Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the numb... | We have $\sqrt{3}+i=2e^{i\pi/6}$ and $\sqrt{3}-i=2e^{-i\pi/6}$. So
\begin{eqnarray*}
z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m} = 2^{n-m} e^{ i \pi (n+m) /6}.
\end{eqnarray*}
So we require $ n+m \equiv 0 \pmod{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
$\int_{/6}^{/2}\sin^3(x+\sin3x+\cos3x)\,dx 4$ Evaluate
$\int_{/6}^{/2}\sin^3(x+\sin3x+\cos3x)\,dx </4$
I tried to use the fact that the $\sin^3x<\sin x$. It does not help. This qesion is proposed by Jalil Hajimir to RMM.
| There is more than likely a much simpler solution.
Use the fact that, over the considered range the cube of the sine is smaller than the cube of its argument. Developing the later and using a couple of integration by parts plus triple angle formulae, you end for
$$I=\int \left(x+\sin3x+\cos3x\right)^3\,dx$$
$$I=\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derivative of sum of two functions I have to find $\frac{dy}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$ but would like to find where I made a mistake in my solution.
Here is my work:
\begin{align}
(f+g)'=& \ f'+g', f = x \sqrt{x}, g= \frac{1}{x^2 \sqrt x}\\ f=& \ x \cdot x^{1/2} = x^{3/2}
\\ f'=& \ \frac{3}{... | $$\frac{d}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$$$$=\frac{d}{dx}\left(x\sqrt{x}\right)+\frac{d}{dx}\ \left(\frac{1}{x^{2}\sqrt{x}}\right)$$$$=\frac{d}{dx}\left(x^{\frac{3}{2}}\right)+\frac{d}{dx}\left(x^{\large-\frac{5}{2}}\right)$$$$=\frac{3}{2}\large x^{\frac{1}{2}}-\frac{5}{2}\large x^{-\frac{7}{2}}$$$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given the result of diagonalization of a matrix, determine the two invertible matrices. Determine $3$ by $3$ invertible matrices P and Q, such that
$$P\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
... | As it stands, the question does not require $P$ and $Q$ to be inverses of each other, or real orthogonal, etc.. So, there is no need to diagonalise the matrix on the left by similarity or by congruence. In fact, since the matrix on the LHS has a zero diagonal and the one on the RHS is positive semidefinite, they are ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $a_n:a_{n+1}$ if $a_n=\int_0^{\pi/2}\cos^nx.\cos nx.dx$
Find $a_n:a_{n+1}$ if $a_n=\int_0^{\pi/2}\cos^nx.\cos nx.dx$
Attempt 1
$$
a_{n+1}=\int_0^{\pi/2}\cos^{n+1}x.\cos(n+1)x.dx=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x-(n+1)\int\cos^{n}x.\frac{\sin (n+1)x}{(n+1)}.dx\\
=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x-\int\cos^{n}... | $\newcommand{\d}[1]{\, \mathrm{d}#1}$
It's easier to start by expanding $\cos{(n+1)x}$.
\begin{align*}
a_{n+1} &= \int_0^\frac{\pi}{2} \cos^{n+1}{x}\cos{(n+1)x} \d{x} \\
&= \int_0^\frac{\pi}{2} \cos^{n+1}{x}(\cos{nx}\cos{x} - \sin{nx}\sin{x}) \d{x} \\
&= \int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} - \int_0^\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $\frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$.
Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $$\large \frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$$
Well, I have provided my solution below, which is clearly not the most str... | We need to prove that $$(x+y)(x-y)^2\geq1-x^2y^2$$ or $$(x+y)(x-y)^2\geq(x-y)^2(x^2+y^2)$$ or $$(x+y)^2(x^2-xy+y^2)\geq(x^2+y^2)^2$$ or $$(x-y)^2xy\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to start solving this Logarithmic problem? If $abc= 2^6$, $a, b, c \ge 0$, $\log_2 (a)\log_2 (bc)+\log_2 (b)\log_2 (c)= 10$, find $\sqrt{((\log_2 (a))^2 + (\log_2 (b))^2 + (\log_2 (c))^2}$
| Using the substitution suggested in the comments gives the equations
\begin{cases}
A+B+C=6\\
AB+AC+BC=10
\end{cases}
We can then square the first equation to give
\begin{align}
(A+B+C)^2
&=A^2+B^2+C^2+2(AB+AC+BC)\\
&=A^2+B^2+C^2+2\cdot10\\
&=A^2+B^2+C^2+20\\
&=6^2\\
&=36\\
\end{align}
Thus we get
$$A^2+B^2+C^2=36-20=16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| Apply the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ to rewrite the equation
$$\frac\pi6=\tan^{-1}\frac{11}{x} -\tan^{-1} \frac{1}{x}
=\tan^{-1}\frac{\frac{10}x}{1+\frac{11}{x^2}}$$
Then, take $\tan(\cdot)$ on both sides along with $\tan\frac\pi6=\frac1{\sqrt3}$,
$$\frac{11}{x^2}+\frac{10\sqrt3}x+1=0$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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probability that two of these boxes contain exactly $2$ and $3$ balls is
If $10$ different balls are to be placed in $4$ distinct boxes at random, then the probability that two of these boxes contain exactly $2$ and $3$ balls is
What I tried:
Total number of ways $\displaystyle 4^{10}$
probability that two of these... | We have the following variants to distribute balls into the boxes:
$$
\underbrace{2+3+0+5}_{A_1}, \quad \underbrace{2+3+1+4}_{A_2}, \quad \underbrace{2+3+2+3}_{A_3}.
$$
And the probability that two of these box contain exactly $2$ and $3$ balls is the sum of probabilities for these variants to distribute balls:
$$
\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding an angle in equilateral triangle
Given $\triangle ABC$ ($AC=AB$). $X$ - the point on side $AC$ such as $AX=BC$. $\angle A = 20^0$. Find $\angle XBC$.
Here is my attempt:
Let side $AB = a$, then side $BC = 2a \sin10^0$.
Construct $B_1X \parallel BC$.
Similar triangles $\triangle BAC \sim \triangle B_1AX$ ... | Let $Y\in CX$,$Z\in AB$ and $X'\in AY$ such that $BY=ZY=ZX'.$
Thus, $$\measuredangle ZX'Y=\measuredangle X'YZ=180^{\circ}-\measuredangle ZYB-\measuredangle BYC=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ},$$
which gives $$\measuredangle AZX'=\measuredangle ZX'Y-\measuredangle A=40^{\circ}-20^{\circ}=20^{\circ},$$ which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $m$ such that $x^4 - (2m - 1)x^2 + 4m -5 = 0$ has real roots Consider the equation:
$$ x ^ 4 - (2m - 1) x^ 2 + 4m -5 = 0 $$
with $m \in \mathbb{R}$. I have to find the values of $m$ such that the given equation has all of its roots real.
This is what I did:
Let $ u = x^2, \hspace{.25cm} u\ge 0$
We get:
$$ u ^ 2 - ... | You need also to consider $\Delta$ to be positive in order for the solutions to be real.
$\Delta = (2m-1)^2-4(4m-5)=4m^2-4m+1-16m+20=4m^2-20m+21$
$m_{1,2}=\frac{10 \pm \sqrt{100-84}}{4}=\frac{10 \pm 4}{4}=\{\frac{3}{2},\frac{7}{2} \}$
Thus $\Delta \geq 0 \Leftrightarrow m \in (-\infty, \frac{3}{2}] \cup [\frac{7}{2},\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
I thought about MVT , what i did is :
let $g(x) = \arctan\left(x\right) $
so by MVT there is $c\in (0,x)$ such that :
$g'(c) = \frac{g\left(x\right)-g\left(0\right)}{x-0} \impli... | Let $f(x)= \arctan x - \frac{x}{\sqrt{1+x^2}}$. Evaluate
$$f'(x)=\frac{\sqrt{1+x^2}-1}{(1+x^2)^{3/2}}>0$$
So, $f(x)$ is strictly increasing for $x>0$. Thus, $f(x) > f(0) = 0$, or
$$\arctan x > \frac{x}{\sqrt{1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$
show that
$$\csc x \cot x=\frac{a^2-b^2}{4ab}$$
(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the an... | Rearrange
$$\frac{a^2-b^2}{4ab}=\frac{(a+b)(a-b)}{(a+b)^2-(a-b)^2}
=\frac{1}{\frac{a+b}{a-b}-\frac{a-b}{a+b}}$$
Then, substitute $\frac{a+b}{a-b}=\frac{\csc x}{\cot x}$,
$$\frac{a^2-b^2}{4ab}=\frac1{\frac{\csc x}{\cot x}-\frac{\cot x}{\csc x}}
=\frac{\csc x\cot x}{\csc^2 x-\cot^2x}=\csc x\cot x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Calculate matrix by using Cayley-Hamilton theorem Calculate matrix $B = A^{10}-3A^9-A^2+4A$ using Cayley-Hamilton theorem on $A$.
$$A = \begin{pmatrix}
2 & 2 & 2 & 5 \\
-1 & -1 & -1 & -5 \\
-2 & -2 & -1 & 0 \\
1 & 1 & 3 & 3
\end{pmatrix}$$
Now, I've calculated the characteristic polynomial of $A$:
$P_A(\lambda... | By polynomial division, we see that
\begin{align}
x^{10}-3x^9-x^2+4x = (x^4-3x^3+x^2-3x)(x^6-x^4+x^2+1)+x
\end{align}
which means
\begin{align}
A^{10}-3A^9-A^2+4A = (A^4-3A^3+A^2-3A)(A^6-A^4+A^2+1)+A = A.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to prove $(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$ where $x, y, z \ge 0$?
For $x, y, z \in \mathbb R^+\cup\{0\} $, prove that
$$(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$$ and
$$(x+y+z)^{(x+y+z)^2}x^{x^2}y^{y^2}z^{z^2} \ge (x+y)^{(x+y)^2}(y+z)^{(y+z)^2}(z+x)^{(z+x)^2... | It should be $x>0$, $y>0$ and $z>0$, otherwise $x^x$ is not defined for $x=0$.
Let $f(z)=\sum\limits_{cyc}(x+y)\ln(x+y)-\sum\limits_{cyc}x\ln{x}-(x+y+z)\ln(x+y+z).$
Thus, $$f'(z)=\ln(x+z)+1+\ln(y+z)+1-\ln{z}-1-\ln(x+y+z)-1=$$
$$=\ln\frac{(x+z)(y+z)}{z(x+y+z)}=\ln\left(1+\frac{xy}{z(x+y+z)}\right)>0.$$
Thus, $$f(z)\geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | There is one trick which is very useful when you face products, quotients, powr,.. : logarithmic differentiation.
$$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}\implies \log(f(x))=\frac 12 \log(3-x^2)-\log(3+x)$$
$$\frac{f'(x)}{f(x)}=\frac 12\frac{-2x}{3-x^2}-\frac 1{3+x}=-\frac{3 (x+1)}{(x+3) \left(3-x^2\right)}$$
Now
$$f'(x)=f(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b... | $$(a+b)(ax^2+by^2)=p^2+ab$$ where $p=\sqrt{a^2x^2+b^2y^2}$
Now assuming $ab>0,$
$$\dfrac{p^2+ab}p\ge2\sqrt{p\cdot\dfrac{ab}p}$$ using AM-GM inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Computation of the $2n$-th order determinant Compute the determinant of the $2n-\text{th}$ order.
$$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\\2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vd... | A recursion can also be helpful here:
*
*$n=1$: $D_2 = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = -5$
*For $n >1$, expanding $D_{2n}$ along the first column gives:
$$D_{2n} = 2\cdot D_{2(n-1)}\cdot 2 - 3 \cdot D_{2(n-1)}\cdot 3 = -5D_{2(n-1)}$$
It follows
$$D_{2n} = (-5)^{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Diagonalizability of a matrix
Show that $$ A :=\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 2 \\ -2 & 0 & 3 \end{pmatrix}$$ is diagonalizable.
What I did:
First, I determined the characteristic polynomial $$\chi_A(X) = \det(X \cdot E_3-A)=(X-3)(X-1)(X-1)=X^3-5X^2+7X-3,$$
so the eigenvalues are $3$ and $1$.
I then determine... | When $X=1$ the matrix becomes $$XI-A=I-A=\begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & 0 & -2\end{bmatrix}$$ which has rank 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$
Consider two sequences $(a_n)_{n\ge 1}$ and $(b_n)_{n\ge 1}$ of
positive real numbers such that
$$\lim_{n\to\infty} \frac {a_{n+1}}{n^2\cdot a_n}=x>0$$
and
$$\lim_{n\to\infty} \frac {b_{n+1}}{n\cdot b... | I think I completed a proof thanks to Gary's idea with the $c_n$ substitution and the ratio limits. Bear with me. First, let's make a little different substitution $c_n = \sqrt[n]{\dfrac{a_n}{b_n}}$ and use the following criterion.
Lemma (Cauchy-d'Alembert): If $x_n > 0,\ (\forall) n \in \mathbb{N}$ and the limit $l =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$ Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$ as $x$, $y$, and $z$ range over all positive real numbers.
My first instinct was to apply AM-GM on each factor in the denominator since every variable is a positive real number. I also t... | By AM-GM
$$\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}=$$
$$=\frac{xyz}{\left(1+3\cdot\frac{5x}{3}\right)(4x+3\cdot y)(5y+3\cdot2z)(z+3\cdot6)}\leq$$
$$\leq \frac{xyz}{4\sqrt[4]{1\cdot\left(\frac{5x}{3}\right)^3}\cdot4\sqrt[4]{4x\cdot y^3}\cdot4\sqrt[4]{5y\cdot(2z)^3}\cdot4\sqrt[4]{z\cdot6^3}}=\frac{1}{5120}.$$
The equality... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \s... | Hint:
$$1=(1-\cos y)^2+(1-\sin y)^2=1-2\cos y-2\sin y+1$$ gives you $y$. And at the same time, $2x$.
Alternatively:
$$\frac{\sin 2x+\sin x}{\cos 2x+\cos x}=\frac{2\sin\dfrac{2x+y}2\cos\dfrac{2x-y}2}{2\cos\dfrac{2x+y}2\cos\dfrac{2x-y}2}=\tan\frac{2x+y}2=1$$ so that
$$2x+y=\frac\pi4+k\pi.$$
Then
$$\cos\left(\frac\pi2+2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Compute$\int_0^{2\pi}\frac{\cos^2\theta }{ |re^{i\theta} -z|^2}d\theta$
$r\gt0$, Compute
$$\int_0^{2\pi}\frac{\cos^2\theta }{ |re^{i\theta} -z|^2}d\theta$$
when $|z|\ne r$
The problem is related to Poisson kernel and harmonic function, but I don't know how to start, $\cos^2\theta =1/2 (1+\cos 2\theta )$.
| Express $z$ also in polar form, $z = s^{i\alpha}$. Then,
$$|re^{i\theta} -z|^2=r^2+s^2-2rs\cos(\theta-\alpha)$$
and, with the variable change $t= \theta - \alpha$, the integral reads,
$$I=\int_0^{2\pi}\frac{\cos^2\theta }{ |re^{i\theta} -z|^2}d\theta
=\int_0^{2\pi}\frac{\cos^2(t+\alpha)}{ r^2+s^2-2rs\cos t}dt$$
Expand... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Generalized version of calculating expected minimum value rolled until $5$ is obtained from a fair die The following is a modified interview question.
Given an $n$-sided fair die where $n\geq 1.$
You roll a die until you get a $m$ where $1\leq m\leq n$.
Calculate the expected value of the minimum rolled.
The o... | A slight variant of angryavian's answer above:
Let $T$ be the number of rolls before the first appearance of $m.$ We have $$P(T = t) = \left(\frac{n-1}{n}\right)^{t}\cdot \frac{1}{n},\ \ t=0,1,2,\ldots.$$
We have $$\mathbb{E}[Y \ | \ T=t] = \sum_{y=0}^{m-1} P(Y> y \ | \ T=t) = \sum_{y=0}^{m-1} \left( \frac{n-y-1}{n-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Greatest common divisor for $x+1$ and $2x+1$? I was just wondering, what would the gcd of $x+1$ and $2x+1$ be?
I know that $x+1$ is not even or odd, whereas, $2x+1$ is odd. Is the gcd $1$?
| If $d(x)|x+1$ and $d(x)|2x+1$ then $d(x)|(2x+1)-(x+1)= x$. And so $d(x)|x$ and $d(x)|x+1$ so $d(x)|(x+1) -x = 1$ so $d(x)|1$. And assuming we are talking $\mathbb Z[x]$ this is as far as we go and $d(x) =1$ is the only (positive) common divisor of $2x+1$ and $x+1$. So the greatest common divisor of $2x + 1$ and $x+1$... | {
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"url": "https://math.stackexchange.com/questions/3534868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=... | The trick is to add and cancel an $x$ term:
$$\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\right) =\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-x +x-\sqrt{x^2-2x}\right)$$
and compute each limit separately:
$$\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-x\right) = \lim_{x\to \infty} \frac{2x}{\sqrt[3]{(x^3+2x)^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
remainder when $a_{1000}$ is divided by $1000$
If $a_{1}=7,a_{2}=7^7,a_{3}=7^{7^{7}}.$ Then the remainder when
$a_{1000}$ is divided by $1000$
what i try
$a_{1}=7=1\mod(1000)$ and
$a_{2}=7^7=1^7\mod(1000)=1\mod(1000)$
from using modulo theorem
$a_{3}=7^{7^{7}}=1^7\mod(1000)=1\mod(7)$
can we say that $a_{1000}= 1\mod... | $a_1=7\equiv\color{red}7\bmod 1000$.
$a_2=7^7=823543\equiv543\bmod 1000$.
$7^4=2401\equiv401\bmod1000,$ so $7^{20}=(7^4)^5\equiv401^5=(400+1)^5\equiv1\bmod1000$.
Since $a_2\equiv543\bmod1000,$ $a_2\equiv3\bmod20$. Therefore, $a_3=7^{a_2}\equiv7^3=343\bmod1000$.
Can you take it from here?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How many pairs of $(x,y,z)$ for $x+y+2z=n$
How many pairs of $(x,y,z)$ for $x+y+2z=n$ when $n$ is an odd integer $\geq 5$
For $n=5$ there are 2 pairs. $n=7$ there are 6 pairs. $n=9$ there are 12 pairs. $n=11$ there are 20 pairs.
$an^2 + bn + c = f(n)$
$25a + 5b + c = 2$
$49a + 7b + c = 6$
$81a + 8b + c = 12$
$f(n) = ... | put z=2 then x+y=n-2 has n-3 ways (x can be anything from 1 to n-3).
then put z=4 then x+y=n-4 has n-5 ways (x can be anything from 1 to n-5).
.
.
.
.
put z =n-3 then x+y =3 have 2 ways.
total number of cases = 2+4+ 6.......n-3 = 2(1+2+3....$\frac{n-3}{2}$) = 2 $\frac{(\frac{n-3}{2})(\frac{n-1}{2})}{2}$ = $\frac{(n-3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Product fractions limit $\lim_{n\to \infty} (\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}})^{1/n^2}$ I need some help with this limit
$$\lim_{n\to \infty} \left(\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}}\right)^{1/n^2}$$
This problem appears in a chapter about Riemann sums, so I think I must ... | $$\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)^{\frac{1}{n^{2}}}=\exp\left(\frac{1}{n^{2}}\ln\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)\right)$$$$=\exp\left(\frac{1}{n^{2}}\ln\left(\prod_{k=1}^{n}\left(\frac{k}{n}\right)^{k}\right)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.
| I found
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2 = xy(x-y)^2+x^2(z+x-y)^2+yz^2(y+z) \geqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solutions of p-Laplace equation I found that for the following problem
\begin{cases}
-\Delta_p u = 1,&x\in B_1(0)\\
u = 0,\quad &x\in\partial B_1(0)
\end{cases}
where $B_1(0)$ is the unitary ball of $\mathbb{R}^N$ and $\Delta_p u = \operatorname{div}(\|\nabla u\|^{p-2}\nabla u) $,
there exists a weak solution of the f... | Assume the given solution $u$ and note that it is smooth almost everywhere.
Compute the partial derivative (define $C:=C_{N,p}$)
\begin{align}
\partial_i u
&= -C \partial_i |x|^\frac{p}{p-1}
\\
&= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \partial_i |x|
\\
&= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \frac{x_i}{|... | {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$ The Function is $$f(x) = \log_3\sqrt{(x+3)}−\log_3(9−x^2)$$
and I need to figure out arguments for which $$ f(x) < 0 $$
So I calculated the domain of function which is $ D: (-3;3)$
However I am still unable to solve $f(x) < 0$
I simplified $ \log_3\sq... | Continue with $ \frac{\sqrt{x+3}}{9-x^2} < 1$, yet with the domain restriction $-3<x<3$. Then, examine the roots of the equation
$$\sqrt{x+3} = 9-x^2$$
Square and factorize to get
$$(x+3)(x^3-3x^2-9x + 26)=0$$
Since $x>-3$, the valid roots come from the second factor. Let $x= t+1$ to depress the second factor as
$$t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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About Harmonic Numbers and a sequence in OEIS I'd been looking for a series in OEIS and the one that fits better (resembles, at least) is one in this link in Examples: http://oeis.org/A082687. There, as we can see, it is equal to
$$H'(2n) = H(2n)-H(n) = \cdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
Also, I know ... | Here is how I would write the expression you give in (1):
$$\overline H_{2n} = H_{2n} - H_n. \tag2$$
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$ while $\overline H_n$ denotes the $n$th skew-harmonic number $\sum_{k = 1}^n (-1)^{k + 1}/k$.
In proving (2), we have
\begin{align}
\overline H_{2n} &= \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate this determinant of a $2\times 2$matrix? This matrix arises from a homework problem which our professor gave.
We need to find the determinant of this matrix.
Does there exist any simple way to find the determinant of this matrix?
$\begin{pmatrix}
x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) && (1-p)(2+\fra... | Call your matrix $M$ and let $s=x+2-n$. Then
\begin{aligned}
M&=\pmatrix{
x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) &(1-p)(2+\frac{l}{x-n+2-l})\\
(1-q)(2+\frac{l}{x-n+2-l}) &x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})}\\
&=\pmatrix{x-pq-p+3 &1-p\\ 1-q &x-pq-q+3}
+\frac{x+2-n}{x+2-n-l}\pmatrix{1-q&1-p\\ 1-q&1-p}\\
&=sI+\pmatrix{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ I am asked to simplify the following expression:
$$\tan \left( \frac{\pi}{2} - \theta \right)$$
The book gives me the answer $\cot \theta$ but, when I try to derive that formula using
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1... | The formula for $\tan(A\pm B)$ is obtained by dividing the formula for $\sin(A\pm B)$ by the one for $\cos(A\pm B)$, cancelling a factor of $\cos A\cos B$ from each to obtain the usual expression. As this is an illegal division by $0$ if $A=\frac{\pi}{2}$, it's easiest just to go back to the tangent's definition, viz.$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A sequence $(a_n)_{n\ge 1}$ such that $a_1>0$ and $a_{n+1}=a_n-\ln(1+a_n)$ Let $(a_n)_{n\ge 1}$ be a sequence such that $a_1>0$ and $$a_{n+1}=a_n-\ln(1+a_n)$$
a) Prove that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$ and $\lim\limits_{n\to \infty} (n^{2019}a_n)=0$.
b) If $a_1<2$, prove that $a_{12} \in (0,10^{-... | From
$ a_{n+1}
<\frac{a_n^2}{2}
$
we get
$ a_{n+2}
<\frac{a_{n+1}^2}{2}
<\frac{(\frac{a_{n}^2}{2})^2}{2}
=\frac{a_{n}^4}{8}
=\frac{a_{n}^4}{2^3}
$
and
$ a_{n+3}
<\frac{a_{n+2}^2}{2}
<\frac{(\frac{a_{n}^4}{2^3})^2}{2}
=\frac{a_{n}^8}{2^7}
$.
By induction
we can show that
$ a_{n+m}
<\frac{a_{n}^{2^m}}{2^{2^{m}-1}}
$.
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove if parallelogram law worked in a quadrilateral that shape is parallelogram Image with my solution
We have a quadrilateral ABCD. And we know: $$a^2+b^2+c^2+d^2=p^2+q^2$$ Prove this shape is parallelogram.
I write law of cosines between sides and diameters. Then I write sum of them and with the formula that the que... | Let $K$, $L$, $M$, $N$, $P$ and $Q$ be a midpoints of $AB,$ $BC,$ $CD,$ $DA$, $AC$ and $BD$ respectively.
Thus, since $PLQN,$ $PKQM$ and $KLMN$ are parallelograms, by your work we obtain:
$$PL^2+LQ^2+QN^2+NP^2=PQ^2+LN^2$$ or
$$AB^2+CD^2=2PQ^2+2LN^2.$$
Also, $$KP^2+PM^2+MQ^2+QK^2=PQ^2+KM^2$$ or
$$BC^2+AD^2=2PQ^2+2KM^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find all triples $(x,y,z)\in \Bbb{R}$ that satisfy the following conditions: The question is to find all real numbers solutions to the system of equations:
*
*$y=\Large\frac{4x^2}{4x^2+1}$,
*$z=\Large\frac{4y^2}{4y^2+1}$,
*$x=\Large\frac{4z^2}{4z^2+1}$,
This seems simple enough, so I tried substituting the v... | $\bullet\; $ Clearly $x=y=z=0$ are the solution of system of eqn
$\bullet\; $ If $x,y,z\neq 0\; $ Then
*
*$\displaystyle y=\frac{4x^2}{4x^2+1}\Longrightarrow \frac{1}{y}=1+\frac{1}{4x^2}$,
*$\displaystyle z=\frac{4y^2}{4y^2+1}\Longrightarrow \frac{1}{z}=1+\frac{1}{4y^2},$
*$\displaystyle x=\frac{4z^2}{4z^2+1}\Long... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How many ways to distribute $m$ distinguishable passengers in $n$ cars when $m>n$? I know there are versions of this question on here, but I'm looking to relate this concept to permutation. Let me explain: If we have $10$ seats and $7$ distinguishable people, and we want to find all ways $7$ people can sit on $10$ seat... |
In how many ways can ten distinct objects be placed in seven distinct boxes if no box is left empty?
Method 1: The number $10$ can be partitioned into seven parts in three ways.
\begin{align*}
10 & = 4 + 1 + 1 + 1 + 1 + 1 + 1\\
& = 3 + 2 + 1 + 1 + 1 + 1 + 1\\
& = 2 + 2 + 2 + 1 + 1 + 1 + 1
\end{align*}
Four obj... | {
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"url": "https://math.stackexchange.com/questions/3553624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve a differential equation $(x^2y+y^5)dx+(x^3-xy^4)dy=0$?
Solve the following differential equation:
$$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
I noticed that it is not exact, since:
$$
(x^2y+y^5)'_y=x^2+5y^4\ne3x^2-y^4=(x^3-xy^4)'_x
$$
Then I tried to express $y'$:
$$
y'=\frac{x^2y+y^5}{xy^4-x^3}
$$
And now I don't... | $$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
Rearrange terms:
$$y^4(ydx-xdy)+x^2(ydx+xdy)=0$$
Divide by $x^2$:
$$-y^4d\left (\frac y x \right ) +d(xy)=0$$
Divide by $(xy)^2$ and integrate:
$$\frac {y^2}{x^2}d\left (\frac y x \right )=\dfrac {d(xy)}{x^2y^2}$$
$$\frac {y^3}{x^3}=-\dfrac {3}{xy}+C$$
Finally, multiply by $x^3y$:
$$ \bo... | {
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"url": "https://math.stackexchange.com/questions/3553915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do I get the inverse of this equation? The equation is the following:
$$y = \frac{x}{x-2}$$
What I did is:
$$y = \frac{x}{x-2}$$
$$y \times (x-2) = x$$
$$\frac{x-2}{x} = \frac{1}{y}$$
$$\frac{x}{x}-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{x}{2} = y$$
I've got a lot of doubts about that ... | \begin{align}
y &= \frac{x}{x-2}
.
\end{align}
\begin{align}
y&=\frac{x+2-2}{x+2}
,\\
y&=1-\frac 2{x+2}
,\\
y-1&=-\frac 2{x+2}
,\\
-\frac{y-1}2&=\frac 1{x+2}
,\\
x+2 &=\frac 2{1-y}
,\\
x &=\frac 2{1-y}-2;
,\\
x &=\frac{2\,y}{1-y}
,
\end{align}
so the answer is indeed
\begin{align}
f^{-1}(x)=\frac{2x}{1-x}
.
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Let a,b and c be the roots of the equation $x^3-9x^2+11x-1=0$ and $s=\sqrt a +\sqrt b +\sqrt c$
Then Find the value of $s^4-18s^2-8s$
$$s=\sqrt a + \sqrt b +\sqrt c$$
$$s^2=a+b+c+2(\sqrt {ab} +\sqrt {bc} +\sqrt {ac})$$
I can’t seem to find a way around this obstacle. Squaring it again gives another $\sqrt {ab}+\sqrt ... | You should use the following properties of the roots of a cubic equation ax^3 + bx^2 + cx + d = 0:
a+b+c = -b/a = -(-9)/1 = 9
abc = -d/a = 1
and
ab + ac + bc = c/a = 11/1 =11
From these you can find:
(a+b+c)^2 = 81 = a^2+b^2+c^2 + 2 (ab + ac + bc) = a^2+b^2+c^2 + 2 * 11
=> a^2+b^2+c^2 = 81 - 22 = 59
Now expand s^2 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Greatest value of $xz$ in expression
Let $x,y,z,t\in\mathbb{R}$ and $x^2+y^2=9,z^2+t^2=4$ and $xt-yz=6$. Then greatest value of $xz$ is
What i try
I am trying to solve without trigonometric substution
$$(xt-yz)^2+(xz+yt)^2=(x^2+y^2)(z^2+t^2)$$
$6^2+(xz+yt)^2=4\cdot 9\Longrightarrow (xz+yt)^2=0$
$$xz=-yt$$
How do i s... | Notice that:
$$4(x^2+y^2) + 9(t^2 + z^2) = 12(xt - yz)$$
and this is equivalent with
$$(2x - 3t)^2+ (2y + 3z)^2 = 0$$
Thus $z=-\frac{2}{3}y$, and we want to maximize $-\frac{2}{3}xy$. This is simple, given that $x^2+y^2=9$ because:
$$0\leq (x+y)^2\Rightarrow -2xy \leq x^2+y^2=9$$
Therefore:
$$zx=-\frac{2}{3}xy\leq 3$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\binom{n}{0} + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n} = \frac{2^{n + 1} - 1}{n + 1}$. I have to prove the identity:
$$\binom{n}{0} + \dfrac{1}{2} \binom{n}{1} + \dfrac{1}{3} \binom{n}{2}
+ ... + \dfrac{1}{n+1} \binom{n}{n} = \dfrac{2^{n + 1} - 1}{n + 1}$$
First ... | Hint: Integrate the binomial theorem
$$(1+x)^n = \sum_{k=0}^n\binom nk x^k$$
from $x=0$ to $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}$ I need to calculate the following sum:
$$
\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}
$$
I've tried adding an imaginary part and differentiating:
$$
f(x) = \sum \limits_{k = 1}^{\infty} \frac{\cos 2xk + i \sin 2xk}{k^2} \\
f(x) = \sum \limits_{k = 1}^{... | Subtract $\sum_n\frac1{n^2}=\frac{\pi^2}6$ to obtain
\begin{eqnarray}
\sum_{n=1}^\infty\frac{\cos2n}{n^2}-\frac{\pi^2}6
&=&
\sum_{n=1}^\infty\frac{1-\cos2n}{n^2}
\\
&=&
-2\sum_{n=1}^\infty\left(\frac{\sin n}n\right)^2\;.
\end{eqnarray}
Now note that $\frac1\pi\frac{\sin n}n$ is the $n$-th coefficient in the Fourier ser... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $a, b \in \mathbb{R}$ such that the polynomial $f(x)=x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$ has a triple root. I am given the following polynomial:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$$
with $a, b \in \mathbb{R}$.
I have to find $a$ and $b$ such that the given polynomial has a triple root.
I know th... | Simplify the equation as
$$10 a^ 3 + 42 a^2 + 57a + 26= 0$$
which may be factorized as follows,
$$(10 a ^ 3 + 20a^2)+ (22 a^2 + 57a + 26 ) $$
$$= 10(a + 2) a^2+ (a + 2)(22a + 13 ) = (a + 2)(10^2+22a + 13 ) =0$$
which yields $a = -2$.
Alternatively, the cubic equation can always be solved with the Cardano's formula. Fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Numerical methods: forward difference operator $${ \text{If }y = \frac{1}{{(4 x + 1)} {(4 x + 5)} {(4 x + 9)}}, }\\ { \text{find } z = \Delta ^ {2} y .}$$
Can someone help with this? I tried it using factorial notation but apparently my answer is partially correct.
Any help would be much appreciated.
| By induction, for $a,b\in \mathbb{R}$ with $a\ne 0$ and denoting forward difference by $\Delta$, we have
$$
{\Delta}^n (ax+b)^{-1} = (-1)^n\cdot n! \cdot a^n \cdot \prod_{k=0}^{n} \left(ax+b +ak\right)^{-1}
$$So, your question can be rewritten as
$$
z = \Delta^2 y = \frac{1}{2!\cdot 4^2}\Delta^4 (4x+1)^{-1} = \frac{4!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stuck on Some Derivative in Derivation of Faà di Bruno's Formula I have no problem with the underlying idea of Faà di Bruno's Formula as generalization of composite function derivatives of single-variable functions. Where I am not sure that there is some recursive operation related to chain rule. Here I will directly d... | The derivation of $\frac{d^2 y}{dx^2}$ is sound.
We obtain
\begin{align*}
\color{blue}{\frac{d^2 y}{dx^2}}
&=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\
&=\frac{d}{dx}\left(\frac{dy}{du}\,\frac{du}{dx}\right)\\
&=\left(\frac{d}{dx}\left(\frac{dy}{du}\right)\right)\frac{du}{dx}+\frac{dy}{du}\,\left(\frac{d}{dx}\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I find the volume of this 4-dimensional ball?
*
*Let $B$ be the $4$-dimensional ball
$$
\left\{(x, y, z, w): x^2 + y^2 + z^2 + w^2 \leq R^2 \right\}
$$
in $\mathbb{R}^4$. I want to make the change of variables
$$
x=r \cos \phi, \quad y=r \sin \phi, \quad z=\rho \cos \theta, \quad w=\rho \sin \theta
$$
to comp... | I've taught this: the main thing is to do induction by two dimensions, using polar coordinates. So: we know the volume of the $n$ ball of radius $R$ is $c_n R^n.$ Next, the unit ball in dimension $n+2.$
So, on the unit disc in dimension 2, at a point $(r, \theta)$ we integrate the $n$ volume of a ball with radius $\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The $x$-coordinate of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$.
The $x$-coordinates of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$. If the tangents at $P$ and $Q$ meet at $T$, then find the distance of $T$ from the focus.
The point of intersection... | Note that
$$x_1+x_2=17,\>\>\>\>\>x_1x_2 =11$$
Then, the intersection coordinates of the two tangents are (GM of abscissa, AM of ordinate),
$$x=\sqrt{x_1x_2} = \sqrt{11}$$
$$y = \frac{y_1+y_2}2=\sqrt2(\sqrt{x_1}+\sqrt{x_2})=\sqrt2\sqrt{x_1+x_2+2\sqrt{x_1x_2} }
=\sqrt{34+4\sqrt{11}}$$
Since the focus of $y^2=8x$ is $(2,0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An interesting question of algebraic manipulation If $a+b+c+d+e=0$ then prove that $a^3+b^3+c^3+d^3+e^3=3(abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde)$ Extend this argument to n integers such that if $a_1+a_2+a_3+\cdots a_n=0$ then $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
My... | Instead of claiming it is "well known", actually prove it.
$0^3 = (a+b+c)^3 =$
$(a^3 + b^3 + c^3)+ 3(a^2b + a^2c + b^2a + b^2 c + c^2a + c^2b) + 6abc=$
$(a^3 + b^2 + c^3) +3((a^2b + a^2c + abc) + (b^2a+b^2c + bac) + (c^2a+c^2b + cab))- 3abc=$
$(a^3 + b^3 + c^2) + 3(a(ab+ac+bc) + b(ba + bc + ac) + c(ac+bc + ab))-3abc=$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$
We have that $$a^2 + 4b^2 + c... | The idea is indeed that of using Cauchy-Schwarz. The problem is you went from a homogeneous expression ($a^2+4b^2+c^2$) to non-homogeneous terms and that may be harder to prove than the original inequality.
Instead, I would apply Cauchy-Schwarz like this:
$$\frac{(a+b+c)^2}{a^2+4b^2+c^2} \leq \frac{a^2}{a^2+b^2}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Given $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ prove $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$ Given a sequence $(a_n)_{n\geq 1}$ such that $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ where $ n\geq1$
prove that $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$
I tried to prove that $a_n$ is increas... | So for given $a_n$, the next term $a_{n+1}$ is one of the roots of
$$X^2-(a_n+1)X-a_n, $$
i.e.,
$$\tag1\begin{align}a_{n+1}&=\frac{a_n+1\pm\sqrt{(a_n+1)^2+4a_n}}{2}\\&=\frac{a_n+1\pm\sqrt{(a_n+3)^2-8}}{2},\end{align}$$
which unfortunately gives us two choices. But there are some restrictions: As we (seem to) want a seq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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${2 x^{1-y}(1 - x^y) \over y (1-x)(1+x^{1-y})} \leq 1$ for $0The title says it all. I'd like to prove the inequality
$${2 x^{1-y}(1 - x^y) \over y (1-x)(1+x^{1-y})} \leq 1$$
for $0<x<1, 0<y<1$. I'd be happy just showing that the left hand side is bounded. Mathematica seems to believe that the inequality holds, and som... | Rewrite our inequality in the following form.
$$\frac{2x+y-xy}{2-y+xy}\geq x^{1-y}$$ or $f(x)\geq0,$ where
$$f(x)=\ln(2x+y-xy)-\ln(2-y+xy)-(1-y)\ln{x}.$$
But $$f'(x)=\frac{2-y}{2x+y-xy}-\frac{y}{2-y+xy}-\frac{1-y}{x}=$$
$$=-\frac{y(1-y)(2-y)(x-1)^2}{x(2-y+xy)(2x+y-xy)}<0.$$
Thus, $$f(x)>f(1)=0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem. I have been losing lots of points for m... | What you did cannot possibly work. Since, for any $x\in\mathbb R$,$$\left\lvert\frac1{2+\sqrt x}\right\rvert+\frac13\geqslant\frac13,$$ if $\varepsilon\in\left(0,\frac13\right)$, then there is no $\delta>0$ such that$$\lvert x-1\rvert<\delta\implies\left\lvert\frac1{2+\sqrt x}\right\rvert+\frac13<\varepsilon.$$Note tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there... | The book solution used the formulas for the sum and difference of two cubes,
$x^3+y^3=(x+y)(x^2-xy+y^2)$ and $x^3-y^3=(x-y)(x^2+xy+y^2),$
with $x=\sqrt5+2$ and $y=\sqrt5-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Can I solve this other than using Newton's sums or Newton's identities?
$\begin{cases}
&x + y + z = 12\\& x^2 + y^2 + z^2 = 12
\\ & x^3 +y^3 + z^3 = 12
\end{cases}$
If $x,$ $y,$ and $z$ satisfy the system of equations above, what is the value of $x^4+y^4+z^4?$
People told me that this can be solved by using Newton... | Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $$u=4,$$
$$3v^2=\frac{12^2-12}{2}=66,$$ which gives $$v^2=22.$$
Also, $$12=x^3+y^3+z^3=27u^3-27uv^2+3w^3,$$ which gives $$w^3=220$$ and use
$$x^4+y^4+z^4=81u^4-108u^2v^2+18v^4+12uw^3.$$
I got $1992.$
I used the known $uvw$'s substitutions: https://artofproblemsolving... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$. If $P_k=(\alpha)^k+(\beta)^k$, $k\ge 1$, then prove that-
a) $P_5=11$$
b) $P_1+P_2+P_3+P_4+P_5 =26$
For the first part
$$\alpha^5+\beta ^5$$
$$=(\alpha^3+\beta ^3)^2-2(\alpha \beta )^3$$
I found the value of $\alpha^3+\beta^3=4$
So $$16-2(-1)=18$$ which doesn’t m... | if $$x^2-x-1=0~~~(1)$$ has roots as $a,b$ then $P_k=a^k+b^k,P_0=2,P_1=a+b=1$ and
$$a^2-a-1=0~~~(2),~~ b^2-b-1=0~~~(3)$$
Multiply Eq.(2) once by $a^k$ and (3) by $b^k$. Adding these two Eqs. you get
$$(a^{k+2}+b^{k+2})-(a^{k+1}+b^{k+1})-(a^k+b^k)=0 \implies P_{k+2}= P_{k+1}+P_k~~~(4)$$
So $P_5=P_4+P_3, P_4=P_3+P_2, P_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Euclid's Lemma in the proof that $\mathbb{Z}[\sqrt{-5}]$ is not euclidean In proving, I induced that $3\gamma + (2+\sqrt{-5})\delta=1$. Textbook says that multiplying both sides by $2-\sqrt{-5}$ implies $2-\sqrt{-5}$ is multiple of 3. I don't know why. Can you explain the calculation.
| If you multiply both sides by $2 - \sqrt{-5}$ you get
$$ \begin{align} (2 - \sqrt{-5}) \cdot 3 \cdot \gamma + (2 - \sqrt{-5}) \cdot (2 + \sqrt{-5}) \delta &= 2 - \sqrt{-5} \\
(2 - \sqrt{-5}) \cdot 3 \cdot \gamma + 3 \cdot 3 \delta &= 2 - \sqrt{-5} \\
3\cdot ((2 - \sqrt{-5}) \cdot \gamma + 3 \delta) &= 2 - \sqrt{-5} \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to prove this inequality with $a,b,c\in [1,3]$ let $a,b,c\in [1,3]$,show that
$$3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{45}{a+b+c}\ge 16\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)$$
I had found this simaler problem,https://artofproblemsolving.com/community/c6h615194
I can't prove t... | $$3\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)+\frac{45}{a+b+c} ≥ 16\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)$$
Let:
$\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)=A$
$\frac{9}{a+b+c}=B$
$2\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)=C$
We use inequality:
$$C=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluate $ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx$
Evaluate
$$ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx,$$
My attempt : $u=\tan\frac{x}{2} \rightarrow x=2\arctan(u) \rightarrow \frac{2}{1+u^2}du=dx$
$$\sin x= \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{2u}{1+u^2... | Hint:
$$2\sin x+\cos x=\frac{4}{5}(\sin x+2\cos x)+\frac{3}{5}(2\sin x-\cos x)$$
This makes the integral $\displaystyle \int\left(\frac{4}{5}\mathrm dx-\frac{3}{5}\mathrm d(\ln|\sin x+2\cos x|)\right)$ with the same limits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$ Evaluate the following integral :
$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$
I was tried use change variable ,
If I use $x=y^2$ integral becomes :
$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}... | Let
$$I(a)=\int_0^{\infty}\frac{\ln(a^8+x^4)}{\sqrt x(1+x)}dx=2\int_0^{\infty}\frac{\ln(a^8+x^8)}{1+x^2}$$
Then
$$I(0)=2\int_0^{\infty}\frac{\ln(x^8)}{1+x^2}dx=2\int_{\infty}^0\frac{\ln(y^{-8})}{1+\frac1{y^2}}\left(-\frac{dy}{y^2}\right)=-2\int_0^{\infty}\frac{\ln(y^8)}{1+y^2}dy=-I(0)=0$$
And
$$\begin{align}I^{\prime}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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If $z=\cos \theta + i \sin \theta$, express $\displaystyle \frac {1}{1-z \cos \theta}$ in the form $a+i\cdot b$. In order to get rid of the $z$, should I substitute $z=\cos\theta + i\sin \theta$ into the complex number, or what conjugate should I multiply the complex number by?
I have tried substituting $z=\cos\theta +... | Note that $z=e^{i\theta}$, $z=e^{-i\theta}$ and $\cos\theta = \frac{z+\bar z}2$
$$\frac {1}{1-z \cos \theta}=\frac {1}{1-z \cos \theta}\cdot\frac {1-\bar z \cos \theta}{1-\bar z \cos \theta}$$
$$=\frac {1-e^{-i\theta}\cos \theta}{1-(\bar z +z)\cos \theta + \cos^2\theta}$$
$$=\frac {1-(\cos\theta - i\sin\theta)\cos \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\sum_{n=0}^{\infty} \frac{n! a^n}{(n+1)^{n+1}}<\pi \sqrt{\frac{2}{e(e-a)}}$ Show $$\sum_{n=0}^{\infty} \frac{n! a^n}{(n+1)^{n+1}}<\pi \sqrt{\frac{2}{e(e-a)}}$$ where $a<e$.
This problem comes from the original version:
$$\int_0^{\infty} \frac{{\rm d}x}{e^x-ax}<\pi \sqrt{\frac{2}{e(e-a)}}(a<e),$$ which can deduce ... | A much tighter bound will be derived:
$$(1) \quad S(a):=\sum_{n=0}^\infty \frac{n! \, a^n}{(n+1)^{n+1}} \le \frac{\sqrt{2}\pi}{a}
\Big(\frac{1}{\sqrt{1-a/e}} - g(a)\Big) \quad ,$$
where
$$ g(a)=1 + 2 c \log \big(\frac{1+\sqrt{1-a/e}}{2}\big) , \quad c= \frac{\sqrt{2}\, e}{\pi} - 1$$
The bound of the OP follows from set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that this function is continuous at $(0,0)$ I need to prove that the function is continuous at $(0,0)$.
$$f(x,y)=\frac{2x^2y^2}{x^2+y^2}, \text{ if } (x,y)\neq(0,0)$$
and $0 \text{ if } (x,y)=(0,0)$
I'm trying to prove that $$\lim_{(x,y)\to (0,0)}\frac{2x^2y^2}{x^2+y^2}=0$$
but I don't know how to relate it to ... | Since $x^2,y^2\leqslant x^2+y^2$,$$0\leqslant\frac{2x^2y^2}{x^2+y^2}\leqslant2\frac{(x^2+y^2)^2}{x^2+y^2}=2(x^2+y^2).\tag1$$So, by the squeeze theorem,$$\lim_{(x,y)\to(0,0)}\frac{2x^2y^2}{x^2+y^2}=0.$$
Or, given $\varepsilon>0$, you can take $\delta=\sqrt{\frac\varepsilon2}$. If $\sqrt{x^2+y^2}<\delta$, then, by $(1)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Is it possible to find where two curves have a common value? Suppose I have;
$$f(x) = x^2 - 42x + 364$$
$$g(y) = y^2 - 35y + 364$$
Computing out values I find they have a common value
$$f(6) = g(8) = g(27) = f(36) = 148$$
Is there a way of finding these values?
Working through based on received help...
(1) $$f(x) = g... | As the abovementioned 2 functions are continuous, you can find infinitely many common values in the intersection of their ranges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculate $\int_{x^2+y^2+z^2
Calculate the integral: $$\int_{x^2+y^2+z^2<x+y+z} (x^2+y^2+z^2)dxdydz$$
Of course calculating the integral $\int_{a}^{b}\int_{c}^{d}\int_{e}^{f} (x^2+y^2+z^2)dxdydz$ would be very easy. However I have a problem because I must find $a,b,c,d,e,f$. I tried to use substitution: $$\begin{cases}... | Note that \begin{align*} x^2 + y^2 + z^2 < x+y+z \,\,\,\,\, &\Longleftrightarrow \,\,\,\,\, (x^2-x) + (y^2-y) + (z^2-z) < 0\\
&\Longleftrightarrow \,\,\,\,\, (x-1/2)^2 + (y-1/2)^2 + (z-1/2)^2 < 3/4.
\end{align*} Thus you are integrating over a sphere centered not at the origin, but at $(1/2,1/2,1/2)$ which should infor... | {
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Lagrange Multipliers - Solving system of equation $$\begin{array}{ll} \text{extremize} & \overbrace{xy + xz + yz}^{=: f(x,y,z)}\\ \text{subject to} & x^2 + y^2 + z^2 = 1\end{array}$$
I get the following Lagrange system of equations
$y + z - 2x\lambda = 0\qquad(1)$
$x + z - 2y\lambda = 0\qquad(2)$
$x + y - 2z\lambda = ... | $$1-(xy+xz+yz)=\sum_{cyc}(x^2-xy)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0,$$
which says that $$xy+xz+yz\leq1.$$
The equality occurs for $x=y=z=\frac{1}{\sqrt3},$ which says that we got a maximal value.
Also, (it's the case, for which you look) $$\frac{1}{2}+xy+xz+yz=\frac{1}{2}\sum_{cyc}(x^2+2xy)=\frac{1}{2}(x+y+z)^2\geq0,$... | {
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How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T
\begin{pmatrix}
10 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25 \\
\en... | Call the matrix $$C'=\begin{bmatrix}\frac{35}{2}&\frac{5\sqrt 3}{2}\\\frac{5\sqrt 3}{2}&\frac{25}{2}\end{bmatrix}=\begin{bmatrix}a&b\\b&d\end{bmatrix}$$ Apply the 'diagonalization of a symmetric matrix by congruence' formula (which comes from 'completing the square' in the corresponding quadratic form). If $$E=\begin{b... | {
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On bounds for the deficiency of $m^2$, where $p^k m^2$ is an odd perfect number with special prime $p$ Hereinafter, call a number $N$ perfect if $N$ satisfies $\sigma(N)=2N$, where
$$\sigma(x)=\sum_{d \mid x}{d}$$
is the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/... | We can get a better bound.
To get a better bound, we need a better inequality than $\sigma(x)-x\ge 1$.
So, let us find a better inequality on $\sigma(m^2)$.
To find a better lower bound, let us consider $m$ of the form $PQ$ where $P\lt Q$ are distinct primes.
Then, we have
$$\begin{align}\sigma(m^2)&\ge (1+P+P^2)(1+Q+... | {
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Problem with polynomial divisilibity proof.
Show that for any natural $k$, $m$, $n$ polynomial $x^{3k}+x^{3m+1}+x^{3m+2}$ is divisible by $x^2+x+1$.
My observation:
$$x^a=(x^{a-2}-x^{a-3})(x^2+x+1)+x^{a-3}$$
so,
$$x^{3k}+x^{3m+1}+x^{3m+2}=(x^{3k-2}-x^{3k-3})(x^2+x+1)+x^{3k-3}+(x^{3m-1}-x^{3m-2})(x^2+x+1)+x^{3m-2}+(x^... | Observe that $q(x)=x^2+x+1$ divides $p(x)=x^{3k}+x^{3m+1}+x^{3n+2}$ if and only if the roots of $q(x)$ are roots of $p(x)$.
Since $x^3-1 = (x-1)(x^2+x+1)$, the roots of $q(x)$ are $\alpha$ and $\beta$ and they are the primitive cubic roots of unity and obviously the following holds:
\begin{gather}
\alpha^3 = \beta^3 =1... | {
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Find delta for the limit I'm having difficulty to solve this problem:
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$.
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displayst... | You should proceed with your approach a little further. Note that in the last step the denominator is not less than $1/\sqrt{2}$ which itself exceeds $1/2$ and therefore $$\left|\dfrac{\dfrac{1}{1+x^2}-\dfrac{1}{2}}{\dfrac{1}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{2}}}\right|<2 \left|\frac{1}{1+x^2}-\frac{1}{2}\right|=\frac{|1-... | {
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Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
I used the fact that
$$(A+B+C)^2=A^2+AB+AC+BA+B^2+BC+CA+CB+C^2$$
and multiply this with $(A+B+C)$ and I got
$$A^3+A^2B+A^2C
+ABA+AB^2+ABC
+ACA+ACB+AC^2
+BA^2+BAB+BAC
+B^2A+B^3+B^2C
+BCA+BCB+BC^2
+CA^2+CAB+CAC
+... | You cannot simplify the full expansion, which has $27$ terms, because matrix multiplication is non-commutative ($AB\ne BA$ in general).
| {
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Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, ... | Just another way is to note for $t \in (0, 1)$,
$$\frac{1+t^2}{1-t^2} \geqslant \frac{27t+11}{16} \iff (3t-1)^2(3t+5)\geqslant 0$$
Hence
$$\sum_{cyc} \frac{1+a^2}{1-a^2}\geqslant \frac1{16}\sum_{cyc}(27a+11) = \frac{15}4 $$
| {
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Prove that $DE\perp EF$.
Question: Point $D$ lies inside $\Delta ABC$ such that $\angle DAC=\angle DCA=30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE\perp EF$.
My approach: Let $\angle CDF=\alpha$ and $\angle EDC=\beta$.
... | Let $AD=DC=p$ and $\measuredangle BAD=\alpha$.
Thus, $$AC=p\sqrt3,$$
$\frac{BD}{\sin\alpha}=\frac{p}{\sin60^{\circ}},$ which gives $$BD=\frac{2p\sin\alpha}{\sqrt3},$$ $\frac{AB}{\sin(60^{\circ}+\alpha)}=\frac{p}{\sin60^{\circ}},$ which gives $$AB=\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}$$ and
$$\vec{DE}\cdot\vec{FE}=\... | {
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Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$. My attempt:
Let $x, y$ be roots of
$$t^2-5t+p=0$$
I got
$$p=\frac{5 \pm \sqrt{-431}}{2}$$
using Vieta's relations.
Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$.
But I got two very hideous complex values of x and y. So I think my answ... | $$x^3+y^3=(x+y)^3-3xy(x+y)=5^3-3xy(5)$$
$$x^2+y^2=(x+y)^2-2xy=5^2-2xy$$
$$(x^3+y^3)(x^2+y^2)=x^5+y^5+x^2y^2(x+y)$$
$$\iff(125-15xy)(25-2xy)=275+5x^2y^2$$
$$\iff(xy)^2-25xy+114=0$$
For the rest, follow How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$?
| {
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What is the solution of the following definite integral? I encountered this integral in my calculations (picture here):
$$
\int_{f=0}^{f=\pi/2} \frac
{\left(n_0 \sqrt{1 - \left(\frac{n_0}{n_1} \sin f \right)^2} - n_1 \cos f\right)^2}
{\left(n_0 \sqrt{1 - \left(\frac{n_0}{n_1} \sin f \right)^2} + n_1 \cos f\right)^2}
2... | This seems to be the Lambertian reflectance for parallelly polarised light passing from medium 0 to medium 1, so let us introduce $n = \frac{n_1}{n_0} > 1$ and write your integral as
\begin{align}
r_\text{p} (n) &= \int \limits_0^{\pi/2} \left(\frac{n^2 \cos(f) - \sqrt{n^2 - \sin^2(f)}}{n^2 \cos(f) + \sqrt{n^2-\sin^2(f... | {
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Expansion of $(a+b+c+d+e+....)^n$, but with all coefficients equal to 1. I'm looking for a formula to calculate the sum of $(a+b+c+d+...)^n$ but with coefficients equal to 1.
For example in $(a+b+c)^2$. I want the sum of $a^2 + b^2 + c^2 + ab + bc + ca$. And for $(a+b+c+d)^3$,
I want the sum of $a^3 + b^3 + c^3 + a^2b ... | For any $k$ variables $x_1, \ldots, x_k$, let $P_n(x_1,\ldots,x_k)$ be the sum of all monomials over them for degree $n$.
$$P_n(x_1,\ldots,x_k) \stackrel{def}{=} \sum_{\sum_{j=1}^{k} e_j = n} \prod_{j=1}^k x_j^{e_j}, \quad\text{where}\quad e_1,\ldots,e_k \in \mathbb{N}$$
Multiply by $s^n$ and sum over $n$, we find
$$\b... | {
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sum of power floor $ S(n,m) = \sum_{i = 1}^{n} \Bigl \lfloor \frac{n}{i^m} \Bigr \rfloor i^m $ How do I calculate
$$ S(n,m) = \sum_{i = 1}^{n} \Bigl \lfloor \frac{n}{i^m} \Bigr \rfloor i^m $$
This can be simplifies to
$$ S(n, m) = \sum_{i = 1}^{\sqrt[m] n} \Bigl \lfloor \frac{n}{i^m} \Bigr \rfloor i^m = \sum_{i = 1}^{... | Hint
Using the division algorithm, we can write $a=kq+r$, where $0 \leq r < k$. Then,
$$\frac{a}{k}=q+\frac{r}{k} \implies q=\lfloor\frac{a}{k}\rfloor.$$
So
$$a=k\lfloor\frac{a}{k}\rfloor+r \implies \lfloor\frac{a}{k}\rfloor k=a-r.$$
| {
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If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4... | $$0 = x^2 + 4 y^2 + 16 z^2 - 2 (4yz + 2zx + xy) = (x-y-2z)^2 + 3 (y-2z)^2 $$
So, as noted $y = 2z$ and $x = y+2z = 4z,$ so
$$ 48 = x^2 + 4 y^2 + 16 z^2 = 16 z^2 + 16 z^2 + 16 z^2 = 48 z^2 $$
and $z = \pm 1.$
| {
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A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.... | Claim 1: $2.3<\ln 10.$
Claim 2: $\ln 1.7<8/15$
Both these claims can be proven easily via Taylor series, etc.
Now, using the above inequalities, we have $1.7^{69}<e^{69\cdot \frac{8}{15}}<10^{16},$ or, multiplying $10^{69}$ on both sides, $17^{69}<10^{85}.$
| {
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System of congruences where $\gcd(m, n)\ne1$ I have to solve this system of congruences:
$$
\begin{cases}
x^2+2x+2\equiv 0\pmod{10}\\
7x\equiv 20\pmod{22}
\end{cases}
$$
after some calculations
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cas... | \begin{cases}
x\equiv 0\;(mod\;2)\\
\hline
x\equiv 1\;(mod\;5)\\
x\equiv 2\;(mod\;5)\\
\hline
x\equiv 6\;(mod\;11)\\
\end{cases}
So you want a solution modulo $2 \cdot 5 \cdot 11 = 110$.
This is how I would solve it.
\begin{array}{r|rrr}
& 2 & 5 & 11 \\
\hline
55 & 1 & 0 & 0\\
22 & 0 & 2 & 0\\
10 & 0 & 0 & -1 \\
\hl... | {
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Mod arithmetic and divisibility The question comes from a textbook
Prove that $3x^5+5x^3+7x$ is divisible by 15 for any integer $x$. I want to do this through mod arithmetic.
The first thing that I did was to try and solve it through mod 3 and 5. If they are congruent to 0 for both, then it is divisible by 15.
So I t... | Note since $5 \equiv 0 \pmod{5}$, then $5x^3 \equiv 0 \pmod{5}$, but in your statement of
$3x^5+5x^3+7x$ is congruent to $3x^5+x^3+2x$
you have that $5x^3 \equiv x^3 \pmod{5}$ instead (but you did correctly simplify $7x \equiv 2x \pmod{5}$). This is only true for $x \equiv 0 \pmod{5}$. Also, it's why your substitutio... | {
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Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$ I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities.
For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^... | $$f(x)=(x^2-(a^2+b^2))^2+(b^2-a^2)^2-(a^2+b^2)^2\ge(b^2-a^2)^2-(a^2+b^2)^2=-4a^2b^2$$
The equality occurs for $x^2=a^2+b^2$, which is always possible.
| {
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"source": "stackexchange",
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Quartic polynomials of a complex variable I want to answer the following:
The equation
$$
a \bigg(z + \frac{1}{z}\bigg)^2
+ b \bigg(z + \frac{1}{z}\bigg)
+ c = 0
$$
has four solutions. Find which quartics can be put in this form after apply a linear change of vars.
| $a(z+\frac{1}{z})^2+b(z+\frac{1}{z})+c = 0$
This is a quartic polynomial, if we multipy by $z^2$ we obtain
$az^4+bz^3+(2a+c)z^2+bz+a = 0$
This quartic can be related to the general quartic polynomial
$f(x) = Ax^4+Bx^3+Cx^2+Dx+E$
Now compare the coefficient of the two polynomials
$az^4+bz^3+(2a+c)z^2+bz+a$ and $Ax^4+Bx^... | {
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Least $x$ Such That $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$ A friend recently texted me the following:
Compute the least $x$ such that $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$.
Is there a way to do what my friend is asking analytically? I graphed it on Desmos, and got $x\approx 5.91608$. I realized that thi... | $$x = I+ F \qquad (I \in \mathbb Z, \quad 0 \le F < 1).$$
$$\lfloor 2IF+F^2\rfloor =10 \tag{A.}$$
A quick check shows that $I$ needs to be positive.
Let's first consider what happens when
$$F^2 + 2IF = 10 \tag{B.}$$
By the quadratic equation,
$$F = \dfrac{-2I+\sqrt{4I^2+40}}{2}
= -I+\sqrt{I^2+10}$$
This clearly ... | {
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Is the number of nonzero digits always at least $m$? if $k,m$ is give postive integers,for any postive integer $p$, if we define
$$p\cdot\dfrac{k^m-1}{k-1}=a_{i}k^{i}+a_{i-1}k^{i-1}+\cdots+a_{1}\cdot k+a_{0}$$
where $a_{i}\in\{0,1,2,\cdots,k-1\}$,and let set $A=\{i|a_{i}\neq 0\}$,show that
$$|A|\ge m$$
this problem i... | Notice that you actually claim that the number:
$$
N=N(p,k,m)=p\cdot\dfrac{k^m-1}{k-1}=p(k^0+k^1+k^2+\dots+k^{m-1})
$$
Has at least $m$ non-zero digits in number base $k$, where $a_{i}\in\{0,1,2,\cdots,k-1\}$ are its digits.
To prove the claim for $p\lt k$ is easy. Notice that for this case, we have:
$$
N=\sum_{i=0}^{... | {
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How to analyze the equation $(x-y)^2=2\big( (x+y)-2\sqrt{xy} \big)$? Suppose that $x,y$ are positive real numbers and that
$$ (x-y)^2=2\big( (x+y)-2\sqrt{xy} \big). \tag{*}$$
Then Mathematica claims that one of the following $3$ options holds:
$$1. \, \, \, x=y.$$
$$2. \, \, \, x = y - 2 \sqrt 2 \sqrt y + 2.$$
$$3. \,... | Factorise to $(x - y)^2 = 2 ( \sqrt{x} - \sqrt{y})^2$. Which in turn gives $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = \pm \sqrt{2} (\sqrt{x} - \sqrt{y})$. So $x=y$ or $\sqrt{x} = \pm \sqrt{2} - \sqrt{y}$.
| {
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In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on bo... | Here's a couple of different methods:
Proof using the cosine angel sum formula:$$
1=\cos(0) =\cos(x + (-x)) =\cos(x)\cos(-x) - \sin(x)\sin(-x) = \cos^2(x) + \sin^2(x)
$$
Proof using differential equations:
Recall the differential equation definitions of $\sin$ and $\cos$: They are the solutions to $f'' = - f$ with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha... | Note
$$\alpha (\alpha^2+2\alpha -4)\beta =-1$$
Thus,
$$\beta =- \frac1{\alpha (\alpha^2+2\alpha -4)}
=- \frac1{\alpha^3+2\alpha^2-4\alpha}
=- \frac1{5\alpha -1 -4\alpha} =\frac1{1-\alpha}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find (ab) if $ \int \frac{15x^2dx}{\sqrt{2x-1}}=\left(ax^2\:+bx\:+b\right)\sqrt{2x-1}+C $ $$
\int \frac{15x^2dx}{\sqrt{2x-1}}=\left(ax^2\:+bx\:+b\right)\sqrt{2x-1}+C
$$
What is the answer written as integer
$$
225\sqrt{2x-1}+C
$$
(excuse my English)
| \begin{eqnarray*}
\int{\frac{15 x^{2}}{\sqrt{2 x-1}} dx} &=& \left(2+2 x+ 3 x^{2} \right) \sqrt{2 x -1} + C
\end{eqnarray*}
which means, $a=3, b=2$ and hence $ab=6$. Is this that you are seeking here?
Note: $y^{2} = 2 x-1$, which means $2 y d y = 2 dx $ or $ \frac{dx}{y}=dy$. The problem then translates to,
\begin{eqna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646839",
"timestamp": "2023-03-29T00:00:00",
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Showing if all eigenvalues of $A$ have negative real parts then our system has a strong Lyapunov function of the form $x^TSx$. Can I please have help solving the problem? I am having a tough time working out the details for $S$ and how to do it without assuming diagonability. Thank you!
Show that if all eigenvalues of ... | Assume all eigenvalues of $A$ have negative real parts. Consider $x'=Ax,$ $$L=x^TSx \text{ where } S=S^T \text{ and } x\ne 0.$$ We must show that $\frac{dL}{dt} < 0$ so $$\frac{dL}{dt} = (x^T)'Sx + x^TSx' = (x')^TSx + x^TSx' = x^TA^TSx + x^TSAx$$ $$=x^T(A^TS+SA)x.$$ Since we know that $A^TS+SA = -I$ so $$\frac{dL}{dt}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Piecewise Laplace transformation The piecewise function is:
$$
\begin{array}{cc}
&
\begin{array}{cc}
t & 0\leq t< 1 \\
2-t & 1\leq x\leq 2 \\
0 & t>2
\end{array}
\end{array}
$$
Now putting the piecewise into LaPlace form:
$$\int_0^1te^{-st}dt +2\int_1^2e^{-st}dt-\int_1^2te^{-st}dt+0$$
The ... | I believe you've made a few small sign slips. Your starting point is correct,
$$\tilde f(s) = \int_0^1 t e^{-st} + \int_1^2 (2-t) e^{-st} \; dt$$
If we're not careful, we'll get far too many terms (like your final answer, which is difficult to figure out whether it's correct), so let's evaluate
\begin{align}
I(t) =\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to find the particular integral? The question is to find the particular integral of the expression
$$ (D^3+1)y = cos(2x-1) $$
I’m not sure how to go about it. How do I find the PI for this expression?
| $\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$This is an interesting differential equation because it yields some really nice graphs when some initial $y(0),y'(0),y''(0)$ are sampled.
As for the solution, first solve the homogenous problem:
$$y''' + y = 0 \xRightarrow{y = e^\lambda} \lambda^3 e^\lambda + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt... | It is already said everything about the solution, but i was curious to see if there is any "straightforward" way to proceed in this and similar cases by working in the appropriate quadratic field, here in $K=\Bbb Q(s)$, with $s=\sqrt {17}$ (with class number one) and asking the computer, here sage, for factorizations o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\frac{x^2 -4}{x-4} \rightarrow -5$ as $x \rightarrow 3$, proving the limit from first principles. ($\delta - \epsilon$). This is a follow up to a previous post. I am sorry of this is considered spamming but I have made significant progress of the question since then. Hopefully I'm on the right track, an tips or advice... | Your choice for $\delta$ does not work.
Let $\varepsilon = \frac{1}{4}$.
Then your formula gives $\delta = \frac{1}{2}$.
If $x = 3.4$ then
$$\quad \bigg|(x-3) \; \frac{x+8}{x-4}\bigg| = 7.6$$
If $x \in [2\frac{1}{2},3\frac{1}{2}]$ then
$$\quad |x+8| \le 12 $$
If $x \in [2\frac{1}{2},3\frac{1}{2}]$ then
$$\quad |x-4| \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$ I have first written $x^4-2x^2+2=(x^2-1)^2+1$ then I choose $t=x^2-1$. From this I got $$\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$$
From this how to proceed?
| HINT:
$$\frac{\text{d}}{\text{dn}}\left(\int_0^\infty\frac{\ln\left(1+\text{n}x^2\right)}{\sqrt{x}\left(1+x\right)}\space\text{d}x\right)=\int_0^\infty\frac{x^\frac{3}{2}}{\left(1+\text{n}x^2\right)\left(1+x\right)}\space\text{d}x=$$
$$\frac{\pi}{2\left(1+\text{n}\right)}\left(2-\frac{\sqrt{2}\left(\sqrt{\text{n}}-1\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$ Show that if $0 \leq x \leq a$ and $n \in \mathbb{N}$, then $$1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a... | By Taylor's theorem,
$$
e^x = 1 + x + \frac{{x^2 }}{{2!}} + \cdots + \frac{{x^{n - 1} }}{{(n - 1)!}} + \frac{{x^n }}{{n!}}e^c
$$
for all $x\geq 0$ with a suitable $0\leq c\leq x$. Now just note that if $0\leq x\leq a$ then
$$
1 \leq e^c \leq e^x \leq e^a .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Vieta's formula for $f(\frac{1}{x})$
Let $f(x)$ be a monic cubic polynomial. Let the solutions of the equation $f(\frac{1}{x})=0$ be $\alpha, \beta$ and $\gamma$. If $\alpha+\beta+\gamma = 10$ and $\alpha\beta\gamma=15$, what is $\left \lceil{f(10)-f(-10)}\right \rceil$?
So since $f(x)$ is a monic cubic we can write ... | First notice that:
\begin{gather}
f(10)-f(-10) = 10^3+10^2b+10c+d - (-10)^3 - (-10)^2b-(-10)c -d =\\
=2000+20c
\end{gather}
So if we find the value of $c$ we are done.
As you noticed $f\left(\frac{1}{x}\right) = \frac{1}{x^3}(dx^3+cx^2+bx+1) = \frac{1}{dx^3}\left(x^3+\frac{c}{d}x^2+\frac{b}{d}x+\frac{1}{d}\right)$ wher... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\r... | $$y=\dfrac{1+x}{\sqrt{1-x}}=\dfrac{2-(1-x)}{\sqrt{1-x}}=2(1-x)^{-1/2}-(1-x)^{+1/2}$$
Now use this by writing $$(1-x)^m=(-1)^m(x-1)^m$$
$$\dfrac{d^n(x-1)^m}{dx^n}=m^n(x-1)^{m-n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Fractions in Questions and Answers
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