Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
| Hint. You may use Taylor expansions, as $x \to 0$,
$$
\cos x =1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)
$$
$$
\sin x =x+O(x^3)
$$ giving
$$
x^2+2\cos x-2=\frac{x^4}{12}+O(x^6)
$$ $$
x\sin^3 x =x^4+O(x^6)
$$ and
$$
\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{\frac{x^4}{12}+O(x^6)}{x^4+O(x^6)}=\frac{1}{12}+O(x^2)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to get the eigenvalue of this matrix Let $\theta \in \mathbb R$, and let $T\in\mathcal L(\mathbb C^2)$ have canoncial matrix
$M(T)$ = $$
\left(
\begin{matrix}
1 & e^{i\theta} \\
e^{-i\theta} & -1 \\
\end{matrix}
\right)
$$
(a) Find the eigenvalues of $T$.
(b) Find an orthonormal basis for $\mathbb C^2$ that consists of eigenvectors for $T$.
I can get the eigenvalues of T and they are $\sqrt 2$ and $-\sqrt 2$. However, I cannot get each eigenvector respect to each eigenvalue. I know how to get eigenvectors by calculating the null space of $(T - \lambda I)$, but it looks like this is not a proper method to solve this problem. So, anyone can help? Thank you!
| Finding the null space of $A - \lambda I$ always work. Let us find for example the eigenvectors of $T$ corresponding to the eigenvalue $\sqrt{2}$. Denote by $v = (x, y)^T$ such an eigenvector. We must have
$$ Av = \begin{pmatrix} 1 & e^{i \theta} \\ e^{-i \theta} & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + e^{i \theta} y \\ e^{-i \theta} x - y \end{pmatrix} = \sqrt{2} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \sqrt{2}x \\ \sqrt{2}y \end{pmatrix}. $$
Thus, we get a system of two linear equations for $v$:
$$ x + e^{i\theta}y = \sqrt{2}x, \\ e^{-i\theta} x - y = \sqrt{2}y. $$
Written differently, we have:
$$ (1 - \sqrt{2})x + e^{i \theta} y = 0, \\ e^{-i \theta} x - (\sqrt{2} + 1)y = 0. $$
Since this homogeneous system must have a non-zero solution, we know that the equations must be linearly dependent and so we can choose to solve only one of the equations. The first equation implies that $x = \frac{e^{i \theta}}{\sqrt{2} - 1} y$. Choosing $y = 1$ we obtain the eigenvector $v = \left( \frac{e^{i \theta}}{\sqrt{2} - 1}, 1 \right)^T$ and the eigenspace associated with $\sqrt{2}$ is
$$ \mathrm{span} \left \{ \begin{pmatrix} \frac{e^{i \theta}}{\sqrt{2} - 1} \\ 1 \end{pmatrix} \right \}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\alpha (1+\alpha/2)^{-1} < \log(1+\alpha) $ for $\alpha > 0$ How does one prove that $\alpha (1+\alpha/2)^{-1} < \log(1+\alpha) $ for $\alpha > 0$?
| By changing variable $x \rightarrow a-x$, we have
$$
I = \int_0^\alpha \frac{dx}{1+x} = \int_0^\alpha \frac{dx}{1+\alpha-x}.
$$
So, by averaging the two expressions, we get
$$
I = \int_0^\alpha \frac{1}{2} \left( \frac{1}{1+x} + \frac{1}{1+\alpha-x} \right) \, dx
$$
But by the convexity of $1/(1+x)$, we have
$$
\frac{1}{2} \left( \frac{1}{1+x} + \frac{1}{1+\alpha-x} \right)
\ge \frac{1}{1+(x + \alpha-x)/2} = \frac{1}{1+\alpha/2}.
$$
The equality is impossible when $x\ne\alpha/2$. So
$$
I > \int_0^\alpha \frac{dx}{1+\alpha/2} = \frac{\alpha}{1+\alpha/2}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all the values: a+b=c+d=ab-cd=4n for n>1 This question was given to my cousin who is in grade 9 in Kazakhstan, hence the solution should not involve any matrices, sets etc. and I cannot think of a simpler solution. Any help would be greatly appreciated!
Find all the values: a+b=c+d=ab-cd=4n for n>1
| The solution is basically the same as the one by Thomas.
WLOG, we can assume $a \le b$ and $c \le d$. So $a,c \le 2n$.
From the equations, we have $b = 4n - a, d = 4n - c$, and
$$
4n = a (4n-a) - c (4n-c) = (4n - a - c) (a - c).
$$
But $a+c$ and $a-c$ share the same parity, so they have to be both even:
$$
\begin{aligned}
a + c &= 2 p, \\
a - c &= 2 q,
\end{aligned}
$$
or
$$
\begin{aligned}
a &= p + q, \\
c &= p - q,
\end{aligned}
$$
and
$$
n = (2n - p) \, q.
$$
This means $q|n$, and $2n-p|n$. Since $p + |q| \le \max\{a, c\} \le 2n$, $2n-p\ge |q|$. This means $q$ is positive, and $n \ge |q| q = q^2$. So the general solution is
$$
\begin{aligned}
a = 2n - \left(\frac{n}{q} - q \right), \\
b = 2n + \left(\frac{n}{q} - q \right), \\
c = 2n - \left(\frac{n}{q} + q \right), \\
d = 2n + \left(\frac{n}{q} + q \right).
\end{aligned}
$$
with $1 \le q \le \sqrt{n}$. The positions of $a$ and $b$ can be exchanged, so are those of $c$ and $d$.
For example, $n = 25$, we have, $q = 1$ or $q = 5$,
$$
\begin{aligned}
q = 1: \quad & a = 26, \; b = 74, \; c = 24, \; d = 76. \\
q = 5: \quad & a = 50, \; b = 50, \; c = 40, \; d = 60.
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I simplify $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ )$? $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ ) =$ ?
The answer is $3$.
My progress so far:
\begin{align}
\sqrt{3}(\cot 70^\circ+4\cos 70^\circ)&= \sqrt{3}(\tan 20^\circ+4\sin 20^\circ) \\
&= \sqrt{3}(\sin 20^\circ)\left(\frac{1}{\cos 20^\circ}+4\right)\\
&= \sqrt{3}\frac{(\sin 20^\circ)(1+4\cos 20^\circ)}{\cos 20^\circ}\\
&= \sqrt{3}\frac{(\sin 20^\circ+2\sin 40^\circ)}{\cos 20^\circ}\\
\end{align}
| $$\cot(70)+4\cos(70) = \frac{\cos(70)}{\sin(70)} + 4\cos(70) =\frac{\cos(70)+4\cos(70)\sin(70)}{\sin(70)} = \frac{\cos(70)+2\sin(140)}{\sin(70)} = \frac{(\cos(70)+\sin(140))+\sin(140)}{\sin(70)} = \frac{(\sin(20)+\sin(140))+\sin(140)}{\sin(70)} = \frac{2\sin(80)\cos(60)+\sin(140)}{\sin(70)} = \frac{2\sin(110)\cos(30)}{\sin(70)} = 2 * \frac{\sqrt3}{2} = \sqrt 3$$
Note $\sin(110)=\sin(70)$ at the end.
$\cos(60)=\frac{1}{2}$ and
$\cos(30)=\frac{\sqrt3}{2}$
Identities used: $\sin(90-x)=\cos(x)$
$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$
$\sin(2x)=2\sin(x)\cos(x)$
$\sin(a)+\sin(b)=2\sin(1/2(a+b))\cos(1/2(a-b))$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the diophantine equation $a = x^p - y^p$ sufficient to find $x$ and $y$ in terms of $a$ and $p$? Let there be a natural number $a,$ that can be expressed as $a = x^p - y^p$ where $x, y$ and $p$ are natural numbers, each two of them being pairwise co-prime and $p$ is an odd prime. Then can $x$ and $y$ be found out uniquely in terms of $a$ and $p$? If yes, how?
| If $x^p-y^p=z^p-w^p$, then $x^p+w^p=y^p+w^p$, so this question is equivalent to asking if there is a number that is the sum of two $p^{th}$ powers in two different ways. It is unknown if there are any examples of this for $n>3$, see:
Guy, Richard K. (2004). Unsolved Problems in Number Theory (Third ed.). New York, New York, USA: Springer-Science+Business Media, Inc. ISBN 0-387-20860-7.
Thus we restrict our attention to $p=3$. It is a result of Euler that all rational solutions are of the form: $(3a^2+5ab−5b^2)^3+(4a^2−4ab+6b^2)^3+(5a^2−5ab−3b^2)^3 = (6a^2−4ab+4b^2)^3$.
This can be rewritten as $(A^2+7AB−9B^2)^3+(2A^2−4AB+12B^2)^3 = (2A^2+10B^2)^3+(A^2−9AB−B^2)^3$ with $a=A+B, b=A-2B.$ Taking the parts of this equation mod $3$ yields $A^2+AB$, $2A^2-AB$, $2A^2+B^2$, $A^2-B^2$. We want all of these to not be $0$ mod $3$, because otherwise $(3,x)=3$. But it's clear that $B\neq 0$, as otherwise $a=b$, so by checking the other possible values of $A$ and $B$ mod $3$, none of them produce four terms none of which are $0$ mod $3$. Thus the answer is yes, it is unique for $p=3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$... Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$.
So far I have the base case completed, and believe I am close to completing the proof itself.
Base case:$(n=1)$
$3^1 + 7^1 - 2 = 8/8 = 1 $
Inductive Hypothesis: Assume that $3^n +7^n −2$ is divisible by 8 for all positive integers n.
Induction step $(n+1)$ case:
$$ 3^{n+1} + 7^{n+1} - 2 $$
$$3(3^{n}) + 7(7^{n}) - 2$$
$$3^n + 7^n = 8x $$
-It seems to me that this could be the end of the proof because whatever the answer is would be a multiple of 8: but I am unsure, any help is appreciated.
| It holds for $n=1,2$.
If it holds for $1,2,\dots,n$, then
\begin{align}
&3^{n+1}+7^{n+1}-2\\
&=3^2\cdot3^{n-1}+7^2\cdot7^{n-1}-2\\
&=(8+1)\cdot3^{n-1}+(48+1)\cdot7^{n-1}-2\\
&=8\cdot(3^{n-1}+6\cdot7^{n-1})+3^{n-1}+7^{n-1}-2\\
\end{align}
Therefore it also holds for $n+1$.
So it holds for all $n\in \mathbb N$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Discrete Random variable, finding pmf of X Q: A fair coin is tossed independently 5 times. Let X denote the difference between the number of heads and the number of tails obtained. find the probability mass function of X
here's my take:
P(X=0) = all heads or all tails = 1/32 + 1/32 = 2/32
P(X=4) = one head or one tail = 5C1 x 1/32 + 5C1 x 1/32 = 10/32
P(X=3) = two heads or two tails = (5C2 x 1/32)^2 = 20/32
P(X=2) = three heads or three tails = (5C3 x 1/32)^2= 20/32
P(X=1) = four heads or four tails = (5C4 x 1/32)^2 x 1/2 = 10/32
but i dont think i am on the right track. did i understand question completely wrong? i know pmf should equal 1 and my answer exceeds that. so i know im wrong for sure
| Not quite. Convince yourself that
$$P(X = 0) = P(X=2) =P(X = 4) = 0.$$
Then for the difference to be 1, then either you got 3 heads in 5 flips, or 3 tails in 5 flips. Thus,
$$P(X = 1) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2
+\binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 2\binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2$$
Similarly,
$$P(X=3) = 2\binom{5}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^1,$$
and
$$P(X=5)= 2\binom{5}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^0.$$
You can verify that indeed the sum of these cases adds to 1.
| {
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"timestamp": "2023-03-29T00:00:00",
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find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the condition $a+b=-c$ are such that $a^2=bc+1$ and $b^2=ca+1$. Also if $a-b=b-a,$ it means $a=b.$ Then both $b-a$ and $a-b$ are $0.$ Hence $c$ can thus be any integer. Hence, there are two groups of integers satisfying the required condition $-a=b$ with any $c$ and $a+b=-c.$
Please let me know whether my solution is correct or not.
| Assuming $b\neq a$ you gained a constraint
$$-(a+b) = c$$
This constriction is not bad, because there is no number $b = c$ such that $a^2 = bc + 1$ for integers.
It's fine to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562396",
"timestamp": "2023-03-29T00:00:00",
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Lagrange Multipliers: find points farthest/closest to a point Find the points of the ellipse:
$$\frac{x^2}{9}+\frac{y^2}{4}=1$$
which are closest to and farthest from the point $(1,1)$.
I use the method of the Lagrange Multipliers by setting:
$$f(x,y)=(x-1)^2+(y-1)^2$$
(no need for the root since maximizing one maximizes the other)
And $$g(x,y)=\frac{x^2}{9}+\frac{y^2}{4}-1=0$$ the constraint.
We get:
$$x-1=λx/9$$ and $$2y-2=λy/2$$
I don't know where to go from there. I tried finding $x$ and $y$ alone and then plugging their values in the constraint $g(x,y)=0$ but I get a really complicated equation.
| Hint: $$\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda g(x,y)=(x-1)^2+(y-1)^2+\lambda \left(\frac{x^2}{9}+\frac{y^2}{4}-1\right) $$
Then:
$\frac{\partial \mathcal{L}}{\partial x}=2(x-1)+\frac{2x\lambda}{9}=0 \implies x=\frac{9}{9+\lambda}$
$\frac{\partial \mathcal{L}}{\partial y}=2(y-1)+\frac{\lambda y}{2}=0 \implies y=\frac{4}{4+\lambda}$
$\frac{\partial \mathcal{L}}{\partial \lambda}= \frac{x^2}{9}+\frac{y^2}{4}-1=0$
so $\implies \frac{\left(\frac{9}{9+\lambda}\right)^2}{9}+\frac{\left(\frac{4}{4+\lambda}\right)^2}{4}=1$
$\implies \frac{9}{(9+\lambda)^2}+\frac{4}{(4+\lambda)^2}=1$
Then solve for $\lambda$ (not easy) then plug in value to obtain $x=\frac{9}{9+\lambda}$ and $y=\frac{4}{4+\lambda}$ that are required
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit problem with roots I'm struggling solving the following limit problem:
$$\lim_{x \to \infty} \left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)$$
At first I thought I could
Multiply by:
$$\frac{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}$$
But that doesn't seem to take me anywhere closer to an answer.
Some help would be appreciated.
| Substitute $t=1/x$, so the limit becomes
$$
\lim_{t\to0^+}\frac{\sqrt[9]{1+t}-\sqrt[9]{1-t}}{t}
$$
Now apply l'Hôpital or Taylor. Surely the conjugate is not
$\sqrt[9]{1+t}+\sqrt[9]{1-t}$ that would be only for the square root.
Actually, this is the derivative at $0$ of the function
$$
f(t)=\sqrt[9]{1+t}-\sqrt[9]{1-t}
$$
and, since
$$
f'(t)=\frac{1}{9}(1+t)^{-8/9}+\frac{1}{9}(1-t)^{-8/9}
$$
we have
$$
f'(0)=\frac{2}{9}
$$
If you have to do it without derivatives, you have to use
$$
a^9-b^9=(a-b)(a^8+a^7b+a^6b^2+a^5b^3+a^4b^4+a^3b^5+a^2b^6+ab^7+b^8)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this sum $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$ Show that
$$\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$$
| Hint Let
$$a_{n}=\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\cdot \binom{2n-k+1}{k}=2^{2n}+\sum_{k=1}^{n}(-1)^k\cdot 2^{2n-2k}\left[\binom{2n-k}{k}+\binom{2n-k}{k-1}\right]$$
other hand we easy to prove $r=k-1$
\begin{align*}&\sum_{k=1}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k}{k-1}=\sum_{r=0}^{n-1}(-1)^{r+1}\cdot 2^{2(n-1)-2r}\binom{2(n-1)-r+1}{r}\\&=
-a_{n-1}
\end{align*}
if we let $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k}{k}=b_{n}$
then we have
$$b_{n}=a_{n}+a_{n-1}$$
and
\begin{align*}b_{n}&=2^{2n}+\sum_{k=1}^{n-1}(-1)^k\cdot 2^{2n-2k}\binom{2n-k}{k}+(-1)^n\\
&=2^{2n}+\sum_{k=1}^{n-1}(-1)^k\cdot 2^{2n-2k}\left[\binom{2n-k-1}{k}+\binom{2n-k-1}{k-1}\right]+(-1)^n\\
&=4a_{n-1}-\sum_{j=0}^{n-1}(-1)^j\cdot 2^{2(n-1)-2j}\binom{2(n-1)-j}{j}\\
&=4a_{n-1}-b_{n-1}
\end{align*}
so
$$a_{n}+a_{n-1}=4a_{n-1}-a_{n-1}-a_{n-2}$$
then we have
$$a_{n}+a_{n-2}=2a_{n-1},a_{0}=1,a_{1}=2$$
so
$$a_{n}=n+1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all ordered triples Problem
Suppose $351_7=aca_b$ for positives $a,b,$ such that $b-1=c$. Compute the ordered triple $(a,b,c).$
Here is what I tried:
$351_7 = 183 = a*b^2+c*b+a = a(b^2+1)+c*b = a(b^2+1)+(b-1)*b = b^2*a+b^2-b+a \implies b^2*a+b^2-b+a -183 = 0 \implies 1-4(a)(a-183) \geq 0.$
I am not sure if this helps or not, though.
| The inequality $1-4a(a-183) \geq 0$ does not help you finding the solution. Based on your computation I suggest the following approach.
The equations $351_{7}=aca_{b}$ and $b-1=c$ imply that $183=ab^{2}+b\left( b-1\right) +a$. Solving for $a$ we obtain $$a=\frac{183-b^{2}+b}{ b^{2}+1}.\tag{1} $$ Since $1\leq a\leq b-1$, we conclude that $b$ must
satisfy the double inequality
\begin{equation*}
1\leq \frac{183-b^{2}+b}{b^{2}+1}\leq b-1;
\end{equation*}
or the following one, after multiplying through by $b^{2}+1$:
\begin{equation*}
b^{2}+1\leq 183-b^{2}+b\leq b^{3}-b^{2}+b-1.\tag{2}
\end{equation*}
The first inequality in $(2) $ is equivalent to $2b^{2}-b-182\leq 0$, which implies
that $b\leq \left( 1+\sqrt{1457}\right) /4\approx 9.793$. The second
inequality is equivalent to $b^{3}\geq 184$, so $b\geq \sqrt[3]{184}\approx 5.688$.
Therefore the possible values of $b$ are $b\in \left\{ 6,7,8,9\right\} $. However none of these values results in $(1) $ being an integer.
| {
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Solve the equation with trigonometry $$\frac{2\cos x}{(1+\cos2x)}=\frac{(1-\cos2x)}{\sin2x}$$
Here's what i did
$$\frac{2\cos x}{ (1+\cos^2x-\sin^2x)}=\frac{1-(\cos^2x-\sin^2x)}{2\sin x \cos x}$$
$$\frac{2\cos x}{2\cos^2x}= \frac{(\sin^2x+\sin^2x)}{2\sin x\cos x}.$$ How do i continue?
| $$2\cos x/(1+\cos 2x)=(1-\cos 2x)/\sin 2x$$
$$2\cos x/(1+\cos^2(x)-\sin^2(x))=(1-(\cos^2(x)-\sin^2(x)))/(2\sin(x)\cos(x))$$
$$2\cos x/(2\cos^2(x))=(2\sin^2(x))/(2\sin(x)\cos(x))$$
$$1/\cos(x) = \sin(x)/\cos(x)$$
$$1 = \sin(x).$$
| {
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The set of real solutions to an equation $$4^x - 7(2^{\frac{x-3}{2}}) = 2^{-x}$$
Set of real solutions is in which interval:
*
*$(-9, -2)$
*$(0, 3]$
*$(-2, 0]$
*$(7, 12]$
*$(3, 7]$
I tried the following. Dividing by $2^{-x}$ I get $2^{3x} - \frac{7\sqrt(2)}{4}2^{\frac{3x}{2}} - 1 = 0$. Substituing
$$
2^{3x} = c
$$ and solving I get $c_1= 2\sqrt(2)$ and $c_2=-\frac{\sqrt(2)}{4}$ and for $x$ I get $\frac{1}{2}$. That's not even in the given line. What is wrong with the calculations. What have I missed?
| $$4^x-7\left(2^{\frac{x-3}{2}}\right)=2^{-x}\Longleftrightarrow$$
$$2^{2x}-7\left(2^{\frac{x-3}{2}}\right)=2^{-x}\Longleftrightarrow$$
$$2^{3x}-7\left(2^{\frac{3x}{2}-\frac{3}{2}}\right)=1\Longleftrightarrow$$
Substitute $y=-7\left(2^{\frac{3x}{2}-\frac{3}{2}}\right)$:
$$\frac{8y^2}{49}+y=1\Longleftrightarrow$$
$$y^2+\frac{49y}{8}=\frac{49}{8}\Longleftrightarrow$$
$$y^2+\frac{49y}{8}+\frac{2401}{256}=\frac{3969}{256}\Longleftrightarrow$$
$$\left(y+\frac{49}{16}\right)^2=\frac{3969}{256}\Longleftrightarrow$$
$$y+\frac{49}{16}=\pm\sqrt{\frac{3969}{256}}\Longleftrightarrow$$
$$y+\frac{49}{16}=\pm\frac{63}{16}\Longleftrightarrow$$
$$y=\pm\frac{63}{16}-\frac{49}{16}\Longleftrightarrow$$
Notice that the positive one from our $y$ soution,
has no solution since for all $z\in\mathbb{R},2^z>0$ and $-\frac{1}{8}<0$:
$$y=-\frac{63}{16}-\frac{49}{16}\Longleftrightarrow$$
$$y=-7\Longleftrightarrow$$
$$-7\left(2^{\frac{3x}{2}-\frac{3}{2}}\right)=-7\Longleftrightarrow$$
$$2^{\frac{3x}{2}-\frac{3}{2}}=1\Longleftrightarrow$$
$$\frac{3x}{2}-\frac{3}{2}=0\Longleftrightarrow$$
$$\frac{3(x-1)}{2}=0\Longleftrightarrow$$
$$x-1=0\Longleftrightarrow$$
$$x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating limit by sandwich theorem: $\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$ $$\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$$
For using Sandwich theorem I need two functions such that $g(x)<f(x)<h(x)$
$$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{n+n^2} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2} $$
But I can't find function greater than the given which will help me evaluate the limit.
| Your expression may be written as $$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{k}{k+n^2}$$ so, you can search for terms $a(k,n)$ and $b(k,n)$ such that $$a(k,n)\le \frac{k}{k+n^2}\le b(k,n)$$ For example $$a(k,n)=\frac{k}{n+n^2} \text{ or } a(k,n)=\frac{k}{2n^2}$$ and $$b(k,n)=\frac{k}{k+k^2} \text{ or } b(k,n)=\frac{k}{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How can one show that $f(0)=0$ for $f$ satisfying certain conditions? Given the functional equation
$$f(x+(1+x)f(y))=y+(1+y)f(x)$$
Such that $f:(-1,\infty) \to (-1,\infty)$ and the function $g(x):=\frac{f(x)}{x}$ is strictly increasing in $I=(-1,0)\cup(0,+\infty),$ how can one show that for every $t \in I$ we have
$f(t)\neq t$?
Plugging $x=y=0,$ that makes $f(f(0))=f(0).$
Does that mean that $f(0)=0$?
Making $x=y,$ $f(x+(1+x)f(x))=x+(1+x)f(x)$ so we make $x+(1+x)f(x)=t$ that makes $f(t)=t$ but we can't make $t=0$ because $t$ is not in $I$? So do I have to use the fact that $\frac{f(x)}{x}$ is increasing?
Then after all this work we have to deduce that $f(0)=0$ and $f(x)=\frac{-x}{x+1}$ for $x \in (-1,\infty).$
| I think that this will do the trick for showing that f(0) = 0:
Let's start by setting $x = 0$. We get
$$f(f(y)) = f(0) + (1+f(0))y.$$
By setting $y = 0$, we get
$$f(x) = f(x + (1+x)f(0)).$$
Apply now $f$ to expression above and use the first equation:
$$f(0) + (1+ f(0))x = f(0) + (1+f(0))(x + (1+x)f(0)).$$
This simplifies to
$$(1+x)(1+f(0))f(0) = 0$$
and thus $f(0) = 0$. We can also conclude that $f(f(y)) = y $, so f is its' own inverse.
Let suppose that there is numbers $a,b$ such that $a < b$ and $f(a) = b$ and $f(b) = a$ ($f$ is its' own inverse). Since function
$$ g(x) = \frac{f(x)}{x}$$
is increasing by the hypothesis. This means that $g(a) \le g(b)$, so
$$\frac{f(a)}{a} \le \frac{f(b)}{b} \iff \frac{b}{a} \le \frac{a}{b} \iff$$
$$\frac{b^2 - a^2}{ab} \le 0 \iff \frac{(b-a)(b+a)}{ab} \le 0 \iff $$
$$ \frac{(b+a)}{ab} \le 0. $$
so we can say that there can't be such $a,b$ with same sign. So $x$ and $f(x)$ must have different signs.
As OP noticed the equation
$$ f(x+(1+x)f(x)) = x + (1+x)f(x) $$
holds. Let denote $x+ (1+x)f(x)$ as $t$. Then the equation says $f(t) = t$ but $f(t)$ and $t$ have different signs, therefore, $t$ must be zero. Thus, we conclude that $x + (1+x)f(x) = 0$ and the function $f$ is:
$$ f(x) = -\frac{x}{1+x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Using FLT for $ n=3$ Using FLT for exponent $3$, I need to show that if n positive integer is divisible by $3$, then there are no $x,y,z$ positive integers such that $x^n+y^n=z^n$.
This is what I did:
if $3/n$ then $n=3.k$, for some $k$ positive integer. Then if $(x^3, y^3,z^3)$ is a solution for the exponent $k$
$(x^3)^k+(y^3)^k=(z^3)^k$
and because $3.k=k.3$ in integers
$(x^k)^3+(y^k)^3= (z^k)^3$
where $x^k=a ,y^k,=b z^k=c$ are positive integers. Then there are no , $a,b,c$ positive integers such that $a^3+b^3=c^3$ .
Could this be remotely correct?
| if $x^{3k} + y^{3k} = z^{3k}$ then set $u=x^k$, $v=y^k$, $w=z^k$ and we have $u^{3} + v^{3} = w^{3}.$
Therefore if $u^{3} + v^{3} = w^{3}$ has no nontrivial solutions, $x^{3k} + y^{3k} = z^{3k}$ has no nontrivial solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$.
Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.
My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $. How will we get that in the integrand?
| The substitution works because you get
$$
x=2\arctan u
$$
so
$$
dx=\frac{2}{1+u^2}\,du
$$
Since
$$
2\sin x-3\cos x=2\frac{2u}{1+u^2}-3\frac{1-u^2}{1+u^2}=
\frac{3u^2- 4u -3}{1+u^2}
$$
the integral becomes
$$
\int\frac{1+u^2}{3u^2- 4u -3}\frac{2}{1+u^2}\,du=
2\int\frac{1}{3u^2-4u -3}\,du
$$
that's solvable by partial fractions.
A different way is to write
$$
2\sin x-3\cos x=A\sin(x-\varphi)
$$
that becomes
$$
2\sin x-3\cos x=A\sin x\cos\varphi-A\cos x\sin\varphi
$$
so we can set
$$
A\cos\varphi=2,\quad A\sin\varphi=3
$$
and so $A^2=13$. There is a unique angle $\varphi\in[0,2\pi)$ such that
$$
\sin\varphi=\frac{3}{\sqrt{13}},\quad
\cos\varphi=\frac{2}{\sqrt{13}}
$$
Now the integral becomes
$$
\sqrt{13}\int\frac{1}{\sin(x-\varphi)}\,dx
$$
and we can do the substitution $x-\varphi=2u$, so $dx=2\,du$ and the integral is
\begin{align}
\sqrt{13}\int\frac{1}{\sin u\cos u}\,du
&=\sqrt{13}\int\frac{\sin^2u+\cos^2u}{\sin u\cos u}\,du\\
&=\sqrt{13}\left(\int\frac{\sin u}{\cos u}\,du
+\int\frac{\cos u}{\sin u}\,du\right)\\
&=\sqrt{13}(\log|\cos u|-\log|\sin u|)+c
\end{align}
and it's just tedious to do the back substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached:
| $u = \ln \left| {x + \sqrt {{x^2} + 16} } \right|$. Then $du = \frac{1}
{{\sqrt {{x^2} + 16} }}dx$. Then
$$\int {\frac{1}{{\sqrt {{x^2} + 16} }}dx} = \int {du} = u + C = \ln \left| {x + \sqrt {{x^2} + 16} } \right| + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Factoring the quartic $\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}$ Define $Q{\left(s;x\right)}$ to be the quartic function of $x$ with real parameter $s$ such that $0\le s\le1$ given as
$$Q{\left(s;x\right)}=\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}.\tag{1}$$
Question: Is there a (relatively) straightforward way of factoring $Q{\left(s;x\right)}$ into a product of two quadratics with real parameters?
In principle, this problem could be solved by brute force using the general methods for finding roots of quartic functions such as the Ferrari method described in the quartic function wiki.
However, I'm having a terrible time actually carrying out the process for my particular quartic $Q$. I'd really like to know if there is a more efficient approach or any cute shortcuts for avoiding some of the cumbersome intermediate expressions.
Note: for reference,the discriminant of $Q$ is
$$\begin{align}
\Delta_{Q}
&=-8\left(s-1\right)\left(128s^{4}-602s^{3}+982s^{2}-508s+5\right).\tag{2}\\
\end{align}$$
| Only the start of some thoughts...
$Q{\left(s;x\right)}=\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}$
$Q{\left(s;x\right)}=x^4 + 2x^3-x^2+x^3+2x^2-x-x^2-2x+1-8sx^3+8sx^2-2sx$
$Q{\left(s;x\right)}=x^4 + (3-8s)x^3+8sx^2+(-3-2s)x+1$
Consider $Q{\left(s;x\right)}=(x^2+ax+p)(x^2+bx+\frac 1p)$
$Q{\left(s;x\right)}=x^4+bx^3+\frac 1p x^2 + ax^3+abx^2+ \frac ap x+px^2+bpx+ 1$
$Q{\left(s;x\right)}=x^4+(a+b)x^3+(p+\frac 1p+ab)x^2+ (\frac ap +bp)x+ 1$
Comparing coefficients:
$(1): a+b=3-8s$
$(2): p+\frac 1p+ab=8s$
$(3): \frac ap +bp=-3-2s$
$(2) + 4 \times (3) \Rightarrow p+\frac 1p+ab+4(\frac ap +bp)=-12 \Rightarrow (1+4b)p^2+(ab+12)p+ (4a+1)=0$
$(1) + (2) \Rightarrow a+b+p+\frac 1p+ab=3 \Rightarrow p^2+(a+ab+b-3)p+1=0$
$\Rightarrow (1+4b)p^2+(1+4b)(a+ab+b-3)p+(1+4b)=0$
Then $1+4b=1+4a \Rightarrow a=b$
And $(1+4b)(a+ab+b-3)=ab+12$
$(1+4a)(2a+a^2-3)=a^2+12$
$2a+a^2-3+8a^2+4a^3-12a=a^2+12$
$4a^3+8a^2+10a-15=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}$
Find $\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}$
Using Taylor series:
$$\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}=\lim\limits_{x\to 0}\frac{(1+x^2+O(x^{2n}))-(1-\frac{x^{2\sqrt{2}}}{2}+O(x^{\sqrt{2}(2n+1)})}{x^2}=\lim\limits_{x\to 0}\frac{\frac{2x^2+x^{2\sqrt{2}}}{2}}{x^2}=\lim\limits_{x\to 0}\frac{2x^2+x^{2\sqrt{2}}}{2x^2}$$
How to evaluate this limit?
| Hint. You may just write, as $x \to 0$,
$$
\frac{e^{x^2}-\cos (x^{\sqrt{2}})}{x^2}=\frac{x^2(1+\frac12x^{2\sqrt{2}-2})+O(x^4)}{x^2}=1+\frac12x^{2\sqrt{2}-2}+O(x^2) \longrightarrow 1.
$$
In fact, $x^{2\sqrt{2}}$ is negligible comparing to $x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Bounds (and range) of a nonlinear difference equation I'm interested in the following set of nonlinear difference equations:
$$x_{n+1} = \frac{c + x_n}{x_{n-1}},\; x_1 = x_0 = 1 \qquad \textrm{for } c > 0$$
For $c=1$ the sequence is periodic with period 5 and range {1,2,3}. For other values of $c$ the sequence appears to be oscillatory. For small $c$, the supremum seems to be slightly more than $1+2c$, while for larger $c$ it seems to approach $3c$.
Is there a simple formula for the bounds of the sequence in this general case? What about the range? Are the sequences periodic? Are they dense within the bounds?
| This is an partial answer. Please consider this as a supplement to merico's excellent answer.
Introduce auxillary sequences $(y_n), (z_n), (w_n)$ by $y_n = x_{n+1}, z_n = x_{n+2}, w_n = x_{n+3}$.
Rewrite the defining recurrence relation of $(x_n)$ to a more symmetric form:
$$x_{n+1} = \frac{c+x_n}{x_{n-1}}\quad \iff\quad x_{n+1}x_{n-1} = c + x_n$$
In terms of the auxillary sequences, this becomes
$$x_n z_n = y_n + c\quad\text{ and }\quad y_n w_n = z_n + c.$$
As a result,
$$
(1+x_n)z_n = z_n + z_n x_n = z_n + (y_n+c) = y_n + (z_n + c) = y_n + y_nw_n = y_n(1+w_n)
$$
Multiply both side by $\frac{(1+y_n)(1+z_n)}{y_nz_n}$, we find
$$\frac{(1+x_n)(1+y_n)(1+z_n)}{y_n}
= \frac{(1+y_n)(1+z_n)(1+w_n)}{z_n}
= \frac{(1+x_{n+1})(1+y_{n+1})(1+z_{n+1})}{y_{n+1}}$$
This means following expression
$$\frac{(1+x_n)(1+y_n)(1+z_n)}{y_n} = \frac{(1+x_n)(1+y_n)(x_n+y_n+c)}{x_ny_n}$$
is independent of $n$ and they are invariant of the system.
Up to an additive offset $c$, the second form is the invariant found by merico.
Since $x_1 = y_1 = 1, z_1 = c+1$, the sequence of points $(x_n, y_n, z_n)$ lies on intersection of following two surfaces in $\mathbb{R}^3$:
$$(1+x)(1+y)(1+z) = 4(c+2)y\quad\text{ and } xz = y + c$$
Parametrize the curves within above interesection by some parameter $s$ and taking logarithm derivatives, we obtain
$$\frac{x'}{1+x} + \frac{y'}{1+y} + \frac{z'}{1+z} = \frac{y'}{y}\quad\text{ and }\quad
\frac{x'}{1+x} + \frac{z'}{1+z} = \frac{y'}{y+c}
$$
At those $s$ where where $y$ is extremal, $y' = 0$.
Above equations implies $x = z$ and hence $x^2 = y + c$. Furthermore,
$$(1+x)^2(1+y) = 4(c+2)y
\iff (1+2x+y+c)(1+y) = 4(c+2)y\\
{\large\Downarrow}\\
x = -\frac{y^2 - 3(c+2)y + (c+1)}{2(y+1)}
$$
Substitute this back into $x^2 = y + c$, we find the extremal values of $y$
are roots of a quartic polynomial.
$$(y^2 - 3(c+2)y + (c+1))^2 - 4(y+1)^2(y+c) = 0 \tag{*1}$$
For $c > 0$, this polynomial has $4$ distinct real roots: $0 \le \lambda_1(c) < \lambda_2(c) < \lambda_3(c) < \lambda_4(c)$.
Start from $x_1 = x_2 = 1$, it is easy to see all $x_n, y_n$ generated are positive. This means the extremal values of $y$ also need to satisfy:
$$y^2 - 3(c+2) + (c+1) \le 0
\iff \frac32(c+2) - \Delta(c) \le y \le \frac32(c+2) + \Delta(c)\tag{*2}
$$
where $\Delta(c) = \frac12\sqrt{9c^2+32(c+1)}$.
If one make an implicit plot of the roots of $(*1)$. One find in general,
$$\lambda_1(c) < \frac32(c+2) - \Delta(c) < \lambda_2(c) < \lambda_3(c) < \frac32(c+2) + \Delta(c) < \lambda_4(c)$$
From this, we can deduce in general, $x_{n+1} = y_n$ are bounded between
the two middle two roots $\lambda_2(c), \lambda_3(c)$ of equation $(*1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Evaluate $\prod \frac {2k - 1} {2k}$ I came across to this: evaluate $$\prod_{k = 1}^{n} \frac {2k - 1} {2k}.$$
I brought it to a closed-form expression in asymptotics but I suspect my asymptotics is bad and I also want to see different solutions, hence the question.
My original solution: Write
$$\begin {eqnarray}
\prod_{k = 1}^{n} \frac {2k - 1} {2k}
& = & \exp \left (\sum_{k = 1}^{n} \log \left (1 - \frac {1} {2k}\right) \right) \nonumber \\
& = & \exp \left (- \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m}\right).
\end {eqnarray}$$
Let $$S = \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} \tag {1}.$$ Multiply both sides of $(1)$ by $2k$ to get $2k S < S + 1$, and by $4k$ to get $4kS > S + 2$. Hence, $$\frac {2} {4k - 1} < S < \frac {1} {2k - 1}.$$ Then, $$\frac {1} {2} \left( \log (4n - 1) - \log 3 \right) - 2 + O \left(\frac {1} {n^2}\right) < \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} < \frac {1} {2} \log (2n - 1) - 1 + O \left(\frac {1} {n^2}\right).$$ Hence, $$\sqrt {\frac {c} {2n - 1}} < \prod_{k = 1}^{n} \frac {2k - 1} {2k} < \sqrt {\frac {3 c^2} {4n - 1}},$$ where $c$ is a constant. Any better approach?
| I tried a different and definitely a far more elementary approach and, surprisingly, it gave a better estimate. We have obviously (which I had overlooked!)
$$\prod_{k = 1}^{n} \frac {2k - 1} {2k} = \frac {(2n)!} {4^n (n!)^2}$$
which, by Stirling formula, is
$$=\frac {1} {\sqrt {\pi n}} \exp \left (\frac {1} {192 n^3} - \frac {1} {8n}\right).$$
This estimate is indeed very precise, since for the very small value $n = 5$ the actual value is $0.24609375$ and the estimated value is $0.2460938695063\cdots$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Arithmetic: Prove that is multiple of 30 Prove that $n^{19}-n^7$ is multiple of $30$
I've seen $6$ can divide it because
$$n^{19}-n^7=n^7(n^{12}-1) = n^7(n^6+1)(n^6-1)=n^4(n^6+1)(n^3-1)n^3(n^3+1)$$
And there are three consecutive numbers, so, at least one is multiple of $3$ and up to two even numbers.
But, how to prove that is multiple of $5$?
| Here is an answer to the actual question: why is $n^{19}-n^7$ a multiple of $5$ ?
Divide $n^{19}-n^7$ by $n^5-n$ as polynomials:
$$
n^{19} - n^7 = (n^{14} + n^{10} + n^6) (n^5 - n)
$$
By Fermat's theorem, $n^5-n$ is always a multiple of $5$ and the result follows.
Somewhat surprisingly, $n^5-n=n (n - 1) (n + 1) (n^2 + 1)
$ is always a multiple of $6$ and so the argument above proves that $n^{19}-n^7$ is always a multiple of $30$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
Let $f(a,b,c,d) = (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2$. For real numbers $a < b < c < d,$ prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
Would the best way to do this be to expand each expression and show directly the inequality which we need to prove?
|
Question
Let $f(a,b,c,d) = (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2$.
Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c), $ if $a < b < c < d$.
Then
$$\begin{align}
f(a,c,b,d)&=(a-c)^2+(c-b)^2+(b-d)^2+(d-a)^2\\
&=(a-c)^2+(b-c)^2+(b-d)^2+(d-a)^2
\end{align}$$
Now
$$\begin{align}
f(a,c,b,d) - f(a,b,c,d)&=(a-c)^2+(b-d)^2-((a-b)^2+(c-d)^2)\\
&=2ab+2cd-2ac-2bd \\
&= 2a(b-c)-2d(b-c)\\
&=2(a-d)(b-c) >0.
\end{align}
$$
(Why is that inequality valid?)
This method works for the second inequality as well.
| {
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How do I find the image of a point in a line in 3D space? The question is,
find the image of the point $(1, 6, 3)$ in the line $$\frac x1 = \frac {y-1}{2} = \frac {z-2}{3}$$
I want to know the general equation to find the image of a point in a line.
EDIT
By image, I mean a point which is diametrically opposite to the point, on the opposite side of the line
| Let $Q(a,b,c)$ be the image of $P(1,6,3)$ in the given line.Then direction ratios of $PQ$ are $(a-1,b-6,c-3)$
As the given line and $PQ$ will be perpendicular to each other.So $a_1a_2+b_1b_2+c_1c_2=0$ is the condition of perpendicularity
$1(a-1)+2(b-6)+3(c-3)=0$
$a+2b+3c=22..........(1)$
Now the mid point of $PQ(\frac{a+1}{2},\frac{b+6}{2},\frac{c+3}{2})$ will lie on the line.
Write the equation of line in parametric form
$x=t,y=2t+1,z=3t+2$
So $\frac{a+1}{2}=t,\frac{b+6}{2}=2t+1,\frac{c+3}{2}=3t+2$
$a=2t-1,b=4t-4,c=6t+1$
Put this in equation $(1)$,find $t$ and hence find $(a,b,c)$
| {
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Subadditivity of square root function Any hint to prove
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$
| I'll try the hard way
and use calculus.
Let
$f(y)
= \sqrt{x+y} - \sqrt{x} - \sqrt{y}
$.
$f(0)
=0
$.
$f'(y)
=\frac1{2\sqrt{x+y}}-\frac1{2\sqrt{y}}
$.
Since
$x+y \ge y$,
$\sqrt{x+y} \ge \sqrt{y}$
so
$\frac1{2\sqrt{x+y}}
\le \frac1{2\sqrt{y}}
$.
Therefore
$f'(y) \le 0$
for all $y$.
Since
$f(0) = 0$,
$f(y)
\le 0
$ for all $y$.
Note that this
can be used to show that
if
$g(y)$
is increasing and
$g'(y)$
is decreasing
then
$g(x+y)
\le g(x)+g(y)
$.
Similarly,
if $g'(y)$
is increasing,
$g(x+y)
\ge g(x)+g(y)
$.
| {
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Probability of random subsets of the same size intersecting $A$ and $B$ are subsets drawn uniformly from a set $C$, such that $|A|=|B|=a$ and $|C|=n$.
For what $a$ is the probability of $A\cap B \ne \varnothing$ equal to $50\%$?
I figured that if $|A|=a$ and $|B|=b$ then the probability that there will be no intersection is
$$\frac{(n-a)!(n-b)!}{n!(n-a-b)!}$$
and for this special case,
$$\frac{(n-a)!(n-a)!}{n!(n-2a)!}$$
If we expand it, we get:
$$\prod_{i=0}^{a-1}{\left(1-\frac{a}{n-i}\right)}$$
and I wasn't able to go further from here to solve when is it equal to $0.5$
| If we use Stirling's Approximation, that is $n!\sim\sqrt{2\pi n}\frac{n^n}{e^n}$, we get that
$$
\frac{\binom{n-a}{a}}{\binom{n}{a}}\sim\frac{\left(1-\frac an\right)^{2n-2a+1}}{\left(1-\frac{2a}n\right)^{n-2a+1/2}}
$$
Using $\left(1+\frac xn\right)^n=e^x\left(1+O\left(\frac{x^2}n\right)\right)$, we get
$$
\begin{align}
\frac{\left(1-\frac an\right)^{2n-2a+1}}{\left(1-\frac{2a}n\right)^{n-2a+1/2}}
&=\frac{\left(1-\frac{2a}n+\frac{a^2}{n^2}\right)^{n-a+1/2}}{\left(1-\frac{2a}n\right)^{n-2a+1/2}}\\
&=\left[1+\frac{\frac{a^2}{n^2}}{1-\frac{2a}n}\right]^{\,n}\frac{\left(1-\frac{2a}n\right)^{2a}}{\left(1-\frac{2a}n+\frac{a^2}{n^2}\right)^a}\left[\frac{1-\frac{2a}n+\frac{a^2}{n^2}}{1-\frac{2a}n}\right]^{1/2}\\
&=\left[1+\frac{a^2}{n^2}+O\left(\frac{a^3}{n^3}\right)\right]^n\left(1-\frac{2a}n+O\left(\frac{a^2}{n^2}\right)\right)^a\left[1+O\left(\frac{a^2}{n^2}\right)\right]\\
&=e^{-a^2/n}\left(1+O\left(\frac{a^3}{n^2}\right)\right)
\end{align}
$$
Finding the $a$ so that this is approximately $\frac12$, we get
$$
a\sim\sqrt{n\log(2)}
$$
| {
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Prove that ${x^7-1 \over x-1}=y^5-1$ has no integer solutions I want to show that $${x^7-1 \over x-1}=y^5-1$$
cannot have any integer solutions. The only observation I have made so far is that the left hand side is the $7$th cyclotomic polynomial
$$\Phi_7(x)= {x^7-1 \over x-1}=x^6+x^5+x^4+x^3+x^2+1$$
If I remember correctly cyclotomic polynomials are irreducible. Now can I use this property to arrive at the conclusion or should I try to approach by contradiction and assume $\Phi_7(a)=b^5-1$ for some integers $a$ and $b$? The only problem is that I don't see where I would look for an easy contradiction. Any hints?
Edit
I also see that the right hand side can be factored as
$$(y-1)(y^4+y^3+y^2+y+1)=(y-1)\Phi_5(y)\implies \frac{\Phi_7(x)}{\Phi_5(y)}=y-1$$
which seems like it could give the result if we prove that the two cyclotomic polynomials have no common factors. How could this be done?
Edit: This is IMO2006 Shortlised Problem N5 (RUS).
| Since this is an IMO2006 shortlisted problem, here is an solution from The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009:
Every prime divisor $p$ of $\frac{x^7-1}{x-1}=x^6+\dots x+1$ is congruent to $0$ or $1$ modulo $7$. Indeed, if $p \mid x-1$, then $\frac{x^7-1}{x-1} \equiv 1+\dots 1\equiv 7 \pmod{p}$, so $p=7$; otherwise the order of $x$ modulo $p$ is $7$ and hence $p\equiv 1 \pmod{7}$. Therefore every positive divisor $d$ of $\frac{x^7-1}{x-1}$ satisfies $d \equiv 0$ or $1\pmod{7}$.
Now suppose $(x,y)$ is a solution of the given equation. Since $y-1$ and $y^4+y^3+y^2+y+1$ divide $\frac{x^7-1}{x-1}=y^5-1$, we have $y\equiv 1$ or $2$ and $y^4+y^3+y^2+y+1\equiv 0$ or $1\pmod{7}$. However, $y\equiv 1$ or $2$ implies that $y^4+y^3+y^2+y+1\equiv 5$ or $3 \pmod{7}$, which is impossible.
| {
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Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
| By AM-GM we have:
$1 + \frac{1}{a} = \frac{1}{a}(a + b + c + a) \ge \frac{1}{a}4\sqrt[4]{{{a^2}bc}}$
$\Rightarrow 1 + \frac{1}{a} \ge \frac{4}{a}\sqrt[4]{{\frac{{{a^4}bc}}{{{a^2}}}}} = 4\sqrt[4]{{\frac{{bc}}{{{a^2}}}}} $
And $1 + \frac{1}{b} \ge 4\sqrt[4]{{\frac{{ca}}{{{b^2}}}}};1 + \frac{1}{c} \ge 4\sqrt[4]{{\frac{{ab}}{{{c^2}}}}}$
$$\Rightarrow \left( {1 + \frac{1}{a}} \right)\left( {1 + \frac{1}{b}} \right)\left( {1 + \frac{1}{c}} \right) \ge 4\sqrt[4]{{\frac{{bc}}{{{a^2}}}}}4\sqrt[4]{{\frac{{ca}}{{{b^2}}}}}4\sqrt[4]{{\frac{{ab}}{{{c^2}}}}} = 64$$
| {
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simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^{x}
$$
The second inequality can be found in Proposition B.3 of "Randomized Algorithms", by Raghaven and Motwani (which itself refers to the book "Analytic Inequalities" by Mitrinović) , and can be proven straight-forwardly by calculus (showing a first derivative is non-negative and such).
While I can also prove the first inequality using familiar calculus methods, it is a bit messy (ultimately requiring that $\frac{1}{y+2}+\frac{1}{3y+2}\ge \frac{1}{y+1}$ for $y\ge 0$).
Does anyone know a "simple" proof of $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$?
| You can do better than the stated inequality starting from $$\log(1+x)<\frac{x(x+6)}{4x+6}$$ for all $x>0$. (Equality for $x=0$ and the difference between the right- and left hand side is strictly increasing as can be shown by taking derivatives.) With this we find $$ \frac{1}{2}\log\left(1+\frac{1}{x+\frac{1}{3}}\right)+x \log\left(1+\frac{1}{x}\right) < \frac38\left(\frac1{3x+1}+\frac1{x+1}\right)+\left(1-\frac3{6x+4}\right)=$$ $$1-\frac{3x}{4(3x+1)(x+1)(3x+2)}<1$$ for all $x>0$. In fact one can show in the same way that for any $\alpha > \tfrac16$ the inequality $$ \frac{1}{2}\log\left(1+\frac{1}{x+\alpha}\right)+x \log\left(1+\frac{1}{x}\right) < 1$$ holds for all $$x>\max\left(0,\frac{2-18\alpha^2}{18\alpha -3}\right).$$
| {
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Elementary number theory equation simplification Let $x\equiv 1 \pmod 3$ and $x\equiv 2 \pmod 5$
first congruence gives $x=1+3k$, substituting this second congruence we get
$1+3k \equiv 2 \pmod 5$...(I), consequently $k \equiv 2 \pmod 5$...(II)
can someone tell how we reach from (I) to (II) ? which rules are applied?
| $$3k\equiv 1\equiv 6\pmod{5}\stackrel{:3}\iff k\equiv 2\pmod{5}$$
Or apply Extended Euclidean Algorithm to find $s,t\in\mathbb Z$ such that $3s+5t=1$, in which case $3s\equiv 1\pmod{5}$, so $k\equiv s\pmod{5}$.
Subtract consecutive equations:
$$5=3(0)+5(1)\\3=3(1)+5(0)\\2=3(-1)+5(1)\\1=3(2)+5(-1)$$
Therefore $3(2)\equiv 1\pmod{5}$, so $k\equiv 2\pmod{5}$.
| {
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Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
| Let's start from the complicated side.
$$\begin{align*}
\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}&=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\times\frac{\cos\theta}{\cos\theta}\quad\text{(Multiply by $1=\frac{\cos\theta}{\cos\theta}$ to simplify)}\\
&=\frac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin\theta-1}{\cos\theta}+1\right)}\quad\text{(Forcefully factoring out the terms we need)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin\theta-1}{\cos\theta}\times\frac{1+\sin\theta}{1+\sin\theta}+1\right)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin^2\theta-1}{\cos\theta(1+\sin\theta)}+1\right)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{-\cos^2\theta}{\cos\theta(1+\sin\theta)}+1\right)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(1-\frac{\cos\theta}{1+\sin\theta}\right)}\\
&=\frac{1+\sin\theta}{\cos\theta}\quad\text{(Cancelling out common terms)}
\end{align*}$$
| {
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Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$ Let $P(x)=4x^2+6x+4$ and $Q(y)=4y^2-12y+25$.
Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$
I can solve this question graphically.
$$P(x)Q(y)=28\implies(4x^2+6x+4)(4y^2-12y+25)=28$$
$$4x^2+6x+4=\frac{28}{4y^2-12y+25}$$
I drew the graph of $4x^2+6x+4$ and found it is a upward parabola with minimum value at $(-\frac{3}{4},\frac{7}{4})$ and I drew the graph of $\frac{28}{4x^2-12x+25}$. I found that it is a bell shaped curve whose maximum value occurs at $(\frac{1}{4},\frac{7}{4})$. So I found that the common abscissa is $y=\frac{7}{4}$ which occurs at $x=-\frac{3}{4}$ for the first curve and at $x=\frac{1}{4}$ for the second curve, so solution should be $(-\tfrac{3}{4},\tfrac{1}{4})$
But I do not know how to solve it algebraically.
| Note that we have
$$P(x)=4x^2+6x+4=4\left(x+\frac 34\right)^2+\frac 74\ge \frac 74$$
and
$$Q(y)=4y^2-12y+25=4\left(y-\frac 32\right)^2+16\ge16$$
| {
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For all nonnegative real numbers $x,y$ and $z$, prove that $\dfrac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$
For all nonnegative real numbers $x,y$ and $z$, prove that $$\frac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$$
It seems like AM-GM works here. We have $\dfrac{(x+y+z)^2}{3} \geq \dfrac{(2\sqrt{xy}+z)(2\sqrt{xz}+y)(2\sqrt{xy}+z)}{3}$. Then I get stuck and don't know what to do next.
| We know that for non negative reals (in fact for all reals) we have $$a^2+b^2+c^2 \ge ab+bc+ca$$
Hence $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) \ge 3(xy+yz+zx)$$
and by applying the same trick on $a=\sqrt{xy}, b=\sqrt{yz}, c=\sqrt{zx}$ we have
$$xy+yz+zx \ge y\sqrt{xz}+z\sqrt{yx}+x\sqrt{yz}$$
This proves the claim.
| {
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Finding the number of real solutions of $2 \cos((x^2+x)/6) = 2^x + 2^{-x}$
The number of real roots of the equation
$$
2\cos\left(\frac{x^2+x}{6}\right) = 2^x + 2^{-x}
$$
is
*
*$0$,
*$1$,
*$2$,
*infinitely many.
I can see that $x=0$ satisfies the equation. But I do not know how to solve the equation explicitly.
Thanks
| we have $$2\cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}\geq 2\sqrt{2^x\cdot 2^{-x}}=2$$ by AM-GM thus we have $$2\cos\left(\frac{x^2+x}{6}\right)\geq 2$$
Can you proceed?
| {
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Determinant problem $n\times n$ with $0$'s and $1$'s $$D_n=\left\vert\begin{matrix}0&1&0&0&\cdots&0&0\\
1&0&1&0&\cdots&0&0\\
0&1&0&1&\cdots&0&0\\
0&0&1&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&0&\cdots&0&1\\
0&0&0&0&\cdots&1&0\\
\end{matrix}\right\vert$$
Can anyone help me with this determinant? I tried to use Laplace and after that recursive formula but that didn't help.
| If you calculate $D_n$ using Laplace's expansion along the first row twice, you get
$$D_n = (-1) \cdot \left\vert\begin{matrix}
1&1&0&\cdots&0&0\\
0&0&1&\cdots&0&0\\
0&1&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\\
0&0&0&\cdots&0&1\\
0&0&0&\cdots&1&0\\
\end{matrix}\right\vert = (-1) \cdot \left( 1 \cdot \left\vert\begin{matrix}
0&1&\cdots&0&0\\
1&0&\cdots&0&0\\
\vdots &\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&0&1\\
0&0&\cdots&1&0\\
\end{matrix}\right\vert \\ + (-1) \cdot
\left\vert\begin{matrix}
0&1&\cdots&0&0\\
0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots& \\
0&0&\cdots&0&1\\
0&0&\cdots&1&0\\
\end{matrix}\right\vert \right) = -D_{n-2} $$
Thus, starting from $D_1 = 0$ and $D_2 = (-1)$ we get
$$ D_{2n} = (-1)^n, \,\,\, D_{2n+1} = 0. $$
| {
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Proof of Multiple angle trigonometri ratios If $\tan^2{A}=1+2\tan^2B$ then prove that $\cos2B=1+2\cos2A$
I could not relate from $\tan$ to $\cos$
| Add $1$ to both sides of the tangent condition.
We get $\tan^2A+1=2+2\tan^2B$.
Simplifying with the identity $\tan^2\theta+1=\sec^2\theta$, we have:
$$\sec^2A=2\sec^2B$$
Taking the reciprocals of both sides and multiplying by $4$, we get:
$$4\cos^2A=2\cos^2B$$
Remembering the identity $2\cos^2\theta-1=\cos2\theta$, we can subtract $1$ from each side to get:
$$2(2\cos^2A-1)+1=2\cos^2B-1$$
which simplifies to:
$$2\cos2A+1=\cos2B$$
| {
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Positive definite binary quadratic forms Please help me to solve this question or introduce references that help me:
Let $f(x,y) = ax^2 + bxy + cy^2$ be a reduced positive definite form.
Suppose that $\gcd(x, y) = 1$ and that $f(x, y) ≤ a + |b| + c$. Show
that $f(x,y)$ must be one of the numbers $a, c, a - |b| + c$ or $a +
|b| + c$.
| I suppose the main thing is that
$$ f \geq \left( \frac{4ac - b^2}{4c} \right) x^2, $$
$$ f \geq \left( \frac{4ac - b^2}{4a} \right) y^2. $$
These just come from completing the square.
Reduced means $$ |b| \leq a \leq c $$ with some additional conditions in case some things are equal. For example, $ac \geq b^2.$ Thus $4ac-b^2 \geq 3ac.$ Thus
$$ f \geq \left( \frac{3a}{4} \right) x^2, $$
$$ f \geq \left( \frac{3c }{4} \right) y^2. $$
Now, suppose $|y|\geq 2,$ so that $y^2 \geq 4$ and $f \geq 3c.$ With the condition $f \leq a + |b| + c$ and reduction this gives $y = \pm 2,$ $a = |b| = c$ and so $a=|b| = c = 1$ because $\gcd(a,b,c) = 1.$ Take $y=2$ and continue with $x^2 + 2x + 4.$
Next, if we do not have $a=|b| = c = 1,$ we have $y = \pm 1$ or $y=0.$ Continue with either $ax^2 + b x + c$ and $y=1$ or $a x^2$ with $y=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve this exercise given:
$\triangle ABC$
$P=20$ Note: P is perimeter
$\cos \alpha = -\frac{1}{3}$
$\cos \beta = \frac{7}{9}$
Find the sides of the triangle
I'm totally lost on this one. I have no idea from where to begin.
The answer given in my textbook is: $a = 9 b = 6 c = 5$
I managed to solve the problem, I case somebody needs to know how:
Find $\sin\alpha$ and $sin \beta$
Then $\gamma = 180 - (\alpha + \beta)$ => $sin \gamma = sin(\alpha + \beta)$ From sin law => a : b : c = $\sin \alpha : \sin \beta : \sin \gamma$ and a + b + c =20 from here it's just arithmetic operations
| I'll assume that $a$ is the side opposite to $\alpha$ and similarly for $b$ and $c$, opposite to $\beta$ and $\gamma$.
The cosine law tells you that
$$
a^2=b^2+c^2-2bc\cos\alpha
$$
so
$$
2bc+2bc\cos\alpha=b^2+2bc+c^2-a^2=(b+c)^2-a^2=(a+b+c)(b+c-a)
$$
which means
$$
\frac{1}{15}bc=b+c-a
$$
Similarly,
$$
2ac(1+\cos\beta)=(a+b+c)(a+c-b)
$$
that is
$$
\frac{8}{45}ac=a+c-b
$$
Summing the two relations we get
$$
\frac{c}{45}(8a+3b)=2c
$$
that implies $8a+3b=90$ (because we assume $c\ne0$ in a triangle).
The sine law tells you that
$$
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}
$$
Since
$$
\sin\alpha=\sqrt{1-\frac{1}{9}}=\frac{2}{3}\sqrt{2}
$$
and
$$
\sin\beta=\sqrt{1-\frac{49}{81}}=\frac{4}{9}\sqrt{2}
$$
we have
$$
\frac{3a}{2\sqrt{2}}=\frac{9b}{4\sqrt{2}}
$$
that simplifies to $2a=3b$.
Now we have
\begin{cases}
8a+3b=90\\[3px]
2a=3b\\[3px]
a+b+c=20
\end{cases}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help me to prove the determinant of given matrix. Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
-x & a_2&a_3&a_4&\cdots &a_n\\
a_{1}+x & -x-a_2 & 0&0&\cdots &0\\
a_1+x&0 & -x-a_3 &0&\cdots &0\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
a_1+x&0 & 0&0&\cdots & -x-a_n\\
\end{array}\end{bmatrix}$, then how to find the $\det (M)$?
The empirical formula I got from considering $n=2,3,4$ in Wolfram Mathematica is
$$
(-1)^n(x^n-\sum_{k=2}^n (k-1)\sigma_k x^{n-k})
$$
where $\sigma_k$ is the $k$-th elementary symmetric polynomial in $a_1,\dots,a_n$.
| I start with the well-known and easily proved fact that the $n\times n$ all $1$'s matrix
$$
\begin{pmatrix}
1 & 1 & \dots &1\\
1& 1 & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1 & 1 &\dots &1
\end{pmatrix}
$$
has one eigenvalue $n$ corresponding to the obvious eigenvector $(1,1,\dots,1)^{T}$ and the eigenvalue $0$ occurring $n-1$ times corresponding to the obvious eigenvectors $(1,-1,0,\dots, 0)^{T}$, $(1,0,-1,\dots, 0)^{T}$, ... $(1,0,0,\dots, -1)^{T}$. We will then also know the eigenvlaues of the matrix got by subtracting the identity matrix from the all-$1$'s matrix, and so we will have that
$$
\det
\begin{pmatrix}
0 & 1 & \dots &1\\
1& 0 & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1 & 1 &\dots &0
\end{pmatrix}
=
(n-1)(-1)^{n-1}.
$$
Now turn to the matrix $M$ of the question. To evaluate the determinant we can replace it by the matrix we get by adding the first row to every other row, that is by
$$
\begin{pmatrix}
-x & a_2 &\dots &a_n\\
a_1& -x & \dots & a_n\\
\vdots & \vdots &\ddots &\vdots\\
a_1 & a_2 &\dots & -x
\end{pmatrix}.
$$
The coefficient of $(-x)^{n-k}$ in this expansion is clearly the sum of all the principal $k\times k$ minors of
$$
\begin{pmatrix}
0 & a_2 &\dots &a_n\\
a_1& 0 & \dots & a_n\\
\vdots & \vdots &\ddots &\vdots\\
a_1 & a_2 &\dots & 0
\end{pmatrix}.
$$
For $k=0$ that sum is $0$; otherwise it is the symmetric sum of terms like
$$
\det
\begin{pmatrix}
0 & a_2 & \dots &a_k\\
a_1& 0 & \dots &a_k\\
\vdots &\vdots &\ddots&\vdots\\
a_1& a_2 &\dots &0
\end{pmatrix}
=
a_1 a_2\dots a_k \det
\begin{pmatrix}
0 & 1 & \dots &1\\
1& 0 & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1& 1 &\dots &0
\end{pmatrix}
=
a_1 a_2\dots a_k (-1)^{k} (k-1).
$$
These calculations will provide a proof of the OP's conjecture.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution ?
$$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$$ Thanks
| $$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)=\lim\limits_{x \to +\infty} \left(\frac{1+x\sin\left(\frac{1}{x}\right)}{\ln\left(\frac{x}{2x+1}\right)-\frac{1}{x^3}} \right)$$
$$=\lim\limits_{x \to +\infty} \left(\frac{1+\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}}{\ln\left(\frac{1}{2+\frac{1}{x}}\right)-\frac{1}{x^3}} \right)$$
So clearly $x \to \infty \Rightarrow\frac{1}{x}\to0$
So $$=\lim\limits_{x \to +\infty} \left(\frac{1+\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}}{\ln\left(\frac{1}{2+\frac{1}{x}}\right)-\frac{1}{x^3}} \right)= \frac{1+1}{\ln\left(\frac{1}{2+0}\right)-0} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the minimum polynomial of $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$? Inspired by a previous question what let $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$. What is the minimal polynomial of $x$ ?
The theory of algebraic extensions says the degree is $4$ since we have the degree of the field extension $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{6}): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] =4$
Does trigonometry help us find the other three conjugate roots?
*
*$+\sqrt{2}-\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_1$
*$-\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_2$
*$-\sqrt{2}-\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot \theta_3$
This problem would be easier if we used $\cos$ instead of $\cot$. If I remember the half-angle identity or... double-angle identity:
$$ \cot \theta = \frac{\cos \theta}{\sin \theta} = \sqrt{\frac{1 - \sin \frac{\theta}{2}}{1 + \sin \frac{\theta}{2}}}$$
Sorry I am forgetting, but I am asking about the relationship between trigonometry and the Galois theory of this number.
| By Galois theory the intermediate fields are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. Your number is not an element of any of those, so it generates the whole 4-d extension. In particular, we know that the minimal polynomial will be a quartic. Therefore any quartic polynomial with integer coefficients with this number as a root is the minimal polynomial.
Let $x=2+\sqrt2+\sqrt3+\sqrt6$. Then
$$
0=(x-2-\sqrt3)^2-(\sqrt2+\sqrt6)^2=x^2-(4+2\sqrt3)x-1.
$$
The idea here is that squaring removes "the $\sqrt2$ content" from $\sqrt2+\sqrt6$ leaving only irrationalities coming from $\sqrt3$.
Using the obvious algebraic conjugate as an extra factor we see that
$$
m(x)=(x^2-(4+2\sqrt3)x-1)(x^2-(4-2\sqrt3)x-1)=x^4-8x^3+2x^2+8x+1
$$
fits the bill.
The algebraic conjugates of values of trig functions (at rational multiples of $\pi$) are of the same type:
$$
\begin{aligned}
2+\sqrt2+\sqrt3+\sqrt6&=\cot\frac{\pi}{24},\\
2+\sqrt2-\sqrt3-\sqrt6&=\cot\frac{17\pi}{24},\\
2-\sqrt2+\sqrt3-\sqrt6&=\cot\frac{13\pi}{24},\\
2-\sqrt2-\sqrt3+\sqrt6&=\cot\frac{5\pi}{24}.
\end{aligned}
$$
I haven't checked the details, but I'm fairly sure that when you dig for the integer multiple angle formulas for cotangent, the polynomial $m(x)$ pops out. After all, we can write $\cot 6x$ as a rational function of $\cot x$ with polynomials of degrees $\le6$ as numerators and denominators. When you use that formula in the l.h.s. of the equation
$$
\cot 6x=1
$$
and clear the denominator, the resulting degree $6$ polynomial equation in $\cot x$ factors as a product of a quartic (obviously $m(\cot x)$) and a quadratic - the latter accounting for solutions $x=3\pi/8$ and $x=7\pi/8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Second order ODE, first derivative missing I have the following second order equation, where the first derivative is missing, and I am asked to find its general solution:
$$6x^{2}yy''=3x(3y^{2}+2)+2(3y^{2}+2)^3$$
I don't know how to solve it. I have tried with a $u(x)=3y^{2}+2$ substitution but it doesn't seem useful...
Is there any method for this kind of equation whithout $y'$?
| Let
$$ u = 3 y^2 +2, u'= 6 y y' $$
where primes are with respect to x. We need to solve
$$ y y''= u/(2 x) + u^3 / (3 x^2) \tag{0} $$
Spoilt earlier erroneous part, substituted by new derivation.
$$ \frac{u'}{6} = \frac{u}{2 x} + \frac{u^3}{3 x^2} ;$$
$$ u' = \frac{u}{ x} ( 3+ 2 \frac{u^2}{x}) ;$$
Unless $3$ is absent, the variables are not separable.
Employing third degree substitution
$$ u = \frac{ x^3}{\sqrt{ c -4 x^5/5 }} = 3 y^2 +2 ; $$
is the integrand with arbitrary constant $c ;$
$$ u = 3 y^2 +2 \tag{1} $$
$$ u' = 6 y y' \tag{2} $$
from $ (1), (2) $
$$ \frac{y}{y'}= \frac{ 2(u-2)}{u'} \tag{3}$$
$$ y \, y'' = \frac{u'}{6} \tag{4}$$
from $(3),(4) $
$$ y =\sqrt{(u-2)/3} \tag {5}$$
$$ y'= \frac {u'/2}{\sqrt{3(u-2)}} \tag {6}$$
Differentiate squared equation of (6) and simplify
$$ 24 y' y'' = \frac {2 (u-2) u'u''-u'^3}{(u-2)^2} \tag {7} $$
Eliminate $y'$ from (6) and (7)
$$ y''= \frac{\sqrt{(u-2)}}{4 {\sqrt 3}u'} \tag{8}$$
Multiply (5) and (8) and simplify
$$ 12 y y''= 2 u''- \frac{u^{'2}}{u-2} \tag{9}$$
that leads to a second order non-linear ODE in $u$ and $x$.
$$ 2 u''= 12\, (\frac{u}{2 x} + \frac {u^3} {3 x^2} ) + \frac{u'^{2}}{(u-2)} \tag {10}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sign the Derivative of $(\frac{1-q^r}{1-q^n})^{1/(n-r)}$ wrt $n$ (where $q \in (0,1)$) For $n, r \in \mathbb{N}$, $n > r$ and $q \in (0,1)$, I'm trying to sign the derivative wrt $n$ of
\begin{eqnarray}
f(n) = \left(\frac{1-q^r}{1-q^n}\right)^{1/(n-r)}.
\end{eqnarray}
I believe the derivative is
\begin{eqnarray}
f'(n) = \left(\frac{1-q^r}{1-q^n}\right)^{\frac{1}{n-r}} \left(\frac{q^n \log (q)}{\left(1-q^n\right) (n-r)}-\frac{\log \left(\frac{1-q^r}{1-q^n}\right)}{(n-r)^2}\right).
\end{eqnarray}
I am very sure that the derivative has the same sign for all admissible values of $x, r,$ and $n$: it should be positive. However proving it is hard. Clearly, $f'(n) > 0$ iff
\begin{eqnarray}
(n-r)q^n \log (q) > (1-q^n)[\log (1-q^r) - \log(1-q^n)],
\end{eqnarray}
iff
\begin{eqnarray}
q^n \log (q^{n-r}) + (1-q^n)\log(1-q^n) > (1-q^n)\log (1-q^r).
\end{eqnarray}
This look on the right track until you realise
\begin{eqnarray}
n > r \iff q^n < q^r \iff 1-q^n > 1-q^r \iff \log(1-q^n) > \log(1-q^r).
\end{eqnarray}
Any ideas what I might be missing?
| I strongly believe the following to be a solution, however, I would be grateful if somebody could quickly go through my reasoning to make sure I didn't go wrong.
--
Let $A \equiv 1-q^r$. Then
\begin{eqnarray*}
f'(n) &=& \frac{\delta}{\delta n}\left( \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \right) A^\frac{1}{n-r} + \frac{\delta}{\delta n}\left( A^\frac{1}{n-r} \right) \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \\
&=& \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \left[ \frac{q^n \ln q}{(1-q^n)(n-r)} + \frac{\ln(1-q^n)}{(n-r)^2}\right] A^\frac{1}{n-r} - \frac{A^\frac{1}{n-r} \ln A}{(n-r)^2} \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \\
&\propto& (n-r) \frac{q^n \ln q}{1-q^n} + \ln(1-q^n) - \ln(A) =: Z(n).
\end{eqnarray*}
Our strategy will be to show that $Z(n) \geq 0$, whence we will know that $f'(n) \geq 0$. To this end, we firstly note
\begin{eqnarray*}
Z''(n) &=& \frac{q^n (n-r) \log ^2(q)}{1-q^n}+\frac{q^{2 n} (n-r) \log ^2(q)}{\left(1-q^n\right)^2} \geq 0.
\end{eqnarray*}
Thus, in order to show $Z'(n) \geq 0$ it suffices to show $Z'(r+1) \geq 0$, i.e.
\begin{eqnarray*}
Z'(r+1) = \frac{q^{r+1} \ln(q)}{1-q^{r+1}} + \ln(1-q^{r+1}) - \ln(1-q^r) \geq 0.
\end{eqnarray*}
Now consider
\begin{eqnarray*}
\frac{\delta}{\delta q} Z'(r+1) = \frac{q^{r-1} \left(q (r+1) \left(q^r-1\right) \log (q)-(q-1) r \left(q^{r+1}-1\right)\right)}{\left(q^r-1\right) \left(q^{r+1}-1\right)^2}.
\end{eqnarray*}
We claim $\frac{\delta}{\delta q} Z'(r+1) \geq 0$ (for $0 < q < 1$). In order to establish this, we solve
\begin{eqnarray*}
q^{r-1} \left(q (r+1) \left(q^r-1\right) \log(q)-(q-1) r \left(q^{r+1}-1\right)\right) &=& 0 \\
\iff \frac{(q-1)(q^{r+1}-1)}{q^r-1} \frac{r}{r+1} = q \log(q) \vee q = 0.
\end{eqnarray*}
Now note
\begin{eqnarray*}
\frac{(q-1)(q^{r+1}-1)}{q^r-1} &=& - \frac{(1-q)^2(1+ \dots + q^r)}{q(1-q)(1+\dots+q^{r-1})} \\
&=& - \frac{(1-q)(1+ \dots + q^r)}{1+\dots+q^{r-1}} \\
&=& -(1-q) \left(1 + \frac{q^r}{1+\dots+q^{r-1}}\right) \leq -(1-q),
\end{eqnarray*}
where the inequality holds on $0 < q < 1$.
On the same domain, we claim $q \log(q) > -(1-q)$. As the LHS tends to $0$ as $q \to 0$ and the RHS tends to $-1$, this is true for at least one point in the domain. Furthermore, the solution to $q \log(q) = -(1-q)$ is given by $q = \exp(1 + W(-e^{-1}))$, where $W(\cdot)$ denotes the Lambert W function. As $W_0(-e^{-1}) = W_{-1}(-e^{-1}) = -1$ (i.e. the two real branches of the Lambert function agree at $-1/e$), we have a unique solution of $q = 1$. Thus, as both functions are continuous, the LHS must lie strictly above the RHS on the domain.
So the only solutions to $\frac{\delta}{\delta q} Z'(r+1) = 0$ on $q \in [0,1]$ are the boundary points. Thus, for any intermediate points $\frac{\delta}{\delta q} Z'(r+1)$ must have the same sign. Now consider
\begin{eqnarray*}
\frac{\delta}{\delta q}|_{q=1/2} Z'(r+1) = \frac{2^{1-r} \left(\frac{1}{2} \left(2^{-r-1}-1\right) r-\frac{1}{2} \left(2^{-r}-1\right) (r+1) \log (2)\right)}{\left(2^{-r-1}-1\right)^2 \left(2^{-r}-1\right)}.
\end{eqnarray*}
It is clear that this is positive iff
\begin{eqnarray*}
\frac{1}{2} \left(2^{-r-1}-1\right) r &\leq& \frac{1}{2} \left(2^{-r}-1\right) (r+1) \log (2) \\
\left(2^{-r-1}-1\right) r &\leq& \left(2^{-r}-1\right) (r+1) \log (2) \\
\frac{2^{-r-1}-1}{2^{-r}-1} &\geq& \frac{r+1}{r} \log(2)\\
\frac{1/2}{1-(1/2)^r} &\geq& \frac{r+1}{r} \log(2)-1.
\end{eqnarray*}
Clearly, the LHS is bounded below by $1/2$ for $r\geq 1$, and the RHS is strictly less than $1/2$ for $r \geq 1$. Thus, this inequality holds for all $r \in \mathbb{N}$. Hence, $\frac{\delta}{\delta q} Z'(r+1) \geq 0$.
Hence, it suffices to show
\begin{eqnarray*}
\lim_{q \to 0} Z'(r+1) &=& \lim_{q \to 0} \frac{q^{r+1} \ln(q)}{1-q^{r+1}} + \ln(1-q^{r+1}) - \ln(1-q^r) \\
&=& 0.
\end{eqnarray*}
in order to establish $Z'(n) \geq 0$. But $f'(n) \geq 0$ iff $Z'(n) \geq 0$, and hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$
If $11 \gamma = \pi$ then prove that
$
\sec(\gamma) \sec(2\gamma) \sec(3\gamma) \sec(4\gamma) \sec(5\gamma) = 32.
$
I could not use the relation $11 \gamma = \pi$.
| Like How to expand $\cos nx$ with $\cos x$?,
we can prove for positive integer $N$:
$$\cos Nx=2^{N-1}\cos^Nx+\cdots$$
If $N=2n+1$ and $\cos(2n+1)x=1, (2n+1)x=2m\pi$ where $m$ is any integer
$x=\dfrac{2m\pi}{2n+1}$ where $m\equiv0,1,2,\cdots,2n-1,2n\pmod{2n+1}$
So, the roots of $$2^{2n}\cos^{2n+1}x+\cdots-1=0$$
are $\cos\dfrac{2m\pi}{2n+1}$ where $m\equiv0,1,2,\cdots,2n-1,2n\pmod{2n+1}$
$$\implies2^{2n}\prod_{m=0}^{2n}\cos\dfrac{2m\pi}{2n+1}=(-1)^n$$
$$\implies2^{2n}\prod_{m=1}^{2n}\cos\dfrac{2m\pi}{2n+1}=(-1)^n$$
Now $\cos(\pi-A)=-\cos A,$
$$\implies2^{2n}\prod_{r=1}^n(-1)^n\cos^2\dfrac{r\pi}{2n+1}=(-1)^n$$
As for $1\le r\le n,0\le\dfrac{r\pi}{2n+1}<\dfrac\pi2\implies\cos\dfrac{r\pi}{2n+1}>0$
$$\implies\prod_{r=1}^n\cos\dfrac{r\pi}{2n+1}=\dfrac1{2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Definite Integral of Polynomial of Sine and Cosines I was wondering if there is a way to compute definite integral
\begin{align}
I(m,n) :=\int_{0}^{\pi} \sin^m (\theta) \cdot \cos^{n}(\theta) \ \mathrm{d} \theta
\end{align}
in general for integer-valued $m$ and $n$. This problem arises when I try to compute integrals of trigonometric functions over a high dimensional sphere.
In fact I am more interested in the asymptotic order of this integral. I was wondering how $I(m,n)$ behaves when $n$ goes to infinity. For example, what is the limit of $\lim _{n\rightarrow \infty} I(2, n)$?
| When at least one of the exponents is odd, substitute $\sin^2 x = (1 - \cos^2 x)$ and $\cos^2 x = (1 - \sin^2 x)$ as in:
$$
\begin{eqnarray}
\int \sin^3 x \cos ^4 x \, \textrm{d}x &=& \int \sin^2 x \cos^4 x(\sin x \, \textrm{d}x) \\
&=& \int (1 - \cos^2 x) cos^4 x (\sin x \, \textrm{d}x) \\
&=& \int \cos ^4 x \sin x \, \textrm{d}x - \int \cos^6 x \sin x \, \textrm{d}x \\
&=& \frac{-\cos^5 x}{5} + \frac{\cos^7 x}{7} + C \\
\end{eqnarray}
$$
When both $m$ and $n$ are even, substitute $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$ as in:
$$
\begin{eqnarray}
\int \sin^2 x \cos^4 x \, \textrm{d}x &=& \int \left(\frac{1 - \cos 2x}{2}\right) \left(\frac{1 + \cos 2x}{2}\right)^2 \, \textrm{d}x \\
&=& \frac{1}{8} \int \, \textrm{d}x + \frac{1}{8} \int \cos 2x \, \textrm{d}x - \frac{1}{8} \int \cos^2 2x \, \textrm{d}x - \frac{1}{8} \int \cos^3 2x \, \textrm{d}x \\
&=& \frac{x}{8} + \frac{1}{16} \sin 2x - \frac{1}{8} \int \frac{1 + \cos 4x}{2} \, \textrm{d}x - \frac{1}{8} \int (1 - \sin^2 2x) \cos 2x \, \textrm{d}x \\
&& ... \\
&=& \frac{x}{16} + \frac{\sin^3 2x}{48} - \frac{\sin 4x}{64} + C
\end{eqnarray}
$$
This requires computing the indefinite integral on which to apply your interval, which might be painful for high values of $m$ or $n$. I hope someone has a better answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find the eccentricity of a conic Find the eccentricity $e$ of the conic $$S \equiv 39x^2+11y^2-96xy+14x+2y-34=0.$$
My try:
Comparing with general second degree conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ we have
$a=39$, $b=11$, $2h=-96$, $2g=14$, $2f=2$ and $c=-34$
The determinant $$\Delta=\begin{vmatrix}
39 & -48&7 \\
-48& 11 & 1\\
7&1 & -34
\end{vmatrix} \ne 0$$ and
$h^2-ab \gt 0$, and hence the conic is a Hyperbola.
Now since there is $xy$ term one way is to rotate the axes to make $xy$ term zero by angle $$\tan(2\theta)=\frac{-96}{28}.$$ But the algebra is very tedious. Is there any other approach?
| The formula derived in this answer is
$$
e^2=\frac{2\sqrt{(A{-}C)^2+B^2}}{\sigma(A{+}C)+\sqrt{(A{-}C)^2+B^2}}
$$
Setting $A=39,B=-96,C=11,D=14,E=2,F=-34$ gives
$$
\begin{align}
e^2&=\frac{2\cdot100}{-50+100}=4\\
e&=2
\end{align}
$$
since $\sigma=\operatorname{sign}\left(F\!\left(B^2-4AC\right)+\left(AE^2{-}BDE{+}CD^2\right)\right)=\operatorname{sign}\left(-250000\right)=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving that $x-\frac{x^{3}}{6} < \sin(x)$ for $0I have to prove that $$x-\frac{x^{3}}{6} < \sin(x) \quad\text{ for }\quad 0<x<\pi $$
I tried to define $f(x) = \sin(x) - (x-\frac{x^{3}}{6})$, and to differentiate it,
but $f'(x) = \cos(x) -1 +\frac{1}{2}x^{2}$. I have no idea how to continue.
Thanks.
| Consider that
$$
\sin(x)=\sum_{n\ge 0} (-1)^n\frac{x^{2n+1}}{(2n+1)!}.
$$
Moreover, for all $x \in [0,\pi]$ and positive $n$ it holds $ (2n+3)(2n+2)\ge 20>\pi^2\ge x^2$, so that
$$
\left|\frac{x^{2n+1}}{(2n+1)!}\right|> \left|\frac{x^{2n+3}}{(2n+3)!}\right|.
$$
Therefore for each integer $m\ge 0$ and $x \in [0,\pi]$ we obtain
$$
\sum_{n=0}^{2m+1} (-1)^n\frac{x^{2n+1}}{(2n+1)!} <\sin(x)<\sum_{n=0}^{2m} (-1)^n\frac{x^{2n+1}}{(2n+1)!}.
$$
In particular, we have
$$
x-\frac{x^3}{6}<\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}.
$$
| {
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"url": "https://math.stackexchange.com/questions/1613919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$? Suppose $\theta=2\pi/5$. Apparently it is true that $1+ \cos^2 \theta + \cos^2(2\theta) + \cos^2(3 \theta) + \cos^2 (4\theta) = 5/2$, or equivalently, $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$. What is the easiest way to see this?
I see that $1 + \cos \theta + \dotsb + \cos 4\theta = 0$ by de Moivre.
This came up while calculating the character of the irreducible two-dimensional representation of the dihedral group of order $10$ (i.e. to prove it is irreducible).
| As $\pi=180^\circ$ $$S=\cos^272^\circ+\cos^2144^\circ$$
$$2S=1+\cos144^\circ+1+\cos288^\circ=2+\cos(180-36)^\circ+\cos(360-72)^\circ$$
As $\cos(180^\circ-A)=-\cos A,\cos(360^\circ-B)=+\cos B,$
$$2S-2=\cos72^\circ-\cos36^\circ$$
Now use Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to isolate variable algebraically in a combinatorics equation? How would I isolate a variable algebraically in a combinatorics equation?
For example, if I'm given:
$$C(k, 2) = 45$$
How would I solve for $k$, without trying random values of $k$? I know that $$C(k, r) = \frac{k!}{r!(k-r)!}$$ but I'm unsure how I'd solve for $$ \frac{k!}{2!(k-2)!} = 45$$ Or if this is even the right approach. A smaller question that might solve the larger would be, how do I get rid of factorials algebraically?
| The number $n!$ can be defined recursively as follows:
*
*$1! = 1$
*$n! = n(n - 1)!$, $n \geq 1$
Using the definition to compute $5!$ yields
\begin{align*}
5! & = 5 \cdot 4!\\
& = 5 \cdot 4 \cdot 3!\\
& = 5 \cdot 4 \cdot 3 \cdot 2!\\
& = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1!\\
& = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\end{align*}
Notice that if $n \geq 2$, we have $n! = n(n - 1)! = n(n - 1)(n - 2)!$. Hence,
$$\binom{n}{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n(n - 1)}{2}$$
Thus,
\begin{align*}
\binom{n}{2} & = 45\\
\frac{n!}{2!(n - 2)!} & = 45\\
\frac{n(n - 1)(n - 2!}{2(n - 2)!} & = 45\\
\frac{n(n - 1)}{2} & = 45\\
n(n - 1) & = 90\\
n^2 - n & = 90\\
n^2 - n - 90 & = 0\\
(n - 10)(n + 9) & = 0
\end{align*}
The quadratic equation has solutions $n = 10$ and $n = -9$. However, it is not possible to select two objects from $-9$ objects, so $n = -9$ is an extraneous solution. Hence, $n = 10$.
If $n \geq 3$, we have $n! = n(n - 1)(n - 2)(n - 3)!$. Thus,
$$\binom{n}{3} = \frac{n!}{3!(n - 3)!} = \frac{n(n - 1)(n - 2)(n - 3)!}{3!(n - 3)!} = \frac{n(n - 1)(n - 2)}{3!}$$
so you could solve the equation
$$\binom{n}{3} = 56$$
by solving the cubic equation
$$\frac{n(n - 1)(n - 2)}{3!} = 20$$
More generally,
\begin{align*}
\binom{n}{k} & = \frac{n!}{k!(n - k)!}\\
& = \frac{n(n - 1)(n - 2) \cdots (n - k + 1)(n - k)!}{k!(n - k)!}\\
& = \frac{n(n - 1)(n - 2) \cdots (n - k + 1)}{k!}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How many non-negative integer solutions does the equation $3x + y + z = 24$ have? If the equation is $x + y + z = 24$ then it is solvable with stars and bars theorem. But what to do if it is $3x + y + z = 24$?
| Put $x=0\;,$ Then $y+z=24$. So we get Total ordered pair of $(y,z)$ is $= 25$
Put $x=1\;,$ Then $y+z=21$. So we get Total ordered pair of $(y,z)$ is $= 22$
Put $x=2\;,$ Then $y+z=18$. So we get Total ordered pair of $(y,z)$ is $= 19$
Put $x=3\;,$ Then $y+z=15$. So we get Total ordered pair of $(y,z)$ is $= 16$
Put $x=4\;,$ Then $y+z=12$. So we get Total ordered pair of $(y,z)$ is $= 13$
Put $x=5\;,$ Then $y+z=9$. So we get Total ordered pair of $(y,z)$ is $= 10$
Put $x=6\;,$ Then $y+z=6$. So we get Total ordered pair of $(y,z)$ is $= 7$
Put $x=7\;,$ Then $y+z=3$. So we get Total ordered pair of $(y,z)$ is $= 4$
Put $x=8\;,$ Then $y+z=0$. So we get Total ordered pair of $(y,z)$ is $= 1$
So total no. of ordered triples $(x,y,z)$ is $ = 25+22+19+16+13+10+7+4+1=$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
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What is the geometry behind $\frac{\tan 10^\circ}{\tan 20^\circ}=\frac{\tan 30^\circ}{\tan 50^\circ}$? This identity is solvable by the help of trigonometry identities, but I guess there is an interesting and simple geometry interpretation behind this identity and I can't find it.
I found it when I was thinking about World's Hardest Easy Geometry Problem
|
Consider a triangle $ABC$ with $\angle A = 10^\circ$, $\angle B = 150^\circ$, and $\angle C=20^\circ$. Let $O$ be the circumcenter of the triangle $ABC$. Then
$$\angle AOC = 360^\circ - 2 \angle B = 60^\circ,$$
so triangle $AOC$ is equilateral.
Let $S$ be the circumcenter of the triangle $OBC$. Then
$$\angle BSC = 2\angle BOC = 4\angle BAC = 40^\circ$$
and
$$\angle SCB = \frac{180^\circ - \angle BSC}2 = 70^\circ,$$
so $\angle SCA = 50^\circ$. Denoting the intersection of $AC$ and $SB$ by $X$ we get
$$\angle CXS = 180^\circ - \angle SCA - \angle BSC = 90^\circ$$ so $AC \perp SB$.
Moreover $\triangle ASC \equiv \triangle ASO$ because $AC=AO$ and $SC=SO$. In particular $\angle CAS = \angle SAO$ and since $\angle CAO=60^\circ$, we have $\angle CAS = 30^\circ$.
Observe that
$$\frac{\tan \angle BAC}{\tan \angle ACB} = \frac{\frac{BX}{AX}}{\frac{BX}{CX}} = \frac{CX}{AX} = \frac{\frac{SX}{AX}}{\frac{SX}{CX}} = \frac{\tan \angle CAS}{\tan \angle SCA},$$
therefore
$$\frac{\tan 10^\circ}{\tan 20^\circ} = \frac{\tan 30^\circ}{\tan 50^\circ}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"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.
| Since $5=2^{\log _2 5}$ we have
$$5^{19}=2^{19 \log_2 5}.$$
Now we get some estimates on that logarithm:
We have $2^{2.25}=4 \sqrt[4]{2}<5$, as $\sqrt[4]{2} < \frac{5}{4}=1.25$. This means $\log_2 5>2.25$ and thus
$$5^{19}=2^{19 \log_2 5}> 2^{19 \times 2.25}=2^{42.75}>2^{39}.$$
Also,
$$52^7=2^{7 \log_2 52} $$
and since $2^{5 \frac{2}{3}}=\frac{64}{\sqrt[3]{2}}<52$ we have
$$52^7=2^{7 \log_2 52}>2^{7\times 5 \frac{2}{3}}=2^{39 \frac{2}{3}}>2^{39}. $$
In summary, we have
$$5^{19}>52^7>2^{39}. $$
| {
"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": 3
} |
Trigonometric Expression for $1 + \cos \alpha + \cos 2\alpha + \cdots + \cos n \alpha$ using complex numbers This question is not a duplicate because I am asked here to use the fact that $1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha = Re (1 + z + z^{2} + \cdots + z^{n})$, where the question this is suspected of being a duplicate of does not use this
I am supposed to use the fact that $1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha = Re (1 + z + z^{2} + \cdots + z^{n})$, where $z = \cos \alpha + i \sin \alpha$, ro find a trigonometric expression for $1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha$.
Now, when $z \neq 1$, we have the geometric sum $\displaystyle 1 + z + z^{2} + \cdots + z^{n} = \frac{1-z^{n+1}}{1-z}$. (When $z = 1$, $1 + z + z^{2} + \cdots + z^{n} = n + 1$, so $Re(1 + z + z^{2} + \cdots + z^{n}) = Re(n+1) = n+1$, finished.)
So, in the case where $z \neq 1$, $\displaystyle 1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha = Re(1 + z + z^{2} + \cdots + z^{n}) = Re\left(\frac{1-z^{n+1}}{1-z} \right) = Re \left(\frac{1-(\cos \alpha + i \sin \alpha)^{n+1}}{1-(\cos \alpha + i \sin \alpha)} \right) = Re \left[ \left(\frac{1-(\cos \alpha + i \sin \alpha)^{n+1}}{(1-\cos \alpha) - i \sin \alpha} \right)\cdot \frac{(1-\cos \alpha) + i \sin \alpha}{(1 - \cos \alpha)+ i \sin \alpha}\right] = Re \left( \frac{(1-\cos \alpha)+i\sin \alpha - ( 1- \cos \alpha)(\cos \alpha + i \sin \alpha)^{n+1}-i \sin \alpha (\cos \alpha + i \sin \alpha)^{n+1}}{(1-\cos \alpha)^{2} + \sin^{2} \alpha}\right)$.
Then, to make a long story short, after an application of De Moivre's Theorem to turn the $(\cos \alpha + i \sin \alpha)^{n+1}$'s into $(\cos(n \alpha + \alpha)+ i \sin (n \alpha + \alpha))$'s, usage of the sine and cosine of sum identities, collecting of like terms and cancelling some things, we obtain
$\displaystyle Re \left(\frac{1-(\cos \alpha + i \sin \alpha)^{n+1}}{1-(\cos \alpha + i \sin \alpha)} \right) = Re \left(\frac{(1-\cos \alpha + \cos (n\alpha)) + i(\sin \alpha - \cos (n\alpha)\sin \alpha + \sin (n \alpha))}{2-2\cos \alpha} \right) = \frac{1-\cos \alpha + \cos (n \alpha)}{2-2\cos \alpha}$.
At least I think it does.
Which brings me to my question: is this correct? If not, where did I go wrong and how do I fix it?
| Since all your work was not shown, I cannot comment on where any problems were, but here is a simple derivation of a few formulas for the sum in the question:
$$
\begin{align}
\sum_{k=0}^n\cos(k\alpha)
&=\frac12\sum_{k=0}^n\left(e^{ik\alpha}+e^{-ik\alpha}\right)\tag{1}\\
&=\frac12\frac{e^{i(n+1)\alpha}-1}{e^{i\alpha}-1}+\frac12\frac{1-e^{-i(n+1)\alpha}}{1-e^{-i\alpha}}\tag{2}\\
&=\frac12\frac{e^{i(n+1/2)\alpha}-e^{-i\alpha/2}}{e^{i\alpha/2}-e^{-i\alpha/2}}+\frac12\frac{e^{i\alpha/2}-e^{-i(n+1/2)\alpha}}{e^{i\alpha/2}-e^{-i\alpha/2}}\tag{3}\\
&=\frac12\left(1+\frac{\sin((2n+1)\alpha/2)}{\sin(\alpha/2)}\right)\tag{4}\\
&=\frac12\left(1+\frac{\sin((n+1)\alpha)+\sin(n\alpha)}{\sin(\alpha)}\right)\tag{5}\\
&=\frac12\left(1+\frac{\cos(n\alpha)-\cos((n+1)\alpha)}{1-\cos(\alpha)}\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$(2)$: sum of a geometric series
$(3)$: multiply left summand by $\frac{e^{-i\alpha/2}}{e^{-i\alpha/2}}$ and the right summand by $\frac{e^{i\alpha/2}}{e^{i\alpha/2}}$
$(4)$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(5)$: multiply fraction in $(4)$ by $\frac{2\cos(\alpha/2)}{2\cos(\alpha/2)}$
$(6)$: multiply fraction in $(4)$ by $\frac{2\sin(\alpha/2)}{2\sin(\alpha/2)}$
| {
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"url": "https://math.stackexchange.com/questions/1627376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$ Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$.
I begin by letting $n=1$ then $\frac{1}{2}<\frac{1}{\sqrt{3}}$. Then assume $\frac{1\cdot 3\cdots(2k-1)}{2\cdot 4 \cdots(2k)}<\frac{1}{\sqrt{2k+1}}$, for some $k\ge 1$. Then if I multiply both sides by $\frac{(2(k+1)-1)}{2(k+1)}$ I get $\frac{1\cdot 3\cdots(2(k+1)-1)}{2\cdot 4 \cdots2(k+1)}<\frac{1}{\sqrt{2k+1}}\cdot \frac{(2(k+1)-1)}{2(k+1)}$. So I have what I need on the left hand side, but I'm not sure how to continue on the right hand side. I realized that if I show that $\frac{(2(k+1)-1)}{2(k+1)}<\frac{\sqrt{2k+1}}{\sqrt{2(k+1)+1}}$ then I would be done. This however leads to the same dilemma in that I can easily make the left the way I need, but not the right. So there must be a trick that I am missing.
| Here is the inductive step: from
$\;\dfrac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\dfrac{1}{\sqrt{2n+1}}$, you deduce
$$\dfrac{1\cdot 3\cdots(2n-1)(2n+1)}{2\cdot 4 \cdots(2n)(2n+2)}<\dfrac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2},$$
hence it is enough to prove $\;\dfrac{1}{\sqrt{2n+1}}\dfrac{2n+1}{2n+2}<\dfrac1{\sqrt{2n+3}}$. This is equivalent to
$$\sqrt{(2n+1)(2n+3)}<2n+2,$$
which is the A.G.M. inequality (case of strict inequality).
| {
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"url": "https://math.stackexchange.com/questions/1627933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $\frac{1}{xy}$ convex for $x,y>0$ Is the function $$f(x,y) = \frac{1}{xy}$$
convex for $x,y>0$. I computed the hessian but it is very complicated and I do not know how to show it is positive semi definite.
| The Hessian is
$$\left( \begin{matrix} \frac{2}{x^3y} & \frac{1}{x^2y^2} \\ \frac{1}{x^2y^2} & \frac{2}{xy^3}\end{matrix} \right).$$
Now fix $x,y \ge 0$ , then we must show that for $a,b \in \mathbb{R}$ we have
$$2g(a,b)=\left( \begin{matrix} a& b \end{matrix} \right) \left( \begin{matrix} \frac{2}{x^3y} & \frac{1}{x^2y^2} \\ \frac{1}{x^2y^2} & \frac{2}{xy^3}\end{matrix} \right)\left( \begin{matrix}a \\ b \end{matrix} \right) = 2 \left( \frac{a^2}{x^3y}+\frac{b^2}{xy^3}+\frac{ab}{x^2y^2} \right) \ge 0.$$
Let $v=ax^{-\frac{3}{2}}y^{-\frac{1}{2}}$ and $w=bx^{-\frac{1}{2}}y^{-\frac{3}{2}}$, then $g(a,b) = v^2+w^2+vw$.
So we can wlog assume $v < 0 < w$, then $-v \le w$ gives $-vw \le w^2$ or $w^2+vw\le 0$. Likewise $-v > w$ gives $v^2 > -vw$ or $v^2+vw>0$.
So we have $g(a,b) \ge 0$ for all $a,b \in \mathbb{R}$.
Thus the Hessian is positive semidefinite, so $\frac{1}{xy}$ is convex.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find derivative of $\sin\sqrt{x}$ using difference quotient? The definition of derivative of a function $f(x)$ is $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$
Using this definition, the derivative of $\sin\sqrt{x}$ will be:
$$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$
$$\lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$
Now i got stuck. How to find the limit or simplify this expression?
I get intuition that we have to use $$\lim_{x\to0}\frac{\sin x}{x} = 1$$
but that too is leading no where. I am unable to remove h from denominator.
NOTE
I know the derivative of $\sin\sqrt{x}$ is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ using chain rule, but this exercise was given to us for practice using division quotient.
| You are right. First multiply and divide by the expression in $\sin()$
$$\lim_{h\rightarrow 0} \dfrac{2\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}\times\dfrac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}$$
Now, as $h \rightarrow 0, \ \ \ \left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)\rightarrow 0$
So we can use $\ \ \lim_{\lambda \rightarrow 0} \dfrac{\sin\lambda}{\lambda} = 1$
So it reduces to:
$$\lim_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\left(\sqrt{x+h}-\sqrt{x}\right)$$
Multiply and divide by $\left(\sqrt{x+h}+\sqrt{x}\right)$
$$\lim_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\dfrac{\left((\sqrt{x+h})^2-(\sqrt{x})^2\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$
$$ = \lim_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\dfrac{h}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$
Apply the limit and you get :
$$\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution to $\sqrt{x^2-2x+1} + 5 = 0$ I asked this question before, I wrote it wrong!
$$\sqrt{x^2-2x+1} + 5 = 0$$
My friends said the the solution could be :
$$|x-1| = -5$$
So the solution is nothing!
But I say the solution is:
$$x^2-2x+1 = 25 $$
so $$x = 6\ |\ x = -4$$
My Question here is which solution is right, and why.
Thanks in advance.
| Your friends are correct.
\begin{align*}
\sqrt{x^2 - 2x + 1} + 5 & = 0\\
\sqrt{x^2 - 2x + 1} & = -5\\
\sqrt{(x - 1)^2} & = -5\\
|x - 1| & = -5
\end{align*}
which has no solutions.
By convention, the expression $\sqrt{u}$ means the non-negative square root of $u$. Therefore, the equation
$$\sqrt{x^2 - 2x + 1} = -5$$
has no solution. When you squared both sides of the equation
$$\sqrt{x^2 - 2x + 1} = -5$$
to obtain
$$x^2 - 2x + 1 = 25$$
you introduced extraneous solutions since the equation $x^2 - 2x + 1 = 25$ has solutions that the equation $\sqrt{x^2 - 2x + 1} + 5 = 0$ does not.
| {
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"answer_count": 4,
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} |
How to prove $\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$ Let $a,b,c>0,a+b+c=1$, show that
$$\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}+\sqrt{\dfrac{ab}{c+ab}} \geq 1$$
I tried
$$\dfrac{bc}{a+bc}=\dfrac{bc}{a(a+b+c)+bc}=\dfrac{bc}{(a+b)(a+c)}$$
It suffice to show that
$$\sqrt{\dfrac{bc}{(a+b)(a+c)}}+\sqrt{\dfrac{ca}{(b+a)(b+c)}}+\sqrt{\dfrac{ab}{(c+a)(c+b)}}\geq 1$$
| As $0<a,b,c<1$, we can find $0<A,B,C<\pi/2$ such that $a=\sin^2A, b=\sin^2B$ and $c=\sin^2C$.
Then, we have $\sin^2A + \sin^2B + \sin^2C=1$.
If $A+B+C<\pi/2$, $\sin(A+B) < \sin(\pi/2 - C) = \cos(C)$. Squaring both sides, $\cos^2C > \sin^2A\cos^2B + \sin^2B\cos^2A + 2\sin A\sin B\cos A\cos B$ $= \sin^2A + \sin^2B -2\sin^2A\sin^2B + 2\sin A\sin B\cos A\cos B$
$= \sin^2A + \sin^2B + 2\sin A \sin B\cos (A+B)$
$\geq \sin^2A + \sin^2B$ as $A+B < A+B+C < \pi/2$ and so $\cos (A+B)>0$
Thus, we have $\sin^2A+\sin^2B+\sin^2C < 1$ contradicting the given condition. Hence, we must have $A+B+C \geq \pi/2$.
Proceeding with the inequality,
$$\sqrt{\frac{bc}{(1-b)(1-c)}}+\sqrt{\frac{ab}{(1-a)(1-b)}}+\sqrt{\frac{ca}{(1-c)(1-a)}}$$
$$=\tan A\tan B + \tan B\tan C + \tan C\tan A$$
$$=\frac{\sin A\sin B\cos C + \sin B\sin C\cos A + \sin C\sin A\cos B}{\cos A\cos B\cos C}$$
$$=\frac{\cos A\cos B\cos C-\cos(A+B+C)}{\cos A\cos B\cos C} = 1 - \frac{\cos(A+B+C)}{\cos A\cos B\cos C} \geq 1$$
because $A+B+C\geq \pi/2$ implies $\cos(A+B+C)\leq 0$. Thus the inequality stands proven.
| {
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Show that $c^2 \equiv -1 \pmod p$ Let $p \equiv 1 \pmod4$ be a prime. Write $p$ in the form $p=a^2+b^2$ where $a$ and $b$ are integers. Let $c \equiv ab^{-1} \pmod p$. Show that $c^2 \equiv -1 \pmod p$
$p = 4n+1$ where $n$ is an integer.
$c = ab^{-1} +pm$ where $m$ is an integer.
Could someone give me a hint?
| Since $c\equiv ab^{-1}$, we have $c^2\equiv a^2b^{-2}$. Also:
$$a^2+b^2=p \implies a^2 + b^2 \equiv 0 \pmod p$$
Now multiply by $b^{-2}$ to get:
$$a^2b^{-2} + 1 \equiv 0\pmod p\implies c\equiv -1\pmod p$$
| {
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} |
solving rectangle Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle.
So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=169$
\begin{align*}
d'^2 & =a'^2+b'^2\\
(d+7)^2 & =(a+4)^2+(b+7)^2\\
20^2 & = a^2+8a+16+b^2+14b+49\\
400 & = 8a+16+14b+49+169\\
\frac{166}{2} & = \frac{8a+14b}{2}\\
83 & =4a+7b
\end{align*}
...
That's what I've done. What should I do next?
| Let us rework the problem; this might make the answer easier to see. Let $a$ and $b$ be the side lengths of the rectangle, in which case $13 = \sqrt{a^{2} + b^{2}}$ is the diagonal. As you have already noted, we have $169 = a^{2} + b^{2}.$
When we enlarge the rectangle, we can use Pythagoras once again to get the following relation:
$$20 = \sqrt{(a + 4)^{2} + (b + 7)^{2}}.$$
This expands to
$$400 = (a + 4)^{2} + (b + 7)^{2},$$
which you have discovered to simplify into $4a + 7b = 83.$
Now, we begin the messy work. Rearranging the previous equation, $a = \frac{83 - 7b}{4}.$ We substitute this into $169 = a^{2} + b^{2}$ and solve as follows:
$$\left(\frac{83 - 7b}{4}\right)^{2} + b^{2} = 169$$
$$\frac{6889 - 1162b + 49b^{2}}{4} + b^{2} = 169,$$
which yields the quadratic form
$$\frac{53}{4}b^{2} - \frac{581}{2}b + \frac{6889}{4} = 0.$$
But the discriminant of the equation is less than $0,$ signaling no real solutions. Looking back at the problem, this makes sense. If you expand the width by $7$ units and the length by a non-zero amount, the diagonal should expand by more than $7$ units.
| {
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Looking for a non-combinatorial proof that $a! \cdot b! \mid (a+b)!$ (I use $a$ and $b$ to denote natural numbers.)
Question. Without appealing to the combinatorial interpretation of $$\frac{(a+b)!}{a! b!}$$ as a multinomial coefficients, is there a proof that for all $a$ and $b$, we have $$a! \cdot b! \mid (a+b)! \qquad?$$
Basically, I want a proof that just uses some clever algebra.
I was thinking that maybe we can use modular arithmetic, and try to understand the value of $(a+b)!$ modulo $a! \cdot b!$, and eventually show that this is $0$.
Ideas, anyone?
| With the settings
\begin{align*}
[k]_q:=\frac{1-q^k}{1-q}\qquad\text{and}\qquad [k]_q!:=\prod_{j=1}^{k}[j]_q=\prod_{j=1}^{k}\frac{1-q^j}{1-q}
\end{align*}
the q-binomial coefficient $\begin{bmatrix}a+b\\a\end{bmatrix}_q$ is defined as
\begin{align*}
\begin{bmatrix}a+b\\a\end{bmatrix}_q:=\frac{[a+b]_q!}{[a]_q![b]_q!}
\end{align*}
Since the following is valid
\begin{align*}
\begin{bmatrix}a+b\\a\end{bmatrix}_q&=\begin{bmatrix}a+b-1\\a\end{bmatrix}_q+q^{b}\begin{bmatrix}a+b-1\\a-1\end{bmatrix}_q\\
\begin{bmatrix}a\\0\end{bmatrix}_q&=\begin{bmatrix}a\\a\end{bmatrix}_q=1
\end{align*}
the $q$-binomial coefficients are polynomials in $q$ with non-negative integer coefficients.
Together with
\begin{align*}
\frac{(a+b)!}{a!b!}=\lim_{q\rightarrow 1}\begin{bmatrix}a+b\\a\end{bmatrix}_q
\end{align*}
the claim $a!b!|(a+b)!$ follows.
| {
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Find unit vector given Roll, Pitch and Yaw Is it possible to find the unit vector with:
Roll € [-90 (banked to right), 90 (banked to left)],
Pitch € [-90 (all the way down), 90 (all the way up)]
Yaw € [0, 360 (N)]
I calculated it without the Roll and it is
\begin{pmatrix}
cos(Pitch) sin(Yaw)\\
cos(Yaw) cos(Pitch)\\
sin(Pitch)
\end{pmatrix}.
How should it be with the Roll rotation and how can I get to this result?
My coordinate system is with +z up, +x right and +y forward
Many thanks!
| There are a lot of questions like this, all slightly different.
My earlier answer to a different question
is closely related to what you need, but the question is different enough
that I thought it better to write an answer a little more applicable
to your case.
I'll use right-handed $x,y,z$ Cartesian coordinates.
I visualize roll, pitch, and yaw using the motion of someone's right hand,
whose thumb and index finger are kept outstretched at right angles.
We start out with index finger of pointing in the direction
$(0,1,0)^T$ (positive $y$ axis)
and thumb pointing in the direction $(0,0,1)^T$ (positive $z$ axis).
We turn the entire hand by the angle $\theta$ (the "yaw" angle)
counterclockwise around the $z$ axis.
We then rotate the hand in the plane now occupied by the index finger
and thumb, turning it by the angle $\phi$ (the "pitch" angle)
with the index finger moving in the direction of the thumb.
At this point the angles $\phi$ and $\theta$ describe the
direction of the index finger in something like geographic coordinates,
where $\phi$ corresponds to latitude and $\theta$ corresponds to longitude.
(Note that these are not the kind of "mathematical" spherical coordinates
where $\phi$ is the angle of a vector relative to the $z$-axis.)
Finally, we rotate the hand using the direction of the index finger as
the axis of rotation, turning the thumb toward the palm of the hand
by an angle $\psi$ (the "roll" angle).
The effect of these motions on a vector "attached" to the hand
can be represented by a $3\times3$ matrix.
We can decompose this matrix into a product of three much simpler matrices.
Each of these three matrices will be a rotation around one of the
principal axes ($x$, $y$, or $z$ axes),
unlike the pitch and roll motions described above,
which were performed around axes relative to an already-rotated hand.
To reproduce the result of the roll, pitch, and yaw on the orientation
of the hand, using rotations only around the principal axes, we
have to do the rotations in reverse order. This is so that we can still
do pitch and roll around the correct axes relative to the hand,
but do them while those axes are still aligned with the principal axes.
We do the "roll" through angle $\psi$ first, using the rotation matrix
$$
R_y(\psi) =
\begin{pmatrix} \cos\psi & 0 & -\sin\psi \\
0 & 1 & 0 \\
\sin\psi & 0 & \cos\psi \end{pmatrix}
$$
(a rotation around the $y$-axis),
then the "pitch" through angle $\phi$, using the matrix
$$
R_x(\phi) =
\begin{pmatrix} 1 & 0 & 0 \\
0 & \cos\phi & -\sin\phi \\
0 & \sin\phi & \cos\phi \end{pmatrix}
$$
(a rotation around the $x$-axis),
and finally the "yaw" through angle $\theta$, using the matrix
$$
R_z(\theta) =
\begin{pmatrix} \cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0 \\
0 & 0 & 1 \end{pmatrix}
$$
(a rotation around the $z$-axis).
These are much like the axis-rotation matrices used in several other places
(including my earlier answer), but with a sequence of axes and signs
of the matrix entries suitable to your system.
The best way to understand how this works may be to try several examples,
using simple pitch, roll, and yaw angles such as $\pi/2$ or $\pi/4$,
and confirm that this sequence of rotations around fixed
principal coordinate axes has the same result as the desired sequence of
rotations around axes defined by the orientation of the hand.
This sequence of rotations is equivalent to the single rotation
performed by the matrix product $R_z(\theta)R_y(\phi)R_x(\psi)$.
For example, here's what this rotation does to the direction of
the index finger, $(0,1,0)^T$:
\begin{align}
R_z(\theta) R_x(\phi) R_y(\psi) \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
&= R_z(\theta) R_x(\phi)
\begin{pmatrix} \cos\psi & 0 & -\sin\psi \\
0 & 1 & 0 \\
\sin\psi & 0 & \cos\psi \end{pmatrix}
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \\
&= R_z(\theta) R_x(\phi) \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \\
&= R_z(\theta)
\begin{pmatrix} 1 & 0 & 0 \\
0 & \cos\phi & -\sin\phi \\
0 & \sin\phi & \cos\phi \end{pmatrix}
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \\
&= R_z(\theta) \begin{pmatrix} 0 \\ \cos\phi \\ \sin\phi \end{pmatrix} \\
&= \begin{pmatrix} \cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0 \\
0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 0 \\ \cos\phi \\ \sin\phi \end{pmatrix} \\
&= \begin{pmatrix} \cos\phi \sin\theta \\
\cos\phi \cos\theta \\
\sin\phi \end{pmatrix}, \\
\end{align}
which agrees with your result.
What this same rotation does to the direction of
the thumb, $(0,0,1)^T$, is
\begin{align}
R_z(\theta) R_x(\phi) R_y(\psi) \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
&= R_z(\theta) R_x(\phi)
\begin{pmatrix} \cos\psi & 0 & -\sin\psi \\
0 & 1 & 0 \\
\sin\psi & 0 & \cos\psi \end{pmatrix}
\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \\
&= R_z(\theta) R_x(\phi)
\begin{pmatrix} -\sin\psi \\ 0 \\ \cos\psi \end{pmatrix} \\
&= R_z(\theta)
\begin{pmatrix} 1 & 0 & 0 \\
0 & \cos\phi & -\sin\phi \\
0 & \sin\phi & \cos\phi \end{pmatrix}
\begin{pmatrix} -\sin\psi \\ 0 \\ \cos\psi \end{pmatrix} \\
&= R_z(\theta) \begin{pmatrix} -\sin\psi \\
-\cos\psi \sin\phi \\
\cos\psi \cos\phi \end{pmatrix} \\
&= \begin{pmatrix} \cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0 \\
0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} -\sin\psi \\
-\cos\psi \sin\phi \\
\cos\psi \cos\phi \end{pmatrix} \\
&= \begin{pmatrix} -\sin\psi \cos\theta - \cos\psi \sin\phi \sin\theta \\
\sin\psi \sin\theta - \cos\psi \sin\phi \cos\theta \\
\cos\psi \cos\phi \end{pmatrix} \\
\end{align}
if I haven't dropped a sign or made some other arithmetic error.
| {
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"answer_count": 1,
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How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the second part: due a change of variable $u=x+1$, I find
$$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$
| Here is one approach to the second integral:
$$\int \frac{x + 1}{ x^2 + x + 1} dx = \frac12 \left(\int \frac{2x + 1}{ x^2 + x + 1} dx + \int \frac{1}{x^2 + x + 1} dx\right)$$
The first integral is identical to the one you integrated. The second is $\arctan$, which can be seen after you complete the square.
$$\int \frac{1}{x^2 + x + 1} dx = \int \frac{1}{x^2 + x + (1/2)^2 - (1/2)^2 + 1} dx$$
$$\int \frac{1}{(x+1/2)^2 + 3/4} dx = \frac{4}{3} \int \frac{1}{\left( \frac{2x+1}{\sqrt 3}\right) + 1} dx = \frac{2}{\sqrt{3}} \int \frac{1}{u^2 + 1} du = \frac{2}{\sqrt{3}} \arctan \left( \frac{2x+1}{\sqrt{3}}\right) + C$$
Where we used the substitution $u = \frac{2x+1}{\sqrt{3}}$.
| {
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"answer_count": 5,
"answer_id": 2
} |
Prove that $16\cos^5A-20\cos^3A+5\cos A=\cos5A$
Prove that
$$16\cos^5A-20\cos^3A+5\cos A=\cos5A$$
My solution begins here;
$$
\begin{align}
\text{RHS} & =\cos5A \\
& =\cos(A+4A) \\
& =\cos A\cos4A-\sin A\sin4A \\
& =\cos A(2\cos^2 2A-1)-\sin A(2\sin2A\cos2A) \\
& =2\cos A\cos^2 2A-\cos A-2\sin A\sin2A\cos2A \\
& =2\cos A\cos^2 2A-\cos A-2\sin A(2\sin A\cos A)\cos2A
\end{align}
$$
Now how do I move on?
| Now use double angle formulae to write $\cos 2A$ in terms of $\cos A$, then turn any $\sin^2 A$ into $1-\cos^2 A$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$ How do I prove the following inequality
$$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$
without the help of induction?
Thanks for any help!!
| You don't need calculus here (although it would make things easier). Let $$S=\frac{1}{n+1}+\frac {1}{2n}+\frac{1}{n+3}+...+\frac{1}{2n}$$
and write it both ways, à la Gauss:
$$\begin{align}
2S &= \frac{1}{n+1}+\frac {1}{n+2}\,\,\,+\,\,\frac{1}{n+3}\,+...+\,\,\,\frac{1}{2n} \\
&\,\,\,\,+\frac{1}{2n}\,\,\;+\frac {1}{2n-1}+\frac{1}{2n-2}+...+\frac{1}{n+1}\\
&=\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-(k-1)}\right)\\
&<\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-k}\right)\\
&=\sum_{k=1}^n\frac{3n}{(n+k)(2n-k)}\\
&\le\sum_{k=1}^n\frac{3n}{2n^2} \,\,\,\,\text{because}\, (n+k)(2n-k) \ge 2n^2 \text{ if } 1 \le k \le n\\
&=\frac{3}{2n}\sum_{k=1}^n1\\
&=\frac32
\end{align}$$
Hence $S < \dfrac{3}{4}$.
| {
"language": "en",
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"source": "stackexchange",
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Conditional probability with dependent events We have 2 dice. One is fair. The other one lands by the following probabilities:
6: 1/2
5: 1/10
4: 1/10
3: 1/10
2: 1/10
1: 1/10
We roll both dice.
What is the probability that the sum is 9 given that exactly one die lands on 6?
Here's what I have so far:
Cases where sum is 9:
Case1: Loaded = 6 | Fair = 3 -> 1/2 * 1/6
Case2: Loaded = 3 | Fair = 6 -> 1/10 * 1/6
Case3: Loaded = 5 | Fair = 4 -> 1/10 * 1/6
Case4: Loaded = 4 | Fair = 5 -> 1/10 * 1/6
------------------------------------
P(sum is 9) = 2/15
P(exactly one die lands on 6) = P(loaded lands on 6) + P(fair lands on 6) - P(both land on 6)
P(exactly one die lands on 6) = 1/2 + 1/6 - (1/2 * 1/6) = 7/12
P(sum is 9 AND exactly one die lands on 6)
= (P(case1) + P(case2)) / P(sum is 9)
= ((1/2 * 1/6) + (1/10 * 1/6)) / (2/15)
= 3/4
P(sum is 9 | exactly one die lands on 6)
= P(sum is 9 AND exactly one die lands on 6) / P(exactly one die lands on 6)
= (3/4) / (7/12)
> 1
What am I doing wrong?
| In short, you erroneously divided by $\Pr\{\text{sum is 9}\}$ when calculating $\Pr\{\text{sum is 9 and exactly one die lands on 6}\}$.
$$
\begin{align*}
\Pr\{\text{sum is 9 and exactly one die lands on 6}\}
& = \Pr\{(6,3),(3,6)\} \\
& = \Pr\{(6,3)\} + \Pr\{(3,6)\} \\
& = \left(\frac{1}{2}\right)\left(\frac{1}{6}\right) + \left(\frac{1}{10}\right)\left(\frac{1}{6}\right) \\
& = \frac{1}{10}
\end{align*}
$$
| {
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Solving $\int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta $ I have the following integration to solve.
$$f(k) = \int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta,\quad0<k<1$$
assuming $\sin\theta = t$ which results $d\theta = \frac{dt}{\sqrt{1-t^2}}$
and when $\theta = 0, t=0$ and $\theta=\frac{\pi}{2},t=1$
so above equation can be rewritten as,
$$f(k) = \int_0^1{t^2\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}}dt$$
I'm stuck in solving this further.
Can somebody help me with some clues/solution to solve this further.
| Possible hints to perfume some kind of calculations.
Leaving apart the extrema of the integra, for the moment.
$$\int\sin^2\theta \sqrt{1 - k^2 \sin^2\theta}\ \text{d}\theta$$
Using the substitution
$$k\sin\theta = \cos\phi ~~~~~~~ \sin\theta = \frac{\cos\phi}{k} ~~~ \to ~~~ \sin^2\theta = \frac{\cos^2\phi}{k^2}$$
$$\phi = \arccos(k\sin\theta)$$
$$\text{d}\phi = \frac{- k\cos\theta}{\sqrt{1 - k^2\sin^2\theta}}\ \text{d}\theta = -\frac{k\sqrt{1 - \sin^2\theta}}{\sqrt{1 - \cos^2\phi}}\ \text{d}\theta = -\frac{k\sqrt{1 - \frac{\cos^2\phi}{k^2}}}{\sin^2\phi}\ \ \text{d}\theta$$
Thence $$\text{d}\theta = -\frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\ \text{d}\phi$$ and the integral becomes
$$- \int \frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\left(\frac{\cos^2\phi}{k^2}\right)\sin^2\phi \text{d}\phi = -\frac{1}{k^2}\int \frac{\sin^4\phi\cos^2\phi}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi$$
Now we can use the trigonometric reduction formula for the numerator of the integrand:
$$\sin^4\phi\cos^2\phi = \frac{1}{32}(2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi))$$
to get
$$
-\frac{1}{32k^2}\int\ \frac{2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi)}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi
$$
Which might be splitter into four parts, and then.. who knows!
The solution, however, lies into Jacobi Elliptic Functions of the First an Second Kind.
More on Jacobi Elliptic Integrals
https://en.wikipedia.org/wiki/Elliptic_integral
| {
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"answer_id": 1
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Find $\int\limits^{\infty}_{0}\int\limits^{\infty}_{0}{\frac{1}{(x+y)^{3/2}}\exp\left\{-\frac{a^2}{2(x+y)}\right\}}\,dy\,dx$. In my posterior probability computation, I got the following integration and I could not figure it out.
$$\int\limits^{\infty}_{0}\int\limits^{\infty}_{0}{\frac{1}{(x+y)^{3/2}}\exp\left\{-\frac{a^2}{2(x+y)}\right\}}\,dy\,dx$$ where $a$ is a positive constant
Any help is appreciated.
| I also think that this integral diverges.
$$\int\frac{e^{-\frac{a^2}{2 (x+y)}}}{(x+y)^{3/2}}\,dy=-\frac{\sqrt{2 \pi } }{a}\text{erf}\left(\frac{a}{\sqrt{2} \sqrt{x+y}}\right)$$ $$\int_0^\infty\frac{e^{-\frac{a^2}{2 (x+y)}}}{(x+y)^{3/2}}\,dy=\frac{\sqrt{2 \pi } }{a}\text{erf}\left(\frac{a}{\sqrt{2x} }\right)$$ provided $\Re(x)>0\lor x\notin \mathbb{R}$ Now, consider $$\int\text{erf}\left(\frac{1}{\sqrt{t} }\right)\,dt=(t+2) \text{erf}\left(\frac{1}{\sqrt{t}}\right)+\frac{2 e^{-1/t} \sqrt{t}}{\sqrt{\pi
}}$$ $$I=\int_0^b\text{erf}\left(\frac{1}{\sqrt{t} }\right)\,dt=(b+2) \text{erf}\left(\frac{1}{\sqrt{b}}\right)+\frac{2 \sqrt{b} e^{-1/b}}{\sqrt{\pi
}}-2$$ Now, for large values of $b$, $$I=-2+\frac{4 \sqrt{b}}{\sqrt{\pi }}+\frac{4 }{3 \sqrt{\pi b
}}+O\left(\frac{1}{b^{3/2}}\right)$$ which would then make $$\frac{\sqrt{2 \pi } }{a}\int_0^b\text{erf}\left(\frac{a}{\sqrt{2x} }\right)=-\sqrt{2 \pi } a+4 \sqrt{b}+\frac{2}{3} a^2
\sqrt{\frac{1}{b}}+O\left(\frac{1}{b^{3/2}}\right)$$
| {
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How to show that $f$ is a straight line if $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$? Let $f:\mathbb R\to\mathbb R$ be continuous such that $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}~\forall~x,y\in\mathbb R.$ How to show that $f$ is a straight line?
| Define a linear function $g \colon \mathbb{R} \rightarrow \mathbb{R}$ by $g(x) = (f(1) - f(0))x + f(0)$. Note that by definition, $g(0) = f(0)$ and $g(1) = f(1)$ and that $g$ also satisfies
$$ g \left( \frac{x + y}{2} \right) = \frac{g(x) + g(y)}{2}. $$
You want to show that $g(x) = f(x)$ for all $x \in \mathbb{R}$. We have
$$ f \left( \frac{x + 0}{2} \right) = \frac{f(x) + f(0)}{2}. $$
Plugging in $x = 1$, we see that
$$ f \left( \frac{1}{2} \right) = \frac{f(1) + f(0)}{2} = \frac{g(1) + g(0)}{2} = g \left( \frac{1}{2} \right) $$
(where we used $f(0) = g(0)$ and $f(1) = g(1)$). Plugging in $x = \frac{1}{2}$ will give you
$$ f \left( \frac{1}{4} \right) = \frac{f \left( \frac{1}{2} \right) + f(0)}{2} = \frac{g \left( \frac{1}{2} \right) + g(0)}{2} = g \left( \frac{1}{4} \right)$$
(where we used $f(0) = g(0)$ and $f \left( \frac{1}{2} \right) = g \left( \frac{1}{2} \right)$). Iterating this inductively, you can show that
$$ f \left( \frac{1}{2^n} \right) = g \left( \frac{1}{2^n} \right) $$
for all $n \in \mathbb{N}$. By a similar inductive argument, you can then show that
$$ f \left( \frac{m}{2^n} \right) = g \left( \frac{m}{2^n} \right) $$
for all $n \in \mathbb{N}$ and $m \in \mathbb{Z}$. Since $f$ and $g$ are two continuous functions that agree on the dense set $ \{ \frac{m}{2^n} \, | \, n \in \mathbb{N}, m \in \mathbb{Z} \}$ of the dyadic rationals, they must agree everywhere and so $f(x) = g(x)$ for all $x \in \mathbb{R}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ by Mathematical Induction Prove by Mathematical Induction:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$$
Now by inductive hypothesis:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+k(k+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+k^2(k+1)}=\frac{3k+5}{3k+1}$$
But verification for $n=k+1$ is creating problem.because $k$ is present in denominator too which is hindering use of equation obtained from inductive hypothesis. Please suggest something.
| $$1\cdot 2^2 + 2\cdot 3^2 +\cdots+ n\cdot (n+1)^2 =2^3 -2^2 +3^3 - 3^2 +\cdots+(n+1)^3 -(n+1)^2 =\left(\frac{(n+2)(n+1)}{2}\right)^2 -\frac{(n+1)(n+2)(2n+3)}{6} =(n+1)(n+2)\left[\frac{n^2 +3n +2}{4} -\frac{2n+3}{6}\right]= (n+1)(n+2)\left[\frac{3n^2 +5n }{12}\right]=\frac{1}{12} n (n+1) (n+2) (3n+5)$$
$$1^2\cdot 2 + 2^2\cdot 3 +\cdots+ n^2\cdot (n+1) =1^3 +2^3 +3^3 +\cdots+n^3 +1^2 +2^2 +\cdots+n^2 =\left(\frac{n(n+1)}{2}\right)^2 +\frac{n(n+1)(2n+1)}{6} =n(n+1)\left[\frac{n^2 +n}{4} +\frac{2n+1}{6}\right]= n(n+1)\left[\frac{3n^2 +7n +2 }{12}\right]=\frac{1}{12} n (n+1) (n+2) (3n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Common tangent line to two functions I have two functions:
$$f(x) = x^2 + 3$$
$$g(x) = -x^2 - 2x - 2$$
This two functions have a common tangent line that its slope is positive.
My approach:
$$f'(x) = 2x$$
$$g'(x) = -2x -2$$
I mark the two tangent points $x=a$ in $f(x)$ and $x = b$ in $g(x)$
$$(a, a^2 + 3)\qquad(b, -b^2 - 2b - 2)$$
I place the chosen points in their respective derivatives and equal:
$$2a = -2b - 2$$
$$a + b = -1$$
For trying to find the slope I do this:
\begin{align}
m = \frac{\Delta y}{\Delta x} & = \frac{a^2 + 3 + b^2 + 2b + 2}{a - b}\\
& = \frac{a^2 + b^2 + 2b + 5}{a - b}\\
\end{align}
I'm stuck here. What can I do to resolve my problem?
| You are almost there.
You have $2a=-2b-2$
You also know that
$$m = \frac{a^2 + 3 + b^2 + 2b + 2}{a - b}
= \frac{a^2 + b^2 + 2b + 5}{a - b}=2a$$
You have two equations in two unknowns. Solve for $a$ and $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results.
| Starting with
$\cos(\dfrac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$
this also means starting with the triangle of trig ratios drawn and Pythagoras theorem:
$\tan(\dfrac{\arctan(\frac{11}{2})}{3}) = \frac{1}{2} = t ,$
Now use the $ \tan 3 \theta = \dfrac{3 t - t^3}{1-3 t^2} \rightarrow \dfrac{11}{2} $ triple angle formula and simplify, done!
| {
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"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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euclidean algorithm iteration problem So if I have a congruence of the form $ad \equiv 1 \mod m$, where $a$, $d$, $m$ are all integers and $m > a$, I should be able to find the integer $d$ satisfying the congruence using the Euclidean algorithm.
For example, $7d \equiv 1 \mod 30$ inserted into the Euclidean algorithm gives
\begin{align*}
30 & = 4 \cdot 7 + 2\\
7 & = 3 \cdot 2 + 1
\end{align*}
Working backwards gives
\begin{align*}
1 & = 7 - 3 \cdot 2\\
& = 7 - 3(30 - 4 \cdot 7)\\
& = 13 \cdot 7 + (-3) \cdot 30
\end{align*}
[easily seen through some rearranging]
And so obviously $d = 13$, as $13 \cdot 7 = 1 \mod 30$ as desired.
I'm trying to execute this for $3d = 1 \mod 100$ But I always end up like this
\begin{align*}
A & = qB + r\\
100 & = 33 \cdot 3 + 1\\
1 & = 3 \cdot (-33) + 100 \cdot 1
\end{align*}
$d$ should turn out to be $67$, but I can't get to it. Help is appreciated
| Your answer is correct, since
$$-33\equiv 67\mod 100$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int{ \left( \frac{1-x}{1+x} \right)^\frac{3}{2}dx}$ Integrate $$\int{ \left(\frac{1-x}{1+x} \right)^\frac{3}{2}dx}$$
I guess that there is sub $x = \cos t$ so integral gets to $$\int{ \left(\tan \frac{t}{2} \right)^3 d\cos t}$$ then I used that $\sin t = \frac{\tan \frac{t}{2}}{(\tan \frac{t}{2})^2 + 1}$ and got $$-2\int {\frac{(sin \frac{t}{2})^4}{(\cos \frac{t}{2})^2}dt}$$but then I stuck with transformations. Please, help.
| HINT:
$$\int\left(\frac{1-x}{1+x}\right)^{\frac{3}{2}}\space\text{d}x=$$
Substitute $u=\frac{1-x}{1+x}$ and $\text{d}u=\left(-\frac{1-x}{(1+x)^2}-\frac{1}{x+1}\right)\space\text{d}x$:
$$-2\int\frac{u^{\frac{3}{2}}}{(-u-1)^2}\space\text{d}u=$$
Substitute $s=\sqrt{u}$ and $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$:
$$-4\int\frac{s^4}{(-s^2-1)^2}\space\text{d}s=-4\int\frac{s^4}{(s^2+1)^2}\space\text{d}s=-4\int\left[1+\frac{1}{(s^2+1)^2}-\frac{2}{s^2+1}\right]\space\text{d}s$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many numbers between $0$ and $999,999$ are there whose digits sum to $r$ How many numbers between $0$ and $999,999$ are there whose digits sum to $r$
In generating a function for the answer, here is what I came to.
We have a maximum of 6 number slots to use to sum to r.
Then $e_1 + e_2 + e_3 + e_4 + e_5 + e_6 = r$
And since each slot has the capacity to be a number between $0$ and $9$, then $e_i = (1 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9) $,
and the generating function is therefore $(1 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9)^6$.
Do I have this generating function correct?
| Note: Your derivation of the generating function is perfectly valid.
The generating function has following representation
\begin{align*}
(1+x^1+x^2+\cdots+x^9)^6&=\left(\sum_{j=0}^9x^j\right)^6\\
&=\left(\frac{1-x^{10}}{1-x}\right)^6\\
&=1+6x+21x^2+56x^3+126x^4+252x^5+\cdots
\end{align*}
The exponents of the terms vary from $0$ which corresponds to the smallest sum of digits $r=0$ and the number $0$ up to $54$ which corresponds to the largest sum of digits $r=54$ and the number $999999$.
We calculate the coefficient of $x^r$ of the generating series. In order to do so we use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain for $0\leq r\leq 54$
\begin{align*}
[x^r]\left(\frac{1-x^{10}}{1-x}\right)^6
&=[x^r](1-x^{10})^6\sum_{k=0}^{\infty}\binom{-6}{k}(-x)^k\tag{1}\\
&=[x^r]\sum_{j=0}^{6}\binom{6}{j}(-1)^jx^{10j}\sum_{k=0}^{\infty}\binom{k+5}{5}x^k\tag{2}\\
&=\sum_{j=0}^{\lfloor\frac{r}{10}\rfloor}\binom{6}{j}(-1)^j[x^{r-10j}]\sum_{k=0}^{\infty}\binom{k+5}{5}x^k\tag{3}\\
&=\sum_{j=0}^{\lfloor\frac{r}{10}\rfloor}(-1)^j\binom{6}{j}\binom{r-10j+5}{5}
\end{align*}
Comment:
*
*In (1) we use the binomial series representation
*In (2) we expand the left binom and use the identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$
*In (3) we use the rule $[x^{r+s}]A(x)=[x^r]x^{-s}A(x)$ and keep care of the upper limit of the sum to not add anything when $r-10j<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to check Cohen-Macaulayness?
Let $R=k[x,y,z]$. Consider the ideal $I=(x^2z^2,xyz,y^2z^4,y^4z^3,x^3y^5,x^4y^3)$. Is $R/I$ Cohen-Macaulay ?
By definition it seems tough to solve this problem. Is there any other way to check this?
| You can take a primary decomposition of the ideal $I$. For monomials $x,y$ and monomial ideal $J$,
$(x + J) \cap (y + J) = (xy) + J$
provided that $\gcd(x,y) \sim 1$.
Then
\begin{align}
I&= (x^2z^2,xyz,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\
&= (xyz,x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\
&= (x,x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \cap (y, x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\
&\qquad \cap (z, x^2z^2,y^2z^4,y^4z^3,x^3y^5,x^4y^3) \\
&= (x,y^2z^4,y^4z^3) \cap (y, x^2z^2) \cap (z,x^3y^5,x^4y^3) \\
&= [ (x, y^2) \cap (x,z^4,y^4z^3)] \cap [(y,x^2) \cap (y,z^2)] \cap [(z,x^3) \cap (z, y^5,x^4y^3)] \\
&= [ (x, y^2) \cap (x,z^4,y^4) \cap (x,z^3)] \cap [(y,x^2) \cap (y,z^2)] \cap [(z,x^3) \cap (z, y^5,x^4) \cap (z,y^3)] \\
&= (x, y^2) \cap (x,z^3) \cap (y,x^2) \cap (y,z^2) \cap (z,x^3) \cap (z,y^3) \cap (z, y^5,x^4) \cap (x,z^4,y^4)
\end{align}
Since $(x,y,z)$ is associated to $I$, $\operatorname{depth} R/I = 0$. But $\dim R/I = 1$; hence $R/I$ is not Cohen-Macaulay. In other words, $I$ is a mixed ideal.
| {
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"timestamp": "2023-03-29T00:00:00",
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For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
For all $x$ which are real numbers, prove that
$$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$
I know that
Let $\lfloor x\rfloor = n$
$n \leq x < n+1$
| Let $ x = i + d $ where $ 0 \le d < 1 $
$ \lfloor 2x \rfloor = \lfloor 2i + 2d \rfloor = 2i + \lfloor 2d \rfloor $
$ \lfloor x \rfloor + \lfloor x + 0.5 \rfloor = 2i + \lfloor d \rfloor + \lfloor d + 0.5 \rfloor $
So we reduced the problem to just show for $ d $
Suppose $ 0 \le d < 0.5 $, then both sides are just $ 2d $
Suppose $ 0.5 \le d < 1 $, then both sides are just $ d + 1 $.
QED $ \blacksquare $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{4x-6}{3})=1-\frac{1}{x^{2}}(\frac{x^{2}}{3})=\frac{2}{3}$.
But according to the book the right answer is option E, i.e, $2$.
Is the book right or me ?
| By solving the quadratic equation,$x = 2+\sqrt 2 *i$ or $x = 2-\sqrt 2 *i$ where $i$ is a unit of image number.
*
*When $x = 2+\sqrt 2 *i$, $1-\frac{4}{3x}+\frac{2}{x^{2}}$ = $1-\frac{4}{3*(2+\sqrt 2 *i)}+\frac{2}{(2+\sqrt 2 *i)^{2}} = 1-\frac{4}{9}+\frac{2*\sqrt 2}{9}*i + \frac{1}{9} - \frac{2*\sqrt 2}{9}*i = \frac{2}{3}$ w
*When $x = 2-\sqrt 2 *i$, $1-\frac{4}{3x}+\frac{2}{x^{2}} = \frac{2}{3}$
reference:
*
*Solving quadratic equation: http://mathworld.wolfram.com/QuadraticEquation.html
*Complex number :http://mathworld.wolfram.com/ComplexNumber.html
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$
$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_{2}^{4} \frac{1-\sin^2u}{\sin u}du=\int_{2}^{4} \frac{1}{\sin u}du-\int_{2}^{4} {\sin u}du$$
$$=\ln\left(\tan\left(\frac{u}{2}\right)\right)+\cos u$$
$\frac{x}{4}=\sin u$
$\tan u=\frac{\sqrt{x}}{x^2-16}$
$\ln\left(\frac{{x}}{8(\sqrt{x^2-16})}\right)+\left(\frac{\sqrt{16-x^2}}{4}\right)$ form $2$ to $4$
but I get a $\frac{4}{0}$ the end result according to Wolfram is $1.80$
| I think u forgot to change the limits and also factor of 4 is missing while putting dx term.
Edited answer :
Sorry the previous answer I made a slight mistake in taking limits
So our final answer will look like as seen in the photo above.
As you can see that we get
$tan(\pi/12)$ in the answer.
By placing value of $tan(\pi/12)$ in different form. Look of the answer can be changed.
So now this is the correct answer. Apologies for slight mistake in limits in the previous answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integers $k$ such that $5^k\equiv 97 \pmod{101}$ Find integers $k$ such that $5^k\equiv 97 \pmod{101}$.
By brutal force, If $k=23$ then $101\vert 5^{23}-97$. Furthermore, by Euler-Fermat theorem, since $gcd(5,101)$ we have, $5^{100}\equiv 1\pmod{101}$, then for integer $r$, $5^{100r}\equiv 1\pmod{101}$, so $5^{100r+23}\equiv 97\pmod{101}$, but I'm not sure if this is true, any other idea of how find all intergers k. Thanks!
| $97 \equiv -4 \mod 101$. $-4 * 25 = -100 \equiv 1 \mod 101$.
So if $5^k \equiv 1 \mod 101$ then $5^{k -2} \equiv 97 \mod 101$.
Now $5^{100} \equiv 1 \mod 101$ as 101 is prime so $\phi (101) = 100$ so $5^{\phi(101)} \equiv 1 \mod 101$.
$5^{100} \equiv 1 \mod 101$. So try $5^{50}$ and $5^{25}$ and we see $5^{25} \equiv 1 \mod 101$. If there is any smaller power so that $5^n \equiv 1 \mod 101$ it would be $5$. But $5^5 \equiv 95 \equiv -6 \mod 101$.
So $5^{25} \equiv 1 \mod 101$ so $5^{25-2} = 5^{23} \equiv 97 \mod 101$.
so $k = 25r + 23$ are all the integers such that $5^k \equiv 97 \mod 101$.
| {
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Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$ $$
\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots
$$
up to $n$ terms. I need help in solving this sum. I tried finding the coefficients of terms after splitting the terms..: it becomes
$$(\frac{1}{1\cdot 6}-\frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}-\frac{1}{6\cdot4}) + (\frac{1}{6\cdot 2} - \frac{1}{3\cdot2} +\frac{1}{4\cdot 2} -\frac{1}{6\cdot5})+\cdots.$$ I tried solving it but am getting nowhere .Someone please help me with this sum.
| Hint: Use partial fractions:
$$
\frac1{(n-3)(n-2)(n-1)n}=-\frac1{2(n-2)}+\frac1{2(n-1)}-\frac1{6n}+\frac1{6(n-3)}$$
and note that it telescopes, so that you can find the partial sums.
| {
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"url": "https://math.stackexchange.com/questions/1662784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is there a mathematical way to solve this problem? The question goes something like this:
How many different ways can you add up 2, 3, and 5 to get a sum of 12. Numbers can be repeated, and all three numbers do not have to be used for each solution. For instance, 5 + 2 + 2 is an answer, and so is 2 + 5 + 2.
The answer is 5 different ways:
*
*5 + 5 + 2
*2 + 2 + 2 + 2 + 2 + 2
*3 + 3 + 3 + 3
*3 + 3 + 2 + 2 + 2
*5 + 3 + 2 + 2
Is there some equation or formula that can be used to solve this problem, or does it have to be done manually?
| Assuming that order of numbers doesn't matter, i.e. $5+5+2$ is considered the same as $5+2+5$ is considered the same as $2+5+5$ (as suggested by your list of ways in which $12$ can be written):
One way of approaching this is via Generating Functions. A rather in-depth treatment of the topic as a whole is given in the book Generatingfunctionology by Herbert Wilf and is available legally for download free at the linked site.
Here, we use the generating function $(1+x^2+x^4+x^6+x^8+\dots)(1+x^3+x^6+x^9+x^{12}+\dots)(1+x^5+x^{10}+x^{15}+\dots) \\= \frac{1}{1-x^2}\cdot \frac{1}{1-x^3}\cdot \frac{1}{1-x^5}$
We pick this generating function so that the term picked from the first parenthesis represents the number of times we used a $2$ in our sum, the term picked from the second parenthesis represents the number of times we used $3$ in our sum, etc...
Although each summation in the parentheses are technically infinite series, we may choose to cut them off at an arbitrary finite point after the desired total in order to make calculations possible. Once expanding, the coefficient of $x^n$ will be the number of different ways to create a sum of $n$ using the available numbers.
Expanding, $$\frac{1}{1-x^2}\cdot \frac{1}{1-x^3}\cdot \frac{1}{1-x^5} = 1+x^2+x^3+x^4+2x^5+2x^6+2x^7+\dots+4x^{11}+5x^{12}+5x^{13}+\dots\\+18x^{28}+19x^{29}+21x^{30}+\dots$$
This shows it is impossible to get a sum of $1$ with the above numbers, there are $5$ ways to write $12$ as a sum of $2$'s, $3$'s, and $5$'s and similarly there are $21$ ways to write $30$ as a sum of $2$'s, $3$'s, and $5$'s.
For the related question of in how many ways a number can be written as a sum where order does matter, your example is incorrect as that would imply that $5+5+2$ is considered different than $5+2+5$ which is also different than $2+5+5$, which would give you a total of $3+1+1+\binom{5}{2}+\binom{4}{1,1,2} = 27$ different ways to write $12$ as a sum of twos, threes, and fives.
This is, as mentioned in the comments above, a different problem and is much more difficult.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Laurent series of $\log(z)\sin(1/(z-1))$ I would like to calculate the Laurent series of $$\log(z)\sin \left(\frac{1}{z-1} \right)$$ I have developed separately $\log z$ et the sinus and tried to multiplied them terms by terms but it is complicated. Is there another simplier way?
| Hints and a bit of example
Use the substitution
$$z - 1 = u$$
so that
$$\sin\left(\frac{1}{z-1}\right) = \sin\left(\frac{1}{u}\right) = \frac{1}{u} - \frac{1}{3!u^3} + \cdot$$
and
$$\log(z) = \log(u+1) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdot $$
In this way you get simply
$$\left(\frac{1}{u} - \frac{1}{3!u^3}\right)\cdot \left(u - \frac{u^2}{2} + \frac{u^3}{3}\right) = -\frac{1}{6u^2} + \frac{1}{12u} + \frac{17}{18} - \frac{u}{2} + \frac{u^2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute this integral... I need to compute $$ \int_0^1 \frac{\arctan x}{(1+x)^2}dx$$
I tried substituting $t= \arctan x$ (check please), getting $$ \int \frac{t}{1+2\sin t \cos t}dt $$ but I still can't handle it. Any help would be useful: I haven't done integrals for years. Thanks
| By parts $f=\arctan x$ and $g=-\frac{1}{x+1}$
$$-\frac{\arctan x}{x+1}+\int\frac{dx}{(x+1)(x^2+1)}$$
Now partial fractions
$$-\frac{\arctan x}{x+1}+\frac 1 2\int \frac{1-x}{x^2+1}dx+\frac 1 2\int\frac{dx}{x+1}\\
=-\frac{\arctan x}{x+1}-\frac 1 2\int \frac{x}{x^2+1}dx+\frac 1 2\int\frac{dx}{x^2+1}+\frac 1 2 \int \frac {dx}{x+1}\\
=\frac{\ln(x+1)}{2}-\frac{\ln(x^2+1)}{4}-\frac{\arctan x}{x+1}+\frac {\arctan x}{2}+\mathcal C$$
$$\left(\frac{\ln(x+1)}{2}-\frac{\ln(x^2+1)}{4}-\frac{\arctan x}{x+1}+\frac {\arctan x}{2}\right)\bigg|_0^1=\frac{\ln(2)}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \times 1 + 8 \times 2) +1, (8 \times 1 + 8 \times 2 + 8 \times 3 + 8 \times 4) +1 ....$
equals to
$1^2, 3^2, 5^2, 7^2 ...$
and my brain stuck here and cannot find out the relationship between two series.
Can someone give me a hint why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$
| $(n+1)^2 = n^2+(2n+1) $ so by induction $n^2= \sum 2k-1$
And for just the odd squares. $(2n+1)^2= 1 + \sum_{k=2;+2}^{2n} [(2k - 1)+(2k+1) ]= 1 +4\sum_{j=1}^{n }2j=1+8\sum j$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I get a wrong determinant - why? I'm trying to calculate the following determinant:
$$\begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & x & a_2 & \dots & a_n \\
a_0 & a_1 & x & \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
a_0 & a_1 & a_2 & \dots & x
\end{vmatrix} = $$ $$ = \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & a_1 & a_2 & \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
a_0 & a_1 & a_2 & \dots & a_n
\end{vmatrix} + \begin{vmatrix}
0 & 0 & 0 & \dots & 0 \\
0 & x - a_1 & 0 & \dots & 0 \\
0 & 0 & x - a_2 & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & x-a_n
\end{vmatrix} = 0 + 0 = 0 $$
Still, experimental results contradict, since for one example I get a non-zero determinant.
What am I doing wrong?
| Determinant is a multilinear map so the linearity works in a different way. Operation that you made corresponds to linearity in a sense
$\det(A+B) = \det(A) + \det(B)$. This is true in general only for 1x1 matrices. The correct way to use the linearity of determinant is
$$
\det(x_1 + \alpha y_1,x_2,\cdots,x_n) = \det(x_1,x_2,\cdots,x_n) + \alpha \det(y_1,x_2,\cdots,x_n)
$$
where $x_1,\cdots,x_n,y_1$ are vectors of rows/columns of the matrix. So in your case the correct manipulation is
$$
\begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & x & a_2 & \dots & a_n \\
a_0 & a_1 & x & \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
a_0 & a_1 & a_2 & \dots & x
\end{vmatrix} = \begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & x - a_1 + a_1 & a_2 & \dots & a_n \\
a_0 & a_1 & x & \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
a_0 & a_1 & a_2 & \dots & x
\end{vmatrix} =
\begin{vmatrix}
a_0 & 0 & a_2 & \dots & a_n \\
a_0 & x - a_1 & a_2 & \dots & a_n \\
a_0 & 0 & x & \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
a_0 & 0 & a_2 & \dots & x
\end{vmatrix} +
\begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & a_1 & a_2 & \dots & a_n \\
a_0 & a_1 & x & \dots & a_n \\
\dots & \dots & \dots & \dots & \dots \\
a_0 & a_1 & a_2 & \dots & x
\end{vmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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The value of $\sum_{k=0}^\infty k^n x^k$ Any known conclusions for $\sum_{k=0}^\infty k^n x^k$ where $n$ is a nonnegative integer and $0<x<1$?
*
*If $n=0$, the sum is simply $\frac{1}{1-x}$.
*If $n=1$, the sum can be computed as shown in this post
*For general $n$, it seems the method by taking derivative is
extremely complicated though I believe the sum would have a
finite value because $x^k$ decreases faster than $k^n$ increases.
| I don't know of any simple solutions for the general summation, but I do know a method that will work. Since $0 < x < 1$ let $x = \frac{1}{y}$ where $y > 1$. Consider $n = 2$
Let $S = \sum_{k = 0}^\infty \frac{k^2}{y^k}$.
\begin{align}
S &= \frac{1}{y} + \frac{4}{y^2} + \frac{9}{y^3} + \frac{16}{y^4} + \cdots\\
\frac{S}{y} &= \frac{1}{y^2} + \frac{4}{y^3} + \frac{9}{y^4} + \cdots\\
S - \frac{S}{y} = \frac{S(y - 1)}{y} &= \frac{1}{y} + \frac{3}{y^2} + \frac{5}{y^3} + \frac{7}{y^4} + \cdots\\
\frac{S(y - 1)}{y^2} &= \frac{1}{y^2} + \frac{3}{y^3} + \frac{5}{y^4} + \cdots\\
\frac{S(y - 1)}{y} - \frac{S(y-1)}{y^2} = \frac{S(y-1)^2}{y^2} &= \frac{1}{y} + \frac{2}{y^2} + \frac{2}{y^3} + \frac{2}{y^4} + \cdots\\
&= \frac{1}{y} + \frac{2}{y^2}(1 + \frac{1}{y} + \frac{1}{y^2} + \cdots)\\
&= \frac{1}{y} + \frac{2}{y^2}(\frac{1}{1-\frac{1}{y}}) = \frac{1}{y} + \frac{2}{y(y-1)}
\end{align}
Therefore $S = \frac{y^2}{(y-1)^2}(\frac{1}{y} + \frac{2}{y(y-1)}) = \frac{y}{(y-1)^2} + \frac{2y}{(y-1)^3}$. Now try this method for $n = 3$.
\begin{align}
S &= \frac{1}{y} + \frac{8}{y^2} + \frac{27}{y^3} + \frac{64}{y^4} + \cdots\\
\frac{S}{y} &= \frac{1}{y^2} + \frac{8}{y^3} + \frac{27}{y^4} + \cdots \\
S - \frac{S}{y} = \frac{S(y-1)}{y} &= \frac{1}{y} + \frac{7}{y^2} + \frac{19}{y^3} + \frac{37}{y^4} + \cdots \\
\frac{S(y-1)}{y^2} &= \frac{1}{y^2} + \frac{7}{y^3} + \frac{19}{y^4} + \cdots \\
\frac{S(y-1)^2}{y^2} &= \frac{1}{y} + \frac{6}{y^2} + \frac{12}{y^3} + \frac{18}{y^4} + \cdots\\
\frac{S(y-1)^2}{y^3} &= \frac{1}{y^2} + \frac{6}{y^3} + \frac{12}{y^4} + \cdots\\
\frac{S(y-1)^3}{y^3} &= \frac{1}{y} + \frac{5}{y^2} + \frac{6}{y^3} + \frac{6}{y^4} + \frac{6}{y^5} + \cdots
\end{align}
Which yields the geometric series.
Basically the procedure consists of taking the difference between the sum and multiplying it by the common ratio $\frac{1}{y}$. This procedure will be done $n$ times in order to yield a finite amount of terms plus an infinite convergent geometric series. This occurs due to the property of finite differences. Quadratics have second differences, Cubics have third, $x^n$ has $n$ differences. This means the numerators eventually disappear into constants while the exponentials on the bottom remain the same. So just repeat this process $n$ times to compute the sum. This method only requires elementary algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find sum of $1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$ Find the sum of the following series :
$$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots$$ and
$$1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$$
For the first series , we have , $$(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}+\cdots$$
$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots=\frac{1}{2}\log 2$$
But I am unable to set the second series to find its sum..Please help.
| For conditionally convergent series, you can move terms around to a limited extent and still get the same sum. Thus
$$1+{1\over3}+{1\over5}-{1\over2}-{1\over4}-{1\over6}+{1\over7}+{1\over9}+{1\over11}-\cdots=1-{1\over2}+{1\over3}-{1\over4}+{1\over5}-{1\over6}+{1\over7}-{1\over8}+\cdots$$
because no term is moving more than $3$ positions. In this case the proof is also utterly straightforward: The partial sum $s_n$ for each series is the same whenever $n$ is a multiple of $6$, so the series (each of which is convegent) must converge to the same sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inverting Modular Exponentiation How can I go about solving the equation $4 = y^4 \bmod{7}$? Do I have to try all of the possible $y$'s in between $1$ and $7-2$ or is there a smarter way that can be generalized for larger numbers?
| First, factor the equivalence as a difference of squares.
\begin{align*}
y^4 & \equiv 4 \pmod{7}\\
y^4 - 4 & \equiv 0 \pmod{7}\\
(y^2 + 2)(y^2 - 2) & \equiv 0 \pmod{7}
\end{align*}
Hence, $$y^2 + 2 \equiv 0 \pmod{7} \implies y^2 \equiv -2 \equiv 5 \pmod{7}$$ or $$y^2 - 2 \equiv 0 \pmod{7} \implies y^2 \equiv 2 \pmod{7}$$
Observe that
\begin{align*}
4 & \equiv -3 \pmod{7}\\
5 & \equiv -2 \pmod{7}\\
6 & \equiv -1 \pmod{7}
\end{align*}
and that
\begin{align*}
0^2 & \equiv 0 \pmod{7}\\
1^2 & \equiv (-1)^2 \equiv 1 \pmod{7}\\
2^2 & \equiv (-2)^2 \equiv 4 \pmod{7}\\
3^2 & \equiv (-3)^2 \equiv 9 \equiv 2 \pmod{7}
\end{align*}
Hence, $y \equiv 3 \pmod{7}$ or $y \equiv -3 \equiv 4 \pmod{7}$, which you can check by direct substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
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The series $\sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k}$ is bounded below by $c \log\log x$ near 0, thus fails to be uniformly convergent The convergent series
$$ \sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k} $$
defines a function in $H^{1/2}([0,2\pi])$. This is an example of such a series for which convergence on is not uniform.
I want to show that the convergence is not uniform on $(0,2\pi)$ by showing that the series is $\geq c\log \log \frac{1}{x}$ as $x\to 0$.
For a fixed positive integer $N$, we can split the series into
$$ \sum_{k=2}^N \frac{ \cos(kx)}{k \ln k} + \sum_{k=N+1}^\infty \frac{ \cos(kx)}{k \ln k} = \sum_{k=2}^N \frac{ \cos(kx)}{k \ln k} + R_N(x),$$
where $R_N(x)$ denotes the tail series evaluated at $x$.
We use continuity of cosine at $x=0$ in an essential way: the first $N$ factors of cosine stay away from 0. If $x < \frac{\pi}{6N}$, then $\cos(kx) > \frac{\sqrt 3}{2}$ for $2\leq k \leq N$. Thus we can bound below the first $N$ terms in the sum by
$$ \frac{\sqrt 3}{2} \sum_{k=2}^N \frac{1}{k\ln k} \geq \frac{\sqrt 3}{2} \int_2^{N+1} \frac{1}{t \ln t}dt = \frac{\sqrt 3}{2}( \ln \ln (N+1) - \ln \ln 2),$$
where we use monotonicity and convexity of $t\to \frac{1}{t\ln t}$ on $[2,\infty)$. Since $x < \frac{1}{N+1}$, $\ln \ln (N+1) \geq \ln \ln \frac{1}{x}.$
This looks almost like what I need to show:
$$ \sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k} \geq \frac{\sqrt 3}{2} \ln \ln \frac{1}{x} - \frac{\sqrt 3}{2} \ln \ln 2 + R_N(x),$$
for all positive integers $N$ and all $x < \frac{\pi}{6N}$.
My question is: how can I get rid of the tail series $R_N(x)$?
If I can show that it is positive or at least negative but very small, then we can simply drop it.
I also see that I might be able to go back in my argument and use that for every $\epsilon >0$, the convergence of the series is uniform on $[\epsilon, 2\pi-\epsilon]$. Then I can pick some $N$ so that $R_N(x)$ is very small, at least on $[\epsilon, 2\pi - \epsilon]$. But the problem is that for a choice of $\epsilon$, the interval for which $\cos(kx) > 0$ for $2\leq k \leq N$ guaranteed by continuity may be contained in $[0,\epsilon)$.
There appear to be two competing forces: continuity is useful close to 0, but the uniform convergence can only be used away from 0.
How can I proceed? Or am I completely off the mark? I would very much appreciate any advice or criticism for this problem.
| The easiest - in some sense - way to show that the series doesn't converge uniformly on $(0,2\pi)$ is to note that all terms of the sequence are continuous functions with common period $2\pi$, and if the series were uniformly convergent on $(0,2\pi)$, it would by continuity be uniformly convergent on $[0,2\pi]$ (and by periodicity on all of $\mathbb{R}$), so the limit function would be a continuous (and $2\pi$-periodic) real-valued function. But the series diverges for $x = 0$, hence the convergence cannot be uniform.
This argument doesn't use any deep theory, the only mildly uncommon thing is to notice that if a sequence/series of continuous functions converges uniformly on $S$, then it also converges uniformly on $\overline{S}$ - since for continuous $f,g$ we have
$$\sup \{ \lvert f(x) - g(x)\rvert : x \in \overline{S}\} = \sup \{ \lvert f(x) - g(x)\rvert : x \in S\}$$
by continuity of $f-g$ on $\overline{S}$.
But we can also continue on the way you started. That has the advantage that we do not only show that the series doesn't converge uniformly, we also get information about the behaviour of the sum function near $0$ (and near $2\pi$). To estimate the size of $R_N(x)$, summation by parts is a useful technique. We start with
\begin{align}
\Biggl\lvert \sum_{k = m+1}^n \cos (kx)\Biggr\rvert
&\leqslant \Biggl\lvert \sum_{k = m+1}^n e^{ikx}\Biggr\rvert \tag{$\lvert \operatorname{Re} z\rvert \leqslant \lvert z\rvert$}\\
&= \Biggl\lvert \sum_{k = 0}^{n-m-1} e^{ikx}\Biggr\rvert\\
&= \biggl\lvert \frac{e^{i(n-m)x}-1}{e^{ix}-1}\biggr\rvert\\
&\leqslant \frac{2}{\lvert e^{ix}-1\rvert}\\
&= \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert}.
\end{align}
Now for fixed $N$ and $n \geqslant N$ we define $S_n(x) := \sum\limits_{k = N+1}^n \cos (kx)$ and find (using $S_N(x) = 0$)
\begin{align}
\sum_{k = N+1}^n \frac{\cos (kx)}{k\log k}
&= \sum_{k = N+1}^n \frac{S_k(x) - S_{k-1}(x)}{k\log k}\\
&= \frac{S_n(x)}{(n+1)\log (n+1)} + \sum_{k = N+1}^n S_k(x)\biggl(\frac{1}{k\log k} - \frac{1}{(k+1)\log (k+1)}\biggr).
\end{align}
Taking absolute values, the inequality $\lvert S_k(x)\rvert \leqslant \frac{1}{\bigl\lvert\sin \frac{x}{2}\bigr\rvert}$ yields
\begin{align}
\Biggl\lvert \sum_{k = N+1}^n \frac{\cos (kx)}{k\log k} \Biggr\rvert &\leqslant
\frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert (n+1)\log (n+1)} + \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert} \sum_{k = N+1}^n \biggl\lvert \frac{1}{k\log k} - \frac{1}{(k+1)\log (k+1)}\biggr\rvert\\
&= \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert (N+1)\log (N+1)}.
\end{align}
Letting $n \to \infty$, we obtain
$$\lvert R_N(x)\rvert \leqslant \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert (N+1)\log(N+1)}.$$
The $\sin \frac{x}{2}$ in the denominator of that bound means that we cannot use this estimate to reach our desired conclusion if we work with a fixed $N$ independent of $x$. But we can choose $N$ depending on $x$ so that the singularity of $\frac{1}{\sin (x/2)}$ is compensated by the vanishing of $\frac{1}{N(x)}$. If for some $a > 0$ we choose $N(x) > \frac{a}{x} - 1$, then for $0 < x \leqslant \pi$ we have
$$(N(x)+1)\sin \frac{x}{2} \geqslant (N(x)+1)\frac{x}{\pi} > \frac{a}{x}\cdot\frac{x}{\pi} = \frac{a}{\pi}$$
and hence
$$\frac{1}{\bigl(\sin \frac{x}{2}\bigr) (N(x)+1)\log (N(x)+1)} < \frac{\pi}{a\log (N(x)+1)} < \frac{\pi}{a\log \frac{a}{x}}$$
for $0 < x < \min \{a,\pi\}$.
Now we have a bound for $\lvert R_{N(x)}(x)\rvert$ that we can use.
If we have $N(x) \leqslant \frac{b}{x}$ for some $b < \frac{\pi}{2}$, your computation shows
$$\sum_{k = 2}^{N(x)} \frac{\cos (kx)}{k\log k} > \cos b \log \log (N(x)+1)$$
(since $\log \log 2 < 0$).
If we choose $N(x) = \bigl\lfloor \frac{b}{x}\bigr\rfloor$ with $0 < b < \frac{\pi}{2}$, then we have $N(x) \leqslant \frac{b}{x} < N(x)+1$, and by the above
\begin{align}
\sum_{k = 2}^{\infty} \frac{\cos (kx)}{k\log k}
&\geqslant \sum_{k = 2}^{N(x)} \frac{\cos (kx)}{k\log k} - \Biggl\lvert \sum_{k = N(x)+1}^{\infty} \frac{\cos (kx)}{k\log k}\Biggr\rvert\\
&> \cos b\cdot \log \log \frac{b}{x} - \frac{\pi}{b \log \frac{b}{x}}
\end{align}
for $0 < x < b$. On the other hand we have
$$\sum_{k = 2}^{N(x)} \frac{\cos (kx)}{k\log k} \leqslant \sum_{k = 2}^{N(x)} \frac{1}{k\log k} \leqslant \log \log N(x) - \log \log 2 + \frac{1}{2\log 2}$$
and so
$$\sum_{k = 2}^{\infty} \frac{\cos (kx)}{k\log k} \leqslant \log \log \frac{b}{x} - \log \log 2 + \frac{1}{2\log 2} + \frac{\pi}{b\log \frac{b}{x}}$$
for $0 < b < x$. Since
$$\log \log \frac{b}{x} = \log \log \frac{1}{x} + \log \biggl(1 + \frac{\log b}{\log \frac{1}{x}}\biggr) = \log \log \frac{1}{x} + O\bigl(\lvert\log x\rvert^{-1}\bigr),$$
it follows that
$$\sum_{k = 2}^{\infty} \frac{\cos (kx)}{k\log k} \sim \log \log \frac{1}{x}$$
as $x\searrow 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2} $ . Using AM-HM inequality we get $t(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2) \geq 2^{2}.$ let $ m =(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2)$ and using AM-GM inequality we get $m \leq 5/3$.
But from this point my inequality signs are getting mixed up. Am I on right track?
| EDIT : The previous solution was utterly mistaken.
Let$$\frac{4}{4-x^2} + \frac{9}{9-y^2} =f(x,y)$$
$$f(x,y)=\frac{36}{36-9x^2}+\frac{36}{36-4y^2} \ge \frac{(6+6)^2}{72-9x^2-4y^2} (\because \text{Cauchy})\ge \frac{144}{60}=\frac{12}{5}(\because \text {AM-GM})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$A, B$ and $C$ can do a piece of work
$A, B$ and $C$ can do a piece of work in $18$, $24$ and $30$ days respectively. $A$ and $B$ work together for a certain number of days. Then $B$ leaves and $C$ joins. They work together for half the number of days for which $A$ and $B$ had worked together. After that $A$ goes away and $B$ rejoins the work. If $B$ and $C$ together complete the remaining work in one third of the number of days for which $A$ and $B$ had previously worked together, for how many days does each of them work?
My attempt:
In $18$ days, $A$ can do $1$ work
In $1$ day, $A$ can do $\frac{1}{18}$ work
In $1$ day, $B$ can do $\frac{1}{24}$ work
In $1$ day, $C$ can do $\frac{1}{30}$ work
| Let the time worked by $A$ and $B$ together be $x$ days.
Then, the fraction of work done by $A$ and $B$ together $=\frac x{18}+\frac x{24}$.
Time worked by $A$ and $C$ will then be $\frac x2$ days.
The fraction of work done by them $=\frac{x}{2\times18}+\frac{x}{2\times30}=\frac x{36} + \frac x{60}$
Time worked by $B$ and $C$ is given to be $\frac x3$ days.
The fraction of work done by the pair $=\frac{x}{3\times24}+\frac{x}{3\times30}=\frac x{72}+\frac x{90}$
But by now, the work is complete.
Thus,
$$\frac x{18}+\frac x{24}+\frac x{36} + \frac x{60}+\frac x{72}+\frac x{90} = 1$$
Solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form for the infinite product $\prod\limits_{k=0}^{\infty} \left( 1-x^{2^k} \right)$ There is a known identity:
$$\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)=\frac{1}{1-x}, ~~~~~|x|<1$$
It's easy to derive it by converting it to a telescoping product as shown in this answer.
However, we can't use the same method here.
$$\left( 1-x^{2^k} \right) \left( 1+x^{2^k} \right)=\left( 1-x^{2^{k+1}} \right)$$
$$\left( 1-x^{2^k} \right) =\frac{\left( 1-x^{2^{k+1}} \right)}{ \left( 1+x^{2^k} \right)}$$
This product will not telescope. We can't even use this to find something new about it:
$$p(x)=\prod_{k=0}^{\infty} \left( 1-x^{2^k} \right)=\frac{\prod_{k=0}^{\infty} \left( 1-x^{2^{k+1}} \right)}{\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)}$$
We only get the obvious recurrence relation:
$$p(x)=(1-x)~p(x^2)$$
Mathematica gives this plot (for 25 terms).
Does this product have a closed form?
| Not a closed form, but we have the following series representations:
$$\begin{align*} p(x) = \prod_{k=0}^{\infty} \left( 1-x^{2^k} \right)
&= \frac1{1-x}\left(1+2\sum_{k\geq0}\frac{(-1)^kx^{2^k}}{1+x^{2^k}}\right) \\
&= \frac{1}{1-x} - \frac{4x}{(1-x)^2} + \frac6{1-x} \sum_{k=0}^\infty \frac{x^{2^{2k}}}{1-x^{2^{2k+1}}}
\end{align*}$$
where the last sum is $$\sum_{\nu_2(n)\text{ even}}x^n$$
Proof. We have: (on the level of formal series; although everything converges absolutely for $|x|<1$)
$$p(x)-1 = \sum_{n=1}^\infty x^n (-1)^{b(n)}$$
where $b(n)$ is the number of $1$'s in the binary expansion of $n$. We can write:
$$(-1)^{b(n)} = 1+2\sum_{\substack{k \text{ for which the }\\k\text{th digit is 1,}\\\text{starting from }k=0}}(-1)^k$$
plug this in the sum and change the order of summation $n \leftrightarrow k$:
$$\begin{align*}p(x)-1
&= \sum_{n=1}^\infty x^n + 2 \sum_{n=1}^\infty x^n\sum_{\substack{k \text{ for which the }\\k\text{th digit in }n\text{ is 1}}}(-1)^k \\
&= \frac x{1-x} + 2 \sum_{k=0}^\infty (-1)^k x^{2^k} \prod_{j \neq k}(1+x^{2^j}) \\
&= \frac x{1-x} + 2 \sum_{k=0}^\infty (-1)^k \frac{x^{2^k}}{(1-x)(1+x^{2^k})} \\
\end{align*}$$
For the series $S = \sum_{k=0}^\infty (-1)^k \frac{x^{2^k}}{1+x^{2^k}}$, we can write each term as a geometric series:
$$S = -\sum_{k=0}^\infty \sum_{n=1}^\infty (-1)^{k+n}x^{2^kn}$$
Grouping the terms with the same power of $x$ gives:
$$\sum_{n=1}^\infty c_nx^n$$
where $c_n=1$ if $\nu_2(n)$ is even, and $c_n=-2$ if $\nu_2(n)$ is odd.
We get $$\begin{align*}S
&= -2\frac x{1-x} + 3\sum_{\nu_2(n)\text{ even}}x^n \\
&= -2\frac x{1-x} + 3\sum_{k=0}^\infty \left( (x^{4^k})^1 + (x^{4^k})^3 + (x^{4^k})^5 + \cdots \right)\\
&= -2\frac x{1-x} + 3\sum_{k=0}^\infty \frac{x^{2^{2k}}}{1-x^{2^{2k+1}}}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Prove that $C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$, for every $n \in N$ EDIT: I forgot to mention $C = 0.5uu^T + 0.33vv^T$ and now if I use it, I solve it easily.
Given: $Cx = \frac{1}{2}au + \frac{1}{3}bv$, $x \in R^2$, $u,v$ are orthonormal vectors in $R^2$, $x = au + bv$ and $a,b \in R$, $C$ is a matrix $2x2$. Note: $a,b$ are scalars.
Prove that $$C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$$, for every $n \in N$
Well, it's trivial that we need to show that with induction, so for $n = 1$ this works because it is given (actually I proved it and it is really true).
Now I assume that it works for $k = n$, and I want to prove for $k = n + 1 $. I use $k$ because it is more comfortable for me that way, and I got stuck with $a^2$ and $u,v$ are gone in the last equation, instead of $a$ and multiplied with $u,v$ as required.
$C^{k+1}x=\frac{1}{2^{k+1}}au + \frac{1}{3^{k+1}}bv$ I need to prove.
Going from the left side of the equation and I try to prove the right side:
$C^{k+1}x = C^k(Cx) = C^k(\frac{1}{2}au + \frac{1}{3}bv) = \frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =
\frac{1}{2^{k+1}}a^2 + \frac{1}{3^{k+1}}b^2$.
Edit: if I take $au$ out, then I get was it required. can I do that?
I mean: $\frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =$ long equation $= au\frac{1}{2^{k+1}} + bu\frac{1}{3^{k+1}}$
| $Cx = \frac{1}{2}au + \frac{1}{3}bv$
$x = au + bv \implies Cx = C(au +bv) = Cau + Cbv$
Thus
$Cau + Cbv = \frac{1}{2}au + \frac{1}{3}bv$
$\implies a(C-\frac{1}{2}I)u + b(C-\frac{1}{3}I)v = 0$
$\implies a(C-\frac{1}{2}I)u.u^T + b(C-\frac{1}{3}I)v.u^T = 0.u^T = 0$
$\implies a(C-\frac{1}{2}I) = 0$ , since u&v are orthanormal.
By the same reason we also have $ b(C-\frac{1}{3}I) = 0$
Therefore either $a = 0$ or $b = 0$
WLOG choose $b = 0$, then
$Cx = \frac{1}{2}au$, and $x = au$, and it is easy to show
$C^nx = \frac{1}{2^n}au$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding all left inverses of a matrix I have to find all left inverses of a matrix
$$A = \begin{bmatrix}
2&-1 \\
5 & 3\\
-2& 1
\end{bmatrix}$$
I created a matrix to the left of $A$,
$$\begin{bmatrix}
a &b &c \\
d &e &f
\end{bmatrix} \begin{bmatrix}
2&-1 \\
5 & 3\\
-2& 1
\end{bmatrix} = \begin{bmatrix}
1&0 \\
0&1
\end{bmatrix}$$
and I got the following system of equations:
\begin{array} {lcl} 2a+5b-2c & = & 1 \\-a+3b+c & = & 0 \\
2d+5e-2f & = & 0 \\ -d+3e+f & = & 1 \end{array}
After this step, I am unsure how to continue or form those equations into a solvable matrix, and create a left inverse matrix from the answers of these equations.
| Alternative less advanced solution. Multiply it with its own transpose:
$A^TA$ is a $2\times2$ matrix:
$\left(
\begin{array}{cc}
33 & 11 \\
11 & 11 \\
\end{array}
\right)$
This has a trivial inverse by swapping the diagonals and finding the determinant:
$\frac{1}{22}\left(
\begin{array}{cc}
1 & -1 \\
-1 & 3 \\
\end{array}
\right)$
If you multiply this with the original matrix:
$(A^TA)^{-1}A^TA=I_2$
And note that this is your inverse:
$(A^TA)^{-1}A^T$
Which gives:
$\begin{pmatrix}\frac{3}{22}&\frac{1}{11}&-\frac{3}{22}\\ -\frac{5}{22}&\frac{2}{11}&\frac{5}{22}\end{pmatrix}$
Even less advanced, your system of linear equations work too. What I think you're not realising is, due to the very fact that there are less equations than variables, that there are an infinite number of inverses to matrices. Due to that fact you can take the liberty of just letting two variables be 0, and solving the rest. If you choose $c=f=0$, you can quite easily solve $b=\frac{1}{11}, a=\frac{3}{11}, e=\frac{2}{11}, d=-\frac{5}{11}$, which gives the solution:
$\begin{pmatrix}
\frac{3}{11}&\frac{1}{11}&0\\
-\frac{5}{11}&\frac{2}{11}&0
\end{pmatrix}$
Whilst this method is superior from a linear algebra perspective (because unlike the first solution above I can find multiple by choosing different values for a pair of unknowns), it's less practical if all you want is to find another inverse.
A third idea I have is to use block matrices. Generally speaking I can write your matrix as a joining of the vector $\mathbf{a}=\begin{pmatrix}-2&1\end{pmatrix}$ with the matrix $\begin{pmatrix}
2&-1 \\
5 & 3
\end{pmatrix}$
Written as a block matrix:
\begin{bmatrix}
\mathbf{a}\\M_a
\end{bmatrix}
Writing the inverse, $B$, as a block matrix of a similar form (albeit with a
vetrical vector), we get:
$\begin{bmatrix}
\mathbf{b}&M_b
\end{bmatrix}
\begin{bmatrix}
\mathbf{a}\\M_a
\end{bmatrix}
=I_2$
By the property of block matrices, this gives us:
$I_2=\mathbb{a}\mathbb{b}+M_aM_b$
Hence for convenience, if $\mathbf{b}$ is a $0$ vector (the equivalent of setting $a=d=0$ in your linear equations), then all you need to do is invert $M_a$:
$M_b=\begin{pmatrix}
5&3\\
-2&1
\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{11}&-\frac{3}{11}\\ \frac{2}{11}&\frac{5}{11}\end{pmatrix}$
Hence your answer would be:
$B=\begin{pmatrix}0&\frac{1}{11}&-\frac{3}{11}\\ 0& \frac{2}{11}&\frac{5}{11}\end{pmatrix}$
For explanatory purposes I wrote everything out. In general, just invert the largest square you can fit (in any position) in the matrix you wish to invert, and just bung it in a $0$ matrix with reversed dimensions (in the same position along the height or width)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
3 x 4 Matrix with Complex Numbers I have a $3 \times4$ matrix $$
\begin{pmatrix}
i & 4c & -1 & 0 \\
2ic & 4 & -2c & 0 \\
-i & -2 & 1 & 6c-3 \\
\end{pmatrix}
$$
to figure out the complex numbers $c$ such that the row space has dimension 2, how would I approach this problem? Also, for each $c$ I would like to try to find the subset of the columns of A that makes it a $\Bbb C$-basis of the column-space A.
So far, i've taken the determinate's of the 3 sub $3 \times3$ matrices. In particular:
$$(A_1) = \begin{pmatrix}
4c & -1 & 0 \\
4 & -2c & 0 \\
-2 & 1 & 6c-3 \\
\end{pmatrix} = det(A_1) -48c^2+48c-12$$
$$(A_2) = \begin{pmatrix}
i & -1 & 0 \\
2ic & -2 & 0 \\
-2 & 1 & 6c-3 \\
\end{pmatrix} = det(A_2) 12ic^2-18ic-12i$$
$$(A_2) = \begin{pmatrix}
i & 4c & -1 \\
2ic & 4 & -2 \\
-i & -2 & 1 \\
\end{pmatrix} = det(A_3) -8c^2i+12ic-4i$$
From here, I'm not sure what to do.I set the determinate equal to zero and got the following (below).
$$c(A_1)=\frac{1}{2}$$ $$c(A_2)= \frac{3i \pm 5i}{4i}$$ $$c(A_3)= \frac{-3i\pm i}{4}$$
Does anyone have any clue how or where I have messed up because I feel as though I have made a mistake somewhere.
| I think reducing by rows can help:$${}$$
$$ \begin{pmatrix}
i & 4c & -1 & 0 \\
2ic & 4 & -2c & 0 \\
-i & -2 & 1 & 6c-3 \\
\end{pmatrix}\longrightarrow\begin{pmatrix}
i & 4c &\!\!-1 & 0 \\
0 & 4-8c^2 & 0 & 0 \\
0 & 4c-2 & 0 & 6c-3 \\
\end{pmatrix}$$
Observe that if $\;6c-3=0\iff c=\frac12\;$ , then it follows that the last row becomes all zeros and none of the two first one is all zeros, and also if $\;8c^2=4\iff c=\pm\frac1{\sqrt2}\;$, and with this value the second row is all zeros.
Thus, only for $\;c=\frac12,\,\pm\frac1{\sqrt2}\;$ the matrix has row dimension equal to two
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple method to solve a geometry question for junior high school student Rencently, my sister asked me a geometry question that came from her mock examination, please see the following graph.
Here,
*
*$\angle DOE=45°$
*the length of $DE$ is constant, and $DE=1$. Namely, $OD,OE$ are changeable.
*$\triangle DEF$ is equilateral triangle.
Q: What is the maximum length of $OF$?
My solution
Denote $OD,OE,\angle ODE$ as $x,y,\theta$, respectively.
Via sine theorem
$$
\begin{cases}
ED^{2} = OE^{2} + OD^{2} - 2OE \times OD\cos \angle EOD \\[6pt]
\cos \theta = \dfrac{EO^{2} + ED^{2} - OD^{2}}{2 EO \times ED}
\end{cases}
$$
$$
\begin{align}
1^{2} &= x^{2} + y^{2} - 2xy\cos 45^{\circ} \\
&= x^{2} + y^{2} - \sqrt{2} xy
\end{align}
$$
$$
\implies
\begin{cases}
\color{red}{xy} = \dfrac{x^{2} + y^{2} - 1}{\sqrt{2}} \color{red}{\leq}
\dfrac{x^{2} + y^{2}}{2} \implies x^{2} + y^{2} \color{red}{\leq}
2 + \sqrt{2} \\[6pt]
\cos \theta = \dfrac{x^{2} + 1 - y^{2}}{2x}
\end{cases}
$$
Via cosine theorem
$$
\frac{y}{\sin \theta} = \frac{DE}{\sin \angle EOD} = \frac{1}{\sin
45^{\circ}} \implies \sin \theta = \frac{y}{\sqrt{2}}
$$
$$\begin{align}
OF^{2} &= EO^{2} + EF^{2} - 2EO \times EF\cos \angle OEF \\
&= x^{2} + 1^{2} - 2x\cos(\theta + 60^{\circ}) \\
&= x^{2} + 1 - 2x(\cos \theta \cos 60^{\circ} - \sin \theta \sin
60^{\circ}) \\
&= x^{2} + 1 - 2x\left(\frac{x^{2} + 1 - y^{2}}{2x} \frac{1}{2} -
\frac{y}{\sqrt{2}} \frac{\sqrt{3}}{2}\right) \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3} xy}{\sqrt{2}} \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3}}{\sqrt{2}}
\frac{x^{2} + y^{2} - 1}{\sqrt{2}} \\
&= \frac{(\sqrt{3} + 1)(x^{2} + y^{2})}{2} + \frac{1 - \sqrt{3}}{2}
\\
&\color{red}{\leq} \frac{(\sqrt{3} + 1)(2 + \sqrt{2})}{2} + \frac{1 -
\sqrt{3}}{2} = \frac{1}{2}(3 + \sqrt{3} + \sqrt{2} + \sqrt{6})
\end{align}
$$
However, for junior high school student, she doesn't learn the following formulae:
*
*sine theorem
*cosine therem
*$\cos(x+y)=\cos x \cos y-\sin x \sin y$
*fundamental inequality $x y\leq \frac{x^2+y^2}{2}$
Question
*
*Is there other simple/elegant method to solve this geometry question?
Update
Thanks for MXYMXY's hint
Here, the line $O'F$ pass the center of the circle. Namely, $O'D=OF$
In Rt $\triangle O'OF$, the inequality $O'F>OF$ holds.
| Practical approach.Take a cardboard in the shape of an equilateral triangle of side 1. Slide two vertices along arms of the $ 45 ^0$ lines,note the other vertex is farthest when the setup is symmetrical, lying along bisector of $45^0 $ corner.
Only new trig ratio to be known is $ \tan 67.5^0$ is $ \sqrt2 +1 $
So the required length is:
$$ \frac12 \; (\tan 60^0 + \tan 67.5^0 ) =\frac {\sqrt3 + \sqrt2 +1}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"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.