Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $A+B+C=0$, then prove that the value of the determinant is $0$. I'll state the question from my textbook below:
If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.
This is how I tried solving the problem:
$LHS = \begin{vmatrix}1 & \cos ... | When at least one of $A,B$ or $C$ is not an integer multiple of $\pi$, the vector $(\sin A,\sin B,\sin C)$ is nonzero. However, as $A+B+C=0$, the sum-of-angle formula implies that
$$
\pmatrix{1&\cos C&\cos B \\ \cos C&1&\cos A\\ \cos B&\cos A& 1}
\pmatrix{\sin A\\ \sin B\\ \sin C}
=0
$$
and hence the matrix is singular... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Jordan basis unique? Is Jordan basis unique? I have a 4x4 matrix, find eigenvectors and one generalized eigenvector, also trying different linearly independent eigenvectors but the matrix P so that PJP^-1 =A only works for certain vector.
\begin{pmatrix}-\frac{1}{2}&0&-\frac{1}{2}&0\\ \frac{3}{2}&0&\frac{1}{2}&0\\ -\fr... | It is not: the Jordan matrix is unique, but there are several basis which will yield that matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Let $a^2<2, b=2(a+1)/(a+2)$. Show $b^2<2$ (assignment) It is a part of my assignment.
$$ \text {Let }a^2<2, \quad b=2\frac {(a+1)}{(a+2)}\quad \text{ Show } b^2<2$$
I already proved that a
But, I am struggling to prove $b^2<2$.
My lecturer said that I need to manipulate $b^2$, which is larger than $b^2$ but less th... | same thing but with more details.
$b = \frac{(2a+1)}{a+2}$
$\frac{b}{2} = \frac{a+1}{a+2} = \frac{a+2-1}{a+2} = = 1 - \frac{1}{a+2}$
since
$a^2 < 2$
$-\sqrt{2} < a < \sqrt{2}$
$2 -\sqrt{2} < 2 + a < 2 + \sqrt{2}$
both sides are positive so
$\frac{1}{2 + \sqrt{2}} < \frac{1}{2+a} < \frac{1}{2 - \sqrt{2}}$
$-\frac{1}{2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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If $\cos2\theta=0$, then $\Delta^2=$? I'll state the question from my textbook here:
If $\cos2\theta=0$, then $\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2=$?
This is how I solved the problem:
$\begin{vmatrix} 0 & \cos \theta & \sin \t... | When in doubt, use the relations in the original problem. Let $\theta=\frac{3\pi}4$. Then $\cos2\theta=0$, while $\sin\theta=\sqrt2/2=a$ and $\cos\theta=-\sqrt2/2=-a$.
The expression in the question is now
$$\begin{vmatrix}0&-a&a\\-a&a&0\\a&0&-a\end{vmatrix}^2$$
Clearly, adding up all three rows of the matrix produces ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Using polar co-ordinates to solve a dynamical system I have the system
$x^\prime = x-y-x(x^{2}+y^{2})$
$y^\prime = x+y-y(4x^{2}+y^{2})$.
I want to show with polar coordinates that I eventually end up with
$r^\prime = r-r^{3}(1+ \frac{3}{4}$sin$^{2}(2 \theta))$.
I have used the following substitutions:
$x = r$cos$(\the... | Actually you are on the right track but you've just made a few slips with your algebra.
You should have $$rr'=x(x-y-x^{3} -xy^{2}) + y(x+y-4x^{2}y^{\color{red}{1}}\color{red}{-}y^{3})$$
this will simplify to $$x^2+y^2-(x^4+y^4)-5x^2y^2$$
$$=r^2-(r^4-2x^2y^2)-5x^2y^2$$
$$=r^2-r^4-3x^2y^2$$
now use the double angle form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of a function tending to a finite number If
$$\lim_{x\to 0} \frac{ae^x - b\cos x +ce^{-x}}{x\sin x} = 2$$
then find the value of $a+b+c$.
My book has given the following solution to the above problem :-
We observe that as $x$ tends to zero , numerator tends to $a-b+c$ whereas the denominator tends to zero. Ther... | If $a-b+c \neq 0$, the limit would be either $+\infty$ or $-\infty$.
Note, as written as Robert Z, that
$$
\frac{ae^x-b\cos x+cx^{-x}}{x\sin x}=\frac{(a-b+c)+x(a-c)+x^2(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}))+O(x^3)}{x^2+O(x^3)}
$$
as $x\to 0$. Since the limit exists, you need $a-b+c=0$, $a=c$ and $\frac{a}{2}+\frac{b}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do you solve this equation $\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$? This is a problem from Prilepko's book on Mathematics for High School, I have tried to solve this equation but cannot succeed. The hint is to put $t =$ $\sqrt{3+log_{0.2}x}$. But then it reduces to some quite complicated expression. If I've a... | Let $t = \sqrt{3+log_{0.2}x} \implies \frac{\ln x}{\ln 0.2} = t^2 - 3$
Now $\sqrt{1+log_{0.04}x} = \sqrt{1 + (t^2 - 3)\ln{0.2}/\ln{0.04}} = \sqrt{1 + \frac{t^2 - 3}{2}} = \sqrt{\frac{t^2 - 1}{2}} $
So we get $\sqrt{\frac{t^2 - 1}{2}} + t = 1 \implies t^2 -1 = 2(t-1)^2$
$ t^2 -1 = 2t^2 - 4t + 2 \implies t^2 - 4t + 3 = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Bounded sequences. We have the following sequences:
$a_n=n+sin(n+3^n)$
$b_n=\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}$
Show that $a_n$ is non bounded and $b_n$ is bounded. Also show that the subsequences $b_{2n}$ and $b_{2n-1}$ are convergent.
My work: $a_n<n+1$ and as $n\rightarrow∞$ $a_n<$∞ , so $a_n$ is non bounded
For $b... | Note that
$$a_n=n+\sin(n+3^n)>n-100 \to \infty$$
thus $a_n$ is not bounded and
$$-\frac{(n+1)}{3\sqrt{n^2+7}}\le b_n=\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}\le \frac{(n+1)}{3\sqrt{n^2+7}}$$
with $\frac{(n+1)}{3\sqrt{n^2+7}}$ bounded since $\frac{(x+1)}{3\sqrt{x^2+7}}$ is continuous and
$$\frac{(n+1)}{3\sqrt{n^2+7}}\to\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to find the value of the $k$ to make $\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=A$ exist? the question described as follow:
$$\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=A$$
the $A$ is constant and $A\not=0$
and find the $k$ to make this limit exist.... | It seems correct indeed
$$\sqrt[5]{x+\sqrt[3]{x}}\sim x^\frac1{15} \left(1+ \frac{x^\frac23}5 \right)=x^\frac1{15} + \frac{x^\frac{11}{15}}5 $$
$$\sqrt[3]{x+\sqrt[5]{x}}\sim x^\frac1{15} \left(1+\frac{x^\frac45}3\right)=x^\frac1{15} +\frac{x^\frac{13}{15}}3$$
thus
$$\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}\sim \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it wrong to factor in this way? I found an exercise for factoring:
$\frac{4x^2-8x+4}{x^2-1} * \frac{x+1}{12}$
I resolve like this:
= $\frac{(2x -2)^2}{(x-1)(x+1)} * \frac{(x+1)}{12}$
= $\frac{(2x -2)^2}{12(x-1)}$
Well up here, because here comes my problem:
$1-.$ If I factor like this:
= $\frac{2(x -1)^2}{12(x-1)}$... | When you factor something out of an expression which is squared, be wary of where the brackets go
$$(2x-2)^2=[2(x-1)]^2=[2]^2[x-1]^2=4(x-1)^2$$
So the first answer is wrong, the second is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
polynomial long division Integration $$
\int \frac{3x^2 +6x +2}{x^2 + 3x +2}dx
$$
I've been trying to solve this integral, my gut tells me to use polynomial long division so I factored and got
$$
\int \frac{3x^2 +6x +2}{(2+x)(1+x)}dx
$$
I have never done a division with two factors in the denominator, would anybody be ... | You do polynomial long division just like you do long division on integers.
$
\require{enclose}
\begin{array}{r}
3 \\
x^2 + 3x + 2 \enclose{longdiv}{3x^2+6x+2} \\
\underline{3x^2 + 9x + 6} \\
-3x - 4
\end{array}
$
$\frac {3x+6x + 2}{x^2+3x + 2} = 3 - \frac {3x+ 4}{x^2+3x+2}$
But ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Compute $I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x$
$$I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x=?$$
I saw the related problem but that didn't help much. I attempted this question by splitting the numerator into $$I_1=\int\frac{\sin x}{\sin^2x+\cos^4x}\mathrm d x \text { and } I_2=\int\frac{\... |
Here is a sketch only that should facilitate leading to the coveted result. To that end, we proceed.
First, noting that
$$\sin^2(x)+\cos^4(x)=\cos^2(x)+\sin^4(x)$$
we can write
$$\begin{align}
\int\frac{\sin(x)+\cos(x)}{\sin^2(x)+\cos^4(x)}\,dx&=\int \frac{\cos(x)}{1-\sin^2(x)+\sin^4(x)}\,dx+\int \frac{\sin(x)}{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Suppose $A \in M_{3,3}$ with eigenvalues $1,2,3$, eigenvectors $b_1, b_2, b_3$. Let $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$. I think I did this right, but someone skim and double check?
Q: Suppose $A$ is a $3 \times 3$ matrix, with eigenvalues $1,2,3$ and corresponding eigenvectors $b_1, b_2, b_3$. Suppose that $v = b... | That's fine, to check note that for each eigenvectors
$$A\vec v=\lambda \vec v\implies A^n(c\vec v)=c\lambda^n \vec v$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all possible values of $p$, $q$ and $r$ such that this matrix is orthogonal
Find all possible values of $p$, $q$ and $r$ such that the following matrix is orthogonal.
$$B= \begin{pmatrix}
\frac1{3}&\frac2{3}&\frac2{3} \\
\frac2{3}&\frac1{3}&-\frac2{3} \\
p&q&r
\end{pmatrix}$$
| If $B$ is your matrix, then$$B^T.B=\begin{pmatrix}p^2+\frac{5}{9} & p q+\frac{4}{9} & p r-\frac{2}{9} \\ p q+\frac{4}{9} & q^2+\frac{5}{9} & q r+\frac{2}{9} \\ p r-\frac{2}{9} & q r+\frac{2}{9} & r^2+\frac{8}{9}\end{pmatrix}$$and you want this to be the identity matrix. So, since you only want to have $1$'s in the m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $\sin(x)=3\cos(x)$, compute $\sin(x)*\cos(x)$ I drew a triangle and was instructed to use a property unknown to me. Apparently the answer is $\frac{3}{\sqrt{10}}*\frac{1}{\sqrt{10}}=\frac{3}{10}$. Is this answer correct, and what theorem/formula is this?
| We have the identity
$1 + \tan^2 \theta = \sec^2 \theta = \dfrac{1}{\cos^2 \theta}; \tag 1$
so with
$\sin x = 3\cos x, \tag 2$
we infer
$\tan x = \dfrac{\sin x}{\cos x} = 3, \tag 3$
whence
$\dfrac{1}{\cos^2 x} = 1 + \tan^2 x = 1 + 3^2 = 10, \tag 4$
yielding
$\cos x = \pm \dfrac{1}{\sqrt{10}}; \tag 5$
since
$\sin^2 x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
reverse three digit subtraction I cannot get this. How does it work? Like what is the concept in this?
Imagine $895 - 598$:
*
*Step 1. $8 - 5 = 3$ (I got $8$ and $5$ from the hundreths place in the numbers I am subtracting)
*Step 2. $3 × 100 = 300$ ($100$ is because $8$ and $5$ were originally from the hundreth... | Simply,
$$895-598=\left(8\times10^2+9\times10^1+5\times10^0\right)-\left(5\times10^2+9\times10^1+8\times10^0\right)=\left(8\times10^2-5\times10^2\right)+\left(9\times10^1-9\times10^1\right)+\left(5\times10^0-8\times10^0\right)=3\times10^2+0\times10^1+\left(-3\times10^0\right)=300-3=297$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Number of ways 6 things can be selected from a group of identical things
In how many ways can a party of 6 men be selected out of 10 Hindus, 8 Muslims, and 6 Christians if the party consists of at least one person of each religion( Consider only the religion of the person?
Now, I know this problem can be solved by fi... | Stars and Bars
Once we pre-select one of each religion, this is the same as asking "how many ways we can choose $3$ men from $9$ Hindus, $7$ Muslims, and $5$ Christians?" This is the same as asking how many ways to solve
$$
x+y+z=3
$$
with $x,y,z\ge0$. Stars and Bars say $\binom{5}{2}=10$.
Generating Functions
$$
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2689421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Fourier Transform of $ e^ {t/2} \operatorname{sech}(t) $ My question is what is the Fourier transform of the function
$f(t)=\exp(t/2)\text{sech}(t)$
Is there any closed form available? One possibility is that
$f(t)^2=\frac{1}{2} \text{sech}(t) + \frac{1}{4} (\dfrac{d}{dt}\operatorname{sech}t)$
and then the Fourier tra... | The best way to attack the problem is to use the definition of the hyperbolic secant, together with the Geometric Series.
$$\text{sech}(t) = \frac{2}{e^t + e^{-t}}$$
We then have:
$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{2}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{t/2} e^{-i kt }\frac{\text{d}t}{e^t + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Summation of Fourier Coefficients in Fourier Series Problem First poster!
I have been posed with a problem that is to find the Fourier Series of the periodic function
$$f(x) =
\begin{cases}
0, & -\pi < x < 0 \\
1, & 0<x<\frac{\pi}{2}\\
0, &\frac{\pi}{2}<x<\pi
\end{cases}$$
$$f(x) = f(x+2\pi)$$
I have found
$$
a_0 = \f... | What you have looks fine, and I likely wouldn't bother trying to simplify it at all (unless you want to take Spencer's suggestion and work with complex numbers). That being said, one might note that the angle addition formula for the sine function might allow some moderate simplification. Recall that
$$ \sin(\alpha +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit (proof). (assignment) This is my assignment question.
Find $\lim_{n \to \infty} (2n^2+10n+5)/n^2$.
My attempt: suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$
Then, $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}$.
Then, let $M \in N$ and $\frac {15}M< \varepsilo... | You could just do $$\lim_{n \to \infty}\frac{2n^2+10n+5}{n^2}=\lim_{n\to\infty}\frac{2+\frac{10}n+\frac5{n^2}}{1}=\frac{2+0+0}{1}=2$$ since both $\dfrac1n$ and $\dfrac1{n^2}$ tend to $0$ as $n\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find derivative of $y=\sin^{-1}\frac{2x}{1+x^2}$
Find $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{2x}{1+x^2}$
The solution is given as $\frac{2}{1+x^2}$. But is it a complete solution ?
My Attempt
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2},\quad |x|\leq 1\\
\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\; x>0\\
-... | Hint
Your answer seems correct to me even I don't really understand the need of tan function
$$y=\sin^{-1}\frac{2x}{1+x^2} \implies \sin y=\frac{2x}{1+x^2}$$
Differentiate
$$y'\cos(y)=\frac {-2(x^2-1)}{(x^2+1)^2}$$
use the fact that $$\sin^2(y)+\cos^2(y)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Alternative for calculating the nth of quadratic sequence Given the quadratic sequence
$$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$
$$f(2)=4a+2b+c$$
$$f(3)=9a+3b+c$$
Now solve the simul... | Yes!
You know $f(1)= 1 \implies f(1) - 1 = 0 \implies f(x) - 1$ has a root at x=1
Now,
$f(x) - 1 = a(x-1)(x-b)$
Put $x = 2$, $a(2-b) = 6$
Put x = 3, $a*2*(3-b) = 18$
Divide both equation, we get $b = 0 $ and a = 3
$f(x) - 1 = 3*(x-1)x \implies f(x) = 3x^2 - 3x + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim _{t\to \infty}\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}$ I am having difficulties evaluating this limit:
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right)$$
I have tried to divide out by $\frac{\sqrt{t}}{\sq... | What I interpret Dr. Graubner's directions to mean.
\begin{align*}
\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}
&= \frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}} \cdot \frac{\sqrt{t+1}\sqrt{t+2}}{\sqrt{t+1}\sqrt{t+2}} \\
&= \frac{\sqrt{t+1}-\sqrt{t} }{2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Prove that the general term of this multinomial expansion is as given
If $(6-12x+12x^2)^n = \displaystyle \sum_{r=0}^{2n} T_r x^r$, prove $ T_r = (-2)^r 3^n \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg]$
I tried by writing $$(6−12x+12x^2)^n= 6^n(1-2x+2x^2)^n = 6^n\dis... | Working with the innermost sum we have for
$$\sum_{q=0}^n {2n-2q\choose r} {n\choose q}$$
that it is
$$\sum_{q=0}^n [w^r] (1+w)^{2n-2q} [z^{n-q}] (1+z)^n
\\ = [w^r] (1+w)^{2n} \sum_{q=0}^n (1+w)^{-2q}
[z^{n}] z^q (1+z)^n.$$
Now we may extend $q$ to infinity as there is no contribution
to the coefficient extractor $[z^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trace of matrices with primes as elements I noticed this curious pattern that matrices composed of primes seem to have traces equal to or close to square numbers:
\begin{align}tr\begin{bmatrix}2&3\\5&7\end{bmatrix}=9=3^2\end{align}
\begin{align}tr\begin{bmatrix}2&3&5\\7&11&13\\17&19&23\end{bmatrix}=36=6^2\end{align}
\b... | I think this pattern exist, because these matrices are relatively small. If you take for example matrix of dimension $n=7$, then sum is equal to $724$, which is $3$ away from $27^2$.
Interesting thing happens when for $n=10$ it gives diagonal value of: $\sum=2,451$ which is in "the middle" of $49^2=2401$ and $50^2=2500... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric series sum involving tangents
$$\frac{1}{4}\tan\bigg(\frac{\pi}{8}\bigg)+\frac{1}{8}\tan\bigg(\frac{\pi}{16}\bigg)+\frac{1}{16}\tan\bigg(\frac{\pi}{32}\bigg)+\cdots\cdots \infty$$
Try: $$\cos x\cdot \cos(x/2)\cdot\cos(x/2^2)\cdots\cdots \cos(x/2^{n-1})=\frac{1}{2^{n-1}}\frac{\cos(x)}{\sin(x/2^{n-1})}$$
... | Using the identity:
$$
\cot x-\tan x = 2\cot 2x \Rightarrow \tan x=\cot x -2\cot 2x,
$$
one obtains a telescoping sum:
$$
S(x):=\sum_{n=0}^\infty \frac{\tan(x/2^n)}{2^n}=\sum_{n=0}^\infty\left( \frac{\cot(x/2^n)}{2^n}-\frac{\cot(x/2^{n-1})}{2^{n-1}}\right)\\=
-2\cot(2x)+\lim_{n\rightarrow\infty}\frac{\cot(x/2^n)}{2^n}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to simplify arc length problems: expressions under the root I am having trouble during my studies of arc length by integration in Calculus.
Arc Length formula: $\int\sqrt{1+('y)^2}dx$
My problem values:
$y=\frac{1}{10}x^5+\frac{1}{6}x^-3$
$y'=\frac{5}{10}x^5+(-\frac{3}{6}x^{-4})$ = $\frac{1}{2}x^4-\frac{1}{2}x^-4$... | $$\begin {align}1+\frac12\left(x^4-x^{-4}\right)^2&=1+\frac12\left(x^8-2\cdot x^4 \cdot x^{-4}+x^{-8}\right)\\&=1+\frac12(x^8-2+x^{-8})\\
&=-1+\frac12(x^8+2+x^{-8})\\
&=-1+\frac12\left(x^4+x^{-4}\right)^2\end {align}$$
Your expression is missing the $-1$ out front.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve: $5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$ Solve to find the general value of $x$:
$$5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$$
My Attempt:
$$5(1-\sin^2 (x))-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$5-5\sin^2 (x)-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$2\sin^2 (x)+4\sin (x)\cos (x)=3$$
| You can divide by $\cos^2 x \quad $ (Since $x=\pi/2$ does not work)
$$\implies 5-4 \tan x+3 \tan^2 x=2 \sec^2 x$$
yet $\sec^2 x=1+\tan^2 x .....$
Continuing you have a quadratic in $\tan x$ which I'm sure you know how to solve...
$$\implies z^2-4z+3=(z-3)(z-1)=0 \quad |\quad z=\tan x $$
So for any $n \in \mathbb{Z}$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluate $\sum_{k=1}^\infty \frac{k-1}{2^{k+1}}$ as a telescoping series? Okay, so my book shows (with absolutely no explanation) an equivalency to
$\lim\limits_{x \to \infty} (\frac{1}{2} - \frac{x+2}{2^x})$
, and I have no idea where this came from. From my understanding, telescoping series generally telescope down... | Apparently there was a typo somewhere: “$\frac{x + 2}{2^x}$” should have being replaced by “$\frac{x}{2^x}$”. That being said, let $s_x \mathop{:=} 1/2 - x \mathbin/ 2^x$ for $x \in \mathbf{N}^*$. Observe that
$$
s_{x + 1} - s_x = \frac{1}{2} - \frac{x + 1}{2^{x + 1}} - \biggl(\frac{1}{2} - \frac{x}{2^x}\biggr) = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How many integers in the range satisfy a given congruence? Specifically how many integers $x$ in the range $1 ≤ x ≤ 60$ satisfy $21x ≡ 24\ (mod\ 60)$? And how many integers $a$ are there in the range $1 ≤ a ≤ 74$ for which the equation $x^2 ≡ a\ (mod\ 37)$ is soluble?
I have the solutions to be $3$ and $38$ respectivel... | \begin{align}
21x &\equiv 24 \pmod{60} \\
7x &\equiv 8 \pmod{20} \\
x &\equiv 4 \pmod{20} \\
x &\in \{4, 24, 44 \}
\end{align}
TOTAL = 3
\begin{array}{|c|c|}
\hline
x & x^2 \pmod{37}\\
\hline
0,37 & 0 \\
1,36 & 1 \\
2,35 & 4 \\
3,34 & 9 \\
4,33 & 16 \\
5,32 & 25 \\
6,31 & 36 \\
7,30 & 12 \\
8,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof by Induction $\frac{n+1}{2n} = \left(1-\frac{1}{2^2}\right)\dotsc \left(1-\frac{1}{n^2}\right)$ proof by induction. I'm pretty close to done with this proof but got stuck at the end
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{n^2}\right) = \frac {n+1}{2n}$$
for all integers ... | You are close.
You want to show that
$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right)
= \frac{(k+1)+1}{2k+2}
$.
Note that the denominator on the right is
$2(k+1) = 2k+2$,
not the $2k+1$ you have.
That is why you were unable
to complete the proof -
you were trying to prove
something that was false.
$\begin{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Explaining $\int_{-1}^1\frac{1}{1+x^2}\,dx = \frac{\pi}{2}$. How would you explain to a student that
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) = \frac{\pi}{2} $$
and not
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) \neq \frac{\pi}{4} - \frac{3\pi}{4} = -\frac{\pi}{2}$$
besides the obv... | We are actually using the substitution $x=\tan\theta$.
$$ \int_{-1}^1\frac{1}{1+x^2}dx=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \left(\frac{1}{1+\tan^2\theta}\right)(\sec^2\theta)d\theta=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} d\theta=\frac{\pi}{2}$$
We take the $\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Number theory and set theory problem
Let $A$ be a finite set. For $0\leq i\leq 2$, let $a_i$ be the number of subsets $B$ of $A$ such that
$$|B|\equiv i \pmod{3}.$$
Prove that $$|a_i - a_j| \leq 1,$$ for all $0\leq i,j\leq 2$.
Firstly I assumed that $|A|=n$. Then $a_0 = {^nC_0} + {^nC_3} + {^nC_6} +\cdots$ and ... | Let $a_0(n),a_1(n),a_2(n)$ be given by
$$
\begin{cases}
a_0(n) = \binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \cdots\\[4pt]
a_1(n) = \binom{n}{1} + \binom{n}{4} + \binom{n}{7} + \cdots\\[4pt]
a_2(n) = \binom{n}{2} + \binom{n}{5} + \binom{n}{8} + \cdots\\
\end{cases}
$$
We want to show that for all nonnegative integers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\sin(z) =\frac{e^{iz}-e^{-iz}}{2i} =\sin(x)\cosh(y) +i\cos(x)\sinh(y)$ For complex $z=x+iy$.
How to prove that
$$\sin(z) =\frac{e^{iz}-e^{-iz}}{2i} =\sin(x)\cosh(y) +i\cos(x)\sinh(y)$$
by using the power series definition
$$\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots$$
and Euler's formula $$e^z =e^x... | Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
\begin{align}
\sin z &=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =\sum_{n=0}^\infty \frac{i^{2n}}{(2n+1)!} z^{2n+1}\\
&=\frac{1}{i} \sum_{n=0}^\infty \frac{i^{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f... | Use summation by parts. Given a sequence $f(n)$ define $(\Delta f)(n)=f(n+1)-f(n)$ and note that
$$
\sum_{n=a}^b(\Delta f)(n)=f(b+1)-f(a).
$$
Further given two sequences $f(n)$ and $g(n)$, note that
$$
(\Delta fg)(n)=f(n+1)\Delta g(n)+g(n)\Delta f(n)
$$
whence
$$
\sum_{n=a}^bg(n)\Delta f(n)=[f(b+1)g(b+1)-g(a)f(a)]-\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Possible values of $x+y+z$ Given that $xyz=11(x+y+z)$, find the possible values of $x+y+z$ such that $x\leq y$ satisifies $x^2+y^2=2018$
I know that $x,y$, or $z$ should be divisible by $11$. I am having trouble going to the next step
| I think the problem also states that $x,y,z$ must be integers.
Assume that $x$ is divisible by $11$, then $x=11k$ ($k$ is an integer) and $(11k)^2+y^2=2018$.
We will have $y^2=2018-121k^2$. Because $y^2\ge 0$, we can prove that $k^2$ must be in this set of numbers: $(1;4;9;16)$.
We have four cases:
*
*$k^2=1\Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Polar form from a complex number? How do I solve for the polar form of $-3\sqrt{2} - 3\sqrt{3}\,i\; ?$
I think I solved for $r$ which is $3\sqrt{5}$ from using $r=\sqrt{a^2+b^2}.$
When I look for theta I use $\tan\theta = \frac ab,$ but when doing so I get $\frac{\sqrt{6}}{ 2}.$ Anyway I looked in the answer key for $... | Same way always:
$z = -3\sqrt{2} + 3\sqrt 3 i = re^{i\theta}$ where $r = |-3\sqrt{2} + 3\sqrt 3 i| = \sqrt{(-3\sqrt{2})^2 + (3\sqrt 3)^2} = \sqrt{9*2 + 9*3}=\sqrt {45} = 3\sqrt 5$.
And $\theta$ is so that $\cos \theta = \frac {-3\sqrt{2}}r=\frac {-3\sqrt{2}}{3\sqrt 5}=\frac {-\sqrt 2}{\sqrt 5}; \sin\theta = \frac {3\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What is the remainder when $ 6^{65} $ is divided by 80? In the solutions it used Chinese remainder theorem in the following way: $6^{65}\equiv 1 \pmod 5 $ and $ 6^{65} \equiv 0 \pmod{16} $. I can see how they got the first congruence but am really struggling to see it for the second one. Any help would be much apprecia... | Note:
$$\begin{align} 6 &\equiv 1\pmod 5 \\
6^{61} &\equiv 1 \pmod 5 \\
6^{61}\cdot 2^4 &\equiv 1\cdot 2^4 \pmod {5\cdot 2^4}\\
3^4 &\equiv 1\pmod{80}\\
6^{61}\cdot 2^4\cdot 3^4 &\equiv 1\cdot 2^4\cdot 1 \pmod{80}\\
6^{65} &\equiv 16 \pmod{80}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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The square of the solution of the equation... The square of the solution of the equation
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
is equal to: ...
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$
$\implies 7x^2 = \left... | Alt. hint: show first that $\;\sqrt{8 \pm 3\sqrt{7}} = \dfrac{1}{\sqrt{2}}\left(3 \pm \sqrt{7}\right)\,$, then no squaring is needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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What is $x$, if $\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots = 1$?
Let
$$S_{n}=\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots \quad\text{($n$ terms)}$$
where $x... | We have
\begin{eqnarray*}
\cot^{-1}(A)-\cot^{-1}(B)=\cot^{-1}\left(\frac{AB+1}{B-A} \right).
\end{eqnarray*}
In your case this gives
\begin{eqnarray*}
\cot^{-1}\left(\frac{i(i+1)x}{2}+\frac{2}{x} \right)=\cot^{-1}\left(\frac{ix}{2} \right)-\cot^{-1}\left(\frac{(i+1)x}{2} \right).
\end{eqnarray*}
So the sum is telescop... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Finding a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ I am trying to find a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ as a function of $x$ and $n$ and for large $n$,
where $x$ changes with $n$ such that $\lim_{n\to\infty}x=0$.
I am not sure wether my approach to solve this is righ... | Start considering $$y=\frac{1+x}{1+\left(\frac x2 \right)^2}$$ and use Taylor around $x=0$; this will give
$$y=1+x-\frac{x^2}{4}-\frac{x^3}{4}+O\left(x^4\right)$$ Now, use the binomial expansion and get
$$y^n=1+nx+\frac n4 (2n-3)x^2+\frac n{12}(2 n^2-9 n+4)x^3+O\left(x^4\right)$$ Define $t=nx$ and you can write as
$$y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What does this connection between addition and multiplication mean? I noticing that some numbers when multiplied together give the same product as their sum. After some trial and error I came up with a simple rule for this.
$$n\times\frac{n}{n-1} = n+\frac{n}{n-1}$$
For example
$$7\times\frac{7}{6} = \frac{49}{6}$$ $$... | Notice $1 + \frac 1{n-1} = \frac {n-1}{n-1} + \frac 1{n-1} = \frac n{n-1}$
So $n\times \frac n{n-1} = n(1 + \frac 1{n-1}) = n + \frac n{n-1}$
Also note: If we were asked to solve $m*n = m + n$ then $m*n -m = n$ and $m(n-1) = n$.
Then if $n -1 = 0$ then $n = 1$ and $m*0 = 1$ which is impossible. So $n\ne 1$ and $n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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General formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$. If $k=1$, we have
\begin{align}
\sum_{n\ge1}\frac{1}{\binom{2n}{n}}&=\frac12\sum_{n\ge1}n\mathrm{B}(n,n)\\
&=\frac12\int _{0}^{1}\sum_{n\ge1}n(t-t^2)^{{n-1}}{\mathrm d}t\\
&=\frac12\int _{0}^{1}\frac{1}{(t-t^2-1)^2}{\mathrm d}t\\
&=\frac13+\frac{2\sqrt3\pi}{27}
\... | Chappers already wrote these series as hypergeometric functions, so I will just outline how to deal with the case $k=2$, exhibiting a relation with elliptic integrals. The key identities are
$$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx,\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$ How should I approach this type of a system equation?
$$
\begin{cases}
\dfrac xy-\dfrac yx=\dfrac{15}4\\
2x-5y=9
\end{cases}
$$
I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I ... | Put $x/y =t$ then $$t-{1\over t} = {15\over 4} \implies 4t^2-15t-4 =0$$
Since $(4t+1)(t-4)=0$ so $t=-1/4$ or $t=4$.
*
*case $y=-4x$ put in to $2x+20x=9$ so $x=9/22$ and $y=-18/11$.
*case $x=4y$ and $y=3$ and $x=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Finding the Equation of the Ellipse By Completing the Square I have the following equation of an ellipse:
$$5x^2+9y^2+40x=100$$
I need to put it in the form:
$$\frac{(x-h)^2}{m^2} + \frac{(y-k)^2}{n^2} =1 $$
I was trying to complete the square with the coefficients that have an $x$ variable.
$$5x^2 + 40x + 9y^2 = 10... | You did a good job just you made a mistake here
$$5(x^2 + 8x + 16) + 9y^2 = 100 + 16$$
Should be
$$5(x^2 + 8x + 16) + 9y^2 = 100 + 80$$
$$5(x^2 +4)^2 + 9y^2 = 180$$
$$\frac {(x^2 +4)^2}{36} + \frac {y^2}{20} = 1$$
$$.....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate without L'Hopital: $\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$ Evaluate the following limit without using L'Hospital method
$$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$
My turn is
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$
$$L=\lim_{x\to1}\le... | Observe that, by the standard binomial expansion,
$$
(1+h)^9-1=9h+36h^2+O(h^3),\qquad (1+h)^5-1=5h+10h^2+O(h^3)
$$ then, by setting $x=1+h$, as $x \to 1$ we have $h \to 0$, we get
$$
\begin{align}
\lim_{x\to1}\left(\frac{9}{x^9-1}-\frac{5}{x^5-1}\right)&=\lim_{h\to0}\left(\frac{9}{(1+h)^9-1}-\frac{5}{(1+h)^5-1}\right)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Probability of forming a triangle given sides to choose from
What is the probability of being able to form a triangle from three segments chosen at random from five line segments of lengths $1,3,5,7$ and $9$?
Denote with $A$ the event of forming a triangle from three segments chosen from the list above. Then, the seg... | You can choose $3$ segments from those $5$ on ${5\choose 3} =10$ ways, but only $3$, you listed, are OK.- So the answer is $0.3$
Order is here irrelevant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\cot(x)=2$ and $\cot(y)=3$ show that $x+y=\frac{\pi}{4}$. If $\cot(x)=2$ and $\cot(y)=3$ show that $x+y=\frac{\pi}{4}$.
I was trying to find x and y but is some weird thing. Also note that $x,y\:\in (0,\pi/2)$
Second idea was to write : $\frac{\cos x}{\sin x}=2$ and $\frac{\cos y}{\sin y}=3$ and adding this two I'l... | From this wikipedia page on trigonmetric identities, we see that
$\tan(\alpha + \beta) = \dfrac{(\tan \alpha + \tan \beta)}{1 - (\tan \alpha)(\tan \beta)}; \tag 1$
with
$\cot x = 2, \; \cot y = 3, \tag 2$
we find
$\tan x= \dfrac{1}{2} = 2^{-1}, \; \tan y = \dfrac{1}{3} = 3^{-1}; \tag 3$
thus
$\tan(x + y) = \dfrac{2^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c... | In the case $3 \times 3$ you might want to use the rule of Sarrus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\... | De Moivre's theorem says
$(\cosθ+i\sinθ)^{10}=\cos{10θ}+i\sin{10θ}$
$$\sin{10θ}=10\sinθ\cos^9θ-_{10}C_3\sin^3θ\cos^7θ+_{10}C_5\cos^5θ\sin^5θ
-_{10}C_7\cos^3θ\sin^7θ+10\cosθ\sin^9θ$$
$=\dfrac{10*3*4^9-10*3*4*3^3*4^7+9*2*7*3^5*4^5-10*3*4*3^7*4^3+10*4*3^9}{5^{10}}$
$$=\dfrac{1476984}{9765625}.$$ The result of wolfram alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Calculating $3^{m-n}=?$ $$9^m + 9^n = 52$$
$$9^m -4 = 2 \cdot 9^n$$
$$3^{m-n}=?$$
Let me show what I've tried
Simpifyling the both equalities.
$$3^{2^m} + 3^{2n} = 2 \cdot 13 \tag{1}$$
$$3^{2m} -2^2 = 2 \cdot 3^{2n} \tag{2}$$
Diving the second equality by $2$ and we have
$$\frac{3^{2m} -2^2}{2} =3^{2n} \tag{3}$$
H... | Putting
$$
x=9^{m}, \quad y=9^{n},
$$
we have
\begin{eqnarray*}
x+y&=&52\\
x-2y&=&4
\end{eqnarray*}
It follows that
$$
9^{m}=x=\frac{3x}{3}=\frac{108}{3}=36, \quad 9^{n}=y=\frac{3y}{3}=\frac{48}{3}=16,
$$
i.e.
$$
3^{m}=\sqrt{9^{m}}=\sqrt{36}=6,\quad 3^{n}=\sqrt{9^{n}}=\sqrt{16}=4.
$$
Hence
$$
3^{m-n}=\frac{3^{m}}{3^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the function is differentiable at $0$? Is the function given by
$\displaystyle f(x) = \begin{cases} \dfrac{1}{x\log(2)} - \dfrac{1}{2^x -1}, \quad &x \neq 0, \\ \dfrac{1}{2} , &x = 0 \end{cases}$
differentiable at $0$ ?
My attempt : $f'(0) = \frac {f(x) -f(0)}{x-0}$
$$\lim _{x\to 0}f'(0) = \lim _{x\to 0}\frac ... | You approach is correct, but the substitution is wrong! You should have proceeded in this manner:-
$$f'\left( 0 \right) = \lim_{x \rightarrow 0}\dfrac{f \left( x \right) - f \left( 0 \right)}{x - 0}$$
$$= \lim_{x \rightarrow 0}{\dfrac{ \left( \dfrac{1}{x \log 2} - \dfrac{1}{2^{x} - 1} \right) - \left( \dfrac{1}{2} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions to parametric polynomial equations The equation $ax^3-x^2+1=0$ has three real solutions if $a^2<4/27$. The three solutions are plotted below against parameter $a$
Is there some intuitive reason why instead of being two continuous functions, both the yellow branch and green branch are discontinuous at $a=0$? ... | One obvious thing is that the three functions given are all odd functions: the contents of the cube root is an even function, so if we write it as $b = \frac{1}{2}(2-27a^2-3\sqrt{3}\sqrt{27a^4-4a^2})$, we have three functions of the form
$$ \frac{1}{3a} \left(1+\omega b^{-1/3}+\omega^{-1} b^{1/3} \right), $$
where $\om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$ This is a question from a past Olympiad paper:
Three positive real numbers $a, b, c$ satisfy the following constraint:
$a+b+c = 1$. Show that the following inequality is true:... | Alt. hint: by AM-GM:
$$
\frac{1}{a}+\frac{1}{bc}=\frac{1}{a}+\frac{a+b+c}{bc} = \frac{1}{a} + \frac{1}{b}+\frac{1}{c}+\frac{a}{bc}\ge 4 \sqrt[4]{\frac{1}{b^2c^2}} = 4 \sqrt{\frac{1}{bc}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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If $x+y+z=2$ prove that $\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\ge2+\frac{1}{\sqrt{2}}$ Let $x,y,z$ be non-negative reals whose sum is $2$. Prove that
$\frac{1}{\sqrt{x^2+y^2}}+\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{z^2+x^2}}\ge2+\frac{1}{\sqrt{2}}$
I have tried bounding them up (assuming that $a\le b\le c$), many inequal... | Without loss of generality, assume that $x \geq y \geq z$. It is easy to verify that using $(x + \frac{z}{2}, y + \frac{z}{2}, 0)$ to replace $(x, y, z)$ would make the LHS smaller, i.e.,
$$
\frac{1}{\sqrt{x^2 + y^2}} +
\frac{1}{\sqrt{y^2 + z^2}} +
\frac{1}{\sqrt{x^2 + z^2}}
\geq
\frac{1}{\sqrt{\left(x + \frac{z}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this binomial sum? I come across the following binomial sum when studying the average-case time complexity of an algorithm:
$$\sum_{i = 1}^{n-k+1} i \binom{n-i}{k-1}$$
How to evaluate this sum?
| \begin{align*}
\sum_{i=1}^{n-k+1} i \binom{n-i}{k-1} &= \sum_{i=0}^{n-k} (i+1) \binom{n-i-1}{k-1} \\
&= \sum_{i=0}^{n-k} \big((n+1) - (n-i)\big) \binom{n-i-1}{k-1} \\
&= \sum_{i=0}^{n-k} (n+1) \binom{n-i-1}{k-1} - \sum_{i=0}^{n-k} (n-i) \binom{n-i-1}{k-1} \\
&= (n+1) \sum_{i=0}^{n-k}\binom{n-i-1}{k-1} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have th... | Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 16,
"answer_id": 0
} |
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
... | $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + \frac {1}{2} \ln C$$
$$ | 1 + 2y | =C(1+x^2)$$
$$ 2y = -1 +C(1+x^2)$$
$$ y = \frac {-1 +C(1+x^2)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\... | You can simplify it like this
$$ y = \sin(\sin^{-1} x + C) = \cos (C) \sin(\sin^{-1}x) + \sin (C)\cos(\sin^{-1}x) \\
= \cos (C)x + \sin(C)\sqrt{1-x^2} $$
The explanation is
$$ \cos(\sin^{-1}x) = \sqrt{1-\sin^2(\sin^{-1}x)} = \sqrt{1-x^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction.
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$.
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus:
$3^{2(k+1)+1}+2^{(k+1)+2}... | $$9(3^{2k+1}) + 2(2^{k+2}) = 7(3^{2k+1}) + 2(3^{2k+1}) + 2(2^{k+2}) = 7(3^{2k+1}) + 2(3^{2k+1} + 2^{k+2}) = 7(3^{2k+1} + 2m)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes. Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$
By induction: $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$. Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{... | It is even true for $n\ge3$. Indeed $\;\dbinom 93=\dfrac{9!}{3!\,6!}=84 >64$.
\begin{align}
&\text{Next, we have }\qquad\qquad
&\binom{3(n+1)}{n+1}&=\binom{3n}n\,\frac{(3n+1)(3n+2)(3n+3)}{(n+1)(2n+1)(2n+2)},&&\qquad\qquad\end{align}
so, by induction, all we have to prove is that
$$\frac{3(3n+1)(3n+2)}{2(n+1)(2n+1)}\ge4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Computing the coefficient of the term of a certain degree in a polynomial Given the polynomial
${1\over8}((1+z)^9 + 3(1-z)^4(1+z)^5 + (1-z)^6(1+z)^3)$
(which is the weight enumerator of a code)
how do I find out the coefficient of $z^2$?
The solution given is ${1 \over 8}(36-12+0) = 3$.
I got $36$ for the $z^2$ coeffic... | Notice that the coefficient of $z^2$ in $(a+bz+cz^2+\cdots)(d+ez+fz^2+\cdots)$ is $af+be+cd$.
After expansion of the powers,
$$1+9z+36z^2+\cdots+\\
3(1-4z+6z^2-\cdots)(1+5z+10z^2+\cdots)+\\
(1-6z+15z^2-\cdots)(1+3z+3z^2+\cdots)
$$
Which gives
$$\frac{36+3(10-20+6)+(3-18+15)}8=\frac{24}8.$$
Using Lord Shark the Unknown... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
quadratic system of three equations with three unknowns If $x,y,z$ are real and positive and they verify the system
$$\begin{cases}
x^2+xy+y^2=36\\
y^2+yz+z^2=64\\
z^2+zx+x^2=100
\end{cases}$$
Find the value of the sum $S=xy+yz+zx$
Of course I tried the obvious, to sum them all up, to bring each equation to the canonic... | If you are willing to use the Wolfram Cloud Sandbox then the answer comes out as
In[1]:= (x y+y z+z x)^2 /. Solve[{x^2+x y+y^2 == 36,
y^2+y z+z^2 == 64, z^2+z x+x^2 == 100}] //
Simplify//Sqrt//InputForm
Out[1]//InputForm=
32*Sqrt[3], 32*Sqrt[3], 32*Sqrt[3], 32*Sqrt[3]}
and so the answer is $\;32\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that
$$A=\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}$$
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Chole... | It is well-known that the eigenvalues of
$$
S=\begin{bmatrix} 0&1&0&0&0\\ 1&0&1&0&0\\ 0&1&0&1&0\\ 0&0&1&0&1\\ 0&0&0&1&0
\end{bmatrix}
$$
are
$$
2\cos\frac{k\pi}{n+1},\ \ \ k=1,\ldots,n.
$$
So the eigenvalues of $4I+2S$ are
$$
4+4\cos\frac{k\pi}{6}>0,\ \ \ k=1,\ldots,5.
$$
Being symmetric, this tells us that $$4I+4S=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 7
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Prove that the level curves of $\log\frac{|z-\sqrt{z^2-1}|}{2}$ are ellipses In a book it was mentioned that the level curves of the complex function $\phi (z) = \log\frac{|z-\sqrt{z^2-1}|}{2}$ is an ellipse with foci $1$ and $-1$. I can not establish a relation between the ellipse equation in the complex plane $|z-a|+... | You can just ignore $\log$ and $2$ in the fraction, and concentrate on $|z-\sqrt{z^2-1}| = r$ for $r>0$.
Let $w = z-\sqrt{z^2-1}$. Then $z - w = \sqrt{z^2-1}$ and $(z-w)^2 = z^2 - 1$. It follows that $z = \frac{w^2+1}{2w}.$ We have
\begin{align}
\left| \frac{w^2+1}{2w} + 1 \right| + \left| \frac{w^2+1}{2w} - 1 \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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prove this inequality with five variables Let $$\begin{cases}S=a^2+b^2+c^2+d^2+e^2\\
ab+bc+cd+de+ea=T_{1}\\
ac+ce+eb+bd+da=T_{2}
\end{cases}$$
Find the range $A$,such any postive real numbers have
$$S\ge AT_{1}+(1-A)T_{2}$$
This problem is creat by Wang yong xi .since I'm using the AM-GM inequality.
$$S=a^2+b^2+c^2+d^... | Indeed, the condition is satisfied for $-0.62 \simeq \frac{1-5^{1/2} }{2} < A< \frac43$ as will be shown below. Two examples are given.
We need to establish $S - AT_{1}-(1-A)T_{2} \ge 0 $ and we have that
$$
2[S - AT_{1}-(1-A)T_{2}] = A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2
$$
where $\sum_{cyc}$ says that the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by t... | You can do it using the concept of odds and evens which is sort of a poor mans version of unique prime factorization but surely acceptable. After all the classic proof of the irrationality of $\sqrt 2$ used even and odds without assuming unique prime factorization.
First note: If $b \in \mathbb Z$ is $b = 2m$ is even... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Without resorting to induction show that $L_n^2=L_{n+1}L_{n-1}+5(-1)^n$,Where $L_n$ is $n^{th}$ Lucas number.
Without resorting to induction show that
$L_n^2=L_{n+1}L_{n-1}+5(-1)^n$,Where $L_n$ is $n^{th}$ Lucas number.
By definition of Lucas number $L_n=L_{n-1}+L_{n-2}\implies L_{n-1}=L_n-L_{n-2}$ and $L_{n+1}=L_{... | The identity
$$
L_n^2=L_{n+1}L_{n-1}+5(-1)^n
$$
for the Lucas numbers seems to be closed related to Cassini's identity for the Fibonacci numbers:
$$
F_n^2 - F_{n+1}F_{n-1} = (-1)^{n-1}
$$
which follows from taking determinants in
$$
\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}
=
\begin{pmatrix} 1 & 1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Eliminate $x,y,z$ from equations Eliminate $x,y,z$ from equations
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=a$$
$$\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=b$$
$$\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right) =c$$
I try this problem with this method.
$$ab=\left(\fr... | Following is an alternate way to derive $c = ab - 1$.
Let $\lambda_1 = \frac{x}{y}$, $\lambda_2 = \frac{y}{z}$ and $\lambda_3 = \frac{z}{x}$. It is trivial to see $\lambda_1\lambda_2\lambda_3 = 1$.
In terms of $\lambda_1,\lambda_2,\lambda_3$, we have
$$\begin{align}
a &= \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = \lambd... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational.
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?
| Since $x\ne 0$ and $x\ne 2$ we have $${x^3-5x\over x^2-2x} = {x^2-5\over x-2}= q\in \mathbb{Q}$$
So $$x^2-5 = xq-2q \;\;\;\;{\rm and }\;\;\;\;x^2 =2x+r\in \mathbb{Q}$$
so $$ 2x+r-5 = xq-2q\implies (q-2)x =-2q-r+5$$ so if $q\ne 2$ we get $$x={-2q-r+5\over q-2}\;\;\;$$
but this can't be since $x$ is irational and left s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\int \frac{2x}{(x^2+x+1)^2}dx$ Solve
$$\int \frac{2x}{\left(x^2+x+1\right)^2}dx$$
I tried using integrating in parts by using the partial fraction method where the rational function is split into
$$\frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{(x^2+x+1)^2}$$
But then I just get the original formula with A=$0$ B=$0$ C=2 an... | Hint:
$\dfrac{d(x^2+x+1)}{dx}=?$
$$\dfrac{2x}{(x^2+x+1)^2}=\dfrac{2x+1}{(x^2+x+1)^2}-\dfrac1{(x^2+x+1)^2}$$
As $4(x^2+x+1)=(2x+1)^2+3$
how about starting with $2x+1=\sqrt3\tan y$
See also : Trigonometric substitutions, wiki
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$ How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
My attempt
$$(3x+5y)^{12}=\left(3x\left(1+\frac{5y}{3x}\right)\right)^{12}$$
$$=3^{12}x^{12}\left(1+\frac{5y}{3... | The $n^{th}$ term of $\left(\frac 32 + \frac{20}{3} \right)^{12}$ is
$t_n =\binom{12}{n}\left(\frac 32\right)^{12-n}\left(\frac{20}{3}\right)^n$ for $n = 0,1,\dots,12$. The ratio of consecutive terms is
$\rho_n = \dfrac{t_{n+1}}{t_n} = \dfrac{40}{9}\left(\dfrac{12-n}{n+1}\right)$.
These ratios are easy to compute
\be... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$ Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$
i got this problem in olympiad mathematics .
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+... | Hint:
You're on the right track. In the new sum you have, you can pair up the terms $1/n$ and $1/(101-n)$:
$$\sum_{n=34}^{50} \frac{1}{n}+\frac{1}{101-n}.$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\s... | $$\int(1+\cos2u)du=u+\dfrac{\sin2u}2+C$$
$$\sin2u=\dfrac{2\tan u}{1+\tan^2u}=?$$
Another way: $$\int\dfrac{dx}{(x^2+1)^n}=\int\dfrac1{2x}\dfrac{2x}{(x^2+1)^n}dx$$
$$=\dfrac1{2x}\int\dfrac{2x}{(x^2+1)^n}dx-\int\left(\dfrac{d(1/2x)}{dx}\int\dfrac{2x}{(x^2+1)^n}dx\right)dx=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to show that $\ln\left(\sin(\pi x)/(\pi x)\right)=\sum_{n=1}^{+\infty}\ln\left(1-x^2/n^2\right)$ I wanted to show for $x \in \left]-1,1\right]$ the beautiful relation
$$\ln\left(\frac{\sin\left(\pi x\right)}{\left(\pi x\right)}\right)=\sum_{n=1}^{+\infty}\ln\left(1-\frac{x^2}{n^2}\right)$$
I want to use it in order... | It's pretty straightforward if you are familiar with the product expansion of $\sin\theta$
$$\frac {\sin\theta}{\theta}=\prod\limits_{n\geq1}\left(1-\frac {\theta^2}{n^2\pi^2}\right)$$
Take the natural log of both sides to immediately see that$$\log\left(\frac {\sin\theta}{\theta}\right)=\sum\limits_{n\geq1}\log\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x... | $$\int\frac{1}{1+sin(x)}dx$$
Apply u-substitution:$u=tan(\frac x2)$
$$=\int \frac{2}{u^2+1+2u}$$
$$=2\int \frac{1}{u^2+1+2u}$$
$$=2\int\frac{1}{(u+1)^2}$$
Again apply u-substitution:$v=u+1$
$$=2\int\frac{1}{v^2}dv$$
$$=2\int\ v^{-2}dv$$
$$=2\frac{v^{-2+1}}{-2+1}$$
Substitute back $v=u+1,u=tan\frac {x}{2}$
$$=-\frac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqr... | $$\int\frac{-y^2}{\sqrt{1-y^2}}dy=\int\frac{1-y^2-1}{\sqrt{1-y^2}}dy
=\int\frac{1-y^2}{\sqrt{1-y^2}}dy-\int\frac{1}{\sqrt{1-y^2}}dy=$$ $$=
\int\sqrt{1-y^2}dy-\arcsin(y)=\left[y=\sin{t}, dy=\cos{t}\ dt\right]=\int\cos^2{t}dt-\arcsin(y)=$$ $$=\frac12\int(1+\cos{2t})dt-\arcsin(y)=\frac12\left[t+\frac12\sin{2t}\right]_{t=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using identity Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$
where $p_k(n)$ is the number of partitions of a... | You have $$P_2(x) = \frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right) = \frac12 (1-x)^{-2} + \frac12 (1-x^2)^{-1}$$
Now, the generalised binomial expansion is $(1+x)^a = 1 + \frac{a}{1!}x + \frac{a(a-1)}{2!}x^2 + \cdots$
Therefore $$\begin{eqnarray}
(1 - x)^{-2} &=& 1 + \frac{(-2)}{1!}(-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the inverse laplace transform of the given problem Find the inverse laplace transform of $ \ Y(s)= \large \frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10} \ $
Answer:
Let $ \ \mathcal{L}^{-1} \ $ be the inverse laplace operator.
Then,
$ y(t)=\mathcal{L}^{-1} [Y(s);t] \\ \Rightarrow y(t)= \mathcal{L}^{-1} \left... | You are on the right track, after you work out the constants $A$, $B$ and $C$ you should end up with
\begin{eqnarray}
Y(s) &=& \frac{1}{s - 6} - \frac{3s + 10}{s^2 + 6s + 10} \\
&=& \frac{1}{s - 6} - \frac{3(s + 3) + 1}{(s + 3)^2 + 1} \\
&=& \frac{1}{s - 6} - 3\frac{(s + 3)}{(s + 3)^2 + 1} - \frac{1}{(s + 3)^2 + 1}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which one of the following are correct?
Let, $A= \left[ {\begin{array}{cc} -1 & 2 \\ 0 & -1 \\
\end{array} } \right]$ , and $B = A + A^2 + A^3 +···+ A^{50}$. Then
$(A) B^2 = I $
$(B) B^2 = 0$
$(C) B^2 = A$
$(D) B^2 = B$
Eigenvalues of $A$ are $-1,-1$. So, Eigenvalues of $B$ are $0,0$. So, $\det(B)=0$. So... | Since $A-I$ is triangular with nonzero diagonal entries, it follows that $A-I$ has nonzero determinant, so is invertible, hence $I-A$ is also invertible.
Note also that any two poynomials in $A$ commute.
\begin{align*}
\text{Then}\;\;(I-A)B&=(I-A)(A+A^2+A^3+\cdots+A^{50})\\[4pt]
&=(A + A^2 + A^3 +···+ A^{50})-(A^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve this system of equations for the value of $a,b$ and $c$ such that $a^3+3ab^2+3ac^2-6abc=1$,$b^3+3ba^2+3bc^2-6abc=1$,$c^3+3cb^2+3ca^2-6abc=1$ Solve this system of equations for the value of $a,b$ and $c$.
$$a^3+3ab^2+3ac^2-6abc=1$$
$$b^3+3ba^2+3bc^2-6abc=1$$
$$c^3+3cb^2+3ca^2-6abc=1$$
This is symmetric equation.... | Let $$a^3+3ab^2+3ac^2-6abc-1=b^3+3a^2b+3bc^2-6abc-1$$ then we get
$$a^3-b^3-3ab(a-b)+3c^2(a-b)=0$$ Now use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Then you will get
$$(a-b)\left((a-b)^2+3c^2)\right)=0$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cd... | $$\int e^{-3x}\cos^3x dx=\frac{1}{8}\int e^{-3x}(e^{ix}+e^{-ix})^3dx\\
=\frac{1}{8}\int e^{-3x}(e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix})dx\\
=\frac{1}{8}e^{-3x}\left(\frac{e^{3ix}}{-3+3i}+\frac{e^{-3ix}}{-3-3i}+\frac{3e^{ix}}{-3+i}+\frac{3e^{-ix}}{-3-i}\right)\\
=-\frac{1}{8}e^{-3x}\left(\frac{3(e^{3ix}+e^{-3ix})+3i(e^{3ix}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
My Attempt
Using Partial fractions
$$
\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\frac{1}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)}+\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=I_1+I_2\\
... | $$I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$$
If $x = \frac ab \tan \theta$, then
*
*$(a^2+b^2x^2)=a^2(1+tan^2 \theta)=a^2 sec^2 \theta$
*$dx = \frac ab sec^2 \theta \ d\theta$
*$\displaystyle
\int \frac{dx}{(a^2+b^2x^2)^2}
= \int \frac{\frac ab sec^2 \theta \ d\theta}{a^4 sec^4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the DE: $y''+1/(2y^3)=0$. \begin{align}
y''+\frac{1}{2y^3}=0\\
\end{align}
This is a second order differential equation which doesn't contain $x$ explicitly.
Let $y'=p(y)$, so that $y''=p(y)*p'(y)$.
\begin{align}
pp'+\frac{1}{2y^3}=0\\
\int{pdp}=\int{-\frac{1}{2y^3}dy}\\
\frac{p^2}{2}=-\frac{1}{2}*\frac{-1y^{-2... | $y' = \frac {\sqrt {Cy^2 + 1}}{\sqrt 2 y}\\
\int \frac {y}{\sqrt {Cy^2 + 1}} \ dy =\int \frac {1}{\sqrt 2} \ dt \\
\frac 1C \sqrt {Cy^2 + 1} = \frac {1}{\sqrt 2} t + D $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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General term of a sequence. So i have the following sequence:
${1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ...}$
Where the number $i$ appears $i + 1$ times.
I would like to know the $n$-th term of this sequence. I tried to analise certain patterns within the sequence, but wasn´t able to conclude anything so far.
| I would like to put a "number pyramid" like this:
$$1-1$$
$$2-2-2$$
$$3-3-3-3$$
$$4-4-4-4-4$$
$$5-5-5-5-5-5$$
$$\cdots$$
$$k-\text{th floor: } k-k-k-k-k-...-k-k\text{ ($k+1$ times)}$$
The number of numbers appear in all the floors from $1$ to $k-1$:
$$2+3+4+5+\cdots+k=\dfrac{(k+2)(k-1)}{2}$$
The number of numbers appea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
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Simplifying $\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$ How to simplify the expression and get a closed form?
$$\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$$
where $\alpha := \frac12(m+1)\theta$ and $\beta:=\frac12(m-1)\theta$; and where $m$... | \begin{align}\frac{\sin\frac{m+1}{2}\theta\cos^2\frac{m-1}{2}\theta}{\sin\frac{m-1}{2}\theta}+\frac{\sin\frac{m-1}{2}\theta\cos^2\frac{m+1}{2}\theta}{\sin\frac{m+1}{2}\theta}&=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$
Evaluate $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$
I tried using the substitution $u=\frac {\pi}{2}-x$ and maybe thought this function might be symmetrical in some way and got: $$\lim_{n\to\infty}\int_... | A heuristic solution. Interchanging the order of integral and limit, we have
$$ \lim_{n\to\infty} \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx = \int_{0}^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx. $$
Now depending on the relative size of $\sin x$ and $\cos x$, we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Another way to find the sum $S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}$ I want to know the sum
$$S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}.$$ This is my way.
First, we find the sum
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}=... | $$S+\frac S2=\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{k-1}}+\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{\color{red}k}}\\
=\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{k-1}}+\sum_\color{red}2^{\color{red}{n+1}}\frac{(-1)^{k-2}(2k-3)}{2^{\color{red}{k-1}}}\\
=1+\sum_2^n\frac{(-1)^{k-2}(\color{red}{2k-1-2k+3})}{2^{k-1}}+\frac{(-1)^{n-1}(2n-1)}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find smallest $A\in \mathbb{R}$ such that $|f'(0)|\leq A.$ Problem
Find smallest $A\in \mathbb{R}$ such that for 3-degree polynomial satisfying the condition $\forall x\in[0;1],\ |f(x)|\le 1$ holds $|f'(0)|\le A$.
Solution
Assume that the 3-degree polynomial is $$ax^3+bx^2+cx+d,(a \neq 0).$$Let $x=0,\dfrac{1}{3},\dfrac... | I think an approach from interval arithmetic could be appropriate. Here the equation
$$
f([0,1])=[0,1]
$$
must be satisfied for your requirements.
$$
f([0, 1]) = a[0,1]^3+b[0, 1]^2+c[0,1]+d
$$
$$
= a[0,1] + b[0,1] +c[0,1] + d
$$
$$
= (a +b + c) [0,1] + d = [0,1]
$$
In order for the last equation to be valid statement $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Same integration with 2 different answers? $$\int x(x^2+2)^4\,dx $$
When we do this integration with u substitution we get
$$\frac{(x^2+2)^5}{10}$$
as $u=x^2+2$
$du=2x\,dx$
$$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$
Although when we expand the fraction and then integrate the answer we get is different:... | Like mentioned in the comments this is all fixed if you remember your constant of integration.
$$\int x(x^2+2)^4\ dx= \frac{(x^2+2)^5}{10}+C$$
Note if you expand
$$
\begin{split}
\frac{(x^2+2)^5}{10}&=\frac{1}{10}\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\right)\\
&=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$
Find the sum of
$$\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$$
I tried to convert all into
$\tan {}^{ - 1}(x) $
using
$\cot {}^{ - 1} (x) = { \tan {}^{ - 1} ( \frac{1}{x} ) }$
and then tried to simplify them using property
$ \tan {}^{ - 1} ( \frac... | After the fix of sign in front of $\frac{5}{12}$ in the question body, the sum becomes pretty simple.
Let $a_{\pm} = 3(r\pm \frac12)$, notice
$$\frac{a_+ - a_-}{1 + a_+ a_-} = \frac{3}{1 + 9(r^2-\frac14)}
= \frac{12}{36r^2 - 5} = \left(3r^2 - \frac{5}{12}\right)^{-1}$$
We have
$$\begin{align}\cot^{-1}\left(3r^2 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Value of $\cos\frac{2k\pi}{n}$ We know that $\cos\frac{2\pi}{5} = \frac{-1 + \sqrt{5}}{4}$ and $\cos\frac{4\pi}{5} = \frac{-1 - \sqrt{5}}{4}$. In general is there any formula for
$$\cos\frac{2k\pi}{n}$$
where $k\leq n/2$.
| Claim: for any $n\geq 3$, $\cos\frac{2\pi}{n}$ is an algebraic number over $\mathbb{Q}$ with degree $\frac{1}{2}\varphi(n)$, hence it can be found by solving $p(x)=0$ where $p(x)\in\mathbb{Z}[x]$ is such that $\deg p = \frac{1}{2}\varphi(n)$.
Proof: we first invoke a well-known lemma, i.e. that the cyclotomic polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $x-\frac{1}{x}$ if $x^{3}-\frac{1}{x^{3}}=76$. If $x$ is a real number that satisfies $x^{3}-\frac{1}{x^{3}}=76$, determine the value of $x-\frac{1}{x}$.
How can I solve this problem?
| $a^3-b^3=(a-b)(a^2+ab+b^2)$ and note that $(x-\frac{1}{x})^2 =x^2+\frac{1}{x^2}-2$ then if we set $(x-\frac{1}{x})=t$ we have $t(t^2+3)=76$ now find $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of given sequence Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$
My try:
$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the lim... | Because $f(x)={1\over 3}(1+x+x^3)$ is a strictly increasing function, compute its derivative, show recursively that that $0<a_n$, remark that if $a_{n-1},a_{n-2}$ are strictly inferior to $1$, $a_n\geq f(max(a_{n-1},a_{n-2}))$ and $f(max(a_{n-1},a_{n-2}))<f(1)=1$. _{n-1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve Differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ with substitution I am trying to solve this differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ using substitution, with the initial value $y(1)=0$ This is what I've done so far:
$$
\begin{align*}
y'(t) &= \frac{t^2+3y(t)^2}{2ty(t)} = \frac{t^2}{2ty(t)} + \frac{... | Or start with setting $v=y^2$,
$$
tv'=t^2+3v
$$
which is linear with integrating factor $1/t^3$ for the normalized equation,
$$
\left(\frac{v}{t^3}\right)'=\frac{tv'-3v}{t^4}=\frac1{t^2}\implies \frac{v}{t^3}=-\frac1t+C,~~v=Ct^3-t^2,\\~\\y=\pm t\sqrt{Ct-1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that w... | Your original statement is equivalent to
$$a_{n+1}^2=a_n(a_{n+1}+a_n)+(-1)^n,$$
that is
$$a_{n+1}^2-a_na_{n+1}-a_n^2=(-1)^n.\tag{1}$$
To prove $(1)$ all you need to do is to prove it for $n=1$ and verify
$$a_{n+2}^2-a_{n+1}a_{n+2}-a_{n+1}^2=-
(a_{n+1}^2-a_na_{n+1}-a_n^2)^.\tag{2}$$
This is an easy induction.
One can pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $a, b$ so that $0\leqq a, b\leqq 2, a^{-1}+ 2b^{-1}\geqq 2$. Prove that $a^{2}+ b^{2}\leqq 5$ .
Given $a, b$ so that $0\leqq a, b\leqq 2, a^{-1}+ 2b^{-1}\geqq 2$. Prove that $a^{2}+ b^{2}\leqq 5$ (equality: $a= 1, b= 2$).
For $a\leqq 1$, the inequality is clearly true!
For $a\geqq 1, F= a^{-1}+ 2b^{-1}- 2\geqq ... | Note that if $a=2$ and $b=4/3$ then $0\leq a,\,b\leq 2$, $\displaystyle \frac{1}{a}+ \frac{2}{b}= 2$, but
$$a^{2}+ b^{2}=\frac{52}{9}> 5.$$
P.S. Actually on the compact set
$$K:=\left\{(a,b)\in [0,2]\times[0,2]: \frac{1}{a}+ \frac{2}{b}\geq 2\right\}$$
the continuous function $f(a,b)=a^2+b^2$ attains its maximum val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt... | Your question was how to determine the value of $x$ in the given equation:
$$
\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}
$$
1.)
$
x+4+2 \sqrt{x+3} = \bigg( \frac{x+8}{3} \bigg)^2
$
2.)
$
x+4+2 \sqrt{x+3} = \frac{x^2+8^2}{3^2}
$
3.)
$
9(x+4+2 \sqrt{x+3}) = (x+8)^2
$
4.)
$
9x+36+18 \sqrt{x+3} = (x+8)(x+8)
$
5.)
$
18 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int... | Split the integral at $\frac{1}{\sqrt{2}}$ and use the substitution $x = \sqrt{1-y^2}$ in the second part to obtain
\begin{align}
I &\equiv \int \limits_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_{\frac{1}{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 0
} |
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Fractions in Questions and Answers
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