Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $(z-2)/(z+2)$ is purely imaginary, find $z$ that satisfies this condition?
If $(z-2)/(z+2)$ is purely imaginary, then what is the value of $z$ that satisfies this condition?
I am troubled solving this equation for a long time. I tried to put $z = a + i b$ and then I got $(a-2+ib)(a+2+ib)$ is imaginary.
Not sure h... | Putting $z=a+ib$, with $z\neq-2$, is a correct way to start. Then, you get \begin{align}\frac{z-2}{z+2}&=\frac{z-2}{(a+2)+ib}=\frac{(a-2)+ib}{(a+2)+ib}\cdot\frac{(a+2)-ib}{(a+2)-ib}\\[0.2cm]&=\frac{(a-2)(a+2)+b^2+ib(a+2-a+2)}{(a+2)^2+b^2}=\underbrace{\frac{(a-2)(a+2)+b^2}{(a+2)^2+b^2}}_{=\text{ real part }=0}+i\frac{4b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods... | Hint: your objective function $x^2+y^2-4x-6y$ is convex, so its maximum is attained at an extreme point of your convex feasible region.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How many real, rational and complex solutions has this system of equations? Let the system $\left\{\begin{aligned}
a+b+c &= 3\\
a^2+b^2+c^2 &=5\\
a^3+b^3+c^3 & =12
\end{aligned}\right. $
How many real, rational and complex solutions has it?
I read System of three variables of simultaneous equations and... | Note
$$ 3^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=5+2(ab+bc+ca). $$
from this, one has
$$ ab+bc+ca=2. $$
Also
\begin{eqnarray}
3^3&=&(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\
&=&12+3ab(a+b+c)+3bc(a+b+c)+3ca(a+b+c)-3abc\\
&=&12+3(ab+bc+ca)(a+b+c)-3abc\\
&=&30-3abc
\end{eqnarray}
from which, one has
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
| $z=(-x-y)$ leads to
$$ x^2+y^2+(x+y)^2 = 6 $$
that is equivalent to $x^2+xy+y^2=3$, i.e. the equation of an ellipse in the $xy$-plane. Such ellipse goes through $(1,1)$, hence by considering the intersections between such ellipse and the lines through $(1,1)$, i.e. by solving $x^2+xy+y^2=3$ and $y=m(x-1)+1$, we get tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Addition of two piecewise function Suppose I have function
$$y(x)=\begin{cases}
x+1\qquad & 0\leq x\leq1 \\
2-x\qquad & 1<x \leq2 \\
0\qquad & \mathrm{elsewhere}
\end{cases}$$
I need to find the function $g(x)=y(x+2)+2y(x+1)$
$$y(x+2)=\begin{cases}
x+3\qquad &-2\leq x\leq-1 \\
-x\qquad & -1<x \leq0 \\
0\qqu... | \begin{eqnarray}
g_1(x)=y(x+2)=\begin{cases}
x+3&\text{ for }-2\le x\le -1\\
-x&\text{ for }-1< x\le0\\
0&\text{ otherwise}
\end{cases}
\end{eqnarray}
and
\begin{eqnarray}
g_2(x)=2y(x+1)=\begin{cases}
2x+4&\text{ for }-1\le x\le0\\
2-2x&\text{ for }0< x\le1\\
0&\text{ otherwise}
\end{cases}
\end{eqnarray}
Giving the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadra... | Solving $$z^2-2i=0$$
we have $$z^2=2i=2\exp\left( \frac{i\pi}{2}\right)$$
$$z=\pm\sqrt2\exp\left( \frac{i\pi}{4}\right)=\pm\sqrt{2}\left(\cos\left( \frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4} \right)\right)=\pm(1+i)$$
Hence, $$z^2-2i=(z-(1+i))(z+(1+i))$$
Try the same trick on $z^2+2i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Epsilon-Delta proof for a limit of a function $\lim\limits_{x \to 2} \frac{x^2+4}{x+2}=2$
I understand the structure of the epsilon delta proof, but I need help with the scratchwork/setup.
Using |$\frac{x^2+4}{x+2}-2$| $<\epsilon$ , you can factor and you're left with |$x-4$|$< \epsilon$. What I'm stuck on is solving f... | $$\left|\frac{x^2+4}{x+2} -2 \right|=\left|\frac{x^2-2x}{x+2} \right|=\frac{|x||x-2|}{|x+2|}$$
WLOG, we can assume that $\delta < 1$,
Hence if $|x-2| < \delta$, then $2-\delta < x < 2+\delta$ which implies that $x$ is between $1$ and $3$ while $x+2$ is in between $3$ and $5$.
Hence, $$\left|\frac{x^2+4}{x+2} -2 \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\frac{x}{\sqrt[3]{x+1}-1}$
Simplify the following fraction:
$$\frac{x}{\sqrt[3]{x+1}-1}$$
How should I approach this? unlike $(a-b)(a+b)=a^2+b^2$?
| Depending what you mean by simplify: presumably, get rid of the denominator.
Let $y = \sqrt[3]{x+1}$ then
$$
\frac{x}{\sqrt[3]{x+1}-1} = \frac{y^3 -1}{y-1} = y^2 + y +1 = (x+1)^{\frac23} + \sqrt[3]{x+1} + 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
| Use the inequality $\dfrac{a^3}{b^2}\geq \dfrac{2a^2}{3b} - \dfrac{a}{9}$ and then use $\dfrac{a^2}{b}\geq \dfrac{2a}{3}-\dfrac{b}{9}$, so $$\sum_{x,y,z}\dfrac{x^3}{(y+2z)^2}\geq\sum_{x,y,z}(\dfrac{x}{3} - \dfrac{2(y+2z)}{27})=\dfrac{3(x+y+z)}{27}=\dfrac{1}{3}.$$
The inequalities used are trivial AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplify: $\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$ I am doing a pretty hard problem:
$$\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
So it is a pretty long a... | Let $$x=\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
The square of numerator is $$2\sqrt[4]{27} -2\sqrt{\sqrt{27}-(\sqrt{3}-1)}=2\sqrt[4]{27}-2\sqrt{2\sqrt{3}+1}$$
The square of denominator is $\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}$.
Hence $x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Show that a function $f(x,y)$ is bijective Show that a function is bijective
$ f: \mathbb{R}_{+} \times \mathbb{R}_{+} \rightarrow \mathbb{R}_{+} \times \mathbb{R}, f(x,y) := (x+y; \frac{1}{x} - \frac{1}{y}) $
I know I have to show that this function is injection and surjection.
My attempts:
*
*Injection
$ f(x,y)... | Injectivity: You have shown that $x-a=b-y$ and $\frac{a-x}{ax}=\frac{b-y}{by}$. If $x-a$ and $b-y$ is not zero, then you have $ax=-by$ (as you got). But this is impossible since $a,b,x,y>0$. Hence, the only possibility is $x-a=b-y=0$. That is $(x,y)=(a,b)$.
Surjectivity: Your approach is fine, but you should separate c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplifying an algebraic fraction The image below is from an online algebra quiz I did recently for uni. I got this question wrong and it indicated what the correct answer was, however I cannot understand how they got to this answer.
Question from my algebra quiz
This is what I came up with when working out the questi... | \begin{eqnarray*}
x^3-4x^2y+6xy^2-3y^3 &=& x^3-y^3 -2y(y^2-3xy+2x^2)\\
&=& (x-y)(x^2+xy+y^2)-2y(y-x)(y-2x)\\
&=& (x-y)(x^2+xy+y^2+2y^2-4xy)\\
&=& (x-y)(x^2-3xy+3y^2)
\end{eqnarray*}
So we have
$$...=\frac{x^3-4x^2y+6xy^2-3y^3}{x-y} ={(x-y)(x^2-3xy+3y^2)\over x-y}=x^2-3xy+3y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $a \sin\theta + b \cos\theta = c$ Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$
I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$
I just can't figure out the right steps to arrive at this sol... | Well, we have:
$$\text{a}\cdot\sin\left(x\right)+\text{b}\cdot\cos\left(x\right)=\text{c}\tag1$$
Substitute $\text{y}=\tan\left(\frac{x}{2}\right)$, so $\sin\left(x\right)=\frac{2\cdot\text{y}}{1+\text{y}^2}$ and $\cos\left(x\right)=\frac{1-\text{y}^2}{1+\text{y}^2}$:
$$\text{y}^2-\frac{\text{b}-\text{c}}{\text{b}+c}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at soluti... | We have $a_k \geq 1$ and $a_{k-1} \leq 2$ ; which gives $ \frac{2}{a_{k-1}} \geq 1$. So
\begin{eqnarray*}
a_{k+1} = \frac{1}{2} \left( a_k +\frac{2}{a_{k-1}} \right) \geq \frac{1}{2} \left( 1+1 \right)=1.
\end{eqnarray*}
We have $a_k \leq 2$ and $a_{k-1} \geq 1$ ; which gives $ \frac{2}{a_{k-1}} \leq 2$. So
\begin{eqna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\sum\limits_{\mathrm{cyc}} \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3} \le 6$ for $a, b, c \ge 1$
I have already post this inequality but I add this condition $a,b,c\geq 1$
$$6\geq \frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b +... | The given inequality can be transformed into
$$\sum \frac {a^2+b^2+ab+2a+3}{a^2+b^2+ab-2a+3}\le 6$$
Which changes to
$$\sum 1+\frac {4a}{a^2+b^2+ab-2a+3}\le 6$$
$$\Rightarrow \sum \frac {4a}{a^2+b^2+ab-2a+3}\le 3$$
Hence it suffices to show that
$$\sum \frac {a}{a^2+b^2+ab-2a+3}\le \frac{3}{4}$$
Hence by AM GM we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why is the dimension of the null space of this matrix 1? Consider this problem from wikipedia:
$$A = \begin{bmatrix}
5 & 4 & 2 & 1 \\
0 & 1 & -1 & -1 \\
-1 & -1 & 3 & 0 \\
1 & 1 & -1 & 2 \end{bmatrix}$$
From the wikipedia link: "Including multiplicity, the eigenvalues of A are λ = 1, 2, 4, 4. The dimens... | Let's do the Gaussian elimination slowly:
\begin{align}
A - 4I =
\begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
-1 & -1 & -1 & 0 \\
1 & 1 & -1 & -2
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 4 & 2 & 1 \\
0 & -3 & -1 & -1 \\
0 & 3 & 1 & 1 \\
0 & -3 & -3 & -3
\end{bmatrix}
&&\begin{aligned} R_3&\gets R_3+R_1\\R_4&\gets... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int_0^\pi x\,\cos^4x\, dx$ Assume that $$\int_{0}^\pi x\,f(\sin(x))dx=\frac{\pi}2 \int_{0}^\pi f(\sin(x))dx$$
and use it to calculate $$\int_{0}^\pi x\,\cos^{4}(x)\, dx$$
Can anyone help me with that? I proved the identity but I am stuck with the rest.
| With different approach. First note:
\begin{align}
I=\int_0^\pi \cos^4(x) dx = 2 \int_0^{\pi/2} \cos^4(x)dx
\end{align}
Substitute $z=\tan(x)$ so that $\cos^4(x)= \frac{1}{(1+z^2)^2}$ and $dx=\frac{dz}{1+z^2}$. So we get:
\begin{align}
I=2\int^\infty_0 \frac{1}{(1+z^2)^3}dz = \int_{-\infty}^\infty \frac{1}{(1+z^2)^3} d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Prove $ ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$
Let $0<\theta<\pi/2$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \r... | Clearly, we only need to prove the case when $\theta \in (0, \pi/4]$.
Let
\begin{align*}
A &= \frac{4}{17}\left(\frac{\sin^2 \theta}{2} + \frac{2}{\cos^2\theta}\right), \\
B &= \frac{4}{17}\left(\frac{\cos^2 \theta}{2} + \frac{2}{\sin^2\theta}\right).
\end{align*}
It suffices to prove that
$$A^{1/4} + B^{1/4} \ge 2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
area of points in a square closer to the center than the edge I'm hoping someone can confirm my solution to the following problem or perhaps provide better alternatives. Cheers.
A shape is defined by consisting of all the points closer to the center than the edge of a 2 inch square. Find the area of the shape.
Use symm... | Since our square is $[{-1},1]^2$ the arc you considered is not characterized by $\sqrt{x^2+y^2}=2-x$, but by $\sqrt{x^2+y^2}=1-x$. It is a parabolic arc given by $$x={1\over2}(1-y^2)\qquad\bigl(0\leq y\leq\sqrt{2}-1\bigr)\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If polynomial is divisible by quadratic, find values of $a$ and $b$ Equation is $z^4+(a+b)z^3+4az^2+(a+b+32)z+45$ which is divisible by $z^2+6z+9$.
We're meant to find values of $a$ and $b$ and then solve the equation itself!
I figured $z^2+6z+9=(z+3)^2$, but was wondering if this automatically implies that it is a dou... | Let $p(z)=z^4+(a+b)z^3+4az^2+(a+b+32)z+45$. Then\begin{multline}p(z)=p\bigl((z+3)-3\bigr)=(z+3)^4+(a+b-12)(z+3)^3+\\+(-5a-9b+54)(z+3)^2+(4a+28b-76)(z+3)+6a-30b+30.\end{multline}On the other hand, $z^2+6z+9=(z+3)^2$. Therefore, $p(z)$ is a multiple of $z^2+6z+9$ if and only if$$\left\{\begin{array}{l}4a+28b-76=0\\6a-30b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Probably that an $80\%$-truthful person actually rolled a $6$ A person, $A$, speaks the truth $4$ out of $5$ times. The person throws a die and reports that he obtained a $6$. What is the probability that he actually rolled a $6$?
I know there is a similar question like this
but my doubts are different from it and als... | Let $D_{6}$ denote the event that the die lands on a $6$ and $R_{6}$ denote the event that the person reports that it landed on a $6$. You know that
\begin{align*}
P(R_{6}|D_{6}) &= 0.8 \\
P(D_{6}) &= 6^{-1}
\end{align*}
In order to obtain the correct answer, you also need to assume that $P(R_{6}|D_{6}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
How many sets of digits are there so each digit would belong to the corresponding set? Given nine sets:
$X_1 \{2,3,9\}, X_2 \{1, 2, 3, 5, 6, 7, 8, 9\}, X_3 \{3, 9\}, X_4 \{1, 2, 3, 7, 9\}, X_5 \{1, 2, 3, 4, 5, 7, 9\}, X_6 \{2, 3, 6, 7\}, X_7 \{1, 6, 7, 9\} , X_8 \{1, 3, 4, 6, 7, 8, 9\}, X_9 \{2, 9\} $
How many sets of ... | Consider only $X_1$, $X_3$, and $X_9$. $a_1$ can be 2, 3, or 9. If $a_1=2$, then $a_9=9$ so $a_3=3$. If $a_1=3$, then $a_3=9$ so $a_2=2$. Finally if If $a_1=9$, then $a_3=3$ and $a_9=2$.
The possibilities for $(a_1, a_3, a_9)$ are $(2,3,9)$, $(3,9,2)$, or $(9,3,2)$.
Since 2, 3, 9 are "used up" you can repeat the pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Locating the third vertex of an equilateral triangle Two vertices of an equilateral triangle are at $A=(10,-4)$ and $B=(0,6)$.
How can one locate the third vertex?
Maybe someone could give me the easy way please.
My attempt:
*
*Find the average of $M=(x , y)$ of $A$ and $B$, which is $(10+0)/2 = 5$, $(-4+6)/2 = 1... | We know that multiplying with $\varepsilon ={1\over 2}+{\sqrt{3}\over 2}$ is rotation around $0$ for $60^{\circ}$ in counterclockwise direction and with $-\varepsilon$ around $0$ for $60^{\circ}$ in clockwise direction.
Let in complex plain $z=10-4i$ correspond to the point $(10,-4)$ and $w=6i$ to the point $(0,6)$. If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the probability that the first head will appear on the even numbered tosses Question
$\text{Consider a coin with probability R to be heads. What is the probability}$
$\text{that the first head will appear on the even numbered tosses?}$
My Approach
let the required Probability$=P$.
Hence we can write our eau... | The probability that the first head will appear on the second toss is $(1 - R)R$.
The probability that the first head will appear on the fourth toss is $(1 - R)^3R$.
The probability that the first head will appear on the sixth toss is $(1 - R)^5R$.
In general, the probability that the first head will appear on the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms. The sum is t... | By induction, it is pretty clear that
$$ x^{2^{n+1}}-1 = (x-1)\prod_{k=0}^{n} \left(x^{2^k}+1\right) \tag{A}$$
and by applying $x\cdot\frac{d}{dx}\log(\cdot)$ to both sides we have:
$$\frac{2^{n+1}x^{2^{n+1}} }{x^{2^{n+1}}-1}=\frac{x}{x-1}+\sum_{k=0}^{n}\frac{2^k x^{2^k}}{x^{2^k}+1} \tag{B}$$
and by replacing $x$ with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Solving $3^x = 2^y + 1$ with $x,y \in \mathbb{N}^2$ I consider the following equation
$$ 3^x = 1 + 2^y \tag{$\star$} $$
with $(x,y) \in \mathbb{N}^2$ and $y \geq 3$.
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$
I assume that $3^x \equiv 1 \; [2^y]$. Writing
$$ 3^x = \... |
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$
To show this, I would use that, for every pair of positive integers $(a,s)$ where $s$ is odd, there exists an odd integer $t$ such that
$$3^{2^as}-1=2^{a+2}t\tag1$$
(the proof are written at the end of the answer)
Proof for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^... | it must be $$\frac{(n^2+n)^2-(n^4+1)}{(\sqrt{n^2+n}+\sqrt[4]{n^4+1})(n^2+n+\sqrt{n^4+1})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$. Determine all solutions of the congruence $y^2≡5x^3\pmod7$ in integers $x$, $y$.
I learned about primitive roots and the theory of indices.
By trial, I check that $3$ is a primitive root of $7$ and $\operatorname{ind}_35≡5\pmod 7$.
| With $p=7$, you have that $6|p-1$, which means there are only a limited number of squares and cubes mod $7$. Whatever $y$ is, $y^2$ can only be $0$, $1$, $4$, or $2$.
And $x^3$ can only be congruent to $0$, $1$, or $6$. So $5x^3$ can only be $0$, $5$, or $2$.
So with the given congruence, for the left side to match th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to generate number in ascending order using the expression $a^2\cdot b^3$ where $a$ and $b$ are distinct primes. How to generate numbers in ascending order using the expression $a^2\cdot b^3$ where $a$ and $b$ are distinct primes.
Here are a couple of examples :
$$\begin{array}{c|ccc}
\text{No}&a&b&a^2\cdot b^3\\
\... | I got Matlab to calculate numbers up to $10^8$, or a hundred billion.
The $N$th number, over that range, was about $9N^{2.5}/\ln N$. The $1000$th number was $42797187 = 3^31259^2$.
The largest prime involved in the first thousand was $2311$, which is the $344$th prime.
According to my formula, the biggest prime $p$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to differentiate a power series? I have a task where I need to write the power series:
$\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^n$
differentiated 2 times where $x=0$.
This is what I have done so far:
$f'(x) = \sum\limits_{n=1}^\infty n\frac{(-1)^n}{(2n)!}x^{n-1} \\
f''(x) = \sum\limits_{n=1}^\infty n(n-1)... | \begin{eqnarray*}
\sum\limits_{n=1}^\infty\frac{(-1)^n n(n - 1) x^{n - 2}}{(2 n)!} \mid _{x=0}
\end{eqnarray*}
Note that the first term is zero, the third & higher terms are all zero, so only the second term contributes & gives the value $\color{red}{1/12}$.
Another way to figure is the function is $y=\cos(\sqrt{x})$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
| We have
$(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz$
With $x=9\sqrt[3]{9},y=-3\sqrt[3]{3},z=-27$ all terms on the right side are rational, try it. So multiply the given numerator and denominator by $x^2+y^2+z^2-xy-xz-yz$ with $x,y,z$ as rendered above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Determine the centroid of the region $ D \ $ within the circle of radius $ \ 2 \ $
Determine the centroid of the region $ D \ $ within the circle of radius $ \ 2 \ $ as follows :
Where $ \ AB \ $ is the line joining $ \ A (-\sqrt 2, -\sqrt 2) \ $ and $ \ (\sqrt 2, \sqrt 2 ) \ $
Step 1: Use the formula $ \ \bar x=... | Note that
$$|D|=\iint_D dxdy= \int_{y=-\sqrt 2}^{2} \int_{x=-\sqrt{4-y^2}}^{x=\sqrt{4-y^2}} dxdy =2\int_{-\sqrt 2}^{2} \sqrt{4-y^2} dy =2+ 3 \pi.$$
Therefore
$$\bar y=\frac{1}{|D|}\iint_D ydxdy=\frac{1}{A}\int_{y=-\sqrt 2}^{2} y\int_{x=-\sqrt{4-y^2}}^{x=\sqrt{4-y^2}} dxdy\\=\frac{2}{|D|}\int_{-\sqrt 2}^{2} y\sqrt{4-y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding function $f(x)$ which satisfy given functional equation
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$
Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some he... | Just as you replaced $x$ by $\frac{1}{x}$, you can replace $x$ by $1-x$. Then you can replace $x$ by $\frac{1}{x}$ in this new equation.
Repeating this process, I believe you will obtain 6 independent equations in the variables $f(z)$ where $z$ takes the values $$x,\frac{1}{x},\frac{1}{1-x},\frac{x}{x-1},\frac{x-1}{x},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding number of circles and area of polygon formed by their centres. Let $S(x,y)=0$ represent a circle with radius $\frac{3}{\sqrt 2}$ such that $S(\lambda -3, \lambda) =0$ has equal roots and $S(\mu, 7- \mu )=0$ also has equal roots then find
1) number of such circles
2) number of circles whose centre lie in first ... | The following is just painful computation from where you left off. I would be very interested to see if there is a slicker way to do this problem; in particular, why did the magic cancelations happen?
The general equation is $$S(x,y) = (x-a)^2 + (y-b)^2 - 9/4.$$
Then
\begin{align}
S(\lambda - 3, \lambda)
&= (\lambda -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
*
*a) Show the monotony of the partial sums
*b) estimate upwards
*c) remember the geometric series (I do not know how to use that here.)
The following... | For a) you have to show, that $\sum_{k=0}^n a_k<\sum_{k=0}^{n+1} a_k$, which is kinda clear, since $a_k>0$ for every $k\in\mathbb{N}$. Then stipulate $\frac{3^k}{5^k+1}<\frac{3^k}{5^k}$ and use the geometric series. With the comparision test we get that the series converges.
But this can be also done immediatly. We ki... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Index Manipulation of Summation in Real Cosine Transform Fast Computation Consider $X_k = \sum_{n=0}^{N-1}x_{n}\cos\frac{(2n+1)k\pi}{2N}$ for $k=0,1,2,...,N-1$. Now we define $g_{n} = x_{n} + x_{N-1-n}$ and $h_{n} = \frac{x_{n}-x_{N-1-n}}{2\cos\frac{(2n+1)k\pi}{2N}}$ for $\{x_{n}\}_{n=0}^{N-1}$ and $\{g_{n}\}_{n=0}^{\f... | Here is the first part. We want to show equality $G_k=X_{2k}$ with
\begin{align*}
G_k&=\sum_{n=0}^{\frac{N}{2}-1}\left(x_n+x_{N-1+n}\right)\cos\frac{(2n+1)k\pi}{N}\\
X_{2k}&=\sum_{n=0}^{N-1}x_n\cos\frac{(2n+1)k\pi}{N}
\end{align*}
We obtain
\begin{align*}
\color{blue}{G_k}&\color{blue}{=\sum_{n=0}^{\frac{N}{2}-1}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}$ by induction? Let $(2n)!!$ be the product of all positive even integers less than or equal to $2n$. Let $(2n − 1)!!$ be the product of all odd positive integers less than or equal to $(2n − 1)$. Prove that
$$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}.$$
... | $$\frac{(2n-1)!!}{(2n)!!}=\frac{2n-1}{2n}\cdot \frac{(2n-3)!!}{(2n-2)!!}\le \frac{2n-1}{2n\sqrt{2(n-1)+1}}=\frac{2n-1}{2n\sqrt{2n-1}}= \frac{\sqrt{2n-1}}{2n}$$
Now to show that $\frac{\sqrt{2n-1}}{2n}\le \frac{1}{\sqrt{2n+1}}$ we show that $\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}\ge 0$:
$$\frac{1}{\sqrt{2n+1}}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to find $\sum_{n=1}^{\infty}\frac{\sin(nx)}{2n+1}$? My attempt was : $f(t) = \sum_{n=1}^{\infty}\frac{\sin(nx)}{2n+1}t^{2n+1}$, where $|t|\le 1$. So we can find $f'(t) = \displaystyle \Im\sum_{n=0}^{\infty}(e^{ix}t^2)^{n} = \Im\frac{1}{1-e^{ix}t^2} = \frac{t^2\sin(x)}{t^4-2\cos(x)t^2+1}$. So $f(t) = \displaystyle \... | Let $z=\rho e^{ix}$ with $\rho\in(0,1)$. Then
$$\sum_{n\geq 0}\frac{\rho^n\sin(nx)}{2n+1}=\text{Im}\sum_{n\geq 0}\frac{z^n}{2n+1} =\text{Im}\left(\frac{\text{arctanh}\sqrt{z}}{\sqrt{z}}\right)$$
and by considering the limit as $\rho\to 1^-$ we get
$$ \sum_{n\geq 0}\frac{\sin(nx)}{2n+1}=\frac{\pi}{4}\cos\left(\frac{x}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating fuctions: $\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$ I'm trying to solve one of my combinatorics exercise but I struggle a bit.
Is the equality correct for all the $n\ge 0$?
$$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}... | This follows directly (more or less) from Vandermonde's identity:
\begin{align*}
\sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} &= \binom{n-1/2}{n} \\
&= \frac{(n-1/2)(n-3/2)\cdots(n-1/2-(n-1))}{n!}\\
&= \frac{(2n-1)(2n-3)\cdots (1)}{2^nn!} \\
&= 2^{-2n}\binom{2n}{n}.
\end{align*}
But also,
$$\binom{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the real solutions for the system: $ x^3+y^3=1$,$x^2y+2xy^2+y^3=2.$
Find the real solutions for the system:
$$\left\{
\begin{array}{l}
x^3+y^3=1\\
x^2y+2xy^2+y^3=2\\
\end{array}
\right.
$$
From a book with exercises for math contests. The solutions provided are: $(x,y)=(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{\sqr... | Well, the second equation is a quadratic equation in $x$:
$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$
Now, in your example $\text{a}=\text{y},\text{b}=2\cdot\text{y}^2,\text{c}=\text{y}^3-2$, so w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is it true that $\sum_{k=0}^m\binom{n-k}k$ outputs the $(n+1)$th Fibonacci number, where $m=\frac{n-1}2$ for odd $n$ and $m=\frac n2$ for even $n$?
Does $$\sum_{k=0}^m\binom{n-k}k=F_{n+1}$$ where $m=\left\{\begin{matrix}
\frac{n-1}{2}, \text{for odd} \,n\\
\frac n2, \text{for even} \,n
\end{matrix}\right.$ hold for a... | Here is more of an algebraic solution through generating functions. We have
\begin{align}
\sum_{n=0}^{\infty}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}x^n
&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\binom{n}{k}x^{n+k+1}\\
&=\sum_{k=0}^{\infty}x^{2k+1}\sum_{n=k}^{\infty}\binom{n}{k}x^{n-k}\\
&=\sum_{k=0}^{\infty}x^{2k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving trigonometric identity: $ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $
Prove that
$$ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $$
This is my working -
$$\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos... | Consider fractions that are equivalent to $2/3$.
$$\frac{2}{3}\cdot\frac{2}{2} = \frac{4}{6}$$
$$\frac{2}{3}\cdot\frac{3}{3} = \frac{6}{9}$$
$$etc.$$
Conclusion ... you're allowed to multiply by one.
Similarly, with the expression $\displaystyle\frac{\sin y\cos x+\cos y \sin x}{\sin y \cos x−\cos y \sin x}$, you're al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the reduction formula for $\int x\cos^nx\,dx$ such that $n \in \mathbb{Z}^+ $? I am pretty bad at deducing reduction formulas but I tried and I reached
$$\frac{1}{2}I_{n-2} + \frac{1}{2}(x\sin x\cos^{n-1}x+\sin^{n-1}x+\frac{1}{n} \cos^nx)$$
Which I tried testing for $n = 2$ and comparing my answer to the integ... | The first step is correct. You made a mistake during integration by parts
$$ v = \frac{\sin 2x}{2} = \sin x\cos x, \quad du = \big( \cos^{n-2}x - x(n-2)\cos^{n-3}x\sin x \big)\ dx $$
Then
$$ \begin{align} \int x\cos^{n-2}x\sin 2x\ dx &= x\cos^{n-1}x\sin x - \int \cos^{n-1}x\sin x\ dx \\ &\quad + (n-2)\int x \cos^{n-2}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is $\operatorname{ord}_{22}(5^6)$?
Find $\operatorname{ord}_{22}(5^6)$.
So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I co... | You have found that $5^5 \equiv 1 \pmod {22}$. So as $5^6 \equiv 5 \pmod {22}$. Then we must have $\text{ord}(5^6) = \text{ord}(5)$. But from the first equation we have that $\text{ord}(5) \mid 5$ and as $5$ is prime and $5^1 \not \equiv 1 \pmod {22}$ we must have that $\text{ord}(5^6) = \text{ord}(5) = 5$
Note that w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integral question (Using Fundamental Theorem) Let's say I have the function
$$ x^2 = \int_{\tan(x)}^{y(x)}\frac{1}{\sqrt{2+t^2}}\,\mathrm dt $$
can I replace the y in the upper limit of the integral with $ x^2 $ ? If not, any steps on how to proceed? How could I find $ y'(0) $ ?
| By the fundamental theorem of calculus, we have
$$ F(x) := \int_{a}^{x} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
\implies F'(x) = \frac{1}{\sqrt{2+t^2}}. $$
Splitting the original integral, we obtain
$$ x^2 = \int_{\tan(x)}^{y(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
= \int_{a}^{y(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the real part of $ e^{e^{i \theta}} ?$ How to find the real part of the complex number (in Euler's form) $ z = e^{e^{i \theta } } $ ?
I got confused on how to proceed.
I am a beginner to complex numbers.
| For Euler's formula:
$z=x+iy=r(cos\theta+isin\theta)$ (It is easy part.)
$cos\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+...$
$sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$| Then
$cos\theta+isin\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2550387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding GCD of two polynomials I am finding the GCD ($x^{24}-1, x^{15}-1$) using Euclidean Algorithm. So far I have
EDIT:
$x^{24}-1=x^9(x^{15}-1)$ w remainder $x^9-1$, then $x^{15}=x^6(x^9-1)$ with the remainder $x^6-1$, then $x^9-1=x^3(x^6-1)$ with remainder $X^3-1$, then $x^6-1=x^3(x^3-1)$ with remainder $x^3-1$. The... | $$x^6-1 = x^3(x^3 -1 ) + x^3 -1$$
$$x^3-1 = 1(x^3-1) + 0$$
$$\gcd{(x^{24}-1, x^{15}-1)}=\gcd{(x^3-1,0)}$$
Therefore, $\gcd{(x^{24}-1, x^{15}-1)}=x^3-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim\limits_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$ In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.
Here is my approach/what I have done so fa... | Use $|\sin(x/n)|\leq|x/n|=x/n$ for $x>0$, $\left|\dfrac{n^{2}\sin(x/n)}{n^{3}x+x(1+x^{3})}\right|\leq\dfrac{nx}{n^{3}x+x(1+x^{3})}\stackrel{\sqrt{ab}\leq \frac{1}{2}(a+b)}{\leq}\dfrac{nx}{2(n^{3}x^{2}(1+x^{3}))^{1/2}}=\dfrac{1}{2n^{1/2}(1+x^{3})}\leq\dfrac{1}{2(1+x^{3})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| Say $$(1-2x+3x^2-4x^3)^{1/2} =a+bx+cx^2+dx^3...$$
then
$$1-2x+3x^2-4x^3 =(a+bx+cx^2+dx^3...)^2$$
but $$(a+bx+cx^2+dx^3...)^2 = a^2+2abx+(2ac+b^2)x^2+2(ad+bc)x^3+...$$
So $a=\pm 1$.
If $a=1$ then $b=-1$ and $c=1$ and $d=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac... | This is a classic telescoping series.
$$\frac1{n\cdot(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$
Thus
$$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}+\cdots$$
$$=\frac12\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\frac19-\frac1{11}+\cdots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the Integral: $\int{\sqrt{8y-x^2}dx}$ I am trying to solve the following integral:
$$\int{\sqrt{8y-x^2}}dx$$
For which I don't understand clearly how to work with root values.
What I've tried:
$$
\int{\sqrt{8y-x^2}dx} = \int{(8y-x^2)^{\frac{1}{2}}dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}\int{(8y-x^2)dx}... | Note that if we use the substitution, $x=\sqrt{8y}\sin u$, then the integrand can be easily reduced to:
$$I=\int \sqrt{8y-x^2} dx = \sqrt{8y} \int \cos u \sqrt{8y-8y\sin^2 u} \,du=8y \int \cos^2 u \,du$$
I hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$ The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$ then find the locus of focus of rolling parabola and also find the locus of vertex of rolling parabola
|
\begin{align}
\text{Static parabola: }f(x)&=-\tfrac1{12}x^2
,\\
f'(x)&=-\tfrac1{6}x
.
\end{align}
Tangent line at $x=x_t$:
\begin{align}
f_t(x,x_t)&=
f'(x_t)\,(x-x_t)+f(x_t)
\\
&=
-\tfrac1{6}x_t\,(x-x_t)-\tfrac1{12}x_t^2
\\
&=
-\tfrac16\,x_t\,x+\tfrac1{12}\,x_t^2
.
\end{align}
\begin{align}
|OD|&=\tfrac12x_t \quad\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Trying to derive expected value of triangularly distributed random variable I'm having trouble deriving the expected value of a triangularly distributed random variable with lower bound $a$ upper bound $b$ and mode $c$ for the case when the distribution is symmetric about the mode. The expected value in this case is kn... | UPDATE
Finally you got
$$E(x)=\frac{1}{6\beta^2}\left(-2c^3+a^3+b^3\right)$$
but notice that since $a=c-\beta$ and $b=c+\beta$, we have
$$
\begin{split}
a^3+b^3-2c^3
&=(c-\beta)^3 + (c+\beta)^3-2c^3\\
&= \left(c^3-3c^2\beta+3c\beta^2-\beta^3\right)
+ \left(c^3+3c^2\beta+3c\beta^2+\beta^3\right)
- 2c^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use Viete's relations to prove the roots of the equation $x^3+ax+b=0$ satisfy $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$ Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.
I know that viete's relat... | If $b=0$, then the roots are $0,\pm\sqrt{-a}$ from which
$$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3$$
follows.
In the following, $b\not=0$.
We have, by Vieta's formulas,
$$x_1+x_2+x_3=0,\quad x_1x_2+x_2x_3+x_3x_1=a,\quad x_1x_2x_3=-b$$
So, we can have $$\begin{align}x_3(x_1-x_2)^2&=x_3((x_1+x_2)^2-4x_1x_2)\\\\&=x_3(-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$ ?
EDIT: So ... | You should see the domain is at first restricted by the square root $\sqrt{1-x^2}$ so that we must have $|x|\le 1$. Also note that $g(x) = 2x\sqrt{1-x^2}$ is an odd function, so we can just analyse for $x> 0$.
$$g'(x) = 2 \sqrt{1-x^2} -\frac{2x^2}{\sqrt{1-x^2}} = \frac{2-4x^2}{\sqrt{1-x^2}}$$
Giving maxima at $x = \tfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do diff... | Hint 1: Since $(4n-2)^2\gt(4n-1)(4n-3)$, we have
$$
\begin{align}
\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}
&\le\sqrt{\frac{1\cdot3\cdots(4n-3)}{3\cdot5\cdots(4n-1)}\cdot\frac{1\cdot3\cdots(4n-3)}{1\cdot3\cdots(4n-3)}}\\
&=\sqrt{\frac1{4n-1}}
\end{align}
$$
Hint 2: Since $(4n-3)^2\gt(4n-2)(4n-4)$, we have
$$
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How do you go about factoring $x^3-2x-4$? How do you factor $x^3-2x-4$?
To me, this polynomial seems unfactorable.
But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.
So I got to solve backward:
$$x(x^2-2)-4$$
And add some terms and subtract them later,
$$x(x^2+2x+2-2x-4)-4=0$$
$$x(x^2+2x+2)-4-2x^... | It is better to go with Rational root theorem as suggested. I just want to write another way to see but clearly will not always work out that nicely.
\begin{equation}
x(x^2-2)-4
\end{equation}
instead of $x^2-2$, write $x^2-4$(because it is $(x-2)(x+2)$) and see if it is possible to factor:
\begin{equation}
x(x^2-4) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
What is the probability of getting at least one pair in Poker? Problem:
A 5-card hand is dealt from a perfectly shuffled deck of playing cards. What is the probability of each
of the hand has at least two cards with the same rank.
Answer:
By a rank, I mean a card like a $2$ or a king. The set of all poker hands is ${52... | The denominator is correct. So the numerator must be wrong.
The probability of not getting a pair is:
$$
\dfrac{48}{51} \dfrac{44}{50}\dfrac{40}{49}\dfrac{36}{48}\approx 0.507
$$
It's complement is $0.493$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.... | The formula is already true for $n=1,2,....,5$ and we know that
$$ \sum_{i=1}^ni= \frac{n\left(n+1\right)}{2}$$
Assume
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4 =\frac{n^4\left(n+1\right)^4}{8}$$
then,
$$\begin{align}\sum_{i=1}^{n+1} i^5+i^7&=\sum_{i=1}^{n} i^5+i^7 +(n+1)^5 +(n+1)^7\\&=\color{blue}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| $$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos nx=1-\frac{n^2x^2}2+O(x^4),$$
$$(\cos nx)^{1/n}=1-\frac{nx^2}2+O(x^4),$$
$$(\cos 2x)^{1/2}(\cos 3x)^{1/3}
=\left(1-\frac{2x^2}{2}+O(x^4)\right)\left(1-\frac{3x^2}{2}+O(x^4)\right)
=1-\frac{5x^2}{2}+O(x^4)$$
and so
$$\cos x-(\cos 2x)^{1/2}(\cos 3x)^{1/3}=2x^2+O(x^4)$$
so the limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
How to compute $\sum^k_{i=0}{k\choose i}\frac{\prod_{j=0}^{i-1}(x+jm)\prod_{j=0}^{k-i-1}(y+jm)}{\prod_{j=0}^{k-1}(x+y+jm)}$? This basic course on probability and statistics is the first course where I feel like a total idiot... especially since I've already forgotten much from the basic courses at discrete mathematics ... | You start off good:
$\operatorname{P}(X_1)=\frac{x}{x+y}$
$\operatorname{P}(Y_1)=\frac{y}{x+y}$
$\operatorname{P}(X_2)=\operatorname{P}(X_1)\operatorname{P}(X_2|X_1)+\operatorname{P}(Y_1)\operatorname{P}(X_2|Y_1)=\frac{x}{x+y}\frac{x+m}{x+m+y}+\frac{y}{x+y}\frac{x}{x+y+m}$
But you can simplify here:
$$\frac{x}{x+y}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
find the range of the function : $y=(3\sin 2x-4\cos 2x)^2-5$ find the range of the function :
$$y=(3\sin 2x-4\cos 2x)^2-5$$
My try :
$$y=9\sin^22x+16\cos^22x-24\sin 2x\cos 2x-5\\y=9+7\cos^22x-12\sin4x-5$$
now what do I do؟
| For real $x,$ $$(a\cos2x-b\sin2x)^2\ge0$$ the equality occurs if $$a\cos2x-b\sin2x=0\iff\dfrac{\cos2x}b=\dfrac{\sin2x}a=\pm\sqrt{\dfrac1{b^2+a^2}}$$
Using Brahmagupta-Fibonacci Identity,
$$(a\cos2x-b\sin2x)^2+(b\cos2x+a\sin2x)^2=(a^2+b^2)(?)$$
$$(a\cos2x-b\sin2x)^2=a^2+b^2-(b\cos2x+a\sin2x)^2\le a^2+b^2$$
the equality... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$ If $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?
Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?
| $$\sqrt{x+14-8\sqrt{x-2}} +\sqrt{x+23-10\sqrt{x-2}} = 3$$
$$\sqrt{(\sqrt{x-2}-4})^2+\sqrt{(\sqrt{x-2}-5)^2}=3$$
Then
$$|\sqrt{x-2}-4|+|\sqrt{x-2}-5|=3$$
Let $\sqrt{x-2}\ge5 (x\ge27)$. Then
$$\sqrt{x-2}-4+\sqrt{x-2}-5=3$$
$$\sqrt{x-2}=6$$
Let $\sqrt{x-2}\le4 (2\le x\le18)$. Then
$$4-\sqrt{x-2}-\sqrt{x-2}+5=3$$
$$\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Invertible 4x4 matrix $$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
$$
Is this matrix invertible? I would like to show that it is invertible but first I should find the det(Matrix) which should not be equal to zero. To find the determinant, ... | A combination of Gaussian elimination and a well-known trick readily gives the answer.
$$\det
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
=\det\begin{pmatrix}
5 & 6 & 6 & 2 \\
2 & 2 & 2 & 6 \\
6 & 6 & 2 & 2 \\
2 & 3 & 6 & 4 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Limit of $h(x)$ as $x \to \infty$ Let $h(x) = \sqrt{x + \sqrt{x}} - \sqrt{x}$ , find $\lim_{x \to \infty } h(x)$ . I've tried substitution , multiplying by conjugate and l'hospital's rule but didn't work .
| Multiplying and dividing by the "conjugate" $\sqrt{x + \sqrt{x}} + \sqrt{x}$ is a good strategy!
Note that for $x>0$,
$$ \sqrt{x + \sqrt{x}} - \sqrt{x}=\frac{ (x + \sqrt{x}) - x}{ \sqrt{x + \sqrt{x}} + \sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}} + \sqrt{x}}=\frac{ 1}{ \sqrt{1 + \frac{1}{\sqrt{x}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\fra... | $$s =\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n} = $$
$$ -\frac{1}{2} + \frac{1}{1} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{5} - \cdots $$
$$\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$$
$$ \sum_{k =1}^\infty \left(\frac{1}{2k-1}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that
$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$
I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\l... | Via Stirling's formula,
$$
\begin{align}
\frac{(n^2+n)!}{n^2!n^{2n}} &= \frac{\sqrt{2\pi(n^2+n)}\left(\frac{n^2+n}{e}\right)^{n^2+n}}{\sqrt{2\pi n^2}\left(\frac{n^2}{e}\right)^{n^2}n^{2n}}(1+o(1)) \\
&= e^{-n}\left(1+\frac{1}{n}\right)^{n^2+n}(1+o(1))
\end{align}
$$
Taking logarithms,
$$
\begin{align}
\log\frac{(n^2+n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum... | It's simpler still with equivalents:
$$\sin \frac{1}{2n} - \sin \frac{1}{2n+1}=2\sin\frac1{4n(2n+1)}\underbrace{\cos\frac{4n+1}{4n(2n+1)}}_{\substack{\downarrow\\\textstyle1}}l\sim_\infty2\,\frac1{4n(2n+1)}\cdot 1\sim_\infty\frac1{4n^2},$$
which converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solut... | If you factor out a $\sqrt{x}$ term from the denominator one has
\begin{align*}
\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} &= \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{x} \sqrt{x + \sqrt{x}}}}\\
&= \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac{1}{x} + \frac{1}{x^{3/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Is there an infinite number of numbers like $1600$? My reputation is at this moment at $1600$.
I did some experimenting with $1600$ and obtained the following:
Evidently, it is a perfect square $1600=40^2$
Also, it is a hypothenuse of a Pythagorean integer-triple triangle $1600=40^2=32^2+24^2$.
Also, it can be written ... | Yes, there are an infinite number of them. Given your result for $1600$ we can say that $1600k^2$ is another one. It is a square, it is $(32k)^2+(24k)^2$ and it is $4\cdot (20k)^2$. We can use the parameterization of Pythagorean triples to get others. If you choose $m,n$ relatively prime and of opposite parity, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of 'a' for Given Condition Let $f(x) $ be a positive differentiable function on $[0,a]$ such that $f(0)=1$ and $f(a)=3^{\frac{1}{4}}$. If $f'(x)\ge(f(x))^3 +(f(x))^{-1} $,what is the maximum value of a?
$ a) \frac{\pi}{12} b) \frac{\pi}{36}c) \frac{\pi}{24} d) \frac{\pi}{48}$
From the given conditions i ... | On the one hand, because$$
f'(x) \geqslant (f(x))^3 + \frac{1}{f(x)}, \quad \forall 0 < x < a
$$
then $f(x) \neq 0$ for $0 < x < a$. Since $f$ is continuous on $[0, a]$ and $f(0) > 0$, then $f(x) > 0$ for $0 < x < a$. Now,\begin{align*}\def\d{\mathrm{d}}
a &= \int_0^a \,\d x \leqslant \int_0^a \frac{f'(x)}{(f(x))^3 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it true that $n = 2t^2-2$ for even $t$ is a congruent number? This post asks for $m$ such that the simultaneous Pythagorean triples,
$$a^2+m^2b^2 = c^2\\b^2+c^2 = d^2\tag1$$
have solutions. Will Jagy found an infinite family given by,
$$m = 2t^2-2 = 0, 6, 16, 30, 48, 70, 96, 126, \dots$$
where,
$$\begin{aligned}
a &... | (This is a partial answer.)
After some persistence and effort, I managed to find a partial answer. It can be proven that
$$n = 2t^2-2$$
is a congruent number for infinitely many $t$ (odd or even).
Proof: If,
$$n = 2(v^2\pm3)^2-2$$
then,
$$p^2+nq^2=r^2\\p^2-nq^2 = s^2$$
has the simple solution,
$$\begin{aligned}p &=v^4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
we need to compute determinant of matrix. If$$
A=\begin{pmatrix}-2 & 9 & -1 \\ 8 & 0 & 7\end{pmatrix}, \quad B=\begin{pmatrix}1 & 5 \\ 6 & 3 \\ 2 & 4\end{pmatrix},
$$
find $|AB|$ without finding $AB$.
| By the Cauchy-Binet formula, we have
\begin{align*}
det(AB) &= \begin{vmatrix}
-2 & 9\\
8 & 0
\end{vmatrix}\begin{vmatrix}
1 & 5\\
6 & 3
\end{vmatrix} + \begin{vmatrix}
9 & -1\\
0 & 7
\end{vmatrix}\begin{vmatrix}
6 & 3\\
2 & 4
\end{vmatrix} + \begin{vmatrix}
-2 & -1\\
8 & 7
\end{vmatrix}\begin{vmatrix}
1 & 2\\
5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that two polynomials are constants if $P(x^2+x+1)=Q(x^2-x+1)$
Let $P$ and $Q$ be two polynomials with complex coefficients. It is known that $P(x^2+x+1)=Q(x^2-x+1)$. How can I prove that $P$ and $Q$ are constants?
I evaluated at various points to deduce $Q(3) = P(1) = Q(1) = P(3)$ but I don't know how to conclu... | Set $x=0$ and you get $P(1)=Q(1)$
With $x=1-0=1$ you get $P(3)=Q(1)$ and $x=-1-0=-1$ gives $P(1)=Q(3)$
With $x=-1-1=-2$ you get $P(3)=Q(7)$ and with $x=1-(-1)=2$ there is $Q(3)=P(7)$
Notice that $x^2+x+1=(-1-x)^2+(-1-x)+1$ and $x^2-x+1=(1-x)^2-(1-x)+1$.
So you can use vieta jumping with the functional relation to show ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How many complex roots? Find the number of solutions, counting multiplicity, in the domain $\{z \in \mathbb{C} : 1 < |z| <2\}$ to the equation
$$z^9+z^5-8z^3+2z+1=0$$
Notice that along $e^{i \theta}$ where $\theta \in [0,2\pi]$ we have $|8z^3| $dominating the polynomial. Thus, by Rouche's theorem, there are 3 zeroes w... | The original question was how to calculate the number of roots, now the question has changed to confirming that the number the OP got was correct.
Sketch:
Use Rouche's theorem to count the number of roots within the disk $|z|<2$ and then use Rouche's theorem to count the number of roots within the disk $|z|<1$. A reve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infinite geometric sum (asking for insight on an easier solution) Let the sequence $F$ be defined as: $F_1=F_2=1$ and $F_n=2F_{n-1}+F_{n-2}$, for $n>2$. Evaluate $\sum_{n=1}^{\infty}\frac{F_n}{10^n}$.
The obvious solution involves solving for the explicit formula for $F$ (using the standard linear recurrence technique)... | \begin{align*}
\sum_{n=1}^\infty\frac{F_n}{10^n}&=\frac{1}{10}+\frac{1}{100}+\sum_{n=3}^\infty\frac{F_n}{10^n}\\
&=0.11+\sum_{n=3}^\infty\frac{2F_{n-1}+F_{n-2}}{10^n}\\
&=0.11+\frac{2}{10}\left(\sum_{n=1}^\infty\frac{F_n}{10^n}-\frac{1}{10}\right)+\frac{1}{100}\sum_{n=1}^\infty\frac{F_n}{10^n}\\
0.79\sum_{n=1}^\infty\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cro... | It's $$(x^2-1)(3(x^2-1)^2+7(x^2-1)+4)=(x^2-1)(3(x^2-1)+4)(x^2-1+1)=$$
$$=x^2(x^2-1)(3x^2+1)=x^2(x-1)(x+1)(3x^2+1).$$
By your way.
It should be $$3p^2+7p+4=(p+1)(3p+4).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
linear algebra- how to find the representation matrix of a linear transformation when given the base of the origin and its transformations as the title states i want to find the matrix representing some linear transformations $T:V\rightarrow W$ but all i know is the base for $V$is $ (b_1,b_2,b_3)$ and $T(b_1),T(b_2),T(... | Your method is correct and the preferable method in this case.
Usually you first try inspection, so you try to find out what happens to the standard basis vectors $e_1,e_2,e_3$, so you try write $e_1$ as a linear combination of $b_1,b_2$ and $b_3$, and then you can directly calculate $T(e_1)$, which is the first column... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I integrate these functions? I deleted my last question, since some of you wanted me to rewrite the question properly. I feel sorry for inconvenience, but please understand that this is the first time I use $\texttt{MathJax}$.
Up to now, I tried every method I know in integration, like substituition, partial fra... | There are lots of different methods to calculate your integrals,
the simplest way is partial fraction expansion [wikipedia]
(https://en.wikipedia.org/wiki/Partial_fraction_decomposition):
$
\frac{1}{(x+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}}}=\frac{x^{%
\frac{4}{3}}}{2\left( 1-x\right) ^{\frac{1}{3}}}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 3
} |
Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
| Let $X_n =(a_n,b_n)$ with $X_0=(1,0)=(a_0,b_0)$
\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases} Then it is equivalent $$X_{n+1} = AX_n \Longleftrightarrow X_n =A^nX_0$$
where, $$A= \begin{pmatrix}1&2\\ 2&1\end{pmatrix}$$
Then prove by induction that,
$$A^n= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.
In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.
I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to bui... | $\mathbf {Hint...}$
$$\cos {3A}+\cos {3B}+ \cos{3C}=1$$
$$\Rightarrow 4 \sin {\frac {3C}{2}} .\sin {\frac{3B}{2}} .\sin{\frac{3A}{2}}=0$$
Hence the largest angle of triangle is $\frac{2\pi}{3}$ which can be either angle $C$ or angle $A$. By applying cosine rule in each of these cases we get the value of $AB$ as $\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expre... | In order to obtain the second expression you have to know that:
$$a^2b^2=(ab)^2\implies(m-1)^2(m+1)^2=((m+1)(m+1))^2=(m^2-1)^2\tag{1}$$
$$\frac{a^x}{a^y}=a^{(x-y)}\tag{2}$$
Hence
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(1)}\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(2)}\\=(m^2+2)(m^2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting f... | Hint:
As $2^{-y}=\dfrac1{2^y}$
$$\dfrac1{2^{3-x}}=2^{-(3-x)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Calculate $\int \frac{1}{x^2+x+1} \, dx$ $$ \int \frac{1}{x^2+x+1}\, dx = \int \frac{1}{(x+\frac 1 2)^{2} + \frac 3 4}\, dx $$
Substitute $x+\frac 1 2 = u$, $dx = du$:
$$\int \frac 4 3 \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du = \frac 4 3 \int \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du$$
Substitute... | Your book is wrong. If you differentiate the answer of the book we get $$\frac{3}{2(x^2+x+1)}$$ which is not the same as the integrand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$
Any point on the ellipse is given by $(a\cos \theta,b\sin... | Let $\alpha$ be the angle for which $\cos\alpha = \frac a{\sqrt{a^2+16 b^2}}$ and $\sin \alpha=\frac {4b}{\sqrt{a^2+16 b^2}}$.
Then $\theta-\alpha$ can be solved from $cos(\theta-\alpha)=\sin \alpha\sin \theta+\cos \alpha \cos \theta=\frac 3{\sqrt {a^2+16 b^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find value of $\lambda$ $S$ is a circle having center at $(0, a)$ and radius $b\lt a$. A variable circle centred at $(\alpha, 0)$ and touching the circle $S$ meets $X$ axis at $ M$ and $N$. A point $P=(0,\pm \lambda\sqrt {a^2-b^2})$ on the $y$ axis such that angle $MNP$ is a constant for any choice of $\alpha$ then fin... | There is no $\lambda$ such that $\angle MNP$ stays constant when $\alpha$ varies, since $\tan(\angle MNP)$ is a ratio of two segments, one of which is constant $\lambda \sqrt{a^2-b^2}\,$, while the other one clearly varies with $\alpha$.
The more interesting question, however, is whether such a $\lambda$ exists so that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all the triangles $ABC$ for which the perpendicular line to AB halves a line segment Let $AA_1$ and $BB_1$ be the bisectors of angles in triangle $ABC$. The bisectors intercept at the point $I$. How do I find all the triangles for which the perpendicular line from $I$ to $AB$ halves the line segment $A_{1}B_1$?
... | The answer is as follows :
$$\text{$\triangle{ABC}$ is either a right triangle with $\angle C=90^\circ$ or an isosceles triangle with $|\overline{CA}|=|\overline{CB}|$}$$
We may suppose that $A(-1,0),B(1,0),C(c,d)$ where $c\ge 0$ and $d\gt 0$.
The equation of the line $AC,BC$ is $dx-(c+1)y+d=0,dx-(c-1)y-d=0$ respectiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Intersection points of $2^x$ and $x^2$ I am trying to find the intersection points of $2^x$ and $x^2$.
To do this, I set the two functions equal to each other: $2^x = x^2$. It seems intuitive to take the log base two of both sides of the equations, and I got: $x = \log_2{x^2}$.
I'm not entirely sure how to proceed from... | All three solutions can be found using an analytic approach based on the Lambert W function.
Starting with the equation $2^x = x^2$, taking the square root of both sides gives
$$x = \pm 2^{x/2}.$$
Rewriting this equation in the form for the defining equation for the Lambert W function, namely
$$\text{W} (x) e^{\text{W}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating the probability of a function being real when it has random variables I'm trying to calculate the probability of
$ax^2+bx+c$ root being a real number when the variables $a,$ $b,$ and $c$ values are all randomized by throwing a standard die.
I got to the point where I can get the probability by calculating... | We need $b^2 - 4ac \ge 0$.
If $b= 6$ then $ac \le 9$. If $a =1$ then $c = 1...6$. If $a=2$ then $c=1...4$. If $c=3; a= 1..3$ if $a=4; c=1,2$ if $a=5,6;c =1$. There are $6+4+3+2+1+1 = 17$ ways to do this.
If $b = 5$ then $ac \le 6$. If $a=1$ then $c = 1...6$ if $a= 2$ then $b=1..3$ and if $a = 3; c=1,2$ and if $a \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Identify $\mathbb{Z}[x]/(x^2-3,2x+4)$. I need help to identify $\mathbb{Z}[x]/(x^2-3,2x+4)$.
I've been solving such problems in an approach like:
$$
2(x^2-3)=2x^2-6, x(2x+4)=2x^2+4x \\
(2x^2+4x)-(2x^2-6)=4x+6, 2(2x+4)=4x+8 \\
(4x+8)-(4x+6)=2
$$
What shall I do next, please? Thank you.
Simon
| we have $2=2(x^2-3)+(2-x)(2x+4)$ and then $(x^2-3,2x+4,2)=(2,x^2-3)$
$\mathbb Z[x]/(x^2-3,2x+4) \cong\mathbb Z[x]/(2,x^2-3)\cong \mathbb Z_2[x]/(x^2-1) \cong \mathbb Z_2[x]/(x-1)^2 \cong \mathbb Z_2[x]/(x)^2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $f\left(x\right)=$ $\sum_{n=0}^{\infty}$ $\frac{x^{3n}}{\left(3n\right)!}$ then prove $f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
QuestionIf $f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$
then prove that
$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
My Appro... | Your solution is generally ok, but for $f''$ we should rewrite it like below:
$f''=(f')'=(\sum_{n=0}^{\infty}\frac{x^{3n+2}}{\left(3n+2\right)!})'=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{\left(3n+1\right)!}$
So finally, we will have:
$f\left(x\right)+f'\left(x\right)+f''\left(x\right)=\sum_{n=0}^{\infty}\frac{x^{3n}}{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k... | Wolfram Alpha says it diverges because you typed in $n=1$ to infinity instead of $k=1$ to infinity. Correct the typo and it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating an integral2 Evaluate the following integral
$$\int\frac{dx}{\tan^2x +\csc x}$$
I tried the sub $$u=\tan{\frac{x}{2}}$$ to be
$$\int\frac{2u(1-u^2)^2}{(1+u^2)(u^6-u^4+3u^2+1)}du$$
Then i used the partial fractions to get
$$\int\frac{-2u}{1+u^2}+\frac{x^5-x^3+2x}{x^6-x^4+3x^2+1}$$
The first integrand is eas... | Try rewriting
$$\begin{align}
\frac{x^5 - x^3 + 2x}{x^6 - x^4 + 3x^2 + 1} &= \frac16 \frac{(6x^5 - 4x^3 + 6x) + (-2x^3 + 6x)}{x^6 - x^4 + 3x^2 + 1} \\
&= \frac16 \ \frac{6x^5 - 4x^3 + 6x}{x^6 - x^4 + 3x^2 + 1} - \frac16 \frac{(x^2-3)2x}{x^6 - x^4 + 3x^2 + 1}
\end{align} $$
The second term can be reduced to a cubic
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$... | Note that
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6...+n^6}{(1^2+2^2+3^2...+n^2)(1^3+2^3+3^3...+n^3)}}=\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\frac{12}{7}$$
indeed by Stolz-Cesaro
$$\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{(n+1)^7-n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{7n^6+...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Real ordered pair $(a,b)$ in equation
If $$a^2+5b^2+2b=6a+2ab-10$$
then all real ordered pair of $(a,b)$ is?
Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.
| The equation can be written as
$$(2b-1)^2+ a^2+b^2+9-2ab-6a+6b=0$$
$$\Rightarrow (2b-1)^2+ (-a)^2+(b)^2+(3)^2+ (2)(-a)(b)+(2)(3)(-a) +(2)(3)(b)=0$$
$$\Rightarrow (2b-1)^2+(a-b-3)^2=0$$
Hence we get $(a, b) = (\frac {7}{2} , \frac {1}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
F... | Let's take the derivative:
$$g'(x)=(-3(1-2x-3x^2)^\frac{2}{3})'=-3 \frac 2 3 (1-2x-3x^2)^{\frac {-1}3}(-6x-2)$$
$$g'(x)=2 (1-2x-3x^2)^{\frac {-1}3}(6x+2)=\frac {12x+4}{(1-2x-3x^2)^{\frac {1}3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $1/x+1/y+1/z=0$
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0$
My thinking was that since the numbers are integers, then there can't be $2$ negative values that cancel out the positive or $2$ positive nu... | Another proof using symmetric polynomials/Vieta's formulas. This generalizes to any field.
Let $\sigma_1=x+y+z$, $\sigma_2=xy+yz+xz$, and $\sigma_3=xyz$ be the elementary symmetric polynomials. In order to make sense of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$, we can assume that $xyz\ne 0$. Then
$\sigma_3(\frac{1}{x}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 2
} |
How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$.
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
| Let $n = 4$ and $m = 2n = 8$.
Then $1 + 2 + 3 + 4 + 5 +6+7+8 = 36$.
$1+2+3+4+5+6+7+8 = \frac {8(8+1)}{2}= \frac {2*4(2*4 + 1)}2 = 36$.
$1 + 2 + 3 + ....... + m = \frac {m(m+1)}{2}$ and if $m = 8$ then $1 + 2 + 3 +....... + 8 = \frac {8(8+1)}{2} = 36$.
And $1+2+3+...... + 2n -1 + 2n = \frac {2n(2n+1)}2$ and if $n= 4$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
B... | Since
$$
(5+10i)^{1/3}=\sqrt5\,e^{i\arctan(2)/3}
$$
and
$$
(5-10i)^{1/3}=\sqrt5\,e^{-i\arctan(2)/3}
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{(5+10i)^{1/3}+(5-10i)^{1/3}=2\sqrt5\cos\left(\frac{\arctan(2)}3\right)}
$$
Another Approach to the Original Question
Let $\alpha=2\sqrt5$. Then, using $\cos(3x)=4\cos^3(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Coefficients of the stirling's series expansion for the factorial. Knowing the Stirling's approximation for the Gamma function (factorial) for integers:
$$\Gamma(n+1)=n!\approx \sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$
Using the above approximation one can write:
$$(n+1)!=\sqrt{2\pi(n+1... |
We expand the series up to terms of $\frac{1}{n^3}$. We obtain for $\left|\frac{1}{n}\right|<1$
\begin{align*}
\color{blue}{\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}e^{-1}}
&=e^{-1}e^{\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)}\\
&=\exp(-1)\exp\left[\left(n+\frac{1}{2}\right)\left(\frac{1}{n}-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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.