Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving for $x$ in $x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$ How would I go about solving for $x$ in this equation?
$$x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$$
| $$x+\frac{s}{s^2+4}=\frac{2x}{s^2+4}$$
$$\frac{s^2x+4x+s}{s^2+4}=\frac{2x}{s^2+4}$$
For real $s$, $s^2+4\ne0$. Hence,
$$s^2x+4x+s=2x$$
$$x(s^2+2)+s=0$$
$$x=-\frac{s}{s^2+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable? Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?
Having some trouble with this one. I tried using the f... | Consider the matrix $B = A - \lambda I$, where $ \lambda \in \text{Spec}(A)$.
Then $\lambda_{1} = n - \lambda$ is an eigenvalue of $B$ with an associated eigenvector $v_{1} = \begin{pmatrix} 1 \\ 1 \\ 1\\ \cdot \\ \cdot \\ \cdot \\ 1 \\1 \\1\\ \end{pmatrix}$.
The remaining eigenvalues of $B$ are $\lambda_{2} = \lambda_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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The sum of k times the kth power of a is given analytically by? I was wondering how would someone derive (Not prove) the result in terms of n and a of the following sum:
$$\sum_{k=1}^n ka^k$$
My question basically is, given that summation how would you tackle the problem in order to get to a solution such as, for exam... | The trick is to eliminate the $k$ in the sum by integrating:
$$\begin{align*}
\sum_{k = 1}^n ka^k & ~=~ a \sum_{k = 1}^n k a^{k - 1} \\
& ~=~ a \sum_{k = 1}^n \frac{d}{da} \left ( \int ka^{k - 1} \ da\right ) \\
& ~=~ a \frac{d}{da} \left ( \sum_{k = 1}^n \int ka^{k - 1} \ da \right ) \\
& ~=~ a \frac{d}{da} \left ( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Quick Question on Zero Matrix Find all $2\times2$ matrices
$$B =\begin{pmatrix} \alpha & \beta\\ \gamma & \delta\\ \end{pmatrix}$$
such that $B^2$ is the zero matrix.
Any help would be greatly appreciated. Thanks
| $B^2=0$ means $\operatorname{im}(B)\subseteq\ker(B)$. In particular, $B$ needs to be of rank at most one: If $\dim\operatorname{im}(B)=2$, then $\dim\ker(B)=0$ and the above inclusion is impossible. Furthermore, $B^2=0$ is equivalent to $(SBS^{-1})^2=0$, so we can change bases as we please. Let us assume $B\ne0$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving Recurrence Relation problem I am trying to solve the recurrence relation:
$$G_n = \frac{1-G_{n-1}}{4}$$
$$G_0 = 0$$
$$G_1 = \frac{1}{4}$$
I am told that the answer is
$$G_n = \frac{1}{5}\left(1+\left(\frac{-1}{4}\right)^{n+1}\right)$$
I have found the characteristic equation to be
$$G_n = B\left(\frac{-1}{4}\r... | By rewriting the recurrence formula we have $G_n=\dfrac{1}{4}-\dfrac{1}{4}G_{n-1}$.
If we write the first elements of the sequence we get:
$$\begin{align*}
G_0&=0, G_1=\dfrac{1}{4} \\
G_2&=\dfrac{1}{4}-\dfrac{1}{4}G_1 =\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2\\
G_3&=\dfrac{1}{4}-\dfrac{1}{4}G_2 =\dfrac{1}{4}-\dfrac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerat... | Use equivalents:
$$\ln(1+u)\sim_0 u, \enspace\sqrt{1+u}-1\sim_0 \dfrac u2, \enspace \sin u\sim_0 u, \enspace \cos u\sim_0 1, \enspace\cot u\sim_0 \dfrac1u, \enspace \tan u\sim_0 u$$
hence $$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\sim_0 \frac{x^3\cfrac1{x^3}x^4}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
$f(z) = \frac{1}{z^2-2z+2}$ - $|f^{n}(0)| \leq n!$ - Cauchy inequality Let the function $f(z) = \frac{1}{z^2-2z+2}$. Show that for each integer $n \geq 0$, we have $|f^{n}(0)| \leq n!$.
Honnestly, I don't know how to do that problem, maybe in using the Cauchy inequality (complex analysis).
Is anyone could help me at th... | We'll use the following result from geometric series to compute the power series of $f$.
Let $c\neq 0$, the power series of $\frac{1}{z-c}$ around $0$ is $\frac{1}{z-c}=\frac{\frac{-1}{c}}{1-\frac{z}{c}}=\frac{-1}{c} (1+(\frac{z}{c})+(\frac{z}{c})^2+(\frac{z}{c})^3+...)$, radius of convergence is $c$.
$f(z)=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a Cartesian equation for the curve and identify it. $ r^2 \cos 2\theta = 1$
Find a Cartesian equation for the curve and identify it. $$ r^2 \cos 2\theta = 1$$
I'm confused by the $2\theta.$
I isolated $r^2$ to get $r^2 = \frac{1}{\cos2\theta}$
Now, normally if it was just a $\cos \theta$ I would multiply both si... | Since $\cos 2 \theta = 2 \cos^2 \theta -1$, then $$r^2 \cos 2 \theta = 2 r^2 \cos^2 \theta - r^2 = 2(r \cos \theta)^2 - r^2 = 2x^2 - (x^2 + y^2) = x^2 - y^2 ,$$ so your curve is $x^2 - y^2 = 1$ which is known to be a hyperbola having the lines $y = \pm x$ as asymptotes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simplifying Fractions involving negative numbers I want to simplify $$\frac{\frac{7}{-10} \times \frac{-15}{6}}{\frac{7}{-19} + \frac{-17}{-8}}$$
I really don't understand how to do this, or even how to start?
Negative numbers make it even harder for me to try and simplify it.
| A few important facts to know are the following. For all complex numbers $a,b, n$:
$$ \frac{-a}{b} = \frac{a}{-b} = -\left(\frac{a}{b}\right) $$
$$ \frac{a}{b} + \frac{c}{b} = \frac{a+c}{b} $$
$$ \frac{n\times a}{n \times b} = \frac{a}{b} $$
$$ \frac{a}{b} = a \times \frac{1}{b} $$
In particular, by the last identity a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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In how many ways can $10$ chocolates be distributed to $3$ people such that no one gets more than $4$? $10$ chocolates are distributed to $3$ people such that no one gets more than $4$ or $A,B,C \leq 4$. How many ways can this be done?
I know how to do this if the condition is every person gets at least one. Here the c... | drhab has provided you with an elegant solution to this problem based on the assumption that the chocolates are indistinguishable. I will make the same assumption.
We wish to distribute ten chocolates to three people so that no person receives more than four chocolates. The number of ways we can do this is equal to t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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positive definite of a matrix(diagonal decay) Is this matrix positive definite:
$$\begin{bmatrix}a & b &c & d\\b & b & c & d\\c & c & c & d\\d&d&d&d\end{bmatrix}$$
when $a>b>c>d>0$?
I tried several $a,b,c,d$, the result seems to be true, but I cannot prove it. Thanks!
| Thanks to @Friedrich Philipp and @Omnomnomnom's comments, I notice huge defects in my original proof. Instead, I would like to present an another
proof followed by @Omnomnomnom's method.
Denote the given matrix by $A$. Observe that
\begin{align}
A=\begin{bmatrix}a & b &c & d\\b & b & c & d\\c & c & c & d\\d&d&d&d\end{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Trigonometric inequality - x domain the following equality is given:
$$2 \sqrt2\sin{}x+\sqrt2\cos{x}=\sqrt{-\sin2x}$$
I managed to solve it, but I have a problem with this inequality to define x domain:$$2 \sqrt2\sin{}x+\sqrt2\cos{x}\ge0$$
Wolfram says:
$$x \in\left\langle 2\arctan(2- \sqrt{5}) + 2k \pi, 2\arctan(2+ \... | Since $\sqrt{(2\sqrt 2)^2+(\sqrt 2)^2}=\sqrt{10}$,
we have$$\begin{align}2\sqrt 2\sin x+\sqrt 2\cos x&=\sqrt{10}\left(\frac{2\sqrt{2}}{\sqrt{10}}\sin x+\frac{\sqrt 2}{\sqrt{10}}\cos x\right)\\&=\sqrt{10}\left(\frac{2}{\sqrt 5}\sin x+\frac{1}{\sqrt 5}\cos x\right)\\&=\sqrt{10}(\cos\theta\sin x+\sin \theta\cos x)\\&=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$ Integrate $$\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$$
I have tried to do Tangent substitution, but there is huge power.
What method is better to use here?
| $$
\begin{aligned}
\int_0^{2 \pi} \frac{2}{\cos^6 x + \sin^6 x} dx
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^6 x + \sin^6 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{(\cos^2 x + \sin^2 x)(\cos^4 x-\cos^2 x\sin^2 x+\sin^4 x)} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^4 x-\cos^2 x\sin^2 x+\sin^4 x} dx\\
&=8\int_0^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do i find sequences using linear algebra? My task is this;
Find two sequences $\{x_n\}$ and $\{y_n\}$ such that $$x_{n+1} = x_n + 3y_n\\ y_{n+1} = 2x_n + 2y_n$$
When $x_0 = 5, y_0 = -5$.
My work so far (not sure if this is the right approach):
We are interested in finding eigenvectors and values to this system. Le... | So now you know
$$A=\begin{pmatrix}1&3\\2&2\end{pmatrix}=\overbrace{\begin{pmatrix}3&1\\\!\!-2&1\end{pmatrix}}^{=P}\begin{pmatrix}\!\!-1&0\\0&4\end{pmatrix}\overbrace{\begin{pmatrix}\frac15&\!\!-\frac15\\\frac25&\frac35\end{pmatrix}}^{=P^{-1}}\implies$$
$$A^n=P\begin{pmatrix}(-1)^n&0\\0&4^n\end{pmatrix}P^{-1}=\frac15\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Finding when $y=1+2\sin(x)$ and $y=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$ are equal How would you solve the following equation?
$$1 +2\sin(x) = 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \quad\text{for }−2\pi\leq x \leq 2\pi.$$
Steps I tried:
\begin{align}
1+2\sin(x) &= 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \\
2\sin(x... | HINT:
Avoid squaring as it immediately introduces extraneous root.
Using $\sin x=\sin2\cdot\dfrac x2=2\sin\dfrac x2\cos\dfrac x2$
$$1+2\sin x=2\left(\sin\dfrac x2+\cos\dfrac x2\right)$$
$$\iff\left(2\sin\dfrac x2-1\right)\left(2\cos\dfrac x2-1\right)=0$$
Hope you can take it from here!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maclaurin serie of $\int_0^x\frac{sin(t)}{t}$ If $f(x)=\int_0^x\frac{\sin(t)}{t}$. Show that
$$f(x)=x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+...$$
Calculate f(1) to three decimal places.
Would you mind showing how to build this Maclaurin serie?
| $$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
Now, we divide by $x$ to get that
$$\frac{\sin x}{x} = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$
Now, we write
$$f(x)=\int_0^x\frac{\sin tdt}{t}=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728891",
"timestamp": "2023-03-29T00:00:00",
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Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
| Here's yet another way. First write: $$\int x^3\sqrt{4-x^2} \, dx = \int x^2 \sqrt{4-x^2} \, \cdot x \, dx$$
Now let $u = 4-x^2$, so then $du = -2x\, dx$ (meaning $x\,dx = -\frac{1}{2}du$), and $x^2 = 4 - u$. Then you get:
\begin{align}
\int x^2 \sqrt{4-x^2} \, \cdot x \, dx
&= \int (4-u)\sqrt{u} \, \cdot \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Possible number of terms in an Arithmetic Progression The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is
$a)$ $3$
$b)$ $4$
$c)$ $5$
$d)$ $6$
My approach : I used the formula for the first $n$ terms of ... | \begin{align}
n^2 +(a-1)n - 153 &= 0 \\
(n+u)(n-v) &= 0 \\
n^2 + (u-v)n - uv &= 0
\end{align}
So you need two positive integers, say $u>v>0$, such that $uv=153$ and $u-v=a-1$. Which will have solution $n=v$.
There aren't that many possibilities
\begin{array}{|rr|r|rr|}
\hline
u & v & u-v & a & n \\
\hline
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How to solve the congruence $y^{31}\equiv 3 \mod{100}$ $\phi (100) = 40$
Hence:
$y^{31}\equiv y^{-9} \equiv 3\mod{100}$
$y^{9}\equiv 67 \mod{100}$
However I do not know where to go from here.
| This is a kind of tedious solution:
$$x^{31}\equiv 3\pmod{100}\iff \begin{cases}x^{31}\equiv 3\pmod{25}\\ x^{31}\equiv 3\pmod{4}\end{cases}$$
$$\iff \begin{cases}x^{31}\equiv 3\pmod{25}\\ x\equiv 3\pmod{4}\end{cases}$$
Clearly $\gcd(x,25)=1$, so by Euler's theorem:
$$\iff \begin{cases}x^{11}\equiv 3\pmod{25}\\ x\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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integral $\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x$
How to prove that $$0.78<\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x<0.93$$
approch: $x^2<\sqrt{1+x^4}<x^2+\frac{1}{2x^2}$
Any hint would be appreciated.
| I know this doesn't constitute as a proof, but I thought the following was interesting. It turns out we can use the Beta function to express your integral in terms of the Gamma function by taking advantage of the fact that
\begin{align*}
B(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dt.
\end{align*}
First, substitu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Prove that $36|n^{12k-6}-n^6$ Prove that if gcd(n,6)=1 and k>0, then $36|n^{12k-6}-n^6$.
Idea: I want to show that $2|n^{12k-6}-n^6$ and $3|n^{12k-6}-n^6$.
For the first part, since gcd(n,6)=1, n must be an odd number, so it is easy to show that $n^{12k-6}-n^6$ is divisible by 2. I am stuck by the second part and i ... | Let $n$ and $k>0$ integers. Notice that
\begin{align*}
n^{12k-6}-n^6&=\left(n^{6k-3}-n^3\right)\left(n^{6k-3}+n^3\right)\\
&=\left(n^{2k-1}-n\right)\left(n^{4k-2}+n^{2k}+n^2\right)\left(n^{2k-1}+n\right)\left(n^{4k-2}-n^{2k}+n^2\right)\\
&=n^6\left(n^{2k-2}-1\right)\left(n^{4k-4}+n^{2k-2}+1\right)\left(n^{2k-2}+1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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None of $3,5,7$ can divide $r^4+1$ Let $n=r^4+1$ for some $r$. Show that none of $3,5,$ and $7$ can divide $n$.
I am thinking to use a corollary that "each prime divisor p of an integer of the form $(2m)^4+1$ has the form $8k+1$", but I failed. Anyone can give some hint?
| Use Fermat little theorem. If $r $ isn't divisible by 3, $r^2 \equiv 1 \mod 3$. If $r$ is divisible by 3 then $r^2 \equiv 0 \mod 3$. So $r^4 \not \equiv -1 \mod 3$ so $r^4 + 1$ is not divisible by 3.
Similarly $r^4 \equiv 1,0 \mod 5$ so $r^4 + 1$ is not divisible by 5.
If $r$ is not divisible by 7 then $r^6 \equiv 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
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How to find whether equation of angle bisector represents the obtuse or acute angle bisector of two given straight lines? Two lines: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are given. I know that the equation of its bisectors is ${a_1x + b_1y + c_1 \over \sqrt{(a_1^2 + b_1^2)}} = \pm {a_2x + b_2y + c_2 \ove... | We have two lines :
$$L_1 : a_1x+b_1y+c_1=0,\quad L_2 : a_2x+b_2y+c_2=0$$
and the angle bisectors :
$$L_{\pm} : \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$
If we let $\theta$ be the (smaller) angle between $L_+$ and $L_1$, then we have
$$\cos\theta=\frac{\left|a_1\left(\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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How to evaluate $\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\... | $x^{\frac{1}{x}}-1 = \exp\left(\frac{\log x}{x}\right)-1 = \frac{\log x}{x}+O\left(\frac{\log^2 x}{x^2}\right)$ for $x\to +\infty$, so the denominator behaves like $x^{1/2}\log(x)+O(\log x)$. On the other hand:
$$ \sqrt{1+\frac{x^3}{x+1}}-x = \frac{1+\frac{x^3}{x+1}-x^2}{x+\sqrt{1+\frac{x^3}{x+1}}}=\frac{x+1+x^3-x^2(x+... | {
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Determine if the set of vectors are linearly independent or linearly dependent $$u=(1,1,1,3)$$
$$v=(1,2,1,3)$$
$$w=(1,2,3,2)$$
I need help understanding the method of how to do solve this type of problem. I understand that the concept is just to find out if the constants $k_n$ in $k_1u+k_2v+k_3w=0$ all equal zero or no... | To answer where those $a$'s come from, and what they represent, when you construct an augmented matrix like you did, what you really have is,
$$[A|0] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 1 & 2 & 2 & 0 \\ 1 & 1 & 3 & 0 \\ 3 & 3 & 2 & 0 \\ \end{array} \right].$$
Which is equivalent to saying,
$$A\vec{x} = \vec{... | {
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Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:
*
*$f(x)-1$ is divisible by $(x-1)^3$
*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f ... | Since $x^2$ and $(x-1)^2$ divide $p'(x)$ we have $$p'(x) =ax^2(x-1)^2 =ax^4-2ax^3+ax^2,$$ thus $$p(x) = {a\over 5}x^5-{a\over 2}x^4+{a\over 3}x^3+c$$
Since $p(0)=0$ we have $c=0$ and since $p(1)=1$ we have ${a\over 5}-{a\over 2}+{a\over 3}=1$ so $a=30$.
| {
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If $x+y+z=0$, prove that $\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}=1$ A problem in my homework had asked me:
When $x+y+z=0$, evaluate$$\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}$$
Without too much difficulty, one can see that the value should be $1$ using $(x,y,z)=(1,0,-1)$.
I dec... | HINT:
$$2x^2+yz=2(y+z)^2+yz=(2y+z)(y+2z)$$
Now $2y+z=x+y+z+y-x=y-x$
| {
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If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are? If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are?
I thought that $(x+1)/(x+2)=-x$.... | Don't forget the possibility $(x+1)/(x+2)=+x$.
Setting $\frac{x+1}{x+2}=-x$ leads to $x+1=-x^2-2x$, i.e., $x^2+3x+1=0$, $x=\frac{-3\pm\sqrt{5}}2$. Setting $\frac{x+1}{x+2}=+x$ leads to $x+1=x^2+2x$, i.e., $x^2+x-1=0$, $x=\frac{-1\pm\sqrt{5}}2$.
All four values are well within $(-5,5)$.
| {
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$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}} < 2$
If ${{a}_{1}},{{a}_{2}},\ldots ,{{a}_{n}}$ are distinct odd natural
numbers not divisible by any prime greater than 5, then show that
$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}}
< 2$.
First I have noted that $a_i... | Look at the product
$$\left(1+\frac{1}{3}+\frac{1}{3^2}+...\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+...\right) = \frac{15}{8}$$
and notice it contains every possible $\frac{1}{3^a 5^b}$, hence it gives you upper bound for all of your finite sums.
| {
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If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $$M=abc+abd+acd+bcd-abcd$$
I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$. ... | Let $\mathcal{T}$ be the regular tetrahedron of side $\sqrt{2}$ with vertices at
$\begin{cases}
A &= (+1,+1,+1),\\
B &= (+1,-1,-1),\\
C &= (-1,+1,-1),\\
D &= (-1,-1,+1).
\end{cases}$
For every point $X \in \mathcal{T}$, its baricentric coordinate is an unique $4$-tuple $(a,b,c,d)$ such that
$$\begin{cases}
a, b, c, d \... | {
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"question_score": "9",
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How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?... | So your problem reduces to
$$\int\frac{x^2dx}{(x^2+1)^2}$$
Integrate by parts: let $u=x$ and $dv=\frac{x}{(x^2+1)^2}dx$. Then $du=dx$ and $v=-\frac{1}{2}(x^2+1)^{-1}$ so we have
$$-\frac{1}{2}x(x^2+1)^{-1}+\frac{1}{2}\int\frac{dx}{x^2+1}$$
and that last integral may jump out at you as a particular trig derivative
| {
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Sum of number-of-divisors function equals $\sum_{j=1}^{n} \lfloor n/j \rfloor$. I am trying to prove the identity
$$t(1) + t(2) + \cdots + t(n) = \Big\lfloor \dfrac{n}{1} \Big\rfloor + \Big\lfloor \dfrac{n}{2} \Big\rfloor + \cdots + \Big\lfloor \dfrac{n}{n} \Big\rfloor,$$
where $t(j)$ is the number of divisors of $j$. ... | For the sake of completeness I would like to point out that this is
usually done by induction. We seek to show that
$$\sum_{k=1}^n \tau(k)
= \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
It holds for $n=1:$ $$\tau(1) = \lfloor 1\rfloor.$$
In the induction step we start from
$$\sum_{k=1}^n \tau(k)
= \sum_{k=... | {
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arithmetic problem concerning equilateral triangle An equilateral triangle exists with vertices $(0,0), (a,11)$ and $(b,37)$. Using the distance formula three times, I eventually arrive to:
$$a^2 + 121 = b^2 + 1369 = a^2 + b^2 - 2ab + 676$$
The only problem is that since none of the equations are actually equal to anyt... | You actually have two equations: for instance,
\begin{align}
a^2+121&=b^2+1369\\
b^2+1369&=a^2+b^2-2ab+676.
\end{align}
It is easy to see that $b=0$ gives no solution: the second equation becomes $121=676$. So we assume $b\ne0$. We rewrite the equations as
\begin{align}
a^2-b^2&=1248\\
a^2-2ab&=693.
\end{align}
If yo... | {
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Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin ... | The general method of solving $A\cos x+B\sin x=C$ when $A^2+B^2\ne 0$ is to take any $y$ such that $$(1)\quad A/\sqrt {A^2+B^2}=\cos y\;\land \;B/\sqrt {A^2+B^2}=\sin y.$$ $$\text {Then } C/\sqrt {A^2+B^2}=\cos y \cos x+\sin y \sin x =\cos (x-y).$$ So let $x=y+z$ for any $z$ such that $\cos z=C/\sqrt {A^2+B^2}.$
If w... | {
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"source": "stackexchange",
"question_score": "2",
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calculate arbitrary points from a plane equation I understand how one can calculate a plane equation (ax+by+cz=d) from three points but how can you go in reverse?
How can you calculate arbitrary points from a plane equation?
| From your comment I finally understood what you are looking for:
If you have a plane defined by $a x + b y + c z = d$ then you also have the following properties:
*
*Plane normal direction:
$$\hat{n} = \begin{pmatrix}
\frac{a}{\sqrt{a^2+b^2+c^2}} \\
\frac{b}{\sqrt{a^2+b^2+c^2}} \\
\frac{c}{\sqrt{a^2+b^2+c^2}... | {
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"source": "stackexchange",
"question_score": "5",
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Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$
$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we ... | Firstly, let's say $f(x) = \sqrt{a+\sqrt{a-x}} - x$. If $f$ has real solutions, then it must hold that $x \in [0,a]$. Now, taking the derivative of $f$, we have:
$$f'(x) = -1-\dfrac{1}{4\sqrt{a+\sqrt{a-x}}\cdot (\sqrt{a-x})}.$$
Clearly, $f'(x)<0, \forall x \in [0,a).$ Thus, $f$ is strictly decreasing in $[0,a]$.
Also,... | {
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Is it true that $log(i) = \frac\pi2i$ ? If so, are both of these legitimate proofs? They seem too beautiful not to be... Sorry if this is a naive question. I have not yet taken any upper level math courses involving complex numbers. However, in preparation for those courses, together with utilizing the knowledge that m... | It depends on how you define "logarithm" in the first place. But the simplest and most straightforward approach would be to define
Definition. $\log x$ means the number $y$ such that $e^y=x$.
With this definition the first two lines of your first proof is a complete, direct proof that $\log i = \frac12\pi i$.
The onl... | {
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Solve $\log_9(x-4) - \log_9(x-8)= \frac{1}2$
Solve $\log_9(x-4) - \log_9(x-8)= \frac{1}2$
$(x-4) - (x-8)= 9^\frac{1}2$
$(x-4) - (x-8)= 3$
The answer is 10 but I am not sure how that was obtained.
| $\log_9(x-4)-\log_9(x-8)=\log_9(\frac{x-4}{x-8})=\frac{1}{2}$
$\frac{x-4}{x-8}=9^{1/2}=3$
$x-4 = 3(x-8)=3x-24$
$x-4=3x-24$
$2x=20$
$x=10$
| {
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| Note this convenient fact: $$(a+b+c)^3-(a^3+b^3+c^3) = 3(a+b)(b+c)(c+a)$$
Since we know the left-hand side is zero, we also know the right-hand side is zero; this tells us that at least one of these is true:
$$a+b=0 \qquad b+c=0 \qquad c+a=0 \tag{$\star$}$$
We'll go ahead and take the middle one (the others will work o... | {
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Find all integral solutions for the Diophantine Equations $x^4 - x^2y^2 + y^4 = z^2$ and $x^4 + x^2y^2 + y^4 = z^2$.
Find all integral solutions for the Diophantine Equations $$x^4 - x^2y^2 + y^4 = z^2$$ and $$x^4 + x^2y^2 + y^4 = z^2$$
I basically think that to solve these equations we need to use the fact that all ... | COMMENT.- I feel but I have no proof that there are only the trivial solutions $(x,y,z)= (t,0,t^2),(0,t,t^2)$. I give here an outline of what could perhaps lead to a proof.
$$x^4-x^2y^2+y^4=z^2\iff (x^2-y^2)^2+x^2y^2=z^2\qquad (1)$$ Hence, as it is well known,
$$\begin{cases}x^2-y^2=t^2-s^2\\xy=2ts\\z=t^2+s^2\end{case... | {
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What is the coefficient and constant term in the following sequence defined recursively? Let $f_n(x)$ be a sequence of polynomials defined inductively as
$f_1(x) = (x - 2)^2$
$f_{n+1}(x) = (f_n(x) - 2)^2$ $; n \ge 1$
Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then
... | A straightforward approach is to see how $a_n$ and $b_n$ change when $n$ is increased by $1$.
The constant term of $f_1(x)$ is $4$. Suppose that $f_n(x)$ for some $n$; then $f_n(x)=xg_n(x)+4$ for some polynomial $g_n(x)$ (why?), so
$$f_{n+1}(x)=\left(\big(xg_n(x)+4\big)-2\right)^2=\big(xg_n(x)+2\big)^2=x^2\big(g_n(x)\b... | {
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If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$ If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$
I substituted $B=k-b,C=k-c,A=k-a$ and plugged them to get a quadratic of $k$ which I had to show positive .SO, the discriminant ... | From
$$
ABC = (k-a)(k-b)(k-c) \\
= k^3 - (a+b+c)k^2 + (ab + bc + ca)k - abc \\
= k \bigl( k^2 - (a+b+c)k + (ab + bc +ca) \bigr) - abc \\
= k \bigl( k^2 - (a B + b C + c A ) \bigr) - abc
$$
it follows that
$$
a B + b C + c A = k^2 - \frac{abc + ABC}{k}
\le k^2
$$
Equality holds if and only if
$$
abc = ABC = 0
$... | {
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What is the probability to fill rows of a cinema hall? This is the problem I'm trying to solve, but I'm not sure I'm on the correct path! would appreciate your feedback guidence and help.
So the problem is: there're 3 rows in a cinema hall. the first one is of 5 chairs, the second one with 7 chairs and the third one wi... | What matters here is which seats are occupied, not who sits in which seat. Therefore, we can take our sample space to be the
$$\binom{5 + 7 + 10}{18} = \binom{22}{18}$$
ways the patrons can occupy $18$ of the $22$ available chairs.
What is the probability the first row is full?
If the first row is full, $5$ of th... | {
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Prove the inequality $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$ $a, b, c, d$ are positive reals.
How would I prove the inequality
$$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4$$
I have tried using the rearrangement inequality with $a\leq b\leq c\leq d$
But it does... | By C-S
$$\sum_{cyc}\frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\geq$$
$$\geq\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{b+c+d+a}=4.$$
| {
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Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps:
*
*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
*${\displays... | This would have been lots easier with $x=\cosh\theta$:
$$\begin{align}\int x^3\sqrt{x^2-1}dx&=\int\cosh^3\theta\sinh^2\theta\,d\theta\\
&=\int(\sinh^2\theta+1)\sinh^2\theta\cosh\theta\,d\theta\\
&=\frac15\sinh^5\theta+\frac13\sinh^3\theta+C\\
&=\frac15(x^2-1)^{5/2}+\frac13(x^2-1)^{3/2}+C\\
&=\left(\frac15x^2-\frac15+\f... | {
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Factoring a 4th degree trinomial I am trying to factor $3x^4-8x^3+16$, but I have no idea how to even start. I put into Wolfram Alpha, and it said that the answer was $(x-2)^2 (3 x^2+4 x+4)$. How would you factor something like this by hand? Is there any way without using the quartic formula in this case?
| You can factor it if you know its roots. In general if $r$ is a root of a polynomial $f(x)$, then $(x-r)$ divides it.
For example, if you see that $f(x)=3 x^4−8 x^3+16$ has the root $2$, that is, $3(2)^4-8(2)^3+16=48-64+16=0$, then you know that $(x-2)$ divides it. You can then use long division to show that $3 x^4−8 ... | {
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Cauchy but not rapidly Cauchy I want to show that the sequence $\{\frac{(-1)^n}{n}\}$ is Cauchy but not rapidly Cauchy. Here is the work I done so far. I am curiously if I made any errors.
Consider the normed linear space $\mathbb{R}$ with the norm of the absolute value. Let
$$
\{f_n\}=\left\{\frac{(-1)^n}{n}\right\}=... | Your argument looks correct, although it's a bit long and drawn out. I might argue this way: From your definition we have $f_k$ rapidly Cauchy iff
$$\sum_{k=1}^{\infty} |f_{k+1} - f_k|^{1/2} < \infty.$$
(Never heard of "rapidly Cauchy" before). Now
$$f_{k+1} - f_k = \frac{(-1)^{k+1}}{k+1} - \frac{(-1)^{k} }{k} = (-1)^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use generating functions to determine the number of ways Use generating functions to determine the number of different ways $12$ identical action figures can be given to $5$ children so that each child receives at most $3$ action figures
So far I have come up with the generating function of: $$(1 + x + x^2 + x^3)^5$$
T... | It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
[x^{12}]&(1+x+x^2+x^3)^5\\
&=[x^{12}]\left(\frac{1-x^4}{1-x}\right)^5\tag{1}\\
&=[x^{12}]\left(\sum_{k=0}^{5}\binom{5}{k}(-1)^kx^{4k}\right)\left(\sum_{n=0}^\infty\binom{-5}{n}(-x)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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finding the number of factors $2^{15}\times3^{10}\times5^6$ The number of factors of $2^{15}\times3^{10}\times5^6$ which are either perfect square or perfect cubes(or both)
I don't know how to start this even!
Plz solve this!
| A factor can be described by a tuple of exponents
$f \in
\{ 0, \dotsc, 15 \}
\times
\{ 0, \dotsc, 10 \}
\times
\{ 0, \dotsc, 6 \}
$.
As
$$
(2^{f_1} 3^{f_2} 5^{f_3})^{1/n} =
2^{f_1/n} \, 3^{f_2/n} \, 5^{f_3/n}
$$
we need $n$ to divide the exponents $f_i$.
The perfect squares ($n=2$) are combined from
$
S = \{ 0, 2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Evaluate the improper integral with residues.
$$\int_0^{\infty} \frac{x^2+1}{x^4+1}dx$$
What i've found are the singularities at: $e^{\pi/4+\pi/2n}$ for $n=0,1,2,3$. But i'm unsure how to calculate the integral since I don't want to include the singularity at $x_2=\frac{-1+i}{\sqrt{2}}$ since then I would be calculat... | Let $I$ be the integral defined by
$$I=\int_0^\infty\frac{x^2+1}{x^4+1}\,dx \tag 1$$
We proceed to evaluate the integral in $(1)$ by analyzing the contour integral $J$
$$J=\oint_C \frac{z^2+1}{z^4+1}\,dz$$
where $C$ is the quarter circle in the first quadrant, centered at the origin, with radius $R$. Then, we have
$$J... | {
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"source": "stackexchange",
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$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
$$x^2y+y^2z+z^2x < \frac12$$
This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.... | Let $x=0$.
Then $y^2+z^3=1 \Rightarrow z^3=1-y^2$
Seek to maximise $p=y^2z=y^2(1-y^2)^{\frac 13}$
WLOG let $u=y^2$ so that $p=u(1-u)^{\frac 13}$
$\frac {dp}{du}=(1-u)^{\frac 13}-u{\frac 13}(1-u)^{-\frac 23}$
$\frac {dp}{du}=\frac 13(3-4u)(1-u)^{-\frac 23}$
Maximum is at $u=\frac 34$
Thus maximum is $p=\frac 34(\frac 14... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a ... | This is too long to fit into a comment. I wanted to ask a question about my proof on this problem. (It might help discover another proof)
This proof has a flaw -- From $AB \ge C$ and $A \ge D$, I wrongly implied that $DB \ge C$.
Is there a way to slightly modify it such that it can prove the statement or is it comple... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
"answer_count": 8,
"answer_id": 5
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If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{... | Minimizing $1 - \frac{1}{(t - \frac 4t +1)^2 + 2}$ is equivalent to minimizing $(t - \frac 4t +1)^2 + 2$. But obviously the minimal value for this is $2$, as the square of a number is always bigger than $0$. To find the value which minimizes it just solve $t- \frac4t + 1 = 0$
| {
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"url": "https://math.stackexchange.com/questions/1776723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Inequality $\sqrt{1-abc}(3-a-b-c)\geq |(1-a)(1-b)(1-c)|$ Let $a,b,c>0$ and $a^2+b^2+c^2+abc=4$. Prove that
$$\sqrt{1-abc}(3-a-b-c)\geq |(1-a)(1-b)(1-c)|.$$
Note that equality holds trivially when $a=b=c=1$, but it is not the only case. It also holds more generally when $a=x, b=x, c=2-x^2$, where $0<x^2<2$.
| squaring the given inequality we get
$$- \left( ac+b-2 \right) \left( bc+a-2 \right) \left( ab+c-2 \right) \geq 0$$
from $$a^2+b^2+c^2+abc=4$$ we get
$$(b-2)(b+2)=-(a^2+c^2+abc)$$
$$b-2=-\frac{a^2+c^2+abc}{b+2}$$
$$ac+b-2=ac-\frac{a^2+c^2+abc}{abc}$$
$$ac+b-2=-\frac{(a-c)^2}{b+2}<0$$
etc. Thus our product above is no... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 20... | A bit algebra shows that the inequality is equivalent to
$$
5 \left(5 a^5 \left(13 b^2+5 c^2\right)+13 a^4 \left(5 b^3-13 b^2 c-5 b c^2-5 c^3\right)+a^3 \left(-65 b^4+144 b^2 c^2+65 c^4\right)+a^2 \left(25 b^5-65 b^4 c+144 b^3 c^2+144 b^2 c^3-169 b c^4+65 c^5\right)-13 a \left(13 b^4 c^2+5 b^2 c^4\right)+5 b^2 c^2 \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that,
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how can I simplify th... | Explanation for your question here
$2arctan(\frac{1}{3})$ = $arctan(\frac{3}{4})$ as you already understand
$arcsin(x) = arctan(\frac{x}{\sqrt{1-x^2}})$Inverse trigonometric functions
This simplifies $arcsin(\frac{1}{5\sqrt2}) = arctan (\frac{1}{7})$ which makes
$2arcsin(\frac{1}{5\sqrt2}) = 2arctan(\frac{1}{7})$
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
How can I prove this sum?
$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$
My try:... | Let's define the sum we are looking for with $S$.
Furthermore, notice that ($k \in [1,4] $)
$$
S_k=\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+k)^2}=\sum_{n=0}^{\infty}\frac{1}{(5(2n)+k)^2}-\sum_{n=0}^{\infty}\frac{1}{(5(2n+1)+k)^2}=\\
\frac{1}{5^22^2}\left[\sum_{n=0}^{\infty}\frac{1}{(n+k/10)^2}-\sum_{n=0}^{\infty}\frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$
I was thinking of using mathematical induction for this. That is,
We prove by induction on $n$. The cas... | If you want to do a proof by induction.
you have covered the base case.
Inductive hypothesis:
$\sum_\limits{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$.
We need to show that:
$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
thoughts:
$\ln(2n) + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve this equation algebraically Solve the following simultaneous equations on the set of real numbers:
\begin{cases}x^2 + y^3 = x+1 \\ x^3+y^2=y+1\end{cases}
Thanks for helping!
| Hint:
Eliminate one of the unknowns in a way to get a polynomial.
From the second equation,
$$y^2-y+1=2-x^3$$ and multiplying by $y+1$ and using the first equation,
$$(2-x^3)(y+1)=(y^2-y+1)(y+1)=y^3+1=2+x-x^2,$$
and
$$y=\frac{2+x-x^2}{2-x^3}-1=\frac{x-x^2+x^3}{2-x^3}.$$
Then
$$\left(x-x^2+x^3\right)^3=(1+x-x^2)(2-x^3)^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\sum_{n=0}^\infty(n+2)x^n$ I am trying to calculate $\sum_{n=0}^\infty(n+2)x^n$.
I was thinking it is like the second derivative of $x^{n+2}/(n+1)$ but I am not sure how to go about calculating it. Any hints?
| Hint: It is sometimes convenient to work with operators. Since
\begin{align*}
\left(x\frac{d}{dx}\right)\sum_{n=0}^\infty a_nx^n=x\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=0}^\infty na_nx^{n}
\end{align*}
we consider $xD:=\left(x\frac{d}{dx}\right)$ as operator and obtain for $k\geq 0$
\begin{align*}
\left(xD\right)^k\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781930",
"timestamp": "2023-03-29T00:00:00",
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Number of polynomials which are divisible by $x+1$ Let $a,b,c,d$ be four integers (not necessarily distinct) in the set ${1,2,3,4,5}$ . The number of polynomials $f(x)=x^4+ax^3+bx^2+cx+d$ which are divisible by $x+1$ are:
$(A)$ Between 55 and 65
$(B)$ Between 65 and 85
$(C)$ Between 86 and 105
$(D)$ More than 105
I see... | Your approach of looking for the number of solutions to $1-a+b-c+d=0$ is excellent.
Let $f(n)$ be the number of ways of choosing two (possibly identical) numbers $h,k$ from $\{1,2,3,4,5\}$ so that $h-k=n$. It is easy to check that $f(n)=1,2,3,4,5,4,3,2,1$ for $n=-4,-3,-2,\dots,3,4$ and 0 for all other $n$.
So $1-a+b$ m... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square.
Since $n^2+n+43$ is odd,if it's a perfect square it can be written as: $8k+1$,then:
$$n^2+n+43=8k+1\Rightarrow\ n^2+n+42=8k\ \Rightarrow\ n(n+1)\equiv6\pmod8$$
So $... | Trying to solve: $n^2+n+43=m^2$, multiply by $4$ and complete the square, and you get:
$$(2n+1)^2+4\cdot 43-1 = (2m)^2.$$
Subtract and you are trying to solve $(2m)^2-(2n+1)^2=171$.
The difference of two squares means you need:
$$(2(m-n)-1)(2(m+n)+1)=171=9\cdot 19$$
So you need to factor $171$ as two odd numbers whose ... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $n^2+11n+2$ is not divisible by $113^2$ ( n is integer) Show that $n^2+11n+2$ is not divisible by $113^2$ ( n is integer)
It's obvious that if we show $113$ doesn't divide $n^2+11n+2$ we are done...
| A good start would be to find if the equation $n^2 + 11n + 2 \equiv 0$ has any solutions modulo 113. We can do this by completing the square, so we want to find $a,b \in \mathbb{Z}/113\mathbb{Z}$ such that
$$
n^2 + 11n + 2 \equiv (n-a)^2 + b \pmod{113}.
$$
By expanding we see that $-2a \equiv 11 \pmod{113}$, so $a \eq... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with t... | Michael's grouping answer works in general. Note that the group with $n$ terms is of the form
$${1\over k+1}+{1\over k+2}+\cdots+{1\over k+n}$$
so the next group is
$${1\over k+n+1}+{1\over k+n+2}+\cdots{1\over k+2n}+{1\over k+2n+1}$$
It's easy to see that
$${1\over k+j}-{1\over k+n+j}={n\over(k+j)(k+n+j)}\gt{n\over(k... | {
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Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equ... | $\mathbb{Z}/(25\mathbb{Z})^*$ is a cyclic group generated by $2$, hence
$$ \sum_{\substack{1\leq n \leq 25\\ 5\nmid n}}n^4 \equiv \sum_{r=1}^{20}2^{4r}\equiv \frac{16^{24}-16^4}{16-1}\equiv 20\pmod{25} $$
and
$$ \sum_{\substack{1\leq n \leq 25\\ 5\nmid n}}n^8 \equiv \sum_{r=1}^{20}2^{8r}\equiv \frac{256^{168}-256^8}{25... | {
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"timestamp": "2023-03-29T00:00:00",
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The function $h(x) = x^3 + bx^2 + d$ has a critical point at $(2, -4)$. Determine the constants $b$ and $d$ and find the equation of $h(x)$. The function $h(x) = x^3 + bx^2 + d$ has a critical point at $(2, -4)$. Determine the constants $b$ and $d$ and find the equation of $h(x)$. Please help me answer this question.
| The critical points occur where the derivative is equal to zero.
\begin{align*}
h(x) & = x^3 + bx^2 + d\\
h'(x) & = 3x^2 + 2bx\\
& = x(3x + 2b)
\end{align*}
Setting the derivative equal to zero yields
\begin{align*}
x & = 0 & 3x + 2b & = 0\\
& & 3x & = -2b\\
& & x & = -\frac{2b}{3}
\end{align*}
We know that a... | {
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Radical equation $\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$ The Question: $\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$
Only thing I can take from this is that $x^2 -1=(x+1)(x-1)$, but I don't think that would help in any way.
I know the answer, but I don't know how to use that to work backwards.
Answer:
$x= \frac54$
| You have already seen a solution which is much more clever. But let us try whether this can be done with repeated squaring - which is usually one of the first thing which comes to mind in problems like this.
Do not forget that when you square an equation, you can get an extraneous solutions. (See here, here or here.)
W... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$? On the Wikipedia Page about Pentagons, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$
My question is: How would you justi... | You surely know that $z=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}$ is the root in the first quadrant of $z^5-1=0$ so it satisfies
$$
z^4+z^3+z^2+z+1=0
$$
that can also be written as
$$
z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0
$$
or, by noting that $z^2+1/z^2=(z+1/z)^2-2$,
$$
\left(z+\frac{1}{z}\right)^2+\left(z+\frac{1}{z}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$ I have this equation from this paper (Q.63)
Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.
My attempt-
$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$
So, $x=6$ and $x=-10$ makes the power $0$ an... | You missed some cases.
The cases are: $\text{number}^0,1^\text{number},{-1}^{\text{even}}$
You have only considered the first case.
*
*For first case, $$x^2+4x-60=0$$ the roots are $6,-10$.
*For the second case, $$x^2-5x+5=1$$ the roots are $1,4$.
*For the third case, $$x^2-5x+5=-1$$ $x$ can be $2$ or $3$. If you... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Find the limit $\lim_{x \to 1} \left(\frac{p}{1-x^p} - \frac{q}{1-x^q}\right) $ $p ,q >0$ I Know series expansion and L'Hospital's rule . But here both of them are not of any help.
| Here's a rather standard series expansion method: rewriting $x$ as $1+(x-1)$, and using Newton's generalised binomial expansion,
\begin{align}
\frac{p}{1-x^p}-\frac{q}{1-x^q} & = \frac{p}{1-(1+(x-1))^p}-\frac{q}{1-(1+(x-1))^q} \\
& = \frac{p}{1-(1+p(x-1)+\frac{p(p-1)}{2}(x-1)^2+O(|x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
How can I compute $\tan(.5\arctan(x))$? The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.
| You have to compute $y=\tan\left( \frac{\arctan{(x)}}{2} \right)$. Let's call $\theta=\arctan(x)$, so you have to compute $y=\tan\left(\frac{\theta}{2}\right)$.
You should write $\tan\left(\frac{\theta}{2}\right)$ in terms of $\tan\left(\theta\right)$, say you use:
$$\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the condition such that the roots of the polynomial are in AP
$f(x)=x^3+3px^2+3qx+r$ has roots in AP.Find the relation between $p,q$ and $r$.
[Answer:$-2p^2-3pq+r=0$]
My attempt:-
Taking $d$ as the common difference of the roots in AP we have $f(x)=(x-(a-d))(x-a)(x-(a+d))=x^3-3x^2a+(3a^2-d^2)x-a(a^2-d^2)$.
Compa... | As you suggest, take the roots as $a-d,a,a+d$, then you get $p=-a,3q=3a^2-d^2,r=-a^3+ad^2$. Substituting the first and third in the second we get $$-2p^3+3pq-r=0$$ Check: take roots 1,2,3. The equation is $(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$, so $p=-2,3q=11,r=-6$. The relation $-2p^3+3pq-r=0$ holds. The relation given in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
how to verify $\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x))}$? How would I verifty the following trig identity?
$$
\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}
$$
I am not sure how to start.
| $$\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}\Longleftrightarrow$$
$$\frac{\sec^2(x)}{\sec^2(x)}\cdot\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}\Longleftrightarrow$$
$$\frac{\tan(x)}{\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)}=\frac{\tan(x)}{1-\tan^2(x)}$$
Now, use $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $f$ is differentiable in $(a,b)$ then $\frac{1}{f}$ is differentiable at $(a,b)$, provided $f(a,b)\neq0$ "Suppose that $f$ is a differentiable function at $(a,b)$. Prove that $\frac{1}{f}$ is differentiable in $(a,b)$, provided $f(a,b)\neq0$"
We were given the following definition of differentiability: a function is... | To be able to prove differentiability, you first need to guess what the linear map is. Usually one does this by using differentiation rules (derivative of a product, chain rule) that have appropriate extensions to the multivariable setting. But let us work this directly.
A decent rule of thumb is to mimic the one-varia... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges.
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1... | $$
\sum_{n=1}^N \frac{1}{n+2} =
\sum_{n=3}^{N+2} \frac{1}{n}
$$
So your series differs just by finite many terms from the harmonic series, which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$
Prove that if $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$.
I am not sure how to prove this statement, but it seems that from $9 \mid 2^b-2^a$ we have $b-a = 6n$. Then what should I do from here to prove the statement?
| $9 \mid 2^b-2^a$
Let $a \leq b$, otherwise just exchange $a$ and $b$ and note that if $ x \mid -y \Rightarrow x \mid y$
$2^b -2^a=2^a(2^{b-a}-1)$
So, $9 \mid (2^{b-a}-1)$, as $9 \nmid 2^a$, because $9=3^2$ and so they don't have any common prime factors.
$2^{b-a} \equiv 1 \pmod 9$ and $\phi(9)=6$
$2$ is a primitive roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$6 \times 6$ and $7 \times 7$ integer matrices
Can one fill a $6 \times 6$ matrix with integers so that the sum of all the numbers in each $3 \times 3$ square equals $2016$ and the sum of all the numbers in each $5 \times 5$ square equals $2015$?
Solve the same problem for the $7 \times 7$ case.
My work so far:
I sol... | Hint: in the $7 \times 7$ case, how many $3 \times 3$ squares and how many $5 \times 5$ squares contain each entry of the table?
EDIT:
OK, time's up... each entry of the $7 \times 7$ table is in the same number of $3 \times 3$ squares as it is in $5 \times 5$ squares. So i you take the sums of the numbers in each $3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such th... | Note that $S_6$ is the symmetric group of $6$ symbols which consists of $6!$ element.
We can use the theorem that two permutations are conjugate iff they have the same cycle type.
As you have already simplified the permutations into the product of disjoint cycles, it is easy to see that the two permutations have the sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric s... | Let
$$a_k := 4 k - 3 \qquad \qquad \qquad b_k := k - 1$$
Using summation by parts,
$$\begin{array}{rl} \displaystyle\sum_{k=1}^n a_k &= \displaystyle\sum_{k=1}^n a_k (b_{k+1} - b_k)\\ &= \left(a_k \, b_k \,\bigg|_1^{n+1}\right) - \displaystyle\sum_{k=1}^n (a_{k+1} - a_k) \, b_{k+1}\\ &= a_{n+1} \, b_{n+1} - a_1 \, b_1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 2
} |
Household reflector or transformation
Let $A\in\mathbb{R}^{n\times k}$, $n\geq k$, and $rank(A) = k$. Consider the use of Household reflectors, $H_i$, $1\leq i\leq k$, to transform $A$ to upper trapezoidal form, i.e., $$H_{k}H_{k-1}\ldots H_2 H_1 A = \begin{pmatrix}R\\0\end{pmatrix}$$
where $R\in\mathbb{R}^{k\times ... | The point of $H_1$ is to make the first column a column with zeros except in the top entry.
To find $x_1$, note that $\|(2,1,1)\| = \sqrt{6}$, so we should aim to get $(2,1,1)$ to $\sqrt{6}(1,0,0)$. Take
$$
x_1 = (2,1,1) - (\sqrt{6},0,0) = (2-\sqrt{6},1,1)
$$
(why should this work? Try to figure it out geometrically).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Curve sketching without a computer program How to sketch the curve x^6 + y^6 = (x^4)*y without using a computer program ? Could someone give me the step by step ?
| $\dfrac{y^6}{x^4} = y - x^2$, so there are no solutions whenever $y - x^2 < 0$. Whenever $y - x^2 = 0$, we find (by substitution) that the only solution is the origin. Notice that the curve is symmetric with respect to the y-axis, that is $(-x,y)$ is a solution if $(x,y)$ is a solution. Therefore, it is only necessary ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral inequality :$\int_0^1(f'(x))^2dx\geq 32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$ Assume $f:[0,1]\to \mathbb{R}$ is differentiable and $f'$ is integrable. Given $f\left(\frac{1}{4}\right)=f\left(\frac{3}{4}\right)=f(1)-f(0)=0$, then prove that $$\int_0^1(f'(x))... | Thanks for mentioning Wirtinger's inequality, i was not aware of it( have not studied fourier analysis yet). I will use two of its applications : $1)$ If $f(a)=f(b)=0$ then $$\frac{(b-a)^2}{\pi^2}\int_a^b(f'(x))^2dx\geq\int_0^a(f(x))^2dx$$ $2)$ If either one of $f(a)$ or $f(b)$ is zero then $$\frac{4(b-a)^2}{\pi^2}\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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How to find the Laplacian Eigen Values of the given graph
Find the Laplacian eigen-values of the of the graph on $N$ vertices whose edge set is given by $\{(i,i+1),1\le i<n\}$ and the edge $(1,n)$ .
The answer is given to be $2-2\cos{\frac{2k\pi}{n}}$ .
My try:
I tried to proceed by induction.
For the case $n=3$ ;I g... | Consider the Laplacian matrix of the cycle on $n$ vertices, given by
\begin{equation*}
L = \begin{bmatrix}
2 & - 1 & 0 & \cdots & 0 & -1\\
-1 & 2 & -1 & \cdots & 0 & 0\\
0 & -1 & 2 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2 & -1\\
-1 & 0 & 0 & \cdots & -1& 2\\
\end{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0$ Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0.$
Let one of the $x,y,z$ be even number.Let $x=2p$
$x^2+y^2+z^2=x^2y^2$
This gives $y^2+z^2$ is also even,which means either $y,z$ are both even or $y... | $$ z^2 + 1 = (x^2 - 1)(y^2 - 1) $$
From quadratic reciprocity, $z^2 + 1$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4.$
If $x$ or $y$ is odd, then $z^2 + 1$ is divisible by $4.$ Nope.
If $x$ or $y$ is $0$ then all $x,y,z = 0.$
Otherwise, if $x$ or $y$ is even, say it is $x,$ then one of $x-1,x+1$ is $1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find dimension from subspace equations In order to calculate its dimension, I need the basis of the subspace. How can I get it from the equations?
Find the dimension of the following subspaces of $\mathbb{R}^5$:
$$ U = \{(x_1,x_2,x_3,x_4,x_5) \ | \ 2x_1 - x_2 - x_3 = 0, x_4-3x_5=0 \}\\
V = \{(x_1,x_2,x_3,x_4,x_5) \ | \... | If $(x_1,x_2,x_3,x_4,x_5) \in U$, then $x_1 = \frac{1}{2}x_2 + \frac{1}{2}x_3$ and $x_4 = 3x_5$.
So
$$\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2}x_2 + \frac{1}{2}x_3\\
x_2 \\
x_3 \\
3x_5 \\
x_5
\end{pmatrix}= x_2\begin{pmatrix}
\frac{1}{2}\\
1\\
0\\
0\\
0\\
\end{pmatrix}+ x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of equation If $f(x) = x^2 - 2ax + a(a+1)$ , $f:[ a, \infty] \to [a,\infty]$ . If one of the solution of the equation $f(x)=f^{-1}(x)$ is $5049$ , then what may be the other solution ?
My WORK:
I found the inverse, but when I equate them , it is very difficult to solve . Please help me with some method to solv... | Your method of equating the function and its inverse directly is perfectly alright.
$f(x) = x^2 - 2ax + a^2 + a$
or $f(x) = (x-a)^2 + a$
or $f(x) - a = (x-a)^2$
$x = a \pm \sqrt{f(x) - a}$
Here the negative sign is rejected as $x \in [a,\infty]$
So, we have $f^{-1}(x) = a + \sqrt{x - a}$
Now, $f(x) = f^{-1}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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To show that two quadratic forms are equivalent over $\mathbb{C}$ but not over $\mathbb{R}$ Consider the quadratic forms $p$ and $q$ given by $$q(x,y,z,w)=x^2+y^2+z^2+bw^2$$ and $$p(x,y,z,w)=x^2+y^2+czw.$$
Then we can write their respective matrices as
$$
A = \pmatrix{1\\&1\\&&1\\&&&b}, \quad
B = \pmatrix{1\\&1\\&&0... | Write the coordinates in $q(x,y,z,w)$ as $\widetilde z:= \frac1{\sqrt c}\cdot (z - i\sqrt bw)$ and $\widetilde w:= \frac1{\sqrt c}\cdot (z+i\sqrt bw)$ (i. e. make a base change). Then you have
\begin{align*}
q(x,y,z,w) &= x^2+y^2+z^2+bw^2\\
&= x^2+y^2 + (z-i\sqrt bw)(z+i\sqrt bw)\\
&= x^2+y^2 + c\cdot \left(\frac{1}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three?
Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three?
Attempt: I feel I'm having trouble formalizing the expres... | I would use a table to first see this intuiutively
$$\begin{array}{|cc|cccccc|}\hline
&&&&&D_1\\
&&1&2&3&4&5&6\\
\hline
&1&2&3&4&5&\color{red}6&7\\
&2&3&4&5&\color{red}6&7&8\\
D_2&3&4&5&\color{blue}6&7&8&9\\
&4&5&\color{red}6&7&8&9&10\\
&5&\color{red}6&7&8&9&10&11\\
&6&7&8&9&10&11&12\\
\hline\end{array}$$
The red numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Diophantine equations using Euclidean algorithm I solved two systems of Diophantine equations using the Euclidean algorithm and I can't figure out where I went wrong because the solutions I test aren't working but I have rechecked my work several times.
a) $56x+72y=40$
using the algorithm:
$72=1*56+16$
$56=3*16+$
$16=... | Let $d = \gcd(a, b)$. The linear Diophantine equation $ax + by = c$ has a solution if and only if $d \mid c$. If $(x_0, y_0)$ is a particular solution of the equation $ax + by = c$, then the general solution is
$$x = x_0 + \frac{b}{d}t \qquad y = y_0 - \frac{a}{d}t$$
where $t \in \mathbb{Z}$.
In both problems, you ... | {
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"url": "https://math.stackexchange.com/questions/1822748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Can $b+c$ in the pythagorean triplets $(a, b, c)$ be a prime number? If $a^2 + b^2 = c^2$, can $b + c$ be a prime number?
| Hint: $a^2+b^2=c^2\implies b^2-c^2=-a^2\implies (b+c)(b-c)=-a^2 \implies b+c=\frac{a^2}{c-b}$ $ \implies b+c=a^2 \times \frac{1}{c-b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Uniform convergence of following series Prove that $\sum_{n=1}^\infty \frac{x^{2n}}{(1 + x + \dots + x^{2n})^2}$ converges uniformly when $x \geq 0$.
| $\sum \dfrac {x^{2n}}{(1+x+...x^{2n})^2}$
Suppose $x< 1$
$\sum \dfrac {(1-x)^2x^{2n}}{(1-x^{2n+1})^2}\\
\sum \big(\dfrac {(1-x)x^{n}}{1-x^{2n+1}}\big)^2\\
\big(\dfrac {(1-x)x^{n}}{1-x^{2n+1}}\big)^2<x^{2n}$
$\sum x^{2n}$ converges when $x<1$ and the series converges by the comparison test.
Suppose $x>1$
$\sum \big(\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
| Assume w.l.o.g that $a \leq b \leq c$.
Then we have : $(a+b+c)^3 \geq 7(a^2b+b^2c+c^2a)+6abc$; thus it is sufficient to prove: $(a+b+c)(a+b+c-abc) \geq \dfrac{2}{7}((a+b+c)^3-6abc)$.
Let $x=a+b+c$ and $y=abc$. The last inequality is then: $x^2-xy \geq \dfrac{2}{7}(x^3-6y)$, which is equivalent to: $x^2+\dfrac{12y}{7} \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof ... | The previous answer was not a full answer, let me offer my thoughts and let me expand a little (using Cauchy-Schwartz):
Using the Cauchy-Schwarz inequality we get
\begin{align}
a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}
\end{align}
From the AM-GM inequality we obtain $a^2+b^2+c^2\g... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Limit of the sequence $a_{n+1}=\frac{1}{2} (a_n+\sqrt{\frac{a_n^2+b_n^2}{2}})$ - can't recognize the pattern Consider the sequence:
$$a_0=x,~~~b_0=y$$
$$a_{n+1}=\frac{1}{2} \left(a_n+\sqrt{\frac{a_n^2+b_n^2}{2}} \right),~b_{n+1}=\frac{1}{2} \left(b_n+\sqrt{\frac{a_n^2+b_n^2}{2}}\right)$$
$$\lim_{n \to \infty} a_n=\lim_... | Let $z_i=a_i + ib_i$. Then your recursion is
$$
z_{n+1}=a_{n+1}+ib_{n+1}=\frac{1}{2}(a_n+ib_n)+\frac{1}{2}(1+i)\sqrt{\frac{a_n^2+b_n^2}{2}}=\frac{1}{2}z_n+\frac{1+i}{2\sqrt{2}}|z_n|=\frac{1}{2}z_n+\frac{1}{2}e^{i\pi/4}|z_n|.
$$
Now, let $z_n=e^{i(\theta_n+\pi/4)}y_n$, where the $y_n$ and $\theta_n$ are real. Then
$$
y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to
$$\Gamma\left ( -k+\frac... | Is this Possible?
Since,
$\Gamma(a)=\frac{\Gamma(a+n)}{(a)_n}$ You can deduce that
\begin{eqnarray*}
\Gamma(-m+\frac{1}{2})&=&\frac{\Gamma(\frac{1}{2})}{(-m+\frac{1}{2})_m}\\
&=&\frac{\sqrt{\pi}}{\left(\frac{-2m+1}{2}\right)\cdot\left(\frac{-2m+3}{2}\right)\cdot\left(\frac{-2m+5}{2}\right)\cdots\left(\frac{1}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of $\frac{1}{5}-\frac{2}{5^2}+\frac{3}{5^3}-\frac{4}{5^4}+....$ Find the sum of following series:
$S=\frac{1}{5}-\frac{2}{5^2}+\frac{3}{5^3}-\frac{4}{5^4}+....$ upto infinite terms
Could someone give me slight hint to solve this question?
| $$S=\frac 15-\frac2{5^2}+\frac 3{5^3}-\dots\tag 1$$
and $$\frac S5=\frac 1{5^2}-\frac2{5^3}+\frac 3{5^4}-\dots\tag 2$$
Now, $(1)+(2)$ yields $$S+\frac S5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots\\\implies \frac{6S}5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots=\frac 16\\\implies S=\frac5{36}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Expected Number of flips for alternating Heads/Tails 10 times What is the expected number of flips needed to flip a coin 10 times and have the outcome be alternating heads/tails (starting with heads, then tails, then heads etc...).
I wrote a c++ program and it gives me 2,730. Is this correct and how would you do this ... | Let $x$ be the expected number of flips to get $(HT)^5$ and $y$ the expected number of further tosses required if we are starting with $XH$ where we cannot use part of $X$ to form the first part of the required run (eg $X$ is empty or ends in $H$).
The first toss is $H$ or $T$, so $x=\frac{1}{2}(1+y)+\frac{1}{2}(1+x)$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.
I took extremely long to solve this
I got
$50= 7^2 + 1^2 ... | Consider the equation $x^2+y^2=z^2+w^2=N.$
This is equivalent to: $x^2-z^2=w^2-y^2=D.$
and we need $D$ to have at least two different factorizations with the factors having the same parity. One approach is thus to find the smallest values of $D$ which satisfy this property. Another approach is to find parametric soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
If $\ln(1+x) \approx A+Bx+Cx^2$, differentiate twice both sides and show that $\ln(1+x) \approx x-\frac{1}{2}x^2$
Question: $\ln(1+x) \approx A+Bx+Cx^2$, for $-1<x\leq1$, where $A,B,C$ are constants.
Differentiate twice both sides of the approximation above and hence show that $$ \ln(1+x) \approx x-\frac{1}{2}x^2$$
U... | @Michael Hardy offered a clear solution to the first part.
Second Part solution
The second part of the problem can be solved by applying the solution from the first part:
\begin{align*}
\left(1+\frac{1}{2n}\right)^{n+3} &< \left(1+\frac{1}{n}\right)^{n-1} \\
\ln \left(1+\frac{1}{2n}\right)^{n+3} &< \ln \left(1+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\f... | You are very close but you have actually made two mistakes that have 'cancelled' each other out to get the right result.
Like you pointed out the inequality you used was wrong. In fact, you have the reverse of this inequality. However, this is exactly what you need since you have to remember that dividing reverses the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
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.