Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find equation of the circle whose diameter is the common chord of two other circles? Circle 1:
$$x^2 + y^2 +6x + 2y +6 = 0$$
Circle 2:
$$x^2 + y^2 + 8x + y + 10 = 0$$
My attampt:
From circle 1 and 2, I found
$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got
$$5x^2 + 26x + 30 = 0$$
here I go... | First, obtain the equations of the intersection points below for both $x$ and $y$,
$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$
It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,
$$x_1+x_2=-\frac{26}{5},\>\>\>x_1x_2=6$$
$$y_1+y_2=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-... | $\!\! \bmod n\!+\!1\!:\,\ \color{#c00}{n\equiv -1}\,\Rightarrow\, 3\,\color{#c00}n+11\equiv 3(\color{#c00}{-1})+11\equiv 8\ $ by Congruence Sum & Product Rules $\ \ \ \ \ $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Prove inequalities $\frac 34 \le I(a,b) \le 1$ Given the expression,
$$ I(a,b) = \frac{a^2}{(1+a)(a+b)} + \frac{b^2}{(1+b)(a+b)} + \frac{1}{(1+a)(1+b)}$$
where $a\ge 0$ and $b\ge 0$, prove the following inequalities:
$$\frac 34 \le I(a,b) \le 1$$
I had trouble figuring it out. Tried some inequity techniques I am aware ... | I think, it means that $a+b>0$, otherwise your inequality is wrong for $a=b=0$.
Let $a+b=2x$.
Thus, by C-S and AM-GM we obtain:
$$I(a,b)=\frac{1}{a+b}\left(\frac{a^2}{1+a}+\frac{b^2}{1+b}\right)+\frac{1}{1+a+b+ab}\geq$$
$$\geq\frac{1}{2x}\cdot\frac{(a+b)^2}{1+a+1+b}+\frac{1}{1+2x+\left(\frac{a+b}{2}\right)^2}=$$
$$=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Parabolic mirror: why is $b^2-1/4 = \tan(2\tan^{-1}(2b)+{\pi}/{2})(b-0)$ for all $b$? Let us take the simple parabola $x^2$.
A ray of light will bounce on it making equal angles to both sides.
The derivative of $x^2$ is $2x$ and its normal is $-1/2x$.
Given light rays coming straight from above we get, given a variable... | Apply the following identities
$ tan(x+ \frac \pi 2)=-cot(x)$
$cot(2x) = \frac{ \cot^2(x)-1 }{ 2\cot(x) }$
$ \cot( \tan^{-1}(x) ) =\frac 1x $
Get $$ \tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)
\\ = -\cot( 2 \tan^{-1}(2b) )
\\ =\frac { 1-\cot^2( \tan^{-1}(2b) ) }{ 2\cot( \tan^{-1}(2b) }
... | {
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"timestamp": "2023-03-29T00:00:00",
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Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$. It is trivial to prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \le
8\sqrt{30}$ using numeric methods. For example by multiplying
$(1+\sqrt{2}) \le 3 $
$(1+\sqrt{3}) \le 3 $
$(1+\sqrt{5}) \le 4 $
We get:
$(1+\sqrt{2})(1+\sqr... | $$(1+\sqrt 2)(1+\sqrt 3)(1+\sqrt 5)=1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}$$
and thus you can use the AM-GM inequality on this
Actually I think I messed up, the AM-GM inequality goes the wrong way and the exponents don't work out. Use the Root-Mean Square - Arithmetic Mean Inequa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Infinite product $\Gamma(\tfrac14)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}$ I saw the following infinite product on Wikipedia:
$$\Gamma\left(\tfrac14\right)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}\tag{1}$... | A short proof.
$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$
$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3 = 2^{7/12}Q_1\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369673",
"timestamp": "2023-03-29T00:00:00",
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A (possibly) difficult epsilon-delta proof So I'm struggling to find an epsilon-delta proof for why the function $f(h) = \dfrac{\sqrt[3]{1+h}-1}{h}$ approaches $\dfrac{1}{3}$ as $x \rightarrow 0$.
I'd like to know how one can show using an epsilon-delta proof or some other means that for $0 < \epsilon < \dfrac{1}{8}$ ... | $\lim_\limits {h\to 0} \frac{\sqrt [3]{1 + h} - 1}{h} = \frac{d}{dx} \sqrt [3] x$ evaluated at $1.$
First, let's write the $\epsilon - \delta$ definition of the limit
$\forall \epsilon > 0, \exists \delta > 0 : |h| < \delta \implies |\frac{(1 + h)^\frac 13 - 1}{h} - \frac 13| < \epsilon$
Use the generalized binomial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371401",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $ This functional equation problem is from the Latvian Baltic Way team selection competition 2019:
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 ... | Suppose we have $(x^2+1)^2+4f(x)\geq 0$ for some $x$. Then there exists some $y_0$, such that $$y_0^2-(x^2+1)y_0-f(x)=0.$$ We may also assume that $y_0\ne 0$, because the roots of the above quadratic can't both be $0$. Plugging $P(x, y_0)$ in the equation we get $$y_0f(x)^2=0,$$ and because $y_0\ne 0$, we must have $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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What is the wrong step in the integral $ \int {\ln x \over x\sqrt{1-4\ln x -\ln^2 x}}dx $?
Evaluate the following integral:
$$
I= \int {\ln x \over x\sqrt{1-4\ln x -\ln^2 x}}dx
$$
I've started with a substitution: $t = \ln x$, then:
$$
dt = {dx \over x} \iff dx = xdt\\
I = \int {tdt\over \sqrt{1 - 4t - t^2}}
$$
Com... | Your computation of $I_1$ is wrong. $\int \frac 1 {\sqrt {5-p}} dp$ is $-2\sqrt {5-p}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution to $x^2-y^{11}=23$ in positive integers? Are there positive integers that make the following equation true? $x^2-y^{11}=23$
I do not think it is possible because I have tried many numbers any none of them seem to work. I was able to solve it with an easier case: $x^2 - y^2 = 23$. We can factor to get: $(x+y)(... | Suppose a solution exists. First note that $x$ has to be even. Indeed, if it was odd, then $y^{11}=x^2-23$ would be even but not divisible by $4$ (look mod $4$), which is impossible.
It follows that $y$ has to be odd. But then, since $y^{11}$ is congruent to $y$ modulo $4$ for odd $y$, we find that $y\equiv 1\pmod 4$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let $ a \in \mathbb{N}^\ast $, prove that $ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $ I have the following problem to solve. It's about convergent sequences.
Let $ a \in \mathbb{N}^\ast $, prove that:
$$ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $$
To solve it, I first sol... | The Maclaurin series of $\sqrt{x^2+1}$:$$\sqrt{x^2+1}=1+\dfrac{1}{2}x^2-\dfrac{1}{8}x^4+\dfrac{1}{16}x^5-\cdots$$ Then, find $\dfrac{\sqrt{x^2+1}-1}{x}$,$$\dfrac{\sqrt{x^2+1}-1}{x}=\dfrac{1}{2}x-\dfrac{1}{8}x^3+\dfrac{1}{16}x^5-\cdots$$ Substitute $x=\dfrac{1}{a}$:$$\dfrac{\sqrt{x^2+1}-1}{x}=\dfrac{\sqrt{\frac{1}{a^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan ... | This is a nice case for composition of Taylor series. Using
$$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\tan(\tan(x))=x+\frac{2 x^3}{3}+\frac{3 x^5}{5}+O\left(x^7\right)$$
$$\sin(\sin(x))=x-\frac{x^3}{3}+\frac{x^5}{10}+O\left(x^7\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show with the epsilon-delta definition that $\lim_{x \to 2} \frac{1}{x - 1} = 1$ I have an assignment about epsilon-delta proofs and I'm having trouble with this one. I have worked it through using some of the methods I've picked up for similar proofs but it's just something about this particular expression that doesn'... | Assuming wlog $\frac32\le x\le \frac52$ we have
$$\left|\frac{2 - x}{x - 1}\right| < \varepsilon \iff \left|2 - x\right| < \varepsilon \left|x - 1\right|\le \frac 32 \varepsilon $$
then it suffices to take $\delta <\frac 32 \varepsilon $.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$ I have this equation
$$x^3-x^2-x+0.5=0$$
Now i have to use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$ which will be the roots of the new equation obatained.
I know that i can use many other ways but how to solve by this method
I end up with t... | We have $$2x^3+1=2x^2+2x$$
Cubing both sides $$8(x^3)^3+1^3+3(2x^3)(2x^3+1)=(2x^2+2x)^3=8(x^3)^2+8(x^3)+3(2x^2)(2x)(2x^2+2x)$$
Now as $x^3=y, 2x^2+2x=2x^3+1=2y+1$
$$8y^3+1+6y(2y+1)=8y^2+8y+12y(2y+1)$$
$$\implies8y^3+y^2(12-8-24)+(\cdots)y+(\cdots)=0$$ whose roots are
$$\alpha^3,\beta^3,\gamma^3$$
Using Vieta's formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to use radical rules for this?
$$\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}=?$$
Which one of the following is a true answer and why?
*
*$\sqrt{7}-7$
*$1-\sqrt{7}$
*$\sqrt{7}-1$
*$\sqrt{7}$
*$\sqrt{7}+7$
| Square it and take root:
\begin{align}
& \sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}} \\
= & \sqrt{\Big[\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}\Big]^2} \\
= & \sqrt{\big(3+\sqrt{2\sqrt{7}+1}\big) + \big(3-\sqrt{2\sqrt{7}+1}\big) - 2\sqrt{(3+\sqrt{2\sqrt{7}+1})(3-\sqrt{2\sqrt{7}+1})}} \\
= & \... | {
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"timestamp": "2023-03-29T00:00:00",
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Tricky epsilon-delta proof $$\lim_{x\to -1}\frac{x-1}{x^2-x+1}=-\frac{2}{3}$$
What I've got so far is that:
$$\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<|x-(-1)|< \delta\implies\left|\frac{x-1}{x^2-x+1}-(-\frac{2}{3})\right|<\epsilon\\
\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<| x+1|< \delta\implies\left|\fr... | Note that you have $0<|x+1|<\delta$, and you have $(x+1)$ in your numerator in the second line. So we make this more explicit:
$$
\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|=|x+1|\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\epsilon
$$
This is what we want to hold. Now, in order to get a handle on this, we have
$$
|x+1|\cd... | {
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Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x} $ using the quadratic formula I am trying to prove the fact that $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ or alternatively $\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$. (I understand that it can be proved using the half-angle identities of $\sin$ and $\cos$ but... | Hint. Note that $\pm|x|=\pm x,$ without loss of any generality. Then split into two cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I find general formula of this sequence? Let be the sequence $(a_n)$ so that $a_1 = 1$, $a_{n+1}=\dfrac{20}{3 a_n+4}$, $\forall n \geqslant 1.$ I can prove that the sequence is an increasing sequence and convert to 2. But, I can find the general formula of this sequence. How can I find it?
| $$A_{n+1}=\frac{20}{3A_n+4} \implies A_{n+1}~(3A_n +4)= 20.~~~(1)$$
Let $$(3A_n+4)=\frac{B_n}{B_{n-1}} \implies A_n=\frac{1}{3} \left(\frac{B_n}{B_{n-1}}-4\right)~~~(2)$$
Then (1) becomes $$B_{n+1}-60 B_{n-1}-4 B_n =0~~~(3)$$
Now let us put $B_n=t^n$ in (3) to get
$$t^2-4t-60=0 \implies t=10,-6.$$
So the solution of (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393152",
"timestamp": "2023-03-29T00:00:00",
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If volume of a sphere increases by 72.8%, what is change of its surface area? If the volume of a sphere is increased by 72.8% what would be the change in surface area?
I'm trying to solve the problem using application of derivatives. I noticed that on differentiating the formula for volume of a sphere we directly end u... | Set up a formula that sets up Volume in terms of Area.
As Volume is determined by $V= \frac 43\pi r^3$ and Surface Area of a Sphere is $SA= 4 \pi r^2$ then $SA = 4\pi r^2 = 4\pi{\sqrt[3]{\frac {3V}{4\pi}}}^2$
So if volume is increased by $1.728$ then surface area is increased from $4\pi{\sqrt[3]{\frac {3r}{4\pi}}}^2$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
M... | Show that $n^3/3 \ge 3n-3$ for $n\ge 3$, or equivalently
$n(n^2-9)\ge -9$, or
$n(n+3)(n-3) \ge -9$.
For $n \ge 3:$
$n(n+3)(n-3) \ge 0$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ ... | Let $\sqrt[3]{x+2}=a$ etc.
So, $a+b+c=0$
and $a^3+c^3=2b^3=2(-a-c)^3=-2(a^3+c^3+3a^2c+3ac^2)$
$$\iff0=a^3+ c^3+2a^2c+2ac^2=(a+c)(a^2-ac+c^2)+2ac(c+a)=(c+a)(a^2+ca+c^2)$$
But $a^2+ca+c^2=0$
$\implies$ either $a=c=0$
or $\dfrac ac=$ imaginary as $\left(\dfrac ac\right)^2+\dfrac ac+1=0$
$\implies a+c=0\implies a^3+c^3=0$... | {
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"timestamp": "2023-03-29T00:00:00",
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No solutions to $x^2+y^2+z^2 = 7t^2$ A problem in my book asks me to show that there are no solutions to
$$x^2+y^2+z^2 = 7t^2$$
in the integers apart from $(x,y,z,t)=(0,0,0,0)$.
The solution states that reducing modulo $4$ we see that $x,y,z,t$ must be even
and dividing through we get a smaller solution.
I don't under... | Hmm.... Well, if $x,y,z$ are even then $t$ is, of course and if, $x,y,z$ are all odd then $t$ is. and then $\mod 4$ we get $x^2 + y^2 + t^2 \equiv 3\mod 4$ and $7t^2 \equiv 3\pmod 4$ so that's not a problem. (Yet.)
Indeed It will always be the case that $(2m+1)^2 + (2n+1)^2 + (2k+1)^2 \equiv 3 \equiv 7(2j+1)^2$.
But... | {
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"url": "https://math.stackexchange.com/questions/3395107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Series using partial sums... Determine if the following series converge by studying the partial sums. If it converges, compute its value.
$\sum_{n=1}^{\infty} \frac{2n+3}{(n+1)(n+2)}$.
We use partial fractions so
\begin{equation*}
\begin{split}
\frac{2n+3}{(n+1)(n+2)} &= \frac{A}{n+1}+\frac{B}{n+2} \\
\Longleftrightarr... | Your idea of looking at the partial sum is good. So, as you wrote,
$$
S_k = \sum_{n=1}^{k} \frac{2n+3}{(n+1)(n+2)} = \sum_{n=1}^{k} \frac{1}{n+1}+\sum_{n=1}^{k} \frac{1}{n+2} $$ Now
$$\sum_{n=1}^{k} \frac{1}{n+a}=H_{a+k}-H_a$$ where appear harmonic numbers. So
$$S_k=H_{k+1}+H_{k+2}-\frac{5}{2}$$ Now, using the asympto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x^5-bx^3+cx^2+dx-e$ can be expressed as the product of a perfect square and a perfect cube then prove following If $x^5-bx^3+cx^2+dx-e$ can be expressed as the product of a perfect square and a perfect cube then prove that
$$\frac{12b}{5}=\frac{9d}{b}=\frac{5e}{c}=\frac{d^2}{c^2}$$
My attempt is as follows:
$$E=x... | Since $$\alpha=-\frac{3}{2}\beta,$$ we obtain:
$$E=\left(x+\frac{3}{2}\beta\right)^2(x-\beta)^3=x^5-\frac{15}{4}\beta^2x^3+\frac{5}{4}\beta^3x^2+\frac{15}{4}\beta^4x-\frac{9}{4}\beta^5,$$ which gives
$$b=\frac{15}{4}\beta^2,$$
$$c=\frac{5}{4}\beta^3,$$ $$d=\frac{15}{4}\beta^4$$ and $$e=\frac{9}{4}\beta^5.$$
Can you end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Formal proof function is continuous I'm a bit stuck in this formal proof that the functions is continuos in all of it's domain.
$f(x)= 3(x^2+1)^3$
$ \epsilon >0 , \delta > 0 $
$|x-c|<\delta \to |3(x^2+1)^3-3(c^2+1)^3| \to 3|x^6-c^6+3x^4+3x^2-3c^4-3c^2| $
that's where I get to by myself. Could somebody help m... | As I said in the comments: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ so we can write this as
\begin{align}
&3|(x^2 + 1) - (y^2 + 1)| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2| \\
={} & 3|x - y| \cdot |x + y| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2|.
\end{align}
Then you just need a crude bound on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding more than one basis for a column space Let's say the matrix A is defined as the following:
$$A=\begin{pmatrix}6 & 3 & -1 & 0 \\ 1 & 1 & 0 & 4 \\ -2 & 5 & 0 & 2\end{pmatrix}$$
Whose RRE form is:
$$A_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
Then I can write a basis for t... | Yes it works fine and all the basis you have found are correct as any triple of linearly independent vectors.
A simpler choice would be: $v_1=(1,1,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For which values of $k$ does the equation $2\cos^{2}\theta +k\sin \theta + k = 2$ have real solutions? So I take A level maths and this question was in our textbook. We solved an inequality for when the discriminant is less than zero and this gave us the same answer that is in the textbook. The problem is that the sol... | Rewriting $2\cos^2\theta$ as $2-2\sin^2\theta$, we can substitute $t = \sin\theta$ and our equation becomes
$$
2t^2-kt-k = 0
$$
with solutions
$$
t = \frac{k\pm\sqrt{k^2+8k}}{4}
$$
Having a non-negative discriminant $k^2+8k$ forces either $k \leq -8$ or $k \geq 0$. But in addition, we require $-1 \leq t \leq 1$ for at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(... | Hint
Let $t=\tan(x)$ to make
$$2t \frac{1-t^2}{2t}-5t+5=0\implies t^2+5t-6=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$ Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$
My attempt is as follows:
Taking LHS:
$$\left(\sin A-\cos A\right)(1+\sin A\cos A)$$
$$\left(\sin^2A-\cos^2A\right)\frac{\left(1+\sin A\cos A\right)}{\left(\sin A+\cos A\right)... | $(\sin^2A-\cos^2A)(1-\sin^2A\cos^2A)$
$=(\sin A-\cos A)(\sin A+\cos A)(\sin^4A+\cos^4A).$
But I think the problem is wrong, just try with $A=\frac{\pi}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Fixed Point Theorems and Contraction Mappings I am trying to solve this following exercise.
Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a) < 0$, $F(b) > 0$, and
\begin{align*}
0 < K_1 \leq F'(x) \leq K_2 \; \; \; (a \leq x \leq b).
\end{align*}
Use Theorem 1 ... | You have
$$0 < \frac{K_1}{K_1+K_2} \leq \frac{F'(x)}{K_1+K_2} \leq \frac{K_2}{K_1+K_2} < 1$$
It follows
$$0 < 1- \frac{K_2}{K_1+K_2} \leq 1- \frac{1}{K_1+K_2}F'(x) \leq 1 - \frac{K_1}{K_1+K_2} < 1$$
So, $\boxed{\lambda = \frac{1}{K_1+K_2}}$ is a good choice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{... | Recognize that the numbers cannot both contain a multiple of $5$ and a multiple of $2$; otherwise they would include a zero. Since $10000=2^4\cdot 5^4$, we know that each of $a,b,c$ is of the form $2^r5^s,$ where $0\leq r,s\leq 4$. Using the above fact we only have three possibilities to consider, though two of them ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a stat... | Let $m:=3n+1$. We have
$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac3m\right)^{2(m-1)/3}=\left(\left(1-\frac3m\right)^m\right)^{2/3}\left(1-\frac3m\right)^{-2/3}.$$
Hence the limit is $e^{-3\cdot2/3}\cdot1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Can someone explain this strange relationship between arithmetic, quadratic, and cubic means? Let $a,b,c$ be positive real numbers. Then, there exist unique positive real $x,y$ such that $\frac{a+b+c}{3}$=$\frac{x+y}{2}$, and $\sqrt{\frac{a^2+b^2+c^2}{3}}$=$\sqrt{\frac{x^2+y^2}{2}}$.
Strangely, if $c=(a+b)/2$, then $\... | A slightly different approach: the system is equivalent to $$\left\{\begin{array}{c}x+y=\frac 2 3(a+b+c):=A\\
x^2+y^2=\frac 23(a^2+b^2+c^2):=B\end{array}\right.$$ By elimination, one clearly has $$xy=\frac 12(A^2-B):=C$$ which together with the first equation can be used to solve for $(x,y)$. Note that in general there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2... | Let $\cos x=u\implies \sin x=\sqrt{1-u^2}$. Then $$\sin^8x+\cos^6x=(1-u^2)^4+u^6=(1-4u^2+6u^4-4u^6+u^8)+u^6=1$$ gives $$u^8-3u^6+6u^4-4u^2=0\implies u^2(u^2-1)(u^4-2u^2+4)=0$$ so $u=\cos x=0,\pm1$ as $\Delta_{v^2-2v+4}<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$ I am stuck on a limit of the indeterminate form $\infty-\infty$. I have tried many approaches, such as multiplying with conjugates etc. and I am unable to find a solution. I suspect that there is an elementary trick that I am plainly missing right here. Can anyb... | If
$f(x)
=\dfrac{x^2}{x+1}-\sqrt{x^2+1}
$
then
$\begin{array}\\
f(x)
&=\dfrac{x^2}{x+1}-\sqrt{x^2+1}\\
&=\dfrac{x^2+x-x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x+1-1}{x+1}-\sqrt{x^2+1}\\
&=x-1+\dfrac{1}{x+1}-\sqrt{x^2+1}\\
\text{so}\\
f(x)
&\lt x-1+\dfrac{1}{x+1}-\sqrt{x^2}\\
&= -1+\dfrac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2... | Perform a parametrization of the form $$x = r \cos \theta, \quad y = r \sin \theta.$$ Then we seek to minimize $(x^2 + y^2)^2 = r^4$ subject to the constraint $$\begin{align*}
6 & = r^2 \cos^2 \theta + 2 r^2 \cos \theta \sin \theta - r^2 \sin^2 \theta \\
&= r^2 \left( \cos^2 \theta - \sin^2 \theta + 2\cos \theta \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Show $(a^2 + b^2 + c^2 + d^2)^3 \ge 3 (a^3 + b^3 + c^3 + d^3)^2$ Let $a,b,c,d$ be real numbers such that $a + b + c +d = 0$.
Show:
$(a^2 + b^2 + c^2 + d^2)^3 \ge 3 (a^3 + b^3 + c^3 + d^3)^2.$
I've applied AM-GM to the left hand side to find $ (a^2 + b^2 + c^2 + d^2) \ge 4 \cdot\sqrt[4]{a^2b^2c^2d^2}, $ so $ (a^2 +... | We need to prove that $$(a^2+b^2+c^2+(a+b+c)^2)^3\geq3(a^3+b^3+c^3-(a+b+c)^3)^2$$ or
$$\left(\sum_{cyc}(a+b)^2\right)^3\geq27(a+b)^2(a+c)^2(b+c)^2,$$ which is just AM-GM:
$$\frac{\sum\limits_{cyc}(a+b)^2}{3}\geq\sqrt[3]{\prod_{cyc}(a+b)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Trouble Calculating for $30 \cos(30x)+14=−16$ I am unable to see how,
$$30\cos(30x)+14=−16$$
is equal to
$$\frac{\pi}{30} + n \frac{\pi}{15}$$
I solved up to this
$$\cos(30x) = -1$$
$$\pi= 30x$$
$$\frac{\pi}{30} = x$$
But I am unsure where the $\frac{\pi}{15}$ came from. Can someone help me understand this?
| The thing is $\cos (x + 2n\pi ) = \cos x$ so if $\cos 30x = -1$ then $\cos 30(x+ \frac {2n\pi}{30})$ will also be equal to $-1$.
So when you got $\cos (30x ) = -1$ then
$30x = \pi$ is ONE of the possible values for $30x$ (because $\cos \pi = -1$).
$30x = 3\pi$ is another one (because $\cos 3\pi = -1)$ and $30x = -pi$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
Finding Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
what I tried:
let $S = 4x^2+4xy+4y^2+x-5$
$\dfrac{dS}{dx}=8x+4y+1$ and $\dfrac{dS}{dy}=4x+8y$
for center $\dfrac{dS}{dx}=0$ and $\dfrac{dS}{dy}=0$
getting center as $ x=-\dfrac{1}{6}$ and $y=\dfrac{1}{12}$
How do I ... | The eccentricity only depends on the ratio of the axis so that only the quadratic terms matter.
By symmetry, it is obvious that one of the axis is parallel to $x=y$ and we apply the orthogonal transformation
$$x=u-v,y=u+v$$
which yields
$$3u^2+v^2.$$
Hence
$$e=\sqrt{1-\dfrac13}.$$
More generally, when the quadratic te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Ideal factorization in cubic extension Let $a:=\sqrt[3]3$
Factoring the ideal $(5)$ in $\Bbb Z[a]$, I used reduction of $X^3-3$ mod $5$ to find gives $(5)=(5,a-2)(5,a^2+2a+4)$
Checking my result under sage math gives $(5)=(a-2)(a^2+2a+4)$
I don't know why this difference?
We have $(5)=(5,a-2)(5,a^2+2a+4)=(5\Bbb Z[a]+(a... | In a commutative ring, the product of two principal ideals $(a_1)(a_2)$ is equal to the ideal $(a_1a_2)$. So in this case, this indeed gives us $(a - 2)(a^2 + 2a + 4) = (a^3 - 8) = (-5) = (5)$.
However, we cannot remove the $p$ from your expression in general. For example, if we are working in $\mathbb{Z}[\sqrt[3]{5}]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$ If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$.
I am confused as square root always gives positive number.
But the irony is that if we have $a^2=9$, then we write $a=\pm3 \text { where $a=\sqrt{a^2}$ }$
| Note that $x^2=9$ has two solutions namely $x=\sqrt 9=3$ and $x=-\sqrt 9 = -3$
When we write $x=\pm 3$ we mean that both $3$ and $-3$ are solutions and it does not mean that they are equal.
Thus we have $x=\pm \sqrt 9$ which means there are two solutions and they are opposite to each other.
In general for a real n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
interpretation on $1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}$ and $1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}$ How to interpret the results
$$
1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}
\\
1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}
$$
I want to find a clear a... | The sum of cubes formula can also be derived by slightly generalizing this answer by Mike Earnest for the sum of squares case.
Consider counting ordered quadruplets of integers $(w,x,y,z)$ s.t.
*
*$0 \le w, x, y < z$
*$1 \le z \le n$
When $z=k$ the number of ways to choose $(w,x,y)\in \{0, 1, \dots, k-1\}^3$ is $k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve binomial coefficient equation My book asks me to solve this equation:
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\begin{pmatrix} 7\\x \end{pmatrix}$$
The solution is $x=3$ and the formula $$\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n\\k \end{pmatrix}=\begin{pmatrix} n+1\\... | The formula should be
$$\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}.$$
It is the fundamental recurrence of the binomial coefficients.
Hence
$$\binom{6}{2}+\binom{6}{x}=\binom{7}{x}=\binom{6}{x-1}+\binom{6}{x}\implies \binom{6}{2}=
\binom{6}{x-1}.$$
and, by symmetry, it follows that we have TWO solutions: $x-1=2$ OR $x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If $h(t)=-16t^2+80t$ then simplify $\frac{h(a)-h(1)}{a-1}$ I am to simplify $\frac{h(a)-h(1)}{a-1}$ given $h(t)=-16t^2+80t$. The solution provided is
$\frac{-64+80a-16a^2}{-1+a}$ = $-16a+64$
I cannot see how this was arrived at. Here's as far as I got:
$\frac{h(a)-h(1)}{a-1}$
$\frac{(-16a^2+80a)-(-16+80)}{a-1}$ # subs... | We have that
$$\frac{-16a^2+80a-64}{a-1}=\frac{-16(a^2-5a+4)}{a-1}$$
and
$$a^2-5a+4=(a-4)(a-1)$$
indeed by quadratic equation for $a^2-5a+4=0$
$$a_{1,2}=\frac{5\pm \sqrt{25-16}}{2}=4,1$$
that is
$$a^2-5a+4=(a-a_1)(a-a_2)=(a-4)(a-1)$$
therefore providing that $a\neq 1$
$$\frac{-16(a^2-5a+4)}{a-1}=\frac{-16(a-4)(a-1)}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove $n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$ for all $n \ge 2$? I'm trying to prove that $$n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$$ for all $n \ge 2$. I try induction on $n$ as follows:
My attempt:
The inequality holds for $n=2$. Let it holds for $n$. Our goal is to show that $$2 {... | We can use that bound for harmonic series
$$\sum_{k=1}^n\frac{1}{k} \leq \ln n + 1$$
therefore
$$\frac{1}{(n+1)^2} + \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k}\le \frac{1}{(n+1)^2} + \frac{\ln n + 1}{n+1} \le 1$$
indeed
$$ \frac{1}{(n+1)^2} + \frac{\ln n + 1}{n+1} \le 1 \iff \frac{1}{n+1} + \ln n + 1 \le n+1 \iff \ln n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate the number of solutions Calculate the number of solutions presented by the equation.
$$\sqrt{1-x}+\sqrt{1-2x}+\sqrt{1-4x}=x^2+2$$
What I thought: The LHS is concave, the RHS is convex
| Hint : If $X= \sqrt{a}+ \sqrt{b}+\sqrt{c}$ ... keep squaring and rearranging ...
\begin{eqnarray*}
X &=& \sqrt{a}+ \sqrt{b}+\sqrt{c} \\
\frac{X^2-a-b-c}{2} &=& \sqrt{ab}+ \sqrt{bc}+\sqrt{ca} \\
\left(\frac{X^2-a-b-c}{2} \right)^2 &=& 2\sqrt{abc}(\sqrt{a}+ \sqrt{b}+\sqrt{c})=2\sqrt{abc}X. \\
\end{eqnarray*}
Now square o... | {
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"url": "https://math.stackexchange.com/questions/3436698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
r... | One:
$$r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \Rightarrow y^2=\frac{1-x^2-(1-x)^2r}{1+r}$$
Two:
$$\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \Rightarrow \\
y^2=\left[\frac1{1+r}-x+\frac{r}{1+r}\right]\cdot \left[\frac1{1+r}+x-\frac{r}{1+r}\right]=\\
(1-x)\cdot\frac{(1+x)-(1-x)r}{1+r}=\frac{1-x^2-(1-x)^2r}{1... | {
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would ca... | We can also use
$$\arctan(u) \pm \arctan(v) = \arctan\left(\frac{u \pm v}{1 \mp uv}\right)$$
to obtain in four steps
$$\frac{\frac15 - \frac1{239}}{1 + \frac1{5\cdot 239}}=\frac{239-5}{5\cdot 239+1}=\frac{234}{5\cdot 239+1}=\frac9{46} \to$$
$$\to \frac{\frac15 + \frac9{46}}{1 - \frac15\frac9{46}}=
\frac7{17} \\\to \fra... | {
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Prove that $ \lim_{x \to 0} \frac{ \sqrt{x + \sqrt{ x + \sqrt{x}}} }{\sqrt[8]{x}}= 1$
Prove that $$ \lim_{x \to 0} \frac{ \sqrt{x + \sqrt{ x + \sqrt{x}}} }{\sqrt[8]{x}}= 1$$
Let $x=y^8$ with $t>0$. Then after we get
$$
\lim \frac{ \sqrt { t^8 + \sqrt{ t^8 + \sqrt{t^8}}}}{ t} = \lim \frac{ \sqrt{ t^8 + \sqrt{ t^8 + ... | We have that
$$ \frac{\sqrt{x + \sqrt{ x + \sqrt x}}}{\sqrt[8] x} = \frac{ \sqrt{x + \sqrt[4] x\sqrt{ \frac{x}{ \sqrt x}+1}}}{\sqrt[8] x} = \frac{\sqrt[8]x\sqrt{\frac{x}{\sqrt[4]x} + \sqrt{ \frac{x}{ \sqrt x}+1}}}{\sqrt[8] x} =$$
$$=\sqrt{\frac{x}{\sqrt[4]x} + \sqrt{ \frac{x}{ \sqrt x}+1}} \to \sqrt{0 + \sqrt{ 0+1}}=1$... | {
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Is my approach accurate to find original position of boat? A boat goes upstream for $3$ hr $30$ min and then goes downstream for $2$ hr $30$ min. If the speed of the current and the speed of the boat
in still water are $\frac{10}{3}$ kmph and $\frac{15}{2}$ kmph respectively, how far
from its original position is the b... | Your method looks great. As an alternative:
With reference to ground, the water travels downstream for $2.5+3.5=6$ hours.
With reference to water, the boat went upstream for $1$ hour. So the boat is
$$ \frac{10}{3} \times 6- \frac{15}{2}\times 1 = 12.5$$
$km$ downstream from the start.
| {
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Find the right positive definite block matrix Let $0<a<b$, and let $A, B \in \mathbb{R}^{n \times n}$ be two positive definite symmetric matrices (that do not commute).
My question is: are the matrices $M, P \in \mathbb{R}^{2n \times 2n}$ defined by
\begin{align*}
M = \begin{bmatrix}
bABA & aAB\\
aBA & bBAB
\end{bma... | First of all, we must assume that $b>0$ for either matrix to be positive definite.
For $M$, we compute the Schur complement
$$
bABA - \frac{a^2}{b}(AB)(BAB)^{-1}(BA)=
bABA - \frac{a^2}{b}A.
$$
Since $bABA$ is positive definite, $M$ will be positive definite if and only if $bABA - \frac{a^2}{b}A$ is positive definite. ... | {
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Where am I going wrong in calculating the projection of a vector onto a subspace? I am currently working my way through Poole's Linear Algebra, 4th Edition, and I am hitting a bit of a wall in regards to a particular example in the chapter on least squares solutions. The line $y=a+bx$ that "best fits" the data points $... | The column space of $A$, namely $U$, is the span of the vectors $\mathbf{a_1}:=(1,1,1)$ and $\mathbf{a_2}:=(1,2,3)$ in $\Bbb R ^3$, and for $\mathbf{b}:=(2,2,4)$ you want to calculate the orthogonal projection of $\mathbf{b}$ in $U$; this is done by
$$
\operatorname{proj}_U \mathbf{b}=\langle \mathbf{b},\mathbf{e_1} \r... | {
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Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statem... | $4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$
How do I substitute the statement where n=2k−1 to the above
By factoring one more power out...
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}=(16)4^{2k-1} + (25)5^{2k-1} + (36)5^{2k-1}$
So this is $[16(4^{2k-1} + 5^{2k-1}+6^{2k-1})] + 9*5^{2k-1} + ... | {
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Prove that $3^{2015}-2^{2015}>2016^2$ I have to prove:$$3^{2015}-2^{2015}>2016^2$$
by induction.
I wanted to use the standard factorizing formula:
$$a^{mn}−b^{mn}=(a^n−b^n)(a^{n(m−1)}+a^{n(m−2)}b^n+...+a^nb^{n(m−2)}+b^{n(m−1)}).$$
$2015=403\cdot 5$
What would be the next step?
| We will prove for a more general case , i.e :
$$3^x - 2^x > (x+1)^2 \quad\quad \text{ For x > 3 }$$
For base case , $3^3 - 2^3 = 19 > 16$
Now let be true for a value $n$ , such that $3^n - 2^n > (n+1)^2$
$$\begin{align}3^n - 2^n + 2n+3& > (n+1)^2 + 2n+3\\
3^n - 2^n + 2n+3& > (n+2)^2 \end{align}$$
Now it is left to s... | {
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"answer_count": 3,
"answer_id": 2
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Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$$
How to consider it?
| Here's my solution. Please correct me if I'm wrong.
First, let's introduce and prove a result, which will be applied soon. That is
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}=2.$$
Consider using the squeeze theorem. Notice that
\begin{align*} \frac{1}{\sqrt{n+1-k}+\sqrt{n-k}}\geq\frac{1}{2\sqrt{n+1-k... | {
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In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$ In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$
My attempt is as follows:-
$$\triangle=\dfrac{1}{2}bc\sin A$$
For having the maximum area,
$$A=90^{\ci... | It's true that a triangle with given sides $b$ and $c$ attains it's maximum area when $\angle A = 90^{\circ}$.
But it's not true that a triangle with given medians attains it's maximum area when one of it's angle is a right angle which you assume in your attempt.
Here's one useful lemma to tackle the problem :
Given a... | {
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Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing ... | We see that we need $x>3$ then let $x=\frac3{\cos y}$ with $y\in\left(0,\frac \pi 2\right)$
$$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4} \iff \frac1{\cos y}+\frac1{\sin y}=\dfrac{35}{12}$$
and by half tangent identities by $t=\tan \frac y2$ we obtain
$$\frac{1+t^2}{1-t^2}+\frac{1+t^2}{2t}=\dfrac{35}{12} \iff (3t-1... | {
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Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x... | Note that, as $x\to 0$,
$$\begin{align}\frac{1-\cos{x}\sqrt{\cos{2x}}\pm\sqrt{\cos{2x}}}{x\sin{x}}&=
\sqrt{\cos{2x}}\cdot \frac{1-\cos{x}}{x^2}\cdot\frac{x}{\sin(x)}\\&\quad\qquad+2\cdot\frac{\sqrt{1-2\sin^2(x)}-1}{-2\sin^2(x)}\cdot\frac{\sin(x)}{x}\\
&\to1\cdot \frac{1}{2}\cdot 1+2\cdot\frac{1}{2}\cdot 1=\frac{3}{2}\e... | {
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Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{... | $$
\cos x = 1 - \frac{x^2} 2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots
$$
So for $x$ near $0$ we have
\begin{align}
(\cos x)^{1/x^2} & \ge \left( 1 - \frac{x^2} 2 \right)^{1/x^2} \\[10pt]
& = \left( 1 + \frac{-1/2}{u} \right)^u \\[8pt]
& \to e^{-1/2} \quad \text{as } u \to+\infty. \\[12pt]
(\cos x)^{1/x^2} & \le \lef... | {
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Problem with derivative of generating function and sequence of Catalan number
$Q^1$: Given that
\begin{align}
a_0 &= 0\\
a_{n+1} &= na_n+1\\
\end{align}
find the closed form$?$
My attempt:
\begin{align}
\sum_{n\geq0}^{\infty}a_{n+1}x^{n+1}&=\sum_{n\geq0}^{\infty}na_nx^{n+1}+\sum_{n\geq0}^{\infty}1.x^{n+1}\\
(G(z)-a_0... | $Q^1$: I've played around a bit with initial conditions and indexing and arrived at the following solution. This is just the OEIS sequence A000522 with a $0$ tacked on in the front.
We will use exponential generating functions. First let $b_n = a_{n+1}$. Then we have
$$b_{n} = nb_{n-1} + 1; \qquad b_0 = 1$$
Let $B(x) ... | {
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Find derivative $\frac{dy}{dx}$, given $y(x)=\sin^{-1}\left(\frac{5\sin x+4\cos x}{\sqrt{41}}\right)$
Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\dfrac{5\sin x+4\cos x}{\sqrt{41}}\bigg]$
My Attempt
Put $\cos\theta=5/\sqrt{41}\implies\sin\theta=4/\sqrt{41}$
$$
y=\sin^{-1}\big[\sin(x+\theta)\big]\implies\sin y=\sin(x+\... | As you correctly pointed out we have
$$y = \arcsin\left[\sin\left(x +\theta\right) \right],$$
where $\theta = \arcsin\frac4{\sqrt{41}}$.
Now observe that ($k\in \Bbb Z$)
$$
\arcsin\sin \alpha = \begin{cases}\alpha-2k\pi & \left(2k\pi-\frac{\pi}2\leq \alpha < 2k\pi+\frac{\pi}2\right)\\
- \alpha - (2k-1)\pi & \left(2k\pi... | {
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"answer_count": 3,
"answer_id": 0
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Show that $2^{y-1}|(\alpha_{0}+\alpha_{1}+...+\alpha_{2^{y-1}-1})$
Problem show that
Let $\alpha_{0},\alpha_{1},\alpha_{2},...,\alpha_{2^{y-1}-1}$ are smallest non negative integers such that
$$2^y|(3^{(2^{y-1}+i)}-\alpha_{i})$$
For $0\le i \le 2^{y-1}-1$
Then $$2^{y-1}|(\alpha_{0}+\alpha_{1}+...+\alpha_{2^{y-1}-1})$$... | Let $m=2^{y-1}$.
Then $\alpha_i\equiv 3^{m+i}\pmod{2m}$.
Since $\varphi(2m)=m$, we have $3^m\equiv 1\pmod{2m}$.
Consequently:
\begin{align}
s
&=\sum_{i=0}^{m-1}\alpha_i\\
&\equiv\sum_{i=0}^{m-1}3^{m+i}\\
&\equiv\sum_{i=0}^{m-1}3^{i}
\pmod{2m}
\end{align}
Then
\begin{align}
3s
&\equiv\sum_{i=0}^{m-1}3^{i+1}\\
&\equiv s\... | {
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Determinant with ones on diagonal and increasing powers of a constant in rows and columns I've found a few determinants of this form and I would like to reach a simple expression for this determinant in terms of a and n.
$$
D_{a,n}=\begin{vmatrix}
1 & a & a^2 &\cdots & a^{n-1} & a^n\\
a & 1 & a &\cdots & a^... | Denote by $M$ the matrix in question (note that there are $n+1$ entries on the first row of $M$; thus $M$ is $(n+1)\times(n+1)$, not $n\times n$). Then
$$
\pmatrix{1\\ -a&1\\ &\ddots&\ddots\\ &&-a&1}M=\pmatrix{1&\ast&\cdots&\ast\\ &1-a^2&\ddots&\vdots\\ &&\ddots&\ast\\ &&&1-a^2}.
$$
Therefore $D_{a,n}=\det M=(1-a^2)^n$... | {
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Prove that $W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ Let $F_{a}(n)$ be the digit sum of $n$ in base $a$,
define $W(a,b)=F_{a}(a^{\lceil\frac{\log{b}}{\log{a}} \rceil}-b)$,
prove that $\displaystyle\ W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ if $n−1 \in 6\mathbb{N} \pm 1$.
| For positive integers $n,k$, let
$$S(n,k)=\sum_{i=1}^{n}i^k$$
and for positive integers $m,b$, with $b>1$, let $D(b,m)$ be the sum of the base-$b$ digits of $m$.
Let $k=2$.
Thus, suppose $a$ is a positive integer such that $a \mid S(a,2)$, and let $b=a+1$.
Identically, we have
$$
S(n,2)
=
\sum_{i=1}^n i^2
=
\frac{n(n+1... | {
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Asymptotic evaluations of a limit: $\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}$ I have this simple limit
$$\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}$$
I have solved this with these steps:
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}=\frac{\ln(1+x^3\cos 3x)(1+\sqrt{1+\c... | We have
*
*$\ln(1+x^3\cos 3x) \sim x^3\cos 3x$
moreover we have $\psi(x) \to 1$ and binomial expansion doesn't apply.
Therefore we obtain
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}\sim \frac{\cos 3x}{2(1-\sqrt{1+\cos 3x)}} \to\frac1{2(1-\sqrt 2)}$$
Anyway I suggest to proceed with caution for the asymptoti... | {
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Solve intial value problem using power series $xy''+y'+2y=0$ with y(1) =2, y'(1) =4.
What I tried:
\begin{equation}
xy'' +y'+2y = 0
\end{equation}
Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$
Then
\begin{equation}
x\sum_{k=2}... | We propose to solve $$xy''+y'+2y=0 ~~~(1)$$ by taylor series about $x=1$ as
$$y(x)=\sum_{k=0}^{\infty}~ y^{(k)}(1) \frac{(x-1)^k}{k!}, y^{(k)}(1)~ \text{denotes
$k$th derivative of $y(x)$ at $x=1$}~~~~(2)$$
solution where we are
given that $y(1)=2, y'(1)=4$. We differentiate (1) $n$ times as
$$xy^{(n+2)}+n y^{(n+1)}+2y... | {
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prove that the series convereges and find its value $\frac{5^{k+1}+(-3)^k}{7^{k+2}}$ $$\sum_{k=0}^{\infty}\frac{5^{k+1}+(-3)^{k}}{7^{k+2}}$$
My first inclination is to use the divergence test.
$$\lim_{k\to\infty} \frac{5^{k+1}+(-3)^{k}}{7^{k+2}}$$
$$ \frac{5^{\infty}+(-3)^{k}}{7^{\infty}} \to \frac{5^{\infty}+(-3)^{\in... | hint
We have
$$|5^{k+1}+(-3)^k | \le 5^{k+1}+3.5^{k+1}\le 5^{k+2}$$
thus
$$|u_n| \le (\frac 57)^{k+2}$$
Its sum is
$$\frac{5}{49}\sum_{k=0}^\infty (\frac 57)^k+\frac{1}{49}\sum_{k=0}^\infty(\frac{-3}{7})^k=$$
$$\frac{5}{49}\frac{1}{1-\frac 57}+\frac{1}{49}\frac{1}{1+\frac{3}{7}}=\frac{13}{35}$$
| {
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Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^... | How to fix your approach
You define $d_n=\sqrt[3n]{n^2+n}-1$. This then implies that $\sqrt[\large\color{#C00}{3}]{n^2+n}=(1+d_n)^n\ge nd_n$ (the $3$ was left out). Thus, $d_n\le\frac{\sqrt[3]{n^2+n}}n=\sqrt[\large3]{\frac{n+1}{n^2}}\to0$.
A Different Bernoulli Approach
Bernoulli's Inequality says
$$
\begin{align}
n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
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Evaluate the following integral: $\int_{0}^{\pi/4} \frac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx$ This is a question for the 2006 MIT Integration Bee that went unanswered by the contestants. I am not sure how to solve it either. I was only able to use the double angle formula to simplify ... | Note that \begin{align}
9+ 16\sin\left(2x\right)&=25+32\sin(x)\cos(x)-16\\
&=25-16\sin^2x-16\cos^2x+32\cos x\sin x\\
&=5^2-4^2(\sin x - \cos x)^2\\
&=(5-4\cos x+4\sin x)(5+4\cos x-4\sin x)
\end{align}
hence
\begin{align}
\frac{\sin x+\cos x}{(5-4\cos x+4\sin x)(5+4\cos x-4\sin x)}&=-\frac{1}{40}\frac{-4\cos x-4\sin x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solution to the following Non linear Ode We have the following ODE:
$x'(t) = \frac{3x(t)ln(x(t))}{t^2-3t}$
$x(t_0) = x_0 ∈ R$
My professor claims that the solution is, for $t_0 = 6$ and $x_0 = 3$, $x(t) = 9e^{\frac{− 6 ln 3}{t}}$. Its derivative is $9e^{\frac{− 6 ln 3}{t}}\frac{6 ln 3}{t^2}$. But if we plug our solutio... | There is at least an error in the work for the Laplace transform of the integral in $t$. Since both of these integrals are fairly simple, let's just do them.
\begin{align*}
\int \frac{\mathrm{d}x}{3 x \ln x} &= \frac{1}{3} \int \frac{\mathrm{d}u}{u} & &\left[u = \ln x, \mathrm{d}u = \frac{\mathrm{d}x}{x} \right] \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Riemann sums for $x^3$ on [a,b] So far, I have this for the n'th right Riemann sum: $\sum_{i=1}^{n} \left(a+\frac{(b-a) i}{n}\right)^3 \left(\frac{b-a}{n}\right)$
The first component is the height of the rectangle and the second is the width
Obviously I know this should eventually become $\frac{b^4}{4}-\frac{a^4}{4}$ ... | First, we divide the area from a to b into n-equal subintervals. So, $\Delta x_i = \frac{(b-a)}{n}$ . From now on, let k = (b-a). So, as we know, the Riemann sum will be found by the summation $\sum_{i=1}^n f(x_i) \Delta x_i$, we need to find $x_i$. See $$x_0 = a$$ $$x_1 = a + \frac{k}{n}$$ $$x_2 = a + \frac{2k}{n}$$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve: $\int \frac{\cos x}{x} dx$ Please help me solve this question. I have tried using series expansion, but I am not getting an finite answer.
| There is no closed form of this integral using elementary functions,but we can show an infinite series using Taylor expansion as an answer. In case you want to know more about the integral, you can find it here
$\int \frac{\cos(x)}{x} dx= \int( \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots}{x})dx = \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive $a$, $b$, $c$, show $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$
Prove that for every three positive numbers $a, b, c$:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$$
I tried using $\sum_{\mathrm{cyc}}$ but I haven't got far. I als... | I'll prove a stronger inequality:
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}.$$
Indeed, by C-S
$$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2}{a^2+ab}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Prove that the sum of the squares of three consecutive odd numbers plus one is divisible by 12 but not by 24. I tried to find a similar problem to help me with this question, but I couldn’t find anything. The only relevant thing I may know is that this may be a quadratic word problem; any ideas how to solve this?
| Let the numbers be $x-2$ , $x$ and $x+2$. Now ,
$$\begin{align}S &= (x-2)^2 + x^2 + (x+2)^2+1\\ &= x^2 +4 -4x+x^2+x^2+4+4x +1\\ & = 3x^2 + 9 \end{align}$$
Since $x$ is odd , we have $x=2k+1$
$$S = 3(2k+1)^2 + 0 \implies \color{#d05}{S = 12k^2 + 12k+12}$$
Clearly $S $ is divisible by $12$ .
Since $(k^2+k)$ is always di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution for $I = \int_1^2 \int_0^\sqrt{1-(1-x)^2} x/(x^2+y^2) \ \mathrm dy\ \mathrm dx$ The given integration is:
$$I = \int_1^2 \int_0^\sqrt{1-(1-x)^2} \dfrac{x}{x^2+y^2} \ \mathrm dy\ \mathrm dx$$
After substituting to polar coordinate, I get:
$$I = \int_0^{\pi/2} \int_0^{2\cos\theta} \cos\theta\ \mathrm dr\ \mathrm... | This is an attempt, I can't really find a simple way with polar coordinates.
Assuming your domain is the circle of radius $1$ with center at $(1,0)$ your integral can be written as
$$
\int_{-1}^1 \int_{1 - \sqrt{1-y^2}}^{1 + \sqrt{1-y^2}} \frac{x}{x^2+y^2}dxdy =
2 \int_{0}^1 \int_{1 - \sqrt{1-y^2}}^{1 + \sqrt{1-y^2}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$ABC$ rectangle in $A$ or $C$ iff $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$ The triangle $ABC$ is rectangle in $A$ or $C$ if and only if $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$, where $\alpha$ is the angle in $A$, $\beta$ is the angle in $B... | $$\cot\frac{\beta}{2}=\frac{\cos\frac{\beta}{2}}{\sin\frac{\beta}{2}}=\frac{2\cos^2\frac{\beta}{2}}{\sin\beta}$$
so if you assume that equality, you get
$$\sin\alpha+\sin\gamma=2\cos^2\frac{\beta}{2}\Leftrightarrow \\
2\sin\frac{\alpha+\gamma}{2}\cos\frac{\alpha-\gamma}{2}=2\cos^2\frac{\beta}{2}\Leftrightarrow \\
2\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
| Add the two to get:
$$y=\frac{44-2x^2}{x}$$
Substitute to the first:
$$x^2-\frac{(44-2x^2)^2}{x^2}=7 \Rightarrow \\
3x^4-169x^2+44^2=0 \Rightarrow \\
x^2=\frac{169\pm \sqrt{169^2-12\cdot 44^2}}{6}=\frac{169\pm 73}{6}=16;\frac{121}{3} \Rightarrow \\
x_{1,2}=\pm 4,x_{3,4}=\pm \frac{11}{\sqrt{3}}\\
y_{1,2}=\pm 3,y_{3,4}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\t... | When $x$ is small, $\sin x \approx x-x^3/6+x^5/120$ and $\sin^{-1} x\approx x+x^3/6+3x^5/40$
Then $$\lim_{x \rightarrow 0} \frac{(x-x^3/6+x^5/120)(x+x^3/6+3x^5/40)-x^2}{x^{6}}=
\lim_{x\rightarrow 0}\frac{x^6/18+(.)x^8}{x^6}=\frac{1}{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Formula for the inverse of a block-matrix Let $A, B, C, D ∈ R^{n×n}$. Show that if $A, B, C − D(B^{−1})A$, and $D − C(A^{−1}B$ are nonsingular then
$$\begin{bmatrix}
\mathbf{A} & \mathbf{B} \\
\mathbf{C} & \mathbf{D}
\end{bmatrix}^{-1} = \begin{bmatrix}
\mathbf{A}^{-1} + \mathbf{A}^{-1}\mathbf{B}\left(\m... | Let $\begin{bmatrix}\mathbf{X}_1 & \mathbf{X}_2 \\\mathbf{X}_3 & \mathbf{X}_4 \end{bmatrix} \in \mathbb{R}^{2n\times2n}$ be a matrix such that,
$$\begin{bmatrix}\mathbf{A} & \mathbf{B} \\\mathbf{C} & \mathbf{D} \end{bmatrix} \begin{bmatrix}\mathbf{X}_1 & \mathbf{X}_2 \\\mathbf{X}_3 & \mathbf{X}_4 \end{bmatrix} = \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Doubts in solving the integral $\int \frac{1}{\sqrt{1+\sin2x}}dx$ I tried to solve the integral $\int \frac{1}{\sqrt{1+\sin2x}}dx$ by using $1+\sin2x=(\sin x+\cos x)^2$ but got stuck. So I referred the solution in my book which is given below:
$$
I=\int \frac{1}{\sqrt{1+\sin2x}}dx
$$
$$
= \int \frac{1}{\sqrt{1-\cos(\fr... | $\sqrt{1+sin2x}=\sqrt{(sinx+cosx)^{2}}=\sqrt{2}\left| sin(x+\frac{\pi}{4}) \right|\\
z=x+\frac{\pi}{4}\Rightarrow dz=dx\\
\displaystyle\int\frac{dx}{\sqrt{1+sin2x}}=\frac{1}{\sqrt{2}}\int\frac{dz}{\left| sinz \right|}\\
1)sinz\gt 0\\
\displaystyle\int\frac{dz}{\left| sinz \right|}=\int\frac{sinzdz}{sin^{2}z}=\int\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Getting co-ordinates from equation of a circle I am given the question:
How many points with integer coordinates lie on the circumference of circle $x^2 + y^2 = 5^2$
I know the general equation of a circle so the radius is 5. How do I go about it now ?
| We want to find the integer solutions of the equation $x^2+y^2=5^2$ .
Observe that we must have $-5\le x,y\le 5$ . WLOG assume $x\lt y.$
We get : $x^2+x^2\lt x^2+y^2\implies 2x^2 \lt 25 \implies x\le3$
Setting $x=0$ , we get $y = \pm5$.
Setting $x=\pm1$ , we get $y = \pm \sqrt{24}$ , which is not a solution.
Setting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+... | By your work $$7(a+b+c)^2-9(a+b+c)(a^2+b^2+c^2)-108abc=$$
$$=\sum_{cyc}(12a^2b+12a^2c-2a^3-22abc)=\sum_{cyc}(a^2b+a^2c-2a^3+11(a^2b+a^2c-2abc))=$$
$$=\sum_{cyc}a^2(c-a-(a-b))+11(b^2c+a^2c-2abc))=$$
$$=\sum_{cyc}((a-b)(b^2-a^2)+11c(a-b)^2))=$$
$$=\sum_{cyc}(a-b)^2(11c-a-b)\geq\sum_{cyc}(a-b)^2(3c-a-b).$$
Now, let $a\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$\frac5x-\frac3y=-4$ and $\frac3{x^2}+\frac6{y^2}=\frac{11}3$
Solve the system:
$$\begin{array}{|l} \dfrac{5}{x}-\dfrac{3}{y}=-4 \\ \dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3} \end{array}$$
First, we have $x,y \ne 0$. If we multiply the first equation by $xy$ and the second by $3x^2y^2$, we get $5y-3x=-4xy$ and $9y... | $$\dfrac{5}{x}-\dfrac{3}{y}=-4 \tag 1$$
$$\dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3}\tag 2$$
Substitute $\frac1y = \frac13(\frac5x+4)$ from (1) into (2) to obtain,
$$\frac{59}{x^2} +\frac{80}x+21=0\implies\left(\frac1x+1\right)\left(\frac{59}{x}+21\right)=0$$
Solve to obtain the solutions $(-1,-3)$ and $(-\frac{59}{21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}$ An interesting infinite product with the closed form,
How can we show show that the formula is correct
$$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=\fra... | As I said in the comments, evaluate the finite product and then take the limit as the number of terms, $N \to \infty$.
First let's state up front that
$$\frac{\Gamma{\left ( n+\frac12 \right )}}{\Gamma{(n)}} = \sqrt{\pi} n \frac{(2 n)!}{2^{2 n} (n!)^2} $$
so that we can write the product out to $N$ terms as follows:
$$... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding a formula for a triangle similar to Pascal's. I need to find a formula expressed in binomial coefficients of the following triangle:
$$D_n^k=\begin{cases}
n \quad\quad\quad\quad\quad\text{ if } n=k \text{ or } k=0 \\
D_{n-1}^k+D_{n-1}^{k-1} \text{ otherwise}
\end{cases}$$
This triangle is the same as the Pascal... | As darij grinberg has pointed out, the formula is $$D_n^k=\binom{n+2}{k+1}-2\binom{n}{k}$$ Here is a simple proof that it fits the definition:
$$D_n^n=\binom{n+2}{n+1}-2\binom{n}{n}=n+2-2=n$$
$$D_n^0=\binom{n+2}{1}-2\binom{n}{0}=n+2-2=n$$
$$D_n^k=\binom{n+2}{k+1}-2\binom{n}{k}\stackrel{(*)}{=}\binom{n+1}{k+1}+\binom{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alterative Sum of Squared Error formula proof The well-known formula of calculating Sum of Squared Error for a cluster is this:
SSE formula
where "c" is the mean and "x" is the value of an observation.
But this formula also brings the same result:
Alternative SSE formula
where "m" is the number of the observations and ... | Working with your numerical example:
$$14=SSE=(6-3)² + (6-7)² + (6-8)² = \\
\left(\frac{3+7+8}{3}-3\right)^2+\left(\frac{3+7+8}{3}-7\right)^2+\left(\frac{3+7+8}{3}-8\right)^2=\\
\frac{(7-3+8-3)^2}{3^2}+\frac{(3-7+8-7)^2}{3^2}+\frac{(3-8+7-8)^2}{3^2}=\\
\frac{2[(7-3)^2+(8-3)^2+(7-8)^2]+2[(7-3)(8-3)+(3-7)(8-7)+(3-8)(7-8)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the remainder when $5^{55}+3^{55}$ is divided by $16$
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.
What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please
| We can simplify the expression as follows: $5^{55} + 3^{55} \equiv (5^4)^{13}\cdot 5^3 + (3^4)^{13}\cdot 3^3 \pmod {16}\\
\equiv (9^2)^{13}\cdot 5^2\cdot 5 + 1^{13}\cdot 11\pmod {16}\\
\equiv 1^{13}\cdot9\cdot 5 + 11\pmod {16}\\
\equiv 8\pmod {16}$.
So the remainder is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 7
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recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $ then $a_n>0 \forall n \in N \rightarrow
$ the succession is positive $\foral... | Note that if $a_0>0$ then $a_n>0$ for all $n\geq 1$ and
$$|a_{n+1}-2/3|=\frac{|a_n-2/3|}{|3a_n+2|}\leq \frac{|a_n-2/3|}{2}.$$
Hence,
$$|a_{n+1}-2/3|\leq \frac{|a_n-2/3|}{2}\leq
\frac{\frac{1}{2}|a_{n-1}-2/3|}{2}=\frac{|a_{n-1}-2/3|}{2^2}\leq \dots\leq \frac{|a_0-2/3|}{2^{n+1}}.$$
Therefore, we have that $\frac{|a_0-2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If n approaches infinity, will the descending numbers in n number system divided by the ascending numbers of n number system be closer to an integer? We know that $$\frac{987654321}{123456789}$$ is very close to $8$. If $n$ is the total amount of numbers in the number system ($10$ being the total amount of numbers in t... | We have that in a base $b$ number system, the numerator is
$$1\cdot b^0 +2\cdot b^1+3\cdot b^2+\cdots +(b-1)b^{b-2}$$
while the denominator is
$$(b-1)\cdot b^0 +(b-2)\cdot b^1+(b-3)\cdot b^2+\cdots +b^{b-2}$$
Thus, you are asking if
$$\frac{\sum_{i=1}^{b-1}i b^{i-1}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}$$
gets closer to an i... | {
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"url": "https://math.stackexchange.com/questions/3494759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can someone please explain why the roots are real as well as imaginary? I have this polynomial $x^3+x^2(3-4n)+x(4n^2-6n)-2n^2+6n-4=0$
If I put $n=3$, I get it as $x^3-9x^2+18x-4$
On computing the roots of $x^3+x^2(3-4n)+x(4n^2-6n)-2n^2+6n-4=0$ for any $n$ using Wolfram Alpha I am getting imaginary roots, here.
But if I... | Let $x_1$, $x_2$ and $x_3$ be roots of our polynomial.
Easy to see that if all roots are reals so $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2\geq0,$$ while if there are two complex roots, so
$$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2<0.$$
Now, let $x_1+x_2+x_3=3u,$ $x_1x_2+x_1x_3+x_2x_3=3v^2,$ where $v^2$ can be negative and $x_1x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a function $f$ such that $\gcd(f(x)-f(y),x-y)\mid 2$ For all integers $x,y$. Find a function $f$ such that $\gcd(f(x)-f(y),x-y)\mid 2$ For all integers $x,y$.
This was a question proposed to me by my teacher to attempt,
I tried to find a function such that $f(x)-f(y)=x-y+1 $ but there are no solutions for that bec... | I claim that no such function exists.
Suppose that $f$ is such a function, and consider the values
$$
\begin{align*}
a & = f(4) - f(1) & b & = f(7) - f(4) & c & = f(10) - f(7).
\end{align*}
$$
If $3 \mid a$ then we would have that $\gcd(f(4) - f(1), 4 - 1) \geq 3$, which would be a contradiction. Thus $3 \nmid a$. Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving recursive function outputs $2^x \cdot {x \choose y}$ Consider following function
$f: \mathbb{N}\times\mathbb{N} \rightarrow \mathbb{N}$:
\begin{align*}
f(x ,y) =
\begin{cases}
0 & \text{if } x < y\\
2^x & \text{if } y = 0\\
2 \cdot (f(x - 1, y - 1) + f(x - 1, y)) & \text{else.}\\
\end{cases}
\end{alig... | Going directly, I would try induction on $x$.
If $x=0$, $$f(0,y) = \begin{cases} 1 & y = 0 \\ 0 & \text{otherwise}\end{cases}$$
as required.
Now fix $x$, and let $F(y) = f(x,y)$. You have $F(0) = 2^{x}$, and (by assumption), $$F(y) = 2\left(2^{x-1}\binom{x-1}{y-1} + 2^{x-1}\binom{x-1}{y}\right) = 2^{x}\binom{x}{y}$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Determine all value of $p,q\in\mathbb{N}$ such that : $2^{5}5^{3}=(p+1)(2q+p)$ Problem :
Determine all value of $p,q\in\mathbb{N}$ such that :
$$2^{4}5^{3}=(p+1)(2q+p)$$
My try :
$$2q+p-p-1=2q$$
So $2p+q$ odd or $p+1$ odd
I'm going to try all divisible :
$$1,4,5,8,25,125,10,50,250,20,200,500,40,200,1000,16,80,40... | Observe that $p+1 \mid 2^{4}5^{3}$, so we can write $p+1 = 2^{a}5^{b}$, where $a$ and $b$ are integers such that $0\le a\le 4$, $0\le b\le 3$.
Now, plug this into the original equation, we then have $2q+p = 2^{4-a}5^{3-b} \Rightarrow 2q = 2^{4-a}5^{3-b} - p \Rightarrow 2q = 2^{4-a}5^{3-b} - 2^a5^b +1$. Now observe that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
... | Hint: $$a+a^2+...+a^n=\frac{a(a^n-1)}{a-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence... | Hint
Rationalizing you get
$$\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1}= (\sqrt{n + 1} - \sqrt{n})-(\sqrt{n} - \sqrt{n - 1})\\=\frac{1}{\sqrt{n+1}+\sqrt{n}}- \frac{1}{\sqrt{n}+\sqrt{n-1}}\\
=\frac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n}+\sqrt{n-1})}$$
Repeat:
$$\sqrt{n-1}-\sqrt{n+1}=\frac{-2}{\sqrt{n-1}+\sqr... | {
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"url": "https://math.stackexchange.com/questions/3510658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the conditions for $n$ and $\theta\neq0+2k\pi$ such that $(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n \in\mathbb{R}$ Let $z=1+\cos\theta+i\sin\theta=|z|(\cos\alpha+i\sin\alpha)$ and $z'=1-\cos\theta+i\sin\theta=|z'|(\cos\alpha'+i\sin\alpha')$.
My line of reasoning is to convert the numerator... | Use the exponential form for complex numbers, it will be much shorter:
$$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}=\frac{1+\mathrm e^{i\theta}}{1-\mathrm e^{-i\theta}}=\frac{\mathrm e^{\tfrac{i\theta}2}}{\mathrm e^{-\tfrac{i\theta}2}}\,\frac{\mathrm e^{\tfrac{i\theta}2}+\mathrm e^{-\tfrac{i\theta}2}}{\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Bernoulli's equation(differential equations) Solve the following equation:
$$\sec^2y\frac{dy}{dx}+x \tan y =x^3$$
I have tried like this:
Let, tan y =z
$$\implies \sec^2 y\frac{dy}{dx}=\frac{dz}{dx}\\$$
$$\therefore \frac{dz}{dx}+zx=x^3\\$$
$$I.F.=e^{\int{xdx}}=e^{\frac{x^2}{2}}\\$$
$$\therefore ze^{\frac{x^2}{2}} =\i... | $$(\tan y )'+x \tan y =x^3$$
Substitute $z=\tan y$
$$z'+xz=x^3$$
$$(ze^{x^2/2})'=x^3e^{x^2/2}$$
$$ze^{x^2/2}=\int x^3e^{x^2/2} dx$$
Substitute $u=\dfrac {x^2}2$
$$ze^{x^2/2}=2\int ue^{u} du$$
$$ze^{x^2/2}=2( ue^{u}.-e^u)+C $$
$$z=2(\frac {x^2}2-1)+Ce^{-x^2/2}$$
Finally:
$$ \boxed {\tan y= {x^2}-2+Ce^{-x^2/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find the limit for this $(a_{n})_{n \in \mathbb{N}}$, and $a_{0}>0 \\$ and
$a_{n+1}=a_{n}+\frac{1}{a_{n}^2+a_{n}+1}$
Find the limit of: $\lim_{n \to \infty} \frac{a_{n}^3}{n}$
please help me! :D
Can i solve this whith Cesaro-Stolz theorem or what else?
| $$a_{n+1}^3-a_n^3 = (a_{n+1}-a_n)(a_{n+1}^2+a_{n+1} a_{n}+a_n^2) = \frac{a_{n+1}^2+a_{n+1}a_n+a_{n}^2}{a_n^2+a_n+1}$$
and by letting $\delta_n=a_{n+1}-a_n$
$$\begin{eqnarray*} a_{n+1}^2+a_{n+1}a_n+a_n^2 &=& (a_n+\delta_n)^2+(a_n+\delta_n)a_n+a_n^2\\&=&3a_n^2+3\delta_n a_n+\delta_n^2\end{eqnarray*} $$
such that
$$ a_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can the quadratic formula be explained intuitively? Most people know what the quadratic formula is so I won’t post it here (Besides, I don’t know how to properly post formulas in general).
I was wondering if there is an intuitive explanation as to why the quadratic formula is structured the way it is.
| The development of the quadratic formula is based on solving the quadratic equation in the form $$ax^2+bx+c=0.$$
By completing the square, we have
\begin{align}
ax^2+bx+c&=0\\
ax^2+bx&=-c\\
x^2+\frac{b}{a}x&=-\frac{c}{a}\\
x^2+\frac{b}{a}x\color{blue}{+\left(\frac{b}{2a}\right)^2}&=\color{blue}{\left(\frac{b}{2a}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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