Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What's the derivative of $f(x)=(\frac{x^2 + 1}{x^2 + 3})^{\sin(2x)}$ I'm trying to calculate the derivative of $f(x)=(\frac{x^2 + 1}{x^2 + 3})^{\sin(2x)}$ using the difference quotient but somehow I don't succeed.
Help and elegant solution appreciated!
| Hint
$$\log f(x) = \sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)$$
$$\begin{align} \implies f'(x) &= f(x)\frac{d}{dx}\left[\sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)\right]\\&=\left(\frac{x^2 + 1}{x^2 + 3}\right)^{\sin(2x)}\frac{d}{dx}\left[\sin(2x) \log\left(\frac{x^2 + 1}{x^2 + 3}\right)\right]\end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I compare the numbers $2^{39}$, $5^{19}$ and $52^7$? I have to compare the numbers $2^{39}$, $5^{19}$ and $52^7$. I don't know how to do that because their exponents don't have anything in common.
| Hint: you can establish the right order between $5^{19}$ and $2^{39}$ as follows
\begin{align}
5^{19} &= 5^{20-1}\\
&=\frac{5^{20}}{5}\\
&=\frac{(5^2)^{10}}{5}\\
&=\frac{25^{10}}{5}\\
&>\frac{16^{10}}{2}\\
&=\frac{(2^4)^{10}}{2}\\
&=2^{39}
\end{align}
then note that
\begin{align}
52^7 &= 52^{10-3}\\
&=\frac{52^{10}}{52... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
least value of expression What is the least value of $\csc^2(x)+36\sec^2(x)$ ?
So I differentiated and when simplifying we get $\tan^4(x)=1/36$, but that is giving me max value i think. Can someone just help me . Or is there any other efficient way. Thanks !
| As you stated, yes taking derivative equal to zero you get $$\tan^4(x)=\frac{1}{36} \rightarrow tan^2(x)=\frac{1}{6} $$
More over $$ \frac{1}{\cos^2(x)}= 1+tan^2(x) = 1+ \frac{1}{6} =\frac{7}{6}$$ so the minimum is at such $x$. Substitute in the original formula :
$$\frac{1}{\sin^2 x} + \frac{36}{\cos^2 x} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $(a^2+b^2+c^2)(a+b-c)(b+c-a)(a+c-b)\leq abc(ab+bc+ac)$ Let $a,b,c$ are $3$ edge of a triangle. Prove $(a^2+b^2+c^2)(a+b-c)(b+c-a)(a+c-b)\leq abc(ab+bc+ac)$. My try: I suppose $c=\min\{a,b,c\}$ but I don't know what next.
| I give a correct answer here (in the last one there was a wrong factor $4$ instead of the correct $16$).
Basically I use here the following two facts:
(a) Stewart’s theorem.-It is applied to triangles with a cevian. In the figure below we apply this theorem twice, with the cevian $d=OG$ in $\triangle DOB$ and with th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts:
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}}
\end{equation}
I really just don’t know where ... | Hint:
Use integration by parts with $\;\displaystyle \int\dfrac{\mathrm d\mkern1mu x}{(x^2+a^2)^{n-1}}$, setting
$$u=\frac1{(x^2+a^2)^{n-1}},\quad\mathrm d\mkern1mu v=\mathrm d\mkern1mu x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Say for what values of $a \in \mathbb {R} $ this matrix system has solutions
Let $a \in \mathbb{R}$ and
$$
A_a =
\begin{pmatrix}
1 & a & 1 \\
a & 2 & 3 \\
2 & 3 & 4
\end{pmatrix}.
$$
*
*Say for each values of $a \in \mathbb{R}$ the system:
$$
A_a
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
=
... | You can do the exercise in one swoop by using Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
1 & a & 1 & 1\\
a & 2 & 3 & 0\\
2 & 3 & 4 & 1
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & a & 1 & 1\\
0 & 2-a^2 & 3-a & -a\\
0 & 3-2a & 2 & -1
\end{array}\right] && \begin{matrix}R_2\gets ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $f(n) = 3n^5 + 5n^3 + 7n$ is divisible by 15 for every integer $n$
*
*So far I have only been able to complete the base case for which I got the following:
$$f(n) = 3n^5 + 5n^3 + 7n$$
$$f(n) = 3(1)^5 = 5(1)^3 + 7(1)$$
$$f(n) = 3 + 5 + 7$$
$$15/15 = 1$$
From here I got a bit confused with the inductive st... | Although you asked for induction, it's much easier to simply split this into cases. We'll first prove that it's divisible by $3$.\begin{align}
n\equiv 0\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 0+0+0\equiv 0\mod 3\\
n\equiv 1\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3+5+7\equiv 0\mod 3\\
n\equiv 2\mod 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
I need some help with Geometry. Is this a correct answer to this problem? Good day,
I have a question regarding geometry. I don't know whether my answer is correct because the answer in my book uses a totally different method for solving this particular problem.
Here's the problem:
Given is a triangle, ABC, in which th... | It is straightforward.
$\angle A+\angle B+\angle C=\pi$
But given that $CM=AM=BM$, hence creating two isocele triangles $AMC$ and $BMC$ you have
$\angle A+\angle B=\angle C$
Hence the same as in your book $2\angle C=\pi$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For which $a$ and $b$ the function $\frac{ax+b}{a^2x+b^2}$ is increasing?
For which $a$ and $b$ the function $$\frac{ax+b}{a^2x+b^2}$$ is increasing?
I know that function is increasing if $x_1 > x_2 \implies f(x_1)>f(x_2)$ but how can I find $a$ and $b$ for which this statement will be true?
| Divide both polynomials:
$$\frac{ax+b}{a^2x+b^2}=\frac{a}{a^2}\left(1+\frac{b/a-b^2/a^2}{x+b^2/a^2}\right)=\frac{a}{a^2}+\frac{{a/a^2(b/a-b^2/a^2)}}{x+{b^2/a^2}}$$
This is function $1/x$ scaled and moved. The scale is
$$k=\frac{a}{a^2}\left(\frac{b}{a}-\frac{b^2}{a^2}\right)$$
You know that $c/x$ is decreasing when $c>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Area stacked between common tangent and circles Is there any way to find area of shaded region?
The radii of circles are $4$ and $12$ units.
|
By similar triangles, $CD=8$.
By Pythagoras' Theorem, $CG=EF=8\sqrt{3}$
Area of trapezoid $ACEF=\frac{(4+12)\times 8\sqrt{3}}{2}=64\sqrt{3}$
$\angle BAF=\cos^{-1} \left( \frac{1}{2} \right)=60^{\circ}$ and
$\angle BCE=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore$ area of sector $BAF
=\pi (12)^{2} \times \frac{60^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $|a\sqrt{1-b^2}+b\sqrt{1-a^2}-\sqrt{3(1-a^2)(1-b^2)} +\sqrt{3}ab| \le2$
Prove for any $a, b \in [-1, 1]$ that $$|a\sqrt{1-b^2}+b\sqrt{1-a^2}-\sqrt{3(1-a^2)(1-b^2)} +\sqrt{3}ab| \le2$$
I'm sure there is a solution using the Cauchy-Swartz inequality. Thus i tried to prove $$\Big(a\sqrt{(1-b)(1+b)}+b\sqrt{(1-... | Hint: if $a=\cos\alpha$, $\sqrt{1-a^2}=\sin\alpha$, $b=\cos\beta$ and $\sqrt{1-b^2}=\sin\beta$, then you need to show that
$$
\lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta|
=\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2,
$$
that is,
$$
\left|\frac12\sin(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$ How do I prove $$\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$$
without using induction? Note that clearly $n\neq 0$
Thanks for any help!!
| The reverse inequality is true for $n≥1$.
If $1≤k≤n$ then $\frac{1}{n+k}≥\frac{1}{n+n}$. Therfore, you get :
$$\frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n} ≥ \frac{1}{n+n}+\frac {1}{n+n}+\frac{1}{n+n}...+\frac{1}{n+n} = n \cdot \frac{1}{n+n} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{n}{n+1} < \frac{n+1}{n+2}$ by induction? I have the inequality
$\frac{n}{n+1} < \frac{n+1}{n+2}$
I'm not sure how to go about proving it. I've started by testing with n = 1, which results in
$\frac{1}{2} < \frac{2}{3}$ which is true
I then assume true for n = k and have to prove that it is true for n = k... | if you need induction:
\begin{gather*}
n(n+2)<(n+1)^2 \\
\begin{aligned}
(n+1)(n+3)={}&n(n+2+1)+(n+3)=(n(n+2)+n+n+3)<(n+1)^2+2n+3={} \\
{}={}&(n^2+4n+4)=(n+2)^2.
\end{aligned}
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
The equation represents the line of intersection of two planes. Using augmented matrix
$$
\begin{bmatrix}
1 & 1 & 1 & 3 \\
2 & 3 & 4 & 6
\end{bmatrix}$$
$R_2\rightarrow ... | Note:
I get
$$
y + 2z = 0 \Rightarrow z = - y/2 \\
3 + z = (x + y + z) + z = x + y + 2z = x
$$
So
$$
\frac{x - 3}{1} = \frac{y-0}{-2} = \frac{z - 0}{1}
$$
or
$$
(x, y, z) = (3,0,0) + t (1, -2, 1)
$$
Hint: The projection onto $z=0$ should affect what coordinate?
Spoiler:
The $z$-coordinate is forced to $0$. So we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
integrate $\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx=\int \frac{3\left(\frac{16}{9}-x^2\right)^{\frac{3}{2}}}{x^6}dx$$
$x=\frac{4}{3}\sin\theta$
$dx=\frac{4}{3}\cos\theta d\theta$
$$\int \frac{3\left(\frac{16}{9}-\frac{16}{9... | Another way using reduction formula:
$$\int(a^2-x^2)^nx^mdx$$
$$=(a^2-x^2)^n\int x^m\ dx-\int\left(\dfrac{d\{(a^2-x^2)^n\}}{dx}\int x^m\ dx\right)dx$$
$$\implies\int(a^2-x^2)^nx^mdx=(a^2-x^2)^n\cdot\dfrac{x^{m+1}}{m+1}+\dfrac{2n}{m+1}\int(a^2-x^2)^{n-1}x^{m+2}\ dx$$
Setting $n=\dfrac32, m=-6$
$$\int(a^2-x^2)^{3/2}x^{-6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $A$ be a $2\times2$ matrix with real entries such that $A$ is invertible. If $Det(A)=k$,and $Det(A+kadj(A))=0$ Let $A$ be a $2\times2$ matrix with real entries such that $A$ is invertible.
If $Det(A)=k$,and
$Det(A+kadj(A))=0$,
then find the value of $Det(A-kadj(A))$
My attempt:
$Det(A+kadj(A))=0$
$Det(A)Det(A+ka... | Since you're given that $A$ is a $2\times 2$ you could brute force it...
Calculating the determinant of
$$
\begin{align*}
A^2 + k^2I =
\begin{pmatrix}
a + k^2 & b \\
c & d + k^2
\end{pmatrix}
\end{align*}
$$
and setting it equal to zero gives that $a+d = -(1+k^2)$. Now solving for the determinant of
$$
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}$, Prove that $x+y+z$ $=0$
Question If $$ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} $$ Prove that $x+y+z$ $=0$
I've attmempted this question by cross multiplying so that
$$ x(c-a)(a-b) = y(b-c)(a-b) = (b-c)(c-a)z $$
but that did not work
I also tried splitting ... | \begin{align}
x+y+z&=\frac{b-c}{a-b}z+\frac{c-a}{a-b}z+z\\
&=z\left(\frac{b-c+c-a+a-b}{a-b}\right)\\
&=z\cdot 0\\
&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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To evaluate integral using Beta function - Which substitution should i use? $$\int_{0}^{1} \frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}dx = \frac{B(m,n)}{(a+b)^ma^n}$$
I have to use some kind of substitution but i do not understand what i use and why ?
Thanks
| Let's try the substitution
$$x=\frac{1-y}{1+cy}$$
so that $1-x=(1+c)\frac{y}{1+cy}$.
Then, when $x=0$, $y=1$ and when $x=1$, $y=0$. We also have $dx=-(1+c)\frac{1}{(1+cy)^2}\,dy$. Then, we can write
$$\begin{align}\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\int_0^1 \frac{\left(\frac{1-y}{1+cy}\right)^{m-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Understanding basis algorithm result I've a matrix ${\bf A}$ defined as
A = \begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
2 & 1 & -3 & 1 & 1\\
\end{pmatrix}
And ${W_1}$ is the solution space associated to the homogenous system ${\bf A}$. I am asked to find a basis for ${W_1}$.
I re-wrote the matrix as li... | A standard Gaussian elimination leads to a reduced row echelon form:
\begin{align}
\begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
2 & 1 & -3 & 1 & 1
\end{pmatrix}
&\to
\begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
0 & 5 & -3 & -5 & -13
\end{pmatrix}
&&R_2\gets R_2-2R_1
\\
&\to
\begin{pmatrix}
1 & -2 & 0 & 3 & 7\\
0 & 1 & -3/5 & -1 & -13/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove the inequality regarding complex numbers If $\theta_i\in [0,\pi/6],i=1,2,3,4,5$.And $$\sin \theta_1\ z^4 + \sin\theta_2 \ z^3 + \sin\theta_3 \ z^2 + \sin\theta_4 \ z + \sin\theta_5=2$$ Prove that $|z|\gt \frac{3}{4}$.
| $\theta_i\in [0,\pi/6]$ implies that $0 \le \sin \theta_i \le \frac 12$.
So it follows from your equation that
$$
2 \le \frac 12 \left( 1 + |z| + |z|^2 + |z|^3 + |z|^4 \right)
$$
or, with $r = |z|$,
$$
f(r) = 1+ r+r^2 +r^3 +r^4 - 4 \ge 0
$$
$f(r)$ is monotonically increasing in $r$ and $f(3/4) < 0$,
therefore a solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\r... | Given limit does not tend to $\infty$. This is the graph of $f(x)=\frac{x^2+9}{\sqrt{x^2+16}-5}$ in WolframAlpha.
As you see,
$$
\lim_{x\to -3+0}f(x)=-\infty
$$
and
$$
\lim_{x\to -3-0}f(x)=\infty.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How many ways we can choose items from different boxes I searched through the internet but couldn't find my answer, which can either be a very simple or a hard one.
Assume there are $3$ boxes, which carry, respectively, $1$, $4$, $2$ items.
My question is how many ways we can select $3$ items from these boxes.
I am loo... | Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Consecutive strings of heads problem So the question asks:
We toss a fair coin $n$ times and record the outcome as a sequence of H and T. We say that there is a run of heads if there is a consecutive string H...H which starts either at the first toss or after the coin lands tails and which ends either at the last toss... | With $z$ representing tails and $w$ representing heads and $u$
counting runs we get the generating function
$$\frac{1}{1-z}
\left(\sum_{q\ge 0}
\left(u \frac{w}{1-w} \frac{z}{1-z}\right)^q\right)
\left(1+u\frac{w}{1-w}\right).$$
Actually we are not interested in the distinction between heads and
tails once r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is there any integral for the Golden Ratio? I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$.
The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:
$$ \p... | Here is a collection of the series with reciprocal binomial coefficients.
$$\sum_{n=0}^\infty (-1)^n \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=\frac{4}{5} \left(1-\frac{\sqrt{5}}{5} \ln \phi \right)$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} \left( \begin{matrix} 2n \\ n \end{matrix} \right)^{-1}=-\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "166",
"answer_count": 42,
"answer_id": 1
} |
sum of floor ("weighted" divisor summary function) is there a closed form for
$$
\sum_{i=1}^N\left\lfloor\frac{N}{i}\right\rfloor i
$$
Or is there any faster way to calculate this for any given N value?
| This answer does not give closed form formula (for reasons mentioned in comments), but it gives a way to calculate the sum more efficiently than just going through all $i\leq N$.
The trick is that values of $\left\lfloor\frac{N}{i}\right\rfloor$ do not change for long ranges of $i$, for example:
$$
\left\lfloor\frac{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration of $\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx$ How can we integrate:
$$
\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx
$$
Using simple algebraic identities I deduced it to
$$
\int\frac{1-2\sin^2x\cdot\cos^2x}{(\sin x+\cos x)(1-\sin x\cdot\cos x)}dx
$$ but can't proceed further. Please provide some dire... | Noting that
$$
\sin ^4 x+\cos ^4 x= \left(\sin ^3 x+\cos ^3 x\right)(\cos x+\sin x)-\sin x \cos x\left(\sin ^2 x+\cos^2x \right),
$$
we have
$$
\begin{aligned}
I&=\int(\cos x+\sin x) d x-\int \frac{\sin x \cos x}{\sin ^3 x+\cos^3 x} d x \\
& =\sin ^2 x-\cos x-\int \frac{\sin x \cos x}{\sin ^3 x+\cos ^3 x} d x \\
& =\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find all functions $f:\mathbb R\to \mathbb R$ such that $f(a^2+b^2)=f(a^2-b^2)+f(2ab)$ for every real $a$,$b$ I guessed $f(a)=a^2$ and $f(a)=0$, but have no idea how to get to the solutions in a good way.
Edit: I did what was suggested:
from $a=b=0$
$f(0)=0$
The function is even, because from $b=-a$
$f(2a^2)=f(-2a^2)$.... | Define $g : \Bbb{R} \to \Bbb{R}$ by $g(x) = f(\sqrt{x})$ and $g(-x) = -g(x)$ for $x \geq 0$. ($g$ is well-defined since $f(0) = 0$.) We claim that
Claim. $g$ solves the Cauchy functional equation
$$ g(x+y) = g(x) + g(y), \quad x, y \in \Bbb{R} \tag{1}. $$
Proof. Let $x, y \in \Bbb{R}$.
*
*If $x, y \geq 0$. then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{... | I think your working is perfect. If the book is correct then
$$1-\frac4{3x}+\frac2{x^2}=2\implies 3x^2-4x+6=6x^2\implies3x^2+4x-6=0$$
whereas according to what is written there it must be $\;3x^2+4x-6=4x^2\;$, so I'd say the book's answer is wrong...or you forgot a coefficient $\;3\;$ for the quadratic term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that if $n \in \mathbb{Z}[\sqrt{2}]$ has an even norm, then $\sqrt{2} \mid n$ Aside from multiplying and dividing some specific numbers in this ring, e.g., $(1 + \sqrt{2})\sqrt{2}$ I have not really done anything productive on this question. I either go around in circles or jump to conclusions, such as that it fo... | If the norm $N(a+b\sqrt 2)=a^2-2b^2$ is even, then $a^2$ is even, and so is $a$. Then $a=2a'$ for some integer $a'$. Therefore :
$$a+b\sqrt 2 = 2a'+b\sqrt 2=\sqrt 2 (b+a'\sqrt 2)$$ From this you conclude that $\sqrt 2 \mid a+b\sqrt 2$ in $\Bbb Z[\sqrt 2]$, provided that the norm of $a+b\sqrt 2$ is even.
More general... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of
$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$
the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but ca... | This technique comes from A Course in Pure Mathematics by Margaret M. Gow:
$E=(a+b+c)^{3}-(b+c-a)^{3}-(c+a-b)^{3}-(a+b-c)^{3}$
$c=0 \implies E=(a+b)^{3}-(b-a)^{3}-(a-b)^{3}-(a+b)^{3}=0$
$\therefore c$ is a factor of $E$.
By symmetry, $a, b$ also are factors of $E$.
Since $E$ is cubic, let $E=kabc$ where $k$ is a consta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $x_1 = 2 $, and define $x_{n+1} = \frac{1} {2} (x_n + \frac {2} {x_n})$
Show that $x^2_n $ is always greater than or equal to $2$. And then use this to prove that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$ Hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$.
$\bullet~$ I can ... | Note the fact that $x_{n+1} =\frac{1}{2}(x_n+\frac{2}{x_{n}}) \le \frac{1}{2}(x_n+x_{n})=x_{n}$ from $x_{n}^2 \ge 2$ (or $x_{n} \ge \frac{2}{x_{n}}$)
Thus, $x_{n}-x_{n+1}\le0$.
This implies that $x_{n}$ is a decreasing sequence that is greater than $\sqrt{2}$, implying that $x_{n}$ is convergent.
Let $\lim{x_n}=a$.
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove the inequality $\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$ For every $a,b,c>0$ such that $abc=1$ prove the inequality
$$\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$$
My work so far:
$abc=1 \Rightarrow \frac 13 (a+b+c)\ge 1$
Then $$c^2+a+b \le ... | As we know:
$$2ab \le a^{2}+b^{2} $$, and
$$3a^{\frac{1}{3}} = 3(a^{2}bc)^{\frac{1}{3}} \le a^{2} + b + c $$ (because of $abc =1$)
Now we have :
$$2(\frac{ab}{3c^{1/3}} + \frac{ac}{3b^{1/3}} +\frac{cb}{3a^{1/3}})$$
Now using Cauchy we could prove our inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$. I am currently preparing for a certain quiz show when I encountered this question:
What is the remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$?
I know for a fact that $x^3-1=(x-1)(x^2+x+1)$.
And I got
$$x^{2016}+x^{2013... | \begin{align}
\sum_{k=1}^{672}x^{3k}&=\sum_{k=1}^{672}\left((x-1)(x^2+x+1)+1\right)^{k}\\
&=\sum_{k=1}^{672}\left(\sum_{j=0}^{k}{k \choose j}\left((x-1)(x^2+x+1)\right)^{j}\right)\\
&=\sum_{k=1}^{672}\left((x^2+x+1)P_k(x)+1\right)\\
&=(x^2+x+1)P(x)+\sum_{k=1}^{672}1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Using the ratio test for the series $\sum \frac1{3^n-2^n}$, I can't compute the limit $$\sum_{n=1}^{\infty} \frac{1}{3^n-2^n}$$
I know this series is convergent and using the ratio test.
But I can't conclude the proving.
$$\begin{align}\lim_{n\to\infty} \left| \frac{\frac{1}{3^{n+1}-2^{n+1}}}{ \frac{1}{3^n-2^n}}\right|... | You've almost got this one solved
$\lim_{n\to\infty} {(\frac{2}{3})}^n = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} .... = 0.\dot{6} \times 0.\dot{6} \times 0.\dot{6} ... = 0$
so your solution will come out to $\frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to find Ratio And proportion If $\frac{4x+3y}{4x-3y}=\frac{7}{5}$.Find the Value of the $\frac{2x^2-11y^2}{2x^2+11y^2}$.
Okay I just want hint how to solve this problem.
| In your first equation $\frac{4x+3y}{4x-3y}=\frac{7}{5}$, try solving x in terms of y or vice versa.
Then you will find that $x=\frac{9}{2}y$ or $y=\frac{2}{9}x$,
so now substitute one of these to $\frac{2x^2-11y^2}{2x^2+11y^2}$
if you choose $x=\frac{9}{2}y$ you'll have,
$$\frac{2x^2-11y^2}{2x^2+11y^2}=\frac{\frac{81... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \right... | Using that $\frac{k}{n^2} \xrightarrow[n\to\infty]{} 0$ for all $1\leq k\leq n$, one can use the Taylor expansion of $\sqrt{1+x}$, $$\sqrt{1+x} = 1+\frac{x}{2} - \frac{x^2}{8} + o(x^2)$$ when $x\to 0$:
$$\begin{align}
x_n &= \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}
= \sum_{k=1}^n \left(1+\frac{k}{2n^2}-\frac{k^2}{8n^4}+o\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int_{0}^{\pi} \frac{\pi}{1-\cos(x)\sin(x)}dx$ How can I evaluate
$$\int_{0}^{\pi} \frac{\pi}{1-\cos(x)\sin(x)}dx$$
I tried it using identity $\sin(x)=\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$ and $\cos(x)=\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$ but it is making question quite calculative. ... | This problem can also solve by complex method. Let $t=2x$, we have
$$
\int_0^\pi\frac{\pi}{1-\cos(x)\sin(x)}dx=\int_0^\pi\frac{\pi}{1-\frac{1}{2}\sin(2x)}dx\\
=\pi\int_0^{2\pi}\frac{dt}{2-\sin(t)}dt:=\pi I
$$
Now, let $z=e^{it}$, then
$$
\sin(t)=\frac{z-\frac{1}{z}}{2i}
$$
and
$$
dt=\frac{dz}{iz}
$$
Plug in, we have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Convergence of a sum, which test should I use? I have to show if this sum
$$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}$$
converges or not.
I tried to do something like
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$$
but I got nothing, because the best I could make it was:
$$\fr... | Three methods, increasingly involved (but increasingly generalizable):
To use a very simple trick:
As for any $n\geq 1$ we have
$$
0 \leq \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
\leq \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
$$
we can immediately conclude by comparison, since the series $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
limit $ \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} $ Calculate the limit $ \displaystyle \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} $
I now the answer, it is $ \displaystyle e^\frac{\log^2z}{2} $, but I don't know how to prove it. It ... | If $L$ is the desired limit then we have
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Radius of an e-circle in terms of triangle's sides and area. Following is the derivation given in my textbook:
I can't figure out that how do the areas of triangles $ABI_1$ and $ACI_1$ can be given by $\frac{1}{2} cr_1$ and $\frac{1}{2} br_1$ when $r_1$ is the radius of e-circle and $b,c$ are sides of triangle $ABC$ o... | Let the excircle at side AB touch at side AC extended at G, and let this excircle's radius be $ r_{c}$ and its center be $ {\displaystyle J_{c}}$.
Then $ {\displaystyle J_{c}G}$ is an altitude of $ {\displaystyle \triangle ACJ_{c}}$, so $ {\displaystyle \triangle ACJ_{c}}$ has area $ {\tfrac {1}{2}}br_{c}$. By a simila... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving correctness of a recursive function of multiplication by induction. Define multiplication of two numbers y and z as:
$$m_c(y,z)=\begin{cases}0&z=0\\m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)+y(z\mod c)&z\neq 0\end{cases}\tag{$\forall c\geq 2$}$$
Now I need to show that this is correct using induction... | Base Case: If $z=0$ then it is true.
Assumption: true for $z\le n,c\ge 2,y\ge 0$
Induction: $$\begin{align}m(y,n+1)&=m\left(cy,\left\lfloor\frac {n+1}c\right\rfloor\right)+y((n+1)\mod c)\\
&=cy.\left\lfloor\frac {n+1}c\right\rfloor+y((n+1)\mod c)\tag{$\left\lfloor\frac{n+1}c\right\rfloor<n+1\forall c\ge 2$}
\\&=y\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$ positive integers $\Rightarrow a+c \le n-1$
and $ad+bc \le bd-1.$
| Since $n>a+c \ge 2$ then $n \ge 3$. Moreover, $a<b$ and $c<d$, because $$\frac ab + \frac cd <1$$.
Consider the following cases.
a) Let $b \ge n$ and $d \ge n$, then $$\frac ab + \frac cd \le \frac an + \frac cn=\frac {a+c}{n} \le \frac {n-1}{n}=1-\frac 1n<1- \frac{1}{n^3}.$$
b) Let $b \le n$ and $d \le n$, then
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I... | There are two errors in your solution.
First, the transformation $$
\frac{\sin x}{\cos x} = 2\sin x \iff \sin x = 2\sin x \cos x
$$ is false. The correct inference is $$
\left( \frac{\sin x}{\cos x} = 2\sin x \text{ and } \cos x \neq 0 \right) \text{ or } \cos x = 0 \iff (\sin x = 2\sin x \cos x \text{ and } \cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Integral $I=\int \frac{dx}{(x^2+1)\sqrt{x^2-4}} $ Frankly, i don't have a solution to this, not even incorrect one, but, this integral looks a lot like that standard type of integral $I=\int\frac{Mx+N}{(x-\alpha)^n\sqrt{ax^2+bx+c}}$ which can be solved using substitution $x-\alpha=\frac{1}{t}$ so i tried to find such s... | The curve $t^2=x^2-4$ is a hyperbola, which can be parametrized by a single value. Rewrite this equation as $(x+t)(x-t)=4$, and set $y=x+t$. Then $4/y=x-t$, and we have
$$
x=\frac{y+\frac{4}{y}}{2},\;\;\;\;t=\frac{y-\frac{4}{y}}{2}.
$$
Compute $dx=(1/2-2 y^{-2})dy$, and the integral becomes
$$
\int \frac{dx}{(x^2+1)\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$
Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\... | Use $\sin\left(\dfrac\pi2-A\right)=\cos A$ and $\sin2B=2\sin B\cos B$
to find the product $$=\dfrac{\sin2t\sin4t\sin6t}8$$ where $7t=\pi$
Now use Method to find $\sin (2\pi/7)$, to find
$$\sin2t\sin4t\sin6t=\sqrt{\dfrac7{64}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Pascal's triangle induction proof I am trying to prove
$$\binom{n}{k} = \binom{n}{k-1}\frac{n-k+1}{k}$$ for each $k \in \{1,...,n\}$ by induction. My professor gave us a hint for the inductive step to use the following four equations:
\begin{align*}
\binom{n + 1}{k} & = \binom{n}{k} + \binom{n}{k - 1}\\
\binom{n + 1}... | Assume that $$\binom{n}{k} = \binom{n}{k-1}\frac{n-k+1}{k}$$
We need to prove:
$$\binom{n}{k+1} = \binom{n}{k}\frac{n-k}{k+1}$$
We will have
$$\begin{equation}\begin{aligned}
\binom{n}{k+1} &= \frac{n!}{(k+1)!(n-k-1)!} \\
&= \frac{n!}{k!\times(k+1)\times1\times2\times...\times{(n-k)\div(n-k)}} \\
&= \frac{n!}{k!(n-k)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the inverse of a matrix given an equation So I've been given this equation:
$A\begin{bmatrix}
2&3&1&5\\
1&0&3&1\\
0&2&-3&2\\
0&2&3&1
\end{bmatrix} = \begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}$
I'm supposed to find the inverse of A using this but I'm not really sure where to start. Any i... | If $AB=C$, then $A^{-1}=BC^{-1}$.
Now if we interpret the right-hand side matrix as a change of basis matrix, such that
$$u_1=e_4,\enspace u_2=e_1,\enspace u_3=e_3, \enspace u_4=e_2,$$
then, conversely
$$ e_1=u_2,\enspace e_2=u_4,\enspace e_3=u_3,\enspace e_4=u_1,$$
so that
$$\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$? How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$
where
$$f(x,y)=x^4+y^4-x^2-2xy-y^2$$
Answer:
$$f_x=4x^3-2x-2y$$
$$f_y=4y^3-2y-2x$$
Critical points are $(0,0), (... | We have on the boundary $x^2+y^2=4$:
$$f=x^4+y^4-(x^2+y^2)-2xy=(x^2+y^2)^2-2x^2y^2-(x^2+y^2)-2xy=12-2((xy)^2+xy)$$
Note that $|xy| \leq \frac{x^2+y^2}{2}=2$, and the extremal values of $t \mapsto t^2+t$ for $|t| \leq 2$ are $6$ (for $t=2$) and $-\frac{1}{4}$ (for $t=-\frac{1}{2}$).
This shows
$$0 \leq f=12-2((xy)^2+xy)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
integral for range in x axis Calculate the area in the shaded region:
${f(x) = x^3 -2x + 7}$
${{{\int_{-1}^2}} f(x) dx = x^4 - x^2 + 7x}$
$= {[(2^4) -(2^2) + 14] - [(-1)^4 - (-1)^2 + 7(-1)]}$
$= [16 - 4 + 14] - [- 7] = 19$
But the answer in the book is $21{3\over 4}$.
| HINT:
$$\int ax^3 + bx^2 + cx^1 + d \, \mathrm{d}x$$
is
$$ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^3 + dx + C $$
You have miscalculated your first term. Everything else is fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Differentiation Calculus: $\tan^{-1} \text{Problem}$ Well, Today at Math Revision exam I have to answer for $\frac{dy}{dx}$
Question:$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
I got the answer $$ \frac{2}{1+(\frac{2x}{1+x})^2}\frac{cos(2\tan^{-1}x)}{1+x^2}$$
i think this not correct.
I have to know the right solution with ste... | By using the chain rule, we have
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\left(\frac{{2x}}{{1+x^2}}\right)'\times\frac1{1+(\frac{2x}{1+x^2})^2}
$$ or
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\frac{2 \left(1-x^2\right)}{\left(1+x^2\right)^2}\times\frac1{1+(\frac{2x}{1+x^2})^2}
$$ giving
$$
\left(\arctan\frac{{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show that if a and b have the same sign then |a| + |b| = |a + b| We start with a, b > 0.
We know that $|a| = \sqrt{a^2}$, $|b| = \sqrt{b^2}$ and $|a+b| = \sqrt{(a+b)^2}$
We do the following :
$|a| + |b| = \sqrt{a^2} + \sqrt{b^2}$
$=a + b$
So : $|a| + |b| = a+b$
We take this result :
$|a| + |b| = a+b$
$(|a| + |b|)^2 =... | If $a\geq 0$ and $b\geq 0$ then
$$|a|+|b|=a+b=|a+b|.$$
If $a\leq 0$ and $b\leq 0$ then
$$|a|+|b|=-a-b=-(a+b)=|a+b|.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ Find the equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ and passes through the point... | write the first equation in the form $$ [x,y,z]=[1,2,3]+t[2,3,4]$$ and the second one in the form $$[x,y,z]=[-2,3,-1]+s[1,2,4]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Work out the area enclosed? I am doing a simple exercice and I think that either the book's solution is wrong or I misunderstood the problem.
Here is the problem,
平面上で次の曲線又は直線で囲まれる図形の面積を求めよ。
極座標系について、曲線 $r = a(1+2\cos \theta)$, $(0 ≦ \theta ≦ \frac{2\pi}{3})$, $(a > 0)$と直線 $\theta = 0$.
and its translation:
In the p... |
This only shows what the book's answer means. It does not mean that the op's answer is wrong because the given curve is $\left[0,\dfrac{2 \pi}{3}\right]$ (blue one), the red one is $\left[\pi,\dfrac{4 \pi}{3}\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Multiple Nested Radicals $\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$
I have no idea how to unnest radicals, can anyone help?
| Work from the inside out.
Let $\sqrt{3 - 2\sqrt{2}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides yields
$$3 - 2\sqrt{2} = a + b - 2\sqrt{ab}$$
Then
\begin{align*}
a + b & = 3 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{2} \tag{2}
\end{align*}
Dividing equation by $-2$ yields
$$\sqrt{ab} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
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Bivariate normal distribution $X$ and $Y$ I need help figuring out the following.
Let $X$ and $Y$ have the bivariate normal distribution
$$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$
Show that $X$ and $Z=\frac{Y−ρX}{\sqrt{1−ρ2}}$ are independent standar... | I assume your initial distribution is
$$
f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right)
$$
Apply the transformation
$Y = Z\sqrt{1-\rho^2} + \rho X$
Including the Jacobian determinant $\sqrt{1-\rho^2}$, the pdf for $X,Z$
$$
f_{XZ}(x,z) = \frac{1}{2\pi } \exp \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Moment generating function, determine distribution If $Y$ is a random variable such that $\mathrm{E}[Y^k] = \frac{1}{4} + 2^{k-1}$ for $k = 1,2,{...},$ I'm supposed to determine the distribution. I know the answer is $P(X=0) = \frac{1}{4}, P(X=1) = \frac{1}{4}, P(X=2) = \frac{1}{2},$ but I don't know how to reach that ... | Knowing those moments, the moment generating function is then $$M_X(t) = 1 + (\tfrac{1}{4} + \tfrac{1}{2} 2^{1})t+(\tfrac{1}{4} +\tfrac{1}{2} 2^{2})\frac{t^2}{2!}+(\tfrac{1}{4} + \tfrac{1}{2}2^{3})\frac{t^3}{3!}+\cdots $$
and rewriting $1 = \tfrac{1}{4} +(\tfrac{1}{4} +\tfrac{1}{2} 2^{0})t^0$ we have $$M_X(t)= \tfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Good substitution for this integral What is $$\int \frac{4t}{1-t^4}dt$$ is there some kind of substitution which might help .Note that here $t=\tan(\theta)$
| $$I=\int \frac{4t}{1-t^4}dt$$
Let $t=\sqrt{\sin\theta}$. Then, $dt=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$.
Then,
$$I=\int \frac{4\sqrt{\sin\theta}\cos\theta d\theta}{2\sqrt{\sin\theta}\cos^2\theta}=2\int\sec\theta d\theta=2\ln|\sec\theta+\tan\theta|+c=2\ln\left|\frac1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$
I found that:
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$
and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{... | Starting from where you left off: $S = \displaystyle \sum_{k=2}^\infty \left(\left(\dfrac{1}{2(k-1)} - \dfrac{1}{2k}\right)+\left(\dfrac{1}{2k} - \dfrac{1}{2(k+1)}\right)\right)$. Can you find the $2$ telescoping sums and their values?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find $ \lim\limits_{\substack{x\to\infty\\y\to0\,\,}}\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}$ I have to solve the following limit:
\begin{align*}
\lim_{\substack{x\to\infty\\y\to0\,\,}}\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}
&=
\lim_{\substack{x\to\infty\\y\to0\,\,}}\left\{\left[\left(1+\frac{1}{x}\right)^x\;... | $$\lim_{x\to\infty}\left[\lim_{y\to0}\left[\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}\right]\right]=\lim_{x\to\infty}\left[\lim_{y\to0}\left[\exp\left[\ln\left(\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}\right)\right]\right]\right]=$$
$$\lim_{x\to\infty}\left[\lim_{y\to0}\left[\exp\left[\frac{x^2\ln\left(1+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute all possible values of $\cot{\theta} - \frac{6}{z}$. Note that $ 0 < \theta < \frac{\pi}{2}$.
Let $\theta = \arg{z}$ and suppose $z$ satisfies $|z - 3i| = 3$.
Compute all possible values of $\displaystyle \cot{\theta} - \frac{6}{z}$. Note that $\displaystyle 0 < \theta < \frac{\pi}{2}$.
$\bf{My\; Try::}$ Let ... | WLOG $z-3i=3(\cos2t+i\sin2t)$
$\iff z=3\{\cos2t+i(1+\sin2t)\}$
$=6\sin\left(\dfrac\pi4-t\right)\left[\cos\left(\dfrac\pi4-t\right)+i\sin\left(\dfrac\pi4-t\right)\right]$
$\dfrac6z=\cdots=\dfrac{\cos\left(\dfrac\pi4-t\right)-i\sin\left(\dfrac\pi4-t\right)}{\sin\left(\dfrac\pi4-t\right)}=\cot\left(\dfrac\pi4-t\right)-i$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove $\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$ Im trying to prove the following equation:
$$\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$$
I have tried to do it several times now but get stuck at different places. I don't know if it's a calculation error or the question is defective. It would ... | The problem is simple but you may have made a calculation mistake. Let's start with the identity $$\sin^{2}x = \frac{1 - \cos 2x}{2}$$ and then cube it to get $$\sin^{6}x = \frac{1}{8}(1 - 3\cos 2x + 3\cos^{2}2x - \cos^{3}2x)$$ and then we need to note that $$\cos^{2}2x = \frac{1 + \cos 4x}{2},\,\cos 6x = 4\cos^{3}2x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Find the transformation matrix given a set of points The map $A:ℝ^3→ℝ^3$ satisfies:
$A(2,−1,2)=(−2,−9,−30)$
$A(−1,2,2)=(−23,12,−9)$
$A(2,2,−1)=(13,6,−6)$
So if I understand it correctly there is a matrix $A$ such that
$ [A] \begin{bmatrix}2\\-1\\2\end{bmatrix} = \begin{bmatrix}-2\\-9\\-30\end{bmatrix}$
and
$ [A] \begin... | An answer for noobs.
You said you had this:
(1)
$ [A] \begin{bmatrix}2\\-1\\2\end{bmatrix} = \begin{bmatrix}-2\\-9\\-30\end{bmatrix}$
(2)
$ [A] \begin{bmatrix}-1\\2\\2\end{bmatrix} = \begin{bmatrix}-23\\12\\-9\end{bmatrix}$
(3)
$ [A] \begin{bmatrix}2\\2\\-1\end{bmatrix} = \begin{bmatrix}13\\6\\-6\end{bmatrix}$
$ A = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rusty at complex integrals, finding error in steps I am looking for some help spotting where I messed up some computations, it's been awhile... I'll show my work to my answer and then say what the correct answer is...
Integrate:
$$\int_0^{2\pi} \frac{\cos(3\theta)}{5-4\cos(\theta)}\mathrm{d}\theta$$
Let $z=e^{i\theta}$... | Community wiki answer so the question can be marked as answered:
As DavidP commented, the residue at $0$ is $\frac{21}4$. Apparently you forgot to multiply the Wolfram|Alpha result by the factor $\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can every parabola be written in the form of a quadratic $y=ax^2+bx+c$ or $x=dy^2+ey+f$? I understand that the graph of any equation of the form $y=ax^2+bx+c$ is a parabola (please correct me if I am mistaken).
My question is about the converse: Can every parabola be written in the form of an equation $y=ax^2+bx+c$ or ... | A parabola is just the locus of points (a collection) that is equidistant from a point and a straight line. Let $(x,y)$ be the set of points, then for let's say you have a fixed point $(p,q)$ and a line $ax+by+c=0$,
The distance between $(x,y)$ and the straight line is given by:
$$\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$$
We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(... | Notice, when $\Re[n]>0$:
*
*$$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^n\space\text{d}x=\frac{\Gamma(n+1)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(1+\frac{2}{n}+n\right)}$$
*$$\int_{0}^{1}\left(1-x^{\frac{n}{2}}\right)^{n+1}\space\text{d}x=\frac{\Gamma(n+2)\Gamma\left(\frac{2+n}{n}\right)}{\Gamma\left(2+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Calculate Tangent Points to Circle Problem
Given a circle with radius $r = 2$ and center $C = (4,2)$ and a point $P = (-1,2)$ outside the circle.
How do I calculate the coordinates of two tangent points to the circle, given that the tangents both have to go through $P$?
My (sad) try
All I can think of is finding the eq... | There is another way to find the that may strike you as funny. Use homogeneous coordinates for the points in the place so the exterior point is located at $P = (-1,2,1)$.
The homogeneous coordinates of a circle located at $(c_x,c_y)$ with radius $r$ are
$$ C = \begin{vmatrix} 1 & 0 & -c_x \\ 0 & 1 & -c_y \\ -c_x & -c_y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let... | $$I_2=\frac{-1}{2} \int_0^a \frac{d(a^2-x^2)}{\sqrt{a^2-x^2}}$$
let $a^2-x^2=t$. If $x=0,$ then $t=a^2$ and if $x=a,$ then $t=0$.
Hence $$I_2=\frac{-1}{2} \int_{a^2}^0 \frac{dt}{\sqrt{t}}$$
$$\Rightarrow I_2= \int_{0}^{a^2} \frac{dt}{2\sqrt{t}}$$
$$\Rightarrow I_2=a-0=a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Using induction, show ${(1+\sqrt{2})}^{2n}+{(1-\sqrt{2})}^{2n}$ is an even integer. I'm having serious difficulties with that task, so it should be nice, if there is someone that can help!
The task says:
Prove that the number
$${(1+\sqrt{2})}^{2n}+{(1-\sqrt{2})}^{2n}$$ is an even integer for all $n\in\mathbb{N}$
| If $a_n=pr^n+qs^n$, with $r\ne s$, the sequence $a_n$ satisfies the recurrence relation
$$
a_0=p+q,\qquad a_1=pr+qs,\qquad a_{n+2}-(r+s)a_{n+1}+rsa_n
$$
Indeed,
\begin{align}
a_{n+2}-(r+s)a_{n+1}+rsa_n
&=pr^{n+2}+qs^{n+2}-(r+s)(pr^{n+1}+qs^{n+1})+rs(pr^n+qs^n) \\
&=p(r^{n+2}-r^{n+2}-sr^{n+1}+sr^{n+1}) \\
&\qquad+q(s^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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When will the sequence $k \mapsto A + Bk + k^2$ yield a perfect square? Consider the following sequence:
$$a(k) = A + Bk + k^2 ,$$
where $A$ and $B$ are both integers, and $A < B$ ($k$ is of course an integer variable, B is even).
Problem:
For which $k^*$ is $a(k^*) = r^2$ (in other words for which $k$ the given sequen... | By repeatedly replacing $k$ by either $k-1,$ or repeatedly by $k+1,$ we can demand that $0 \leq B < 2.$ This will alter $A,$ let us now call it $C.$ This is simply Gauss reduction, if you think in terms of the quadratic form $f(k,j)=k^2 + B k j + A j^2.$
Two sub-problems only:
$$k^2 + C = u^2$$
$$ k^2 + k + C = v^2 $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Point on a plane closest to the origin Consider the plane with equation $x-y+z=1.$ Find the point on the plane closest to the origin.
I think I need to find a function for the distance between the plane and the origin and then find it's derivative and set it equal to zero. I'm not sure exactly how to go about doing th... | The point on the plane $ax+by+cz=d$ nearest the origin is $$\biggl(\frac{da}{\sqrt{a^2+b^2+c^2}}, \frac{db}{\sqrt{a^2+b^2+c^2}}, \frac{dc}{\sqrt{a^2+b^2+c^2}}\biggr)$$
Here, the point on the plane $x-y+z=1$ nearest the origin is $$\Biggl(\frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times-1}{\sqrt{1^2+(-1)^2+1^2}}, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Critical points of $g(x, y) = x^3+3xy^2-12x-6y$
Find the critical points of $g(x, y) = x^3+3xy^2-12x-6y$.
$\nabla g(x, y) = (3x^2+3y^2-12, 6xy-6)$ so $\nabla g(x, y) = 0 \implies (3x^2+3y^2-12, 6xy-6) = (0,0)$
So $x^2+y^2 -4 = 0$ and $xy = 1$; combining and writing it in two different ways we get:
$\begin{aligned} ... | Your method can't find two other critical points. You successfully found that $x+y=\pm \sqrt{6}$ and $x-y=\pm \sqrt{2}$, but plus and minus signs don't need to coincide. Then four possibilities occur:
*
*$x=\dfrac{\sqrt{6}+\sqrt{2}}{2}$
*$x=\dfrac{\sqrt{6}-\sqrt{2}}{2}$
*$x=\dfrac{-\sqrt{6}+\sqrt{2}}{2}$
*$x=\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3? It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In som... | The case $n=1$ is uninteresting, so let $n\gt 1$. We show that $n^2+n+1$ has a prime factor congruent to $1$ modulo $3$, by showing that $4(n^2+n+1)$ has such a prime factor.
Note that $n^2+n+1$ is odd. We first show that $3$ is not the only odd prime that divides $n^2+n+1$. For suppose to the contrary that $(2n+1)^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Differential forms should be invariant under coordinate transformations I am wondering why, if we transform the following differential form, it does not seem to be invariant under the coordinate transformation. The $1$-form on $\mathbb R^2$ is
$$
\omega = \sqrt{x^2 + y^2} dx + 0\cdot dy
$$
and as the coefficient funct... | The problem with your computation is that the polar coordinates of the basis tangent vector $\partial_x$ at $(1,1)$ are $(\sqrt 2 /2,-1/2)$:
$$v = \partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding when $y=1+2\sin(x)$ and $y=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$ are equal How would you solve the following equation?
$$1 +2\sin(x) = 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \quad\text{for }−2\pi\leq x \leq 2\pi.$$
Steps I tried:
\begin{align}
1+2\sin(x) &= 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \\
2\sin(x... | Examining your attempt
Your first five lines do nothing but divide the original equation by $2$.
$$1+2\sin(x)=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$$
$2\sin(x)-2\sin(\frac{x}{2})=2\cos(\frac{x}{2})-1$
$2[\sin(x)-\sin(\frac{x}{2})]=2\cos(\frac{x}{2})-1$
$\sin(x)-\sin(\frac{x}{2})=\cos(\frac{x}{2})-\frac{1}{2}$
$$\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
| $$\int x^3\sqrt{4-x^2}\space\text{d}x=$$
Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:
$$\frac{1}{2}\int u\sqrt{4-u}\space\text{d}u=$$
Substitute $s=4-u$ and $\text{d}s=-\space\text{d}u$:
$$\frac{1}{2}\int(s-4)\sqrt{s}\space\text{d}s=\frac{1}{2}\int\left[s^{\frac{3}{2}}-4\sqrt{s}\right]\space\text{d}s=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
What does the diagonalized matrix say about a Transformation? I have a matrix given:
$$A=\begin{pmatrix}
7 & -2 \\
-1 & 8
\end{pmatrix}
$$
I have found its characteristic polynomial: $\lambda^2 - 15\lambda +54 = 0$,
which gave me $\lambda = 6, 9$.
Now, I used that to find the Diagonal matrix.
$$\begin{pmatrix}
6 & 0\\
... | Did you do the computation $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}2 & 1 \end{pmatrix}$? That gives $\begin{pmatrix}7(2)-2(1) \\ -1(2)+ 8(1)\end{pmatrix}= \begin{pmatrix}12 \\ 6 \end{pmatrix}$ not $\begin{pmatrix}12 \\ 9 \end{pmatrix}$. You can multiply the matrix $\begin{pmatrix}6 & 0 \\ 0 & 9 \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplification of surds $\frac{x}{\sqrt{x^2 - x^4}}$
$$\frac{x}{\sqrt{x^2 - x^4}}$$
I believe that I can factor out the $x^2$ in the square root to get
$$\frac{1}{\sqrt{1-x^2}} .$$
However, Wolfram Alpha doesn't do the simplification, hence my confusion.
| $$\frac{x}{\sqrt{x^2 - x^4}}=
\Bigg\{\begin{array}{c}
\frac{1}{\sqrt{1-x^2}}, \ \ 0<x<1 \\
-\frac{1}{\sqrt{1-x^2}}, \ \ -1<x<0
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Roots of quadratic equation are given by $b \pm \sqrt{b^2 - c}$ I was reading slides about the cancellation error in quadratic equations and it's written:
The roots of the quadratic equation:
$$x^2 - 2bx + c = 0$$
with $b^2 > c$ are given by $b \pm \sqrt{b^2 - c}$.
Fact that let me perplexed, since I always thought t... | Never mind, they can be found by that formula only for this quadratic equation $$x^2 - 2bx + c = 0$$
Indeed if we use the usual formula:
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
to the equation above we obtain:
$$\frac{-(-2b) \pm \sqrt{(-2b)^2 - 4*1*c}}{2*1}$$
$$\frac{2b \pm \sqrt{4b^2 - 4c}}{2}$$
$$\frac{2b \pm \sqrt{4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Proving a congrent modular How do I show that $3^{1974}+5^{1974}\equiv 0 \pmod {13}$? I have tried feeding the values onto a calculator but they are so big to be computed. What is the best approach?
| since Fermat's little theorem, implies $3^{12}\equiv5^{12}\equiv 1\pmod{13}$ we have:
$$
3^{1974}+5^{1974}=3^{12*164+6}+5^{12*164+6}=(3^{12})^{164}3^6+(5^{12})^{164}5^6\equiv 3^6+5^6 \pmod{13}
$$
Now, you can easily see that:
$$3^3=27\equiv 1\pmod{13}\Rightarrow 3^6=1\pmod{13}$$
and:
$$5^2=25\equiv -1\pmod{13}\Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:
*
*$f(x)-1$ is divisible by $(x-1)^3$
*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f ... | Because $f$ is divisible by $x^3$, we must have $f(x)=r_3x^3+r_4x^4+r_5x^5$.
The second condition gives, comparing coefficients of powers of $x$, that
$$
f(x)=6x^5 - 15x^4 + 10x^3.
$$
Edit: Comparing coefficients gives me the linear equations $r_4 + 3r_5 - 3=0, r_3 - 3r_5 + 8$ and $r_5=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $$M=abc+abd+acd+bcd-abcd$$
I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$. ... | The continuous function
$$
f(a, b, c, d) = abc+abd+acd+bcd-abcd
$$
has a maximum on the compact set
$$
\{ (a, b, c, d) \in \Bbb R^4\mid a, b, c, d \ge 0, a+b+c+d = 1 \}
$$
which is obtained at some point $(A, B, C, D)$. If we can show
that $A=B=C=D$ then it follows that the maximum is
$$
M = f(\frac 14,\frac 14,\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?... | To find $$\int\frac{x^2dx}{(x^2+1)^2}$$ you may want consider the trigonometric subsitution $x=\tan t$ and $dx=\sec^2t dt$
This gives:
$$\int\frac{\tan^2t \sec^2t dt}{(\tan^2t+1)^2}=\int\frac{\tan^2t \sec^2t dt}{\sec^4t}=\int\frac{\tan^2t dt}{\sec^2t}=\int \frac{\sin^2t\cos^2t}{\cos^2t}dt=\int \sin^2tdt$$
The last int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Sum of number-of-divisors function equals $\sum_{j=1}^{n} \lfloor n/j \rfloor$. I am trying to prove the identity
$$t(1) + t(2) + \cdots + t(n) = \Big\lfloor \dfrac{n}{1} \Big\rfloor + \Big\lfloor \dfrac{n}{2} \Big\rfloor + \cdots + \Big\lfloor \dfrac{n}{n} \Big\rfloor,$$
where $t(j)$ is the number of divisors of $j$. ... | It really helps to write $t(1)+t(2)+\dotsb+t(n)$ in a triangular array:
1 +
1 + 1 +
1 + 0 + 1 +
1 + 1 + 0 + 1 +
1 + 0 + 0 + 0 + 1 +
1 + 1 + 1 + 0 + 0 + 1 +
1 + 0 + 0 + 0 + 0 + 0 + 1 +
1 + 1 + 0 + 1 + 0 + 0 + 0 + 1 +
1 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 1 + ...
Row $k$ is $t(k)$ written in a way that indicates which num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve $\lfloor x \rfloor = ax+1$ for integral $a$ Solve $\lfloor x \rfloor = ax+1$, where $a$ is an integer.
I have found the values of $x$ for $a=0$, $a=1$ and $a=-1$. But I don't know how to continue. How can I find the solutions for integral $a\leq-2$ and $a\geq2$ ?
| When $a=0$ there are infinite solutions , they are clearly $x\in [1,2)$. We now assume $a\neq 0$.
Since $\lfloor x \rfloor$ is an integer we conclude $ax+1$ is an integer, so $x=\frac{k}{a}$ for some $k$.
We must therefore find all the integer values for $k$ such that $\lfloor \frac{k}{a} \rfloor=k+1$.
It is clear that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin ... |
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$?
It is the latter.
So, having $\tan\alpha=-6/8=-3/4$ gives $\alpha=\arctan(-3/4)\approx -36.9^\circ$.
Hence, for $n\in\mathbb Z$,
$$\begin{align}10\cos(x-(-36.9^\circ))=5&\Rightarrow \cos(x+36.9^\circ)=1/2\\&\Rightarrow x+36.9^\circ=\pm 60^\circ+360^\circ n\\&\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$
$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we ... | It factors!
$x^4-2ax^2+x+(a^2-a)=(x^2+x-a)(x^2-x+(1-a))$
How do I get that? Well, there will be an extraneous root with the wrong signs for the inner square root, thus:
$x=\sqrt{a-\sqrt{a-x}}$
$-x=-\sqrt{a-\sqrt{a+(-x)}}$
The second extraneous equation is consistent with the simpler recursion
$y=-\sqrt{a+y}$
with $y=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Block matrix of order $m$ with three block matrices How to find eigenvalues of following block matrices?
$M=\begin{bmatrix}
A & B & O & O & O & O & O & \cdots & O & O\\
B & A & B & O & O & O & O & \cdots & O & O\\
O & B & A & B & O & O & O & \cdots & O & O\\
O & O & B & A & B & O & O & \cdots & O & O\\
O & O & O &... | You can find a computation method for the eigenvalues and eigenvectors in the following recent document: "Eigendecomposition of Block Tridiagonal Matrices" by
A. Sandryhaila and J.M.F. Moura (arxiv.org/pdf/1306.0217)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A seemingly-trivial divisibility conjecture While working on another problem, I stumbled on the following divisibility claim.
Conjecture: No integers $a,b,c,d$ satisfy all of the following conditions:
*
*$a^2+b^2-c^2-d^2 = 2(ad-bc)-1$;
*$\gcd(ac+bd,a^2+b^2-c^2-d^2) > 1$;
*$\gcd(ac+bd+1,ad-bc) > 1$;
*$(ac+bd)-(ad-... | Not an answer, but just to share if it could be a way, on the track of your edit2.
Put $z = a + i\,b\quad w = c + i\,d$
then:
$$
a^{\,2} + b^{\,2} - c^{\,2} - d^{\,2} = \tilde z\,z - \tilde w\,w
$$
$$
\left( {\tilde z + \,\tilde w} \right)\left( {z - w} \right) = \tilde z\,z - \tilde z\,w + \tilde w\,z - \tilde w\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Problem about find the extreme of a function (Multipliers of Lagrange) Good morning. I have a problem with this:
Find the maximum and minimum distances from the origin to the curve* $$g\left(x,y\right)=5x^{2}+6xy+5y^{2}$$
I have done this:
Function to optimize:$f\left(x,y\right)=x^{2}+y^{2}$
Restriction: $g\left(x,y\ri... | It turns out that this system of Lagrange equations doesn't really warrant the use of Mathematica. We are faced with
$$ \ 2x \ = \ (2x \ + \ \frac{6}{5}y)\lambda \ \ , \ \ 2y \ = \ (2y \ + \ \frac{6}{5}x)\lambda \ \ . $$
Since bringing all the terms to one side in each equation won't help us to factor the expressions,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Logarithm question with base change If $\log_{12} 27 = a$ then find the value of $\log_6 16$.
| $a=\log_{12}27$ is equivalent to $3^3=12^a=2^{2a}3^a$. So $2^{2a}=3^{3-a}$. Hence $2^{3+a}=2^{3-a}3^{3-a}=6^{3-a}$. So $16=2^4=6^b$ where $b=\frac{4(3-a)}{3+a}$. Hence $\log_616=b=\frac{4(3-a)}{3+a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Please prove the following: Given $ƒ(x) = e^x$, verify that $\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$. Given $ƒ(x) = e^x$, verify that
$$\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$$ and explain how this illustrates that $f'(x) = \ln e \cdot f(x) = f(x)$.
| Recall that all we need to show is that $\lim_{h\to 0} \frac{e^h - 1}{h} = 1$. The only things we need in this case are the definition of $e$ given by $e^h = \lim_{n\to \infty}\left ( 1 + \frac{h}{n} \right )^n$, and the binomial theorem. Observe that
\begin{eqnarray*}
\lim_{h\to 0} \frac{e^h -1}{h} & = & \lim_{h\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve the first order ordinary differential equation $y'(x)=2x \cos^2 y(x)$
Solve $$y'(x) =2x \cos^2 y(x) .$$
\begin{align}
\frac{dy}{dx} &= \ 2x \cos^2 y(x) \\
\frac{dy}{\cos^2 y(x)} &=2x \, dx \\
\tan y(x) &=x^2+k, \qquad\qquad k \in \mathbb{R} \\
y(x) &=\arctan (x^2+k)
\end{align}
Is is correct?
Thanks!
| You can check if your proposed solution actually solves the equation. Let's calculate the derivative:
$$
y(x) = \arctan(x^2 + k)\\
y'(x) = \frac{2x}{(x^2+k)^2+1}
$$
This should equal $2x\cos^2(y(x))$ if $y(x)$ solved the original equation:
$$
2x\cos^2(y(x)) = 2x\cos^2(\arctan(x^2 + k)) = 2x\left(\frac{1}{\sqrt{(x^2+k)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
$$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$
My thoughts were turning the right hand side to $abc$ as $abc=1$ ho... | Note that $$\sum_{cyc}\frac{a}{bc+2}=\sum_{cyc}\frac{a^2}{2a+abc}=\sum_{cyc}\frac{a^2}{2a+1} \ge \frac{(a+b+c)^2}{2a+2b+2c+3} $$
Now one can use that $$\sum_{cyc}a \ge 3\sqrt[3]{abc}=3$$This establishes that $$(a+b+c-1)^2 \ge 4 \Leftrightarrow (a+b+c)^2 \ge 2(a+b+c)+3$$Thus, our proof is done, with equality at $a=b=c$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps:
*
*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
*${\displays... | How about this:
\begin{align}
u & = \sqrt{x^2-1} \\
u^2 & = x^2 - 1 \\
u^2+1 & = x^2 \\
2u\,du & = 2x\,dx \\
\end{align}
\begin{align}
\int x^3 \sqrt{x^2-1} \, dx & = \int x^2 \sqrt{x^2-1} \Big( x\,dx\Big) \\
& = \int (u^2+1) u \left( \frac 1 2 \, du \right) \\
& = \frac 1 2 \int (u^4 + u^2) \, du \\
& \phantom{{}={}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the $\lim_{n\to \infty} {\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}$ Because $(1+\frac{k}{n})\leq (1+\frac{1}{n})^k$ $$ {\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}\leq\sqrt[n]{(1+\frac{1}{n})^{n(n+1)/2}}=(1+\frac{1}{n})^{(n+1)/2}$$ and so $$\lim_{n\to \infty}{\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}<\sqrt{e}$$
But what is the e... | You can apply Cauchy's limit theorem
$$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n},$$
with
$$a_n = (1+1/n)(1+2/n)…(1+n/n) = \frac{(2n)!}{n! n^n}.$$
Then
$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!} \\ = \frac{(2n+2)(2n+1)n^n}{(n+1)(n+1)^{n+1}} \\ = 2\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Induction Proof $2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{ 2}$ I am looking for an induction proof...
$$2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{2}$$
when $n \geq 1$.
I know there are $3$ steps to this.
1) Check
2) Do $n = k$
3) Do $n = k + 1$
Problem is, I can't seem to get past the first ... | For $n=1$, Ok. It's easy to check
Now assume it works for $n$, the we have to show it works for $n+1$.
(note that $9(n+1)-1=9n+8$)
we have to show the sentence below is true:
$$2+5+8+...+9n-1+9n+2+9n+5+9n+8=\frac{3(n+1)(9(n+1)+1)}{2}=\frac{27n^2+27n+30}{2}$$
Knowing it works for $n$, then we have:
$$2+5+8+...+9n-1+9n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the value of integral $ \int_{ \mid z \mid =1} \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } $ Integral to get the result?
$$ \int_{ \mid z \mid =1} \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } $$
with Solutions ,tnx.
| With partial fraction decomposition you have :
$$f(z)=\frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } = 2\frac{1}{z-\frac{1}{3}}+\frac{1}{2}\frac{1}{z-\frac{1}{2}}+\frac{1}{2}\frac{1}{(z-\frac{1}{2})^2}$$
We will try to use Residue theorem now :
$Res(f,\frac{1}{3})=Res(2\frac{1}{z-\frac{1}{3}})=2$
you should also find $Res(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
$$x^2y+y^2z+z^2x < \frac12$$
This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.... | I don't think the best bound can be solved for analytically.
using Lagrange multipliers , I tried to maximize $x^2y+y^2z+z^2x$ $\quad$ subject to the constraint $x+y^2+z^3=1$
maximize $g=x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$
$\frac {\partial g}{\partial x}=0:2xy+z^2+\lambda=0$
$\frac {\partial g}{\partial y}=0:x^2+2yz+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 2
} |
If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{... | Write
$$
\frac{a^2+1}{a^2+2}=1-\frac{1}{a^2+2}
$$
Minimizing this is the same as maximizing $1/(a^2+2)$ which, in turn, is the same as minimizing $a^2+2$ or, as well, minimizing $a^2$.
Since
$$
a=-\frac{x^2-3x-2}{x-1}
$$
the minimum value for $a^2$ is obtained when $x^2-3x-2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 20... | Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2\geq \frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$\frac{a}{13b} \sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2+\frac{1}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{b}{c}))^2+\frac{c}{13b}\sin(\arctan(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 5,
"answer_id": 0
} |
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.