Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Inverse Fourier transform with a $\delta$-function integrand I'm self-studying math and trying to find the inverse Fourier transform of $\frac{4+w^2}{1+w^2}(4\pi * (\delta(w-2)+\delta(w+2)))$
Based on wolframalpha, the result is $32/5\sqrt{2\pi}\cos(2t)$.
But I can't even find the fourier transform of $\frac{4+w^2}{1+w... | You don't need to know, just apply the definition
\begin{eqnarray}
f(t) &=& \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2}[4\pi\delta(\omega - 2) + 4\pi\delta(\omega + 2)] e^{i\omega t}{\rm d}\omega \\
&=& 2\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2}4\pi\delta(\omega - 2) e^{i\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I... | Alternatively, use Formulas for generating Pythagorean triples
WLOG $a=2pqk, b=(p^2-q^2)k,c=(p^2+q^2)k$
$c-a=k(p-q)^2$
$c-b=2kq^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Seeking Methods to solve $F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$ I'm looking for different methods to solve the following integral.
$$ F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$$
For $\alpha > 0$
Here the method I took was to employ integration by parts and then call to special ... | Here is a method that relies on using a double integral.
Noting the integral converges for $\alpha > -2$, recognising
$$\arcsin x = \int_0^x \frac{du}{\sqrt{1 - u^2}},$$
the integral can be rewritten as
$$I = \int_0^1 \int_0^x \frac{x^\alpha}{\sqrt{1 - u^2}} \, du dx.$$
On changing the order of integration, one has
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Minimizing distance from a point to a parabola Problem: The point on the curve $x^2 + 2y = 0$ that is nearest the point $\left(0, -\frac{1}{2}\right)$ occurs at what value of y?
Using the distance formula, I get my primary equation: $L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
However, when using the se... | The correct solution is $(0,0)$. Your first approach is correct.
For the second approach, note that the domain of $D(y)$ is $y \le 0$, so the solution $y= \frac12$ cannot be correct.
Also note that $D$ is always decreasing for $\forall y \le 0$, therefore $D(y)$ attains its minimum at $y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that point does not belong to a plane Problem
Show that point $\textbf{q}$ does not belong to plane defined by these 3 points:
$$ \textbf{p}_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \textbf{p}_2=\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \textbf{p}_3 = \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} $$
Point $\... | Let me check your solution:
$$(-2) \left( -\frac{15}{19}\right)+(-2) \left( \frac{1}{19}\right)= \left( \frac{2}{19}\right)(15-1) \ne 1$$
Hence you made a mistake in solving for your $s$ and $t$.
Here is the RREF, we can see that the augmented matrix is not consistent.
octave:1> rref([-2 -2 1; 0 -3 0 ; -2 -1 2])
ans =
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trouble to find the solution to a linear system where a matrix is not invertible Hello community I am new here and I have a question which might be pretty basic.
So I am trying to solve an equation. I have 3 matrices
A = \begin{pmatrix}1&0&0&0&0&0&0&0\\0&1&1&0&0&0&0&0\end{pmatrix}
B = \begin{pmatrix}0&0&0&0&0&0&1&0\\0... | For the sake of simpler notations, let us rename $$L:=\begin{pmatrix}H\\K\end{pmatrix} \ \ \text{and} \ \ P:=\begin{pmatrix}0\\Z\end{pmatrix} $$
giving :
$$Lv=P \tag{1}$$
which is a $7 \times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).
Instead of your $$ v = (L^T L)^{-1}L^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the Range of $5|\sin x|+12|\cos x|$ What is the Range of $5|\sin x|+12|\cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $\sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
| If $f(x) = 5|\sin x| + 12 |\cos x|$, then
\begin{align*}
f(x) &= \sqrt{f(x)^2} \\
&= \sqrt{25 \sin^2 x + 144 \cos^2 x + 60 |\sin x \cos x|} \\
&= \sqrt{25 + (144 - 25) \cos^2 x + 60 |\sin x \cos x|} \\
&\ge 5
\end{align*}
with equality obtained when $\cos x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How many $n$-digit sequences of $0,1 \ \text{or} \ 2$s contain an odd numbers of $0$s? Problem : There are $3^n$ n-digit sequences in which each digit is $0$, $1$ or $2$. How many of these sequences have an odd number of $0$'s ?
Let $o(n)$ = the number of n-digit sequences which have an odd number of $0$'s and
$e(n)$ =... | Assume that $n$ is odd, $n = 2s+1$
Note that number of $n$-digit sequences with $k$ number of zeroes is ${N_k}(n) =$ $n \choose k$$2^{n-k}$.
Thus, $$o(n) = \sum_{m=0}^s{N_{2m+1}}(n) = \sum_{m=0}^s{n \choose 2m+1}{2^{n-2m-1}} = \sum_{m=0}^s{n \choose 2m+1}{1^{2m+1} 2^{n-2m-1}} \\ = \frac{(2+1)^n-(2-1)^n}{2} = \frac{3^n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3041858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\big... | Hint: your fraction is $(x^2+1)^{-1}+\frac{x^2}{x^4+1}$, and $\frac{x^2}{x^4+1}=\frac{ax+b}{x^2-\sqrt{2}x+1}+\frac{-ax-b}{x^2+\sqrt{2}x+1}$ for some real numbers $a,b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Simplify the determinant of a $4 \times 4$ matrix. I have to find the determinant of the following 4x4 matrix:
$\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bma... | Just expand by the first row:
\begin{align}
\begin{vmatrix}3&0&1&0 \\
0&2&0&0 \\
1&0&3&0 \\
0&0&0&-4
\end{vmatrix}&=
3\,\begin{vmatrix}
2&0&0 \\
0&3&0 \\
0&0&-4
\end{vmatrix}+
1\,\begin{vmatrix}
0&2&0 \\
1&0&0 \\
0&0&-4
\end{vmatrix}
=3(2\cdot3\cdot(-4))+1(-1)\begin{vmatrix}
2&0 \\
0&-4
\end{vmatrix}\\
&=-72+8=-64.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^... | Just to give another approach, let $x^2+1=u^2$, in which case $xdx=udu$ and we have
$$\int{x^3\over\sqrt{x^2+1}}dx=\int{x^2\cdot xdx\over\sqrt{x^2+1}}=\int{(u^2-1)udu\over u}=\int(u^2-1)du={1\over3}u^3-u+C\\={1\over3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C$$
The main virtue (if any) of this approach is that it gets rid of the sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
exponential equation has non positive roots
Find real values of $a$ for which the equation
$4^x-(a-3)\cdot 2^x+(a+4)=0$
has non positive roots
Try: Let $2^x=y\in (0,1].$ Then equation convert into
$y^2-(a-3)y+(a+4)=0$
For real roots its discriminant $\geq 0$
So $$(a-3)^2-4(a+4)\geq 0$$
$$a^2-10a-7\geq 0$$
$$a\in \big... | You want $f(y) = y^2 - (a-3) y + (a+4)$ to have at least one root in $(0,1]$.
Note that $f(0) = a+4$ and $f(1) = 8$, while the minimum value of $f(y)$, occurring at
$y = (a-3)/2$, is $(-a^2 + 10 a + 7)/4$. There are several cases to consider.
*
*If $a < -4$, $f(0) < 0 < f(1)$ so there is a root in $(0,1]$.
*For $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
what went wrong in solving this singular ODE? I want to solve $$x(2-x) y'' +(1+x)y' -\frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r \sum \limits_{n=0}^{\infty} a_n x^n$. i found that $r = -1, \frac{3}{2}$. substituting the value $r=-1$ and the standar... | You have
$$
a_{n + 1} = a_n\frac{n^2 - 4n + 3}{2n^2 - n - 3}
$$
Using $a_0 = \color{blue}{1}$ you get
$$
a_1 = 1 \frac{3}{(-3)} = \color{red}{-1}
$$
and
$$
a_2 = 0 = a_3 = \cdots
$$
So one of the solutions is
$$
y_1(x) = x^{-1}\left(a_0 + a_1 x + a_2 x^2 + \cdots \right) = x^{-1}\left[\color{blue}{1} + (\color{red}{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $T(n) = T(2n/3) + \lg^2 n$ by substitution? My solution through substitution is as follows:
$$T(n) = T(2n/3) + \lg^2 (n)$$
$$T(2n/3) = T(4n/9) + \lg^2 (2n/3)$$
$$T(4n/9) = T(8n/27) + \lg^2 (4n/9)$$
And so on...
But my actual problem is how can I calculate the below step which cause to obtain order of the a... | $$T(n)=T\left(\frac{2n}{3}\right)+\lg^2 n$$
$$T\left(\frac{2n}{3}\right)=T\left(\frac{2\cdot\frac{2n}{3}}{3}\right)+\lg^2\left(\frac{2n}{3}\right)=T\left(\frac{2^2n}{3^2}\right)+\lg^2\left(\frac{2n}{3}\right)$$
$$T\left(\frac{2^2n}{3^2}\right)=T\left(\frac{2^3n}{3^3}\right)+\lg^2\left(\frac{2^2n}{3^2}\right)$$
$$\dots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,c\in R,a\ne0$)
$(1)\;5a$
$(2)\;3b$
$(3)\;2c$
$(4)\;0$
As $x=1$ is the root of t... | The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, \quad \frac14 \left(-3-i\sqrt{31} \right), \quad \mbox{and} \quad \frac14 \left(-3+i\sqrt{31} \right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=\frac14 \left(-3-i\sqrt{31} \right)$ (to check your computation maybe use wolf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Equivalence of two antiderivatives involving trigonometric/hyperbolic functions I am struggling to see how two antiderivatives of the same function—obtained in two different ways—are equivalent (what I mean by equivalent is that they differ from just a constant), if they even are equivalent.
The function in question is... | Whenever you face this problem, you can check the answer by differentiating. If
$$
F(x)=\frac{2}{3} \arctan \left(\frac{e^{x} + \sqrt{e^{2x}-9}}{3}\right) + C
$$
then
\begin{align}
F'(x)
&=\frac{2}{3}\frac{1}{1+\left(\dfrac{e^{x} + \sqrt{e^{2x}-9}}{3}\right)^2}
\frac{1}{3}\left(e^x+\frac{e^{2x}}{\sqrt{e^{2x}-9}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimizing area of ellipse Find minimum area of an ellipse that can pack three unit circles such that all three touch the ellipse internally:
I took a point H as shown in the diagram and used the fact that the radius of the circle is 1, and that the circle touches the ellipse at point H. I am getting four equations fo... | Let me try to help with the discriminant.
Start with $$\frac{x^2}{a^2}+ \frac{y^2}{(1+2\cosθ)^2}=1$$
Put $$y^2=1−(x−2 \sinθ)^2$$
So the equation becomes
$$\frac{x^2}{a^2}+ \frac{1−(x−2 \sinθ)^2}{(1+2\cosθ)^2}=1$$
Compare it with the standard quadratic equation $px^2 + qx + r = 0$. If the discriminant is zero, then the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$ I'm trying to evaluate the following integral:
$$\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$$
I was playing around trying to numerically approximate the answer with known constants and found that the integral is almost exactly $\frac{\pi^4}{64}$. The int... | As MrTaurho pointed out in the comments, we can rewrite $\,\displaystyle{\operatorname{arctanh}x=\frac12 \ln\left(\frac{1+x}{1-x}\right)}$, this gives:
$$I=\int_{0}^1 \frac{\operatorname{arctanh}(x)^3}{x}dx=-\frac18\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx$$
And we will substitute $\displaystyle{\frac{1-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$
\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\rig... | Show by induction that
$a_1=1!/0!$
$a_2=2!/0!+2!/1!$
$a_3=3!/0!+3!/1!+3!/2!$
...
$\color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$
Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of $\sqrt {(x+2)^2+(y+2)^2}+\sqrt {(x+1)^2+(y-1)^2}+\sqrt{ (x-1)^2+(y+1)^2}$ Let $x;y\in R$. Find Minimum value of the function $$\sqrt {(x+2)^2+(y+2)^2}+\sqrt {(x+1)^2+(y-1)^2}+\sqrt{ (x-1)^2+(y+1)^2}$$
My try: By Minkowski inequality:
$LHS=\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+2\sqrt{\dfrac{1}{... | FERMAT POINT
I get, from the $120^\circ$ construction, that the minimizing point is at $$ \left(\frac{-1}{\sqrt 3},\frac{-1}{\sqrt 3} \right)$$
The main calculation I did was $45^\circ + 60^\circ = 105^\circ$ and $\tan 105^\circ = -2-\sqrt 3.$ The sum of three distances is
$$ \sqrt 6 + \sqrt 8 \approx 5.2779168675293... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve this ode $xy''-(1+x)y'+y=x^2$? Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.
I have found that $a=1$ or $a=1/x$.
| The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
$$
\frac{xy''-y'}{x^2} + \frac{y-xy'}{x^2} = 1,
$$
which can be rewritten as
$$
\Big( \frac{y'}{x} \Big)' - \Big( \frac{y}{x} \Big)' = 1.
$$
Taking the antiderivative gives
$$
\frac{y'}{x} - \frac{y}{x} = x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\... | Using the method I employed here: we observe that
\begin{equation}
\int_{0}^{\infty} \frac{x^2}{\left(x^4 + 1\right)^2}\:dx = \frac{1}{4} \cdot 1^{\frac{2 + 1}{4} - 2} \cdot B\left(2 - \frac{2 + 1}{4}, \frac{2 + 1}{4} \right) = \frac{1}{4}B\left(\frac{5}{4}, \frac{3}{4}\right)
\end{equation}
Using the relationship betw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\le... | We have
\begin{align}
\sum_{n=1}^\infty\frac{H_n}{n^2+n}
&=\sum_{n=1}^\infty H_n\left(\frac1n-\frac1{n+1}\right)
\\&=\sum_{n=1}^\infty\sum_{m=1}^n\frac1m\left(\frac1n-\frac1{n+1}\right)
\\&=\sum_{m=1}^\infty\frac1m\sum_{n=m}^\infty\left(\frac1n-\frac1{n+1}\right)
\\&=\sum_{m=1}^\infty\frac1m\cdot\frac1m
\\&=\frac{\pi^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\c... | A slightly different approach using de Moivre's theorem would be: notice that $1 + i \tan \theta = \sec \theta \cdot e^{i \theta}$. Therefore, cubing both sides, $(1 + i \tan \theta)^3 = \sec^3 \theta \cdot e^{3 i \theta}$. Now, if you take the "slope" of this complex number, i.e. the imaginary part divided by the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
The closed form representations of Integrals of logarithm functions I wish to find a closed form representations of the following integral
$$\int\limits_{0}^1\frac{\log^p(x)\log^r\left(\frac{1-x}{1+x}\right)}{x}dx=?$$
Here $p\ge 1$ and $r\ge 0$ are nonnegative integers. It can be expressed in terms of a linear combinat... | Here is an alternative approach for the case $r = 1, p \in \mathbb{N}$ that has already been given by @Ininterrompue.
Let
$$I_p = \int_0^1 \ln^p x \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{x}.$$
Since
$$\ln \left (\frac{1 - x}{1 + x} \right ) = - 2 \sum_{n = 0}^\infty \frac{x^{2n+1}}{2n + 1}, \qquad |x| < 1,$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to find a recurrence relation for counting the number of solutions? Consider the diophantine equation
$$x_1+3x_2+5x_3 = n$$
where $x_i\geq 0$ and $n\geq 1.$
Let $P_n(1,3,5)$ denote the number of solutions to this equation. I want to express $P_n(1,3,5)$ in terms of $P_{1}(1,3,5), P_{2}(1,3,5),\cdots, P_{n-1}(1,3,... | One way to solve this is with generating functions. For $x_1$ you get a factor $(1 - z)^{-1}$, $x_2$ gives $(1 - z^2)^{-1}$, $x_3$ adds $(1 - z^3)^{-1}$. Pulling all together:
$\begin{align*}
[z^n] \frac{1}{(1 - z) (1 - z^2) (1 - z^3)}
&= \frac{1}{24} [z^n] \frac{11 + 3 z + 3 z^2}{1 - x^3}
+ \frac{1}{8} [z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to evaluate analytically the following infinite double sum? Consider the following double sum
$$
S = \sum_{n=1}^\infty \sum_{m=1}^\infty
\frac{1}{a (2n-1)^2 - b (2m-1)^2} \, ,
$$
where $a$ and $b$ are both positive real numbers given by
\begin{align}
a &= \frac{1}{2} - \frac{\sqrt{2}}{32} \, , \\
b ... | Let
$$q=\sqrt{\dfrac{b}{a}},$$
then attack via digamma-function leads to
$$\begin{align}
&S = \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{(2n-1)^2-q^2(2m-1)^2} \\[4pt]
&= \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{2(2n-1)}\left(\dfrac1{2n-1-q(2m-1)}+\dfrac1{2n-1+q(2m-1)}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$ Let $m,n\in\mathbb N-\{1,2\}$ such that $m\ne n.$ How to show that
$$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}?$$
Please help me. I am clueless.
| I am going to prove this for cases. First if $m < n$:
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{m}-\frac{1}{n}
$$
For all $n\in \mathbb{N}$
$$
\frac{2}{n+2} > 0
$$
Then
\begin{eqnarray}
(n-1) + \frac{2}{n+2} &>& (n-1) \\
\frac{n^2+n-2 +2}{n+2} &>& n-1 \\
\frac{n^2+n}{n+2} &>& n-1 \\
\end{eqnarray}
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \d... | Set $1/x=h$
$$\lim_{h\to0^+}\dfrac{2h+(3h+1)\{\ln(1+ah)-\ln(1+bh)\}}{h^2}$$
$$=3\left(a\lim_{h\to0^+}\dfrac{\ln(1+ah)-1}{ah}-b\lim_{h\to0^+}\dfrac{\ln(1+bh)-1}{bh}\right)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$
$$=3(a-b)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2\lim_{h\to0^+}\dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding remainder when unknown $f(x)$ is divided to $g(x)$ When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.
Using the... | No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+\frac{6}{5}x + \frac{13}{5}$$
with any polynomial $q(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Finding all roots of $x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$, knowing that one root is $x=-1+i$.
Find all roots of
$$V(x) = x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$$
knowing that one root is $x=-1+i$.
Sorry for the picture. I found two roots of the polynomial and also found an equation which can help me to find all... | To solve $$x^2+2x+2=0$$ use $$x^2+2x+1=-1$$ hence $$(x+1)^2=-1$$ which gives the solutions $\ -1-i\ $ and $\ -1+i\ $
To solve $$x^4+3=0$$ first note that the absolute value of the solutions must be $3^{\frac{1}{4}}$ and then use the $4$ fourth roots of $-1$ , being $$\pm \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}i$$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $a, b\in\mathbb{R}$ so that $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite.
Find $a, b\in\mathbb{R}$ so that
$$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.
I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?
| Hint:
Use the following lemma:
Lemma: $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.
Proof: If $ax^6+bx^5+1 = p(x)\cdot (x-1)^2$ then $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving a strict inequality in the limit I want to prove that
$$ \lim_{k \to \infty} \left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < e .$$
Using the $AM-GM$ inequality we arrive at
$$\left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k... | $$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\frac{1}{2^k}=1,$$
then exponentiate both sides. There is also a simple strengthening, namely
$$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\left(\frac{1}{2^k}-\frac{1-\log 2}{4^k}\right)=\frac{2+\log 2}{3},$$
leading to
$$ \prod_{k\geq 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to 1}\frac{^5x-^4x}{(1-x)^5}$, where $^nx$ is the repeated exponent ("tetration") operation
$$\lim_{x\rightarrow1}\frac{x^{x^{x^{x^x}}}-{x^{x^{x^x}}}}{(1-x)^5}$$
My friends told that it appeared on Instagram (some social media network). I tried various methods but failed. (I tried using Taylor ser... | Start using composition of Taylor series using mainly $t=e^{\log(t)}$. Then
$$x^x=1+(x-1)+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+\frac{1}{12}
(x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^x}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{4}{3} (x-1)^4+\frac{3}{2}
(x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^{x^x}}=1+(x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Cannot find angle for trigonometry problem
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $\me... | We have
$$2R=\frac{r}{\tan(\alpha/2)}+\frac{r}{\tan(\beta/2)},$$
hence
$$\frac{13}{2}=\frac{1}{\tan(\alpha/2)}+\frac{1}{\tan(\beta/2)}.$$
We know that $\alpha+\beta=\pi/2$, so $\beta/2=\pi/4-\alpha/2$. Let $t=\tan(\alpha/2)$. From the addition theorems for $\sin$ and $\cos$ derive that
$$\tan(\pi/4-\alpha/2)=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\sum_{n=1}^\infty \frac{\sin(nx)}{n}$ without integrating $\sum_{n=1}^\infty e^{nx}$ I am looking for alternative solutions for finding this sum
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n} $$
My solution proceeds by integrating $$\sum_{n=1}^\infty e^{nx}=\frac{e^{ix}}{1-e^{ix}}$$
With suitable limits and then t... | Note that\begin{align}\sum_{n=1}^\infty\frac{\sin(nx)}n&=\operatorname{Im}\left(\sum_{n=1}^\infty\frac{e^{inx}}n\right)\\&=\operatorname{Im}\left(\sum_{n=1}^\infty\frac{\left(e^{ix}\right)^n}n\right).\end{align}If $x\in\mathbb{R}\setminus\{0\}$, then $e^{ix}\in S^1\setminus\{1\}$, and therefore$$\sum_{n=1}^\infty\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $n$ such that polynomial is divisible Find $n \in N$, such that:
$$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$
If I let $Q(x) = x^2 + x + 1$, I have that $x^3 \equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
I suppo... | Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $\omega$ is a root of $x^2 + x + 1$, we want
$\omega$ to be a root of $P_n(x)$ of multiplicity $\ge 2$,
which means both $P_n(\omega) = \omega^n + (\omega+1)^n + 1 = 0$ and
$P_n'(\omega) = n \omega^{n-1} + n (\omega+1)^{n-1} = 0$. This last
s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to find the value this sum converges to?$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $ How to find the value this sum converges to?$$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$\sum_{n=2}^{\infty}\frac{1/2}{(2n+1)}+\frac{3/2}{(2n-1)}-\frac{1}{n}$$
and writing some terms and I ge... | Proceeding on the partial fraction decomposition you already got
$$
\eqalign{
& S = \sum\limits_{2\, \le \,n} {{{n + 1} \over {n\left( {2n - 1} \right)\left( {2n + 1} \right)}}} = \cr
& = \sum\limits_{2\, \le \,n} {{{1/2} \over {\left( {2n + 1} \right)}} + {{3/2} \over {\left( {2n - 1} \right)}} + {1 \over n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of series: 9 + 16 + 29 + 54 + 103 + ... The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + ... T_n$
$S = 0 + 9 + 16 + 29 + 54 + ... T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1}) - T_n$
... | From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + \sum_{k=1}^n (6(2^{n-2}) + 1) \\
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 \cdot 2^n$$
The sum is then
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the set of the real numbers $x$ satisfying the inequalities $|x+4|<|2x-1|$ and $|x|+|x+1|<3$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
1) $|x+4|<|2x-1|$
*
*If $x<-4$, we have $x+4<0$ and $2x-1<0$, Then $|x+4|=-x-4$ and $|2x-1|=1-2x$
hence, in this c... | Problem 1.
If you square it you get $$x^2+8x+16<4x^2-4x+1$$
so $$ 3x^2-12x-15>0\implies 3(x-5)(x+1)>0$$
so $x\in(-\infty,-1)\cup (5,\infty)$
And a solution to the second problem you solved is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Closed form expression for infinite series I was given the following function:
$$ f(x) = x + \frac{2x^3}{1\cdot3} + \frac{2\cdot4x^5}{1\cdot3\cdot5} + \frac{2\cdot4\cdot6x^7}{1\cdot3\cdot5\cdot7}... $$ $$ \forall x \in [0,1) $$
And then I was asked to find the value of $ f(\frac{1}{\sqrt{2}}) $, which obviously ... | I would suggest the following representation
$$\sum_{n=0}^{\infty}\frac{x^{2n+1}(2n)!!}{(2n+1)!!}=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}$$
Plug in $x=\frac{1}{\sqrt{2}}$. The result will be
$$\frac{\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{1-\frac{1}{2}}}=\frac{\sqrt{2}\pi}{4}$$
Update:
Your question is similar to pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 1... | Completing the square;
$$4x^2+8x+5=(2x+2)^2+1,$$
so solving the congruence $\;4x^2+8x+5\equiv 0\mod 13$ amounts to solving
$$\bigl(2(x+1)\bigr)^2\equiv -1\mod 13,$$
and ultimately to finding the square roots $y$ of $-1\bmod 13$ (we know this is possible because $13\equiv 1\mod 4$).
Note that $5^2\equiv -1\mod 13$, hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides ... | It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2\sqrt{a(b-a)}>0$, hence
$$2\sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please
| Use substitution method:
$$(x-4)^2+(y^2-3)^2=1 \Rightarrow y^2=\pm\sqrt{1-(x-4)^2}+3\\
f(x,y)=x^2+y^4 \Rightarrow \\
f(x)=x^2+1-(x-4)^2\pm6\sqrt{1-(x-4)^2}+9\\
f'(x)=8+\frac{\mp6(x-4)}{\sqrt{1-(x-4)^2}}=0 \Rightarrow \\
4\sqrt{1-(x-4)^2}=\pm3(x-4) \Rightarrow \\
16(-x^2+8x-15)=9(x^2-8x+16) \Rightarrow \\
25x^2-200x+384... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
equating coefficients in algebraic expansion If $\displaystyle \bigg(\frac{1+x}{1-x}\bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots +\infty,$ then value of
$(1)\; \displaystyle \frac{3b_{3}-b_{1}}{b_{2}}$
$(2)\; \displaystyle \frac{2b_{4}-b_{2}}{b_{3}}$
$(3)\; \displaystyle \frac{3b_{6}-2b_{4}}{b_{5}}$
$(4)\; \displaystyle... | You made a mistake. It is not $\displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=\frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$.
By brut force :
$b_{4}=\frac23(2n^2+n^4)$
$b_{5}=\frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=\frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$ for natural number $n$
Prove that
$$n^n\left(\frac{n+1}{2}\right)^{2n}\geq (n!)^3$$ for a natural number $n$.
what i try
i have use AM GM inequality
$$\frac{1^3+2^3+3^3+\cdots +n^3}{n}\geq ((n!))^{\frac{1}{3n}}$$
how i prove question inequality help me please... | First of all, by expanding the terms we obtain$$n^n\bigg(\frac{n+1}{2}\bigg)^{2n}{={1\over n^n}\cdot \bigg(\frac{n(n+1)}{2}\bigg)^{2n}\\={1\over n^n}\cdot (1+2^3+\cdots +n^3)^n.}$$By substitution and taking the $n$-th root, we need to prove that $${1+2^3+\cdots+n^3\over n}\ge \sqrt[n]{n!^3}=\sqrt[n]{1^3\cdot 2^3 \cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the $n$th derivative of $\sin(x)$? I don't understand why for $f(x) = \sin(x)$, $\;f^{(n)}(x) = \sin(n\pi/2 + x)$
It does not make sense to me; can anyone help with the reasoning?
Thank you.
| Part I:
$\sin'(x) = \cos (x)$ and $\cos'x = -\sin x$.
If we iterate these $n$ times then for $f(x) = \sin x$ if
$n = 4k$ for some $k$ then $f^n(x) =\sin x$.
If $n = 4k + 1$ then $f^n(x) = \cos x$
If $n = 4k + 2$ then $f^n(x) = -\sin x$
And if $n = 4k + 3$ then $f^n(x) = -\cos x$.
.....
Now lets walk away and notice.
P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find permutations whose third power is known I have to find permutations $a$ such that $a^3=(1 \ 2)(3 \ 4)(5 \ 6)(7 \ 8 \ 9 \ 10)$
and I have to find at least 3 solutions.
So first I must find disjoint cycles.
Those are:
1 [2,2,2][4] and 2 [4,2][4] and 3[6][4]
are there any more cycles?... | Yes for $p=(1 \ 2)(3 \ 4)(5 \ 6)(10 \ 9 \ 8 \ 7)$.
No for $p=(1 \ 3 \ 2 \ 4)(5 \ 6)(7 \ 10 \ 9 \ 8)$ ; it isn't a solution because $p^3$ would send $1$ onto $4$ instead of $2$.
Here is a way to find many solutions :
First, as you have well seen it, we must take for the last cycle $\color{red}{the \ reversed \ cycle \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$ I am a beginner at trigonometry, I want to know the answer to this question.
If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} =... | Hint:
$\cos\dfrac{14\pi}{15}=\cos\left(\pi-\dfrac\pi{15}\right)=?$
Use $\sin2x=2\sin x\cos x\iff\cos x=\dfrac{\sin2x}{2\sin x}$ repeatedly
Finally here $\sin\dfrac{16\pi}{15}=\sin\left(\pi+\dfrac\pi{15}\right)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem,
I started with the 'hig... | It just jumped out at me that if I define $y=x+1$ the first two become
$$3y \equiv 0 \pmod 5\\3y \equiv 0 \pmod 7$$
which clearly calls for $y$ to be a multiple of $35$. Now we have $$3y \equiv 9 \pmod {11}\\y \equiv 3 \pmod {11}$$
We note that $35 \equiv 2 \pmod {11}$ so $7 \cdot 35 = 245 \equiv 7\cdot 2 \equiv 14\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $a,b$ at $f(x)=\frac{x^2+x-12}{x^2-ax+b}$ An High school question:
Given :$$f(x)=\frac{x^2+x-12}{x^2-ax+b}$$ it's given that $x=3$ is a vertical asymptote find $a$ and $b$.
I tried:
Since $x=3$ is a vertical asymptote then $3^2-3a+b=0$, but now what
| You can try also like that: Since $b= 3a-9$ we have $$f(x) = {(x+4)(x-3)\over x^2-ax+3a-9}= {(x+4)(x-3)\over (x-3)(x+3-a)} ={x+4\over x+3-a}$$
so $3-a=-3\implies a = 6$ and $b= 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Evaluate the indefinite integral $ I = \int (x^2 + 2x)\cos(x) dx $
Integration by Parts, choose $u$:
$$\begin{align}
u &= \cos(x) \\
dv &= (x^2 + 2x)dx \\
du &= -\sin(x) \\
v &= \frac{1}{3}x^3 + x^2
\end{align}
$$
Substitute into formula:
$$
\begin{align}
\int udu &= uv - \int vdu \\
&= \cos(x)\left(\frac{1}{3}x^3 + x^... | I would do this way
$$\int(x^2+2x)\cos x\ dx$$
By Integration By Parts: $u=(x^2+2x),v^{\prime}=\cos x$
$$=(x^2+2x)\sin x-\int(2x+2)\sin xdx$$
Again apply Integration By Parts for $\int(2x+2)\sin xdx$ $u=(2x+2), v^{\prime}=\sin x$
and we get
$\int(2x+2)\sin xdx=2\sin x-\cos x(2x+2)$
So, we finally get,
$$=(x^2+2x)\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion of $\int_0^1 \frac{\operatorname{K}(r x)}{\sqrt{(1-r^2 x^2)(1-x^2)}} \, \mathrm{d} x $ Notation: For $\varphi \in [0,\frac{\pi}{2}]$ and $k \in [0,1)$ the definitions
$$ \operatorname{F}(\varphi,k) = \int \limits_0^\varphi \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin^2(\theta)}} = \int \limits_0^{\sin... | EllipitcElena's answer and Maxim's correction show that we have
$$ \psi(k) = - \frac{1}{4} \operatorname{K}(k) \log(1-k^2) + 2 \log(2) \operatorname{K}(k) + \chi(k) \, ,$$
where ($\psi_0$ is the digamma function, so $\psi$ was not the best choice)
$$ \chi (k) = \sum \limits_{m=1}^\infty \frac{\left(\frac{1}{2}\right)_m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving that two determinants are equal without expanding them So I need to prove that
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
$$
= \begin{vmatrix}
... | $$\begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\
\sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\
\sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\
\end{vmatrix}=\begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin\alpha\color{red}{\cos\delta}+\color{red}{\sin\delta}\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Asymptotic behavior of the integral $H(\beta)=\frac{2}{\pi}\beta\int_{0}^{\infty}\exp\left(-x^{3/2}\right)\sin(\beta x)xdx$ I found the integral $H(\beta)$ (which is called Holtsmark distribution) in Holtsmark's theory of ion field in plasma. In a book there is its asymptotic representation at small and great $\beta$:
... | Well, we are looking (in a more general sense) at the following integral:
$$\mathcal{H}_\text{n}\left(\beta\right):=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty x\cdot\sin\left(\beta\cdot x\right)\cdot\exp\left(-x^\text{n}\right)\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
| In the worst case possible you'd get $$a = b = c = \frac{1}{3} \Longrightarrow a^2 + b^2 + c^2 = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \geq \frac{1}{3} $$
In the best case possible you'd get $$a = 1, b = c = 0 \Longrightarrow 1^2 + 0^2 + 0^2 = 1 \geq 1/3 $$
Therefore the inequality holds. Didn't use th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x = a \cos t^3 , y = b \sin t^3$ then what is $d^3y/dx^3$? If $ x = a \cos t^3 $, $ y = b \sin t^3 $, then what is $ \frac{d^3y}{dx^3} $?
I tried doing this problem by dividing $ \frac{d^3y}{dt^3} $ by $ \frac{d^3x}{dt^3} $ and got $ \frac{b}{a} $.
However my book says the third derivative doesn't exist. Why is ... | Given $x = a \cos t^3,y=b\sin t^3$,
the first derivative:
$$y'_x=\frac{y_t'}{x_t'}=\frac{(b\sin t^3)'}{(a\cos t^3)'}=\frac{b\cos t^3\cdot 3t^2}{-a\sin t^3\cdot 3t^2}=-\frac{b}{a}\cot t^3,$$
which does not exist at $t=\pi k,k\in\mathbb Z$.
The second derivative:
$$y''_{xx}=\frac{(y_x')'_t}{x_t'}=\frac{(-\frac ba\cot t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\frac34$ The original exercise is to
Prove that
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ)$$
Dividing both sides by $\cos20^\circ+\sin10^\circ$ leads me to the problem in the question title.
I've tried rewriting the left side in te... | A complex numbers proof of the original exercise.
By Euler's formula, we have that
$$\text{Re}\left((\cos 20^\circ+i\sin 20^\circ)^3\right)=\cos(60^\circ)=\frac{1}{2}=\sin(30^\circ)=\text{Im}\left((\cos 10^\circ+i\sin 10^\circ)^3\right)$$
Hence
$$\cos^3 20^\circ-3\cos 20^\circ \sin^2 20^\circ
=3\cos^2 10^\circ\sin 10^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - ... | Don't forget that when you add two logarithms with the same base, the expressions under the logarithm sign are multiplied: $\log_b{x}+\log_b{y}=\log_b{(xy)}$. When two logarithms with the same base are subtracted, you divide: $\log_b{x}-\log_b{y}=\log_b{\frac{x}{y}}$. And also don't forget that the power on the express... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to evaluate $\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$
How to evaluate
$$\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$$
My attempt
$$\begin{align}\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx&=\frac{1}{4}\int_{0}^{1}{\ln(1+x)}d(\ln^2(1+x^2))\\ &=\frac{1}{4}(\ln(1+x)\ln^2(1+x^2)|_0^1-\int_{0}^{... | You already used contour integration to split the integral into four parts:
$$\begin {align}
&\int_{0}^{1} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x}dx\\
=&\int_{0}^{1} \frac{x \ln ^{2}\left(1-x^{2}\right)}{1+x^{2}} dx+\frac{1}{2} \int_{0}^{\pi/2} \tan \left(\frac{x}{2}\right) \ln ^{2}(2 \cos x)dx+\int_{0}^{\pi/2} x \ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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"answer_id": 1
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Find shortest distance from point on ellipse to focus of ellipse.
Find the point $(x,y)$ on the ellipse $b^2x^2 + a^2y^2 = a^2b^2$ such that the distance to the focus $(c,0)$ is a minimum.
My book I got this problem out of gave a suggestion saying to express the distance as a function of $x$ and work the problem and ... | The hint.
We need to find a a values of $d>0$, for which the system
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and
$$(x-c)^2+y^2=d^2$$ has solutions.
We obtain:
$$\frac{x^2}{a^2}+\frac{d^2-(x-c)^2}{b^2}=1$$ or
$$\left(\frac{1}{a^2}-\frac{1}{b^2}\right)x^2+\frac{2cx}{b^2}+\frac{d^2-c^2}{b^2}-1=0$$ or
$$-\frac{c^2x^2}{a^2b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin... | To answer the actual question (“How did the book”, etc.): probably just by inspection (chain rule backwards), or else via the substitution $u=\sin x$, $du = \cos x \, dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$? I want to solve the following equation for $x$ :
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$
My approach:
Let the given eq.:
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$
On rearranging, we get:
$$\sqrt{x-6} \, +... | Hint:
Clearly,the LHS is increasing function of $x$
so,we cannot have multiple roots
For real solution, $x\ge6$
Also, $3\sqrt{x+6}>9>3\sqrt{x-6}$
$\implies x+6>9>x-6\iff3< x<15\implies6\le x<15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Is this sequence following some rule: $(1+3)^2 + (1+2)^2=(2+3)^2$ , $(1+3+5)^2 + (3+4+5)^2=(4+5+6)^2$ , ...? I found its interesting but don't know whether there is any rule that is governing its next term.
The sequence is given as follows:
$$(1+3)^2 + (1+2)^2=(2+3)^2$$
$$(1+3+5)^2 + (3+4+5)^2=(4+5+6)^2$$
$$(1+3+5+7)^2... | A simple induction gives $\left(\frac{n^3+n}{2}\right)^2$ for the right-hand side. On the left-hand side we have $(n^2)^2 + \left(\frac{n^3+n}{2}- n\right)^2$. Since
$$(n^2)^2 + \left(\frac{n^3+n}{2}- n\right)^2 = n^4 + \left(\frac{n^3+n}{2}\right)^2 - 2\frac{n^4+n^2}{2} + n^2 = \left(\frac{n^3+n}{2}\right)^2,$$
they a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find $\max\{y-x\}$
If $x+y+z=3, $ and $x^2+y^2+z^2=9$ , find $\max\{y-x\}$.
I tried to do this geometrically, $x+y+z=3$ is a plane in $\Bbb{R}^3$ and $x^2+y^2+z^2=9$ is a ball with radius 3 and center of origin . So the candidate points for $y-x$ are on the intersection of the plane and the ball. But now I am confu... | Note that $y^2=(3-x-z)^2=9-x^2-z^2\implies z^2+(x-3)z+(x^2-3x)=0$ so $$2z=3-x\pm\sqrt3\sqrt{3+x-x^2}\implies y=3-x-z=\frac{3-x}2\mp\frac{\sqrt3}2\sqrt{3+2x-x^2}$$ giving $$\max\{y-x\}=\max\left\{\frac32-\frac32x+\frac{\sqrt3}2\sqrt{3+2x-x^2}\right\}$$ and differentiating gives $$-\frac32+\frac{\sqrt3}4\cdot\frac{2-2x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Leibniz integral rule for solving integrals I am giving as homework to solve these integral
1) $\int \limits_{0}^{1} \frac{\arctan(y x)}{x \sqrt{1-x^2}}dx$
2)$\int \limits_{0}^{\frac{\pi}{2}} \frac{x}{\tan x}dx $ with the hint of $0\leq y \leq 1$ and $\int \limits_{0}^{\frac{\pi}{2}} \frac{\arctan( y \tan x)}{\tan x}... | $$I_1(y)=\int_0^1 \frac{\arctan(yx)}{x\sqrt{1-x^2}}dx\Rightarrow I_1(0)=0$$
$$I_2(y)=\int_0^{\pi/2}\frac{\arctan(y\tan x)}{\tan x}dx\Rightarrow I_2(0)=0$$
First for $I_1$, take $\frac{d}{dy}$ on both sides (and use the Leibniz integral rule)
$$I_1'(y)=\int_0^1\frac1{x\sqrt{1-x^2}}\frac{\partial}{\partial y}\arctan(yx)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integrals involving multi-valued functions Show that
$$\int_{0}^{\infty}{\frac{\cos{x}}{x^\alpha}dx}=\mathrm{\Gamma}\left(1-a\right)\sin{\left(\frac{\mathrm{\pi\alpha}}{2}\right)}\ ,\ \ \ \ 0<\alpha<1$$
what would be the contour of this integration if I want to solve this integration using residue theorem
| Assuming you wish to show
$$\int_0^\infty \frac{\cos x}{x^\alpha} \, dx = \Gamma (1 - \alpha) \sin \left (\frac{\pi \alpha}{2} \right ), \qquad 0 < \alpha < 1,$$
what follows is a real method that can be used.
Let
$$I = \int_0^\infty \frac{\cos x}{x^\alpha} \, dx.$$
By employing the following useful property for the La... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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solve $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$ $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$
trig sub
$y = \sqrt{3}\sin(u)$
$dy = \sqrt{3}\cos(u)du$
\begin{align}\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}& 4 \sqrt{3-3\sin^2(u)}\sqrt{3}\cos(u)\,du = 12 \int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sq... | Say $x(y) = \sqrt{3-y^2}$. This is a formula for a semi-circle of radius $\sqrt{3}$, valid from $y=-\sqrt{3}$ to $y = \sqrt{3}$.
The area of this semi-circle is $\pi r^2 / 2 = 3\pi/2$, which is $\int_{-\sqrt{3}}^{\sqrt{3}}x(y)\,dy$.
Your integral has an extra factor of four in it, so you need to multiply the result ... | {
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Finding $\int\frac{\sin x+\tan x}{\cos x+\csc x}dx$ Finding $\displaystyle \int\frac{\sin x+\tan x}{\cos x+\csc x}dx$
what i try
$\displaystyle \Lambda =\int\frac{\sin^2 x(1+\cos x)}{\cos x(\sin x\cos x+1)}dx$
$\displaystyle \Lambda=\int\frac{\sin^4 x}{\cos x(1-\cos x)(\sin x\cos x+1)}dx$
How do i solve it help me plea... | Hint:
$$\text{Define }I=\underbrace{\int\dfrac{\tan x}{\cos x+\csc x}\mathrm dx}_{I_1}+\underbrace{\int\dfrac{\sin x}{\cos x+\csc x}\mathrm dx}_{I_2}$$
For $I_1$, substitute $u=\tan(x/2)$ (Weierstrass substitution) and perform Partial Fraction Decomposition twice.
For $I_2$, rewrite everything in terms of $\tan x$ an... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the roots of $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$? Find all $x$ in the interval $(0,\pi/2)$ such that $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$.
The options are (i)$\pi/9,2\pi/7$, (ii)$\pi/36,11\pi/12$ (iii)$\pi/12,11\pi/36$ (iv) All
I have been able to find ... | Use $$\sin15^{\circ}=\frac{\sqrt3-1}{2\sqrt2}$$ and
$$\cos15^{\circ}=\frac{\sqrt3+1}{2\sqrt2}.$$
We obtain:
$$\sin(15^{\circ}+x)=\sin2x.$$
Thus, $$15^{\circ}+x=2x+360^{\circ}k,$$ where $k$ is an integer number, or
$$15^{\circ}+x=180^{\circ}-2x+360^{\circ}k.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
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Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the midd... | Prompted by an additional point in a duplicate question asking whether the construction is always possible for arbitrary concentric circles, the following is about the general case of radii $\,A,B,C\,$ (where OP's question here is about $\,A=1,B=2,C=3\,$).
Let $\,a,b,c\,$ be points in the complex plane on the respectiv... | {
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"timestamp": "2023-03-29T00:00:00",
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How to approximate $49^{4}81^{5}$? I am doing this exercise for the GMAT test
Which of the following option is closest to $49^{4}81^{5}$?
A. $8^{18}$
B. $8^{19}$
C. $8^{20}$
D. $8^{21}$
E. $8^{22}$
My attempt:
$49^{4}81^{5} = (7^2)^4(9^2)^5 = 7^{8}9^{10} = (8-1)^{8}(8+1)^{10} \approx 8^{8}8^{10} = 8^{18}$
But I am ... | Let's take it exactly for a few more steps:
$$
(8-1)^{8}(8+1)^{10} =(8-1)^8(8+1)^8(8+1)^2\\
=(8^2-1)^8\cdot9^2<8^{16}\cdot 9^2
$$
And $9^2$ is much closer to $8^2$ than to $8^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A pizza parlor has 5 meat and 5 veggie toppings and 3 different sizes. How many pizzas are there with at least 1 meat and at least 1 veggie topping? I managed to get
$$3(2^{10} -1) - 3(2^{5}-1) - 3(2^{5}-1) $$
or
$$3(2^{10} -1) - 3(2^{5} + 2^{5}-1) $$
However, I am a little confused why the answer isn't
$$3(2^{10} ... | Let's use the Inclusion-Exclusion Principle to solve the problem.
There are $3$ ways to choose the size of the pizza and $2^{10}$ ways to choose which subset of toppings will be placed on the pizza. Hence, if there were no restrictions, we could select a pizza in $3 \cdot 2^{10}$ ways.
From these, we must subtract tho... | {
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Prove that for $a,p,q \in \Bbb R$ the solutions of: $\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$ are real numbers. Prove that for $a,p,q \in \Bbb R$ the solutions of: $$\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$$
are real numbers.
I tried manipulating the expression, getting rid of the denominators, but i can'... | Observe that your equation is equivalent to
$$ a^2(x-q)+a^2(x-p)=(x-p)(x-q)$$ $$\iff a^2x-a^2q+a^2x-a^2p-x^2+(p+q)x-pq=0$$ $$\iff-x^2+(2a^2+p+q)x-(a^2q+a^2p+pq)=0$$
Which is a quadratic equation with discriminant $$\color{red}{D=(2a^2+p+q)^2-4(a^2p+a^2q+pq)=4a^4+p^2+q^2-2pq>0}$$
| {
"language": "en",
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If $f(a)=f(b)=0$ and $|f''(x)|\le M$ prove $|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$ If $f(a)=f(b)=0$ and $|f''(x)|\le M$. Prove
$$|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$$
I have thought about that since $f(a) = f(b) = 0 $ there is $\xi$ such that $f'(\xi) = 0$. Then when $x \le \xi$, $|f'(x)| \le... | For simplicity let $a = -1, b = 1$. Let $\varepsilon > 0$. Consider the two functions $g_+(x) = \frac{M + \varepsilon}{2}(1-x^2)$ and $g_-(x) = -g_+(x)$. Then
$$
g_\pm(-1) = g_\pm(1) = 0, \; g_+''(x) = -M - \varepsilon, \; g_-''(x) = +M + \varepsilon
$$
and
$$
\int_{-1}^1 g_+(x) = 2\frac{M+ \varepsilon}{3} = - \int_{-... | {
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"source": "stackexchange",
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What is the closed form of infinite sum $x+x^4+x^7+x^{10}+...$? I know that $1+x+x^2+x^3+...=\frac{1}{1-x}$, $1+x^2+x^4+...=\frac{1}{1-x^2}$. It seems to follow some rules of geometric series. But I'm confused how to get the title from here. Thanks
| It's always nice, if we want to see if the sum is geometric, to do factoring to ensure that the first term of the sum is $1$, so factor out an $x$:
$$x + x^4 + x^7 + x^{10} + ... = x(1 + x^3 + x^6 + x^9 + ...)$$
The parenthetical expression is a geometric series in which the ratio is $x^3$. Then as a result, by the for... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Is $0$ an eigen value of $A$
Consider the following matrix:
$A=\begin{bmatrix} 9&1&1&1&1&1&2&2\\1& 9&1&1&1&1&2&2\\1&1&9&1&1&1&2&2\\1&1&1&9&1&1&2&2\\1&1&1&1&9&1&2&2\\1&1&1&1&1&9&2&2\\2&2&2&2&2&2&13&1\\2&2&2&2&2&2&1&13\end{bmatrix}$.
I got the eigen values of the matrix
\begin{bmatrix} 9&1&1&1&1&1\\1& 9&1&1&1&1\\1&1&9&... | Subtracting $8$ and $12$ from the main diagonal, it is easily seen that $8$ is an eigenvalue of multiplicity $5$ and $12$ is an eigenvalue of multiplicity $1$. Hence we are still missing two eigenvalues. The trace tells you that the sum of the remaining two eigenvalues has to be $80-52=28$. If one is $0$, then the othe... | {
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Convergence of $\sqrt[k]{z+\sqrt[k]{z+\sqrt[k]{z+\cdots}}}$, where $z=(1+x)^k-(1+x)$ If one writes
$$1+x=\sqrt{(1+x)^2}=\sqrt{1+2x+x^2}=\sqrt{x+x^2+(1+x)}$$
then one has a recursive definition of the function $1+x$ which can be used to write $1+x$ as the infinite nested radical:
$$1+x=\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{x+x^... | The expression
$$
y=\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\cdots}}
$$
where $x\ge 0$, represents the limit of the recursive sequence
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)}, \quad a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n},
\quad n\in\mathbb N.
$$
if such a limit exists.
Clearly, the sequence $\{a_n\}$ is increasing. (The fac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
... | You're almost there. You just have to use your hypothesis in your final case for $k+1$.
You found for $n=k$ that
$$2^0+2^1+2^2+...+2^{k-2}+1=2^{k-1}.$$
You want to prove for $n=k+1$ that
$$2^0+2^1+2^2+...2^{k-2}+2^{k-1}+1=2^{k},$$
Fill in the first equation into the second
\begin{align*}
2^0+2^1+2^2+...+ 2^{k-2}+&(2^0+... | {
"language": "en",
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How to determine the smallest value of $N=n^4+6n^3+11n^2+6n$ if 13 and 19 both divide N? I tried to solve for an integer solution by making N equal to multiples of 247 but this is not leading me anywhere. I then tried using the tests for divisibility which did not seem to lead me anywhere either.
| $N=n^4 + 6n^3 + 11n^2 + 6n = n(n+1)(n+2)(n+3)$ so $n, n+1, n+2$ and $n+3$ all divide $N$.
$19$ and $13$ are prime so we need $19|K$ and $13|M$ where $K$ and $M$ are each one of $n,n+1,n+2,$ or $n+3$
So basically we need $|13a + 19b| \le 3$ and the smallest value that is so.
Do a hobbled Euclid's Algorithm.
$19 - 13 = ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| Hint:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}\iff \frac{a+b+c}{b+c}+\frac{b+a+c}{a+c}+\frac{c+a+b}{a+b}\ge \frac{3}{2}+3$$
$$\iff \big(a+b+c\big)\cdot \bigg(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\bigg)\ge \frac{9}{2}\iff \color{blue}{\frac{2\cdot (a+b+c)}{3}\ge \frac{3}{\frac{1}{b+c}+\frac{1}{a+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3... | Part of the issue seems to be mixing up $x$ and $y$. Usually once you identify a substitution it's easiest to bite the bullet and change the variable over completely. Here, $y=\sqrt{x+3}$ implies $x=y^2-3$ and then
$13(y^2-3)+2(3y^2-7)y+42=0$
$6y^3+13y^2-14y+3=0$
Trying out rational root candidates with the Rational ... | {
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Find (a,b,c) if $ax^2-bx+c=0$ have roots lying in $(0,1)$ where $a,b,c\in\mathbb{Z_+}$
Let the equation $ax^2-bx+c=0$ have distinct real roots both lying in the open interval $(0,1)$ where $a,b,c$ are given to be positive integers. Then the value of the ordered triplet $(a,b,c)$ can be
$$
a)\quad (5,3,1)\;,\quad b)\... | You have that $b^2-4ac\gt0$, since the roots are real. Only $(5,5,1)$ meets this condition.
| {
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Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
| Let $x\geq0$ and $f(x)=x^5+x^4.$
Thus, since $f$ increases, continuous, $f(0)<1$ and $f(1)>1$, we obtain that our equation has an unique root.
Let $x<0$.
Thus, $-x>0$.
We'll prove that the equation $-x^5+x^4=1$ has no roots for $x>0$.
Indeed, by AM-GM we obtain:
$$x^5+1-x^4=4\cdot\frac{x^5}{4}+1-x^4\geq5\sqrt[5]{\left(... | {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry AS I've been trying this question for a while but still don't understand it. Do you use trigonometry to form a equation for the sides and then add them up as you know the perimeter is 40?
| Construct a perpendicular from $B$ to $AC$ such that it touches $AC$ at point $D$.
Using the fact that Perimeter is $40cm$
$$a+b+14=40 \\ \Rightarrow a=26-b\\ $$
Using Pythagoras Theorem on Triangle $ABD$, $$\ \ \ \ \ (26-b)^2=BD^2+AD^2 \\\Rightarrow (26-b)^2 -AD^2=BD^2 \\ \\ \\ \\ \\ \\$$
Using Pythagoras Theorem on T... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I solve this absolute value equation? This is the equation:
$|\sqrt{x-1} - 2| + |\sqrt{x-1} - 3| = 1$
Any help would be appreciated. Thanks!
| Let $x$ be a solution of the equation. Notice $x \geq 1$, since $\sqrt{x-1}$ has its domain as $x \geq 1$.
If $x \geq 10$, then each of the absolute value is just the term inside (i.e.$|\sqrt{x-1}-2| = \sqrt{x-1}-2$ and similarly $|\sqrt{x-1}-3| =\sqrt{x-1}-3$) so that the given equation in this case becomes $2\sqrt{x... | {
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"timestamp": "2023-03-29T00:00:00",
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Roots of a Complex Number expression If $z_1$, $z_2$, $z_3$, $z_4$ are roots of the equation $z^4+z^3+z^2+z+1=0$,
then what is the least value of $\lfloor mod(z_1 + z_2)\rfloor + 1$?
($\lfloor.\rfloor$ denotes Greatest Integer Function)
| The roots are
$$w_k = e^{\frac{2\pi k}{5}i} \quad(k = 1,2,3,4)$$
$$\begin{align*}
\min\left\lfloor\left|w_m+w_n\right|\right\rfloor+1
&= \min\left\lfloor\left|e^{\frac{2\pi m}{5}i}+e^{\frac{2\pi n}{5}i}\right|\right\rfloor+1\\
&= \min \left\lfloor\left|e^{\frac{2\pi m}{5}i}\right|\left|1+e^{\frac{2\pi (n-m)}{5}i}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3178747",
"timestamp": "2023-03-29T00:00:00",
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How to get an equation system from a Simplex table Let's assume I already have a simplex table (with an optimal solution):
$$\left(\begin{array}{ccccc|c}
& x_1 & x_2 & S_1 & S_2 & \\
S_1 & 0 & 2 & 0 & 1 & 2 \\
S_2 & 1 & 0 & \frac{1}{5} & 0 & 4 \\
\hline
P & 5 & 0 & \frac{1}{2} & 1 & 1
\end{array}\right)$$
note: the ... | If you write a better and correct example, you can find it easier. Just make sure that:
1) The matrix in front of your basic variables should be the identity matrix.
2) The objective function row in the optimal tableau will have 0s for basic variables.
I'll change your example to the following and show you the steps to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and $x^2 + 5x - 10^{2013} = 0$ have common roots.
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and
$x^2 + 5x - 10^{2013} = 0$ have common roots.
Let $a,b $ be the solutions of second equation, then by Vieta we have ... | Your proof is entirely correct, though you might want to clarify two things:
*
*First, that if the two polynomials have a common root, then they have two common roots.
*Second, that $a^3-b^3\neq0$ when dividing by $a^3-b^3$ to deduce that $-p=a^3+b^3$.
| {
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"url": "https://math.stackexchange.com/questions/3182637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$
Then, $G(x) = a_0 + a_1x + a_2x^2 + \su... | Given that $G(x) = \sum_{n=0}^{\infty} a_nx^n$, you can write $a_n$ as $[x^n]G(x)$. Then the recurrence becomes $$[x^n]G(x) = 3[x^{n-1}]G(x) - 2[x^{n-2}]G(x) + 2^n + (n+1)3^n$$
If we introduce functions $P(x) = \sum_{n=0}^\infty 2^n x^n$ and $Q(x) = \sum_{n=0}^\infty (n+1)3^n x^n$ then, by the linearity of coefficient ... | {
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Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
I know that I should use Taylor's theorem and create power series. However I don't have idea how I can find $a_{n}$ such that $f(x)=\sum_{n=1}^{+\infty} a_{n}x^{n}$. I know only that $f(x)=\frac{x}{(x-1)(... | $$
\begin{split}
f(x) &= \frac{x}{x^3-x^2+2x-2}
= \frac{x}{x^2(x-1)+2(x-1)}
= \frac{x}{(x^2+2)(x-1)}
\end{split}
$$
and use partial fractions. Then use
$$
\frac{1}{1-u} = 1 + u + u^2 + \ldots
$$
and
$$
\frac{1}{1+u} = \frac{1}{1-(-u)} = 1 -u + u^2-u^3 \pm \ldots
$$
UPDATE
From partial fractions,
$$
\frac{A... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
I know that when I calculate derivative then I get true answer. However I want to know why the way that I will present soon does not work. My try:$$\tan(x)=x+r_{1}(x), \quad r_{1}(x)=o(x)$$ $$\tan(\sin(x))=\si... | The problem statement contains a sign error.
In any case you want to compute $\tan \sin x$ modulo $o(x^3)$, so you should compute $\sin x$ and $\tan x$ modulo $o(x^3)$, not $o(x)$. Indeed,
$$\sin x = x - \frac{1}{6} x^3 + o(x^3), \qquad \tan y = y + \frac{1}{3} y^3 + o(y^3) ,$$
so substituting gives $$\color{#df0000}{\... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How does this infinite series $1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots$ simplify to an integral $\int_0^1\frac{dx}{1+x^3}$? How does the infinite series below simplify to that integral?
$$1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots=\int_0^1\frac{dx}{1+x^3}$$
I thought of simplifying the series to the sum to inf... | for $x$ real, $n\geq 0$ integer
\begin{align}\frac{1}{1+x^3}&=\frac{1-(-x^3)^{n+1}}{1-(-x^3)}+\frac{(-x^3)^{n+1}}{1-(-x^3)}\\
&=\frac{1-(-x^3)^{n+1}}{1-(-x^3)}+\frac{(-x^3)^{n+1}}{1+x^3}\\
\end{align}
For $x\neq 1$, $n\geq 0$ integer, \begin{align}\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}\end{align}
Therefore,
\begin{alig... | {
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"answer_count": 3,
"answer_id": 0
} |
Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\s... | As $0\le x\le1$
WLOG $x=\sin^2t;0\le t\le\dfrac\pi2,dx=?$
So $\sqrt x=+\sin t,\sqrt{1-x}=?$
Alternatively $4x(1-x)=1-(2x-1)^2$
Set $2x-1=\sin y$ or $\cos y$
Observe that $2x-1=-\cos2t$ in the first method
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How many columns of this matrix are linearly independent? \begin{matrix}
1 & 2 & 0 & 1 \\
2 & 4 & 1 & 4 \\
3 & 6 & 3 & 9 \\
\end{matrix}
I have tried to transpose it and then reduce it by row echelon form and i get zeros on the last two rows. But i can't grasp if i should be doing that or doing it anoth... | If you look at your matrix
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 &4 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$
The rank can be no more than $3$. Do the following
Subtract $2 R_{1} $ from $R_{2}$
$$ R_{2} - 2 R_{1} = \begin{bmatrix} 2 & 4 & 1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 0 & 2\end{bmatrix} = \begin{b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Combinatorics problem on counting. How many positive integers n are there such that all of the following take place:
1) n has 1000 digits.
2) all of the digits are odd.
3) the absolute value of the difference of any two consecutive (neighboring) digits is equal to 2.
Please help. I don’t even know how to start.
| Define $n_i=2i-1$ (so a bijection between 1,2,3,4,5 with 1,3,5,7,9).
Consider the 5x5 matrix $A=(a_{i,j})$ with $a_{i,j}=1$ if $n_i$ and $n_j$ differ by 2 and $a_{i,j}=0$ otherwise. Then, the number of positive integers with "m" digits satisfying your properties is the sum of entries of $A^{m-1}$. So you want to find ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
| First, we rearrange the equation into the equivalent equation
$$
\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}} \overset?= \sqrt{2-2b}
$$
which we want to prove. Both sides are positive (keep in mind we have not yet proven the equality), so we can square both sides, and equivalence still holds:
$$
4+2b-2\sqrt{(2+b)^... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugat... | Here is your error $1-i^2=1-(-1)=2\ne 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How can I find the P matrix in $P^{-1}AP $= D? I have the following exercise:
Let a matrix A = $\begin{pmatrix}
4 & 0 & 1\\
2 & 2 & 1\\
0 & 4 & 1
\end{pmatrix}$
1) Determine its eigenvalues and their multiplicity.
2) Give a basis of the eigen spaces associated with each of the
distinct eigenvalues of A.
3) fin... | You got the last eigenvector wrong.
If you take $A$ and multiply with $(-1,1,1)$ you will get $(-3,1,5)$.
I believe you need to take $(1,1,1)$ instead. Then everything will work.
Good Luck!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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