Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the linear transformation of a matrix knowing 4 linear transformations I'm stuck with an exercise where they give me 4 linear transformations $T:M_{2\times2}\to\Bbb R$ for 4 matrices 2x2 and then they ask me for the linear transformation of a fifth matrix.
$$ T
\begin{pmatrix}
1 &0\\
0 & 0\\
\end{pmatr... | What is given can be translated as follows. Denote, in a standard way,
$$E_1=\begin{pmatrix}1&0\\0&0 \end{pmatrix},\quad E_2=\begin{pmatrix}0&1\\0&0 \end{pmatrix}, \quad E_3=\begin{pmatrix}0&0\\1&0 \end{pmatrix},\quad E_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$
We have $\:M=\begin{pmatrix}a&b\\c&d \end{pmatrix}=aE_1+bE... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$ There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent seri... | It is not clear for what you are using $w = z-1$ and the link between $f(z)$ and the remainder of your argument is unclear. However, you have nearly got the right answer.
Instead, you could justify the result by writing,
\begin{align}
f(z) &= \frac{1}{z(z-1)} \\
&= \frac{1}{(z-1)^2(1+\frac{1}{z-1})}
\end{align}
and w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$ I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$
So far I have:
*
*Base case: true
*Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$... | In the induction step you want to prove the increment from $n$ to $n+1$ on both sides are the same, namely,
$$(n+1)3^{n+1} = \dfrac{3}{4}(3^{n+1}(2n+1)-3^n(2n-1))\tag1$$
By looking at the increment you got rid of the annoying constant $1$ from the RHS. Then you notice immediately that you can divide $3^{n+1}$ from both... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$ Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$
My attempt: multiply both side... | I know the question has already been answered, and the answer is quite nice, but I was thinking along a different line:
$$\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)} \geq \frac{2}{n(n+1)}\left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right)$ holds on a smooth surface.
Problem: Given a infinitely small geodesic triangle $\triangle ABC$ on a smooth surface, denote the corresponding edges as $a,b,c.$ Prove: the area of $\triangle ABC$ (denote as $S$) and the Gauss curvature on $C$ ( denote as $K$) s... | Comment @ robjohn : Can we check for both signs of const $K?$
So for constant $K$ approximation we have $ \cos \tilde C$
$$= \cos (C+ \dfrac{KS}{3})= \cos (C\pm \dfrac{A+B+C-\pi}{3}) $$
K>0
For an equilateral spherical triangle $( a=b=c,A=B=C=\pi/2) $
$$\text{RHS}=\cos 2 \pi/3= -\dfrac12$$
$$ c^2= a^2+b^2+ab = 3 a^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Polar to cartesian form of $r=\cos(4θ)$?
Consider an equation in polar coordinates, $r = \cos(4θ)$. Find the equation of the curve in the first quadrant in Cartesian coordinates.
This is for an assignment and this is what help I have received so far from user170231-
$$\begin{align}
r(\theta)&=\cos(4\theta)\\[1ex]
&=\... | Welcome to MSE! Here you can see how $r=cos(4\theta)$ is negative and positive and the importance of solving for both $\pm \sqrt{x^2+y^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there any way to calculate $\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$ faster? During solving a problem I get this expression:
$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$$
To calculate it normally, I do this way:
$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\bi... | $$(1+i)^{10} = {10\choose 0} + {10\choose 1}i -{10\choose 2}-{10\choose 3}i +{10\choose 4}+ ... -{10\choose 10} $$
$$(1-i)^{10} = {10\choose 0} - {10\choose 1}i -{10\choose 2}+{10\choose 3}i +{10\choose 4}+...-{10\choose 10}$$
So $$...={(1+i)^{10}-(1-i)^{10}\over 2i} = {1\over 2i}((2i)^5-(-2i)^5) = 32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Range of error in truncated division $\lfloor\frac{a\pm 1}{b\pm 1}\rfloor$ With $a,b\in\mathbb{N}^+$, $b > 1$ and $2\lfloor\log_2(b+\Delta_b)\rfloor\ge\lfloor\log_2(a+\Delta_a)\rfloor$ where $\Delta_n\in\{-1,0,1\}$ is my guess that
$$x + \Delta_x = \left\lfloor\frac{a+\Delta_a}{b+\Delta_b}\right\rfloor$$
correct? If th... | The largest and smallest numbers that can be expressed as $\frac{a+\Delta_1}{b+\Delta_2}$ with $\Delta_1,\Delta_2\in\{-1,0,1\}$ are $\frac{a+1}{b-1}$ and $\frac{a-1}{b+1}$, respectively. We can calculate that
$$\frac{a+1}{b-1}-\frac{a}{b}=\frac{a+b}{b(b-1)}$$
and
$$\frac{a}{b}-\frac{a-1}{b+1}=\frac{a+b}{b(b+1)}.$$
If t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ I need to find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ so that I may find the relationship between $I_4$ and $I_3$, I have done the following:
Setting up for Integration by parts
$$u=(1+x^3)^n \implies u^{'}=3nx^2(1+x^3)^{n-1... | For completeness, by $$x^{6}\left(1+x^{3}\right)^{n}-x^{6}\left(1+x^{3}\right)^{n-1}=x^{6}\left(1+x^{3}\right)^{n-1}\left(1+x^{3}-1\right)= x^{9}\left(1+x^{3}\right)^{n-1}, $$ we have $$
\begin{aligned}
I_{n} &=\frac{2^{n}}{7}-\frac{3 n}{7}\left(I_{n}-I_{n-1}\right) \\
7I_{n} &=2^{n}-3 n I_{n}+3 n I_{n-1} \\
I_{n} &=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$ Let $N$ be a Poisson random variable with the mean parameter $m$.
We are interested in finding the following limit
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N... | We will show that $lim_m m E\ln \frac{N+ \frac12}{m+1} = -1$.
We have $$m E\ln \frac{N+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + Em\ln \frac{m+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + m \ln (1 - \frac{0.5}{m+1})$$
and thus
$$lim_m m E\ln \frac{N+ \frac12}{m+1} = \lim_m E\ln \frac{N+ \frac12}{m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the double integral $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$ $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$
I considered changing the order of integration but that I do not think it does anything as it just leads me to evaluating dy still.
Another possibility I considere... | I think changing the order of the integration is best:
$$\sqrt[4]{x}\leq y \leq 2$$
$$0 \leq x \leq 16$$
is the same as
$$0\leq x \leq y^4$$
$$0 \leq y \leq 2$$
from which the integral becomes $\int_0^2\int_0^{y^4}\sqrt{2y^5 +1}dxdy=\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy$
which is easy to solve given the substitution me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ ... | Let $t = \sqrt{ab}$ and
$$f(a,b,c) = a^2+b^2+c^2 + 6 - 3(a+b+c).$$
Suppose $c = \min\{a,\,b,\,c\},$ then $c \leqslant 1,$ so
$$\sqrt{a}+\sqrt{b} \geqslant 2\sqrt{t} \geqslant 2.$$
We have
$$f(a,b,c) - f(t,t,c) = a^2+b^2-2t^2 - 3(a+b-2t) = \left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\right]\left(\sqrt{a} - \sqrt{b}\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Trying to show that $r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right)$ is an ellipse I am trying to prove that planets move in ellipses,
I watched this video: https://www.youtube.com/watch?v=DurLVHPc1Iw
and read this: https://arxiv.org/pdf/1009.1738.pdf .
But both sources end up with this as an equ... | We have
$$
\begin{split}
r&= \dfrac{1}{C} \left(\dfrac{1}{1+e \cos(\theta+\omega)}\right) \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \cdot \dfrac{1/C}{1/C} \\
&= \dfrac{1/C}{1+ e\cos(\theta+\omega)}
\end{split}
$$
Write $P= 1/C$. Then we have
$$
\begin{split}
r= \dfrac{P}{1+e \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
determinant of matrix in polynomial Suppose $X=\sqrt{k^2 - k_1^2}, Y= \sqrt{k^2 - k_2^2},Z=\sqrt{k^2 - k_3^2}, S=\sqrt{k^2 - k_4^2}, T= \sqrt{k^2 - k_5^2}$. \
And there is $5 \times 5 $ matrix as given below.
\begin{equation}
\begin{vmatrix}
a_{11} & a_{12} & a_{13} & a_{14}*S & a_{15}*T\\
a_{21}*X & a_{22}*Y & a_{23... | Here is a special case of your matrix
$$
\begin{vmatrix}
1 & 0 & 0 & 0 & 0\\
0& 0 & 0 & 1 & 0\\
0 & 0 & Z & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0& 1\\
\end{vmatrix}.
$$
It evaluates to $Z=\sqrt{k^2 - k_3^2}$. (Or is it $-Z$? No matter.) You can't achieve what you are asking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that integral converges How can I prove that this integral converges?
$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$
I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when serie... | Observe that for $x\geq 2$ we have $\frac{x}{2}\geq 1$ and $\frac{x^2}{2}\geq 2\geq 1$ . Hence,
\begin{align}
(x-1)\cdot \sqrt{x^2-1}&\geq (x-\frac{x}{2})\cdot \sqrt{x^2-\frac{x^2}{2}}\\
&=\frac{x}{2}\cdot \sqrt{\frac{x^2}{2}}\\
&=\frac{x^2}{2\sqrt{2}}
\end{align}
Hence for every $x\geq 2$
$$\frac{1}{(x-1)\cdot \sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$ For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:
*
*$\sin x \approx x$
*$\cos x \approx 1-\frac{x^2}{2}$
*$\tan x \approx x +\frac{x^3}{3}$
I am looking for a... | Look at the trigonometric circle and approximate the chord $s$ with the angle $\theta$.
Here $x=\cos \theta$ and $y=\sin \theta$.
From the big triangle, you have $x^2+y^2=1$ which is obvious, and from the small circle, you have $s^2 = y^2 + (1-x)^2$. But with approximation $s \approx \theta$ the above is
$$\theta^2 \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $(1+x+x^2)^n = C_0 + C_1x + C_2 x^2...$, then find $C_0C_1 -C_1C_2 + C_2C_3...$ $$(1+x+x^2)^n = C_0 + C_1 x + C_2 x^2...C_n x^n$$
And
$$(1-\frac 1x +\frac{1}{x^2})^n = C_0 -\frac{C_1}{x}+\frac{C_2}{x^2}...$$
So if we multiply, the two equations, the given series will be obtained by the coefficient of $x^{-1}$ on the... | $$
(1+x+x^2)(1-\frac 1x +\frac{1}{x^2}) = x^2 + 1 + \frac{1}{x^2}
$$
so that
$$
(1+x+x^2)^n(1-\frac 1x +\frac{1}{x^2})^n = (x^2 + 1 + \frac{1}{x^2})^n
$$
contains only even powers of $x$. In particular, the coefficient of $x^{-1}$ is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3971436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$ The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____
My approach is as follow
Already this question is solved Maximum value of argument but I... | Hint:
Arg$\left(\dfrac12+i\cdot\dfrac{\cot\dfrac\theta2}2\right)=$Arg$\left(\cot\dfrac\theta2\right)=$Arg$\left(\tan\left(\dfrac{\pi-\theta}2\right)\right)$
Use this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both ... | The provided solution first employs the angle addition identity $$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,$$ with the choices $$\alpha = \arcsin \frac{3}{5}, \quad \beta = \arctan \frac{7}{24}.$$ That is to say, $$\sin \alpha = \frac{3}{5}, \quad \tan \beta = \frac{7}{24}.$$ In order t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove the roots of $z^7+7z^4+4z+1=0$ lie inside a circle of radius $2$. Problem: Prove that all the roots of the equation $z^7+7z^4+4z+1=0$ lie inside the circle of radius $2$ centered at the origin.
This question came in the roots of unity section of a summer course. My attempts have been to factor and hope something ... | Assume that $z$ is a root of the equation with $|z| \ge 2$. Then
$$
1 = \left|\frac{7z^4+4z+1}{z^7} \right|
\le \left|\frac{7}{z^3}\right| + \left|\frac{4}{z^6}\right| +\left| \frac{1}{z^7}
\right| \\
\le \frac{7}{8} + \frac{4}{64} + \frac{1}{128} = \frac{121}{128} < 1
$$
gives a contradiction.
The idea is to show th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove $H$ is a normal subgroup of $K$ $G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1 & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i & ... | a. $H=\{I,-I\}$, so, for all $k\in K$, $kH=\{kI,-kI\}=\{k,-k\}$,
and also $Hk=\{Ik,-Ik\}=\{k,-k\}$,
so indeed each left coset of $H$ in $K$ is a right coset of $H$ in $K$, so $H\lhd K$.
b. One element of $G/H$, for example,
is $\left\{I\begin{bmatrix}0 & i \\ -i & 0\end{bmatrix}, -I\begin{bmatrix}0 & i \\ -i & 0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$ Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).
My Work :-
Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\fra... | $T_n = \dfrac{1}{(2n-1)(2n+1)(2n+3)}$
Let $V_n= \dfrac{1}{(2n-1)(2n+1)}=(2n+3)T_n$
$V_{n} - V_{n+1}=\dfrac{1}{(2n-1)(2n+1)}-\dfrac{1}{(2n+1)(2n+3)}=\dfrac{2}{(2n-1)(2n+1)(2n+3)}=2T_n$
Let $\sum_{n=1}^{\infty}T_n=S$
Therefore,
$\sum_{n=1}^{\infty}(V_n - V_{n+1})=(V_1-V_{\infty})=2S$
$V_1=\frac{1}{3},V_{\infty}=0$
Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1}, ~~f_k=k/n$ Very often the summations at pre-degree level are done by using differencing-telescoping and by some symmetry properties. The following summation is fabricated keeping one thing in mind, I may tell it later. $$\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-... | There a many ways to do it, I presume.
$$S_n=\sum_{k=1}^n \frac{k^3}{3 k^2 n-3 n^2k+n^3}=\sum_{k=1}^n \frac{k^3}{3(k-a)(k-b) }$$ where
$$a=\frac{n}{6} \left(3-i \sqrt{3}\right)\qquad \text{and} \qquad b=\frac{n}{6} \left(3+i \sqrt{3}\right)$$
$$S_n=\sum_{k=1}^n \Bigg[\frac{a^3}{3 (a-b) (k-a)}-\frac{b^3}{3 (a-b) (k-b)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Positive coefficients after the substitution? Let $n>1$ be an integer and $\Phi_n(x)$ be the $n$ th cyclotomic polynomial.
If we subsititute $x$ by $x+1$ , do we always get a polynomial with positive coefficients ? If yes, how can this be proven ? I also want to prove that no
coeffcient upto the $\varphi(n)$ th is mis... | Deal with the case $n=1,$ $ n = 2$ separately.
Notice that $ \Phi_1(x+1) = (x+1) - 1 = x$, which has a non-positive constant term.
This will be the only coefficient of $ \Phi_n (x+1)$ which is non-positive.
For $ n > 2$,
Let $A_n$ denote the set of integers $ 0 < k < n$ such that $ \gcd( n,k) = 1$.
Let $A_n^*$ denote ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$ For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$
I immediately thought of saying that from symmetry we have that $x\le y \le z$.
Also, $y^2+z... | this is called a CW answer; recommend beginning with
Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.
Here are a number of posts about
Vieta Jumping/ Hurwitz/Markov Grundlösung
Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$.
Diophantine quartic equation in four variables
Is it true that $f(x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $ The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :
$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$
But is there any other way to solve this? Any hints would be appreciated... | Just for fun, try letting $x=\tan\theta=\sin\theta/\cos\theta$ with $\theta\to\pi/2^-$, and use
$$\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=\sec\theta={1\over\cos\theta}$$
so that
$$\begin{align}
(\sqrt{x^2+1}-x)(x+1)
&=\left({1\over\cos\theta}-{\sin\theta\over\cos\theta}\right)\left({\sin\theta\over\cos\theta}+1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please... | Showing that $a+b > - \frac73$:
$$(a-2)^2+(b-2)^2+(a-b)^2\geq(a-2)^2+(b-2)^2 \geq\frac{(a+b)^2}{2}-4(a+b)+8$$
Now using the previous result $a + b < -2$ we get:
$$\frac{(a+b)^2}{2}-4(a+b)+8> 18$$
From OP's equation in the question, this gives
$$a+b=-\frac{6}{(a-2)^2+(b-2)^2+(a-b)^2}-2>-\frac{6}{18} -2 = -\frac{7}{3}. \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Integral of $1/x^2$ without power rule I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents
\begin{equation*}
\int \frac{1}{x^2} \; dx
\end{equation*}
When using integration by parts, you end up with the same integral in the rhs so it seems out of l... | Let $x=\cos \theta +i\sin \theta$. By De Moivre's Theorem, $$\frac{1}{x^2}=x^{-2}=\cos (-2\theta)+i\sin (-2\theta)=\cos (2\theta)-i\sin(2\theta)$$
Also $$dx=(-\sin \theta+i\cos \theta) d\theta$$
The integral becomes $$\int( \cos 2\theta-i\sin 2\theta)(-\sin \theta+i\cos\theta)d\theta\\
=\int (-\sin \theta\cos 2\theta+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 11,
"answer_id": 0
} |
Determine if there's a linear map such that... Today in the class we were solving a following problem:
Determine if there's a linear map $\phi : R^3 \rightarrow R^3$ such that $\phi(2,1,0)=(3,2,1), \phi(4,2,1) = (4,1,0), \phi(2,2,1) = (2,3,3), \phi(2,2,2) = (0,0,1)$. It was then solved by creating an augmented matrix w... | $$A=\left(
\begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i \\
\end{array}
\right)$$
$$A\cdot (2,1,0)=(3,2,1),A\cdot (4,2,1)=(4,1,0),A\cdot (2,2,1)=(2,3,3),A\cdot (2,2,2)=(0,0,1)$$
$$\begin{cases}
2 a + b=3\\
2 d + e=2\\
2 g + h=1\\
4 a + 2 b + c=4\\
4 d + 2 e + f=2\\
4 g + 2 h + i=1\\
2 a + 2 b + c=2\\
2 d +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solving the integral of $\frac {dx} {1+x^3}$
Compute integral $$\int_0^{\infty} \frac{dx}{1+x^3}$$.
I used the formula from complex variables by fisher (2.6 formula (9)) that $$\int_0^{\infty} \frac{dx}{1+x^a}=\frac {\pi} {a\sin(\pi/a)}$$ This means $$\int_0^{\infty} \frac{dx}{1+x^3}=\frac {\pi} {3\sin(\pi/3)}$$ Is t... | Next, simplify
$$
F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1|
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|}
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Nested equilateral triangles Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.
Firstly, I would be very grateful if someone can explain to me how I am su... | WLOG, $AB = 8, DF = 5$. Say $\angle DEC = \theta$
If $CD = x$ then $CE = 8 - x$
Applying sine law in $\triangle CDE$,
$\displaystyle \frac{\sin 60^0}{5} = \frac{\sin \theta}{x} = \frac{\sin(60^0+\theta)}{8-x}$
From first two, we have $\sin \theta = \frac{x \sqrt3}{10}$
From first and third, $\sin 60^0 \cos \theta + \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Transformation of the integrand of an elliptic integral to the general form of an elliptic curve First, some background.
An ellipse is defined by the following equation:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies y=b\sqrt{1-\frac{x^2}{a^2}}.$
To get the arclength of the ellipse, we use calculus to obtain:
$4\int_{0}^{a... | $y^2=(1+e^2t^2)/(1+t^2)$
If $2\ne 0$,
$y=z/(1+t^2)$ gives $z^2=(1+e^2t^2)(1+t^2)=e^2(t^2+a)^2+b$
$z=w+ie(t^2+a)$ gives $w^2 +2wie(t^2+a)=b$
$t=u/w$ gives $w^2 +2ieu^2/w+2wiea=b$
or $$w^3+2w^2iea-bw=-2ieu^2$$
If $e\ne 0,3\ne 0$ it can be transformed into $U^2= W^3+AW+B$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denomin... | The conjugation/rationalization approach works, but is somewhat tedious. The key is to introduce the right terms in order to force the numerator and denominator into a differences of integer powers:
$$\begin{align}
\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}&=\frac{\left((19-x)^2\right)^{\frac14}-\left(2^4(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.
I wanted to calculate $2 \cdot Im(x) = ... | Consider $(a+b+c+ab+ac+bc)(1+\overline {abc})$.
When you multiply out remember that $a\overline {a}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Link between thetrahedral numbers and combinatorics The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1... | I found an analogous explaintion using people.
Let's assume we want to calculate $\binom{5}{3}$, i.e. the number of subsets of order 3 with 5 elements / people {A,B,C,D,E}.
*
*Subgroups of size 3 with A are all possible pairs using {B,C,D,E} and
A itself, i.e. $\binom{4}{2}$.
*Subsets of size 3 with B (and without ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find in closed form.$\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx$ Find in closed form .
$\displaystyle\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx$
My work
We put $t=\cos(x)$ we find $\displaystyle\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\... | I did not check your calculations.
$$\frac{1}{2n^2+3n}=\frac{1}{3 n}-\frac{2}{3 (2 n+3)}$$
$$S_p=\sum_{n=1}^{p}\frac{1}{2n^2+3n}=\frac{1}{3}H_p-\frac{1}{3} \left(\psi \left(p+\frac{5}{2}\right)-\psi
\left(\frac{5}{2}\right)\right)$$Using asymptotics
$$S_p=\left(\frac{8}{9}-\frac{2 \log (2)}{3}\right)-\frac{1}{2 p}+O\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$ How can I approach:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$$
I tried the usual differentiation under the integral sign but it didn't work so well.
I also tried to rewrite it the following 2 ways:
$$\int _0^1\frac{\ln \left(1+\sqr... | I do not think that I should arrive to the result but here are my attempts.
Trying to work the antiderivative
$$\int \frac{\log \left(1+\sqrt{x}\right)}{1+x^2}\,dx=2\int \frac{ t \log (t+1)}{t^4+1}\,dt$$ Using partial fractions
$$\frac{ t }{t^4+1}=\frac{ t }{(t-a)(t-b)(t-c)(t-d)}$$ where $(a,b,c,d)$ are the roots of un... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$ let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$
maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x... | Since at the point $x=y$, equality holds, and multiple derivatives are $0$, it appears likely that a solution with derivatives is needed. Here is one:
Following @River Li's calculations, it suffices to prove, for $x \ge 1$, that (let $n$ be a parameter)
$$
f(x) = \ln (x^{n+1} + 1) - \ln (x^n + 1)- \frac{1}{2n+1}\ln \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Mistake proving that $3 = 0$ I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:
*
*Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.
*Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $... | You made a mistake When you said that if $$x^3=1$$
then $$x=1$$
In fact,
$$x^3=1\iff x^3-1=0\iff (x-1)(x^2+x+1)=0$$
$$\implies x=1 \text{ xor } x^2+x+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$ How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as
$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfra... | Since,
for $|x| < 1$,
$\arctan(x)
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{2n+1}
$,
so
$\arctan(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n}{2n+1}
$.
Similarly,
for $|x| < 1$,
$arctanh(x)
=\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1}
$,
so
$arctanh(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$
How to evaluate $$\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$$
Attempt:
$$f(x) = \frac{x^8-1}{x^{10}+1}$$
$$I = \int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$$
Substituting $t = \frac{1}{x}$,
$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{... | with $x = e^u$ and $dx = e^u du,$ I got
$$ \int_{- \infty}^\infty \frac{e^{4u} - e^{-4u}}{e^{5u} + e^{-5u}} du = \int_{- \infty}^\infty \; \frac{ \sinh 4u \; }{ \cosh 5u \;} du$$
which is an odd integrand over symmetric endpoints
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that for every triangle $ABC$ $\frac{1}{h_b}+\frac{1}{h_c}-\frac{1}{h_a}=\frac{1}{r_a}$ Show that for every triangle $ABC$ $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{1}{r_a}.$$
Let $S_{ABC}=S$. Then $$S=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}\Rightarrow h_a=\dfrac{2S}{a};h_b=\dfrac{2S}{b};h_c=\d... | If $p$ is the semiperimeter $\frac {a+b+c}2$:
$$\frac1{h_b}+\frac1{h_c}-\frac1{h_a}=\frac {b+c-a}{2\color{red}S} = \frac {a+b+c-2a}{2\color{red}S} = \frac{\frac {a+b+c}2-\frac {2a}2}{\color{red}S} = \frac {p-a}{\color{red}S} = \frac1{r_a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$ This is a math Olympiad problem.
my attempt :
By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100... | From the equation, we have
$$(x-4)^2+y^2 = 25$$
Hence, we can deduce that $(|x-4|,y) = (0,5),(3,4),(4,3)$ or $(5,0)$.
The solution is then
$$(x,y) = (4,5),(7,4),(1,4),(0,3),(8,3),(9,0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
PDF of a rectangle If I want to find the $c$ of a $PDF$ when it's given:
$f_{X,Y}\left(x,y\right)=c\:\:\:\:\left(The\:area\:in\:blue\right),\:otherwise:\:0$
I try to do that:
$$\int _{\frac{1}{2}}^1\:\int _{-x+\frac{3}{2}}^1\:c\,dy\,dx+\int _0^1\:\int _{-x+\frac{1}{2}}^{-x+1}\:c\,dy\,dx=1$$
But I got $c=\frac{8}{5}$ a... | $$
\begin{aligned}
\iint\limits_{\text{square}}f(x, y)dxdy &= \iint\limits_{\text{white triangle}}0dxdy+
\iint\limits_{\text{blue trapezoid}}cdxdy+
\iint\limits_{\text{white trapezoid}}0dxdy+
\iint\limits_{\text{blue triangle}}cdxdy = \\
&= c\left(\iint\limits_{\text{blue trapezoid}}dxdy+
\iint\limits_{\text{blue trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
sum of floor function simplification Let $n \in \mathbb{N}$.
I'm trying to show $\sum_{i=1}^{\infty} \left(\left \lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}}\right \rfloor \right) = 2^k-1$.
I know that when $2^i > n-1$, the right floor function will become zero.
I'm not sure how I cancel ... | What you're trying to show is not always true. For example, let $n = 4$, so $n - 1 = 3$, and $k = 1$, so $2^k = 2$. This means $n + 2^k - 1 = 5$. However, using that each summation term is non-negative, the first few terms then become
$$\begin{equation}\begin{aligned}
& \sum_{i=1}^{\infty} \left(\left\lfloor{\frac{n+2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$ I was solving a question and saw a pattern. Can someone prove it, please?
We know
$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$
So we... | You have
\begin{align}
k(k+1)(k+2)\cdots(k+r) &= k(k+1)(k+2)\cdots(k+r)\frac{r+2}{r+2}\\
&= k(k+1)(k+2)\cdots(k+r)\frac{(k+r+1)-(k-1)}{r+2}\\
&= \frac{k(k+1)(k+2)\cdots(k+r)(k+r+1)}{r+2}-\frac{(k-1)k(k+1)(k+2)\cdots(k+r)}{r+2}\\
\end{align}
Make the sum, you deduce easily that
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Integral solutions to the diophantine equation $y^2=x^3+2017$. Case 1: $y^2=0 \mod{3}$
\begin{align*}
& y^2 = 3b \quad (b=3k^2) \implies x^3=3a +2 \\
& 3b = 3a + 2 + 2017 \implies b-a = 673 \\
& k^2 = \frac{a-2}{3} + 225 \\
\end{align*}
Then I just took $a =2$ since 225 is already a perfect square. Thus I got the solu... | Comment:
You can try this method:
Take floor of $2017^{\frac 13}=12$
You can check 12 and less to see it gives a solution:
$2017=12^3+17^2=y^2-x^3$
Which gives $y=\pm 17$ and $x=-12$
Among other numbers bellow 12 only 2 gives integer for y:
$2017=45^2-2^3=y^2-x^3$
Which give what you found; $x=2$ and $y=\pm 45$
This me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding length of diagonal in parallelogram given two sides and other diagonal A parallelogram has sides of length 11 cm and 13 cm and has one diagonal 20 cm long the length of the other diagonal is what I am supposed to find. Now my answer came out to be 13.2 cm.
Finding the area of the first triangle:
Perimeter = $11... | Obviously, both diagonals cannot be $20$, otherwise you would have a rectangle, and this would imply $11^2 + 13^2 = 20^2$, which is false.
One solution that is accessible to you is to employ Heron's formula as you have done, but extend the computation. You already found that the area of the parallelogram is $66(2) = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$ How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... ... | Hint:
The $n$th term $T(n)$ is $$(2n-1)(2n)(2n+1)(2n+2) =16n^4+16n^3-4n^2-4n$$
WLOG $$T(n)=P(n+1)-P(n)$$ where $P(m)=\sum_{r=0}^ua_rm^r$
$$\sum_{r=0}^5a_r((n+1)^r-n^r)$$
$$=a_1+a_2(2n+1)+a_3(3n^3+3n+1)+a_4(4n^3+6n^2+4n+1)a+a_5(5n^4+10n^3+10n^2+5n+1)+\cdots$$
Clearly, $a_r=0$ for $r\ge6$
Now compare the coefficients of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2... | Or just integration by parts:
\begin{eqnarray*}\int_0^1 r^2 \frac{r}{\sqrt{1-r^2}}dr
& = & \left.-\sqrt{1-r^2}\cdot r^2\right|_0^1 + \int_0^1 (2r)\sqrt{1-r^2}\;dr \\
& = & \left.-\frac 23\left(1-r^2\right)^{\frac 32}\right|_0^1 \\
& = & \frac 23
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$ It is visible that the result is 0, but I can't calculate it.
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$
It occurred to me to express it as a product and take a ratio test, but I'm not sure
| $$
f(n)=\frac{3^n}{(1+3)(1+3^2)\cdots(1+3^n)}
=\frac{\frac{3^n}{3^1 3^2 \cdots 3^n}}{\frac{1+3^1}{3^1}\frac{1+3^2}{3^2}\cdots\frac{1+3^n}{3^n}}\\
=\frac{\frac{3^n}{3^{\frac{n(n+1)}2}}}{(1+3^{-1})(1+3^{-2})\cdots(1+3^{-n})}
=\frac{2\cdot 3^{-\frac{n(n-1)}{2}}}{\left(-1;\frac{1}{3}\right)_{n+1}}
$$
in terms of a $q$-Poch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Symmetry in the quadratic form matrix The following is an excerpt from Greene's Econometric Analysis (7th Edition). Therein, the author states the matrix in quadratic forms must be symmetric. I would like to know why. What if, for instance, $a_{12} \neq a_{21}$?
| $$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+(b+c)xy+dy^2$$ can be symmetrized as
$$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&\frac{b+c}2\\\frac{b+c}2&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+2\frac{b+c}2xy+dy^2.$$
So a quadratic form has a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the fami... | We will prove the last statement.
Let $\alpha$ be a root of $p_1:= X^3 + X^2-1$, i.e., $\alpha^3 +\alpha^2-1 = 0$. We will prove that $\beta := \alpha^2$ is a root of $p_2:= X^3-X^2+2X-1$.
We have the following identities:
$$\beta = \alpha^2,$$
$$\beta^2 = \alpha^4 = \alpha\cdot \alpha^3 = \alpha (-\alpha^2 +1) = -\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.
The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
x &= 1\\
x &= -\sqr... | The multiplicative group of $9$-th roots is isomorphic to the additive group $\mathbf Z/9\mathbf Z$, under the isomorphism:
$$\mathbf Z/9\mathbf Z\longrightarrow U_9,\quad k\bmod9\longmapsto\mathrm e^{\tfrac{2ik\pi}9}.$$
Under this isomorphism, generators of $\mathbf Z/9\mathbf Z$ map onto generators of $U_9$, i.e. pri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the meaning of $A =\prod_{jI am trying to parse this proof in Hungerford's book:
When he gets into computing this:
$$A =\prod_{\substack{j<k\\ j,k \neq c,d}}^{}(i_j-i_k)$$
What does this means?
*
*Does it means that $j<k$ and $j\neq c$ and $k\neq d$; or
*Does it means that $j<k$ and $j\neq c$ and $k\neq... | I feel like Hungerford may have painted himself into a corner with that notation, so let me offer you a different way of doing exactly what the proof is doing.
For a fixed $n$, consider commuting variables $x_1,\ldots,x_n$ and form the Vandermonde matrix $V=V(x_1,\ldots,x_n)$ with $V(x)_{ij} = x_i^{j-1}$. It is a pleas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$ When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a... | Original problem : $$\color{green}{(a-b)}\color{red}{(2b-a+2)}\color{green}{(2b+1)} - (b+1)\color{blue}{(2a-2b-1)}(2b-a)$$
*
*) Notice how
$\color{green}{(a-b)(2b+1)}=-2b^2+2ab-b+a=b(-2b+2a-1)+a=b\color{blue}{(2a-2b-1)}+a$
$$\Rightarrow\color{blue}{(2a-2b-1)}=\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}$$
*) We find th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$ I'm struggling to grasp this particular question:
When a polynomial $f(x)$ is divided by $x+3$, the quotient is $2x^2-x+7$ and the remainder is $10$. What is $f(x)$?
This is what I did:
$$\begin{align}
f(x) &= (x+3)(2... | In general, when you divide a function $p(x)$ by $q(x)$ and you get $f(x)$ as the quotient with a remainder of $r(x)$, you can alyways represent the answer as: $\frac{r(x)}{q(x)} + f(x)$
So you can rewrite your question algebraically as $\frac{10}{x+3} + 2x^2-x+7$ and then can simplify as follows:
$\frac{10}{x+3} + 2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verif... | Note that
$$f(x)=\tan\left(\frac{\pi}{2}\frac{(1+x)^2}{3+x^2}\right)\tan\left(\frac{\pi}{2}\frac{(1-x)^2}{3+x^2}\right)=\frac{2\cos(2\pi/(3+x^2))}{\cos(2\pi x/(3+x^2))-\cos(2\pi/(3+x^2)} +1,$$
which is slightly easier to take the derivative:
$$f'(x)=\frac{8 \pi x \cos(\frac{2 \pi x}{3 + x^2}) \sin(\frac{2 \pi}{3 + x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Complex Quadratic, what did I do wrong? $$z^2-2(1-2i)z-8i=0$$
Here is my working:
$$\begin{align}
x_{1,2} &= \frac{2-4i \pm \sqrt{(4i-2)^2 + 4\cdot 8i}}{2} \\
&=\frac{2-4i \pm \sqrt{-16+4-16i + 32i}}{2} \\
&= \frac{2-4i \pm \sqrt{-12+16i}}{2} \\
&= \frac{2-4i \pm \sqrt{4(-3+4i)}}{2} \\
&= \frac{2-4i \pm 2\sqrt{(4i-3)}}... | The mistake is when you square $a+bi$, you should get $a^2-b^2+\color{red}2abi$ and not $a^2-b^2+abi$.
$$a^4+3a^2-4=0$$
$$(a^2+4)(a^2-1)=0$$
$$a=\pm 1, b = \pm 2$$
Hence,
$$x_{1,2}=1-2i\pm(1+2i)$$
which gives you the desired solution of $2$ and $-4i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to? What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?
Sorry for the basic question. I am a little confused when solving such problems. They are very easy, I know, but still... Which is the easiest algorithm and how would you solve it? We are supposed... | $\sqrt{8\cdot32\cdot(-3)^2} = \sqrt{8\cdot32\cdot9} = 3 \cdot \sqrt{8\cdot32} = 3 \cdot \sqrt{16\cdot16} = 3 \cdot 16 = 48$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How should I finish this by completing to square? I am trying to calculate the integral:
$$\int_{-\infty}^{+\infty}\exp\left[-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t\right]dk$$
I thought on maybe mark $k-k_0=u$ but I am not see how it gonna help me
P.S the integral is only with respect to $k$. $k_0$ is constant ... | $$
-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t
= -\frac12 a^2 k^2 + a^2 k_0 k - \frac12 a^2 k_0^2 + ikx - i\frac12 \frac1m ht k^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) k^2 + (a^2 k_0 + ix) k - \frac12 a^2 k_0^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( k^2 - 2 \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} k + \frac{a^2 k_0^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$ It's an inequality based on two found on the website MSE (see the reference):
Let $a,b,c>0$ then we have:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(... | The given inequality is homogenious by $\;(a,b,c)\;$ and symmetric by $\;(a,b).\;$
Let WLOG
$$a+b+c = 6,\quad a+b=s,\quad ab=p,\tag1$$
then it suffices to prove the inequality in the form of
$$\dfrac a{6-a}+\dfrac b{6-b}+\dfrac{6-a-b}{a+b} \ge \sqrt{\dfrac94
+9\,\dfrac{(a-b)^2}{(a+b)(6-a)(6-b)}},\tag2$$
with the both p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In p... | Here's something you may use:
Consider $p \leq 0$ first. In this case, $\frac{1}{n} = n^{-1} \geq 0$. The limit of this implies that it doesn't converge to $0$ as $n \rightarrow \infty$ and so $\sum_{n = 1}^{\infty} \frac{1}{n^p}$ does not converge.
Now assume $p > 0$. Then $\frac{1}{n^p}$ is a decreasing sequence of p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve one system of equations Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$
I tried solving this using cross multiplication method but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in ... | Here is a systematic method that brings back your issue to solving a linear system.
I assume $abc \ne 0$. Setting
$$X=x/a, Y=y/b, Z=z/c \ \ \ \text{and} \ \ \ A=a^2,B=b^2,C=c^2,$$
then replacing $x,y,z$ by $aX,bY,cZ$ resp., and inverting equ. 1, with equ. 2, your system can be written under the form:
$$\begin{pmatrix}1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Tennis match probability - is my logic incorrect? A tennis match consists of sets, until one player wins three sets. Player $A$ has a two-thirds chance of winning. Player $B$ has a one-third chance of winning. If they play a match, what is the probability of $A$ winning?
My attempt:
The minimum games for $A$ victory is... | Your working: $\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
Please see the second term. You are allowing any $1$ of the $4$ sets to be won by player B. But if there are $4$ sets,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When i... | The statement is equivalent to proving $$10^{3^k}-1\equiv_{3^{k+2}}0$$
Notice that $\phi(3^{k+2})=3^{k+2}-3^{k+1}=2\cdot 3^{k+1}$, so that for a unit $a$ we have $$a^{2\cdot 3^{k+1}}\equiv_{3^{k+2}}1$$ We also have
Claim: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}}$$
Proof: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}-2\cdot 3^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Limit of a continued product $\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $ as $n \to \infty$ Evaluate $\lim_{n\to \infty }\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $
I don't have any ideas on how to approach the problem. Hints or solutions are appreciated.
| If you factor the general term, you get:
$$\frac{r^2 \left(r^2+3\right)^2}{(r-2) (r+2)
\left(r^2-2 r+4\right) \left(r^2+2 r+4\right)}$$
Notice that $r^2-2r + 4 = (r-1)^2 + 3,\ r^2+2r + 4 = (r+1)^2+3,$
So the product telescopes: and almost everything cancels. (the final answer is $\frac{72}7.)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are there random variables $X,Y$ such that $X,Y,X+Y \sim N(0,1)$? Let's suppose we have two, not necessarily independent, random variables $X,Y \sim N(0,1)$.
Then, is it possible that $X+Y \sim N(0,1)$?
What I have managed to show is that because $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$ it would be necessary that $Cov(... | The pdf of the sum of two absolutely continuous random variables is simply given by the convolution of the pdf's of the variables. In our case, $f_X(x) = f_Y(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, so $f_{X + Y}(x) = \int_{- \infty}^\infty f_X(t)f_Y(x - t) dt = \frac{1}{4} \int_{- \infty}^\infty \exp(- \frac{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end wit... | $0$ is last digit. Note that there are only 2 numbers from the set whose mod 3 is 1.
Case 1: Pick $0$ $x$ s.t. $x \mod 3 = 1$: Not possible
Case 2: Pick exactly 1 digit x s.t. $x \mod 3 = 1$. We must pick exactly 1 digit x s.t. $x \mod 3 = 2$. Remaining are filled with $x \mod 3 = 0$.
Sum = $2 * 2 * (4!) = 96$
case 3: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
A biquadratic equation to have at least two real roots Which of the equations have at least two real roots?
\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}
I wasn't able to notice something clever, so I solved each of the equations. The first one has $2$ r... | $$ x^4 -13 x^2 + 36 = (x^2 + 6)^2 - 25 x^2 = (x^2 + 5x+6) (x^2 - 5x+6) $$
and both quadratic factors have real roots(positive discriminants).
$$ 4 x^4 -10 x^2 + 25 = (2x^2 + 5)^2 - 30 x^2 = (2x^2 + x \sqrt{30}+5) (2x^2 - x \sqrt{30}+5) $$
and both quadratic factors have negative discriminants. OR
$$ 4 x^4 -10 x^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find $\mathbb{E}(X)$ and Var$(X)$ from the c.d.f. Suppose a child plays outside in the yard. On their own, they come back inside at a random time uniformly distributed on the interval [0,1] (Take the units to be hours.) However, if the child is not back in 50 minutes, their mother brings them in. Let X be the time when... | Another way to solve this is by ysing the heaviside step function $H(x)$ and the dirac delta function $\delta(x)$:
\begin{align*}
F(s)=
\begin{cases}
0, & s<0\\
s, & 0\leq s<5/6\\
\frac{5}{6} + \frac{1}{6}\cdot H(x-\frac{5}{6}), & s\geq 5/6
\end{cases}
\end{align*}
We can then write the pdf in terms of the dirac delta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help understanding the solution of $m^2 = n^3 -4$. The material I'm studying has the solution for $m^2 = n^3 + 4$ but I can't quite understand it. It starts like this:
First suppose, for a contradiction, that $m$ is odd. Then $n^3 = m^2 − 4 = (m + 2)(m − 2)$. Any common factors of $m + 2$ and $m − 2$ divide $(m + 2) − ... | a) You do know $m \neq 10$ because you have assumed $m$ odd. From "any common factors of $m+2$ and $m-2$ divide $(m+2)-(m-2)=4$," we can conclude that the only possible common factors of $m+2$ and $m-2$ are $1$, $2$, or $4$. But if $m$ is odd, then so are $m+2$ and $m-2$. So $2$ and $4$ cannot be divisors. This leaves ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can one simplify $\sum_{n=1}^8\frac{\sin10n^\circ}{\cos5^\circ\cos10^\circ\cos20^\circ}$? I have trouble solving the following question:
Consider the following expression:
$$\sum_{n=1}^8\frac{\sin10n^\circ}{\cos5^\circ\cos10^\circ\cos20^\circ}$$
The value of the above expression can be fully simplified into the fo... | $\textbf{Hint:}$
Use the formula that, $\sum_{r=0}^{n-1}\sin(a+rh) = \frac{\sin(\frac{nh}{2})\sin(a+(n-1)\frac{h}{2})}{\sin(\frac{h}{2})}$
so, Let $$P = \sum_{n=1}^{8}\frac{\sin(10n^\circ)}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}$$
$$=\frac{1}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}\sum_{n=1}^{8}\sin(10n^\circ)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x^{x^{x+1}}=\sqrt{2}$, then evaluate $x^{x^{p}}$, where $p = 2x^{x+1}+x+1$ I can't figure out how to give a proper form to this expression to use the root of two.
If
$$x^{x^{x+1}}=\sqrt{2}$$
find the value of $W$ if
$$W=x^{x^{p}} \quad\text{where}\; p = 2x^{x+1}+x+1$$
EDIT: This is an algebraic manipulation probl... | Assuming $x^{x^{x+1}} = \sqrt 2$,
\begin{align*}
x^{x^{2x^{x+1} + x + 1}} &= x^{x^{2x^{x+1}} \cdot x^{x+1}} \\
&= x^{\left(x^{x^{x+1}}\right)^{2} \cdot x^{x+1}} \\
&= x^{\left(\sqrt{2}\right)^{2} \cdot x^{x+1}} \\
&= x^{2 \cdot x^{x+1}} \\
&= \left(x^{x^{x+1}}\right)^{2} \\
&= \left(\sqrt{2}\right)^{2} \\
&= 2
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Power of little o - asymptotic - series Consider the following, when $n \to + \infty$:
$$\log\left(1 + \frac{-1}{q}n^{\frac{1-q}{q}} +o(n^{-1}) \right)= R_1(n^{\frac{-1}{q}}) + o(n^{-1}).$$
And I am trying to determine the polynomial $R_1$. From a lemma, I know that this polynomial satisfies $R_1(0)=0$ and is of degree... | If you expand the logarithm using $\log(1+w)=w+\mathcal{O}(w^2)$, you find
\begin{align*}
& - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)\left( { - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right)} \right)^2
\\ &
= - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate integral $\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$ I recently saw the integral problem
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$$
and tried to solve it. Below is what I did.
Interesting to look at other easier solutions.
$$\int\limits_{0}^{2\pi}\frac{dx}{\l... | Just for the fun !
When there is a square in denominator, we never know ! Trying
$$\int\frac{dx}{\left ( 1+n^2\sin^2 (x) \right )^2}=\frac{P(x)}{ 1+n^2\sin^2 (x)} $$ Differentiate both sides and remove the denominators
$$(1+n^2 \sin ^2(x))P'(x)-n^2 \sin(2x) P(x)=1$$ which is not very difficult to integrate. So, by the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Misprint(s) In "Combinatorial Identities" by Riordan? On page 38 in the exercises I was able to confirm that
$$a_{kj}(x)=(x+1)\dots(x+k-1)a_{k-1,j}(x)+x^{k-1}a_{k-1,j-1}(x+1).$$
For $k\ge{}j$ the author defines $b_{kj}(x)$ via
$$a_{kj}(x)=(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\dots(x+k-1)b_{kj}(x),a_{kk}(x)=b_{kk}(x),$$
where ... | First question: Your assumption is correct. There are two typos as indicated by you.
We have the recurrence relation
\begin{align*}
a_{k,0}(x)&=(x+1)\cdots(x+k-1)a_{k-1,0}(x)\\
&=(x+1)^{k-1}(x+1)^{k-2}\cdots(x+k-1)\\
a_{k,j}(x)&=(x+1)\cdots(x+k-1)a_{k-1,j}(x)\\
&\qquad+x^{k-1}a_{k-1,j-1}(x+1)\qquad\qquad\qquad\qquad 0<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix Multiplication - Express a row as a linear combination $$ Let \ A \ = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 3 & 6 \\ \end{bmatrix} and
\ let \ B = \begin{bmatrix} 0 & 1 & -3 \\ -3 & 1 & 4 \end{bmatrix} $$
Express the third row of AB as a linear combination of the rows of B
$$ AB \ = \ \begin{bmatrix} -6 & 3 & 5 \\... | \begin{align} \textrm{third row of } AB
&= \begin{bmatrix}
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} 0 \\ -3 \end{bmatrix} &
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} &
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} -3 \\ 4 \end{bmatrix}
\end{bmatrix} \\
&= \begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor Expansion of the first four non-zero terms $$f(x) = \sqrt{1+x}\sin(4x)$$
Okay, so at this point, I am able to find the answer to these kinds of problems when the outside term is x to the power of a whole number, but this is two terms added together then to the 1/2 power.
The sin expansion looks like this:
$$\si... | Near $0$, you have$$\sin(4x)=4x-\frac{4^3}{3!}x^3+\cdots$$and$$\sqrt{1+x}=(1+x)^{1/2}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots,$$and therefore$$\sqrt{1+x}\sin(4x)=4x+2x^2-\frac{67x^3}6-\frac{61x^4}{12}+\cdots$$The RHS of this equality is what one gets after ignoring the terms of the product$$\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the rank of $\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$ Found this exercise in Serge Lang's Introduction to Linear Algebra:
Find the rank of the matrix $$\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$$
So my process to solve it is as follows. First, I set a syst... | The first two rows are independent (by inspection: they're not multiples of each other).
Then $r_3=2r_1+r_2$, as you can see.
So the row rank is $2$. So the rank is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that the product of four consecutive natural numbers are not the square of an integer
Prove that the product of four consecutive natural numbers are not the square of an integer
Would appreciate any thoughts and feedback on my suggested proof, which is as follows:
Let $f(n) = n(n+1)(n+2)(n+3) $.
Multiplying out... | Your solution is perfect and detailed.
Another alternative shorter approach based on observation is if we can consider $n(n+2)=k$ and then redefine the function as:
$$\begin{align*}f(k) &= k(k+2)\\ f(k) &= (k+1)^2-1\end{align*}$$
Now, we need to prove that $f(k)$ is not the square of an integer.
Well, there exists no t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find infimum and closure of $A$ Let $$
\mathbf{A}=\left\{\frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}: \mathbf{m}, \mathbf{n} \geq \mathbf{3}\right\}
$$
Now find :
*
*$inf(A)$
*$int(A)$ , interior points
*$cl(A)$ , closure of $A$
With euclidean metric on $\mathbb{R}$
We have $int(A) =\emptyset $ bec... | $$\dfrac{3m + 2n}{5m + 7n} = k$$
$$(3m + 2n) = k(5m + 7n)$$
$$(7k -2)n + (5k-3)m = 0$$
$$-\dfrac{7k-2}{5k-3}= \dfrac{m}{n} > 0$$
so $\dfrac{7k-2}{5k-3} < 0$ and $\dfrac{2}{7} \leq k \leq \dfrac{3}{5}$. You can prove that $A$ is dense in $[2/7, 3/5]$, so $inf(A) = \dfrac{2}{7}$, $int(A) =\{\}$ and $cl(A) = [2/7, 3/5]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Scalar "exchange" in determinants Consider the following:
Find
$$
\begin{vmatrix}
x^2+1 & xy & xz \\
xy & y^2+1 & yz \\
xz & yz & z^2+1 \\
\end{vmatrix}
$$
Multiplying $R_1,R_2,R_3$ with $x,y,z$ and dividing $C_1,C_2,C_3$ by $x,y,z$ respectively, we get
$$
\begin{vmatrix}
x^2+1 & x^2 & x^2 ... | *
*Multiplying the $k$th row by $c$ is accomplished by multiplying on the left by the matrix $M(k,c)$ given by
$$ M(k,c)_{ij}=\begin{cases} c, & i=j=k,\\ 1, &i=j\neq k, \\ 0 &\text{otherwise} \end{cases}$$
*Multiplying the $k$th column by $c$ is done by multiplying by $M(k,c)$ on the right.
*Determinants of produc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction that $3^{n-2} \geq n^5$ for $n\geq20$ I am new to induction proofs and wanted to know if my reasoning behind the following proof is correct. Here are my steps :
*
*$n=18 \Rightarrow 3^{18} \geq 20^5$
*Given that $3^{n-2} \geq n^5$ is true for n, show $3^{n-1} \geq (n+1)^5$
$3^{n-1} = 3\times(3)^{n... | It seems like the limit is less ambitious than it could be. Let's aim for $n=15$.
Here is a $n=15$ base case to show $15^5 < 3^{13} $:
$$\begin{align}
15^5 &= 3^5\cdot 5^5\\
&<3^{5}\cdot 27^2\cdot 5\\
&=3^{11}\cdot 5\\
&<3^{13}\\
\end{align}$$
Then for $n\ge 15$,
$$\begin{align}
(n+1)^5 &= n^5 + 5n^4 + 10n^3+10n^2+5n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involvin... | This is a partial answer and not very rigorous (I don't know how to make it rigorous) but i think this way might be able to produce some results.
$$ \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} - \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2} = 4\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Now Consider the function
$$f(a) = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 1
} |
Divisibility of an expression by 11 How to prove that $5^{5n+1}+4^{5n+2}+3^{5n}$ where $n\in \mathbb{N}$ is divisible by $11$ using mathematical induction?
I have tried and got to this $$5 \cdot 25 \cdot 25 \cdot 5^{5k+1}+4\cdot 16 \cdot 16 \cdot 4^{5k+2}+3\cdot 9 \cdot 9 \cdot 3^{5k}$$
and I even skipped the first ste... | Let $f(n)=5^{5n+1}+4^{5n+2}+3^{5n}$
$$f(m+1)-3^5\cdot f(m)=5^{5(m+1)+1}+4^{5(m+1)+2}-3^5(5^{5m+1}+4^{5m+2})$$
$$=5^{5m+1}(5^5-3^5)+4^{5m+2}(4^5-3^5)$$ which is divisible by $11$
as $3^5=9(22+5)\equiv1\pmod{11}, 5^5=5^2\cdot5^3\equiv3\cdot4\equiv1\pmod{11}$
Similarly, $4^5\equiv1\pmod{11}$
$$\implies11|f(m)\iff11\mid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4125397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$
This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
... | Let
$$
S_n:=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}
$$
Obviously $S_0=1$. Assume that equality
$$S_{n-1}=1\tag1$$
is valid for some $n$. Then it is valid for $n+1$ as well:
$$
\begin{align}
S_n
&=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}\\
&=\sum_{k=0}^n\frac{\binom{n+k-1}{k-1}+\binom{n-1+k}{k}}{2^{n+k}}\\
&=\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 0
} |
How to integrate $\int_0^{\infty}\frac{1}{(1+x)(1+x^2)}$ What is the method of integrating the following:
$$\int_0^{\infty}\frac{1}{(1+x)(1+x^2)}$$
I tried doing it via using partial fractions and deduce that:
$$\frac{\ln(x+1)}{2} - \frac{\ln(x^2+1)}{4} + \frac{\arctan(x)}{2} + C$$
However I am unsure how to use the l... | HINT:
Note that we have
$$\begin{align}
\frac12\log(x+1)-\frac14\log(x^2+1)&=\frac14 \log\left(\frac{x^2+2x+1}{x^2+1}\right)\\\\
&=\frac14 \log\left(1+\frac{2x}{x^2+1}\right)
\end{align}$$
And we have the estimates for $x>0$
$$0\le \log\left(1+\frac{2x}{x^2+1}\right)\le \frac{2x}{x^2+1}$$
Can you finish now?
ALTERNAT... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $ I=\int_{0}^{+\infty} \frac{1}{\sqrt{t}} \cdot \sin \left(t+\frac{1}{t}\right) \, dt $ What have i done?
$I$ is improper at $0$ and $+\infty$. The function inside is continuous on $(0 ;+\infty)$ and then integrable on any closed interval contained in $(0 ;+\infty).$
$$
I=\int_0^1 \frac{1}{\sqrt{t}} \sin... | We can go ahead and actually compute the value of this integral by recognizing
$$I = \int_0^\infty\frac{2\,dt}{2\sqrt{t}}\sin\left((\sqrt{t})^2+\frac{1}{(\sqrt{t})^2}\right)$$
which suggests using the substitution $s = \sqrt{t}$
$$I = \int_0^\infty 2\sin\left(s^2+ \frac{1}{s^2}\right) = \int_{-\infty}^\infty \sin\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding generating function and closed formula Find generating function and closed formula $1,0,1,0,1,0,1,0,1,...$
Solution Attempt)
$$\begin{align}G(x) &= 1 + 0x + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1 x^6 + 0 x^7 + 1 x^8 + \dots\\
&= 1 + x^2 + x^4 + x^6 + x^8 + \dots\end{align}$$
$$G(x) = \sum_{k = 0}^{∞} x^{2k} = \su... | Following Ethan remark, we have
$$u_{2n}=1=|\sin((2n+1)\frac{\pi}{2})|$$
and
$$u_{2n+1}=0=|\sin((2n+2)\frac{\pi}{2})|$$
So, we can take
$$f(x)=|\sin((x+1)\frac{\pi}{2})|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\left(y+\frac23\right)\mathrm{ln}\left(\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$ We consider the limit for large $y$ of the following expression :
$$\left(y+\frac23\right)\mathrm{ln}\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}.$$
Many references state that the large $y$ beha... | Considering $$A=\left(y+\frac23\right)\log\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$$ first simplify
$$A=\left(y+\frac{2}{3}\right) \log \left(\frac{y+2 \sqrt{y+1}+2}{y}\right)-2
\sqrt{y+1}$$ For large values of $y$, you could start using $y=\frac 1x$ to make
$$A=\frac{(2 x+3) \log \left(2 x+2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving for integers $m>8, n > 0$ for $2^m - 3^n = 13$ Let $m,n$ be integers such that:
$$2^m - 3^n = 13$$
$m > 8$ since $2^8 - 3^5 = 13$.
I am trying to either find a solution or prove that no solution exists.
I tried to use an argument similar to this one for $2^m - 3^n = 5$ where $m > 5$.
$3^n \equiv -13 \pmod {512}... | we have $$ 256(2^x - 1) = 243(3^y - 1) $$
and we assume $x,y \geq 1$
since $2^x \equiv 1 \pmod {243}$ we calculate that $162 | x.$
Next, $2^{162} - 1$ is divisible by the prime $262657.$
since $3^y \equiv 1 \pmod {262657}$ we calculate that $14592 | y.$ In particular, $256|y.$
Well, $3^{256} - 1$ is divisible by 102... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Complete the Matrix The eigenvectors of A (corresponding to the eigenvalue 2) = \begin{bmatrix}
1 & 2 \\
-1 & 4 \\
\end{bmatrix}
can be solved by solving the matrix equation:
Here is my working:
My answers were $4$ and $-1/2$ which were apparently wrong, though I am wondering if I made a typo when I put ... | I don't know where the matrix $\begin{pmatrix} 4 & -2 \\ 1 & -1/2 \end{pmatrix}$ comes from.
The way I have seen it done is that $\begin{pmatrix} x \\ y \end{pmatrix}$ is a $\lambda$-eigenvector if
$\begin{pmatrix} 1-\lambda & 2 \\ -1 & 4-\lambda \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 0 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
(rectified) proof by induction - Fibonacci Sequence Define the sequence $ (a_n) $ by induction, putting $ a_1 = a_2 = 1 $ and $ a_ {n + 2} = a_ {n + 1} + a_n $, $ \forall n \in \mathbb{N} $ , thus obtaining the Fibonacci sequence $ (1,1,2,3,5,8,13,...)$.
Write $ x_n = \frac{a_n}{a_ {n + 1}} $ and prove that $ \lim x_n ... | There are several mistakes/typos in your proof, and I suggest you go over your proof much more carefully. Once you have reached the equation
$$
\frac{1}{x_{n+1}} = 1+x_n
$$
you can simply apply the limit as $n \to \infty$ from both sides as the limits being finite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2 $ Hi can someone help me please simplify the following showing the working out step by step?
$$
\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2
$$
I can't get the... | Using $a^2-b^2 = (a+b)(a-b)$ we get
$$\left(\sqrt{x} + \frac{1}{\sqrt{x}} + \sqrt{x} - \frac{1}{\sqrt{x}}\right)\left(\sqrt{x}+\frac{1}{\sqrt{x}} - \sqrt{x}+\frac{1}{\sqrt{x}}\right) = \left(2\sqrt{x}\right)\left(\frac{2}{\sqrt{x}}\right) = 4$$
Hence, we get our answer as $4$.
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations a... | If you could show that it is an increasing function of real positive $N$ and its limit is $1$ then that would be enough.
Showing the limit is $1$ as $N \to \infty$ looks easy enough: each component heads towards $1$
while the derivative seems to be $\mathcal P'(x)=\frac{{{\left( \frac{x}{2x+1}\right) }^{\frac{2x+1}{2x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}=2-2\ln(2)$ The question is simple, to prove that $\sum_{n=1}^\infty{\frac{2}{2n}-\frac{2}{2n+1}}=2-2\ln(2)$.
I did it by wirtting down some terms of the series, rearranging as follows:
$$\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}$$
$$2\sum_{n=1}^\infty \frac... | Well, you asked for an elegant way, which depends on your perspective for elegance. I came up with a solution using integrals.
Simplifying the sum,
$$ S =2\sum_{n=1}^{\infty} \frac{1}{2n}-\frac{1}{2n+1} = \sum_{n=1}^{\infty} \dfrac{1}{n(2n+1)} $$
Also,
$$ \int_0^1 x^{2n}\ \mathrm{d} x = \dfrac{1}{2n+1} $$
Substituting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solution Verification: Find Extrema points of the function $f(x,y)=2^{3x+8y}$ that are on $x^2+y^2=1$.
$f(x,y)=2^{3x+8y}$
$x^2+y^2=1$
Polar Coordinates approach:
Let $x=cos(t), y=sin(t), 0\le t \le 2\pi$ (We can see that this solved the circle equation).
Substituting them into my function:
$f = 2^{3\cos(t)+8\sin... | You could have made your work much simpler by noticing that $2^z$ is monotonically increasing – maximising/minimising $2^{3x+8y}$ is equivalent to maximising/minimising $3x+8y$. Then Lagrange multipliers (or any other reasonable method, like plotting the objective function and constrant and eyeballing) will work just f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prob. 2.10 in Bourne & Kendall: These four points lie on a circle Here is Prob. 2.10 in the book Vector Analysis and Cartesian Tensors by D. E. Bourne and P. C. Kendall, 3rd edition:
Show that the four points with position vectors
$$
\mathbf{r}_1, \mathbf{r}_2, \frac{r_2}{r_1} \mathbf{r}_1, \frac{r_1}{r_2} \mathbf{r}_... | You can just observe that the origin has the same power with respect to the two pairs of points collinear with it:
$$
|\mathbf{r}_1|\cdot \left|\frac{r_2}{r_1} \mathbf{r}_1\right|
=r_1r_2=
|\mathbf{r}_2|\cdot \left|\frac{r_1}{r_2} \mathbf{r}_2\right|.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $|f(x)-f(y)| \leq A |x-y|^\alpha$, then $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$ Let $f$ be a $2\pi$ periodic continuous function and let
\begin{align}
\hat{f}(n) = \frac{1}{2\pi} \int_0^{2\pi} e^{-inx} f(x) dx
\end{align}
I want to show if
\begin{align}
|f(x)-f(y)| \leq A |x-y|^\alpha, \quad0< \alpha \leq 1
\end{ali... | Comment from @Medo
\begin{align}
|\hat{f}(n) - \hat{f}(1)| &= \frac{1}{2\pi} \left| \int_0^{2 \pi} (e^{-inx} f(x) - e^{-ix} f(x) ) dx \right| \\
&
= \frac{1}{2\pi} \left| \int_0^{2\pi} \left( e^{-inx} f(x) - f(x+\frac{1}{n} ) - f(x+\frac{1}{n}) - e^{-ix} f(x) \right) dx \right| \\
& \leq \frac{1}{2\pi} \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.