Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $ y = x^{\frac{1}{n}} \Rightarrow y^n = x $ Tao, Analysis I, exercise 5.6.1
I have to prove the following where $x,y \in \Bbb R^+, \ n \in \Bbb Z^+$
$$ y = x^{\frac{1}{n}} \Rightarrow y^n = x $$
Hints: review the proof of Proposition
5.5.12. Also, you will find proof by contradiction a useful tool, espec... | Let $a := x^{1/n}$.
Then $a^n > x$ does not conflict with the definition of the set, since it merely states that $a^n$ is an upper bound.
Here is my version of the proof following the same mechanism than for
Proposition 5.5.12. There exists a positive real number $x$ such that
$x^2 = 2$.
and using the result of
E... | {
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"timestamp": "2023-03-29T00:00:00",
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How to tackle this squaring of inequality problem
If the roots of quadratic equation $$x^2 − 2ax + a^2 + a – 3 = 0$$
are real and less than $3$, find the range of $a$.
The roots are $a \pm \sqrt {3 – a}$
For the roots to be real, we must have a < 3.
Also, for the roots to be less than 3, we must have $\pm \... | $y=(x-a)^2+(a-3)$;
$a-3 >0:$ This parabola does not cut the $x-$axis ($(x-a)^\ge 0)$.
Hence $a-3\le 0$.
Roots:
$y=(x-a)^2 +(a-3) = 0;$
$x_{1,2}=a\pm \sqrt{3-a} <3$.
1)Let $0\le a \lt 3$ ($a=3$ is ruled out)
$\sqrt{3-a} <3-a=\sqrt{3-a}\sqrt{3-a}$.
This is true for $\sqrt{3-a} >1$, or $ 3-a >1$.
Hence $2>a \ge 0$.
2) Let... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 8
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Evaluate $\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$ I would like to evaluate this integral.
$$\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$$
$1+x+x^2=(x+\frac{1}{2})^2+1-\frac{1}{4}=(x+\frac{1}{2})^2-\frac{3}{4}$
$$\int_{0}^{1}\frac{x(1-x)^2}{(x+\frac{1}{2})^2-\frac{1}{4}}\frac{\math... | Consider the function
$$ I(s) = \int_{0}^{1} \frac{x^{s+1} (1 - x)^2}{1+x+x^2}\,\frac{\mathrm{d}x}{\log x}. $$
Then
\begin{align*}
I'(s)
&= \int_{0}^{1} \frac{x^{s+1} (1 - x)^3}{1-x^3}\,\mathrm{d}x \\
&= \frac{1}{3}\int_{0}^{1} \frac{u^{(s-1)/3} (1 - u^{1/3})^3}{1-u}\,\mathrm{d}u \tag{$x=u^{1/3}$} \\
&= \frac{1}{3} \su... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$
Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t... | $K=\frac{1+t^2}{1+t+4t^2}$ To find min and max, set $K'(t)=0$ or $t^2-6t-1=0$ and solve for $t$. Answer $t=3\pm \sqrt{10}$ leading to max and min $K=\frac{20\pm 6\sqrt{10}}{80\pm 25\sqrt{10}}$. or $1.088303688022443$ and $0.245029645310883$
Corrected (stupid error in solving quadratic).
| {
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Find residues of a function $f(z) = z^{10}e^{\frac{1}{2z}} + \frac{z+4\pi}{z-4\pi}\cot{\frac{z}{2}}$ I am trying to find residues at singularity points of this function, but I cannot wrap my head around its Laurent series. Singularity points are $2k\pi, k \in \mathbb{Z}.$ $f$ is a sum of two functions and $0$ is an ess... | Actually, $0$ is an essential singularity of $z^{10}e^{1/2z}$, but not of $\frac{z+4\pi}{z-4\pi}\cot\left(\frac z2\right)$. So, $0$ is an essential singularity of $f$.
Anyway,$$\cot\left(\frac z2\right)=\frac2{z-4\pi}-\frac16(z-4\pi)+\cdots,$$and so$$\frac{\cot\left(\frac z2\right)}{z-4\pi}=\frac2{(z-4\pi)^2}-\frac16+\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{(m!)^{1/m}}{m/e}$ is a decreasing function of $m$ Show that
$\dfrac{(m!)^{1/m}}{m/e}$
is a decreasing function of $m$.
Here is my proof.
I would like to see others,
preferably simpler.
I have shown in
Proof explanation $\lim\limits_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=e$
that,
if
$r_m
=\dfrac{(m!)^{1/m}... | I guess the following is "quicker", but relies on some heavy machinery. We have
\begin{align*}
\frac{d}{dm} \log r_m &= -\frac{1}{m^2} \log m! + \frac{1}{m} \frac{\Gamma'(m+1)}{m!} - \frac{1}{m} \\
&\le -\frac{\log\sqrt{2\pi}-m+(m+\frac{1}{2})\log m}{m^2} + \frac{H_m - \gamma - 1}{m}
\end{align*}
where we used the fact... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that a following equation has no solution in integers: $x^3-x+9=5y^2$
Show that a following equation has no solution in integers: $$x^3-x+9=5y^2$$
Clearly we see that $y$ is odd, so $y^2\equiv_8 1$ and thus $8\mid x^3-x-4$.
So if $x$ is odd, then $x-1$ and $x+1$ one is divisible by $2$ and other by $4$ so $8\mid... | Aqua, here it is.
As you've obtained, $x\equiv 2\pmod{5}$. Now, $x(x-1)(x+1)=5y^2-9$. Note that, $x+1\equiv 3\pmod{5}$, and is odd. I now claim that, there is a prime $p\mid x(x-1)(x+1)$ such that $p\equiv 2,3\pmod{5}$, and $p\neq 3$. If $x\equiv 1,2\pmod{3}$, then the object $x$ clearly has such a prime divisor (inde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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U~$(0,i)$ where $i$ is a fair dice roll.
If I throw a fair dice and the result is $i$ then I choose a point $X$~$(0,i)$
1st) What is the expected value and standard deviation of $X$?
2nd) if $X > 3$ then what is the probability that the result of the dice was $6$?
I think I got the 1st part correct but the second par... | a)
Note that, for any $\sigma$-field $\mathcal G$, we have $E[X] = E[E[X|\mathcal G]]$
Let us consider $\mathcal G$ = $\sigma(Y)$, where Y is rv. that characterises the number we've rolled. Due to Y being discrete, we only need to find $E[X|Y=k]$ to know $E[X|Y]$.
So take $k\in\{1,2,3,4,5,6\}$
$E[X|Y=k]$ = $\frac{E[X\c... | {
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Find minimum value of $P=\sum \frac{a}{a^2+b^3}$ If $a,b,c$ be non-negative numbers and $a+b+c=3$, then find minima value of function $$P=\frac{a}{a^2+b^3}+\frac{b}{b^2+c^3}+\frac{c}{c^2+a^3}$$
Here is my idea.
By WA i see that if the equality occurs $a=b=c=1$, we will get $\text {Min}_P=1.5$. Indeed i will prove $P\... | By the Andreas's hint it remains to prove that
$$\sum_{cyc}\frac{a}{a^2+b^3}\geq\frac{1}{3}.$$
Now, by C-S
$$\sum_{cyc}\frac{a}{a^2+b^3}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^3+b^3a)}.$$
Id est, it's enough to prove that
$$(a+b+c)^4\geq(a+b+c)(a^3+b^3+c^3)+3(a^3c+b^3a+c^3b)$$ or
$$\sum_{cyc}(3a^3b+6a^2b^2+12a^2bc)\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan x=3$, then what is the value of ${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$?
If $\tan x=3$, then what is the value of
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$$
So what I did is change all the $\sin{2x}$ and $cos{2x}$ with double angle formulas, getting
$${3\cos^2{x}-3\sin^2{x}-4\sin{x}\cos{x}\o... | Since $\tan x = {\sin x \over \cos x}$ we have $\sin x = 3\cos x$ so
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}= {3\cos^2x-3\sin^2x-4\sin x \cos x\over 8\sin x \cos x +5\cos^2-5\sin^2 x}$$
$$ {3-27-12\over 24+5-45} = {9\over 4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | In addition to the possible typo $\sqrt{3x+7}=1+\sqrt{x+2}$ not $1-\sqrt{x+2}$ in the RHS, you made an arithmetic error later.
From $4x^2+16x+16=4(x+2)$, you should get $4x^2+12x+8=0$, not $4x^2+12x+14=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Find $ \frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1} $ Find the following sum
$$
\frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1}
$$
I am not able to find any short trick for it.
Is there any short trick or do we have to simplify and add it?
| $$\sum_{n=1}^{10}\frac{1}{4n^2-1}=\frac12\sum_{n=1}^{10}\frac{1}{2n-1}-\frac{1}{2n+1}=\frac12\left(1-\frac{1}{21}\right)=\frac{10}{21}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$
$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able t... | Look at the sequence of numerators and their differences
\begin{array}
n &1 & 2 &3 &4& 5 \\
u_n & 9 & 19 &35&57&85 \\
1st diff. & 10&16&22&28& \\
2nd diff. &6&6&6&&
\end{array}
The first differences show the sequence is not of the form $an+b$, i.e., not a linear sequence. Working out the $2$nd differences gives all the... | {
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For which of the following functions $f$ is $f(x)=f(2-x)$ for all $x$? The options are:
$A)$ $f(x) = x(x+2)$
$B)$ $f(x) = x-2$
$C)$ $f(x) = 2-x$
$D)$ $f(x) = 3x(x-2)$
$E)$ $f(x) = x^2(2-x)^2$
The answer given in my worksheet is $E)$ and I got it. My doubt is why it can't be $D)$ also? Solving thus:
$f(x) = 3x(x-2)$
$f(... | The functions such that $f(x)=f(2-x)$ for all $x$ are the functions with a symmetry around the vertical line $x=1$ (we can see this solving $x=2-x$, giving us $x=1$).
Therefore we can exclude $B)$ and $C)$ since the graph is a line and they can't have this kind of symmetry (unless they are horizontal lines, which is no... | {
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"url": "https://math.stackexchange.com/questions/3277723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a second-order homogeneous differential equation I got into this ODE which looks simple but I have a hard time figuring out how to solve:
$$a\frac{d^2 y}{dx^2}+\frac{da}{dx}\frac{dy}{dx}-\frac{l(l+1)}{x^2}y(x)=0$$
where $a=(1-k/x)^2$, $l\geq 0$, and $k>0$. Can anyone help me? I just want the general solution. I... | Let $z'(x)=y'(x)\,a(x).$ Then $z''(x)=y''(x)\,a(x)+y'(x)\,a'(x).$ This allows us to write the DE as $z''-l(l+1)y/x^2=0.$ Now
$a(x)=(1-k/x)^2=(x-k)^2/x^2;$ our goal is to write $y/x^2$ in terms of $z$ and $x,$ if possible. We have
\begin{align*}
z'&=y'a \\
\frac{z'}{a}&=y' \\
\int\frac{z'}{a}\,dx&=y \\
z''-l(l+1)\,\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the rate of convergence of nested radicals The sequence of nested radicals $\sqrt {a+\sqrt{a+\sqrt{a+...}}}$ is defined by
$x_1=\sqrt a$, $x_{n+1}=\sqrt{a+x_n}$, where $a>0$.
Here are three questions:
(a) Show that $x=\displaystyle\lim_{n\to\infty} x_n=\frac{1+\sqrt{1+4a}}{2}$.
(b) Prove $\displaystyle x-x_n\... | Let $x_n:=x-h_n$, so that $$(x-h_{n+1})^2=a+x-h_n$$ which simplifies to $$h_{n+1}^2-2xh_{n+1}+h_n=0$$ \begin{align*}h_{n+1}&=x-\sqrt{x^2-h_n}\\
&=\frac{h_n}{2x}(1+\frac{h_n}{4x^2}+\frac{h_n^2}{8x^4}+\cdots)\\
\end{align*}
So multiplying out gives $$h_n=\frac{h_1}{(2x)^n}\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{... | $$\dfrac{\binom n2}{(n+1)!}=\dfrac{n(n-1)}{2(n+1)!}$$
As $n(n-1)=(n+1)n-2(n+1)+2$
$$ \dfrac{\binom n2}{(n+1)!}=\dfrac12\cdot\dfrac1{(n-1)!}-\left(\underbrace{\dfrac1{n!}-\dfrac1{(n+1)!}}\right)$$
The terms under brace telescopes
Use $e^y=\displaystyle\sum_{r=0}^\infty\dfrac{y^r}{r!}$ for the first term
See also : Evalu... | {
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Finding zero divisors in a polynomial quotient ring
Is $x^2+x+I$ a zero divisor in $\mathbb{Z}_7[x]/I$, where $I=(x^3+5x^2+2x+5)$?
I know that $\gcd(x^3+5x^2+2x+5,x^2+x)=x+1$, and that it means that $x^2+x$ is indeed a zero divisor.
What I struggle with is finding $g(x)$ such that $$(f(x)+\langle x^3+5x^2+2x+5\rangl... | The long division of $x^3+5x^2+2x+5$ by the GCD $x+1$ yields $x^2+4x+5$, so we have (in the given ring)
$$(x^2+x)(x^2+4x+5)=x(x+1)(x^2+4x+5)=x(x^3+5x^2+2x+5)=x\cdot0=0$$
Thus if $f(x)=x^2+x$, $g(x)$ may be $x^2+4x+5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\sum_{n=0}^\infty\frac{S_n}{(n+1)!}$ if $S_n$ is the sum of the products of the first $n$ natural numbers taken two at a time
If $S_n$ denotes the sum of the products of the first $n$ natural numbers taken two at a time, then find $$\sum_{n=0}^\infty\frac{S_n}{(n+1)!}$$
$$
S_n=\sum_{1\leq i<j\leq n} a_ia_j=\fra... | You are almost correct, just pay attention to the starting index of each sum at each step:
$$\begin{align}
\sum_{n=0}^\infty\frac{S_n}{(n+1)!}&=\frac{1}{24}\sum_{n=1}^\infty\frac{3n(n+1)-2(2n+1)}{(n-1)!}\\
&=\frac{1}{24}\sum_{n=1}^\infty\frac{3(n-1)(n-2)+8(n-1)}{(n-1)!}\\
&=\frac{1}{24}\sum_{n=3}^\infty\frac{3}{(n-3)!}... | {
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"url": "https://math.stackexchange.com/questions/3284347",
"timestamp": "2023-03-29T00:00:00",
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Show that $\sum_{cyc} \sqrt{8a+b^3}\ge 9$ Prove that $$\sqrt{8a+b^3}+\sqrt{8b+c^3}+\sqrt{8c+a^3}\ge 9 \text{
for } a,b,c\ge0 \text{
and } a+b+c=3.$$
By Holder $$\left(\sum _{cyc}\sqrt{8a+b^3}\right)^2\left(\sum _{cyc}\frac{1}{8a+b^3}\right)\ge 27$$
Or $$\sum _{cyc}\frac{1}{8a+b^3}\le \frac{1}{3} \text{ WLOG } ... | By Holder $$\left(\sum_{cyc}\sqrt{8a+b^3}\right)^2\sum_{cyc}\frac{(2a^2+2b^2+3ab+4ac+bc)^3}{8a+b^3}\geq$$
$$\geq\left(\sum_{cyc}(2a^2+2b^2+3ab+4ac+bc)\right)^3=64(a+b+c)^6.$$
Thus, it's enough to prove that
$$64(a+b+c)^3\geq27\sum_{cyc}\frac{(2a^2+2b^2+3ab+4ac+bc)^3}{8(a+b+c)^2a+9b^3},$$
which is obviously true by BW a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving equation with two unknown degrees I sticked at this problem:
$3^x-7^y=2$ where it should be solved for natural $x$ and $y$.
I made the conclusion that $x=12a+2$ and $y=6b+1$ after playing with different mods. I am sure that the only solution is $x=2$ and $y=1$ as wolfram alpha confirmed it. Can you help with pr... | Let $x>2$ and $y>1$.
Thus, our equation it's $$3^x-9=7^x-7$$ or
$$9\left(3^{x-2}-1\right)=7\left(7^{y-1}-1\right),$$
which says that $3^{x-2}-1$ is divisible by $7$, which gives $x-2$ is divisible by $6,$
which gives $3^{x-2}-1$ is divisible by $3^6-1=8\cdot7\cdot13,$ which gives $7^{y-1}-1$ is divisible by $13$,
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3287777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a Limit Problem if the Limit Exists My question is, consider $\lim [(x^{2} +2x +1) / (x^{4} -1)]$ for $x \to -1$ (just arrow sign to the right)
What I have done so far is
$(x+1)^{2}/[(x^{2} +1)(x+1)(x-1)]$
$(x+1)/[(x^{2}+1)(x-1)]$
I'm not too sure what to do at this point. If I substitute in $x=-1$, I would get... | Note that $x^2+2x+1=(x+1)^2$ and $x^4-1=(x^2+1)(x+1)(x-1),$
so that $$\frac{x^2+2x+1}{(x^2+1)(x+1)(x-1)} = \frac{x+1}{x^2+1)(x-1)},$$ so $$\lim_{x \to -1} \frac{x^2+2x+1}{(x^2+1)(x+1)(x-1)} = \frac{0}{2 \cdot -2} = 0.$$
$0$ is a perfectly fine answer. You may be confusing this with getting 0 in the denominator, which... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| Let $x=y=-1$.
Thus, we obtain the value $16$.
We'll prove that it's a minimal value.
Indeed,
$$(5+x)(5+y)-16=xy+5(x+y)+9=3x^2+4xy+3y^2+5(x+y)\sqrt{\frac{x^2+y^2+xy}{3}}\geq$$
$$\geq3x^2+4xy+3y^2-5\sqrt{\frac{(x+y)^2(x^2+y^2+xy)}{3}}=$$
$$=\frac{3(3x^2+4xy+3y^2)^2-25(x+y)^2(x^2+y^2+xy)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqr... | {
"language": "en",
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"source": "stackexchange",
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Discuss a set of three numbers $x, y, z ∈ N$ such that $x^2+5^4=5^y+z$ Discuss a set of three numbers $x, y, z ∈ N$ such that $x^2+5^4=5^y+z$ .
What about the possible pairs of numbers $x, y ∈ N$, $y$ being an even number, such that $x^2+5^4=5^y$ ? What if $y$ being odd? What if we have another prime number instead of ... | Consider $x^2+5^4 = 5^y+z \implies z = x^2-5^y+5^4$ sub to $x,y,z \in N$. There is one equation and three unknowns, which means degree of freedom is two (possibly infinite solutions). Choose any two variables and the third one will be constrained by the equation. Since $x, y$ are appear in square and exponential terms,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe... | Using the Reflection property of Beta function, we have
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \tan ^a x d x=\int_0^{\frac{\pi}{2}} \sin ^a x \cos ^{-a} x d x =& \frac{1}{2} B\left(\frac{a+1}{2}, \frac{1-a}{2}\right)=\frac{\pi}{2 \sin \left(\frac{a+1}{2} \pi\right)}=\frac{1}{2} \sec \frac{a \pi}{2}
\end{aligned}
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294446",
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"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. Could you please give me a hint on how to prove the inequality below for $k \geq 3$?
$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$
Thank you in advance.
| If $k \geq 3, $ then...
$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} &=\frac{\sqrt{k^2+k}}{k-1} \\
&\lt \frac{\sqrt{k^2+k+\frac 1 4}}{k-1}= \frac{k+ \frac 1 2}{k-1}=1+\frac{3}{2k-2} \\
\end{align}$$
Have to compare $\frac{3}{2k-2}$ with $\frac{8k+3}{4k^2}$ when $ \ \ k \geq 3$
$$\begin{align}\frac{3}{2k-2}&=\frac{12k^2}{... | {
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"source": "stackexchange",
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) ... | The given sequence is equivalent to
$$ 1 + 2(2) + 4(2^{2}) + 8(2^{3}) + 16(2^{4}) + \ldots $$
$$ = 1 + (2^{2})^{1} + (2^{2})^{2} + (2^{3})^{3} + \ldots + (2^{2})^{k-1} + \ldots $$
Now, since this is a geometric series, we may solve
$$ s_{k} = \frac{a(1 - r^{k})}{1-r} = \frac{1(1-4^{k})}{1-4} = 1025 $$
which easily ca... | {
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"source": "stackexchange",
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Alternative methods for solving a system of one linear one non linear simultaneous equations Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious... | The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find range of $m$ such that $|x^2-3x+2 | = mx$ has 4 distinct solutions
$|x^2-3x+2 | = mx$
I think that $(x^2-3x+2) = mx$ or $-(x^2-3x+2) = mx$
For $-(x^2-3x+2) = mx$, $D > 0$,
$m = 3 \pm \sqrt{8}$
For $(x^2-3x+2) = mx$, $D > 0$,
$m = -3 \pm \sqrt{8}$
Now. What do you think it's the range?
Note that $ |x^2-3x+2 | = m... | If $f(x) = x^2-(3+m)x+2$, the discriminant is
$D = (3+m)^2-4\cdot 2 = (m+3)^2-8.$
This gives you a quadratic polynomial in $m$. If you have two solutions for $m$, you get in total four solutions of $f(x)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
|
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}=\sqrt{\frac{(1+\sqrt3)^2}{2}}=\frac{1+\sqrt3}{\sqrt2}$$
Can you end it now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\sqrt{a\pm bi\sqrt c}$ I recently encountered this problem
$$\sqrt{10-4i\sqrt{6}}$$
To witch I set the solution equal to $a+bi$ squaring both sides leaves
$${10-4i\sqrt{6}}=a^2-b^2+2abi$$
Obviously $a^2-b^2=10$ and $2abi=-4i\sqrt{6}$, using geuss and check, the solution is $a=\sqrt12, b=\sqrt2$
But I was wo... | One thing right off the bat. Every non-zero complex number will have two square roots so there wont be just $x + yi$ there will also be $-x -yi$
" using geuss and check" Why use guess and check when you can use the quadratic formula?
$a^2 - b^2 = 10$ and $2abi = -4\sqrt 6i$
$a=\frac{-2\sqrt 6}b$
$\frac {4*6}{b^2} - b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h... | $$\frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}=\frac{\left((x-a)-(x-a-h)\right)\left((x-a)+(x-a-h)\right)}{h(x-a-h)^2(x-a)^2}=$$
$$=\frac{h(2x-2a-h)}{h(x-a-h)^2(x-a)^2}=\frac{(2x-2a-h)}{(x-a-h)^2(x-a)^2}$$
If now you want to take the limit when $\;h\to0\;$ , for example, the above equals
$$\frac2{(x-a)^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to calculate $\lim_{x \to - 1} \frac{2}{(x+1)^4}$ Calculate $$\lim_{x \to - 1} \frac{2}{(x+1)^4}$$
a) $0$
b) $\infty$
c) $-\infty$
d) $2$
I am able to see that it is equivalent the limit as $x$ approaches $-1$ of $\frac{2}{(x^2+2x+1)^2}$.
I know that when doing limits to infinity this would be $0$ because the deno... | Let's construct tables of values.
$$
\begin{array}{c c}
x & f(x) = \dfrac{2}{(x + 1)^4}\\ \hline
0 & 2\\
-0.9 & 20,000\\
-0.99 & 200,000,000\\
-0.999 & 2,000,000,000,000
\end{array}
\qquad
\begin{array}{c c}
x & f(x) = \dfrac{2}{(x + 1)^4}\\ \hline
-2 & 2\\
-1.1 & 20,000\\
-1.01 & 200,000,000\\
-1.001 & 2,000,000,000,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 1... | Find the intersection of the curves $y=\frac{x^2}{2}$ with $x^2+y^2=8$
From the parabola $y=\frac{x^2}{2}$ we have $2y=x^2$. Pluging this result $x^2=2y$ in the equation function of the circle gives $2y+y^2=8$. The equation $2y+y^2=8$ is a quadratic equation $y^2+2y-8=0$ and can be solved by many methods. One of these ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| By equating coefficients, you have the system of equations
$$A+B = 1,$$
$$2A-3B = 6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|$. Find the minimum value of $E$. Can any one help me finding the minimum value of the following expression:
$$
E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|
$$
where $x$ is a real number.
| Note that $$\begin{align}E(x+2)&=|x+2|-|x+3|
\mp\cdots+|x+2016|-|x+2017|+|x+2018|\\&=E(x)-|x|+|x+1|-|x+2017|+|x+2018|\end{align} $$
and $$|x+1|-|x|=\begin{cases}1&x\ge 0\\-1&x\le -1\\2x+1&-1\le x\le0\end{cases} $$
$$|x+2017|-|x+2018|=\begin{cases}1&x\ge -2017\\-1&x\le -2018\\2(x+2017)+1&-2018\le x\le2017\end{cases} $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Determining multiplicity of 1 as an eigenvalue for a certain matrix By Matlab, I know that the eigenvalues of the matrix $B^{-1}A$ are 2.457, 0.542, and 1 (multiplicity 3) where $A$ and $B$ are defined as:
\begin{equation}
A=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 1 & 1 \\
0 & 0 & 1 & 2 &... | Here's one.
Since $A$ and $B$ must always be invertible, the following conditions are equivalent:
*
*$1$ is an eigenvalue of $B^{-1}A$ for eigenvector $x$
*$B^{-1}Ax=x$
*$Ax = Bx$
*$(A-B)x=0$
*$x \in \ker(A-B)$
So the multiplicity of the eigenvalue $1$ of $B^{-1}A$ is equal to the dimension of $\ker(A-B)$.
Edit:... | {
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"source": "stackexchange",
"question_score": "4",
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Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$
Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3... | You have the identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$
As far as I know, this is the best you can do, though I'd be happy if someone proves me wrong.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Modular arithmetic $(3n - 2)x + 5n \equiv 0 \pmod{9n-9}$ $(3n - 2)x + 5n \equiv 0 \pmod{9n-9}$
I know that the congruence has a solution if $gcd((3n - 2), (9n -9)) \mid -5n$ And it seems to have solution because $gcd((3n - 2), (9n -9)) = 1$
I tried with several $x$’s in the equation $(3n-2)x +5n$ but couldn’t found one... | Okay... this is a little symbol heavy but you are correct.
This will have a solution if and only if $\gcd(3n-2,9n-9)|-5n$.
Let's work with that first. Remember Euclids algorithm for finding $\gcd$s: $\gcd(a, b) = \gcd(a, b - ka)$ and repeat until you can go any further. Also bear in mind $\gcd(a,b) = \gcd (b,a)$ and... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator ... | \begin{align*}
\frac{2x+1}{(x-1)(x-2)} & = \frac{(2x-2) + 3}{(x-1)(x-2)} = \frac{2(x-1)}{(x-1)(x-2)} + \frac{3}{(x-1)(x-2)}\\\\
& = \frac{2}{x-2} + 3\times\frac{(x-1) - (x-2)}{(x-1)(x-2)} = \frac{2}{x-2} + \frac{3}{x-2} - \frac{3}{x-1}\\\\
& = \frac{5}{x-2} - \frac{3}{x-1}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Inscribed Equilateral Triangle Trig - Expression for P/A Problem: The vertices of an equilateral triangle, with perimeter P and area A, lie on a circle with radius r. Find an expression for $\frac{P}{A}$ in the form $\frac{r}{k}$, where k ∈ Z+.
Hi, I'm having a bit of trouble solving this problem. What I've tried is us... | I get that $A = \frac{P^2\sqrt{3}}{36}$. Also, $r = \frac{P}{3\sqrt{3}} \implies P = 3\sqrt{3}r$. This effectively matches your results. Thus,
$$\frac{P}{A} = \frac{36}{\sqrt{3}P} = \frac{36}{9r} = \frac{4}{r} \tag{1}\label{eq1A}$$
In your calculations, since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, then $P = 3(2r\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequality $\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$ I'm interested by the following problem :
Let $a,b,c>0$ such that $abc=a+b+c$ then we have :
$$\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt... | Using the fact that $\ln x \le \ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(x-8\sqrt{3})$ for $x > 0$
(the proof is simple and thus omitted),
it suffices to prove that
$$\sum_{\mathrm{cyc}}\frac{\ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(7a+b-8\sqrt{3})}{7a+b}
\le \frac{3\ln (8\sqrt{3})}{8\sqrt{3}}$$
or
$$\frac{1}{7a+b} + \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$
Given $a$, $b$, $p$, $q$, $r \in\mathbb{R_{>0}}$ s.t.
$$\begin {cases}\phantom{b}a=p+q+r=pqr \\ab =pq+qr+rp\end{cases} $$
Find the maximum of $$\dfrac{a^2+6b+1}{a^2+a}$$
This question is terrifying that I ... | Hint.
Using the AM-GM we get
$$
a \ge 3\sqrt[3]{pqr}=3\sqrt[3]{a}\Rightarrow a\ge \sqrt{27}
$$
$$
ab\ge 3\sqrt[3]{a^2}\ge 3\sqrt[3]{27}
$$
now we can solve
$$
\max \frac{a^2+6b+1}{a^2+a}\ \ \ \text{s. t.}\ \ \ (a\ge \sqrt{27})\cap (ab \ge 9)
$$
and the solution is
$$
a = 3\sqrt 3,\ \ b = \sqrt 3
$$
with
$$
\frac{a^2+6b... | {
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"url": "https://math.stackexchange.com/questions/3319843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Determine points on line with specific distance from plane There is a line $$p: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1}$$
and plane $$\pi : x+y+2z-3=0.$$
I need to find points $T_1, T_2 \in p$. Requirement when finding those points are that they have a set distance from plane $\pi$, with distance being $\sqrt... | Let $ \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1} = t$ for some point $(x,y,z)$ on the line $p$.
Then $p(t) = (1+2t, -1+3t, -1-t)$
Using your distance formula, we get
\begin{align}
\dfrac{|1(1+2t) + 1(-1+3t) + 2(-1-t) - 3|}{\sqrt{(1^2 + 1^2 + 2^2)}} &= \sqrt 6 \\
|3t-5| &= 6 \\
3t - 5 &= \pm 6 \\
t &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/... | We have $5y-1+12=5y+11$ and not $=5y-11.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Line$ 3x-4y+k$ touches a circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$, then find $k+a+b =?$
The line $3x-4y+\lambda=0, (\lambda > 0)$ touches the circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$. Find the value of $\lambda+a+b$.
I tried solving in the following manner:
Clearly the circle has center at $(2,4)$ with a radius of $5... | Here is the picture for the problem, could not insert it as a comment, so it became an answer:
$5$ through the points $(2,-1)$, $(5,0)$, $(6,1)$, $(7,4)$, $(6,7)$, $(5,8)$, $(2,9)$, $(-1,8)$, $(-2,7)$, $(-3,4)$, $(-2,1)$, and $(-1,0)$">
In the picture, we see the circle with center $C(2,4)$ and radius $5$, it passes th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r... | Apply the following identity involving an orthogonal array of sign patterns:
$4(a^2+b^2+c^2+d^2)=(a+b+c+d)^2+(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2$
Put in the given values and simplify giving
$3=(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2$
The condition $a\ge b\ge c\ge d>0$ forces $a+b-c-d$ to have an absolute value greater than... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:
If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$
I tried many times, and finally I used Muirhea... | By AM-GM you have
$$(1 - a)^{1\over 2}(1 - b)^{1 \over 2} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$(1 - b)^{1\over 2}(1 - c)^{1 \over 2} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$(1 - c)^{1\over 2}(1 - a)^{1 \over 2} \leq {1 - c + 1 - a \over 2} = {b \over 2}$$
Multiplying these three inequalities together g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Study the convergence of the series I need to study the convergence of the following series:
$$\sum_{n=1}^{\infty}(\frac{1\cdot3\cdot5\cdot\cdots\cdot(2n+1)}{1\cdot4\cdot7\cdot\cdots\cdot(3n+1)}x^n)^2$$ I tried to expand that term and reduce some terms but didn't succeed, I also tried to apply the ratio test but still ... | The ratio test gives you
$${\left(\frac{1 \cdot 3\cdot 5\cdot ...\cdot (2n+1)\cdot (2n+3)}{1\cdot 4\cdot 7\cdot ...\cdot (3n+1)\cdot (3n+4)}x^{n+1}\right)^2\over \left(\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n+1)}{1\cdot 4\cdot 7\cdot ...\cdot (3n+1)}x^n\right)^2}=\frac{(2n+3)^2}{(3n+4)^2} x^2$$
and here you need to take... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to refactor $\pi / 12$ into $\pi /4 - \pi /3$? How to refactor $\pi / 12$ into $\pi /4 - \pi /3$ (standard calculus angles)? How do you know that $\pi / 12$ converts exactly into $\pi /4 - \pi /3$? Do I need to memorize that or is there a simple procedure to follow in order to refactor any angle into combination of... | A good place to start is by expressing each of the standard angles as multiples of $\frac{\pi}{12}$.
\begin{align*}
\frac{\pi}{6} & = \frac{2\pi}{12}\\
\frac{\pi}{4} & = \frac{3\pi}{12}\\
\frac{\pi}{3} & = \frac{4\pi}{12}
\end{align*}
From there, you can immediately see that
\begin{align*}
\frac{\pi}{12} & = \frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A problem on inequality If real numbers $a >b >1$ satisfy the inequality
$$ ( ab+1)^{2} +( a+b)^{2} \leqslant 2( a+b)\left( a^{2} -ab+b^{2} +1\right) $$
What is the minimum possible value of $\frac{\sqrt{a-b}}{b-1}$?
My approach:
$$ ( a+b)\left( a^{2} -ab+b^{2} +1\right) =( a+b)\left( a^{2} -ab+b^{2}\right) +( a+b) =a^... | The condition gives
$$2a^3-(b^2+1)a^2-2a(2b-1)+2b^3-b^2+2b-1\geq0$$ or
$$a^2(2a-b^2-1)-2a(2b-1)+(2b-1)(b^2+1)\geq0$$ or
$$(2a-b^2-1)(a^2-2b+1)\geq0$$ and since $$a^2-2b+1>a^2-2a+1=(a-1)^2>0,$$ we obtain $$2a\geq b^2+1$$ or
$$2(a-b)\geq(b-1)^2$$ or
$$\frac{\sqrt{a-b}}{b-1}\geq\frac{1}{\sqrt2}.$$
The equality occurs for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find real numbers c and d from $\frac{1}{a+bi}$
Suppose $a$ and $b$ are real numbers, both not 0. Find real numbers $c$ and $d$ such that
$$ \frac{1}{a+bi} = c + di$$
I am not really sure what the question is asking me to do. Am I supposed to represent $c$ and $di$ both in terms of $a$ and $b$ since $a$ and $b$ are ... | Notice that
$$
\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di.
$$
Hence
$$
c=\frac{a}{a^2+b^2}\,\, \text{ and }\,\,d=-\frac{b}{a^2+b^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use formal definition of a limit to find the largest value of $\delta > 0$ Given $f(x)= {\sqrt x}$, L = 2, $\epsilon = {\frac 1 5}$ and C=4.
I know that I start with
$$|{\sqrt x} -2| \le {\frac 1 5}$$
If I rewrite that as
$$-{\frac 1 5} \le {\sqrt x} -2 \le {\frac 1 5}$$
I can then add 2 to both sides and square eve... | Nothing wrong with the approach so far, aside from mixing up $<$ with $\le$ sometimes. What you have shown so far is
$$\frac{81}{25} < x < \frac{121}{25} \iff |\sqrt{x} - 2| < \frac{1}{5}.$$
Further, we can say,
$$-\frac{19}{25} < x - 4 < \frac{21}{25} \iff |\sqrt{x} - 2| < \frac{1}{5}.$$
Our largest possible value of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to calculate this definite integral $\int_0^1\frac{{\ln^4 x }}{1+x^2}dx=\frac{5\pi^5}{64}$ $$\displaystyle\int_0^1\dfrac{{\ln^4 x}}{1+x^2}\text{d}x=\dfrac{5\pi^5}{64}$$
let $x=e^{-t}$，
$$
\displaystyle\int_0^1\dfrac{({\ln x})^4}{1+x^2}\text{d}x=\displaystyle\int_0^{+\infty}\dfrac{t^4\text{e}^{-t}}{1+\text{e}^{-2t}}... | \begin{align}
J&=\int_0^1 \frac{\ln^4 x}{1+x^2}\,dx\\
&=\frac{1}{2}\int_0^\infty \frac{\ln^4 x}{1+x^2}\,dx\\
J_n&=\int_0^\infty \frac{\ln^{2n} x}{1+x^2}\,dx\\
J&=\frac{1}{2}J_2\\
K_n&=\int_0^\infty \int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&\overset{u=yx}=\int_0^\infty\left(\int_0^\infty\frac{y\ln^{2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Stem and leaf diagrams I have the following data:
$2.6$ $ $ $3.3$ $ $ $2.4$ $ $ $1.1$ $ $ $0.8$ $ $ $3.5$ $ $ $3.9$ $ $ $1.6$ $ $ $2.8$ $ $ $2.6$ $ $ $3.4$ $ $ $4.1$ $ $ $2.0$ $ $ $1.7$ $ $ $2.9$ $ $ $1.9$ $ $ $2.9$ $ $ $2.5$ $ $ $4.5$ $ $ $5.0$
Built stem and leaf plot:
$Stem$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $Leaf$ $ ... | It's important to note that (for example) $2\le X\le 3$ translates to "$2$ is less than or equal to $X,$ and $X$ is less than or equal to $3.$" Thus, $2\le X\le 3$ will be true if and only if both $2\le X$ and $X\le 3$ are true. Put another way, $2\le X\le 3$ will be true if and only if neither $X<2$ nor $3<X$ is true.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Quartic polynomial roots The equation
$x^4 - x^3-1=0$
has roots α, β, γ, δ. By using the substitution $y=x^3$ find the exact value of $α^6+β^6+γ^6+δ^6$ .
The solution is
$x=y$ (1/3)
$y^4=(1+y)^3$
$y^4 -y^3 -3y^2-3y^2-1=0$
$S$N+4 $=$ $S$N + $S$N+3
$S$-1 = $\frac {0}{1} =0$
$S$2 = $ 1^2 -2*0 =1$
$S$3 = $0+1 =1$
$S$4 = $... | This looks like an application of the Newton identities which are a generalization of Viete's identities connecting roots and coefficients. If $S_k=\sum_{j=1}^d α_j^k$ is the degree $k$ power sum of the roots of the polynomial equation $0=x^4+c_1x^3+c_2x^2+c_3x+c_4$, then
$$
c_1+S_1=0\\
2c_2+c_1S_1+S_2=0\\
3c_3+c_2S_1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ramanujan's series for the exponential integral According to Wikipedia, Ramanujan came up with the following series expansion of the exponential integral:
$$\operatorname{Ei}(x)=\gamma+\ln|x|+e^{x/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1}$$
My first instinct was to... | You can subtract $\frac{1}{x}$ and multiply by $xe^{-x/2}$ to avoid the Cauchy product and instead use the known series expansions of $e^x$. You would get $$e^{x/2} -e^{-x/2} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is my solution of $\lim_{(x, y)\rightarrow (0, 0)} \left( \frac{\cos (x^2+y)}{1-x^2y} \right) ^{\frac{1}{x^4+y^2}}$ correct? Is this solution correct?
If not, why?
Alternative (especially less wordy) proofs are also welcome.
$$\lim_{(x, y)\rightarrow (0, 0)} \left( \frac{\cos (x^2+y)}{1-x^2y} \right) ^{\frac{1}{x^4+y^2... | You have approached this by looking at one curve, namely $y=x^2.$ And that curve should have yielded the limit $e^{-1/2}.$ So there's that mistake. The bigger mistake is: One curve is never enough.
We can write
$$\tag 1 \cos (x^2+y) = 1-\frac{(x^2+y)^2}{2} +o(x^4+y^2)$$
and
$$\tag 2 \frac{1}{1-x^2y} = 1 + x^2y + o(x^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$
$a$, $b$, $c$ are sides of a triangle, prove:
$$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$
What I have tried:
$$
⇔2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use
$$\sum a^3+3abc\geq \sum ab(a+b)$$
| Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to solve $x(3x+3)(x+5)(2x+12)+576 = 0$?
How to solve $$x(3x+3)(x+5)(2x+12)+576 = 0?$$
This was a question on a test i recently took, And i wasn't able to solve it. I later tried to solve it using online calculators and it turns out this doesn't have any real solutions.
I know there's a general formula for quartic... | Another possible way is using the symmetry of $x(3x+3)(x+5)(2x+12) = 6x(x+1)(x+5)(x+6)$ around $x=\color{blue}{3}$:
$$6x(x+1)(x+5)(x+6) = 6(x+\color{blue}{3}-3)(x+\color{blue}{3}+3)(x+\color{blue}{3}-2)(x+\color{blue}{3}+2)$$ $$=6(\underbrace{(x+\color{blue}{3})^2}_{y:=}-9)((x+\color{blue}{3})^2-4)$$
The minimum value ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Prove that $0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$ Prove
$$0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$$
where $g(k)$ is the greatest odd divisor of k
Please Find Holes in my Proof.
Let $k=2m+1$ if we show that the right hand side of the equation is true for odd numbers, then it is t... | $$g(2^m (2k+1))=2k+1$$ Let $$h(n) = \sum_{2k+1 \le n} 1=\lfloor (n+1)/2\rfloor$$ then $$\sum_{k=1}^n \frac{g(k)}{k} =\sum_{m\ge 0} \frac{h(n/2^m)}{2^m}=\sum_{m\ge 0} \frac{\lfloor (n/2^m+1)/2\rfloor}{2^m}=\sum_{m\ge 0} \frac{ (n/2^m+1)/2-O(1)}{2^m}\\ = \frac{n/2}{1-2^{-2}}+\frac{1/2}{1-2^{-1}}-\frac{O(1)}{1-2^{-1}}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Number of words with no adjacent vowels I am faced with the following problem:
Given $C = \{B,C,D,F\}$ and $V = \{A, E, I, O, U\}$ find the number of 9-letter words with elements from $C$ and $V$ such that no two vowels (elements of V) are adjacent.
Following this answer about a very similar question I get that I sho... | You boil your problem down to a linear recurrence relation, which is nice!
There's a little issue with $a_0$ and $b_0$, though : how can the empty word start by anything? One should start the recurrence at $n=1$, we have $a_1 = 5$ and $b_1 = 4$.
Call
$A = \begin{pmatrix} 0 & 5 \\ 4 & 4 \end{pmatrix}$, then
$$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Ways to prove the area of an ellipse formula One can prove the ellipse area formula $A=\pi a b$ ($a$, $b$ the major and minor semi-axis) either by integration or by the stretched-circle argument. See for instance here: https://proofwiki.org/wiki/Area_of_Ellipse
Are there any other proofs of this formula?
| Let's get creative. For any $R>1$ we have
$$ f_R(\theta)=\frac{1}{R+e^{i\theta}} = \sum_{n\geq 0}(-1)^n \frac{e^{ni\theta}}{R^{n+1}} $$
hence by Parseval's identity
$$\int_{0}^{2\pi}\frac{d\theta}{(R^2+1)+2R\cos\theta}= \int_{0}^{2\pi}f_R(\theta)\overline{f_R}(\theta)\,d\theta = 2\pi\sum_{n\geq 0}\frac{1}{R^{2n+2}}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Olympiad Inequality $\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$ with proof (long post) It takes me a lot a time to solve this :
Let $a,b,c>0$ with $a\geq b\geq c$ then we have :
$$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$$
My proof :
We need some lemmas to begin :
First lemma :
Let $x,... | As an alternative, while not shorter, it can be brute-forced with just elementary algebra . . .
Since the $\text{LHS}$ is homogeneous, we can assume $c=1$, and $a\ge b \ge 1$.
Replacing $c$ by $1$, and simplifying, we get
$$
1-\text{LHS}
=
\frac
{-a^2b+2a^3b-a^2-ab^2-a^2b^2-ab+2b^3-b^2+2a}
{(a+b+1)(2a+b)(2b+1)(2+a)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$... | Here's an elementary way to do it: Since $x\equiv 2\mod 4$ then $x$ is even. Since $x\equiv 1\mod 5$ then the decimal expansion of $x$ has $1$ or $6$ in the units. But $x$ is even, so it must be $6$.
So $x=10k+6$ for some $k$. But also, $x\equiv 8\mod 9$, so the sum of the algarisms of $x$ is of the form $9p+8$. But si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now ... | Note that we have $$n^2(n^2-1)(n^2+1)=n^2(n^2-1)(n^2-4)+5n^2(n^2-1)$$ and that $n^2$ is divisible by $4$ or $n^2-1$ is divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find all solutions of $a,b,c,d$ if $a+b+c+d=12$ and $abcd = 27 +ab+ac+ad+bc+bd+cd$ Find all solutions of $a,b,c,d>0$ if
$$a+b+c+d=12$$
$$abcd = 27 +ab+ac+ad+bc+bd+cd$$
Attempt:
$a + b + c + d = 12 \implies 3 \ge \sqrt[4]{abcd} \implies 81 \ge abcd$ by AM-GM, eqyality when $a=b=c=d$. Also
$$ abcd - 27 = ab + ac + ad ... | I think it's better from $$abcd-27\geq6\sqrt{abcd}$$ to write
$$abcd-6\sqrt{abcd}-27\geq0$$ or
$$(\sqrt{abcd}-9)(\sqrt{abcd}+3)\geq0$$ or
$$abcd\geq81$$ and since by your work
$$abcd\leq81,$$ it's enough to check an equality case.
Another solution.
By AM-GM we obtain:
$$abcd=27\left(\frac{a+b+c+d}{12}\right)^4+(ab+ac+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$ $a,b,c > 0$ and $a+b+c=3$, prove
$$ \frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2 $$
Attempt:
Notice that by AM-Gm
$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3\frac{\sqrt[3]{abc}}{\sqrt[3]{... | $$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{b^2+1}\geq$$
$$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}(ab+ac+bc).$$
Can you end it now?
Since by your work $$3-\frac{1}{2}(ab+ac+bc)=3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2},$$ it's enough to prove that
$$3-\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Prove that $\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1}$ Given $x, y,z \ge 0$. Prove that
$$\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} $$
Attempt
Notice that $$(x^{3} + y^{3}+1) ... | The inequality $$\sum_{cyc}\sqrt{a+1}\geq2+\sqrt{a+b+c+1}$$ we can prove also by the following way:
$$\sum_{cyc}\sqrt{a+1}-2-\sqrt{a+b+c+1}=$$
$$=\sqrt{a+1}+\sqrt{b+1}-2-\left(\sqrt{a+b+c+1}-\sqrt{c+1}\right)=$$
$$=\sqrt{a+1}+\sqrt{b+1}-2-\frac{a+b}{\sqrt{a+b+c+1}+\sqrt{c+1}}\geq$$
$$\geq\sqrt{a+1}+\sqrt{b+1}-2-\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find equation of the circle whose diameter is the common chord of two other circles? Circle 1:
$$x^2 + y^2 +6x + 2y +6 = 0$$
Circle 2:
$$x^2 + y^2 + 8x + y + 10 = 0$$
My attampt:
From circle 1 and 2, I found
$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got
$$5x^2 + 26x + 30 = 0$$
here I go... | You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 \Rightarrow x_1=\frac{-13-\sqrt{19}}{5},x_2=\frac{-13+\sqrt{19}}{5}\\
y_1=\frac{-6-2\sqrt{19}}{5}, y_2=\frac{-6+2\sqrt{19}}{5}$$
The center of the new circle:
$$\frac{x_1+x_2}{2}=-\frac{13}{5},\frac{y_1+y_2}{2}=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$why $A = \frac{1}{2} A + \frac{1}{2} A^\top $ is not always true? If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$,
It is possible to prove that
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top(\frac{1}{2} A + \frac{1}{2} A^\top)... | In the product $\mathbf{x}^\top A \mathbf{x}$ we get terms like $x_1 A_{12} x_2 + x_2 A_{21} x_1 = (A_{12}+A_{21}) x_1 x_2$ so that for example $\begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}$, $\begin{bmatrix}1&5\\ 0&4 \end{bmatrix}$ and $\begin{bmatrix}1&3\\ 2&4 \end{bmatrix}$ give the same result since $2.5+2.5 = 5+0 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equations and find $x,y,z$ Given:
$$x^3+y^3+z^3=x+y+z$$
And:
$$x^2+y^2+z^2=xyz$$
Find all real and positive solutions to these equations, if any.
So most probably, we'll factorise in this way:
$$x^3+y^3+z^3-3xyz=x+y+z-3xyz
\rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)-3(x^2+y^2+z^2)$$
But I'm having trou... | You're given
$$x^3+y^3+z^3=x+y+z \tag{1}\label{eq1A}$$
$$x^2+y^2+z^2=xyz \tag{2}\label{eq2A}$$
As you've shown, you have
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) \tag{3}\label{eq3A}$$
Next, note
$$(x-y)^2 + (y-z)^2 + (x-z)^2 = 2(x^2 + y^2 + z^2 - xy - yz - zx) \tag{4}\label{eq4A}$$
Since $x,y,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculating the value of a triple integral I would like to evaluate the following integral
$$
I=\int_{\mathbb{R}^3}\frac{x^2+y^2+1-z^2}{|(x,y,z-1)|^3|(x,y,z+1)|^3}\,dx\,dy\,dz.
$$
My attempt: with $R^2=x^2+y^2$ we have
$$
I=2\pi\int_{\mathbb{R}_+\times\mathbb{R}}\frac{R^2+1-z^2}{[(R^2+(z-1)^2)(R^2+(z+1)^2)]^{3/2}}R\,dR... | Probably the fastest method: Do not rewrite the denominator and integrate with respect to $D$ directly:
\begin{align}
I &= \pi \int \limits_{-\infty}^{\infty} \int \limits_0^\infty \frac{D + 1 - z^2}{[(D + (1-z)^2)(D + (1 + z)^2)]^{3/2}} \, \mathrm{d} D \, \mathrm{d} z \\
&= \pi \int \limits_{-\infty}^{\infty} \left[\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Polynomial of 4th degree I would like to ask if someone could help me with the following equation.
\begin{equation}
x^4+ax^3+(a+b)x^2+2bx+b=0
\end{equation}
Could you first solve in general then $a=11$ and $b=28$.
I get it to this form but I stuck.
\begin{equation}
1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac... | The polynomial equation for $(a, b)=(11,28)$ is given by
$$
x^4+11x^3+39x^2+56x+28=(x^2 + 7x + 7)(x + 2)^2=0.
$$
"How do we come up wit this layout"? By the rational root test, we find the factor $x-2$ twice, and dividing gives the factor $x^2+7x+7$. Hence the roots are $x=-2,-2,\frac{\sqrt{21}-7}{2},\frac{-\sqrt{21}-7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Diophantine equation from the Latvian Baltic Way team selection competition 2019 So here is the problem statement:
Find all integer triples $(a, b, c)$ such that
$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$
The only things I have so far figured out is that (-1, 0, 0) and (0, 1, 0) are solution, gcd((a-b), c) = 1 and that c must... | Consider the equation already found:-
$$(a-b)\left((a^2-b^2)^2-2\right)=c^2+1,$$
where $c^2+1$ is odd or singly even.
If $a-b$ is even then the LHS would be divisible by 4. Therefore $a-b$ is odd and $c$ is even.
Then $(a^2-b^2)^2-2$ is of the form $4k-1$ and it is therefore either $-1$ or has a prime divisor of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all real matrices such that $X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$ The following question come from the 1998 Romanian Mathematical Competition:
Find all matrices in $M_2(\mathbb R)$ such that $$X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$$
Can you guy please help me... | There is a matrix $U$ such that
$$U^{-1}\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix} U=\begin{pmatrix} 20 & 0 \\ 0&0 \end{pmatrix}.$$
Let $Y=U^{-1}XU$, then $Y^{3}-4Y^{2}+5Y=\begin{pmatrix} 20 & 0 \\0 & 0 \end{pmatrix}$. The matrix $Y$ then commutes with $\begin{pmatrix} 20 & 0 \\ 0&0 \end{pmatrix}$ and so is diag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Simplifying $i \left[ \ln(x+i)-\ln(x-i)-\ln(1+ix)+\ln(1-ix) \right]$ Can the following expression further be simplified and expressed in terms of usual functions such as inverse hyperbolic or inverse trigonometric functions?
$$
f(x) = i \left[ \ln(x+i)-\ln(x-i)-\ln(1+ix)+\ln(1-ix) \right] \, ,
$$
where $x\ge 0$ is a r... | Another approach to simplifying $f(x)$ is to simply use the expansion in power series of $x$.
Specifically,
\begin{align}
i \left( \ln(x+i)-\ln(x-i) \right) &=-\pi+ 2\left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) , \\
i \left(-\ln(1+ix)+\ln(1-ix) \right) &= 2\left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $ This functional equation problem is from the Latvian Baltic Way team selection competition 2019:
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 ... | You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is the constant zero function, by continuing your own argument. You've established showing
$$ f ( - x ) ^ 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan ... | Your second step is invalid. In general one can't replace a sub-expression by its limit while evaluating limit of a bigger expression in step by step fashion. You can get more details in this answer.
You can approach this problem by adding and subtracting $\tan(\sin x) $ in numerator. The numerator can thus be expresse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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$\sqrt[3]{2}$ satisfies $x^3-2=0$ Show that there is no polynomial $P(x)$ of degree less than 3 with $P(\sqrt[3]{2})=0$ $\sqrt[3]{2}$ satisfies $x^3-2=0$
Show that there is no polynomial $P(x)$ of degree less than $3$ with $P(\sqrt[3]{2})=0$ All coefficients are rational numbers.
Is it by induction? Say, if $x$ has deg... | You didn't precised what are the nature of the coefficients of P, so we can build a polynomial $P(x) = x - 3\sqrt{2}$ such that $P(3\sqrt{2}) = 0$.
And, we can build an infinity of polynomials of degree 2 such that $P(2 \sqrt{3}) = 0$, these polynomials are : $$P(x) = (x - 2 \sqrt{3})(x - k), k \in \mathbb{C}$$
We all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is the average area of all triangles that can be inscribed in a unit circle? There is only one circle that passes through any three given points. Hence by suitable scaling, we can inscribe every triangle inside a unit circle of radius $1$. We define distinct triangles as triangles which have different sides regard... | Comment:
As can be seen in figure the area and perimeter of right angled triangle in half circle is maximum when the height is maximum , i.e. $h=r$ where h is height of triangle and r is the radius of circle. The area and perimeter are minimum when $h → 0$. So we can write:
$a/2 = r= 1$
$S_{max}=(\frac{(\sqrt 2)^2}{2})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x} $ using the quadratic formula I am trying to prove the fact that $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ or alternatively $\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$. (I understand that it can be proved using the half-angle identities of $\sin$ and $\cos$ but... | Because of the $\pm$, the absolute value is superfluous.
$\begin{align}
B &= \frac{-1\pm |\sec x|}{\tan x} \\
B &= \frac{-1\pm \sec x}{\tan x} \\
B &= \frac{(-1\pm \sec x)(\cos x)}{(\tan x)(\cos x)} \\
B &= \frac{-\cos x\pm 1}{\sin x} \\
\end{align}$
We know that $\dfrac{1 - \cos x}{\sin x} = \tan \dfrac x2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} +... | The equality that you have used is not correct.$$x^{4}-4x^{3}+2x^{2}+4 \ne (x-1)^{4}-4(x-1)^{2} + 7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of 6-character passwords containing at least one number - why is my answer wrong? Here's the question:
How many six-character passwords can be built with lowercase letters and numbers, given that at least one of its characters is a number?
Here's my answer:
$$10\dbinom{6}{1}36^5$$
*
*$10$ for the number pos... | The number of passwords with exactly $k$ digits is
$$\binom{6}{k}10^k26^{6 - k}$$
since there are $10$ choices for each of the $k$ digits, $26$ choices for each of the $6 - k$ lowercase letters, and $\binom{6}{k}$ ways to select $k$ of the $6$ positions in the password for the letters to appear. Hence, the number of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ ... | $$\sqrt[3]{x+2}+\sqrt[3]{x+4}=-\sqrt[3]{x+3}$$
Take Cube in both sides
$$2(x+3)-3\sqrt[3]{(x+2)(x+4)}\sqrt[3]{x+3}=-(x+3)$$
Set $\sqrt[3]{x+3}=y\implies x+3=y^3, (x+2)(x+4)=(y^3-1)(y^3+1)$
$$2y^3-3y\sqrt[3]{y^6-1}=-y^3$$
$$y(y^2-\sqrt[3]{y^6-1})=0$$
If $y\ne0$ $$y^2=\sqrt[3]{y^6-1}$$
Take cube in both sides $$y^6=y^6-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Evaluate $\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}$ I was given that
$$\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}=\frac{-9 \left(-\frac{2 C}{3}-1+\frac{1}{3} \pi \log \left(\sqrt{3}+2\right)\right)}{2 \pi }$$
Here $\beta$ denotes Dirichlet Beta. Unfortu... | Here is a sketch. Using, say, the link with Euler numbers, one gets
$$\sum_{n=0}^{\infty}\beta(2n+1)x^{2n}=\frac{\pi}{4}\sec\frac{x\pi}{2},$$
which implies
$$\sum_{n=0}^{\infty}\frac{\beta(2n+1)}{2n+2}x^{2n+2}=\frac{1}{\pi}f\left(\frac{x\pi}{2}\right),\quad f(y)=\int_0^y\frac{x\,dx}{\cos x}.$$
The given sum is then equ... | {
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"url": "https://math.stackexchange.com/questions/3397355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$ Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$. Here generalized harmonic numbers should be used.
$$H^{(2)}_n = 1^2 + \frac{1}{2}^2 + \dots + \frac{1}{N}^2.$$
For now, I was able to find $[z^N]$ for $\left(\ln \frac{1}{1-z}\right)^2$ ... | Using the generalization:
$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$
Let $a_{n}=H_n^2$ to have
\begin{align}
\sum_{n=1}^\infty H_n^2x^n&=\frac1{1-x}\sum_{n=1}^\infty \left(H_n^2-H_{n-1}^2\right)x^n\\
&=\frac1{1-x}\sum_{n=1}^\infty \left(\frac{2H_n}{n}-\frac1{n^2}\right)x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2 = \frac{2}{n+1}\binom{2n}{n}$
Show that $$\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \dfrac{2}{n+1}\dbinom{2n}{n}$$ where $n \in \mathbb{N}$
Consider $\dbinom{n}{r}=0 $ for $r<0 $ and $r>n$. Using $\displaystyle\su... | The way you simplify $\displaystyle{\sum_{k=0}^{n+1}\binom nk^2}$ also works for $\displaystyle{\sum_{k=0}^{n+1}\binom nk \binom n{k-1}}$.
Rewriting $\binom n{k-1} = \binom n{n-k+1}$ we can turn the second sum into $$\sum_{i+j=n+1} \binom ni \binom nj.$$ This is the coefficient of $x^{n+1}$ in $(1+x)^n (1+x)^n$, or $\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Ramanujan's Nested Radicals: evaluating $\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{\cdots}}}}$
Find the exact value of
$$\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{\cdots}}}}$$
My approach:
Suppose
$$\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = p \tag{1}$$
By multiplying each side by $2$, we have
$$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqr... | The mistake is:
$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} \neq \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} $
$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = \sqrt{4^2 + \sqrt{4^4 + \sqrt{4^7 + \sqrt{\cdots}}}} $
I think that an a way is:
$$2^n+1=\sqrt{(2^n+1)^2}$$
$$2^n+1=\sqrt{4^n+2\cdot2^n+1}$$
$$2^n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(... | Convert all to $\tan$:
$$\frac{2 \sin(x) \cos(2x)}{\cos(x) \sin(2x)} - 5 \tan(x) +5 = \frac{2 \tan(x)}{\tan(2x)} - 5 \tan(x) +5 = (1 - \tan^2(x)) - 5 \tan(x) + 5$$
$$ \Longrightarrow \tan^2(x) + 5 \tan(x) - 6 = 0$$
Now we have:
$$\tan(x_1 + x_2) = \frac{\tan(x_1) + \tan(x_2)}{1 - \tan(x_1)\tan(x_2)} = \frac{-b/a}{1-c/a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the value of C in this function If $$x^2 + (c-2)x -c^2 -3c + 5$$ is divided by $x + c$, the remainder is $-1$. find the value of c
I replaced all the x value to -c and set it to an equation which equated to $-1$
I am confused what to do after that, show me the steps how I can retrieve the value of $c$
| Given that:
$$\begin{align}x^2 + (c-2)x -c^2 -3c + 5 & = (-c)^2 + (c-2)(-c) -c^2 -3c + 5 \\
& = c^2 - c^2 +2c -c^2 -3c + 5 \\
c^2 +c -5& = 1 \\
c^2 +c-6 &= 0 \\
(c+3)(c-2) & = 0 \implies \color {blue}{ \boxed {\text { c = 2 , -3}} }\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the given inequality if $a+b+c=1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that
$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$
Please provide some hint to proceed. I have used Arithmetic Mean-Geometric Mean inequality to proceed in such qu... | We need to prove that:
$$\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\frac{a+b+c}{5abc}$$ or
$$\sum_{cyc}\left(\frac{a}{5abc}-\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^3+b^3+c^3+a^2b+a^2c-5abc)}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or
$$\sum_{cyc}\frac{a((a-b)(a^2+5ac-2ab-2b^2)-(c-a)(a^2+5ab-2ac-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that integer Let $a, b,$ and $c$ be real such that $ a ^ 2b + b ^ 2 c + c ^ 2a - ab ^ 2 - bc ^ 2 - ca ^ 2 = 6 $ and $ a ^ 2 + b ^ 2 + c ^ 2 -ab - ac - bc = 7$. Prove that if a is integer, then $b$ and $c$ are also integers.
What I tried:
$ ab (a-b) + bc (b-c) + ca (c-a) = 6 $
$a (a-b) -c (c-a) + b (b-c) = 7 $
$c... | Put $x = a - b$, $y = b - c$, $z = c - a$. Then:
*
*$x + y + z = 0$;
*$x^2 + y^2 + z^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca) = 14$;
*$xyz = -(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2) = -6$.
From 1. and 2., we get:
*$xy + yz + zx = \frac{1}{2}((x + y + z)^2 - (x^2 + y^2 + z^2)) = -7$.
Therefore, the equations 1. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{... | Recognize that you cannot have a factor that includes both a power of $2$ and $5$ at the same time, and recognize that (obviously) $5^2+5^2<5^4$. We would try to cover all the $2$s required, then split up the $5$s as evenly as possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Sum of floor function of general term While doing some solution to get a general term for some series I encountered at the end with this.
$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$
, where $i> \dfrac{n}{2}$. How can I get the ge... | What's your problem with floor function?
However you can't get rid of it, but this may help:
$$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \dfrac{n^2+n-2i}{2n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} -... | HINT
Divide by $\sqrt{x}$ if $x \ne 0$ to get
$$
\sqrt{x} + 1 - \sqrt{x-1} = m
$$
move $1$ to the other side and square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a stat... | We can use that
$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac{3}{3n+1}\right)^{2n}=\left[\left(1-\frac{3}{3n+1}\right)^{-\frac{3n+1}3}\right]^{-\frac{6n}{3n+1}}\to e^{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Is there any. elementary formula for the sequence$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}$ Is there any formula for the following sequence which does not use any derivative and also is less
advanced:
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}$$
I've calculated the general formula... | It's a fairly simple proof by induction.
$$\mathcal{H}_n \ : \ \left\{ \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3\right\}$$
Base case is obvious, $\mathcal{H}_1$ is true.
Induction. Suppose $\mathcal{H}_n$ true. Now
$$\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A pair of limits involving sums Find the following pair of limits:
$$\lim_{r \to \infty} 2 \lfloor r \rfloor \sum_{n=2\lfloor r \rfloor+1}^\infty \frac{(-1)^{n+1}}{n}$$
$$\lim_{r \to \infty} (2 \lfloor r \rfloor +1) \sum_{n=2 \lfloor r \rfloor+2}^\infty \frac{(-1)^{n+1}}{n}$$
Plugging in large values of $r$, I would gu... | For $r\in \Bbb N$ we have $$\sum_{n=2r+1}^{\infty}(-1)^{n+1}/n=$$ $$=(\frac {1}{2r+1}-\frac {1}{2r+2})+(\frac {1}{2r+3}-\frac {1}{2r+4})+(\frac {1}{2r+5}-\frac {1}{2r+6})+...=$$ $$=\frac {1}{(2r+1)(2r+2)}+\frac {1}{(2r+3)(2r+4)}+\frac {1}{(2r+5)(2r+6)}+...$$
Now for $n>1$ we have $$\frac {1}{4}\int_{n+1}^{n+2}(1/x^2)... | {
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"url": "https://math.stackexchange.com/questions/3407889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2... | The function to be minimized is the fourth power of the distance from the origin, so will have its extrema at the same points at which the distance is also an extremum. The constraint is a hyperbola, so the minimum distance occurs at its vertices, therefore this problem is equivalent to computing the fourth power of a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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