Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers
If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but se... | Here I shall find all critical points without verifying what kind of critical points they are (which seems to be your question). My notations are similar to yours, but a bit different, so I shall work from scratch but my answer borrows a lot of your ideas (great attempt, by the way). However, I do not expect that any... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + ... | As we know that $$\lim_{x\to 0}{x\over \sin x}=1$$therefore $$\lim_{x\to 0}{x^2\cos {1\over x}\over \sin x}=\lim_{x\to 0}x\cos{1\over x}=\lim_{u\to \infty}{\cos u\over u}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$ I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1... | Set $1/x=h\implies h\to0^+$
and $\sqrt{x^2+1}=\dfrac{\sqrt{1+h^2}}{|h|}=\dfrac{\sqrt{1+h^2}}h$ as $h>0$ as $h\to0^+$
So, we have $$ \lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}$$
$$=1+\lim_{h\to0^+}\dfrac{\sqrt{1+h^2}-1}h$$
$$=1+\lim_{h\to0^+}\dfrac{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many sequences can be made with 5 digits so that the difference between any two consecutive digits is $1$? Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten-digit sequences can be written so that the difference between any two consecutive digits is $1$?
I was wondering if my solution is right.
Let $a(n)$ ... | Your approach is correct, with the above 2 relations mentioned by you:
$$a(n) = b(n-1) = c(n)$$
$$b(n) = a(n-1) + 2c(n-1)$$
we can easily get the relation $$x(n)=3x(n-2)$$
Hence $x(10) = 3x(8) = 3^2 x(6) =...= 3^4 x(2)$, where $x(2) = 8$.
So, the answer to your question would be $2^3 3^4 = \boldsymbol{648}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Finding the Laurent series (complex numbers) I have
$$
f(z)={\frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=\infty$.
I did
$$
{\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n
$$
But in the b... | Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = \frac{1}{z(1-z)} = \frac{1}{z} + \frac{1}{1-z} = \frac{1}{1+(z-1)} + \frac{1}{1-z} $$
$$=$$
$$\frac{1}{(z-1)\left(\frac{1}{z-1} + 1\right)} - \frac{1}{z-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to find the other asymptote of $y=\sqrt{x^2+x}$ The task is to find the asymptotes of $y=\sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $\lim_{x \to \pm}y= \infty$.
Next, to find $m_{1,2}$: $$m_1=\lim_{x \to +\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_1$$
and $$m_2=... | You have the wrong value for $m_2.$
If $x \leq -1,$ then $\frac{\sqrt{x^2+x}}{x} \neq \sqrt{1+ \frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \frac{-1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$ Task
Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = -\frac{1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$
About
Hi, I have been trying to solve this task sin... | Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$\frac{1}{n (n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
$$\frac{1}{(n-1) n} = \frac{1}{n-1} - \frac{1}{n}$$
$$\frac{1}{(n-1)(n+1)} = \frac{1/2}{n-1} - \frac{1/2}{n+1}$$
$$\frac{1}{(n-1)n(n+1)} = \frac{1}{(n-1)n} - \frac{1}{(n-1)(n+1)} = \frac{1}{n-1} - \frac{1}{n} - \frac{1/2}{n-1} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $p, q, r$ prime numbers such that $r^2 - q^2 - p^2 = n^2$. Find $p, q, r$ prime numbers $p < q < r$ such that there exists natural number $n$ such that $r^2 - q^2 - p^2 = n^2$.
What I have done so far is obtaining the smallest number $p$ which is $2$.
The reason is that if we assume that $p$ is odd, as $p$ is the ... | Assume $q$ and $r$ are primes bigger than $3$. Then they are not a multiple of $3$, so their squares are $1$ mod $3$. Then $r^2-q^2-p^2$ will be $2$ mod $3$, so not a square.
So we need $q=3$. Then we need $r^2-13=n^2$ and there are only finitely many pairs of squares differing by $13$, and as you found $7^2-13=6^2$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Radius of convergence of $\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n$ I want to determine the convergence of the following series in dependency of $x$:
$\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n=\frac{3}{4}x+\frac{2}{5}x^2+\frac{1}{4}x^3+\frac{3}{17}x^4+ ... $
How can I solve this?
EDIT:
@Winther said, I... | Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
\to 1$).
Therefore
$f(x)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$\sum\limits_{n=1}^{\infty} x^n$
which is
$-1 < x < -1$.
At $x=1$,
the series is
$f(1)
=\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers
\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}... | Let $a=1/x, b=1/y, c=1/z$.
$$a-pb-pc=0$$
$$pa-8b+pc=0$$
$$pa+pb-13c=0$$
From here, you can get $$(p+1)a=(13+p)c\\\text{and}$$
$$(p+1)a=(p+8)b$$
Substitute those values for $b$ and $c$ in last equation to get
$$p^3+11p^2-52=0$$
It'll give you p, and then you can find the corresponding values of $x,y$ and $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Rationalizing denominator of $\frac{18}{\sqrt{162}}$. Cannot match textbook solution I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{18}{\sqrt{162}}$$
The solution is provided:
$\sqrt{2}$
I arrived at:
$$\frac{\sqrt{162}}{9}$$
Here is my thought process to arrive at this incor... | $$\require{cancel}\frac{18}{\sqrt{162}}=\frac{2\cdot3^2}{\sqrt{2\cdot3^4}}=\frac{2\cdot\cancel{3^2}}{\sqrt{2}\cdot\cancel{3^2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\cancel2\sqrt{2}}{\cancel{2}}=\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that a group with $3$ elements is cyclic?
Prove that a group with $3$ elements is cyclic.
I tried the case where $G=\{e,a,b\}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.
Are there any other methods
?
I have another question :
... | Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
\begin{array}{c|ccc}
& e & a & b \\ \hline
e & e & a & b \\
a & a & \\
b & b & \\
\end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find all $p \in \mathbb{Z}$ such that $ p^2+ 4p + 16 $ is a perfect square I've tried some things involving modular residues but they don't seem to work. Anyone know how to do this?
| So, we need $(p+2)^2+12=m^2$ for some integer $m$
$-12=(p+2-m)(p+2+m)$
As $p+2-m+p+2+m$ is even, the multiplicands have the same parity $\implies $ both must be ven
Consequently, $$\dfrac{p+2-m}2\cdot\dfrac{p+2+m}2=-3=1(-3)=(-1)3$$
Alternatively let $p^2+4p+16=(p+a)^2$ for some integer $a$
$\implies p=\dfrac{a^2-16}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\le... | Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^n \, dy = \frac{1}{n + 1},$$
the sum can be written as
$$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = \int_0^1 \int_0^1 \frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly... | A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 0
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Find the locus of the foot of perpendicular from the centre of the ellipse. Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
Consider two points $P$ and $Q$ such tha... | Let the centre of the ellipse be O and the foot be R. Also let point P is $(acos\theta, bsin\theta)$ then Q is $(acos(\theta+\frac{\pi}{2}), bsin(\theta+\frac{\pi}{2})$ or $(-asin\theta, bcos\theta)$ then PQ is
$$\frac{y - bcos\theta}{x + asin\theta} = \frac{b(sin\theta - cos\theta)}{a(sin\theta + cos\theta)} or \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Expected area of triangle inscribed in a circle On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .
| For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2\phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = \sqrt{(1 - (\frac{2}{\pi})^2}$ and $sin\phi = \frac{2}{\pi}$
Let the line joi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix demonstration $A^k$ Given a matrix $A = \begin{bmatrix} 7 & 4\\ -9 & -5 \end{bmatrix}$ $\in \mathcal{M2\times2}\, (\mathbb{R}) $
Show that $A^k = \begin{bmatrix} 1+6k & 4k\\ -9k & 1-6k \end{bmatrix} $
for every $k \in \mathbb{N}$
| Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = \begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 1 + 6 \cdot 1 & 4 \cdot 1 \\ -9 \cdot 1 & 1 - 6 \cdot 1 \end{bmatrix}; \tag 1$
then assuming that for some $k$
$A^k = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate $f ( x ) = \frac { \ln \left( x ^ { 2 } \cos ( x ) \right) } { \sqrt { 1 - x ^ { 2 } } }$ could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for m... | I would use logarithmic differentiation. Setting $u(x)=x^2\cos x$, we have
\begin{align}
\frac{f'(x)}{f(x)}&=\frac{u'(x)}{u(x)\ln u(x)}-\frac12\frac{-2x}{1-x^2} =\frac{2x\cos x-x^2\sin x}{x^2\cos x\,\ln(x^2\cos x)}-\frac12\frac{-2x}{1-x^2}\\[1ex]
&=\frac{2\cos x-x\sin x}{x\cos x\,\ln(x^2\cos x)}+\frac x{1-x^2},
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
minimum value of $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2\geq 3\cdot 2\sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})\geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c... | Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 \geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Extremum value of $f(x,y)=ax^2+2hxy+by^2$ subject to constraints $g(x,y)=x^2+y^2-c^2=0$. Find the extremum value of $f(x,y)=ax^2+2hxy+by^2$ subject to constraints $g(x,y)=x^2+y^2-c^2=0$, where $abc \neq 0$ and $(a+1)^2+4(a^2-b^2) \geq 0$.
My attempt: I have use Lagrange Multipliers to solve this. Let us assume $$\phi... | An alternative approach is using polar coordinates. WLOG, let $c=1$. Then you need to find the extrema of
$$a\cos^2t+2h\cos t\sin t+b\sin^2t=\frac{a-b}2\cos2t+\frac{a+b}2+h\sin2t.$$
As this expression describes a sinusoid, the values of the extrema are immediate:
$$\frac{a+b\pm\sqrt{(a-b)^2+4h^2}}2.$$
(If $c\ne1$, time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I procee... | Note that$$(x+1)^{n+m}=(x+1)^n(x+1)^m$$
Then by binomial theorem and collecting terms
\begin{align}
\sum_{r=0}^{n+m}\binom{n+m}{r}x^r &=
\sum_{i=0}^{n}\binom{n}{i}x^i\sum_{j=0}^{m}\binom{m}{j}x^j \\&=
\sum_{r=0}^{n+m}\sum_{i+j=r}\binom{n}{i}\binom{m}{j}x^r \\ &=
\sum_{r=0}^{n+m}\sum_{i=0}^r\binom{n}{i}\binom{m}{r-i}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Compute $ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$
Compute $$ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$$
I did the following:
$$ \lim\limits_{n \to \infty}\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}} = \... | By dividing the numerator and the denominator by $n$, you should have
$$\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}}=\frac{\sqrt{3+\frac{1}{n}-\frac{1}{n^2}}}{1+\sqrt{1-\frac{1}{n^2}}}.$$
Instead at your denominator we have $\frac{n}{n^2}+\sqrt{\frac{n^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.
When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately ... | It cannot occur between term $n$ and term $n+1$ if $\frac{x}{n} - \frac{x}{n+1} \ge 1$, equivalently $x \ge n^2 + n$, equivalently $n \le -\frac{1}{2} + \frac{\sqrt{1+4x}}{2}$.
It must occur, either between term $n$ and $n+1$, or between term $n+1$ and $n+2$, if $\frac{x}{n} - \frac{x}{n+1} \le \frac{1}{2}$, equivalent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
} |
Why does this math trick work? 35 by 11 is 385 because 3+5 is 8, so it's the digit in the middle.
Same for:
72 by 11 is 792 because 7+2 is 9, so it's the digit in the middle.
I see it works because 35 by 10 is 350, or 72 by 10 is 720. The 0 is replaced with the extra digit. The last digit is 5 by 1 or 2 by 1, so it sta... | Let "$ab$" be a two digit number.
Then $ab = 10a + b$ and $ab\times 11 = (10a+b)(10 + 1) =$
$ 10a(10 + 1) + b(10+1) =$
$ (100a + 10a)+ (10b + b) =$
$100a + (10a + 10b)+b =$
$100a + 10(a+b) + b$.
And if $a+b < 10$ we get $100a + 10(a+b) + b = a(a+b)b$.
Not it doesn't work if $a+b \ge 10$. Example $84\times 11= 924$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\sigma_n - e$ is decreasing faster than $e-S_n$ given specific $\sigma_n$ and $S_n$.
Let $\sigma_n$ and $S_n$ be defined as:
$$
\sigma_n = 3 - \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!} \\
S_n = 1 + \sum_{k=1}^n\frac{1}{k!}
$$
Show that $\sigma_n - e$ is decreasing faster than $e-S_n$.
I may use anything befo... | Presuming my cursory review of your work leading up to the inequality $\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!} \le e$ didn't miss an error and assuming $n \in \mathbb N$, we may proceed as follows:
\begin{aligned}\\
e-\sum_{k=0}^{n} \frac{1}{k!}
&= \sum_{k=n+1}^{\infty} {1\over k!}\\
&\geq \frac1{(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Complex analysis proof triangle inequality: Given: $|z+w|^2=|z|^2+|w|^2+2Re(z\bar w)$
Prove:$|z+w|\leq |z|+|w|$
Work done so far:
Let $z=x+iy$ and $w=a+bi$, then:
$$|x+iy+a+ib|=|z+w|=\sqrt{(x+a)^2+(y+b)^2}$$
$$\sqrt{x^2+y^2}+\sqrt{a^2+b^2}=|z|+|w|$$
Squaring it I get,
$$x^2+y^2+2|z||w|+a^2+y^2$$
After this I am lost, ... | You want to show that
$$\tag1
\sqrt{(x+a)^2+(y+b)^2}\leq \sqrt{a^2+b^2}+\sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$\tag2
{(x+a)^2+(y+b)^2}\leq a^2+b^2+x^2+y^2+2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2},
$$
which reduces to
$$\tag3
2ax+2yb\leq 2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2},
$$
and squaring again this is (a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Making $x$ the subject of $x^3-2x+y=1$ How to I make x the subject of this formula?
$$x^3-2x+y=1\tag1$$
My attempt:
I apply complete the square
$$(x-1)^3+y-1=1\tag2$$
$$(x-1)^3=2-y\tag3$$
$$x-1=\sqrt[3]{2-y}\tag4$$
$$x=1+\sqrt[3]{2-y}\tag5$$
but my teacher said it is wrong!
Can you please help? Thank in advance!
| Use the formula for a cubic and find three solutions for $x$:
*
*$\frac{\sqrt[3]{2} \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}+4 \sqrt[3]{3}}{6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$
*$\frac{i \sqrt[3]{2} \left(\sqrt{3}+i\right) \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}-4 \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of extracting a ball after two balls were swapped We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the bal... | The probability of ending up with an additional white ball equals:
$$\frac{11}{21} \frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$\frac{10}{21} \frac{13}{26}$$
We thus find an overall probability of:
$$\frac{11}{21} \frac{12}{26} \frac{11}{21} + \frac{10}{21} \frac{13}{26} \frac{9}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.
| Well, if you hit a snag that means either... the thing you are trying to prove is false... or it's not actually a snag.
You say $6^{n+5}\equiv 76*6^n \pmod {100}$ does not give the correct result.
How do you know $76*6^n\not \equiv 6^n \pmod {100}$?
...
$76*6^n \equiv 6^n \pmod {100} \iff$
$75*6^n \equiv 0 \pmod {100}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\se... | $$\begin{align}
\int t^2 \sqrt{(1+4t^2)}dt&=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx\\
&=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8\times 32}\left(\sinh(4x)-4x\right) +C \\
&= \frac{1}{256}\left((4\sinh(x)\cosh^3(x) + 4\sinh^3(x)\cosh(x)) - 4x\right) + C\\
&=\left[\frac{1}{256}\left((4(2t)(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Simplifying $\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$ In simplifying
$$\frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
| $\frac{(2x+1)(x-3)}{2x^3(3-x)}= -\frac{(2x+1)(x-3)}{2x^3(x-3)}= -\frac{(2x+1)}{2x^3} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| Let $\arctan x=y\implies x=\tan y,x^2-1=\dfrac{\sin^2y}{\cos^2y}-1=-(1+x^2)\cos2y$
$$2x\cos(2x)+(x^2-1)\sin2x=(1+x^2)\sin2(\arctan x-x)$$
$$I=\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} =\int\dfrac{2x^2}{(1+x^2)\sin2(\arctan x-x)}$$
Now set $u=\arctan x-x,du=-\dfrac{x^2\ dx}{1+x^2}$
$$I=-2\int\dfrac{du}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$
Calculate $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$
$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)... | This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$\begin{align*}
\int_0^\infty\frac{\ln^2(x)}{(1-x^2)^2}\,dx
&=\int_{-\infty}^\infty\frac{t^2e^{t}}{(1-e^{2t})^2}\,dt\\
&=\int_{0}^{+\infty}\frac{t^2e^{-t}}{(1-e^{-2t})^2}\,dt+\int_{0}^\infty\frac{t^2e^{-3t}}{(1-e^{-2t})^2}\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$ Show $\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
from LS
$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(3/2) + \log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2\big(... | Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.
1)$$
\log_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding weight in Megagrams (Mg) given circumference and density Glaciers often deposit large rocks called erratics. The granite rock has a circumference of 9.5 m. Assuming it conforms to the shape of a sphere, what would be its weight in Megagrams (Mg), where 1 Mg = 1,000 Kg ≈ 1 US ton. The average density of granite ... | HINT
*
*Given a circumference of the sphere, find its volume $V$.
*With volume and density, find the mass.
*Convert the mass to the desired units.
UPDATE
Since the circumference is $$C=2\pi r \iff r = \frac{C}{2\pi}$$ and $V = 4\pi r^3/2$, you have
$$
\begin{split}
V &= \frac43\pi r^3
= \frac43 \pi \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$
I would like to prove using Mean Value theorem for $n \ge 1$
$$\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$$
RHS can be proved by rationalizing the square root term, not sure about the LHS.
| $n^2+n=(n+\frac12)^2-\frac14<(n+\frac12)^2\Rightarrow \sqrt{n^2+n}-n<\frac12$
$\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}>\frac{n}{n+\frac12+n}>\frac{n}{2n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Problem solving a 4 by 4 linear system with infinite solutions I have a question here. The following is a 4 by 4 system of linear equations which has infinitely many solutions.
$$
\left\{
\begin{array}{c}
x+y+3z+t=0 \\
x-y-z-t=0 \\
3x+y+5z+3t=0 \\
x+5y+11z+8t=0
\end{array}
\right.
$$
An online calculator gave me t... | Use the matrix of the linear system and put it in reduced row echelon form:
\begin{align}
&\left[\begin{array}{rrrr}
1&1&3&1\\1&-1&-1&-1\\3&1&5&3\\1&5&11&8
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrrr}
1&1&3&1\\0&-2&-4&-2\\0&-2&-4&0\\0&4&8&7
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrrr}
1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ wi... | Just factorise:
$$\frac{x^2+x}{x^2-x-2} \equiv \frac{x(x+1)}{(x-2)(x+1)}$$
Assuming $x \neq -1$, we can divide the numerator and denominator by $x+1$. For all $x \neq -1$, we have
$$\frac{x^2+x}{x^2-x-2} = \frac{x}{x-2}$$
As $x \to -1$, we see that
$$\frac{x^2+x}{x^2-x-2} \to \frac{-1}{-1-2} = \frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Evaluate $\frac{2a}{a^2-4} - \frac{1}{a-2}-\frac{1}{a^2+2a}$
Evaluate
$$\dfrac{2a}{a^2-4} - \dfrac{1}{a-2}-\dfrac{1}{a^2+2a}$$
We have to see what their common term is. Therefrom, we can evaluate the simplified expression by canceling out.
$$a^2 - 4 = (a)^2 - (2)^2 = (a-2)(a+2) \tag {1}$$
$$a^2 +2a = a(a+2)\tag{2}$$... | \begin{align}
\dfrac{1}{a+2}\biggr(\dfrac{2a}{a-2} - \dfrac{a+2}{a-2}-\dfrac{1}{a}\biggr)&=\dfrac{1}{a+2}\biggr (\dfrac{2a-a-2}{a-2}-\dfrac{1}{a}\biggr)\\
&=\dfrac{1}{a+2}\biggr(\dfrac{a-2}{a-2}-\dfrac{1}{a}\biggr )\\
&=\dfrac{1}{a+2}\biggr(1-\dfrac{1}{a}\biggr)\\
&=\dfrac{1}{a+2}\biggr(\dfrac{a-1}{a}\biggr)\\
&=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Functional equation $ (n+1) f(n+1)= (a n+b) f(n) $ for $n=0,1,...$ I am looking for a solution to the following functional equation:
\begin{align}
(n+1) f(n+1)= (a n+b) f(n), n=0,1,...
\end{align}
where $a$ and $b$ are some positive constants.
Moreover, $f(n)$ is positive and
\begin{align}
0<f(n) &\le 1,\\
\sum_{n=0}... | One way is to use generating functions. Define
\begin{align*}
g(x) = \sum_{n=0}^{\infty} f(n) x^n
\end{align*}
Henceforth, I will use the notation
\begin{align*}
g(x) \leftrightarrow\{f(n)\}
\end{align*}
to indicate that $g(x)$ has coefficients of $f(n)$ for $x^n$ in its series expansion. Then we have
\begin{align*}
g'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the area of $\triangle ABC$ where $\triangle ADC$ is cyclic, point $P$ is on the circumference and $AD = AP$?
$\triangle ABC$ is a right angled triangle. The perpendicular drawn form $A$ on $BC$ intersects $BC$ at point $D$. A point $P$ is chosen on the circle drawn through the vertices of $\triangle ADC$ such... | Since $ADCP$ is square, we obtain $AD=BD=DC=x$
because $\measuredangle ACD=45^{\circ}$ and from here also $\measuredangle ABD=45^{\circ}.$
Thus, $BC=2x$ and $$350=PB^2=(2x)^2+x^2,$$ which gives $x^2=70$ and
$$S_{\Delta ABC}=\frac{2x\cdot x}{2}=70.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that |z|=1. I am trying to prove that
If $z\in \mathbb{C}-\mathbb{R}$ such that $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$. Show that $|z|=1$.
1 method , through which I approached this problem is to assume $z=a+ib$ and to see that $$\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$$.
So problem reduces to show that $... | We are given that the imaginary part of $\frac{z^2+z+1}{z^2-z+1}$ is zero.
Therefore $\frac{z^2+z+1}{z^2-z+1}$ is equal to its own conjugate:
$$\frac{z^2+z+1}{z^2-z+1} = \frac{\bar z^2+\bar z+1}{\bar z^2-\bar z+1}.$$
Cross-multiply (multiply both sides by $(z^2-z+1)(\bar z^2-\bar z+1)$)
and cancel all terms on the righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $\lim\limits_{(x,y)\to(0,0)}\frac{x^3y-xy^3}{x^4+2y^4}$ does not exist.
Show that $$\lim_{(x,y)\to(0,0)}\frac{x^3y-xy^3}{x^4+2y^4}$$ does not exist.
I'm not even sure how to approach this. I tried factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$, but I don't think that gets me anywhere with the de... | Doing the change to polar coordinates:
$$
\frac{x^3y - xy^3}{x^4 + 2y^4} =
\frac
{r^3\cos^3\theta\,r\sin\theta - r\cos\theta\,r^3\sin^3\theta}
{r^4\cos^4\theta + 2r^4\sin^4\theta}
= \frac
{\cos^3\theta\sin\theta - \cos\theta\sin^3\theta}
{\cos^4\theta + 2\sin^4\theta},
$$
dependent of $\theta$ (and independent of $r$),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Show that $\cot{142\frac{1}{2}^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
Show that $\cot{142\frac{1}{2} ^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
What I have tried:
Let $\theta = 142\frac{1}{2}^\circ \text{ and } 2\theta = 285^\circ$.
$$\cos 285^\circ = \cos 75^\circ$$
$$\cos 75^\circ = \frac{\sqrt3 - 1}{2\sqrt2}$$
$$\... | Good start. You got $$\cot \theta= \sqrt{\frac{1+\frac{\sqrt3-1}{2\sqrt2}}{1-\frac{\sqrt3-1}{2\sqrt2}}}=\sqrt{\frac{2\sqrt2+\sqrt3-1}{2\sqrt2-(\sqrt3-1)}}\; .$$
To simplify, multiply numerator and denominator by $2\sqrt2+\sqrt3-1.$
Note that the denominator then becomes $8-(\sqrt3-1)^2=(\sqrt3+1)^2. $
Thus $$\cot \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Does $\triangle ABC$ exist such that $\triangle ABC \sim \triangle DEF$, with $D, E, F$ being the incentre, centroid, orthocentre of $\triangle ABC$? Question:
Does $\triangle ABC$ exist such that $\triangle ABC \sim \triangle DEF$, with $D, E, F$ being the incentre, centroid, orthocentre of $\triangle ABC$, resp.?
F... | Here's what I mean by a brute force approach：
Without loss of generality, let's choose an appropriate scale such that $AB=1$.
Set
$$A=(0,0)$$ $$B=(1,0)$$ $$C=(x,y)$$
Then the coordinate of the centroid $E$ is
$$E=\left(\frac {x+1}{3},\frac {y}{3}\right)$$
To find out the coordinate of the incentre $D$, let's calcul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
For what values of $a$ and $b$ is the function $\frac{x^ay^b}{x^2+y^2}$ continuous at $(0,0)$? I have the function $$f(x,y)=\begin{cases}\dfrac{x^ay^b}{x^2+y^2} &(x,y)\neq(0,0)\\ 0 &(x,y)=(0,0) \end{cases}$$ I am trying to figure out what constants $a$ and $b$ will make the function continuous at $(0,0)$. I know that t... | $0\leq (x-y)^2=x^2+y^2-2xy$
so
$xy\leq \frac{1}{2}(x^2+y^2)$
then
$|\frac{x^ay^b}{x^2+y^2}|\leq|\frac{x^{a-1}y^{b-1}\frac{1}{2}(x^2+y^2)}{x^2+y^2}|=\frac{1}{2}|x^{a-1}y^{b-1}|\to 0$
if $a>1 $ and $b>1$
with the polar coordinate you get a better condition on $a,b$ :
$a+b>2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can anybody help with the integral: $\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} dx$ please? I have searched online. I have got a solution from Wolfram but I don't understand it, or how to reach it. If anyone has a method that would be fantastic, thanks!
| This is an elliptic integral, but it is interesting to point out what is the explicit value of the integral over $(0,1)$.
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-x}{x(1+x)}}\,dx=\frac{1}{2}\int_{0}^{1}\sqrt{\frac{x}{(1-x)(2-x)}}\,dx$$
equals
$$ \int_{0}^{1}\frac{x^2}{\sqrt{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
easy riemann sum problem goes hard Ok, so here it is the integral
$$\int_{0}^{\pi/2}\left[1 + \sin\left(x\right)\right]\,\mathrm{d}x
$$
which I must compute with the definition. And here is my atempt
$$
\int_{0}^{\pi/2}\left[1 + \sin\left(x\right)\right]\,\mathrm{d}x =
\lim_{n \to \infty}\sum_{k = 1}^{n}\mathrm{f}\left... | You are looking to evaluate the limit $$\lim_{n \to \infty}\sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\frac{\pi}{2n}.$$ Time to simplify! [Warning: this is longer than I expected.] The sum can be split into two parts so that
$$\sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factor $X^4 + 3$ into irreducible factors in $F_7[X]$ I am not quite sure where to start with this problem. I am new to polynomial rings and want to learn how to factor polynomials in polynomial rings made of fields.
Factor $X^4 + 3$ into irreducible factors in $F_7[X]$.
| You are working in $\Bbb F_7[x]$, the ring of polynomials with coefficients in the finite field $\Bbb F_7$. As Will Jagy also suggested, $3\equiv -4$ mod $7$, hence your polynomial is basically $x^4-4$.
$$x^4-4=(x^2+2)(x^2-2)$$
We have other two factors to check. Notice that $3^2=9\equiv 2$ mod $7$, hence the second f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $|\alpha-i|=|\beta-i|$ if $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2e^{i\theta}$
If $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2(\cos\theta+i\sin\theta),$ $0<\theta<\pi$, then prove that $|\alpha-i|=|\beta-i|$
My Attempt
$$
z^2-2e^{i\theta}z+1=0\implies\alpha+\beta=2e^{i\theta}\quad\&\qu... | This problem can be worked out using polar coordinates, but it seems to be a bit simpler to use rectangular coordinates.
That $|\alpha-i|=|\beta-i|$ follows from proving this claim: If $z$ is any root of $z+\frac{1}{z}=2e^{i\theta}$ with $0 \lt \theta \lt \pi$, then $|z-i| = \sqrt{2}$.
Let $z = x + yi$ and let $2e^{i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Error in fun problem of series and the binary representation of n The problem is as follows:
Let $a_n = :$ number of digits of $n$ in binary notation, and $$A_n = \frac{1}{n} \sum_{k=1}^{n} (-1)^{a_n} $$.
Show that $\limsup A_n = 1/3$.
From previous questions, I have asserted that
$$a_n =
\left\{
\begin{array}{111111}... | In the last line, instead of
$$A_i = \frac{1}{2^{k+1} - 1} \times \frac{1 - {2^{2}}^{k/2}}{1 - 2^2} $$
it should be
$$A_i = \frac{1}{2^{k+1} - 1} \times \frac{1 - 2^{k+1}}{1 - 2^2}=\frac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Real matrix satisfying $A^3=4I_n-3A$
Let $A\in M_n(\mathbb{R}) $ so that $A^3=4I_n-3A$. Prove that $\det(A+I_n) =2^n$.
My work : $A$ 's eigenvalues are the roots of $x^3+3x-4=0$, so one of the eigenvalues is $1$ and the others are $\lambda_1$ and $\lambda_2$, the roots of $x^2+x+4=0$.
Hence,
$$\det(A+I_n) =2(\lam... | Careful, you only know that the spectrum of $A$ is contained in the set of zeroes of $x^3+3x-4$.
Since $A$ is a real matrix, its minimal polynomial has real coefficients and divides $x^3+3x-4$, so it is one of
$$x-1, x^2+x+4, (x-1)(x^2+x+4)$$
Hence the characteristic polynomial is of the form $\chi_A(x) = (x-1)^{n-2r}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $3\mid p^3 \implies 3\mid p$ I want to prove $3\mid p^3 \implies 3\mid p$ (Does it?)
The contrapositive would be $3 \nmid p \implies 3 \nmid p^3$ I believe.
$3\nmid p \implies p = 3q + r$ ($0<r<3$), so $p^3 = 27q^3+27q^2r+9qr^2+r^3$
Dividing by $3$ we get $3(9q^3 + 9q^2r + 3qr^2) + r^3$
Is this correct so far? ... | Since $0\lt r\lt3$ so either $r=1$ or $r=2$.
If $r=1$ then $r^3=1$ . So $3(9q^3+9q^2r+3qr^2)+r^3=3(9q^3+9q^2r+3qr^2)+1$ which leaves remainder $1$ when divided by $3$
And similarly, if $r=2$ then $r^3=8$ . So
$3(9q^3+9q^2r+3qr^2)+r^3=3(9q^3+9q^2r+3qr^2)+8=3(9q^3+9q^2r+3qr^2)+3\cdot2+2=3(9q^3+9q^2r+3qr^2+2)+2$
which le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Show that the following sequence converges. Please Critique my proof. The problem is as follows:
Let $\{a_n\}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}\leq a_n+\frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.
My (wrong) proof:
Notice that
$$
|a_{n+1}-a_n|\leq \left|\frac{(-1)^n}{n}\right|\leq\fr... | Define $b_k := a_{2k+1}$. Then
$$b_k \leq a_{2k} + (-1)^{2k}\frac{1}{2k} \leq b_{k-1} + (\frac{1}{2k} - \frac{1}{2k-1}) \leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k \to b$.
Suppose $a_n \nrightarrow b$. Then there exists an $\varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > \va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove that in every triangle the inequality $a^3r_a + b^3r_b + c^3r_c \ge 8S(2R-r)^2 $ takes place Prove that in every triangle the inequality $$a^3r_a + b^3r_b + c^3r_c \ge 8S(2R-r)^2 $$ takes place, with the usual notations ($a,b,c$ lengths of sides, $r_a, r_b, r_c$ radii of corresponding excircles, $R$ radius of cir... | Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x>0$, $y>0$ and $z>0$ and in the standard notation we need to prove that
$$\sum_{cyc}\frac{a^3\cdot2S}{b+c-a}\geq8S\left(\frac{abc}{2S}-\frac{2S}{a+b+c}\right)^2$$ or
$$\sum_{cyc}\frac{a^3}{b+c-a}\geq\frac{(4abc(a+b+c)-16S^2)^2}{16S^2(a+b+c)^2}$$ or
$$\sum_{cyc}\frac{a^3}{b+c-a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Harmonic series in probability mass function problem Suppose $X$ is a discrete random variable with possible values $\{1, 2, 3,\dots\}.$ Further, suppose the p.m.f is $$c\left(\frac{1}{x}-\frac{1}{x+1}\right)\enspace\text{s.t. $c > 0$}$$
Find c and $E[X].$
Idea:
We have $$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1... | You can write $$\sum a_n-b_n=\sum a_n-\sum b_n$$only if at least one of $\sum a_n$ or $\sum b_n$ is bounded. In this case$$\sum_{x=1}^{\infty}{1\over x}-{1\over x+1}{=\left(1-{1\over 2}\right)\\+\left({1\over 2}-{1\over 3}\right)\\+\left({1\over 3}-{1\over 4}\right)\\+\left({1\over 4}-{1\over 5}\right)\\+\cdots\\=1}$$t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A binomial double sum I was doing some numerical experiment and I found that
$$\sum_{k=2}^{\infty}\frac{1}{k(k-1)}\sum_{n=k}^{\infty}\frac{1}{\binom{n}{k}(n+1)}=\sum_{k=2}^{\infty}\frac{1}{(k-1)k^2}=2-\zeta(2).$$
I wonder if the following variation
$$\sum_{k=2}^{\infty}\frac{1}{k(k-1)}\sum_{n=k}^{\infty}\frac{1}{\binom... | We have that the inner sum equals
$$
\eqalign{
& \sum\limits_{k\, \le \,n} {{1 \over {\binom{n}{k}\left( {n - 1} \right)}}} = \cr
& = \sum\limits_{0\, \le \,n} {{1 \over {\binom{n+k}{k}\left( {n + k - 1} \right)}}} = \cr
& = k!\sum\limits_{0\, \le \,n} {{1 \over {\left( {n + k} \right)^{\,\underline {\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve these trigonometry equations? I have to work with the following 5 equations:
*
*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$
*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$
*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$
*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfra... | $2. 2\cos2y+1=2(1-2\sin^2y)+1=\dfrac{\sin3y}{\sin y}$ for $\sin y\ne0$
$3.\dfrac{\cos2y}{\sin3y}=\dfrac{2\cos2y\sin y}{2\sin3y\sin y}=?$
$4. \dfrac{\cos x}{\sin3x}=\dfrac{\sin(3x-x)}{2\sin3x\sin x}=?$
$5.\sin x=\sin((n+1)x-nx)=?$
Set $n=1,2,3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of $(x^2+y^2)^2$ if $x,y$ are real number such that $x^2+2xy-y^2=6$ Then find minimum value of $(x^2+y^2)^2$
what i try : $x^2+2xy+y^2-2y^2=6$ or $(x+y)^2-\bigg(\sqrt{2} y\bigg)^2=6$
put $\displaystyle (x+y)=\sqrt{6}\cos \alpha$ and $\displaystyle \sqrt{2}y=\sqrt{6}\sin \alpha$
$\displaystyle x=\sqrt{6}\c... | I think a fun way to do the problem similar to what you've done is by using polar co-ordinates. Sub $x=r \cos \theta$ and $y=r \sin \theta$. Then you want to minimise $(x^2+y^2)^2=r^4$. Note that the constraint simplifies massively.
$$ x^2+2xy-y^2=6$$
$$r^2( \cos ^2 \theta +2 \sin \theta \cos \theta -\sin ^2 \theta )=6... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I'm having a hard time rationalizing the proof that $2^n > n^2$ if $n >4$ I'm not sure how to apply $n > 4$ in the proof, when I work it out I get that $n >2$...
Basis Step $P(5) = 2^5 > 5^2 = 32 > 25 $ which is True
Hypothesis assume $P(k) = 2^k > k^2$ is true. Prove that $P(k+1)$ is true.
Substituting $k+1$ into the... | First of all, this is false when $n=3$. $2^3 = 8 < 9 = 3^2$. This becomes clear if you properly write out your inequality:
$$2^k > k^2 >\frac{k^2 + 2k + 1}{2}.$$
The right inequality holds for $k > 2$ but the left one only holds for $k =5$, and $k \geq 5$ once you prove the inequality. When you plug in $n=3$, you get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i so... | In fact, you are given that $$ \int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{px^3+qx^8}{(x^5-1)^2}+c$$ Because of the denominator in the integrand, rewrite the rhs as
$$\frac{P_n(x)}{x^5-1}$$ Differentiate both sides to get
$$\frac{2x^7+3x^2}{(x^5-1)^2}=\frac{\left(x^5-1\right) P_n'(x)-5 x^4 P_n(x)}{\left(x^5-1\right)^2}$$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How to simplify $\sqrt{2+\sqrt{3}}$ $?$
Simplify $\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$
The answer to this question is $\frac{4}{3}$ in a workbook.
How would I simplify $\sqrt{2+\sqrt3}$ $?$ If it was something like $\sqrt{3 + 2\sqrt2}$ , I would have simplified it as follows:
$\sqrt{3 + 2\sqrt2... | Hints:
$$\sqrt 2+\sqrt 6=\sqrt 2(1+\sqrt 3)=\sqrt{2(1+\sqrt 3)^2}=\sqrt{2(4+2\sqrt 3)}=2\sqrt{2+\sqrt 3}$$
$$\sqrt{2+\sqrt 3}=\frac 1{\sqrt 2}\sqrt{4+2\sqrt 3}=\frac 1{\sqrt 2}\sqrt{(\sqrt 3+1)^2}=\frac{\sqrt 3+1}{\sqrt 2}=\frac{\sqrt 6+\sqrt 2}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Prove:$$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
I tried to use the Laplace expansion, but it seems useless.
| Hint:
$C_1'=C_1+C_2+C_3$
$$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}$$
$$=\begin{vmatrix}a+b+c&b&c\\b+c+a&c&a\\c+a+b&a&b\end{vmatrix}$$
$$=(a+b+c)\begin{vmatrix}1&b&c\\1&c&a\\1&a&b\end{vmatrix}$$
Now either expand or
use $R_2'=R_2-R_1, R_3'=R_3-R_1$
| {
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"url": "https://math.stackexchange.com/questions/3151461",
"timestamp": "2023-03-29T00:00:00",
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$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$ Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$
Answer: $K=(ab+bc+ca)$
My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatr... | Note that we can rewrite the determinant as a sum:
$$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}b^2&c^2\\b^3&c^3\end{vmatrix} - \begin{vmatrix}a^2&c^2\\a^3&c^3\end{vmatrix} + \begin{vmatrix}a^2&b^2\\a^3&b^3\end{vmatrix} = (b^2c^3 - b^3c^2) - (a^2c^3 - a^3c^2) + (a^2b^3 - a^3b^2)$$
Now, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of the absence of isolated vertices. We have a graph with n vertices. Each edge can appear with probability $\frac{1}{n\ln(n)}$. Prove that $\lim\limits_{n\rightarrow\infty}Pr[$there is no isolated vertices$] = 0$.
I think that first of all we should connect vecrtices so there won't be isolated vertices. We... | I assume that the edge choices are independent. Put $p=\frac{1}{n\ln(n)}$. A probability that a given vertex is isolated is $$(1-p)^{n-1}\ge 1-(n-1)p\ge 1-\tfrac{1}{\ln n}$$ by Bernoulli's inequality, and the last value tends to $1$.
| {
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"url": "https://math.stackexchange.com/questions/3151921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a ... | You can prove that $$\color{red}{(x-1)^3\geq {3\over 4}x-1}$$ for all nonegative $x$; it is equivalent to $x(2x-3)^2\ge 0$, so $$(a-1)^3+(b-1)^3+(c-1)^3\geq {3\over 4}(a+b+c)-3= -{3\over 4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve tan(x)+cos(x)=1/2 Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
\frac{\sqrt{1-c^2}}{c}+c=\frac{1}{2}
$$
$$
\sqrt{1-c^2}+c^2=\frac{1}{2}c
$$
$$
1-c^2=c^2(\frac{1}{2}-c)^2=c^2(\frac{1}{4}-c+c^2)
... | If we put $t=x/2$ then we get $${2\tan t\over 1-\tan ^2t} +2\cos ^2 t -1={1\over 2}$$
Let $y= \tan t$. Since $\cos ^2t = {1\over 1+y^2}$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the type of the equation Given
$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0$
I need to find the type of the equation, for that i tried to get the canonical form
So, turn into lambdas
$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3... | The equation
$$\sum_{i,j=1}^na_{ij}u_{x_ix_j}+a(x,u,\nabla u)=0$$
can be reduced to canonical form by applying the non singular linear transformation $\xi=B^Tx$, where B is a matrix such that the transformation $y=B\eta$ reduces the quadratic form $\sum_{i,j=1}^na_{ij}y_iy_j$ to canonical form.
Example. Reduce to cano... | {
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Finding the value of $\lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$
Finding the value of $\displaystyle \lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$
What I tried
Let $\displaystyle I_{k} =\int^{\frac{\pi}{4}}_{0}\cos^{k}(x)\mat... | Result
Let
$$f(n) = \int_0^\frac{\pi}{4} \cos(x)^n\,dx$$
then
$$\lim_{n\to \infty } \, \sqrt{n} f(n)=\sqrt{\frac{\pi }{2}} \simeq 1.2533141373155001\tag{1}$$
and
$$\lim_{n\to \infty } \, \sqrt{n} f(2n)=\frac{\sqrt{\pi }}{2} \simeq 0.8862269254527579\tag{2}$$
Here $(2)$ is the limit asked for in the OP.
Derivation
Decom... | {
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Meaning of linear independence with row vectors So far I have understood that a set of vectors $S = {v_1, v_2, . . . , v_k }$ in a vector space V is linearly independent
when the vector equation
$c_1v_1 + c_2v_2 + . . . + c_kv_k = 0$
has only the trivial solution$c_1 = 0, c_2 = 0, . . . , c_k = 0.$
An example in matrix... | Yes, it is "exactly like the second example" you gave. In that example, you were showing that the original four vectors were dependent. But there will always be some subset of those vectors that is independent. (That subset might consist of a single vector. A set containing just one vector is always "independent".)... | {
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An integration that wolfram cannot help me. $$\int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx$$
I noted the fact that $\frac{d(x\cos x)}{dx}=-x\sin x+\cos x$ but I cannot apply the substitution on it.
| Let's write an antiderivative Ansatz $f(x)\exp (x\sin x+\cos x)$ so $$f^\prime(x) + f(x)x\cos x=x^2\cos x-\frac{1}{x}\sec x\tan x+\frac{1}{x^2}\sec x\\=x^2\cos x-\left(\frac{1}{x}\sec x\right)^\prime=1+x\cos^2 x-\left(\frac{1}{x}\sec x\right)^\prime-\frac{1}{x}\sec x\cdot x\cos x.$$But by inspection, this has solution ... | {
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Using transformation to evaluate double integral
Given the transformation $T(x, y) = (x - y, x + y)$, evaluate the double integral
$\iint_R (x^2+y^2) dA$, where $R$ is the rectangle in the $xy$-plane with vertices $A(1, 1)$, $B(2, 2)$, $C(-1, 5)$ and $D(-2, 4)$.
Computing the Jacobian of the transformation, I get $... | HINT
One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
\begin{split}
A &= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[\left(\frac{u+v}{2}\right)^2 + \left(\frac{v-u}{2}\right)^2\right] 2\ du\ dv \\
&= \frac12
\int_{v=2}^{v=4} \int_{u=-6}^{u=0}... | {
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Find $\lim_{n \rightarrow\infty } P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$. $\lim_{n \rightarrow\infty } P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$.
This is a past problem of a high-school level math compettion.
M... | There's some telescoping going on here. We have
$$\frac{n^3-1}{n^3+1}=\frac{n-1}{n+1}\frac{n^2+n+1}{n^2-n+1}.$$
Then
$$\prod_{n=2}^N\frac{n-1}{n+1}=\frac13\frac24\frac35\frac46\cdots\frac{N-1}{N+2}$$
and
$$\prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}=\frac73\frac{13}7\frac{21}{13}\frac{31}{21}\cdots\frac{N^2+N+1}{N^2-N+1}.$$
B... | {
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Geometric proof of $\sin x \geq x - x^3 /6 $? We know (from Taylor expansion for example) that if $x \geq 0$, then $\sin x \geq x - \frac{x^3}{6}$.
In Prove that: $\sin(x) \cos(x) \geq x-x^3$ a geometric proof of the inequality $\sin x \geq x - \frac{x^3}{4}$ is given. Is there any geometric proof of the first one (whi... | You meant this holds for acute $x$. Take a radius-$\sqrt{2}$ centre-$O$ circle of radii $OA,\,OB$ with $\angle AOB=x$. We'll work in Cartesian coordinates $X,\,Y$; rotate the diagram so the line segment $AB$, which is of length $2\sqrt{2}\sin\frac{x}{2}$, has endpoints at $$X=\pm\sqrt{2}\sin\frac{x}{2},\,Y=0,$$and let ... | {
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"source": "stackexchange",
"question_score": "2",
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Partial fractions, disagreement with Wolfram Alpha On my math homework, I have this problem, and WolframAlpha says that $A=-\frac{1}{7}$ and $B=\frac{1}{7}$. However, while solving the problem on my own, I found a restriction that $A \neq -B$. How is it that this answer works? The problem in question is:
$$\dfrac{1}{x^... | To find $A$ and $B$:
Note that
$(x + 2)(x - 5) = x^2 -3x - 10; \tag 1$
then from
$\dfrac{1}{x^2 -3x - 10} = \dfrac{A}{x + 2} + \dfrac{B}{x - 5} \tag 2$
we have
$A(x - 5) + B(x + 2) = \dfrac{(x + 2)(x - 5)}{x^2 -3x - 10} = 1; \tag 3$
this may be written
$(A + B)x + (2B - 5A) = 1, \tag 4$
whence
$A + B = 0 \Longrightarro... | {
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Calculate $\sum_{n=1}^\infty \frac{n+1}{n(n+2)^2}$ using Basel Problem sum I have this series I need to calculate:
$$\sum_{n=1}^\infty \frac{n+1}{n(n+2)^2}$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$\sum_{n=... | Write (for example using Partial-fraction decomposition)
$$\frac{n+1}{n(n+2)^2} = \frac{1}{2(n+2)^2}+\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+2}\right).$$
The second bracket summed is a telescoping sum, specifically
$$\sum_{n=1}^k\frac{1}{n}-\frac{1}{n+2}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{... | {
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Integral roots of cubic equation $x^3-27x+k=0$ The number of integers $k$ for which the equation $x^3-27x+k=0$ has atleast two distinct integer roots is
(A)$1$
(B)$2$
(C)$3 $
(D)$4$
My Attempt: The condition for cubic $x^3+ax+b=0$to have $3$ real roots happens to be $4a^3+27b^2\leq0$. But how to go about finding cond... | If $x=b$ is one of the solutions
$$k=27b-b^3$$
Now if $b$ is a repeated root and $c$ is the third one,
$0=b+b+c\iff c=-2b$
$$\implies x^3-27x+k=(x-b)^2(x+2b)=x^3+x(2b+b^2)-2b^3$$
$\implies b^2+2b=-27\iff b^2+2b+27=0$ which does not have an integer solution.
So, we can not have repeated roots.
The rest two solutions wil... | {
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"question_score": "2",
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Find the value of D.
The complex numbers $1+i$ and $1+2i$ are both roots of the equation $x^5-6x^4+Ax^3+Bx^2+Cx+D=0$, where $A, B, C, D \in R$ Find the value of D.
My attempt: The given equation will have 5 roots (distinct or undistinct) since it is a polynomial equation of degree 5. The coefficients are all real. Si... | Hint:
Use Vieta's formula
$6=1+i+1-i+1-2i+1+2i+t$
where $t$ is the fifth root
Can you complete the solution now?
| {
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Determine the real numbers $a$, $b$, $c$ such that $1$, $\frac1{1+\omega}$ and $\frac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$ I am stuck on this question:
Let $1$, $\omega$ and $\omega^*$ be the cube root of unity.
a. Show that $\dfrac1{1+\omega}=-\omega$ and $\dfrac1{1+\omega^*}=-\omega^*$.
b... | We use the familiar fact that
$$
(x - \alpha) (x - \beta) = x^{2} - (\alpha + \beta) x + \alpha \beta.
$$
For part a, note that $1, \omega, \omega^{*}$ are the three distinct roots of
$$
x^{3} - 1 = (x -1) (x^{2} + x + 1).
$$
Therefore $\omega, \omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals th... | {
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Numbers of roots in Quadratic Equations If $a,b,c,d$ are real numbers, then show that the equation $$(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$$ has at least two real roots.
| Adding the discriminants of the factors we get $(a^2+12b)+(c^2-4b)+(d^2-8b)=a^2+c^2+d^2.$
If the sum is positive then at least one of discriminants is positive and so at least one of three factors has two distinct roots.
Let $a^2+c^2+d^2=0$ then the equation is $(x^2-3b)(x^2+b)(x^2+2b)=0$ and evidantly has 2 or 4 dist... | {
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Alternate complex binomial series sum
Calculation of $\displaystyle \sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}$ is
My Try: Using $$\int^{1}_{0}x^m(1-x)^ndx = \frac{1}{(m+n+1)}\cdot \frac{1}{\binom{m+n}{n}}$$
So $\displaystyle \int^{1}x^{2n-r}(1-x)^r=\frac{1}{2n}\cdot \frac{1}{\binom{2n}{r}}$
Sum ... | $\text{Lemma: } \sum_{k=0}^{n-1} k x^k=\frac{(n-1)x^n+1}{(x-1)}-\frac{(x^n-1)}{(x-1)^2}
$$\text{Proof: }\sum_{k=0}^{n-1} x^k=\frac{x^n-1}{x-1}\implies\sum_{k=0}^{n-1} k x^{k-1}=\frac{n(x-1)x^{n-1}-(x^n-1)}{(x-1)^2}$
$$\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}=\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdo... | {
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"source": "stackexchange",
"question_score": "8",
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Is $x=0$ the inflection point of $f$?
Is $x=0$ the inflection point of $f= \begin{cases} \sin \frac{1}{x}\cdot e^{\frac{-1}{x^2}}, x \neq 0 \\ 0, x=0 \end{cases}?$
I know that the infection point exist when $f''$ changes the sign in this point.I examine that $f$ has $f', f''$ in $x=0$ but from Mathematica I know that... | $f$ is of the form
$$\tag1f(x)=\begin{cases}\left(p(\frac1x)\sin\frac1x+q(\frac1x)\cos\frac1x\right)e^{-1/{x^2}},&x\ne0\\0,&x=0 \end{cases} $$
where $p,q$ are polynomials. Show that every function of form $(1)$ is differentiable with derivative also of form $(1)$ (but of course with diffferent polynomials $p,q$).
Also... | {
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$y_0 \ge 2$, $y_n = y_{n-1}^2 -2$ $\Rightarrow$ $\frac{1}{y_0}+\frac{1}{y_0y_1}+\cdots = \frac{y_0 - \sqrt{y_0^2 - 4}}{2}$ $y_0 \ge 2$ and $y_n = y_{n-1}^2 - 2$. Let $S_n = \frac{1}{y_0} + \frac{1}{y_0 y_1}+\cdots + \frac{1}{y_0 y_1 \cdots y_{n}}$. Prove that
$$\lim_{n \rightarrow \infty} S_n = \frac{y_0 - \sqrt{y_0^2 ... | $\text{let} P_n=y_0 y_1\text{...} y_n$, now I will show that $S_n^2-y_0 S_n+1=\frac{1}{P_n^2}$ by MI.
Base case $n=0$ is obvious, assume $S_k^2-y_0 S_k+1=\frac{1}{P_k^2}$, when $n=k+1$
\begin{align}
S_{k+1}^2-y_0 S_{k+1}+1 &=
\left(S_k+\frac{1}{P_{k+1}}\right)^2-y_0 \left(S_k+\frac{1}{P_{k+1}}\right)+1 \\&=
S_k^2-y_0S... | {
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Integral $\int_0^1 \frac{\ln(1+x)}{1+x^3}dx$ Earlier today I saw this integral around here and gave it a try without success, unfortunately it got taken down so it didn't receive to much attention, but I think it's a nice integral (although it seems quite hard) and finding a closed form it's worth trying.
Evaluate $$I... | The 'sister integral' approach works for quite a few other integrals, but I do not know how to proceed in this particular case as well (note, by the way, that you have used the geometric series outside its radius of convergence in your calculations), so here's a sketch of the somewhat laborious brute-force method:
Afte... | {
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"source": "stackexchange",
"question_score": "10",
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searching for a formula for $T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$ So we have a number $n$, which is a power of two.
And we have the following recursion:
$$ T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$$
I solved some exercises like this, but I have a problem with this one.
I don't see the structure:
for the first... | For $n=2^N$ let $F(N)=T(n)$. Then
$$T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$$
implies that
$$F(N) = F(N-1) + F(N-2) + 2^N.$$
which is a linear recurrence with characteristic equation $z^2-z-1=0$, the same of the Fibonacci sequence $1,1,2,3,5,8,13,\dots$. Hence the solution is
$$F(N)=A\varphi^N+B(-\varphi)^{-N}+2^{N+... | {
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 ... | We have $$11x\equiv -9\equiv 28\mod 37$$ so we get
$$x\equiv \frac{28}{11}\equiv \frac{65}{11}\equiv\frac{102}{11}\equiv \frac{139}{11}\equiv \frac{176}{11}\equiv 16\mod 11$$
| {
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"url": "https://math.stackexchange.com/questions/3198371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I prove this combinatorial identity Show that
$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$
One way that I did it was to use the idea of generating functions.
For ... | Using your functions, consider
$$
3^n f_2(\frac13) = 3^n \frac{1}{(1-\frac13)^{n+1}} = \frac32 (\frac92)^n\\ = {n \choose n}3^n + {n+1 \choose n}3^{n-1} + \cdots + {2n \choose n} + {\color{red}{ {2n +1 \choose n} 3^{-1}+ \cdots}}
$$
and further
$$
2^n f_4 (\frac12) = 2^n (\frac32)^{2n+1} = \frac32 (\frac92)^n \\= {2n... | {
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Number of ways to pick a team of 4 with at least 1 girl and 1 boy from 4 girls and 4 boys. To find the number of ways to pick a team of $4$ with at least $1$ girl and $1$ boy from $4$ girls and $4$ boys, I thought to manually assign the team with $1$ girl and $1$ boy to begin with.
$$G, \quad B, \quad G \textrm{ or } B... | The counting
$${4 \choose 1} \cdot {4 \choose 1}\cdot {6 \choose 2}=240$$
is the number of teams with 1 "labelled" girl and 1 "labelled" boy. In order to "unlabel" them split ${6 \choose 2}$ into ${3 \choose 2}{3 \choose 0}+{3 \choose 0}{3 \choose 2}+{3 \choose 1}{3 \choose 1}$ and divide each term by the total numbers... | {
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"source": "stackexchange",
"question_score": "2",
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Show that $a^3 - b^3 = c! - 18$ does not have a solution
Let $a, b,$ and $c$ be positive integers and $c \gt 6$.
Show that the equation $$a^3 - b^3 = c! - 18$$ does not have a solution for all positive integers $a, b,$ and $c$.
What I have realized so far is that if $a^3 - b^3 = c! - 18$, then it must also be true t... | Consider modulo $3$. When $c$ is sufficiently large, arbitrarily large (in particular, $>2$) powers of $3$ divide $c$. Suppose $c$ is sufficiently large in this sense, and suppose $a,b$ existed so that the equation is satisfied. The RHS is $0$ modulo $3$, so we have $a^3=b^3$ mod $3$, which you can easily check implies... | {
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"source": "stackexchange",
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Solve for $x$ and $y$, The equations are $x \cos^{3} y+3x \cos y \sin^{2} y =14 $ and $ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $ Consider the system of equations
$$x \cos^{3} y+3x \cos y \sin^{2} y =14 $$
$$ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $$
$1)$ the values of $x$ is /are..
Answer is $ \pm\sqrt 5 $
The number ... | I think your approach is quite good.
\begin{align*}
\frac{x(\sin^3y+3\cos^2y\sin y)}{x(\cos^3y+3\cos y\sin^2y)}&=\frac{13}{14}\\
\frac{\tan^3y+3\tan y}{1+3\tan^2y}&=\frac{13}{14}\\
14\tan^3y-39\tan^2y+42\tan y-13&=0\\
(2\tan y-1)(7\tan^2y-16\tan y+13)&=0
\end{align*}
$\tan y=\frac12$.
$y$ has $6$ values in $(0,6\pi)$.
... | {
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can re... | Let $\tan\beta=t$.
Thus, $t\neq1$, otherwise $\gamma=\frac{\pi}{2},$ which is impossible.
Also, $$0<\beta<\frac{3\pi}{4},$$ which is
$$0<\beta<\frac{\pi}{4}$$ or
$$\frac{\pi}{4}<\beta<\frac{\pi}{2}$$ or $$\frac{\pi}{2}<\beta<\frac{3\pi}{4},$$ which is
$$0<t<1$$ or
$$t>1$$ or
$$t<-1.$$
Consider three cases.
*
*$t>1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove $\cos(x+y)+\sin(x-y)=2 \sin(x+\frac{\pi}{4}) \sin(\frac{\pi}{4}-y)$ I spend some time trying to figure out how to prove the following identity:
$$ \cos(x+y)+\sin(x-y)=2 \sin\left(x+\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}-y\right) $$
I tried to use the following identities:
$$ \cos(x+y)=\cos(x) \sin(y)... | Note that
$$\sin(x+y)+\sin(x-y)=2\sin x \cos y$$
i.e.
$$\sin(A)+\sin(B)=2\sin \frac{A+B}{2} \cos \frac{A-B}{2}$$
Then you can say
\begin{align}
\cos(x+y)+\sin(x-y)
&= \sin(x+y+\frac{\pi}{2})+\sin(x-y)\\
&= 2 \sin(x+\frac{\pi}{4})\cos(y+\frac{\pi}{4})\\
&= 2 \sin(x+\frac{\pi}{4})\sin(\frac{\pi}{4}-y)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $2^{2^{2^{\cdot^{\cdot^{2}}}}} \mod 9 = 7$
Prove that $\underbrace{2^{2^{2^{\cdot^{\cdot^{2}}}}}}_{2016 \mbox{
times}} \mod 9 = 7$
I think that it can be done by induction:
Base:
$2^{2^{2^{2}}} \equiv 2^{16} \equiv 2^8 \cdot 2^8 \equiv 2^4 \cdot 2^4 \cdot2^4 \cdot2^4 \equiv 7^2 \cdot 7^2 \equiv 4 \cdot ... | Where you fail, is in assuming exponents mod the modulus works. It wouldn't generally. the exponent is mod 6, when doing mod 9. So you get it equivalent to: $$128\cdot 2^{3k}=128\cdot8^k$$ Which then depends on parity of k (because 8's order mod 9 is 2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Need help to understand this solution of $2^mp^2 +1=q^5$ The Question-
Find all triples $(m,p,q)$ where $ m $ is a positive integer and $ p ,
> q $ are primes.
****$2^mp^2 +1=q^5$****
After trying my best to solve this problem, and progressing a little bit, I decided to look up the solution and I am unable to quite u... | I'm going to add some words to the solution.
$$2^mp^2=(q-1)A\tag1$$
where $A=q^4+q^3+q^2+q+1$.
If:$d=5\large \Rightarrow p=5\Rightarrow A=5$ since$ A$ is odd , But
$A>1+1+1+1+1=5$
If $d=5$, then the right hand of $(1)$ is divisible by $5$, so it follows that $p=5$. Since $A$ is odd, we see that $2^m$ divides $q-1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Combinatorics generating function question Find the generating function for the sequence $a_n$ which counts the number of bags containing $n$ pieces of fruit in which there is an odd number of apples, at most $1$ banana, and at least $1$ orange. Find $a_{40}$.
I came up with the equation $(x+x^3+x^5...)(1+x)(x+x^2+x^... | No, $a_{40}$ is the coefficient of $x^{40}$ of the Taylor expansion of $\frac{x^2}{(1-x)^2}$. You can find this by starting $\frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+\dots$ and taking the derivative to get $$\frac{1}{(1-x)^2}=\sum_{n=0}^\infty nx^{n-1}$$
So,
$$\frac{x^2}{(1-x)^2}=\sum_{n=0}^\infty nx^{n+1}$$
What i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises.
Theme in the book is mathematical induction. There were examples, where w... | You like to prove $$q+q^2+\dots+q^n=\frac{q(1-q^n)}{1-q}.$$ using Mathematical induction.
The expression is true for $n=1$ because it is simply $q=q$
If true for n, you want to show that it is also true for $n+1$
Note that $$ q+q^2+\dots+q^n+q^{n+1} = \frac{q(1-q^n)}{1-q} + q^{n+1} = $$
$$ \frac {q(1-q^n)+(1-q)q^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find inverse Laplace transform of : $\ln(\frac{s^2+a^2}{s^2+b^2})$ Question :
Find inverse Laplace transform of :
$$\ln \left(\frac{s^2+a^2}{s^2+b^2}\right)$$
My try :
I'm trying use this identity :
$f(t)=-\frac{\mathcal{L}^{-1}(\frac{dF(s)}{ds})}{t}$
Let $F(s)=\ln(\frac{s^2+a^2}{s^2+b^2})$
Then :
$\frac{dF(s)}{ds}=\... | Using the change of variable $s=\frac{1}{x}$
$$F\left( s \right)=\ln \left( \frac{{{s}^{2}}+{{a}^{2}}}{{{s}^{2}}+{{b}^{2}}} \right)=\ln \left( \frac{1+{{a}^{2}}{{x}^{2}}}{1+{{b}^{2}}{{x}^{2}}} \right)=\ln \left( 1+{{a}^{2}}{{x}^{2}} \right)-\ln \left( 1+{{b}^{2}}{{x}^{2}} \right)$$
and the power series expansion:
$$\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Chebyshev polynomial property
I want to prove inequality (5.13) but I have a problem with (5.16). I have: $$ \sin(n\theta) = \sin\theta \cos(n-1)\theta + \sin(n-1)\theta \cos\theta = $$ $$ = \sin\theta \cos(n-1)\theta + \cos\theta [\sin\theta \cos(n-2)\theta + \sin(n-2)\theta \cos\theta] = $$ $$ = \sin\theta \cos(n-1... | Note that
\begin{align*}
\frac{\sin n\theta}{\sin\theta}
&=\frac{e^{in\theta}-e^{-in\theta}}{e^{i\theta}-e^{-i\theta}}\\
&=\frac{(e^{i\theta}-e^{-i\theta})[e^{i(n-1)\theta}+e^{i(n-3)\theta}+\dots+e^{-i(n-1)\theta}]}{e^{i\theta}-e^{-i\theta}}\\
&=e^{i(n-1)\theta}+e^{i(n-3)\theta}+\dots+e^{-i(n-1)\theta}\\
&=\begin{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \lim_{x\to \frac{1}{{\sqrt 2}^+}} \frac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\frac{1}{\sqrt{2}}}$
$\displaystyle \lim_{x\to {1\over \sqrt{2}^+}} \dfrac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\dfrac{1}{\sqrt{2}}}$
I have tried substituting $x$ for $\sin \theta$, doing the calculations and ended up with -$2... | Let $x=t+\frac1{\sqrt 2}$ to make
$$A=\dfrac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\dfrac{1}{\sqrt{2}}}=\dfrac{\cos ^{-1}(T)}t\qquad \text{with} \qquad T=2 \left(t+\frac{1}{\sqrt{2}}\right) \sqrt{1-\left(t+\frac{1}{\sqrt{2}}\right)^2}$$
Now, using Taylor expansions around $t=0^+$
$$T=1-4 t^2-4 \sqrt{2} t^3+O\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"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.