Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Closed form of $\int_0^2\frac{1}{2+\sqrt{3\,e^x+3\,e^{-x}-2}}dx$ Could you please help me to solve this integration problem?
$$\int_0^2\frac{1}{2+\sqrt{3\,e^x+3\,e^{-x}-2}}dx$$
Its approximate numeric value is $0.419197813818367...$, but I could not find an exact symbolic expression for it.
| $\int_0^2\dfrac{1}{2+\sqrt{3e^x+3e^{-x}-2}}dx$
$=\int_1^{e^2}\dfrac{1}{2+\sqrt{3y+\dfrac{3}{y}-2}}d(\ln y)$
$=\int_1^{e^2}\dfrac{1}{y\left(2+\sqrt{3y+\dfrac{3}{y}-2}\right)}dy$
$=\int_1^{e^2}\dfrac{1}{2y+y\sqrt{3y+\dfrac{3}{y}-2}}dy$
$=\int_1^{e^2}\dfrac{1}{\sqrt{3y^3-2y^2+3y}+2y}dy$
$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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How to express sum of even numbers in sigma notation Given a positive integer number n, how can I express the sum of all positive even numbers up to n in sigma notation?
| Given a real number $x$ we can define $\lfloor x \rfloor$ to be the greatest integer less than or equal to $x$. For instance, $\lfloor 3 \rfloor = 3$ and $\lfloor 7.81 \rfloor = 7$.
Thus we can write the sum of even integers less than or equal to $n$ as
$$\sum_{k=0}^{\lfloor n/2 \rfloor} 2k$$
If you stare hard enough a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\lim_{n\to\infty}{\frac{3n+1}{2n+5}=\frac{3}{2}}$ Problem Prove $$\lim_{n\to\infty}{\frac{3n+1}{2n+5}=\frac{3}{2}}$$ using a epsilon proof
Scratchwork
We see that $\mid{\frac{3n+1}{2n+5}-\frac{3}{2}}\mid=\mid{\frac{6n+2}{4n+10}-\frac{6n+15}{4n+10}}\mid= \mid\frac{-13}{4n+10}\mid=\frac{13}{4n+10}<\frac{13}{4n}<\f... | Yes, absolutely :)
Would post this as a comment, but then this question remains unanswered, so...
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac {1}{\cos 0^\circ \cdot \cos 1^\circ} + \ldots +\frac {1}{\cos 88^\circ \cdot \cos 89^\circ}= \frac{\cos 1^\circ}{\sin 1^\circ}$ Prove the following identity:
$$\frac {1}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {1}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$... | I think it's a typo.
$$
\frac{1}{\cos k \cos (k+1)}=\frac{1}{\sin 1}\frac{\sin 1}{\cos k \cos (k+1)}=\frac{1}{\sin 1} \frac{\sin(k+1)\cos(k)-\sin(k)\cos(k+1)}{\cos k \cos (k+1)} \\ =\frac{1}{\sin 1} \left(\tan(k+1)-\tan (k)\right)
$$
Then
$$
\sin 1\; \sum_{k=0}^{88}\frac{1}{\cos k \cos (k+1)}=\tan(89)-\tan(0)=\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real... | Here's what I've got so far.
It's not a complete solution,
but the ideas might be useful
to someone else.
$2x^2-6x+9
=x^2+(x-3)^2
$.
If
$f(x)
=2x^2-6x+9
$,
$f'(x)
=4x-6
$
is zero at $x = 3/2$.
Since $f(3/2)
=9/2
$,
$f(x) \ge 9/2$
for all $x$.
This means that
$\dfrac1{f(x)}
\le \dfrac{2}{9}
$
for all $x$,
so
$\dfrac1{f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}... | One of the easiest ways to deal with infinite sums/products, is to stop at a finite value, say $N$, and look at what happens to the finite sum/product and then let $N \to \infty$. (In fact, this is the typical way infinite sums/products are to be understood/interpreted.)
Hence, in your case, let us look at
\begin{align... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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Are there identities which show that every odd square is the sum of three squares? I am looking for algebraic identities of the form
$$
(2n+1)^2 = f(n)^2 + g(n)^2 + h(n)^2,
$$
where the functions are polynomials in $n$.
EDIT: Evidently $(6k)^2 = 36k^2$ is trivially the sum of three squares when $k$ is odd. We also ha... | That's not really going to work, the highest degree terms do not cancel here, so all you get is $(2n+1)^2 = (2n+1)^2 + 0^2 + 0^2.$
What does work is due to Gordon Pall; every number is the sum of four squares, so write $$ 2n+1 = a^2 + b^2 + c^2 + d^2.$$ Then you get nontrivial expressions
$$ (2n+1)^2 = \left(a^2 + b^2 ... | {
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Any 'odd unit fraction' whose denominator is not $1$ can be represented as the sum of three different 'odd unit fractions'? Let us call a fraction whose denominator is odd 'odd fraction'. Also, let us call an odd fraction whose numerator is 1 'odd unit fraction'.
Then, here is my question.
Question : Is the following... | If $n$ is not a multiple of 3, then $$\frac{1}{n}=\frac{1}{n+2}+\frac{3}{(n+2)(n+4)}+\frac{1}{n(n+2)(n+4)}$$
Second try: You have the first fraction is $1/(n+2)$. Then you want to solve $$\frac{2}{n(n+2)}=\frac{1}{a}+\frac{1}{b}$$
Rearrange that into $(2a-n(n+2))(2b-n(n+2))=n^2(n+2)^2$ You can make the right-hand sid... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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finding range of function of three variables Three real numbers $x$, $y$, $z$ satisfy the following conditions.
$x^{2}+y^{2}+z^{2}=1~$, $~y+z=1$
Find the range of $~x^{3}+y^{3}+z^{3}~$ without calculus.
I solved this problem only with Lagrange-Multiplier and wonder if there exist other methods.
| Let $z=1-y$ to get $x^2+y^2+(1-y)^2 = x^2 + 2 y^2 -2y +1 = 1$, or $x^2 + 2 y^2 -2y = 0$.
This gives $x = \pm\sqrt{2y(1-y)}$ and $y \in [0,1]$.
Substituting gives:
$x^3+y^3+z^3 = \pm (2y(1-y))^\frac{3}{2} +3 y(y-1) +1$, with $y \in [0,1]$.
Let $\phi_\pm(y) = \pm (2y(1-y))^\frac{3}{2} -3 y(1-y) +1$, with $y \in [0,1]$.
L... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$ Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$.
Since $x^6 + x^3 + x + 1 = (x^2 + 1)(x^4 - x^2 + x + 1)$, $\mathrm{gcd}[f(x),g(x)] = x^2 + 1$.
My question is how could I JUSTIFY that the ... | To justify, you could say that it divides g(x) and f(x) to the rational polynomials such that they have no other common divisor, or common factor. So the answer is proved.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Induction $(1+\frac{1}{x^n})(1+\frac{1}{y^n}) \geq (1+2^n)^2$ How to prove this inequality using Induction (or any simpler method):
Let (x,y) be real positive numbers, so that x+y=1; and n an integer:
Prove this:
$\begin{align}(1+\frac{1}{x^n})(1+\frac{1}{y^n}) \geq (1+2^n)^2\end{align}$
| Replace $ y$ with $1-x$.
Hint: Show that $ \frac{1}{4} \geq x(1-x)$.
Hint: Show that $\frac{1}{x^n} \times \frac{1}{(1-x)^n} \geq 2^{2n} $.
Hint: Show that $ \frac{ 1}{x^n} + \frac{1}{(1-x)^n} \geq 2 \times 2^n$ using the technique AM-GM.
Expand and compare.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$? I'm trying to find the orthogonal trajectories of the family of circles through the points $(1,1)$ and $(-1, -1)$. Now such a family can be given by an equation of the form $$ x^2 + y^2 + 2g(x-y) - 2 = 0, ... | Assuming your calculations are correct up to the last part, I begin with:
$$ y^\prime = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}. $$
Observe that:
$$ y^\prime = \frac{1 + 2y/x - (y/x)^2 - 2/x^2}{1 - 2y/x - (y/x)^2 + 2/x^2}. $$
Thus, let $v = y/x$ for which $y=vx$ and $y' = xv'+v$ and the given problem changes ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Ellipse: product of the distance from foci to a tangent is a constant I am supposed to determine what is the result of said product. Given $P(x_0,y_0)$, I need to calculate the distance from the foci of an ellipse to the tangent line that passes through $P$, and then multiply the distances.
In essence it is quite simpl... | $$
\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac{y_0^2}{b^4}}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac1{b^2}\left(1-\frac{x_0^2}{a^2}\right)}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac1{b^2}\left(1-(a^2-b^2)\frac{x_0^2}{a^4}\right)}
=b^2,
$$
where in the final step we used the identity $a^2-b^2=c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the vector field $X(x, y, z)=(xy-z^2, yz-x^2, x^2+z^2+xz-1)$ is tangent to the set $x^2 + y^2 + z^2 = 1$ I know I need to find functions $F(t)$, $G(t)$, and $H(t)$ such that $F(0)=x$, $G(0)=y$, and $H(0)=z$ and $F'(0)=xy-z^2$, $G'(0)=yz-x^2$, and $H'(0)=x^2+z^2+xz-1$. It's also necessary that $(F(t))^2 +(G(t... | We will show that the vector field is perpendicular to any point $(x,y,z)$ on sphere by considering the dot product of the vector field and the normal to the sphere at that point.
$(xy-z^2,yz-x^2,x^2+z^2+xz-1)\cdot(x,y,z)=x^2y-xz^2+zy^2-yx^2+zx^2+z^3+xz^2-z=zy^2+zx^2+z^3-z=z(x^2+y^2+z^2-1).$ We know on the surface of t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form,... | Hint: Try matching first and last terms: $A^3 = x^3$ and $B^3=-\frac{64}{x^3}$ and check if it fits the other terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Limit of a Lebesgue integral What is the value of:
$$\lim_{n\to\infty}\sqrt{n}\int_0^{1}(1-t^2)^ndt$$
I think I have to use the Theorem of dominated convergence
| From my comment:
$$\begin{align*}
A_n & = \int_0^1 (1-t^2)^n dt = \int_0^1 (1-t^2)^{n-1} dt - \int_0^1 t^2 (1-t^2)^{n-1} dt \\
& = A_{n-1} - \left( \underbrace{\frac{1}{n} \left[ t^2 (1-t^2)^n \right]_0^1}_{=0} - \frac{2}{n} \int_0^1 t (1-t^2)^n dt \right) \\
& = A_{n-1} + \frac{2}{n} \left( \underbrace{\frac 1 {n+1} \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{... | Another way: You can bound it by $$\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n+1} \ge \int_{n+1}^{3n+1} \frac{1}{x} = \log \Big(\frac{3n+1}{n+1}\Big)$$ where $\frac{3n+1}{n+1}$ is monotone increasing in $n$, and $\log(\frac{3*7 + 1}{7+1}) > 1$ already. Then check the cases $n=1$ through $6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Matrix multiplication - Express a column as a linear combination Let $A = \begin{bmatrix}
3 & -2 & 7\\
6 & 5 & 4\\
0 & 4 & 9
\end{bmatrix} $ and $B = \begin{bmatrix}
6 & -2 & 4\\
0 & 1 & 3\\
7 & 7 & 5
\end{bmatrix} $
Express the third column matrix of $AB$ as a linear combination of the column matrices of $A$
I do... | The third column of $AB$ is $$A\begin{bmatrix}
4\\
3\\
5
\end{bmatrix}$$
and if we denote the columns of $A=[C_1 C_2 C_3]$ then the third column of $AB$ is
$$A\begin{bmatrix}
4\\
3\\
5
\end{bmatrix}=[C_1 C_2 C_3]\begin{bmatrix}
4\\
3\\
5
\end{bmatrix}=4C_1+3C_2+5C_3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$? I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework probl... | $$\sum_{n=1}^\infty\frac{(-1)^n}{z^2-n^2}=2\sum_{n=1}^\infty\frac{1}{z^2-(2n)^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
$$=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
Now since,
$$\pi \cot(\pi z)=\frac{1}{z}+2z\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
We get that, $$\pi\coth(\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/535116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to prove the inequality $\sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $ for $n\in\mathbb{Z}^+$? I have to prove this inequality:
$$
\forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}
$$
So far, I have done the base cases and assumed the inequality is true for some integer k, and ... | I am not a fan of using (in)equality that you have to prove, since you have to be really careful what you do with that. So, let's try this:
\begin{align*}
\frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} &> \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+4} = \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2(k+2)} \\
&> \frac{\sqrt{k}}{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find this inequality find the maximum $z_{5}$ let $z_{1},z_{2},z_{3},z_{4},z_{5}\in C$,such
$$\begin{cases}
|z_{1}|\le 1,|z_{2}|\le 1\\
|2z_{3}-(z_{1}+z_{2})|\le|z_{1}-z_{2}|\\
|2z_{4}-(z_{1}+z_{2})|\le|z_{1}-z_{2}|\\
|2z_{5}-(z_{3}+z_{4})|\le|z_{3}-z_{4}|
\end{cases}$$
Find the maximum $|z_{5}|$
My ugly soluti... | It seems the following.
Your solution is nice but I shall use a geometric approach.
The points $z_1$ and $z_2$ belong to a disk $D_1$ of radius $R_1=1$ centered at the point $o_1=0$. The points $z_3$ and $z_4$ belong to a disk $D_2$ of radius $R_2=|z_1-z_2|/2$ centered at the point $o_2=(z_1+z_2)/2$. The point $z_5$ b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of $(\forall x)(x^2+4x+5 \geqslant 0)$ $(\forall x)(x^2+4x+5\geqslant 0)$ universe is $\Re$
I went about it this way
$x^2+4x \geqslant -5$
$x(x+4) \geqslant -5$
And then I deduce that if $x$ is positive, then $x(x+4)$ is positive, so it's $\geqslant 5$
If $ 0 \geqslant x \geqslant -4$, then $x(x+4)$ is also $\geq... | \begin{align}
x^2+4x+5
&= x^2 + 2x(2) +(2)^2 +1\\
&= (x+2)^2 +1
\end{align}
$(x+2)^2\ge0$ for any $x\in\mathbb{R}$. Adding $1$ for both sides of inequalities, we have
$$
(x+2)^2+1\ge 1$$
for any $x\in \mathbb{R}$.
As $x^2+4x+5=(x+2)^2+1$,
$$
x^2+4x+5\ge 1>0$$
for any $x\in \mathbb{R}$.
Finally,
$$
x^2+4x+5>0$$
for any... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539375",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $n^9 \equiv n \pmod{30}$ if $(30,n) > 1$ I'm looking at the following number theory problem:
Prove that $n^9 \equiv n \pmod{30}$ for all positive integers $n$ if $(30,n) > 1$.
It is easy to show that $n^9 \equiv n \pmod{30}$ if $(30,n)=1$ by using Euler's Theorem. Is there any way to prove the above easily w... | You know that $n^9-n=0\mod 30$ if, and only if $n^9-n=0\mod 2$, $n^9-n=0\mod 3$ and $n^9-n=0\mod 5$. But modulo $2$; we have $n^2=n$, so $n^9=n$ immediately. If $n\neq 0$, $n^2= 1$ modulo $3$, so $n^3=n$, so $n^9=n^3=n$ so $n^9=n$. Modulo $5$, if $n\neq 0$, $n^2=\pm 1$ so $n^4=1$ so $n^5=n$ so $n^9=n^5n^4=n^5=n$ and th... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How come when $2^{k} | (x-1)(x+1)$ one of the terms is divisible by $2$ and not by $4$ when $k \in \mathbb{N} $ and $3 \leq k$ So I'm reading Knuth's 'Discrete Mathematics' at the moment and there's a paragraph detailing how many solutions are there for $x^{2} \equiv 1 \pmod{p}$.
So other cases (when $p$ is an odd prim... | In general, there are cases in which neither $x+1$ nor $x-1$ is divisible by $2$ (take $x$ to be an even number).
It is given that $2^k | (x+1)(x-1) $.
Suppose neither $x+1$ nor $x-1$ is divided by $2$, then $2$ does not divide $(x+1)(x-1)$. So, there is no point of considering $2^k | (x+1)(x-1)$.
So, without loss of ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\sum_{n=0}^\infty \frac{n}{3^n}$ How can i prove that $ \sum\limits_{n=0}^\infty \frac{n}{3^n}=\frac{3}{4}$?
I tried to find some pattern but didn't succeed
| Another approach: As the series converges, let L denote its limit. Now you have $$L=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty \frac{n}{3^{n-1}}= \frac{1}{3}\sum_{n=0}^\infty \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty \frac{n}{3^{n}}+\sum_{n=0}^\infty... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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we need to find the points where $|f(z)|$ has maximum and minimum value $f(z)=(z+1)^2$ and $R$ be the triangle with vertices $(0,0),(0,1),(2,0)$, we need to find the points where $|f(z)|$ has maximum and minimum value
so here $z=2$ is the point of maximum and $z=0$ is minimmum (intuitively), as $f$ is analytic so it m... | On the lower edge of the triangle, we have $z = x$, and
$$|f(z)| = (x + 1)^2$$
This is a strictly increasing function of $x$, and so its minimum is at $0$ and maximum at $1$.
On the vertical edge, we have $z = iy$ and
$$|f(z)| = |(iy + 1)^2| = 1 + y^2$$
Again, this is strictly increasing in $y$ (take a derivative), a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\cos x+\cos 3x+\cos 5x+\cos 7x=0$, Any quick methods? How to solve the following equation by a quick method?
\begin{eqnarray}
\\\cos x+\cos 3x+\cos 5x+\cos 7x=0\\
\end{eqnarray}
If I normally solve the equation, it takes so long time for me.
I have typed it into a solution generator to see the steps. One of the step... | In case you want to crack the question through solving polynomial equation, let $cos(x)=y$
You've found $$64y^7-96y^5+10y^2-4y=0$$
It can be factorized to
$$4y(2y^2-1)(8y^4-8y^2+1)=0$$
Furthermore
$$2y^2-1=(\sqrt{2}y+1)(\sqrt{2}y-1)$$
and
$$(8y^4-8y^2+1)=(2\sqrt{2}y^2+1)^2-(8+4\sqrt{2})y^2=(2\sqrt{2}y^2+\sqrt{8+4\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
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Sum of squares of the quadratic nonresidues modulo $p$ is divisible by $p$ Let $p$ be a prime number with $p > 5$. Prove that the sum of the squares of the quadratic nonresidues modulo $p$ is divisible by $p$.
My idea is to use the fact that any quadratic residue is congruent modulo $p$ to one integer in the set $\{1^2... | Using primitive roots finishes the problem quickly: Let $g$ be a primitive root, so that the sum of the squares of the quadratic non-residues is just
$$\sum_{i=1}^{\frac{p-1}{2}}{(g^{2i-1})^2} \equiv g^2\sum_{i=0}^{\frac{p-3}{2}}{g^{4i}} \equiv g^2\frac{1-(g^4)^{\frac{p-1}{2}}}{1-g^4} \equiv \frac{g^2(1-(g^{p-1})^2)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$a^2-b^2=bc$ and $b^2-c^2=ac \Rightarrow a^2-c^2=ab$ Some weeks ago our math teacher asked the following question and gave us a week to solve it:
If $a^2-b^2=bc$ and $b^2-c^2=ac ,$ Prove $a^2-c^2=ab$, Where $a,b,c$ are non-zero real numbers.
This seemed really easy at the first, but when i tried to prove it i just ... | You can chug through this by eliminating the $a$ variable and following your nose.
$$b(b + c) = b^2 + bc = a^2 = {(b^2 - c^2)^2 \over c^2}$$
We can assume that $b \neq -c$, since if $b = -c$ the first equation implies $a = 0$, contradicting that $a,b,c$ are all nonzero. So we can divide by $b+c$ to get
$$b = {(b^2 - c... | {
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"source": "stackexchange",
"question_score": "8",
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How find this $\int_{0}^{\pi}\frac{\cos{(nx)}}{\cos{x}+a}dx$ Fin the integral
$$I_{n}=\int_{0}^{\pi}\dfrac{\cos{(nx)}}{\cos{x}+a}dx$$
where $n\in {\mathbb N}\,,\ a>1$
My try:
let
$$I_{n}-I_{n-1}=\int_{0}^{\pi}\dfrac{\cos{(nx)}-\cos{(n-1)x}}{\cos{x}+a}dx$$
and note
$$\cos{x}-\cos{y}=-2\sin{\dfrac{x+y}{2}}\sin{\dfrac{x-... | A little complex analysis helps. First, by the evenness of $\cos$, we have
$$I_n = \frac12 \int_{-\pi}^\pi \frac{\cos (nx)}{a+\cos x}\,dx,$$
and by Euler's formula, we have $\cos (nx) = \operatorname{Re} e^{inx}$, so
$$I_n = \frac12 \operatorname{Re} \int_{-\pi}^\pi \frac{e^{inx}}{a+\cos x}\,dx.$$
Since the imaginary p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/547684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding a,b,c,d in a quartic expression Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far.
\begin{align}
a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51
\end{align}
Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)... | (2012's answer is correct, but the algebraic manipulations doesn't reveal what is happening. This answer explains why we could calculate the expression, despite not having enough information.)
By the remainder factor theorem, since $p(x) - 10x$ is a monic quartic polynomial with roots 1, 2, and 3, hence
$$p(x) - 10x = ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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How to solve the diophantine equation:$ xa^3+yb^3=c^3$ Let $a,b,c,x,y \in \mathbb{Z}> 1$. Any hint on how to solve of the diophantine equation $ xa^3+yb^3=c^3$?
| Consider following relations:
$1^3+6^3+8^3=9^3$
$3^3+4^3+5^3=6^3$
$⇒1^3+3^3+4^3+5^3+8^3=9^3$
or: $1^3 + 3^3+5^3+ 4^3(1+2^3)=9^3$
or: $(1^3 + 3^3+5^3). 1^3+ 4^3(1+2^3)=9^3$
or: $153 . 1^3 +4^3 . 9 = 9^3$
Compare this with equation:
$x a^3 + y b^3=c^3$
you find: $x=153, a=1, y=9, b=4, c=9$
It can be seen that there are i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx\leqslant\frac{a+b}{\sin a+\sin b}$ $a,b\in(0,\frac{\pi}{2}),aFor $a,b\in(0,\frac{\pi}{2}),a<b$, prove
$$\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx\leqslant\frac{a+b}{\sin a+\sin b}.$$
By mean value theorem, there is $c\in(a,b)$ s.t.
$$\frac{c}{\sin c}=\frac{1}{b-a}\int_a^b\fr... | From the Hermite-Hadamard inequality, since $\frac{x}{\sin x}$ is convex on $[a,b]$, you have that $$\frac{1}{b-a}\int_a^b\frac{x}{\sin x}\,dx\leq\frac{1}{2}\left(\frac{a}{\sin a}+\frac{b}{\sin b}\right).$$ So it's enough to prove the following:
$$\frac{a\sin b+b\sin a}{2\sin a\sin a b}\leq\frac{a+b}{\sin a+\sin b}\Lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$? For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$?
Well,
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1}$$
and,
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1}$$
Notice that $x^{m+1} - 1|(x^{m+1})^n - 1$, ... | In your second formula you need $x^n$ in denominator.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Direct proof that $\sum_{n\geq 0}\frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$ Is there a direct proof that $$\sum_{n\geq 0}\frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n?$$
We dont know what logarithms or exponentials are.
| Applying the Binomial theorem,
\begin{equation*}
\lim_{n\to\infty}(1+\frac{x}{n})^n = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\frac{(n-2)(n-1)x^3}{6n^2}+\frac{(n-3)(n-2)(n-1)x^4}{24n^3}+\frac{(n-4)(n-3)(n-2)(n-1)x^5}{120n^4}+... \\
=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+... \\
=\sum_{n\geq 0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/553253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$. Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$.
I did
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{9+h}-3}{h} \frac{\sqrt(9+h)+3}{\sqrt(9+h)+3} = \frac{9+h-3}{h\sqrt{9+h}+3}= \frac{6}{\sqrt{9+h}+3}= \... | let $f(x) = x^a$
$f'(x) = a*x^{a-1}$
for example:
$f(x) = x^2$ -> $f'(x) = 2*x^1$
$f(x) = x^3$ -> $f'(x) = 3*x^2$
So...
if $f(x)=x^\frac{1}{2}$ what is $f'(x)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/553494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
| EDITED. Some simplifications were made.
Here is a solution.
1. Basic facts on the dilogarithm. Let $\mathrm{Li}_{2}(z)$ be the dilogarithm function defined by
$$ \operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx. $$
Here the branch cut of $\log $ is chosen to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 8,
"answer_id": 2
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Laplace differential equation Can somebody help me work out $2y''+y'-y=\mathrm{e}^{3t}$, y(0)=2 and y'(0)=0 with the method of Laplace? I got
\begin{align*}
Y(s)&=\frac{1}{(s-3)(2s^2+s-1)}+ \frac{2+4s}{(s-3)(2s^2+s-1)}\\
&=-\frac{4}{15}\frac{1}{2s-1}+\frac{1}{12}\frac{1}{s+1}+\frac{1}{20}\frac{1}{s-3}-\frac{1}{6}\frac{... | Given:
$$\tag 1 2 y''+ y'- y = e^{3t},~ y(0)=2, ~ y'(0)=0$$
Taking the Laplace Transform of $(1)$, we have:
$$\mathcal{L}(2 y''+ y'- y) = 2(s^2y(s) - sy(0)-y'0) + (sy(s)-y(0)) - y(s) = \dfrac{1}{s-3}$$
Using the initial conditions from $(1)$, this reduces to:
$$y(s)(2s^2 + s - 1) -4s - 2 = \dfrac{1}{s-3}$$
Solving for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the first three terms of the Maclaurin Series
Determine using multiplication/division of power series (and not via WolframAlpha!) the first three terms in the Maclaurin series for $y=\sec x$.
I tried to do it for $\tan(x)$ but then got kind of stuck. For our homework we have to do it for the $\sec(x)$. It is kin... | Let's write
$$1 = \cos{x} \sec{x} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\right)(a_0 + a_1 x + a_2x^2 + \dots)$$
Expand the right hand side to find that
$$1 = 1 (a_0) + x (a_1) + x^2 \left(a_2 - a_0 \frac{1}{2!}\right) + x^3 \left(a_3 - a_1 \frac{1}{2!}\right) + x^4 \left(a_4 - a_2 \frac{1}{2!} + a_0 \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer.
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!... | $$
\binom{n}{r} + \binom{n}{r+1} \\
\frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} \\
\frac{n!}{(n-r)(n-r-1)!r!} + \frac{n!}{(n-r-1)!r!(r+1)} \\
\frac{n!}{(n-r-1)!r!}\left(\frac{1}{n-r} + \frac{1}{r+1}\right) \\
\frac{n!}{(n-r-1)!r!}\left(\frac{n+1}{(n-r)(r+1)}\right) \\
\frac{(n+1)!}{(n-r)!(r+1)!}\\
\binom{n+1}{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/558156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Uniform continuity of $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0.$ For the $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0,$ my text book asks the following questions.
(b) Why is $f$ uniformly continuous on any bounded subset of $\mathbb{R}$?
(c) Is $f$ uniformly continuous on $\mathbb{R}$??
T... | Would anyone be so kind as to critique the following direct demonstration that $\displaystyle x \sin \frac{1}{x}$ is uniformly continuous on $(0,1)$ ?
Let $\epsilon > 0$ and let $x, y \in (0,1)$.
Then
\begin{align*}
x\sin\frac{1}{x} - y \sin\frac{1}{y}
&= x\sin\frac{1}{x} - y\sin\frac{1}{x} + y\sin\frac{1}{x} - y \sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
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Check my solution: Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$ Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$
First I rationalized the numerator,
$$
\begin{align*}
&\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x) \cdot \frac{\sqrt{(x+a_1)(x+a_2)}+x}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{x(a_1+a_2... | You do it right until the last line $\lim_{x\to+\infty}\frac {x(a_1+a_2)+a_1a_2}{\sqrt {(x+a_1)(x+a_2)}+x}=\lim_{x\to+\infty}\frac {x(a_1+a_2)+a_1a_2}{\sqrt {x^2+(a_1+a_2)x+a_1a_2} +x}$=$\lim_{x\to+\infty}\frac {x((a_1+a_2)+a_1a_2x^{-1})}{|x| \left( \sqrt {1+(a_1+a_2)x^{-1}+a_1a_2x^{-2}}+1 \right)} $
For $x>0$ we have ... | {
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"url": "https://math.stackexchange.com/questions/560938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(... | using de moivres theorem we can generalise.$$\cos(n\theta)=\cos^n(\theta)-\frac{n(n-1)}{2!}\cos^{n-2}(\theta)\sin^2(\theta)+\frac{n(n-1)(n-2)(n-3)}{4!} \cos^{n-4}(\theta)\sin^4(\theta)........$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/564174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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how to find the asymptotic expansion of the following sum: I need to determine an asymptotic expansion when $q \rightarrow 1$ of the sum
$$S(q)=\sum_{n=0}^{\infty} \frac{q^n}{ (q^n + 1)^2 }.$$
Numerical computations suggest that $S(q)\sim\frac{c}{|q-1|}$ with $c \approx 0.5$.
| Suppose that $q>1$ and re-write the sum as
$$S(q) = \frac{1}{4} + \sum_{n\ge 1} \frac{q^n}{(q^n+1)^2}$$
The sum term is harmonic and may be re-written as
$$T(x) = \sum_{n\ge 1} \frac{e^{nx}}{(e^{nx}+1)^2}$$
which we will evaluate at $x=\log q.$
We will evaluate $T(x)$ by inverting its Mellin transform.
Recall the harm... | {
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"url": "https://math.stackexchange.com/questions/569447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Arithmetic series relationship with difference of two consecutive cubes. Is this a thing? Excuse my dodgy notation and my write-up in general, this is the first proof I've done since leaving school a while back. Anywho, has anyone come across anything like this before? Read the whole thing, not just the arithmetic seri... | I think you can prove it directly :
$$(n+1)^3 - n^3 = 3 n^2 + 3 n + 1 = 3 n (n + 1) + 1 = 6 \frac{n (n + 1)} 2 + 1 $$
$\frac{n (n + 1)} 2$ being the sum of all integers from $1$ to $n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Approximating the square root of two with fractions I would like to prove that there exist only finitely many $m, n \in \mathbb{N}$ satisfying $$\left | \sqrt{2} - \frac{m}{n} \right | < \frac{1}{4n^2}.$$
Any thoughts?
Thank you for your help.
| $$|\sqrt{2} - \frac{m}{n}| < \frac{1}{4n^2}\\\implies
|\sqrt{2}n-m|< \frac{1}{4n}\\\implies
\sqrt{2}n-\frac{1}{4n}<m<\sqrt{2}n+\frac{1}{4n}\\\implies
0<|(\sqrt{2}n-m)(\sqrt{2}n+m)|<\frac{1}{4n}(2\sqrt{2}n+\frac{1}{4n})=\frac{\sqrt{2}}{2}+\frac{1}{16n^2}<1.$$
But $|(\sqrt{2}n-m)(\sqrt{2}n+m)|=|2n^2-m^2|$ is an integer, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd per... | Let $3n+1=K^2$ and $4n+1=P^2$, where $P$ and $K$ are integers. Therefore, we know that $4(K^2)-3(P^2)=1$. Rearranging terms gives us $K^2= \frac {1+3P^2}4$. By testing values and guesswork we know that $P=13$ and $K=15$. Therefore $3n+1=169$, and then solving for $n$ gives us $n=56$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Limit with Stolz-Cesàro theorem: $\lim\limits_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}$ $$\lim_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}= \text{?}$$
Book has no answers.
It's on Stolz-Cesàro theorem lesson, if that helps. Can't find a solution.
| This is the direct application of the Stolz-Cesàro theorem
$$
\begin{align}
&\lim_{n\to \infty} \frac{1+2\sqrt{2}+...+n\sqrt{n}}{n^2 \sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(1+2\sqrt{2}+...+(n+1)\sqrt{n+1})-(1+2\sqrt{2}+...+n\sqrt{n})}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}}{(n+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding determinant using properties of determinant without expanding show that determinant
$$\left|\matrix{
x^2+L & xy & xz \\
xy & y^2+L & yz \\
xz & yz & z^2+L \\
}\right|
= L^2(x^2+y^2+z^2+L)$$
without expanding by using the appropriate properties of determinant.
All i can do is LHS
$$x^2y^2z... | $$\left|\matrix{
x^2+L & xy & xz \\
xy & y^2+L & yz \\
xz & yz & z^2+L \\
}\right|
= $$
$$xyz\left|\matrix{
x+L/x & y & z \\
x & y+L/y & z \\
x & y & z+L/z \\
}\right|
= $$
$$=\left|\matrix{
x^2+L & y^2 & z^2 \\
x^2 & y^2+L & z^2 \\
x^2 & y^2 & z^2+L \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine one triple of positive integer Note that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and $\frac{1}{6}-\frac{1}{7}=\frac{1}{6\times 7}$ and $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1$.
Determine one triple $(x,y,z)$ of positive integers with $1000<x<y<z<2000$ and $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\fra... | There are three solutions for
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} + \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{45} = 1$
in the required range, I
found them by exhaustive search:
$$x=1771,\;y=1932,\;z=1980$$
$$x=1806,\;y=1892,\;z=1980$$
$$x=1830,\;y=1891,\;z=1953$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/580290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Generalizing $\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}$ The following integral
\begin{align*}
\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} = \frac{5\pi^{2}}{96}
\tag{1}
\end{align*}
is called the Ahmed'... | The following is only a partial answer, but it might be useful.
Assuming that all the parameters are positive, the integral $$ I(p,q,r) = \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx $$ can be expressed in terms of $I \left(\frac{1}{q}, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "193",
"answer_count": 1,
"answer_id": 0
} |
Compute a sum with partial fraction decomposition and generating functions I am trying to compute $$\sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$ using a term by term partial fraction decomposition and also with generating functions, but I'm stuck.
| Note that:
$$
\frac{1}{k (n - k)} = \frac{1}{k n} + \frac{1}{n (n - k)}
$$
Thus your sum is:
$\begin{align}
\sum_{1 \le k \le n - 1} \frac{1}{k (n - k)}
&= \sum_{1 \le k \le n - 1} \frac{1}{k n}
+ \sum_{1 \le k \le n - 1} \frac{1}{n (n - k)} \\
&= \frac{1}{n} \sum_{1 \le k \le n - 1} \frac{1}{k}
+ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all solutions for a complex equation: $(1+i)z^2 - (6+i)z + 9+7i=0$ There is this math assignment that we've been given to find all the answers for some diffrent math problems.
The problem is: $(1+i)z^2 - (6+i)z + 9+7i=0$, find all the solutions and answer in geometric form.
I've tried the following:
$(1+i)z^2 - (6... | $$(1+i)z^2-(6+i)z+9+7i=0\Longleftrightarrow$$
Use the abc-formula:
$$z=\frac{-\left(-(6+i)\right)\pm\sqrt{\left(-(6+i)\right)^2-4\cdot\left(1+i\right)\cdot\left(9+7i\right)}}{2\cdot\left(1+i\right)}\Longleftrightarrow$$
$$z=\frac{-\left(-6-i\right)\pm\sqrt{\left(-6-i\right)^2-4\cdot\left(1+i\right)\cdot\left(9+7i\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the residue of $\frac{e^{iz}}{(z^2+1)^5}$ at $z = i$ and evaluate $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ I know the evaluation of $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ requires that I solve the first part, but for some reason I'm stumped. I get that I should use $\lim_{z \to i}\frac{1}{24}\frac{d^4}{dz^4}((z-i)^5\f... | $$\frac{d^4}{dz^4}\left((z-i)^5\frac{e^{iz}}{(z-i)^5(z+i)^5}\right)= \frac{d^4}{dz^4}\left(e^{iz}(z+i)^{-5}\right).$$
By Leibnitz Rule
$$ \frac{d^4}{dz^4}\left(e^{iz}(z+i)^{-5}\right) = \binom{4}{0} \left(\frac{d^4}{dz^4} e^{iz}\right)(z+i)^{-5}+\binom{4}{1} \left( \frac{d^3}{dz^3}e^{iz}\right) \left( \frac{d}{dz}(z+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -... | $f(n) = 2^{3n}-1 $
$f(0) = 0$ and $7|0$
Suppose that $7|f(n)$, let's say $f(n) = 7k$, $\Rightarrow f(n+1) = 2^{3n+3}-1 = 8\cdot2^{3n}-1 = 8\cdot2^{3n}-1 + 8 - 8 = 8(2^{3n}-1) - 7 = 8\cdot(7k) - 7 = 7\cdot(8k-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/584686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
| Replace $x^2=\tan(\alpha)$. We have $$\text{Slope} = \frac{\frac{x^3}{1+x^4}-\frac{-x^3}{1+x^4}}{2x}= \frac{x^2}{1+x^4} = \frac{\sin(2\alpha)}{2}\le\frac{1}{2}$$ with equality exactly when $\alpha=n\pi+\frac{\pi}{4}, x^2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
logarithms equation Hello I have following problem: solve equation $\log{(x-5)^2}+\log{(x+6)^2}=2$
and I rewrited this equation as
$2\log{(x-5)}+2\log{(x+6)}=\log{100} \implies 2(\log{(x-5)(x+6))=\log{100}} \implies \log{x^2+x-30}=\log10 \implies x^2+x-40=0 $
and I solved this equation, but I obtained only two soluti... | As the logarithm term contains squares, we can allow $x-5,x+6$ to be negative
In fact, $\log b^2=2\log |b|$ for real $b$
As $\log A+\log B=\log (AB),$
$$\implies \log_{10}\{(x-5)(x+6)\}^2=2$$
$$\implies \{(x-5)(x+6)\}^2=10^2=100$$
$$\implies (x-5)(x+6)=\pm10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding a Jordan canonical form $J$ and an invertible matrix $Q$ such that $J = Q^{-1}AQ$ $$
A = \begin{bmatrix}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1
\end{bmatrix}
$$
The characteristic polynomial can be found to be: $p(t)= -(t-1)(t-2)^2$.
I found $E_{\lambda_1} = N(A - I) = (1, 2, 1)$.
I'm having problems finding ... | We asked to find the Jordan Normal Form of the matrix:
$$
A = \begin{bmatrix}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1
\end{bmatrix}
$$
We start out by solving $|A - \lambda I| = 0$ to find the eigenvalues, which gives:
$$-(\lambda - 2)^2(\lambda-1) = 0 \rightarrow \lambda_1 = 1,~ \lambda_2 = 2, ~\lambda_3 = 2$$
We have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to calculate such sums of Legendre symbols? How to calculate such sums as $\sum_{x\in\mathbb{F}_p} \left(\frac{x^2+ax+b}{p} \right)$
If $x^2+ax+b$has a root, $b$ may be eliminated and the sum is evaluated to be $0+\sum_{x\in\mathbb{F}_p^*} \left(\frac{1+ax^{-1}}{p} \right)=-1$. Otherwise to linearize it, you need t... | We have in general for $p>2$ an odd prime
$$\sum_{x=0}^{p-1}{\left(\frac{x^2+ax+b}{p}\right)}=\begin{cases} -1 & \text{if} \; p\nmid a^2-4b \\ p-1 & \text{if} \; p\mid a^2-4b \end{cases}$$
We begin by completing the square, using that $(\frac{4}{p})=1$:
\begin{align}
\sum_{x=0}^{p-1}{\left(\frac{x^2+ax+b}{p}\right)}& =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square? How would I prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square?
| Short solutions: $2^{1000} + 15 \equiv 3 \pmod{4}$, $2^{1000} + 16 \equiv 2 \pmod{3}$, neither of which are quadratic residues.
More elementary solution: $2^{1000} = (2^{500})^2$. The next perfect square is $$(2^{500} + 1)^2 = 2^{1000} + 2(2^{500}) + 1$$ which is clearly more than both $2^{1000} + 15$ and $2^{1000} + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 10,
"answer_id": 5
} |
Simpson Rule special case of Gauss quadrature Prove that the Simpson Rule for integrating the function $f$ on the interval $[a,b]$: $$I_2(f)=\frac{b-a}{6}(f(a)+4f(\frac{a+b}{2})+f(b))$$ is actually the Gauss quadrature for the weight $g=1$ and $3$ nodes.
I can not get why when we have $3$ nodes in the Gauss quadrature ... | Start from the definition of a Gauss quadrature:
$$\int_{-1}^1 f(x)\ dx \approx \sum_{i=1}^3 g_i f(x_i).$$
We perform a change of variables to change the range of integration, so
$$\int_a^b f(x)\ dx = \frac{b-a}{2}\int_{-1}^1 f\left(\frac{b-a}{2}z+\frac{b+a}{2}\right)\ dz.$$
Apply gaussian quadrature to the new integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Don't understand simple logarithm problem with fractional base $$3\log_{\frac{4}{9}}\sqrt[4]{\frac{27}{8}}$$
$$\log_{\frac{3}{2}}\frac{16}{81}$$
I understand using the expansion property to expand the division into a subtraction but how do I proceed from there?
| First observe that
$$
3\log_\frac{4}{9}\sqrt[4]{\frac{27}{8}} =\frac{3}{4}\log_\frac{4}{9}\frac{27}{8}
$$
So all we need now is to compute the log term. By definition $\log_ab=y$ means that $a^y=b$ so we need to find $y$ such that
$$
\left(\frac{4}{9}\right)^y=\frac{27}{8}
$$
Expanding this, we have
$$
\left(\frac{2^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $x$ for $\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12}...$ $$\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12} + \dots +\frac1{100\times110}$$
Find x
My younger sist... | $$\frac{1}{k(k+10)}=\frac{a}{k}+\frac{b}{k+10}$$
$$\begin{array}{l}
\frac{a}{k}+\frac{b}{k+10}=\frac{k(a+b)+10a}{k(k+10)}
\end{array}$$
$$k\leftarrow 0, a=\frac{1}{10}$$
$$k\leftarrow 1, b=-\frac{1}{10}$$
$$\boxed{\cfrac{1}{k(k+10)}=\cfrac{1}{10}\left(\cfrac{1}{k}-\cfrac{1}{k+10}\right)}$$
$$\begin{array}{l}
\sum_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Expected value of a negative binomial that has finite $n: n \lt \infty$?
Cards from an ordinary deck are turned face up one at a time. Compute
the expected number of cards that need to be turned face up in order
to obtain
(a) 2 aces;
(c) all 13 hearts.
This is a homework problem straight from a chapter on Expec... | Using the identity
$$
\sum_{k=a}^{n-b}\binom{k}{a}\binom{n-k}{b}=\binom{n+1}{a+b+1}\tag{1}
$$
here is how I would approach (c):
The number of arrangements which get all $13$ hearts in exactly $k$ draws is $\binom{k-1}{12}\binom{52-k}{0}$; that is, the number of arrangements, with the $k^\text{th}$ draw being a heart, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/594960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Asymptotic expansion of $J(t) = \int^{\infty}_{0}{\exp(-t(x + 4/(x+1)))}\, dx$ I want to derive an asymptotic expansion for the following Bessel function. I think I need to rewrite it in another form, from which I can integrate it by parts. I am interested in obtaining the expansion up to second order as $t$ approaches... | First make the (cosmetic) change of variables $x = y+1$, so that
$$
\begin{align}
J(t) &= \int^{\infty}_{0} \exp\left[-t\left(x + \dfrac{4}{x+1} \right) \right]\, dx \\
&= \int_{-1}^{\infty} \exp\left[-t\left(y + 1 + \dfrac{4}{y+2} \right) \right]\, dy \\
&= \int_{-1}^{\infty} \exp\Bigl[-t f(y) \Bigr]\,dy.
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$
Let $a$, $b$ and $c$ be positive real numbers.
$(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
$(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$
For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab... | for (i);
notice that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This is the standard factorization for the sum of cubes. Then $4(a^2-ab+b^2)\ge(a+b)^2$.
Thus,
$$4a^2-4ab+4b^2\ge (a+b)^2$$
$$a^2+2ab+b^2+3a^2-6ab+3b^2\ge (a+b)^2$$
$$(a+b)^2+3(a-b)^2\ge(a+b)^2$$
Since $3(a-b)^2$ is always positive or $0$, the inequality is proven.... | {
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"url": "https://math.stackexchange.com/questions/596257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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what is the limit of $\displaystyle \lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$
find the limit of:
$\displaystyle\lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$
$ a,b>0$
i know that:
$\lfloor\frac{b}{x}\rfloor\le \frac{b}{x} + 1$; and
$\lfloor\frac{b}{x}\rfloor\ge \frac{b}{x} -1$
but i c... | Consider two cases: $x>0$ and $x<0$.
In each case multiply your inequalities by $\frac{x}{a}$.
Then, say for $x<0$, you get:
$\frac{b}{a} -\frac{x}{a} \geq \frac{x}{a} \lfloor \frac{b}{x} \rfloor \geq \frac{b}{a}+ \frac{x}{a}.$
Now use the squeeze theorem.
| {
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"url": "https://math.stackexchange.com/questions/605897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Calculate the lim $\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}$ $$
\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}
$$
I multiply it with:
$$
\frac{ \sqrt{x} + 1}{\sqrt{x} + 1}
$$
And I get :
$$
\lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)}
$$
But the solution is still division b... | Hint: $$\frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1} =\frac{3}{2} \cdot \frac{ e^\frac{\ln x}{2} - 1}{\frac{\ln x}{2}} \cdot \left( \frac{ e^\frac{\ln x}{3} - 1}{\frac{\ln x}{3}}\right)^{-1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Counting Combinations of a List of Numbers Let's say I have the following list of 9 (including repetitions) single-digit numbers: 7,7,7,5,5,3,2,2,2. I want to know, for an arbitrary list of natural numbers, how many ways I can select one, two,...,all of them without the obvious double counting. For example, for the li... | Suppose we have $N$ balls colored $n$ different colors. For each color $k$, let $d_k$ be the number of balls of color $k$. Our goal is to find the number $a_r$ of ways to choose $r\leq N$ balls without repetition and ignoring order, where balls of the same color are considered identical. Now, any such choice of $r$ bal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Transforming from cartesian to cylindrical Here is the question:
Transform $\textbf{A} = \hat{\mathbf{x}} 2 - \hat{\mathbf{y}}5 + \hat{\mathbf{z}}3$ into cylindrical coordinates at point ($x=-2, y=3, z=1$).
What I have tried is this: consider the point the exercise has given and find $r$ and $\theta$ with it. This le... | When you substituted in for the trig functions in the $A_{\theta}$ line, you put the values in wrong. You should have $-2\sin \theta - 5 \cos \theta=\frac 1{\sqrt{13}}(2(-2)-5(-2))=\frac 4{\sqrt{13}}\approx 1.11$
Now we have an overall sign error relative to the book. As I draw the picture, the book is wrong. The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Showing that this IVP has a solution for all of $\mathbb{R}$ I'm preparing for an exam and came across this problem:
Given the IVP
$f'(x) = \frac{f(x)}{1+x^2+(f(x))^2}$
$f(x_0) = y_0$
-show this has an unique solution $f \in C(\mathbb{R})$ defined on all of $\mathbb{R}$.
-What is the solution if $y_0=0$?
We're usin... | You have
$$
\left|\frac{y}{1+x^2+y^2}-\frac{y_1}{1+x^2+y_1^2}\right|
\leq
\left|\frac{y}{1+x^2+y^2}-\frac{y_1}{1+x^2+y^2}\right|
+\left|\frac{y_1}{1+x^2+y^2}-\frac{y_1}{1+x^2+y_1^2}\right|\\
=\frac{|y-y_1|}{1+x^2+y^2}+\left|\frac{|y_1|(1+x^2+y_1^2)-|y_1|(1+x^2+y^2)}{(1+x^2+y^2)(1+x^2+y_1^2)}\right|\\
=\frac{|y-y_1|}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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squares which are not the sum of a square and twice a triangular number I'm trying to determine conditions on integer squares which cannot be written as a square and twice a triangular [all numbers positive], i.e. integers $n \ge 1$ where there are no integers $a,b \ge 1$ such that
$$ n^2 = a^2 + b^2+b.$$
For example, ... | In fact every odd square can be written as $n^2= 4T_{k} + a^2 + (a+1)^2$. Every odd square can be decomposed into:
$n^2 = [(n-1)/2 + (n+1)/2]^2$ with $(n-1)/2=4T_{k}$ and $(n+1)/2 = a^2 + (a+1)^2$ or equivalently $n^2 = (a+b)^2 = a^2 +2ab +b^2$.
Your example $9= 2^2 +5$ can in fact be rewritten as:
$9=(9-1)/2 + (9+1)/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/614519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, then $n-4m$?
If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, reduced fraction, then what is $n-4m$?
Th... | The following is not particularly clever but works.
Note that $$1\cdot 3 \cdot 5 \cdot \cdots \cdot (2N-1) = \frac{(2N)!}{2^N N!}$$
Using this (or just considering ratios of adjacent terms as suggested in comments), each term in the series is
$$\frac{120(2n)! (n+3)!}{n!(2n+6)!} = \frac{15}{(2n+1)(2n+3)(2n+5)} = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy... | $$(x+y+1)(x+y-1)-1(x^2+y^2-1)=2xy$$
If we assume $x\equiv y\pmod 2$, then $\text{lcm}(x+y+1,x+y-1)=(x+y)^2-1$ since the expressions are coprime. If $x+y-1\mid x^2+y^2-1$, the above equality would imply that $x+y-1\mid 2xy$, but since $2xy<(x+y)^2-1$ ($x,y>0$), this implies $x+y+1\not \mid 2xy$.
If $x\not\equiv y\pmod2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Help with derivative of $y=x^2\sin^5x+x\cos^{-5}x$
Find $y^{\prime}$ of $y=x^2\sin^5x+x\cos^{-5}x$
My try:
$\dfrac{d}{dx}(x^2\sin^5x)=x^2(-5\sin^4x)+(2x\sin^5x)$
$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}x)+1(\cos^{-5}x)$
This doesn't seem right. Can you please show how to do it correctly?
The answer is $y^{\prim... |
$$y=x^2\sin^5x+x\cos^{-5}x$$
When taking the derivative of each summand of $y$, you left out a component of the chain rule for each of the trigonometric functions: $$[g(x)]^n = n[g(x)]^{n - 1} \cdot \color{blue}{g'(x)}$$ See the text highlighted in blue:
$$\dfrac{d}{dx}(x^2\sin^5x)=x^2(5\sin^4x){\bf\color{blue}{ \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove that if $10|A$ then $100|A$. For : $A=a^2+ab+b^2$ with $a,b \in \mathbb{N}$.
We known that $10|A$ prove that $100|A$.
| Solution 1: Suppose $10|A$. If either $a$ or $b$ is odd, then $A$ is also odd, so $a$ and $b$ must be even and hence $4|A$. Now consider $A$ modulo $5$. If either $a$ or $b$ is divisible by $5$, the other must be as well, and so $25|A$. Otherwise, $a^2$ and $b^2$ can be either of $\pm 1$ modulo $5$. If one is $1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$ $n$ is a positive integer, then
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$
please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.
I want to find a better proof.
My stupid method:
$$1+\frac1{2^2}+\fra... | We show by induction that if $n\gt 1$ then
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}\lt \frac{5}{3}-\frac{2}{2n+1}.$$
The result is true at $n=2$. Suppose that the result holds at $n=k$. We show it holds at $n=k+1$.
By the induction assumption,
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{k^2}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 3,
"answer_id": 0
} |
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
... | Hint $\ \ \sqrt{z+1}+\sqrt{z}\, =:\, y\ $ times $\,(\sqrt{z+1}-\sqrt{z}\,=\,-1) \ \Rightarrow\ 1 = -y\,$ contra $\, y > 0\ \ \ $ QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
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Hermitian matrix such that $4M^5+2M^3+M=7I_n$ $n$ is a positive integer. Besides the identity matrix $I_n$, does there exist other $n\times n$ Hermitian matrix $M$, such that the following equality
$$4M^5+2M^3+M=7I_n $$
hold?
I try this: Since $4M^5+2M^3+M=7I_n$, then
$$(M-I_n)(4M^4+4M^3+6M^2+6M+7I_n)=0$$
but, Wha... | Hermitian matrices are diagonalizable and they have only real eigenvalues. Let
$$
M=H^{-1}\mathrm{diag}(d_1,\ldots,d_n)H,
$$
where $D=\mathrm{diag}(d_1,\ldots,d_n)$ is the diagonal matrix which contains the eigenvalues of $M$. Note that
$$
4x^4+4x^3+6x^2+6x+7= (2x^2+1)^2+3(x+1)^2+2x^2+4 \ge 5,
$$
for all real $x$. This... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| $\begin{eqnarray}{\bf Hint} &&\ \ \underbrace{2^{11}} \\
\,&=&\ \ \overbrace{2^{10} + \underbrace{2^{10}}}\\
\,&=&\ \ 2^{10} + \overbrace{{2^9+\underbrace{2^9}}}\\
\,&=&\ \ 2^{10} + 2^9+\overbrace{2^8+\underbrace{2^8}}\\
\,&=&\ \ 2^{10} + 2^9+2^8+\overbrace{2^7+2^7}\\
\,&&\qquad\qquad\ \ \vdots\qquad\qquad\qquad\ddot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
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How would you explain to a 9th grader the negative exponent rule? Let us assume that the students haven't been exposed to these two rules: $a^{x+y} = a^{x}a^{y}$ and $\frac{a^x}{a^y} = a^{x-y}$. They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^... | *
*Help for confusing $$ 2^{-3}\neq (-2)(-2)(-2)$$
Advise them to watch out for the exact position of the - sign,
and tell them since it isn't next to 2, it simply does not belong to the 2, instead it belongs to the 3.
Example would be
$$2^{-3} = \frac{1}{2^3} = \frac{1}{2\times2\times2} = \frac{1}{8}$$
as already p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 24,
"answer_id": 4
} |
The equation $3a+5b=n$.
For a given number $n$, how can we find out whether we have any non-negative values for $a$ and $b$ for the equation
$$3a+5b=n,$$
where $1\le n\le 100,000$.
For example: If $n=5$, then $a=0, b=1$.
| Your question is a special case of the Frobenius problem or coin problem. You want to know for which $n$ it is possible to find nonnegative integers $a,b$ such that $3a+5b=n$. Take a look at the two-variable case on the Wikipedia page: It is always possible if $n>3\cdot5-3-5$, that is, $n>7$.
For values smaller then or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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What's the explanation for why n^2+1 is never divisible by 3? What's the explanation for why $n^2+1$ is never divisible by $3$?
There are proofs on this site, but they are either wrong or overcomplicated.
It can be proved very easily by imagining 3 consecutive numbers, $n-1$, $n$, and $n+1$. We know that exactly one of... | Another approach is to write $n=3k+r$, with $r=0$, or $r=1$, or $r=2$, which is always possible: it's a simple division with the remainder. Now, the case $r=0$ is easily settled:
$$
n^2+1=9k^2+1
$$
which can't be divisible by $3$.
If $r=1$, then
$$
n^2+1=(3k+1)^2+1=9k^2+6k+1+1=3(3k^2+2k)+2
$$
which is not divisible by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)... | You should, through looking at only the leading terms of
$$\dfrac{20 n + 6}{\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}}$$
see
$$\dfrac{20 n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{20 n}{2 n^2}=\dfrac{10}{n}$$
which is clearly going to go to $0$ as $n\rightarrow \infty $. That's what we want to work towards.
So we try to get rid of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Derivative of trigonometric function help I am having trouble computing derivatives at $x=2k \pi$.
Let $f(x)=x \sqrt{1-\cos(x)}$.
Then $\displaystyle f'(x)=\frac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}$.
Then $\displaystyle f'(2k\pi)=\lim_{x \rightarrow 2k\pi}\frac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}$.
I know lim... | As $x$ approaches $2k\pi$ from the negative side, $\sin x$ is negative. From the positive side, $\sin x$ is positive. We have
$$\lim_{x\rightarrow2k\pi^-}\dfrac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}=\lim_{x\rightarrow2k\pi^-}\dfrac{2(1-\cos x)-x\sqrt{1-\cos^2x}}{2\sqrt{1-\cos x}}=$$
$$\lim_{x\rightarrow2k\pi^-}\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integral of $\sqrt{16x^{2}+9}$ with respect to x Find the Integral of $\sqrt{16x^{2}+9}$ with respect to x
My try: $\int\sqrt{16x^{2}+9} dx =\int(16x^{2}+9)^{1/2}dx$ then make u substitution $u=16x^{2}+9$
but then have no idea how to continue
| HINT:
$$I=\int\sqrt{x^2+a^2}dx=\int dx \sqrt{x^2+a^2}-\int\left(\int dx \frac{d \sqrt{x^2+a^2}}{dx}\right)dx$$
$$=x\sqrt{x^2+a^2}-\int \frac{x^2}{\sqrt{x^2+a^2}}dx$$
$$=x\sqrt{x^2+a^2}-\int \frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx$$
$$=x\sqrt{x^2+a^2}-I+a^2\int \frac1{\sqrt{x^2+a^2}}dx$$
For $\displaystyle \int \frac1{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$ Let $a,b,c\in R$ and satisfying $a^2+b^2+c^2=1$
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$
| We need to consider only non-negative $a, b, c$ as: $P(a, b, c)=P(-a, -b, -c)$ means if all three are negative we may replace them with positive numbers, if exactly two are negative then we can replace them with a triple having only one negative, and finally if only one is negative (say $a$), $P(a, b, c) \ge P(-a, b, c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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specify the limit of the following equation? What is the limit of the following $\lim_{n \to \infty}$ $\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2}$ and how to calculate it?
| Write
$$\begin{array}{rcl}
\lim_{n \to \infty}\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2} &=& \lim_{n \to \infty}\frac{(n-3)\cdot(n+3)\cdot(3n+1)^2}{(n+3)\cdot(n+1)^3}\\
&=& \lim_{n \to \infty}\frac{(n-3)\cdot(3n+1)^2}{(n+1)^3}\\
&=& \lim_{n \to \infty}\frac{9n^3 -21n^2-17n-3}{n^3+3n^2+3n+1} \\
&=& \lim_{n \to \infty}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/638067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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mixed numbers subtraction vertically In the following subtraction we are subtracting $2$ mixed numbers vertically. I know how it works except the last step.
$$ 7 \frac{1}{3} - 4 \frac{1}{2} = 3 + \frac{-1}{6} = 2 + \frac{5}{6} = 2 \frac{5}{6}$$
I am confused about this step: $ 3 + \frac{-1}{6} = 2 + \frac{5}{6}$
How do... | 3+-1/6=2+(1-1/6)=2+(6/6-1/6)=2+5/6
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Fractions in convergence proof Sequence: $a_n = \sqrt{2+ \frac{3}{n}}$
To prove convergence, want to show that $\left|\sqrt{2+ \frac{3}{n}} - \sqrt{2}\right| \le \varepsilon$
Simplifying, we get that $\sqrt{2+ \frac{3}{n}} - \sqrt{2}= \frac{3/n}{\sqrt{2 + \frac{3}{n}} + \sqrt{2}}$.
I understand up to this part. But the... | $\sqrt{2 + \frac{3}{n}}$ is larger than $\sqrt{2}$; then $\sqrt{2 +\frac{3}{n}} + \sqrt{2}$ is larger than $2\sqrt{2}$.
So, $\frac{1}{\sqrt{2+\frac{3}{n}}+\sqrt{2}}$ is smaller than $\frac{1}{2\sqrt{2}}$ . So, now, you have an upper bound expressed in a simpler manner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/639962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Clarification Regarding evaluation of $\lim_{n\rightarrow \infty} n\sin(2\pi e n!)$- NBHM-$2009$ Question is to evaluate $$\lim_{n\rightarrow \infty} n\sin(2\pi e n!)$$
We have $e = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots$
$$n!e=n!(1 + \dfrac1{1!} ... | your answer seems ok to me. my back-of-the-envelope calc using Landau's "big-O" was:
$$
n \sin (2\pi e n) = n \sin(2\pi E_n +2\pi e_n) = n \sin 2\pi e_n \\
= n \sin \left(\frac{2\pi}{n+1} + O(n^{-2})\right) = 2\pi (1+n^{-1})^{-1} + nO(n^{-2}) \\
= 2\pi + O(n^{-1})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/640110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Infinite Geometric Series Issue i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio
here is the series
$\... | $$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{2n}{2^n}-\sum_{n=1}^{\infty}\frac{1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1.$$
We use Maclaurin series for function $\frac{1}{(1-x)}$
$$\frac{1}{1-x}=1+x+x^2+x^3+\dots = \sum_{n=0}^{\infty} x^{n}$$
Differentiating both sides of this equation we get t... | {
"language": "en",
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"source": "stackexchange",
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How many zero elements are there in the inverse of the $n\times n$ matrix How many zero elements are there in the inverse of the $n\times n$ matrix
$A=\begin{bmatrix}
1&1&1&1&\cdots&1\\
1&2&2&2&\cdots&2\\
1&2&1&1&\cdots&1\\
1&2&1&2&\cdots&2\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
1&2&1&2&\cdots&\cdots
\end{bmatri... | Investigation with Maple indicates that for $n>1$, the inverse of $A$ has the following form:
(1) the main diagonal starts with $2$, finishes with $\pm1$, has zeros in between;
(2) the next diagonal each way has $1$ and $-1$ alternating;
(3) everything more than "one step away" from the main diagonal is $0$.
For exampl... | {
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Find the minimum value of $\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(c+a)}}{\sqrt{ab+bc+ca}}$ Let $a,b,c\ge0$ such that: $(a+b)(b+c)(c+a)=1$.
Find the minimum value of:
$$P=\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(c+a)}}{\sqrt{ab+bc+ca}}$$
I've tried many things but all failed. Please help. Thank you.
| If $c=0$ and $a=b=\frac{1}{\sqrt[3]2}$ then $P=\sqrt[3]2$.
We'll prove that it's a minimal value of $P$.
Indeed, we need to prove that $$\frac{\sum\limits_{cyc}\sqrt{ab(a+b)}}{\sqrt{ab+ac+bc}}\geq\sqrt[6]{4(a+b)(a+c)(b+c)}.$$
Now, by AM-GM
$$\sum\limits_{cyc}\sqrt{ab(a+b)}=\sqrt{\sum_{cyc}\left(a^2b+a^2c+2\sqrt{a^2bc(a... | {
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"question_score": "3",
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Prove trigonometric relation Recently, I found this identities in a sheet of paper I was given as studying material:
$$\prod^n_{k=1}\sin\left(\frac{k\pi}{2n+1}\right)=\frac{\sqrt{2n+1}}{2^n}\tag1$$
$$\prod^n_{k=1}\cos\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^n}\tag2$$
$$\prod^n_{k=1}\tan\left(\frac{k\pi}{2n+1}\right)=\... | Consider
$$\prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right).$$
By symmetry, it's the square of your first product, and expanding it with Euler's formula, we find
$$\begin{align}
\prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right) &= \frac{1}{(2i)^{2n}}\prod_{k=1}^{2n} \left(\exp \frac{k\pi i}{2n+1} - \exp \frac{-k... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solve the error in the simultaneous equation. The question is:
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1):
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
[(3)^2 - 2\cdot 3\c... | your equation (3) is wrong. it will be 2x=3+3y and then try.
the solving process is alternatively done by using the formula a^2-b^2 in the LHS of (2) and then use eq (1).
This particular type of problem follow always this type method.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving a property of about the Fermat numbers
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.
For our base step we let $n=2$. Now we have $2^{2^2}=16$. So the assertion holds for our base case. Then we assume it holds for $n$. For the $n+1$ case, is there a way to de... | $$ (2^{2^n}+1)^2 = 2^{2^{n+1}} + 2\cdot 2^{2^n}+1$$
Modulo 10 the left-hand side is 9 by the inductive hypothesis. Also
$$2\cdot 2^{2^n} = 2(2^{2^n}+1)-2$$
Modulo 10, this is 2.
What we have is:
$$(2^{2^n}+1)^2 - 2\cdot 2^{2^n}\,\,\, (\text{mod 10}) = 9-2=7$$
But that's the same as
$$2^{2^{n+1}} + 1 \,\,\, (\text{mod... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Maximum and minimum distance from the origin
Find the maximum and minimum distances from the origin to the curve $5x^3+6xy+5y^2-8=0$
My attempt:
We have to maximise and minimise the following function $x^2+y^2$ with the constraint that $5x^3+6xy+5y^2-8=0$.
Let
$$F(x,y)=x^2+y^2+\lambda(5x^2+6xy+5y^2-8)$$
$$\frac{\del... | if the case is $5x^2+6xy+5y^2-8=0$, it could be solved in a easy way:
$5x^2+6xy+5y^2=8,2xy \le x^2+y^2 \implies 5x^2+5y^2 +3(x^2+y^2) \ge 8 \implies x^2+y^2 \ge 1$ when $x=y$
$2xy \ge -(x^2+y^2) \implies 2(x^2+y^2) \le 8 \implies x^2+y^2 \le 4$ when $x=-y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/655607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin... | That is exactly correct! Just two things: First, $\tan,\sin,\cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $\tan x$, $\cos x$ etc rather than just $\tan$ or $\cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
\frac{1}{\cos x\sin x}=\frac{1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
How does $\frac1{n^4} =\frac1{(2n-1)^4} + \frac1{(2n)^4}$ I need to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4}.$
In the solution document it written:
$$\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4}.$$
My question is, how did he get ... | Since the even and odd natural numbers form a complete residue system modulo two the author can separate the sum into two sums of even and odd terms. This could be done with any modulus for example instead of modulo two, we could separate the sum using a modulus of three:
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{n=1}^\in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"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.