Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute $A^5-4A^4-7A^3+11A^2-A-10I$ for $A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$ by Cayley-Hamilton Question: Use Cayley-Hamilton theorem to express $A^5-4A^4-7A^3+11A^2-A-10I$, where $$A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$$
Attempt: I only know how to get the characteristic polynomial equation which is $\lambda^2-... | Continuing your analysis,
$$A^2 - 4A -5I=0\implies A^2 = 4A+5I$$
Thus,
$$\begin{array}{cc}
A = A\\
A^2 = 4A + 5I\\
A^3 = 4A^2 + 5A &= 4(4A+5I) + 5A = 21A +20I\\
\vdots
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $A: l_2 \to l_2$ is a bounded operator Let us consider the following linear operator acting on $l_2$:
$$ A(x_1,x_2,x_3,\ldots) ~\colon=~ \left(x_1,\frac{x_1+x_2}{2},\frac{x_1+x_2+x_3}{3},\ldots\right) $$
I need to show that $A$ is a bounded operator, that is $||Ax|| \leq C~||x||$ for some constant $C$ and all $... | *
*We show that $\left|\sum_{j=1}^nx_j\right|\leq \left(\sum_{j=1}^nx_j^2\sqrt j\right)^{1/2}\left(\sum_{j=1}^n\frac 1{\sqrt j}\right)^{1/2}$, applying Cauchy-Schwarz inequality $\sum_j a_jb_j\leq \sqrt{\sum_ja_j}\sqrt{\sum_jb_j}$ to $a_j=x_j\sqrt j$ and $b_j=\frac 1{\sqrt j}$.
*We have $$\sum_{j=1}^n\frac 1{\sqrt j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Finding root of a polynomial Let's say I've got $ax^3 + bx^2 + cx + d = 0$
I want to find x for a general-case.
Mathematica gave me a very long solution, and even longer for one which has $kx^4$.
Can you please help me reproduce the algorithm used by Mathematica to generate formula of:
$$P(x) = a_0x^0 + a_1x^1 + a_2x^2... | Here's a (standard) method to treat the cubic case, though it hides some of what is going on, which is basically Galois Theory. You might as well suppose that $a \neq 0$ (otherwise you would be dealing with a quadratic at worse). Hence you can divide through and assume that $a = 1.$ So let's work instead with the cubic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\int \sin^n{x} \ dx$ I am working on trying to solve this problem:
Prove: $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$
Here are the steps that I follow in the example that I am reading:
$u = \sin^{n - 1}{x}$
$du = (n - 1) \cdot \sin^{n - ... | First let us denote
$$I_n = \int \sin^n{x} \ dx $$
$$ \int u(x) v'(x) dx = u(x) v(x) - \int v(x) u' (x) dx $$
Here $u(x) = \sin^{n-1}{x} \hspace{3pt}$ and $\hspace{3pt} v'(x) = \sin x $
$ \Rightarrow v(x) = -\cos x$
Therefore
$$
\begin{align*}
I_n &= -\cos x \hspace{3pt} \sin^{n-1}x + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a^2 + b^2 + c^2 $ is not a prime number I am having difficulty solving this problem:
Let $a, b, c \in\mathbb{Z}$, $abc \neq 0$ and $a\neq c$ be such that
$$\frac{a}{c} = \frac{a^2+b^2}{c^2+b^2}.$$
Prove that $a^2 + b^2 + c^2$ is not a prime number.
Thanks in advance!
| W.l.o.g we may assume that $a,c>0$. The equation
$$
\frac{a}{c}=\frac{a^2+b^2}{c^2+b^2}
$$
together with the assumption $a\neq c$ quickly gives us $ac=b^2$ as a corollary. Therefore we have
$$
a^2+b^2+c^2=a^2+ac+c^2
$$
and the extra condition that $ac=b^2$ must be a perfect square.
There are two main cases. If $gcd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n=... | A trick I learnt from Bill Dubuque.
Write it as
$$\frac{n^5 - n}{5} + \frac{n^4 - n}{2} + \frac{n^3 - n}{3} + n$$
(See Bill's answer here: Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 3
} |
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, ... | Since powers of x and y is always greater than or equal to zero, You can prove it by mathematical induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 2
} |
Chinese Remainder theorem with non-pairwise coprime moduli Let $n_1,...,n_k \in \mathbb{N}$ and let $a_1,...,a_k \in \mathbb{Z}$.
How to prove the following version of the Chinese remainder theorem (see here):
There exists a $x \in \mathbb{Z}$ satisfying system of equations:
$$x=a_1 \pmod {n_1}$$
$$x=a_2 \pmod {n_2}$... | If we factor $n_k$ into primes, $n_k = p_{1}^{b_{1}}\cdots p_r^{b_{r}}$, then the Chinese Remainder Theorem tells us that $x\equiv a_k\pmod{n_k}$ is equivalent to the system of congruences
$$\begin{align*}
x&\equiv a_k\pmod{p_1^{b_{1}}}\\
x&\equiv a_k\pmod{p_2^{b_{2}}}\\
&\vdots\\
x&\equiv a_k\pmod{p_r^{b_{r}}}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 1,
"answer_id": 0
} |
Exercise on irreducible polynomials I know that this is not the right place for questions like that, but I need someone that explain me step-by-step how can I resolve this exercise (I've exam in the next days):
Write as products of irreducible factors the polynomial $f=x^3-3x^2+x-3 \in\mathbb{R}[x]$ and $f=x^3-\overlin... | This polynomial factors, and that can be seen using the grouping method. If your polynomial has four terms like this one, the grouping method might work. You group terms together two at a time and factor out a Greatest Common Factor:
$$
\begin{align*}
x^3-3x^2+x -3 & = x^2(x-3)+1(x -3)\\
& = (x^2+1)(x-3)
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find the integral $\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$ Could somebody tell me how to find the integral
$$\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$$
for constants $a$ and $b$?
Thanks!
| I will assume that $a, b$ are positive constants, in order to circumvent singularity issues. By the substitution $z = \sqrt{b} \, x$, the integral in question is equal to
$$ \frac{1}{\sqrt{b}} \int_{-\infty}^{\infty} \frac{e^{-z^2}}{a + e^{-z^2}} \; dz.$$
From the identity
$$ 1 + x^{2n+1} = (1 + x)(1 - x + \cdots - x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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outer automorphisms of $S_6$ $$
\begin{array}{|l|c|c|}
\hline
\text{cycle structure} & \text{number of permutations} & \text{order} \\
\hline
6 & 120 & 6 \\
5+1 & 144 & 5 \\
4+2 & 90 & 4 \\
4+1+1 & 90 & 4 \\
3+3 & 40 & 3 \\
3 + 2 + 1 & 120 & 6 \\
3 + 1 + 1 + 1 & 40 & 3 \\
2 + 2 + 2 & 15 & 2 \\
2 + 2 + 1 + ... | I think the best source for understanding the outer automorphisms of $S_6$ is the following paper:
Combinatorial Structure of the automorphism group of $\mathbf{S_6}$ by T.Y. Lam and David B. Leep, Expositiones Mathematicae 11 (1993), no. 4, pp. 289-903.
It describes exactly how to think about the exchange map betwee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
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Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under ... | Edit: Note that when the answer below was posted, and for a long time thereafter, the expression on the right was $\dfrac{1}{\sqrt{1+ab}}$.
I am puzzled. Suppose without loss of generality that $a\le b$. Then $1+a^2\le 1+ab$, and therefore $\dots$. So the desired inequality is true for any positive $a$, $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Expressing an algebraic function as a single expression I have the following question in my text book:
Express as a single fraction
$$ \frac{3}{x-4} - \frac{2}{(x-4)^2} $$
The answer the book gives is this:
$$ \frac{3x-14}{(x-4)^2} $$
I understand how they get to this through these steps:
$$ \frac{3(x-4)-2}{(x-4)^2} = ... | It is never allowed to cancel across a plus or minus in a fractional expression. If it were possible, strange things would happen. Consider your expression when $x = 6$:
$$
\frac{3(x-4) - 2}{(x-4)^2} = \frac{3(6-4)-2}{(6-4)^2} = \frac{3(2) - 2}{2^2}
= \frac{4}{4} = 1.
$$
But if you tried canceling the $x-4$ firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Induction Proof: Proof of Strict Inequality involving Exponents of 3 in the Denominator. I have been trying to do this problem by using induction but I became stuck halfway through:
Use induction to show that
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{n^3}\right) \lt 3 - \frac{1}{n^2}$$
for $n\geq... | Your sum converges to
$$
\sum_{n=1}^\infty \frac{1}{n^3} = \zeta(3),
$$
because the series $ \sum_{n=1}^\infty \frac1{n^{1+\varepsilon}} $ (cf. Riemann zeta function) converges for every ε > 0, because
$$
\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Proving the maximum of $(x^2-y^2-1)^2+4x^2y^2$ occurs at $x=0$, $y=\pm 1$. I'm trying to get some practice using the Maximum Modulus theorem, and want to use it to conclude that the maximum of $(x^2-y^2-1)^2+4x^2y^2$ occurs at $x=0$, $y=\pm 1$, supposing $x^2+y^2\leq 1$.
My thinking is I want to find some suitalbe com... | $x^2+y^2=1$
$(x^2-y^2-1)^2+4x^2y^2=(x^2+y^2)^2-2(x^2-y^2)+1=4-4x^2\leq 4$
$x=0,y=±1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$
Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$
I was just wondering what I am doing wrong, as I don't seem to be arriving at the correct answer for $\sin(\frac{\pi}{8})$
What I did:
Let $\theta = \frac{\pi}{8}$
$\cos(2\theta) = 2\cos^2(\theta) - 1$
$\... | The trick is to rationalize the denominator by multiplying by the conjugate, along the following lines:
$$\begin{align}\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}&=\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}\cdot\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\\&=\dfrac{\sqrt{2}\sqrt{2-\sqrt{2}}}{2 \sqrt {4-2}}=\dfrac{\sqrt{\not 2}\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$. I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome
Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble by $2^{n+1}$.
here n belongs to natural numbe... | This is similar to Henry's answer, but things are simpler if we use $4+2\sqrt{3}=(1+\sqrt{3})^2$ and $4-2\sqrt{3}=(1-\sqrt{3})^2$ instead. These satisfy $x^2-8x+4=0$.
The sequence defined by
$$
a_0=2\text{, }a_1=8\text{, and }a_n=8a_{n-1}-4a_{n-2}\tag{1}
$$
has the solution
$$
\begin{align}
a_n
&=(4+2\sqrt{3})^n+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/127893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$ The question I am working on: Evaluate
$$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$$
So using integration by parts to solve:
(letting $u=(1-x)^4$ and $dv=x^4$)
$$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x... | Hint :
Rewrite integral into form :
$$I=\int \frac{x^8+x^6}{1+x^2} \,dx + \int \frac{x^6+x^4}{1+x^2} \,dx +4\cdot \int \frac{x^6}{1+x^2} \,dx -4\cdot \int \frac{x^7+x^5}{1+x^2} \,dx$$
for third integral do long division .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Probabilistic method: what is the probability of the event that randomly selected composition of $n$ Can anyone help me out with the following question?
Q. What is the probability of the event that randomly selected composition of $n$ has a second part and that second part is $1$?
I know that the expected number, $E(X)... | Obviously for $n=1$ the probability is $0$. I suggest that you do some experimenting with small values of $n$, say $n=2,3,4$, and $5$. The compositions that meet the requirement are as follows.
$n=2$: $$1+1$$
$n=3$: $$\begin{align*}&1+1+1\\&2+1\end{align*}$$
$n=4$: $$\begin{align*}&1+1+1+1\\&1+1+2\\&2+1+1\\&3+1\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/129768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Basketball Team Combinatorial Question: A basketball team has 5 players, 3 in forward position (which includes a center) and 2 in guard position. In how many ways can we make a team if there are 6 forwards, 4 guards and 2 people who can play forward or guard?
The way I am approaching this question is as follows:
We nee... | When you write $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$, you consider the possibility of not using the free positioned player at all, thrice. So, I think you should subtract $2\times \binom{6}{3}\times \binom{4}{2}$ from $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$ How to prove: $$\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$$ Help please. Don't know where to start.
| $$\sin2\theta$$
$$2\cdot\sin\theta\cdot \cos\theta$$
multiply and divide by $\cos\theta$
$$ 2\cdot \dfrac {\sin\theta}{\cos\theta}\cdot \cos^2\theta$$
$$2\cdot\tan\theta\cdot\cos^2\theta$$
$$\dfrac{2\cdot\tan\theta}{\sec^2\theta}$$
$$\dfrac{2\cdot\tan\theta}{1+\tan^2\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear?
Anyways, I did:
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} +... | For these types of problems, try to substitute the variable $x$ so that the integral reduces into a form that is easier to integrate.
For this integral, I would start by completing the square in the integrand:
$\sqrt{10x-x^2} = \sqrt{-(x^2-10x)}= \sqrt{-((x-5)^2-25)}=\sqrt{25-(x-5)^2}$.
Note that, so far, I have made n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Simplify a trigonometric expression I've been simplying a slew of trigonometric expressions and most of them fall out pretty clearly, but this one has been giving me fits:
$$\frac{(\sec x - \tan x)^{2} + 1}{\sec x\csc x - \tan x\csc x}$$
Here's my best attempt so far:
$$\begin{align*}
&=\frac{\left(\frac{1}{\cos x}-\fr... | You have
$$\begin{align*}
&=\frac{(\frac{1}{\cos x}-\frac{\sin x}{\cos x})^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\
&=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}}
\end{align*}$$
Your next steps might be to note that on top, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/142045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Show that $\sum\limits_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\frac{N+1}{2}$ I have recently found this excercise and was not able to solve it so far.
Show that
$$\sum_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\frac{N+1}{2}\;,$$ where $m \in \lbrace1,2,...,N\rbrace$.
This was one of my attempts:
$$\sum_{i=1}^N\sin^2\frac{m\pi i}{N+1}=\... | $$
\sum_{k=1}^{n} \cos\left( \frac{2 m \pi k}{n+1}\right) = \Re \sum_{k=1}^{n} \exp\left( i \frac{2 m \pi k}{n+1}\right) = \Re\left( \sum_{k=1}^n z^k \right) = \Re\left( \frac{1-z^n}{1-z} z \right)
$$
where $z= \exp\left( i \frac{2 m \pi}{n+1}\right)$. Since $z^{n+1} = 1$:
$$
\Re\left(\frac{z-1}{1-z}\right) = -1
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Root equation - What am I missing? There's a problem of which I know the solution but not the solving process:
$(\sqrt{x} + 7)(\sqrt{x} - 1) = \frac{105}{4}$
I'm convinced that up to:
$x + 6\sqrt{x} - 7 = \frac{105}{4}$
everything is correct.
But afterwards, I never seem to be able to get to the solution of $x = \f... | Hint $\ $ Put $\:z = \sqrt{x},\:$ clear denominators and rearrange to obtain the quadratic equation$\:0 = 4 z^2 + 24 z - 133 = (2z-7)(2z+19),\:$ solvable by the quadratic formula, or the Rational Root Test or AC method, etc.
Alternatively, purely arithmetically, if it has rational root $\:\sqrt{x} = a/b\:$ in lowest t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/145498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Square root of negative numbers If:
$$a = \sqrt{ b^2 - b }$$
The problem I have is that for values of:
$0 < b < 1$
the result of:
$b^2 - b$
Is a negative number which gives rise to an error on Excel and my calculator.
I understand that negative numbers don't have square roots (I read it on Wikipedia at least), so how ... | If $b^2-b<0$, then $b-b^2>0$ and $$a = \pm i\sqrt{ b - b^2 },$$ where $i$ is the imaginary unit, which by definition is the unique complex number that satisfies $$i^2=-1\Leftrightarrow i=\pm\sqrt{-1}.$$
The complex numbers are numbers of the form $a+bi$, where $a$ and $b$ are real numbers. They appear e.g. in the sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Is there a quicker way of doing this integral? I have an integral $\int_0^\infty {x^2\over 1+x^4} dx$. I gave it a go and it turned out quite messy, so I consulted Wolfram Alpha but the steps given there seem rather long winded too. Is there is a faster way of doing the integral?
| $$\int_0^\infty\frac{x^2}{1+x^4}dx$$
Let, $t=\frac{1}{x}$, and, $dt=\frac{-dx}{x^2}$
$$\int_\infty^0 \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$
$$=\int_\infty^0\frac{-dt}{1+t^4} =\int_0^\infty \frac{dt}{1+t^4}$$
Follow Norberts solution after that.
Another way to do this is
$$
I=\int_0^\infty\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \... | You reached the point where you want to prove $\,\displaystyle{\frac{3n^2+n+1}{n^3+n^2}<\epsilon}\,$ , but $$\frac{3n^2+n+1}{n^3+n^2}\leq \frac{5n^2}{n^3}\leq \frac{5}{n}$$ so $\,\displaystyle{\frac{5}{n}<\epsilon \Longrightarrow n>\frac{5}{\epsilon}}\,$ , and thus it is enough to choose $\displaystyle{\,N_\epsilon:=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/147109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Determinant of matrix composition Considering a matrix given by composition of square matrices like:
$$
M = \begin{pmatrix}
A & B\\
C & D
\end{pmatrix}
$$
I want to calculate its determinant $|M|$. Consider that all components are square matrices of course.
Can this be related somehow to the components $A$, $B$, $C$ an... | Note that $$\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \begin{pmatrix} A & 0 \\ C & I \end{pmatrix} \begin{pmatrix} I & A^{-1}B \\ 0 & D - CA^{-1}B \end{pmatrix} = \begin{pmatrix} A - BD^{-1}C & BD^{-1} \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ C & D\end{pmatrix}$$
The middle factorization is valid assuming $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/147227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_... | *
*The determinant of a block diagonal matrix is equal to the product of the determinants of the diagonal blocks. In your case, you have a block diagonal matrix of the form
$$A=\left(\begin{array}{ccc}
A_1 & 0 & 0\\
0 & A_2 & 0\\
0 & 0 & A_3
\end{array}\right)$$
so $\det(A) = \det(A_1)\det(A_2)\det(A_3)$.
*You can, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
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Analyze the convergence or divergence of $\{1/n^2\}$
Analyze the convergence or divergence of $\{1/n^2\}$
$$\left|\frac{1}{m^2} - \frac{1}{n^2}\right| < \left|\frac{1}{m^2}\right| < \frac{1}{N^2} < \varepsilon$$
whenever $N > \dfrac{1}{\sqrt{\varepsilon}}$ (first step because $n^2 > 0$ and hence $\dfrac{1}{n^2} > 0$ ... | The reasoning is not quite correct, but it is in fact a Cauchy sequence that converges to $0$.
$\frac{1}{n^2}>0$ is true when $n$ is a positive integer, and this does imply that $\frac{1}{m^2}-\frac{1}{n^2}<\frac{1}{m^2}$. However, that does not mean that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|<\frac{1}{m^2}$. For ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to compute this integral involving a cdf? $\int_0^\infty\Phi(\frac{-x}{\sqrt{2}})d\Phi(x)=?$ where $\Phi(x)$ is the cumulative distribution function of a standard normal random variable.
| Consider $I(a) = \int_0^\infty \Phi(a x) \mathrm{d} \Phi(x)$. Differentiate with respect to $a$, and denote $\phi(x) = \Phi^\prime(x)$:
$$
I^\prime(a) = \int_0^\infty x \phi(a x) \phi(x) \mathrm{d} x = \frac{1}{2 \pi} \int_0^\infty x \mathrm{e}^{-\frac{(1+a^2) x^2}{2}} \mathrm{d} x = \frac{1}{2 \pi} \frac{1}{1+ a^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculate sums of inverses of binomial coefficients How to calculate the sum of sequence $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？$$ How about its limit?
| As an alternative, starting from this result by robjohn's answer [tag (7)]
$$\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}=\frac{\sum_{k=1}^{n+1}\frac{2^k}{k}}{\frac{2^{n+1}}{n+1}}$$
by Stolz-Cesaro we obtain
$$\frac{\frac{2^{n+2}}{n+2}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/151441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 5,
"answer_id": 4
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Simultaneous equations My question is: Solve simultaneously:(anwers are in integers)
$$\begin{align} y^3 - 9x^2 + 27x - 27 &= 0 \\
z^3-9y^2+27y-27 &= 0 \\
x^3-9z^2+27z-27 &= 0 \end{align}$$
Any hints to solve this question would be greatly appreciated.
| Edit: It turns out that it is only integer solutions that are being asked about. Then there is a simpler approach that involves play with inequalities. Below is the original answer, followed by a solution of the integer version of the problem.
A start: Note that the first equation can be rewritten as
$$y^3+(x-3)^3=x^3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrices with $A^3+B^3=C^3$ Problem: Find infinitely many triples of nonzero $3\times 3$ matrices $(A,B,C)$ over the nonnegative integers with
$$A^3+B^3=C^3.$$
My proposed solution is in the answers.
| OR OR OR, given
$$ x, y > 0, $$
let
$$ R \; = \;
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
x & 0 & 0
\end{array}
\right) , \; \;
S \; = \;
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
y & 0 & 0
\end{array}
\right) , \; \;
T \; = \;
\left( \begin{array}{ccc}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Mod operator and divisibility by $3$? I wanted to prove that $n(n+1)(2n+1)$ is always divisible by three, for this I used the following approach:
$n = 0, 1\pmod2$
for $n = 0$;
$$n(n+1)(2n+1) \equiv 0\pmod3$$
for $n = 1$;
$$n(n+1)(2n+1) \equiv 0\pmod3$$
But now I am unsure if this the correct way of doing because of the... | Every number $n$ is congruent to $0$, $1$, or $2$ mod $3$.
If $n \equiv 0 \pmod 3$, then $3 \mid n$.
If $n \equiv 1 \pmod 3$, then $2n \equiv 2 \pmod 3$ which implies $2n + 1 \equiv 3 \pmod 3$ which implies $2n + 1 \equiv 0 \pmod 3$.
If $n \equiv 2 \pmod 3$, then $n + 1 \equiv 3 \pmod 3$ which implies $n + 1 \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Integral of $\int \tan^4 x \, \sec^6 x dx$ I am trying to find the integral of $$\int \tan^4 x \, \sec^6 x dx$$
I tried to rewrite as trig identities using $\sec^2 - \tan^2 = 1$ but that got me nowhere so I wrote it like this.
$$\int \frac{\sin^4x}{\cos^4 x} \frac{1}{\cos^6 x} dx$$
$$\int \frac{\sin^4x}{\cos^{10} x} dx... | Here is a relatively easier way. Remember that $\sec^2(x) = 1 + \tan^2(x)$. Hence the integral $$I = \int \tan^4(x) \sec^6(x) dx = \int \tan^4(x) \sec^4(x) \sec^2(x) dx\\ = \int \tan^4(x) \left(1 + \tan^2(x) \right)^2 \sec^2(x) dx $$
Let $\tan(x) = t$. Hence, $\sec^2(x) dx = dt$. Hence, we get that
$$I = \int t^4 \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$ My question is: Solve $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$
I deduced that:$LHS= x+1-(x-2)$
I am unable to solve this equation. I would like to get some hints to solve it.
| $\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|=3$
$|x+1|-|x-2|=3$
1) $x\in(-\infty, -1)$$\Rightarrow$$|x+1|=-(x+1)=-x-1$, $|x-2|=-(x-2)=2-x$.
$|x+1|-|x-2|=3$$\Rightarrow$ $-x-1-2+x=3$$\Rightarrow$$-3=3$, this is a contradiction.
In this interval equation has no solution.
2) $x\in[-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Strategies to find the set of functions $f:\mathbb R\to\mathbb R$ satisfing the functional equation: $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$ My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case wher... | Using the equation, we can see that $f(0) = 0$ is necessary. Let $y = -x$, then
\begin{gather}
f(x^3) + f(-x^3) = 0
\end{gather}
This implies that $f$ must be an odd function. Also letting $y = x$ we see that
\begin{equation}
f(x^3) = x f(x^2)
\end{equation}
Plugging this relationship into the LHS gives
\begin{align}
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find the vertical and horizontal asymptotes of the function. I am asked to find the vertical and horizontal asymptotes of the equation:
$$f(x)=(a^{-1}+x^{-1})^{-1}$$
I simplify this to
$$f(x)=\frac{1}{a^{-1}+x^{-1}}$$
$$f(x)=a^1+x^1$$$$f(x)=a+x$$Which is some constant, graphed as horizontal line - that will not have a ... | \begin{align}
f(x) & = (a^{-1} + x^{-1})^{-1} = \dfrac1{a^{-1} + x^{-1}}\\
& = \dfrac1{\dfrac1a + \dfrac1x} = \dfrac1{\dfrac{x+a}{ax}}\\
& = \dfrac{ax}{x+a} = \dfrac{ax + a^2 - a^2}{x+a} \text{(Adding and subtracting $a^2$ to the numerator)}\\
& = \dfrac{ax+a^2}{x+a} - \dfrac{a^2}{x+a} = \dfrac{a(x + a)}{x+a} - \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| $$\int\frac 1{1+x^4}dx=\frac12\int\frac{1+x^2+1-x^2}{1+x^4}dx$$
$$\int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{\left(x-\frac1x\right)^2+2}dx$$
Set $x-\frac1x=\sqrt2\tan\phi$
$$\int\frac{1-x^2}{1+x^4}dx=-\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}dx$$
Set $x+\frac1x=\sqrt2\sec\psi$
Reference: Trigonometri... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 0
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When does $q^b = (p^a - 1)/(p - 1)$? $\DeclareMathOperator{\Ord}{Ord}$When does $q^b = (p^a - 1)/(p - 1)$ for $p, q$ odd primes and $a, b$ odd integers $> 1$ ? If no examples are possible, please give a simple proof.
A proof of this for $q=3, p=5$ might be
Assume $3^b = (5^a - 1)/4$; then
\begin{equation}
5^a - 4 ... | q^b = (p^a - 1)/(p - 1)
Now for q^b to be defined p is ne 1
p is an odd prime number therefore p = 3 or p = (6k - 1) or p = (6k + 1) where k is a positive integer.
when p = 3
q^b = (3^a - 1)/(3 - 1)
when p = 6k - 1
q^b = ((6k - 1)^a - 1)/(6k - 2)
= [(6k - 1 - 1)(6n + 1)]/(6k - 2) in atleast one of the instances.
= [(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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a representation of $f''$ only using terms of $f$ and constants. Using Taylor formulare How can I prove the following result?
$
f''\left( x \right) \approx Af\left( x \right) + Bf\left( {x + h} \right) + Cf\left( {x + 2h} \right) + Df\left( {x + 3h} \right)\,\,h > 0
$
Obviously I have to consider the taylor expansion ... | As for $1 \le i \le 3$
\[ f(x+ih) = f(x) + ihf'(x) + \frac 12 {i^2}{h^2}f''(x) + \frac 16{i^3}{h^3}f'''(x) + O(h^4)
\]
So we have for your right hand side
\begin{align*}
f(x)\bigl(A + B + C + D)\\
{} + f'(x)\bigl(B + 2C + 3D\bigr)h\\
{} + f''(x)\frac 12\bigl(B + 4C + 9D\bigr)h^2\\
{} + f'''(x)\frac 16 \bigl(B + 8C +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that
$$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$
I have reduced it the form
$$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$
I am not sure what to do next to get snappy solution. Any ideas?
| This refers to the original question, which had the left hand side equal to $\frac{1}{2}$ instead of $\frac{1}{12}$.
They are not equal.
Your simplification is correct. Then we can rewrite the left hand side as
$$\left(\frac{4^{\ln(4)}}{3^{\ln(3)}}\right)^{1/\ln(3/4)}$$
so raising both sides of the equation to the $\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2... | The standard way is to find one solution to $x^2-2y^2=2$, e.g., $x=2$, $y=1$, and find the fundamental solution to $x^2-2y^2=1$, which is $x=3$, $y=2$, and then go $(2+\sqrt2)(3+2\sqrt2)^n$, etc., etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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For prime $p>2: 1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
Possible Duplicate:
Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}\pmod p$?
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$
... | $$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} \pmod p$$
is the same as
$$1^2 2^2 3^2 \cdots
\left(\frac{p-1}{2}\right)^2
\equiv (-1)^{\frac{p-1}{2}} \pmod p,$$
since we are working modulo $p$.
Now we can split each square itself as $ i^2 = i(p-i)(-1) $
$$(-1)^{\frac{p-1}{2}} 123\cdot\cdot\cdot(p-1)\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Computing the value of $(\frac{p-1}{2})!$ modulo $p$. I want to prove that for $p \geq 3$, and for $a=(\frac{p-1}{2})!$, if $p \equiv1\pmod 4$, then $a^2\equiv -1 \pmod p$, and if $p \equiv 3\pmod4$, then $a \equiv +/-1 \pmod p$.
For the first part, I used Wilson's theorem which says that for prime $p$, $(p-1)!=-1 \pmo... | You got the hard part. To summarize, we have $a^2\equiv (-1)^{(p+1)/2}\pmod{p}$.
If $p$ is of the form $4k+1$, then $(p+1)/2$ is odd, and therefore $a^2\equiv -1\pmod{p}$.
If $p$ is of the form $4k+3$, then $(p+1)/2$ is even, and therefore $a^2\equiv 1\pmod{p}$. It follows that $a\equiv 1\pmod{p}$ or $a\equiv -1\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
This seems clear, but I don't know how to prove this..
I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
| Lemma: $\gcd(a,b)=1\implies\gcd\left(a^2,b^2\right)=1$
Proof: By Bezout, $\gcd(a,b)=1$ implies
$$
\begin{align}
ax+by&=1\\
a^2x^2&=1-2by+b^2y^2\\
\left(2y-by^2\right)b&=1-a^2x^2\\
\left(2y-by^2\right)^2b^2&=1-2a^2x^2+a^4x^4\\
\left(2x^2-a^2x^4\right)a^2+\left(2y-by^2\right)^2b^2&=1
\end{align}
$$
Therefore, $\gcd\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 13,
"answer_id": 8
} |
Infinite product How do I solve the infinite product of $$\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}?$$
I know that I have to factorise to $$\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)},$$
but how do I do the partial product?
Thanks a lot in advance.
If I'm not mistaken the Answer is 2/3
| After factorization, the product looks like $(\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}\frac{5}{7}\cdots)(\frac{7}{3}\frac{13}{7}\frac{21}{13}\frac{31}{21}\frac{43}{31}\cdots)=(2)(1/3)=2/3$.$$$$ Here terms in first () are from expression $\frac{n-1}{n+1}$ and terms in second () from expression $\frac{n^2+n+1}{n^2-n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b... | On what grounds did you move $b$ to the top on (c)? It's incorrect. You cannot just multiply by $\frac{b}{1}$ because it pleases you to do so. And, after you suddenly create a factor of $\frac{b}{1}$ ex nihilo, it would have cancelled with the denominator. So both the penultimate and antepenultimate equality signs are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Verify trigonometry equation $\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A$ Sorry for asking so many of these type of questions.
How would I verify the following trigonometry identity:
$$\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A.$$
My work is
$$\frac{\sin A + \frac{\sin A}{\cos A}}{\frac{\cos A}{\sin A} +... | $\begin{align}
\dfrac{\sin A + \tan A}{\cot A + \csc A} &= \dfrac{\sin A + \dfrac{\sin A}{\cos A}}{\dfrac{\cos A}{\sin A} + \dfrac{1}{\sin A}}\\
&= \dfrac{\dfrac{\sin A}{\cos A}\color{green}{\left(\cos A + 1\right)}}{\dfrac{1}{\sin A}\color{green}{(\cos A + 1)}}\\
&=\dfrac{\sin A}{\cos A} \div \dfrac{1}{\sin A}\\
&= \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$
Let $x,y,z >0$. Prove
that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}}
\ge \sqrt{3}$$
My solution:
By Hölder,
$$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\... | The function $f(x) = \frac 1 {\sqrt x}$ is convex, by Jensen's inequality
$$
\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}} \geq \\ \frac {(x + y + z)^{3/2}} { \sqrt{x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2)} }
$$
Our inequality becomes
$$
x(y^2 + yz + z^2) + y(z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
evaluate the integral: $\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$ $$\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$$
I got $4(5^\frac{1}{4})$ but I'm not sure if thats right
| Your computation is incorrect.
The substitution $u=x^2+4$ gives $du = 2x\,dx$. When $x=0$, we have $u=4$; when $x=\sqrt{5}$, we get $u=9$. So
$$\begin{align*}
\int_0^{\sqrt{5}}\frac{4x}{\sqrt{x^2+4}}\,dx &= 2\int_0^{\sqrt{5}}\frac{2x\,dx}{\sqrt{x^2+4}}\\
&= 2\int_4^9\frac{du}{\sqrt{u}}\\
&= 2 \int_4^9 u^{-1/2}\,du\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to conceptualize conditional expectation inductively? Attempting the solve the following problem
A fair die is successively rolled. Let $X$ and $Y$ denote,
respectively, the number of rolls necessary to obtain a 6 and a 5.
Find
$E[ X|Y = 5]$
I am confused about how to conceptualize such a problem?
I know... | I am not quite sure what kind of induction you are looking for, so I will just try giving a more general solution to this problem, i.e. $E[X|Y=y]$.
The point is to notice that random variable X distributed differently before and after the y-th roll, for tosses before the first one which yield 5 cannot have 5 as result,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\... | We have
$$
\tan(A+B+C)=\tan(A+(B+C))=\frac{\tan A+\tan(B+C)}{1-\tan A \tan(B+C)}=
$$
$$
\frac{\tan A+\frac{\tan B+\tan C}{1-\tan B \tan C}}{1-\tan A\frac{\tan B+\tan C}{1-\tan B\tan C}}=
\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}
$$
In the last step we multiplied the numera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| For a more geometric proof:
Let $C$ be a cuboid whose edges are $a$, $b$ and $c$. If $\text{diam}(C)=\sqrt{a^2+b^2+c^2}$ is fixed, then $a+b+c$ is at most $\sqrt{3}\text{diam}(C)$. So $a+b+c \leq \sqrt{3(a^2+b^2+c^2)}$.
Then, if $\text{diam}(C)$ is fixed, $\text{Vol}(C)=abc$ is at most $\displaystyle \left( \frac{a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 3
} |
Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\... | $$\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}=\frac1{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
} |
Is this 3D curve a circle? The following is a curve in $3$ dimensions:
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/3) \\ z & = & \cos(\theta - 2\pi/3) \end{eqnarray}$$
Is the curve a circle?
If it is, what about this curve in $4$ dimensions?
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \... | $\cos3\theta=cos3(\theta)$
$\cos3(\theta-\frac{\pi}{3})=cos(3\theta-\pi)=-\cos3\theta$
As $-y=-\cos(\theta-\frac{\pi}{3})=cos(\theta+\frac{2\pi}{3})$,
$\cos3(\theta+\frac{\pi}{3})=\cos(2\pi+3\theta)=\cos3\theta$
$\cos3(\theta-\frac{2\pi}{3})=cos(3\theta-2\pi)=\cos3\theta$
Now, $\cos3\theta=4cos^3\theta-3\cos\theta$
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 3
} |
The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$ How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
| Let $f: \mathbb R \to \mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $\tilde f := f-g$ is also a solution and we have $\tilde f(-1) = \tilde f(1) = 0$. We will show $\tilde f \equiv 0$, so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots. Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots.
for real roots, $a \gt 0$, $b^2-4{a}{c} \ge 0$
Here we h... | You have basically answered the question yourself. You just need $a>0$ which is automatic and $b^2\geq 4ac=4a$. So for $a=1$, $b$ can be $2,3,4$, for $a=2$, $b$ can be $3,4$ and for $a=3$ or $4$, $b$ can be only $4$.
Thus there are $7$ possible polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can a partial differential equation have two different solutions? Consider:
$$x^2p+y^2q=(x+y)z$$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Thus by Lagrange's Method
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dy}{y^2}
\Rightarrow \frac{1}{x... | Both are not right.
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{dz}{dt}=(x+y)z=\left(-\dfrac{1}{t}-\dfrac{1}{t+y_0}\right)z$ , we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding $a_n$ for very large $n$ where $a_n = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3} $ I have a recurrence relation,
$$ a_n = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3} $$
for $n>3$ and $a_1 = 0, a_2 = 0, a_3 = 1$
I have to find the value of $a_n$ for very large values of n. I tried an approach similar to that which we u... | For $n \in \mathbb N$, we let $b_n = a_n - 2^n$. We get by using the equation for $(a_n)$:
\begin{align*}
b_n &= a_n - 2^n\\
&= a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3} - 2^n\\
&= b_{n-1} + 2^{n-1} + b_{n-2} + 2^{n-2} + b_{n-3} + 2^{n-3} + 2^{n-3} - 2^n\\
&= b_{n-1} + b_{n-2} + b_{n-3} + 2^{n-3}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Inequality. $(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $
Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality)
$$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$
Thanks :)
| The inequality is obviously true if $a = b$, because we have $2(n+1)a^{n+1}\geq 2na^{n+1}$, which is true for $a\geq 0$.
Now, if $a\neq b$, we have $a^n+a^{n-1}b+\dots+b^n=\displaystyle\frac{a^{n+1}-b^{n+1}}{a-b}$. Let's assume $a>b$ (otherwise we can simply inverse the roles of $a$ and $b$). The inequality to prove is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outlin... | I guess your problem is step four:
$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$
Note that
$$\begin{align}\int(1-\cos^2 x) \ \cos^{n-2} x \ dx & = \int \cos^{n-2} x \ dx-\int \cos^2 x\; \cos^{n-2} x \ dx \\ & =\int \cos^{n-2} x \ dx-\int \cos^{n} x \ dx \end{align}$$
so we get
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
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Finding $\sum\limits_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $ How we can find
$$\sum_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $$
| Hint: Compute $(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n+1}+\sqrt[4]{n})(\sqrt[4]{n+1}-\sqrt[4]{n})$.
*
*Warm up: Simplify the expression $(a^2+b^2)(a+b)(a-b)$.
*Pre-warm up: Simplify the expression $(a+b)(a-b)$ and deduce that $\sum\limits_{n=1}^{99}\frac1{\sqrt{n+1}+\sqrt{n}}=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/196156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Rational number solution for an equation Does there exist $v=(a,b,c)\in\mathbb{Q^3}$ with none of $v$'s terms being zero s.t. $ a+b\sqrt[3]2+c\sqrt[3]4=0$ ?
And I was doing undergraduate algebra 2 homework when I encountered it in my head. At first It seemed like it can be proved there can be no such $v$ like how $\sq... | I'll try to show some more elementary approach, not using field theory. Suppose there is such a $v$. We have $c \ne 0$ (as $\sqrt[3]2$ is irrational). Rewriting the equation, we find $\alpha, \beta \in \mathbb Q$ with
\[ \sqrt[3]4 = \alpha + \beta \sqrt[3]2 \]
and $\alpha, \beta \ne 0$ (as $\sqrt[3]2, \sqrt[3]4 \not\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Trigonometry Table Problem Evaluate the given trigonometric expression:
\[
\frac{5\sin^2 30° + \cos^245° - 4\tan^2 30°}{2\sin30°\cdot\cos 30° + \tan 45°}
\]
| HINT: Posted below are some identities/axioms. $$\begin{aligned}\sin 30^{\circ} &= 0.5 \\ \\ \sin 45^{\circ} &= {1 \over \sqrt 2} \\ \\ \cos 30^{\circ} & = {\sqrt3 \over 2} \\ \\ \cos 45^{\circ}& = {1 \over \sqrt 2} \\ \\ \tan \theta &= {\sin \theta \over \cos\theta } \\ \\ \sin^2\theta &= (\sin\theta)^2 \\ \\ \tan^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Exponential modular equation I am having some trouble in proving that the only solutions to
$$ -2^{m-1} \equiv m \pmod{7} $$
are $m \equiv 3,5, 13 \pmod{42}$.
What I tried to use:
If $-2^{m-1} \equiv m \pmod{7}$, then $2^{m-1} \equiv 6m \pmod{7}$ and we know that $2^6 \equiv 1 \pmod{7}$.
But now I can't go on. Could s... | Note that modulo $7$, $2^m$ is periodic with period $3$. So $2^{m-1}+m$, modulo $7$, is periodic with period at most $21$.
Now one only needs to make a somewhat long table, starting at $m=1$. The powers of $2$, modulo $7$, are $1,2,4,1,2,4,1,2,\dots$. And $m$ is congruent to $1,2,3,4,5,6,0,1,\dots$. Add, check where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof the logarithmic identity $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$ Please help me proof $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$, for $a,b,c>0$ and $a^2+b^2=c^2$. Thanks.
| Note that $\log_xy=\displaystyle\frac{\ln y}{\ln x}$, here $\ln$ is the natural log. Therefore, we have
$$\log_{b+c} a+\log_{c-b} a=\frac{\ln a}{\ln(b+c)}+\frac{\ln a}{\ln(c-b)}
=\ln a\left(\frac{\ln(b+c)+\ln(c-b)}{\ln(b+c)\ln(c-b)}\right)$$
$$=\ln a\cdot\frac{\ln[(b+c)(c-b)]}{\ln(b+c)\ln(c-b)}
=\frac{\ln a\ln(c^2-b^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving an equation with three quadratic radicals in the set of real numbers How do we solve the following equation in the set of real numbers?
$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$
I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.
| If you put $a=\sqrt{5x-1}$ and $b = \sqrt{5-2x}$, the equation says
$$ (26-x) a - (13x+14) b + 12 a b = 18 x + 32$$
The resultant of $(26-x) a - (13x+14) b + 12 a b - 18 x - 32$ and $5x-1-a^2$ with respect to $a$ is
$$ -169\,{x}^{2}{b}^{2}+5\,{x}^{3}-588\,{x}^{2}b+356\,x{b}^{2}-585\,{x}^{
2}+1808\,xb-340\,{b}^{2}+2280\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/201457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Prove the trigonometric identity Please help me prove the identity:
$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$$
| Implement the formula:
1) $1-\cos^2\alpha=\sin^2\alpha$
2) $\cos2\alpha=\cos^2\alpha-\sin\alpha$
3) $1=\sin^2\alpha+\cos^2\alpha$
Now turn the proof given identity.
$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$
$\cos^2\alpha(1-\cos^2\alpha)+\sin^4\alpha=\frac{1}{2}(1-\cos2\alpha)$
$\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the solution of this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
Possible Duplicate:
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Help me solve this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
| Let $t_1=(\sqrt{2+\sqrt3})^x$, and $t_2=(\sqrt{2-\sqrt3})^x$. Now the given equation is:
$t_1+t_2=4$..........(1)
Above it follows that:
$t_1\cdot t_2=(\sqrt{2+\sqrt 3})^x\cdot(\sqrt{2-\sqrt3})^x=(\sqrt{({2+\sqrt3})({2-\sqrt3}})^x=(\sqrt{2^2-(\sqrt3)^2})^x=(\sqrt{4-\sqrt 3^2})^x=(\sqrt{4-3})^x=1^x=1$ $\Rightarrow$ $t_1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Multivariable limit - Two variables $ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$ How can I calculate the following limit and show that it equals $0$:
$$ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$$
Thanks in advance
| First I would change coordinates to $(x,z)$ where $z=\pi-y$. The limit becomes
$$\lim_{(x,z) \to (0, 0 ) } \frac{x^2 (\pi-z) \sin (\pi-z) } {\sin^2x + z^2 }$$
which we can evaluate by changing to polar coordinates and using the fact that near $0$, $\sin x=x+O(x^3)$. If you aren't familiar with big O notation, you can r... | {
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"answer_count": 3,
"answer_id": 2
} |
Summation manipulation. How can this be true?
\begin{align*}
\sum_{n=1}^\infty \sum_{k=n}^\infty \frac 1{k^3}
&= \sum_{k=1}^\infty \sum_{n=1}^{k} \frac 1{k^3}\\
\end{align*}
| On the left-hand side, imagine the following infinite sum of infinite sums:
$$\begin{aligned}
\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\
\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\
\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\
\frac{1}{4^3}+\cdots
\end{aligned}$$
This is equal to the infinite sum
$$... | {
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"answer_count": 2,
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Is this composition of polynomial correct? Suppose $p(x)=x^2+5x+3$ and $q(x)=3x^3-3x+7$. Write the expression $(q \circ p)(x)$ as a sum of terms, each which is a constant times the power of $x$.
Here is my work for the problem:
$(q\circ p)(x)=3(x^2+5x+3)^3-3(x^2+5x+3)+7$
$(q\circ p)(x)=3(x^6+125x^3+27)-3x^2-15x-9+7$
$(... | The mistake you made is assuming that $(a+b+c)^3=a^3+b^3+c^3$. Note that $(a+b+c)(a+b+c)=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2$. Simplify and multiply by $a+b+c$ again to get the cube. Alternatively you can rewrite $q(x)=3x(x^2-1)+7$ and work from there.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Fundamental Set of Complex Solutions I am confused on how to do this problem, it states:
Find a fundamental set of solutions and put it in general form for the given system.
$x' = \begin{pmatrix} -1/2 & 1\\ -1 & -1/2 \end{pmatrix}x$
I got the eigen values to be $\lambda_1$ = [ (-1/2) + i ] , $\lambda_2$ = [ (-1/2) - i... | Based on your derivations, the general solution has the form
$$ x(t) = c_1 \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{(-\frac{1}{2}+i)t} + c_2 \begin{pmatrix} 1\\ -i \end{pmatrix} {\rm e}^{(-\frac{1}{2}-i)t}\,, $$
where $c_1,c_2$ are arbitrary constants. Simplifying further
$$ x(t) = c_1 {\rm e}^{-\frac{1}{2}t} \be... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Twist on log of sine and cosine integral $\int_{0}^{\frac{\pi}{2}}x\ln(\sin x)\ln(\cos x)dx$ I ran across this integral and have not been able to evaluate it.
$$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$$
I had some ideas. Perhaps some how arr... | Here is an approach that avoids using the Fourier series expansions for $\ln (\cos x)$ and $\ln (\sin x)$.
Consider
$$I = \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx.$$
Enforcing a substitution of $x \mapsto \pi/2 - x$ gives
\begin{align*}
I &= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln \left [\sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equation... | You can approach this with homogeneous coordinates. A circle with equation $(x-x_c)^2+(y-y_c)^2 - r^2 = 0$ is represented by a 3x3 matrix as
$$ {\rm Circle}(x_c,y_c,r) = \begin{bmatrix} 1 & 0 & -x_c \\ 0 & 1 & -y_c \\ -x_c & -y_c & x_c^2+y_x^2-r^2 \end{bmatrix} $$
This means that the equation for a circle $C_1 = \begin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
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Circle Least Squares Fit So my question is this:
Find the equation of the circle that gives the best least squares circle fit to the points $(-1,-2), (0,2.4), (1.1,-4),$ and $(2.4,-1.6).$
So far I have this general equation:
$2xc_1+2yc_2+(r^2-c_1^2-c_2^2)=x^2+y^2$ where $r^2-c_1^2-c_2^2 = c_3$
So then I think I creat... | Just multiply both sides of the equation by $A^{T}\,,$ where $A$ is your matrix, and then solve the resulting system.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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A definite integral involving a parameter Does this integral seem to have a nice closed form (at least for a subset of values of $a > 0$)
$$
\int_{\sqrt{a}}^\infty \frac{y^2}{(y^2-a+1)^2} dy
$$
Using a symbolic math software and for $a$ being some small integers, I end up getting different answers in terms of $\sinh^{-... | Partial fractions does it. The partial fraction expansion of the integrand is
$$
\frac{1}{4(y-\sqrt{a-1})^2} + \frac{1}{4(y+\sqrt{a-1})^2}
+ \frac{1}{4 \sqrt{a-1} (y - \sqrt{a-1})} - \frac{1}{4 \sqrt{a-1} (y + \sqrt{a-1})}
$$
so an antiderivative is
$$ -\frac{1}{4(y-\sqrt{a-1})} - \frac{1}{4(y+\sqrt{a-1})} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/215019",
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"source": "stackexchange",
"question_score": "1",
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How to integrate $\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? How to integrate $\displaystyle \int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? I Tried using partial fractions but i got lost, thanks.
| From $\frac{2x^2+ x}{(x+ 1)(x^2+ 1)}= \frac{A}{x+ 1}+ \frac{Bx+ C}{x^2+ 1}$ you can add the fractions on the right, then compare coefficients of "like terms". But we can also use the fact that this is to be true for all x. Setting x to any three values gives three equations to solve for A, B, and C.
First multiplying... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Continued fractions with rational functions Express the following rational function in continued-fraction form:
$${4x^2+3x-7\over 2x^3+x^2-x+5}$$
The answer is :
$${4 \over 2x- \frac{1}{2}} + { \frac{23}{8} \over x-\frac{63}{92}}-{\frac{406}{529} \over x+\frac{33}{23}}\tag{inline continued fraction}$$
which means
$$
\c... | Here is the solution using integer coefficients
$$
\begin{align}
\cfrac{4x^2+3x-7}{2x^3+x^2-x+5}
&=\cfrac1{\cfrac{2x^3+x^2-x+5}{4x^2+3x-7}}\\
&=\cfrac{8}{4x-1+\cfrac{23x+33}{4x^2+3x-7}}\\
&=\cfrac{8}{4x-1+\cfrac{529}{92x-63-\cfrac{1624}{23x+33}}}\\
\end{align}
$$
The book's answer can be shown to be equal by cancelling... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How did my teacher gets this result? Algebra/Limits $$\begin{align*}&\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}=\\
&\lim_{x\to0}\left(\frac{(1+x)-(1-x)}{(1+x)-(1-x)}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}}\right)
\end{align*}$$
My teacher was... | You should use the following identities:
$$a^2-b^2=(a-b)(a+b)$$
$$c^3-d^3=(c-d)(c^2+cd+d^2)$$
(prove by opening brackets) for
$$a=\sqrt{1+x}, \hspace{5pt} b=\sqrt{1-x}, \hspace{5pt} c=\sqrt[3]{1+x}, \hspace{5pt} d=\sqrt[3]{1-x}$$
Then you multiply your function in the limit by $\frac{a+b}{c^2+cd+d^2}\cdot \frac{c^2+cd... | {
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"source": "stackexchange",
"question_score": "1",
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Range of a trignonmetric function I came across this in an Engineering entrance book,
What is the range of this: $a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x$
What is the method to find it? I tried the graph approach but didn't know how to proceed.
| The double angle formulas yield
$$
a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x=\frac{\color{#C00000}{b\sin(2x) + (c-a^2)\cos(2x)}+(\color{#00A000}{c+a^2})}{2}\tag{1}
$$
Since the range of $a\cos(x)+b\sin(x)$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$, we get that the range of $(1)$ is
$$
\left[\frac{\color{#00A000}{c+a^2}+\... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Prove that $4w^2 + 5z^2 = 4z^2 + 5w^2$ if $|z| = |w|$ and $4z^2+5w^2=azw$ I know that:
$$|z| = |w| = p$$
$$4z^2 + 5w^2 = azw ,\qquad a \in R$$
I need to prove that $4w^2 + 5z^2 = azw$
How I solved it is:
$$|z| = p \implies |z|^2 = p^2 \implies zz^* = p^2$$
then solved as $z$ and replaced it to the original equation.
Th... | This's strongly inspired by the other answer.
Let's denote $\cos C+i\sin C=cis C$
If $z=p(cis A),w=p(cis B)$ with $p\ne 0$
So, $\frac z w=cis(A-B),4cis 2(A-B)+acis(A-B)+5=0$
Equating real & imaginary parts(using the fact that $a$ is real ),
$4\cos2(A-B)+a\cos(A-B)+5=0--->(1)$
and $4\sin2(A-B)+a\sin(A-B)=--->(2)$
$(2)... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluate: $\int{\frac{x^{5}-x}{x^{8}+1}}\:\mathrm dx.$
Evaluate: $$\int{\frac{x^{5}-x}{x^{8}+1}\:\mathrm dx}.$$
I am unable to see a decent starting point for this integral, there are no radicals so trigonometric substitution isn't helpful; there is no nice partial fraction decomposition to simplify the integrand, in... | Letting $y=x^2$ reduces the powers of $x$ and then dividing both the denominator and numerator of the integrand by $y^2$ yields
$$
\begin{aligned}
\int \frac{x^5-x}{x^8+1} d x&=\frac{1}{2} \int \frac{y^2-1}{y^4+1} d y \\
=& \frac{1}{2} \int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}} d y \\
=& \frac{1}{2} \int \frac{d\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/226996",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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How to calculate the maximum value of: $\frac{25x}{x^2+1600x+640000}$? Wolfram says it's 800, but how to calculate it?
$$
\frac{25x}{x^2+1600x+640000}
$$
| It’s straightforward as a calculus problem. To solve it without calculus, note that
$$\frac{25x}{(x^2+1600x+640000)}=\frac{25x}{(x+800)^2}\;,\tag{1}$$
so the denominator is always positive, the the function has its maximum at some positive value of $x$. That maximum will occur where
$$\frac{(x+800)^2}{25x}=\frac1{25}\... | {
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"source": "stackexchange",
"question_score": "2",
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explain why $3|n$ with the following conditions.. If $ord_ma=3$ and if $a^n\equiv 1 \pmod{m}$ for some $n\ge1$, explain why $3|n$.
Attempt at solution:
We know that $ord_ma=3$ means $a^3\equiv 1 \pmod{m}$. Therefore, $n$ is a multiple of $3$, because $a^{3n}\equiv 1 \pmod{m}$.
| The reasoning is incomplete. First note that $\text{ord}_m(a)=3$ means that $a^3\equiv 1\pmod{m}$ and for any positive integer $k \lt 3$, we have $a^k\not\equiv 1\pmod{m}$.
Now we proceed to show that if $a^n\equiv 1\pmod{m}$, where $n\ge 1$, then $3$ divides $n$.
Suppose to the contrary that $3$ does not divide $n$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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wave equation with initial values and boundary condititon I have a homogenious 2-dimensional wave equation:
$$ - \frac{\partial^2 u}{\partial x^2} (x, y, t) - \frac{\partial^2 u}{\partial y^2} (x, y, t) + \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} (x, y, t) = 0$$
With:
$$ 0 < x < a, 0 < y < b, t > 0$$
$$ u(x, y, 0... | Let $u(x,y,t)=\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}$ so that it automatically satisfies $u(0,y,t)=u(a,y,t)=u(x,0,t)=u(x,b,t)=0$ ,
Then $\sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty\dfrac{m^2\pi^2}{a^2}C(m,n,t)\sin\dfrac{m\pi x}{a}\sin\dfrac{n\pi y}{b}+\s... | {
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"source": "stackexchange",
"question_score": "2",
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How to prove $\lim_{n \to +\infty} \sqrt{n}\int_0^\pi{\cos(\frac{t}{2})^n}dt>0$ I want to prove
$$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$
First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller th... | HINT
In fact, it is not hard to compute the limit exactly. Let $I_n = \displaystyle \int_0^{\pi} \cos^n(t/2) dt$. Compute $I_n$ and use Stirling's formula to obtain asymptotics of $I_n$.
Move your cursor over the gray area for the complete solution.
Setting $t = 2x$, we get $$I_n = 2 \int_{0}^{\pi/2} \cos^n(x) dx$$ Fo... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Bijective Proof of a Fibonacci Identity Prove (Using bijections):
$F_{1}+F_{3}+\cdots+F_{2n-1}=F_{2n}$
Where $F_{i}$ is the $i$th Fibonacci number.
Apparently you use monomers and dimers to prove this, but I don't really know what to do.
| HINT: A monomer is a $1\times 1$ square, and a dimer is a $2\times 1$ rectangle, a domino. The number of ways to tile a $1\times n$ strip using monomers and dimers is $F_{2n+1}$. (For example, a $1\times 4$ strip can be tiled in $F_5=5$ ways: $1+1+1+1,1+1+2,1+2+1,2+1+1$, and $2+2$. The sum
$$F_1+F_3+F_5+\ldots+F_{2n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232633",
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"source": "stackexchange",
"question_score": "3",
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Encode the message $[1,1,0,1,1,0,1]$ in BCH code based on the field $\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1}$ So here's what I understand so far:
$\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1} = GF(16)$
The code is written as $[x^{14},x^{13},x^{12},x^{11},x^{10},x^{9},x^{8}$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2}... | Let $\alpha$ be a root of $m_{\alpha}(x)=x^4+x+1$ in the field $GF(16)$. As this polynomial is primitive, $\alpha$ is of order $15$. Therefore $\alpha^3$ is of order five. Thus the minimal polynomial of $\alpha^3$ is $m_{\alpha^3}(x)=x^4+x^3+x^2+x+1$. The generator polynomial of a double-error-correcting BCH-code is
$... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Find k $\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$ Find k with
$$\frac{1}{(3a-b)^2}+\frac{1}{(3b-c)^2}+\frac{1}{(3c-a)^2} \ge \frac{k}{a^2+b^2+c^2}$$
| In other words, you want to minimize $$f(a,b,c) = (a^2 + b^2 + c^2)\left(\frac{1}{(3a-b)^2} + \frac{1}{(3b-c)^2} + \frac{1}{(3c-a)^2}\right)$$
(presumably for $a,b,c$ real numbers).
By homogeneity, we may restrict to the sphere $a^2 + b^2 + c^2 = 1$.
Using a Lagrange multiplier, we look for critical points of $F(a,b,c,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Exponential and power functions through two points I have a problem where I'm asked to determine the constants of exponential and power functions that go through both points (5, 50) and (10, 1600). I have tried to solve them below, but would appreciate it if someone could check. Would also appreciate feedback on how I ... | The steps seem to be good. In both cases, you could divide your first equation by the second one (or vice versa) and then take ln on both sides. It would save you some time.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometric identity, cannot show equivalence using trigonometric identities
clearly
$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$
expanding and using the Weierstrass substitution we find that
$$\theta= 2 \arctan \frac{\left( 2ay- \sqrt{ 4a^2y^2 - ( (x-a)^2+y^2-b^2)( (x+a)^2+y^2-b^2) }\right)}{(x-a)^2+y^2-b^2} $$
if w... | You could look at the angles at the vertex between sides $c$ and $a$. The arctan comes from the $x,y$ triangle and the arccos from the $abc$ one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/236510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding sum $\sum_{i=1}^n \frac1{4i^2-1}$ I have been having problem with calculating the following summation:
$$
\sum_{i=1}^n {1\over 4i^2-1} = {1\over3} + {1\over15} + {1\over35} + \cdots + {1\over 4n^2-1}
$$
I do know the answer, but just can not find the way to get it.
Thank you in advance.
| You have,
$$
\sum_{i=1}^n {1\over 4i^2-1} = \sum_{i=1}^n\frac{1}{{(2i + 1)(2i - 1)}}
$$
$$
= \sum_{i=1}^n \left(\frac{1}{{2(2i - 1)}} - \frac{1}{{2(2i + 1)}}\right)
= \frac{1}{2}\left(\sum_{1\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n}\frac{1}{{(2i + 1)}}\right)
$$
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Derivatives and graphs $f(x) = \sin(x)+ \cos(x) $ for $ 0≤x≤2\pi$
I have to do the following
1) Find the intervals on which $f$ is increasing or decreasing
2) Find the local maximum and minimum values of $f$
3) Find the intervals of concavity and the inflections points
I got until this point when trying to solve probl... | (1) $f$ is increasing or decreasing accordingly as $f'(x)>0$ or $<0$
Now, $f'(x)=\cos x-\sin x=\sqrt2\cos(x+\frac \pi 4)$
$f'(x)>0$ if $\cos(x+\frac \pi 4)>0$ if $ x+\frac \pi 4$ lies in the 1st or 4th quadrant.
As,$0\le x\le 2\pi,$ for $f'(x)>0, 0\le x<\frac \pi 4$ or $\frac{3\pi}2-\frac \pi4<x\le 2\pi$
Similarly for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Period of linear congruential generator How can you calculate the probability distribution of the period length of a linear congruential generator? That is $X_{n+1} = (aX_n + c) \bmod m$ where $a$ is chosen uniformly at random from $\{1,\dots, m-1\}$ and $c$ is chosen uniformly at random from $\{0,\dots, m-1\}$ and $m... | This may not address the question exactly, but the results derived indicate that the final answer may depend on the factors common to $a-1$ and $m$.
A preliminary lemma and theorem
Lemma: Suppose $p$ is prime and $j\ge2$. Then, unless $p=j=2$,
$$
p^k\,|\,n\implies\left.p^{k-j+2}\,\middle|\,\binom{n}{j}\right.\tag{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Probability problem (withdraw balls from the urn)
Problem:
An urn contains 3 red and 7 black balls. Player A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls... | Well, the second entry on top is wrong (there should be another factor of $7$), so that could be the source of your confusion. Let's assume it's just a typo, and see how they might have gone about it.
There are $4$ mutually exclusive possibilities to consider for ways Player A can draw the red ball:
(i) The first ball... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to calcualate how many unique set of 6 can i have in a given set. Hello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\{1,2,3,4,5,6,7,8,9,10\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set.
In other words for the number... | It exactly the number of ways to choose $6$ elements out of $10$, i,e. the binomial coefficient$$\binom{10}{6}=\frac{10!}{6!4!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.