Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cd... | If $f(m)=1^2-3^2+5^2-7^2+\dots+(2m-3)^2-(2m-1)^2$
$m=2\implies f(2)=1^2-3^2=-8=-2.2^2$
$m=4\implies f(4)=1^2-3^2+5^2-7^2=1-9+25-49=-8-24=-2\cdot4^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$ let $g(x)\in \Bbb R$ and for any $x\in \Bbb R$ such that
$$g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0, g(0)=0$$
find $g(x)$
my idea
let $x\longrightarrow x+1$, then we have
$$g^2(x+1)-g(x+2)-\dfrac{(x+1)^2+2(x+1)-6}{4}=0$$
$$g^2(x+2)-g(x+3)-\dfrac{(x+2)^2+2(... | HINT:
Let $g(x)=\sum_{n\ge i\ge 0} a_ix^i$
So, $\displaystyle \frac{x^2+2x-6}4= g^2(x)-g(x+1)=(\sum_{n\ge i\ge 0} a_ix^i)^2-\sum_{n\ge i\ge 0} a_i(x+1)^i=a^2_nx^{2n}+\cdots$
So comparing the coefficients of the highest power of $x$, $n=1$ and $a_1=\frac12$
$\implies g(x)=\frac x2+a_0$
As $g(0)=0$ and $g(0)=a_0\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Question involving approximation, taylor series and proving Question:
Consider the approximation $$\ln(2)\approx 2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}} \right )$$
Prove that the error in this approximation is less than $$\frac{1}{7\times 2^{2} \times 3^{5}}$$
Attempt:
It looks like the exp... | We can say that $2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right)<2 \left (\frac{1}{7\times 3^{7}}+\frac{1}{7\times 3^{9}}+\frac{1}{7\times 3^{11}}\cdots\right)= 2 \left(\frac{1}{7\times 3^7}\right)
\div \left(1-\frac{1}{3^2}\right)=\frac{1}{7\times 2^2\times3^5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true? Can someone explain?
| Let's start with the power series definitions:
$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots, $$
$$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots. $$
If you expand the product $$\sin x \cos x= \left( x-\frac{x^3}{3!}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 8,
"answer_id": 3
} |
Calculate the integer part I have to calculate the integer part of this:
$$[(\sqrt{2}+\sqrt{5})^2]$$
I tried to write it like this:
$$[2+5+2\sqrt{10}]=[7+2\sqrt{10}]=7+[2\sqrt{10}]$$
Any ideas?
| A direct way of solving this problem is to realize that
$$ 0 < \sqrt{ 5} - \sqrt{2} < 1 $$
This can be seen by squaring both sides, and using that $ 6 < 2 \sqrt{10} < 7 $ which again is true by squaring.
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = 5 + 2\sqrt{10} + 2 + 5 - 2 \sqrt{10} + 2 = 14 $, hence th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Using partial fractions to find an antiderivative of $(x^2+2x)/(x+1)^2$
Evaluate $$
\int\frac{x^2+2x}{(x+1)^2}dx
$$
My solution
Let $u =x+1$,
$
du=dx
$.
Then
$
du(x^2+2x)=(x^2+2x)dx
$ and
$
x=u-1
$.
We get
$$
\int\frac{(u-1)^2+2(u-1)}{u^2}du
=
\int\frac{u^2-2u+1+2u-2}{u^2}du
=
\int\frac{u^2-1}{u^2}du
$$
which sim... | The question has been well-answered, this is just a comment. Note that $x^2+2x=(x+1)^2-1$.
So we are integrating $\dfrac{(x+1)^2-1}{(x+1)^2}$, that is, $1-\dfrac{1}{(x+1)^2}$. Easier!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/460865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Simple question about antiderivative? So this has been confusing me a lot. Let $f(x)=x^2$ and let $F(x)=\displaystyle \int_{1}^{x} f(x) \, \mathrm{d}x$. Then $F(1)=0$, obviously, but the antiderivative of $f$ (which is the same as $\displaystyle \int_{1}^{x} f(x) \, \mathrm{d}x$) is $F(x)=x^3/3$, so $F(1)=1/3$, not $0$... | It does not matter what C is equal to.
The antiderivative of f is $F(x) = \int _0 ^x f(t) dt = \int _0 ^x t^2 dt = (\frac{t^3}{3} + C) |_0^x = (\frac {x^3}{3}+ C) - (\frac {0^3}{3} +C) = \frac{x^3}{3}$
$\therefore F(1) = \int _0 ^1 f(t) dt = 1/3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Given three integers in $\{0,\ldots,100\}$ which sum up to $100$. What is the probabilty that two of them are the same? We pick $3$ numbers (one by one) from set $\{0,1,...,100\}$.
What is probabilty that two numbers are the same if sum of those $3$ numbers is $100$?
My solution:
Which two are the same we can pick in $... | First, we count the triples $(a,b,c)$ such that $a + b + c = 100$.
For the choice of $a,b$ we need $a + b \leq 100$. Then $c = 100 - a - b$ is uniquely determined.
For $a + b \leq 100$, the number of possibilities is
$$101 + 100 + 99 + \ldots + 2 + 1 = \frac{102\cdot 101}{2} = 5151.$$
Now we count the triples with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {... | At $x=0$, both numerator and denominator are $0$, so one way to do this is to apply L'Hospital's rule :
$$
\lim_{x\to 0} \frac{\sqrt{x+49}-7}{3-\sqrt{x+9}} = \lim_{x\to 0} \frac{1/2}{-1/2}\frac{(x+49)^{-1/2}}{(x+9)^{-1/2}}
$$
Now just plug in $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Easy trigonometry question How is
$$\frac{2 \sin x}{(1+ \cos x)^2}= \tan\left(\frac{x}{2}\right)\sec^2\left(\frac{x}{2}\right)\;?$$
It should be easy.
But somehow I don't get it.
Can you help me with this?
| $$\frac{2 \sin x}{(1+ \cos x)^2}=\frac{4 \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\left[1+ \cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)\right]^2}=\frac{4 \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\left[2 \cos^2\left(\frac{x}{2}\right)\right]^2}=\tan\left(\frac{x}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{... | Hint: Switch the functions you originally set equal to $u$ and $dv$.
General Strategy: Assume you're trying to compute $\int p(x) f(x) dx$ where $p(x) = a_0 + a_1 x + \ldots + a_nx^n$ is a polynomial and $f(x)$ is a function that you know how to integrate $n$ times. (For example, in your problem $p(x) = x^2$, $f(x) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finesse vs. brute force in solving quadratic equations In Higher Algebra by Hall and Knight, the following "artifice" for solving a certain type of equations is given:
Solve: $\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x-11} = 9$
They make use of the fact that $(3x^2-4x+34) - (3x^2-4x-11) = 45$, and utilizing the formula $a^2 ... | Assuming the real, actual question has a plus sign between the square roots and $\,-11\,$ in the second square root, and doing your substitution:
$$\sqrt{3x^2-4x+34}+\sqrt{3x^2-4x-11}=9\;,\;\;\color{red}{t:=3x^2-4x}\implies$$
$$\sqrt{t+34}+\sqrt{t-11}=9\implies 2t+23+2\sqrt{(t+34)(t-11)}=81\implies$$
$$29-t=\sqrt{(t+34... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopi... | EDIT: The original proof contained an error. It is (hopefully) fixed now.
Consider the function $f(x) = 10x^3 - 9x^5$. Then our goal is to show that $f(a) + f(b) + f(c) \leq 9/4$.
We will need the following two claims
Claim1: $f(x) + f(1-x) \leq 9/4$ for all $x \in [0,1]$.
Proof: A straightforward calculation says tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$4^\text{th}$ power of a $2\times 2$ matrix $$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.
I trie... | Here is a hint:
We have
$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$
Use matrix multiplication to compute (I'll do it for you) ...
$B=A^2 = \left(\begin{array}{cc}\cos^2 x -\sin^2 x& -2\sin x \cos x \\ 2\sin x \cos x & \cos^2 x -\sin^2 x\end{array}\right)$
Now simplify using the tr... | {
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"url": "https://math.stackexchange.com/questions/470541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and ... | I would do it this way:
We have $x^n=y^n$ and $n=2k$ so we have $(y^2)^k=(x^2)^k$, taking $k$-th root of the both sides we obtain $y^2=x^2$ (beacuse $k$-th root function is an injection) so it follows $(x-y)(x+y)=0$ from which it follows that either $x=y$ or $x=-y$, short enough?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
What is the remainder? What is the remainder when $7^{2002}+3^{2002}+2002$ is divided by $29$ ?
I tried out the problem using congruent modulo but cannot help my cause.
| These kind of problems can always be solved in a similar matter. The goal is to find powers that are congruent $1$ or $-1$ modulo $29$. So you start with $7$ and calculate it's powers modulo $29$:
$$\begin{align*}
7^1&\equiv 7\\
7^2&\equiv 7\cdot 7\equiv 49 \equiv 20\\
7^3&\equiv 20\cdot 7\equiv 140\equiv 24\\
7^3&\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determining the general form of $10^x \bmod 210$ While solving a problem I came across solving $10^x\bmod 210$ for various values of $x$. It seems that the values repeat after an interval of 6 for $x\geq4$. Can any one explain how can solve this equation for some value of $x$ and show that it repeats at an interval of ... | Using Carmitcheal function, $\lambda(21)=$lcm$(\lambda(3),\lambda(7))=$lcm$(2,6)=6$
and $(10,21)=1,10^6\equiv1\pmod{21}$
$10^0\equiv1\pmod{210}$
As $a\equiv b\pmod m\implies a\cdot k\equiv b\cdot k\pmod{m\cdot k} $
$10^0\equiv1\pmod{21}\implies 10\equiv10\pmod{210}$
$10^1\equiv10\pmod{21}\implies 10^2\equiv100\pmod{21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the domain of $\sqrt{x^2-9}$ $$\sqrt{x^2-9}$$
I know that the domain of square root is greater than or equal to zero. I solve for when $x^2-9<0$ and get $x^2<9$. Now I get $x<-3$ and $x<3$. I know that the domain is $(-\infty,-3] \cup [3,\infty)$ The problem is I do not understand how the $x<3$ gets flipped to $x>... | You need to have $x^2-9\ge 0$, not $x^2-9<0$. This is $(x-3)(x+3)\ge 0$, which is the case when
*
*one of $x-3$ and $x+3$ is $0$,
*both $x-3$ and $x+3$ are positive, or
*both $x-3$ and $x+3$ are negative.
In all other cases $(x-3)(x+3)$ is negative, which is precisely what you don’t want.
*
*The first of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the give... | $${ 1 }^{ 2 }-{ 2 }^{ 2 }=-\left( 1+2 \right) \\ { 3 }^{ 2 }-{ 4 }^{ 2 }=-\left( 3+4 \right) \\ { 5 }^{ 2 }-{ 6 }^{ 2 }=-\left( 5+6 \right) \\ \vdots \\ { \left( 2n-1 \right) }^{ 2 }-{ \left( 2n \right) }^{ 2 }=-\left( 2n-1+2n \right) $$
Sum up, you get the answer that O.L. has written.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you evaluate $\int \frac{y^2}{y^2+d^2}dy$? $\color{green}{question}$:
How do you evaluate this integral?
$$\int \frac{y^2}{y^2+d^2}dy=y-d\,{\tan}^{-1}\left ( \frac{y}{d} \right )+\mathrm{constant}$$
$\color{green}{I~know}$ I should use the change of variables, But I do not know how to do.
Thank you for any hin... | Hint
*
*$$\frac{y^2}{y^2+d^2}=\frac{y^2+d^2-d^2}{y^2+d^2}=1-\frac{d^2}{y^2+d^2}$$
*$$\frac{d^2}{y^2+d^2}=\frac{1}{(y/d)^2+1}=\frac{1}{t^2+1}$$
*$$\int\frac{dt}{t^2+1}=\arctan t+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/475210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\ta... | Here's another solution for the identity of Antonio Cagnoli :
We want to show that :
$\tan A + \tan B + \tan C = \tan A\times \tan B \times \tan C\ $ with $A+B+C=180^\circ=\pi$.
By definition we have : $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ so here, we want to prove that :
$\frac{\sin A}{\cos A}+\frac{\sin B}{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 8
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Finding the minimum value of a rational function. Prove that if $x$ is real and $a>c$ & $b>c$ the minimum value of $$\frac{(a+x)(b+x)}{(c+x)} ;Given\space( x>-c)$$ is $$({\sqrt{a-c}+\sqrt{b-c \space}})^2$$
I tried using minima condition but the expression was too complicated to arrange and solve .How can this be solved... | Expand and simplify (or rather, do partial fraction decomposition), we get that
$$ \frac{ (a+x)(b+x)}{c+x} = (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c $$
Apply AM-GM to the first two terms to conclude that
$$ (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c \geq 2 \sqrt{ (a-c)(b-c) } + a+b-2c = \left( \sqrt{a-c} + \sqrt{b-c} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$
reduces to $2p^2+p-10=0$.
I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and findin... | $$
\begin{align}
2x^4+x^3-6x^2+x+2 & = x^2\Big(2x^2 + x - 6 + \frac1x + \frac{2}{x^2}\Big) \\[12pt]
& = x^2 \Big(2\left(x+\frac1x\right)^2 + \left(x+\frac1x\right) - 10 \Big) \\[12pt]
& = x^2(2p^2 +p-10).
\end{align}
$$
This equals $0$ only if $x=0$ or the second factor equals $0$. But clearly $x=0$ is not one of the ... | {
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"url": "https://math.stackexchange.com/questions/480102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$ I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.
Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\fra... | Hint: Using the Binomial Theorem, prove by induction that
$$
(1+x)^{-1/2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}(-x)^k
$$
Truncate as needed. Note that $n!!$ is the Double Factorial:
$$
\begin{array}{rll}
(2k-1)!!&=1\cdot3\cdot5\cdots(2k-1)&=\frac{(2k)!}{2^kk!}\\[6pt]
(2k)!!&=2\cdot4\cdot6\cdot8\cdots(2k)&={2^kk!}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
| The values of $2(\frac 12 +\frac 12 \sqrt{-7})^{n-2}$ give $x+y\sqrt{-7}$, which are a solution to the equation $x^2+7y^2 = 2^n$, where $x$ and $y$ are always odd.
These solutions are thence unique, because all representations of $x^2+7y^2$ reduce to a complex number $x+y\sqrt{-7}$, and that $\frac 12 \pm \frac 12 \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Showing that $\int_{a}^{b} \frac{(x-x_1)(x-x_2)}{(x_0 - x_1)(x_0-x_2)} dx = \int_{-1}^{1} \frac{t(t-1)}{2} \frac{b-a}{2} dt$ I'm a second year math student, and I have some trouble understanding this equality in my book. I would appreciate a little help :)
Let $x_0=a,x_1=(a+b)/2,x_2=b$. \begin{align*} \int_{a}^{b}
\f... | The substitution $t = \frac{x-x_1}{x_1-x_0}$ converts the left integral into the right:
We have $\frac{dx}{dt} = x_1 - x_0 = \frac{b-a}{2}$, and $\frac{x - x_2}{x_2 - x_0} = \frac{(x-x_1) - (x_2-x_1)}{2(x_1-x_0)} = \frac{1}{2}\left(t - \frac{x_2-x_1}{x_1-x_0}\right) = \frac{t-1}{2}$ since $x_2 - x_1 = x_1 - x_0$. Furth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Factoring an element in $\mathbb{Z}[\sqrt6]$ I want to know if $1+\sqrt6 $ factors in $\mathbb{Z}[\sqrt6]$.
$1+\sqrt6=(a+b\sqrt6)(c+d\sqrt6)$ implies $a^2-6b^2\not=0,c=\frac{a-6b}{a^2-6b^2},a\not=0, d=\frac{a-b}{a^2-6b^2}$
How do I determine if there are integer solutions?
| Use the multiplicativity of the norm : if you define
$N(a+b\sqrt{6})=a^2-6b^2$, then $N(w_1w_2)=N(w_1)N(w_2)$ for any $w_1,w_2$.
So if $1+\sqrt{6}=(a+b\sqrt{6})(c+d\sqrt{6})$, then
$-5=(a^2-6b^2)(c^2-6d^2)$, so that one of $a^2-6b^2$ or $c^2-6d^2$ is $\pm 1$.
If, say, $a^2-6b^2= \pm 1$, then $a+b\sqrt{6}$ is a unit in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I find integer values for which a given expression gives a perfect square? Find the integer values for which $x^2+19x+92$ is a perfect square.
Also, How to proceed if you have to find values ( not necessarily integer)?
| Complete the square:
$$\begin{align}
x^2 + 19x + 92 &= m^2\\
\iff (2x)^2 + 2\cdot 19\cdot (2x) + 368 &= (2m)^2\\
\iff (2x+19)^2 + 7 &= (2m)^2\\
\iff (2m)^2 - (2x+19)^2 &= 7.
\end{align}$$
Since $7$ is prime, that means $2m = \pm 4$ and $2x+19 = \pm 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| An answer using the Stolz–Cesàro theorem: $$\lim_{n\to\infty} \frac{ \sum_{k=1}^n 1/\sqrt{k} }{\sqrt{n}} = \lim_{n\to\infty} \frac{1/\sqrt{n} }{\sqrt{n} - \sqrt{n-1}} = \lim_{n\to\infty} \frac{\sqrt{n}+\sqrt{n-1} }{\sqrt{n}}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| As $x=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$
Using de Moivre's formula for positive integer $n$
$$x^n=\left(\cos\frac{2\pi}5+i\sin\frac{2\pi}5\right)^n=\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5$$
$$\implies x^5=\cos2\pi=1\text{ and }x^{-n}=\frac1{x^n}=\frac1{\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5}=\cos\frac{2n\pi}5-i\sin \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Smallest number with specific number of divisors Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:
Sinc... | This approach is not fishy at all, but can be laborious. Note that you can't use $5+3 \lt 7+2$ to conclude that $2^5*3^3 \lt 2^7*3^2$, although it is true, you have to take the ratio and use $3 \lt 2^2$. For example, if you were looking for $36$ factors, one comparison would be $2^8*3^3$ versus $2^5*3^5$. Despite th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
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Solve a congruence linear equation. Solve the following congruence:
$19x\equiv 1\;(\text{mod}\;36)$
My work:
I found an inverse of $19$ and $36$ which is $9$.
$9\cdot 19x\equiv 9\cdot 1\;(\text{mod}\;36)$
$171x\equiv 9\;(\text{mod}\;36)$
$9x\equiv 9\;(\text{mod}\; 36)$
$x\equiv 1\;(\text{mod}\;4)$ is my final answer..... | Method $1:$
We need $x\equiv19^{-1}\pmod{36}$
Using Carmichael function $\lambda(36)=$lcm$(6,2)=6\implies a^6\equiv1\pmod{36}$ if $(a,6)=1$
$\implies$ord$_{36}(a)$ must divide $6$
$19^1\equiv19\pmod{36},19^2=361\equiv1\pmod{36}\implies 19\equiv19^{-1}\pmod{36}$
Method $2:$
Like this, expressing as Continued Fraction,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluating the area between the curves $r=2\sin\theta$ and $r=\sin\theta+\cos\theta$
So the problem asked me to find the area of the region that lies inside both of the circles
$$r=2\sin\theta, \quad r=\sin\theta +\cos\theta $$
I know that $r=2\sin\theta$ is $x^2+(y-1)^2=1,$but the second one is a little bit harder ... | Hint: Try multiplying both sides of $r=\sin\theta + \cos\theta$ by $r$, and using the identities $r^2=x^2+y^2$, $x=r\cos\theta$, and $y=r\sin\theta$, to find an equation in terms of $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2} $ by induction
Prove by induction that for all $n > 0$,
$$
\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots
+ \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}
$$
I have done the ba... | By cross-multiplying, it can easily be seen that
$$
\begin{align}
\frac{\sqrt{k+1}}{2k}
&\gt\frac1{\sqrt{k+1}+\sqrt{k\,}}\\[6pt]
&=\sqrt{k+1}-\sqrt{k\,}
\end{align}
$$
Summing gives
$$
\sum_{k=1}^n\frac{\sqrt{k+1}}{2k}
\gt\sqrt{n+1}-1
$$
This is not the inequality given, but for $n\ge2$, $\sqrt{n+1}-1\gt\frac{\sqrt{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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How do I determine whether $18 \notin A$ with these premises? With $A \subseteq \mathbb{N}$, given that
A1. $(x \in A \land y \in A) \implies x^2 + y ^2 \in A$
A2. $1 \in A$
A3. $3 \notin A$
Determine whether:
$18 \notin A$
I've been unable to prove this. First, I tried to demonstrate $18 \in A$, but no matter what... | \begin{align}
\phantom{\left(1 \in A\quad \wedge\quad 1 \in A\right)}
&\phantom{\quad\Longrightarrow\quad
1^{2} + 1^{2} =} 1 \in A
\\
\left(1 \in A\quad \wedge\quad 1 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 1^{2} = 2 \in A
\\
\phantom{\left(1 \in A\quad \wedge\quad 1 \in A\right)}
&\phantom{\quad\Longright... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\lim_{n\to \infty} \prod_{i=0}^{n-1} $ $(2+\cos \frac{i \pi}{n})^{\frac{\pi}{n}}$=$\sqrt{3}$ by the sum form of a integration? The integral seems like
$$
\exp\left(\int_{0}^{\pi}\left[\ln(2+\cos\left(x\right)\right)]\,{\rm d}x\right)
$$
but the above will be a number bigger than $5$ by appraisement.
| Taking log,
\begin{align*}
\log \left( \prod_{i=0}^{n-1} \left( 2 + \cos \left( \frac{i \pi}{ n } \right ) \right )^{\frac \pi n}\right ) &=
\sum_{i=0}^{n-1} \frac \pi n \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right ) \\
&= \pi \cdot \frac {1}{n } \sum_{i=0}^{n-1} \log \left( 2 + \cos \left( \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integer $x$ for which $x^4+x^3+x^2+x+1$ is perfect square. Integer values of $x$ for which $\bf{x^4+x^3+x^2+x+1}$ is a Perfect Square.
$\underline{\bf{My\; Try}}$:: Let $\bf{x^4+x^3+x^2+x+1 = k^2}$, where $k\in \mathbb{Z}$
$4x^4+4x^3+4x^2+4x+1 = 4k^2 = (2k)^2$
Now How can I proceed after that
Help Required,
Thanks
| We can just change this into a quadratic equation (or cubic equation, but that becomes very highly complicated).
$$\because x^4 + x^3 + x^2 + x + 1 = k^2$$
then this means that $x^2 + x + (1 - k^2 - x^3 - x^4) = 0$. This means that:
$$x = \frac{-1 \pm \sqrt {1 - 4(1 - k^2 - x^3 - x^4)}}{2}$$
And this means that:
$2x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Elementary manipulation with elements of group Let $(G,*)$ be a group with identity $e$ , let $a,b∈G$ such that $a*b^3*a^{-1}=b^2$ and $b^{-1}*a^2*b=a^3$ , then how do we prove that $a=b=e$ ?
| Given $$ab^3a^{-1}=b^2$$ thus we get $$ab^6a^{-1}=b^4\;,\;\;ab^9a^{-1}=b^6$$
Replaceing $b^6$ in $ab^6a^{-1}=b^4$, we get $a^2b^9(a^{-1})^2=b^4$. Thus $a^2b^{18}(a^{-1})^2=b^8$. Again $ab^3a^{-1}=b^2$ gives $ab^{27}a^{-1}=b^{18}$. Replacing $b^{18}$, we get $a^3b^{27}(a^{-1})^3=b^8$. But $b^{-1}a^2b=a^3$. Thus $(b^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it... | Assuming that $4\times10^{2n}+9\times10^{2n-1}+5=99k$,
$$
4\times10^{2(n+1)}+9\times10^{2(n+1)-1}+5=4\times10^{2n+2}+9\times10^{2n+1}+5=100\times(4\times10^{2n})+100\times(9\times10^{2n-1})+5=100\times(4\times10^{2n}+9\times10^{2n-1})+5=100\times(4\times10^{2n}+9\times10^{2n-1}+5-5)+5=100\times(4\times10^{2n}+9\times10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Implicit differentiation of a lemniscate at a point: So here's the problem:
Find the slope of the tangent line of :
$2(x^2 +y^2)^2 = 25(x^2 - y^2)$ at the point (3,1)
Cool:
So here's what I did:
Simplification step:
$2(x^2 +y^2)^2 = 25(x^2 - y^2) \Rightarrow 2(x^2+y^2)^2 = 25x^2 - 25y^2$
Differentiate both sides:... | The answers above were wrong since the equation is wrong. It should be the following -
y1= 8(x^2+y^2)-50)x/-(8(x^2+y^2)+50)y
Then you get the slope as -9/13.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Pattern matching puzzles Problems: I wish to have an equation for below data:
$x = 2, y = 5$
$x = 2, y = 6 $
$x = 2, y = 7 $
$x = 2, y = 8 $
$x = 1, y = 9$
$x = 0, y = 10$
And Will you show me the steps to acquire that equation?
| First sqap the values for $x$ and $y$. Then apply the Lagrange interpolation. The first 4 points lie on the the line $y=2$ Then do the Lagrange interpolation explained here:
After the application, the first 5 points then lie on the function
$$y = 2 - \frac{(x - 5) (x - 6) (x - 7) (x - 8)}{24}$$
Do this method again and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c... | It is a bit unclear what you want to generalise. $x^3+x+c$ is a strictly increasing function of $x$ - a strictly increasing (or decreasing) polynomial of odd order with real coefficients will always have a single real root.
Or you may be saying that the polynomial can be written as $p(x)=(x-a)q(x)$ where $q(x)$ is a po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find hypotenuse given acute angle bisectors
In a right triangle $ABC$ (right-angled at $B$), $D$ and $E$ are points of $\overline{AB}$ and $\overline{BC}$ respectively such that $\overline{CD}$ and $\overline{AE}$ are the angle bisectors of the acute angles of the triangle.
Given that $AE=9$ and $CD=8\sqrt{2}$, find t... | Let $\alpha = \angle BAE$. Let $\beta = \angle BCD$. What we want is $AC = \sqrt{AB^2 + BC^2}$. In other words, we want to solve for $\alpha, \beta, AB, BC$ such that
$$AB = 9 \cos\alpha \quad\quad BC = 8\sqrt{2} \cos\beta \quad\quad \alpha + \beta = \frac{\pi}{4} \quad\quad \frac{BC}{AB} = \tan 2\alpha$$
From the last... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Logarithm problem : Prove that $log_{3^2} \frac{1}{2} > 0$ Logarithm problem :
Prove that $log_{3^2} \frac{1}{2} > 0$
My approach :
$log_{3^2} \frac{1}{2} > 0$
$\Rightarrow \frac{1}{2} log_3 \frac{1}{2} >0$
$\Rightarrow \frac{1}{2} [ log_3 1 -log_3 2] >0 $
$\Rightarrow \frac{1}{2} [ 0 - log_3 2] >0 ... | To evaluate $log_{3^2}\frac{1}{2}$ you solve $(3^2)^x=\frac{1}{2} \rightarrow xlog(3^2)=log(\frac{1}{2}) \rightarrow x =\frac{log(\frac{1}{2})}{log(3^2)}$, but $log(\frac{1}{2}) = -ln(2)$. Hence, $log_{3^2}\frac{1}{2}$ is not greater than 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof that:$ \sum\limits_{n=1}^{p} \left\lfloor \frac{n(n+1)}{p} \right\rfloor= \frac{2p^2+3p+7}{6} $ where $p$ is a prime and $p \equiv 7 \mod{8}$ Proof that:$ \sum\limits_{n=1}^{p} \left\lfloor \frac{n(n+1)}{p} \right\rfloor= \frac{2p^2+3p+7}{6} $
where $p$ is a prime number such that $p \equiv 7 \mod{8}$
I trie... | $$\sum_{i=1}^p \frac{n(n+1)}p = \frac1 p \frac {p(p+1)(p+2)}3 = \frac{p^2+3p+2}3$$, and
$$\frac{p^2+3p+2}3 - \frac{2p^2+3p+7}6 = \frac{p-1}2$$
So you are asking to prove that $$\sum_{n=1}^p \frac{n(n+1)}p - \lfloor\frac{n(n+1)}p\rfloor = \frac{p-1}2$$.
The term being summed is $\dfrac 1 p$ times the residue of $n(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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A closed form of $\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$ Is it possible to evaluate the following integral in a closed form?
$$\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx,$$
where $\phi$ is the golden ratio:
$$\phi=\frac{1+\sqrt{5}}{2}.$$
| Since $\frac1\phi=\phi-1$, we get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\int_0^\infty\frac{x^\phi\arctan(x)}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{1}\\
&=\int_0^\infty\frac{x^\phi(\frac\pi2-\arctan(x))}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/520494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 2,
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Writing u as a linear combination of the vectors in S. Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$
as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\ri... | You're heading in the right direction, you've found the augmented matrix:
$$
\left[\begin{array}{ccc|c}
1 & 4 & 5 & 2\\
2 & 2 & 4 & 10\\
2 & 1 & 1 & 1\\
\end{array}\right].
$$
The solutions $(x,y,z)$ to this system of linear equations are precisely the values for which $$x\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix}+y\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Fibonacci Sequence Exercise I need some help checking the following solution.
The Fib sequence is defined by $a_1 = 1, a_2 = 1$ and for all $n\geq 2$, $a_{n+1} = a_n + a_{n-1}$. Thus, the sequence begins:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...
Prove that for all $n\geq 1$, $a_n <\left(\frac{5}{3}\right)^n$.
So far I h... | We need Strong induction here
Let $a_n<\left(\frac53\right)^n$ for integer $1\le n\le m$
$\implies \displaystyle a_{m+1}=a_m+a_{m-1}<\left(\frac53\right)^m+\left(\frac53\right)^{m-1}=\left(\frac53\right)^{m-1}\left(\frac53+1\right)$
which will be $<\left(\frac53\right)^{m+1}$ if $\displaystyle \left(\frac53+1\right... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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finding factors for gcd To compute $gcd(25, 11)$, Euclid's algorithm would proceed as follows:
$$\underline{25} = 2 \cdot \underline{11}+3$$
$$\underline{11} = 3 \cdot \underline{3}+2$$
$$\underline{3} = 1 \cdot \underline{2}+1$$
$$\underline{2} = 2 \cdot \underline{1}+0$$
Thus $gcd(25,11)$ = $gcd(11,3)$ = $gcd(3,2)$ =... | Perhaps it would actually be easier to see what’s going on in a symbolic example. Suppose that we applied the algorithm to $m$ and $n$ and get:
$$\begin{align*}
n&=q_0m+r_0\tag{1}\\
m&=q_1r_0+r_1\tag{2}\\
r_0&=q_2r_1+r_2\tag{3}\\
r_1&=q_3r_2+0\tag{4}\\
\end{align*}$$
We know then that $\gcd(n,m)=\gcd(m,r_0)=\gcd(r_0,r_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Calculation of real values of $x$ in $\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1$
Calculate the real solutions $x\in\mathbb{R}$ to
$$
\tag1\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1
$$
My Attempt:
Let $2^x = a$ and $3^x = b$ . Then $(1)$ becomes
$$
\sqrt{a^2-a\cdot b+b^2}+\... | Hint:
$$a^2+b^2\ge 2ab$$
$$\implies 4a^2+4b^2-4ab\ge a^2+b^2+2ab$$
$$\implies \sqrt{a^2+b^2-ab}\ge \frac{a+b}{2}$$
With equality iff $a=b$
You see where this is going?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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How to solve the inequality $\frac {5x+1}{4x-1}\geq1$ Please help me solve the following inequality.
\begin{eqnarray}
\\\frac {5x+1}{4x-1}\geq1\\
\end{eqnarray}
I have tried the following method but it is wrong. Why?
\begin{eqnarray}
\\\frac {5x+1}{4x-1}&\geq&1\\
\\5x+1&\geq& 4x-1\\
\\x &\geq& -2
\end{eqnarray}
Th... | As Carlos Eugenio Thompson Pinzón has commented, we need $\displaystyle4x-1\ne0$
Method $1:$
If $\displaystyle4x-1\ne0, (4x-1)^2>0$
Multiplying either sides of $\displaystyle\frac{5x+1}{4x-1}\ge1$ by $(4x-1)^2$
we get $\displaystyle(5x+1)(4x-1)\ge(4x-1)^2$
$\displaystyle\implies (4x-1)\{5x+1-(4x-1)\}\ge0 $
$\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ in which every number is made by inserting $48$ in the middle of previous as indicated, each number is the square of an integer.
| Without words:
$$\begin{align}
\left(6\frac{10^k-1}{9}+1\right)^2 &= 36 \frac{10^{2k} - 2\cdot 10^k + 1}{81} + 12\frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k - 4 \frac{10^k-1}{9} + 12 \frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k + 8 \frac{10^k-1}{9} + 1.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:
$$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$
for some integers $m, n$
$$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 ... | As you have proved, $p^2+q^2\in\Bbb Q$ and $pq\in \Bbb Q$, so
$$p-q=\frac{a-b}{p^2+pq+q^2}\in \Bbb Q.$$
Combining this with $p+q\in \Bbb Q$, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 0
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How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.
My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't
| IN $\bigtriangleup$ABC we take AB=c;AC=b &BC=a
*
*NOW $a^2$ = $6^2$+$7^2$ -2.6.7.cos$\angle$BPC
*NOW cos$\theta$ $\geq$ -1$\longrightarrow$ -cos$\theta$$\leq$1. from this we get $a^2$$\leq$13
*$c^2$= $10^2$+$6^2$ -2.10.6.cos$\angle$APB.similarly from this we get [applying the inequality] c$\leq$16
*now the area... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using pen and paper How can you calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using nothing more than a pen and pencil? Simply typing this in any symbolic calculator will give us $1/81$. I could also possibly find this formula if I was actually looking at given numbers ... | We know that
$$
\begin{align}
A&=\hphantom{1+}\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots\\
10A&=1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots\\[4pt]
9A&=1
\end{align}
$$
So $\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots=A=\frac19$.
Likewise,
$$
\begin{align}
\color{#C00000}{\frac1{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Finding minimum $\sqrt{x^2-3x+3}+\sqrt{x^2-3x+2}$ I would appreciate if somebody could help me with the following problem
Q. Finding minimum $f(x)$
$$f(x)=\sqrt{x^2-3x+3}+\sqrt{x^2-3x+2}$$
| Hint: Since
$$f(x)=\sqrt{x^2-3x+3}+\sqrt{x^2-3x+2} = \sqrt{\left(x-\dfrac32\right)^2 + \dfrac34} + \sqrt{\left(x-\dfrac32\right)^2 - \dfrac14},$$
we have $\left(x-\dfrac32\right)^2 + \dfrac34 \ge \dfrac14+\dfrac34 = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?
| Suppose that $4k+1 = a^2$ and $9k+1 = b^2$, where $a, b$ are positive integers.
Then, we get that $ 9a^2 - 4b^2 = 5 $.
Since we have $(3a-2b)(3a+2b) = 5$, and both terms are integers,$3a+2b > 0$, $3a+2b>3a-2b$, so we must have $3a+2b = 5, 3a-2b = 1$. This gives us $a=1, b= 1$ and hence $k=0$ is the only solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How find series $\sum_{n=1}^{\infty}\frac{(-1)^{[\sqrt[m]{n}]}}{n^a}$?
let $m$ is give a positive integers,
Determine for which values of $a$,the series $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}$$ converges
where $[x]$ is the largest integer not greater than $x$.
and I have see this not hard problem... | Let's start with the observation that if $a > 1$, then the sum converges absolutely, and if $a \leqslant 0$, the terms of the sum don't converge to $0$, hence the sum cannot converge. So in the following, we always assume $0 < a$.
The case $m = 1$ is immediate by Leibniz' criterion, $\sum\limits_{n=1}^\infty \dfrac{(-1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{2}^{\infty}\frac{dx}{x^2-1}$ converge or diverge How can I find whether the integral converge or diverge.
I did the following.
$\int_{2}^{\infty}\frac{dx}{x^2-1}$
I did the following
$\frac{a}{x-1}+\frac{b}{x+1}$
$A(x+1)+B(x-1)=1$
$t\rightarrow\infty$
$\int_{2}^{t}\frac{1/2}{x-1}-\frac{1/2}{x+1}$
$\frac{1}{2}[\l... | One can prove convergence, without evaluating, by noting that for $x\ge 2$, we have $x^2-1\gt x^2-\frac{x^2}{2}=\frac{x^2}{2}$. Thus
$$0\lt \frac{1}{x^2-1}\lt \frac{2}{x^2}.$$
It is a standard fact that $\int_2^\infty \frac{1}{x^2}\,dx$ converges, so our integral does.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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What are the generators of $\mathbb{Z}_9^*$? I am to understand that $\mathbb{Z}_9^*$ is cyclic because $9=3^2$, where $3^2$ is of the form $p^{\alpha}$, with $p$ an odd prime... but I can't find any generators for the set...
$|\mathbb{Z}_9^*|=6$, and $\mathbb{Z}_9^*=\{\overline{1},\overline{2},\overline{4},\overline{5... | When you are looking for generators, a trick to reduce the work is to note that, once you know that $a$ is not a generator, then neither is any power $a^i$. So in your case you'd start by testing $2$ rather than $2^2=4$ or $2^3=8$.
A good way to check $a$ is a generator module ${\mathbb Z}_{p^k}^\times$ is to check whe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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$x^2+xy+y^2$ and $x^2-xy+y^2$ are not both perfect squares
Prove that $x^2+xy+y^2$ and $x^2-xy+y^2$ cannot be both perfect squares.
Surely $x$ and $y$ are natural numbers. If $x^2+xy+y^2 =a^2$ and $x^2-xy+y^2=b^2$ simultaneously then we have to show that there are no such integers $a$ and $b$.
I have tried that:
Supp... | The question amounts to showing that the elliptic curve
$Y^2 = X^2+X+1$, $Z^2 = X^2-X+1$
has no rational points other than those with $X=0$ and $X=\infty$.
That can be done by elementary means but requires a Fermat-style descent.
According to Dickson's History of the Theory of Numbers
(Volume II, Chapter XVI, bottom of... | {
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How can I determine this function? If i know that$$ f(3)-f(-1/2) = 7$$ and $$ f(2)-f(1/2) = 3 $$ and $$f(3/2)-f(-2) =7$$ How can I determine the function?
| $$
f(x) = \left\{
\begin{array}{ll}
110004 & \quad x = -2 \\
68 & \quad x = -\frac{1}{2} \\
4 & \quad x = \frac{1}{2} \\
110011 & \quad x = \frac{3}{2} \\
7 & \quad x = 2 \\
75 & \quad x = 3 \\
0 & \quad x\neq-2,-\frac{1}{2},\frac{1}{2},\frac{3}{2},2,3
\end{array}
\right.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Solving trigonometry equation Please help me understand how to solve this for $0\leq x\leq360 $
I seem to have a problem with equations with powers.
$$3\sin^2 x-3\cos^2x+\cos x-1=0 $$
thinking that I would start by simplifying:
$$3 (\sin^2 x- \cos^2x)+\cos x - 1=0 $$
How I wish the equation in the bracket was in fo... | $3\sin^2{x} - 3\cos^2{x} + \cos{x} - 1 = 0$
$6\cos^2{x} - \cos{x} - 2 = 0$
$\cos{x} = -\frac{1}{2}\:\:\:\cos{x} = \frac{2}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $... | $2ab = (a+b)^2 - (a^2+b^2)$, hence we need to find the minimum value of $a + b + \frac{2}{(a+b)^2 - (a^2+b^2)}$ $ = a + b + \frac{2}{(a+b)^2 - 1}$.
Let us find the maximum value (k) of $a+b$ given $a^2+b^2=1$. Thinking graphically, when $a+b = k$ is tangent to the circle, its gradient is $-1$ and hence the gradient of ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
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Finding the limit of $(\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$ Any Ideas/Hints?
$$\lim\limits_{n \to \infty } (\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$$
| $$\lim_{n\to\infty}{\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}=\lim_{n\to\infty}\left({\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}\right)\cdot 1$$
$$=\lim_{n\to\infty}\left({\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}\right)\cdot \frac{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all positive integers m, n, p such that $(m+n)(mn+1)=2^p$ Find all positive integers m, n, p such that
$$(m+n)(mn+1)=2^p$$
Please give me some hints
Thanks
| We have $m+n=2^a$ and $mn+1=2^b$ with $a+b=p$. First note that both $m$ and $n$ need to be odd.
First case: Suppose $m=1$ or $n=1$, then the other is equal to $2^{p/2}-1$ and this is a solution for even $p$.
Now suppose $m>1$ so $b>a>1$. Adding the two equations above we get
$(m+1)(n+1)=mn+1+m+n=2^a(2^{b-a}+1)$. Let $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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binomial theorem with conditional probability There are four men in a room, 1 pair of brothers, and 2 unrelated men. The
probability that any man has blood-group X is
1
4
. The probability that if one
brother has blood-group X, the other brother also has X is
3
4
, otherwise the
blood-groups are independent. Find the p... | $$\frac{3}{4}\cdot\left({3 \choose 2}\cdot\left(\frac{1}{4}\right)^2\cdot\frac{3}{4}\right)+\frac{1}{4}\cdot\left(\frac{3}{4}\cdot \left({2\choose 1}\frac{1}{4}\cdot\frac{3}{4}\right)+\frac{1}{4}\cdot\left(\frac{1}{4}\right)^2\right)=\frac{23}{128}$$
Chance of 1$^{\mathrm{st}}$ brother having X and not having X on the ... | {
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"timestamp": "2023-03-29T00:00:00",
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How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 - 3x + ... | Note that the expression can be written as $f(x)+g(y)$, where $f(x) = x^2-3x+2 $ and $g(y) = 2y(y+2)$.
The two parts are independent of each other, so we can minimize them separately.
Setting $f'(x) = 0$ gives $2x-3 = 0$, so we see that $x = \frac{3}{2}$ is the minimizer. Noting that $(x-\frac{3}{2})^2 = x^2-3x+\frac{9... | {
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"timestamp": "2023-03-29T00:00:00",
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If : $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1\dots$ Problem :
If $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1$
Then find the value of $\sin^2\alpha + \sin^2\... | HINT:
If $\sin^2\alpha=x$ etc,
$\displaystyle\tan^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}=\frac x{1-x}$
Just simplify to find $x+y+z=1$ assuming $(1-x)(1-y)(1-z)\ne0$ i.e, $\tan\alpha$ etc. are finite
| {
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How to solve the equation $x^2 + 4 |x| - 4 = 0$ How do I get the value of x from the equation that is provided
| Putting $x=a+ib$ where $a,b$ are real
we get $\sqrt{a^2+b^2}=4-(a+ib)^2=4+b^2-a^2-2abi$
Equating the imaginary parts $ b=0$
Now for real $x,$
$$|x|= \begin{cases} +x &\mbox{if } x\ge0 \\
-x & \mbox{if } x<0 \end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $x_1=1, x_{n+1}=x_n+\frac1{x_n^2}$
Given that $x_1 = 1$ and $x_{n+1}=x_{n}+ 1/x_{n}^{2}$. Find the limit of the sequence.
Let $ c $ be the limit of the sequence, then $ c=c+\frac{1}{c^2} $, that means $ \frac{1}{c^2}=0 $. That can't be like that.What is wrong???
| Notice $$x_{n+1}^3 = x_n^3 (1 + \frac{1}{x_n^3})^3 = x_n^3 + 3 + \frac{3}{x_n^3} + \frac{1}{x_n^6}$$
We have $x_{n+1}^3 - x_n^3 > 3$ for all $n$ and hence we can bound $x_n$ from below
$$x_n^3 \ge x_1^3 + 3(n-1) = 3n-2 \quad\implies\quad x_n \ge \sqrt[3]{3n-2}$$
As a result, the sequence $x_n$ diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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verify this limit of radical function $$\lim_{x \to -\infty} \frac{\sqrt[3]{x}- \sqrt[5]{x}}{4\sqrt[3]{x}+\sqrt[5]{x}} $$
$$= \lim_{x \to -\infty} \frac{x^{1/3}-x^{1/5}}{4x^{1/3}+x^{1/5}} $$
divide numerator and denominator by highest power of $x$ in the denominator, $x^{1/3},$ to get:
$$= \lim_{x \to -\infty} \frac... | Your work is correct
$$\lim_{x\to-\infty}\frac{\sqrt[3]{x}-\sqrt[5]{x}}{4\sqrt[3]{x}+\sqrt[5]{x}}=$$
$$\lim_{x\to-\infty}\frac{\sqrt[15]{x^5}-\sqrt[15]{x^3}}{4\sqrt[15]{x^5}+\sqrt[15]{x^3}}=$$
$$\lim_{x\to-\infty}\frac{\frac{\sqrt[15]{x^5}-\sqrt[15]{x^3}}{\sqrt[15]{x^5}}}{4\frac{\sqrt[15]{x^5}+\sqrt[15]{x^3}}{\sqrt[15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ let $a,b,c>0$ ,and such $a+b+c=3$.prove that
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$
if this problem find this minimum,then
$$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$
where $a=f'(1),a+b=f(1)$
since $$f'(x)=-\dfrac{2x}{(1+x^2)^... | By symmetry, it just suffices to look at the case $0 \le a \le b \le c$ where $a + b + c = 3$.
Let $f(x) = \frac{1}{1+x^2}$ and $F(a,b,c) = f(a)+f(b)+f(c)$. Notice
$$\frac{d^2}{dx^2} f(x) = \frac{2(3x^2-1)}{(1+x^2)^3} \longrightarrow
\begin{cases} > 0,&\text{ for } x > \frac{1}{\sqrt{3}}\\< 0,&\text{ for } x <\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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problem with comparing inequality Hello I just wondering if it's possible to prove this inequality: there are positive, various $a,b,c$ and
$ \frac{3a-b}{3} \ge x \ge \frac{3(a^2-b^2)}{3a+b}$
$ \frac{3b-c}{3} \ge y \ge \frac{3(b^2-c^2)}{3b+c}$
$ \frac{3c-a}{3} \ge z \ge \frac{3(c^2-a^2)}{3c+a}$
I want to prove that $x... | Hint:
To prove that $x+y\ge z$ we need to prove that minimum value of $x$ plus minimum value of $y$ is $\ge$ maximum value of $z$.
So you have:
$x=\frac{3(a^2-b^2)}{3a+b}; y=\frac{3(b^2-c^2)}{3b+c}; z=\frac{3c-a}{3}$
Just prove that $\frac{3(a^2-b^2)}{3a+b}+\frac{3(b^2-c^2)}{3b+c}\ge\frac{3c-a}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/566925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation formula for $x^2+x$ Since I learned easier ways of calculating summations I've been curious as to how I could find formulas for as many equations as possible. I came across the equation $x^2+x$, I've spent quite some time on this problem and could not find a solution. If someone has maybe already done this or... | Note that
$$k^2+k=\frac{1}{3}\left((k+1)^3-k^3-1\right).$$
Thus our sum $\sum_{k=1}^n (k^2+k)$ is equal to
$$\frac{1}{3}\left((2^3-1^3-1)+(3^3-2^3-1)+(4^3-3^3-1)+(5^3-4^3-1)+\cdots +((n+1)^3-n^3-1)\right).$$
Observe the nice almost total cancellations. We end up with
$$\frac{1}{3}\left((n+1)^3-1^3-n\right).$$
Remarks:... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $ S_{20} $ if $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n Define $ a_{n} $ is a sequence and all terms of $ a_{n} $ are positive. $ S_{n} $ is the summation of the first n terms. If $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n, then find $ S_{20} $
| $S_n=a_1+a_2+\cdots+a_n=S_{n-1}+a_n=S_{n-1}+2(\sqrt{2S_n}-1)\Rightarrow$ $S_{19}+2(\sqrt{2S_{20}}-1)=S_{20}$. Also we have $S_{20}-S_{19}=a_{20}$. Adding these two we get
$2\sqrt{2S_{20}}=a_{20}\Rightarrow S_{20}=\frac{a_{20}^2}{8}$.
| {
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Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.
My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases... | We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation.
Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,... | Actually you have shown that there is a number $b$ that if you divide it by $15$ you will get $16^n-1$. But this does not mean that $16^n-1$ is divisible by $15$. For example $23=\frac{345}{15}$ but it does not mean that $23$ is divisible by $15$ . To show that $a$ is divisible by $b$ you have to show that $a=kb$ ($k$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$ How can I evaluate the following integral?
$$
\int_0^{\pi/2}
\log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\,
\frac{1}{\cos\left(x\right)}\,{\rm d}x\,,
\qquad\left\vert\,a\,\right... | $$\begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1+ b\cos x}\right)\frac{1}{\cos x}dx &= \int_{0}^{\pi/2} \int_{b}^{a} \frac{1}{1+t \cos x} \ dt \ dx \\ &= \int_{b}^{a} \int_{0}^{\pi/2}\frac{1}{1+t \cos x} \ dx \ dt \end{align}$$
Let $ \displaystyle u = \tan \frac{x}{2}$.
$$\begin{align} &= \int_{b}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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show continuity of $\frac{xy}{\sqrt{x^2+y^2}}$ How to show this function's continuity?
My book's 'Hint' says $|xy| \leq \frac12(x^2+y^2)$ can be used.
$ f(x,y) = \left\{
\begin{array}{l l}
\frac{xy}{\sqrt{x^2+y^2}} & \quad , \quad(x,y)\neq(0,0)\\
0 & \quad , \quad(x,y)=(0,0)
\end{array} \right.$
| Here is another way to prove the continuity of $f(x,y)$ at $(0,0)$.
$$\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right|={{|x||y|}\over{\sqrt{x^2+y^2}}} <{{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\over{\sqrt{x^2+y^2}}}=\sqrt{x^2+y^2} < \varepsilon \\\,\,(\text{where}\,\,\, \varepsilon \,\text{is a preassigned positive number})$$ if $x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/572098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \... | HINT
Let $$u = \tan(\frac{x}{2})$$
Using the trig. identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$ (1)
$$u = \tan(\frac{x}{2}) = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$$
Squaring both sides $$u^2 \cdot \cos^2(\frac{x}{2}) = \sin^2(\frac{x}{2})$$
Using the trig. identity $\sin^2(x) + cos^2(x) = 1$ (2) This becomes
$$u^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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why $x^3=11^3$ mod $5083$ has only one solution? Why $x^3=11^3$ mod $5083$ has only one solution? (The only answer is $x=11$)
I know it has at most 3 roots but how to find that there isn't another answer?
| As $5083=13\cdot 17\cdot 23$
We have $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{13}\ \ \ \ (1)$
As all primes have primitive roots, taking discrete logarithm wrt some primitive root $g$ of $13,$(say $2$)
$3$ind$\displaystyle_g \left(\frac x{11}\right)\equiv0\pmod {12}\ \ \ \ (2)$
Using Linear Congruence Theo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Triangle inequality problem. Prove that from sections $x,y,z$ where
$x= \sqrt[3]{(a-b)^2(a+b)}, y=\sqrt[3]{(b-c)^2(b+c)}, z=\sqrt[3]{(c-a)^2(c+a)}$ and $a,b,c>0$, $a\neq\ b \neq c$
it is possible to construct a triangle.
I started the limitation $x,y,z$ from cauchy inequality, but I am not sure if inequality I need to... | $x^3+y^3-z^3+3xyz=(x+y-z)(x^2+y^2+z^2-xy+yz+xz)$
sinc $(x^2+y^2+z^2-xy+yz+xz) >0 \implies x^3+y^3-z^3+3xyz>0 \iff x+y-z>0$
$x^3+y^3-z^3+3xyz>0 \iff -(a-b)(b-c)(a+c+2b)+3xyz>0 \iff 27(a-b)^2(b-c)^2(c-a)^2(a+b)(b+c)(a+c)> (a-b)^3(b-c)^3(a+c+2b)^3 \iff 27(c-a)^2(a+b)(b+c)(a+c) >(a-b)(b-c)(a+c+2b)^3$
if$(a-b)(b-c)<0$, then... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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FoxTrot Bill Amend Problems So I found this on the Wolfram website today:
So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:
*
*$ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $
*$ \displaystyle\... | My take on $\#3$: $$ \begin {align*} \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] &= \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot (n+1) \ln u - \dfrac {u^{n+1}}{(n+1)^2} \right] \\&= \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
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Expressing $\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$ in closed form I want to express $$\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.
I know that $$\pi z\cot(\pi z)=1+2z^2\sum_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.
| First rewrite it as:
$$\frac{1}{z^3} + \sum_{n=1}^\infty \left(\frac{1}{z^3-n^3} + \frac{1}{z^3+n^3}\right) = \frac{1}{z^3} +\sum_1^\infty \frac{2z^3}{z^6-n^6}$$
so we only need to find:
$$\sum_{n=1}^\infty \frac{1}{z^6-n^6}$$
Now, if $\omega$ is a primitive cube root of unity, there are functions $a(u),b(u),c(u)$ such... | {
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"url": "https://math.stackexchange.com/questions/576343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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All roots of the quartic equation $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real Problem
Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$.
Source
This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time ... | Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic
$ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots
iff $f$ does. But then the same is true of
$f(x-\frac14) = x^4 + \frac58 x^2 + Bx + A$ for some $B$ and $A$
(we don't need the formula).
If this is $(x-a)(x-b)(x-c)(x-d)$ for some $a,b,c,d$
t... | {
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"url": "https://math.stackexchange.com/questions/578360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Proving that $(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$, where $a$, $b$, $c$ are the sides, and $\Delta$ the area, of a triangle Let $a$, $b$, $c$ be the sides of $\triangle ABC$.
Prove $$(abc)^2\geq\frac{4\Delta}{\sqrt{3}}$$
where $\Delta$ is the area of the triangle.
(Editor's note: As observed in the... | I think you mean to the following:
$$(abc)^{\frac{2}{3}}\geq\frac{4\Delta}{\sqrt3},$$ which is
$$16\Delta^2\leq3\sqrt[3]{a^4b^4c^4}$$ or
$$\sum_{cyc}(2a^2b^2-a^4)\leq\sqrt[3]{a^4b^4c^4}$$ or
$$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)\geq0,$$ which is true by Schur and AM-GM:
$$\sum_{cyc}\left(a^4-2a^2b^2+... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Limit at infinity involving $e$ I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$
I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.
| $$
\begin{align}
\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^{\frac{x^3+8}{x}}
&=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{x^2+8/x}\\
&=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)\frac{x^2+8/x}{5x^2+x+8/x}}\\
&=\left(\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
... | We count the ways in which Team A can win the series, and double the result. To count the ways A can win the series, we make a list like yours.
A wins in $4$: There is $1$ way this can happen.
A wins in $5$: A has to win $3$ of the first $4$, and then win. There are $\binom{4}{3}$ ways this can happen.
A wins in $6$: A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\in... | $\dfrac{1}{3}+\dfrac{1}{4}\dfrac{1}{2!}+\dfrac{1}{5}\dfrac{1}{3!}+\dots\\=\displaystyle\sum_3^\infty\dfrac{1}{n}\times\dfrac{1}{(n-2)!}\\=\displaystyle\sum_3^\infty\dfrac{n-1}{n!}\\=\displaystyle\sum_3^\infty\left[\dfrac{1}{(n-1)!}-\dfrac{1}{n!}\right]\\=\dfrac{1}{2!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I al... | $$
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^2\sin^2x}=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}\frac{x^2}{\sin^2x}
$$
Apply l'Hôpital or Taylor expansion (better) to the first fraction.
For the Taylor expansion, it's easier to do it in pieces:
$$
\sin x... | {
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"timestamp": "2023-03-29T00:00:00",
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$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} +\cdots+{1 \over {\sqrt n }}\right)$ $$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }}+\cdots+{1 \over {\sqrt n }}\right)$$
... | $$ \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{{\sqrt{n} + \sqrt{n}}} < \frac{1}{2\sqrt{n}}$$
add inequalities from 1 to n,
$$\sqrt{n+1} - 1 > \frac{1}{2}\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$
also for lower bound note that,
$$\l... | {
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"timestamp": "2023-03-29T00:00:00",
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Exponentials in complex numbers If $\displaystyle z-\frac1z=i$, then find $\displaystyle z^{2014}+\frac{1}{z^{2014}}$.
The answer should be in terms of $1, -1,\;i\;or\;-i$. I am not able to understand how to simplify the given expression so that its $2014^{th}$ power can be found out easily.
| $(1)$
We have $\displaystyle z^2+\frac1{z^2}=1$
$\displaystyle\implies z^4-z^2+1=0$
Using $\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2),$
$z^6+1=(z^2+1)(z^4-z^2+1)=0\implies z^6=-1$
$(2A)$ We have $\displaystyle z^2=iz+1$
$\displaystyle \implies z^3=z\cdot z^2=z(iz+1)$
$\displaystyle=iz^2+z$
$\displaystyle=i(iz+1)+z=i+z(i^... | {
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Suppose that we flip a coin until either it comes up tails twice or we have flipped six times. What is the expected number of times we flip the coin? So this is the answer that I got (where $N$ represents the number of coin flips after which you can stop) but can someone tell me if this looks right?
$P(N=2) = P(TT) = \... | What you are looking at is a right-censored negative binomial distribution. That is, let $Y = X \wedge 6 = \min\{X, 6\}$ where $X \sim {\rm NegBinomial}(r = 2, p = 1/2)$, where the particular choice of parametrization for $X$ is $$\Pr[X = k] = \binom{k-1}{k-r} (1-p)^r p^{k-r}, \quad k = r, r+1, r+2, \ldots.$$ That is... | {
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$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| Let $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=x $$
Clearly, $x>0$
$$\implies x^2=7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7x$$
Now left is the proof of converge(as conversed with Abdulh Khazzak Gustav ElFakiri)
Observe that the $r$th term $T_r$ of this infinite product is $\displaystyle7^{\left(\frac1{2^r}\rig... | {
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Product of a matrix and its Hermitian transpose? Suppose I need to find a matrix B such that
$B^H B = A$
and $A = \begin{bmatrix}4 &0& 0\\ 0 &1 &i\\ 0 &-i& 1\\\end{bmatrix}$
How do I proceed with a Product of a matrix and its Hermitian transpose ?
| This will illustrate how to handle your specific, small matrix, but the other post points to a general and practical method.
Diagonalize A. It must have non-negative eigenvalues or $B^{H}B=A$ is not possible. The eigenvalues of $A$ are $4,2,0$. Check that $A$ has a basis of eigenvectors:
$$
A\left[\begin{array}{c}1... | {
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The $\sum\limits_{n = 1}^\infty {\frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}}$ series is absolutely convergent? $$\sum\limits_{n=1}^\infty\frac{\sin(\sqrt n)}{n^{\frac{3}{2}}}$$
Let ${a_n} = \frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}$, then, using the criterion of quotient I must prove that $\mathop {\lim }\limits... | Let us take absolute series of given series,
ie.$$\sum_{n=1}^\infty |\frac {\sin \sqrt n}{n^{3/2}}|$$.since we know $|\frac {\sin \sqrt n}{n^{3/2}}|$$\leq$ $|\frac {1}{n^{3/2}}|$.But $\sum_{n=1}^{\infty} |\frac {1}{n^{3/2}}|$ is converging.Hence by comparison theorem, $\sum_{n=1}^\infty |\frac {\sin \sqrt n}{n^{3/2}}|$... | {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar... | HINT. The second part is the triangle inequality in reverse. So instead of starting with $|a+b|$ and getting $|a+b|\leq |a|+|b|$. You have $|a|+|b|$ and want to write something like $|a+b|\leq |a|+|b|$.
| {
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy... | let
$$a=\dfrac{2(x+y)}{x},b=\dfrac{x+y}{y},x,y>0$$
then
$$C=a+b+\sqrt{a^2+b^2}=\dfrac{(x+y)(x+2y+\sqrt{4y^2+x^2})}{xy}$$
so
$$C-10=\dfrac{x^2-7xy+2y^2+(x+y)\sqrt{x^2+4y^2}}{xy}$$
and note
$$(x+y)^2(x^2+4y^2)-(x^2-7xy+2y^2)^2=4xy(2x-3y)^2\ge 0$$
so
$$a+b+\sqrt{a^2+b^2}\ge 10$$
| {
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"source": "stackexchange",
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finding minimum of function Can you please give me some hints finding minimum of this function:
$ (r-1)^2 + (\frac{s}{r} -1)^2 + (\frac{t}{s}-1)^2 + (\frac{4}{t}-1)^2$
where $ 1 \le r \le s \le t \le 4 $,
$r,s,t \in \Bbb R $
| Let us write $a = \frac{r}{1}$, $b = \frac{s}{r}$, $c = \frac{t}{s}$, and $d = \frac{4}{t}$. Then the function you want to minimise is
$$f(a,b,c,d) = (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2,$$
subject to the constraints $a,b,c,d \geqslant 1$ and $p(a,b,c,d) = a\cdot b\cdot c\cdot d = 4$.
That looks like using the method ... | {
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