Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding range of a log function with variable base How can I find the range of $\log_{x-7}(x-5)$?
Here is my attempt.
Let $$y=\log_{x-7}(x-5)$$
Let $z=x-7$, $y=\log_z(z+2)$
then $y$ satisfies
$$z^y-z-2=0, z>0, z \ne 1.$$
$y$ be written as $\frac{\ln (x-5)}{\ln (x-7)}$.
| The domain of the function is $x>7$ and $x \ne 8$. We note that $\ln (x-5)>0$. $\ln(x-7)$ takes negative value when $x<8$ and it takes positive value when $x>8$.
Let's differentiate it,
$$f'(x) = \frac{\frac{\ln (x-7)}{x-5}-\frac{\ln (x-5)}{x-7}}{(\ln (x-7))^2}=\frac{(x-7)\ln (x-7) - (x-5) \ln (x-5)}{(x-5)(x-7)(\ln (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$
Prove
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$$
Proving right hand side to left hand side:
$$\begin{align}\csc^2x+2\csc x \cot x+\cot^2x
&= \frac{1}{\sin^2x}+\dfrac{2\cos x}{\sin^2x}+\dfrac{\cos^2x}{\si... | The following relations would be used in the answer:$$1-\cos^2x = \sin^2x$$ $$\sin 2x=2\sin x\cos x$$
Taking LHS,
$$\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x} = \frac{2\sin x(1+\cos x)}{2\sin x(1-\cos x)}$$
Cancelling $\sin 2x$
$$=\frac{1+\cos x}{1-\cos x}$$
Multiplying $1+\cos x$ to numerator and denominator
$$=\frac{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Roots of An Equation - Finding Equations in other degrees of roots (Link: https://pastpapers.papacambridge.com/viewer/caie/cambridge-advanced-as-and-a-level-mathematics-9231-2021-may-june-9231s21gtpdf-9231s21ms11pdf-9231s21ms12pdf-9231s21ms13pdf-9231s21ms21pdf-9231s21ms22pdf-9231s21ms23pdf-9231s21ms31pdf-9231s21ms32pdf... | (From dxiv)
Original Equation $x^4 - 2x^3 - 1 = 0$
Therefore $x^4 = 2x^3 +1 $
Using each of the roots instead of x, we get
$\alpha^4 = 2\alpha^3 +1$
.....
.....
$\delta^4 = 2\delta^3 +1$
Add these all together to get $\alpha^4+\beta^4+\gamma^4+\delta^4 = 2(\alpha^3+\beta^3+\gamma^3+\delta^3) + 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$10$ circles ($2$ large of radius $R$, $6$ small of radius $r$ and 2 small of radius $t$) are enclosed in a square. How we find $r$ in terms of $t$? Let us embed $2$ large intersecting circles of radius $R$ into a square as depicted by the figure below. These two circles are highlighted green. Into these $2$ circles we... | Let the centres of the green circles be $O_1$ (lower) and $O_2$ (upper).
Let $P$ be the midpoint of $O_1O_2$ which is also the centre of the square.
Let the uppermost two orange circles above the line $O_1O_2$ have centres $C_1$ (lower) and $C_2$ (upper).
Let $Q$ be the point where the circle centre $C_2$ touches the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove that $\sqrt1+\sqrt2+\sqrt3+....+\sqrt n
$Q.$ Prove that $$\underbrace{\sqrt1+\sqrt2+\sqrt3+....+\sqrt n}_{\alpha}<n\sqrt{\underbrace{\frac{n(n+1)}{2}}_\beta}$$
Basically we need to prove that $$\alpha^2<n^2\beta$$
And $$\alpha^2=\beta+2(\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+......+\sqrt{(n-1)n})$$
using AM-GM on $(n-... | Here's what I did , $$\alpha^2=(\sqrt{1}+\sqrt{2}+\sqrt{3}+......+\sqrt{n})^2\Rightarrow$$ $$\color{red}{\sqrt1.\sqrt1}+\sqrt1.\sqrt2+\sqrt1.\sqrt3+.....+\sqrt1.\sqrt n\\\sqrt2.\sqrt1+\color{red}{\sqrt2.\sqrt2}+\sqrt2.\sqrt3+.....+\sqrt2.\sqrt n\\\sqrt3.\sqrt1+\sqrt3.\sqrt2+\color{red}{\sqrt3.\sqrt3}+.....+\sqrt3.\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve the following limit $\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n}$ can someone try to solve this limit? $$\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times \sqrt{n}$$
I tried to solve it like $\lim _{n\to \infty }\left( e^{\sqrt{n+1} -\sqrt{n}} -1\right) \times ... | Note that $$ \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} +\sqrt{n}} $$
Let us denote this expression $x_n$. We have $\lim_{n\to\infty} x_n = 0$. Knowing that $\lim_{x\to0} \frac{e^x-1}{x} =1$, we get
\begin{align} & \lim_{n\to\infty} (e^{\sqrt{n+1} - \sqrt{n}} - 1)\sqrt{n} = \\
&=\lim_{n\to\infty} \frac{e^{x_n} - 1}{x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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f(xy)=f(x) * f(y) Let f be a function such that $f(mn) = f(m) f(n)$ for every positive integers m and n. If $f(1), f(2)$ and $f(3)$ are positive integers, $f(1) < f(2),$ and $f(24) = 54$, then $f(18)$ equals ?
Process:-
I attempted to solve this question using 2 approaches , but couldn't reach to the answer using appro... | With $f(2)=3$ and $f(3)=2$ the form of $f(x)=x^t$ is not the correct form.
Note that $2^t=3 \implies t=\log_2 3$ while $3^t=2 \implies t=\log_3 2$
We do not necessarily get a closed form for this function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Probability all number will appear when rolling k n-sided dice If k balanced n-sided dice are rolled, what is the probability that each of the n different numbers will appear at least once?
A case of this was discussed here, but I’m not sure how to extend this. Specifically, I’m not sure how the to calculate the number... | (Please allow me to use $n$ in place of your $k$, and $m$ in place of your $n$)
So we have $n$ fair $m$-face dice.
If you consider the dies to be distinct, by color or by launching them in sequence, then
the space of events is given by $m^n$ equiprobable words (strings , m-tuples) of length $n$
formed out of the alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Eliminating $r$ from $6\tan(r+x)=3\tan(r+y)=2\tan(r+z)$ Here is the question:
If $$\boldsymbol{6\tan (r+x)=3\tan(r+y)=2\tan(r+z)}$$
show that
$$\boldsymbol{3\sin^2(x-y)+5\sin^2(y-z)-2\sin^2(z-x)=0}$$
It seems easy but the natural approach goes off the rails.
What I have so far,
Take the two equalities and cross multipl... | Just as you derive,
$$\sin(2r+x+y)=3\sin(y-x)\tag1$$
$$\sin(2r+y+z)=5\sin(z-y)\tag2$$
you can get, $$\sin(2r+z+x)=-2\sin(z-x)\tag3$$
From (1), can you get, $$\begin{align}\sin(2r+x+y)\sin(y-x)&=3\sin^2(y-x)\\ \frac12\left[\cos(2r+2x)-\cos(2r+2y)\right]&=3\sin^2(y-x)\tag{1´}\end{align}$$?
From (2), can you get, $$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.
| Make life easier writing
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}=x\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^2\sin(x^2)}$$ Now, let $t=x^2$ to make
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^2\sin(x^2)}=\frac{(t+1) \ln(t+1)-\sin(t)}{t\sin(t)}$$ Using equivalent $\ln(t+1)\sim t$ and $\sin(t)\sim t$
$$\frac{(t+1) \ln(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Determining the period of $ \frac{\sin(2x)}{\cos(3x)}$ I would like to compute the period of this function which is a fraction of two trigonometric functions. $$ \frac{\sin(2x)}{\cos(3x)}$$
Is there a theorem for this? what trick to use to easily find the period?
I started by reducing the fraction but I'm stuck on the ... | Suppose the period is $p$, and suppose the domain of the function is suitably defined, then for all values of $x$ in the domain, we must have $$\frac{\sin 2x}{\cos 3x}=\frac{\sin(2(x+p))}{\cos(3(x+p))}$$
$$\implies \sin 2x\cos(3x+3p)=\cos 3x\sin(2x+2p)$$
If we set $x=0$, we have $$\sin(2p)=0\implies 2p=k\pi.k\in\mathbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
If $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$, then find the value of $\frac{(1+x)^3}{1+x^3}$
Let $x$ be a real number such that $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$. What is the value of $\frac{(1+x)^3}{1+x^3}$?
Of course, one way is that to solve for $x$ from the quadratic $37(1+x)^2=13(1+x^2)$ which gives the value $x... | Given
$$\frac{(1+x)^2}{1+x^2}=a,$$
we have
$$\frac{(1+x)^3}{1+x^3}=\frac{(1+x)^3}{(1+x)(1-x+x^2)}=\frac{(1+x)^2}{1-x+x^2}.$$
Now,
$$\frac{1-x+x^2}{(1+x)^2}=\frac{\frac32(1+x^2)-\frac12(1+x)^2}{(1+x)^2}=\frac{3}{2a}-\frac12,$$
so the answer is
$$\frac1{\frac{3}{2a}-\frac12}=\frac{2a}{3-a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofco... | We deal with the case $c > 0, h > 0, n\ge 1$.
The equation is written as
$$b = \frac{2n\pi}{h} + \frac{\arctan\frac{c}{b}}{h}.$$
Clearly, the equation has a unique real solution $b$. More precisely,
this solution is located in $ (\frac{2n\pi}{h}, \frac{2n\pi + \pi/2}{h})$.
Using the famous Lagrange inversion theorem, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integration operator appears inside an integration operator as I try $~\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right)~$ $$ A:=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1} $$
$$=\lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\cdot\sin^{}\left(x\right)\right)\cdot\exp\... | Just proceed calculations using equations of laplace transformations of $~ \sin^{}\left(x_{}\right) ~~,~~ \cos^{}\left(x\right) ~$
$$ A=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1} $$
$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Quadratic Recurrences is there any way to solve 'simple' recurrences like $a_0=2$, $a_{n+1}=(2a_n-1)^2$ or even $a_0=1$, $a_{n+1}=(2a_n+1)^2$ ? Any help much welcomed !
Regards, Knud
| Let us denote $b_n = 2a_n -1$, then
$$b_0 = 3$$
$$b_{n+1} = 2b_n^2 -1$$
We seek $x$ satisfying $b_0 = \text{cosh}(x) := \frac{e^x+e^{-x}}{2}$ then
$$x = \text{arcosh}(3) = \ln(3+2\sqrt{2})=2\ln(\sqrt{2}+1)$$
Now, it is easy to prove that $b_n = \text{cosh}(2^nx)$, indeed
$$b_{n+1} = 2\left(\frac{e^{2^n x} +e^{-2^n x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+...+x_4 \leq 54, x_1+...+x_5 \leq 90$. Find the maximum of ...
$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+...+x_4 \leq 54, x_1+...+x_5 \leq 90 \text{ for } x_1, ..., x_5 \in R_0^+. \\ \ \\ \text{Find the maximum of } \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\s... | $$x_1 \leq 4 \tag {1}$$ $$x_1+x_2 \leq 13\tag{2}$$ $$x_1+x_2+x_3 \leq 29\tag{3}$$ $$x_1+...+x_4 \leq 54\tag {4}$$ $$x_1+...+x_5 \leq 90\tag 5$$
now $$(5)\times 10+(4)\times 2+(3)\times 3+(2)\times 5+(1)\times 10 \implies$$ $$10x_5+12x_4+15x_3+20x_2+30x_1\le 1200$$ also by C-S $${\left(10x_5+12x_4+15x_3+20x_2+30x_1 \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve: $\frac{1}{\sin(\pi/n)}=\frac{1}{\sin(2\pi/n)}+\frac{1}{\sin(3\pi/n)}$
The positve integer satisfying value of $n: n>3$ satisfying the equation: $\dfrac{1}{\sin(\pi/n)}=\dfrac{1}{\sin(2\pi/n)}+\dfrac{1}{\sin(3\pi/n)}$ is
This question is from a Practice Book for CBSE Term$-1$ Maths Preparation Class $10$.
What ... | Using the formula
$$\sin A-\sin B=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
we have
\begin{align*}
\sin \dfrac{4\pi}{n}&=\sin \dfrac{3\pi}{n}\\
\sin \dfrac{4\pi}{n}-\sin \dfrac{3\pi}{n}&=0\\
2\sin\left(\frac{\pi}{2n}\right)\cos\left(\frac{7\pi}{2n}\right)&=0
\end{align*}
then
$$
\begin{array}{rcl}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294678",
"timestamp": "2023-03-29T00:00:00",
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Find all integer solutions of $(a+b+3)^2+2ab=3ab(a+2)(b+2)$ Here is another number theory problem that I am not able to do
Find all integer solutions of $$(a+b+3)^2+2ab=3ab(a+2)(b+2)$$
My attempt$:$
On expanding we get,
$$b^2+4ab+6b+a^2+6a+9=3a^2b^2+6a^2b+6ab^2+12ab$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9=3a^2b^2+6a^2b$$ or ... | Let x=a+1, y = b+1, so
$(x+y+1)^2 + 2(x-1)(y-1) = 3(x^2-1)(y^2-1)$.
The right hand side is quite obviously larger when |x|, |y| >= 3. Finding solutions for x=-2, -1, 0, 1, 2 are easy to find.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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An integral identity with a complex parameter Conjecture: For $z\in\mathbb{C}\setminus[-1,1]$, we have
$$
\int_{0}^{\pi}\frac{dx}{|z-\cos x|}=\frac{\pi}{\sqrt{|z^{2}-1}|}.
$$
I know it is true for $z\in\mathbb{R}\setminus[-1,1]$ but not in general.
| First note that the function
$$f(z) = \frac{dx}{|z-\cos x|}$$
is an even function. Hence,
$$
I = \int_{0}^{\pi}\frac{dx}{|z-\cos x|} = \frac{1}{2}\int_{-\pi}^{\pi}\frac{dx}{|z-\cos x|}
$$
If you do the following substitution:
$$ \cos x = \frac{w+w^{-1}}{2} = \frac{w^2+1}{2w}$$
$$dx = \frac{dw}{wi}$$
$$ I = \frac{1}{2}... | {
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"url": "https://math.stackexchange.com/questions/4303061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it possible to prove the formula to factorize $(a^n + b^n)$ using the formula that holds for $(a^n - b^n) $? ( with $n\geq 3$) Is it possible to prove the factorization of $a^3 + b^3$ such that $a^3 + b^3= (a+b) ( a^2 - ab + b^2)$ using the formula for the factorization of
$ (a^n - b^n) = (a-b) ( a^{n-1} + a^{n-2}b^... | Just let $c=-b$. Then $a^3+b^3=a^3-c^3=(a-c)(a^2+ac+c^2)=(a+b)(a^2-ab+b^2)$.
Note that this only works because $3$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that $\lim\limits_{n\to\infty} \frac{a_1+\cdots + a_n}n = 1$
Let $(a_n)$ be a bounded sequence of positive real numbers such that $\lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n} = 1.$ Prove that $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n = 1.$
By the AM-GM inequality, $ \sqrt[n]{a_1\cdots a_n} \leq \d... | This is false. Consider the sequence
$$a_{j} = \begin{cases}2 && \text{if }j\text{ is odd}\\
\frac{1}{2} && \text{if }j \text{ is even} \end{cases}$$
we have for all $n$
$$1 \leq (a_{1}...a_{n})^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$
Thus
$$\lim_{n \rightarrow \infty}(a_{1}...a_{n})^{\frac{1}{n}} = 1$$
But
$$\lim_{n \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove the sign of $1+2x-e^x$ around 0 I want to prove that the sign of $1+2x-e^x$ around $0$ is negative if $x\to 0^-$ and positive if $x\to 0^+$. If I replace $e^x$ by its equivalent function (the tangent at 0) I conclude that the sign of this expression depends on the sign of $x$ as I have $1+2x - 1-x $.
I need this... | You should already know from the definition that:
$$
e^x=\lim_{n\to\infty}(1+\frac xn)^n=1+x+\frac{x^2}{2!}+...+\frac{x^k}{k!}+...
$$
Obviously for all $x>0$: $e^x>1+x$. To consider negative $x$'s we write:
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}(1-\frac x4)+...+\frac{x^{2m+1}}{(2m+1)!}(1-\frac x{2m+2})+...
$$
Thus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this method of showing that $\frac{2n^2}{n^3+3} \rightarrow 0$ correct? Given
$$
a_n = \frac{2n^2}{n^3+3}
$$
I want to show that
$$
a_n \rightarrow 0
$$
My solution proceeds as follows.
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert &< \epsilon \\
\frac{2n^2}{n^3+3} &< \epsilon
\end{align}
since
\begi... | Two points:
*
*You set $N:=\tfrac{2}{\varepsilon}$. But what is $\varepsilon$? Also $N$ need not be an integer. That's fine, but the symbol $N$ does suggest an integer value.
*Next you 'fix $\varepsilon$ such that $\forall n>N$...'.
But to prove that the limit equals $0$, you want to show that $$(\forall\varepsilon>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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show this $(f(x))''+2\ge 0$,for any real numbers, where $f(x)=\frac{x^2-1}{x^{2n}-1}$ When I did a question today, I turned to the following question
let $n$ be postive integer,and $x\in R$, let $f(x)=\dfrac{x^2-1}{x^{2n}-1}$,show this
$$(f(x))''+2\ge 0 \tag{1}$$
For example $n=2$
then
$$(f(x))''+2=\dfrac{2x^2(x^4+3x^... | As stated above in a comment, we can write
$$f(x)=\frac{1}{1+x^2...+x^{2n-2}}=\frac{1}{g(x)}$$
which yields that
$$f''(x)+2=\frac{2g'^2-gg''+2g^3}{g^3}$$
so we only have to decide the sign of the numerator, given that $g(x)>0$ for all x. Since $f(x)=f(-x)$ it also follows that the second derivative is even, which allow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less th... | Thanks for hermites identity that @mymolecules showed...
$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{6}}\right]=\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\: \\ $
$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right]+\left[\frac{\mathrm{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4311152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ways of Changing 85/90 dollars The number of ways of changing 85 dollars with 1,5,10 and 20 dollar bills should be the coefficient of $x^{85}$ in $(1+x+x^2+\ldots+x^{85})(1+x^5+\ldots+x^{85})(1+x^{10}+\ldots+x^{80})(1+x^{20}+\ldots+x^{80})$, that is, $\frac{(1-x^{81})^2}{(1-x)(1-x^5)}\cdot\frac{(1-x^{86})^2}{(1-x^{10})... | First thing, your formula for finite geometric sums is incorrect. It should be
$$1+x^a+x^{2a}+x^{3a}+x^{4a}+\ldots +x^{na}=\frac{x^{(n+1)a}-1}{x-1}$$
So what you actually should have is that in the $85$ case, it is the coefficient of $x^{85}$ in
$$\frac{(1-x^{86})(1-x^{90})(1-x^{90})(1-x^{100})}{(1-x)(1-x^5)(1-x^{10})(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4311473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\sqrt{\frac{\sqrt[4]{x^3}-8}{\sqrt[4]{x}-2}+2\sqrt[4]{x}}\left(\frac{\sqrt[4]{x^3}+8}{\sqrt[4]{x}+2}-\sqrt{x}\right)$ Simplify $$\sqrt{\dfrac{\sqrt[4]{x^3}-8}{\sqrt[4]{x}-2}+2\sqrt[4]{x}}\left(\dfrac{\sqrt[4]{x^3}+8}{\sqrt[4]{x}+2}-\sqrt{x}\right)$$
Is it a good idea to simplify the square root with the commo... | Start with setting $\sqrt[4]{x} = u$ (also mentioned by $\textit{Achilles hui}$ in the comments).
We get
$$L= \sqrt{\frac{u^3-8}{u-2}+2u}\left(\frac{u^3+8}{u+2}-u^2\right)$$
Using the identities for sum and differences of cubes we have
$$L=\sqrt{(u^2+2u+4)+2u}((u^2-2u+4)-u^2)$$
$$L=(u+2)\cdot ( -2(u-2))=-2(u^2-4)=-2(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the solution of $x^2\equiv 25\pmod{32}$? I am trying to find square root of $57$ modulo $32\times 49$. For that I need to find the solutions of $x^2\equiv 57\pmod{32}$ and $x^2\equiv 57\pmod{49}$ which are $x^2\equiv 25\pmod{32}$ and $x^2\equiv 8 \pmod{49}$. Now I could find the solution of the second one ... | Well, the easiest way to find a solution of $x^2\equiv 25\pmod{32}$ is to observe that $25$ is a perfect square, so $5^2=25$ which means that certainly $5^2\equiv 25\pmod{32}$ as well. If it holds in the good old integers, it certainly holds mod whatever.
You quickly earn another solution for free: $(-5)^2 \equiv 27^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solid of revolution problem I would like to check with you guys if this exercise is right. The problem is:
"Let be $R$ be the region bounded by the graph of $y=2x$", $y=\frac{x^2}{4}$ and $y=2$. Find the volume of the solid generated by the rotation of the region $R$ about the $y$-axis. Make a sketch of the region. Sho... | Yes, this is exactly correct.
You can also verify your solution by using the method of cylindrical shells. For instance, the volume of the parabolic region including the cone is just $$\int_{x=0}^{2 \sqrt{2}} 2\pi x \left(2 - \frac{x^2}{4}\right) \, dx = \left[2x^2 - \frac{x^4}{8}\right]_{x=0}^{2\sqrt{2}} \pi = 8\pi.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Angles satisfying $(\cos\theta,\sin\theta)=(\frac{\sqrt a }{\sqrt{a+b}},\frac{\sqrt b}{\sqrt{a+b}})$ for integer $a$, $b$ If we list the commonly used angles in the first quadrant $\theta_0=0$, $\theta_1=\dfrac{\pi}{6}$, $\theta_2=\dfrac{\pi}{4}$, $\theta_3=\dfrac{\pi}{3}$, and $\theta_4=\dfrac{\pi}{2}$,
then $$(\cos \... | Regarding the second question, there are no other possible values of $\theta\in[0,\pi/2]$. Squaring, we have
$$\cos^2\theta=\frac{a}{a+b}\Leftrightarrow\frac{1+\cos 2\theta}{2}=\frac{a}{a+b}\Leftrightarrow\cos 2\theta=\frac{a-b}{a+b} $$
This means that $\cos2\theta$ is rational and $2\theta$ is a rational multiple of $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Epsilon - delta proof $\lim_{x \to \frac{\pi}{4}} \tan(x)=1$ I need to prove the limit using the definition of limit $$\lim_{x\to c}f(x)=L \leftrightarrow \forall \epsilon >0 \hspace{0.5 cm} \exists \delta >0 : 0<|x-c|<\delta \rightarrow |f(x)-L|<\epsilon $$
My attempt
$$|\tan(x)-1|<\epsilon\\ -\epsilon <\tan(x)-1 < \... | Here is a more explicit way to do it, without using arctan of epsilon.
Since $\tan(x)$ has an asymptote at $x=\frac{\pi}{2}$, we should stipulate at the outset a maximum value of delta which is less than the distance from the point we are working at and the asymptote (this distance is $\frac{\pi}{4}$), for instance $\d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to solve this linear system of three equations using Cramer's rule? I have a 3-by-3 matrix,
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
1 & 0 & 1 \\
1 & 1 & -1\\
\end{matrix} \right]$
the known terms are (-6, 2, -5), at the right of "=" symbol.
(1) I've calculated the determinant,
(2) I've used Cramer'... | The determinant of the matrix representing $x$ is, expanding down the middle column (which you definitely can do, no row operations needed!):
$$\det\begin{pmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{pmatrix}=(-2)\cdot\det\begin{pmatrix}2&1\\-5&-1\end{pmatrix}+0+(-1)\cdot\det\begin{pmatrix}-6&3\\2&1\end{pmatrix}$$
Which is: $$(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometry problem with triangle Given a triangle $ABC$ with side lengths $a$, $b$, and $c$. There exists a unique point $P$ such that
$$d^2 = AB^2+AP^2+BP^2 = AC^2+AP^2+CP^2=BC^2+BP^2+CP^2.$$
Question: How to express $d^2$ in terms of $a$, $b$, and $c$, and the circumradius $R$ of the triangle $ABC$?
What I tried: $R$ ... | $\begin{array}{} AB=c & BC=a & AC=b \\ AP=f & BP=g & CP=h \end{array}$
$\begin{array}{} d^2=c^2+f^2+g^2 & (1) \\ d^2=b^2+f^2 +h^2 & (2) \\ d^2=a^2+g^2+h^2 & (3) \ \end{array}$
$\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) \end{array}$
where:
$\begin{array}{} p=\frac{a^2+b^2-c^2}{2a} & q=\frac{2S}{a} & S=\sqrt{s(s-a)(s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$. I have written an attempted proof of the theorem on the title, and I need help verifying it.
I have used the following theorems to proof the theorem on the title.
Theorem 5.14)a) Let $x$ be a real number. $-|x| \leq x \leq |x|$.
Theorem 5.14)b) Let $... | To answer the solution-verification part of the question, this step is wrong.
one obtains
\begin{align}
(1+z)|(1+w)| \leq |(1+z)||(1+w)|
\end{align}
Since $(1+w) \leq |(1+w)|$, it follows that
\begin{align}
(1+z)(1+w) \leq |(1+z)||(1+w)|
\end{align}
The above is of the form $\, a \cdot |b| \le c \implies a \cdot b \l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
The answer is:
the number of four-digit integers that are divisible by $5\;-\;$ the number of four-digit integers ... | You are over counting a few cases. For example, $1330$ is counted in the first case when $x\in\{1,…,9\},y=3,z\in\{0,…,9\},w\in\{0,5\}$ and it is counted again in the second case when $x\in\{1,…,9\},y\in\{0,…,9\},z=3,w\in\{0,5\}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove inequality using a given inequality Question: It is given that $$a^r(a-b)(a-c)+b^r(b-c)(b-a)+c^r(c-a)(c-b)\geq 0$$ for positive integers $a,b,c,r$. We are supposed to show that $$\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}+\frac{a+b+c}{a^2b^2c^2}\geq \frac{b^2+c^2}{a^3b^2c^2}+\frac{c^2+a^2}{a^2b^3c^2}+\frac{a^2+b^2... | Alternative (brute force/buffalo way) solution: Your inequality is upon full expansion (using symmetric sum notation) $$\frac{\frac{\sum_{\text{sym}} a^5 b^5}2+\frac{\sum_{\text{sym}}a^4 b^3 c^3}2-\sum_{\text{sym}}a^5 b^3 c^2}{a^5b^5c^5}\ge 0.$$
Since $a,b,c>0$, you just need to prove $$\frac{\sum_{\text{sym}} a^5 b^5}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Partial differential equation - integral surface I have to find the integral surface of the PDE
$$x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$$
containing the straight line $x+y=0, z=1$.
From the auxiliary equations, I can obtain $xyz=C_{1}$ for some constant $C_{1}$. To proceed further, I need to find another relation involvin... | $x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$ is ambiguous because the symbols are not defined. I suppose that the PDE is :
$$x(y^2+z)\frac{\partial z}{\partial x} - y(x^2+z)\frac{\partial z}{\partial y} = (x^2-y^2)z(x,y)\tag 1$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326024",
"timestamp": "2023-03-29T00:00:00",
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Composed series expansion understanding Recently I'm studying series expansion and there is something I don't understand but it's hard to tell what... For example, imagine I need to calculate the series expansion of $\sqrt{1+\sin(x)}$ at $ a=0$.
I want to do it as fast as possible, so my idea is to have the series expa... | Your approach is correct. Until you've got it mastered I'd use two different variables: so
$$
\sqrt{1+y} = 1 + \frac{1}{2}y - \frac{1}{8}y^2 + \frac{1}{16}y^3 + O(y^4)
$$
Then you want to substitute
$$
y = x - \frac{1}{3!}x^3 + O(x^5)
$$
So
$$
\sqrt{1+y} = 1 + \frac{1}{2}\left(x - \frac{1}{3!}x^3 + O(x^5)\right) - \fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all positive integers $(a,b)$ such that $\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}$ is an integer
Find all positive integers $(a,b)$ such that
$\displaystyle\frac{a^{2^{b}}+b^{2^{a}}+11}{2^{a}+2^{b}}$ is an
integer.
The machine/code says that $a=2$ and $b=3$ are suitable up to symmetry. And I bet t... | As noted by the comments one of $a,b$, say $b$ must be $2$ and the parity of $a$ and $b$ must differ. So let us assume that $a$ is odd.
We first make the following claim:
Claim 1: Let $c$ and $a'$ be nonnegative integers. Then for some nonegative integer $d\le a'$ the equation holds:
$$2^c \equiv_{2^{a'}+1} \pm 2^d.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Where did I go wrong in finding the sum of $1+2k+3k^2+...+nk^{n-1}$ using Abel's formula The version of Abel's formula I'm using is $a_1b_1+a_2b_2+...+a_nb_n=(a_1-a_2)B_1+(a_2-a_3)B_2+...+(a_{n-1}-a_n)B_{n-1}+a_nB_n$ where $B_1=b_1$, $B_2=b_1+b_2$, $B_3=b_1+b_2+b_3$...$B_n=b_1+b_2+b_3+...+b_n$ and $a_n$ and $b_n$ are t... | Hint:
With the correct expression $b_n=k^{n-1}$ instead of $b_n=k^n$ you will obtain the desired result.
Observe also that you wrongly simplified the last term $a_nB_n$ in your expression (these two errors have eliminated each other to give correctly the first term in the final expression).
| {
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"timestamp": "2023-03-29T00:00:00",
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$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n ... | There is a slight mistake
From this I need to subtract the sum of squares of even terms.
No. You have to subtract twice of above because
$1^2-2^2+3^2+\cdots+1999^2=(1^2+2^2+3^2+\cdots+1999^2)-2(2^2+4^2+6^2+1998^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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What is the transformation matrix of the linear mapping? $ f:M_{2.2}(\mathbb{R}) \rightarrow \mathbb{R^3}, $$
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\mapsto
\begin{pmatrix}
2a - 4b \\
-6d \\
8a-16b+2d
\end{pmatrix}\beta =\big
(\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}
,\begin{pmatrix}
0 & 1 \\
0 & 0
\end{p... | Your idea is not correct, since the domain of $f$ is $\Bbb R^{2\times2}$; so it makes no sense to talk about, say$$f\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right).$$
Note that\begin{align}f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)&=\begin{bmatrix}2\\0\\8\end{bmatrix}\\&=0\times e_2+1\times(2e_1)-8(-e_3).\end{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A die is rolled once. Call the result N. Then, the die is rolled N times, and those rolls which are equal to or greater than N are summed
A die is rolled once. Call the result N. Then, the die is rolled N times, and those rolls which are
equal to or greater than N are summed (other rolls are not summed). What is the d... | I will expand slightly on my and Jaap's comments.
Denote by $S$ the random variable of the resulting sum. By the law of total probability, you can find its distribution by
$$ P(S=s) = \sum_{n=1}^6 P(S=s|N=n) P(N=n) = \frac16 \sum_{n=1}^6 P(S=s|N=n). $$
As the OP described, the probabilities $P(S=s|N=n)$ can be found vi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question abut the ambiguity of a maths problem. I am trying to prove by induction that:
$$1^3-2^3 +\cdots+n^3=(1+2+\cdots+n)^2 $$
This was a problem from a practice worksheet, but I don't understand how to interpret the LHS.
Is the following correct:
$$1^3-2^3+3^3-4^3+5^3-6^3+\cdots+n^3$$
Or this:
$$1^3-2^3+3^3+4^3+5^3... | The true formula is that
$1^3+2^3+...+n^3=(1+2+...+n)^2$,
and you can prove this by induction. Assume that
$1^3+2^3+\cdots +(n-1)^3=(1+2+\cdots +(n-1))^2$,
then
$1^3+2^3+\cdots +n^3=(1+2+...+n-1)^2+n^3=\frac{n^2(n-1)^2}{4}+n^3=(1+2+\cdots n)^2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate residue of pole with order 5 I am trying to evaluate the residue of $\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}$ at z=0.
Is there a way to do this without having to do the long derivative calculation?
I know we have the formula $Res(f,0)=\lim_{z\to0}\frac{1}{(5-1)!}\frac{\mathrm{d}^{5-1}f}{\mathrm{d}z^{5-1}}z^5f(z)$.... | For practical calculations, I have never used that derivative formula to calculate any residue. What is useful is learning how to perform a limited Laurent expansion. One of the most useful "tricks" is the simple geometric series formula $\frac{1}{1-\zeta}=\sum_{n=0}^{\infty}\zeta^n$ provided $|\zeta|<1$. You should al... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Volume of a container with unknown side and known surface area We have to make a closed container. All side surfaces (walls, ceiling and bottom) must be rectangles and stand perpendicular to each other. One of the sides should be $3 \,m$ long. The surface area of the container shall be $32 \,m^2$. Let one of the ot... | Unless we let $x=y$ the formula does not have enough information to identify volume. If, however, we let $x=y$, then we can show that the formula works, with the restriction that $x=-8$ or $x=2$.
$$
2(3\cdot x)+2(3\cdot y) + 2(x\cdot y) = 32 \\
2 x y + 6 x + 6 y = 32\\
x = \dfrac{16 - 3 y}{y + 3},
y=x\implies x,y\in\b... | {
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Find parameters $a,b$ such that $x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1-\left(x^3-x^2+a x+b\right)^2>0$ The probrem is to prove that
$$x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1>0.$$
(the minimum value is about 0.02, tested by wolframalpha.)
I use sos(sum of squares) method, my idea is to reduce the degree of the polynomial gradually.
F... | Let $p(x)=x^6-2x^5+2x^4+2x^3-x^2-2x+1$. A possible way to show $p(x)>0$ for all $x\in\mathbb{R}$ follows. Consider 3 cases.
Case 1: $x<-1$. Let $r(x)=p(x-1)$. Then $r(x)=x^6-8x^5+27x^4-46x^3+40x^2-18x+5$. Clearly, $r(x)>0$ for $x<0$, so $p(x)=r(x+1)>0$ for $x<-1$.
Case 2: $x>\frac 12$. Let $q(x)=p\left(x+\frac 12\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Convergence of the series with $a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$ I need to determine the convergence of the series whose general term is given by:
$$a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot ... | Let $N \in \mathbb{N}$. One has
\begin{align*}
\left( \sum_{k=1}^N \frac{1}{\sqrt{k}}\right)\left( \sum_{j=1}^N \frac{1}{j^2}\right) &= \sum_{k=1}^N \sum_{j=1}^N \frac{1}{j^2\sqrt{k}}\\
&\leq \sum_{n=2}^{2N} \sum_{j+k=n} \frac{1}{j^2\sqrt{k}}\\
&= \sum_{n=1}^{2N-1} \sum_{j+k=n+1} \frac{1}{j^2\sqrt{k}}\\
&= \sum_{n=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How far can I go with the integral $\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x, $ where $n\in N$? Latest Edit
By the aid of my recent post, a closed form for its definite integral is obtained as below:
$$
\int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta d \theta= \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)... | Let’s find a more general result and simplify with standard functions. The first step is a geometric series and the double angle formula:
$$\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x = \int \frac{\frac1{2^n}\sin ^{n}(2 x) }{1-\frac12\sin(2x)} d x\mathop=^{\big|\frac12\sin(2x)\big|<1} 2^{-n}\int \sin ^{n}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to grasp the relationship between Laurent series depending on the region they are developed at in trying to understand how to set up the Laurent series for a fractional expression, I have the given function
\begin{equation}
\frac{1}{(z-2)}-\frac{1}{(z-1)}
\end{equation}
The Laurent series of each of the two fractio... | The function
\begin{align*}
&f:\mathbb{C}\setminus\{1,2\}\to\mathbb{C}\\
&f(z)=\frac{1}{z-2}-\frac{1}{z-1}
\end{align*}
has simple poles at $z=1$ and $z=2$. Let's assume we want a Laurent expansion of $f$ at $z=0$. We distinguish three regions of convergence
\begin{align*}
\color{blue}{D_1:}&\color{blue}{\quad 0\leq |... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4348500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do we handle the integral $I_{4}=\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{4} d \theta$? After struggling with $I_2,I_3$ in the post, I dare to tackle $I_4$ now.
We first rewrite the integral
$$I_{4}=\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{4} d \theta =\int \frac{\sec ^{2} \theta}{\l... | In general, for any even power, substitute $\tan^2\theta = \frac12\tan^2 x$ to transform the integral to a manageable one
$$I_{2m}= \int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{2m} d \theta
= \frac1{2^{m-\frac12}}\int \cos^{2m}x\>(1+\cos^2x)^{m-1}dx
$$
which, for $m=2$, reduces to
\begin{align}
I_4
= &\>\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4349163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Density function of average of max and min of uniform random variables Suppose $X_1, \ldots, X_n \sim U(\theta - \frac{1}{2}, \theta + \frac{1}{2})$. I want to find the density of $Z = \frac{X_{(1)} + X_{(n)}}{2}$. My strategy is to use the transformation $F(x, y) = (\frac{x + y}{2}, y)$ and the joint density of $X_{(1... | The mistake is in the integral to find density function of $Z$.
$ \displaystyle \theta - \frac{1}{2} < 2z - y < y < \theta + \frac{1}{2}$ would mean
For $\theta - \frac 12 \lt z \lt \theta, z \lt y \lt 2z + \frac 12 - \theta$
And for $\theta \lt z \lt \theta + \frac 12, z \lt y \lt \theta + \frac 12$
Integrating you ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area of a circle passing through two vertices of a parallelogram touching one edge. For reference:
Let $ABCD$ be a parallelogram, $AB = 6, BC= 10$ and $AC = 14$ ; traces a circle passing through the vertices $C$ and $D$ with $AD$ being tangent and this circle is $BC$ a secant. Calculate the area of the circle. (Answ... | The center $(x,y)$ of the circle has obvious $x=10$; the ordinate $y$ is given by the intersection of lines $x=10$ and the perpendicular to the segment $DC$ at its midpoint (and this ordinate is clearly the radius also by the tangency).
$► \cos(\angle{ABC})=-\dfrac12\Rightarrow \angle{BAD}=60^{\circ}$.
It follows:
$►$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Identity involving double sum with binomials (Edit: This question has now been answered in MathOverflow, here)
In the course of a calculation, I have met the following complicated identity. Let $A$ and $a$ be positive integers. Then I believe that
$$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-... | In the case $A = 1, a = 1$, and letting $n = B + b -1$, this reduces to
$$-\sum_{n=1}^\infty\sum_{b=1}^n (-1)^bb\dfrac{(n-b)!b!}{(n+1)(x-b)^{(n+1)}} = \dfrac 1{2x(x+1)}$$
But that cannot be. The residue at $0$ can be found by multiplying both sides by $x$ and taking the limit as $x \to 0$. The right-hand side easily gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$ Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to u... | Square the equation rearrange & square again. This will allow you to remove the radicals & leave you with a polynomial.
\begin{eqnarray*}
(3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12)=8(x+1)\sqrt{2(x^2+1)(6x^2+18x+12)} \\
\end{eqnarray*}
\begin{eqnarray*}
((3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12))^2-128(x+1)^2(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$ I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$
$$2^{\frac{y^2}{2... | Using the hyperbolic method when there is only one real root as for$$u^3-u^2+14u-126=0$$ the solution is not ugly at all. It write
$$u=\frac{1}{3} \left(1+2 \sqrt{41} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{1639}{41 \sqrt{41}}\right)\right)\right)$$ while using Cardano method, it write
$$\frac{1}{3} \left(1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On Sylvester’s criterion for $2 \times 2$ matrices In my electrical engineering text, I have an expression of the form:
$$E_m(x_1, x_2) = \frac{1}{2}(A_{11}x_{1}^2+2Mx_1x_2+A_{22}x_{2}^2) > 0$$
where $$A = \begin{bmatrix}
A_{11} & M \\
M & A_{22}
\end{bmatrix}$$
and so the expression can be written in matrix form ... | (In this answer, I assume that $x_1$ and $x_2$ are not zero simultaneously.)
Let us first exclude the case where both $A_{11}$ and $A_{22}$ are both zero because $M x_1 x_2 >0$ cannot hold for arbitrary values of $x_1$ and $x_2$.
Without loss of generality, since the expression is symmetric with respect to $x_1$ and $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why is $\left ( 1+\frac{1}{k} \right )^kIn Artin's Gamma function (page 20), he says in footnote
(..) we consider the elementary inequalities
$$
\left ( 1+\frac{1}{k} \right )^k<e<\left ( 1+\frac{1}{k} \right )^{k+1}
$$
for $k=1,2,\dots, n-1$.
Where does the inequalities come from? From what I know is that
$$
e=\lim_... | Partial answer.
Let $t_k = \left( 1 + \frac{1}{k} \right) ^k.$
"Borrowing" from Rudin's PMA, in Theorem $3.31$ he gives the nice formula, due to the Binomial theorem:
$$t_k = 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k} \right) \left( 1 - \frac{2}{k} \right) + \ldots$$
$$ + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\frac{a^3+b^3+c^3-3abc}{a^4+b^4+c^4-4abc}$, given $a+b+c=a^2+b^2+c^2=1$ How do I solve this problem?
Says so:
If $$a+b+c=a^2+b^2+c^2=1$$
find the value of
$$M=\frac{a^3+b^3+c^3-3abc}{a^4+b^4+c^4-4abc}$$
edit:
I know these identities
$$1)(a + b + c)^2 = a^2 + + c^2 + b^2+2 a b + + 2 a c + 2 b c$$
$$2)(a ... | Expanding on the comments, given that $\,a+b+c=1\,$ and $\,ab+bc+ca=0\,$:
$$
\begin{align}
p(x) = (x-a)(x-b)(x-c) &= x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc
\\ &= x^3 - x^2 - abc
\end{align}
$$
Since $\,p(a)=p(b)=p(c)=0\,$:
$$
\begin{cases}
a^3 = a^2 + abc
\\ b^3 = b^2 + abc
\\ c^3 = c^2 + abc
\end{cases}
$$
Adding the th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$ I want to find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$. I have plugged this into a calculator which tells us that if we solve for $a$, without loss of generality, we obtain
$$a = - \frac{b}{2} \pm \frac{\sqrt{3}}{2}i$$
and vice versa. I am not sure how to s... | To begin with, let us assume that $b\neq 0$ and $a\neq -b$.
Now we can multiply both sides by $b(a+b)$ in order to obtain
\begin{align*}
(a + b)^{2} = ab & \Longleftrightarrow a^{2} + 2ab + b^{2} = ab\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = 0\\\\
& \Longleftrightarrow \left(a + \frac{b}{2}\right)^{2} + \frac{3b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove $\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}=\frac n4\csc\frac{2\pi}n$ I would like to evaluate the sum
$$\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}
$$
which is said to reduce to the simple close-form $\frac n4\csc\frac{2\pi}n$. I have verified it numerically for a large number ... | To get rid of the $k$, I use the following identity:
$$\sum_{k=1}^{n-1} kz^k=z\frac{d}{dz}\sum_{k=0}^{n-1}z^k=z\frac{d}{dz}\frac{1-z^n}{1-z}=\frac{(n-1)z^{n+1}-nz^n+z}{(1-z)^2}$$
Thus using $z=e^{\frac {4i\pi}n}$,
$$\begin{split}
\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n} &= \frac 1 2\sum_{k=1}^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$ I have this equation to solve:
$$\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$$
Since $\overline{z}\cdot z = |z|^2$ and utilizing the de Moivre's formula this can be simpl... | From the beginning:
Since $i^8 = 1$, we can multiply by $i^8$ to get $(- \cos(\pi/5) - i \sin (\pi/5))^8$ which is the same as $(\cos \pi/5 + i \sin \pi/5)^8$.
So now you have:
$$|z|^3z^4=8\sqrt{2} e^{8i \pi/5}$$
which means that $z$ has modulus $(8 \sqrt{2})^{1/7} = \sqrt{2}$. Now since $|z|^3$ is real, you have:
$$\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Using Lagrange Mult. to find maximum of $2x + y^2$ with constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ I was trying to maximize the function $2x + y^2$ with the following constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ using Lagrange multipliers. I already did all the work of proving that the maximu... | Using the two constraint-multiplier equations you wrote,
$$ 2 \ - \ 2\alpha x \ - \ \beta \ \ = \ \ 0 \ \ \ , \ \ \ 2y \ - \ 2\alpha y \ - \ 2 \beta y \ \ = \ \ 0 \ \ \ , \ \ \ 0 \ - \ 2\alpha z \ - \ \beta \ \ = \ \ 0 \ \ , $$
we could eliminate $ \ \beta \ $ between the first and third equations to write $ \ 2·(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Why is $ab$ likely to have more divisors than $(a-b)(a+b)$? Consider the two numbers $ab$ and $(a-b)(a+b), \gcd(a,b) = 1, 1 \le b < a$. On an average, which of these two numbers has more distinct prime factors? All the prime factors of $a$ and $b$ divide $ab$ and similarly all the prime factors of $a-b$ and $a+b$ divid... | On one hand,
\begin{align*}
\sum_{a,b\le x} \omega(ab) &= \sum_{a,b\le x} \sum_{\substack{p\le x \\ p\mid ab}} 1 = \sum_{p\le x} \sum_{\substack{a,b\le x \\ p\mid ab}} 1 \\
&= \sum_{p\le x} \biggl( \sum_{\substack{a,b\le x \\ p\mid a}} 1 + \sum_{\substack{a,b\le x \\ p\mid b}} 1 - \sum_{\substack{a,b\le x \\ p\mid a,\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
Finding the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$
Find the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$
My Attempt: Let $z=x+iy$, so, $$\sqrt{x^2+y^2}=\text{max}\{\sqrt{(x-1)^2+y^2},\sqrt{(x+1)^2+y^2}\}$$
Case I: $\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\implies \pm x=x... | It follows from
$$
|z-1|^2 = (x-1)^2 + y^2 = x^2 + y^2 - 2x + 1 \\
|z+1|^2 = (x-1)^2 + y^2 = x^2 + y^2 + 2x + 1
$$
that
$$
\max(|z-1|, |z+1|)^2 = x^2 + y^2 + 2|x| + 1 > x^2 + y^2 = |z|^2
$$
which shows that the equation has no solution.
Geometrically:
$$
\max(|z-1|, |z+1|) = |z+1|
$$
means that the distance from $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$ab+ac+bc=2\quad ,\quad\min(10a^2+10b^2+c^2)=?$
If $ab+ac+bc=2$, find minimum value of $10a^2+10b^2+c^2$
$1)3\qquad\qquad2)4\qquad\qquad3)8\qquad\qquad4)10$
I used AM-GM inequality for three variables:
$$ab+ac+bc\ge3(abc)^{\frac23}\quad\Rightarrow\quad 2\ge3(abc)^{\frac23}$$
$$10a^2+10b^2+c^2\ge 3(10abc)^{\frac23}$$
... | Hint: $$10a^2+10b^2+c^2-4(ab+bc+ca)=\frac{2}{5}(5a-b-c)^2+\frac{3}{5}(4b-c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I prove: $xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{2(xy+yz+xz)}{3}\right)^3,\ x,y,z>0$? I have a question which askes to prove:
$\displaystyle \tag*{} xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{2(xy+yz+xz)}{3}\right)^3,\ x,y,z>0$?
My approach: I tried AM-GM inequality, didn't work well. So I tried substituting $2(xy+yz+xz)=... | We want $x+y+z$ largest when $xyz,xy+yz+zx$ fixed. By 3.83 in the Mathematical Inequalities (Volume 1) by Vasile Cîrtoaje, we can assume $x=y\le z$. Now we rewrite the inequality as $x=y$:
$$ x^2z(x-2)^2(z-2)\leq\left(1+\frac{2(x^2+2xz)}{3}\right)^3$$
We must have $z\ge 2$ otherwise left is smaller than zero, and right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^\infty\frac{x^2\ln x}{x^4+x^2+1}dx$ by the residue theorem The result should be $\frac{\pi^2}{12}$.
Edit: I have tried to reproduce the image, but limitations of MathJax required some reformatting. Here is the original image.
$$
\begin{align}
\int_0^\infty\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}&=\int_0... | You can use Feynman's method to evaluate.
Let
$$I(a)=\int_0^\infty\frac{x^a}{x^4+x^2+1}dx.$$
Then
\begin{eqnarray}
I(a)&=&\int_0^1\frac{x^a+x^{2-a}}{x^4+x^2+1}dx\\
&=&\int_0^1\frac{(1-x^2)(x^a+x^{2-a})}{1-x^6}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x^2)(x^a+x^{2-a})x^{6n}dx\\
&=&\sum_{n=0}^\infty\int_0^1(1-x^2)(x^a+x^{2-a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Seeing if a polynomial is irreducible over a field extension of $\Bbb Q$. Exercise. Determine $m_{i\sqrt[3]{2}}$ over $\Bbb Q(i)$, where $m_{i\sqrt[3]{2}}$ represents the minimal polynomial of ${i\sqrt[3]{2}}$.
My attempt. We know that $m_i = x^2 +1$ over $\Bbb Q$ (this is obvious) and thus $\deg(i) = \deg(m_i) = 2$ ov... | What you've done is correct. There are a few ways to do it by hand, without using a computer.
The most elementary way is to take a general element $a + bi \in \mathbb{Q}(i)$ and show that it can't be a root of $f(x)$. That is, let $a, b \in \mathbb{Q}$ and suppose that $f(a + bi) = 0$. Then
$$
(a + bi)^3 = -2i,
$$
whic... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}}$ How to solve the limits without using L-hospital law, like using rationalisation
L-hospital method is taking too long
The final answer I got was
$$\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}} = 1$... | $$\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}{((n^4+1)^{1/4}-\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}{(n^{4}+1)^{\frac{1}{2}}-... | {
"language": "en",
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"source": "stackexchange",
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How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right... | Let $r^2=x^2+y^2$
$|\frac{r^2}{\sqrt{r^2+1}-1}-2|<\epsilon$
$2-\epsilon<\sqrt{r^2+1}+1<2+\epsilon$ by rationalizing the denominator, cancelling terms and pushing $2$ to the outsides of the compound inequality.
$ \sqrt{r^2+1}\le1+\frac{r^2}{2} \implies -\epsilon< r^2/2<\epsilon$ by Taylor's Theorem, then cancelling, th... | {
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"source": "stackexchange",
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What's the ratio between the segments $\frac{AF.BG}{FG}$ in the figure below? For reference: In the figure below the trapezoid has height $13$ and is inscribed in a circle of radius $15$. Point $E$ is on the minor arc determined by $A$ and $B$, the points $F$ and $G$ intersect with $ED$, $E$C and $AB$. Then the ratio b... |
Notice that points $H$ and $I$ in this answer are different from those in the original post.
Let $\frac{AF\cdot BG}{FG}=R$. First of all, $$AB=AF+FG+BG=2\sqrt{15^2-13^2}=4\sqrt{14}$$
Power of a point gives us
\begin{align*}
AF\cdot FB&=AF\cdot\left(4\sqrt{14}-AF\right)=EF\cdot DF&&(1)\\
BG\cdot AG&=BG\cdot\left(4\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Confusions in Holder's Inequality Holder's Inequality states that for nonnegative real numbers $a_1,...,a_n$ and $b_1,...,b_n$ we have $$\left(\sum_{i=1}^na_i\right)^p\left(\sum_{i=1}^nb_i\right)^q\ge \left(\sum_{i=1}\sqrt[p+q]{a_i^pb_i^q}\right)^{p+q}$$
Where $p$ and $q$ are positive real numbers.
Here is my problem :... | I would think Holder's inequality implies
$$(a^3+2)(b^3+2)(c^3+2)\geq \left((a^3b^3c^3)^{1/3}+ (2^3)^{1/3} \right)^3=(abc+2)^3$$
So it would be good to know how your last inequality follows from Holder's.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Markov Chain Contest Problem Five cards labeled 1, 3, 5, 7, 9 are laid in a row in that order, forming the five-digit number 13579 when
read from left to right. A swap consists of picking two distinct cards, and then swapping them. After
three swaps, the cards form a new five-digit number n when read from left to right... | Here's a slightly simpler way of calculating the exact expected value of $\ n\ $. After the OP's correction to my original calculation (in which I made an arithmetical error), it gives the same answer as his.
After $3$ swaps, there will be a probability $\ p\ \ \big($which turns out to be $\ \frac{3}{10}\ \big)
$ tha... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ ... | There is a Riemann sum in disguise.
Note, however, that the function to integrate is unbounded. It still works with the improper integral, as is explained here: Convergence of Riemann sums for improper integrals.
$$t_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt k}=\frac{1}{n}\sum_{k=1}^n\frac{1/\sqrt{n}}{\sqrt {k/n}}=\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve the recurrence relation $a_n = 7a_{n-1} - 16a_{n-2} +12a_{n-3} + n4^n $ Solve the recurrence relation
$$a_n = 7a_{n-1} - 16a_{n-2} +12a_{n-3} + n4^n $$
With initial conditions:
$a_0 = −2$
$a_1 = 0$
$a_2 = 5$
I solve the homogenous part like that:
$$a_n - 7a_{n-1} + 16a_{n-2} -12a_{n-3} = 0 $$
$$ p(r) = r^3 -7r^2... | You can find a particular solution using the $Z$ transform:
$$
Z[a_n-7a_{n-1}+16a_{n-2}-12a_{n-3}]=Z[n 4^n]
$$
then
$$
\left(1-\frac 7z+\frac {16}{z^2}-\frac{12}{z^3}\right)Z[a_n]=\frac{4z}{(z-4)^2}+7a_{-1}-16 \left(a_{-2}+\frac{a_{-1}}{z}\right)+12\left(a_{-3}+\frac{a_{-2}}{z}+\frac{a_{-1}}{z^2}\right)
$$
now taking t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $6^{50-x} = 2^{50}$ I'm being asked to solve an equation for $x$, giving the answer in the form below: $$a \log_{b} c$$
One of them is: $$6^{50-x} = 2^{50}$$
So I started by taking the $\log_6$ of both sides to give:
$$\log_{6}6^{50-x}=\log_{6}2^{50} \tag1$$
Leading to:
$$\begin{align}
50-x &=\log_{6}2^{50} \ta... | Your last line:
$x= 50 -50\log_6 2$;
$x=50-50\log_6 ((2 \cdot 3)/ 3));$
Can you finish?
Option:
$(2 \cdot 3)^{50}=2^{50}6^x ;$
$2^{50}3^{50}=2^{50}6^x;$
$6^x=3^{50};$
$x=50 \log_6 3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use induction to prove $ S(n) = \sum_{k=1}^n \sum_{i=1}^k 2^{n-k} = 2^{n+1} - (n+2)$ Use induction to prove $$ S(n) = \sum_{k=1}^n \sum_{i=1}^k 2^{n-k} = 2^{n+1} - (n+2)$$
So far I worked out that you can convert the two summations into one:
$$\sum_{k=1}^n k\cdot 2^{n-k} $$
My problem comes when I try to write $S(n+1)$... | $$S(n+1)=\sum_{k=1}^{n+1} k \, 2^{n+1-k} = 2 \sum_{k=1}^n k \, 2^{n-k} + (n+1) \\ \stackrel{\text{I.H.}}{=} 2\left(2^{n+1}-n-2\right) + n + 1 = 2^{n+2} - n - 3$$
which is what you would expect.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $ T: \mathbb{R^3} -> \mathbb{R^2} : T(x,y,z) = (x+1, y+z)$ a linear transformation? Need help for my proof So I have
$$ T: \mathbb{R^3} \rightarrow\mathbb{R^2} $$
$$ T(x,y,z) = (x+1, y+z),$$
and I know that it is indeed a linear transformation if I show both cases:
$$ \forall v \in T : F(a+b) = F(a)+F(b),$$
and
$... | Any linear transformation can be written in the form $$T(\overline{v})=M \overline{v}$$ where $M$ is a matrix of transformation. The transformation cannot be written in this form because
$$T(x,y,z)= \begin{pmatrix} x+1 \\ x+z
\end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\
1 & 0 & 1
\end{pmatrix} \begin{pmatrix} x \\ y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a Rational Expression is Surjective I need to prove that the function $f:(-1,1)\rightarrow\mathbb{R}$ defined by $f(x) = \frac{x}{x^2 - 1}$ is surjective.
My work.
$b = \frac{a}{a^2 - 1} \iff b(a^2 -1)=a \iff ba^2 - b - a=0$
From here I did a few cases:
Case 1) $b=0$. Then $a=0$.
Using the quadratic formula: $a... | It is correct. But you should justify the assertions that $\frac{1+\sqrt{1+4b^2}}{2b}\notin(-1,1)$ and that $\frac{1-\sqrt{1+4b^2}}{2b}\in(-1,1)$. This follows easily from the fact that$$\frac{1+\sqrt{1+4b^2}}{2b}\times\frac{1-\sqrt{1+4b^2}}{2b}=-1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sum_{n=1}^\infty (-1)^n\tan\left(\frac{n+1}{n^2}\right)$ converges according to the Leibniz test I need to prove the following series converges:
$$\sum_{n=1}^\infty (-1)^n\tan\left(\frac{n+1}{n^2}\right)$$
I'm trying to understand a solution that uses Leibniz's test.
I understand why $\tan\left(\frac{n+1}{n^2... | HINT
To begin with, notice the $\tan$ function is strictly increasing.
Moreover, its argument is decreasing, as the following reasoning proves:
\begin{align*}
a_{n + 1} - a_{n} & = \frac{(n + 1) + 1}{(n + 1)^{2}} - \frac{n + 1}{n^{2}}\\\\
& = \frac{n^{2}(n + 1) + n^{2} - (n + 1)^{3}}{n^{2}(n + 1)^{2}}\\\\
& = \frac{n^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assume $\sum \frac{1}{a_n} < \infty$, How about $\sum \frac{1}{a_n - 1}$? Problem :
Given sequence $\left\{a_n\right\}$ s. t. $\forall n \in \mathbb{N},a_n(a_n-1)\neq 0$.
If $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n}$ converge, then how about $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n-1}$?
Rewrite $$\frac{1}{a_n-1}... | For the same example given by Lorago (assuming $n>1$), notice that $\dfrac{1}{a_n(a_n-1)}$ would be positive for sufficiently large $n$ (in fact positive for every $n>1$). Specifically, $$\dfrac{1}{a_n(a_n-1)} = \dfrac{1}{n+(-1)^{n+1}\sqrt{n}}.$$
Thus $\displaystyle\sum \dfrac{1}{a_n(a_n-1)} $(considering from the stag... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show this $\left(\left(\frac{p-1}{3}\right)!\right)^3+1\not\equiv 0\pmod p$ let $p\equiv 1\pmod 3$ is prime number,show that
$$\left(\left(\dfrac{p-1}{3}\right)!\right)^3+1\not\equiv 0\pmod p$$
I know this well
Show that $[(\frac{p-1}{2})!]^2+1\equiv0 \pmod p$
and I try use same idea to
$$-1\equiv (p-1)!=\left(\dfrac{p... | Let $X$ be the elliptic curve $u^3+v^3+w^3= 0$, which has the Weierstrass form $y^2 = x^3 - 432$. Now one can confirm the following facts (not necessarily easily).
*
*If $p \equiv 1 \bmod 3$, the number of points on $X$ has the form $1 + p - a_p$, where
$|a_p| < 2 \sqrt{p}$ (Weil bounds).
*If $p \equiv 1 \bmod 3$,... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\dfrac{nm^2-n+1}{2mn-2n+1} \notin\mathbb{Z}$ when $m \geq 2$ Given: $n$ and $m$ are positive integers with
$m≥2$, show that:
$$\frac {nm²-n+1}{2mn - 2n+1}\notin \mathbb Z$$
My attempt:
$n, m \in \mathbb{Z}$
$nm^2 \in \mathbb{Z}$
$nm^2 - n \in \mathbb{Z}$
$2mn - 2n \in \mathbb{Z}$
So $\dfrac{nm^2 - n }{2mn -... | If $\frac{2nm^2-2n+2}{2mn-2n+1}$ is not an integer, then the original fraction $\frac{nm^2-n+1}{2mn-2n+1}$ cannot be an integer either. So we show that $\frac{2nm^2-2n+2}{2mn-2n+1}$ is not an integer.
\begin{align*}
\frac{2nm^2-2n+2}{2mn-2n+1} &= m + \frac{2mn-2n-m+2}{2mn-2n+1} \\[6pt]
&=m + 1 + \frac{1-m}{2mn-2n+1} \\... | {
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"url": "https://math.stackexchange.com/questions/4420902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating $\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}$ Maybe someone could explain the following to me:
So I want to calculate $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}$$
It is pretty much obvious that $$\lim _{x\rightarrow 0^{+}}\le... | You didn't correctly use the known limit though, because one cannot here.
To be able to use the result $\lim_{z \rightarrow 0^+}(1+z)^{\frac{1}{z}} = e$ to evaluate $\lim_{x \rightarrow 0^+} \left(1+\frac{1}{-x-2}\right)^{\frac{1-\sqrt{x}}{1-x}}$, you would need both [among other things]
*
*the equation $\lim_{x \rig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rhombus rotates in a circle. Two vertexes of the rhombus is on the circle, rotate the rhombus $ABCD$ clockwise around the point A to the rhombus $AB'C'D'$, where the point $B'$ falls on the circle, link $B'D,C'C$.
If $B'D:CC'=4:3$, then value of $tan\angle BAD$ will be?
It seems $B',B,D'$ or $D,D',C'$ are in a straight... |
Let $\angle AOD=\angle DOC=\alpha, \angle EOC=\beta, \angle ODC=\gamma$
We have that $\sin \alpha: \sin \beta=4:3$ or $3\sin \alpha=4\sin \beta$, also $\sin(2\alpha+\beta)=0$Thus,
$$\sin 2\alpha\cos \beta+\cos2 \alpha\sin\beta=0$$
$$2\sin \alpha\cos \alpha \cos \beta+\frac{3}{4}\cos2 \alpha\sin\alpha=0$$
$$2\cos \alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle Find the angle $\alpha$ of a triangle with sides $a,b$ and $c$ for which the equality $$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}$$ holds.
My idea is to use the law of cosines: $$\cos\alpha=\dfrac{b^2+c^2-a^2}{2bc}$$ after simplifying the given equality a... | You are doing great, expand the last line, you get $b^2+c^2=a^2+bc$, so the $\cos \alpha = 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ten digit numbers have the sum of their digits equal to $4$? How many ten digit numbers have the sum of their digits equal to $4$?
Well, my solution goes like this:
The first digit of the $10$ digit number cannot be $0$.
It can be $1,2,3$ or $4$ only .
If the first digit is $1$ then we can have $3$ other digi... | As Mark Saving pointed out in the comments, the error you made was not taking into account the order of the digits $1$ and $2$ in numbers in the case in which the leading digit is $1$ and the other nonzero digits are $1$ and $2$. In that case, there are $9$ ways to place the second $1$ and eight ways to place the $2$.... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $x,y,z$ satisfying $x(y+z-x)=68-2x^2$, $y(z+x-y)=102-2y^2$, $z(x+y-z)=119-2z^2$ Solve for $x,y,z$:
$$x(y+z-x)=68-2x^2$$
$$y(z+x-y)=102-2y^2$$
$$z(x+y-z)=119-2z^2$$
After some manipulation, I obtain
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
After combining equations, I get
$$y=\frac{-51-x^2+z^2}{x-z}$$
T... | If you add
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
you get $(x+y+z)^2=17^2$, and so $x+y+z=\pm 17$. If you substitute $x+y=\pm 17-z$ in the third of your equation, you get $\pm 17 z=119$, and so $z=7$ or $z=-7$; now it is easy to get the two solution $x=4$, $y=6$, $z=7$ and $x=-4$, $y=-6$, $z=-7$.
| {
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"url": "https://math.stackexchange.com/questions/4434581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the sum of all the numbers less than $10000$ that can be formed using the digits $1, 2, 3$ and $4$ without repetition. I was not able to solve this question on my own and then when I looked up on the web, I found that there is a formula somewhat related to this problem which states that
If all the possible $n$-di... | Let's do some case work.
One-digit numbers: There are four such numbers: $1, 2, 3, 4$. They add to $10$.
Two-digit numbers: Since repetition is not permitted, there are four ways to choose the tens digit and three ways to choose the units digit. Hence, there are $4 \cdot 3 = 12$ such numbers. By symmetry, each of ... | {
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"timestamp": "2023-03-29T00:00:00",
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Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an ... | This does not answer the question asked in title.
If the goal is to investigate
$$S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}=\frac 1{4^{2n+1}}\Bigg[\sum_{k=0}^{\infty} \frac 1 {\left(k+\frac{1}{4}\right)^{2 n+1} } -\sum_{k=0}^{\infty}\frac 1 {\left(k+\frac{3}{4}\right)^{2 n+1} }\Bigg]$$ it could be sim... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $? I tried this:
$\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $=
= $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2}{n\sqrt{n}}) $ + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{4}{n\sqrt{n}... | Call the required limit $L$. Use the Taylor expansion
$$\cos t=1-t^2/2+O(t^4)$$ to deduce that
$$L=\lim_n \sum_{k=1}^n [2(k/n)^2 n^{-1}+O(n^{-2})] =\int_0^1 2x^2 \,dx=2/3 \,,$$
since the sum is a Riemann sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Question from isi previous years (a) Show that $\left(\begin{array}{l}n \\ k\end{array}\right)=\sum_{m=k}^{n}\left(\begin{array}{c}m-1 \\ k-1\end{array}\right)$.
(b) Prove that
$$
\left(\begin{array}{l}
n \\
1
\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}
n \\
2
\end{array}\right)+\frac{1}{3}\left(\begin{array}{... | Sketch of a proof using Mathematical Induction:
Recall the identity (well-known, easy to prove)
$$\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}. \tag{1}$$
For (a):
Using (1), we have
$$\sum_{m=k}^{n + 1} \binom{m-1}{k-1} = \sum_{m=k}^{n} \binom{m-1}{k-1} +
\binom{n}{k-1} = \binom{n}{k} + \binom{n}{k-1} = \binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Cross covariance and trace identity This may be a simple answer but I can't find any proof. I know that the following identities are true
$$ E\left \{ \left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right )^T\mathbf{Q}\left ( \mathbf{x} - E\left ( \mathbf{x} \right ) \right ) \right \}=tr\left ( \mathbf{Q}cov\lef... | The answer to both is yes.
For the first one, let $\mathbf{c}=\mathbf{x}-E\left ( \mathbf{x} \right)$, $\mathbf{d}=\mathbf{y}-E\left ( \mathbf{y} \right)$ and $Q=\text{diag}(q_1,\ldots,q_n)$, then you can easily check
$$ E\left[ \mathbf{c}^TQ\mathbf{d}\right]=E\left[\sum_i q_i\mathbf{c}_i \mathbf{d}_i\right]=\sum_i q_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Tschirnhaus transformation for cubic equation - "Galois Theory" by Stewart, 3rd edition Consider the following cubic equation:
$$x^3 + ax^2 + bx + c = 0$$
where $a, b, c \in \mathbb{C}$.
Suppose $x \in \mathbb{C}$ satisfies this equation.
Define $y := x + \frac{a}{3}$, so that $x = y - \frac{a}{3}$.
We now substitute t... | Nevermind, I checked the 4th edition of the book and there the author writes $p = \frac{-a^2 + 3b}{3}$, so it must have been a mistake in the 3rd edition that has now been corrected.
I'm leaving this thread up so that other people who encounter this same issue while reading the 3rd edition can resolve it quickly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this eliptic equation? I want to find a solution to
$$ \frac{x_1}{x_2 + x_3} +\frac{x_2}{x_1 + x_3} + \frac{x_3}{x_1+x_2} = 4 $$
for $x_1,x_2,x_3>0$, and $x_1+x_2+x_3=1$.
We have
$$\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}+\frac{1-x_1-x_2}{x_1+x_2}=4.$$
Let $\dfrac{x_1}{1-x_1}=s,\dfrac{x_2}{1-x_2}=t,$then $x_1=... | First notice that x=0,y=52 is a solution. Take a tangent to the curve at this point and it will necessarily intersect at another rational point. By successively taking intersections of the curve with either tangents of rational points or the lines passing through two different rational points you can find more ratio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $a^3 + 15ab^2 = 9$, $\;\frac 35 a^2b + b^3 = \frac 45$ I have problem solving the following system of two cubic equations.
$$\begin{cases}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{cases}
$$
I don't have any idea how to approach this kind of problem.
I'm looking for solutions included i... | Let's convert this to a nice single-variable cubic. Let $a = kb$. Then we have
$$
b^3(k^3 + 15k) = 9
\\ b^3(3k^2 + 5) = 4
$$
Now multiply top by $4$, bottom by $9$, and subtract to get
$$
b^3(4k^3 - 27k^2 + 60k - 45) = 0.
$$
We definitely don't have $b = 0$, so it's the cubic in $k$ that must be $0$. The discriminant i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does this proof of the binomial expansion (a+b)^2 work? I was rereading Terence Tao's Analysis 1 and found this question in the section:
Exercise $2.3.4.$ Prove the identity $(a + b)^2 = a^2 + 2ab + b^2$ for all natural
numbers a, b.
Prior to this we already have proved:
$1.a\cdot b=b\cdot a\\2.a\cdot b=0\implies a=0... | Basically correct, but here are two remarks:
*
*From $(x + y)(x + y)$ to $x(x + y) + y(x + y)$ you need the result $(a + b)\cdot c = a \cdot c + b \cdot c$ which can be proved from 1. 4. but you haven't done it.
*From $x(x + y) + y(x + y)$ you can only get $(x^2 + xy) + (yx + y^2)$. In order to be able to write it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to Evaluate $\int_1^4 (\frac{1}{2t}+i)^2 dt$ Question:
$$\int \limits _1^4\left (\frac{1}{2t}+i \right )^2\,dt.$$
How can I solve this? I believe its an indefinite integral and I can probably expand it by using $(a+b)^2=a^2+2ab+b^2$ to get something like$$\int \frac{1}{4t^2}+\frac{2i}{2t}+i^2\,dt$$but I'm not eve... | We have
\begin{align*}\int \limits _1^4\left (\frac{1}{2t}+i\right )^2\,dt & =\int \limits _1^4\left (\frac{i}{t}+\frac{1}{4t^2}-1\right )\,dt \\
& =i\int \limits _1^4\frac{1}{t}\,dt+\frac{1}{4}\int \limits _1^4\frac{1}{t^2}\,dt-\int \limits _1^41\,dt \\
& =\left .i\ln |t|-t-\frac{1}{4t}\right |_1^4 \\
& =\frac{16i\ln ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Doubts on asymptotic criterion for $\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$ with $a>0$. I have to valuate the character of the following series:
$$\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$$
with $a>0$.
I have thought tha... | First note that
$$
\arctan x = x - \frac{{x^3 }}{3} + \mathcal{O}(x^5 )\quad \text{ and }\quad e^x = 1 + x + \mathcal{O}(x^2 )
$$
as $x\to 0$. Thus,
\begin{align*}
a_n &= n^a \left( {\frac{1}{{n^a }} - \frac{1}{{3n^{3a} }} + \mathcal{O}\!\left( {\frac{1}{{n^{5a} }}} \right)} \right) - \left( {1 + \frac{1}{n} + \mathc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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