Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Implicit differentiations of same equation with two different answers:Why? I've an equation that I like to use for implicit differentiation. The equation is:$x^2 = \frac{(x+2y)}{(x-2y)}$
I used two different methods but got two different answers for same equation.
Can anyone kindly tell me where I am wrong? Why am I ge... | They may not be different. Remember, there's an equation relating $x$ and $y$. If you use that equation, you might be able to get one of the answers to look like the other one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| If $x,y,z$ are nonnegative reals, then
$x^2+y^2+z^2\ge xy+yz+zx$ (with equality iff $x=y=z$), hence
$3(x^2+y^2+z^2)\ge x(x+y+z)+y(y+z+x)+z(z+x+y) = (x+y+z)^2$
(with equality iff $x=y=z$).
Letting $x=a+b, y=b+c, z=a+c$, we find $x+y+z=6$ and
$3(a+b)^2+3(b+c)^2+3(c+a)^2 \ge 36$.
Note that $(a+b)^2+(b+c)^2+(c+a)^2= 2(a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
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System of 3 linear congruences
Find all solutions: $$\begin{cases} x\equiv 39 \mod(189) \\ x\equiv 25 \mod(539) \\ x\equiv 399 \mod(1089) \end{cases}$$
But $189=3^3\cdot7$, $539=11\cdot7^2$ and $1089=3^2\cdot 11^2$, so I can't use here Chinese remainder theorem.
| You can find a system of congruences equivalent to the original system, but with pairwise relatively prime moduli.
For example, the first congruence is equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 39\pmod{7}$.
The second original congruence is equivalent to the system $x\equiv 25\pmod{11}$ $x\equiv 25\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$ I need to find an expression for $n$th derivative of $f(x) = e^{x^2}$. Really need help.
| Let $y=e^{x^2}$. We want to find $\frac{d^n y}{dx^n}$.
Note that $\ln y = x^2$ and so $\frac{1}{y} \frac{dy}{dx} = 2x\iff \frac{dy}{dx}=2xy$.
Differentiating again yields:
$\frac{d^2y}{dx^2}=2x\frac{dy}{dx}+2y$,
$\frac{d^3 y}{dx^3} = 2x \frac{d^2 y}{dx^2} + 4\frac{dy}{dx}$,
$\frac{d^4 y}{dx^4} = 2x \frac{d^3 y}{dx^3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 9,
"answer_id": 0
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Positive Definite Matrix Problem Suppose that the Symmetric Matrix
$$B=\left(
\begin{array}{cc}
\alpha & a^T \\
a & A
\end{array}
\right)$$
of order $n+1$ is positive definitive.
(a) Show that the scalar $\alpha$ must be positive and the $n*n$ matrix $A$ must be positive definitive.
(b) What is the Cholesky factoriz... | I also find a solution for my problem:
$$B=\left(
\begin{array}{cc}
\alpha & a^T \\
a & A
\end{array}
\right)$$
is a symmetric matrix of order $n+1$ positive definite.
We will show that $\alpha$ is positive and $A$ is a $n\times n$ positive definite matrix.
$B$ is positive definite, so
$$\exists\;v=\left(
\begin{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Compute the limit of $\sum\limits_{k=1}^{n} \left(\frac{k}{n^2}\right)^{1+k/n^2}$ when $n\to\infty$
Compute the limit
$$\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\frac{k}{n^2} + 1}$$
At a first look, I only thought of Riemann sums, but I don't see how I may apply it. What else could I do? I need ... | Let $x: = \frac {k} {n^2}$. Then, we have $$\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)} = \lim_{n\to\infty} n^2 \int_{1/n^2}^{1/n} x^{x + 1}\, dx.$$ Trapezoidal rule (in quadrature) gives $$\int_{1/n^2}^{1/n} x^{x + 1}\, dx = \frac {1} {2} \left(\frac {1} {n} - \frac {1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$? Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$
| I think that if this exercise can be solved in a simple fashion (without heavy
computation), then this should be the way to approach it (otherwise, it is
just a mindless computation which just requires to apply some algorithm like
the, so-called, Weierstrass substitution and teaches you nothing).
So, since
$$
\sin^{3}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/198083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Proof the logaritmic identity Please help me proving the basic logarithmic identity $\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$
| Let $\gcd(a,b) = 1$ and $x = a^nb$ where $a,b,n \in \mathbb{Z}$, then
$$\begin{aligned}
\log_a x &= n + \frac{\ln b}{\ln a} = n + \frac{\ln b}{\ln(b+1)}\frac{\ln(b+1)}{\ln(b+2)}\cdots \frac{\ln(b+(m-1))}{\ln(b+m)}\frac{\ln (b+m)}{\ln a} \newline
&= n + \log_{b+1} b\log_{b+2}{b+1}\cdots \log_{b+m} (b + (m-1)) \ln_{a} ... | {
"language": "en",
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Evaluating $\int \cos(x) \sqrt{\sin(2 x)} dx$
Evaluate the following indefinite integral:
$$\int \cos(x) \sqrt{\sin(2 x)} dx$$
Only hint I have is from W|A that expresses the integral in terms of a hypergeometric function and it looks rather ugly.
Can we solve it in a simpler way and get a nicer form? Thanks.
| Alternatively, rewrite
$$I=\sqrt{2}\int(1-(\sin x)^2)^\frac{1}{4}(\sin{x})^\frac{1}{2}d(\sin{x})\\=\sqrt{2}\int(1-t^2)^{\frac{1}{4}}t^\frac{1}{2}dt\\=\sqrt{2}\int\left(\frac{1}{t^2}-1\right)^\frac{1}{4}tdt \\=\frac{1}{\sqrt{2}}\int\left(\frac{1}{z}-1\right)^\frac{1}{4}dz$$
Now let
$$\frac{1}{z}-1=u^2$$
$$-\frac{dz}{z^2... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
| You know that $$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$ and $$(a^2-b^2)=(a-b)(a+b)$$ so if $a=\sqrt{3+x}$ and $b=2$, we have $(a^2-b^2)=(\sqrt{3+x}-2)(\sqrt{3+x}+2)=3+x-4$. The same calculation can be done for another identity by taking $x=\sqrt[3]{7+x}$ and $y=2$. In fact you have $$(\sqrt[3]{7+x}-2)\big((\sqrt[3]{7+x})^2+2\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a spanning set for a null space I have a matrix $A$ that is like this:
\begin{equation}
A = \pmatrix{
1 & 2 & -3 & 1 & 5 \\
1 & 3 & -1 & 4 & -2 \\
1 & 1 & -5 & -2 & 12 \\
1 & 4 & 1 & 7 & -7
}
\end{equation}
The question is: Find a set of $5\times 1$ matrices whose linear span is the null space of $A$.
I did Gau... | Your reduced matrix is correct. First you need to characterize the set of vectors $x$ that satisfy $A x = 0$. This set is called the null space or kernel, and I use the standard notation $\ker A$.
The reduction process above corresponds to pre-multiplying $A$ by an invertible matrix $G$ such that $G A = \tilde{A}$, whe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Arithmetic Mean & Geometric Mean Question 1: if the arithmetic mean of two numbers is twice of their geometric mean, their ratio of sum of numbers to the difference of numbers equals?
Question 2: if the quadratic equation:
$(b^2+c^2)x2-2(a+b)cx+(c^2+a^2)=0$
has equal roots then?what is its AP & GP?
Question 3: If the... | $1.$ We can even do it without the quadratic formula. We have $a+b=4\sqrt{ab}$ and therefore
$$(a+b)^2=16ab.$$
Also,
$$(a-b)^2=(a+b)^2-4ab=(a+b)^2-\frac{1}{4}(a+b)^2=\frac{3}{4}(a+b)^2.$$
Thus
$$\frac{(a+b)^2}{(a-b)^2}=\frac{4}{3},$$
and therefore
$$\frac{a+b}{a-b}=\pm\frac{2}{\sqrt{3}}.$$
$2.$ The roots are equal pr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
| Let $s_n=\sum_{k=0}^nk^3$; your generating function for these numbers will be $$f(x)=\sum_{n\ge 0}s_nx^n\;.$$
You know that the sequence satisfies the recurrence $s_n=s_{n-1}+n^3$. Multiply this recurrence by $x^n$ and sum over $n\ge 0$:
$$\sum_{n\ge 0}s_nx^n=\sum_{n\ge 0}s_{n-1}x^n+\sum_{n\ge 0}n^3x^n\tag{1}\;.$$
The ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 2
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A Differential Equation with Trigonometric Coefficients Suppose we have the following second-order differential equation:
$\cos^2(x)y'' -\sin(x)y' + y = 0$
How do we determine its general solution? I couldn't even guess a particular solution; all my efforts led nowhere. I started off with something like $y = Ae^{B\cos(... | This is a linear ODE of trigonometric function coefficients. The current approach of solving it is to transform it to a linear ODE of polynomial function coefficients first.
Let $u=\cos x$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-(\sin x)\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-(\sin x)\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/209947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin... | I won't pretend that I would have thought of this without seeing the answer but we can look at the problem geometrically. This is more of a nice illustration than a genuine 'proof by pictures'.
We know firstly that the curve $x^4+y^4=a$ is symmetric about the $x$- and $y$-axes:
Suppose that $f(\theta)=\sin^4\theta+\co... | {
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"timestamp": "2023-03-29T00:00:00",
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When is $1^5 + 2^5 + \ldots + n^5$ a square? When is $1^5 + 2^5 + \ldots + n^5$ a square? I found that this happens sometimes: $n=13$ gives $1001^2$, $n=133$ gives $9712992^2$ and $n=1321$ gives $942162299^2$.
I feel that the identity$$\displaystyle\sum_{i=1}^n i^5 = \tfrac{1}{12}[2n^6+6n^5+5n^4-n^2]$$ will be useful, ... | If $$a=1+2+3+\cdots+n$$ then
$$
1^5+2^5+3^5+\cdots+n^5 = \frac{4a^3-a^2}{3},
$$
so now the question is: when is that a square?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/217016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Probability of rolling 5 dice? Say you picked $\{4 5 2 6 5\}$ to show up in the dice rolls. You win if you have 3, 4, or 5 numbers in the correct sequence.
*
*3 numbers in correct sequence would be; $\{4 5 2\}$ OR $\{5 2 6\}$ OR $\{2 6 5\}$
*4 numbers in the correct sequence would be; $\{4 5 2 6\}$ OR $\{5 2 6 5\}$... | 3 numbers correct: $(\frac{5}{6})^2\cdot\frac{1}{6^3}\cdot\binom{5}{2}$
4 numbers correct: $\frac{5}{6}\cdot\frac{1}{6^4}\cdot\binom{5}{1}$
5 numbers correct: $\frac{1}{6^5}$
total: $\frac{1}{6^5}\bigg(1+5\cdot\binom{5}{1}+5^2\cdot\binom{5}{2}\bigg)=\frac{276}{7776}=\frac{23}{648}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/217556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected Number of Successes in a Sample
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $1... | The answer should be
$\left({20 \choose 1}{180 \choose 9} + 2{20 \choose 2}{180 \choose 8} + \cdots
+ 10{20 \choose 10}{180 \choose 0} \right) / {200 \choose 10}$, where ${n \choose k}$ denotes the binomial coefficient ${n \choose k} = \frac{n!}{k!(n-k)!}$.
This follows because the probability $p_k$ that you your sampl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/218086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $\int\frac{dx}{x+\sqrt{1-x^2}}$?
Possible Duplicate:
Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$
$$\int\frac{dx}{x+\sqrt{1-x^2}}$$
I used $x = \sin(a),\ dx = \cos (a)\,da$.
However, I am getting suck after that.
I can't seem to find what to do after $$ \int \frac{\cos(a)\,da}{\sin(a)+\cos(a)}$$... | Hint or starting point:
Nothing immediate jumps out to me when I see the integral
$$\int \frac{\cos \theta}{\sin \theta + \cos\theta} d\theta$$
However, when integrating rational function of trig functions, we can always use the Weierstrass substitution.
So, let $t = \tan \left(\frac{\theta}{2}\right), dt = \frac{1}{2}... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Process for $(k+1)^3$? I've mentioned in previous questions that I have a hard time with simplifying algebraic equations.
For this equation, I assumed we would first put the equation into easier to read form:
$(k+1)(k+1)(k+1)$
Then apply FOIL to the first two:
EDIT: $(k^2+2k+1)(k+1)$
Although this is where I'm stuck. ... | Use the distributive law (after correcting your first multiplication):
$$(k^2+2k+1)(\color{red}{k}+\color{blue}{1})=(k^2+2k+1)\cdot\color{red}{k}+(k^2+2k+1)\cdot\color{blue}{1}\;.$$
Now $(k^2+2k+1)\cdot k=k^3+2k^2+k$, and $(k^2+2k+1)\cdot 1=k^2+2k+1$, so all that’s left is to add the two polynomials:
$$\begin{align*}
(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Concecutive last zeroes in expansion of $100!$
Possible Duplicate:
Highest power of a prime $p$ dividing $N!$
In decimal form, the number $100!$ ends in how many consecutive zeroes. I am thinking of the factorization of $100!$ but I am stuck. I try to count them and since there are 10, 20, 30,..., 100, there are at ... | \begin{align}
2\text{ goes into }100 & & 50\text{ times} \\
2^2\text{ goes into }100 & & 25\text{ times} \\
2^3\text{ goes into }100 & & 12\text{ times} \\
2^4\text{ goes into }100 & & 6\text{ times} \\
2^5\text{ goes into }100 & & 2\text{ times} \\
2^6\text{ goes into }100 & & 1\text{ time} \\
\end{align}
$$
50+25+12+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find solutions of $x^2-3y^2=-2$? According to MathWorld,
Pentagonal Triangular Number: A number which is simultaneously a pentagonal number $P_n$ and triangular number $T_m$. Such numbers exist when
$$\frac{1}{2}n(3n-1)=\frac{1}{2}m(m+1).$$
Completing the square gives
$$(6n-1)^2-3(2m+1)^2=-2.$$
Substitu... | $$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
-1 \\
1
\end{array}
\right) \; = \;
\left( \begin{array}{c}
1 \\
1
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 4
} |
Find monic gcd($x^4+x^3+x+1$, $x^6+x^5+x^4+x+1$) in $\mathbb Z_{2}$ My working so far using the Euclidean algorithm and polynomial long division (which I won't fully show here)
$x^6+x^5+x^4x+1$ = $(x^2+1) \times (x^4+x^3+x+1) + (-2x^3-x^2)$
and $(-2x^3-x^2) \equiv (x^2)$ in $\mathbb Z_{2}[x]$
So a non-monic gcd would b... | As wj32 stated, all I needed to do was to keep going one more step:
$x^2=(x+1)(x+1)+1$
$(x+1)^2$ is $(x^2+2x+1)$ but of course $2x \equiv 0$ in $\mathbb Z_{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integration of a trigonometric function I am having some difficulties with the calculation of the following integral. Can somebody help me please?
$$\int \frac{dx}{1+a\cos x},\text{ for }0<a<1$$
Thank you in advance
| Substitute, $t = \tan \left(\dfrac x2\right)$. So $x = 2\tan^{-1}t$ and $dx = \dfrac{2dt}{1+t^2}$. And the integral becomes,
$$\begin{align*}\int\dfrac{\dfrac{2}{1+t^2}}{1+\dfrac{a(1-t^2)}{1+t^2}}dt
&=\dfrac2{(1+a)}\int\dfrac1{1+\left[t\dfrac{\sqrt{1-a}}{\sqrt{1+a}}\right]^2}dt
&\color{blue}{u =\left[t\frac{\sqrt{1-a}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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some problems related to primes I would like to learn the following:
a) Prove that the equation $1 + x + x^2 = py$ has integer solutions for infinitely many primes $p$.
b) Twin primes are those difference by 2. Show that 5 is the only prime belonging to two such pairs. Show also that there is a one-to-one correspondenc... | a) The proof is a modification of the "Euclid" proof that there are infinitely many primes. Let $p_1=3$. Putting $x=1$ and $y=1$, we see that $1+x+x^2=py$ has a solution.
Now suppose that we have found $n$ primes $p_1,p_2,\dots,p_n$ such that $1+x+x^2=p_iy$ has a solution for any $i$ from $1$ to $n$. We exhibit a new ... | {
"language": "en",
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"source": "stackexchange",
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How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives:
\begin{align*}
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
3^7 \mod 5 = 2 \\\\
3^8 \mod 5 = 1 \\
3^9 \mod 5 = 3 \\
3^{10} \mod 5 = 4 \\
3^{11} \mod 5 = 2 \... | The pattern you've seen here is that $3^{4k+1}\;\text{mod}\:5=3$ for integers $k\ge 0$. You can prove this inductively from the base case, and the fact that $3^4=81$ is equal to $1$ modulo $5$. Hence, since $4\cdot 11339+1=45357$, then your prediction holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/234698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake... | $\displaystyle \lim_{n \rightarrow \infty} \left( \sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 + 1} \right) = \lim_{n \rightarrow \infty} \left\{
n \left[
\left(
1 + \frac 1n
\right)^{\frac 13} - \left(
1 + \frac 1{n^3}
\right)^{\frac 13}
\right]
\right\} = \\
\displaystyle \lim_{n \rightarrow \infty} \left[
n \left(
1 + \frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/236901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Geometry Prove - two perpendicular lines in a circle In a circle of radius r, two lines (AB and CD) are perpendicular to each other and meet at X.
Show that:
| Let $O$ be the center of the circle, and let $M$ and $N$ be the midpoints of segments $AB$ and $CD$, respectively. Write $a:=|XA|$, $b:=|XB|$, $c:=|XC|$, $d:=|XD|$, $m := |OM|$, $n:=|ON|$.
Then (noting that $OMXN$ is a rectangle),
$$\begin{align}
|AM| = \frac{1}{2}(a+b) &\qquad |ON|=|MX| = \frac{1}{2}|a-b| \\[5pt]
|CN|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/237541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find $x$ such that $\arctan(3/2)+\dots=\arctan x$ Find $x$ such that
$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$
| You may use the fact if $x > 0$, the argument of the complex number $x+iy$ is $\tan^{-1} \left(\frac{x}{y}\right)$. So if you compute the product
$$(2+3i)(4+5i)(2-5i)(3-8i) = 920 - 531i$$
you get that
$$\tan^{-1}\left(\frac{3}{2}\right) + \tan^{-1}\left(\frac{5}{4}\right) + \tan^{-1}\left(-\frac{5}{2}\right) + \tan^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the maxium $\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$ 1) $a, b, c$ are triangle edges's length such that $abc = 1$. Find max:
$$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$$
My idea: $$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}=\f... | For $0 < \epsilon < 1$ and $a = \epsilon^2$, $b = c = \frac 1 \epsilon$ we have
$$
abc = 1\\
a = \epsilon^2 \leq \frac 2 \epsilon = b + c \\
b = \frac 1 \epsilon \leq \frac 1 \epsilon + \epsilon^2 = a + c \\
c = \frac 1 \epsilon \leq \frac 1 \epsilon + \epsilon^2 = a + b
$$
All question conditions are satisfied.
Now
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/243077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is this series convergent or divergent? Kindly asking, what can I do about series
$$
\left(\frac{1}{3}\right)^2+\left(\frac{1\times 4}{3\times 6}\right)^2+\left(\frac{1\times 4\times 7}{3\times 6\times 9}\right)^2+...+\left(\frac{1\times 4\times 7\times...\times (3n-2)}{3\times 6\times 9\times...\times3n}\right)^2+...$... | Let $a_n$ be the $n$th term.
The ratio of successive terms is
$$\frac{a_{n+1}}{a_n} = \frac{(3n+1)^2}{9(n+1)^2} = 1 - \frac{4}{3n} + O\left(\frac{1}{n^2}\right).$$
Thus, the series converges by Raabe's test, which tells us that if
$$\left|\frac{a_{n+1}}{a_{n}}\right| \sim 1 - \frac{s}{n} \hspace{5ex}(n\to\infty),$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Factor $1+x+x^2+x^3+...+x^{14}$ In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving
$1+x+x^2+x^3 = 0 \to $
$1+x(1+x+x^2) = 0 \to $
$x(1+x+x^2) = -1$
which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$... | Using the fact
$$ x^n-1 = (x-1)(1+x+x^2+\dots+x^{n-1}),$$
our polynomial can be written in the form
$$ 1+x+x^2+x^3+…+x^{14} = \frac{x^{15}-1}{x-1}. $$
Now, we can find the roots of $ x^{15} - 1 $ using the complex variable tecniques
$$ x^{15}=1=e^{i2k\pi} \implies x = e^{\frac{i2k\pi}{15}},\quad k=0,1,2,\dots,14. $$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$
Possible Duplicate:
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?
How can I calculate precise value of that series: $\sum\limits_{i=0}^{n-1} i2^i$ ?
So far, I tried to differentiate the $ \sum\limits_{i=0}^{n} 2^i = 2^{i-1} - 1 $ series, b... | Here is another way to do this
Consider the polynomial $$\begin{align}&P(x)=\sum^{n-1}_{i=0} \ i\ \cdot \ x^i= 0x^0 +1x^1+2x^2+3x^3+\cdots +(n-1)\ x^{n-1}\\&Q(x)=\cfrac{P(x)}{x}=1x^0+2x^1+3x^2+\cdots+(n-1)\ x^{n-2}, \quad \quad x \ne 0\\ &\int Q(x)\ \text d x = cx^0+x^1+x^2+x^3+\cdots+ x^{n-1}=c+\sum^{n-1}_{i=1}x^i \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/250746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Combinations of 5-digit numbers I'm just studying for finals here and wanted some confirmation that I'm doing things right.
So if we have 5-digit decimal numbers, there are:
*
*$10^5 - 9*8*7*6*5$ numbers with no $0$,
*$10^5 - 9*8*7*6*5$ numbers with no $1$, and
*$10^5-2(9*8*7*6*5)+8*7*6*5*4$ numbers with no $0$ ... | *
*A $5$-digit number with no $0$ can have any of the other $9$ digits in each of its $5$ places, so there are $9^5$ such numbers. Your calculation would be right if the digits of the number were required to be distinct, but they’re not. The number of $5$-digit numbers with at least one $0$ is therefore $9\cdot10^5-9^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the direct expression of $f(t)$ I am trying to find the direct expression of the function $f(t)$ given by $$f(t)= \int_{1}^\infty \frac{\arctan (tx)}{x^2\sqrt{x^2-1}}dx$$
It's hard for me to calculate the integration directly.Should I try the method of interchanging $\frac{d}{dx}$ with$\int$ or$\int$ with $\int$ ?... | Presumably this was intended to be done with pen and paper, so here is what they might have had in mind. First note that convergence is excellent, so we may differentiate with respect to $t$ to obtain
$$ f'(t) = \int_1^\infty \frac{x}{1+t^2x^2} \frac{1}{x^2\sqrt{x^2-1}} dx.$$
Now let $x^2-1 = u^2$ so that $x\, dx = u\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I wa... | Here $\prod_{cyc}$ refers to the cyclic product of $x, y, z$.
Let $a=x^2, b=y^2, c=z^2$ for positives $x, y, z$. Then by the Cauchy-Schwarz inequality, we have:
$$LHS^2=\prod_{cyc}[(x^4+y^2z^2)(x^2z^2+y^4)]\ge \prod_{cyc}(x^3z+y^3z)^2=x^2y^2z^2\prod_{cyc}(x^3+y^3)^2$$
Then, by Power-Mean and AM-GM:
$$x^2y^2z^2\prod_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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Algebraic manipulation of binomial theorem Prove, by algebraic manipulation, that:
\[ {{2n} \choose {n}} + {{2n} \choose {n+1}}={1\over2} {{2n+2} \choose {n+1}} \]
| $$\binom m r+\binom m{r+1}=\frac{m!}{(m-r)! r!}+\frac{m!}{(m-r-1)! (r+1)!}=\frac{r+1}{m+1}\frac{(m+1)!}{\{(m+1)-(r+1)\}! (r+1)!}+\frac{m-r}{m+1}\frac{(m+1)!}{\{m-(r+1)\}! (r+1)!}$$
$$=\binom{m+1}{r+1}\cdot\left(\frac{r+1}{m+1}+\frac{m-r}{m+1}\right)=\binom{m+1}{r+1}$$
Putting $m=2n,r=n,$
$$\binom {2n} n+\binom {2n}{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded.
Thank you!
| Let $\arctan x=y\Leftrightarrow x=\tan y$. Then,
$$\sin^2 y+\cos^2 y=1\Leftrightarrow \tan^2 y+1=\frac{1}{\cos^2 y}\Leftrightarrow \frac{1}{x^2+1}=1-\sin^2 y\Leftrightarrow \sin^2 y=\frac{x^2}{x^2+1}$$
and so $$\sin y= \frac{x}{\sqrt{1+x^2}}\Rightarrow \arctan x=y=\arcsin \frac{x}{\sqrt{1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 3
} |
Prove that the equation $y^2=x^3-73$ has no integer solutions Prove that there are no integers $x,y$ such that $y^2=x^3-73$.
Thank you.
| You should send the equation into the congruence modulo $4$, for that according to Division Algorithm, (or Complete Residue System modulo 4) we have 4 possibilities for arbitrary number $x$.
$$ \text{If} \ x \equiv 0 \ \text{or} \ 2 \pmod{4} \ \ \text{then} \ \ y^2\equiv 0 - 73 \equiv 3 \equiv -1 \pmod{4} $$
So as we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| We have $\gcd(a+b,a^2-ab+b^2)=\gcd(a+b,3ab)$.
If $p$ is a prime with $p|\gcd(a+b,3ab)$, then $p=3$ or $p|a$ or $p|b$.
If $p|a$, then $p\nmid b$, hence $p\nmid a+b$. Similarly, if $p|b$, then $b\nmid a+b$.
Thus we conclude $p=3$.
But then if $9|\gcd(a+b,3ab)$, clearly $3|a$ or $3|b$ and again $3\nmid a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 1
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simplify summation of factorial (random walk) I suspect that the expression
$$\sum_{n=0}^N \frac{(N-2n)^2}{n!(N-n)!}$$
simplifies to
$$\frac{2^N}{(N-1)!}$$
But I cannot find the intermediate steps. Can someone give me a hint how I can deduce this result?
(The expression comes up when calculating the average final posi... | In the course of this post, you can see how to prove that $$\sum_{n=0}^Nn\binom{N}{n}=N2^{N-1}.\tag{1}$$ Two other nice facts we'll use are $$\sum_{n=0}^N\binom{N}{n}=2^N\tag{2}$$ and $$\binom{N+1}{n}=\binom{N}{n}+\binom{N}{n-1}.\tag{3}$$ From these facts, along with $\binom{N}{N+1}=0$, we see that $$\begin{align}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/258882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
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Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
| One reason the result looks difficult is because it seems to involve two variables.
But if you divide both sides by $y^n$, it can be rewritten as
$$ \eqalign{\frac{x^n}{y^n} - 1 &= \left(\frac{x - y}{y}\right) \left( \frac{x^{n-1} + x^{n-2} y + \ldots + x y^{n-2} + y^{n-1}}{y^{n-1}}\right)\cr
&= \left(\frac{x}{y} - 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
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Help with Inequality Given that $x, y, z$ are nonnegative real numbers such that :
$$x^2 + y^2 + z^2 + xyz = 4$$
Prove that
$0 ≤ xy + yz + zx − xyz ≤ 2$
| use this $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
and $$x^2+y^2+z^2+xyz=4$$
then we set
$$x=2\cos{A},y=2\cos{B},z=2\cos{C}$$
$$\Longleftrightarrow 0\le 4(\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{A}\cos{C})-8\cos{A}\cos{B}\cos{C}\le 2$$
and since
$$\cos{A}\cos{B}\cos{C}=\dfrac{s^2-(2R+r)^2}{4R^2},\cos{A}\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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A property of finite field of order $2^n$ Suppose $a$ and $b$ are elements of a finite field of order $2^n$ with $n$ odd and $a^2+ab+b^2=0$. Is it necessary that both $a$ and $b$ must be zero ?
I understand that the field has characteristic $2$ but don't know how to use the fact that $n$ is odd, please help.
| If $a=b$,
then $a^2+ab+b^2=3a^2=a^2=0$ and $a=0=b$,
we are done.
Suppose that $a\neq b$.
Observe that $0=(a-b)(a^2+ab+b^2)=a^3-b^3$.
Thus, $a^3=b^3$.
We claim that $a=0$ and $b=0$.
If $a\neq 0$,
then $(a^{-1}b)^3=1$ and the multiplicative order of $a^{-1}b$
in the multiplicative group $F-\{0\}$ is $1$
because $3\not\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Finding Eccentricity from the rotating ellipse formula I see that from a normal ellipse formula, we can acquire the eccentricity via this formula here.
However, for this formula (1):
$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$
When parameter $B = 0$, we would have normal ellipse, and the formula from the link abo... | If $S(p,q)$ be a focus, $L: ax+by+c=0$ be the directrix and $e$ be the eccentricity of a conic with $P(x,y)$ is any point on the conic,
$$e=\frac{\sqrt{(x-p)^2+(y-q)^2}}{\frac{|ax+by+c|}{\sqrt{a^2+b^2}}}$$
On simplification, $$\{a^2(1-e^2)+b^2\}x^2-2abe^2xy+\{a^2+(1-e^2)b^2\}y^2+()x+()y+()=0$$
Comparing with the given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
How to integrate, $\int_{0}^{\pi/2}\sin (\tan\theta) \mathrm{d\theta}$? Integrate, $$\int_{0}^{\frac{\pi}{2}}\sin (\tan\theta) \mathrm{d\theta}$$
| Making a change of variables $u=\tan(\theta)$:
$$
\int_0^{\pi/2} \sin\left(\tan \theta\right) \mathrm{d} \theta = \int_0^\infty \frac{\sin(u)}{1+u^2} \mathrm{d} u \tag{1}
$$
In order to evaluate this we use the technique of Mellin transform.
*
*Evaluate the Mellin transforms of $\sin(u)$ and $\left(1+u^2\right)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Solving an Inequality Involving a Fraction I have the following inequality:
$$ \frac{x-1}{x+2} \geq 0.$$
I solved it pretty fast:
$$\begin{align}
\frac{x-1}{x+2} +1 & \geq 1\\\\
\left(\frac{x-1}{x+2} + 1\right)\cdot(x+2) & \geq 1 \cdot (x+2)\\\\
x-1 + 1\cdot(x+2) & \geq 1\cdot (x+2)\\\\
2x + 1 & \geq x+2\\\\
x + 1 & \... | First note that when you multiply by a negative number the inequality changes in sign. In general, if $\dfrac{a}{b} > 0$, we have $a>0, b>0$ or $a<0, b<0$. Hence, we get that
$$(x-1) > 0, \,\,\,\,\, (x+2) > 0 \text{ i.e. } x>1$$
or
$$(x-1) < 0, \,\,\,\,\, (x+2) < 0 \text{ i.e. }x < -2$$
If $\dfrac{x-1}{x+2} = 0$, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How is $\sqrt{n^2+2n}-\lfloor\sqrt{n^2+2n}\rfloor=\frac{2}{\sqrt{1+\frac{2}{n}}+1}$. I have trouble seeing how
$$\sqrt{n^2+2n}-\lfloor\sqrt{n^2+2n}\rfloor=\frac{2}{\sqrt{1+\frac{2}{n}}+1}.$$
I can't see where to start even.
| Assuming that $n$ is a non-negative integer, we know that
$$ n^2 \leq n^2 + 2n < n^2 + 2n + 1 = (n+1)^2$$
Hence, this show that $\lfloor \sqrt{n^2 + 2n} \rfloor = n$, which simplifies the LHS to
$$ \sqrt{n^2+2n} - n.$$
Now let's work on simplifying the RHS, multiplying the numerator and denominator by $n$, we get
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Proof of a trigonometric inequality Does anyone know the proof of the following inequality
$$\sin(A)\sin(B)\sin(C)\le\left(\frac{3\sqrt{3}}{2\pi}\right)^3ABC$$
where $A,B,C$ are the vertex angles of a triangle.
| First note that $\log(\sin(x)/x)$ is concave for $0 \leq x,y,z < \pi$. Hence, we have that
$$\dfrac{\log(\sin(x)/x) + \log(\sin(y)/y) + \log(\sin(z)/z)}3 \leq \log \left( \dfrac{\sin((x+y+z)/3)}{(x+y+z)/3} \right)$$
Now you should be able to finish it off. Essentially same as Jensen's inequality if we consider the func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/273188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
If $r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\}$ , find $ r_{\max}$ Studying differential equations I came cross through this:
Let $ \displaystyle{ r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\} } $, where $ a,b >0$. Find $ r_{ \max} $.
Here is what I did:
Fix $ a> 0$ and define $ \displaystyle{ g(b) = \fr... | The maximum is when $$a=(4/27)^{1/7}\;\; \text{or}\;\; a=.76125127398793390138765377686933423...$$
$a$ is increasing as a increases. The other formula decreases as $a$ increases
$$a = 2^{\frac{2}7}/3^{\frac{3}7}$$ is when both sides are equal. Any change in a in either direction will cause one of them (and therefore th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Compute $\sum_{m>n=1}^{\infty} \frac{1}{m!n!}$ Compute the series
$$\sum_{m>n=1}^{\infty} \frac{1}{m!n!}$$
| We have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{ m!\,n!} = \left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)\left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)=(e-1)^2$$
We must substract from that the "diagonal" terms: $$\sum_{n=1}^{\infty} \left(\frac{1}{n!}\right)^2 = I_0(2) - 1 \app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/277063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Prove that the only root of the equation $z-\sin(z)$ in the unit disk is zero. Prove that the only root of the equation $z-\sin(z)$ in the unit disk is $z=0$.
My first thought is Rouche's Theorem, but I don't know any bounds on $|\sin(z)|$. Suggestions?
| Not the most elegant solution, but here's something.
First of all, $f(z) = z - \sin z$ has a triple zero at $z=0$. Define
$$g(z) = \frac{z-\sin z}{z^3} = \frac{1}{3!} + \underbrace{\left(- \frac{z^2}{5!} + \frac{z^4}{7!} - \cdots \right)}_{h(z)}. $$
If $|z| < 1$, then by the triangle inequality,
\begin{align}
|h(z)| &\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/278875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Differential equation involving matrices I want to show that
$ Y(t)=\cos(At)$ and $Y(t)=\sin(At)$ satisfy the equation
$$Y''+A^2Y=0 $$
subjected to the initial conditions $Y(0)=I, Y'(0)=0$ and $Y(0)=0, Y'(0)=I$ respictively
where $A$ is an constant $n \times n$ matrix..
| I found the shifting of indices to be highly confusing so I propose a method that does not rely on that.
We will first show that $Y(t)=\cos (At)$ solves the initial value problem $Y''+A^2 Y=0$ where $Y(0)=I$ and $Y'(0)=0$. Because $\cos At$ is defined as the infinite sum $\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/278938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
gambling probability problem We are given a fair coin. We start out with 5 dollars. We keep tossing the coin. If the outcome is different than the previous one, we are awarded another 5 dollars. However, we do not get anything if the outcome is the same as the previous one. Let's say we toss the coin X times in the lon... | This problem may be solved using generating functions which is admittedly more powerful than strictly necessary in this case, but it does illustrate the method, which is worth knowing. Let $p_A(n)$ be a polynomial in $u$ such that $[u^k] p_A(n)$ is the probability of having a capital of $k$ dollars and the last coin ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Number of strings of length n formed by $\{0,1,2\}$ such that $1$ and $2$ do not occur successively. What is the number of ways of forming a string of length $n$ from the set $\{0,1,2\}$ such that $1$ and $2$ do not occur successively.
| Let's use generating functions (Wilf's "generationfunctinology explains the technique)to solve this one. Call $a_n$, $b_n$, $c_n$ the number of strings of length $n$ ending in 0, 1, 2 respectively. By considering the ways of getting strings of length $n + 1$ by appending a digit to sequences of length $n$, this leads t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Which of the following statements are true for the sides of a triangle? Which of the following statements are true?
(a) if $a,b$ and $c$ are the sides of a triangle , then $(ab+bc+ca)/(a^2+b^2+c^2)≥(1/2)$
(b) if $a,b$ and $c$ are the sides of a triangle , then $(ab+bc+ca)/(a^2+b^2+c^2)≤ 1$
(c) both statements above a... | $$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
$$\implies a^2+b^2+c^2\ge ab+bc+ca\iff \frac{ab+bc+ca}{a^2+b^2+c^2}\le1$$
If $c\ge a-b, c^2\ge (a-b)^2=a^2+b^2-2ab$ which is true for any triangle
Similarly, $a^2\ge b^2+c^2-2bc$
and $b^2\ge c^2+a^2-2ca$
Adding we get, $$a^2+b^2+c^2\ge 2(a^2+b^2+c^2)-2(ab+bc+ca)$$
$$\iff 2(ab+bc+ca)\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| It is enough to show that for $t\geqslant0$,
$$f(t) = \log(3)-\log\left(t^3+t+1\right)+\tfrac23\left(t^2-1\right)
\geqslant 0$$
as the inequality in question is merely $f(a)+f(b)+f(c)\geqslant0$.
But $f’(t)=0$ only when $t=1$ where it has a minimum, so $f(t)\geqslant f(1)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 8,
"answer_id": 4
} |
What is wrong with this problem We know that:
$(a^n)^m=a^{nm}$
From this we have:
$-3^3=[(-3)^2]^\frac{3}{2}=(3^2)^\frac{3}{2}=27$
Find what's wrong
| The "law of exponents" that you cite:
$$\large (a^n)^m=a^{nm}$$
applies PROVIDED $\bf{a \gt 0}$.
Here, though, we have $\,\bf{a = -3 \lt 0}$, and hence:
$$-3^3=[(-3)^2]^{\large\frac{3}{2}}=(3^2)^{\large\frac{3}{2}}=27\quad \Longleftarrow \;\text{ False}.$$
$$-3^3 = -(3^3) = -[(3^2)^{\large \frac{3}{2}}] = -(9^{\large... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Improper integral with log and absolute value: $\int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx$ How do you show that $$ \int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx = \frac{\pi}{2} \log (\tanh 1)\, ?$$
I know that the integral converges since $\log |\tan x| = \frac{1}{2} \log(\tan^{2}x)$, and $\log (\ta... | Denote the evaluated integral as $I$, then $I$ may be rewritten as
$$I=\frac{1}{2}\int_0^\infty \frac{\ln \sin^2 x}{1+x^2}\,dx-\frac{1}{2}\int_0^\infty \frac{\ln \cos^2 x}{1+x^2}\,dx$$
Using Fourier series representations of $\ln \sin^2 \theta$ and $\ln \cos^2 \theta$,
$$\ln \sin^2 \theta=-2\ln2-2\sum_{k=1}^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 0
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Real part of complex z must be zero
Show that the real part of any solution of $(z+1)^{100}=(z-1)^{100}$
must be zero.
I know that I must use the fact $e^{2\pi ki}$ somewhere but not sure exactly how or where to use it.
| If $z=x+iy$ and if we put the value of $z$ in $|z+1|=|z-1|$ which has been derived in other answers,
$$|x+iy+1|=|x+iy-1|\implies (x+1)^2+y^2=(x-1)^2+y^2$$
$$\implies (x+1)^2-(x-1)^2=0\implies4\cdot x\cdot1=0\implies x=0$$
A little generalization:
Let $|z-w|=|z+w|$
If $z=x+iy,w=a+ib$ where $x,y,a,b$ are real and $a\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| As $-1\leq\sin x\leq1\implies |\sin x-1|=1-\sin x$, $|\sin x-2|=2-\sin x$, $|\sin x-3|=3-\sin x$, $|\sin x+1|=1+\sin x$
Thus the given expression becomes, $1-\sin x+2-\sin x+3-\sin x+1+\sin x=7-2\sin x$ which is minimum when $\sin x $ is maximum and equal to $1$ which gives minimum value equal to $5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Solve without using L'Hôpital's rule? Is it possible to solve this without using L'Hôpital's rule?
$$
\lim_{x\to 0} \Big(\frac{3x+1}{x}-\frac{1}{\sin x}\Big)
$$
I tried to solve it but I got stuck at the $\frac{1}{\sin x}$ part.
$$
=\lim_{x\to 0}\frac{3+x^{-1}}{1}-\lim_{x\to 0}\frac{1}{\sin x}
$$
since $\lim_{x\to 0}\f... | Let $\displaystyle L=\lim_{x \to 0} \frac{\sin x - x}{x^2}$. Replacing $x$ by $2y$, we obtain:
$
\begin{align*}
L & = \lim_{y \to 0} \frac{\sin 2y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2 \sin y + 2 \sin y - 2y}{4y^2} \\
& = \lim_{y \to 0} \left( \frac{2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/289465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
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Definite integral $\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{\sin 2x}}\,\mathrm{d} x$ $$\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{\sin 2x}}\,\mathrm{d} x$$
I'm pretty sure I can finish it after finding the anti derivative. I tried changing the denominator to $2\sin x \cos x$ and subbing $u$ as $\sin x$.
$$\int_... | After a little algebra, we get
$$ \int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} dx \: \frac{\sin{x}}{\sqrt{\cos{x} \sqrt{1-\cos^2{x}}}}$$
Substitute $u=\cos{x}$:
$$\int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} = \frac{1}{\sqrt{2}} \int_0^1 d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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First week in Linear Algebra, need some help on this simple problem If you have
$5x + 2y + z = 0$
$2x + y = 0$ and you're asked to solve using back-substitution how would you go about doing it?
Initially I thought just simply the following:
$x + \frac{2}{5} y + \frac{1}{5} z = 0$ (divide by $5$)
$2x + y ... | I will write up what I wrote above into a solution:
Note that we begin with
$$5x + 2y + z = 0 $$
$$2x + y = 0$$
Subtracting twice the second from the first we get $x+z=0$, therefore, $z=-x$. Therefore, we get solutions of the form $y=-2x$ and $z=-x$ where $x$ can be anything. This means that all solutions are on the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is my Fourier Series computation done correctly? See my fourier series calculation of this function if you please!
$
f(t)=\left\{\begin{array}{ll} 0, & \text{for } \ -\pi<t<0 \\
1, & \text{for } \ 0 < t < \pi\end{array}\right. .
$
I start by calculation of the coefficient $a_0,a_k,b_k$: $$a_0=\frac{1}{\... | $$\begin{equation*} % a_0
f(t)=\begin{cases}0,&-\pi<t<0\\1,&0<t<\pi \end{cases}
\end{equation*}$$
\begin{equation*}a=-\pi\qquad l=\pi\qquad a+2l=\pi\qquad\qquad\qquad\end{equation*}
$$\begin{equation*}S(t)=\frac{a_0}2+\sum\limits_{k=1}^\infty a_k\cos\left(\frac{k\pi x}l\right)+b_k\sin\left(\frac{k\pi t}l\right)\\
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find real roots? $4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$ How to find real roots?
$$4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$$
I try LMVT but vary difficult
| Trying to piece together the above comments. First, the equation can be rewritten:
$$
\frac{5^x - 4^x}{5 - 4} = \frac{6^{x^2} - 5^{x^2}}{6 - 5}
$$
By LMVT, there exist constants $4 \le \alpha \le 5$ and $5 \le \beta \le 6$
such that:
$$
\begin{align*}
\frac{5^x - 4^x}{5 - 4} &= x \alpha^{x - 1} \\
\frac{6^{x^2} - 5^{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/293323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding all matrices $B$ such that $AB=BA$ for a fixed matrix $A$ Let
$$ A=\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 2
\end{pmatrix}
$$
Find all matrices $B$ such that $AB=BA$.
Attempt at solution: I can show that $A$ is invertible so its inverse must be one of the elements. But how do I go about showing there... | $$A=\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 2
\end{pmatrix}$$
$$A=\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
-3 & -1 & 1
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
0 & 0 & 2
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 1
\end{pmatrix}$$
This is the diagonalization of $A$, $A=PDP^{-1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/296857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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Determinant of block matrix Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.
\begin{align}
D=\begin{vmatrix}
a_{11} &a_{12} & \cdots & a_{1k} &0 &\cdots &0 \\
a_{21}& a_{22}& \cdots & a_{2k} & 0 &\cdots &0 \\
\vdots& \vdots & & \vdots & \vdots & &\vdots\\
a_{k1} & a_{k2} & \cdots & a_{kk} ... | Apply Laplace expansion http://en.wikipedia.org/wiki/Laplace_expansion.
you can directly calculate $D_1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solve for $ x$, $-\frac{1}{2}x^2 + 2x + 5 = 0$ I'm having trouble solving this equation for $x$:
$$-\frac{1}{2}x^2 + 2x + 5 = 0$$
What's the steps to take to solve it?
Thanks.
| If you multiply
$$-\frac{1}{2}x^2 + 2x + 5 = 0\tag{1}$$
by $-2$ you get $$x^2-4x-10=0$$ Using the quadratic formula: $$ax^2 + bx + c = 0 \iff x = \frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right),$$ where in this case, $a = 1,\;b= -4,\; c = -10$ we have $$x=\frac 12 (4 \pm \sqrt {16+40})=2 \pm \sqrt {14}\tag{2}$$
Walki... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to find the exact value of $ \cos(36^\circ) $? The problem reads as follows:
Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of
$$\cos(36^\circ)$$
it says that you may find useful the following identities:
$$\cos^2 t+\sin^2 t = 1,\\
\sin 2t = 2\sin t\cos t,\\
\sin 3t = 3\sin t - 4\sin... | Let $t=\frac{\pi}{5}$ (so $t$ is $36^\circ$). Since $108=180-72$, we have $3t=\pi-2t$ and therefore
$$\sin(3t)=\sin(\pi-2t).$$
But $\sin(\pi-2t)=\sin(2t)=2\sin t\cos t$.
Also, by the identity you were given, $\sin(3t)=3\sin t-4\sin^3 t$. Thus
$$3\sin t-4\sin^3 t=2\sin t\cos t.$$
But $\sin t\ne 0$, so we can cancel a $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Is there a closed form or a better solution? This is another integral in the book "irresistible integral"
I can find that:
$$\int_0^\infty \frac{1}{\left( x^4 +2ax^2+1 \right)^{m+1}} \, \text{d}x =\frac{{{\left( -1 \right)}^m}\sqrt{2}\pi }{4\left( m! \right)}\left. \left( \frac{\text{d}^m}{\text{d}x^m} \sqrt{x}\sqrt{a+... | For every $a\gt-1$, consider
$$
I_m(a)=\int_0^\infty \frac{\mathrm dx}{\left(x^4+2ax^2+1\right)^{m+1}},$$
then, for every $|t|$ small enough,
$$
\sum_{m=0}^{+\infty}I_m(a)t^m=\frac{\pi}{2\sqrt2\sqrt{1-t}\sqrt{a+\sqrt{1-t}}}.
$$
To deduce a closed form formula for each $I_m(a)$ from this expression does not seem obvious... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| Hint: calculate the remainder each of the first few terms contributes. You should be able to make a reasonable hypothesis, then confirm it. What is the remainder when $(2k+1)^2$ is divided by $8?$ How many terms are in the sum?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Theory of Partial Fraction Decomposition The function in question that I want to decompose is $$\dfrac{8x^3 + 7}{(x+1)(2x+1)^3}$$
I had the idea to to break this down into:
$$\dfrac{A}{x+1} + \dfrac{Bx^2 +Cx + D}{(2x+1)^3} + \dfrac{Ex + F}{(2x+1)^2} + \dfrac{G}{2x+1}$$
Well this turns into a really messy system of... | You only need to break it down into $\displaystyle\frac{A}{x+1}$ and $\displaystyle\frac{Bx^2 + Cx + D}{(2x+1)^3}.$ What you end up with is
$\begin{align*}
8A + B &= 8\\
12A + B +C &= 0\\
6A + C + D &= 0\\
A + D &= 7
\end{align*}$
Solving gives $A = 1, B = 0, C = -12, D =6.$
Note that this is equivalent to lab bhattac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/304969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Calculating simple limit: $\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}$ I am trying to calculate this limit:
$$
\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}
$$
What I get is $\frac{4}{3}$, however, according to Wolfram Alpha, it should be 1. What am I missing here?
| $$ \frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\frac{(x+2)(3x-1)}{(x+2)(4x+1)}=\frac{(3x-1)}{(4x+1)}\text{ if }x+2\ne0$$
As $x\to-2,x\ne-2\implies x+2\ne0$
So, $$\lim_{x\to-2}\frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\lim_{x\to-2}\frac{(3x-1)}{(4x+1)}=\frac{3(-2)-1}{4(-2)+1}=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{a}{b^2+5}+ \frac{b}{c^2+5} + \frac{c}{a^2+5} \le \frac 12$ Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$
I have an ugly solution for this solution.
| First ,we can get rid of denominators, by noticing that
$\frac{1}{x^2+5} \leq \frac{4-x}{18}$ for any $x\in [0,2]$
(indeed, we have
$$
(4-x)(x^2+5)-18=((x-1)^2)(2-x) \geq 0 \tag{1}
$$
So it will suffice to show that the number
$$
f(a,b,c)=a(4-b)+b(4-c)+c(4-a) \tag{2}
$$
is $\leq 9$. Note that $f(a,b,c)=4(a+b+c)-(ab+a... | {
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"source": "stackexchange",
"question_score": "12",
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How to evaluate this rational limit Evaluate:
$$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6}$$
| Exploiting the Taylor series of a function at the point $x=0$, we have
$$ \frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln(1+x)+x^4-x^6}$$
$$=\frac{(-3\sqrt{x}+x^3+(x^6-\frac{x^{12}}{3!}-\dots )))( (1+4\sqrt{x}+\dots)-1)}{4 (x+\frac{x^2}{2}\dots)+x^4-x^6}$$
$$ \sim \frac{(-3\sqrt{x})(4\sqrt{x})}{4x} = -3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
| $$\begin{align}
\gcd(a+b, a^2 + b^2)
&= \gcd(a+b, a^2 + b^2 - a(a+b))
\\&= \gcd(a+b, b^2 - ab)
\\&= \gcd(a+b, b^2 - ab + b(a+b))
\\&= \gcd(a+b, 2b^2)
\end{align}
$$
Now, $\gcd(a+b,b) = \gcd(a,b) = 1$, so we can get rid of the factors of $b$ and have
$$ \gcd(a+b, a^2 + b^2) = \gcd(a+b, 2) $$
The strategy I used was stil... | {
"language": "en",
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"source": "stackexchange",
"question_score": "14",
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"answer_id": 5
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Complex Integration : $\int_1^{1+i}\frac{1}{1+z^2}dz$
Integrate alonf the line segment from $z=1$ to $z=1+i$ :
$$\int_1^{1+i}\frac{1}{1+z^2}dz$$
If I integrate, it is just the identity $tan^{-1}z$, but the answer to this question is $$\frac{\pi}{4}-\frac{1}{2}\arctan2+\frac{i}{4}\log5$$ which I don't understand how... | The integral can be parameterized by writing $z=1+it$:
$$\int_0^1 \frac{i}{(1+ti)^2+1}dt = \int_0^1 \frac{2 i + 2 t - i t^2}{4+t^4}dt$$
$$ = \int_0^1 \frac{2 t}{t^4+4} + i\int_0^1 \frac{2-t^2}{t^4+4}dt$$
$$ = \frac{i}{2} \arctan(1/2) - \frac{\log 5}{4} =
\frac{\pi}{4}-\frac{1}{2}\arctan2+\frac{i}{4}\log5$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$ How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$?
For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is $(-\infty, -... | Consider the families of quadratic functions $f_a(x)=(a+4)x^2-2ax+2a-6 $ we needs to find functions with their graph below $x$ axis these functions has maximum if $a+4<0$ and don't intersect the $x$ axis if discriminant is negative $D<0$ so we nedd to find solution of system of innequalities
$$\begin{matrix}
a+4<0\\... | {
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"source": "stackexchange",
"question_score": "8",
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help for proving an equation by induction For this equation:
$$-1^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)$$
how can I prove this by induction?
When I set $n = 1$ for the base case I got:
$$-1^3 + (-3)^3 + (-5)^3 + \ldots + (-3)^3 = -28$$
but am having trouble with the following inductive steps
| First, as the power of each summand is odd, I'd write the equation as
$$-1^3-3^3-5^3-\ldots-(2n+1)^3=-(n+1)^2(2n^2+4n+1)\Longleftrightarrow$$
$$1^3+3^3+5^3+\ldots+(2n+1)^3=(n+1)^2(2n^2+4n+1)$$
The base case is
$$n=1:\;\;\;\;\;1^3+3^3=28\stackrel ?=(1+1)^2(2+4+1)=28\ldots\ldots good$$
Assume for $\,n\,$ and show for $\,... | {
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"source": "stackexchange",
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Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable... | Hint: use Parseval/Plancherel theorem on $(\sin{x}/x)^2$.
That is, the FT of $(\sin{x}/x)^2$ is
$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$
Plancherel/Parseval says that
$$\int_{-\infty}^{\infty} dx \... | {
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Evaluating $\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$ $$I(a)=\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$$ I'd like to evaluate the integral by differentiating with respect to parameter $a$ but no success yet. Seems impossible. What would the other options?
Edit: A hypothetical closed form solut... | A special case is easy, using $4 \sin^3(x) = 3 \sin(x) - \sin(3x)$:
$$
I(0) = \int_0^\infty \frac{\sin^3(x)}{x} \mathrm{d} x = \frac{3}{4} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x - \frac{1}{4} \int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x = \frac{\pi}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/322704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$F:= \{ a+b\sqrt{7} \mid a,b \in \mathbb{Q} \}$ closed under addition, subtraction, multiplication, and division I am in my math class and I came across this problem on my past midterm. How can we prove that $F:=\{ a+b\sqrt{7} \mid a,b \in \mathbb{Q} \} $ is closed under addition, subtraction, multiplication, and divis... | The only somewhat tricky one is division. Let $x=a+b\sqrt{7}$, where $a+b\sqrt{7}\ne 0$. Note that $a-b\sqrt{7}\ne 0$, since $\sqrt{7}$ is irrational. (This needs proof.)
Then
$$\frac{1}{x}=\frac{a-b\sqrt{7}}{(a-b\sqrt{7})(a+b\sqrt{7})}=\frac{a}{a^2-7b^2}+\frac{-b}{a^2-7b^2}\sqrt{7}.$$
Remark: To show that if $a$ and ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Infinite Series :$ \sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2}$ Prove that:
$$\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2} = \frac{-\pi^{\frac{3}{2}}}{12}\left... | Consider the series
\begin{align}
S(x,y) = \sum_{n=0}^{\infty} \frac{\Gamma(n+x)}{n! \ (n+y+1)^{2}}.
\end{align}
This series can be evaluated as follows
\begin{align}
S(x,y) &= \sum_{n=0}^{\infty} \frac{\Gamma(x) \ (x)_{n}}{\Gamma(2) \ n!} \ \int_{0}^{\infty} e^{-(n+y+1)t} \ t \ dt \\
&= \Gamma(x) \ \int_{0}^{\infty} e... | {
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"timestamp": "2023-03-29T00:00:00",
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Exponential equation, $(3+2\sqrt2)^x+1=6(\sqrt2+1)^x$ http://i.stack.imgur.com/YP2Ha.png
$$(3+2\sqrt2)^x+1=6(\sqrt2+1)^x \qquad\qquad x\in\mathbb{R}$$
I managed to find one of the solutions (x=2), but I got stuck. I would really appreciate a step by step solution. Thanks in advance :)
| $$(3+2^{1/2})^x+1=6((2^{1/2}+1)^x)$$
$$(2^{1/2}+1)^{2x}+1=6((2^{1/2}+1)^x)$$ taking$(2^{1/2}+1)^x=y$ we get
$$y^2-6y+1=0$$ from there
$$y_{1,2}=\frac{6\pm\sqrt{32}}{2}=\frac{6\pm4\sqrt{2}}{2}=3\pm 2\sqrt2$$ or
$$(2^{1/2}+1)^x=3+2\sqrt2=(\sqrt2+1)^2\Rightarrow x_1=2$$
$$(2^{1/2}+1)^x=3-2\sqrt2=(\sqrt2+1)^{-2}\Rightarr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof: Two non identical circles have at most 2 same points I'm struggeling with an analytic proof for the fact, that two different circles have at most 2 same points. (I try to solve it analytical, because geometrical I already prooved it).
I tried to start with the equations
$r_1^2=(x-a_1)^2+(y-b_1)^2$ and $r_2^2=(x-... | Write the circle equation as:
$$
x^2 + y^2 + \alpha x + \beta y + \gamma = 0
$$
This equation has three unknowns, so three points are sufficient to uniquely identify a circle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the double integral $$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x^2+y^2+\sin(\pi(x^2+y^2)))\,dy\,dx$$
*sorry if the mathjax is off, I'm new at it.
Anyways, I can use the properties of double integrals to make it
$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x^2+y^2 \,dy\,dx + \int_0^3 \int_{-\sqrt{9-x^2}}... | Convert the integral into polar coordinates. Note that the region you are integrating is the semi-circle of radius lying on the right half plane. Hence, the integral becomes
$$\int_{-\pi/2}^{\pi/2} \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr \,d \theta = \pi \cdot \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$$
All we need to ev... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
Consider an equilateral triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Rotate $P$ clockwise along $A$ by $60^o$ to $B'$, along $B$ by $60^o$ to $C'$, along $C$ by $60^o$ to $A'$
We note that,
$${\triangle APB} \cong ${\triangle ACB'}$$
$${\triangle APC} \cong... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to simplify square root $$
\sqrt[3]{a}(\sqrt[3]{a^2}-\sqrt[3]{a^5})
$$
How can this be simplified? I can't find anything for doing the subtraction.
| *
*Distribute $\;\sqrt[\large 3]{a}\;$ over the difference,
*use the property that $\;\sqrt[\large a]{b}\cdot \sqrt[\large a]{c} =
\sqrt[\large a]{b\cdot c},\;$ and
*remember that $\sqrt[\Large a]{b^a} = b$
$$
\begin{align} \sqrt[\Large 3]{a}\left(\sqrt[\Large 3]{a^2}-\sqrt[\Large 3]{a^5}\right) & = \sqrt[... | {
"language": "en",
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For $x > -1$ proof that $ \arctan x + \arctan\frac{1-x}{1+x} = \frac{\pi}{4} $ For $x > -1$ proof that $\arctan x + \arctan\dfrac{1-x}{1+x} = \dfrac{\pi}{4} $
I have no idea how to approach this, some kind of help would be greatly appreciated!
edit: Thank you all!
| Let $\arctan(x) = a$. We then have
\begin{align}
\arctan(x) + \arctan \left(\dfrac{1-x}{1+x}\right) & = a + \arctan \left(\dfrac{1-\tan(a)}{1+\tan(a)}\right)\\
& = a + \arctan\left(\dfrac{\tan(\pi/4) - \tan(a)}{1+\tan(\pi/4) \cdot \tan(a)}\right)\\
& = a + \arctan \left(\tan(\pi/4-a)\right)
\end{align}
Note that since ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that :
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ .
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $
What to do... | As $A+B=\pi-C\implies \frac{A+B}2=\frac\pi2-\frac C2$
So, $$\tan\left(\frac{A+B}2\right)=\tan\left(\frac\pi2-\frac C2\right)$$
$$\frac{\tan \frac A2+\tan \frac B2}{1-\tan \frac A2\tan \frac B2}=\cot\frac C2=\frac1{\tan\frac C2} $$
$$\sum \tan \frac A2\tan \frac B2=1$$
Now, $$\sum (\tan \frac A2-\tan \frac B2)^2\ge0$$
$... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve in the positive real number the system of equation. Solve in the positive real numbers, the system of equations :
$$(2x)^{2013} + (2y)^{2013} + (2z)^{2013} =3 $$
and $$xy + yz + zx + 2xyz = 1.$$
| Let $x=\frac{a}{2}, y=\frac{b}{2}, z=\frac{c}{2}$, so
$$ab+bc+ca+abc=4, a^{2013}+b^{2013}+c^{2013}=3$$
Now by A.M.$\geq$G.M. $4=ab+bc+ca+abc \geq 3\sqrt[3]{a^2b^2c^2}+abc$. Let $x=\sqrt[3]{abc}$, so that $4 \geq 3x^2+x^3$.
Factorising, we get $(x-1)(x+2)^2 \leq 0$, so $x \leq 1$. Thus $abc \leq 1$, so $ab+ac+bc=4-abc ... | {
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Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4... | We shall use the combinatorial identity
$$\sum_{j=0}^{k}{(-1)^j\binom{n}{j}}=(-1)^k\binom{n-1}{k}$$
This can be proven easily by induction, and there is also probably some combinatorial argument why it holds. We shall use the equivalent form
$$\sum_{j=0}^{k}{(-1)^{k-j}\binom{n}{j}}=\binom{n-1}{k}$$
Now $(r-k+1)(r-k+2)=... | {
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Find the number of pairwise coprime triples of positive integers (a,b,c) with aFind the number of pairwise coprime triples of positive integers $a,b,c$ with $a\lt b\lt c$ such that
a|bc−31, b|ca−31, c|ab−31
Details and assumptions:
The notation n∣m means that $n$ is a divisor of $m$.
Clarifica... | Hint: $a, b, c$ all divide $ab+ac+bc-31$, so since they are pairwise coprime, $$abc \mid (ab+ac+bc-31)$$
Continuation: Note that the above is equivalent to the given conditions.
Consider 3 cases, where $ab+ac+bc-31$ is $=0, <0, >0$.
Case 1: $ab+ac+bc-31=0$, then $31=ab+ac+bc \geq a(a+1)+a(a+2)+(a+1)(a+2)=3a^2+6a+2$, so... | {
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Factorize $3m^4-6m^3+14m^2-6m+11$ I have this expression:
$3m^4-6m^3+14m^2-6m+11=0$ and I want to factorize it in $(m^2+1)(3m^2-6m+11)$.
How can I do it? Thanks for any help!
| $3m^4-6m^3+14m^2-6m+11$
$=3m^4-6m^3+11m^2+3m^2-6m+11$
$=m^2(3m^2-6m+11)+(3m^2-6m+11)$
$=(m^2+1)(3m^2-6m+11)$
| {
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Proof: How to prove $n$ is odd if $n^2 + 3$ is even New to the whole proof thing.
Trying to figure out that, for all integers $n$, if $n^2 + 3$ is even, then $n$ is odd.
Thank you for the help.
| If $n^2 + 3$ is even then $n^2 + 3 = 2k$ , where $k$ is an integer.
Thus, $n^2 = 2k - 3 = 2k - 4 + 1 = 2(k-2) + 1$, where $k$ is an integer, which means $k-2$ is an integer.
Now, this implies that $n^2$ is odd.
Assume $n$ is even.
Then $n$ can be written in the form $2m$, where $m$ is an integer.
Thus, $n^2 = (2m)^2 = ... | {
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How do I determine the intersection of span A and span B? Consider the following 2 sets of vectors in $\mathbb R^4$: $A = \{v_1, v_2, v_3\}, B = \{w_1, w_2, w_3\}$. You are given that $A$ is a set of linearly independent vectors and that $B$ is a set of linearly independent vectors.
Let
$v_1 = (3,1,4,1), v_2 = (5,9,2,6... | The intersection will have dimension 2 or 3 so we can control our results. As your vectors doesn't look very nice we will use the gauss algorithm to make them sweeter. Sweet vectors are vectors with a lot of zeroes. When we calculate $$a_1=v_2-v_3=\begin{pmatrix} 5-5 \\ 9-3 \\ 2-5 \\ 6-8\\ \end{pmatrix} = \begin{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Complete induction:figure out the last step I tried to solve this task on complete induction. I am sorry for the form, if you see a $6$
it means: a divisible through six. Could anyone please check if I have done right? Because at the end, I don´t know how to show that 4n³+2n²+3n is divisible through six.
Please show... | If you want to use induction, you are looking to assume $n^4-n^3-n^2+n$ is a multiple of $6$ and prove $(n+1)^4-(n+1)^3-(n+1)^2+n$ is. So $$((n+1)^4-(n+1)^3-(n+1)^2+n+1)-(n^4-n^3-n^2+n)=4(n+1)^3+6(n+1)^2+4(n+1)+1-3(n+1)^2-3(n+1)-1-2(n+1)-1+1=4n^3+3n^2-n+=n(n+1)(4n-1)$$
One of the first two terms is even and one of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/342046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$ Show that
$$\begin{aligned} \int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2}
\right)dx &= \frac{7\pi^3}{108} \\ \int_0^{\pi/3}x\log^2
\left(2\sin\frac{x}{2} \right)dx &= \frac{17\pi^4}{6480}\end{aligned}$$
*
*... | Here's another approach for evaluating the one without the $x$ in front.
First notice that it's equivalent to showing that $$\int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx = \frac{7 \pi^{3}}{216}. $$
Using the principal branch of the logarithm and assuming that $0 < x < \pi$, we have $$ \begin{align} \log(1-e^{2ix}) &= \lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/342777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 3,
"answer_id": 0
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logarithm problem - four tuple How many distinct four tuple (a,b,c,d) of rational numbers are there with $a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7=2005$
Can we proceed like this :
Using $\log a +\log b = \log(ab)$ and $m\log a = \log a^m$
$\Rightarrow \log_{10}2^a \cdot 3^b \cdot 5^c \cdot 7^d = 2005$
Please gu... | Hint: $10^{2005}=2^a \cdot 3^b \cdot 5^c \cdot 7^d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/343811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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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.