\frac{\pi}{2}\times \frac{(2p)!(2q)!}{2^{2(p+q)}(p+q)!p!q!}
\end{align}
Therefore, for $n\geq 0$, integer,
\begin{align}C_n&=\sum_{j=0}^n\frac{(-1) ^j}{(2j+1)!(2(n-j))!}\int_0^{\frac{\pi}{2}}\sin^{2j+2}x\cos^{2(n-j)}x\,dx\\
&=\frac{\pi}{2^{2(n+1)}(n+1)!}\sum_{j=0}^n\frac{(-1)^j}{j!(n-j)!}\\
&=\frac{\pi}{2^{2(n+1)}(n+1)!n!}\sum_{j=0}^n\frac{(-1)^jn!}{j!(n-j)!}\\
&=\frac{\pi}{2^{2(n+1)}(n+1)!n!}\sum_{j=0}^n (-1)^j\binom{n}{j}
\end{align}
Observe that, $C_0=\dfrac{\pi}{4}$ and, for $n>0$, integer,
\begin{align}C_n&=\frac{\pi}{2^{2(n+1)}(n+1)!n!}(-1+1)^n\\
&=0
\end{align}
Therefore,
\begin{align}\int_0^\pi \text{e}^{\cos x}\cos(x-\sin x) dx&=4\times \frac{\pi}{4}\\
&=\boxed{\pi}
\end{align}
NB:
For $n\geq 0$, integer,
\begin{align}\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n)!}{2^{2n}n!}\sqrt{\pi}\end{align}
For $x$ real,
\begin{align}\cos(\pi-x)=-\cos x\\
\sin(\pi-x)=-\sin x\\
\end{align}
$\text{B}$ is the Euler Beta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| For a start, $$ x^9 - 1 = (x-1) (x^2+x+1)(x^6 + x^3 + 1)$$
This leads us to
$$ a^9 - b^9 = (a-b)(a^2 + ab + b^2)(a^6 + a^3 b^3 + b^6) $$
Now take $a,b$ to be the two cube roots in your fraction
After you cancel the $(a-b)$ factors, you can simply set $a,b$ equal, so the denominator becomes $3 b^2 3 b^6 = 9 b^8$ with $b = 2^{1/3}.$ Now, to get all the way to Mohammad's answer, note $b^8 = b^6 b^2 = 4 b^2,$ while $b^2 = 2^{2/3}= 4^{1/3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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Eigenvalues of a Hermitian tridiagonal matrix. I want to know how to calculate the eigenvalues of the following Hermitian tridiagonal $(N+1)\times(N+1)$ matrix,
$$
A=\begin{pmatrix}
N+1&i\sqrt{N}\\
-i\sqrt{N}&N+1&i\sqrt{2(N-1)}\\
&-i\sqrt{2(N-1)}&\ddots&i\sqrt{3(N-2)}\\
&&-i\sqrt{3(N-2)}&\ddots&\ddots\\
&&&\ddots&N+1&i\sqrt{N}\\
&&&&-i\sqrt{N}&N+1
\end{pmatrix}
$$
that is $a_{kk} = N+1$, $a_{k,k+1} = i\sqrt{k(N-k+1)}$.
From another method to treat the problem (it is from physics), I get the eigenvalues are
$$
\lambda = 1,3,5,\ldots,2N+1.
$$
Any help is appreciated!
My answer: I find $A-(N+1)I$ is the same as the matrix of $J_y$ with using eigenstates of $J_z$ as basis ($J_y,J_z$ are angular momentum operators in quantum mechanics). Then use the result of representations of $\mathfrak{su}_2$, we can get the eigenvalues of $A-(N+1)I$ are $-N,-N+2,\dots,N-2,N$.
Really thanks for help!!!
| (For convenience, all matrices below are zero-indexed.)
Let $B=A-(N+1)I$. Then $B$ is a tridiagonal complex skew-symmetric matrix. When $0\le k<N$, we have
\begin{align}
b_{k,k+1}&=i\sqrt{(k+1)(N-k)},\\
b_{k+1,k}&=-b_{k,k+1}.
\end{align}
(Note that the above formula is different from the one in the OP because our matrix is now zero-indexed.) Let $D=\operatorname{diag}(d_0,d_1,\ldots,d_N)$ where
$$
d_k=\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}.
$$
The product of super-diagonal entries of $B$ in the above is considered empty when $k=0$, so that $d_0=1$ by convention. Then $C=DBD^{-1}$ is a tridiagonal matrix such that
\begin{aligned}
c_{k,k+1}
=\frac{d_k}{d_{k+1}}b_{k,k+1}
&=\frac{\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}}
{\frac{1}{(k+1)!}\prod_{i=0}^kb_{i,i+1}}
b_{k,k+1}
=k+1,\\
c_{k+1,k}
=\frac{d_{k+1}}{d_k}b_{k+1,k}
&=\frac{\frac{1}{(k+1)!}\prod_{i=0}^kb_{i,i+1}}
{\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}}
b_{k+1,k}\\
&=\frac{\frac{1}{(k+1)!}\prod_{i=0}^kb_{i,i+1}}
{\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}}
(-b_{k,k+1})\\
&=\frac{-b_{k,k+1}^2}{k+1}=N-k.
\end{aligned}
In other words, $C$ is the Kac matrix
$$
C=\begin{pmatrix}
0&1\\
N&0&2\\
&N-1&\ddots&3\\
&&N-2&\ddots&\ddots\\
&&&\ddots&0&N\\
&&&&1&0
\end{pmatrix}.
$$
The eigenvalue problem for Kac matrix was solved in this answer. The spectrum of $C$ is $S=\{-N,\,-(N-2),\,\ldots,\,N-2,\,N\}$. As $B$ is similar to $C$ and $A=B+(N+1)I$, the eigenvalues fo $A$ are given by $N+1+S=\{1,3,5,\ldots,2N+1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solve : $2\sin^3 (x) = \cos (x)$
Solve : $2\sin^3 (x) = \cos (x)$
How to solve the above equation?
I tried, but failed to succeed.
Graph of the following equation
But how to solve the equation without graph?
| Here's how I solved it by hand:
Notice that if we square both sides, we get $$4 \sin^6 x = \cos^2 x$$
Using the Pythagorean identity,
$$4 \sin^6 x + \sin^2 x - 1 = 0$$
Let $y = \sin^2 x$. Our equation becomes $$4y^3 + y - 1 = 0$$ Normally, we would have to use the cubic equation formula, but according to the Rational Root Theorem, we can brute force our way through the eligible rational roots: $\pm 1$,$\pm \frac{1}{2}$, and $\pm \frac{1}{4}$. We notice that $\frac{1}{2}$ is indeed a solution. Dividing out, we see that the remaining roots are roots of $2y^2 + y + 1$, which are complex. Thus, all we have to do is find the numbers for which $$\sin^2 x = \frac{1}{2}$$ You should be able to get that $\pm \frac{\pi}{4}$ and $\pm \frac{5\pi}{4}$ are solutions. We see $-\frac{\pi}{4}$ and $-\frac{5\pi}{4}$ are extraneous solutions. Thus, our solution set is all $\frac{\pi}{4} + k\pi, k\in \mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Computing $\sum \frac{1}{n^2(n^2+a^2)}$ using the residue theorem I have to find a closed form for $\sum_{n=1}^{\infty}\frac{1}{n^2(n^2+a^2)}$ (with $a\in \mathbb{R}$).
My idea was to use the residue theorem on $f(z)=\frac{\pi \cot(\pi z)}{z^2(z^2+a^2)}$, thus obtaining:
\begin{equation}
\text{Res}_{0}(f)+\text{Res}_{\pm ia}(f)+2\sum_{n=1}^{\infty}\frac{1}{n^2(n^2+a^2)}
=
\lim_{n\to \infty}\frac{1}{2\pi i}\oint_{c_n}f(z)dz
\end{equation}
($c_n$ is a circle with $r=n+\frac{1}{2}$, in order to avoid poles on the contour.)
Going on:
\begin{equation}
2\sum_{n=1}^{\infty}\frac{1}{n^2(n^2+a^2)}=\frac{\pi^2}{3a^2}-\frac{\pi\coth(\pi a)}{a^3}+\lim_{n\to \infty}\frac{1}{2\pi i}\oint_{c_n}f(z)dz
\end{equation}
At this point, I was quite sure that the integral was $0$, but this does not match the closed form I know to be true.
My questions are:
*
*Is my approach correct? (maybe using
Poisson summation formula, even if it seems to me quite long to deal with the singularity of the series in $n=0$)
*What kind of tecniques should I apply to compute the integral?
Note:
As Denklo noted, we can calculate the series using $\frac{1}{n^2}-\frac{1}{n^2+a^2}=\frac{a^2}{n^2(n^2+a^2)}$. The closed form is thus $\frac{1}{2a^4}-\frac{\pi \coth(\pi a)}{2a^3}+\frac{\pi^2}{6a^2}$, which is consistent with my results so far and should imply $\frac{1}{2\pi i}\oint f=\frac{1}{a^4}$
| We first decompose the series in two parts:
$\sum_{n=1}^{\infty} \dfrac{1}{n^{2}(n^{2}+a^{2})} = \dfrac{1}{a^{2}} \sum_{n=1}^{\infty} \dfrac{1}{n^{2}} - \dfrac{1}{a^{2}} \sum_{n=1}^{\infty} \dfrac{1}{n^{2} + a^{2}}$.
For the first series we use $\sum_{n=1}^{\infty} \dfrac{1}{n^{2}} = \dfrac{\pi^{2}}{6}$.
For the second series we use the identity $\sum_{n = -\infty}^{\infty} \dfrac{1}{n^{2} + a^{2}} = \dfrac{\pi}{a} coth(\pi a)$ in order to get $\sum_{n = 1}^{\infty} \dfrac{1}{n^{2} + a^{2}} = \dfrac{\pi}{2a} coth(\pi a) - \dfrac{1}{2a^{2}}$.
The final result is given by:
$\sum_{n=1}^{\infty} \dfrac{1}{n^{2}(n^{2}+a^{2})} = \dfrac{\pi^{2}}{6a^{2}} - \dfrac{\pi}{2a^{3}} coth(\pi a) - \dfrac{1}{2a^{4}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $a,b,c \in \mathbb{R^+}$ such that $a^2+b^2+c^2+abc=4$ and $a+b+c=3$
Find $a,b,c \in \mathbb{R^+}$ such that $$a^2+b^2+c^2+abc=4\;\;\;{\rm and}\;\;\;a+b+c=3$$
Seeing $a^2+b^2+c^2+abc=4$,I substituted $a=2\cos{A}$,$b=2\cos{B},c=\cos{C}$ with $A+B+C=180°$.
Therefore,$cosA+cosB+cosC=\frac 32$ have thousands of solution. What to do now? Please help me.
| It is $$\cos(C)+\cos(B)+\cos(A)\le \frac{3}{2}$$ and this Comes from
$$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$?
How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$
I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.
| The only real root is
$$
\alpha=-\frac{1}{\sqrt{2}}=- 0.707106781187\cdots
$$
The polynomial divided by $x-\alpha$ then has degree $4$ and has roots $\alpha\pm i\beta$, $\alpha\pm i \gamma$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the ratio of the area $EDPQ$ to the area of ABC In triangle $ABC$,points $E$ and $D$ are on side $AC$ and point $F$ is on side $BC$ such that AE=ED=DC and $BF:FC$ =2:3. $AF$ intersects $BD$ and $BE$ at points $P$ and $Q$, respectively. Find the ratio of the area $EDPQ$ to the area of $ABC$
taken from the 2017 IMC held in India
I assumed ABD was equilateral so that the area of ABE would be equal to BED but didn’t gain anything from that
| Let $k\in DC$ such that $FK||BD$.
Thus, $DK:KC=2:3.$
Let $DC=5x$.
Thus, $DK=2x$ and $AD=10x$, which gives
$$\frac{AP}{PF}=\frac{AD}{DK}=\frac{10x}{2x}=5.$$
Similarly, let $M\in EC$ such that $FM||BE.$
Thus, $EM:MC=2:3,$ which gives $EM=4x$ and
$$\frac{AQ}{QF}=\frac{AE}{EM}=\frac{5x}{4x}=\frac{5}{4}.$$
From here we obtain:
$$AQ:QP:PF=10:5:3.$$
Now, $$S_{\Delta BPQ}=\frac{5}{18}S_{\Delta ABF}=\frac{5}{18}\cdot\frac{2}{5}S_{\Delta ABC}=\frac{1}{9}S_{\Delta ABC}=\frac{1}{3}S_{\Delta BED},$$
which gives
$$S_{EDPQ}=\frac{2}{3}S_{\Delta BED}=\frac{2}{9}S_{\Delta ABC}$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$? I encountered the following hard problem in a math olympiad book:
Evaluate $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}.$$
I tried to evaluate $\sum^{k}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$ where $k=1$ to $10$ and got $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{5}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \frac{10}{5}$ respectively. How can I prove that the pattern continues?
| I will use the following Lemma (that you can prove by induction):
Lemma. For all $N\in\Bbb N:$ $$\sum_{n=1}^N n(n+1)(n+2)(n+3)=\frac15 N(N+1)(N+2)(N+3)(N+4).$$
In our case, we have $$\sum_{n=1}^{2016} n(n+1)(n+2)(n+3) = 2016\cdot2017\cdot2018\cdot2019\cdot2020\cdot\frac15.$$
So $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}=\frac{2020}5=404.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Rearranging the formula Transpose this formula to make $y$ the subject.
$$x=\sqrt{x^2y^2+1-y}$$
My try:
$$x^2=x^2y^2+1-y$$
$$x^2-x^2y^2=1-y$$
$$x^2(1-y^2)=1-y$$
Here I got 2 $y$ terms, but I am not sure what to do next.
| If you can't see the factorization, treat the thing as a quadratic in $y$:
$$
x^2y^2-y+(1-x^2)=0
$$
The discriminant is
$$
1-4x^2(1-x^2)=4x^4-4x^2+1=(2x^2-1)^2
$$
Apply the quadratic formula:
$$
y=\frac{1\pm(2x^2-1)}{2x^2}
$$
and you find either $y=1$ or
$$
y=\frac{1-2x^2+1}{2x^2}=\frac{1-x^2}{x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Regularization of $\sum_{n=2}^\infty (-1)^n \log n$ I accidentally stumbled on the following regularization of this divergent series:
$$\sum_{n=2}^\infty (-1)^n \log n "=" \frac{1}{2} \log \frac{\pi}{2}$$
I'm not familiar enough with regularization, so I wanted to ask if this result agrees with any other known regularization method?
I derived this result in the following way:
$$\log n=\int_0^\infty \frac{dx}{x} (e^{-x}-e^{-n x})$$
Now consider the function:
$$\sum_{n=2}^\infty (-1)^n (\log n) s^n=\int_0^\infty \frac{dx}{x} \left(e^{-x} \frac{s^2}{1+s}-\frac{s^2 e^{-2 x}}{1+s e^{-x}} \right)$$
$$\sum_{n=2}^\infty (-1)^n (\log n) s^n= \frac{s^2}{1+s}\int_0^\infty \frac{e^{-x}dx}{x} \frac{e^x-1}{e^x+s} $$
The right hand side converges for any $s>0$, and in particular, for $s=1$ we have:
$$\frac{1}{2}\int_0^\infty \frac{e^{-x}dx}{x} \frac{e^x-1}{e^x+1}=\frac{1}{2} \log \frac{\pi}{2}$$
I got the result with Wolfram Alpha, but I'm sure there's a proof somewhere on this site.
There's an interesting corollary here. If we write:
$$\sum_{n=2}^\infty (-1)^n (\log n) s^{n-1}= \frac{s}{1+s}\int_0^\infty \frac{e^{-x}dx}{x} \frac{e^x-1}{e^x+s} $$
And then integrate w.r.t. $s$ from $0$ to $1$, we obtain:
$$\sum_{n=2}^\infty (-1)^n \frac{\log n}{n}= \int_0^\infty \frac{\log(1+e^{-x})-e^{-x} \log 2}{x} dx= \gamma \log 2- \frac{\log^2 2}{2}$$
| We can "calculate" the sum using the Wallis product formula.
Denoting our sum by
$$f = \sum_{n=1}^\infty (-1)^n \log(n)$$
we have
$$f= \log(2) - \log(3) + \log(4) \mp ...$$
and
$$e^f = \frac{2}{3} \frac{4}{5} \frac{6}{7} ...$$
Taking the square gives
$$e^{2f} = \frac{2}{3} \frac{2}{3} \frac{4}{5} \frac{4}{5} \frac{6}{7}...$$
Adding a factor $1$ in the denominator we arrive at
$$e^{2f} = \frac{2}{1} \frac{2}{3} \frac{4}{3} \frac{4}{5} \frac{6}{5} ...$$
But this is the famous Wallis product which has the value $\frac{\pi}{2}$.
Hence we have derived
$$f = \frac{1}{2} \log(\frac{\pi}{2})$$
The "regularization" is hidden somewhere in the sloppy derivation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$. Question : If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$.
My Attempt
Corresponding Equation of Inequality, $f(x) =5x^2 - 2kx + 1 = 0$. Let $\alpha,\beta$ be the roots. $f(x)<0$ for exactly one integral value of $x$.
Possible Graphs
Conditions Required
*
*$D>0 \\ 4k^2-20>0 \\k^2-5>0\\k \in(-\infty,-\sqrt5) \;\cup \;(\sqrt5,\infty) \qquad\text{(1)}$
*$|\alpha-\beta|\leq2 \\ |\frac{\sqrt D}a|\leq 2 \\\\ 2\frac{\sqrt{k^2-5}}{5}\leq2 \;\;\&\;\; 2\frac{\sqrt{k^2-5}}{5}\geq-2 \\\sqrt{k^2-5}\leq{5}\;\;\&\;\;\sqrt{k^2-5}\geq{-5}\\$
Solving we get $k\in[-\sqrt{30},\sqrt{30}]\qquad{(2)}\\$
and $k\in(-\infty,-\sqrt5] \; \cup \;[\sqrt5,\infty) \qquad\text{(3)}$
From $(1),(2),(3)$, we get $\boxed{k\in[-\sqrt{30},-\sqrt5) \; \cup \;(\sqrt5,\sqrt{30}]}$
Hence, sum of all positive integral values of $x$ is $3+4+5=12$
Issue: I think I have solved this problem correctly. But the answer at the back of my book is: $4+5=9$.
I mean, $\{3\}$ does not belong to the solution set, which I find strange. I might have made a mistake somewhere while solving it, but couldn't find it. I am just asking why $\{3\}$ does not belong to the solution set. Because it should, from what I see. Please help me with it. Thanks.
| Using your notations:
$$\alpha=\frac{k-\sqrt{k^2-5}}{5}; \beta=\frac{k+\sqrt{k^2-5}}{5}.$$
The integral solution of $5x^2 - 2kx + 1 < 0$ must be in $(\alpha,\beta)$.
Consider only $k>0$ by the requirement and note that:
$$\\
0<\alpha<1 \iff 0<k-\sqrt{k^2-5}<5 \iff k-5<\sqrt{k^2-5} \Rightarrow k>0
$$
Hence:
$$\beta>1 \Rightarrow k+\sqrt{k^2-5}>5 \Rightarrow \sqrt{k^2-5}>5-k \Rightarrow k>3.$$
This is an additional condition that is missed by your $|\alpha-\beta|<2$.
For $k=3$, the inequality $5x^2 - 6x + 1 < 0$ has a solution $(0.2,1)$, which does not have an integral solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Evaluating $\sum_{r=1}^n \frac{\tan(x/2^r)}{2^{r-1}\cos(x/2^{r-1})}$
I was asked to find the sum of
$$\sum_{r=1}^n \dfrac{\tan \dfrac{x}{2^r}}{2^{r-1}\cos\dfrac{x}{2^{r-1}}}$$
I proceeded by breaking $\tan x$ into $\sin x$ and $\cos x$ and writing numerator as follows
$$\sin\left(\frac{x}{2^{r-1}}-\frac{x}{2^r} \right)$$
then by opening brackets and simplifying I got
$$\frac{1}{2^{r-1}}\left(\tan\frac{x}{2^{r-1}}-\tan\frac{x}{2^r}\right)$$
But I couldn't proceed from here.
Any help will be appreciated
| $$\sum^{n}_{k=1}\frac{\tan(x/2^k)}{2^{k-1} \cos(x/2^{k-1})} = \sum^{n}_{k=1}\frac{\sin^2(x/2^k)}{2^{k-1}\sin (x/2^k)\cos(x/2^k)\cos(x/2^{k-1})}$$
$$=2\sum^{n}_{k=1}\frac{1-\cos(x/2^{k-1})}{2^{k-1}\sin(x/2^{k-1})\cos(x/2^{k-1})}$$
$$= 2\sum^{n}_{k=1}\bigg[\frac{1}{2^{k-2}\sin(x/2^{k-2})}-\frac{1}{2^{k-1}\sin(x/2^{k-1})}\bigg]$$
$$ = \bigg[\frac{2}{\sin x}-\frac{2}{2^n \sin (2x/2^n)}\bigg].$$
| {
"language": "en",
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"source": "stackexchange",
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ambiguity in solving algebraic equations. Given $x+\frac{1}{x}=2 \tag{1}$
$x^3+\frac{1}{x^4}=1 \tag{2}$
Find the value of $$x^4+\frac{1}{x^3}$$
If equation $\bf{2}$ is multiplied by $\bf x$ we get $x^4+\frac{1}{x^3}=x$ now, we just need to find the value of $x$ which from $\bf{1}$ is equal to $1$.
But the answer to this question is $\bf3$.
Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.
Another approach
Given, $(x + 1/x) = 2$
Squaring both sides,
$$x^2 + 1/x^2 + 2 = 4$$
$$ x^2 + 1/x^2 = 2 $$
Let,
$A = x^3 + 1/x^4$
$B = x^4 + 1/x^3$
Now, add A and B,
$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$
$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$
$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$
$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2 $
$A + B = 2 + (2)2 – 2$
$ A + B = 4$
Given that $A = 1$
Then, $B = 4 – 1 = 3$
$ x^4 + 1/x^3 = 3$
| If $x=3$ and $x$ is solution to $(1)$, we have
$$
x + \frac{1}{x} = 2 \iff 3 + \frac 1 3 = 2
$$
which is obviously wrong. There must be some mistake in the solution.
| {
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"source": "stackexchange",
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} |
Arc length of an implicit function Find the arc length of $$y=\frac{x^4}{8} + \frac{1}{4y^2}$$ for the region between $~x=1~$ to $~x=2~$.
I have tried implicitly deriving it to find $~\frac{dy}{dx}~$, but since $~\frac{dy}{dx}~$ is dependent on two variables, you cant integrate through the normal definition of $$\int_1^2 \sqrt{1+(dy/dx)^2}dx$$
| Why not to look at $x$ as a function of $y$ and then integrate along $y$ ?
For the range of interest, we have
$$x=\frac{\left(8 y^3-2\right)^{1/4}}{\sqrt{y}}$$
$$x'=\frac{2 y^3+1}{y^{3/2} \left(8 y^3-2\right)^{3/4}}$$ and we face the problem of
$$L=\int_a^b \sqrt{1+\frac{\left(2 y^3+1\right)^2}{y^3 \left(8 y^3-2\right)^{3/2}}}\,dy$$ $a$ being the solution of $y=\frac{1}{8} + \frac{1}{4y^2}$ and $b$ the solution of $y=2 + \frac{1}{4y^2}$ that is to say
$$a=\frac{1}{24} \left(1+\sqrt[3]{1729-24 \sqrt{5190}}+\sqrt[3]{1729+24
\sqrt{5190}}\right)\approx 0.674504$$
$$b=\frac{1}{12} \left(8+\sqrt[3]{728-24 \sqrt{465}}+2 \sqrt[3]{91+3 \sqrt{465}}\right)\approx 2.05897$$ and now numerical integration.
| {
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"source": "stackexchange",
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\
&= 11\cdot10^{2n}-\left(100^{n}-1\right) \\
&= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\
&= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\
&= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\
&= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right)
\end{align}
Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!
| Not as easy as the other answers, but this is a classic induction question.
For $n = 0$, we have $10^{2n + 1} + 1 = 11$, which $11$ divides, confirming the base case.
Suppose $11$ divides $10^{2n + 1} + 1$ for some $n$. Then some $k$ exists such that $10^{2n + 1} + 1 = 11k$. Then,
$$10^{2(n + 1) + 1} + 1 = 100 \cdot 10^{2n + 1} + 1 = 100 (11k - 1) + 1 = 11(100k - 9),$$
thus $11$ divides $10^{2(n+1) + 1} + 1$. By induction, $11$ divides $10^{2n + 1} + 1$ for all $n$.
| {
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Conjecture: ${_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)=\prod_{n=2}^\infty \frac{(2n-2+x)(2n-1+x)}{2n (2n-3+2x)}$
Is the following conjecture for $0<x<1$ true and how do we prove it?
$${_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)=\prod_{n=2}^\infty \frac{(2n-2+x)(2n-1+x)}{2n (2n-3+2x)}$$
I encountered this function when answering a question a while back, where I have shown that:
$$S=\sum_{k=1}^\infty \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1}=\frac{\sqrt{\pi}}{2} \int_0^1 x ~{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right) dx$$
Where $G_k$ are so called Gregory coefficients.
Now I returned to this problem and considered the ratio:
$$\frac{{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};n;1 \right)}{{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};n+1;1 \right)}$$
For several cases for $n \geq 2$ Wolfram Alpha shown me the terms from the product above. This, together with an obvious identity:
$$\lim_{n \to \infty} {_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};n;1 \right)=1$$
Makes me believe that the infinite product converges.
However, I don't know how to prove the ratio. Gauss's contiguous relations connect the two functions and the derivative, but I'm not sure how to use them in this case.
The product is not telescoping (which is obvious, otherwise this hypergeometric function would be a rational function).
Another question: are there more infinite product identities like this one for hypergeometric functions?
We can rewrite the product in a more convenient way:
$$\prod_{n=2}^\infty \frac{\left(1-\frac{2-x}{2n}\right)\left(1-\frac{1-x}{2n}\right)}{\left(1-\frac{3-2x}{2n}\right)}$$
Which looks a lot like a more complicated Gamma function.
Claude Leibovici proposed a closed form for the product which means that the following holds:
$$\sum_{k=1}^\infty \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1}=\int_0^1 2^x \frac{x}{1+x} \frac{\Gamma \left(x+ \frac{1}{2} \right)}{\Gamma \left( x+1 \right)}~ dx$$
The series converges extremely slow, for example:
$$\sum_{k=1}^{500} \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1}=0.50667610440857923696$$
While the integral gives:
$$\int_0^1 2^x \frac{x}{1+x} \frac{\Gamma \left(x+ \frac{1}{2} \right)}{\Gamma \left( x+1 \right)}~ dx=0.50718128906684564110$$
Which is the correct numerical value.
Unfortunately, there's no in-built Gregory coefficient list in Mathematica, so it's hard to compute them for large $k$.
| The conjecture is indeed true for all $x\in\Bbb C$ when $\Re\, x>-1/2$.
If $\Re\, x>-1/2$ then DLMF $15.4.20$ tells us that
$$
F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right)=\frac{\Gamma(2)\Gamma(x+\frac{1}{2})}{\Gamma(\frac{x}{2}+\frac{3}{2})\Gamma(\frac{x}{2}+1)}.
$$
Now notice that the sum of the arguments in the numerator equals the sum of arguments in the denominator, that is,
$$
2+x+\tfrac{1}{2}=\tfrac{x}{2}+\tfrac{3}{2}+\tfrac{x}{2}+1=x+\tfrac{5}{2};
$$
thus, according to DLMF $5.8.5$ we have the following infinite product expansion
$$
F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right)=\prod_{k=0}^\infty\frac{(k+\frac{x}{2}+\frac{3}{2})(k+\frac{x}{2}+1)}{(k+2)(k+x+\frac{1}{2})},
$$
which can be easily derived from the product expansion of the gamma function. Substituting $n=k+2$ and simplifying then gives
$$
\begin{aligned}
F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right)
&=\prod_{n=2}^\infty\frac{(n-\frac{1}{2}+\frac{x}{2})(n-1+\frac{x}{2})}{n(n-\frac{3}{2}+x)}\\
&=\prod_{n=2}^\infty\frac{(2n-1+x)(2n-2+x)}{2n(2n-3+2x)},
\end{aligned}
$$
which is the conjectured product expansion.
| {
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"source": "stackexchange",
"question_score": "8",
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What is the ratio the areas of △$PST$ and quadrilateral $STRQ$? I tried to answer this question:
What is the ratio of the areas of △$PST$ and quadrilateral $STRQ$ if $∠1\cong ∠2$
Since $∠1\cong ∠2$, △$PST\sim$ △$PQR$. And since the two triangles are similar, the ratio of their areas is $\frac{1}{25}$. The area of the whole triangle is 1, so the area of quadrilateral $STRQ$ is $1-\frac{1}{25}=\frac{24}{25}$. The ratio between the two areas now is $\frac{\frac{1}{25}}{\frac{24}{25}}=\frac{1}{24}$.
Is my answer correct?
| $\triangle PST$ and $\triangle PQR$ are similar due to AA (two angles congruent). I will assume you can prove that.
Given that the area of a triangle is $A=\frac{1}{2}ab\sin{C}$, we can calculate the area for both triangles, then find the ratio. Let the side $\overline {PT}=y$. Then we need to calculate $\overline {PR}$.
By similar triangles,
\begin{align}
\frac{\overline{PS}}{\overline{PQ}}&=\frac{\overline{PT}}{\overline{PR}}\\
\frac{3}{15}&=\frac{y}{\overline{PR}} \\
\implies \overline{PR}&=5y \\
\end{align}
Then apply the area formula:
\begin{align}
A_{\triangle {PST}} &=\frac{1}{2}\cdot3\cdot y \cdot \sin{C} \\
&=\frac{3y}{2}\cdot \sin{C} \\
\end{align}
\begin{align}
A_{\triangle {PQR}} &=\frac{1}{2}\cdot15\cdot 5y \cdot \sin{C} \\
&=\frac{75y}{2}\cdot \sin{C} \\
\end{align}
The area of the quadrilateral is:
\begin{align}
A_{STRQ}&=A_{\triangle {PQR}}-A_{\triangle {PST}}\\
&=\frac{72y}{2}\sin{C}
\end{align}
Therefore the ratio is $\frac{A_{\triangle {PQR}}}{A_{STRQ}}=\frac{\frac{3y}{2}\sin{C}}{\frac{72y}{2}\sin{C}}=\frac{1}{24}$
You are correct!
| {
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Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left( x + \frac{1}{x} \right),$$ but I don't know how to proceed. I tried something like $$\left(x -\frac{1}{x} \right) = a$$ and $$\left( x + \frac{1}{x} \right) = a + \frac{2}{x},$$ trying to apply $h(a)$ but it doesn't seem to work.
Any hints?
| instead of writing x+1/x as a+2/x,
$ \sqrt{(x-1/x)^2+4} = \sqrt{a^2+4} = |x+1/x| = x+1/x$ for x>0 and $-(x+1/x)$ for x<0$
$h(a)=a \sqrt{a^2+4}$ for x>0 and $h(a)= -a \sqrt{a^2+4}$ for x<0, where a=x-1/x.
Try simplifying this into a function in terms of x, considering cases for x>0 and x<0.
| {
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"source": "stackexchange",
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Prove that if $x$ is odd, then $x^2$ is odd
Prove that if $x$ is odd, then $x^2$ is odd
Suppose $x$ is odd. Dividing $x^2$ by 2, we get:
$$\frac{x^2}{2} = x \cdot \frac{x}{2}$$
$\frac{x}{2}$ can be rewritten as $\frac{x}{2} = a + 0.5$ where $a \in \mathbb Z$. Now, $x\cdot\frac{x}{2}$ can be rewritten as:
$$x\cdot\frac{x}{2} = x(a+0.5) = xa + \frac{x}{2}$$
$xa \in \mathbb Z$ and $\frac{x}{2} \notin \mathbb Z$, hence $xa + \frac{x}{2}$ is not a integer. And since $xa + \frac{x}{2} = \frac{x^2}{2}$, it follows that $x^2$ is not divisible by two, and thus $x^2$ is odd.
Is it correct?
| Let $x = 2k + 1 \in \mathbb{Z}$ be odd. So, $$ x^{2} = (2k + 1)^2 = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$$ Let $t = 2k^{2} + 2k \in \mathbb{Z}$, thus: $ x^{2} = 2t + 1$.
| {
"language": "en",
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"source": "stackexchange",
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Where is the mistake in the equality of roots? I have this statement:
If $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}}, g(x) =
> \frac{\sqrt[6]{8x^2}}{2}$, with $x < 0$ is $f(x) = g(x)$ ?
My attempt was:
$(1)$ $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$
$(2)$ $g(x) = \frac{\sqrt[6]{8x^2}}{2}$ using the property of $\sqrt[n]{k^m} = k^{\frac{m}{n}}$ and since $x^2 \geq 0$, thus:
$(3)$ $g(x) = \frac{\sqrt[6]{2^3}\sqrt[6]{x^2}}{2} = \frac{2^{\frac{3}{6}}\cdot (x^2)^{\frac{1}{6}}}{2} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$, that is equal to $f(x)$ from the step $(1)$
But according to the guide, this is false. So, what is wrong?
| with $x<0$, $\sqrt[3]x<0$, but $\sqrt[6]{x^2}>0$
| {
"language": "en",
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Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| Your solution is incorrect, as others pointed out. The reason is because you've made two "irreversible" operations, and did not took care of the branches they create.
The first one is:
$$
\sin x \cos x = \frac{1}{2} \Rightarrow \sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}
$$
This is is simply wrong, because if your are working on real numbers, the outcome of the square root must be a non-negative number. The correct statement would actually be:
$$
\sin x |\cos x| = \frac{1}{2} \Rightarrow \sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}
$$
The second one is:
$$
\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2} \Rightarrow \bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2
$$
This is actually right, but you needed the arrow to go both ways, and it doesn't because:
$$
\sin x \sqrt{1 - \sin^2 x} = -\frac{1}{2} \Rightarrow \bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(-\frac{1}{2}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2
$$
Because of this step, you've introduced 2 additional solutions.
You could have checked all the answers you've obtained to verify which remained valid, but the advantage of the double-arc method is that you circumvent those irreversible operations, making it easier to reach the final answer.
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Floor function simplification identities I can't seem to find any identity(if any)for division/multiplication involving floor functions: for example
$$\lfloor{\frac{n-1}{2}}\rfloor\cdot 2$$
I know does not simplify down to $$n-1$$.
| You could consider separately what happens when $n$ is an even number and when $n$ is an odd number.
If $n=2k$, then
\begin{align}
2\left\lfloor \dfrac{n-1}{2} \right\rfloor
&= 2\left\lfloor \dfrac{2k-1}{2} \right\rfloor\\
&= 2\left\lfloor k - \dfrac 12 \right\rfloor\\
&= 2(k-1) \\
&= 2\left( \dfrac n2 - 1 \right) \\
&= n - 2
\end{align}
If $n=2k+1$, then
\begin{align}
2\left\lfloor \dfrac{n-1}{2} \right\rfloor
&= 2\left\lfloor \dfrac{2k+1-1}{2} \right\rfloor\\
&= 2\left\lfloor k \right\rfloor\\
&= 2k \\
&= n - 1
\end{align}
| {
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Formula for the first base-$n$ digit of $n^{d+1}-\sum_{i=1}^{n}i^d$ Let $k,n,d \in \mathbb{N}$ and $n>1$.
Converting $k$ in a base $n$,
$$k = (n_{ l} ~\cdots~ n_2~ n_1)_{n} \quad\text{where}\quad n_{l} \ne 0$$
Show that, if
$$k = n^{d+1} - \sum_{i=1}^n i^d$$ then
$$
n_l=n - \left\lceil\frac{\sum_{i=1}^n i^d}{n^d}\right\rceil
$$
Example
Let $n=3$ and $d=2$.
$$3^3-(1^2+2^2+3^2)= 13 = (111)_{3}$$
So $n_l = 1$.
| Your conjecture doesn't always hold based on the stated conditions. For example, $n = 2, d = 3$ results in
$$\sum_{i=1}^{n}i^d = \sum_{i=1}^{2}i^3 = 1 + 2^3 = 9 \tag{1}\label{eq1A}$$
$$k = n^{d+1} - \sum_{i=1}^{n}i^d = 2^4 - 9 = 7 = (111)_2 \tag{2}\label{eq2B}$$
Thus, $l = 3$ and $n_l = n_3 = 1$. However,
$$n - \left\lceil\frac{\sum_{i=1}^n i^d}{n^d}\right\rceil = 2 - \left\lceil\frac{9}{2^3}\right\rceil = 2 - 2 = 0 \neq n_l = n_3 = 1 \tag{3}\label{eq3A}$$
This fails because $\frac{\sum_{i=1}^n i^d}{n^d} \gt 1$, so the ceiling result is always $\ge 2$, meaning the final result is always $\le n - 2$, but this gives $0$ for $n = 2$. However, your conjecture does work by changing the restriction to be $n \ge 3$.
Note that to prove your conjecture, it's sufficient to prove
$$\sum_{i=1}^n i^d \le n^{d+1} - n^d \; \iff \; \sum_{i=1}^{n-1} i^d \le n^{d+1} - 2n^d \tag{4}\label{eq4A}$$
The left inequality with $d = 1$ gives $\frac{n(n+1)}{2} \le n^2 - n \iff n + 1 \le 2(n - 1) \iff 3 \le n$, so it holds.
Next, consider the remaining cases of $d \ge 2$ with $n \ge 3$. Note that $i^d = \int_{i}^{i+1}i^d dx$. Since $\lfloor x \rfloor = i$ for $x \in [i, i + 1)$, we also have $i^d = \int_{i}^{i+1}(\lfloor x \rfloor)^d dx$. In addition, since for non-integral $x$ means that $\lfloor x \rfloor \lt x$, we have $\int_{i}^{i+1}(\lfloor x \rfloor)^d dx \lt \int_{i}^{i+1}x^d dx$. Using these results and summing over the various values of $i$ gives
$$\sum_{i=1}^{n-1} i^d = \int_{1}^{n}(\lfloor x \rfloor)^d dx \lt \int_{1}^{n}x^d dx = \left. \frac{x^{d+1}}{d+1} \right|_{1}^{n} = \frac{n^{d+1} - 1}{d+1} \tag{5}\label{eq5A}$$
To prove the right side of \eqref{eq4A}, it's sufficient to prove
$$\begin{equation}\begin{aligned}
\frac{n^{d+1} - 1}{d+1} & \le n^{d+1} - 2n^d \\
n^{d+1} - 1 & \le (d + 1)n^{d+1} - 2(d + 1)n^d \\
-1 & \le dn^{d+1} - 2(d + 1)n^d \\
-1 & \le n^d(dn - 2d - 2) \\
-1 & \le n^d(d(n-2) - 2)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Since $d \ge 2$ and $n \ge 3$ means $d(n-2) - 2 \ge 0$, the inequality above holds, so \eqref{eq4A} also holds. This means \eqref{eq4A} is true for all $d \ge 1$ and $n \ge 3$. Note this gives $-\sum_{i=1}^n i^d \ge - n^{d+1} + n^d \; \to \; n^{d+1} - \sum_{i=1}^n i^d \ge n^d$. Thus, $n^d \le k \lt n^{d+1}$ so, expressed in base $n$, we have $l = d + 1$ and
$$\begin{equation}\begin{aligned}
n_l & = \left\lfloor \frac{k}{n^d} \right\rfloor \\
& = \left\lfloor \frac{n^{d+1} - \sum_{i=1}^n i^d}{n^d} \right\rfloor \\
& = \left\lfloor n - \frac{\sum_{i=1}^n i^d}{n^d} \right\rfloor \\
& = n + \left\lfloor -\frac{\sum_{i=1}^n i^d}{n^d} \right\rfloor \\
& = n - \left\lceil \frac{\sum_{i=1}^n i^d}{n^d} \right\rceil
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
For going to the last step, note if $x$ is any integer then $\lfloor -x \rfloor = -x = -\lceil x \rceil$, while if $x$ is non-integral so $x = m + r$ with $m \in \mathbb{Z}$ and $0 \lt r \lt 1$, then $\lfloor -x \rfloor = \lfloor -m - r \rfloor = -m - 1 = -(m + 1) = -\lceil x \rceil$. Thus, for all real $x$, we have $\lfloor -x \rfloor = -\lceil x \rceil$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $a_1,a_2,a_3....$ be an arithmetic progression. Question is
Let $a_1,a_2,a_3....$ be an arithmetic progression. If $\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2},p\not=q$, then find $\frac{a_6}{a_{21}}$
Answer: $\frac{11}{41}$
All options: $\frac{7}{2}$, $\frac{2}{7}$, $\frac{11}{41}$, $\frac{41}{11}$
My approach to the question is
Let $S_r$ be sum of first $r$ terms of this AP
Method 1
we know
$\frac{S_p}{S_q}=\frac{p^2}{q^2}$
So,
$\frac{\frac{p}{2}(a_1+a_p)}{\frac{q}{2}(a_1+a_q)}=\frac{p^2}{q^2}$
$\frac{a_1+a_p}{a_1+a_q}=\frac{p}{q}$
After that I am stuck.
Method 2
But after taking another look at the question, I observed
$\frac{S_6}{S_{21}}=\frac{6^2}{21^2}=\frac{36}{441}$
Also,
$\frac{S_5}{S_{20}}=\frac{5^2}{20^2}=\frac{25}{400}$
And we know $S_r-S_{r-1}=a_r$
Then subtracting numerators and denominators of both
$\frac{S_6-S_5}{S_{21}-S_{20}}=\frac{a_6}{a_{21}}=\frac{36-25}{441-400}=\frac{11}{41}$
which we can't always do in maths, I cannot understand why Method 2 gave correct answer.
| We have that
$$\frac{a_1+a_2}{a_1}=\frac{2^2}{1^2}=4$$
$$a_1+a_2=4a_1$$
$$a_2=3a_1$$
So the common difference is
$$d=a_2-a_1=2a_1$$
Hence our final answer is
$$\frac{a_6}{a_{21}}=\frac{a_1+(6-1)(2a_1)}{a_1+(21-1)(2a_1)}=\frac{11}{41}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Area boundes by two circles
Find the area bounded by $r \le 3\sin(t)$ and $r \le -3 \sqrt3 \cos(t)$.
Well, I've found the intersection point, which is $t=\frac{2 \pi}{3}$ . Then, I set up two integrals
$$
A=\frac{1}{2} \int_{\pi/2}^{2\pi/3} (3\sin(t))^2 dt
$$
and
$$
A=\frac{1}{2} \int_{2\pi/3}^{\pi}(-3 \sqrt3 \cos(t))^2 dt
$$
Is that correct?
|
$$
\text{Area}_1=\frac12\,\frac\pi3\left(\frac{3\sqrt3}2\right)^2=\frac{9\pi}8
$$
$$
\text{Area}_2=\frac12\,\frac{2\pi}3\left(\frac32\right)^2=\frac{3\pi}4
$$
$$
\text{Area}_3=\frac32\,\frac{3\sqrt3}2=\frac{9\sqrt3}4
$$
$$
\begin{align}
\text{Area}
&=\text{Area}_1+\text{Area}_2-\text{Area}_3\\[3pt]
&=\frac{9\pi}8+\frac{3\pi}4-\frac{9\sqrt3}4\\
&=\frac{15\pi}8-\frac{9\sqrt3}4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317764",
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How to prove that $(\sqrt{3} + 1)b/2 < a < 1 + (\sqrt{3} + 1)b/2$? I'm starting with the assumption that $a, b > 0$, and $$\frac{a - 1}{a + b - 1} \cdot \frac{a}{a + b} = \frac{1}{3}$$
I am trying to show that $$(\sqrt{3} + 1)b/2 < a < 1 + (\sqrt{3} + 1)b/2.$$
Since $$\frac{a - 1}{a + b - 1} < \frac{a}{a + b}$$ I can write that $$\left(\frac{a - 1}{a + b - 1}\right)^2 < \frac{1}{3} < \left(\frac{a}{a + b}\right)^2$$
so $$\frac{a - 1}{a + b - 1} < \frac{1}{\sqrt{3}} < \frac{a}{a + b}$$
If I write $$\frac{a - 1}{a + b - 1} < \frac{1}{\sqrt{3}} < \frac{a}{a + b} < \frac{a}{a + b - 1}$$
I can derive the worse bound $(\sqrt{3} + 1)(b - 1)/2 < a < (\sqrt{3} + 1)(b + 2)/2$. If I could have a hint on how I can derive the one stated in the question, that would be much appreciated. And, if possible, what are some general strategies I should consider when trying to solve these kinds of problems?
Thank you!
Edit: I got this from Probability, Random Variables, and Stochastic Processes by Papoulis. The problem is the following -
If we assume that $a$ and $b$ must be integers greater than or equal to 1 does this resolve the problem with the bound?
| The condition gives:
$$2a^2-2a(b+1)-b^2+b=0,$$ which gives
$$a=\frac{b+1+\sqrt{3b^2+1}}{2}$$ or
$$a=\frac{b+1-\sqrt{3b^2+1}}{2}.$$
In the first case we need to prove that
$$\frac{(1+\sqrt3)b}{2}<\frac{b+1+\sqrt{3b^2+1}}{2}<1+\frac{(1+\sqrt3)b}{2}$$ or
$$\sqrt3b<1+\sqrt{1+3b^2}<2+\sqrt3b,$$ which is obvious.
In the second case we need $$\frac{b+1-\sqrt{3b^2+1}}{2}>0,$$ which gives $$0<b<1.$$
Can you end it now?
I got that there is a problem with $$\frac{(1+\sqrt3)b}{2}<\frac{b+1-\sqrt{3b^2+1}}{2},$$ which is wrong for $b\rightarrow1^-.$
Because in this case we need to prove that:
$$\frac{1+\sqrt3}{2}\leq\frac{2-2}{2},$$ which is wrong.
If $a\geq1$ and $b\geq1$ so the second case is impossible because the second root is negative and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
| We have that
$$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
therefore
\begin{align}(1+3x)\ln(1+x)&=\ln(1+x)+3x\ln(1+x) \\&= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}~+~\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(3x)x^n}{n}\\&=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(3x+1)x^n}{n}\\&=(3x+1)x-\frac{(3x+1)x^2}{2}+\frac{(3x+1)x^3}{3}-\frac{(3x+1)x^4}{4}+\cdots \\&=
x + \big(3-\frac{1}{2}\big)x^2-\big(\frac{3}{2}-\frac{1}{3}\big)x^3+\big(\frac{3}{3}-\frac{1}{4}\big)x^4-\big(\frac{3}{4}-\frac{1}{5}\big)x^5+\dots \\&=x+\frac{5}{2}x - \frac{7}{6}x^3+\frac{3}{4}x^4 - \frac{11}{20}x^5 +\dots
\\&= x + \sum_{n=2}^{\infty}(-1)^n\frac{2n+1}{(n-1) n}x^n
\end{align}
| {
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Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.
Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.
What I have tried.
$3^9+3^{12}+3^{15}+3^n$
$=3^9(757+3^{n-9})$
Let $757+3^{n-9}=a^3$
Taking modulo $3$:
$a^3\equiv 1 \pmod 3$
$\implies a\equiv 1 \pmod 3$
Also, $a^3>757>729=9^3$
$\therefore a =10+3k$
$a=10$ gives $n=14$
Now, how can I know if there is any other $n>14$ satisfying the given condition.
PS: Please not use computer programmes to answer. I want pure mathematical solution.
| Check manually if there exist any solution with $n \le 9$.
Now assume $n>9$, and look at the equation $3^m+757=a^3$ modulo 7 (with $m=n-9$):
$3^m+1 \equiv 3^m+757 \equiv a^3 \equiv \{0,-1,1\} \pmod 7$
$3^m \equiv \{0,-1,-2\} \pmod 7$
$n \equiv \{3,5\} \pmod 6$
If $m=6k+3$,
$ 757 = a^3-(3^{2k+1})^3 = (a-b)(a^2+ab+b^2)$
where $b=3^{2k+1}$. Factoring 757, you can check there is no solutions.
Else, if $m=6k+5$,
$757 = a^3-3^5(3^{2k})^3 = a^3-3^5b^3$
where $b=3^{2k}$. This is a Thue equation, effectively solvable:
using PARI/GP
tnf = thueinit(x^3-243)
thue(tnf, 757)
[[10, 1]]
you can check the only solution is $(a,b)=(10,1)$, hence $(a,n)=(10,14)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
simplifying $\frac{15^4+105^2-225\cdot210}{130^2-50^2}$ $x=\frac{15^4+105^2-225\cdot210}{130^2-50^2}$
Since $225 = 15^2$ and $210 = 2\cdot105$ the numerator is $(15^2-105)^2$ which is $120^2 = 14400$.
The denominator is $(130-50)(130+50)$ which is $14400$. The answer is $1$. The solutions for the exam have proven to be wrong from time to time and in this case they state that $x=10$. This is wrong answer, or?
| $\dfrac{15^4+105^2-225\cdot210}{130^2-50^2}=\dfrac{(15^2-105)^2}{(130+50)(130-50)}=\dfrac{120^2}{180\times80}=\dfrac{12^2}{18\times8}=1$
| {
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"url": "https://math.stackexchange.com/questions/3338614",
"timestamp": "2023-03-29T00:00:00",
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How to get the following summation of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$ I am trying to find the sum $$\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$$ I had factorized the sum as $$\frac{1}{2n(n!)}\left(\frac{1}{n(n-1)+1}-\frac{1}{n(n+1)+1}\right)$$ From this step, how to proceed?
| Given your decomposition of the main term, summation by parts ensures that the series equals
$$1+\frac{1}{2}\sum_{n\geq 1}\left(\frac{1}{n\cdot n!}-\frac{1}{(n+1)\cdot(n+1)!}\right)\left(1-\frac{1}{n^2+n+1}\right)$$
or
$$ 1+\frac{1}{2}\sum_{n\geq 1}\frac{n^2+n+1}{n(n+1)(n+1)!}\cdot\frac{n(n+1)}{n^2+n+1}=1+\frac{1}{2}\sum_{n\geq 1}\frac{1}{(n+1)!}=1+\frac{1}{2}\sum_{m\geq 2}\frac{1}{m!}=1+\frac{e-2}{2}=\color{red}{\frac{e}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the recursive formula for $a_{n}={b^n}$ for $n>2$ How can I solve it,
and can anybody help me to find recursive formula for $a_{n}=3n^3$ for $n\geq 0$.
| Set $b_n=a_{n+1}-a_n=3(n+1)^3-3n^3=9n^2+9n+3$
Set $c_n=b_{n+1}-b_n=9(n+1)^2+9(n+1)+3-(9n^2+9n+3)=18n+18$
So $c_{n+1}=c_n+18$ and $c_0=18$
Then $b_{n+2}-b_{n+1}=b_{n+1}-b_n+18$
then $a_{n+3}-a_{n+2}-(a_{n+2}-a_{n+1})=a_{n+2}-a_{n+1}-(a_{n+1}-a_n)+18$
So $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n+18$, with $a_0=0, a_1=3, a_2=24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340944",
"timestamp": "2023-03-29T00:00:00",
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Equations of the sides an equilateral triangle with centroid at the origin and one side is $x+y=1$
An equilateral triangle has its centroid at the origin and one side is $x+y=1$. Find the equations of the other sides.
My Attempt
$$
OD=\frac{1}{\sqrt{2}}\implies OC=\sqrt{2}\implies C=(-1,-1)\\
m_{AB}=m_1=-1\implies\tan60=\sqrt{3}=|\frac{m+1}{1-m}|\\
\sqrt{3}-m\sqrt{3}=m+1\quad\text{or}\quad m\sqrt{3}-\sqrt{3}=m+1\\
m(1+\sqrt{3})=\sqrt{3}-1\quad\text{or}\quad m(\sqrt{3}-1)=\sqrt{3}+1\\
m=\frac{\sqrt{3}-1}{\sqrt{3}+1}\quad\text{or}\quad m=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\
y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)\quad\text{or}\quad y+1=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x+1)
$$
But, my reference gives the solutions
$$y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)$$
and
$$y\pm1=3+\sqrt{3}(x-1)$$
So, what are the actual solution to the problem and the easiest way to solve it?
| The coordinates of the vertices of the triangle are found to be $$A=(\frac {1+\sqrt 3}{2}, \frac {1-\sqrt 3}{2})$$
$$B=(\frac {1-\sqrt 3}{2}, \frac {1+\sqrt 3}{2})$$
$$C= (-1,-1)$$
The equations of lines passing through these points are
$$ CA: y+1=\frac {\sqrt 3 -1}{\sqrt 3+1}(x+1)$$
$$ CB: y+1=\frac {\sqrt 3 +1}{\sqrt 3-1}(x+1)$$
$$ AB: x+y=1$$
Thus your calculations are correct.
| {
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"url": "https://math.stackexchange.com/questions/3342078",
"timestamp": "2023-03-29T00:00:00",
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Geometry - Find angle ∠C of a triangle given angle bisector and two segment ratios. In the triangle ABC, one has $∠A = 70^o$.
The point $D$ is chosen on the segment
$AC$ such that the angle bisector $AE$ intersects the segment $BD$ at the point $H$, with the following ratios
$$\dfrac{AH}{HE}=\dfrac 31, \>\>\> \dfrac{BH}{HD}=\dfrac 53.$$
Find the angle $∠C$.
|
Examine the two ratios as follows,
$$\frac{5}{3}= \frac{BH}{HD} = \frac{\triangle ABE} {\triangle AED} $$
$$=\frac{\frac{BE}{BC}\triangle ABC}{\frac{EC}{BC}\frac{AD}{AC}\triangle ABC}= \frac{BE}{EC}\left(\frac{CD}{AD}+1\right)\tag{1}$$
Similarly,
$$\frac{3}{1} = \frac{AD}{CD}\left(\frac{EC}{BE}+1\right)\tag{2}$$
Combine (1) and (2) to obtain
$$ \frac{BE}{EC} = 1$$
Since AE is the angle bisector, $\triangle ABE$ and $\triangle AEC$ are congruent, and therefore $\triangle ABC$ is an isosceles triangle, which yields,
$$∠C= \frac{1}{2}(180^\circ -70^\circ) = 55^\circ$$
| {
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Evaluating $ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $
I need help with this limit:
$$ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $$
I've tried using the $$ \tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b} $$ with $(x+a)$ as $a$ and $(x)$ as $b$ but having difficulty getting beyond that.
| $$\lim_{a \to 0} \frac{\tan(x+a)- \tan x}{a}= \lim_{a \to 0} \frac{\frac{\sin(x+a)}{\cos(x+a)}- \frac{\sin x}{\cos x}}{a}\\
=\lim_{a \to 0} \frac{\cos x \sin(x+a)-\sin x \cos (x+a)}{a(\cos x \cos (x+a))}\\
= \lim_{a \to 0} \frac{\cos x (\sin x \cos a + \cos x \sin a) - \sin x (\cos x \cos a - \sin x \sin a)}{a\cos x \cos (x+a)}\\
= \lim_{a \to 0} \frac{\sin a (\cos^2 x+ \sin^2 x)}{a \cos x \cos (x+a)}= \lim_{a \to 0} \frac{\sin a}{a} \frac {1}{\cos x \cos (x+a)}= \frac {1}{\cos^2 x}= \sec^2 x.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The polynomial $x^3 + 2ax^2 + (2a^2 + b)x + c$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
The polynomial $$\large x^3 + 2ax^2 + (2a^2 + b)x + c$$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
I'm uncertain of how to prove this.
If $b$ is a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$(2a + b)x^2 + 2a^2x + (b^2 + c) = 0$$
which means the above polynomial has at least one root $\implies (a^2)^2 - (2a + b)(b^2 + c) \ge 0$
$\iff a^4 - 2ab^2 - 2ca - b^3 - bc \ge 0$
And $b$ is also a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$2ax^2 + (2a^2 + b)x + (b^3 + c) = 0$$
which means the above polynomial has at least one root $\implies (2a^2 + b)^2 - 4 \cdot 2a(b^3 + c) \ge 0$
$\iff 4a^4 + 4a^2b - 8ab^3 - 8ca + b^2 \ge 0$
But that's all I got.
| That $b$ is a root implies $-c = b^3+2ab^2+b^2+2a^2b$. Then $x^3+2ax^2+(2a^2+b)x+c = (x-b)(x^2+(2a+b)x+(2a^2+b+2ab+b^2)$ is valid. Since both roots of the quadratic are real, the discriminant must be non-negative: $(2a+b)^2-4(2a^2+b+2ab+b^2) \ge 0$, which is equivalent to $a \in [\frac{-b-\sqrt{-2b^2-4b}}{2},\frac{-b+\sqrt{-2b^2-4b}}{2}]$. We therefore must have $b \le 0$. $\bf\text{For ease, replace $b$ by $-b$}$.
Our problem comes down to maximizing $$(ab^3-2a^2b^2-ab^2+2a^3b)^2$$ given that $$a \in [\frac{b-\sqrt{4b-2b^2}}{2},\frac{b+\sqrt{4b-2b^2}}{2}].$$ Note that we in particular must have $b \in [0,2]$. Let $f(a) = ab^3-2a^2b^2-ab^2+2a^3b$. Then $f'(a) \ge 0$ if and only if $6a^2-4ba+b^2-b \ge 0$, which has roots $\frac{2b\pm\sqrt{6b-2b^2}}{6}$. Now, for any $b \in [0,2]$, $\frac{2b-\sqrt{6b-2b^2}}{6} < \frac{2b+\sqrt{6b-2b^2}}{6} < \frac{b+\sqrt{4b-2b^2}}{2}$, and it holds that $\frac{b-\sqrt{4b-2b^2}}{2} < \frac{2b-\sqrt{6b-2b^2}}{6}$ if and only if $ b \in [0,\frac{50}{33})$. Since we want to maximize $f(a)^2$, it suffices to check extremal $a$ and $a$ for which $f'(a) = 0$, since $(f^2)' = 2ff'$ and $f(a)=0$ implies $(ac)^2 = 0 \le 3$. That is, we just have to show $f(a)^2 \le \sqrt{3}$ for $a = \frac{b+\sqrt{4b-2b^2}}{2},\frac{2b+\sqrt{6b-2b^2}}{6},\frac{2b-\sqrt{6b-2b^2}}{6}$, and for $a = \frac{b-\sqrt{4b-2b^2}}{2}$ when $b \in (\frac{50}{33},2]$.
And this is easily doable. For example, at $a = \frac{b+\sqrt{4b-2b^2}}{2}$, $f(a) = \frac{1}{4}(2-b)b^2(b+\sqrt{2b(2-b)})$. Since $(2-b)b \le 1$, $f(a) \le \frac{1}{4}b(b+\sqrt{2})$, which is $\le \sqrt{3}$. The other values of $a$ can also be handled easily.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $ \lim_{x\to \infty} x \left({{\left(\frac{x}{x+1}\right)}^{x}-\frac{1}{e}}\right)$
Evaluate the following:
$$
\lim_{x\to \infty}
x \left({{\left(\frac{x}{x+1}\right)}^{x}-\frac{1}{e}}\right)$$
I tried to first solve the interior portion as $1^\infty$ indeterminate form but ended up getting a different indeterminate form of $\infty\cdot 0$.
| Let $$y={\left(\frac{x}{x+1}\right)}^{x}\implies \log(y)=x \log\left(\frac{x}{x+1}\right)=-x \log\left(1+\frac 1 x\right)$$
Now, using Taylor expansion
$$\log\left(1+\frac 1 x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\frac{1}{x^4}\right)$$
$$\log(y)=-1+\frac{1}{2 x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right)$$
$$y=e^{\log(y)}=\frac{1}{e}+\frac{1}{2 e x}-\frac{5}{24 e
x^2}+O\left(\frac{1}{x^3}\right)$$
$$x\left(y- \frac 1e\right)=\frac{1}{2 e}-\frac{5}{24 e x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.
Use the above formula for $x=10$ (quite small). You should get $\frac{23}{48 e}\approx 0.176276$ while, using your pocket calculator, the result would be $\approx 0.176638$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the value of an infinite sum from a definite integral: It can be computed that $$\int_0^1 \frac{x^2+1}{x^4+x^2+1}dx = \frac{\pi}{2\sqrt{3}}.$$
The integral can be done using $$\frac{2(x^2+1)}{x^4+x^2+1} = \frac{1}{x^2-x+1}+\frac{1}{x^2+x+1}.$$
But I got stuck in the following part:
We need to show that $$1-\frac{1}{5}+\frac{1}{7} - \frac{1}{11}+\frac{1}{13}- \cdots = \frac{\pi}{2\sqrt{3}}$$ using the above definite integral. The sum of the above series can be computed using Fourier series. Life becomes hard for me when the above sum is to be shown using the above integral. Any suggestions of help is much appreciated.
| $$\frac{x^2+1}{x^4+x^2+1}
=\frac{(1-x^2)(x^2+1)}{(1-x^2)(x^4+x^2+1)}
=\frac{1-x^4}{1-x^6}=1-x^4+x^6-x^{10}+x^{12}-x^{16}+\cdots.$$
Now integrate this termwise.
| {
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"timestamp": "2023-03-29T00:00:00",
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Adding $\left(\frac{b}{c}\right)^{1/2}+\left(\frac{c}{b}\right)^{3/2}$ I am trying to refresh my algebra skills by reading the book Elements Of Algebra by Leonhard Euler. I came across a problem:
$$\text{Add}\;\left(\frac{b}{c}\right)^{1/2}\;\text{to}\;\left(\frac{c}{b}\right)^{3/2}. \qquad\qquad \text{Ans.}\;\frac{b^2+c^2}{b\sqrt{bc}}$$
I don't know how he did it. My understanding is that you can't add radicals that are not the same; for example, $\sqrt{5} + \sqrt{3}$
Could someone enlighten me? Or did he use a some properties of radicals?
| \begin{align*}
\left(\frac{b}{c}\right)^{1/2} + \left(\frac{c}{b}\right)^{3/2} = \frac{\sqrt{b}}{\sqrt{c}} + \frac{\sqrt{c^{3}}}{\sqrt{b^{3}}} = \frac{\sqrt{b}\times\sqrt{b^{3}} + \sqrt{c}\times\sqrt{c^{3}}}{\sqrt{cb^{3}}} = \frac{b^{2}+c^{2}}{b\sqrt{bc}}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proper way to solve $\lim\limits_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}$ I made a problem for my friends, but not everyone agreed with my answer.
$$\lim_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}+\lim_{x\rightarrow\infty}\frac{6}{x}\left[\frac{x}{3}\right]+\lim_{x\rightarrow0}\left[\frac{6}{x}\right]\frac{x}{3}$$
$[x]$ is the biggest integer less than or equal to $x$
My answer is $0+2+2=4$.
Everyone agreed that the second limit should be $2$.
However, I said the first limit should be $0$ because $\left[\frac{6}{x}\right]=0$ as $x$ approaches infinity.
so I said $0$ multiplied by anything should equal to $0$ even when $x$ approaches infinity.
Then my friend said $\left[\frac{6}{x}\right]$ can be expressed as $\frac{6}{x}-k$ where $0\leq k<1$.
Therefore, $$\lim_{x\rightarrow\infty}\left(\frac{6}{x}-k\right)\frac{x}{3}=\lim_{x\rightarrow\infty}2-\frac{kx}{3}=-\infty$$
Then I said we don't know what $k$ approaches to, so we can't conclude the answer here.
Then my friend said $0\times\infty$ can't be 0.
When I put it on WolframAlpha, $$\displaystyle\lim_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}=0$$
But I am not sure which answer is correct.
Can someone explain which is correct and why?
I didn't learn L'Hôpital's rule yet.
| Note that when $x=6,\left \lfloor \frac{6}{x} \right \rfloor\frac{x}{3} = \left \lfloor \frac{6}{6} \right \rfloor\frac{6}{3} =2$.
While when $x>6$, say $x=8, \left \lfloor \frac{6}{x} \right \rfloor\frac{x}{3} = \left \lfloor \frac{6}{8} \right \rfloor\frac{8}{3} =0\cdot \frac{8}{3}=0$.
Similarly for any value of $x \in (6,\infty)$.
What your friend said: "$0×\infty$ can't be 0." is not true.
However, $0 \times \infty$ is indeterminate, it can be $0$, it can be $\infty$, and it can be any other value.
| {
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logarithm inequality: 2log5(x) - logx(125) I was given the following problem:
$2\log_{5}x$ - $\log_{x}125$ $< 1$
I tried to use the exponent law and the change of base law and I got this.
$\frac{\log_{}x^2}{\log_{}5}$ - $\frac{3log_{}5}{\log_{}x}$ $< 1$
Which I then changed back to:
$\frac{2log_{}x}{\log_{}5}$ - $\frac{3log_{}5}{\log_{}x}$ $< 1$
I then made
$\frac{(log_{}x^2)(log_{}x)}{(log_{}5)(log_{}x)}$ - $\frac{(log_{}5)(log_{}125)}{(log_{}5)(log_{}x)}$ $< 1$
When I combined these, I got $\frac{(log_{}\frac{x^3}{625})}{log_{}5x}$ $< 1$, which I thought I could combine to:
${log_{}\frac{x^2}{3125}}$ $< 1$
I got from here $x^2$ < 3,1250, which means x < $\sqrt31250$. However, the answer given is:
0 < x < 0.2 and 1 < x < 5$\sqrt5$.
Where did I go wrong? Does it have to do with original different basis, one being a constant and one being a variable? Can someone please help me solve this problem?
| $2\log_{5}x - \log_{x}125 < 1$
$2\log_{5}x - 3\log_{x}5 < 1$
$2\log_{5}x - 3\frac{1}{\log_{5}x} < 1$At this point you need to be careful as $\log_5 x$ can be negative (which is where your mistake is).
If $\log_5 x > 0$ we have $2(\log_{5}x)^2 - \log_5 x -3 <0 \rightarrow 0 < \log_5 x < 1.5 \rightarrow 1 < x < 5\sqrt{5}$
If $\log_5 x < 0$ we have $2(\log_{5}x)^2 - \log_5 x -3 >0 \rightarrow \log_5 x < -1 \rightarrow 0 < x < 1/5$
| {
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For $x\geq 2$,prove that$(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right)>1$ For $x\geq 2$,prove that$$(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right)>1$$
My Attempt:
Let $f(x)=x\cos\left(\frac{\pi}{x}\right)$
$$f'(x)=\cos\left(\frac{\pi}{x}\right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)>0$$for all $x\in[2,\infty)$
$f''(x)=-\frac{{\pi}^2}{x^3}\cos\left(\frac{\pi}{x}\right)<0$ for all $x\in[2,\infty)$
By LMVT on $[x,x+1]$,we have
$$f(x+1)-f(x)=\cos\left(\frac{\pi}{c_{x}}\right)+\frac{\pi}{c_{x}}\sin\left(\frac{\pi}{c_{x}}\right)$$for some $c_{x}\in(x,x+1)$
After this I am not able to go ahead.
| Another possible way of proving your inequality is rearranging and then applying the MVT as follows:
\begin{eqnarray*}
(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right) & > & 1\\
& \Leftrightarrow & \\
x\left( \cos \frac{\pi}{x+1} - \cos \frac{\pi}{x} \right) & > & 1 - \cos \frac{\pi}{x+1} \\
& \stackrel{MVT:\, \color{blue}{\frac{\pi}{x+1}<\xi < \frac{\pi}{x},\, 0< \zeta < \frac{\pi}{x+1}}}{\Longleftrightarrow} & \\
x\sin \xi \cdot \left(\frac{\pi}{x} - \frac{\pi}{x+1} \right) = \color{blue}{\sin \xi} \cdot \frac{\pi}{x+1} & > & \color{blue}{\sin \zeta} \cdot \frac{\pi}{x+1} \\
\end{eqnarray*}
Now note that $\sin x$ is strictly increasing on $[0,\frac{\pi}{2}]$. Hence, for $x\geq 2 \Rightarrow \color{blue}{\frac{\pi}{2}> \xi > \zeta > 0}\Rightarrow \color{blue}{\sin \xi > \sin \zeta}$. So, the last inequality is $\color{blue}{\mbox{true}}$ for $x\geq 2$.
| {
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Simplifying $\frac{a^x}{a^x+\sqrt{a}} + \frac{a^{1-x}}{a^{1-x}+\sqrt{a}}$ While solving a problem related to functions, I came across the following algebraic expression:
$$\frac{a^x}{a^x+\sqrt{a}} + \frac{a^{1-x}}{a^{1-x}+\sqrt{a}}$$
Although I tried, I couldn't find an easy way to simplify it.
Can anyone simplify the stated equation in the easiest way so that anyone can understand it?
| $\ {a ^ x \over a^x + \sqrt a} + {a ^ {1-x} \over
a ^ {1-x} + \sqrt a}$
Well, often it's a good idea to put over common denominator:
$\frac {a^x(a^{1-x} + \sqrt a) + a^{1-x}(a^x+\sqrt a)}{(a^x + \sqrt a)(a^{1-x} + \sqrt a)} =$
$\frac {(a^{x + 1-x} + a^x\sqrt a) + (a^{1-x+x} + a^{1-x}\sqrt a)}{a^{x+1-z} + a^x\sqrt a + a^{1-x}\sqrt a + a}=$
$\frac {2a^1 + (a^x + a^{1-x})\sqrt a}{2a+(a^x + a^{1-x})\sqrt a} = 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $a$ and $b$, so the linear system stays consistant I was wondering if someone could take a look at my solution and tell me if i'm wrong. I feel like my solution is incorrect.
Given
$ \left[
\begin{array}{ccc|c}
1&1&2&2a-b+1\\
2&1&1&3a-b+1\\
1&2&5&3a-2b+6
\end{array}
\right] $ find the values of a and b so the system stays consistent.
I tried to put the system into RREF. So, the first thing I did was R1-R2
$$ \left[
\begin{array}{ccc|c}
-1&0&1&-a\\
2&1&1&3a-b+1\\
1&2&5&3a-2b+6
\end{array}
\right] $$
R2-2R1
$$ \left[
\begin{array}{ccc|c}
-1&0&1&-a\\
0&1&3&a-b+1\\
1&2&5&3a-2b+6
\end{array}
\right] $$
R3+R1
$$ \left[
\begin{array}{ccc|c}
-1&0&1&-a\\
0&1&3&a-b+1\\
0&2&6&2a-2b+6
\end{array}
\right] $$
R3-2R2
$$ \left[
\begin{array}{ccc|c}
-1&0&1&-a\\
0&1&3&a-b+1\\
0&0&0&-a-4b+6
\end{array}
\right] $$
Lastly, -R1
$$ \left[
\begin{array}{ccc|c}
1&0&-1&a\\
0&1&3&a-b+1\\
0&0&0&-a-4b+6
\end{array}
\right] $$
The last equations gives $0=-a-4b+6$. This means that $a$ and $b$ can be any values as long as it follows that equation.
| Multiplying the first equation by-2 and adding to the second we get
$$-x_2-3x_3=-a+b-1$$
and the first by -1 and adding to the third
$$x_2+3x_3=a-b+5$$
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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Does each positive integer can be represent in difference of power sum? Let $a,b,m,n$ are positive integers with, if $m= 1$ then $n\ne 1$ and vice versa.
Let define
$S_m(a)=1^m+2^m+...+a^m$
and $S_n(b)=1^n+2^n+...+b^n$
Question
Show that every $k\in \mathbb{N}$ can be represented as
$$ k = S_m(a)-S_n(b)$$
Example
$1=1^2+2^2-(1+2)$
$2= 1+2-(1^2)$
$3= 1^4+2^4-(1^2+2^2+3^2)$
$4= 1^2+2^2-(1^2)$
$5= 1^2+2^2+3^2-(1^3+2^3)$
$6= 1^3+2^3+3^3-(1^2+2^2+3^2+4^2)$
$7= 1^4+2^4-(1+2+3+4)$
$8= 1^4+2^4-(1^3+2^3)$
$9=1^2+2^2+3^2-(1^2+2^2)$
$10= 1+2+3+4+5-(1^2+2^2)$
$\vdots$
| Probably, most positive integers $k$ cannot be represented in this way, though proving such a result is, I believe, rather out of range of current technology. Perhaps $k=163$ is the first non-representable $k$, though I didn't try very hard to verify this (actually proving this for even one fixed $k$ is likely extremely hard).
| {
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"timestamp": "2023-03-29T00:00:00",
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If $~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ $, find $~ a+b=~?$ If $$~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ ,$$ find $~ a+b=~?$
I get $a-16b = 24$,
But how to get $a+b$?
a,b is positive integer
| $\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32$
Define $k(x) \equiv
\sqrt{64x^2 + ax + 7} - 8x$
$k(x)= (\sqrt{64x^2 + ax + 7} - 8x) \frac{\sqrt{64x^2 + ax + 7} + 8x}{\sqrt{64x^2 + ax + 7} + 8x}$
$$=\frac{64x^2 + ax + 7 - 64x^2}{\sqrt{64x^2 + ax + 7} + 8x}$$
$$=\frac{a + \frac7x}{\sqrt{64 + \frac ax + \frac 7 {x^2}} + 8}$$
So $\lim_{x\rightarrow\infty}k(x) = \frac a {16}$
And by assumption, $\lim_{x\rightarrow \infty} k(x) + b = \frac 3 2$, so
$$\frac a {16} + b = \frac 3 2$$
This is a necessary and sufficient condition for the conditions stated, meaning there are many possible solutions. For example, setting $a = 0, b = \frac 3 2$ works and then $a + b = \frac 3 2$.
| {
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Limit of $f(x, y)$ at $(0, 0)$ I need to show
$$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0
$$
Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.
Edit: I do have the inequality:
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} \le \frac{xy(x^2+y^2)}{(x^2+y^2)^{3/2}} = \frac{xy}{(x^2+y^2)} $$
| Let $y=tx$. Then
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}=\frac{x^2tx^2(1-t^2)}{x^3(1+t^2)^{3/2}}=x\frac{t(1-t^2)}{(1+t^2)^{3/2}}.$$
As the fraction remains bounded for all values of $t$, the limit is $0$.
| {
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how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$
right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$
left $\ne$ right ,what to do next?
$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$
$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $
so that $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$
this solution is right.
| Following your solution from here
$$...= \lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=...$$
we have that
$$\sqrt{1+\frac{1}{n}}-1=\frac1{2n}-\frac1{8n^2}+O\left(\frac1{n^3}\right)$$
and therefore
$$\lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=\lim_{n\to\infty}{n\ln{\left(1-\frac1{4n}+O\left(\frac1{n^2}\right)\right)}}=-\frac14$$
| {
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"answer_id": 3
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Proof of the identity $F_{(n+1)k}=\frac{F_{2k}}{F_k}F_{nk}+(-1)^{k+1}F_{(n-1)k}\quad\forall k,n\in\mathbb{N}$ I'm reading the solution to an exercise I have been given in which they use the identity
$$F_{(n+1)k}=\frac{F_{2k}}{F_k}F_{nk}+(-1)^{k+1}F_{(n-1)k}\quad\forall k,n\in\mathbb{N}$$
where $F_n$ denotes the $n$th Fibonacci number. Is this a commonly known identity? Is there any way to obtain this identity in a relatively straightforward manner?
| $$ \varphi = \frac{1+\sqrt5}{2}$$
$$ F_n = \frac{\varphi^n +(-1)^{n+1}\cdot \varphi^{-n}}{\sqrt5} $$
Example: $$ F_{13} = \frac{\varphi^{13} + \varphi^{-13}}{\sqrt5} = 233 $$
Then: $$ \frac{F_{2k}}{F_k}=\frac{\frac{\varphi^{2k} +(-1)^{2k+1}\cdot \varphi^{-2k}}{\sqrt5}}{\frac{\varphi^k +(-1)^{k+1}\cdot \varphi^{-k}}{\sqrt5}} = \frac{\varphi^{2k} +(-1)^{2k+1}\cdot \varphi^{-2k}}{\varphi^{k} +(-1)^{k+1}\cdot \varphi^{-k}} =$$
$$ \frac{\varphi^{2k} - (-\varphi)^{-2k}}{\varphi^{k} - (-\varphi)^{-k}} = \varphi^{k} + (-\varphi^{-1})^{k}$$
and set:
$$ \varphi^k = a $$
$$ (-\varphi^{-1})^k = b $$
$$ G_n = F_{kn} = \frac{a^n - b^n}{\sqrt5} $$
Test:
$$ G_{n+1} = \frac {G_2}{G_1}G_n-(-1)^kG_{n-1}$$
$$ G_{n+1} + (-1)^kG_{n-1} = \frac {G_2}{G_1}G_n$$
$$ \frac{a^{n+1} - b^{n+1}}{\sqrt5} + (-1)^k\frac{a^{n-1} - b^{n-1}}{\sqrt5} = (a+b)\frac{a^n - b^n}{\sqrt5}$$
$$ a^{n+1} - b^{n+1}+ (-1)^k(a^{n-1} - b^{n-1}) = (a+b)(a^n - b^n)$$
$$ a^{n+1} - b^{n+1}+ (-1)^k(a^{n-1} - b^{n-1}) = a^{n+1} - b^{n+1} + ba^n - ab^n$$
$$ (-1)^k(a^{n-1} - b^{n-1}) = ba^n - ab^n $$
$$ (-1)^k(a^{n-1} - b^{n-1}) = ab(a^{n-1} - b^{n-1}) $$
$$ (-1)^k = ab \lor a^{n-1} - b^{n-1} = 0 $$
But:
$$ ab = \varphi^k \cdot (-\varphi^{-1})^k = \varphi^k \cdot \varphi^{-k} \cdot (-1)^k = (-1)^k $$
Demostrated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$, find $x$ If $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$, find $x$.
My working: if $x$ is positive then by estimation it must be in $(6,7)$ and for this interval I : $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor<2003$. So no solution for positive $x$.
For negative $x$, I got $x \in (-7,-6)$ and using this I got $x=\frac{2018}{k}, k$ can be $-336,-335,\cdots,-289,$
Now using excel sheet I got $x=-\frac{2018}{305}$.
| $$x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$$
Let $f(x) = x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor$
Let $\xi = -\dfrac{6862735}{1037232} = -6 \dfrac{639343}{1037232}$
Then $f(\xi) - 2018 = -\dfrac{1}{1037232} = -\dfrac{1}{2^4 3^3 7^4}$
Let $x = \xi + \delta$. I want to solve $f(x)-2018 = 0$ for $\delta$.
I will assume that $\delta$ is very small. What I mean by that will make itself clear as I progress.
$\lfloor x \rfloor = -7$
$x \lfloor x \rfloor
= -7\xi - 7\delta
= 46 \dfrac{46639}{148176} - 7\delta$
$\lfloor x \lfloor x \rfloor \rfloor = 46$
$x \lfloor x \lfloor x \rfloor \rfloor
= 46\xi + 46 \delta
= -304 \dfrac{183641}{518616} + 46 \delta$
$\lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor = -305$
$x \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor
= -305\xi -305\delta
= 2017 \dfrac{1037231}{1037232} - 305 \delta $
\begin{align}
f(x) &= 0 \\
\left(2017 \dfrac{1037231}{1037232} - 305 \delta \right) - 2018 &= 0 \\
305 \delta &= -1/1037232 \\
\delta &= -\dfrac{1}{316355760}
\end{align}
$\color{red}{x = \xi + \delta = -\dfrac{2018}{305} = -6 \dfrac{188}{305}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$ I have this to solve :
Let $x,y,z>0$ such that $x+y+z=3$ then we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$
I try to use Jensen's inequality but the function $f(x)=\frac{x^2}{4x^3+3}$ is neither concave or convex on the interval $[0,3]$
I can't use Karamata's inequality too .
Maybe brute force is the only way to solve it .
I try also to use the derivative but it becomes a little bit difficult .
In fact my idea was to use rearrangment inequality we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{x^3}{4x^3+3}+\frac{y^3}{4y^3+3}+\frac{z^3}{4z^3+3}$$
An use the inequality of Jensen's on $[0.8,1.2]$ with $f(x)=\frac{x^3}{4x^3+3}$
So it's a partial answer .
My question is how to complete my answer or can you provide an other answer ?
Thanks a lot for sharing your knowledge and your time .
| Using AM-GM, we have
$$4y^3+3=y^3+y^3+y^3+y^3+1+1+1\geq 7\left((y^3)^41^3\right)^{1/7}=7y^{12/7}.$$
So
$$\frac{xy^2}{4y^3+3}\leq\frac{xy^2}{7y^{12/7}}=\frac{xy^{2/7}}{7}.$$
Now, note by Holder's inequality that
$$\sum_{cyc}xy^{2/7}\leq (x+y+z)^{5/7}(xy+yz+zx)^{2/7}.$$
By Cauchy, $xy+yz+zx\leq \frac{(x+y+z)^2}{3}$. Thus,
$$\sum_{cyc}xy^{2/7}\leq \frac{(x+y+z)^{9/7}}{3^{2/7}}.$$ So
$$\sum_{cyc}\frac{xy^2}{4y^3+3}\leq \frac{\sum_{cyc}xy^{2/7}}{7}\leq \frac{(x+y+z)^{9/7}}{7\cdot 3^{2/7}}.$$
Thus, $\sum_{cyc}\frac{xy^2}{4y^3+3}\le\frac{(x+y+z)^{9/7}}{7\cdot 3^{2/7}}$ for all $x,y,z\ge 0$, where the equality case is $x=y=z=1$. When $x+y+z=3$, we get
$$\sum_{cyc}\frac{xy^2}{4y^3+3}\leq \frac{3^{9/7}}{7\cdot 3^{2/7}}=\frac{3}{7}.$$
The equality holds iff $x=y=z=1$.
More generally, for non-negative real numbers $x,y,z$, for parameters $a,b>0$, and for real exponents $m,n$ such that $$(m-1)(a+b)\le an\le m(a+b),$$ we have $$\sum_{cyc}\frac{xy^m}{ay^n+b}\leq \frac{\sum_{cyc}xy^{\frac{m(a+b)-an}{a+b}}}{a+b}\le \frac{\left(\sum_{cyc}x\right)^{\frac{an-(m-1)(a+b)}{a+b}}\left(\sum_{cyc}xy\right)^{\frac{m(a+b)-an}{a+b}}}{a+b}\leq \frac{\left(\sum_{cyc}x\right)^{\frac{(m+1)(a+b)-an}{a+b}}}{3^{\frac{m(a+b)-an}{a+b}}(a+b)}.$$
The equality case is $x=y=z=1$. In particular if, in addition, $x+y+z=3$, we get
$$\sum_{cyc}\frac{xy^m}{ay^n+b}\leq \frac{3}{a+b}.$$
The equality holds iff $x=y=z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Partial derivatives problem, find $\frac{\partial z}{\partial x}$ Assume that $z$ and $w$ are differentiable functions of $x$ and $y$, satisfying the equations $xw^3+yz^2+z^3=-1$ and $zw^3-xz^3+y^2w=1.$ Then, at $\left(x,y,z,w\right)=\left(1,-1,-1,1\right)$, $\ \frac{\partial z}{\partial x}=$?
My solution
$xw^3+yz^2+z^3=-1$
$\frac{\partial}{\partial x}\left(xw^3+yz^2+z^3\right)=0
\Rightarrow w^3+3xw^2\frac{\partial w}{\partial x}+2yz\frac{\partial z}{\partial x}+3z^2\frac{\partial z}{\partial x}=0$
$zw^3-xz^3+y^2w=1$
$\frac{\partial }{\partial x}\left(zw^3-xz^3+y^2w\right)=0
\Rightarrow w^3\frac{\partial z}{\partial x}+3zw^2\frac{\partial w}{\partial x}-z^3-3xz^2\frac{\partial z}{\partial x}+y^2\frac{\partial w}{\partial x}=0$
$\Rightarrow
\begin{cases}
3xw^2\frac{\partial w}{\partial x}+\left(2yz+3z^2\right)\frac{\partial z}{\partial x}=-w^3 \\
\left(3zw^2+y^2\right)\frac{\partial w}{\partial x}+\left(w^3-3xz^2\right)\frac{\partial z}{\partial x}=z^3
\end{cases}$
When $\left(x,y,z,w\right)=\left(1,-1,-1,1\right)$, we have
$
\begin{cases}
3\frac{\partial w}{\partial x}+5\frac{\partial z}{\partial x}=-1 \\
-2\frac{\partial w}{\partial x}-2\frac{\partial z}{\partial x}=-1
\end{cases}
$.
Therefore $\frac{\partial z}{\partial x}=-\frac{5}{4}$
Is there a problem with this solution? Or is this a better solution?
| Bavo, your solution is good. The method is correct and simple.
If an even simpler method is wanted, Mr.Mathematician asks his friend Mr.Physicist. But the purists will not be fully happy :
$$\begin{cases}
xw^3+yz^2+z^3=-1\\
zw^3-xz^3+y^2w=1
\end{cases}\quad\to\quad
\begin{cases}
(1+dx)(1+dw)^3+(-1)(-1+dz)^2+(-1+dz)^3=-1\\
(-1+dz)(1+dw)^3-(-1+dx)(-1+dz)^3+(-1)^2(1+dw)=1
\end{cases}$$
After simplification with keeping only the first order variational terms :
\begin{cases}
-5dz-3dw=dx\\
2dz+2dw=dx
\end{cases}
Solving the linear system for $dz$ and $dw$ :
$$\left(\begin{matrix}
dz\\
dw
\end{matrix}\right)=
\left(\begin{matrix}
-5 & -3\\
2 & 2
\end{matrix}\right)^{-1}
\left(\begin{matrix}
1\\
1
\end{matrix}\right)dx =
\left(\begin{matrix}
-\frac54\\
\frac74
\end{matrix}\right)dx $$
$$\quad\frac{\partial z}{\partial x}=-\frac54\quad \text{and} \quad\frac{\partial w}{\partial x}=\frac74$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$
Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
| Use that $$\cos^2(x)=1-\sin^2(x)$$ and your integral will be
$$\int\sin^4(x)-\sin^6(x)dx$$
For your work:
$$\int \sin(x)^4dx=-1/4\, \left( \left( \sin \left( x \right) \right) ^{3}+3/2\,\sin
\left( x \right) \right) \cos \left( x \right) +3/8\,x
+C$$
and
$$\int\sin(x)^6dx=-1/6\, \left( \left( \sin \left( x \right) \right) ^{5}+5/4\,
\left( \sin \left( x \right) \right) ^{3}+{\frac {15\,\sin \left( x
\right) }{8}} \right) \cos \left( x \right) +{\frac {5\,x}{16}}
+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Proving $\sum _{k=1}^{\infty } J_k(k z)^2=\frac{1}{2 \sqrt{1-z^2}}-\frac{1}{2}$ How to prove
$$\sum _{k=1}^{\infty } J_k(k z){}^2=\frac{1}{2 \sqrt{1-z^2}}-\frac{1}{2},\ |z|<1$$
The supplementary one $\sum _{k=1}^{\infty } J_k(k z)=\frac{z}{2 (1-z)}$ might be useful. Any help will be appreciated.
| We can use the integral representation for the product of Bessel functions DLMF
\begin{equation}
J_{\mu}\left(x\right)J_{\nu}\left(x\right)=\frac{2}{\pi}\int_{0}^{\pi/2}J_{\mu+\nu}\left(2x\cos\theta\right)\cos\left((\mu-\nu)\theta\right)\mathrm{d}\theta
\end{equation}
with $\mu=\nu=k$ and $x=kz$:
\begin{equation}
J_{k}^2\left(kz\right)=\frac{2}{\pi}\int_{0}^{\pi/2}J_{2k}\left(2kz\cos\theta\right)\mathrm{d}\theta
\end{equation}
From the quoted series, and using parity properties of the Bessel function,
\begin{equation}
\sum _{k=1}^{\infty } J_k(k x)=\frac{x}{2 (1-x)}
\end{equation}
gives
\begin{align}
2\sum _{k=1}^{\infty } J_{2k}(2k x)&=\sum _{k=1}^{\infty } J_k(k x)+\sum _{k=1}^{\infty } J_k(-k x)\\
&=\frac{x}{2 (1-x)}-\frac{x}{2 (1+x)}\\
&=\frac{x^2}{1-x^2}
\end{align}
Then
\begin{align}
\sum_{k\ge1}J_{k}^2\left(kz\right)&=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k\ge1}J_{2k}\left(2kz\cos\theta\right)\mathrm{d}\theta\\
&=\frac{1}{\pi}\int_{0}^{\pi/2}\frac{z^2\cos^2\theta}{1-z^2\cos^2\theta}\mathrm{d}\theta\\
&=\frac{1}{\pi}\left[-\frac{\pi}{2}+\int_{0}^{\pi/2}\frac{1}{1-z^2\cos^2\theta}\mathrm{d}\theta\right]\\
&=-\frac12+\frac1\pi\int_0^\infty \frac{dt}{t^2+1-z^2} \text{ ...by changing }t=\tan\theta\\
&=\frac{1}{2\sqrt{1-z^2}}-\frac{1}{2}
\end{align}
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Verifying the integral solution: $I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx$
Evaluate:
$$
I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx
$$
I got an answer which does not match the keys section of the book I'm solving and want to verify which one is correct.
Start by recalling:
$$
\cos 2x = \cos^2 x-\sin^2 x
$$
Therefore:
$$
\sqrt{1+2\cos x} =\sqrt{1+\cos^2 x - \sin^2 x} = \sqrt{2\cos^2 x} = \sqrt2|\cos x|
$$
Thus:
$$
I = \int \frac{\sin x}{\sqrt2 |\cos x|}dx
$$
Substitute $t=\cos x$, then $dt = -\sin x dx$, $dx = - {dt\over \sin x}$, hence:
$$
I = -{1\over \sqrt 2}\int \frac{dt}{|t|} = \boxed{-{\text{sgn}(\cos x)\over \sqrt 2}\ln(|\cos x|)+C}
$$
While the answer section suggests:
$$
I = -\sqrt{1+2\cos x} + C
$$
Which one is correct?
| Take $u=1+2\cos{x}$
$\frac{-1}{2}du=\sin{x}dx$
So the integral becomes $$\frac{-1}{2}\int \frac{1}{\sqrt{u}}du=-\sqrt{u}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be equal to $\frac{\sqrt 2}2-i \frac{\sqrt 2}2$?
I used Moivre's formula. But I dont know. How can I must choose value of $k$ in Moivre's formula? I chose $k=0$.
$$z=r\left(\cos x+i\sin x\right)$$
$$z^{\frac 1n}=r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n}\right)$$
| Let $\sqrt{-i} = a + bi$, so $-i = (a+bi)^2 = a^2 - b^2 + 2abi$.
Equating real values, $0 = a^2 - b^2$, and equating imaginery parts, $-1 = 2ab$.
Solving these simultaneously gives us either $a=\frac{1}{\sqrt 2}, b=\frac{-1}{\sqrt 2}$ or $a=\frac{-1}{\sqrt 2}, b=\frac{1}{\sqrt 2}$.
Thus, $\sqrt{-i} = \pm \frac{1}{\sqrt 2}(1-i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Is $\sum_{k=1}^{n} \cos(k^ {1/3} )$ bounded by a constant M? I understand what $\sum_{k=1}^{n} \cos(k)$ is bounded by a constant, but I don't have any idea how to prove if $\sum_{k=1}^{n} \cos(k^ {1/3} )$ is bounded or not.
| Not bounded. The blocks where $\cos(^{1/3})>1/2$
have length going to $\infty$.
First, $\cos(z)>1/2$ if $-\pi/3 < z < \pi/3$.
For each $n \in \mathbb Z$ we have $\cos(z)>1/2$ if $2\pi n-\pi/3 < z < 2\pi n+\pi/3$.
$\cos(k^{1/3}) > 1/2$ if $2\pi n-\pi/3 < k^{1/3} < 2\pi n+\pi/3$
$\cos(k^{1/3}) > 1/2$ if $(2\pi n-\pi/3)^3 < k < (2\pi n+\pi/3)^3$
The interval from $(2\pi n-\pi/3)^3$ to $(2\pi n+\pi/3)^3$ has length
$$
\left(2\pi n+\frac{\pi}{3}\right)^3 - \left(2\pi n-\frac{\pi}{3}\right)^3 =
8\pi^3 n^2 + \frac{2\pi^3}{27}
$$
The number of integers in the inverval from $(2\pi n-\pi/3)^3$ to $(2\pi n+\pi/3)^3$ is at least
$$
8\pi^3 n^2 + \frac{2\pi^3}{27} - 2.
$$
So: consider the difference between $\sum_{k=1}^{N_1}\cos(k^{1/3})$
and $\sum_{k=1}^{N_2}\cos(k^{1/3})$ with these two values of $N$:
$$
N_1 = \left\lfloor\left(2\pi n-\frac{\pi}{3} \right)^3\right\rfloor\quad
\text{and}\quad
N_2 = \left\lfloor\left(2\pi n+\frac{\pi}{3}\right)^3\right\rfloor
$$
The difference will be
$$
\sum_{1}^{N_2} \cos(k^{1/3}) - \sum_{1}^{N_1} \cos(k^{1/3}) =
\sum_{k=N_1+1}^{N_2} \cos(k^{1/3}) > \sum_{k=N_1+1}^{N_2} \frac{1}{2} \ge 4\pi^3 n^2+\frac{\pi^3}{27}-1
$$
And this goes to $\infty$. Therefore,
$$
\text{either}\quad
\limsup_{N \to \infty}\sum_{1}^{N} \cos(k^{1/3}) = +\infty
\quad\text{or}\quad
\liminf_{N \to \infty}\sum_{1}^{N} \cos(k^{1/3}) = -\infty
$$
or, likely, both.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find number of distinct terms in a multinomial expansion?
Find the number of distinct terms in the expansion of
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}$$
(with respect to powers of $x$)
I saw that the formula for the number of distinct terms (or dissimilar) in a multinomial expansion $(x_1+x_2+x_3+...+x_k)^n$ is $$\binom{n+k-1}{k-1}$$
But applying that here means $$\binom{15+4-1}{4-1}= \binom{18}{3} = 816$$
But the answer says 61.
Is there a difference between the situations for which that formula is meant to be used and that in which I am using??
| Write it as:
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}=\frac{(x^4+x^3+x+1)^{15}}{x^{30}}$$
So, it is equivalent to:
$$(x^4+x^3+x+1)^{15}=(x^3+1)^{15}(x+1)^{15}=\\
(x^{45}+15x^{42}+\cdots+15x^3+1)(x^{15}+15x^{14}+\cdots+15x+1)=\\
x^{60}+30x^{59}+\cdots15x+1.$$
Hence, there are $61$ distinct powers of $x$.
Note: You can not use the formula you mentioned directly, because the given polynomial is lacking one term:
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}=\left(x^{-2}+x^{-1}+\color{red}{\require{cancel} \cancel{x^0}}+x+x^2\right)^{15}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$ Consider S=$\{(x,y,z)\mid| 3x^2-4y+2y^2+2z-3=0\}$
How can I determine the volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$? That is the part of $S$ over the x-y axis.
My wrong approach was to integrate $z^2\pi=(\frac{1}{2}(3-2y^2+4y-3x^2))^2\pi$ from $z=0$ to the top $z=5/2$ but as you can see this thing isn't circular. Maybe using cylindrical coordinates?
Furthermore I am looking for equations to describe the bottom and the lateral surface of it.
For the bottom I have
$\{(x,y,z)\mid 3x^2-4y+2y^2-3\le 0, z=0\}$
And the surface:
$\{(x,y,z)\mid\frac{1}{2}(3-2y^2+4y-3x^2)=z,0<z<5/2\}$
| The first thing I would do is "complete the square" (that's the second time today I have written that!). $2y^2- 4y= 2(y^2- 2y)= 2(y^2- 2y+ 1- 1)= 2(y^2- 2y+ 1)- 2= 2(y- 1)^2- 2$. So we can write the formula as $3x^2+ 2(y- 1)^2- 2+ 2z- 3= 0$, $z= 5/2- (3/2)x^2- 2(y- 1)^2$. Now let u= y- 1. We have $z= 5/2- (3/2)x^2- u^2$. In the Xu-plane that is the ellipse, $(3/2)x^2+ u^2= 5/2$. When u= 0, $(3/2)x^2= 5/2$ so $x^2= 5/3$. x goes from $-\sqrt{5/3}$ to $\sqrt{5/3}$. And $u^2= 5/2- (3/2)x^2$ so for each x, u goes from $-\sqrt{5/2- (3/2)x^2}$ to $\sqrt{5/2- (3/3)x^2}$.
The volume is given by $\int_{-\sqrt{5}{3}}^{\sqrt{5/3}}\int_{-\sqrt{5/2- (3/2)x^2}}^{\sqrt{5/2- (3/2)x^2}} \left(5/2- (3/2)x^2- u^2\right) dudx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the coefficient of $x^9$ in the expansion of $(1+9x+27x^2+27x^3)^6$ Find the coefficient of $x^9$ in the expansion of $(1+9x+27x^2+27x^3)^6$.
Consider,
$$(1+9x+27x^2+27x^3)^6 = \sum \frac{6!}{a!\cdot b!\cdot c!\cdot d!}(1^{a}\cdot9x^{b} \cdot (27x^2)^{c} \cdot (27^3)^d)$$
Power of $x=x^{b+2c+d}$
So,
*
*$b +2c +3d = 9$
Also,
*
*$a +b+c+d = 6$
Are we supposed to find different combination of $a,b,c $ and $d$ in such a way that the above $2$ equations are satisfied and then add the coefficients of the different combinations?
I don't know how to proceed further. Any Hint would be appreciated.
| First note that $1+9x+27x^2+27x^3=(1+3x)^3$. So the problem is the same as calculating the coefficient of $x^9$ in $(1+3x)^{18}$. Which is much easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a parametric representation for the line of intersection of the planes
Find a parametric representation for the line of intersection of the planes with equations $x+y+z = 3$ and $2x-y-z = 5$, and find the point where the line intersects the plane with equation $x-y = 2$.
First I get the two norm vectors of the given planes, $v = (1,1,1)$ and $u = (2,-1,-1).
$$\begin{align*}
u\times v &= \begin{pmatrix} i&j&k \\ 1&1&1 \\ 2&-1&-2 \end{pmatrix} = \begin{pmatrix} 1&1 \\ -1&-2\end{pmatrix}i - \begin{pmatrix}1&1 \\ 2&-2 \end{pmatrix}j + \begin{pmatrix}1&1 \\ 2& -1 \end{pmatrix}k \\
&= -1i + 4j - 3k \\
&= (-1,4,-3)
\end{align*}$$
Now after getting the cross product, I have the direction of the line of intersection. Now I need to find the point where the line intersects the plane, so taking the norm vector of $x - y = 2$ gives $(1,-1,0)$ so this can serve as my point of intersection?
Then the parametric representation is $= t(-1,4,-3) + (1,-1,0)$. Did I use all the information correctly or am I understanding parts of this problem wrong?
| We are working with
\begin{cases}
\pi_1: & x+y+z=3\\
\pi_2: & 2x-y-z=5\\
\end{cases}
Since their normal vectors are non-parallel, we know that the planes are not parallel. As such, we know they intersect in a line at some point with some direction vector.
Suppose the direction vector can be written as some scalar multiple of a vector whose $z$ component is $1$. So, we can let $\boxed{z=t}$. Substituting this into the two equations we get
\begin{align}
x+y+t&=3\\
2x-y-t&=5\\
\end{align}
With these two equations, we can solve for $x$ and $y$ in terms of $t$, and those will become the parametric equations of our line. I will leave the solution as an exercise (use either the substitution or elimination method to solve the linear system in terms of $t$). The equations we end up with are
$\boxed{x=\frac83}$ and $\boxed{y=\frac13-t}$
These three boxed equations are the parametric equations for the line of intersection.
$$l:\begin{cases}
x=\frac83\\
y=\frac13-t\\
z=t\end{cases}
$$
Since we want the point of intersection between $l$ and the given plane $x-y=2$, simply substitute the parametric equations of $l$ in and solve for $t$, and then finally substitute that $t$ value back into the parametric equations.
\begin{align}
x-y&=2\\
(\frac83)-(\frac13-t)&=2\\
\frac73+t&=2\\
t&=-\frac13
\end{align}
Therefore, we have the point $(x,y,z)=(\dfrac83,\dfrac13-t,t)=\boxed{(\frac83,\dfrac23,-\frac13)}$.
To check: we know that the point $(\frac83,\frac23,-\frac13)$ has to be a point that exists on all three planes. As we expect, it satisfies all three equations. This hints that we could have solved the planar system without first solving for the line of intersection of the first two planes!
Simply put, we want to find the solution for the system of linear equations
\begin{cases}
x+y+z&=3\\
2x-y-z&=5\\
x-y&=2
\end{cases}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Substitution integral with $t=x^3$
$$
\int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3}
$$
$$
\int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2
$$
$$
\int \frac{9}{x(3+{x})}
$$
$$
9\int \frac{1}{3x+{x^2}}
$$
Now i'm stuck
my calculation so far
| $$\int \frac{3}{t(3 + t^{1/3})} \, dt = \int \frac{1}{3 + t^{1/3}} \cdot 9t^{-1/3} \cdot \frac{1}{3} t^{-2/3} \, dt,$$ so with the substitution $$u = 3 + t^{1/3}, \quad du = \frac{1}{3} t^{-2/3} \, dt, \\ t^{-1/3} = \frac{1}{u-3},$$ we obtain $$\int \frac{3}{t(3+t^{1/3})} \, dt = \int \frac{1}{u} \cdot \frac{9}{u-3} \, du = 3 \int \left( \frac{1}{u-3} - \frac{1}{u} \right) \, du = 3 \log\left|\frac{t}{(3+t^{1/3})^3}\right| + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
On the exponentiation of three-dimensional numbers. There exists a general formula to calculate $({a+bi})^{c+di}$. However, it there a formula to calculate exponentiation with three-dimensional numbers, such as $({a+bi+cj})^{d+ei+fj}$ ?
| Yes. But what exactly 3-dimensional algebra are you talking about? There are several. In the linked post are descriptions of 3 3-dimensional commutative number systems.
If we define
$1=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$
$j=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{array}
\right)$
$k=\left(
\begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{array}
\right)$
then
$(a_1+b_1j+c_1k)^{(a_2+b_2j+c_2k)}=a_2 \left(b_1 k+c_1 j\right)+a_1 \left(a_2+b_2 k+c_2 j\right)+b_2 \left(b_1 j+c_1\right)+c_2 \left(b_1+c_1 k\right)$
If we define
$1=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$
$j=\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$
$k=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \\
\end{array}
\right)$
then
$(a_1+b_1j+c_1k)^{(a_2+b_2j+c_2k)}=\frac{1}{2} (1-j) \left(\left(a_1-b_1+c_1\right){}^{a_2-b_2+c_2}-\left(a_1+b_1+c_1\right){}^{a_2+b_2+c_2}\right)+\frac{1}{2} (1-k) \left(\left(a_1+b_1-c_1\right){}^{a_2+b_2-c_2}-\left(a_1+b_1+c_1\right){}^{a_2+b_2+c_2}\right)+\left(a_1+b_1+c_1\right){}^{a_2+b_2+c_2}$
If we define
$1=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$
$j=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}
\right)$
$k=\left(
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)$
then the formula is
$(a_1+b_1j+c_1k)^{(a_2+b_2j+c_2k)}=\frac{1}{2} a_2^{a_1-2} \left(a_2^2 \left(\log \left(a_2\right) \left(b_1^2 k \log \left(a_2\right)+2 b_1 j+2 c_1 k\right)+2\right)+2 a_2 \left(a_1 \left(b_2 \left(b_1 k \log \left(a_2\right)+j\right)+c_2 k\right)+b_1 b_2 k\right)+\left(a_1-1\right) a_1 b_2^2 k\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difference quotient of differentiable functions $Q. \text { For a differentiable function } f : \mathbb { R } \rightarrow \mathbb { R } \text { define the differential quotient}$
\begin{array} { l }{ \left( D _ { x } f \right) ( h ) = \frac { f ( x + h ) - f ( x ) } { h } ; h > 0 } \\ { \text { consider numbers of the form } \hat { h } = h ( 1 + \epsilon ) \text { for a } } \\ { \text { fixed } \epsilon > 0 \text { and let } } \end{array}
\begin{array} { l } { e _ { 1 } ( h ) = f ^ { \prime } ( x ) - \left( D _ { x } f \right) ( h ) } \\ { e _ { 2 } ( h ) = \left( D _ { x } f \right) ( h ) - \left( D _ { x } f \right) ( \hat { h } ) } \\ { e ( h ) = e _ { 1 } ( h ) + e _ { 2 } ( h ) } \\ { \text { If } f ( x + \hat { h } ) = f ( x + h ) , \text { then, which are true, } } \end{array}
\begin{array} { l } { \text { (1) } e _ { 1 } ( h ) \rightarrow 0 \text { as } h \rightarrow 0 } \\ { \text { (2) } e _ { 2 } ( h ) \rightarrow 0 \text { as } h \rightarrow 0 } \\ { \text { (3) } e _ { 2 } ( h ) \rightarrow \epsilon f ^ { \prime } ( x ) / ( 1 + \epsilon ) \text { as } h \rightarrow 0 } \\ { \text { (4) } e(h)\rightarrow 0 \text { as } h \rightarrow 0 } \end{array}
My work:
\begin{array} { l } { \text { Here } f \left( x + \hat h \right) = f ( x + h ) , \quad \hat h = h ( 1 + \epsilon ) ,
\quad \epsilon >0 \text { fixed. } } \\ { \qquad \begin{aligned} \Rightarrow D _ { x } f ( \hat h ) & = \frac { f ( x + \hat h ) - f ( x ) } { \hat h } \\ & = \frac { f ( x + h ) - f ( x ) } { h ( 1 + \epsilon ) } \\ & \rightarrow \frac { f ^ { \prime } ( x ) } { 1 + \epsilon } \text { as } h \rightarrow 0 \end{aligned} } \end{array}
\begin{array} { l } { \text { Also } h \rightarrow 0 \Rightarrow \hat h \rightarrow 0 } \\ { \begin{aligned} \Rightarrow & D _ { x } f ( \hat { h } ) = \frac { f ( x + \hat { h } ) - f ( x ) } { \hat h } \rightarrow f ^ { \prime } ( x ) \\ & \text { as } h \rightarrow 0 \\ & \text { Hence } \frac { f ^ { \prime } ( x ) } { 1 + \varepsilon } = f ^ { \prime } ( x ) \end{aligned} } \end{array}
$\Rightarrow \epsilon = 0 , \text { Where the mistake happened? }$
| Not sure if this is what your looking for, but your last implication is incorrect since
$\frac{f'(x)}{1+ \varepsilon} = f'(x)$ and $\varepsilon >0$ implies $$f'(x) = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt{2} +\sqrt{n}$ is irrational Suppose $\sqrt{2} + \sqrt{n}$ is rational,
so it can be represented like so: $\sqrt{2} + \sqrt{n} = {x \over y}$
then $\sqrt{2} = {{x \over y} - \sqrt{n}}$
so $\sqrt{2} = {x - y\sqrt{n}\over y}$
but $\sqrt{2}$ is irrational, which leads to a contradiction.
Does this proof make sense?
edit: fixed some mistakes
| The polynomial $$\left(x^2-\left(\sqrt{2}-\sqrt{n}\right)^2\right)
\left(x^2-\left(\sqrt{2}+\sqrt{n}\right)^2\right) = x^4-2(n+2)\,x^2+(n-2)^2$$
has $\sqrt2+\sqrt n$ as one of its roots. Any rational divisor would have to be an integer factor of $(n-2)^2$ by the rational roots theorem.
But $\sqrt 2+\sqrt n$ is not an integer, indeed, if $\sqrt 2+\sqrt n=a\in\mathbb Z$, then $n=a^2+2-2a\sqrt 2$, implying that $2a\sqrt 2\in\mathbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
reducing $\cos{x} + 2\sin{x}$ to $R\cos{x - a}$ finds strange phase-shift So, I'm analysing the function $\cos{x} + 2\sin{x} (*)$ and determining whether it is convertible to the single function $R\cos{x - a}$ for $a \in (0, \frac{\pi}{2}); R > 0$; so I begin by observing the maximum/minimum of $(*)$ and determining its max./min. at $\pm\sqrt{5}$, so then I apparently have that $R = \sqrt{5}$ as the radius/magnitude of the corresponding function.
Next, I seek to find the phase-shift difference between my $\sqrt{5}\cos{x}$ and $\cos{x} + 2\sin{x}$, so, I take two reference points at their relative roots, i.e. $\cos{x} + 2\sin{x} = 0 \iff x = -\arctan{\frac{1}{2}} + n\pi, n \in \mathbb{Z}$, and respectively, $\sqrt{5}\cos{x} = 0 \iff x = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$. So I figure that if I take the x's matching with the roots of the functions in the same period, then I should have the phase-shift and my function mutation should be done.
I take $n = 1$ for both functions, so I have $x_{\sqrt{5}\cos{x}} = \pi + \frac{\pi}{2}$ and $x_{\cos{x} + 2\sin{x}} = \pi - \arctan{\frac{1}{2}}$. Then I take their absolute difference, and theoretically this should be the shift. So, $|x_{\sqrt{5}\cos{x}} - x_{\cos{x} + 2\sin{x}}| = |[\pi + \frac{\pi}{2}] - [\pi - \arctan{\frac{1}{2}}]| = |\frac{\pi}{2} + \arctan{\frac{1}{2}}| \approx 2.034$.
So immediately this evaluation seems incorrect; 2.034 radians is well over half a period, and according to plots of these functions the difference does not seem this drastic.
Observe the plot for $y = \sqrt{5}\cos{x - (\frac{\pi}{2} + \arctan{\frac{1}{2}})}$ and $y = \cos{x} + 2\sin{x}$
Evidently incorrect, so I then plot $x = -\arctan{\frac{1}{2}}$ and $y = \cos{x} + 2\sin{x}$ to actually determine where this x-value was being matched:
For some reason, the $-\arctan{\frac{1}{2}}$ was in a completely separate period relative to $\sqrt{5}\cos{x}$ at $n=1$. And here my question(s) arises: why is this root observed in a separate period? How can I find where the start of these periods are? Is there a better method to finding the difference between these trigonometric functions besides at the roots?
| Let's say you want to write, with $r\ge0$:
$$a\cos \theta+b\sin\theta=r\cos(\theta-\phi)=r\cos \phi\cos \theta+r\sin \phi\sin \theta$$
We are done if $a=r\cos\phi$ and $b=r\sin\phi$, hence $a^2+b^2=r^2$, which gives $r=\sqrt{a^2+b^2}$.
Then any $\phi$ such that $\cos\phi=\dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi=\dfrac{b}{\sqrt{a^2+b^2}}$.
If $a>0$, it's equivalent to find $\phi$ such that $\tan\phi=\dfrac ba$, hence $\phi=\arctan(\frac ba)$.
The other cases, to get $\phi\in(-\pi,\pi]$:
*
*If $a<0$ and $b<0$, $\phi=\arctan(\frac ba)-\pi$
*If $a<0$ and $b\ge0$, $\phi=\arctan(\frac ba)+\pi$
*If $a=0$ and $b\ne0$, $\phi=\mathrm{sign}(b)\dfrac \pi2$
*If $a=b=0$, $\phi$ is not well defined.
In your case, $a=1$ and $b=2$, so $r=\sqrt5$ and $\phi=\arctan2\in(0,\pi/2)$.
Notice that this is equivalent to convert the complex number $a+ib$ to polar form $r\exp(i\phi)$. The value of $\phi$ given here is the principal value of the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we simply further the expression? We have the set $\left \{z\in \mathbb{C}\mid z\neq 1, \ \left |\frac{z}{z-1}\right |\leq 1\right \}$
Let $z=x+yi$.
We have that \begin{align*}\frac{z}{z-1}&=\frac{x+yi}{(x-1)+yi}=\frac{\left (x+yi\right )\left ((x-1)-yi\right )}{\left ((x-1)+yi\right )\left ((x-1)-yi\right )} \\ & =\frac{x(x-1)+y^2-yi}{(x-1)^2+y^2} =\frac{x(x-1)+y^2}{(x-1)^2+y^2}+\frac{-y}{(x-1)^2+y^2}i\end{align*}
Then we get \begin{align*}\left |\frac{z}{z-1}\right |&=\sqrt{\left (\frac{x(x-1)+y^2}{(x-1)^2+y^2}\right )^2+\left (\frac{-y}{(x-1)^2+y^2}\right )^2} \\ & =\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4}{[(x-1)^2+y^2]^2}+\frac{y^2}{[(x-1)^2+y^2]^2}} \\ & = \sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}} \end{align*}
can we simplify that further?
Is not, we have to solve the inequality from that point. So we have from $\left |\frac{z}{z-1}\right |\leq 1$ the following:
\begin{align*}&\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}}\leq 1 \\ & \Rightarrow 0\leq \frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}\leq 1 \\ & \Rightarrow 0\leq x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq [(x-1)^2+y^2]^2 \end{align*}
From the second inequality we have \begin{align*}&x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq (x-1)^4+2(x-1)^2y^2+y^4 \\ & \Rightarrow x^2(x-1)^2+2x(x-1)y^2+y^2\leq (x-1)^4+2(x-1)^2y^2\end{align*} How do we continue from here? I got stuck right now.
| We have
$$\left |\frac{z}{z-1}\right |=\frac{|z|}{|z-1|}=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2-2x+1}}\leq 1 \iff x\le \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
General formula for the partial sum of the series $\sum_{i=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$
How would I find the formula for the partial sum $S_n $ of this series? $$\sum_{k=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$$
I know that $S_1=\ln(3/4) $, $S_2=\ln(2/3) $, $S_3 =\ln(5/8)$, and $S_4=\ln(3/5)$, but I don't see any pattern that would allow me to write the partial sum formula.
| Consider the partial sum,
$$S_{n} = \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) $$
which can be evaluated in two ways. The first being:
\begin{align}
S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) \\
&= \sum_{k=1}^{n} ( \ln(k) + \ln(k+2) - 2 \, \ln(k+1) ) \\
&= \sum_{k=2}^{n} \ln(k) + \sum_{k=3}^{n+2} \ln(k) - 2 \, \sum_{k=2}^{n+1} \ln(k) \\
&= \sum_{k=3}^{n+2} \ln(k) - \sum_{k=2}^{n+1} \ln(k) - \ln(n+1) \\
&= \ln(n+2) - \ln(n+1) - \ln(2) = \ln\left(\frac{n+2}{2 \, (n+1)} \right).
\end{align}
The second is:
\begin{align}
S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^2}\right) = \ln\left(\prod_{k=1}^{n} \frac{k(k+2)}{(k+1)^2}\right) \\
&= \ln\left(\frac{n! \, \prod_{k=3}^{n+2} k}{\prod_{k=2}^{n+1} k^2} \right) \\
&= \ln\left(\frac{n! \, (n+2)!}{2! \, ((n+1)!)^2}\right) = \ln\left(\frac{n! \, (n+2)}{2 \, (n+1)!} \right) \\
&= \ln\left(\frac{n+2}{2(n+1)}\right).
\end{align}
In both cases
$$S_{n} = \ln\left(\frac{1 + \frac{2}{n}}{2 \, \left(1 + \frac{1}{n}\right)} \right)$$
such that as $n \to \infty$ the result becomes $- \ln(2)$ which yields
$$\sum_{k=1}^{\infty} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) = - \ln(2).$$
| {
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"url": "https://math.stackexchange.com/questions/3395563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$ QUESTION: Let $ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$
(A) Minimum value of $ƒ(x,y)= 5+\sqrt2$
(B) Minimum value of $ƒ(x,y)= 5-\sqrt2$
(C) Minimum value occurs of $ƒ(x,y)$ for $x=\frac{3}{7}$
(D) Minimum value occurs of $ƒ(x,y)$ for $y=\frac{4}{7}$
My approach , all values in the square roots needs to be positive.
${x^2+y^2\ge 0}$ hence it encloses the whole graph
${x^2+y^2-2x+1}$,${(x-1)^2+y^2\ge 0}$
$x^2+y^2-2y+1$, $ x^2+(y-1)^2\ge 0$
$x^2+y^2-6x-8y+25$, $(x-3)^2+(y-4)^2\ge 0$
Not able to approach from here
| Hint. Note that the centers of the four circles are:
$$O(0,0),A(0,1),B(1,0),C(3,4)$$
For the shortest distance, the point $(x,y)$ must lie on the intersection of the line passing through $(0,0)$ and $(3,4)$ and the line passing through $(0,1)$ and $(1,0)$. Hence:
$$|OC|+|AB|=5+\sqrt2.$$
| {
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Three polynomials which have the same value for a variable
Let $P_1(x)= ax^2-bx-c$, $P_2(x)=bx^2-cx-a$, and $P_3(x)= cx^2-ax-b$ be three quadratic polynomials, where $a,b$, and $c$ are non-zero real numbers. Suppose that there exists a real number $k$ such that $$ P_1(k) = P_2(k)= P_3(k) $$ Prove that $a=b=c$.
Hello everybody! The above is a question I got stuck on. This problem is from an Indian Olympiad. Here is what I did:
After manipulating the given information, we derive at $(a-b)k^2-(b-c)k-(c-a)=0$ $(b-c)k^2-(c-a)k-(a-b)=0 $ and $(c-a)k^2 - (a-b)k - (b-c)$. Now the solution involves some addition and subtraction to factor it and get $a=b=c$.
But my question is: Can’t I just compare the coefficients of the three equations because they are all zero from which we get $b-c = c-a$, $c-a = a-b $ and $a-b = b-c$ and get $a=b=c$? If not, why?
Thanks
| You can't do that because we need to show that this is the unique solution.
As an alternative, since the conditions
*
*$(a-b)k^2-(b-c)k-(c-a)=0$
*$(b-c)k^2-(c-a)k-(a-b)=0 $
*$(c-a)k^2 - (a-b)k - (b-c)=0$
are equivalent to the system $Mx=0$
$$\begin{bmatrix}A&-B&-C\\B&-C&-A\\C&-A&-B\end{bmatrix}\begin{bmatrix}k^2\\k\\1\end{bmatrix} =0$$
which, since $A+B+C=0$, is equivalent to
$$\begin{bmatrix}A&-B&-C\\B&-C&-A\\0&0&0\end{bmatrix}\begin{bmatrix}k^2\\k\\1\end{bmatrix} =0$$
Since $(k^2,k,1)$ and $(1,-1,-1)$ are linearly independent solutions we have that $\ker(M)\ge2$ and therefore
$$\det\begin{bmatrix}A&-B\\B&-C\end{bmatrix} =-AC+B^2=A^2+AB+B^2=(A-B)^2+3AB=0$$
and assuming wlog $a\ge b\ge c$ that is $A\ge B\ge 0$ we have that
$$(A-B)^2+3AB=0 \implies A=B=0 \implies C=0$$
that is $a=b=c$.
| {
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"source": "stackexchange",
"question_score": "3",
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} |
Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3+b^3)$ for $a,b\in\mathbb{N}$. In other words, solutions to the Diophantine Equation:
$a^3+b^3=dc^3$ where $a,b,$ and $c$ are pairwise co-prime and $a,b,c>0$
Obviously there are some trivial solutions. If $c=1$, for example, $a$ and $b$ can be any integers, and $d$ can be chosen to simply be $a^3+b^3$.
By requiring that $a,b,$ and $c$ are pairwise co-prime and that $c\not=1$, we eliminate the trivial solutions, and what remains is of much more interest.
For $a,b,c,d\le20$, there are 5 solutions:
$4^3+5^3=7*3^3$
$2^3+7^3=13*3^3$
$1^3+8^3=19*3^3$
$3^3+5^3=19*2^3$
$1^3+19^3=20*7^3$
Under $100$ there are $16$ solutions, as found by Mathematica.
My question about this equation: Has it been studied previously? Are there infinitely many primitive solutions (which it seems like there are)? If so, can they be parametrized?
| It's actually very easy to identify an infinite number of solutions. Suppose that $m^3+1$ is divisible by $3^3=27$, whichnisctrue for $m=8$. Then
$(m+9)^3+1=(m^3+1)+(27m^2+243m+729)$
is also divisible by $27$. Having $1$ as one of the cubes guarantees relative primarily for all these solutions, but we can reasonably expect infinitely many relatively prime solutions to arise similarly from $m^3+n^3$ for other values of $n$.
| {
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"source": "stackexchange",
"question_score": "8",
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Expected hitting time of a certain state in a Markov chain
Let's consider the following Markov chain. If we start at state $2$, compute the probability to hit $4$ and the expected time until it happens.
The probability to hit $4$ in $n$ steps starting at $2$ is $Pr(X_n=4|X_0=2)=(\frac{1}{2})^n$ if $n$ is even, and $0$ if $n$ is odd. Then, the total probability of hitting 4 if we start at 2 will be$\sum_{n\geq1}Pr(X_n=4|X_0=2)=\sum_{i\geq1}(\frac{1}{2})^{2i}=\sum_{i\geq1}\frac{1}{4^i}=\frac{1}{3}$
How can I calculate $\mathbb{E}(T_{24})?$ I have tried considering the mean hitting times $m_{ij}$, so $m_{24}=1+\frac{1}{2}m_{14}+\frac{1}{2}m_{34}$ and $m_{34}=1+\frac{1}{2}m_{44}+\frac{1}{2}m_{24}=1+\frac{1}{2}m_{24}$.
The solution in my book is $m_{24}=m_{34}=3$, which does not seem intuitive to me.
| The general way for computing mean hitting times of any absorbing state is well known, rather easily proven and in this Wikipedia article.
To get the mean hitting time of a specific state (or subset of states), we simply condition the outcome on hitting that state. For this question, the modified chain looks like:
This gives us the transition matrix $\Pi$, where $\Pi_{ij}$ gives the probability of transitioning to state j from state i:
$$
\begin{pmatrix}
0 & 1 & 0\\
\frac{1}{2} & 0 & \frac{1}{2}\\
0 & 0 & 1
\end{pmatrix}
$$
And Q:
$$
\begin{pmatrix}
0 & 1 \\
\frac{1}{2} & 0
\end{pmatrix}
$$
So that
$$
\begin{align}
m &= (I-Q)^{-1}1 \\
&=
\begin{pmatrix}
2 & 2 \\
1 & 2
\end{pmatrix}1 \\
&=
\begin{pmatrix}
4 \\
3
\end{pmatrix}
\end{align}
$$
Indeed, $m_{24} = 4$ and $m_{34} = 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the closed form of this sum? What is the closed form of this sum?
$$
S = \sum_{k\ge1, r>s\ge 1}\frac{1}{k^2(r^2 + s^2)^2}
$$
Note: Though originally posted for Pythagorean triplets, their was an flaw in the question which changed the meaning of the question and was answered accordingly. I will post a separate question with the original question on Pythagorean triangles.
Update: I have posted the related question of Pythagorean triangles in the link belwo
What is the sum of the reciprocal of the square of hypotenuse of Pythagorean triangles?
| \begin{align*}
S &= \sum_{k \geq 1} \frac{1}{k^2} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2} \\
&= \frac{\pi^2}{6} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2}
\end{align*}
Note that $\frac{1}{(r^2 + s^2)^2}$ is unchanged by $\{r \mapsto -r\}$, $\{s \mapsto -s\}$, and $\{r \leftrightarrow s\}$. Applying these symmetries, we obtain eight copies of the region $r > s \geq 1$ in the $r$-$s$ plane. It lacks the four diagonals,
each with value $\displaystyle \sum_{s \geq 1} \frac{1}{(s^2 + s^2)^2} = \frac{\pi^4}{360}$, the four coordinate rays starting at $\pm 1$, each with value $\displaystyle \sum_{s \geq 1} \frac{1}{(0^2 + s^2)^2} = \frac{\pi^4}{90}$, and the origin, which we will continue to exclude. Placing a prime on the summation sign to indicate that we omit the origin,
$$ \sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 8 \sum_{r > s \geq 1} \frac{1}{(r^2 + s^2)^2} + 4 \cdot \frac{\pi^4}{360} + 4 \cdot \frac{\pi^4}{90} $$
Borwein and Borwein [1] (p. 291) (see also (38) at Double Series on MathWorld) show
$$ \sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 4 \beta(2) \zeta(2) = \frac{2}{3} \pi^2 K \text{,} $$
where $\beta$ is Dirichlet's Beta function, $\zeta$ is the Riemann Zeta function, and $K$ is Catalan's constant.
Utilizing these facts,
$$ S = \frac{\pi^2}{6} \cdot \frac{12 K \pi^2 - \pi^4}{144} = 0.1264945807\dots \text{.} $$
[1] Borwein, Jonathan M. and Peter B. Borwein, "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity", New York: Wiley, 1987.
P.S. Mathematica 11.3 does not evaluate the triple sum symbolically. It numerically estimates $0.126494$, which is comfortably close to the above result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Evaluate $\sum_{n=2}^{\infty}\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ Evaluate $\sum_{n=2}^{\infty}\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$
I am not able to convert it into telescopic function.
| So the series
$$
\sum_{n=2}^{\infty}a_n\tag{1}
$$
has to be written in the form
$$
\sum_{n=2}^{\infty} (b_{n} - b_{n-1});\tag{2}
$$
therefore
$$
b_1 = \sum_{n=2}^{\infty} a_n, \\
b_2 = \sum_{n=3}^{\infty} a_n, \\
b_3 = \sum_{n=4}^{\infty} a_n, \\
\cdots \tag{3}
$$
If search $b_n$ in the form $$b_n = \dfrac{1}{2^n}\dfrac{P_4(n)}{Q_4(n)},$$
where $P_4(n), Q_4(n) $ are polynomials of $4$th degree; then one can figure out that
$$
b_n = \dfrac{1}{2^n}\left(1 + \dfrac{3n^2+4n+5}{n^4+2n^3+3n^2+2n+2}\right) \\
= \dfrac{1}{2^n}\left(1 + \dfrac{3n^2+4n+5}{(n^2+1)\Bigl((n+1)^2+1\Bigr)}\right).
$$
Then $$
b_n - b_{n-1} = (weird \;expression) = \dfrac{n^4+3n^2+10n+10}{2^n(n^3+4)} = a_n.
$$
So
$$
\sum_{n=2}^{\infty} a_n = b_{1} = \dfrac{1}{2}\cdot \dfrac{22}{10}=\dfrac{11}{10}.
$$
WolframAlpha checking link .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Diophantine equations with cubic roots.
Find all positive, integral $x, y$ such that
$$x^3-y^3=xy+61$$
Little work:
$$(x-y)(x^2+xy+y^2)-xy=61$$
Also $x^3-xy-y^3$ seems to to be unable to be "completed by the square," or factored, so I'm stuck. Thanks!
| More generally,
suppose
$x^3-y^3=xy+m
$
where $x, y, m > 0$.
I will show that
$m \ge x^2+y^2$,
$x \ge y+1$,
$x \lt \sqrt{m}$
and
$y
\le \frac12(\sqrt{2m-1}-1)
$.
If $m=61$ then
$x \le 7$
and
$y
\le \frac12(\sqrt{120}-1)
\approx 4.97
$
so
$y \le 4
$
and
$x \le \sqrt{61-y^2}
$.
We must have
$x > y$
so that
$x \ge y+1$
and
$xy+m
=(x-y)(x^2+xy+y^2)
\ge x^2+xy+y^2
$
so
$m
\ge x^2+y^2
$.
Therefore
$m \gt x^2
$
and
$x \ge y+1$
so that
$m
\ge (y+1)^2+y^2
=2y^2+2y+1
$
or
$2m-1
\ge 4y^2+4y+1
=(2y+1)^2
$
so
$y
\le \frac12(\sqrt{2m-1}-1)
$.
| {
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"url": "https://math.stackexchange.com/questions/3403526",
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"source": "stackexchange",
"question_score": "4",
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Please help with this limit of a sequence: $\lim\limits_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg)$ I have this limit of a sequence: $$\lim_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg).$$
Can I modify this expression to this expression by knowing that fractions goes to $0$?: $\lim n-n$. How to modify this solution to get the limit?
| Let $a=\sqrt[3]{n^3+n^2},\,b=\sqrt[3]{n^3+n}$ so $a-b=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{n^2-n}{a^2+ab+b^2}\approx\frac{n^2}{n^2+n^2+n^2}=\frac13$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $8(\arctan(\sqrt2-1)) = \pi$ My friend is saying $$8(\arctan(\sqrt2-1)) = \pi$$
I know $4\arctan(1) = \pi$
I thought of $4\arctan(\sqrt2-1) + 4\arctan(\sqrt2-1)$
But I got stuck somewhere
How to solve?
| After taking the $ \tan $ on both sides you get:
$$ \tan\frac{\pi}{8} = \sqrt{2}-1 $$
Note these two double angle formulas $ \sin2x = 2\sin x \cos x $ and $ \cos 2x = 2\cos^2x-1 $. Now we have:
$$ \tan\frac{\pi}{8} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} \cdot \frac{\cos\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{4}}{1+\cos\frac{\pi}{4}} = \frac{\sqrt{2}/2}{1+\sqrt{2}/2} = \sqrt{2} - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Counting the ways on grid if one can move from $(x,y)$ to $(x+a, x+b)$ for arbitrary $x,y,a,b\geq 0$. On 2 dimensional grid, consider the situation that one can move from $(p,q)$ to $(p+α,q+β)$ at once for arbitrary integer $p,q,α,β\geq 0 \land (α,β)\neq(0,0)$.
I want to count how many ways are there to move from (0,0) to (x,y).
I proved there is $\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}$ by combinatorial view.
Then, can we derive this using formal power series?
I've tried to derive this, however different formula appear and I cannot get the combinatorial interpretation of that formula.
The number of ways to get $(x,y)$ by $n$ moves is
\begin{align}
&[s^x t^y]\left(\frac{1}{1-s}\frac{1}{1-t}-1 \right)^n \\
=&[s^x t^y]\left(\frac{s+t-st}{(1-s)(1-t)}\right)^n
\end{align}
Note that $[s^x t^y] f(s,t)$ is the coefficient of $s^x t^y$ term of $f(s,t)$.
Summing up for $n=1,2,...,$ we can get the number of ways to go to $(x, y)$ by arbitrary number of moves.
\begin{align}
&[s^x t^y]\sum_{n=1}^\infty \left(\frac{s+t-st}{(1-s)(1-t)}\right)^n \\
=&[s^x t^y]\frac{s+t-st}{1-2(s+t-st)} \\
=&[s^x t^y]\sum_{i=0}^{\min(x,y)}2^{x+y-i-1} (s+t-st)^{x+y-i} \\
=&\sum_{i=0}^{\min(x,y)}2^{x+y-i-1} (-1)^i \frac{(x+y-i)!}{(x-i)!(y-i)!i!}
\end{align}
However, this seems different from $\sum\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}$.
Also, I cannot come up with the combinatorial interpretation of the formula we get.
UPDATE
I want to explain in details for the following.
\begin{align}
&[s^x t^y]\sum_{n=1}^\infty \left(\frac{s+t-st}{(1-s)(1-t)}\right)^n \\
=&[s^x t^y]\left(\frac{s+t-st}{1-2(s+t-st)} - \frac{(1-s)(1-t)\lim_{N\to\infty}\left(\frac{s+t-st}{(1-s)(1-t)}\right)^N}{1-2(s+t-st)} \right)\\
\end{align}
Here, I suppose the term, $-\frac{(1-s)(1-t)\lim_{N\to\infty}\left(\frac{s+t-st}{(1-s)(1-t)}\right)^N}{1-2(s+t-st)}$ can be treated as $0$ because if we put $s=0$ and $t=0$, $\frac{s+t-st}{(1-s)(1-t)}=0$ which means the degree of this term will go $\infty$ if we take power of $\infty$. Thus this term have nothing to do with the $s^x t^y$ term and it's ok to treat it as $0$.
| We consider non-negative integers $x,y$ and to get a first impression we start calculating the first few values of
\begin{align*}
\sum_{j\geq 0}\binom{x}{j}\binom{y}{j}2^{x+y-{j+1}}\tag{1}
\end{align*}
We write $j\geq 0$ and recall $\binom{p}{q}=0$ if $q>p$. The values of (1) are given in the picture below and we observe the sequence is archived in OEIS as A059576.
The values in OEIS coincide with (1) besides $(x,y)=(0,0)$ which is set to $1$, so that the value of $(x,y)$ is the sum of the values with smaller $x$ or smaller $y$ (an example marked in blue).
We now assume $x,y\geq 0, x+y\geq 1$ and obtain
\begin{align*}
\color{blue}{[s^xt^y]}&\color{blue}{\sum_{n=1}^\infty\left(\frac{s+t-st}{(1-s)(1-t)}\right)^n}\\
&=[s^xt^y]\left(\frac{1}{1-\frac{s+t-st}{(1-s)(1-t)}}-1\right)\\
&=[s^xt^y]\frac{s+t-st}{1-2(s+t-st)}\\
&=\frac{1}{2}[s^xt^y]\frac{1}{1-2(s+t-st)}\tag{2}\\
&=\frac{1}{2}[s^xt^y]\sum_{j=0}^\infty 2^j(s+t-st)^j\\
&=\frac{1}{2}[s^xt^y]\sum_{j=0}^\infty2^j \sum_{k=0}^j\binom{j}{k}s^k(1-t)^kt^{j-k}\\
&=\frac{1}{2}[s^xt^y]\sum_{k=0}^\infty\sum_{j=k}^\infty 2^j\binom{j}{k}s^k(1-t)^kt^{j-k}\tag{3}\\
&=\frac{1}{2}[t^y]\sum_{j=x}^\infty 2^j\binom{j}{x}(1-t)^xt^{j-x}\tag{4}\\
&=\frac{1}{2}[t^y]\sum_{j=0}^\infty 2^{j+x}\binom{x+j}{j}t^j(1-t)^x\\
&=\frac{1}{2}\sum_{j=0}^y2^{j+x}\binom{x+j}{j}[t^{y-j}](1-t)^x\\
&=\frac{1}{2}\sum_{j=0}^y2^{j+x}\binom{x+j}{j}\binom{x}{y-j}(-1)^{y-j}\tag{5}\\
&=\sum_{j=0}^y\binom{x+y-j}{y-j}\binom{x}{j}2^{x+y-j-1}(-1)^{y-j}\tag{6}\\
&=2^{x+y-1}\sum_{j\geq 0}\binom{x}{j}\left(-\frac{1}{2}\right)^j[z^{y-j}](1+z)^{x+y-j}\\
&=2^{x+y-1}[z^y](1+z)^{x+y}\sum_{j\geq 0}\binom{x}{j}\left(-\frac{z}{2(1+z)}\right)^j\\
&=2^{x+y-1}[z^y](1+z)^{x+y}\left(1-\frac{z}{2(1+z)}\right)^x\\
&=2^{x+y-1}[z^y](1+z)^{y}\left(1+\frac{z}{2}\right)^x\\
&=2^{x+y-1}[z^y](1+z)^{y}\sum_{j\geq 0}\binom{x}{j}\left(\frac{z}{2}\right)^j\\
&=\sum_{j\geq 0}\binom{x}{j}[z^{y-j}](1+z)^y2^{x+y-j-1}\\
&=\sum_{j\geq 0}\binom{x}{j}\binom{y}{y-j}2^{x+y-j-1}\\
&\,\,\color{blue}{=\sum_{j\geq 0}\binom{x}{j}\binom{y}{j}2^{x+y-j-1}}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we use $\frac{2(s+t-st)}{1-2(s+t-st)}=\frac{1}{1-2(s+t-st)}-1$. We can ignore the term $1$ which does not contribute to $[s^xt^y]$ since $x+y\geq 1$.
*In (3) we exchange the summation of series.
*In (4) we select the coefficient of $s^x$.
*In (5) we select the coefficient of $t^{y-j}$.
*In (6) we change the order of summation $j\to y-j$.
Note: The expression with the exponent $\infty$ is mathematically not sound and should be avoided.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the "four" variables problem Given x, y, z, and w are real numbers which satisfy these three equation:
$$x^{2} + 5z^{2} = 10$$
$$yz - xw = 5$$
$$xy + 5zw = \sqrt{105}$$
Find the value of $y^{2} + 5w^{2}$
Can anyone give a hint? I don't even know where to start
| Method$\#1:$
Solve the two simultaneous equations in $x,z$
$$z=\dfrac{aw+5y}{y^2+5w^2},x=\dfrac{ay-25w}{y^2+5w^2}$$
Replace the values of $z,w$ in $$10=x^2+5z^2$$
$$10=\dfrac{a^2y^2+625w^2-50ayw+5(a^2w^2+25y^2+10ayw)}{(y^2+5w^2)^2}$$
$$=\dfrac{(a^2+125)y^2+(625+5a^2)w^2}{(y^2+5w^2)^2}=\dfrac{a^2+125}{y^2+5w^2}$$
where $a^2=105$
See also : Brahmagupta-Fibonacci Identity
Method$\#2:$
WLOG $x=\sqrt{10}\cos t, z=\sqrt2\sin t$
$$yz - xw = 5\implies y(\sqrt2\sin t) - (\sqrt{10}\cos t)w = 5$$
$$xy + 5zw = \sqrt{105}\implies 5(\sqrt2\sin t)w+ (\sqrt{10}\cos t)y = \sqrt{105} $$
Solve for $\cos t,\sin t$ and use $\cos^2t+\sin^2t=1$ to eliminate $t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$ $$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$$
I tried using this substitute:
$$(\tan x=u),(sin2x= \frac{2\tan x}{1+\tan^2x} =\frac{2u}{u^2+1}), (dx=\frac{1}
{u^2+1}du)$$
and after long answer I get:
$$\frac{6+3\sqrt{3}}{6+4\sqrt{3}}.\ln(\frac{\tan x+2
-\sqrt{3}}{\tan x+2+\sqrt{3}})$$
and now I can not use $$F(\frac\pi2)-F(\frac\pi6)$$ because $$\tan(\frac\pi2)=\infty$$
| Since
$$\lim_{x\to\frac{\pi}{2}}\frac{\tan x +2-\sqrt{3}}{\tan x +2+\sqrt{3}}=\lim_{x\to\frac{\pi}{2}}\frac{1+\frac{2-\sqrt{3}}{\tan x}}{1+\frac{2+\sqrt{3}}{\tan x}}=1.$$
Then you can get what you want.
| {
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} |
Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}= \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0} \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=
\lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
| Using $\operatorname{sinc}y:=\frac{\sin y}{y}$ so $\lim_{y\to0}\operatorname{sinc}y=1$, the last expression you obtained is$$\frac12\lim_{x\to0}\frac{x\operatorname{sinc}^2\frac{\sqrt{x^2}}{2}}{\operatorname{sinc}^2\frac{\sqrt{x}}{2}}=\frac12\lim_{x\to0}\frac{x\cdot1}{1}=0.$$
| {
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What are the roots of this equation? (Quadratic Equation) $\frac{x}{\sqrt{1-x^2}}=A \Rightarrow \frac{x^2}{1-x^2}=A^2$
$\Rightarrow x^2=A^2(1-x^2)=A^2-A^2x^2$
$\Rightarrow x^2+A^2x^2=A^2$
$\Rightarrow x^2(1+A^2)=A^2$
$\Rightarrow x^2(1+A^2)-A^2=0$
I have tried this;
$\Delta=b^2-4ac=-4(1+A^2)(-A^2)=4(1+A^2)(A^2)$
$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{\pm2A\sqrt{1+A^2}}{{2(1+A^2)}}$
$x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$
But I got nothing.
| Your algebra is fine. Another way of writing your answer (in a more simplified form) would be
$$x=\pm\frac{A}{\sqrt{1+A^2}}$$
since $$\frac{\sqrt{t}}t=\frac1{\sqrt{t}}.$$
| {
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} |
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}$ using the definition of $e$
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}.$
I know that $e^x=\lim\limits_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n,$ so I need to somehow convert the limits to this form. I also noticed that $\left(1+\frac{1}{n+1}\right)=\left(\frac{n+2}{n+1}\right)=\left(\frac{n(n+2)}{(n+1)^2}\right)\cdot\left(1+\frac{1}{n}\right).$
Thus, the limit can be rewritten as $$\lim\limits_{n\to\infty}\dfrac{\left(1+\frac{1}{n}\right)^{n^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}\cdot\left(\dfrac{(n+1)^2}{n(n+2)}\right)^{(n+1)^2}\\
=\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^{-2n}\cdot\left(1+\dfrac{1}{n}\right)^{-1}\cdot\left(1+\dfrac{1}{n^2+2n}\right)^{n^2+2n}\cdot\left(1+\dfrac{1}{n^2+2n}\right)$$
$$=\dfrac{1}{e^2}\cdot(1)\cdot e\cdot(1)=\dfrac{1}{e}$$
| Your proof is nice, we have indeed
$$\left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}=\\=\left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}=$$
$$=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}=$$
$$=\left[\left(1+\frac{1}{n^2+2n}\right)^{n^2+2n}\right]^{\frac{n^2}{n^2+2n}}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}\to e^1 \cdot e^{-2}\cdot 1 =\frac1e$$
| {
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Calculating the exponential of an upper triangular matrix
Find the matrix exponential $e^A$ for
$$ A = \begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix}.$$
I think we should use the proberty
If $AB = BA$ then $e^{A+B} = e^A e^B$.
We can use that
$$\begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}
+\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{bmatrix}$$
Both matrices obviously commute. But I dont know how to calculate the exponential of
$$\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{bmatrix}.$$
Could you help me?
| Edit: As pointed out by @XTChen there is a much easier way to do this but I detail evaluation of the requested exponential nonetheless.
Note that
$$\pmatrix{1&1&1\\0&1&1\\0&0&1}^n=\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}$$
\begin{align}
\sum_{n=0}^\infty & \frac1{n!}\pmatrix{1&1&1\\0&1&1\\0&0&1}^n \\
&=\sum_{n=0}^\infty\frac1{n!}\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}\\
&=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\sum_{n=1}^\infty\pmatrix{1/n!&1/(n-1)!&(n+1)/(2(n-1)!)\\0&1/n!&1/(n-1)!\\0&0&1/n!}\\
&=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\pmatrix{e-1&e&3e/2\\0&e-1&e\\0&0&e-1}\\
&=\pmatrix{e&e&3e/2\\0&e&e\\0&0&e}\\
&=\frac12e\pmatrix{2&2&3\\0&2&2\\0&0&2}\\
\end{align}
| {
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} |
Find $p, q$ such that polynomial $P(x) = 6x^4 - 7x^3 + px^2 + 3x + 2$ is divisible by $x^2 - x + q$ I'm having difficulties with this problem:
Find $p, q$ such that polynomial $P(x) = 6x^4 - 7x^3 + px^2 + 3x + 2$ is divisible by $x^2 - x + q$.
I'm aware of the Bezout's Theorem, but I don't know how to use it in this problem optimally. I've tried finding "solutions" for $x^2 - x + q$ and representing it as $(x-x_1)(x-x_2)$ where $x_1, x_2$ are solutions to this equation, and the only thing left is to check if $P(x_1) = 0$ and $P(x_2) = 0$, but I'm not sure if it's the most optimal (or even a correct) way to solve this problem.
| We can write
$$6x^4 - 7x^3 + px^2 + 3x + 2=(ax^2+bx+c)(x^2-x+q)$$
Comparing $x^4$ coefficients we see that $a=6$. Now comparing $x^3$ coefficients we get $b=-1$:
$$6x^4 - 7x^3 + px^2 + 3x + 2=(6x^2-x+c)(x^2-x+q)$$
Looking at the $x^0$ and $x^1$ coefficients we have $cq=2$ and $q+c=-3$. By Viète's relations $c,q$ are the roots of $x^2+3x+2$, or $\{c,q\}=\{-1,-2\}$. If $c=-1$ then $p=c+1+6q=-12$; if $c=-2$ then $p=-7$. Hence the possible pairs $(p,q)$ are $(-12,-2)$ and $(-7,-1)$.
| {
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Proof of equality of complex numbers with conditions
I have the next question:
Proof that $\operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=|z_{1}|\cdot |z_{2}|$ only if: $\arg z_{1} -\arg z_{2} =2\cdot \pi \cdot n$ ( $\;(n\in \mathbb{Z})$.
This is my solution:
$z_{1}=x+iy, z_{2}=a+ib$
$z_{1}\cdot \bar{z_{2}}=(x+iy)\cdot (a-ib)=xa+yb+i(ya-bx)$
$\rightarrow \operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=xa+yb$
$|z_{1}|=\sqrt{x^{2}+y^{2}}, |z_{2}|=\sqrt{a^{2}+b^{2}}$$|z_{1}|\cdot |z_{2}|=\sqrt{(x^{2}+y^{2})\cdot (a^{2}+b^{2})}=\sqrt{x^{2}a^{2}+x^{2}b^{2}+y^{2}a^{2}+y^{2}b^{2}}$
$\Rightarrow (xa+yb)=\sqrt{x^{2}a^{2}+x^{2}b^{2}+y^{2}a^{2}+y^{2}b^{2}}$
$\overset{()^{2}}{\rightarrow}x^{2}a^{2}+2xayb+y^{2}b^{2}=x^{2}a^{2}+x^{2}b^{2}+y^{2}a^{2}+y^{2}b^{2}$
$\rightarrow2xyab=x^{2}b^{2}+y^{2}a^{2}\rightarrow x^{2}b^{2}-2xyab+y^{2}a^{2}=0\rightarrow (xb-ya)^2=0$
$x=r_{1}\cos\theta , y=r_{1}\sin\theta$
$a=r_{2}\cos\varphi , b=r_{2}\sin\varphi $
$(r_{1}\cos\theta \cdot r_{2}\sin\varphi-r_{1}\sin\theta\cdot r_{2}\cos\varphi)^2=0$
$(r_{1} \cdot r_{2} (\cos\theta \sin\varphi-\sin\theta \cos\varphi))^2=0$
$\sin^2(\varphi-\theta)=0$
For every $n$ that we put in the condition, the equation is true but the angle $\pi$ is also a solution for this equation but it is not created from the condition.
Where have I done a mistake?
Thank you all for the help.
| There is a hidden thing you should be careful of when you square both sides you assuming that $ax+by>0$ so the equality make sense . But this thing means $\cos \phi \cos \theta +\sin \phi \sin \theta >0$ hence $\cos(\theta -\phi)>0$ so the difference is either in first or fourth quarter but if the difference is multiple of $\pi$ then it could happen one solution in first then the other would be in the third.
| {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 3
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How can I find the eigenvalues of $\alpha$? The question is given below:
My trial is:
Let $\lambda $ be an eigenvalue of $\alpha,$ then
$$ \alpha (
\left[ {\begin{array}{cc}
a \\
b \\
c \\
d \\
\end{array} } \right]
) = \lambda
\left[ {\begin{array}{cc}
a \\
b \\
c \\
d \\
\end{array} } \right]
=
\left[ {\begin{array}{cc}
\lambda a \\
\lambda b \\
\lambda c \\
\lambda d
\end{array} } \right]
$$
$$i.e. \left[ {\begin{array}{cc}
b + c \\
c \\
0 \\
0 \\
\end{array} } \right] = \left[ {\begin{array}{cc}
\lambda a \\
\lambda b \\
\lambda c \\
\lambda d
\end{array} } \right]
\rightarrow (1.)$$
Now, for $\left[ {\begin{array}{cc}
a \\
b \\
c \\
d \\
\end{array} } \right] $ to be an eigenvector, at least one of $a, b, c, d$ must be nonzero. WLOG, let $a \neq 0.$ Therefore from $(1.)$ we have $$\lambda = \frac {b + c}{a}, \lambda b = c \quad i.e. \quad b^2 = c(a-b), \lambda c = 0 \quad i.e. c^2 = (-bc).$$ but then what?
Could anyone help me in solving this, please?
| Another way to do:
$$\alpha(a,b,c,d)=(b+c,c,0,0).$$
Assuming that $e_1,\dots, e_4$ is the canonical basis of $\mathbb{R}^4$. Then,
$$\alpha(e_1)=(0,0,0,0)$$
$$\alpha(e_2)=(1,0,0,0)$$
$$\alpha(e_3)=(1,1,0,0)$$
$$\alpha(e_4)=(0,0,0,0)$$
The matrix associated to $\alpha$ is:
$$[\alpha]=\begin{pmatrix}
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$$
Then, the charascterisc polynomail of $[\alpha]$ is
$$p(\lambda)= \det(\lambda I - [\alpha])= \det\left[\begin{pmatrix}
\lambda & 0 & 0 & 0 \\
0 & \lambda & 0 & 0 \\
0 & 0 & \lambda & 0 \\
0 & 0 & 0 & \lambda
\end{pmatrix} - \begin{pmatrix}
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}\right] = \det\left[\begin{pmatrix}
\lambda & -1 & -1 & 0 \\
0 & \lambda & -1 & 0 \\
0 & 0 & \lambda & 0 \\
0 & 0 & 0 & \lambda
\end{pmatrix}\right]$$
Can you proceed from here?
| {
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A general formula to generate functions of power series I have been playing with Maclaurin series lately, I have been able to come across this:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$
$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$
I found out by accident that:
$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^2x^5...$
I found in an old paper of Euler that this can continue on with these functions:
$\dfrac{1-4x+x^2}{(1+x)^4}=1-2^3x+3^3x^2-4^3x^3+5^3x^4+6^3x^5...$
$\dfrac{1-11x+11x^2-x^3}{(1+x)^5}=1-2^4x+3^4x^4-4^4x^3+5^4x^4+6^4x^5...$
$\dfrac{1-26x+66x^2-26x^3+x^4}{(1+x)^5}=1-2^5x+3^5x^4-4^5x^3+5^5x^4+6^5x^5...$
$\dfrac{1-57x+320x^2-302x^3+57x^4-x^5}{(1+x)^5}=1-2^6x+3^6x^4-4^6x^3+5^6x^4+6^6x^5...$
and so on...
Is there a general formula to generate the functions on the right hand side? How did Euler calculate these series? I have to say I deeply respect him since only he and Ramanujan know how to play with series.
| Given any analytic function $f ,f(x)=\sum_{n=0}^{\infty}f^n(0)x^n/n!$.You are guaranteed to get a series representation iff $R_n(x)=f(x)-\sum_{k=0}^{n}f^k(0)x^k/k! \to 0$ pointwise as $n\to \infty$.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
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Methods of evaluating $ \sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}$? I am interested in ways of evaluating the following infinite seris:
$$
\sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}.
$$
I already know the answer from Wolfram Alpha but I would like to see some methods of evaluating it as I haven't been able to find many (any?) examples involving an infinite series with the $(m+k)!$ in the numerator and the $k!$ in the denominator, it seems that it is more common to find $k!$ in the numerator and $(m+k)!$ in the denominator.
So what are some methods that can be used to evaluate this series?
| This is the non-calculus way. Let the sum be $S_m$. Then,
\begin{align}
& 5S_m-S_m =\sum_{k=0}^{\infty}\frac{(m+k+1)\cdots(k+2)}{5^k}-\sum_{k=1}^{\infty}\frac{(m+k)\cdots(k+1)}{5^k}\\
=&(m+1)!+m\sum_{k=1}^{\infty}\frac{(m+k)\cdots(k+2)}{5^k}\\
=&(m+1)!+5m\sum_{k=2}^{\infty}\frac{(m-1+k)\cdots(k+1)}{5^k}\\
=& (m+1)!+5m\left(S_{m-1}-\frac{m!}{5}\right)=m!+5mS_{m-1}\\
\implies& S_m=\frac{m!}{4}+\frac{5m}{4}S_{m-1}\tag{1}\\
\implies&S_m=\frac{m!}{4}+\frac{5m}{4}\left(\frac{(m-1)!}{4}+\frac{5(m-1)}{4}S_{m-2}\right)\\
=&\frac{m!}{4}+\frac{5m!}{4^2}+\frac{5^2m(m-1)}{4^2}S_{m-2}\\
=&\frac{m!}{4}+\frac{5m!}{4^2}+\frac{5^2m!}{4^3}+\frac{5^3m(m-1)(m-2)}{4^3}S_{m-3}\\
=&\frac{m!}{4}\sum_{k=0}^n\left(\frac{5}{4}\right)^k+\frac{5^{n+1}m\cdots(m-n)}{4^{n+1}}S_{m-n-1}
\end{align}
Setting $n=m-1$ gives us,
$$
S_m=\frac{m!}{4}\sum_{k=0}^{m-1}\left(\frac{5}{4}\right)^k+m!\left(\frac{5}{4}\right)^mS_0
$$
As $S_0$ is a geometric series with value $\frac{1}{4}$ our expression becomes,
$$
S_m=\frac{m!}{4}\sum_{k=0}^{m}\left(\frac{5}{4}\right)^k=m!\left(\left(\frac{5}{4}\right)^{m+1}-1\right)
$$
| {
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"question_score": "1",
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Compute $\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx$ How to prove
$$I=\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac16\left(\frac{\pi^3}{12}-\pi\operatorname{Li}_2\left(\frac13\right)\right)$$
This problem is proposed by Cornel which can be found here where he suggested that the problem can be solved with and without harmonic series.
Here is my approach but I got stuck at the blue integral:
Using the common identity
$$ \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$$
Set $p=-\cos(x)$ we get
$$ \sum_{n=1}^{\infty}(-1)^n \cos^n(x) \cos(nx)=-\frac{2\cos^2(x)}{1+3\cos^2(x)}=-\frac23+\frac23\frac1{1+3\cos^2(x)}$$
Multiply both sides by $-x^2$ then integrate from $x=0$ to $\pi/2$ we get
$$\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac23\int_0^{\pi/2} x^2dx-\frac23\color{blue}{\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx}\\=\frac{\pi^3}{36}-\frac23\left(\color{blue}{\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)}\right)=\frac{\pi^3}{72}-\frac{\pi}{6}\operatorname{Li}_2\left(\frac13\right)$$
I have two Questions:
1) Can we evaluate $I$ in a different way?
2) How to finish the blue integral?
My try to the blue integral is using integration by parts
$$\int\frac{dx}{1+3\cos^2(x)}=\frac12\tan^{-1}\left(\frac{\tan(x)}{2}\right)=-\frac12\tan^{-1}\left(2\cot(x)\right)$$
which gives us
$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\frac{\pi^3}{16}-\int_0^{\pi/2}x\tan^{-1}\left(\frac{\tan(x)}{2}\right)dx$$
Or
$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\int_0^{\pi/2}x\tan^{-1}\left(2\cot(x)\right)dx$$
I also tried the trick $x\to \pi/2-x$ but got complicated
Proof of the identity:
\begin{align}
\sum_{n=0}^\infty p^ne^{inx}&=\sum_{n=0}^\infty\left(p e^{ix}\right)^n=\frac{1}{1-pe^{ix}},\quad |p|<1\\&=\frac{1}{1-p\cos(x)-ip\sin(x)}=\frac{1-p\cos(x)+ip\sin(x)}{1-2p\cos(x)+p^2}\\
&=\frac{1-p\cos(x)}{1-2p\cos(x)+p^2}+i\frac{p\sin(x)}{1-2p\cos(x)+p^2}
\end{align}
By comparing the real and imaginary parts, we get
$$\sum_{n=\color{blue}{0}}^\infty p^n \cos(nx)=\frac{1-p\cos(x)}{1-2p\cos(x)+p^2}\Longrightarrow \sum_{n=\color{blue}{1}}^\infty p^{n-1} \cos(nx)=\frac{\cos(x)-p}{1-2p\cos(x)+p^2}$$
and
$$\sum_{n=\color{red}{0}}^\infty p^n \sin(nx)=\frac{p\sin(x)}{1-2p\cos(x)+p^2}\Longrightarrow \sum_{n=\color{red}{1}}^\infty p^n \sin(nx)=\frac{p\sin(x)}{1-2p\cos(x)+p^2}$$
| Evaluating the blue integral:
First we write
$$\frac1{1+3\cos^2(x)}=\frac{1}{5+3\cos(2x)}$$
Using the same identity in the post body
$$\sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1\tag1$$
But lets manipulate the denominator to have it in the form of $\frac1{5+3\cos(x)}$:
$$\frac1{1-2p\cos(x)+p^2}=\frac{-\frac{3}{2p}}{-\frac{3(1+p^2)}{2p}+3\cos(x)}$$
Now set $$-\frac{3(1+p^2)}{2p}=5\Longrightarrow p=-3,-\frac13$$
and since $|p|<1$, so we take $p=-\frac13$. Plug this value in (1) and replace $x$ by $2x$ we get
$$\frac{1}{5+3\cos(2x)}=\frac{1}{4}+\frac12\sum_{n=1}^\infty (-1)^n\left(\frac{1}{3}\right)^n\cos(2nx)\tag2$$
Multiply both sides of (2) by $x^2$ and integrate between $0$ and $\pi/2$ we get
$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)$$
Addendum:
The identity used by @Zacky above:
$$\frac{1}{a+b\cos(x)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nx)},\ a>b\tag{3}$$
can be derived the same way:
$$\frac1{1-2p\cos(x)+p^2}=\frac{-\frac{b}{2p}}{-\frac{b(1+p^2)}{2p}+b\cos(x)}$$
If we set $$-\frac{b(1+p^2)}{2p}=a\tag{4}$$
we can write
$$\frac1{1-2p\cos(x)+p^2}=\frac{\frac{a}{1+p^2}}{a+b\cos x}$$
We proved above that
$$\sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}=-\frac12-\frac12\frac{p^2-1}{1-2p\cos(x)+p^2}$$
$$=-\frac12-\frac12 \color{red}{\frac{p^2-1}{p^2+1}}\frac{\color{red}{a}}{a+b\cos(x)}\tag5$$
From $(4)$ we find $p=\frac{\sqrt{a^2-b^2}-a}{b}$. Note that we ignored $p=\frac{\sqrt{a^2-b^2}+a}{b}$ as $|p|<1$.
Substitute this root in $(5)$ we get
$$\sum_{n=1}^\infty\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos(nx)=-\frac12-\frac12\cdot\frac{\color{red}{-\sqrt{a^2-b^2}}}{a+b\cos(x)}$$
or
$$\frac{1}{a+b\cos(x)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nx)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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How can we continue at the induction step? I want to show by induction that $$\sum_{i=1}^n i^2\geq 3n+2$$ for all $n\in \mathbb{N}$ with $n\geq 3$.
I have done the following:
Base Case: For $n=3$ we have that $\displaystyle{\sum_{i=1}^3i^2=1^2+2^2+3^2=1+2+9=12}$ and $\displaystyle{3n+2=3\cdot 3+2=9+2=11}$. It holds that $12\geq 11$ uand so the given inequality holds.
Inductive hypothesis: We assume that it holds for $n=k$, so $\displaystyle{\sum_{i=1}^ki^2\geq 3k+2}$. (IH)
Induction Step: We want to show that the inequality holds for $n=k+1$, so $\displaystyle{\sum_{i=1}^{k+1}i^2\geq 3(k+1)+2}$.
We have that \begin{equation*}\sum_{i=1}^{k+1}i^2=\sum_{i=1}^{k}i^2+(k+1)^2\overset{(IH)}{\geq}3k+2+(k+1)^2=3k+2+k^2+2k+1=k^2+5k+3 \end{equation*} How can we continue? Can we just say that this is greater than $3k+5=3(k+1)+2$ ?
| We just need to show that
$$\sum_{i=1}^{k+1}i^2=\sum_{i=1}^{k}i^2+(k+1)^2\overset{(IH)}{\geq}3k+2+(k+1)^2\overset{(?)}\ge 3(k+1)+2$$
which is true indeed
$$3k+2+(k+1)^2\ge 3(k+1)+2$$
$$3k+2+(k+1)^2\ge 3k+5$$
$$(k+1)^2\ge 3$$
for $k\ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$ Evaluate :$$ \int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
using residue theorem. The integral is even and it can be written as :
$$ \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
The only pole with a positive imaginary part is $ \frac{i}{\sqrt2} $ and its degree is $4$. How can I evaluate this without differentiating 3 times when calculating the residue?
I also tried calculating the infinite residue but that didn't help much.
| Note first that$$\frac{x}{1+2x^2}=\frac14\left(\frac{1}{x-i/\sqrt{2}}+\frac{1}{x+i/\sqrt{2}}\right).$$Hence$$\left(\frac{x}{1+2x^2}\right)^2=\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+\frac{2}{x^2+1/2}\right)\\\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+i\sqrt{2}\left(\frac{1}{x-i/\sqrt{2}}-\frac{1}{x+i/\sqrt{2}}\right)\right).$$Squaring this again, none of the diagonal terms matter, and many of the cross terms don't either. Let $f\sim g$ denote the condition that $f-g$ has zero residue at $\frac{i}{\sqrt{2}}$, so$$\left(\frac{x}{1+2x^2}\right)^4\sim\frac{i\sqrt{2}}{128}\frac{1}{(x+i/\sqrt{2})^2(x-i/\sqrt{2})}\\\implies\operatorname{Res}_{-i/\sqrt{2}}\left(\frac{x}{1+2x^2}\right)^4=\frac{-i\sqrt{2}}{256}.$$Now we just need to multiply by $\frac122\pi i$ to give $\frac{\pi\sqrt{2}}{256}$ as the value of the integral.
Just to double-check, let's solve the problem with $x=\frac{1}{\sqrt{2}}\tan t$ so the original integral is$$\frac{\sqrt{2}}{8}\int_0^{\pi/2}\sin^4t\cos^2tdt=\frac{\sqrt{2}}{16}\operatorname{B}\left(\frac52,\,\frac32\right)=\frac{\sqrt{2}}{16}\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)}{\Gamma(4)}=\frac{\pi\sqrt{2}}{256}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $4x \equiv 7 \pmod{15}$ and $3x \equiv 5 \pmod{16}$ (different exercises) This is how I solved each:
$$4x \equiv 7\pmod {15} \\
4x - 7 \equiv 0\pmod{15} \\
4x-7 = 15k \Leftrightarrow 4x-15k= 7 \\
$$
$$15 = 4*3+3 \\
4=3*1+1\\
3=3*1+0$$
$$1 = 4-3*1 \\
3 = 15-4*3 \\
1 = 4 - (15-4*3)*1 \\
1 = 4-15+4*3 \\
1 = 4*4-15*1 \\
$$
$$7 = (7*4)*4-(7*1)*15$$
$7*4*4 = 112$ which is congruent with 7 in mod 15. Yet my book says the answer is 13. What went wrong?
The second one:
$$3x - 5 \equiv 0 \pmod{16} \\
3x-5=16k\\
3x-16k=5 \\$$
$$16=3*5+1\\
3=1*3+0$$
$$1 = 16-3*5 \\
5=16*5-3*5*5$$
Yet $3*5*5-5$ is not congruent with zero in mod 16. What went wrong?
| With such small coefficients ($4$ and $3$), here is a quick and easy way to solve the congruences:
Mod $15$: $4x\equiv7\equiv22\equiv37\equiv\color{red}{52}=4\times13\implies x\equiv 13$
Mod $16$: $3x\equiv5\equiv\color{red}{21}=3\times7\implies x\equiv 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find two elementary matrices $E_1$ and $E_2$ such that $E_2E_1A=B$. I have two matrices $A$ and $B$, and I need to find two matrices $E_1$ and $E_2$ that satisfy the question stated in the title.
\begin{align*}
A&=\begin{bmatrix}
1&2&-1\\
1&1&1\\
1&-1&0\\
\end{bmatrix} &
B&=\begin{bmatrix}
1&-1&0\\
1&1&1\\
4&-1&-1
\end{bmatrix}.
\end{align*}
I know I need to find two elementary row operations that will turn $A$ into $B$, I think the first operation is switching the first and third row in $A$, but I don't know what the second operation would be.
| The aim is to find $E_1$ and $E_2$ such that $B = E_2E_1A$. The first thing is to interchange the first and the last row of the matrix $A$, so $$E_1 = \begin{bmatrix}0 & 0 & 1\\0 & 1 & 0\\ 1 & 0 & 0\end{bmatrix}, \quad A_1 := E_1A = \begin{bmatrix}1 & -1 & 0\\1 & 1 & 1\\ 1 & 2 & -1\end{bmatrix}.$$ Then, $B$ can be obtained by multiplying the first row of $A_1$ and adding it to the third row: $$E_2 = \begin{bmatrix}1 & 0 & 0\\0 & 1 &0\\3&0&1\end{bmatrix}, \quad B = E_2A_1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve this: I was doing my homework and I stumbled over this particular exercise. I would've known how to solve it, if it had been the same thing under square root in both cases.
$$
\lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)
$$
| You can multiply both numerator and denominator by $\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}$. Then you will find $$\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1}=\frac{4n^{2}+3n+2-4n^{2}-n+1}{\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}} = \frac{2n(1+3/2n)}{2n\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} =\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} $$
Thus, we have $$\lim_{n\to \infty}\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1} = \lim_{n\to\infty}\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} = \frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
My process:
\begin{align}
\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=\frac{\sqrt{2 a^{3} a-a^{4}}-a \sqrt[3]{a^{2} a}}{a-\sqrt[4]{a a^{3}}}=\frac{\sqrt{a^4}-a\cdot \sqrt[3]{a^3}}{a-\sqrt[4]{a^4}}=\frac{a^2-a\cdot a}{a-a}=\frac{a^2-a^2}{a-a}
\end{align}
which would be undefined, but Wolfram Alpha calculated this:
$\lim_{x\to a} \, \frac{\sqrt{2 a^3 x-x^4}-a \sqrt[3]{a^2 x}}{a-\sqrt[4]{a x^3}}=
\begin{array}{cc}
\{ &
\begin{array}{cc}
\frac{16 a}{9} & a\geq 0 \\
0 & (\text{otherwise}) \\
\end{array}
\\
\end{array}$
What have I done wrong? Is there a way to calculate the limit with L'Hospitals Rule?
| We have that by $y=x/a\to 1$
$$\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=a \frac{\sqrt{2 y-y^{4}}-\sqrt[3]{y}}{1-\sqrt[4]{y^3}}=$$
$$= -a \frac{y-1}{\sqrt[4]{y^3}-1}\left( \frac{\sqrt{2 y-y^{4}}-1}{y-1}- \frac{\sqrt[3]{y}-1}{y-1}\right)=f(y)$$
then use the definition of derivative to obtain
$$f(y)\to -a\cdot \frac43\left(-1-\frac13\right)=\frac{16a}9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that a sequence decreases past a certain point
I would like to try to prove that the sequence
$$
\frac{\sqrt{6}^{x}}{x!}
$$
for $x=0, 1, 2, \cdots$, is strictly decreasing when $x\geq 2$.
It appears to be the case after listing out a few terms. Here's my attempt at a proof:
To the contrary, suppose $\frac{\sqrt{6}^{x}}{x!} \leq \frac{\sqrt{6}^{x+1}}{(x+1)!}$ for $x \geq 2$. Then we have
\begin{align*}
\begin{split}
&\frac{x\text{ln}\sqrt{6}}{x!} \leq \frac{(x+1)\text{ln}\sqrt{6}}{(x+1)!}\\
&\Rightarrow \frac{1}{(x-1)!} \leq \frac{1}{x!}\\
&\Rightarrow \frac{x!}{(x-1)!} \leq 1\\
&\Rightarrow x \leq 1.
\end{split}
\end{align*}
But this contradicts our assumption on $x$, so the inequality above must in fact be ">". Thus the sequence is strictly decreasing for $x \geq 2$.
Any corrects/input appreciated.
| The given sequence IS strictly decreasing for $x\geq 2$. If we follow your approach in a correct way, we find that the inequality
$$\frac{\sqrt{6}^{x}}{x!} > \frac{\sqrt{6}^{x+1}}{(x+1)!},$$ is equivalent to
$$\frac{(x+1)!}{x!} > \frac{\sqrt{6}^{x+1}}{\sqrt{6}^{x}}$$
which reduces to
$$x+1> \sqrt{6}$$
which is satisfied when $x> \sqrt{6}-1\approx 1.449$.
P.S. Note that $\sqrt{6}^{x}=\exp(x\ln(\sqrt{6}))$ (not $x\ln(\sqrt{6})$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ I want to show
\begin{align}
\frac{(p-1)(p+1)}{24} \in \mathbb N \quad \text{for all primes} \quad p \geq 5 \tag{1}.
\end{align}
I can show $(1)$, if the following statement is true.
Let $a,b,c,d \in \mathbb N$ and $a \geq b \cdot c \cdot d$.
\begin{align}
\text{If} \quad \frac{a}{b},\frac{a}{c},\frac{a}{d} \in \mathbb N, \quad \text{then} \quad \frac{a}{b \cdot c \cdot d} \in \mathbb N \tag{2}.
\end{align}
Given $(2)$ we show that $(1)$ is true for $a = (p-1)(p+1)$, $b = 2$, $c = 3$ and $d = 4$. Since $p$ is a prime $(p-1)$ and $(p + 1)$ are even, implying $(p-1)/2 \in \mathbb N$, $(p+1)/2 \in \mathbb N$ and thus $(p-1)(p+1)/2 \in \mathbb N$ and $(p-1)(p+1)/4 \in \mathbb N$. One of the three numbers $(p-1)$, $p$ and $(p+1)$ must be divisible by 3. Since $p$ is a prime either $(p-1)$ or $(p+1)$ is divisible by 3, implying $(p-1)(p+1)/3 \in \mathbb N$.
Question Is $(2)$ true?
| Primes greater than or equal to $5$ can't be divisible by $3$, so must have the form $3k+1$ or $3k+2$ , so one of $p-1$ or $p+1$ has a factor of $3$. Also, $p-1$ and $p+1$ are consecutive even numbers and one of them is divisible by $4$, so $2$ divides $(p-1)(p+1)$ with at least multiplicity $3.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the 4 roots of the equation: $z^{4}+4=0$ Find the 4 roots of the equation: $z^{4}+4=0$
Note: I can't use the form: $e^{i\theta}$
My attempt:
Note we have: $z=(-4)^{\frac{1}{4}}$, consider $w=-4$, then $w^{1/4}=z$ this implies $w=z^4$.
Consider the polar form of $z$ and $w$:
$|w|=r=4$,
$Arg(w)=\pi=\theta$
Then,
$$w=4\cos\pi+i\sin\pi$$
$$z=p\cos\phi+i\sin\phi$$
This implies: $w=z^4$ iff $4=p^4$ and $4\phi=\pi+2k\pi$ iff $p=4^{1/4}$ and $\phi=\frac{\pi+2k\pi}{4}$ with $k=0,1,2,3$
Then the roots are:
$$z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$
with $k=0,1,2,3$
is correct this?
| Consider the factorization of $z^4+4$:
$$z^4+4=z^4\color{blue}{+4z^2}+4\color{blue}{-4z^2}=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z-2)$$
Then we get $z^2+2z+2=0$ or $z^2-2z+2=0$, then finish it by using quadratic formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the greatest $a \in \mathbb{Z}$ such that $x^2-ax-\ln x+e^{\sin x}-1>0$ holding for every $x>0$.
Find the greatest $a \in \mathbb{Z}$ such that $$x^2-ax-\ln x+e^{\sin
x}-1>0$$ holding for every $x>0$. (Reference data: $\sin 1 \approx 0.84, \ln2 \approx 0.693)$
Notice that the inequality is equivalent to
$$f(x):=\frac{x^2-\ln x+e^{\sin x}-1}{x}>a.$$
If we can find the globle minimum of $f(x)$, then the problem is solved. But $f(x)$ is a transcendent function, it's not so easy to obtain the exact minimum. How to enlarge or contract?
| Answer: greatest $a=2$. Proof.
If $a=3$ and $x=e$ then $x^2-ax-\ln x+e^{\sin x}-1<0$.
Let $f(x)=x^2-2x-\ln x+e^{\sin x}-1$. We need to prove that $f(x)>0$ for all $x$.
Cases:
I. Let $x \ge e$. Let $g(x)=x^2-2x-\ln x-1$. Then $g'(x)=2x-2-\frac{1}{x}=\frac{2}{x} \left( x-\frac{\sqrt3+1}{2}\right)\left( x+\frac{\sqrt3-1}{2}\right)$. Since $e>\frac{\sqrt3+1}{2}$ then $\min \limits_{x \ge e} g(x)=g(e)$. Then $f(x)=g(x)+e^{\sin x}\ge e^2-2e-2+e^{\sin x}\ge e^2-2e-2+e^{-1}>0$.
II. Let $\sqrt2+1 \le x < e$. Then $x^2-2x-1=(x-1)^2-2 \ge 0$ and $\ln x<1$ and $e^{\sin x}>1$ (because $e<\pi$, $\sin x>0$). Then $f(x)>0$.
III. Let $\pi-1 \le x < \sqrt2+1$. Then $e^{\sin x}\ge e^{\sin(\sqrt2+1)}>1.94$. Then $f(x)=x(x-2)-\ln x+e^{\sin x}-1>(\pi-1)(\pi-3)-\ln(\sqrt2+1)+1.94-1>0$.
IV. Let $1 \le x < \pi-1$. Then $e^{\sin x} \ge e^{\sin 1}>2.3$. Then $f(x)>x^2-2x-\ln x+1.3$. Let $g(x)=x^2-2x-\ln x+1.3$. Then $g'(x)=2x-2-\frac{1}{x}=\frac{2}{x} \left( x-\frac{\sqrt3+1}{2}\right)\left( x+\frac{\sqrt3-1}{2}\right)$.Then $\min \limits_{1 \le x<\pi-1} g(x)=g(\frac{\sqrt3+1}{2})=\left( \frac{\sqrt3+1}{2}\right)^2-2\left( \frac{\sqrt3+1}{2}\right)-\ln \left( \frac{\sqrt3+1}{2}\right)+1.3>0$. Then $f(x)>g(x)>0$.
V. Let $\frac{\pi}{4} \le x <1$. Then $e^{\sin x} \ge e^{\sin \frac{\pi}{4}}>2$. Then $e^{\sin x}-2-\ln x >0$. Then $f(x)=(x-1)^2+e^{\sin x}-2-\ln x >0$.
VI. Let $0<x<\frac{\pi}{4}$. It hard case. $$f'(x)=2x-2-\frac{1}{x}+\cos x \cdot e^{\sin x}$$ $$f''(x)=2+\frac{1}{x^2}-\sin x \cdot e^{\sin x}+\cos^2 x \cdot e^{\sin x}$$ $$f'''(x)=-\frac{2}{x^3}-\cos x \sin^2 x \cdot e^{\sin x}-3\sin x \cos x \cdot e^{\sin x}$$ Ease to see that $f'''(x)<0$. Then $\min \limits_{0<x<\frac{\pi}{4}} f''(x)=f''(\frac{\pi}{4})=2+\frac{16}{\pi^2}-\frac{\sqrt2}{2}e^{\frac{\sqrt2}{2}}+0.5e^{\frac{\sqrt2}{2}}>0$. Then $f''(x)>0$. Then $\max \limits_{0<x<\frac{\pi}{4}} f'(x)=f'(\frac{\pi}{4})=\frac{\pi}{2}-2-\frac{4}{\pi}+\frac{\sqrt2}{2}e^{\frac{\sqrt2}{2}}<0$. Then $f'(x)<0$. Then $\min \limits_{0<x<\frac{\pi}{4}} f(x)=f(\frac{\pi}{4})=\left(\frac{\pi}{4} \right)^2-\frac{\pi}{2}-\ln \frac{\pi}{4}+e^{\frac{\sqrt2}{2}}-1>0$.
Proof is complete.
| {
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"timestamp": "2023-03-29T00:00:00",
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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.