Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $... | If $x$ is positive, then it must be at most $4,$ so $\lfloor x \rfloor$ is at most $3.$ If $\lfloor x \rfloor = 0,$ then your equation is $x^2 - 3 x +2 = 0,$ so $(x-1)(x-2) = 0,$ so no solutions. If $\lfloor x \rfloor = 1,$ then if $y$ is the fractional part of $x,$ we have:
$\lfloor (y+1)^2 \rfloor - 1 = 0,$ so $\lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| let $S= 1+x+x^2+.......+x^{n-1}$ where $x\ne1$
Multiply this equation by $x$ on both sides and get:
$xS= x+x^2+...........+x^{n}$
now subtract the second equality from the first one and you will end up with:
$(1-x)S= 1+x+x^2+.....+x^{n-1}-x-x^2-.......-x^n$
simplify:
$(1-x)S=1-x^n$ which implies that $S=\frac{1-x^n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
} |
How prove this inequality $a+b+\sqrt{a^2+b^2}\ge 2(m+n+\sqrt{2mn})$ let $m,n$ is give positive numbers,and such
$$\dfrac{m}{a}+\dfrac{n}{b}=1$$
show that
$$a+b+\sqrt{a^2+b^2}\ge 2(m+n+\sqrt{2mn})$$
use this methods:How to find the minimum of $a+b+\sqrt{a^2+b^2}$
let
$$a=\dfrac{m(x+y)}{x}.b=\dfrac{n(x+y)}{y}$$
then
we h... | let $$a=\dfrac{m(x+y)}{x},b=\dfrac{n(x+y)}{y}$$
then
$$\Longleftrightarrow \dfrac{m(x+y)}{x}+\dfrac{n(x+y)}{y}+(x+y)\sqrt{\dfrac{m^2}{x^2}+\dfrac{n^2}{y^2}}\ge 2(m+n+\sqrt{2mn})$$
$$\Longleftrightarrow (x+y)(my+nx+\sqrt{(my)^2+(nx)^2})\ge 2xy(m+n+\sqrt{2mn})$$
$$\Longleftrightarrow(x+y)\sqrt{m^2y^2+n^2x^2}\ge (x-y)(my-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know... | First observe that the sum converges (by, say, the root test).
We already know that $\displaystyle R := \sum_{n=0}^\infty \frac{1}{2^n} = 2$.
Let $S$ be the given sum. Then $\displaystyle S = 2S - S = \sum_{n=0}^\infty \frac{2n+1}{2^n}$.
Now use the same trick to compute $\displaystyle T := \sum_{n=0}^\infty \frac{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 8,
"answer_id": 0
} |
solving a trigonometric identity: $\frac{\sec^2 x + 2 \tan x}{1 + \tan x}=1+\tan x$. $$\dfrac{\sec^2 x + 2 \tan x}{1 + \tan x} = 1 +\tan x$$ I started by making $1/\cos^2 x$ the
$2\tan x/\cos x$ making them but I can't came the the right answer.
| We use the identity $$\sec^2 x = 1 + \tan^2 x$$
This gives us $$\dfrac{\sec^2 x + 2 \tan x}{1 + \tan x} = \dfrac{\overbrace{(1 + \tan^2 x )+ 2\tan x}^{(1 + \tan x)^2}}{1 + \tan x} = \dfrac{(1 + \tan x)^2}{1 + \tan x} = 1 +\tan x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show $(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots$ Last month I was calculating $\displaystyle \int_0^\infty \frac{1}{1+x^4}\, dx$ when I stumbled on the surprising identity:
$$\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\r... | Let's have a try.
$$\sum_{n=0}^{+\infty}\frac{(-1)^n}{4n+1}=\int_{0}^{1}\frac{dx}{1+x^4},\qquad S=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)=\int_{0}^{1}\frac{1+x^2}{1+x^4}dx,$$
$$ S = \int_{0}^{1}\frac{x+x^{-1}}{x^{-2}+x^2}\frac{dx}{x}=\int_{1}^{+\infty}\frac{z}{(2z^2-1)\sqrt{1-z^2}}\,dz = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 2,
"answer_id": 0
} |
Indefinite Integral Inverse Trigonometric Function $$\int \frac{2}{9w^2+25}dw$$
I already know this will be equal to $\frac{1}{a} \arctan(x/a)$,
but I don't know how to factor out the $9$. I only know how to take out the $2$.
| $$
\begin{align}
& \phantom{={}} 2\int \frac{dw}{9w^2+25} = 2\int\frac{dw}{25(\frac{9}{25} w^2 + 1)} = \frac{2}{25}\int\frac{dw}{\left(\frac35 w\right)^2 + 1} \\[12pt]
& = \frac{2}{3\cdot5} \int \frac{\frac35\, dw}{\left(\frac35 w\right)^2 + 1} = \frac{2}{15} \int \frac{du}{u^2+1}\text{ etc.}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sin^2\theta + \cos^2\theta = 1$ How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
| Here is a nice proof by squaring the infinite Taylor series of $\sin x$ and $\cos x$.
The proof:
$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 16,
"answer_id": 6
} |
How many terms of the progression $3,6,9,12,\dots$ must be taken to have a sum not less than $2000$? How many terms of the progression $3,6,9,12,\dots$ must be taken to have a sum not less than $2000$?
I tried the calculation, say $n$ terms, as they are in AP so are they want this?
${n\over 2}[2\times 3 (n-1)3]\ge 2000... | The progression $3,6,9,12, \ldots $ is a sequence $(a_n)_{n \in \mathbb{N}}$, $a_n = 3n$. The sum of this sequence can be written as $$\sum\limits_{i=1}^{n} 3i = 3\cdot\sum_{i=1}^{n}i$$
And we have the sum of a sequence formula that states $\sum\limits_{i=1}^{n}i = \frac{n(n+1)}{2}$. It follows that $$\sum_{i=1}^{n}3i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Get the last two digits of $16^{100}$ and $17^{100}$ I try to get the last two digits of $16^{100}$ and $17^{100}$
I started with:
$6^{1} = 6$
$6^{2} = 36$
$6^{3} = 216$ means last digit for $6$ is always $6$
That is why the last digit for $16^{100}$ is $6$
And for $17^{100}$ i calculated:
$7^{1} = 7$
$7^{2} = 49$
$7^{... | Another way for $16$
We have already found $\displaystyle16^{100}=2^{400}$
$\displaystyle2^{10}=1024\equiv24\pmod{100}\implies 2^{20}\equiv24^2=576\equiv76$
Now, $\displaystyle2^{20a+b}-2^b=2^b(2^{20a}-1)\equiv0\pmod{100}$
if $4$ divides $2^b\iff$ integer $b\ge2$
and $2^{20a}-1$ is divisible by $2^{20}-1$ for any pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Which is larger :: $y!$ or $x^y$, for numbers $x,y$. This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?.
No explicit general solution was presented there and I'm just curious :D
Thank-you.
Edit :: I want a most-general solution lfor arbitrary $x$ and $y$; not some specfic cases which can ... | $$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} =
\left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What is $\lim_{x\to0} \frac{(\cos x + \cos 2x + \dots+ \cos nx - n)}{\sin x^2}$? What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$
| We have
$$\sum_{k=1}^n \cos(kx) = \dfrac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)}$$
Hence, we have
$$\dfrac{\dfrac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)} - n}{\sin(x^2)} = \dfrac{\sin(nx/2) \cos((n+1)x/2) - n\sin(x/2)}{\sin(x^2)\sin(x/2)}$$
Expanding around $0$ gives us
$$\dfrac{\left(nx/2 - \dfrac{(nx/2)^3}{3!} + \mathcal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ In the following thread
I arrived at the following result
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after lo... | We are going to evaluate our sum by establishing a system of two relations.
Lets establish the first relation and using the derivative of beta function ( see here) , we have
$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3}{n}+3\frac{H_nH_n^{(2)}}{n}+2\frac{H_n^{(3)}}{n}$$
divide both sides by $n$ then take the sum with r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/612181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 0
} |
Difficult integral involving tangent and square root functions How do I go about integrating this function:
$$\int{\sqrt{\tan{x}}} \,dx$$
| Subbing $x=\arctan{u^2}$, $dx = (2 u)/(1+u^4) du$ produces
$$2 \int du \frac{u^2}{1+u^4}$$
It turns out that
$$1+u^4=(1+\sqrt{2} u+u^2)(1-\sqrt{2} u+u^2)$$
so that we may invoke partial fractions. The result is that the integral becomes
$$\frac1{\sqrt{2}} \int du \left (\frac{u}{1-\sqrt{2} u+u^2} - \frac{u}{1+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
two short doubts about the inverse function in a point the function is $F(x,y,z)=(y^2+z^2, z^2+x^2, x^2+y^2)$ the point is (-1,1,-1)
task: find the local inverse of F in that point.
I have already proved that F is actually invertible there. then i solved the system:
$$
\begin{cases}
y^2+z^2=a\\
z^2+x^2=b\\
x^2+y^2=c
\e... | The function $F:\>(x,y,z)\mapsto(a,b,c)$ is smooth, and $F(-1,1,-1)=(2,2,2)$. The Jacobian $J_F(x,y,z)=16xyz$ is $\ne 0$ at $(-1,1,-1)$; therefore "by general principles" $F$ maps any sufficiently small neighborhood $U$ of $(-1,1,-1)$ bijectively onto a neighborhood $V$ of $(2,2,2)$. The "local inverse"
$G: \>V\to U$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the ma... | This is a classic application of the so-called Lagrange Multiplier. We want to maximise the value of $2x-\tfrac{9}{2}y$ subject to the constraint $x^2 + 9 y^2-4 x+6 y+4=0$. We define
$$\Lambda(x,y,\lambda) = \left(2x-\tfrac{9}{2}y\right)+\lambda\left(x^2 + 9 y^2-4 x+6 y+4\right)$$
The point(s) you are looking for come ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I ... | I don't know if there is an easier solution but I think that an alternative solution for a,b,c positive real numbers, can be derived using the law of tangents.
http://en.wikipedia.org/wiki/Law_of_tangents
Of course this will not cover all cases since for a,b,c lengths of sides of a triangle we have $a+b>c$, $b+c>a$, $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Prove that $A + B = C$ I drew the diagram here
I honestly do not see how $A$ and $B$ could possibly equal $C$.
| $\sin A = \dfrac{1}{\sqrt{10}}$, $\cos A = \dfrac{3}{\sqrt{10}}$, $\sin B = \dfrac{1}{\sqrt5}$, $\cos B = \dfrac{2}{\sqrt5}$, $\sin C = \cos C = \dfrac1{\sqrt2}$.
$\sin(A + B)= \sin A \cos B + \cos A \sin B = \dfrac2{5\sqrt{2}} + \dfrac3{5\sqrt{2}} = \dfrac1{\sqrt{2}}= \sin C$
$\cos(A + B) = \cos A \cos B - \sin A \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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$(1+1/x)(1+1/y)(1+1/z) = 3$ Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers. Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers
and
$$(1+1/x)(1+1/y)(1+1/z) = 3.$$
I know
$(x+1)(y+1)(z+1) = 3xyz$ which is no big deal. I can't move forward ... | Hint: Suppose that $x, y, z \ge 3$. Then $$1 + \frac 1 x \le \frac{4}{3}$$
and likewise for the other two. Then
$$\left(1 + \frac 1 x\right)\left(1 + \frac 1 y\right)\left(1 + \frac 1 z\right) \le \frac{64}{27} < 3$$
So one of the numbers has to be pretty small; now consider cases with $x = 1$ and $x = 2$.
Something e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$ Calculation of remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$
$\bf{My\; Try}::$ Using Division Algorithm:: $p(x) = q(x)\cdot g(x)+r(x)$
Now Let $r(x) = ax^2+bx+c$
So $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c.................... | Another way : for integer $n\ge0$
$$(x+1)^n=(x-1+2)^n$$
$$\equiv2^n+\binom n12^{n-1}(x-1)+\binom n22^{n-2}(x-1)^2\left(\text{mod}{(x-1)^3}\right)$$
Clearly, the remainder reduces to $(x+1)^n$ for $0\le n\le2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $ Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $. I provide the answer below.
| The tag says "induction," so step one: prove that
$$\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$$
by induction.
Step two: Given that
$$f(n)=\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$$
we have
$$f(2n-1)=\frac {2n(2n-1)(4n-1)}6=\frac {n(2n-1)(4n-1)}3$$
$$4f(n-1)=2\frac {n(n-1)(2n-1)}3=4\sum_{i=1}^{n-1} i^2=\sum_{i=1}^{n-1} (2i)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$
| The equation
$$y^2 = x^4 + x^3 + x^2 + x = x(x+1)(x^2+1)\tag{*1}$$
has four trivial solutions over $\mathbb{Z}\times\mathbb{Z}$:
$$(x,y) = (-1,0), (0, 0),(1,\pm 2)$$
To determine whether there are other non-trivial solutions, let's look at an equivalent problem:
$$y^2 = \begin{cases}
z(z + 1)(z^2 + 1), &\text{ for } z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solve the following equation system using Gaussian Elimination. Solve the following equation system using Gaussian Elimination.
$x_1+2x_2-x_3-x_4+x_5=0$
$x_1+2x_2-2x_4+4x_5=0$
$2x_1+4x_2-2x_3-2x_4+2x_5=0$
$-2x_1-4x_2+4x_3+4x_5=0$
My working so far
Putting the equation system into a coefficient matrix:
$$ \left[
\... | Note in all combinations the last column will stay zero so we will just drop it.
$$
\begin{pmatrix}
1 & 2 & -1 & -1 & 1 \\
1 & 2 & 0 & -2 & 4 \\
2 & 4 & -2 & -2 & 2 \\
-2 & -4 & 4 & 0 & 4 \\
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 & -1 & -1 & 1 \\
0 & 0 & 1 & -1 & 3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$? I am stuck on this one; Its a sum and don't know how to calculate the denominator.
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$$
| It should be possible to work out the following carefully, but at any rate, using comparisons with integrals, you should be easily able to see that the limit is zero. I could do it in my head.
We are considering
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}.$$
First, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determinant of a special $n\times n$ matrix Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$
For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$
Then $det = ... | Yet another way to do it: note that the matrices in question are of the form $I + J$, where $J$ is the matrix every entry of which is $1$. We have $J^2 = nJ$ by an easy calculation; thus the eigenvalues of $J$ are $0$ and $n$. The eigenspace corresponding to $n$ is the one dimensional subspace spanned by the vector ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Is it possible to prove that $\lim_{x\to 0} {ln(x)x} = 0$ without L'Hospital's rule? Could somebody give me the answer:
Is it possible to prove that $\lim_{x\to 0} {\ln(x)x} = 0$ without L'Hospital's rule?
| One can also use the following power series, which converge for $-1 < x \leq 1$:
$$\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n$$
Then $x \ln x$ becomes
$$x \ln x = (x-1) \ln x + \ln x = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^{n+1} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n = (x-1) + \sum_{n=1}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Length of chord in circle
*In Figure 3, arc CD is a semicircle. AB is perpendicular to CD, BC = 3, BD = 4. Then the length of AB =
a) 3.25 b) 4.56 c) 3.46 d) 7.00
| HINT :
Note that $\angle CAD=90^\circ$, and so $AC^2+AD^2=CD^2$.
Letting $AB=x$, since
$$AC^2=x^2+3^2, AD^2=x^2+4^2, CD^2=7^2$$
you'll get an equation of $x$ as
$$7^2=9+x^2+x^2+16\iff x^2=12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares. If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.
My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2... | This will give the answer : http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares
You need to prove that $n+1=\frac{x^2+2}{3}$ is not of the form $4^k(8m+7)$ for $k,m\in \mathbb{N}$ and its not that difficult to show this.
For Square modulo 8 we know that $x^2 =0,1,4 \mod{8}$, so $x^2 +2 =2,3,6 \mod{8}$.
Since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
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} |
Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1... | Since you only need to prove it converges, and you don't need to find what it converges to, yes, your argument is concise and valid.
Refer to lab bhattacharjee's answer for the actual value it converges to.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Third order ODE initial value problem,solution obeys $y(x) \rightarrow 0 $ as $x \rightarrow \infty$ ??? $y''' + y'' -y' -y=0$
$y(0)=7,y'(0)=-3,y''(0)=\alpha$
Find all values of $\alpha$ for which the solution obeys $y(x) \rightarrow 0 $ as $x
\rightarrow \infty$
Here is my work
I used the cubic characteristic equatio... | As $x \to \infty$, $e^x \to \infty$, so we must set $A = 0$ in order to guarantee that $y(x) \to 0$ as $x \to \infty$. Then from $y(0) = 7 = A + B$, we see that $A = 0$ implies that $B = 7$.
From $y'(0) = -3 = A - B + C$, we get $-3 = -B + C = -7 + C$, so $C = 4$.
From $y''(0) = \alpha = B - 2C$, we get $\alpha = 7 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the equation $x^3+7x-14(n^2+1)$ has no integral root for any integer $n$. Show that the equation
$$x^3+7x-14(n^2+1)=0$$
has no integral root for any integer $n$.
My work:
I consider the contraposition that there are integer roots.
Assume that the roots are $\alpha,\beta,\gamma$
We have, $\alpha\beta\gamma=14(... | Consider $x^3+7x-14(n^2+1)\equiv0 \mod (7)$, we get $x^3\equiv0 \mod(7)$ and therefore $x\equiv0 \mod (7)$.
Assume $x=7t$, then $(7t)^3+7\cdot 7t-14(n^2+1)=0$. i.e., $7^2t^3+7t-2(n^2+1)=0$. Similar, $7^2t^3+7t-2(n^2+1)\equiv0\mod (7)$, one has $2(n^2+1)\equiv0 \mod (7)$, which is impossible for integer $n$. So the equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove that the limit definition of the exponential function implies its infinite series definition. Here's the problem: Let $x$ be any real number. Show that
$$
\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \sum_{n=0}^ \infty \frac{x^n}{n!}
$$
I'm sure there are many ways of pulling this off, but there are 3 ... | The answer from robjohn linked in his comment is a must visit. It deals with the definition of $e^{x}$ as $\lim_{n \to \infty}(1 + 1/n)^{nx}$. Another direct approach starting with $(1 + x/n)^{n}$ is given below.
Let $$F_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}$$ so that $\lim_{n \to \infty}F_{n}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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pre algebraic factoring with polynomials I really need help solving this particular problem.
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.
| Let's see, we have
$$
\frac{1}{4}x^2y(x-1)^3-\frac{5}{4}xy(x-1)^2
$$
Notice the first term has three $x-1$'s and the second term has two of them. So they share two of them, so we can pull out two of them leaving
$$
(x-1)^2\left(\frac{1}{4}x^2y(x-1)^1-\frac{5}{4}xy\right)
$$
Notice, we 'used' two $x-1$ in the first ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$n^2+3n+2$ is even for any integer $n$ I am trying to prove the following:
If n is an integer, then $n^{2}+3n+2$ is even.
I tried proving this by contradiction, assuming that $n_{2}+3n+2=2k+1$, but I couldn't continue from that point.
Thanks!
| If $n$ is even it can be written as $n=2k$
$$\begin{array}{rcl}n^2 + 3n + 2 &=& (2k)^2 + 3(2k) + 2 \\&=& 2\cdot(2k^2+3k+1)\end{array}$$
If $n$ is odd it can be written as $n=2k+1$
$$\begin{array}{rcl}n^2 + 3n + 2 &=& (2k+1)^2 + 3(2k+1) + 2\\ &=& 4k^2+4k+1+6k+3+2 \\&=& 2\cdot(2k^2+5k+3)\end{array}$$
Hence $n^2 + 3n + 2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as sayi... | $\dfrac{1}{2^x}= \dfrac{5^x\cdot2^x}{2^x\cdot10^x}=\dfrac{5^x}{10^x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find $\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$ I want to find the sum:
$$\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$$
I start with finding the antiderivative of the integrant, which is:
$$\frac{1}{2\sqrt{2}}[\ln(x^2-\sqrt{2}x+1)-\ln(x^2+\sqrt{2}x+1)+2\arctan(\sqrt{2}x-... | A more manageable way should be the following:
Let us make the change of the variable: $x=\frac{y}{\sqrt{n}}$
$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=\frac{2}{n^{\frac{3}{2}}}\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy$$
Now, let's expand the integrand:
$$\frac{y^2}{1+\frac{y^4}{n^2}}=\sum_{k=0}^\inft... | {
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"url": "https://math.stackexchange.com/questions/647168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdo... | Let $P(n) = 6\cdot 7^n - 2\cdot 3^n$. If $4| P(1)$ and we assume $P(k)$ to be true for some arbitrary natural number $k$ and we can show $4| P(k+1) - P(k)$ then we have proved that $4|P(n) \ \forall n\in\mathbb{N}$.
$$\begin{align*} P(k+1) - P(k) & = 6\cdot 7^{k+1} - 2\cdot 3^{k+1} - 6\cdot 7^k + 2\cdot 3^k \\ & = 7\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Formula help with this equation I don't know what the answer to this formula is, can someone please help me.
I've tried lots of things but getting no where.
If $x=\dfrac56+\dfrac{15}{18}-\dfrac{10}{12}$, then $(x-1)3=$ ?
| \begin{align}
x&=5/6+15/18-10/12\\
x-1&=5/6+15/18-10/12- 3/3\\
(x-1)3&=(5/6+15/18-10/12- 3/3)3\\
(x-1)3&=3*5/6+3*15/18-3*10/12- 3*3/3\\
(x-1)3&=5/2+15/6-10/4- 3\\
(x-1)3&=5/2+5/2-5/2- 3\\
(x-1)3&=5/2- 3\\
(x-1)3&=5/2- 6/2\\
(x-1)3&=-1/2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Magnitude relation of fractions for four real numbers with a condition I would like to know if the following proposition holds or not.
For all $a, b, c, d$ such that $a, b, c, d$ are positive real numbers,
IF $a < b$ and $a - c > b - d \geq 0$ THEN $\frac{c}{a} < \frac{d}{b}$ .
Could you kindly advise?
It is really app... | It's false. $a=1$ ; $b=2$ ; $c = 3$ ; $d=5$ is a counterexample.
We have $a<b$, $a-c=-2$ and $b-d = -3$ so $a-c>b-d$.
$\frac{c}{a}=3$ and $\frac{d}{b}=2.5$ so we don't have $\frac{c}{a} <\frac{d}{b}$
EDIT : With your new constraints, I believe this is true.
$a-c>b-d$ so $\frac{a}{c}-1> \frac{b-d}{c}$.
$d>c$ (because $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $(n-k)! \cdot n^k$ as $n$ approaches infinity Is it true that $(n-k)! \cdot n^k$ tends to $n!$ as $n \to \infty$?
I think it is correct but can't think of a satisfying proof.
| Sterling's approximation says $\ln (x!) = x \ln x - x + \frac12\ln(2\pi x) + O(\frac1x)$.
$$\ln \left((n - k)! n^k\right) = k \ln n + \ln \left((n - k)!\right)$$
$$ = k \ln n + (n - k)\ln(n - k) - (n - k) + \frac12 \ln (2 \pi (n - k)) + O(\frac1{n-k})$$
Since $\ln (n - k) = \ln\left(n(1 - \frac{k}n)\right) = \ln n + \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/662241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine for what value of $x$ the series converges $∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $ Determine for what value of $x$ the series converges
$∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $
Here is what I got
Using the ratiotest, I got
$D_n =\frac{\frac{(3^{n+1}+(-2)^{n+1} )}{n+1}(x+1)^{n+1}}{\frac {(3^n+(-2)^n )... | First way. The radius of convergence can be found using the ratio test, since
$$
\frac{a_{n+1}}{a_{n}}=\frac{n}{n+1}\cdot \frac{3^{n+1}+(-2)^{n+1}}{3^{n}+(-2)^{n}}=3\cdot
\frac{1}{1+\frac{1}{n}}\cdot \frac{1+\left(-\frac{2}{3}\right)^{n}}{1+\left(-\frac{2}{3}\right)^{n+1}}\,\to\, 3.
$$
Second way. Here you can use the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^5-x^2+1$ is separable over all fields
Prove that $p(x)=x^5-x^2+1$ is separable over all fields.
When the field is finite or of characteristic zero it is automatically true, since any polynomial is separable. The definition of separability requires to look at the irreducible components, which seems pretty hard.
T... | That idea leads you to the solution.
If the characteristic is $5$ you get $\gcd$ $=1$.
Now assume that characteristic is not $5$. Then you can divide by $5$ (cleaner to multiply by $5$) and compute the gcd.
We get, $\gcd(x^5-x^2+1,5x^3-2)=\gcd(5x^5-5x^2+5,5x^3-2)=\gcd(3x^2-5,5x^3-2)$
If characteristic is $3$ we get th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int x^3\sqrt{9-x^2} dx$ using a trigonometric substitution The Integral is $$\int x^3\sqrt{9-x^2} dx$$
My method: $$x = 3\sin\theta$$ $$dx = 3\cos\theta d\theta$$
$$\sqrt{9-9\sin^2\theta} = 3\cos\theta$$
$$= \int 27\sin^3\theta (9\cos^2\theta) d\theta$$
$$=243\int (\cos^2\theta - \cos^4\theta)\sin\theta d\... | Perhaps easier, shorter and/or simpler: by parts
$$u=x^2\;\;,\;\;u'=2x\\v'=x\sqrt{9-x^2}\;\;,\;\;v=-\frac13(9-x^2)^{3/2}$$
so
$$\int x^3\sqrt{9-x^2}\;dx=-\frac13x^2(9-x^2)^{3/2}+\frac23\int x(9-x^2)^{3/2}dx=$$
$$=-\frac13x^2(9-x^2)^{3/2}-\frac2{15}(9-x^2)^{5/2}+C$$
Two things: this answers negatively your last question... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Transform $\int \frac{x^2}{x - 2}$ to $ \int x +\frac{4}{x-2} + 2 $ I was trying to solve the following integral and I failed so I went to wolframalpha to see the step by step solution, but the following transformation is confusing me.
$\int \frac{x^2}{x - 2} = \int x +\frac{4}{x-2} + 2 $
I am unable to do the transfo... | You can rewrite the expression in exactly the same way you normally complete the square: $x^2 = (x^2 - 4x + 4) + 4x - 4 = (x-2)^2 + 4x-4$.
Then work with the linear terms in the same way: $4x - 4 = 4(x-2) + 8 - 4 = 4(x-2) + 4$.
Put it together to get $x^2 = (x-2)^2 + 4(x-2) + 4$.
Then you can throw in your denominator ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How find this $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^6}\right)$ How find this
$$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^6}\right)$$
I think we can find this value have closed form
$$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^{2k}}\right)$$
since
$$1-\dfrac{1}{n^6}=\left(1-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n^3}\righ... | As Mhenni Benghorbal mentioned, there is a similar problem which has been treated yesterday and I shall use a similar approach to the one he proposed.
$$\prod_{n=2}^{m}\left(1-\dfrac{1}{n^3}\right)=\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i
\sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Agai... | Here's a geometric method, just for variety.
First (as in Yiyuan Lee's solution) we consider cases: how many of the factors on the RHS are negative?
Case 3: All three are negative; then the RHS is negative and we are done.
Case 2: Exactly two are negative; wlog, $3-2a<0$ and $3-2b<0$. This is impossible because $3-2a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Prove a $n \times n $ matrix has rank 3 I have been examining a problem dealing with finding the rank of a $n \times n $ matrix $M$ as follows:
\begin{bmatrix}
0&1&4&9&16&\cdots &(n-1)^2\\
1&0&1&4&9&\cdots&(n-2)^2\\
4&1&0&1&4&\cdots&(n-3)^2\\
9&4&1&0&1&\cdots&(n-4)^2\\
16&9&4&1&0&\cdots&(n-5)^2\\
\vdots&\vdots&\vdots... | When $n\ge4$, let $P=\pmatrix{1&-1\\ &\ddots&\ddots\\ &&\ddots&-1\\ &&&1}^3
=\pmatrix{1&-3&3&-1\\
&\ddots&\ddots&\ddots&\ddots\\
&&\ddots&\ddots&\ddots&-1\\
&&&\ddots&\ddots&3\\
&&&&\ddots&-3\\
&&&&&1}$. Then $P^TMP=\pmatrix{0&1&1\\ 1&-6&1\\ 1&1&0}\oplus 0_{(n-3)\times(n-3)}$. Hence $M$ has rank 3.
Remark.
Knowing tha... | {
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"answer_id": 0
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How to solve $x^3 - 2x^2 -16x+16=0$? How to solve $x^3 - 2x^2 -16x+16=0$ ?
I tried to factor out $x^2$ but it doesn't work..any hints?
Ok I tried to use rational root theorem and have possible answer $\pm 1, \pm 2, \pm 4, \pm 8 , \pm 16$, but my calculator gives out some decimal numbers...
| To solve the cubic, we first perform a transformation to eliminate the $x^2$ term: by letting $x = y + 2/3$, we then obtain the 'depressed' cubic $$y^3 - \frac{52}{3} y + \frac{128}{27} = 0.$$ Next, recall the triple angle cosine identity: $$4 \cos^3 \theta - 3 \cos \theta = \cos 3\theta.$$ This suggests letting $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Flux and Gauss theorem I have a problem; There seems to be something wrong with my understanding of gauss theorem.
Let's say $F = [y ; x^2y; y^2z]$. I want to calculate the flux of $F$ going out of $$D = \{1 \le z \le 2 - x^2 - y^2\}$$
Method number 1
LEt's use Gauss theorem.
So $\nabla \cdot F = x^2 + y^2$, and $$\int... | You have the associated form $$G=ydy\wedge dz+x^2ydz\wedge dx+y^2z dx\wedge dy$$ Then $$dG=({\rm div}\; F)dx\wedge dy\wedge dz=(x^2+y^2)dx\wedge dy\wedge dz$$
Now the flat part of the surface, $S_1$; can be parametrized by $(r\cos t,r\sin t,1)$ with $0\leqslant r \leqslant 1$ and $0\leqslant t\leqslant 2\pi$. The upper... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
| Another way.
By Rearrangement
$$\sum_{cyc}\left(\frac{a^3}{bc}-a\right)=\sum_{cyc}\frac{a}{bc}(a^2-bc)\geq\frac{1}{3}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a^2-bc)=\frac{1}{6}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to find the minimum value of this function? How to find the minimum value of
$$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$.
It seems to be hard if we use calculus methods. Are there another method? I have no idea.
Thank you.
| To find a minimum of the function I'm going to find a lower bound and show that the lower bound is attained. For that I'm going to use nothing harder than Cauchy-Schwarz inequality.
Let $f(x, y, z)$ denote your function. Using mentioned Cauchy-Schwarz inequality we get:
$$f(x, y, z)\cdot\big(x(3y^2+3z^2+3yz+1) + y(3x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Proving that $\left(\frac{\pi}{3} \right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k+1)\zeta(2k+2)}{3^{2k+2}}$. I have asked a question or two like this one before and I've tryed to use similar methods to prove this identity(?), but I failed. By using WA it seems that numerically the LHS=RHS
$$
\left(\frac{\pi}{3} \right)^{2}... | The sum
$$\sum_{k=1}^{\infty} \frac{(2 k+1) \zeta(2 k+2)}{3^{2 k+2}} = \sum_{n=1}^{\infty} \frac1{9 n^2}\sum_{k=1}^{\infty} \frac{2 k+1}{(9 n^2)^k}$$
from the definition of the zeta function. The inner sum is a geometrical sum and its derivative. In essence, it is simple to derive the following:
$$\sum_{k=1}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/682282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can 11 be represented by $a^2 - 3b^2$ where a and b are integers? I know the answer is no, just wan't to know how. From a similar question on the site I got that $a^2 - 3b^2$ should always equal a square modulo 3 which 11 is not.
But I don't understand how to get to this conclusion.
Please help
| There are two indefinite binary quadratic forms (well, classes) of discriminant $12.$ These are $x^2 - 3 y^2$ and $3 x^2 - y^2.$ The primes $2,3$ divide $12$ and are given separate consideration; both can be written as $3 x^2 - y^2.$
For any larger (positive) prime with Legendre symbol $(3|p) = 1,$ we have two choices.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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I cannot find the derivatives of these functions using the product rule? I need help finding the derivative of $y=x^3(3x+7)^2$ at $x=-2$
I tried to simplify the function to $y=x^3 (3x+7)(3x+7)$ and the simplify it into two terms and the derivatives of those terms using the product rule, but that doesn't work.
Since I d... | The easiest solution (by this I mean no fancy theorems needed) is to calculate the product. So we have to way to do so. The first way:
\begin{align}
y&=x^3(3x+7)^2\\&=x^3(9x^2+42x+49), \\
\Longrightarrow y'&=3x^2(9x^2+42x+49)+x^3(18x+42)\\&=45x^4+168x^3+147x^2,
\end{align}
The second way:
\begin{align}
y&=x^3(3x+7)^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$
In a triangle $ABC$, it's given that the following two equations are satisfied:
$$ 3\sin A + 4\cos B = 6 $$
$$ 3\cos A + 4\sin B = 1 $$
Source: ISI B-math UGA 2017
We have to find the angle $ C$. Now, it's easy to see that $ \... | when $C=\dfrac{\pi}{6}, A $ can be large enough,it is different from the case $C=\dfrac{5\pi}{6}$as $A\le \dfrac{\pi}{6}$ that you can't find solution for $A,B$
Now in case $C=\dfrac{\pi}{6}$, it is possible to find solution but you can't use same method to check as $\sin{A} $ can large than $\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to move from powers to simple logarithms? I'm following a book that briefly moves from
$$16000 \times 2^{\displaystyle \left (-\frac{x}{24} \right )} = 1600$$
to
$$x = \frac{24 (\log(2) + \log(5))}{\log(2)}$$
adding the comments that
$$\log(1600) = 6\log(2) + 2\log(5) \\
\log(16000) = 7\log(2) + 3\log(5)$$
What gen... | In general:
\begin{align*}
\log a^b &= b \log a\\
\log (a \cdot b) &= \log a + \log b
\end{align*}
Specific to the above problem:
$$\log(1600) = \log(2^6 \cdot 5^2) = \log(2^6) + \log (5^2) = 6 \log 2 + 2 \log 5$$
$$\log(16000) = \log(2^7 \cdot 5^3) = 7 \log 2 + 3 \log 5$$
So,
\begin{align*}
16000 \times 2^{-x/24} &= 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
| Set up an equation like:
$$
u = \sqrt{31 -8 \sqrt{15}} + \sqrt{31 + 8 \sqrt{15}}
$$
Square to get it down to:
$$
u^2 = 31 - 8 \sqrt{15} + 31 + 8 \sqrt{15} + 2 \sqrt{31^2 - 8^2 \cdot 15}
= 62 + 2 \cdot 1 = 64
$$
Thus you have $u = \pm 8$. The original expression is clearly positive, so:
$$
\sqrt{31 -8 \sqrt{15}} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ So as title says I wanna show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$
So I know that the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ is $x^{4}-6x^{2}+7$ the zeroes of which are $\pm \sqrt{3-\sqrt2},... | Since $2,7$ and $2\times 7$ are all nonsquares
in $\mathbb Q$, we see that $K={\mathbb Q}[\sqrt{2},\sqrt{7}]$ has
degree $4$ over $\mathbb Q$. If we put
$\theta=\sqrt{7}+\sqrt{3+\sqrt{2}}$, we have
$$
\theta^2+4=14+\sqrt{2}+2\sqrt{7\times(3+\sqrt{2})}=
\sqrt{2}+2\sqrt{7}\theta \tag{1}
$$
So $\theta$ is a root of the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that if the $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$ I need to prove that:
If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1.$
So far I used what's given so I have:
$ax+by=1$ (I wrote the gcd of a and b as a linear combination)
and
$(a+b)u+ab(v)=1$ (I also wrote this as a linear combination)
where do I go from here?
| We're given $\gcd(a,b) = 1$, and from Bézout's identity we have $\gcd(a,b) = 1 \iff \exists x,y: ax + by = 1$ for integer $x$ and $y$.
$1 = ax + by = ax + bx - bx + by = (a + b)x + b(y - x)$, so $\gcd(a+b,b) = 1$. Likewise, $\gcd(a+b,a)=1$ because $(a + b)y + a(x - y) = 1$
$$ \begin{align}
1 &= ((a + b)x + b(y - x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
} |
solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
| A difficulty of this equation is that it contains both $\sin(x)$ and $\cos(x)$. You can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to express $\sin(x) = \pm \sqrt{1 - \cos^2(x)}$ (or $\cos(x)$ as a function of $\sin(x)$), but it is very unwieldy because it introduces square roots (another difficulty) an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Proof: $ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $. I need some help with the following proof:
$ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $.
I got:
(1) $[ \sqrt{x} ] \le \sqrt{x} < [\sqrt{x}] + 1 $ (by definition?).
(2) $[ \sqrt{x} ]^2 \le x < ([\sqrt{x}] + 1)^2 $.
(... | Note that every nonnegative number $x$ is between the squares of two consecutive nonnegative integers. Say $n^2 \leq x < (n+1)^2$. Then $n \leq \sqrt{x} < n + 1$, so $\lfloor \sqrt{x} \rfloor = n$ in this situation.
So you just have to make sure the left hand side is also $n$. Since $n^2$ is an integer, one has
$n^2 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/705597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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The normal line intersects a curve at two points. What is the other point? The line that is normal to the curve $\displaystyle x^2 + xy - 2y^2 = 0 $ at $\displaystyle (4,4)$ intersects the curve at what other point?
I can not find an example of how to do this equation. Can someone help me out?
| A key to
easily solving this
is lab bhattacharjee's observation that
$x^2+xy-2y^2=(x+2y)(x-y)
$.
Therefore,
the curve is the union
of the lines
$L_1: x+2y = 0$
and
$L_2: x-y = 0$.
Since $(4, 4)$ is on
$L_2$, and not $L_1$,
and the slope of $L_2$
is $1$,
the slope of the normal is $-1$.
The equation of the line through
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/707716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Factor $x^4 + 64$ I found a page where this problem was solved but his english is broken so its difficult to understand his explanation. His first step was to divide the constant, 64, by the exponent 4. What is his reasoning behind this step?
http://www.wyzant.com/resources/answers/163/factor_x_4_64
| 1)He added and then subtracted $16x^2$, so the value wouldn't change
$$x^4 + 64 = x^4 + 16x^2 - 16x^2 + 64$$
2)He used the fact that $(A+B)^2=A^2+2AB+B^2$, in your case $A=x^2$ and $B=8$ and the formula was used "backwards" to get $(A+B)^2$ back
$$x^4 + 64 = x^4 + 16x^2 + 64- 16x^2=(x+8)^2-(4x)^2$$
3)$A^2-B^2=(A-B)(A+B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/711108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Sum of a simple infinite series
Evaluate:
$$\sum\limits_{n=1}^\infty \frac{n^2}{3^n}.$$
By the ratio test, $\displaystyle\lim_{n\to\infty} \frac{(n+1)^2}{3^{n+1}}\cdot\frac{3^n}{n^2}=1/3,$ which is less than 1, therefore the series is convergent.
Now I am stuck on how to evaluate this series, without the $n^2$ on t... | Just to expand on David's comment:
\begin{align*}
\frac{1}{1-x} &= 1 + x + x^2 + \cdots + x^n + \cdots \\
\frac{d}{dx} \left( \frac{1}{1-x} \right) =
\frac{1}{(1-x)^2} &= 1 + 2x + 3x^2 + \cdots + nx^{n-1} + \cdots \\
\frac{x}{(1-x)^2} &= x + 2x^2 + 3x^3 + \cdots + nx^{n} + \cdots \\
\frac{d}{dx} \left( \frac{x}{(1-x)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 ... | Here's how you could have found that identity without having known it before hand.
First, notice that setting $x = -1$ gives $0$. Thus, $(x - (-1)) = x+1$ is a root.
Using polynomial long division, we find that we can reduce it to $(x^4 + x^2 + 1)(x+1)$
Now, what to do with the $x^4 + x^2 + 1$ term? Recall that $x^3 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Series Expansion Of An Integral. I want to find the first 6 terms for the series expansion of this integral:
$$\int x^x~dx$$
My idea was to let: $$x^x=e^{x\ln x}$$
From that we have: $$\int e^{x\ln x}~dx$$
The series expansion of $e^x$ is: $$\sum\limits_{n=0}^\infty\frac{x^n}{n!}$$
Then we have:
$$\int e^{x\ln x}~dx=\i... | You are in the right approach. Then you have to face $\int x^n(\ln x)^n~dx$ for non-negative integers $n$ :
$\int x^n(\ln x)^n~dx$
$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Solving for $3^x - 1 = 2^y$ Besides $x=2, y=3$, are there any other solutions?
I know that if there is another solution:
*
*$y$ is odd since $2^y \equiv -1 \pmod 3$
*$x$ is even since $3^x - 1 \equiv 0 \pmod 8$
*$3 | y$ since $-1 \equiv 2^y \pmod 9$
Are there any other solutions? If not, what is the argument f... | You don't need the full strength of Catalan's conjecture here.
Two solutions are found easily. $3^1-1=2^1$ gives $x=y=1$. $3^2-1=2^3$ gives $x=2$ and $y=3$.
To prove that these are the only solutions, assume $x > 2$. As you have noted, $x$ must be even: $x = 2z$. Now we have
$$
3^x-1 = 3^{2z}-1 = (3^z-1)(3^z+1) = 2^y.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find any affine transformation that swaps affine lines The task is to find any affine transformation that will swap the following two lines:
$$L_1:(1,1,1) + span((1,0,2))$$
$$L_2:(1,0,1) + span((1,0,-1))$$
From what I understand there is a number of equations I can make:
$$f((1,1,1))=(1,0,1)$$
$$f((1,0,1))=(1,1,1)$$
I ... | \begin{pmatrix} 2\\ 0\\ 0 \end{pmatrix}
\rightarrow
\begin{pmatrix} 2\\ 1\\ 3 \end{pmatrix}.
$$
I will use general approach from "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely" to solve the latter problem. It is shown there, the transformation can be presented as
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Calculus I: Tangent line at certain point, second derivative of the curve given. How to solve? I have never seen this problem in my homework.. I need help right before final!!
Let $\frac{d^2y}{dx^2}=y''=-x^3$ at every point on a curve. The equation of the tangent line at $(1,1)$ is $y = 3-2x$. Find the equation of the ... | Since
$y'' = -x^3, \tag{1}$
the general form of $y'$ is, by integration,
$y' = -\dfrac{1}{4}x^4 + a, \tag{2}$
and so the general form of $y(x)$ must be
$y = -\dfrac{1}{20}x^5 + ax + b; \tag{3}$
since the curve (3) passes through the point $(1, 1)$ we have
$1 = -\dfrac{1}{20} + a + b; \tag{4}$
an since the slope at this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How prove this $\ln{(x+\sqrt{x^2+1})}<\frac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$ let $0<a<1,x<0$,show that
$$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
My idea:
$$\Longleftrightarrow \ln{(\sqrt{x^2+1}+x)}<\dfrac{x(a^x-1)\ln{a}}{(a^x+1)\ln{(\sqrt{x^2+1}-x)}}$$
Then following I f... | If $x<0$, then $x+\sqrt{x^2+1}<x+\sqrt{x^2-2x+1}=x+|x-1|=|x-1|-|x|\leq1$, so
$$\ln(x+\sqrt{x^2+1})<0$$
and $\sqrt{x^2+1}-x>1+0=1$, so
$$\ln(\sqrt{x^2+1}-x)>0$$
Additionally if $0<a<1$, then $a^x-1>0$, $a^x+1>0$ and $\ln a<0$, therefore
$$\frac{x(a^x-1)\ln a}{(a^x+1)\ln(\sqrt{x^2+1}-x)}=\frac{-+-}{++}>0$$
So the inequal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Remainder of the polynomial A polynomial function $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2$. Then the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
What I have thought-The remainder must be of the form $ax^2+bx+c$. Now app... | Hint $\ f(x) = 2x\!+\!1+ (x\!-\!1)^2 (c + (x-3)g(x))\,$ and $\,15 = f(3) = 7+4c\ $ so $\ c = \,\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/724875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $89|(2^{44})-1$ Show that $89|(2^{44})-1$
My teacher proved this problem using mod can someone explain the process step by step? Thank you so much!
| With a bit of calculation.
Need to prove $2^{44}\equiv 1\pmod{89}$.
$2^8=256=2\cdot 89+78=3\cdot 89-11$, so $2^8\equiv 78\equiv -11\pmod{89}$ $-$ it is a definition of $\pmod{}$ operation.
Then $2^3\cdot 2^8\equiv 8\cdot 2^8\equiv8\cdot 11\equiv-88 \equiv 1\pmod{89}$ $-$ because $1=1\cdot 89-88$.
So, you have $2^{11}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/726899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your ... | My more general method of calculation. Enables us to solve and other factors. Find out whether or when given coefficients solutions and immediately write the formula.
For example, consider the equation:
$4(a^2+b^2+c^2+d^2)=3(a+b+c+d)^2$
Then the solutions are of the form:
$a=-(p^2+4(k+s)p+2k^2+2ks+2s^2)$
$b=p^2+4(k+s)p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find all ordered pairs $(m,n)$ such $\frac{n^3-1}{mn-1}$ is an integer Question:
Find all ordered pairs $(m,n)$ where $m$ and $n$ are positive integers, such that $$\dfrac{n^3-1}{mn-1}$$ is an integer.
I know this problem is very simaler this 1994 IMO,problem 4,But My problem is different.
This IMO problem is :
F... | Note that if $(m, n)$ is a solution then $(mn-1) \mid (n^3-1)$ so $(mn-1) \mid m^3(n^3-1)$. Also $(mn-1)\mid m^3n^3-1$ so $(mn-1) \mid m^3-1$, so $(n, m)$ is also a solution.
We may thus WLOG assume $m \geq n$. If $n=1$ then $\frac{n^3-1}{mn-1}=0$ so $(m, 1)$ is a solution for all positive integers $m>1$. I would rejec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$ If $a+b+c+d = 2$, prove that
$$\dfrac{a^2}{(a^2+1)^2}+\dfrac{b^2}{(b^2+1)^2}+\dfrac{c^2}{(c^2+1)^2}+\dfrac{d^2}{(d^2+1)^2}\le \dfrac{16}{25}$$
Also $a,b,c,d \ge 0$.
| Assume $0 \le a \le b \le c \le d$, with $a+b+c+d=2$,
Then $(48a-4)(a^2+1)^2-125a^2 = (2a-1)^2(12a^3+11a^2+32a - 4) \ge 0$, for $a \ge \frac{1}{8}$;
(See that $32a-4 \ge 0$ and $12a^3+11a^2$ is positive).
That is $\dfrac{a^2}{(a^2+1)^2} \le \dfrac{48a-4}{125}$, and similarly for $b,c,d$ and adding them we have $\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Linear Transformation from $\alpha$ to $\beta$ T: $R^3$ $\to$ $R^2$
$$[T]_{\beta\alpha} =
\begin{matrix}
2 & 3 & 1 \\
1 & 2 & 1 \\
\end{matrix}
$$
$\alpha$ = {(1, -1, 1), (0, 1, 0), (1, 0, 0)}
$\beta$ = {(3, 2), (2, 1)}
Find: T((x, y, z)) for any x, y, z in $R^3$
My approach:
$[T(v)]... | it is correct in my humble opinion except rhe last step.
finally,
$$(x + 3y + 4z)\begin{pmatrix}2\\1\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}3\\2\\\end{pmatrix} =...$$
instead of
$$(x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the coefficient of a generating function Given $f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4$. This is the generating function $f(x)$ of $a_n$, which is the number of ways to get $n$ as the sum of the upper faces of four thrown dice.
How do I calculate a coefficient from said generatin... | $$f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4=x^4(1+x+^2+x^3+x^4+x^5)^4=\\
=x^4\left(\sum_{k=0}^{5}x^k\right)^4=x^4\left(\frac{1-x^6}{1-x}\right)^4=x^4(1-x^6)^4(1-x)^{-4}=\\=x^4\sum_{j=0}^{4}(-1)^j\binom{4}{j}x^{6j}\sum_{h=0}^{\infty}(-1)^h\binom{-4}{h}x^h$$
Note that $(-n)!=(-n)(-n-1)(-n-2)\... | {
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"timestamp": "2023-03-29T00:00:00",
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How do I integrate $\frac{1}{x^6+1}$ My technique so far was substitution with the intent of getting to a sum of three fractions with squares in their denominators.
$t = x^2 \\
\frac{1}{x^6 + 1} = \frac{1}{t^3+1} = \frac{1}{(t+1)(t^2-t+1)}$
Then I try to reduce this fraction into a sum of two fractions
$\frac{A}{t+1} +... | Observe that $$x^6 + 1 = (x^2)^3 + 1 = (x^2 + 1)(x^4 - x^2 + 1).$$ To factor the quartic term, suppose there is a factorization of the form $$\begin{align*} x^4 - x^2 + 1 &= (x^2 + ax + 1)(x^2 + bx + 1) \\ &= x^4 + (a+b)x^3 + (ab+2)x^2 + (a+b)x + 1. \end{align*}$$ We thus require $a + b = 0$, and $ab + 2 = -1$, from ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is
The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is?
I tried this: $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}=2\binom{50}0\binom{50}1... | $$(1+x)^{50}=\binom{50}{0}+\binom{50}{1}x+\cdots+\binom{50}{49}x^{49}+\binom{50}{50}x^{50}$$
$$(x+1)^{50}=\binom{50}{0}x^{50}+\binom{50}{1}x^{49}+\cdots+\binom{50}{49}x+\binom{50}{50}$$
Multiplying we observe that our required sum is the coefficient of $x^{49}$ in $(1+x)^{100}$
$$S=\binom{100}{49}=\binom{100}{51}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/738169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Hard Olympiad Inequality Let x,y,z be positive real numbers such that $xy+xz+yz=1$. Prove that $$\sqrt{x^3+x}+ \sqrt{y^3+y}+ \sqrt{z^3+z} \geq 2 \cdot \sqrt{x+y+z}$$.
I tried to square expand homogenize then majorize. But I couldn't make it work. Any help would be much appreciated.
| By C-S and Schur we obtain:
$$\sum_{cyc}\sqrt{a^3+a}=\sqrt{\left(\sum_{cyc}\sqrt{a^3+a^2b+a^2c+abc}\right)^2}=$$
$$=\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}(a^3+a^2b+a^2c+abc+2(ab(a+b+c)+abc))}=$$
$$=\sqrt{\sum_{cyc}(a^3+3a^2b+3a^2c+5abc)}=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Area of triangle A triangle is inscribed in a circle. The vertices of triangle divide the circle into three arcs of length 3, 4 and 5 units, then find the area of triangle.
|
Study the above diagram which describes the problem.
From the calculations of the arc lengths, we have:
$$3 = rA, \qquad4 = rB, \qquad 5 = rC$$
$$3 + 4 + 5 = rA + rB + rC = r(A + B + C)$$
$$r = \frac{12}{A + B + C} = \frac{6}{\pi}$$
We also know that the angles $A,B,C$ are in proportions corresponding to the respectiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $y = e^{x(y-1)}$ then $y \approx 1-2(x-1)$ when $0Assume $y = e^{x(y-1)}$.
Then $y \approx 1-2(x-1)$ when $0<x-1<<1$
I thought of something like that:
$$ e^{x(y-1)} = e^{-2(x-1)}e^{xy+x-2}=(1-2(x-1)+O(x-1)^2)e^{xy+x-2}$$
But I failed to prove that $e^{xy+x-2} \approx 1$
Any ideas?
Thanks
| This isn't rigorous, but will give you a starting point: since $e^z \approx 1 + z + \dfrac 12 z^2$ you have $y = e^{x(y-1)} \approx 1 +x(y-1) + \dfrac 12 x^2 (y-1)^2$. That is, $$y-1 \approx x(y-1) + \frac 12 x^2 (y-1)^2$$ so that (after canceling the $y-1$) $$1 \approx x + \dfrac 12 x^2 (y-1)$$ and $$y \approx \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverge? $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverges, and I have no idea why? It would seem using L'hop, $\frac{n^3 + n^2 + n}{n^3 + 1}$ goes to 1. So it should end up just being $1\cdot1\cdot1\cdots$, which makes me feel it converges. Is the p... | Notice that for $n\ge1$ you have
$$\frac{n^3+n^2+n}{n^3+1}=
1+\frac{n^2+n-1}{n^3+1} \ge
1+ \frac{n^2-n+1}{n^3+1} =
1+\frac1{n+1} = \frac{n+2}{n+1}.$$
Therefore
$$\prod_{k=1}^n \frac{k^3+k^2+k}{k^3+1} \ge \prod_{k=1}^n \frac{k+2}{k+1} = \frac32 \cdot \frac 43 \cdots \frac{n+2}{n+1} = \frac{n+2}2.$$
And for $n\to\infty$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove that if b is coprime to 6 then $b^2 \equiv 1 $ (mod 24) Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24).
Now I have that as $\gcd(b,6) = 1$, we know that $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.)
So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd.
Then I'm not sure where to... | Starting with $b$ is odd, so $b = 2k + 1$, and $b^2 - 1 = 4k^2 + 4k = 4k(k+1)$ clearly divisible by $8$. So we need to show that $b^2 - 1$ is divisible by$3$. Since $b$ is not divisible by $3$, $b = 3k + 1$ or $b = 3k +2$. If $b = 3k + 1$, then $b - 1 = 3k$ and $3 | b - 1$, so $3 | b^2 - 1$. And if $b = 3k + 2$, then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/747166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \fra... | Use the mean value theorem. ${}{}{}{}{}{}{}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$... | Let $\alpha,\beta$ be two distinct roots of $ax^2-bx+c=0,$ Where $a,b,c\in \mathbb{N}$
So here $1<\alpha,\beta <2,$ So here $\displaystyle \alpha+\beta = \frac{b}{a}$ and $\displaystyle \alpha \cdot \beta = \frac{c}{a}$
Now Using $\bf{A.M\geq G.M}$ For $0<\alpha, 1-\alpha,\beta, 1-\beta<1$
So $$\frac{1-\alpha+\alpha}{2... | {
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"url": "https://math.stackexchange.com/questions/748229",
"timestamp": "2023-03-29T00:00:00",
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If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $a^4c+b^4d\ge cd$.
If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $$a^4c+b^4d\ge cd$$
It kind of seems useful to begin with a division of both sides by $cd$:
$$\frac{a^4}{d}+\frac{b^4}{c}\ge1$$
It seems like a simple Cauchy-Schwarz would... | $\frac{A^4}{d} >= 1-\frac{B^4}{c}$
$\frac{A^4}{d} >= \frac{(c-B^4)}{c}$
$\frac{C\times A^4}{d} >= C-B^4$
$C\times A^4 >= CD -D\times B^4$
$C\times A^4 + D\times B^4 >= CD$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/748413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integrate $\int_0^\pi \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)d \theta$ Hello I am trying to integrate
$$
I=\int_0^\pi \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)d \theta
$$
which is similar to Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$., however we have hyperbolic functions in... | $$\int_0^\pi\theta^2\ln^2\left(2\cosh\dfrac{\theta}{2}\right)d\theta$$
$$=\int_0^\pi\theta^2(\ln(e^\frac{\theta}{2}+e^{-\frac{\theta}{2}}))^2~d\theta$$
$$=\int_0^\pi\theta^2(\ln(e^\frac{\theta}{2}~(1+e^{-\theta})))^2~d\theta$$
$$=\int_0^\pi\theta^2\left(\dfrac{\theta}{2}+\ln(1+e^{-\theta})\right)^2~d\theta$$
$$=\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/748956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Calculation of $\int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$ Calculate
$$ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$$
$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\frac{\pi... | Proposition :
\begin{equation}
\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]\quad,\quad\text{for }\,\mu\ge0
\end{equation}
Proof :
Let
\begin{equation}
I(\mu)=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/749493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Solving this equation for $\theta$ $$ 0 = x\cos\theta\cos\phi+y\sin\theta\cos\phi+z\sin\phi$$
Here's what I've tried doing.
$$\begin{align}
x\cos\theta\cos\phi+y\sin\theta\cos\phi & = -z\sin\phi \\
x\cos\theta+y\sin\theta & = -z\tan\phi \\
x\cos\theta+y\sqrt{1-\cos^2\theta} & = -z\tan\phi \\
y\sqrt{1-\cos^2\the... | HINT :
$$
x\cos\theta+y\sin\theta=r\cos(\theta-\alpha),
$$
where $r^2=x^2+y^2$ and $\tan\alpha=\dfrac{y}{x}$. You may refer to this link.
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
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"question_score": "2",
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Find an integer $x$ satisfying the congruence: $$x \equiv \ 1 \pmod3$$
$$x \equiv \ 2 \pmod5$$
$$x \equiv \ 8 \pmod{11}$$
From the first, I have $x=3k+1$, $x=5j+2$ from the second and $x=11l+8$ from the third.
Subbing the third into the second I get
$11l+8 \equiv 2 \pmod5$
$l \equiv -6\pmod5$
$l \equiv -1\pmod5$
So... | For small moduli, I came up with this method. Make the following table.
\begin{array}{r|ccc}
& \mod 3 & \mod 5 & \mod{11}\\
\hline
55 & 1 & 0 & 0\\
33 & 0 & 3 & 0\\
15 & 0 & 0 & 4\\
\hline
103 & 1 & 3 & 4
\end{array}
Let $N = 3 \times 5 \times 11 = 165$. Then
$55=\dfrac N3,\;
33= \dfrac N... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$ let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$
My Try $$A=\sqrt{a^2+4a+4}+\sqrt{b^2+2b+1}=$$then I can't
Please just a hint.
| Hint: You need to know that $a\leqslant -2$ iff $a+2\leqslant 0$, and $b\leqslant -1$ iff $b+1\leqslant 0$. And the most important thing: $$ \color{lightblue}{\boxed{\color{white}{\boxed{\color{white}{\overline{\underline{\color{black}{\displaystyle\sqrt{\rm X^2}=|\,\rm X|\,}}}}}}}}$$
I hope this helps.
Best wishes, $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the series $\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6$
Prove the series $$\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6.$$
I've tried various methods for proving the series:
The series is not geometric, but I see that $\frac 1 6 n (\frac 5 6)^{n-1} \rightarrow 0$. Also the series is n... | Here's a calculus approach:
We know that:
$$\frac{1}{1 - x} = 1 + x + x^2 + \dots$$
Differentiating both sides,
$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \dots$$
Now, substitute $x = \frac{5}{6}$. Then,
$$\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{\left(1 - \frac{5}{6}\right)^2}$$
So we have:
$$\frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
| $\dfrac{x^3-4x^2+5x-6}{x^2-x-6}=\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}$
Now we are going to use Factor Theorem, to determine one of the factors of the numerator, which I call call $h(x)$.
(Note: Just in case you're not familiar with the Factor Theorem, it says that a polynomial $p(x)$ has a factor $(x-c) \iff p(c)=0$ )
We n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \fr... |
I believe this surface can be described by two networks of lines. ( Shown in Red and Cyan). Make the substitution $nt$ for $u$ , and separately $nt$ for $t$ ,
so that the new parametrization might look like : $<t,nt,a(1/n + n/1)> $ and/or $<nt,t,a(1/n+ n)>$ these are the asymptotic lines. Then switch the sign of n to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/761350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove the matrix satisfies the equation $A^2 -4A-5I=0$ How to prove that
$$
A=\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{bmatrix}
$$
satisfies the equation $A^2 -4A-5I=0$?
| If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach.
Let's take the equation
$$ A^2 - 4A - 5I = 0.$$
And now forget about the matrices. What we have is simple quadratic equation in $x$ written as
$$ x^2 - 4x - 5 = 0,$$
now just find the roots
$$ (x-5)(x+1) = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How I can refine the proof of $\mathop {\lim }\limits_{(x,y)\, \to \,(3, - 1)} \left( {{x^2} + {y^2} - 4x + 2y} \right) = - \,4$ To prove that $$\mathop {\lim }\limits_{(x,y)\, \to \,(3, - 1)} \left( {{x^2} + {y^2} - 4x + 2y} \right) = - \,4$$ I followed the following process:
Because the hypothesis and the definitio... | From your last step, you want that: $\delta^2 + 2\delta < \epsilon$, but this means $(\delta + 1)^2 < 1 + \epsilon$, and taking square root $\delta + 1 < \sqrt{1 + \epsilon}$, so $\delta < -1 + \sqrt{1 + \epsilon}$. You can take $\delta = \dfrac{-1 + \sqrt{1 + \epsilon}}{2}$, and it does the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing coefficient of $x^n$ Find the coefficient of $x^n$ in the expansion of $$\left(1 + \frac{x}{1!} + \frac{x^2}{2!}+\cdots +\frac{x^n}{n!} \right)^2$$
How do you even start this problem? Do you use multinomial theorem or binomial theorem?
Could anyone please help? I found this in a textbook of mine. What I feel ... | To do this, you need to look at every combination of terms that gives $x^n = x^j \cdot x^{n-j}$. We can do this using the multinomial theorem, or just multiply it out.
\begin{align*}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)^2 &= \left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\left(1+\frac{x}{1!} + \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)$ How to solve this limit
$$
\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)
$$
without using L'Hospital's rule?
| By the formulas
$$
\cos 2x=\cos x ^2-\sin x ^2=1-2\sin x ^2, \sin 2x=2 \sin x \cos x,
$$
we have
$$
1-\cos x=2\sin^2 {\frac{x}{2}}, \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}.
$$
Then,
$$\begin{align}
\lim_{n\rightarrow \infty} (\frac{1}{\sin x}-\frac{1}{\tan x})
&=\lim_{n\rightarrow \infty} (\frac{1}{\sin x}-\frac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/764365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
} |
Bernoulli numbers identity with binomial coefficient The generating function for the Bernoulli numbers $B_k$ is given by $f(z) = \frac{z}{e^z -1}= \sum_{k=0}^{\infty} \frac{B_k}{k!} z^k$. Applying the identity
$$1 = \frac{e^z -1}{z} \cdot \sum_{k=0}^{\infty} \frac{B_k}{k!} z^k$$ yields $\sum_{n=0}^{k} \binom{k+1}{n} B_... | If we don't have to use the equation
$$\frac{e^z -1}{z} \cdot \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}= 1- \frac{e^z -1}{z}(1- \frac{z}{2}),$$
then we can simply use the already proven
$$\sum_{n=0}^k \binom{k+1}{n}B_n = 0$$
for $k > 0$ and insert $k = 2m$ to obtain
$$\sum_{n=0}^m \binom{2m+1}{2n}B_{2n} = - \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/764535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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