Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Integral $\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} d x$ Consider the indefinite integral
$$F_{\pm}(x):=\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} \mathbb{d} x.$$
For all $x\in \mathbb{R}$ it is
$$ F_{+}(x)= \log(-x + \sqrt{1 + 2 x^2}).$$
But what is $F_{-}$?
Note that numerical studies suggest that... | The problem is $$F_\pm(x)=\int\frac{2x^2-1}{\pm x(2x^2-1)+2x+\sqrt{1+2x^2}}\,\mathrm{d}x.$$ Let $x=2t/(2-t^2)$, then $$\begin{aligned}F_+(t)&=2\int\bigg(\frac{t}{2-t^2}+\frac{1+t}{t^2+2t+2}\bigg)\,\mathrm{d}t,\\F_-(t)&=2\int\bigg(\frac{t}{t^2-2}-\frac{t^3-t^2-6t-2}{t^4-2t^3-12t^2-4t+4}\bigg)\,\mathrm{d}t.\end{aligned}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Values of $a$ such that $x^5-x-a$ has quadratic factor I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.
My Attempt
Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:
$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\e... | Another way is to do a long division of $x ^ 5-x-a$ by the arbitrary trinomial $x ^ 2 + bx + c$ and set the remainder to zero. This gives the remainder
$$(c ^ 2-3b ^ 2c + b ^ 4-1) x + (cb ^ 3-2bc ^ 2-a) = 0$$ from where we have
$a = cb (b ^ 2-2c)$ and $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$.
This means that for every solution of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
We know that $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. Prove ${x_n}$ is convergent Suppose $a>0, x_0>0$ and $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. How to prove ${x_n}$ is convergent? Can someone helps me?
I tried to prove that $x_n$ is monotonically decreasing and have a lower bound: $a^\... | Easy to see that all terms of $(x_n)$ are positive.
Now by AM-GM, we have the lower bound:
$$x_{n+1} = \frac 13\left(x_n + x_n + \frac{a}{x_n^2}\right)\ge \sqrt[3] a, \quad (\forall n \in \mathbb N) \tag{1}$$
Also, from the given formula we have
$$x_{n+1}-x_n = -\frac{1}{3}x_n + \frac {a}{3x_n^2} = \frac{-x^3_n + a}{3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| \begin{align}
5^{6n+6}-2^{2n+2} &\equiv 2^{6n+6}-2^{2n+2} \pmod{3} \\
&\equiv 2^{2n+2}(2^{(2n+2)2}-1) \pmod{3}\\
&\equiv2^{2n+2}(1-1) \pmod{3}\\
&\equiv 0 \pmod{3}
\end{align}
Notice that I used $\gcd(3, 2^{2n+2})=1$ and Fermat's theorem to conclude that $2^{(2n+2)2}\equiv1 \pmod{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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△ with altitude from and angle bisector of intersecting at $P$. Find $BP$ Let $\triangle ABC$ be a triangle with side lengths $AB = 17, BC = 28, AC = 25$. Let the altitude from $A$ to $BC$ and the angle bisector of angle $B$ meet at $P$. Given the length of $BP$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive ... | Let $\angle B=2\beta$ and $D$ be the foot of the altitude from $A$. Then
$$\cos\beta = \frac{BD}{BP}, \>\>\> \cos 2\beta = \frac{BD}{17}
$$
leading to
$$BP= \frac{17}{\cos\beta}(2\cos^2\beta-1)\tag1$$
From the cosine rule
$$\cos2\beta= \frac{ 17^2+28^2-25^2}{2\cdot 17\cdot 28}=2\cos\beta^2-1
$$
which yields $\cos^2\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$.
Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$. If $AB = \sqrt{1 + \sqrt{3}}$ and $[\Delta ABF]... | Let's use $AB=k=\sqrt{1+\sqrt 3}$. Then $$\sin 75^\circ=\frac{1+\sqrt 3}{2\sqrt 2}=\frac {k^2}{2\sqrt 2}$$
Area of triangle $ABF$ is then:
$$[\triangle ABF]=\frac12 AB\cdot AF\sin60^\circ=\frac12 k\frac {k\sin 45^\circ}{\sin 75^\circ}\sin 60^\circ\\=\frac 12 k^2\frac{\frac{\sqrt 2}2}{\frac{k^2}{2\sqrt 2}}\frac{\sqrt 3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Combinatorics with repetition - is my textbook right? I am suspicious that my textbook is incorrect.
Here's the question:
In how many ways can we distribute 20 candies among 6 children so that the youngest gets at most 2 candies?
My solution:
There are 20 candies (identical) and 6 children.
This is a combinations with ... | You are correct. The flaw in the textbook's solution is that if the youngest child gets $0$, $1$, or $2$ candies, only the number of candies the other children may receive can vary. In the textbook's solution, $y_1$ should be $0$ in each case.
That said, there is another way to do this problem.
If we ignore the restr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\e... | The truth is that $\sqrt{1+\sqrt2}$ cannot be expressed as $a+b\sqrt2$ where $a,b$ are integers or rationals. If they were reals then the problem becomes trivial.
(Another way to see the first fact above is that the minimal polynomial of $\sqrt{1+\sqrt2}$ is degree-$4$, whereas if it were expressible as $a+b\sqrt2$ wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Show that $4x^2-yz$ is a perfect square Here is my problem.
$A=xy+yz+zx$, where $x,y,z\in\mathbb{Z}$. It is known that if we add $1$ to $x$, and subtract $2$ from both $y$ and $z$, the value $A$ won't change. Prove that $-A$ is a square of whole number.
My attempt:
\begin{align}
A=xy+yz+zx
\end{align}
and
\begin{align... | If we expand, we get
$$A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)\quad =\quad x y + x z - 4 x + y z - y - z$$
Subtracting the original from this equation should be zero
$$(x y + x z - 4 x + y z - y - z)-(xy+yz+xz)=0\\
\implies 4x+y+z=0\implies z = -4 x - y$$
Substituting $z$ now allows the equation to be viewed more simply
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim_{n\to \infty} (n+1)^{\frac 23} -(n-1)^{\frac 23}$ Clearly, $\infty -\infty \not =0$
I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?
| Let $f:[n-1,n+1]\to\mathbb{R}$ and $f(x)=x^{\frac{2}{3}}$. Then from Mean value theorem (Lagrange’a) exist $\xi\in (n-1,n+1)$ such that:
$$\frac{f(n+1)-f(n-1)}{n+1-(n-1)}=f'(\xi)$$
but (check it out)
$$\frac{2}{3 \sqrt[3]{n+1} }\le f'(\xi)\le \frac{2}{3 \sqrt[3]{n-1} } $$
so
$$\frac{4}{3 \sqrt[3]{n+1} }\le f(n+1)-f(n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3930223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the average area as the point $X$ varies We consider a fixed triangle $ABC$ with side lengths $a = BC$, $b = CA$, $c = AB$, and
a variable point $X$ in the interior. The lines through $X$ parallel to $AB$ and $AC$,
together with line $BC$, determine a triangle $T_a$. The triangles $T_b$ and $T_c$ are define... | Let $[ABC]=S_0$ , its perimeter $= P_0$
Let bases (parallel to $a$) of ${T_a, T_b, T_c}$ be ${x,y,z}$ respectively.
Note each $T_i \sim \triangle ABC$ with similarity ratios ${x/a, y/a, z/a}$ respectively.
These three triangles leave three parallelograms out of $\triangle ABC$. Equating base lengths, we get $x+y+z=a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3930378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square Find all $b>5$ so that $x_n = \frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square for all sufficiently so large integers n.
I think the only value of $b$ is 10.
If there is $p \in \mathbb{P}$ (prime), $p \neq 3, p \mid b-1$, then $v_p(b-1)=1$. If $v_p(b-1) > 1$, we can choose n.
If $... | COMMENT: May be using these ideas help:
1-Suppose $\frac{b^{2n}+b^{n+1}+3b-5}{b-1}= (b+a)^2$
If polynomial $f(b)=b^{2n}+b^{n+1}+3b-5$ must have a double root like $-a$ then we must have:
$f(-a)=f'(-a)=0$
This gives following system of equations:
$\begin{cases}\frac{(-a)^{2n}+(-a)^{n+1}-3a-5}{-a-1}=0\\2n(-a)^{2n-1}+(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3931550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $1^3 + 2^3 + \ldots n^3 = \frac{n^2(n + 1)^2}{4}$ with mathematical induction Hello guys I'm trying to prove this equation with mathematical induction method.
So for n = 1 I know that :
1 = 1
Now I know that it works for some k
I wanna prove it for k + 1:
Since :
I can rewrite it as :
So I'm stuc... | Note: It's enough to prove $\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4 = n^3$.
That is to say, if we assume Equation 1: $1^3 + 2^3 + ...... + (n-1)^3= \frac {(n-1)^2n^2}4$
And we want to prove Equation 2: $1^3 + 2^3 +.... + (n-1)^3 + n^3 = \frac {n^2(n+1)^2}4$
That is a matter of proving by subtraction Equation 2 fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability Q on rolling two 8-sided dice Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?
So this is my answer and please correct me if I am wrong:
Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%
Prob of getting... | This answer assumes the person wins if there is a sum of $\, 7\,$ or $\,11\,$ in any toss before a sum of $\,9\,$ appears twice.
Probability of winning in first toss $ = \frac{12}{64}$
a) If the person gets a sum of $9$ once, the chance of winning from there
$\displaystyle P(W1) = \frac{12}{64} + \frac{44}{64} \times (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to put $(-\sqrt{3}-i)^{\frac{5}{7}}$ into polar form How to put $(-\sqrt{3}-i)^{\frac{5}{7}}$ into polar form and find all roots.
What I tried:
$$w = -\sqrt{3}-i$$
$$\arg(w)=\arctan(\frac{1}{\sqrt 3})-\pi = \frac{\pi}{6} - \pi = \frac{-5\pi}{6}$$
$$w = 2(\cos(\frac{-5\pi}{6})+ i\sin(\frac{-5\pi}{6})) $$
$$w^5 = 2^5... | you have :
$$(-\sqrt{3}-i)^{\frac{5}{7}}=\frac{1}{2^{5/7}}\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}i\right)^{5/7}=\frac{1}{2^{5/7}} \left(\exp{\frac{i7\pi}{6}}\right)^{5/7}=\frac{1}{2^{5/7}}\exp{\frac{i5\pi}{6}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate of $\int_{0}^{\infty}\frac{e^{x\pi}-1}{e^{x\pi}+1}\cdot \frac{x}{(1+x^2)(9+x^2)} dx$ Would like to know how would you evaluate $(1)$:
$$\int_{0}^{\infty}\frac{e^{x\pi}-1}{e^{x\pi}+1}\cdot \frac{x}{(1+x^2)(9+x^2)} dx=\frac{1}{8}?\tag1$$
Notices that, $$\tanh\left(\frac{x\pi}{2}\right)=\frac{e^{x\pi}-1}{e^{x\pi}... | Basically, we note
$$
\int_{0}^{\infty} e^{-at}\sin(xt) \,\mathrm{d}t = \frac{x}{a^2+x^2}
$$
$$
\int_{0}^{\infty} {\frac{\cos(\alpha x)}{\cosh(\beta x)} \mathrm{d}x} = \frac{\pi}{2\beta} \operatorname{sech}\left(\frac{\pi\alpha}{2\beta}\right)
$$
then denote desired integral with $I(a,b)$ that is
$$
\begin{aligned}
I(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to write this rational sequence : $(u_n=\frac{2^{n-1}+1}{2^n})_{n \ge 1}$ using only the rational numbers : $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{4}$ and the operation $+,-,\times$ ?
I get : $u_1 = 1 = \dfrac{1}{2} +\dfrac{1}{2}= \dfrac{1}{2}\left(1+1\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\rig... | As mentioned in the comments, you can use a recurrence, which needs only $\frac{1}{2},+$ and $\times$:
Since $u_n=\frac{1}{2}+\frac{1}{2^n} \Rightarrow \frac{1}{2^n}=u_n-\frac{1}{2}$, we have $u_{n+1}=\frac{1}{2}+\frac{1}{2^{n+1}}=\frac{1}{2}(1+\frac{1}{2^n})$ and the recurrence relation $u_{n+1}=\frac{1}{2}(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential equation system solution: do I get the right solution? I am very stuck with differential equation systems.
For example:
$
Y'(x) =
\begin{pmatrix}
2 & 0 & 1 \\
0 & 2 & 0 \\
0 & 1 & 3 \\
\end{pmatrix}
Y(x)
$
I get the eigenvalues and eigenvectors:
$ \lambda = 2 (double) \rightarrow \... | The eignevalues and eigenvectors looks correect to me.
$$y(x) = C_{1}
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{1}x
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
0 \\
-1 \\
1 \\
\end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$
This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\a... | If $A(z_A),B(z_B),M(z_M)$ are distinct points in Argand plane, then $\arg \left(\frac{z_M-z_B}{z_M-z_A}\right)$ is an oriented angle between vectors $(\vec{AM},\vec{BM}).$
In the present case $z_M=z,z_A=2,z_B=8.$
$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ says that the vectors are orthogonal, and $M$ lies on an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x\to+∞}\frac{(\sum_{n=0}^∞{(\frac{x^n}{n!})^2})^2}{(\sum_{n=0}^∞{(\frac{x^n}{n!})^1}) (\sum_{n=0}^∞{(\frac{x^n}{n!})^3})}$ Prove the following limit:
$$
\lim_{x\rightarrow +\infty} \frac{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^2} \right) ^2}{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}... | We investigate a heuristic idea that lead the correct answer. Consider a Poisson random variable $N$ with rate $x$, so that
$$ \mathbb{P}(N = n) = \frac{x^n}{n!} e^{-x} $$
for $n \geq 0$. Then by the (local) central limit theorem, we know that $Z = \frac{N-x}{\sqrt{x}}$ approximates the standard normal distribution, lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations? First, let us recall some trigonometric values
$\sin(0)=0$
$\sin(\pi/6)=\frac{1}{2}$
$\sin(\pi/3)=\frac{\sqrt{3}}{2}$
$\sin(\pi/2)=1$
$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$
$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Here, we can obse... | $8x^3−4x^2−4x+1=0$
Where does this polynomial come from?
Suppose you take a regular polynomial and plot the coordinates.
Consider a n-gon, with one vertex on $(1,0)$ all the vertices one unit from the center, and the center at $(0,0).$
$(1,0)\\
(\cos \frac {2\pi}{n},\sin \frac {2\pi}{n})\\
(\cos \frac {4\pi}{n},\sin \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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The limit of a series I need help with this exercise!
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$
I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
But I dont understand why!
So following that then,
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_... | Note $\sum_{k=1}^nk^4$ must be divisible by $n(n+1)$, so that the obvious extension to negative $n$ achieves $0-0^4=0$ at $n=-1$. The large-$n$ behaviour is asymptotic to $\int_0^nx^4dx=\tfrac15n^5$. We can determine coefficients in $\sum_{k=1}^nk^4=\tfrac15n(n+1)(n^3+An^2+Bn+C)$ from the sum at three values of $n$, sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perf... | I have somewhat weird way to see this. Consider the system
\begin{align*}
a + b p^{1/3} + c p^{2/3} & = 0\\
cp + a p^{1/3} + b p^{2/3} & = 0\\
bp + cp p^{1/3} + a p^{2/3} & = 0
\end{align*}
or
$$\begin{pmatrix} a & b & c \\
cp & a & b \\
bp & cp & a \end{pmatrix} \begin{pmatrix} 1 \\ p^{1/3} \\ p^{2/3} \end{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number,... | Start from the second line,
$$
\begin{aligned}
C_n=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}\\
\text{line 3 } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$ I am going around in circles on this system
$$x^2 - yz = 1\\
y^2 - xz = 2\\
z^2 - xy = 3$$
I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get:
$x^2 + y^2 + z^2 - yz - xz - xy =6$. And also t... | This type of the system is well known. Its general form reads
$$x^2 - yz = a,\>\>\>y^2 - xz = b,\>\>\>z^2 - xy = c$$
and has the solutions given in the symmetric expression below
$$(x,y,z)=\pm \frac{(a^2-bc, b^2-ca, c^2 -ab)}{\sqrt{a^3+b^3+c^3-3abc}}
$$
So, substitute $a=1,\>b=2,\>c=3$ to obtain
$\pm(\frac{-5}{3 \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If a, b and c are positive integers such that $ab = 432$, $bc = 96$ and $c < 9$, then the smallest possible value of $a + b + c$ is Below are the steps I have done so far.
Find $a$ in terms of $c$
$$
ab = 432 \\
bc = 96 \\
\frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c.
$$
Find $b$ in terms of $c$
$$
ab = 432 ... | The factors of $96$ that are less than $9$ are $\{1,2,3,4,6,8\}$
Plug these values of $c$ into your equation above to find the set of $a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $9x \equiv 21 \bmod 30$
Solving $9x \equiv 21 \bmod 30$
My approach:
Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}
So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$... | if $9x = 21 + 30 t,$ we see that $3x = 7 + 10 t,$ the reverse also holding. Solve
$$ 3x \equiv 7 \pmod {10} $$
for which you need a reciprocal of $3.$ Since $21 \equiv 1 \pmod {10} $ we see that
$$ \frac{1}{3} \equiv 7 \pmod {10} , $$
and.......
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Identity with the floor function I'm struggling to complete proof of the following identity:
$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.
Then
\begin{align}
\frac{... | Notice that $\forall x \in \mathbb Z, \left\lfloor \frac x2 \right\rfloor = \frac x2 - \left\lfloor \frac{x \mod 2}{2}\right\rfloor$, therefore
$$\left\lfloor \frac{m+n}{2}\right\rfloor + \left\lfloor \frac{m-n+1}{2}\right\rfloor\\
=\frac{m+n}{2} - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor + \frac{m-n+1}{2}- \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution... | Since $x+y=a$,
$$10a^3=x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3axy\,,$$
and $xy=-3a^2$.
$$\begin{align*}
x^4+y^4&=(x+y)^4-2xy\left(2x^2+3xy+2y^2\right)\\
&=(x+y)^4-2xy\left(2(x+y)^2-xy\right)\\
&=a^4+6a^2\left(2a^2+3a^2\right)\\
&=a^4+30a^4\\
&=31a^4\,.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x... | Since:
$f(x) \equiv x+2 \ (mod \ x^2+2x-3)$
$f(x) \equiv 1 \ (mod \ x+2)$
$x^2+2x-3 \equiv -3 \ (mod \ x+2)$
We have that:
$f(x)=(-1/3x^2-2/3x+1)+(x+2)+k \ (x^3+4^2x+x-6)$
so the remainder is: $(-1/3x^2-2/3x+1)+(x+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\fra... | $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n} \implies \lim_{n\rightarrow \infty} \dfrac{1}{1+\frac{1}{n}}-1-\dfrac{1}{n} $ equals?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. Am I correct? I have to find this limit:
\begin{align} \lim_{n \rightarrow \infty}
\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}
My attempt:
Since we know that $-1\leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$ we have:
\begin{align}
\lim_... | The first inequality is not true in general: if $a\leqslant b$ and $c<0$, then $ac\color{red}{\geqslant}bc$.
But note that$$\left|\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}\right|\leqslant\frac{\sqrt n}{n+1}\to0,$$and therefore your limit is $0$ indeed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Is it possible for $x_{n+1}-x_{n}$ to converge to $0$, but the limit of $x_n$ to not exist? I was reading about subsequences today and I came across an example which said the if $x_{n+1}-x_{n}$ converges to $0$, it's possible for $x_n$ to not have a finite limit (for example if we take $x_n$ to be the harmonic series).... | The answer is yes!
Not only you can find examples were the limit is not finite but is for example $\infty$. You can also have examples were a full interval is a set of limit points of the sequence.
Look at a sequence that will slide back and forth from $0$ to $1$ with steps decreasing to zero for example. Let's for exa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\co... | You got $x=y\tan\theta$. Now substitute back into the original equations to get $y=\cos\theta$ and $x=y\tan\theta=\sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Question about proving differentiability I'm asked to prove $f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}$ is differentiable for every x > 1
and calculate it's deriviate.
I have no idea how to approach this question.. Do I need to define a function g that will be the inverse of f?
i.e. $g\left(x\right)=\left(\frac{... | if $x \ge e$ then $\ln x \ge 1$ so $x\cdot \ln x > 0$ and $\ln (x\cdot \ln x)$ exists and $x\cdot \ln x = e^{\ln (x\cdot \ln x)}$.
And then $f(x) = (x\cdot \ln x)^{\ln x}= [e^{\ln (x\cdot \ln x)}]^{\ln x} = e^{\ln x\dot \ln(x\cdot \ln x)}$.
And that can be differentiated by the using the chain rule and the product ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\fra... | $$\int x\sqrt{1-x^2}sin^{-1}x dx$$
$$x = sin\theta$$
$$cos\theta= \frac{dx}{d\theta}$$
$$dx= cos\theta d\theta$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx= \int sin\theta(\sqrt{cos^2\theta})\theta (cos\theta d\theta)$$
$$=\int \theta sin\theta cos^2\theta d\theta$$
$$=\int \theta sin\theta (1-sin^2\theta) d\theta$$
$$=\int \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Time period of a nonuniform oscillator I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the va... | Note that
$$
T=
\int_0^{2\pi}\frac{2e^{-i\theta}}{2i\omega-a(e^{i\theta}-e^{-i\theta}) }ie^{i\theta}\,d\theta=
\int_0^{2\pi}\frac{2}{2i\omega e^{i\theta}-a(e^{2i\theta}-1) }ie^{i\theta}\,d\theta=\int_\gamma f,
$$
with $\gamma\colon[0,2\pi]\to\mathbb C$ given by $\gamma(\theta)=e^{i\theta}$ and $f(z)=2/(-az^2+2i\omega z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Partial fractions of $\frac{x^4}{(x^2-1)^3}$ (is it possible to find coefficients without expanding?) I want to decompose the fraction $\frac{x^4}{(x^2-1)^3}$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac D{x+1}+\frac E{(x+1)^2}+\frac F{(x+1)^3}$$
Because $f(x)=\frac{x^4}{(x^2-1)^3}$ is... | Following my comments, we may try two simple values of $x$ to reduce computations. Here I believe $x=0$ and $x=2$ are simple enough.
Plugging in $x=0$ gives $$A-B+C-(-A)+(-B)-(-C)=0\implies A-B=-\frac 1 8$$
and plugging in $x=2$ gives $$A\cdot 3^3 - B\cdot 3^3 + C\cdot 3^3 - A\cdot 3^2 + B\cdot 3 - C = 2^4\implies A + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$.
Integrate the following integral:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$$
I did this question a while ago and the answer is correct. However, when I ... | $ \sin{(2x)} = 2 \cos{(x)} \sin{(x)}$ if that's what you are asking for...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove or disprove $(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}$ The following question I read in a book, but the book does not give proof. I doubt the correctness of the result
let $p>3$ be prime number. prove or disprove
$$(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}\tag{1}$$... | You got it right. By Lucas correspondence the congruence (1) would be correct if it were modulo $p$ only. If we did this modulo $p$ we would be doing arithmetic in the ring $\Bbb{Z}_p[x]$ — a commutative ring of characteristic $p$. This means that we can apply Freshman's Dream twice, and get
$$
\begin{aligned}
(x+1)^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?
| If $t:=\tan x$ then $\frac{t(3-t^2)}{1-t^2}=\tfrac{2}{\sqrt{3}}$ so $t^3-\tfrac{2}{\sqrt{3}}t^2+3t+\tfrac{2}{\sqrt{3}}=0$. This has one real root, $t=\tfrac{2+33/a-a}{3\sqrt{3}}$ with $a:=\sqrt[3]{154+9\sqrt{443}}$. Now just calculate $\frac{1-t^2}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\fra... | $$\text{Let $y^2 = f(x)=\sin^2\left(\frac{2x+1}{3}\right)$}$$
So $y \in [-1,1]$ and
$y=\sin\left(\dfrac{2x+1}{3}\right)$
where $\dfrac{2x+1}{3} = 2n\pi + \theta $ and $-\pi \le \theta \le \pi$
for some $n \in \mathbb Z$.
We compute
\begin{align}
\theta &= \arcsin y \\
\dfrac{2x+1}{3} &= 2n \pi + \arcsin y \\
2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Probability of finding a non-repeating combination This is a real life problem, which I'm turning into balls just to make it easier to understand.
Suppose balls can be numbered from 1 to 12.
Suppose I have 10 boxes. Each box has 3 distinct balls, distributed at random.
A valid combination is if I can find, from those 1... | Consider a simpler case: 4 boxes and balls numbered from 1 to 6. We still put 3 balls in each box. I have chosen this case, since it is possible to write out all combinations and check corresponding probabilities.
For each box, we now have $6\choose3$ possible combinations.
In the first box, we can select any ball, yie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given a sequence $\{u_{n}\}$ so that $\left\{\begin{matrix} u_0=a>1\\u_{n+1}=\sqrt{1+\frac{1}{1+u_n}}\end{matrix}\right..$ Find $\lim u_n.$
Given a sequence $\left \{ u_{n} \right \}$ so that $\left\{\begin{matrix} u_{0}= a> 1\\ u_{n+ 1}= \sqrt{1+ \frac{1}{1+ u_{n}}} \end{matrix}\right..$ Find $\lim u_{n}.$
If there ... | Hint: Let $f(x) = \sqrt {1 + \frac{1}{{1 + x}}}$ for $x>0$. Then, for all $x>0$, $y>0$,
\begin{align*}
&\left| {f(x) - f(y)} \right| = \left| {\sqrt {1 + \frac{1}{{1 + x}}} - \sqrt {1 + \frac{1}{{1 + y}}} } \right| \\ &= \frac{{\left| {x - y} \right|}}{{(1 + x)(1 + y)\left( {\sqrt {1 + \frac{1}{{1 + x}}} + \sqrt {1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sum_{n=0}^\infty\frac{x^n}{n!}>0$ for all $x\in\Bbb R$ without invoking any Taylor series knowledge base. While studying basic Taylor expansions, I came across this expansion of $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$$We already know this function is positive for all $x$. But let us assume we don't know about... | To elaborate on some of the hints in the comments, let $x,y \in \mathbb{R}$. Then:
\begin{align*}
e^{x+y} &= \sum_{n=0}^{\infty}\frac{(x+y)^n}{n!} \\
&= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}\frac{x^ky^{n-k}}{n!} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} \\
&= 1 + \left(x+y\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Probability of a certain roll with six dice, each with a single marked side There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".
I calculate the probability of rolling three sets of two with 6 dice:... | Actually I didn't get how you calculated the number of possibilities for $2,3$ gold sides (though I have the same answers) but anyways this is my method (we'll solve for $2,3$ gold sides and then go for $1,0$ gold sides):
First of all, let's partition our $6$ dices into $3$ pairs of dices. Clearly there are $\frac{\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3... | Working mod $x^5-2$,
$$
\scriptsize\frac{3x^3}{1+x}=x^3\frac{1+x^5}{1+x}=x^3-x^4+x^5-x^6+x^7=2-2x+2x^2+x^3-x^4\tag1
$$
$$
\scriptsize\frac{5x}{1+x^2}=x\frac{1+x^{10}}{1+x^2}=x-x^3+x^5-x^7+x^9=2+x-2x^2-x^3+2x^4\tag2
$$
$$
\scriptsize\frac{9x^4}{1+x^3}=x^4\frac{1+x^{15}}{1+x^3}=x^4-x^7+x^{10}-x^{13}+x^{16}=4+8x-2x^2-4x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Puzzling solution for $1/x = 3 - 2\sqrt{x}$ What is the solution for
$$\frac{1}{x} = 3 - 2\sqrt{x}$$
When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.
However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.
What am I missing?
| If we let $y = \sqrt{x}$, we get
\begin{align*}
\frac{1}{y^{2}} = 3 - 2y & \Longleftrightarrow 1 = 3y^{2} - 2y^{3}\\
& \Longleftrightarrow 2y^{3} - 3y^{2} + 1 = 0\\\\
& \Longleftrightarrow (2y^{3} - 2y^{2}) - (y^{2} - 1) = 0\\\\
& \Longleftrightarrow 2y^{2}(y-1) - (y-1)(y+1) = 0\\\\
& \Longleftrightarrow (2y^{2} - y - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish t... | The idea in your attempt definitely works here (although it is not the shortest proof, see the answer by Narasimham): Let $a$ be the sidelength of the square. Then we have \begin{align*}
AE^2&=AD^2+DE^2=a^2+\left(\frac{1}{2}a\right)^2\\
EF^2&=EC^2+CF^2=\left(\frac{1}{2}a\right)^2+\left(\frac{1}{4}a\right)^2\\
AF^2&=AB^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$? For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?
I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2... | You are almost there: Note that by Cauchy-Schwarz you did
$$25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\,.$$
Thus
$$\left|\frac{8x-3y}{\sqrt{4x^2+y^2}}\right|\le 5\tag {*}\\[5ex]
\implies -5\le \frac{8x-3y}{\sqrt{4x^2+y^2}}\le 5$$
The important part was in step (*). The modulus was crucial as $8x-3y$ could be negative to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Radius of convergence of $f(z)=\frac{z^4e^z}{(6-z)(2i-z)} $ The Question is:
Compute the radius of convergence for the expansion at the point
$z=4+3i$ of the function $$f(z)=\frac{z^4e^z}{(6-z)(2i-z)}. $$
In my text book the solution is written like this:
$\sqrt{(4-6)^2+3^2}$, but I don't understand how they calcula... | The function $f$ is holomorphic on $\mathbb{C} \setminus \{6, 2i\}$. The Radius of Convergence is given by the minimum of the distances from $4+3i$ to $z = 6$ or $z = 2i$. That's due to the fact that we can write $f$ as a power series on $D_R(4+3i)$ (meaning the open disc of radius $R$ around $4+3i$). You also might fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$ If $r$ and $s$ are roots of $x^3-cx+d=0$, then
$$r^3-cr+d=s^3-cs+d , $$
for $r \neq s$. Solving the equation for $c$ gives me
$$ c=r^2+rs+s^2$$
Now to find an expression for $d$
$$ r^3-cr+d=0 $$
$$ d=r(c-r^2) ... | Let $p(x) = x^3-xc-d$ then $p(s)=p(r)=0$.
This is the part you already made
$$s^3-cs =d = r^3-rc$$ so
$$ (s-r)(r^2+rs+s^2)= (s-r)c$$ so
$$r^2+rs+s^2=c$$
Let $q(x)=x^3-cx+d$.
Now we have \begin{align}q(s+r)&=(s+r)^3-(r+s)c+d\\
&= p(s)+p(r)+3d +3sr(r+s)\\
&= 0+0+3d-3s(c-s^2)\\
&= 3p(s)\\
&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Leading term asymptotic integral containing logarithm I want to get the leading term in the asymptotic of the integral $$\int_{a}^{+\infty}\frac{x^3}{(1+x^2)^{\alpha}}\ln\frac{x+a}{x-a} \,dx$$ for $a\to +\infty$. Here $a>0$ and $\alpha>\frac{3}{2}$ to ensure integrability at infinity.
My main concern is the integrable ... | First, we make a change $x=ta$. The integral becomes $I(a, \alpha)=a^4\int_{1}^{\infty}\frac{t^3}{(1+a^2t^2)^{\alpha}}\ln\frac{t+1}{t-1} \,dt=a^{4-2\alpha}\int_{1}^{\infty}\frac{t^{3-2\alpha}}{(1+\frac{1}{a^2t^2})^{\alpha}}\ln\frac{t+1}{t-1} \,dt$.
Integral converges; the only (integrable) singularity is at $t=1$.
We c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$ Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$
I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.
| Note that $I(a) = \int_0^\infty \frac{\sin at} {t(t^2+1)}dt
= \frac\pi2 (1-e^{-a}) $, which can be obtain by solving
$$I’’(a)-I(a) = -\int_0^\infty \frac{\sin at}t dt= -\frac\pi2$$
Then
\begin{align}
\int_{0}^{\infty}\frac{\cos 3x}{x^4+x^2+1} dx
= &\frac14\int_{-\infty}^{\infty}
\overset{x=\frac{\sqrt3}2 t-\frac12}{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\inf... | Writing
$$\frac 1{x^3-1}=\frac{1}{(a-1) (a-b) (x-a)}+\frac{1}{(b-1) (b-a) (x-b)}+\frac{1}{(a-1) (b-1) (x-1)}$$ where
$$a=-\frac{1}{2}-\frac{i \sqrt{3}}{2} \qquad \text{and} \qquad b=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$$ we face three integrals
$$I(c)=\int \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{x}{c}\right)+\log (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
} |
Factorisation of a given polynomial into another with integer roots Source: Challenge and Thrill of Pre-College Mathematics
"Find all integers $a$ such that
$$(x-a)(x-12)+2$$
can be factored into $(x-b)(x-c)$, such that $b$ and $c$ are integers."
My attempt: Simplifying the given polynomial and by the given condition, ... | We could also look at the properties of the family of "upward-opening" parabolas described by $ \ y \ = \ (x-a)·(x-12) + 2 \ \ . $ Each member of this set must pass through the point $ \ (12 \ , \ 2 ) \ \ , $ which we shall see imposes an important restriction on its zeroes $ \ b \ $ and $ \ c \ \ . $ This forces the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$ $x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$... | $$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
$$\Leftrightarrow (x^2 + y^2 + z^2) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3(x+y+z)$$
Note that
$$(x^2 + y^2 + z^2) \geq \frac{1}{3} (x+y+z)^2 $$ (Cauchy-Schwarz ineq., equality: x = y = z)
and
$$(x+y+z) \left(\frac{1}{x} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Question on solving one algebra relation to obtain another If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:
If $(1-2x)(1-2y)(1-2z)... | let me continue from your work. Note that $$\frac{1}{x}-1=\frac{a}{b} $$ so $$\frac{(1-x)(1-y)(1-z)}{xyz}=1$$
$$\Rightarrow 1-(x+y+z)+xy+yz+xz-2xyz=0$$
$$\Rightarrow xy+xz+yz-2xyz=(x+y+z)-1$$
Also $$(1-2x)(1-2y)(1-2z)=-27$$
$$\Rightarrow 1-2(x+y+z)+4(xy+yz+xz)-8xyz=-27$$
$$1-2(x+y+z)+4(x+y+z)-4=-27$$
$$\Rightarrow x+y+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Binomial identity arising from Catalan recurrence. The following identity features right at the end of the answer to another question I asked: https://math.stackexchange.com/a/4019598/155881.
$$F(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{1- \sqrt{1-4x}}{2x}\right)^n = \sum\limits_{k=0}^{\infty}{n+2k \choose k}x^k$$
Haven'... | We seek to show that with
$$Q(z) = \frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n$$
we have
$$[z^k] Q(z) = {n+2k\choose k}.$$
Now with the branch cut on $[1/4, \infty)$ for $\sqrt{1-4z}$ we have
analyticity of $Q(z)$ in a neighborhood of the origin (note that the
exponentiated term does not in fact have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove equality using sum of squares and determinants Let $a$, $b$, $c$, $d$ be four positive real numbers satisfying that $a^2+b^2 = S = c^2+d^2$. I want to show that, under that hypotheses, it also holds:
$$ (ab+cd)S = (ac+bd)(ad+bc) $$
It seems to me that this equality can be deduced using suitable determinants of ma... | You can simply calculate this algebraically, no need to consider matrices:
$$\begin{align*}(ac+bd)(ad+bc) &= a^2cd + abc^2 + abd^2 + b^2cd \\
&= abc^2 + abd^2 +a^2cd + b^2cd \\ &= ab(c^2+d^2) + cd(a^2+b^2)\\
&= abS+cdS \\
&= (ab+cd)S\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all complex roots of $(z+1+i)^4 - 1 + i =0$. Find all complex roots of $(z+1+i)^4 - 1 + i =0$.
Attempt:
I got
\begin{align*}
(z+1+i)^4 &= 1 - i \\
(z + \sqrt{2} e^{i(\frac{\pi}{4})})^4 &= \sqrt{2} e^{i(-\frac{\pi}{4})} \\
z + \sqrt{2} e^{i(\frac{\pi}{4})} &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} ... | I got
\begin{align*}
(z+1+i)^4 &= 1 - i \\
(z + \sqrt{2} e^{i(\frac{\pi}{4})})^4 &= \sqrt{2} e^{i(-\frac{\pi}{4})} \\
z + \sqrt{2} e^{i(\frac{\pi}{4})} &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} \right) \\
z &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} \right) - \sqrt{2}e^{i(\frac{\pi}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help with inverse trigonometric substitutions $ \int x^2\sqrt{a^2+x^2}\,dx $. how would I go about integrating this? It is a lecture exercise.
$$
\int x^2\sqrt{a^2+x^2}\,dx
$$
I used a substitution of $x=a\tan\theta$ and ended up with $$\int a^4\tan^2\theta \sec^3\theta \,d\theta.$$ I was thinking by parts but then I w... | Integrate by parts directly without substitutions
\begin{align}
\int x^2\sqrt{a^2+x^2}dx
= &\frac14 \int \frac x{\sqrt{a^2+x^2}}d[(a^2+x^2)^2]\\
= &\frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}4\int \frac{\sqrt{a^2+x^2}}{2x}d(x^2)\\
=& \frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}8x{\sqrt{a^2+x^2}}-\frac{a^4}8\int \frac{dx}{\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \i... | Here's the inevitable 'expand and see if everything works out' solution.
$$
\frac{a^2}{a^2 + bc} + \frac{b^2}{b^2 + ac}+\frac{c^2}{c^2 + ab} < 2 \\
\iff a^2 (b^2 + ac)(c^2 + ab) + b^2 (a^2 + bc)(c^2 + ab) + c^2(a^2 + bc)(b^2 + ac) < 2 (a^2 + bc)(b^2 + ac)(c^2 + ab) \\
\iff 3a^2b^2c^2 + 2(a^3b^3 + b^3c^3 + c^3a^3) + a^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 4
} |
Difference between the maximum and minimum values of $a+b$ that satisfy $a+b+\frac{1}{a}+\frac{9}{b}=10, (a,b\in\mathbb{R}^+)$
Find the difference between the maximum and minimum values of $a+b$
that satisfy
$$a+b+\frac{1}{a}+\frac{9}{b}=10,\quad(a,b\in\mathbb{R}^+)$$
I'm trying to use Cauchy–Schwarz inequality, ... | Using the CS-inequality "Engels" form we have: $10 = a+b+\dfrac{1^2}{a}+\dfrac{3^2}{b} \ge a+b+\dfrac{(1+3)^2}{a+b}=a+b+\dfrac{16}{a+b}=x+\dfrac{16}{x}\implies 10x\ge x^2+16\implies x^2-10x+16 \le 0\implies (x-2)(x-8) \le 0\implies 2 \le x = a+b \le 8 \implies \text{max - min} = 8 - 2 = 6$. The minimum value of $a+b$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proof that a certain fraction is always an integer Prove that $$\frac{(n+1)(n)^2(n-1)^2...(n-k+2)^2(n-k+1)}{(k+1)(k)^2(k-1)^2...(2)^2(1)}$$ is or is not an integer for $0\leq k \leq n$, where $k$ and $n$ are integer values.
This looks like $\frac{(n+1)!(n)!}{(n-k+1)(n-k)(k+1)!(k)!}$. The above statement is true for k=1... | I’m going to give the answer more or less as I worked through the problem, since these numbers turn out to be rather interesting; if you just want a quick and easy computational proof, skip to the end.
I rewrote it as
$$\frac{n!}{k!(n-k)!}\cdot\frac{(n+1)!}{(k+1)!(n-k+1)!}=\frac1{k+1}\binom{n}k\binom{n+1}k\,,\tag{1}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Using $z = x + yi$ does not reveal all solutions The question is to find all solutions in $\mathbb{C}$ of the equation: $$z^2 - (i+1)z + i = 0$$
After expanding with $z = x + yi$, I get $(x^2 - y^2+y-x) + (2xy -x-y+1)i = 0$.
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.
Now, we plug that into $(2xy -x-y+1) = ... |
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.
That is your mistake here. $0=x^2-y^2+y-x=(x-y)(x+y-1)$ so $x=y$ or $x=1-y$, not $y-1$.
Then substituting into the second equation $2xy-x-y+1=0$ gives $(x,y)=(0,1)$ or $(1,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a^3b-ab^2$ is divisible by $3$ then either $a$ is divisible by $3$ or $b$ is divisible by $3$ or $(b-1)$ is divisible by $3$ Question:
Prove for all integers $a,b$ that if $a^3b-ab^2$ is divisible by $3$ then either $a$ is divisible by $3$ or $b$ is divisible by $3$ or $(b-1)$ is divisible by $3$
Attempt:
Suppose b... | Your error is that if $3\not \mid b$ (So $b\not \equiv 0\pmod 3$) and $3\not \mid b-1$ (So $b-1\not \equiv 0 \pmod 3$ and $b\not \equiv 1 \pmod 3$ then $b \equiv 2 \pmod 3$ (and not $b \equiv 1 \pmod 3$). So $b = 3j + 2\ne 3j + 1$.
(If $b = 3k$ then $3\mid b$. If $b = 3j+ 1$ then $3\mid b-1$. So to have $3\not \mid b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use Newton's Binomial Theorem to prove series Verify that for all integer $n\ge 2$,
\begin{eqnarray*}
\sum_{k=0}^n\binom{\frac{1}{2}}{k}\binom{\frac{1}{2}}{n-k}=0.
\end{eqnarray*}
(Hint: Consider Newton's Binomial Theorem.)
=================================
EDIT:
From Newton's Binomial Theorem:
\begin{eqnarray*}
(1+x)^... | (1+x)^\frac{1}{2} &=& \sum_{k=0}^\infty \begin{pmatrix}
\frac{1}{2} \\k
\end{pmatrix} x^k\\
B: (1+x)^\frac{1}{2} &=& \sum_{k=n}^\infty \begin{pmatrix}
\frac{1}{2} \\{n-k}
\end{pmatrix} x^{n-k}
\end{eqnarray*}
I assume finding $A-B$ will lead to the answer $0$ but I'm not sure how to continue from here
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve in the set of integers $2^x +5^x = 3^x + 4^x$
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x + 4^x$$
With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+... | Re write the Eq. as $$3^x-2^x=5^x-4^x....(1)$$
Use LMVT on $f(t)=t^x$ in $(2,3)$ and $(4,5)$
to write $$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, t_1\in(2,3)~~~~(2)$$
and
$$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, t_2\in (4,5)~~~~(3)$$
so $t_1 \ne t_2$, By (1), we can equate $xt_1^{x-1}=xt_2^{x-1} \implies x=0~or~ x=1.$
So (1) has only t... | {
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Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac... | So , we need to solve for : $$\sum_{k=1}^{\frac{n+1}{2}}\frac{1}{2k-1}-\sum_{k=1}^{\frac{n+1}{2}}\frac{1}{2k}=\psi$$ Which is also , $$\psi=\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{n}\right)-\frac{1}{2}S_{n+1}$$ $$\psi=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{n}+\frac{1}{n+1}\right)-S_{n+1}$$ Hence , ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Supremum of $x^2y^2(x^2+y^2)$ Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is
*
*$3n^2$
*$2n^6$
*$4n^6$
*$n^6$
My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express ... | This is a good opportunity to use Lagrange's multipliers method:
Let $g(x,y)=x+y-2n$. We have
$$\nabla g=\langle 1,1\rangle \textrm{ and }\nabla f = \langle 2xy^2(y^2+2x^2), 2yx^2(x^2+2y^2)\rangle$$
At a maximum, $\nabla f$ and $\nabla g$ are parallel, so $xy^2(y^2+2x^2)=yx^2(x^2+2y^2)$. If $x=0$ or $y=0$, we have $f(x... | {
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"source": "stackexchange",
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Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & ... | for the case $\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)$
also LU decopmostion
$$\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Squeeze theorem 2 bounded functions $$\lim_{(x,y) \to (0,0)} \frac{x^2 \sin (y)}{x^2+y^2} $$
Squeeze Theorem:
$$g(x) \leq f(x) \leq h(x)$$
$$-1 \leq \sin(y) \leq 1$$
$$- \frac{x^2}{x^2+y^2}\leq \frac{x^2 \sin (y)}{x^2+y^2} \leq \frac{x^2}{x^2+y^2} $$
Also
$$ 0 \leq \frac{x^2}{x^2+y^2} \leq 1 $$
What should be my next s... | I would use polar coordinates and equivalence of functions:
$$\frac{x^2 \sin (y)}{x^2+y^2}=\frac{r^2\cos^2\theta\sin(r\sin\theta)}{r^2}\sim_{r\to 0}r\sin\theta\cos^2\theta.$$
| {
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any physical interpretation for these two matrices? I know that the physical interpretation of $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin\theta & \cos \theta \end{pmatrix}$ is the rotation matrix.
But what are the physical interpretations of the matrices $\begin{pmatrix} \sin \theta & \cos \theta \\ -\cos\thet... | $$\begin{pmatrix} \sin \theta & \cos \theta \\ -\cos\theta & \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \left(\theta-\frac{\pi}{2}\right) & -\sin \left(\theta-\frac{\pi}{2}\right) \\ \sin \left(\theta-\frac{\pi}{2}\right) & \cos \left(\theta-\frac{\pi}{2}\right) \end{pmatrix}$$
So it is also a rotation matrix.
A ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the class of a function Let $$f(x) = \begin{cases}
x^3\sin(\frac{1}{x}); & \text{ if, } x \neq 0 \\
0; & \text{ $x=0$ }
\end{cases}$$
How do I find the greatest value of n such that $f \in C^n ([-1,1])$?
I know to find n, I must find the greatest n such that $f^{(n)}(x)$ is differentiable in the interval... | You have to check if $f^{(n)}$ is continuous $n$ by $n$.
Start with $n=0$.
The function is continuous because
$$\lim_{x \to 0^+} x^3 \sin{\left(\tfrac{1}{x}\right)} = \lim_{x \to 0^-} x^3 \sin{\left(\tfrac{1}{x}\right)}=0.$$
For $n=1$,
$f'(x) = \begin{cases} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1... | {
"language": "en",
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What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and... | If the upper vertex of the triangle is $(x,y)$, the area of the triangle formed by $(0,0)$, $(3,0)$, and $(x,y)$ is ${1 \over 2}bh$ or ${3h \over 2}$.
So the probability the area is at least $2$ is the probability that ${3h \over 2} > 2$, or equivalently the probability that $h > {4 \over 3}$. Since $h$ is uniformly di... | {
"language": "en",
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"source": "stackexchange",
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Prove that the sequence $ \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1) } { (2^1-1)(2^2-1)...(2^k-1) }$ is an integer For any given integer $n\ge 1$ and $k \in\{1, 2, \dots, n\},$
$$ F_n(k) = \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1) } { (2^1-1)(2^2-1)...(2^k-1) } $$
For example, if $n = 1,$ then $k \in \{1\}$ and $$F_1(... | Cobsider $(n)_q:=\frac{q^n-1}{q-1} $, the so called q-analogue of $n$. Observe that this is a polynomial of degree n-1 in the variable q.
Define now the q-factorial and the q-binomial as follow:
$$(n)_q!:= \prod_{i=1}^n (i)_q \qquad \binom{n}{k}_q:=\frac{(n)_q!}{(k)_q! \,(n-k)_q!}$$
It is easy to prove that both are p... | {
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Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.
Let $\triangle ABC$ be a triangle in the complex plane and let $a$,$b$ and $c$, respectively, be the complex coordinates of its vertices. Suppose that the tiangle is inscribed in the circle $... | Given equilateral triangle, i.e.
$$a= e^{i\alpha}, \>\>\>\>\>b = e^{i(\alpha+\frac{2\pi} 3)}, \>\>\>\>\>c= e^{i(\alpha-\frac{2\pi} 3)}
$$
it is straightforward to verify $a^2+b^2+c^2+3ab+3bc+3ca=0$.
Conversely, given
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$
rewrite the equation in term of $b =a e^{i x}$, $c =a ... | {
"language": "en",
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Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$
With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$
where $s$ is the semi-perimeter.
I started by applying the rule of sines:
*
*$a=k\sin A$
*$b=k\sin B$
*$c=k \sin C$
where $k$... | The semi-perimeter formula you're trying to use here is not true
$$a\cos{A}+b\cos{B}+c\cos{C} \ne s $$
If each symbol has its usual meaning in $\Delta ABC$
$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$
$$a \frac{b^2+c^2-a^2}{2bc} + b \frac{a^2+c^2-b^2}{2ac} + c \frac{a^2+b^2-c^2}{2ab} $$
$$\frac{ a^2(b^2+c^2-a^2)+b^2(a^2+c^2... | {
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Find a power series representation for $\frac{1}{2+x}$
Find a power series representation for $\frac{1}{2+x}$
The question did not specify a real number $c$ and an expansion on the form
$$f(x)=\sum_{n\geq 0} a_n(x-c)^n.$$
I did this problem as follows, but I seem to be off by a factor of $1/3$.
$$\frac{1}{2+x} = \fra... | Here is another answer: For this rational function it is possible to give an elementary construction of a powerseries and an "equality"
$$ \frac{1}{2+x}=\sum_{n \geq 0} a_n(x-a)^n$$
for any $a\neq -2$:
$$\frac{1}{2+x}=\frac{1}{2(1-(-x/2))} =\frac{1}{2}\sum_{n\geq 0}(-\frac{x}{2})^n=$$
$$ \sum_{n\geq 0}(-1)^n\frac{x^n}{... | {
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"timestamp": "2023-03-29T00:00:00",
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Logic behind base to number conversion I am trying to understand and visualize the logic behind the base conversion method.
By "base" I mean how many numbers in a number system:
*
*The decimal number system we use every day has 10 digits
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and so it is Base 10
*A binary digit can only be... | Because it works.
We need a way to express every possible whole numbers.
By having a base $b$ (assuming $b$ is a whole number larger than one) we can simply use the $b$ different digits to list the first $b$ whole numbers from $0$ to $(b-1)$.
(You might ask why $0$ to $b-1$ rather than $1$ to $b$. Technically it is ar... | {
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Evaluating $\int_0^1 (1-x^2)^n dx$ In an exercise I'm asked the following:
a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.
b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$
I used the binomial theorem in $a$ and got:
$$\int (1-x^2)^n dx = \sum_{k... | Part b) can be obtained just by induction. The result is immediate for $n=1$ and
\begin{align*}
\int_0^1 (1-x^2)^{n+1}dx = & \int_0^1 (1-x^2)(1-x^2)^ndx = \int_0^1(1-x^2)^n dx-\int_0^1 x^2 (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!} + \frac{1}{2} \int_0^1 x (-2x) (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!}-\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?
| $$1 = a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2-3ab)$$
implies $a+b=\pm 1$ and so $(a+b)^2 - 3ab = 1 - 3ab \not\in \{-1,1\}$ if $ab \ne 0$
| {
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"question_score": "5",
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Show that $a^2 - 2021b^2 = 13$ has no integer solutions
Show that $a^2 - 2021b^2 = 13$ has no integer solutions.
I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$
A s... | $a^2=13+2021b^2 \Rightarrow a^2 \equiv 6b^2 \mod 13 \iff \Big(\frac{a}{b}\Big)^2 \equiv 6 \mod 13 $ for $b\neq0$. But as $\genfrac(){}{}{6}{13}=-1$ then $6$ is not a quadratic residue so the equation does not have solutions unless $a \equiv b \mod 13$ and $a \equiv 0 \mod 13$. But even in this case, $13^2$ divides the ... | {
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Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005) Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.
Source : Vietnamese TST 2005
I'm sure that the only solutions are the zero function, the identity, and the negative... | As requested I'm posting the above comment as an answer here.
In order to show that every $n\geq 8$ can be written as sum of five cubes with absolute value less than $n$:
*
*we can write for odd numbers $n=2k+1$ with $k\geq 4$
$$(2k+1)^3 = (2k-1)^3 + (k+4)^3 - (k-4)^3 - 5^3 - 1^3$$
*for even numbers $n\geq 8$, we ca... | {
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"timestamp": "2023-03-29T00:00:00",
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(intergation) area of region bounded by a curve and a line with unknown a Let $S1$ be the area of the region bounded by the curve $y = x^2$ and the lines $y = ax, (0 < a < 1)$, and let $S2$ be the area of the region bounded by the curve $y = x^2$, the lines $y = ax, (0 < a < 1)$ and the line $x = 1$, where $a$ is the s... | You have calculated value of $\small S_1$ and $\small S_2$ correctly. Now to find value of $a$ that minimizes $ \small S = S_1+S_2$, take the derivative of $S$ with respect to $a$ and equate to zero.
$\small S = S_1 + S_2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$
$\small S' = \displaystyle a^2 - \frac{... | {
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Where did I go wrong with my differentiation here? This question involves the product rule and power of a function rule for differentiation. I often make silly algebra errors, but here I cannot find them. The answer is supposed to be y'=22. Thanks!
When $x=-1$, $y=(2x+1)^5(3x+2)^4$.
Hence, the derivative,
$$y'=5(2x+1)^... | We are computing the derivative at $x=-1$ when $y=(2x+1)^5 (3x+2)^4$. To find the derivative, we will be using chain rule and the product rule in correlation.
The product rule: $f'g+fg'$
In our case, Let $f=(2x+1)^5$ and $g=(3x+2)^4$
Hence, $f'=5(2x+1)^{5-1}\cdot(2x+1)^{'}\\ \;\;\;=5(2x+1)^4\cdot 2\\ \;\;\;=10(2x+1)^4$... | {
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How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_... | Your sum can be put as
$$(1+\frac 23)\Bigl(\frac 13+\frac{1}{3^3}+\frac{1}{3^5}...\Bigr)=$$
$$\frac 53\frac 13\Bigl(1+\frac 19+\frac{1}{9^2}+...\Bigr)=$$
$$\frac 53\frac 13 \frac{1}{1-\frac 19}=\frac 58$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the limit $\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}$ Find the limit, $$\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}.$$
My attempt:
I have tried several paths for evaluating the limit. For instance, $y=0$, $x=0$, $y=x$, $x=\tan y$, etc. I always obtain $0$. But I... | You may use the following limits which you can easily show using Taylor or L'Hosp.:
*
*$\lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac 12$, hence, $\lim_{t\to 0}\frac{1-\cos t}{t}=0$
*$\lim_{x\to 0}\frac{x-\arctan x}{x^2}=0$
Now, just rewrite the given expression as follows:
\begin{eqnarray*}\frac{x-\cos(x^2+y^2)-\arctan(x... | {
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Messed up on the inverse of $\cosh(x)$ but I don't know where I went wrong If someone can tell me where I went wrong, that would be extremely appreciated :)
The function $f:[0,\infty)\mapsto[1,\infty)$ defined by $f(x)=\cosh x$ is bijective. Find a formula for $f^{-1}.$
If $y=\cosh x$, then
\begin{align}
& y = \frac{... | We have
\begin{align}
e^x &= \frac{-2y\pm\sqrt{(2y)^2-4}}{-2} \\[5pt]
&= \color{red}{y\pm\frac{\sqrt{4y^2-4}}{-2}} \\[5pt]
&= y\pm\frac{\sqrt{4}\sqrt{y^2-1}}{-2} \\[5pt]
&= y\pm \frac{2\sqrt{y^2-1}}{-2} \\[5pt]
&= y \mp\sqrt{y^2-1} \, .
\end{align}
Since $x$ is nonnegative, we know that $e^x \geq 1$. Hence, $y\mp\sqrt{... | {
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My sol... | Problem with the Approach in the Question
One must take extra care when taking the limit of the sum of an increasing number of terms, each of which is tending to $0$. For example,
$$
\overbrace{\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\cdots+\frac1{n+n}}^{\text{$n$ terms, each of which tends to $0$}}
$$
If we sum the limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
solve $|\sin(a)|=\cos(3a)$ $|\sin(a)|=\cos(3a)$ is an alternative version of an equation $\sqrt{1-x^2}=4x^3-3x$, where I made a substitution $x=\cos(a)$ for $x \in [-1, 1]$. Unfortunately, I have no idea how to solve trigonometric equations.
| Consider
Case 1: $0 \le a \le \frac \pi 6$ then $0 \le 3a \le \frac \pi 2$. Than $\sin a \ge 0$ and $\cos 3a \ge 0$.
Then $a$ and $3a$ are both in the first quandrant. And the identity that will apply is $\sin a = \cos (\frac \pi 2 - a)$.
So we must have $\frac\pi 2 - a= 3a$ or $a =\frac {\pi} 8$.
Case 2: $\frac \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with arcsin integral I have the following integral:
$$ I =\int x^2\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) dx$$
Where $a$ and $b$ are non-zero positive integers, and $x<a$. I have started by integration by parts:
$$ I = uv - \int vu'$$
$$u = \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right )$$
$$u' = \l... | Make the substitution
$$u = \frac{\sqrt{a^2-x^2}}b.$$
Then $x = \sqrt{a^2 - u^2 b^2},$ and so $d x = - \frac{b^2 du}{2 \sqrt{a^2-u^2}},$
and so the integral becomes
$$\frac{b^2}2\int \sqrt{a^2 - u^2 b^2} \arcsin(u) du= \frac{b^2 a}{2} \int \sqrt{1- (\frac{ub}a)^2} \arcsin u du$$ Now integrate by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the matrix of the linear transformation $T$ Let $A=\begin{pmatrix}0\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},C=\begin{pmatrix}0\\1\end{pmatrix}$ and $D=\begin{pmatrix}1\\1\end{pmatrix}$. The images of these vectors under the linear transformation is $A^*=\begin{pmatrix}0\\0\end{pmatrix},B^*=\begin{pmatri... | It's correct! Note $D^*=\begin{pmatrix}2\\3\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
There is a shortcut for finding the equation of a tangent to a conic. To what other curves can this shortcut be applied? The conics can be written in Cartesian and parametric form:
Conic
Cartesian equation
Parametric equation
Ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$x=a \cos t, y=b \sin t$
Parabola
$y^2=4... | For unified expressions, let $(x_1, y_1)$ be the point on the ellipse $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$, the hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$, and the parabola $y=ax^2$. Then, their tangent lines are respectively
\begin{align}
\text{Ellipse:} &\>\>\>\>\> \frac{xx_1}{a^2}+ \frac{yy_1}{b^2}=1\\
\text{Hy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right)... | Taking the $\log$ of both sides for big enough $x$
$$\begin{align*}\log \frac{x^{2x}}{(x+1)^{x+1}}&=2x\log x-(x+1)\log(x+1)\\&=x\log \frac{x^2}{x+1}-\log(x+1)\\&\geq x\log \frac{x^2}{2x}-\log(2x)\\&=\log \frac{x^{x-1}}{2^{x+1}}.\end{align*}$$
Now take $e$ to both sides to get, for large enough $x$
$$\frac{x^{x}}{(x+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. than... | For any $z = a + ib \in \mathbb{C}$,
$$
z = \rho\left(\cos{\varphi} + i\sin{\varphi}\right),
$$
where
$$
\rho = \sqrt{a^2 + b^2}, \text{ and }\varphi = \arctan\left(\frac{b}{a}\right).
$$
Then,
$$
\sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos{\left(\frac{\varphi + 2\pi k}{n}\right)} + i\sin{\left(\frac{\varphi + 2\pi k}{n}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$... | Another shortcut: Let $I=\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}=\int\frac{1+\frac 1{x^2}}{\frac 1x-x}\frac{dx}{\sqrt{x^2+\frac 1{x^2}}}$ (By dividing $N^r$ and $D^r$ by $x^2$)
$I= -\int\frac{1+\frac 1{x^2}}{-x+\frac 1x}\frac{dx}{\sqrt{(-x+\frac 1x)^2+2}}$
Now substitute $-x+\frac 1x=t$ so that $(1+\frac 1{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $\frac{3^x+2^x}{3^x-2^x}=7$: More than one answer? How to solve for all $x$?
I am trying to solve this problem (math for fun):
$$\frac{3^x+2^x}{3^x-2^x}=7$$
Step 1. Let$\:a=3^x\:and\:b=2^x$
Step 2. $\frac{a+b}{a-b}=7$
Step 3. $a+b=7\left(a-b\right)=7a-7b$
Step 4. $6a-8b=0$
Step 5. $6a=8b=3\cdot 2\cdot a=4\cdo... | To answer you question how do you know how many solutions there are and how to you find them all: By keeping track of your steps and assuring you never add extraneous solution and that you steps are all one to one reversable so you don't lose or add solutions.
So $\frac {3^x + 2^x}{3^x -2^x} = 7$ gives us the conditio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
F... | You alos could use the binomial expansion or Taylor series
$$\sqrt[3]{a x+b}=\sqrt[3]{b}+\frac{a x}{3 b^{2/3}}-\frac{a^2 x^2}{9 b^{5/3}}+O\left(x^3\right)$$
$$\frac{\sqrt[3]{a x+b} - 2}{x}=\frac{\sqrt[3]{b}-2}{x}+\frac{a}{3 b^{2/3}}-\frac{a^2 x}{9
b^{5/3}}+O\left(x^2\right)$$ So, $$\sqrt[3]{b}-2=0 \implies b=8$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
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.