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
$$
\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{\frac{x^4}{12}+O(x^6)}{x^4+O(x^6)}=\frac{1}{12}+O(x^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
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 ... | 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
$$ Av = \begin{pmatrix} 1 & e^{i \theta} \\ e^{-i \theta} & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\alpha (1+\alpha/2)^{-1} < \log(1+\alpha) $ for $\alpha > 0$ How does one prove that $\alpha (1+\alpha/2)^{-1} < \log(1+\alpha) $ for $\alpha > 0$?
| By changing variable $x \rightarrow a-x$, we have
$$
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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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:
$$
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How do I simplify $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ )$? $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ ) =$ ?
The answer is $3$.
My progress so far:
\begin{align}
\sqrt{3}(\cot 70^\circ+4\cos 70^\circ)&= \sqrt{3}(\tan 20^\circ+4\sin 20^\circ) \\
&= \sqrt{3}(\sin 20^\circ)\left(\frac{1}{\cos 20^\circ}+4\right)\\
&=... | $$\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)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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 un... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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 $
I... | 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",
"url": "https://math.stackexchange.com/questions/1561614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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... | 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,
$$P(X = 1) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2
+\binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the conditi... | Assuming $b\neq a$ you gained a constraint
$$-(a+b) = c$$
This constriction is not bad, because there is no number $b = c$ such that $a^2 = bc + 1$ for integers.
It's fine to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Lagrange Multipliers: find points farthest/closest to a point Find the points of the ellipse:
$$\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$$
(no need for the root since maximizing one maximiz... | Hint: $$\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) $$
Then:
$\frac{\partial \mathcal{L}}{\partial x}=2(x-1)+\frac{2x\lambda}{9}=0 \implies x=\frac{9}{9+\lambda}$
$\frac{\partial \mathcal{L}}{\partial y}=2(y-1)+\frac{\lambda y}{2}=0 \implies y=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit problem with roots I'm struggling solving the following limit problem:
$$\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}}}$$
But that doesn'... | Substitute $t=1/x$, so the limit becomes
$$
\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}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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}\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
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 ... | 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-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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?
| $$2\cos x/(1+\cos 2x)=(1-\cos 2x)/\sin 2x$$
$$2\cos x/(1+\cos^2(x)-\sin^2(x))=(1-(\cos^2(x)-\sin^2(x)))/(2\sin(x)\cos(x))$$
$$2\cos x/(2\cos^2(x))=(2\sin^2(x))/(2\sin(x)\cos(x))$$
$$1/\cos(x) = \sin(x)/\cos(x)$$
$$1 = \sin(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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
$$
2^{3x} =... | $$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... | {
"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
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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 $g(x)<f(x)<h(x)$
$$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{... | Your expression may be written as $$\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",
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"answer_id": 1
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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... | 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 simpl... | {
"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
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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 $k$
$(... | 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 $... | 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 f... | {
"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
$$Q{\left(s;x\right)}=\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}.\tag{1}$$
Qu... | 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{\l... | {
"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:
$$\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}=\lim\limits_{x\to 0}\frac{(1+x^2+O(x^{2n}))-(1-\frac{x^{2\sqrt{2}}}{2}+O(x^{\sqrt{2}(2n+1)})}{x^2}=\lim\lim... | Hint. You may just write, as $x \to 0$,
$$
\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$$
For $c=1$ the sequence is periodic with period 5 and range {1,2,3}. For other values of $c$ the sequence appears... | This is an partial answer. Please consider this as a supplement to merico's excellent answer.
Introduce auxillary sequences $(y_n), (z_n), (w_n)$ by $y_n = x_{n+1}, z_n = x_{n+2}, w_n = x_{n+3}$.
Rewrite the defining recurrence relation of $(x_n)$ to a more symmetric form:
$$x_{n+1} = \frac{c+x_n}{x_{n-1}}\quad \iff\qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 2
} |
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
$$\begin {eqnarray}
\pr... | I tried a different and definitely a far more elementary approach and, surprisingly, it gave a better estimate. We have obviously (which I had overlooked!)
$$\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} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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 pr... | 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 + ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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
$$\begin{align}
f(a,c,b,d) - f(a,b,c,d)&=(a-c)^2+(b-d)^2-((a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How do I find the image of a point in a line in 3D space? The question is,
find the image of the point $(1, 6, 3)$ in the line $$\frac x1 = \frac {y-1}{2} = \frac {z-2}{3}$$
I want to know the general equation to find the image of a point in a line.
EDIT
By image, I mean a point which is diametrically opposite to th... | Let $Q(a,b,c)$ be the image of $P(1,6,3)$ in the given line.Then direction ratios of $PQ$ are $(a-1,b-6,c-3)$
As the given line and $PQ$ will be perpendicular to each other.So $a_1a_2+b_1b_2+c_1c_2=0$ is the condition of perpendicularity
$1(a-1)+2(b-6)+3(c-3)=0$
$a+2b+3c=22..........(1)$
Now the mid point of $PQ(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Subadditivity of square root function Any hint to prove
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$
| I'll try the hard way
and use calculus.
Let
$f(y)
= \sqrt{x+y} - \sqrt{x} - \sqrt{y}
$.
$f(0)
=0
$.
$f'(y)
=\frac1{2\sqrt{x+y}}-\frac1{2\sqrt{y}}
$.
Since
$x+y \ge y$,
$\sqrt{x+y} \ge \sqrt{y}$
so
$\frac1{2\sqrt{x+y}}
\le \frac1{2\sqrt{y}}
$.
Therefore
$f'(y) \le 0$
for all $y$.
Since
$f(0) = 0$,
$f(y)
\le 0
$ for all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Probability of random subsets of the same size intersecting $A$ and $B$ are subsets drawn uniformly from a set $C$, such that $|A|=|B|=a$ and $|C|=n$.
For what $a$ is the probability of $A\cap B \ne \varnothing$ equal to $50\%$?
I figured that if $|A|=a$ and $|B|=b$ then the probability that there will be no intersect... | If we use Stirling's Approximation, that is $n!\sim\sqrt{2\pi n}\frac{n^n}{e^n}$, we get that
$$
\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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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 co... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 0
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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
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}... | You can do better than the stated inequality starting from $$\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}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
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)\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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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.
$$\begin{align*}
\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}&=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\times\frac{\cos\theta}{\cos\theta}\quad\text{(Multiply by $1=\frac{\cos\theta}{\cos\theta}$ to simplify)}\\
&=\frac{\sin\theta+1-\cos\theta}{\sin\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$ Let $P(x)=4x^2+6x+4$ and $Q(y)=4y^2-12y+25$.
Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$
I can solve this question graphically.
$$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}$$
I d... | Note that we have
$$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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For all nonnegative real numbers $x,y$ and $z$, prove that $\dfrac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$
For all nonnegative real numbers $x,y$ and $z$, prove that $$\frac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$$
It seems like AM-GM works here. We have $\dfrac{(x+y+z)^2}{3} \geq \dfrac{(... | We know that for non negative reals (in fact for all reals) we have $$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}$$
This proves the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the number of real solutions of $2 \cos((x^2+x)/6) = 2^x + 2^{-x}$
The number of real roots of the equation
$$
2\cos\left(\frac{x^2+x}{6}\right) = 2^x + 2^{-x}
$$
is
*
*$0$,
*$1$,
*$2$,
*infinitely many.
I can see that $x=0$ satisfies the equation. But I do not know how to solve the ... | we have $$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$$
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Determinant problem $n\times n$ with $0$'s and $1$'s $$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$$
Can anyone help me with this deter... | If you calculate $D_n$ using Laplace's expansion along the first row twice, you get
$$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 ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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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 $a, c, a - |b| + c$ ... | 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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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... | 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,
$$
2ac(1+\cos\beta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help me to prove the determinant of given matrix. Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
-x & a_2&a_3&a_4&\cdots &a_n\\
a_{1}+x & -x-a_2 & 0&0&\cdots &0\\
a_1+x&0 & -x-a_3 &0&\cdots &0\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
a_1+x&0 & 0&0&\cdots & -x-a_n\\
\end{arr... | I start with the well-known and easily proved fact that the $n\times n$ all $1$'s matrix
$$
\begin{pmatrix}
1 & 1 & \dots &1\\
1& 1 & \dots &1\\
\vdots &\vdots &\ddots&\vdots\\
1 & 1 &\dots &1
\end{pmatrix}
$$
has one eigenvalue $n$ corresponding to the obvious eigenvector $(1,1,\dots,1)^{T}$ and the eigenvalue $0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How to compute $\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution ?
$$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}... | $$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)=\lim\limits_{x \to +\infty} \left(\frac{1+x\sin\left(\frac{1}{x}\right)}{\ln\left(\frac{x}{2x+1}\right)-\frac{1}{x^3}} \right)$$
$$=\lim\limits_{x \to +\infty} \left(\frac{1+\frac{\sin\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the minimum polynomial of $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$? Inspired by a previous question what let $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$. What is the minimal polynomial of $x$ ?
The theory of algebraic extensions says the degree is $4$ since we have the deg... | By Galois theory the intermediate fields are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. Your number is not an element of any of those, so it generates the whole 4-d extension. In particular, we know that the minimal polynomial will be a quartic. Therefore any quartic polynomial with integer coefficient... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Second order ODE, first derivative missing I have the following second order equation, where the first derivative is missing, and I am asked to find its general solution:
$$6x^{2}yy''=3x(3y^{2}+2)+2(3y^{2}+2)^3$$
I don't know how to solve it. I have tried with a $u(x)=3y^{2}+2$ substitution but it doesn't seem useful..... | Let
$$ 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}) ;$$
Unless $3$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Sign the Derivative of $(\frac{1-q^r}{1-q^n})^{1/(n-r)}$ wrt $n$ (where $q \in (0,1)$) For $n, r \in \mathbb{N}$, $n > r$ and $q \in (0,1)$, I'm trying to sign the derivative wrt $n$ of
\begin{eqnarray}
f(n) = \left(\frac{1-q^r}{1-q^n}\right)^{1/(n-r)}.
\end{eqnarray}
I believe the derivative is
\begin{eqnarray}
f'(n)... | I strongly believe the following to be a solution, however, I would be grateful if somebody could quickly go through my reasoning to make sure I didn't go wrong.
--
Let $A \equiv 1-q^r$. Then
\begin{eqnarray*}
f'(n) &=& \frac{\delta}{\delta n}\left( \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \right) A^\frac{1}{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$
If $11 \gamma = \pi$ then prove that
$
\sec(\gamma) \sec(2\gamma) \sec(3\gamma) \sec(4\gamma) \sec(5\gamma) = 32.
$
I could not use the relation $11 \gamma = \pi$.
| Like How to expand $\cos nx$ with $\cos x$?,
we can prove for positive integer $N$:
$$\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$$
are $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Definite Integral of Polynomial of Sine and Cosines I was wondering if there is a way to compute definite integral
\begin{align}
I(m,n) :=\int_{0}^{\pi} \sin^m (\theta) \cdot \cos^{n}(\theta) \ \mathrm{d} \theta
\end{align}
in general for integer-valued $m$ and $n$. This problem arises when I try to compute integrals ... | When at least one of the exponents is odd, substitute $\sin^2 x = (1 - \cos^2 x)$ and $\cos^2 x = (1 - \sin^2 x)$ as in:
$$
\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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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 $$\Delta=\begin{vmatrix}
39 & -48&7 \\
-48... | The formula derived in this answer is
$$
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}
$$
since $\sigma=\operatorname{sign}\left(F\!\left(B^2-4AC\right)+\left(AE^2{-}BDE{+}CD^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving that $x-\frac{x^{3}}{6} < \sin(x)$ for $0I have to prove that $$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|.
$$
Therefore for each integer $m\ge 0$ and $x \in [0,\pi]$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Why is $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$? Suppose $\theta=2\pi/5$. Apparently it is true that $1+ \cos^2 \theta + \cos^2(2\theta) + \cos^2(3 \theta) + \cos^2 (4\theta) = 5/2$, or equivalently, $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$. What is the easiest way to see this?
I see that $1 + \cos \theta + \dotsb + \cos 4\t... | As $\pi=180^\circ$ $$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)!}$$ but I'm unsure how I'd s... | The number $n!$ can be defined recursively as follows:
*
*$1! = 1$
*$n! = n(n - 1)!$, $n \geq 1$
Using the definition to compute $5!$ yields
\begin{align*}
5! & = 5 \cdot 4!\\
& = 5 \cdot 4 \cdot 3!\\
& = 5 \cdot 4 \cdot 3 \cdot 2!\\
& = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1!\\
& = 5 \cdot 4 \cdot 3 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many non-negative integer solutions does the equation $3x + y + z = 24$ have? If the equation is $x + y + z = 24$ then it is solvable with stars and bars theorem. But what to do if it is $3x + y + z = 24$?
| Put $x=0\;,$ Then $y+z=24$. So we get Total ordered pair of $(y,z)$ is $= 25$
Put $x=1\;,$ Then $y+z=21$. So we get Total ordered pair of $(y,z)$ is $= 22$
Put $x=2\;,$ Then $y+z=18$. So we get Total ordered pair of $(y,z)$ is $= 19$
Put $x=3\;,$ Then $y+z=15$. So we get Total ordered pair of $(y,z)$ is $= 16$
Put $x=4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
What is the geometry behind $\frac{\tan 10^\circ}{\tan 20^\circ}=\frac{\tan 30^\circ}{\tan 50^\circ}$? This identity is solvable by the help of trigonometry identities, but I guess there is an interesting and simple geometry interpretation behind this identity and I can't find it.
I found it when I was thinking abou... |
Consider a triangle $ABC$ with $\angle A = 10^\circ$, $\angle B = 150^\circ$, and $\angle C=20^\circ$. Let $O$ be the circumcenter of the triangle $ABC$. Then
$$\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... | {
"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... | {
"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 suspect... | 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:
$$
\begin{align}
\sum_{k=0}^n\cos(k\alpha)
&=\frac12\sum_{k=0}^n\left(e^{ik\alpha}+e^{-ik\alpha}\right)\tag{1}\\
&=\frac12\frac{e^{i(n+1)\alpha}-1}{e^{i\alpha}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$ Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$.
I begin by letting $n=1$ then $\frac{1}{2}<\frac{1}{\sqrt{3}}$. Then assume $\frac{1\cdot 3\cdots(2k-1)}{2\cdot 4 \cdots(2k)}<\frac{1}{\sqr... | Here is the inductive step: from
$\;\dfrac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\dfrac{1}{\sqrt{2n+1}}$, you deduce
$$\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},$$
hence it is enough to prove $\;\dfrac{1}{\sqrt{2n+1}}\dfrac{2n+1}{2n+2}<\dfrac1{\sqrt{2n+3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is $\frac{1}{xy}$ convex for $x,y>0$ Is the function $$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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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:
$$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$
$$\lim_{h\to 0} \frac{\cos\left(\frac{\s... | You are right. First multiply and divide by the expression in $\sin()$
$$\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)}$$
Now, as $h \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solution to $\sqrt{x^2-2x+1} + 5 = 0$ I asked this question before, I wrote it wrong!
$$\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$$
My Question here is which solution is right, ... | Your friends are correct.
\begin{align*}
\sqrt{x^2 - 2x + 1} + 5 & = 0\\
\sqrt{x^2 - 2x + 1} & = -5\\
\sqrt{(x - 1)^2} & = -5\\
|x - 1| & = -5
\end{align*}
which has no solutions.
By convention, the expression $\sqrt{u}$ means the non-negative square root of $u$. Therefore, the equation
$$\sqrt{x^2 - 2x + 1} = -5$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to prove $\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$ Let $a,b,c>0,a+b+c=1$, show that
$$\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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
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:
$$a^2b^{-2} + 1 \equiv 0\pmod p\implies c\equiv -1\pmod p$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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=1... | 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 follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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 $$a! \cdot b! \mid (a+b)! \qq... | With the settings
\begin{align*}
[k]_q:=\frac{1-q^k}{1-q}\qquad\text{and}\qquad [k]_q!:=\prod_{j=1}^{k}[j]_q=\prod_{j=1}^{k}\frac{1-q^j}{1-q}
\end{align*}
the q-binomial coefficient $\begin{bmatrix}a+b\\a\end{bmatrix}_q$ is defined as
\begin{align*}
\begin{bmatrix}a+b\\a\end{bmatrix}_q:=\frac{[a+b]_q!}{[a]_q![b]_q!}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Find unit vector given Roll, Pitch and Yaw Is it possible to find the unit vector with:
Roll € [-90 (banked to right), 90 (banked to left)],
Pitch € [-90 (all the way down), 90 (all the way up)]
Yaw € [0, 360 (N)]
I calculated it without the Roll and it is
\begin{pmatrix}
cos(Pitch) sin(Yaw)\\
cos(Yaw) cos(Pitch)\\ ... | There are a lot of questions like this, all slightly different.
My earlier answer to a different question
is closely related to what you need, but the question is different enough
that I thought it better to write an answer a little more applicable
to your case.
I'll use right-handed $x,y,z$ Cartesian coordinates.
I vi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$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$$
For the s... | Here is one approach to the second integral:
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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;
$$
\begin{align}
\text{RHS} & =\cos5A \\
& =\cos(A+4A) \\
& =\cos A\cos4A-\sin A\sin4A \\
& =\cos A(2\cos^2 2A-1)-\sin A(2\sin2A\cos2A) \\
& =2\cos A\cos^2 2A-\cos A-2\sin A\sin2A\cos2A \\
& =... | Now use double angle formulae to write $\cos 2A$ in terms of $\cos A$, then turn any $\sin^2 A$ into $1-\cos^2 A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$ How do I prove the following inequality
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
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 i... | In short, you erroneously divided by $\Pr\{\text{sum is 9}\}$ when calculating $\Pr\{\text{sum is 9 and exactly one die lands on 6}\}$.
$$
\begin{align*}
\Pr\{\text{sum is 9 and exactly one die lands on 6}\}
& = \Pr\{(6,3),(3,6)\} \\
& = \Pr\{(6,3)\} + \Pr\{(3,6)\} \\
& = \left(\frac{1}{2}\right)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta $ I have the following integration to solve.
$$f(k) = \int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta,\quad0<k<1$$
assuming $\sin\theta = t$ which results $d\theta = \frac{dt}{\sqrt{1-t^2}}$
and when $\theta = 0, t=0$ and $\theta=\frac{\... | Possible hints to perfume some kind of calculations.
Leaving apart the extrema of the integra, for the moment.
$$\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}$$
$$\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"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.
$$\int\limits^{\infty}_{0}\int\limits^{\infty}_{0}{\frac{1}{(x+y)^{3/2}}\exp\left\{-\frac... | I also think that this integral diverges.
$$\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)$$ provided ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to show that $f$ is a straight line if $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$? Let $f:\mathbb R\to\mathbb R$ be continuous such that $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}~\forall~x,y\in\mathbb R.$ How to show that $f$ is a straight line?
| Define a linear function $g \colon \mathbb{R} \rightarrow \mathbb{R}$ by $g(x) = (f(1) - f(0))x + f(0)$. Note that by definition, $g(0) = f(0)$ and $g(1) = f(1)$ and that $g$ also satisfies
$$ g \left( \frac{x + y}{2} \right) = \frac{g(x) + g(y)}{2}. $$
You want to show that $g(x) = f(x)$ for all $x \in \mathbb{R}$. We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ by Mathematical Induction Prove by Mathematical Induction:
$$\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+\c... | $$1\cdot 2^2 + 2\cdot 3^2 +\cdots+ n\cdot (n+1)^2 =2^3 -2^2 +3^3 - 3^2 +\cdots+(n+1)^3 -(n+1)^2 =\left(\frac{(n+2)(n+1)}{2}\right)^2 -\frac{(n+1)(n+2)(2n+3)}{6} =(n+1)(n+2)\left[\frac{n^2 +3n +2}{4} -\frac{2n+3}{6}\right]= (n+1)(n+2)\left[\frac{3n^2 +5n }{12}\right]=\frac{1}{12} n (n+1) (n+2) (3n+5)$$
$$1^2\cdot 2 + 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Common tangent line to two functions I have two functions:
$$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^... | 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 $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
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 ... | Your answer is correct, since
$$-33\equiv 67\mod 100$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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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 $\sin t = \frac{\tan \frac{t}{2}}{(\tan \frac{t}{2})^2 + 1}$ an... | 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 $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$... | {
"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 $e_1 + e_2 + e_3 + e_4 + e_5 + e_6 = r$
... | 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 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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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: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{... | By solving the quadratic equation,$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 = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$
$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_... | I think u forgot to change the limits and also factor of 4 is missing while putting dx term.
Edited answer :
Sorry the previous answer I made a slight mistake in taking limits
So our final answer will look like as seen in the photo above.
As you can see that we get
$tan(\pi/12)$ in the answer.
By placing value of $ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find all integers $k$ such that $5^k\equiv 97 \pmod{101}$ Find integers $k$ such that $5^k\equiv 97 \pmod{101}$.
By brutal force, If $k=23$ then $101\vert 5^{23}-97$. Furthermore, by Euler-Fermat theorem, since $gcd(5,101)$ we have, $5^{100}\equiv 1\pmod{101}$, then for integer $r$, $5^{100r}\equiv 1\pmod{101}$, so $5^... | $97 \equiv -4 \mod 101$. $-4 * 25 = -100 \equiv 1 \mod 101$.
So if $5^k \equiv 1 \mod 101$ then $5^{k -2} \equiv 97 \mod 101$.
Now $5^{100} \equiv 1 \mod 101$ as 101 is prime so $\phi (101) = 100$ so $5^{\phi(101)} \equiv 1 \mod 101$.
$5^{100} \equiv 1 \mod 101$. So try $5^{50}$ and $5^{25}$ and we see $5^{25} \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$ $$
\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
$$
up to $n$ terms. I need help in solving this sum. I trie... | Hint: Use partial fractions:
$$
\frac1{(n-3)(n-2)(n-1)n}=-\frac1{2(n-2)}+\frac1{2(n-1)}-\frac1{6n}+\frac1{6(n-3)}$$
and note that it telescopes, so that you can find the partial sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Is there a mathematical way to solve this problem? The question goes something like this:
How many different ways can you add up 2, 3, and 5 to get a sum of 12. Numbers can be repeated, and all three numbers do not have to be used for each solution. For instance, 5 + 2 + 2 is an answer, and so is 2 + 5 + 2.
The answe... | Assuming that order of numbers doesn't matter, i.e. $5+5+2$ is considered the same as $5+2+5$ is considered the same as $2+5+5$ (as suggested by your list of ways in which $12$ can be written):
One way of approaching this is via Generating Functions. A rather in-depth treatment of the topic as a whole is given in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Laurent series of $\log(z)\sin(1/(z-1))$ I would like to calculate the Laurent series of $$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 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
$$-\frac{\arctan x}{x+1}+\frac 1 2\int \frac{1-x}{x^2+1}dx+\frac 1 2\int\frac{dx}{x+1}\\
=-\frac{\arctan x}{x+1}-\frac 1 2\int \frac{x}{x^2+1}dx+\frac 1 2\int\frac{dx}{x^2+1}+\frac 1 2 \int \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \... | $(n+1)^2 = n^2+(2n+1) $ so by induction $n^2= \sum 2k-1$
And for just the odd squares. $(2n+1)^2= 1 + \sum_{k=2;+2}^{2n} [(2k - 1)+(2k+1) ]= 1 +4\sum_{j=1}^{n }2j=1+8\sum j$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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I get a wrong determinant - why? I'm trying to calculate the following determinant:
$$\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_... | 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
$$
\det(x_1 + \alpha y_1,x_2,\cdots,x_n) = \det(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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The value of $\sum_{k=0}^\infty k^n x^k$ Any known conclusions for $\sum_{k=0}^\infty k^n x^k$ where $n$ is a nonnegative integer and $0<x<1$?
*
*If $n=0$, the sum is simply $\frac{1}{1-x}$.
*If $n=1$, the sum can be computed as shown in this post
*For general $n$, it seems the method by taking derivative is
extre... | I don't know of any simple solutions for the general summation, but I do know a method that will work. Since $0 < x < 1$ let $x = \frac{1}{y}$ where $y > 1$. Consider $n = 2$
Let $S = \sum_{k = 0}^\infty \frac{k^2}{y^k}$.
\begin{align}
S &= \frac{1}{y} + \frac{4}{y^2} + \frac{9}{y^3} + \frac{16}{y^4} + \cdots\\
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Find sum of $1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$ Find the sum of the following series :
$$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}-\fra... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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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 $$y^2 - 2 \equiv 0 \pmod{7} \implies y^2 \equiv 2 \pmod{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The series $\sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k}$ is bounded below by $c \log\log x$ near 0, thus fails to be uniformly convergent The convergent series
$$ \sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k} $$
defines a function in $H^{1/2}([0,2\pi])$. This is an example of such a series for which convergence on is not... | The easiest - in some sense - way to show that the series doesn't converge uniformly on $(0,2\pi)$ is to note that all terms of the sequence are continuous functions with common period $2\pi$, and if the series were uniformly convergent on $(0,2\pi)$, it would by continuity be uniformly convergent on $[0,2\pi]$ (and by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2... | EDIT : The previous solution was utterly mistaken.
Let$$\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})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$A, B$ and $C$ can do a piece of work
$A, B$ and $C$ can do a piece of work in $18$, $24$ and $30$ days respectively. $A$ and $B$ work together for a certain number of days. Then $B$ leaves and $C$ joins. They work together for half the number of days for which $A$ and $B$ had worked together. After that $A$ goes away... | Let the time worked by $A$ and $B$ together be $x$ days.
Then, the fraction of work done by $A$ and $B$ together $=\frac x{18}+\frac x{24}$.
Time worked by $A$ and $C$ will then be $\frac x2$ days.
The fraction of work done by them $=\frac{x}{2\times18}+\frac{x}{2\times30}=\frac x{36} + \frac x{60}$
Time worked by $B$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form for the infinite product $\prod\limits_{k=0}^{\infty} \left( 1-x^{2^k} \right)$ There is a known identity:
$$\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)=\frac{1}{1-x}, ~~~~~|x|<1$$
It's easy to derive it by converting it to a telescoping product as shown in this answer.
However, we can't use the same meth... | Not a closed form, but we have the following series representations:
$$\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}}... | {
"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 $2x2$. N... | $Cx = \frac{1}{2}au + \frac{1}{3}bv$
$x = au + bv \implies Cx = C(au +bv) = Cau + Cbv$
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 orthanorm... | {
"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 $A$,
$$\begin{bmatrix}
a &b &c \\
d &e &f
\end{bmatrix} \begin{bmatrix}
2&-1 \\
5 & 3\\
-2& 1
\end{bmatrix} = \begin{bmatrix}
1&... | Alternative less advanced solution. Multiply it with its own transpose:
$A^TA$ is a $2\times2$ matrix:
$\left(
\begin{array}{cc}
33 & 11 \\
11 & 11 \\
\end{array}
\right)$
This has a trivial inverse by swapping the diagonals and finding the determinant:
$\frac{1}{22}\left(
\begin{array}{cc}
1 & -1 \\
-1 & 3 \\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
3 x 4 Matrix with Complex Numbers I have a $3 \times4$ matrix $$
\begin{pmatrix}
i & 4c & -1 & 0 \\
2ic & 4 & -2c & 0 \\
-i & -2 & 1 & 6c-3 \\
\end{pmatrix}
$$
to figure out the complex numbers $c$ such that the row space has dimension 2, how would I approach this problem?... | I think reducing by rows can help:$${}$$
$$ \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}$$... | {
"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.
*$\triangle DEF$ i... | 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 re... | {
"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.