Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$ Yes, I know this is the equation of an ellipse.
This is the starting point of the problem:
$$
r_1+r_2=2a,
$$
where
$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$
I want to establish certain properties prior to deriv... | There is a very simple and short answer. Using the definition of monotonicity given in Fundamentals of Mathematics, Volume 1: Foundations of Mathematics: The Real Number System and Algebra Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle, which is a monotonic function is such that
$$x<y\implies f\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$x,y$ and $z$ are consecutive integers, $\frac {1}{x}+\frac {1}{y}+\frac {1}{z}$... $x,y$ and $z$ are consecutive integers, such that $\frac {1}{x}+ \frac {1}{y}+ \frac {1}{z} \gt \frac {1}{45} $, what is the biggest value of $x+y+z$ ?.
I assumed that $x$ was the smallest number so that I could express the other number... | Write $x = y-1$ and $z = y+1$. Then we have $$\frac{1}{y-1} + \frac{1}{y} + \frac{1}{y+1} = \frac{2y}{y^2 - 1} + \frac{1}{y} = \frac{3y^2 - 1}{y^3 - y} > \frac{1}{45}$$ Since $3y^2 - 1< 3y^2$, this implies $$\frac{3y}{y^2 - 1} = \frac{3y^2}{y^3 - y} > \frac{1}{45} \iff 135y > y^2- 1 \iff y-\frac{1}{y} < 135$$ As $y$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all possible Jordan forms Let $A \in M_3(\mathbb{C})$, and:
$\frac1{12}A=[A^2-7A+16I_3]^{-1}$
Find all possible Jordan Forms of A
(no need to show different block orders)
I was thinking of Algebraic manipulations such as:
$12A^{-1}=A^2-7A+16I_3 \space/*A$
$12I_3=A(A^2-7A+16I_3)\space/*A$
$A(A^2-7A+16I_3)-12I_3=0 \... | You obtained $A^3-7A^2+16A-12I_3=0$ therefore the minimal polynomial $m_A$ divides the polynomial $x^3-7x^2+16x-12=(x-2)^2(x-3)$.
The options for the minimal polynomial and respective Jordan forms are:
*
*$m_A(x) = x-2$
$$\pmatrix{2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}$$
*$m_A(x) = (x-2)^2$
$$\pmatrix{2 & 1 & 0 \\ 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve the recurrence $T(n) = T(n-1) + 2T(n-2) + 2$ $T(0) = 1,
T(1) = 0$
I ain't able to get answer from any of the methods.
Substitution:
$T(n) = x^n $
\begin{align}
& x^n = x^{n-1} + 2x^{n-2} + 2 \\
& x^2 = x + 2 + 2x^2\\
& x^2 + x + 2 =0
\end{align}
solving this I will get a complex root.
\begin{align}
x & = \frac{-... | Hint try to make it homogeneous, from
$$T(n) = T(n-1) + 2T(n-2) + 2$$
$$T(n+1) = T(n) + 2T(n-1) + 2$$
we have
$$T(n)-T(n+1)=T(n-1) + 2T(n-2) - T(n) - 2T(n-1) \iff \\
T(n+1)-2T(n)-T(n-1)+2T(n-2)=0$$
leading to characteristic polynomial
$$x^3-2x^2-x+2=0 \iff (x-2)(x-1)(x+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequalities with absolute values My question is:
Show that for all $|x-1|+|x-2|+\dots+|x-10| > 23$
I have solved above problem as below,
If $x-1 > 0$ and $x-2 > 0$ and ......$x-10 > 0$ then
LHS $= x-1+x-2+x-3+\dots+x-10 = 10x-55 > 23$ (because $x>10$)
If $x-1<0$ and $x-2 < 0$ and ......$x-10 < 0$ then
LHS $= -x+1-x+... | Just use triangle inequality:
$$|x-1|+|x-2|+\dots+|x-10| = |1-x|+|2-x|+\dots+|5-x|+ |x-6|+|x-7|\dots+|x-10| >$$ $$|1-x+2-x+...+x-10| = |15-6-7-8-9-10| = |-25|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Specific case of Sherman–Morrison formula I want to find an easy solution to a matrix inverse for a very special case.
My matrix is of the form: $$D+vv^{T}$$
I know that I can use Sherman–Morrison formula to get
$$(D+vv^{T})^{-1}=D^{-1}-{D^{-1} vv^{T} D^{-1} \over 1+v^{T}D^{-1}v}$$
My D is actually of a very simple dia... | \begin{align}
\frac1{1-s^2}\cdot I-\frac{s^2}{({1-s^2)(1+s^2(n-1))}}\cdot\mathbf{1}
,
\end{align}
that is, all diagonal elements are
\begin{align}
\frac{1+s^2(n-2)}{(1-s^2)(1+s^2(n-1))}
,
\end{align}
and all the other elements are
\begin{align}
-\frac{s^2}{(1-s^2)(1+s^2(n-1))}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
General solution of the ordinary differential equation $(D^4+D^2+1)y=0$ $(D^4+D^2+1)y=0$ where $D=\frac{d}{dx}$
$$D^4+D^2+1 =0 \Rightarrow D^4 + D^2 + \frac14 +\frac{{3}}{4} = 0 \Rightarrow (D^2 + \frac12) = \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D^2 = \frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D= \pm \sqrt{... | The roots of $\lambda^4+\lambda^2+1$ are $$ \lambda_1=\frac{1}{2} + i\frac{\sqrt{3}}{2}\\ \lambda_2=\frac{1}{2} - i\frac{\sqrt{3}}{2}\\ \lambda_3=-\frac{1}{2} + i\frac{\sqrt{3}}{2}\\ \lambda_4=-\frac{1}{2} - i\frac{\sqrt{3}}{2} $$
Thus, the solution to the ODE is
$$ y(x) = \sum_{k=1}^4 a_k\exp(x\lambda_k) $$
If we wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m... | This is another way of looking at @jwg's answer.
Let the four consecutive numbers be $a,b,c,d$ and let $t$ be the number half-way between $b$ and $c$. Then
\begin{align}
abcd + 1
&= \bigg(t-\frac 32\bigg)\bigg(t-\frac 12\bigg)
\bigg(t+\frac 12\bigg)\bigg(t+\frac 32\bigg) + 1\\
&= \bigg(t^2-\frac 94\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 12
} |
Convergence of sum of reciprocals of binomial coefficients I am interested in evaluating the sums
$$
\sum_{k=1}^{\infty}{\binom {2k}{k}^{-n}},
$$
where $n$ is a positive integer. It is already known that for $n=1$ we have
$$
\sum_{k=1}^{\infty}\frac{1}{\binom {2k}{k}}=\frac{9+2\sqrt 3 \pi}{27}.
$$
Several papers analyz... | I think that, for $n>1$, you are entering in the world of hypergeometric functions.
Just have a look to the table and notice the patterns
$$\left(
\begin{array}{cc}
n & S_n \\
2 & \frac{1}{4} \, _3F_2\left(1,2,2;\frac{3}{2},\frac{3}{2};\frac{1}{16}\right) \\
3 & \frac{1}{8} \,
_4F_3\left(1,2,2,2;\frac{3}{2},\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization... | There is a matrix $S$ such that
$$
J=S\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}S^{-1}
$$
If you set $Y=S^{-1}XS$, then $X=SYS^{-1}$ and the equation becomes
$$
S(Y^2+Y)S^{-1}=S\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}S^{-1}
$$
that simplifies to
$$
Y^2+Y=\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}
$$
We can try ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
On the property $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$ for complex numbers Suppose we have complex numbers, $x$ and $y$. My question is: we know, from the triangle inequality, that $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$. It makes sense, geometrical... | Let $x=a+bi, y=c+di,$ where $a,b,c,d \in \mathbb{R}.$ Then $$|x+y|=|(a+c)+(b+d)i|=\sqrt{(a+c)^2+(b+d)^2},$$ and $$|x|+|y|=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}.$$ Thus, the inequality is equivalent to $$\sqrt{(a+c)^2+(b+d)^2} \leq \sqrt{a^2+b^2}+\sqrt{c^2+d^2},$$by squaring both sides, namely,$$ac+bd \leq \sqrt{(a^2+b^2)(c^2+d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trig function bounded on interval (without calculus), prove that $x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$.
If $0 \lt x \lt \dfrac{\pi}{2}$, prove that
$$x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$$
This question must be done without calculus. First, I tried splitting it into the intervals $(0,\pi/4)$ and $(\pi/4... | By Cauchy Schwarz $$ 9 = \left(x^3+(9-x^3) \right) \left(\sin^2 x + \cos^2 x\right) \ge \left(x^{3/2}\sin x + \sqrt{9-x^3}\cos x\right)^2$$ and the result follows
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\... | You know
$$\sin^4x+\cos^4x=1-\dfrac12\sin^22x=1-\dfrac12\left(\dfrac{1-\cos4x}{2}\right)$$
is periodic with period $T=\dfrac{\pi}{2}$, so write
$$\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx=4\int_0^\frac{\pi}{2}\dfrac{(1+\tan^2x)^2}{1+\tan^4x}dx=4\int_0^\infty\dfrac{1+t^2}{1+t^4}dt=4\dfrac{\pi}{\sqrt{2}}=\color{blue}{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
} |
Prove by induction $\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$ In this problem I have to prove the following equation using mathematical induction:
$$\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$$
So far I've proved that $P_1$ is true, and written $P_k$ as $$15+36+63+...+3k(k+4)= \frac{(k)(k+1)(2k+13)}{2}$$
... | There are three silly mistakes here: First off, your expression on the RHS should be $\frac{(k+1)(k+2)(2k+15)}{2}$ since.
Secondly you somehow made the addition into multiplication here
$$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$
And finally as Bernard said you should have a $3(k+1)=3k+3$ ins... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Complex Expression is Real
If I have the a complex number $z \in \mathbb{C}$ with absolute value $|z| = 1$, how do I show that $-i \frac {z-1}{z+1}$ is real?
| A number $w$ is real iff $\operatorname{Im}(w) = \frac{1}{2i}(w - \bar w) = 0\,$. In this case:
$$\require{cancel}
\begin{align}
-i \frac {z-1}{z+1} - \overline{\left(-i \frac {z-1}{z+1}\right)} &= -i \frac {z-1}{z+1} - \left(+i \frac{\bar z - 1}{\bar z + 1}\right) \\
&= -i \cdot\frac{(z-1)(\bar z + 1)+ (\bar z -1)(z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
ellipsoid of greatest volume is a sphere The equation of an ellipsoid is $$f(x,y,x)=(\frac xa)^2+(\frac yb)^2+(\frac zc)^2=1$$.
Given that the volume of an ellipsoid is $$V=\frac43\pi abc$$ and the constraint $$L=a+b+c$$ L some positive constant. Show that the ellipsoid with greatest volume is a sphere.
I should use La... | Here the lagrangian reads
$$
L(a,b,c,x,y,z,\lambda,\mu) = \frac 43 \pi a b c +\lambda\left(\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2-1\right)+\mu(a+b+c-L_0)
$$
The stationary conditions give
$$
\nabla L = \left\{
\begin{array}{rcl}
-\frac{2 \lambda x^2}{a^3}+\mu +\frac{4 b c \pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Hints with this Integration Problem $$\int_{0}^{\infty}\frac{(x \cos x-\sin x)\cos (x/2)}{x^3}\mathrm d x $$
I tried solving it by substitution, and I did come close to the answer. But that was really long and tedious, I don't even know whether or not what I did was even correct.
Can anyone suggest me some other way? H... | $\begin{align}J&=\int_{0}^{\infty}\frac{(x \cos x-\sin x)\cos (x/2)}{x^3}\mathrm d x\end{align}$
Perform integration by parts,
$\begin{align}J&=\left[-\frac{1}{2x^2}(x \cos x-\sin x)\cos (x/2)\right]_{0}^{\infty}+\\
&\int_0^{\infty}\frac{-\frac{1}{2}\sin\left( \frac{1}{2}x\right)(x \cos x-\sin x)-x\sin x\cos\left( \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Diophantine equation $a^2+b^2+c^2=a^2b^2$ I am trying to find all non trivial integers for which $a^2+b^2+c^2=a^2b^2$.
As suggested I have tried working (mod 4).
This is what I've gotten so far:
Squares can have a remainder of 0 or 1 (mod 4).
So the rhs can be 0 or 1 (mod 4). It is 1 if and only if both $a^2$ and $b^2$... | Just work from parity.
If $a,b$ are both odd then $a^2+b^2\equiv 2 \pmod 4$ and $a^2b^2\equiv 1 \pmod 4$ This would mean that $c^2\equiv -1\pmod 4$ which is impossible.
If $a=2A$ and $b$ is odd then we get $1+c^2\equiv 0\pmod 4$ which is, again, impossible.
If $a,b$ are both even then $c$ is also even. Writing them ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solutions of $f(x+y^{n})=f(x)+[f(y)]^{n}$. Consider the functional equation $f(x+y^{n})=f(x)+[f(y)]^{n}$ where $f:\mathbb R \to \mathbb R$ and $n$ is given integer $>1$. This equation was discussed yesterday and it was shown that $f$ is necessarily additive. Assuming continuity it was concluded that $f(x)\equiv cx$ for... | This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\in\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you.
As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\in \mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer sho... | $\sqrt{1+x^n}=1+x^n/2+O(x^{2n})$ as $x\to0$.
In particular $\sqrt{1+x}=1+x/2+O(x^2)$, $\sqrt{1+x^2}=1+O(x^2)$
and $\sqrt{1+x^3}=1+O(x^2)$. Then your fraction is
$$\frac{1+O(x^2)-(1+x/2+O(x^2))}{1+O(x^2)-(1+x/2+O(x^2))}
=\frac{-x/2+O(x^2)}{-x/2+O(x^2)}$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Evaluating definite integral of $p(x)$
Let $p(x)$ be fifth degree polynomial such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1 $is divisible by $(x+1)^3 $. Then find the value of the definite integral $$\int _{-10}^{10}p(x)dx$$
Attempt:
$p(x)-1 = (x+1)^3 Q(x)$
$p(x)+1 = (x-1)^3 H(x)$
Where $Q(x)$ and $H(x)$ ... | Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^{\prime}(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^{\prime}(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^{\prime}(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $t\in\mathbb{R}$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
$\frac{x^2+y^2}{x-y}|1995$. Find positive integer $x,y$
$x,y$ are positive integer . Find all $x,y$ such that
$\frac{x^2+y^2}{x-y}|1995$.
My answer: ${x^2+y^2}|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1\,995$.
I observed that o... | Your first assertion that $x^2+y^2 \mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Finding a holomorphic function with real part $u=\frac{x(1+x^2+y^2)}{1+2x^2-2y^2+(x^2+y^2)^2}$. I am asked to
find the holomorphic function $f(z)=f(x\pm iy)$ with real part
$$u=\frac{x(1+x^2+y^2)}{1+2x^2-2y^2+(x^2+y^2)^2}$$
and such that $f(0)=0$.
I have tried to apply Cauchy-Riemann equations directly but I don... | Hint: In general, use these substitutions
\begin{align}
x=\frac{z+\bar{z}}{2}\\
y=\frac{z-\bar{z}}{2i}\\
|z|^2=x^2+y^2\\
=z\bar{z}
\end{align}
Edit: After substitution
\begin{align}
u
&= \dfrac{\frac{z+\bar{z}}{2}\left(1+z\overline{z}\right)}{1+z^2+\overline{z}^2+z^2\overline{z}^2}\\
&= \dfrac12\dfrac{z(1+\overline{z}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the probability that atleast one valve is defective. A factory A produces $10$% defective valves and another factory $B$ produces 20% defective valves.A bag contains $4$ valves of factory $A$ and $5$ valves of factory B.If two valves are drawn at random from the bag,find the probability that at least one valve is ... | The answer is indeed $\frac{517}{1800}$, because:
$$P(\text{at least one of two defective})=1-P(\text{both normal})=\\
1-[P(ANAN)+P(ANBN)+P(BNAN)+P(BNBN)]=\\
1-\left[\frac{4}{9}\cdot \frac9{10}\cdot \frac38\cdot \frac9{10}+
\frac{4}{9}\cdot \frac9{10}\cdot \frac58\cdot \frac8{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\frac{1}{x-y+z}$ from given fraction equations. $$\begin{align}
\frac{3}{x}\,-\,\frac{4}{y}\,+\,\frac{2}{z}\quad&=\quad3\\
\frac{2}{x}\,-\,\frac{8}{y}\,-\,\frac{1}{z}\quad&=\,-\,8\\
\frac{4}{x}\,-\,\frac{6}{y}\,-\,\frac{3}{z}\quad&=\quad1\\
\frac{1}{x-y+z}\quad&=\quad?
\end{align}$$
I tried to find solution by eq... | Go on, go on…
$$x=\frac{43}{133}\qquad y=\frac{86}{149}\qquad z=\frac{43}{14}$$
$$\frac1{x-y+z}=\frac{5662}{15953}=0.3549\dots$$
Beauty is not a requirement of mathematical problems – especially in applied problems, and even in pure problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and... | Multiply both sides by $4$ and rearrange:
$$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl) \Rightarrow \\
\left(a+b-\frac{4ab}{a+b}\right)+\left(b+c-\frac{4bc}{b+c}\right)+\left(c+a-\frac{4ca}{c+a}\right)\ge 0 \Rightarrow \\
\frac{(a-b)^2}{a+b}+\frac{(b-c)^2}{b+c}+\frac{(c-a)^2}{c+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Common roots of recursive defined polynomial I have a series of polynomials $P_j(x)$ given by the recursive formula
$$P_{j+1}=\frac{e_j}{c_j}xP_{j}-\frac{f_j}{c_j}P_{j-1}
$$
with $P_{-1} \equiv 0$, $P_0 \equiv 1$, where
$$c_j = (j+1)(j+2\kappa+1),\\
e_j = (2j+2\kappa+1)(j+\kappa+1),\\
f_j = (j+\kappa)(j+\kappa+1),\\
j=... | Define $Q_n(x) = \dfrac{(2n)!!}{(2n - 1)!!} P_{n - 1}(x)$ for $n \geqslant 0$, where $(-1)!! = 0!! = 1$, then $Q_0(x) = 0$, $Q_1(x) = 2$, and for $n \geqslant 0$,$$
(n + 1)(n + 2) P_{n + 1}(x) = (2n + 2)\left( n + \frac{3}{2} \right) xP_n(x) - \left( n + \frac{1}{2} \right)\left( n + \frac{3}{2} \right) P_{n - 1}(x)\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
exercise on double integration I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:
given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$
$E[X,Y] = \int \int 24x^2y^2dxdy$ =
$24 \int_0^xx^2 (\int_0^{1-x}y... | I believe you either misread '8' as '3', or there is a typo on your prof's part.
\begin{align}
E &= \int_{x=0}^{x=1}\int_{y=0}^{y=1-x} 24x^2y^2 \, dy \, dx \\
&= 24\int_{x=0}^{x=1}x^2\int_{y=0}^{y=1-x} y^2 \, dy \,dx \\
&= 24\int_{x=0}^{x=1}x^2\left[\frac{y^3}{3}\right]_{y=0}^{y=1-x} \,dx \\
&= 8\int_{0}^{1}x^2(1-x)^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear algebra inequality (1) Note that if $\| \mathbf{A} \mathbf{x} \|_{2} = 1$, then
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x} \|_{2} \cdot \| \mathbf{x} \|_{2} \ge \langle \mathbf{A}^{T} \mathbf{A} \mathbf{x}, \mathbf{x} \rangle = \| \mathbf{A} \mathbf{x} \|_{2}^{2} = 1$$
and thus we have $\| \mathbf{A}^{T} \mathbf... | Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $\{ \mathbf{A}\mathbf{x}_{1}, \cdots, \mathbf{A}\mathbf{x}_{r} \}$ is orthonormal, but :
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{1} \|_{2}^{2} + \cdots + \| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{r} \|_{2}^{2} =r= \frac{1}{\| \mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
derivative of an integral coming from an expectation I just wanted to check on the following result from this paper (a very rough sketch of the proof is given in the supplementary material, starting from eq 8).
$$\left.\frac{\partial}{\partial c} \left(\frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi\left(\sqrt{q} u\righ... | I have found my own mistake: I was a little to fast in using the identity $\int f(x) x e^{\frac{-x^2}{2}} = \int f'(x) e^{\frac{- x^2}{2}}$ in the last line. Acutally, the last line should have been
$$\begin{align}
& = \frac{1}{2 \pi} \int \left[\phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) + c \phi(\sqrt{q} z_1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is the... | Let $a\geq 1$ be fixed. Then, $g:[a,\infty)\to\mathbb{R}$ defined by $$g(t):=t+\frac{1}{t}\text{ for all }t\in[a,\infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)\left(1-\frac{1}{at}\right)\,.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2\,\left(\frac{t}{a^2}\right)+\frac{1}{t}\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Showing that $\left|\frac{z}{z^3+1}\right|\leq\frac{R}{R^3-1}$
I am trying to show that if $|z|=R>1$, then
$$\left|\frac{z}{z^3+1}\right|\leq\frac{R}{R^3-1}$$
I've been playing around with the above inequality and this is where I am at.
\begin{align}
\text{Consider} \ \left|z^3+1\right|&\leq\left|z^3\right|+1 \ \... | WLOG $z=R(\cos t+i\sin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3\cos3t\ge R^6+1-2R^3= (R^3-1)^2$$
as $\cos3t\ge-1$ and $R^3>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $AD=2AP$ if and only if $BP \perp CP$. Let the incircle of $\triangle ABC$ contact $AB,AC,BC$ at $D,E,F$ respectively, and intersect $AF$ at a second point $P$. Show that $AD=2AP$ if and only if $BP \perp CP$.
| Here is a not-so-pretty partial solution. I shall prove that, if $AD=2\cdot AP$, then $BP\perp CP$. This is a trigonometric solution, so it is quite boring.
Without loss of generality, suppose that $\angle B\geq \angle C$. Write $a:=BC$, $b:=CA$, $c:=AB$, and $s:=\dfrac{a+b+c}{2}$. Note that $$AP\cdot AF=AD^2\,.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality inside a triangle "Let $ABC$ be a triangle with centroid $G$. A line $PQ$ is drawed in the triangle such that it passes through $G$ and intersects the sides $AB$ and $AC$ in $P$ and $Q$ respectively."
Prove that:
$$\frac {PB \cdot QC}{PA\cdot QA} \le \frac {1}{4}$$
I made an aproximate drawing of this... | Let $M\in AC$ and $N\in AB$ such that $MN||BC$, and $G\in MN.$
Also, let $\measuredangle MGQ=\theta$.
Thus, in the standard notation $$MG=GN=\frac{1}{3}a$$ and by law of sines we obtain:
$$\frac{PN}{\sin\theta}=\frac{\frac{1}{3}a}{\sin\beta}$$ or
$$PN=\frac{a\sin\theta}{3\sin\beta}.$$
Similarly,
$$MQ=\frac{a\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that: $\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$ Given three positive numbers a,b,c satisfying $$a^2+b^2+c^2=1$$ Prove that: $$\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$$
The things I have done so far:
$$\sum \limits_{cyc}\frac{bc}{a^2+1}=\sum \limits_{cyc}\frac{... | By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{bc}{a^2+1}=\sum_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq\frac{1}{4}\sum_{cyc}\frac{(b+c)^2}{a^2+b^2+a^2+c^2}\leq$$
$$\leq\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{a^2}{b^2+a^2}\right)=\frac{3}{4}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
First Order Separable differential Equation
Problem:
Solve the following differential equation:
\begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}
Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, ... | $$\ln (x^3+1)^2=c_1+ \ln |y|$$
For $y > 0$
$$\ln (x^3+1)^2=c_1+ \ln y$$
$$(x^3+1)^2=e^{c_1} y$$
For $y < 0$
$$\ln (x^3+1)^2=c_1+ \ln (-y)$$
$$(x^3+1)^2=-e^{c_1} y$$
so we have that
$$(x^3+1)^2=|e^{c_1} y|=e^{c_1}|y|$$
On the other hand you have that
$$|ab| = |a||b|$$
so
$$|cy| = |c||y|$$
When in your book they choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}... | The original inequality cannot be proved directly by induction but by the solution given by Jack D'Aurizio is interesting to note that we can also proceed by induction proving the stronger result
$$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac12}}<2\sqrt{2}$$
indeed ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Area of a triangle with ratio of a side In the drawing, $AM=MN=NC$ and $\frac {BP}{PC}=\frac {5}{3}$, if the area of the gray region is 8, whats the area of $\triangle ABC$?
I saw that $\triangle ABN$ and $\triangle BNC$ have the same height but their bases are in ratio $2:1$, and the sum of the areas of those triangl... | Add an auxiliary segment through P parallel to AC intersecting BN at Q and AB at S. Call the intersection of BN and MP R.
Let $h_1$ be the altitude of triangle QRP
Let $h_2$ be the altitude of triangle MRN
Let $h_3$ be the altitude of triangle PMC
Let $h_4$ be the altitude of triangle BSP
Let $h_5$ be the altitude of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the equation of the common tangents Find the equation of the common tangents to the parabola $y^2=4ax$ and $x^2=4by$.
My Attempt :
$y^2=4ax$ is the equation of parabola with focus at $(a,0)$
$x^2=4by$ is the equation of parabola with focus at $(0,b)$
The equation of the tangent to the parabola $y^2=4ax$ is
$$y=... | Let $A(at^2,2at)$ and $B(2bu,bu^2)$ be the points on $C_1:y^2=4ax$ and $C_2:x^2=4by$ respectively.
Equation of tangent of $C_1$ at $A$:
$$\frac{x}{-at^2}+\frac{y}{at}=1 \quad \cdots \cdots \: (1)$$
Equation of tangent of $C_2$ at $B$:
$$\frac{x}{bu}+\frac{y}{-bu^2}=1 \quad \cdots \cdots \: (2)$$
Comparing $(1)$ and $(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Are there functions satisfying $\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx$ Are there functions satisfying $\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx$
I came up with one which is
$f(x)=e^{x \sec ^2 \alpha}$ and $g(x)=e^{x \csc^2 \alpha}$
$$\int f(x)g(x)dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha... | Let
$F = \int f,
G = \int g
$.
I'll try to get $f$
in terms of $g$.
Differentiating
$\int fg
=\int f \int g
$,
$fg
=g\int f+f\int g
=g\int f+fG
$
or,
in terms of $F$,
$F'(g-G)
=gF$
so
$\dfrac{F'}{F}
=\dfrac{g}{g-G}
$
or
$(\ln(F))'
=\dfrac{g}{g-G}
$
or
$\ln F
=\int(\dfrac{g}{g-G})
$
so,
$ F
=e^{\int(\dfrac{g}{g-G})}
$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Smallest cube ending in $2017$
What is the smallest cube ending in $2017$?
My try: I know that the only possible units digit is $3$,
$$(a+3)^3 = 2017 \mod 10^4\;$$
and
$$a^3 + 9a^2 + 27a = 1990\mod 10^4$$
I don't know how to proceed, I tried factoring and adding $10^5$ and $1990$ but when I saw the answer, this appr... | $x^3\equiv 7 \pmod {10}$. So $x\equiv 3 \pmod {10}$.
Put $x=10y+3$, and compute $\pmod {100}$.
This yields $1000y^3+900y^2+270y+27 \equiv 17 \pmod {100}$. Thus $70y \equiv -10 \pmod {100}$, or equivalently, $7y \equiv -1 \pmod {10}$. Divide by $7$: $y \equiv 7 \pmod {10}$.
Put $y=10z+7$, hence $x=100z+73$. I think y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
Find $x$, given: $x^2 + \frac{9x^2}{(x+3)^2} = 16$ Here's an equation:
$$x^2 + \frac{9x^2}{(x+3)^2} = 16$$
First, I subtracted 16 from both sides and factored $x^2$ so I would get a quadratic equation, but with no success. Also, I can see that the equation can be rewritten as:
$$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)... | You can rewrite it as:
\begin{align*}
\left(\frac{x}{4}\right)^2 + \left(\frac{3x}{4(x+3)}\right)^2 & = 1.
\end{align*}
Let $\frac{x}{4}=\cos A$, then $\frac{3x}{4(x+3)}=\sin A$. Using $x=4 \cos A$ in the expression for $\sin A$, we get
\begin{align*}
12 (\cos A -\sin A) & = 8 \sin 2A\\
9(\cos A -\sin A)^2 & = 4 \sin ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $
Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4 +y^2 $
$x^2 = y^4 +y^2 - 41$
$x^2 -49 = y^4 +y^2 - 41 ... | Continuing your calculation,
$(\dfrac{30-y^2}y)^2=y^4+y^2-41,y≠0 $
$⇒y^4-60y^2+900=y^2(y^4+y^2-41)$
$⇒y^6+19y^2-900=0$
$⇒y^2(y^4+19)=900$
This give integer solution $y^2=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Finding the least sum of digits possible for an outcome of a function in prime numbers. Let $f(n)=p^4-5p^2+13$ simplified as $f(n)=(p^2-{5\over2})^2+{27\over4}$ where $p$ is an odd prime.
Find the least possible sum of digit of $f(n)$.
My findings:
After putting $p=(3n,3n+1,3n+2)$ in $p(n)$ and dividing $p(n)$ by $3$... | Let $p \not = 5$, then by using Fermat's Little Theorem we get that $p^4 - 5p^2 + 13 \equiv 1 + 3 \equiv 4 \pmod 5$. So the last digit is either $4$ or $9$. However $f(p)$ is always odd number, so the last digit must be $9$. Thus the sum of digits of $f(p)$ for $p \not = 5$ is at least $9$. On the other side we have th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can I solve $x^\sqrt{y} +y^\sqrt{x} =\dfrac{49}{48}$ and $\sqrt{x}+\sqrt{y} =\dfrac72$? I have tried Wolfram Alpha and Mathematica to get the solution of the below system, but no result , I have used variable change $z=\sqrt{x}+\sqrt{y}$ for simplification but no result ,
$$
\left\{
\begin{array}{ll}
x^\sqrt{y} +y... | Doing the same as Jack D'Aurizio (eliminating $y$ from the second equation) and using a Taylor expansion around $x=0$, the first equation is
$$-\frac{1}{48}+\sqrt{x} \log \left(\frac{49}{4}\right)+x \left(\frac{1}{2} \log
^2\left(\frac{49}{4}\right)-\frac{4}{7}\right)+O\left(x^{3/2}\right)$$ the solution of which be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sum of the series $\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$. Sum to n terms and also to infinity of the following series:
$$\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$$
the solution provided by the book is
$$S_n=\frac{17}{96}-[\frac{1}{... | $$S_n=\dfrac{n}{(n+1)(n+3)(n+5)}=\dfrac{A}{n+1}+\dfrac{B}{n+3}+\dfrac{C}{n+5}$$
then $A=-\dfrac18$, $B=\dfrac68$ and $C=-\dfrac58$ then
$$S_n=\dfrac{n}{(n+1)(n+3)(n+5)}=\dfrac18\left(\dfrac{1}{n+3}-\dfrac{1}{n+1}\right) - \dfrac58\left(\dfrac{1}{n+5}-\dfrac{1}{n+3}\right)$$
for using telescopic method we need add some ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A finite alternate binomial sum with finite value (modulo 6) We use Iverson symbol ( [expr] is $1$ when expr is true else $0$), and $n\%6$ for $n$ modulo $6$. .
For any $n >= 2 $.
Let $ F(n) := \sum_{j>=1} (-1)^j {\binom {n-j-2} j } $
Prove that $F(n) = [n\%6 \in \{0,5\}] - [n\%6 \in \{2,3\}]$
In other words:
IF $n$... | Working with the following sum
$$\sum_{q=0}^n (-1)^q {n-q\choose q}$$
we have
$$\sum_{q=0}^n (-1)^q {n-q\choose n-2q}
= \sum_{q=0}^n (-1)^q [z^{n-2q}] (1+z)^{n-q}
\\ = [z^n] (1+z)^n \sum_{q=0}^n (-1)^q z^{2q} (1+z)^{-q}.$$
Now we get zero contribution when $2q\gt n$ hence we may write
$$[z^n] (1+z)^n \sum_{q\ge 0} (-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Range of a rational function with radicals Find the range of the function
$$\frac{6}{5\sqrt{x^2-10x+29} - 2}$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$\frac{6}{5\sqrt{x^2-10x+29} - 2} = y$
getting the inverse,
$\frac{6}{5\sqrt{y^2-10y+29} - 2} = x... | $\frac{6}{5\sqrt(x^2-10x+29)-2}=\frac{6}{5\sqrt((x-5)^2+4)-2}$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $\frac{3}{4}$ and you have a range $]0,\frac{3}{4}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find $\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$ Find the limit without using L'hopital
$$\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$$
I have tried multiply by the conjugate of the denominator and numerator but didn't work
Any hint would be appreciated
Thank you
| Using that $t^2-1=(t-1)(t+1)$ and $t^3-1=(t-1)(t^2+t+1)$ we have
$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}\cdot \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\cdot \dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}\right]$$ and thus
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Non linear first order ODE in in $\mathbb{R}^2$ 2 Let $a, b \in \mathbb{R}$ and consider the system of ODEs
$$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0 \end{matrix} \right. $$
where $x, y :\mathbb{R} \to \mathbb{C}.$
I wonder whether there is, in general, $F$ such that $F(x,y) =... | $$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0 \end{matrix} \right.$$
Multiply first equation by $x^2$ and second equation by $y^2$
$$\left\{ \begin{matrix} x^2x' = ax^3+ by^2x^2 & . \\ y'y^2 = ay^3 +bx^2y^2 & . \end{matrix} \right.$$
$$x'x^2-ax^3=y'y^2-ay^3$$
$$\frac 13 (x^3)'-ax^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x... | So far so good.
You may continue to see the geometry of the solution as well.
$$3x^2+3y^2+32x+8y+48=0$$
$$x^2+y^2+(32/3)x+(8/3)y+16=0$$
Which is the equation of a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Can we factor multivariate polynomial $x+xy+y$? It seems to me we can only get $x(1+y)+y$ or $x+(x+1)y$, but maybe something better is possible with complex numbers?
It has zeros at $y=-\frac{x}{x+1}$, but how to use that fact?
| A possible factorization is :
$\begin{align}x+xy+y &=x+xy+y+1-1\\ &=x(y+1)+(y+1)-1 \\ &=(x+1)(y+1)-1 \\ &=\left(\sqrt{(x+1)(y+1)}\right)^2-1^2 ,\text{assuming $x,y \gt -1$} \\ &=\left(\sqrt{(x+1)(y+1)}+1\right)\cdot\left(\sqrt{(x+1)(y+1)}-1\right)\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find an affine map for given points I am trying to understand affine maps. In the book I am using there is the following example, which I don't understand. Given are the following points:
\begin{array}{lll}
p_0 = \begin{pmatrix} 1 \\ 1 \end{pmatrix},
&
p_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix},
&
p_2 ... | Let $A=
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}$
Now, apply this matrix to your vectors $p_0p_1$ and $p_0p_2$
The first one gives:
$q_0q_1=A*p_0p_1$
So
$\begin{pmatrix}2 \\0 \end{pmatrix}=A* \begin{pmatrix}0 \\1 \end{pmatrix} = \begin{pmatrix} a*0 + b*1 \\ c*0 +d*1\\ \end{pmatrix}= \begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$
WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2... | For $(a,b,c)\rightarrow(0,6,0)$ we have $A\rightarrow72$ and since
$$72-a^2bc-a^2-2b^2-2c^2\geq0$$ it's
$$2(a+b+c)^4-(a+b+c)^2(a^2+2b^2+2c^2)-36a^2bc\geq0$$ or
$$a^4+6a^3b+6a^3c+9a^2b^2+9a^2c^2-14a^2bc+4ab^3+4ac^3+20b^2ac+20c^2ab+4ab^3+4ac^3+8b^2c^2\geq0,$$
which is obvious, we see that the maximum does not exist, but ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate the integral $\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$ $$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$$
My attempt:
$$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2-2x(1-x)+(x+1-x)^2}}dx$$
$$=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{(1-x(1-x))^2}}dx=\i... | Here is another method:
\begin{eqnarray}
I&=&\int_{0}^{1} \frac{(\log {x})^2}{x^2-x+1}dx=\int_{0}^{1} \frac{(1+x)(\log {x})^2}{1+x^3}dx\\
&=&\int_{0}^{1}(1+x)\log^2x\sum_{k=0}^\infty(-1)^kx^{3k}dx\\
&=&\sum_{k=0}^\infty(-1)^k\int_0^1(x^{3k}+x^{3k+1})\log^2xdx\\
&=&2\sum_{k=0}^\infty(-1)^k\left[\frac{1}{(3k+1)^3}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Continued fraction of $ \sqrt{3} $ How can I solve the continued fraction of $\sqrt{3}$?
I know that a continued fraction is [1;1,2,1,2,1,2,....] -> this is the continued fraction of $\sqrt{3}$, but how can I solve? What are the steps?
I don't understand this proof.
Thank you so much.
| We know $\sqrt{3}$ lies between 1 and 2 (because $1^2 < 3 < 2^2$). So the first digit of the continued fraction expansion is 1. So
$\sqrt{3} = 1 + (\sqrt{3}-1)$
and $\sqrt{3}-1$ must be between $0$ and $1$. So next we want to find the reciprocal of $\sqrt{3}-1$:
$\frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{(\sqrt{3}-1)(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The priority of limits How are the two expressions different?
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$
and $$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$
If limit is inside the floor function, Do I have to apply the limits first?
If this is the case, then, $$\lim_{x\to0}\bigg\lfloor\frac{\si... | Yes you are right indeed since
$$\frac{\sin x}{x}<1 \implies \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0 \implies \lim_{x\to0} \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
and since
$$\lim_{x\to0} \frac{\sin{x}}{x}=1 \implies \bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1
$$
For the latter for $x$ sufficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a+b+c\le 0$ $a,b,c$ are integer numbers different from zero and
$$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$
Prove that $a+b+c\le 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
| We have $$ab=(a+c^2)(b+c^2)\\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2\le -c$$or equivalently $$a+b+c\le 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proof of onto under interval $(0,1)$ How do I prove that the function $$f:(0,1)\rightarrow \mathbb R$$ defined by:
$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$
is onto?
| Alternatively, set $t = 2x-1$ and for each $r \in \mathbb{R}$ we show we can find a number $t$ and then an $x \in (0,1)$ such that$f(x) = r$. We have: $\dfrac{t}{1-t^2}=r$. So this gives $t = r - rt^2\implies rt^2+t -r = 0$. If $r = 0 \implies t = 0\implies x = \dfrac{1}{2}$. If $r \neq 0$, $t = \dfrac{-1\pm \sqrt{1+4r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$ I want to find the following limit:
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$$
This is what I do. I change the variable $t=-x$ and I have the following limit:
$$\lim_{t\rightarrow +\infty}{e^{\frac ... | Let set $\frac {1}{2-x}=y \to 0$
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x=\lim_{y\rightarrow 0}{e^{y}\cdot\frac{\frac1{y^2}-\frac 6 y+7}{-\frac1y}}-2+\frac1y$$$$=\lim_{y\rightarrow 0}e^{y}\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$
then use $e^y=1+y+o(y)$ to obtain
$$(1+y+o(y))\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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General formula for Differentiation Operator I was considering the Operator
$$
x\,\frac{\rm d}{{\rm d}x}
$$
and applying it $n$ times to an arbitrary function $f(x)$.
Is there a general formula for it?
I started with the first few
\begin{align}
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{1} f(x) &= x f' \\
\left(x\,\frac{\... | The coefficients are the Stirling numbers of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semester... | If $\sin x+\cos x=a,a^2=1+2\sin x\cos x\iff\sin x\cos x=\dfrac{a^2-1}2 $
$$\sin^{n+2}x+\cos^{n+2}x=\sin^nx(1-\cos^2x)+\cos^nx(1-\sin^2x)$$
If $I_m=\sin^mx+\cos^mx,I_2=1,I_0=2$
$$I_{n+2}=I_n-(\sin x\cos x)^2I_{n-2}$$
Here $n=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$x... | Consider the equation: $A^3-10A^2+32A-xyz=0$, whose roots are $x,y,z$. We need to find the maximum and minimum values of:
$$xyz=A^3-10A^2+32A=f(A).$$
FOC:
$$f'(A)=3A^2-20A+32=0 \Rightarrow A_1=\frac83; \ \ A_2=4.$$
SOC:
$$\begin{align}f''(A)&=6A-20; \\
f''\left(\frac83\right)&=-4<0 \Rightarrow f\left(\frac83\right)=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solving $\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$
Given
$$\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$$
What is $x$?
I just can do this with that equation
$$\left(\sqrt{2+1+2\sqrt{2.1}}\right)^x - \left(\sqrt{2+1-2\sqrt{2.1}}\right)^x = \fr... | Use the fact that $$\sqrt{3-2\sqrt{2}}=\frac{1}{\sqrt{3+2\sqrt{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Separate f(z) into real and imaginary parts I'm doing some practice problems about complex numbers. I want to break $f(z)=(1-z)/(1+z)$ into real and imaginary parts $u(x,y)$ and $v(x,y)$ such that $f(z)=u+iv$. I did a different problem like this correctly, so I think it's more the algebra than the complex number aspe... | You just need to realise the denominator:
$$f(z) = \frac{(1-x) - iy}{(1+x) + iy} \times \frac{(1+x) - iy}{(1+x)-iy}$$ $$= \frac{(1-x)(1+x) -y^2}{(1+x)^2+y^2} + \frac{i(-y(1-x) - y(1+x))}{(1+x)^2 + y^2}$$ $$= \frac{1-x^2 - y^2}{(1+x)^2 + y^2} + i \frac{-y + yx - y - yx}{(1+x)^2 + y^2} $$ and so it is clear that $u = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | $$\lim_{x\to0}\frac{(\sin{x}+x)(\sin{x}-x)}{x\sin{x}\cdot x\sin{x}}$$
Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)
$$\lim_{x\to0}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}$$
$$\lim_{x\to0}\frac{x-\tan{x}}{x^3}=-\frac{1}{3}$$
$$\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$
So using t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
} |
If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$
But could not proceed further please help.
| Let $c=cosx$ and $s=sinx$ :
$s^8+c^8= \frac{48}{2^7}*\frac{2}{2}$
$s^8+c^8= \frac{96}{2^8} $
$(\frac{G}{2^4})^2+(\frac{H}{2^4})^2=\frac{96}{2^8}\implies G^2+H^2=96 $
$\implies s=\pm\frac{\sqrt[4]{G}}{2} \wedge c=\pm\frac{\sqrt[4]{96-G^2}}{2}$
We gonna use: $s^2+c^2=1\rightarrow c^2=1-s^2$:
$\frac{\sqrt{96-G^2}}{4}=1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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Show that $\frac{3^n}{n!}$ converges to $0$ I was wondering if this proof is correct.
$$\left|\frac{3^n}{n!} - 0\right| = \frac{3^n}{n!} \lt \frac{3^n}{2^n} \lt \varepsilon$$
So then
$$\frac{2^n}{3^n} \gt \frac{1}{\varepsilon}$$
$$\left(\frac{2}{3}\right)^n \gt \frac{1}{\varepsilon}$$
$$\log\left(\frac{2^n}{3^n}\right)... | You can modify your argument as follows:
You can assume that $n \geq 4$ because shifting the terms of a sequence does not affect its convergence or the value of convergence.
Then
$$\left|\frac{3^n}{n!} - 0\right| = \frac{3^n}{n!} =\frac{27}{3\times 2\times 1}
\frac{3^{n-3}}{n \times(n-1)\times\cdots\times 4}\lt 5 \ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Find the domain of x $\left \lfloor x \right \rfloor + \left \lfloor x+\frac{1}{2} \right \rfloor + \left \lfloor x-\frac{1}{3} \right \rfloor =8$ Find the domain of x
$$\left \lfloor x \right \rfloor + \left \lfloor x+\frac{1}{2} \right \rfloor + \left \lfloor x-\frac{1}{3} \right \rfloor =8$$
My approach
W... | Let $x = k + y$, where $k$ is an integer part of $x$ ($k=\left \lfloor x \right \rfloor$) and $y$ is its fractional part. Consider $y$-s belonging to the following intervals
$\left[0, \frac{1}{3}\right)$, $\left[\frac{1}{3}, \frac{1}{2}\right)$ and $\left[\frac{1}{2}, 1\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Find all non-negative integer $n$ that satisfy $f(x+1)+f(x-1)=\sqrt nf(x)$
Find all non-negative integer $n$ that there exists a non-periodic function $f:\Bbb R->\Bbb R\quad f(x+1)+f(x-1)=\sqrt nf(x)\forall x$.
My attempt:
$f(x+2)+f(x)=\sqrt n\cdot f(x+1)$
$\sqrt n \cdot (f(x+1)+f(x+3))=n\cdot f(x+2)$
$f(x+2)+f(x+4... | This is not actually an answer, but more than a comment following the previous nice answers. If $a,b$ are the roots (real roots case where $ n\geq 4$ ) of the trionym $t^2-\sqrt{n}\,t+1$ then all the functions of the form : $f(x)=c_1a^x+c_2b^x$ have the desired attribute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n. Q: Prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n.
In fact, I already have a method, but it seems too long. So I would first (1)is there any problem with my logic? (2) Is there any better/faster ... | Modulo $3$, your numerator is:
$$3n^5+5n^3+7n \equiv -n(n^2-1) $$
$$\equiv -n(n-1)(n+1)\pmod{3}.$$
The last expression is a product of three consecutive integers and hence divisible by $3$.
Modulo $5$, you have
$$3n^5+5n^3+7n \equiv -2n^5+2n $$ $$\equiv -2n(n^4-1) \pmod{5}.$$
If $n$ is divisible by $5$, then so is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Greatest term of this expansion? If in the expansion of $(x+a)^{15}$ the $11^{\text{th}}$ term is G. M. i.e. geometric mean of $8^{\text{th}}$ and $12^{\text{th}}$ terms then which term is the greatest term among the terms in expansion of $(x+a)^{15}$?
| Applying the binomial theorem we have
\begin{align*}
(x+a)^{15}=\sum_{j=0}^{15}\binom{15}{j}x^ja^{15-j}
\end{align*}
Since the $11^{\text{th}}$ term is the geometric mean of the the $8^{\text{th}}$ and $12^{\text{th}}$ term, we have
\begin{align*}
\binom{15}{10}x^{10}a^5&=\sqrt{\binom{15}{7}x^7a^8\binom{15}{11}x^{11}a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $x^4 -10x^3 + 26x^2 -10x +1 = 0$. Find the root of the equation
$$x^4 -10x^3 + 26x^2 -10x +1 = 0$$
My attempt:
If $a,b,c$ and $d$ are a roots then $a+b+c+d = 10,$
$(a+b)(c+d) + ab +cd = 26$
$(a+b)cd +ab(c+d) = 10$
$abcd =1$
Now i'm not able procceed further.
| You can proceede also like this:
\begin{eqnarray}x^4 -10x^3 + 26x^2 -10x +1 &=& (x^4 -4x^3 + 6x^2 -4x +1)-6x(x^2-2x+1)+8x^2\\
&=& (x-1)^4-6x(x-1)^2+8x^2\\
&=& \Big((x-1)^2-4x\Big)\Big((x-1)^2-2x\Big)\\
&=& \Big(x^2-6x+1\Big)\Big(x^2-4x+1\Big)\\
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving a recurrence relation with floor function I'm having trouble solving this recurrence relation:
\begin{align}
T(n) &=
\begin{cases}
2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2} &\text{if } n > 7 \\
1 &\text{otherwise}
\end{cases}
\end{align}
where $n \in \mathbb{N}$. I would pre... | Actually, after some manipulations, you can use the master theorem! Let us see how. First, let us prove the following lemma:
Lemma: The function $T$ is non-decreasing, i.e. $T(n) \leq T(n+1)$ for all $n \in \mathbb{N}$.
Proof. By strong induction on $n \in \mathbb{N}$.
Base cases: For all $0 \leq n \leq 6$, one has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Resolution of first order differential equation I have difficulties to solve these two differential equations:
1) $ y'(x)=\frac{x-y(x)}{x+y(x)} $ with the initial condition $ y(1)=1 $ .I'm arrived to prove that $$ y=x(\sqrt{2-e^{-2(\ln x+c)}}-1) $$ but I don't know if it's correct. If it's right, how do I find the cons... | HINT
First of all, notice that
\begin{align*}
y^{\prime} = \frac{x-y}{x+y} \Longleftrightarrow y^{\prime} = \frac{1 - \frac{y}{x}}{1 + \frac{y}{x}}
\end{align*}
Then, if we make $y = ux$, we get:
\begin{align*}
u + u^{\prime}x = \frac{1-u}{1+u} \Longleftrightarrow u^{\prime}x = \frac{1-u}{1+u} - u = \frac{1-2u-u^{2}}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$ Problem
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
Atte... | One point here is that there is only a removable singularity in such a function if the numerator and denominator have a common root. Common factors of polynomials, which of course have identical roots, can be found by Euclid's extended division algorithm (here just dividing through gives what you need, but if there wer... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the value of $r$ such that $\frac{1}{\binom{9}{r}} - \frac{1}{\binom{10}{r}} = \frac{11}{6\binom{11}{r}}$? Find the value of $r$:
$$\frac{1}{\dbinom{9}{r}} - \frac{1}{\dbinom{10}{r}} = \frac{11}{6\dbinom{11}{r}}$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many fact... | Making the problem more general
$$\frac{1}{\dbinom{n}{r}} - \frac{1}{\dbinom{n+1}{r}} = \frac{k}{\dbinom{n+2}{r}}$$ and using the same approach as N. F. Taussig in his/her answer, you should end with a quadratic equation in $r$
$$k r^2- \left(k(2 n+3 )+n+2\right)r+k\left( n^2+3 n+2\right)=0$$ If $k >0$, there are two... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to find $a$ and $b$ in $\lim\limits_{x→0}\frac{\sqrt{ax+b}-2}{x}=1$? How to find $a$ and $b$ such that $$\lim_{x→0}\frac{\sqrt{ax+b}-2}{x}=1?$$
By setting the equation equal to one directly I was able to find out that $b = 4$. However, when I try to solve for $a$ I am stuck and left in circles. How would I go about... | You could also do it using binomial expansion or Taylor series (built at $x=0$)
$$\sqrt{a x+b}=\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right)$$ making
$$\frac{\sqrt{ax+b}-2}{x}=\frac{-2+\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right) }x=\frac{\sqrt{b}-2}{x}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other.
Answer:
If I follow algebra, then I get complicated results to solve it as follows:
Let $u=(u_1,u_2,u_3,u_4), \ v=(v_1,... | as columns
$$
\left(
\begin{array}{rrrr}
1&-1&-1&-1 \\
1& 1&-1&-1 \\
1&0 &2&-1 \\
1&0&0&3
\end{array}
\right)
$$
Pattern, done correctly, works in any dimension
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Regarding solving the partial differential equation using the method of characteristics I would like to solve the p.d.e of $(a+b)f_a + (a-b)f_b -f =2$ subject to the initial condition $f(a,0)=0$ with $a > 0$.
So the characteristic equations will be $a_t(t,s)=a+b$, $b_t(t,s)=a-b$ and $f_t(t,s)=f+2$.
Also the parametric... | I didn't check your calculus. Supposing that it is correct, the solution would be :
$$a(t,s)=\frac{1}{4}se^{-\sqrt{2}t}((2+\sqrt{2})e^{2\sqrt{2}t}+2-\sqrt{2})$$
$$b(t,s) = \frac{1}{2\sqrt{2}}se^{-\sqrt{2}t}(e^{2\sqrt{2}t}-1)$$
$$f(t,s) = 2e^t - 2$$
My answer is only about the method to find $f(a,b)$ explicitly.
Let $\q... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The lines that connect the intersections of the three circles around a triangle meet at the same point
For any $\Delta ABC$, construct three circles with radius $AB,CB,AC$ centered at $A,B,C$ respectively. Then, connect the intersections of the circles as shown in the picture. The lines are concurrent and meet at poin... | The lines through two points of intersection of two circles is called the radical axis; the axis is actually defined for any two non-concentric circles, and the radical axis theorem states that the axes for each pair of three given circles are concurrent. So, that property of the configuration in question is unsurprisi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to transform the graph $\sin^{-1}(3x+1)$? So there is this question in which you have to transform the graph of $\sin^{-1}(3x+1)$ and so is the following:
$\sin^{-1}(3(x+\frac{1}{3}))$ with $D: -1 \leq x \leq 1$
However, if one applies the transformation of moving the graph $\frac{1}{3}$ to the left, it results in ... | Here's why ...
$$\begin{align}
y &= \sin^{-1}\bigg(3(x-(-\frac{1}{3}))\bigg)\\
\sin y &= 3(x-(-\frac{1}{3}))\\
\frac{1}{3}\sin y &= x-(-\frac{1}{3})\\
\frac{1}{3}\sin y + (-\frac{1}{3}) &= x\\
\end{align}$$
so after algebraically flipping the graph $45^{\circ}$ we can easily see that we have a compression along the x-a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked t... | Recall (in your question) the line ...
$$=\frac{(n^2+n)(2n+1)+6(n+1)^2}{6}$$
What followed, however was a missed opportunity, for $n^2+n$ factors as $n(n+1)$, so your expression becomes
$$\begin{align}
\frac{(n^2+n)(2n+1)+6(n+1)^2}{6} &= \frac{n(n+1)(2n+1)+6(n+1)^2}{6}\\
&= \frac{(n+1)\bigg(n(2n+1)+6(n+1)\bigg)}{6}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve this set of definite integrals? I'm looking to solve the integrals
$$
I_n=\int_0^\infty \frac{x^ne^{-2ax}}{\sqrt{x^4+1}}dx.
$$
for $a>0$.
This should be reduced to finding $I_0$ as numerically I find that
$$
I_n=\left(-2\right)^{-n}\frac{d^nI_0}{da^n}.
$$
Mathematica gives a result in terms of the MeijerG ... | One way to get the result is to use the Mellin transform of the integral considered as a function of $a$:
\begin{align}
I_0&=\int_0^\infty \frac{e^{-2ax}}{\sqrt{x^4+1}}\,dx\\
\mathcal{M}\left[ I_0\right]&=\int_0^\infty a^{s-1}\,da\int_0^\infty \frac{e^{-2ax}}{\sqrt{x^4+1}}\,dx\\
&=\int_0^\infty \frac{1}{\sqrt{x^4+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find a basis for the range of a transformation Supposed $T(v) = Av$ and say I have a matrix like this:
$$\begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 0 \\ 1 & 7 & -6 \end{bmatrix}$$
and I want to find the basis for the range.
Is this right?
$$\begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 0 \\ 1 & 7 & 6 \end{bmatrix}$$
$$\begin{bmat... | Your reasoning is sound. Elementary row operations don't chance the linearly independentness of columns, so from the reduced row echelon form you arrived at you can see the two first columns are linearly independent, while the third can (allegedly) be written as a sum of the two first. In the original matrix then $\be... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof: For all integers $x$ there exists an integer $y$ so that $3$ divides $x+y$ and $3$ divides $x-y$ We start of a proof by finding out if it is true or false, I am having trouble with this and I think it is true, because the negation (there exists and integer $x$ for all integers $y$ so that $3$ does not divides $x... | Remember that $(a+b)\pmod n = [a\pmod n + b\pmod n] \pmod n$, for any $a$ and $b$ positive or negative.
So, if $x \pmod 3 = 0$, we are done, and $y=0$ for instance.
If $x \pmod 3 = 1$, we may set $y=2$ (or any other $y$ such that $y \pmod 3 = 2$) so that $(x+y) \pmod 3 = 3 \pmod 3 = 0$. But $(x-y) \pmod 3 = (1 + 1) \pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqr... | Since $$\sqrt{2+\sqrt{3}} = \sqrt{4+2\sqrt{3}\over 2} = {1+\sqrt{3}\over \sqrt{2}}$$
\begin{eqnarray}\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}&=& \frac{\sqrt{2}+{1+\sqrt{3}\over \sqrt{2}}}{\sqrt{2}-{1+\sqrt{3}\over \sqrt{2}}}\\
&=& \frac{2+1+\sqrt{3}}{2-1-\sqrt{3}}\\
&=& \frac{3+\sqrt{3}}{1-\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that Ratio Test gives no information for $3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots$
Q: Examine the series:
\begin{align*}
\frac{1}{3} + \frac{1}{5} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \\
\end{align*}
Prove that the Root Tes... | Given the series is $$3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots=\dfrac13+\dfrac15+\dfrac{1}{3^2}+\dfrac1{5^2}+\dfrac1{3^3}+\dfrac1{5^3}+.......$$
Notice that from the above series we get
$$a_{2n-1}=\dfrac{1}{3^n}\mbox{ and }a_{2n}=\dfrac{1}{5^n}$$
Now consider taking $\dfrac{a_{2n+1}}{a_{2n}}$ and $\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Formalization/Verification of a beginner combinatorics problem I have the following task:
To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?
*
*No defective component... | What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12\times11\times10\times 9$ on the bottom and $10\times9\times8\times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, go... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $A^2 +B^2+C^2=D^2$ using the following diagram (tetrahedron) Slicing a corner off a square gives a right-angled triangle, as shown in the diagram below.
The lengths of the sides of this triangle are related by Pythagoras’s theorem: $a^
2 + b^
2 = c^
2$
. Show
that this two-dimensional setup generalises to thr... | Let $XYZT$ be our tetrahedron, where $TX\perp TY$, $TX\perp TZ$, $TY\perp TZ$, $TX=x$, $TY=y$ and $TZ=z$.
Thus, $XY=\sqrt{x^2+y^2}$, $XZ=\sqrt{x^2+z^2},$ $YZ=\sqrt{y^2+z^2}$ and
$$S_{\Delta XYZ}=\frac{1}{4}\sqrt{\sum_{cyc}(2(x^2+y^2)(x^2+z^2)-(x^2+y^2)^2)}=$$
$$=\frac{1}{4}\sqrt{\sum_{cyc}(2x^4+6x^2y^2-2x^4-2x^2y^2)}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Generating Pythagorean triples I'm asked to generate Pythagorean triples from the polynomial identity:
$$(X^2-1)^2 + (2X)^2=(X^2+1)^2$$
By substituting rational numbers $\frac p q$ for $X$. However, Pythagorean triples are just as the name says, it, three numbers. If I would substitute this number I get:
$$(\left(\frac... | If we take the equation asis, it is not true unless $x=0$ or$X=1$.
$$(x-1)^2+2x=(x+1)^2$$
$$(x^2-2X+1)+4X^2=(x^2+2X+1)$$
Subracting $x^2+2x+1$ from both sides we get
$$4X^2-4X=0\Rightarrow4x(X-1)=0$$
A variation of the identity will work for all $n$ in generating triples where $C-A)=2.$
$$\mathbf{A=4n^2-1}$$
$$\mathbf{... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\sin(\pi \sqrt{4n^2+n})$ We can notice that $\sqrt{4n^2 + n} = \sqrt{4n^2(1+ \frac{1}{4n})} = 2n\sqrt{1+\frac{1}{4n}}$. Therefore
$$\lim_{n \to \infty} \sin (2n\pi \sqrt{1+ \frac{1}{4n}}) \text{ will be an even number}$$
Because the square root becomes $1$ and we end up with an even number: $\sin(\text{even n... | When $n\to\infty$, we have
$$
\sqrt{4n^2+n} = 2n\sqrt{1+\frac{1}{4n}}
= 2n\left(1+\frac{1}{8n} + o\!\left(\frac{1}{n}\right)\right)
= 2n + \frac{1}{4} + o(1) \tag{1}
$$
using the Taylor expansion of $\sqrt{1+x}$ for $x$ around $0$. Recalling that $$\forall a,b,\qquad \sin(a+b)=\sin a \cos b + \cos a \sin b \tag{2}$$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integration of $\int^{\pi/45}_0 \frac{x^2 \ln (1-x)}{1-x^3}\, dx$ How would you integrate:
$$I=\int^{\pi/45}_0 \frac{x^2 \ln (1-x)}{1-x^3} dx$$
My attempt:
$$du=\frac{x^2}{1-x^3}dx ⇒ u=\frac{-1}{3}\ln (1-x^3)$$
$$v= \ln (1-x) ⇒ dv=\frac{-1}{1-x}$$
$$I=\frac{-1}{3}\ln (1-x^3) \ln (1-x)-\int \frac{\ln (1-x^3)}{1-x}\,dx$$... | Hint: A better method
Using the partial fraction:
$$\frac{{{x}^{2}}}{1-{{x}^{3}}}=\frac{1}{3}\frac{1}{1-x}-\frac{1}{3}\frac{2x+1}{{{x}^{2}}+x+1}=\frac{1}{3}\frac{1}{1-x}-\frac{2}{3}\frac{1}{2x-i\sqrt{3}+1}-\frac{2}{3}\frac{1}{2x+i\sqrt{3}+1}$$
The integral becomes
$$\begin{align}
& I=\frac{1}{3}\int{\frac{\ln \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{... | One Approach
Simpler than Cauchy-Schwarz is to note that for $x\in(n,n+1)$,
$$
\frac1{n+1}\lt\frac1x\lt\frac1n\tag1
$$
which, when integrated in $x$ over $(n,n+1)$, gives
$$
\frac1{n+1}\lt\log\left(\frac{n+1}n\right)\lt\frac1n\tag2
$$
then
$$
\begin{align}
\sum_{k=n}^{3n}\frac1{k+1}
&\le\sum_{k=n}^{3n}\log\left(\frac{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the remainder when $13^{13}$ is divided by $25$. Find the remainder when $13^{13}$ is divided by $25$.
Here is my attempt, which I think is too tedious:
Since $13^{2} \equiv 19 (\text{mod} \ 25),$ we have $13^{4} \equiv 19^{2} \equiv 11 (\text{mod} \ 25)$ and $13^{8} \equiv 121 \equiv 21 (\text{mod} \ 25).$
Finall... | We have
*
*$2\cdot 13 = 26 \equiv 1 \mod 25$
*$\Rightarrow 13^{13} \cdot \color{blue}{2^{13}} \equiv 1 \mod 25$
*$\color{blue}{2^{13}} \equiv 2^{10}\cdot 2^3 \equiv -8 \color{blue}{\equiv 17 \mod 25}$
$$\Rightarrow \boxed{13^{13} \equiv 17^{-1} \equiv 3 \mod 25}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the smallest positive integer $X$ such that $478^{870} \equiv X \ (\text{mod} \ 273)$ Appreciate if one could advise if my solution is correct. Here is my attempt of the problem:
Since $(273, 478) =1,$ by Euler's theorem, $478^{\phi(273)}=478^{144} \equiv 1 \ \ (\text{mod} \ 273) \implies 478^{864} \equiv 1 \ \ (... | Recall Carmichael's theorem: $a^{\lambda(m)} \equiv 1 \bmod m$ for all $a$ coprime with $m$.
Now $\lambda(273)=lcm(\phi(3),\phi(7),\phi(13))=lcm(2,6,12)=12$ and $870 \equiv 6 \bmod 12$.
Therefore
$
478^{870} \equiv 205^{6} \equiv (-68)^{6} \equiv 4^6 \cdot 17^6 \equiv 1 \bmod 273
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational?
The number $x$ defined below is rational or irrational?
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$
From: IMO 1973 - Longlist
My attempt (my real question is at the end):
the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-ac-bc)$ whe... | Alternatively
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}=\\
\sqrt[3]{\frac{8\sqrt{5}+16}{8}}+\sqrt[3]{\frac{8\sqrt{5}-16}{8}}=\\
\sqrt[3]{\frac{5\sqrt{5}+3\cdot5+3\sqrt{5}+1}{8}}+\sqrt[3]{\frac{5\sqrt{5}-3\cdot5+3\sqrt{5}-1}{8}}=\\
\sqrt[3]{\frac{\left(\sqrt{5}+1\right)^3}{2^3}}+\sqrt[3]{\frac{\left(\sqrt{5}-1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Average height of raindrops from height $h$
Raindrops drip from a spout at the edge of a roof and fall to the ground. Assume that the drops drip at a steady rate of $n$ drops per second (where $n$ is large) and that the height of the roof is $h$.
(c) What is the average height of these raindrops as $n$ grows larger wh... | $\def\peq{\mathrel{\phantom{=}}{}}$Assume instead that there are $n$ drops per $T$ seconds so that $n$ becomes dimensionless, thus the time interval between two drops is $Δt = \dfrac{T}{n}$.
It takes $t_1 = \sqrt{\dfrac{2h}{g}}$ for the first drop to hit the ground, and when it hits, there are $N = \left[ \dfrac{t_1}{Δ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find maximum value Given $0 \leq a,b,c \leq \dfrac{3}{2}$ satisfying $a+b+c=3$. Find the maximum value of $$N=a^3+b^3+c^3+4abc.$$
I think the equality does not occur when $a=b=c=1$ as usual. I get stuck in finding the strategy, especially I don't know what to do with $4abc$. Thank you a lot.
| For $a=b=\frac{3}{4}$ and $c=\frac{3}{2}$ we get a value $\frac{243}{32}.$
We'll prove that it's a maximal value.
Indeed, $a+b-c=3-2c\geq0,$ which says that there are non-negatives $x$, $y$ and $z$ for which
$a=y+z$, $b=x+z$ and $c=x+y$.
Id est, we need to prove that
$$\sum_{cyc}(x+y)^3+4\prod_{cyc}(x+y)\leq\frac{9(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$
Evaluate :
$$\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$$
This is my attempt in below ... but what I want is simplify more t... | I really do not see how you could simplify the monster. I really wonder if a closed form expression could exist. I would try numerical integration.
We can get an approximate solution using the following series expansions at $x=0$.
$$\log (\Gamma (x))=-\log (x)-\gamma x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.