Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Using Residue theorem to evaluate the integral Using Residue theorem to evaluate the integral:
$$\int_0^{\infty} \frac{x^2}{x^4 + 5x^2 +6}dx$$
I am using partial fraction to get:
$$\int_0^{\infty} \left( \frac{3}{x^2 +3} - \frac{2}{x^2+2} \right)dx$$
Then, next step, can someone show me how to use Residue theorem to ... | You just use partial fraction expansion again, for each of the quadratic terms. We have
$$\frac{1}{x^2+3}=\frac{1}{i2\sqrt{3}}\left(\frac{1}{x-i\sqrt{3}}-\frac{1}{x+i\sqrt{3}}\right)$$
and
$$\frac{1}{x^2+2}=\frac{1}{i2\sqrt{2}}\left(\frac{1}{x-i\sqrt{2}}-\frac{1}{x+i\sqrt{2}}\right)$$
Can you finish from here?
SPOLIER... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| Note that,
$$
\lim_{x \to 0}\dfrac{\sqrt{x + a^2} - a}{x} = \lim_{x \to 0}\dfrac{1}{\sqrt{x + a^2} + a} = \dfrac{1}{2a}
$$
for $a > 0$. Thus,
\begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4} = \dfrac{\lim_{x \to 0}\dfrac{\sqrt{x + 25} - 5}{x}}{\lim_{x \to 0}\dfrac{\sqrt{x + 16} - 4}{x}} = \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Gauss Jordan elimination method. Solve by Gauss-Jordan elimination method.
$$x + y + z = 2 \\
x + 3y + 3z = 0 \\
x + 3y + 5z = 2$$
Putting this in reduced row-echelon form:
$$\left(\begin{array}{ccc|r}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & -2 \\
0 & 0 & 1 & 1
\end{array}\right)
$$
So then
$\begin{pmatrix}
x \\ y \\ z
\end{... | $$\begin{cases} x+y+z=2 \\ x+3y+3z=0 \\ x+3y+5z=2 \end{cases}\Rightarrow \begin{cases} x+y+z=2 \\ x+3y+3z=0 \\ 2z=2 \end{cases}\Rightarrow \begin{cases} x+y=1 \\ x+3y=-3 \\ z=1 \end{cases}\Rightarrow \begin{cases} x+y=1 \\ 2y=-4 \\ z=1 \end{cases}\begin{cases} x=3 \\ y=-2 \\ z=1 \end{cases}\\ \quad \quad \quad \quad \q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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algebra problem, Solve the equation
a nice problem: Solve the equation $$\left|2x-57-2\sqrt{x-55}+\frac{1}{x-54-2\sqrt{x-55}}\right|=|x-1|.$$
It's just for sharing a new ideas, thanks:)
| The left hand side of expression, by completing the square, is:
\begin{align*}
& \left|(\sqrt{x - 55})^2 - 2\sqrt{x - 55} + 1 + x - 3 + \frac{1}{(\sqrt{x - 55})^2 - 2\sqrt{x - 55} + 1}\right| \\
= & \left|(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} + x - 3\right|.
\end{align*}
Since $x \geq 55$, the express... | {
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"url": "https://math.stackexchange.com/questions/1393881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac{1}{1999} < \prod_{i=1}^{999}{\frac{2i−1}{2i}} < \frac{1}{44}$
Prove that $$\dfrac{1}{1999} < \prod_{i=1}^{999}{\dfrac{2i−1}{2i}} < \dfrac{1}{44}$$
from the 1997 Canada National Olympiad.
I have been able to prove the left half of the inequality using induction. Need help with the second part.
Prove... | You have made the problem much hard than it is suppose to be.
For the left hand side, simply observe that $\frac{1}{2}>\frac{1}{3}$ and $\frac{3}{4}>\frac{3}{5}$ and so on... to $\frac{1997}{1998}>\frac{1997}{1999}$. Now, take the product of these inequalities to get the desired result.
For the other inequality, it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluating a function in a closed form Let $$g(x)=\lim_{n\rightarrow\infty}\sqrt{n}\int_0^x z^2e^{n(\cos^2z-1)}\ \mathsf dz$$
Evaluate $g$ in closed form. The answer is right here: http://math.nyu.edu/student_resources/wwiki/index.php/Advanced_Calculus:_2004_January:_Problem_2
However, in the last part of the answer. I... | Actually, that limit for $x = k\pi$ that you're interested in is not as $\delta \to 0$ but as $n \to \infty$.
Heuristically that does follow from symmetry. It was shown in the line above that
$$
\int_{k\pi-\delta}^{k\pi+\delta} \sqrt{n}z^2e^{n(\cos^2z-1)}\ \mathsf dz\stackrel{n\to\infty}\longrightarrow k^2\pi^{5/2},
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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The sum of the lengths of the hypotenuse and another side of a right angled triangle is given. The sum of the lengths of the hypotenuse and another side of a right angled triangle is given.The area of the triangle will be maximum if the angle between them is:
$(A)\frac{\pi}{6}\hspace{1cm}(B)\frac{\pi}{4}\hspace{1cm}(C)... | Let $\theta$ be the angle between the hypotenuse & one side of length say $x$ of a right angled triangle such that $0<\theta<\frac{\pi}{2}$ then we have $$\text{hypotenuse}=x\sec \theta$$
$$\text{another side}=x\tan \theta$$
Now, according to the given condition, we have $$x\sec\theta+x=c\ \ \text{(any constant)}$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using right-hand Riemann sum to evaluate the limit of $ \frac{n}{n^2+1}+ \cdots+\frac{n}{n^2+n^2}$ I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:
\begin{a... | Notice, we have $$\lim_{n\to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\dots +\frac{n}{n^2+n^2}\right)=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}$$
$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}$$
Let, $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Why does this "miracle method" for matrix inversion work? Recently, I answered this question about matrix invertibility using a solution technique I called a "miracle method." The question and answer are reproduced below:
Problem: Let $A$ be a matrix satisfying $A^3 = 2I$. Show that $B = A^2 - 2A + 2I$ is invertible.
... | Many (but not all!) things about scalar functions work with matrices through power series. Everything is easier when $A^n=A$ (I don't remember the name of this property) or the matrix is nilpotent ($A^n=0$). For the more numerically oriented people, I suggest "Computing matrix functions": http://dx.doi.org/10.1017/S096... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "71",
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Application of Poincaré-Bendixson theorem Consider the system
$$x' = 3xy^2-x^2y \\
y' = 5x^2y - xy^2$$
Show that the system has no periodic solutions.
This is a tricky example. Linearization leads nowhere and I'm having a hard time constructing a Lyapunov function that does the trick. $V = 1/2(x^2+y^2)$ gives
$$V'(x,y)... | I'll give the solution to your non-linear system $$\frac{dx}{dt}=3xy^2-x^2y$$
and $$\frac{dy}{dt}=5x^2y-xy^2$$
Divivding eq 2 by eq 1, we get $$\frac{dy}{dx}=\frac{5x^2y-xy^2}{3xy^2-x^2y}$$
Now, dividing by $x^3$ both the numenator and the denumenator we get that $$\frac{dy}{dx}=\frac{5\frac{y}{x}-(\frac{y}{x})^2}{3(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Another integral $\int \frac{3 x^2+2 x+1}{ \left(x^3+x^2+x+2\right) \sqrt{1+\sqrt{x^3+x^2+x+2}}} \, dx$ Here is an indefinite integral that is similar to an integral I wanna propose for a contest. Apart from
using CAS, do you see any very easy way of calculating it?
$$\int \frac{1+2x +3 x^2}{\left(2+x+x^2+x^3\right) \... | Let $$2+x+x^2+x^3=t^2\implies (1+2x+3x^2)dx=2tdt$$ $$\int \frac{2tdt}{t^2\sqrt{1+t}}$$ $$=2\int \frac{dt}{t\sqrt{1+t}}$$
Let $1+t=x^2\implies dt=2xdx$
$$=2\int \frac{2xdx}{(x^2-1)x}$$
$$=4\int \frac{dx}{x^2-1}$$
$$=4\int \frac{dx}{(x-1)(x+1)}$$
$$=4\frac{1}{2}\int \left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx$$
$$=4\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Find the longest side of the triangle. The sides $a,b,c$ of a $\triangle ABC$ are in $GP$ whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.
I used law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
I took $a=a,b... | If the side lengths are $l,\frac{2}{3}l,\frac{4}{9}l$ then the circumradius is given by:
$$ R=\frac{abc}{4\Delta}=\frac{\frac{8}{27}l^3}{l^2\sqrt{\left(1+\frac{2}{3}+\frac{4}{9}\right)\left(1-\frac{2}{3}+\frac{4}{9}\right)\left(1+\frac{2}{3}-\frac{4}{9}\right)\left(-1+\frac{2}{3}+\frac{4}{9}\right)}}$$
by Heron's formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions? This following Problem is from Pell equation chapters exercise
Let $y>3$ positive integer numbers, show that following diophantine equation
$$x^2+x+1=7^y\tag{1}$$ has no integer solutions.
I tried write the equation
$$(2x+1)^2+3=4\cdot 7^y$$... | An elementary solution [albeit not using Pellian methods] is reasonably simple to obtain.
As you show, there is no solution when $y$ is even. Now assume $y$ is odd. Observe that $7 \mid (x^2+x+1)$ implies $x \equiv 2\!\pmod{7}$ or $x \equiv 4\!\pmod{7}$.
In the first case, we substitute $x=7z+2$ into the equation and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 3
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is th... | It works, indeed. If we set $x=\frac{1}{2}\left(y+\sqrt{4+y^2}\right)$ we get:
$$ I = \int_{0}^{+\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2}\left(1+\frac{y}{\sqrt{y^2+4}}\right) $$
hence:
$$ I = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2} = \frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Trigonometric equations $\tan\theta-\sec\theta=\sqrt3$ For the following problem(s) I cannot get any answer(s). I would appreciate your help very much.
$$\tan { \theta -\sec { \theta } =\sqrt { 3 } } $$
TI get 30 degrees as the reference angle. What am I doing wrong because the answer is 210 degrees.
Link for my work... | $$\tan { \theta -\sec { \theta } =\sqrt { 3 } } \\ \frac { \sin { \theta } }{ \cos { \theta } } -\frac { 1 }{ \cos { \theta } } =\sqrt { 3 } \\ \sin { \theta -\sqrt { 3 } \cos { \theta =1 } } \\$$ divide both side to $2$
$$ \frac { 1 }{ 2 } \sin { \theta -\frac { \sqrt { 3 } }{ 2 } } \cos { \theta } =\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $a, b,c,d$ non-negative and $a+b+c+d=4$. Prove that $a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$
This is a problem on Mathlinks.ro and it have had no solution. So I hope it would have a nice answer and as simply as possible.
Given $a, b,c,d$ are non-negative numbers and $a+b+c+d=4$. Prove that $$a^3b+b^3c+c^3d+d^3a+23ab... | Given the problem, we use
$$
\begin{eqnarray}
a &=& u^2\\
b &=& v^2\\
c &=& w^2\\
d &=& x^2
\end{eqnarray} \tag 1
$$
So the boundary condition is given by
$$
u^2 + v^2 + w^2 + x^2 = 4, \tag 2
$$
for any $u$, $v$, $w$ and $x$.
We define
$$
S(u,v,w,x) = u^6 v^2 + v^6 w^2 + w^6 x^2 + x^6 u^2 + 23 u^2 v^2 w^2 x^2.
\ta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,... | Suppose that $x>1$, to get $5$ its needed to add two positive integers bcz $\displaystyle \forall x>1,\left(\lfloor \frac{3}{x}\rfloor < \lfloor \frac{4}{x}\rfloor \right)$.
So its obvious that only integer that satisfy the condition $(2,3)$ or $(1,4),$ However for the
Second couple its not find Corrosponding $x,$ bcz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}....\cos \alpha_{n}.$
Maximum value of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}.$ If it is given that
$\cot \alpha_{1}\cdot \cot \alpha_{2}\cdot \cot \alpha_{3}.......\cot \alpha_{n} = 1$ a... | Apply AM-GM inequalities twice: Let $x_i = \cos \alpha_i \to 0 \leq x_i \leq 1 \to P = x_1x_2\cdots x_n \to P^2 = (1-x_1^2)(1-x_2^2)\cdots (1-x_n^2)\leq \left(\dfrac{n-(x_1^2+x_2^2+\cdots x_n^2)}{n}\right)^n\leq \left(\dfrac{n-n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}}{n}\right)^n= \left(1-\sqrt[n]{P^2}\right)^n\to \sqrt[n]{P^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How many 3 digit numbers can be formed using digits 1,2,3,4 and 5 such that the number is divisible by 6
How many $3$ digit numbers can be formed using digits $1,2,3,4$ and $5$ without repetition such that the number is divisible by $6$
First Approach:
A number is divisible by $6$ if it is divisible by $2$ and $3$.
N... | A refinement of your first approach:
In numbers which end with $2$, the sum of the other two digits must be $4$ or $7$:
*
*$132$ the sum of the other two digits is 4
*$312$ the sum of the other two digits is 4
*$342$ the sum of the other two digits is 7
*$432$ the sum of the other two digits is 7
In numbers whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\frac{n^n}{3^n} < n!$ for $n\ge6$ by induction How would I prove this using mathematical induction:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$.
Here is what I have tried:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$
Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true.
Assume for $k$ it's true so $\dfrac{k^k}{3... | we have to show that $$(n+1)!>\frac{(n+1)^{n+1}}{3^{n+1}}$$ multiplying $$n!>\frac{n^n}{3^n}$$ by $$n+1$$ we get
$$(n+1)!>\frac{n^n(n+1)}{3^n}$$ this must be greater as $$\frac{(n+1)^{n+1}}{3^{n+1}}$$ this is true since $$\left(1+\frac{1}{n}\right)^n<3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
My attempt:
Let $z = x + iy$
$\displaystyle |z+1|=2 \iff | (x + iy)+1|=2 \iff |(x+1) +iy|=2 \iff (x+1)^2 + y^2 = 4$
Let $w = u + iv$
Now let $\displaystyle w = ... | Another approach, using inversive geometry: Observe that $f(z)$ almost produces the inversion of a figure, with respect to the unit circle; in fact, it produces the complex conjugate of the figure's inversion. However, in our case, the circle $|z+1| = 2$ is already symmetric with respect to the real axis, so $f(z)$ wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles? What are the dihedral angles in a disphenoid with four identical triangles, each having one edge of length $2$ and two edges of length $\sqrt{3}$? Tried to look it up, but couldn't find it...
| Here's a hedronometric approach ...
Let $a$ and $d$ be the (necessarily opposite) edges of length $2$. These edges are orthogonal to each other in space; moreover, their projections into a mutually-parallel plane form the diagonals of a square of side-length $\sqrt{2}$. This square is a "pseudo-face" of the tetrahedron... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2.$
| Let $\omega = e^{2 \pi i/11}$, a primitive $11^{\text{th}}$ root of unity. We can assume that the circle is centered at the origin. We can also assume that $A_k$ is associated with the complex number $2 \omega^k$.
Let $p$ be complex number associated with the point $P$. Then
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \sum_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve this integral:$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx$ I occasionally found that $\displaystyle\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\dfrac{\pi}{2}\ln 2$.
I tried that
$$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^{\Large\frac{\pi}{2}}x \ \mathrm d(\ln \sin x)=-\int_0^{\Large\frac{\... | $\begin{align} J&=\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx\tag1\\
&=\int_0^1\frac{\arctan x}{x(x^2+1)}\mathrm dx+\int_1^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align} J&=\int_0^1\frac{\arctan x}{x(x^2+1)}\mathrm dx+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Range of a Rational Function How to find the Range of function $$f(x)= \frac{x^2-3x-4}{x^2 - 3x +4}$$
I tried to equate the expression to $y$, then cross multiplied
$$ y= \frac{x^2-3x-4}{x^2 - 3x +4}$$
$$ y(x^2 - 3x +4)= x^2-3x-4 $$
bought the terms to one side so it becomes a quadratic and made Discriminant to zero ... | Given $$\displaystyle y = \frac{x^2-3x-4}{x^2-3x+4}\Rightarrow yx^2-3yx+4y = x^2-3x-4$$
so we get $$\displaystyle (y-1)x^2-3(y-1)x+4(y+1) = 0$$
For real values of $y\;,$ Then given equation has real roots. So $\bf{Discriminant\geq 0}$
So $$9(y-1)^2-16(y-1)\cdot (y+1)\geq 0$$
So we get $$(y-1)\left[9(y-1)-16(y+1)\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to solve differential equation $3p^2e^y-px+1=0$ ,$p =\frac{dy}{dx}$ How to solve differential equation
$$3p^2e^y-px+1=0$$
where $$p =\frac{dy}{dx}$$
I have tried to solve for p and for x, but i am not getting anywhere. Can someone help me with this
Thanks
| By using the quadratic formula we get
\begin{align*}
3p^2e^y-px+1&=0\\
3p^2-pe^{-y}x+e^{-y}&=0\\
p&=\frac{xe^{-y}\pm\sqrt{x^2e^{-2y}-12e^{-y}}}{6}\\
pe^y&=\frac{x\pm\sqrt{x^2-12e^{y}}}{6}\\
6pe^y-x&=\pm\sqrt{x^2-12e^{y}}
\end{align*}
Let $u=-6e^y+\frac{1}{2}x^2$, then we have
\begin{align*}
-u'&=\pm\sqrt{2u}\\
u^{-1/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x$
Show that $$\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x,\forall x\in \left ( 0;\frac\pi2 \right )$$
Firstly, I had use the differentiation of $f(x)=\left (\frac{\sin x}{x} \right )^3- \cos^{2}x$,but I think it very complex: $$f'(x)=3\left ( \fra... | Start with
$$
1-\cos(x)\ge0\tag{1}
$$
For $x\ge0$, successive integration yields
$$
x-\sin(x)\ge0\tag{2}
$$
and
$$
\frac{x^2}2-1+\cos(x)\ge0\tag{3}
$$
and
$$
\frac{x^3}6-x+\sin(x)\ge0\tag{4}
$$
and
$$
\frac{x^4}{24}-\frac{x^2}2+1-\cos(x)\ge0\tag{5}
$$
From $(4)$, we get
$$
\frac{\sin^3(x)}{x^3}\ge1-\frac{x^2}2+\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How many ways to write $2010$?
Let $ N$ be the number of ways to write $ 2010$ in the form $ 2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $ a_i$'s are integers, and $ 0 \le a_i \le 99$. An example of such a representation is $ 1\cdot10^3 + 3\cdot10^2 + 67\cdot10^1 + 40\cdot10^0$. Find $ N$.
... | The generating function for these representations is
$$
\frac{x^{100\cdot1000}-1}{x^{1000}-1}\cdot\frac{x^{100\cdot100}-1}{x^{100}-1}\cdot\frac{x^{100\cdot10}-1}{x^{10}-1}\cdot\frac{x^{100}-1}{x-1}=\frac{x^{100000}-1}{x^{10}-1}\cdot\frac{x^{10000}-1}{x-1}\;,
$$
which is also the generating function for representations ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$? \begin{align}
& \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt]
= {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)}
\en... | I cannot comment, so I shall answer instead.
The first thing I tend to do with quadratics like is, is rather than use the quadratic formula, I would find the discriminant. If this is a square number, then you know your quadratic will factorise nicely. You can then either do quick trial and error, or use the formula to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What are solutions of the functional equation $f\big(f(x)\big)=-4f(x)+3x$? How to find all the functions $f:[0,\infty)\rightarrow [0,\infty)$ satisfying the functional equation
$$f\big(f(x)\big)=-4f(x)+3x\text?$$
I deduced $f(x) \le \frac 3 4 x.$
| Fix some $x$, and consider the sequence given by
$$T_0 = x$$
and
$$T_{k+1} = f(T_k)$$
for all $k \geq 0$.
The functional equation then tells us that the sequence $(T_n)$ satisfies the recurrence relation
$$T_{k+2}+4T_{k+1}-3T_k=0$$
The roots of the characteristic equation $\lambda^2+4\lambda-3$ are $\phi_1=-2+\sqrt{7}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$min\{ \sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{a+b}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}\}$ There is an Olympiad problem: $$a,b,c \in \mathbb{R}^+,M=\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{a+b}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqr... | Hint: As the expression is homogeneous, we may set $a+b+c=1$, say. Then
$$M = \sum_{cyc}\left( \sqrt[4]{\frac{a}{1-a}}+\sqrt{\frac{1-a}a}\right)$$
Now you are summing a convex function, so use Jensen's inequality to find the minimum...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lagrange Differential equation I have problems proceeding in solving the following differential equation $$xy' + y + (y')^2 = 0.$$
After solving for $\frac{dy}{dx}$ in the quadratic and using the substitution $u^2 = x^2 - 4y$ in the discriminant, I obtain $\frac{du}{dx} = 2 \frac{x}{u} -1$. Please how can I proceed?
| $$xy+y+y’^2=0$$
Let : $y’=p$
$$y=-xy’-y’^2=-xp-p^2$$
$$ \frac{dy}{dp} =-p\frac{dx}{dp}-x-2p$$
$$p=y’=\frac{dy}{dx}=\frac{dy}{dp} \frac{dp}{dx} =-p-(x+2p) \frac{dp}{dx}$$
$$2p=-(x+2p) \frac{dp}{dx}$$
$$2p\frac{dx}{dp}+x=-2p$$
The solution of this first order linear ODE is :
$$x=-\frac{2}{3}p+\frac{C}{2\sqrt{p}}$$
$$y=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$? How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$ ?
I'm not being able to approach the problem.Hints please!
| You might want to prove by induction that the sum
$$\sum_{i=1}^n \frac{i}{i^4+i^2+1} = \frac{\frac{1}{2} n(n+1)}{n^2+n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Length of tangent common to two semicircles which are also tangent to a larger semicircle. In the diagram,the semicircles centered at ${P}$ and $Q$ are tangent to each other and to the large semicircle ,and their radii are $6$ and $4$ respectively.Line $LM $ is tangent to semicircles $P$ and $Q$ .Find $LM$
Efforts ma... | DISCALIMER : This is a brute force method using coordinate geometry. A better method should exists.
Take the center of the large circle be $(0, 0)$. Then $P = (-4, 0)$ and $Q = (6, 0)$.
Now let $y = mx + c$ denote the line extending $LM$ on both sides, where $m, c \in \mathbb{R}$ are constants such that both the syst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Straight lines divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$ Find the equation of straight lines which pass through $(7,1)$,and divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$
My attempt:
As the required li... | HINT:
Let the equation of the line be $y=mx+c$ passing through the point $(7, 1)$ then we have $$1=m(7)+c$$ $$7m+c=1\tag 1$$
Substituting $y=mx+c$ in the equation of circle $x^2+y^2=100$, we get $$x^2+(mx+c)^2=100$$ $$(1+m^2)x^2+2mc x+c^2-100=0\tag 2$$
Let, the roots of the above equation be $x_1$ & $x_2$ then
$$x_1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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What is the limit of $\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}$? $$\lim_{n\to \infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)=? $$
I tried with the squeeze theorem, though I got the upper bound, but I couldn't... | Notice, we have $$\lim_{n\to \infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 1+\sqrt 3}+\frac{1}{\sqrt 3+\sqrt 5}+\cdots +\frac{1}{\sqrt {2n-1}+\sqrt {2n+1}}\right)$$
$$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt {2r-1}+\sqrt {2r+1}}\right)$$
$$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 2
} |
Diophantine-like equations So I was solving a problem and encountered a specific system of equations that I don't know if a solutions exists for it or not.
$$\begin{align}
4ny&=d^2-a^2\\
-4nx+4ny&=d^2-b^2\\
4nx&=d^2-c^2\\
\end{align}\tag1$$
where $0\leq x,y\leq n$ and $a,c,d\geq n$ and $0\leq b\leq n\sqrt{2}$ and $a,b... | I think you made a typo somewhere as $(1)$ and $(2)$ do not have equivalent solutions. Let's assume $(2)$ has no typos. Then what you can do is treat it as four equations in four unknowns $a,b,c,d$.
$$\begin{align}
b^2+c^2+d^2-3a^2&=-8n(x-y)\\
a^2+c^2+d^2-3b^2&=8n(x+y)\\
a^2+b^2+d^2-3c^2&=-8n(y-x)\\
a^2+b^2+c^2-3d^2&=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$. Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$.
When i solved this question,i got the answer $x<\frac{-28}{9}$ but the answer given in the book is $x\leq-3$.
This is how i solved.
$\sqrt{x^2-5x-24}-x-2>0\Rightarrow \sqrt{x^2-5x-24}>x+2$
$$\therefore x^2-5x-24>x^2+4x+4\Rig... | Be careful when squaring inequalities.
The condition is indeed $\sqrt{x^2-5x-24}>x+2$, but this translates into two different sets of conditions:
\begin{align}
\textbf{Set 1}\quad
&\begin{cases}
x^2-5x-24\ge0\\
x+2<0
\end{cases}
\\[10px]
\textbf{Set 2}\quad
&\begin{cases}
x^2-5x-24\ge0\quad (*)\\
x^2-5x-24\ge(x+2)^2\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Proof that there exists an integer $k$ such that $(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)k$ Without using the identity:
$$(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)(x^2+xy+x z+y^2+y z+z^2)$$ I am trying to find an argument to prove that for all $x,y,z$ there exists a nonzero integer $k$ such that:
$$(x+y+z)^5-x^5-y^5-z^5=5(... | Using the factor theorem: the polynomial $P(x)$ is divisible by $x-a$ if an only if $P(a) = 0$.
Then $P(x)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $x+y$ given that
$$P(-y) = (-y+y+z)^5-(-y)^5-y^5-z^5 =z^5+y^5-y^5-z^5 = 0$$
Similarly, $P(x)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $x+z$ given that $P(-z)=0$.
Finally, $P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the possible real values of $(x,y)$ satisfying $\sin x+\cos x=\frac{y^2-2y+4}{y^2+2y+4}$ Find the possible real values of $(x,y)$ satisfying $\sin x+\cos x=\frac{y^2-2y+4}{y^2+2y+4}$.
I could not solve this question after trying many times.How can two variables be solved with one equation?Please help me.
Its answe... | HINT:
Let $u=\dfrac{y^2-2y+4}{y^2+2y+4}$
$\iff y^2(u-1)+2y(u+1)+4u-4=0$
As $y$ is real, the discriminant $4\{(u+1)^2-4(u-1)^2\}\ge0$
$\iff3u^2-10u+3\le0\iff (u-3)\left(u-\dfrac13\right)\le0\iff\dfrac13\le u\le3$
Now $\cos x+\sin x=\sqrt2\sin\left(\dfrac\pi4+x\right)\implies-\sqrt2\le\cos x+\sin x\le\sqrt2$
So, we need... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
$\bf{My\; try::}$ We can write $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldo... | To expand the answer posted by @Arentino, we have
$$\begin{align}
S&=3\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+4)}\\\\
&=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k+4/5}\right)\\\\
&=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k}\right)+\frac15\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Integral by the given $u$-substitution Original equation $$\int\sqrt{e^t-5}dt \qquad u=\sqrt{e^t-5}$$
This is what I've done so far$$\int{u}dt$$ $$du=\frac{e^t}{2\sqrt{e^t-5}}dt$$$$dt=\frac{2\sqrt{e^t-5}}{e^t}du$$$$\int{u}{\frac{2\sqrt{e^t-5}}{e^t}}du$$ From this point on what would i need to do?
| Try making a substitution of the inside function, that often works better. In this case, for the equation
$$\int\sqrt{e^t-5}dt$$
make the substitution $u=e^t$, $\,\,du = e^tdt$
$$=\int\frac{\sqrt{u-5}}{u}du$$
substitute $s = \sqrt{u-5}$, $\,\,ds = \frac{1}{2\sqrt{u-5}}$
$$=2\int\frac{s^2}{s^2 + 5}ds$$
$$=2\int\ \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Complex numbers - how to solve $(\sqrt{3}-i)z^6+16=0$ When $z = x + yi$ (or $a + bi$), I need to solve:
$$(\sqrt{3}-i)z^6+16=0$$
Here is how I started:
$(\sqrt{3}-i)z^6=-16$
$z^6=\frac{-16}{\sqrt{3}-i}$
$z=\sqrt[6]\frac{-16}{\sqrt{3}-i}$
In other cases I get a normal complex number under the root sign in the right side... | $$(\sqrt{3}-i)z^6+16 = 0 \Longleftrightarrow$$
$$(\sqrt{3}-i)z^6 = -16 \Longleftrightarrow$$
$$z^6 = \frac{-16}{\sqrt{3}-i} \Longleftrightarrow$$
$$z^6 = -4\sqrt{3}-4i \Longleftrightarrow$$
$$z^6 = 8e^{-\frac{5\pi}{6}i} \Longleftrightarrow$$
$$z^6 = 8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving that $\alpha:\mathbb{R}\to\mathbb{R}$ where $\alpha(x)=\frac{x^{3}}{x^{2}+1}$ is bijective Using the usual method that I was taught, in order to find that something is injective, I assume $f(x)=f(y)$ for some $x,y\in\mathbb{R}$, and then I demonstrate that $x=y$. This is not working out so well for me, using th... | Assuming that $f(x) = f(y)$ and showing that $x = y$ is a correct method to show that $f$ is injective. You have one subtle error in your work.
$\begin{align} \frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
x^3(y^2+1)&=y^3(x^2+1)\\ x^3y^2+x^3&=x^2y^3+y^3\\
x^3-y^3&=x^2y^3-x^3y^2\\ (x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Parametric equations - stating values. A curve is defined by the parametric equations: $x=\cos2t, y=\sin2t, 0<t<π.$
a) Use parametric equations to find $\frac{dy}{dx}$. Hence find the equation of the tangent when $t=\frac{π}{8}$.
b) Obtain an expression for $\frac{d^2y}{dx^2}$ and hence show that: $\sin2t\left(\frac{d^... | Using $$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
So here $x=\cos 2t\;,$ Then $$\displaystyle \frac{dx}{dt} = -2\sin 2t$$ and $y=\sin 2t\;,$ Then $$\displaystyle \frac{dy}{dt} = 2\cos 2t$$
So $$\displaystyle \frac{dy}{dx} = -\frac{2\cos 2t}{2\sin 2t} = -\cot 2t\;,$$ Now $$\displaystyle \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Fundamental theorem of Calculus: $\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt$ Possible textbook mistake using the FTC we are supposed to evaluate
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
$$
and the answer, the textbook says, should be a constant. When I evaluate that derivative I get
$$
\frac{d}{dx} \int... | $$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
= 1\frac{1}{3+x^2}+1\frac{1}{3+(-x)^2}=\frac{2}{3+x^2}$$
Note that $$
\frac{d}{dx} \int_{g(x)}^{f(x)} h(t) \ dt=f'(x)h(f(x))-g'(x)h(g(x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that the probability that the sum of the digits used leaves the remainder 2 when divided by $4$ is $\frac{1}{4}.$ A number of 5 digits is written down at random.Prove that the probability that the sum of the digits used leaves the remainder 2 when divided by $4$ is $\frac{1}{4}.$
The sum of digits must be like 6,... | While waiting for some smarter solution (if any), I'll give a direct count of how many 5-digit numbers are congruent to $2$ (mod $4$).
Let's start with 1-digit numbers (from $0$ to $9$): we have $3$ of them congruent to $0$ (they are $0$, $4$ and $8$), $3$ congruent to $1$ ($1$, $5$ and $9$), $2$ congruent to $2$ ($2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expected value of size of subset Given a set $S$ such that $|S|=n$,
A random item is chosen randomly from $S$, and being appended to a new set $T$.
This process is being repeated $n$ times (with repetition), what is the expected value of $|T|$ ?
| This question can also be answered by total enumeration using Stirling
numbers of the second kind. Observe that the expectation is given by
$$\mathrm{E}[X] = n^{-n} \sum_{k=0}^n k \times {n\choose k}
k! {n\brace k}.$$
Here we multiply the value of the random variable by the probability,
which is obtained from the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
How to Compute $\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}$
How to compute
$$\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=-\dfrac{3}{2}$$
I'm interested in more ways of computing limit for this expression.
... | You must use power series development at order $4$:
*
*$\sin x= x+o(x^2)$, hence $\sin x^2=x^2+o(x^4)$
*$\ln(1+u)=u-\dfrac{u^2}2+o(u^2) $, hence
$$\ln(1+\sin x^2)=\ln\bigl(1+ x^2+o(x^4)\bigr)=x^2-\frac{x^4}2+o(x^4)$$
*$\arcsin x=x+\dfrac12\dfrac{x^3}3+o(x^4)$, hence $(\arcsin x)^2=x^2+\dfrac{x^4}3+o(x^4)$
Grouping... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the cube roots of $-11-2i$. How do I find the roots of $\sqrt[3]{ - 11 - 2i}$ ?
Tried to use Moivre's theorem, but can not find the solutions by using the polar form:
$z_k=\sqrt{5}[\cos(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})+i.\sin(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})]$, for $k=0, 1$ and $2$.
Also resorted to ... | Let's try to find a solution $z=a+bi$ with $a,b \in \mathbb Z$.
We must have $-11=a^3-3ab^2 = a(a^2-3b^2)$ and so $a=\pm 1$ or $\pm 11$.
We must have $-2=3a^2b - b^3 = b(3a^2-b^2)$ and so $b=\pm 1$ or $\pm 2$.
Testing all cases, we see that $a=1$ and $b=2$ works.
Having found one solution $z=1+2i$, the others are $z\om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
A... | $$
\underbrace{\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x}}_\text{I have no idea what you did here!}
$$
Let's recall some simple algebra:
$$
\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x}
= \overbrace{\frac{x(\cdots\cdots)}{x(\cdots\cdots)} = \frac{\sqrt{3+x}+\sqrt{3-x}}{2} \vphantom{\frac\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Showing that the sequence $a_n=[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}]-\ln(n)$ converges How do you show that
the sequence
$$a_n=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}\right)-\ln(n)$$ is convergent.
Give a convincing argument, that $a_n$ converges to a number called gamma. Usin... | $a_{n+1} - a_n = \dfrac{1}{n+1} - \ln(n+1)+\ln n = \dfrac{1}{n+1} - \ln\left(1+\dfrac{1}{n}\right)= \dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}-\ln\left(1+\dfrac{1}{n}\right)= \dfrac{x}{1+x} - \ln(1+x)=1-\dfrac{1}{x+1} -\ln(1+x)=f(x), 0< x = \dfrac{1}{n}<1\Rightarrow f'(x) = \dfrac{1}{(x+1)^2}-\dfrac{1}{x+1}<0\Rightarrow a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Tricky trig question from GRE
Please evaluate
\begin{align}
1-\sin^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
What I've tried so far is to use Pythagorean identity and I got
\begin{align}
\cos^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
If \begin{align}
\arccos \frac{\pi}{12}=y
\end{align}
then
\b... | Once you get to this stage:
$\cos^2 (\arccos \frac{\pi}{12})$,
Use the facts that $\cos(\arccos x) = x$ and $\cos^2 y = (\cos y)^2$ to get to the final answer.
In this case, that's simply $(\frac{\pi}{12})^2 = \frac{\pi^2}{144}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \fra... | A possible short proof comes from Parseval's identity.
It is well known that the Fourier sine series of the $\text{sign}$ function over $(-\pi,\pi)$ is given by:
$$ \text{sign}(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} \tag{1}$$
hence by Parseval's identity:
$$ 2\pi = \int_{-\pi}^{\pi} 1\,dx = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Find all conditions of a and b for when matrix has determinant zero Find all conditions on a and b for which A =
\begin{bmatrix} a & 0 &a^2 \\ 2 & a & 3a \\ a & -1 & ba+b\end{bmatrix}
has determinant 0.
| Subtract $a$ times first column from third to obtain
$$\begin{pmatrix}a&0&0\\2&a&a\\a&-1&ba+b-a^2\end{pmatrix} $$
which has determinant $0$ iff the original matrix has detrminant $0$.
If $a=0$, the first row is zero, hence $\det A=0$.
If $a\ne 0$, any linear combinations of the columns that results in zero must not use... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm. Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm.
I let two non-parallel sides and one o... | Notice, let the unknown side be $x$ which is parallel to one of three equal sides each of length $k$ then the normal distance between the parallel sides can be determined as follows (using pytagorean theorem in a right triangle)
$$\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}$$
the area of the trapezium is given as
$$A=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin... | first :
$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^3}=\frac{-1}{2}$$
proof :$\lim_{x\to 0} \frac{\tan ^nx - \sin ^m x}{x^r}=?$without l'Hôpital's rule.
now :
$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}=\lim_{x\to 0}\frac{\sin x-\tan x}{x^3}.\frac{2x}{2\sin 2x}=?$$
since :
$$\lim_{x\to 0}\frac{2x}{\sin 2x}=1$$
so :
$$\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the derivative of the function $f(x)=x/(x^2+1)$ at a point a $$\frac{x}{x^2+1}$$
I have to find the derivative and set a point by myself.
So I just set $x=a$ ( is it correct?)
and using the limit,
$$ \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{\frac{a+h}{(a+h)^2 + 1} - \frac{a}{a^2+1}}{h}$$
therefor... | Your answer is correct.
Notice, $$f(x)=\frac{x}{x^2+1}$$
$$\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$$
$$=\lim_{h\to 0}\frac{(x+h)(x^2+1)-x((x+h)^2+1)}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{x^3+hx^2+x+h-x^3-2hx^2-h^2x-x}{h((x+h)^2+1)(x^2+1)}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How do we know that $x^2 + \frac{1}{x^2}$ is greater or equal to $2$? For one problem, we were supposed to know that:
$$x^2 + \frac{1}{x^2}\geq 2.$$
How do you deduce this instantly when looking at the expression above?
| $x^2 + \frac{1}{x^2} = \frac{x^4 + 1}{x^2}$
$ = \frac{x^4 - 2x^2 + 1}{x^2} + 2$
$ = \frac{(x^2 - 1)^2}{x^2} + 2$
$ \geq 0 + 2 = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 0
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How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\... | By replacing $x$ with $\arctan t$,
$$ I = \int_{0}^{\pi/2}\frac{dx}{4\sin^2 x+5\cos^2 x} = \int_{0}^{+\infty}\frac{dt}{4t^2+5}=\color{red}{\frac{\pi}{4\sqrt{5}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
How to multiply radicals? The problem is $$0 = -(\sqrt{7 -2 b} - 2) * -\sqrt{7-2 b} - 2) $$
the original question was this
$$-2 = \sqrt{7 -2 b} - \sqrt{2 b + 3}$$
and than i found out the you had to isolate one radical version so i substracted $$-(\sqrt{7 -2 b} - 2)$$ and than you get $$(-\sqrt{7 -2 b} -2)^2 = (-\sq... | $$\begin{align}(x + y)^2 &= (x + y)(x+y) \\ &= x(x +y) + y(x +y) \\ &= (x^2 + xy) + (xy + y^2) \\ &= x^2 + 2xy + y^2\end{align}$$
Now in your case $x = -\sqrt{7 - 2b}$ and $y = -2$, so
$$\begin{align}(-\sqrt{7 - 2b} + -2)^2 &= (-\sqrt{7 - 2b})^2 + 2(-\sqrt{7 - 2b})(-2) + (-2)^2\\&=(7 - 2b) + 4\sqrt{7 - 2b} + 4\\&=11 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:
*
*the denominator has to be one,
*the numerator has to be a multiple of the denominator or
*the numerator and denominator have to be the same.... | If $a,b$ are not integers, nothing stops $x+\frac1x$ being an integer. In fact, that happens infinitely many times as the function is continuous.
Let $d=gcd(a,b)$ and $a=xd, b=yd$ with $gcd(x,y)=1$. Then
$$\frac{a}{b}+\frac{b}{a}=\frac{x^2+y^2}{xy}.$$
We will show that if $p$ is a common prime divisor of $x^2+y^2$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Find the maximun $n\in N^{+}$, such $x^n+y^n+z^n$ is divisible by $x+y+z$ A prime number $p$ and integers $x,y,z$ with $0<x<y<z<p$ are given. Show that if the numbers $x^3,y^3,z^3$ give the same remainder when divided by $p$ , Find the maximun $n\in N^{+}$, such $x^n+y^n+z^n$ is divisible by $x+y+z$
when $n=2$ case be ... | I think you have to edit your post.
Are you sure you don’t ask about the maximum $n$ such that all of the $x^m+y^m+z^m$ with $m> n$ are not divisible by $x+y+z$?
In your hyperlink, Poland 2003, it is not question of maximum n but just for $n=2$.
See at the following example: $0<3<5<6<7$ and $3^3, 5^3, 6^3$ have the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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square root of $\frac{2+\sqrt{3}}{4}$
Find the square root of $\frac{2+\sqrt{3}}{4}$.
Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?
| $\frac{2+\sqrt{3}}{4}$ is a positive number so it has a real square root.
Notice first that:
$$\sqrt{\dfrac{2+\sqrt{3}}{4}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$$
so what you have to find is $\sqrt{2+\sqrt{3}}$. A technique is to try to find what are $a,b\in\mathbb{R}$ where $2+\sqrt{3}=(a+b)^2$ so that $2+\sqrt{3}=|a+b|$.
Yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?
The options are:
*
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
*$$\frac{1-x^2}{1-y^2}$$
*$$\frac{1-y^2}{1-x^2}$$
I trie... | It looks like there's supposed to be a way to eliminate $a$ from the formula.
Let's try to tackle the term $a+\frac{x}{\sqrt{1-x^2}}$ first, using the fact that
$$a(x-y) = \sqrt{1-x^2} + \sqrt{1-y^2}.$$
In order to avoid dealing with a lot of fractions, I would
multiply $a + \frac{x}{\sqrt{1-x^2}}$ by the factor $x-y$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})?$ What is the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots (1+x^{100})?$
I manually expanded $(1+x)(1+x^2)(1+x^3)...(1+x^{10})$ and calculated the coefficient of $x^9$ as $8$ but i dont know how to solve it withou... | In every factor of your product, you will need to pick either the $1$ or the $x^i$, so immediatly we see that for all factors with $i > 9$ we need to pick the $1$, otherwise we will get $x^k$ with $k > 9$. So we can just look at the product up till $(1 + x^9)$. Now we look at the number of ways to pick $1$ and $x^i$ he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric equation: $\cos3x+\cos x-\cos2x=0$ $\cos3x+\cos x-\cos2x=0$ find general solution
My answer:
$\cos 3x+\cos x-\cos2x=0$
$\cos2x(2\cos x-1)=0$
$\cos2x=0$ or $2\cos x-1=0$
When $\cos2x=0$
$x=(n+1)\pi/2$
When $\cos x=1/2$
Solving...$x=\pi/3$
Then $x=2n\pi\pm \pi/3$
Did I go wrong?
Is it correct?
| $$\cos(x)-\cos(2x)+\cos(3x)=0\Longleftrightarrow$$
$$1-2\cos(x)-2\cos^2(x)+4\cos^3(x)=0\Longleftrightarrow$$
$$\left(2\cos(x)-1\right)\cdot\left(2\cos^2(x)-1\right)=0\Longleftrightarrow$$
$$2\cos(x)-1=0\Longleftrightarrow \vee 2\cos^2(x)-1=0\Longleftrightarrow$$
$$2\cos(x)=1\Longleftrightarrow \vee 2\cos^2(x)=1\Longlef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How many numbers less than $1000$ with digit sum to $11$ and divisible by $11$ How many positive (integers) numbers less than $1000$ with digit sum to $11$ and divisible by $11$?
There are $\lfloor 1000/11 \rfloor = 90$ numbers less than $1000$ divisible by $11$.
$N = 100a + 10b + c$ where $a + b + c = 11$ and $0 \le ... | $N=a+10b+100c$. $a-b+c=0$ or $a-b+c=11$. Also we have $a+b+c=11$. We get two cases:
*
*$a-b+c=0,\;a+b+c=11$ from which we get $2a+2b=11$; impossible.
*$a-b+c=11,\;a+b+c=11$ from which we get $a+c=11,\; b=0$. They are $209, 308, 407, 506, 605, 704, 803, 902$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Checking primality for $2 \uparrow \uparrow n + 3 \uparrow \uparrow n$ Is there a clever way to test primality for
$$2 \uparrow \uparrow n \quad + \quad 3 \uparrow \uparrow n$$
where $n \gt 3$? (Not surprisingly, I got stuck after that)
For $n \le 3$ we get:
$$
\begin{matrix}
2 \uparrow \uparrow 0 \quad + \quad 3 \upa... | For $n$ large enough ($n \ge 6$) ,$ 2\uparrow \uparrow n \equiv 2197 \pmod {4423}$ and $3 \uparrow \uparrow n \equiv 2226 \pmod {4423}$, and so $2 \uparrow \uparrow n + 3\uparrow \uparrow n$ is always a multiple of $4423$.
Assuming the limit of $a \uparrow \uparrow n$ modulo $p$ is "random" in $\Bbb F_p^*$, you can exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Generalizations of the pentagonal number theorem Euler's pentagonal number theorem (see also the original paper and review by Jordan Bell) states
$$
\prod_{n=1}^\infty (1 - q^n) = \sum_{k=-\infty}^{\infty} (-1)^{k} q^{(3k^2 - k)/2},
$$
where $k \, (3k - 1)/2$ is the $k$th pentagon number.
It seems that we can generaliz... | It turns out that it follows directly from the Jacobi triple product identity:
\begin{align}
\prod_{n=0}^\infty
(1 - x^{2n+2}) (1 + x^{2n+1} z) \left(1 + x^{2n+1}z^{-1}\right)
=
\sum_{n=-\infty}^{\infty}
x^{n^2} \, z^n.
\tag{1}
\end{align}
With $x = q^{t/2}$ and $z = -q^{1-t/2}$, the left hand side becomes
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the integer values of c Find all possible positive integer values of c that $\frac {a^2+b^2+ab} {ab-1}$=c can take in $\mathbb N$.
I know that the solutions are c=7 and c=4 but I don't know how to prove this with Vieta Jumping method.
| THIS IS VIETA JUMPING. I RARELY LIE ABOUT MATHEMATICS.
$$ \frac{x^2 + xy + y^2}{xy-1} = c, $$
$$ x^2 + xy + y^2 = c xy - c, $$
$$ x^2 + (1-c)xy + y^2 = -c. $$
We want integers with $$ x^2 + (1-c)xy + y^2 = -c $$ and $c \geq 3,$ because otherwise the quadratic form on the left is positive definite or semidefinite.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the derivative for a product of two polynomial functions? In my problem, I am attempting to find $f'(x)$ when $f(x)=(5x^2-2x+8)(4x^2+7x-3)$. For my work I have:
\begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16... | \begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt]
= {} & 80x^3 + 73x^2+6x+62
\end{align}
In the penultimate line of calculation, the 8th term from left should be $$4x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Linear algebra Plane and line I have a plane with points: $ A = (-5, 1, 3), B = (-2, 6, -3), C = (-4, 3, 4)$
I also have a line with points: $ D = (-2, 7, 6), E = (-1, 8, -2)$
The vector equation form of the line:
$$
\left\{
\begin{array}{c|}
\begin{pmatrix}
-2\\
7\\
6
\end{pmat... | Hint: We have that $\vec{x}$ lies in the intersection iff:
$\vec{x}=\begin{pmatrix}
-2\\
7\\
6
\end{pmatrix}
+ r \begin{pmatrix}
1\\
1\\
8
\end{pmatrix}$ and $\vec{x}=\begin{pmatrix}
-5\\
1\\
3
\end{pmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of trigonometric relationship of angle bisector I have to prove the following equation:
$$CD=\frac{2ab\cos\frac{C}{2}}{a+b}$$ where $CD$ is the angle bisector of $C$ in $\Delta ABC $.
My attempt:
I've started by rearranging the equation in terms of $\cos \frac{C}{2}$ ,then by squaring and subtracting $\sin^2 \fr... | HINT:
$$(s-a)(s-b)=\frac{ (-b^2-a^2+c^2+2 a b)}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the half-range Fourier series expansion of $f(x) = \cos(x)$ I am stuck on the problem of calculating the half-range Fourier series expansion of $$f(x) = \cos(x),$$ $$0 < x < \frac{\pi}{2}$$
I am at the point where I have calculated the definite integral of $b_n$.
$$b_n = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}} \cos... | for $n=1,5,9,....$ the $\sin(\frac{n\pi}{2})=1$
$$\frac{n-\sin(\frac{n\pi}{2})}{n^2-1}=\frac{n-1}{n^2-1}=\frac{1}{n+1}$$
for $n=2,4,6,8....$ all terms equal zero
for $n=3,7,11,...$ the $\sin(\frac{n\pi}{2})=-1$
$$\frac{n-\sin(\frac{n\pi}{2})}{n^2-1}=\frac{n+1}{n^2-1}=\frac{1}{n-1}$$
now add the $\frac{1}{n+1}$ with $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The limit of $(1-\cos^{1/3} x)/(1-\cos x^{1/3})$ as $x\to 0$ I have a problem to find answer for this limit.
$$\lim_{x\to 0}\frac{1-{\cos^{1/3} x}}{1-\cos(x^{1/3})}$$
I did it like this $$\dfrac{\dfrac{\sin x}{3×\sqrt[3]{\cos x²}}}{\sin \sqrt[3]x×\dfrac 1{3×\sqrt[3]{x²}}}$$
| Applying l'Hopital's rule:
$$\lim_{x\to 0} \frac{1-\sqrt[3]{\cos(x)}}{1-\cos\left(\sqrt[3]{x}\right)}=$$
$$\lim_{x\to 0} \frac{\frac{\text{d}}{\text{d}x}\left(1-\sqrt[3]{\cos(x)}\right)}{\frac{\text{d}}{\text{d}x}\left(1-\cos\left(\sqrt[3]{x}\right)\right)}=$$
$$\lim_{x\to 0} \frac{\frac{\sin(x)}{3\cos^{\frac{2}{3}}(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more si... | $$z^5+z^4+z^3+z^2+z+1 = 0 \Longleftrightarrow$$
$$(z+1)(z^2-z+1)(z^2+z+1) = 0 \Longleftrightarrow$$
$$z+1=0 \vee z^2-z+1=0 \vee z^2+z+1=0 \Longleftrightarrow$$
$$z=-1 \vee z=\frac{1\pm\sqrt{-3}}{2} \vee z=\frac{-1\pm\sqrt{-3}}{2} \Longleftrightarrow$$
$$z=-1 \vee z=\frac{1\pm i\sqrt{3}}{2} \vee z=\frac{-1\pm i\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 4
} |
Remainder of $7^{7^{7}}$ divided by $32$ The problem is to find remainder of $7^{7^{7}}$ divided by $32$ the quickest way.
My attempt:
$32 = 8.4$
Mod 8: $7^{7^{7}}\equiv(-1)^{7^{7}}\equiv-1$
Mod 4: $7^{7^{7}}\equiv(-1)^{7^{7}}\equiv-1$
Hence in mod $32$: $7^{7^{7}}\equiv(-1)$ mod $8.(-1)$mod$ 4 (?) \equiv 7.3\equiv21... | Fact: $a^{2^{k-2}}\equiv 1\pmod{2^k}$ for all $k\ge 3$, odd $a$.
Proof: We use induction. $a^2\equiv 1\pmod{8}$ for all odd $a$.
If $a^{2^{m-2}}\equiv 1\pmod{2^m}$, we want to prove $a^{2^{m-1}}\equiv 1\pmod{2^{m+1}}$.
I.e. if $a^{2^{m-2}}=2^mk+1$, we want to prove $\left(2^mk+1\right)^2\equiv 1\pmod{2^{m+1}}$.
$$\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How does $\sqrt{y^2 + y^2} = 1$ give $y = -1/\sqrt{2}$ here? I'm studying this problem:
and I cannot see the last step, how we go from $$\sqrt{y^2 + y^2} = 1$$ to $$y = -1/\sqrt{2}$$
When I try to solve this, I end up with:
$$y^2 + y^2 = 1^2 = 1$$
$$2y^2 = 1$$
$$y^2 = 2$$
$$y = \sqrt{2}$$
| There are basically two interpretations here: either solve $\sqrt{y^2+y^2}=1$ to get $-\frac{1}{\sqrt{2}}$.
Or, the other interpretation is: They didn't need to solve the equation $\sqrt{y^2+y^2}=1$. They just made the observation that $$\cos\left(\frac{5\pi}{4}\right)=\sin\left(\frac{5\pi}{4}\right)=-\frac{1}{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
2nd degree matrix equation Let $X$ be a matrix with 2 rows and 2 columns.
Solve the following equation:
$$ X^2 = \begin{pmatrix}
3 & 5\\
-5 & 8
\end{pmatrix} $$
Here is what I did:
Let $ X = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} $. After multiplying I got the following system:
$$ \left\{\begin{matrix}
a^2 + bc... | A nonsingular $n \times n$ matrix $M$ will have some square roots that are polynomials in $M$ of degree $\le n-1$. Thus in this case we can look for solutions of the form $X = s M + t I$. By the Cayley-Hamilton theorem,
a matrix satisfies its characteristic polynomial: in this case the characteristic polynomial is $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge? Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge?
| \begin{align}
& \underbrace{1+\frac{1}{2}-\frac{1}{3}}_{\ge 1} + \underbrace{\frac{1}{4}+\frac{1}{5}-\frac{1}{6}}_{\ge 1/4} + \underbrace{\frac{1}{7}+\frac{1}{8}-\frac{1}{9}}_{\ge 1/7} + \underbrace{\frac 1 {10} + \frac 1 {11} - \frac 1 {12}}_{\ge 1/10} + \cdots \\[10pt]
\ge {} & 1 + \frac 1 4 + \frac 1 7 + \frac 1 {10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $p$ be prime not equal to 2 or 5. Show $p^2+1$ or $p^2-1$ is divisible by 10. I can do half the proof but can not think of a way to finish. If $p$ is prime then both $p-1$ and $p+1$ are even. In cases where either $p-1$ or $p+1$ is divisible by 5 that implies $(p-1)(p+1)=p^2-1=10n$ for some positive integer $n$. I ... | If $p \equiv 3$ or $7 \pmod{10}$, then $p^2 \equiv 9 \pmod{10}$, so $p^2 + 1 \equiv 0 \pmod{10}$ as desired. But if $p \equiv 1$ or $9 \pmod{10}$, then $p^2 \equiv 1 \pmod{10}$, so $p^2 - 1 \equiv 0 \pmod{10}$ as desired.
Examples: $3^2 + 1 = 10$, $7^2 + 1 = 50$, $11^2 - 1 = 120$, $19^2 - 1 = 360$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
How do I prove that the inequality $\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $ holds for every $x,y>0$ Prove that for every $x,y>0$, the following inequality holds. Begin by proving the right side first and assume that $(\sqrt { x } -\sqrt { y } )^2\ge 0$
$$\frac { 2xy }{ x+y } \le \sqrt { xy } \le ... | Since $x,y>0$, you can divide by any of them (or by their square roots) on both sides of the inequality. Thus,
$$\frac{2xy}{x+y}\le\sqrt{xy} \quad \Leftrightarrow \quad \frac{2\sqrt{xy}}{x+y}\le1 \quad \Leftrightarrow \quad 2\sqrt{xy}\le x+y.$$
You have already proved the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule
Find the following limit
$$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x... | We can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\frac{1}{x^{3}}\left(\left(\frac{2 + \cos x}{3}\right)^{x} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{3}}\left(\exp\left(x\log\left(\frac{2 + \cos x}{3}\right)\right) - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{3}}\cdot x\log\left(\dfrac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Using Hensel's Lifting Lemma to Solve $x^2 + x + 34 \equiv 0 \pmod{81}$ As in the title, I'm trying to solve $$x^2 + x + 34 \equiv 0 \pmod{81}.$$
Let $f(x) = x^2 + x + 34$ throughout.
I'm using Hensel's lemma, but it's a bit dense and I'm not sure my interpretation is correct. My method involves checking for solutions... | You probably can't use Hensel's Lemma:
we need $f'(x) = 2x + 1 \not \equiv 0 \mod 3$ so we need $x \equiv 0, 2 \mod 3$ and that just isn't likely to solve $x^2 + x + 7 + 27 \equiv 0 \mod 3^k$
But $x^2 + x + 34 \equiv x^2 - 80x + (27 + 7) \equiv$
$x^2 -80 x + 40^2 - 40^2 + (27+7) \equiv $
$(x-40)^2 -(27 + 13)^2 + (27 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find limit "1 ^ infinity" $\displaystyle \lim_\limits{x \to \frac{\pi}{4}}
\left[\frac{1 + \sin(\pi/4 - x)\sin(2x)}{1 + 0.5\cos(2x)}\right]^{\cot^3(x - \frac{\pi}{4})}$
I've tried to use $\exp$, but found very difficult to simplify the expression.
| $$\displaystyle \lim_\limits{x \to \frac{\pi}{4}}
\left(\frac{1 + \sin\left(\frac{\pi}{4} - x\right)\sin(2x)}{1 + \frac{1}{2}\cos(2x)}\right)^{\cot^3(x - \frac{\pi}{4})}=$$
$$\displaystyle \lim_\limits{x \to \frac{\pi}{4}}
\left(\frac{2+\sqrt{2}(\cos(x)-\sin(x))\sin(2x)}{2+\cos(2x)}\right)^{-\tan^3\left(\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of powers of sine
Find $\displaystyle \sum_{n=1}^{89} \sin^6(n) = \frac{m}{n}$
Let $x = \sin(n)$ and let $y = \cos(n)$.
Since $\cos(n) = \sin(90 - n)$ it follows that
$= \sin^6(1) +\sin^6(1) + ... + \sin^6(45) + \cos^6(1) + \cos^6(2) + ... + \cos^6(44)$.
EDIT:
$x^6 + y^6 = (x^2 + y^2)(x^4 - (xy)^2 + y^4)$.
$(x... | Let $$S = \sum^{89}_{n=1}\sin^6(n)\;,$$ as you have used $\sin(90^\circ-n) = \cos (n)$.
So we get $$S = \sum^{89}_{n=1}\sin^6(90^{\circ}-n) = \sum^{89}_{n=1}\cos^6(n)\;,$$ as you have used $\sin(90-n) = \cos (n)$.
Now Add these two equation, We get $$2S = \sum^{89}_{n=1}\left[\sin^6(n)+\cos^{6}(n)\right] = \sum^{89}_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Is there a difference between $(x)^{\frac{1}{n}} $ and $\sqrt[n]{x}$? Is there a difference between $(x)^{\frac{1}{n}}$ and $ \sqrt[n]{x}$ ?
I'm confused with this topic. Any ideas or examples ?
If $(x)^{\frac{1}{n}} = \sqrt[n]{x}$
Consider $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . Is it the same if I write $ x=\frac{-b ... | There is no difference. In the same way as I write $e^x$ when $x$ is simple enough, and $\exp(x)$ otherwise, I also write $\sqrt{x}$ when $x$ is simple enough, and $(x)^{1/2}$ otherwise.
For example, I write
$$
\left(\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}\right)^{1/2}
\qquad\text{and not the ugly}\qquad
\sqrt{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$
$|(1+i\sqrt{3})(1+i)|=8$
U... | Hint:
Since you can write
$(1+i\sqrt{3})(1+i)e^{i \phi}=2\sqrt{2}e^{i \phi} e^{i {7 \pi \over 12}}$, so
you can concentrate on showing
$(1+i\sqrt{3})(1+i)=2\sqrt{2} e^{i {7 \pi \over 12}}$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
min polynomial of $\sqrt{6+2\sqrt{5}}$ min polynomial of $\sqrt{6+2\sqrt{5}}$ over $\mathbb{Q}$
I found $\alpha^4 -12\alpha^2 +16 = 0$ but wolframalpha does not agree with me. furthermore, would anything change is it was over $\mathbb{R}$?
| Since $\sqrt{6+ 2\sqrt{5}}$ is a real number, if it were "over R" then the minimum polynomial would be $x- \sqrt{6+ 2\sqrt{5}}$!
The way I would do this problem would be to note that $(x- \sqrt{6+ 2\sqrt{5}})(x+ \sqrt{6+ 2\sqrt{5}})= x^2- (6+ 2\sqrt{5})= (x^2- 6)- 2\sqrt{5}$.
Then $((x^2- 6)- 2\sqrt{5})((x^2- 6)+ 2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Where am I going wrong when trying to solve this system of equations using Gaussian Elimination? $$3x-y+z=5 \\ 2x+y-z=1 \\ x-y+z=2 \\ 4x+4y+z=3$$
Steps I took:
$$\left[\begin{array}{rrr|r}
3 & -1 & 1 & 5 \\
2 & 1 & -1 & 1 \\
1 & -1 & 1 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$\Rightarrow { R }_{ 1 }... | $\left[\begin{array}{rrr|r}
3 & -1 & 1 & 5 \\
2 & 1 & -1 & 1 \\
1 & -1 & 1 & 2\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{ 1 }={ R }_{ 1 }+{ R }_{ 2 }\Rightarrow
\left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
2 & 1 & -1 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cd... | Permit me to contribute an algebraic proof that does not use Beta
functions.
Suppose we seek to verify that
$$S_n = \sum_{k=0}^n \frac{(-1)^k}{2k+1} {n\choose k} =
\frac{2^{2n} (n!)^2}{(2n+1) (2n)!}.$$
We have by inspection that
$$S_n = \sum_{k=0}^n \mathrm{Res}(f(z); z=k)$$
where $$f(z) =
(-1)^n n! \frac{1}{2z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Mathematical Induction Divisibility Problem Prove that if $n \ge 1$ is a positive integer, then $13^n − 6^n$ is divisible by $7$.
In proving the $n = k+1$ case, I get to $133k + 6^k\cdot13 - 6\cdot13^k = 7M$, where $M$ is a positive integer. $133k$ is divisible by $7$. How do I show the $6^k\cdot13 - 6\cdot13^k$ part i... | $13^{k+1} - 6^{k+1} = 13^k* 13 - 6^k * 6 = 13^k(7 + 6) - 6^k*6 = 7*13^k + 6*13^k - 6^k*6 = 7*13^k + 6(13^k - 6^k) = 7*13^k + 6(7*M) = 7*(13^k + k)$ is a multiple of 7.
| {
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"url": "https://math.stackexchange.com/questions/1503990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use mathematical induction to show that $ H_{2^n} \geq 1+ \frac{n}{2} $ An Inequality for Harmonic Numbers.
The harmonic numbers $H_j, j=1,2,3,...,$ are defined by $$H_j = 1 + \cfrac{1}{2}+\cfrac{1}{3}+...+\cfrac{1}{j}$$
Use mathematical induction to show that $$ H_{2^n} \geq 1+ \frac{n}{2} $$
whenever $n$ is a nonnega... | The first inequality is the inductive hypothesis.
As for the second, note that for all $j \in \{ 1, \dots 2^k\}$, $$\frac{1}{2^k+j} \ge \frac{1}{2^{k+1}}$$
So that
$$\sum_{j=1}^{2^k} \frac{1}{2^k+j} \ge \sum_{j=1}^{2^k} \frac{1}{2^{k+1}} = 2^k \frac{1}{2^{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$P$ is a prime number not equal to $2$ or $3$. Show that $p - 6$ is a quadratic residue $\pmod p$ whenever... $P$ is a prime number not equal to $2$ or $3$.
Show that $p - 6$ is a quadratic residue $\pmod p$ whenever $p \equiv 1, 5, 7, 11\pmod {24}$
And show that $p - 6$ is a quadratic non - residue $\pmod p$ whenever... | I.e. prove: if $p\ge 5$, then $\left(\frac{-6}{p}\right)=1\iff p\equiv \{1,5,7,11\}\pmod{24}$. Use Quadratic Reciprocity.
$$\left(\frac{-6}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)=1$$
$$\iff \begin{cases}\begin{cases}p\equiv 1\pmod{4}\\p\equiv \pm 1\pmod{8}\\p\equiv \pm 1\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
If one has the result it is not difficult to verify it by induction, but how can I solve it without induction ?
| We have :
$$
\begin{split}
\require{cancel}
\dfrac{1\cdot2\cdot\bcancel{3\cdot4\cdot5\cdots k}}{\bcancel{3\cdot4\cdot5\cdots k}\cdot k+1\cdot k+2}&=\dfrac{1\cdot2}{k+1\cdot k+2}\\&=\dfrac{2}{k+1}-\dfrac{2}{k+2}
\end{split}
$$
So the summation $S_j=\sum_{k=0}^{j-1}\dfrac{1\cdot2\cdots k}{3\cdot4\cdots k+2}$ is a telesc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove $x=y$, given $\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$ Given $x,y\in(0,\frac{\pi}{2}]$, $\alpha>0$ is a constant satisfying $0<x+y+\alpha<\pi$,
Also it is known that
$$\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$$
How can we prove... | $$\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$$
or, $$\frac{1}{\sin x}-\frac{1}{\sin y}=\frac{1}{\sin(y+\alpha)}-\frac{1}{\sin(x+\alpha)}$$
or, $$\frac{\sin y-\sin x}{\sin x \sin y}=\frac{\sin(x+\alpha)-\sin(y+\alpha)}{\sin(y+\alpha)\sin(x+\alpha)}$$
or, $$-\frac{2\cos(\frac{x+y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to Find the value of a trigonometric function if other very complicated trigonometric equation is given?
Q)If $\operatorname{sin}\alpha+\operatorname{cos}\alpha=\frac{\sqrt7}2$, $0 \lt \alpha \lt\frac{\pi}{6}, then \operatorname{tan}\frac{\alpha}2 $ is:(1)$\sqrt{7}-2$(2)$(\sqrt{7}-2)/3$(3)$-\sqrt{7}+2$(4)$(-\sqrt{... | If $t=\tan\fracα2$ then
$$
\cosα = \frac{1-t^2}{1+t^2}\text{ and }\sinα = \frac{2t}{1+t^2}
$$
so that
$$
a=\cosα+\sinα=\frac{1-t^2+2t}{1+t^2}\iff (1+a)t^2 - 2t +a-1=0\\
\iff ((a+1)t-1)^2 -1 + (a^2-1) \\
\iff t=\frac{1\pm\sqrt{2-a^2}}{a+1}=\frac{a-1}{1\mp\sqrt{2-a^2}}
$$
and with $a=\frac{\sqrt7}2$
$$
t=\frac{2a-2}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Proving a summation inequality with induction The exact question:
Prove:
$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)$
I have looked at similar problems but still don't understand how to prove this inequality by induction. So far I have this:
Induction basis:
Let n=1
$\displaystyle\sum_{k=1}^n \frac{... | The induction step assumes
$$\sum_{k = 1}^n \frac{1}{\sqrt{k}} > 2\sqrt{n + 1} - 2.$$
Using it, it follows that
\begin{align}
& \sum_{k = 1}^{n + 1} \frac{1}{\sqrt{k}} \\
= & \sum_{k = 1}^n \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{n + 1}} \\
> &2\sqrt{n + 1} - 2 + \frac{1}{\sqrt{n + 1}} \\
= & 2 \sqrt{n + 2} - 2 + \left[2(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Computing the basis of a subspace of matrices with the given nullspace. Compute the dimension and find a basis for the subspace of matrices $H ⊂ M_{3×5} (\mathbb{R})$
such that $[0,2,−3,0,1]^T$ is in the nullspace of A ∈ H.
I have said that, if we want $[0,2,−3,0,1]^T$ to be in the nullspace of A, then we must be able ... | $$\{\left( \begin{array}{ccc}
0 & -1 & 1 & 0 & 5\\
1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\end{array} \right),\left( \begin{array}{ccc}
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\end{array} \right),\left( \begin{array}{ccc}
0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 1\\
1 & 0 & 0 & 0 & 0\end{array} \right)
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.