Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the second part: due a change of variable $u=x+1$, I find
$$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$
| Hint:
There's no need to split the integral that way (IMO), we can instead write, $$\begin{align}
\int \frac{3x+2}{x^2+x+1}\,\mathrm dx&=\frac{3}{2}\int\frac{2x+\tfrac{4}{3}}{x^2+x+1}\,\mathrm dx\\
&=\frac32\int\dfrac{2x+1+\tfrac13}{x^2+x+1}\,\mathrm dx\\
&=\dfrac32\left(\int\dfrac{2x+1}{x^2+x+1}\,\mathrm dx+\int\dfrac{{\small 1/3}}{x^2+x+1}\,\mathrm dx\right).
\end{align}$$ The first one is obvious. For the second one we should complete the square in the denominator, like this, $$\int\dfrac{1}{x^2+x+1}\,\mathrm dx=\int\dfrac{1}{\color{royalblue}{x^2+x+\tfrac14}+1-\tfrac14}\,\mathrm dx=\int\dfrac1{\color{royalblue}{\left(x+\tfrac1{2}\right)^2}+\tfrac34}\,\mathrm dx.$$ Now try to proceed further by rewriting that last expression so as to exploit the fact that the integral of $\frac{u'(x)}{u(x)^2+1}$ is $\arctan u(x)+\rm C$, you may have to factor something, and do one or more substitutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $16\cos^5A-20\cos^3A+5\cos A=\cos5A$
Prove that
$$16\cos^5A-20\cos^3A+5\cos A=\cos5A$$
My solution begins here;
$$
\begin{align}
\text{RHS} & =\cos5A \\
& =\cos(A+4A) \\
& =\cos A\cos4A-\sin A\sin4A \\
& =\cos A(2\cos^2 2A-1)-\sin A(2\sin2A\cos2A) \\
& =2\cos A\cos^2 2A-\cos A-2\sin A\sin2A\cos2A \\
& =2\cos A\cos^2 2A-\cos A-2\sin A(2\sin A\cos A)\cos2A
\end{align}
$$
Now how do I move on?
| Use
$$
2\cos x=e^{ix}+e^{-ix}
$$
Then
$$
32\cos^5x=e^{5ix}+5e^{3ix}+10e^{ix}+10e^{-ix}+5e^{-3ix}+e^{-5ix}=
2\cos5x+10\cos3x+20\cos x
$$
Moreover
$$
8\cos^3x=e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}=
2\cos3x+6\cos x
$$
Therefore
\begin{align}
16\cos^5x-20\cos^3x+5\cos x
&=(\cos5x+5\cos3x+10\cos x)-5(\cos3x+3\cos x)+5\cos x\\
&=\cos5x
\end{align}
This is actually backwards, so here's a different proof:
\begin{align}
\cos 5x+i\sin 5x
&=(\cos x+i\sin x)^5\\
&=\cos^5x+5i\cos^4x\sin x-10\cos^3x\sin^2x\\
&\qquad-10i\cos^2x\sin^3x+5\cos x\sin^4x
+i\sin^5x
\end{align}
Taking the real part
$$
\cos5x=\cos^5x-10\cos^3x(1-\cos^2x)+5\cos x(1-\cos^2x)^2
$$
and you can finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the value of :Solving $\lim_{n \to \infty} \sqrt{n} \sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)$ I have trouble finding the value of the following limit:
$$\lim_{n \to \infty} \sqrt{n} \sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)$$
For now I have rewritten the term into:
$$ \lim_{n \to \infty} \dfrac{\sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)}{\large \frac{1}{\sqrt{n}}}$$
Now I have a limit of type $\large \frac{0}{0}$ so I think I could use L'Hopital's rule. But I would like to know if there is a way you can solve this without using L'Hopital's rule.
Thanks for your answers.
| $\sqrt{n + 3} - \sqrt{n - 2} = \frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}$
This yields the following term:
$$\sqrt{n}\sin(\sqrt{n + 3} - \sqrt{n - 2}) = \frac{5 \sqrt{n}}{\sqrt{n + 3} + \sqrt{n - 2}} \frac{\sin\left(\frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}\right)}{\frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}}$$
Now use the fact that $\frac{\sin(h)}{h} \to 1$ for $h \to 0$ and $\frac{5 \sqrt{n}}{\sqrt{n + 3} + \sqrt{n - 2}} = \frac{5}{\sqrt{1 + 3/n} + \sqrt{1 - 2/n}} \to \frac{5}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the minimum of $x^2+y^2$ for $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$. If $x,y \in \mathbb {R}$, find the minimum of $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$.
This problem was inspired by a problem which asked if $x,y \in \mathbb {R}$ and $xy \neq 0$, find the minimum of $x^2+y^2$ when $xy(x^2-y^2)=x^2+y^2$.
By setting $x=a\sin\theta$, $y=a\cos\theta$, the equation can be simplified to $a^2=\frac{1}{\sin\theta\cos\theta(\sin\theta^2-\cos\theta^2)}$
However, notice that $\sin\theta\cos\theta=\frac{\sin2\theta}{2}$, $\sin\theta^2-\cos\theta^2=-\cos2\theta$.
This implies that $a^2=\frac{2}{-\sin2\theta\cos2\theta}$, thus that $a^2=\frac{4}{-\sin4\theta}\ge 4$.
Thus the minimum of $x^2+y^2$ is $4$, with the equality holding when $x=\sqrt{2-\sqrt{2}}$, $y=\sqrt{2+\sqrt{2}}$.
However, since there are no formulas I know of where $\sin^3 x+\cos^3 x$, I did not know how to find the minimum of $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$.
Graphing it seems to imply that such a minimum exists, but I am not aware of how to find it( and when such a minimum exists).
Any help would be appreciated.
| EDIT: Just realized $xy≠0$. Here is the sadly tough solution.
Instead of writing out the minutiae, I will just sketch out the solution method.
To begin, rewrite the initial expression as follows:
$xy(x-y)^2(x^2+xy+y^2)=(x+y)(x^2+y^2-xy)$
We did this by factoring the sum and the difference of cubes. (For which there are established formulas, but which also may be derived).
Now, simply make the substitutions you used, though I would call $x=r\cos(\theta)$ and $y=r\sin(\theta)$ because of the Pythagorean definitions of the trigonometric functions. ("Opposite" the angle corresponding to the $y$ length and "adjacent" corresponding to the abscissa ($x$-axis) length)
You will, solving for $r$ or $a$ or whatever you call it, find an expression in $\theta$ for the cube of that number. Letting $r=\sqrt{x^2+y^2}$, we find that:
$r^3=\frac{2(\sin(\theta)+\cos(\theta))(2-\sin(2\theta))}{(sin(2\theta))(1-\sin(\theta))(2+\sin(2\theta))}$
And that,
$r^3 = -\frac{2(\sin(2x)-2)\cot(2x)}{(\sin(2x)+2)(\cos(x)-\sin(x))^3}$
Take the cube root of both sides, square booth sides, then find the minimum of the resulting expression in $\theta$ on the RHS.
This will be the minimum value. Writing the last sentence was much easier than doing out the math. Because trig functions are periodic, if you consider an interval $0<\theta<2π$ you need to locate the relative minima, of which it becomes apparent that $0$ is the lowest function value taken.
In a world without WolframAlpha we could use some serious single-variable calculus to determine the relative minima in the given interval by locating where the function's derivatives are zero and applying the usual test.
However, you say that $xy≠0$. Aww, man. But you are right, that would have been too easy.
We now want to look at the other relative minima. Computationally, the answer is 3.182. I do not believe there are ways to compute it by hand.
$r=\left(-\frac{2\left(\sin \left(2\theta\right)-2\right)\cot \left(2\theta\right)}{\left(\sin \left(2\theta\right)+2\right)\left(\cos \left(\theta\right)-\sin \left(\theta\right)\right)^3}\right)^{\frac{2}{3}}$
You want the minima of that function that is nonzero with x and y greater than zero in the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Find the limit $\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)$ $$\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)\;=\;?\quad(n\in I) \\ \text{where $\lfloor\cdot\rfloor$ is the greatest integer function.}$$
This is what I did:
Since $[x] = x - \{x\}$ we get our limit equal to
$$\lim_{n\to\infty}\left\{\sqrt{n^2+n+1}\right\}$$
Moving the limit inside the fractional part function and replacing $n=\frac 1h \; \text {where } h\to0^+$ we get
$$\left\{\lim_{h\to0^+} \frac{\sqrt{h^2+h+1}}h\right\}$$
Applying L'Hospital Rule, we get our limit equal to $\left\{\frac 12\right\}$ which is $0$.
The problem:
The answer in the answer key is $\frac12$. So here, the only problem I seem to find in my solution is that $n\in I$ and simply assuming $n = \frac 1h$ doesn't ensure our $n$ to be an integer.
Can anyone provide a way to either correctly assume a new value for $n$ or any alternate way to solve this?
| For every natural number, we have
$$\lfloor \sqrt{n^2+n+1} \rfloor =n$$
because $n^2\leq n^2+n+1< n^2+2n+1=(n+1)^2$. So we get
$$\begin{align}\lim_{n\to\infty} (\sqrt{n^2+n+1}- \lfloor \sqrt{n^2+n+1} \rfloor)&=\lim_{n\to\infty}(\sqrt{n^2+n+1}-n)\\
&=\lim_{n\to\infty}\frac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\\
&=\lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n+1}+n}\\
&=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}\\
&=\frac{1}{2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ by Mathematical Induction Prove by Mathematical Induction:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$$
Now by inductive hypothesis:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+k(k+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+k^2(k+1)}=\frac{3k+5}{3k+1}$$
But verification for $n=k+1$ is creating problem.because $k$ is present in denominator too which is hindering use of equation obtained from inductive hypothesis. Please suggest something.
| This is also a proof by induction and the result is more general.
The identity
$$\sum_{m=0}^{n}\binom{m}{k}=\binom{n+1}{k+1} \tag{1}$$
can be easily proofed by induction using the well known identity of the Pascal Triangle
$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1} \tag{2}$$
$(1)$ says that the sum of this $k$-th degree polynomials in $m$ is a $(k+1)$-st degree polynomial in $n$.
From this it is also simple to proof by induction that for an arbitrary polynomial $p(m)$ of degree $k \in \mathbb{N}$ there is a polynomial $q(m)$ of degree $k+1$ such that
$$\sum_{m=0}^{n}p(m)=q(m) \tag{3}$$
So
$$(1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2)({3n+1}) \tag{4})$$
and
$$(1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1))(3n+5) \tag{5}$$
are both polynomials of degree 5. So to see that the polynomials $(4)$ and $(5)$ are equal it is sufficient to show that $(4)$ and $(5)$ have the same values for $n=1,2,3,4,5,6$. These values are $16$, $154$, $700$, $2210$, $5600$ and $12236$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results.
| Let $\cos\theta=\frac2{\sqrt5}$. Then
$$\cos3\theta=4\cos^3\theta-3\cos\theta=\frac2{5\sqrt5}$$
and so
$$\tan3\theta=\frac{\sqrt{(5\sqrt5)^2-2^2}}2=\frac{11}2\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
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Why is ln(x) bigger than log(x)? I couldn't find any results. I would like to know why $\ln x$ stays higher than $\log x$. Does it have to do with ase $10$ of $\log$? base $e$ with $\ln x$?
| Take two different bases: $2$ and $4$. Now let's look at what exponent is required to equal some numbers (this is what the logarithm computes):
\begin{align}
2^? = 16 \rightarrow 2^4 = 16 && 4^? = 16 \rightarrow 4^2 = 16\\
2^? = 64 \rightarrow 2^6 = 64 && 4^? = 64 \rightarrow 4^3 = 64 \\
2^? = 1024 \rightarrow 2^{10} = 1024 && 4^? = 1024 \rightarrow 4^5 = 1024
\end{align}
Clearly the smaller base requires a larger exponent to make the same value. Thus since $e < 10$, if $e^a = 10^b$ then $a > b$ and thus $\ln(x) > \log_{10}(x)$ (for some values of $x$).
Let's look at some different numbers with the same two bases above:
\begin{align}
2^? = \frac{1}{16} \rightarrow 2^{-4} = \frac{1}{16} && 4^? = \frac{1}{16} \rightarrow 4^{-2} = \frac{1}{16}\\
2^? = \frac{1}{64} \rightarrow 2^{-6} = \frac{1}{64} && 4^? = \frac{1}{64} \rightarrow 4^{-3} = \frac{1}{64} \\
2^? = \frac{1}{1024} \rightarrow 2^{-10} = \frac{1}{1024} && 4^? = \frac{1}{1024} \rightarrow 4^{-5} = \frac{1}{1024}
\end{align}
In this case, first, notice the absolute value of the exponent is still greater, but since they are negative, the smaller number requires a "smaller" exponent (a "more negative" exponent).
So when the exponent is negative, meaning $\log_a(x) < 0$ and thus $x < 1$, the smaller base requires a "smaller" exponent (really a more negative exponent). The general rule of :
$$
\left|\log_a(x)\right| \geq \left|\log_b(x)\right|, a > 1, b > 1, a \leq b
$$
If $a$ and/or $b$ is between $0$ and $1$ then it gets tricky since:
$$
0 < a < 1 \rightarrow \begin{cases}
\log_a(x) < 0 & x > 1 \\
\log_a(x) > 0 & x < 1
\end{cases}
$$
To figure out how the absolute values would compare requires looking at the inverse of the bases. Let's continue to assume that $\frac{1}{a} < b$ and $a, b > 0$:
*
*If $a < b$ then the same rules would apply:
$$
\left|\log_{\frac{1}{a}}(x)\right| \geq \left|\log_b(x)\right|
$$
*If $a > b$ then the opposite rules would apply:
$$
\left|\log_{\frac{1}{a}}(x)\right| \leq \left|\log_b(x)\right|
$$
If both $a$ and $b$ are between $0$ and $1$ then, since the smaller number would mean the larger inverse (i.e. $\frac{1}{a} < \frac{1}{b} \rightarrow b < a$), the smaller number would make the smaller logarithm.
Here is a summarization, assuming $a, b > 1$ and $a < b$:
\begin{align}
\left|\log_a(x)\right| \geq& \left|\log_b(x)\right| \\
\left|\log_{\frac{1}{a}}(x)\right| \geq& \left|\log_b(x)\right| \\
\left|\log_a(x)\right| \geq& \left|\log_{\frac{1}{b}}(x)\right| \\
\left|\log_{\frac{1}{a}}(x)\right| \geq& \left|\log_{\frac{1}{b}}(x)\right|
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Pythagorean triplets of the form $a^2+(a+1)^2=c^2$ and the space between them I was searching for pythagorean triples where $b=a+1$, and I found using a python program I made the first 10 integer solutions:
*
*$0^2+1^2=1^2$
*$3^2+4^2=5^2$
*$20^2+21^2=29^2$
*$119^2+120^2=169^2$
*$696^2+697^2=985^2$
*$4059^2+4060^2=5741^2$
*$23660^2+23661^2=33461^2$
*$137903^2+137904^2=195025^2$
*$803760^2+803761^2=1136689^2$
*$4684659^2+4684660^2=6625109^2$
Now what's so interesting? I discovered that any $c$, divided by the previous (for example $5/1$ or $29/5$) limits to $5.828427...=\left(\frac{1}{\sqrt2-1}\right)^2=\sqrt8+3$. My question: why?
| The values $\; 1,5,29,169,\cdots\;$ are subset of pell number as shown in the OEIS sequence A000129. You are in fact using the odd-value numbers in the sequence.
If we define Euclid's formula for generating Pythagorean triples as:
$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$
the $(m,k)$ values needed to generate these are also Pell numbers and the $n^{th}$ triple can be generated by
\begin{equation}
m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad
k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}
\end{equation}
If you plug these into $\quad C=m^2+k^2\quad $ and take the ratios between adjacent $C$-value polynomials, it should explain the result you have observed, much as ratios of consecutive Fibonacci number converge to $\phi$.
\begin{align*}
\frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 \qquad&
\frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1 \\
&\qquad\qquad F(2,1)=(3,4,5)\\
\frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 \qquad &
\frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2 \\
&\qquad\qquad F(5,2)=(21,20,29)\\
\frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 \qquad &
\frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5 \\
&\qquad\qquad F(12,5)=(119,120,169)\\
\frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 \qquad &
\frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12 \\
&\qquad\qquad F(29,12)=(697,696,985)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 2
} |
Finding the shortest distance between two Parabolas Recently, a problem asked me to find the minimum distance between the parabolas $y=x^2$ and $y=-x^2-16x-65$.
I proceeded with the problem as thus.
Let $P(a,a^2), Q(b, -b^2-16b-65), a-b=x$.
$\therefore PQ^2=x^2+(2a^2+2ax+16a+x^2+16x+65)^2$.
$PQ^2=x^2+(2(a+\frac{x+8}{2})^2+\frac{(x+8)^2+2}{2})^2 \ge (x^2+(\frac{(x+8)^2+2}{2})^2)(1+\frac{1}{4}) \times \frac{4}{5}$
Applying Cauchy gives us that
$PQ^2 \ge (\frac{1}{4}x^2+3x+\frac{33}{2})^2 \times \frac{4}{5} \ge (\frac{15}{2})^2 \times \frac{4}{5}=75$
This implies that the answer is $\sqrt{75}$.
However, it took me a long time to find the values for Cauchy, and the calculations proved tedious.
What are other approaches to this problem?
EDIT: $(\frac{15}{2})^2 \times \frac{4}{5} \neq 75$, it`s $45$ actually!
| $$y_1 = x_1^2, \qquad y_2 = -x_2^2 - 16x_2 -65$$
$$d = \sqrt{(x_2-x_1)^2 + (-x_2^2 - 16x_2 -65-x_1^2)^2}$$
Thus we want to minimize the following: (Note the change of variables, simply to make the notation easier to follow)
$$f(u,v) = (u-v)^2 + (-u^2 - 16u -65-v^2)^2$$
$$f_u(u,v) = 4u^3 + 96u^2 + u(4v^2 + 774) + 32v^2 - 2v + 2080$$
$$f_v(u,v) = 262 v + 4 u^2 v + 4 v^3 + u (-2 + 64 v)$$
Solving this, we get the real solution $u=-7, v = -1$ (it's quite tedious here, so I ended up using Wolfram Alpha). Thus our minimum distance is
$$\sqrt{((-7)-(-1))^2 + (-(-7)^2 - 16(-7) -65-(-1)^2)^2} = \color{red}{\sqrt{45}} \approx \color{red}{6.7082039325}$$
Addendum
While my answer differs from yours, this Desmos demonstration I created seems to numerically support my answer, as long as I interpreted the question correctly; feel free to mess around with it. I have attached a screenshot from the program showing the minimum distance value. Also note the green line in the image represent the normal line to the graphs at the corresponding points, and both points share a tangent line with a slope of $-2$: this is to be expected, and should hold for a number of problems of this type. Accordingly, this could provide another way to solve this problem with fewer steps and calculations than my solution. Nevertheless, I leave this work up to you!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Number of terms in the expansion of $\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$
Number of terms in the expansion of $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$$
$\bf{My\; Try::}$ We can write $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n=\frac{1}{x^{2n}}\left(1+x+x^2\right)^n$$
Now we have to calculate number of terms in $$(1+x+x^2)^n$$ and divided each term by $x^{2n}$
So $$(1+x+x^2)^n = \binom{n}{0}+\binom{n}{1}(x+x^2)+\binom{n}{2}(x+x^2)^2+......+\binom{n}{n}(x+x^2)^n$$
and divided each term by $x^{2n}$
So we get number of terms $\displaystyle = 1+2+3+.....+n+1 = \frac{(n+1)(n+2)}{2}$
Is my solution is Right, If not then how can i calculate it
actually i don,t have solution of that question.
Thanks
| When you expand $(1 + x + x^2)^n$, the highest degree is $2n$, and the lowest is 0. None of them is vanished, and it can be seen easily that there is one term of any degree between 0 and $2n$. So there should be $2n + 1$ terms.
Cheers,
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Can you multiply a natural log term by a constant to take power inside log term? For Calculus 2 homework I must prove that $\displaystyle\int \frac{du}{u^2-a^2} = \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C$
The instructor wants us to use trigonometric substitution to solve.
My question is, may I take the term (acquired along the way):
\begin{align}
\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|
&=\frac{2}{2}\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|\\[3px]
&=\frac12\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|^2\\[3px]
&=\frac12\ln\left|\frac{(a+u)^2}{(\sqrt{u^2-a^2})^2}\right|
\end{align}
I realize the rules of logarithms state that $\log_ax^p = p\log_ax$
However, I can't tell if this is a valid operation or I merely have flawed logic in this case. Thank you!
| The substitution you can use in this case is
$$
u=\frac{a}{\cos2t}
$$
so
$$
\frac{1}{u^2-a^2}=\frac{\cos^22t}{a^2\sin^22t},
\qquad
du=\frac{2a\sin2t}{\cos^22t}
$$
and the integral becomes
$$
\frac{2}{a}\int\frac{1}{\sin2t}\,dt=
\frac{1}{a}\int\frac{\cos^2t+\sin^2t}{\sin t\cos t}\,dt=
\frac{1}{a}\int
\left(\frac{\cos t}{\sin t}+\frac{\sin t}{\cos t}\right)\,dt=
\frac{1}{a}\ln\left|\frac{\sin t}{\cos t}\right|+C
$$
Now we can divide by $2$ and square inside the logarithm:
$$
\frac{1}{2a}\ln\left|\frac{\sin^2t}{\cos^2t}\right|=
\frac{1}{2a}\ln\left|\frac{1-\cos2t}{1+\cos2t}\right|=
\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|
$$
In your computation you have
$$
\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|=
\ln\sqrt{\frac{(u+a)^2}{u^2-a^2}}=
\ln\sqrt{\frac{u+a}{u-a}}=\frac{1}{2}\ln\frac{u+a}{u-a}
$$
but there seems to be something wrong in how you arrived to this. You're missing an absolute value under the square root, to begin with.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
integrate $\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$
$$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$$
$(2-x)^2(4+x^2)=x^4-4x^3+8x^2-16x+16$
$$\int \frac{x^4+4x+6}{x^4-4x^3+8x^2-16x+16}$$
Is there is anything to do else than long division?
| I know that you asked without long division, but
$$\int \frac{x^4+4x+6}{(x-2)^2(x^2+4)}\stackrel{\text{L.div}}{=}\frac 1 8 \int\frac{11x-8}{x^2+4}dx+\frac{21}{8}\int\frac{dx}{x-2}dx+\frac{15}{4}\int \frac{dx}{(x-2)^2}+\int1 dx=\dots$$
| {
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For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
For all $x$ which are real numbers, prove that
$$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$
I know that
Let $\lfloor x\rfloor = n$
$n \leq x < n+1$
| I will try to show
$\lfloor nx \rfloor = \sum_{k=0}^{n-1} \lfloor x+\frac{k}{n} \rfloor$.
Let
$m = \lfloor x \rfloor$
and
$d = x - m$,
so
$0 \le d < 1$.
Let
$j
=\lfloor nd \rfloor
$,
so
$0 \le j \le n-1
$
and
$\frac{j}{n}
\le d
< \frac{j+1}{n}
$.
If
$0 \le k \le n-j-1$,
$\begin{array}\\
m
&\le m+d+\frac{k}{n}\\
&< m+\frac{n-j}{n}+\frac{j}{n}\\
&= m+\frac{n}{n}\\
&= m+1\\
\end{array}
$
so
$\lfloor m+d+\frac{k}{n} \rfloor
=m
$.
If
$n-j \le k \le n-1$,
$\begin{array}\\
m+d+\frac{k}{n}
&\ge m+\frac{n-j}{n}+\frac{j}{n}\\
&= m+\frac{n}{n}\\
&= m+1\\
\end{array}
$
and
$\begin{array}\\
m+d+\frac{k}{n}
&\lt m+\frac{n-j+1}{n}+\frac{n-1}{n}\\
&= m+\frac{2n-j}{n}\\
&= m+2-\frac{j}{n}\\
\end{array}
$
so
$\lfloor m+d+\frac{k}{n} \rfloor
=m+1
$.
Therefore
$\begin{array}\\
\sum_{k=0}^{n-1} \lfloor x+\frac{k}{n} \rfloor
&=\sum_{k=0}^{n-1} \lfloor m+d+\frac{k}{n} \rfloor\\
&= \sum_{k=0}^{n-j-1} \lfloor m+d+\frac{k}{n} \rfloor
+\sum_{k=n-j}^{n-1} \lfloor m+d+\frac{k}{n} \rfloor\\
&= (n-j)m+j(m+1)\\
&= nm+j\\
\end{array}
$
and
$\lfloor nx \rfloor
=\lfloor n(m+d) \rfloor
=nm+\lfloor nd \rfloor
=nm+j
$.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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} |
Prove by induction that: $\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i}$ Basis step:
For $n=1$, equation holds.
Inductive step: Now,
$$\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i} \tag{i. h.}$$
Now we want to show that
$$\sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^{n+1}_{i=1}\frac{1}{n+1+i}.$$
So,
$$ \sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i} = \sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i} + \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2} = \text{(by inductive hypothesis)}= \sum^n_{i=1}\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2} $$
This is where I'm stuck.
| Change the indices in the sum: $$\sum_{i=1}^n\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2}=\sum_{i=0}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\
=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{n+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{i=1}^{n+1}\frac{1}{n+i+1}.$$
| {
"language": "en",
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Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$ $$
\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots
$$
up to $n$ terms. I need help in solving this sum. I tried finding the coefficients of terms after splitting the terms..: it becomes
$$(\frac{1}{1\cdot 6}-\frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}-\frac{1}{6\cdot4}) + (\frac{1}{6\cdot 2} - \frac{1}{3\cdot2} +\frac{1}{4\cdot 2} -\frac{1}{6\cdot5})+\cdots.$$ I tried solving it but am getting nowhere .Someone please help me with this sum.
| HINT:
The $r(\ge1)$th term
$$=\dfrac1{r(r+1)(r+2)(r+3)} =\dfrac{r+3-r}{3r(r+1)(r+2)(r+3)}=v_r-v_{r+1}$$
where $v_m=\dfrac1{3m(m+1)(m+2)}$
$$\sum_{r=1}^n u_r=\sum_{r=1}^n(v_r-v_{r+1})=v_1-v_{n+1}$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Induction Example Here's a (possibly sloppily written) Induction proof that has me stumped. It's the first example from here and supposedly the statement holds?
(*) For n > 1, 2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^n+1 – 2
Let n = 1. Then:
2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^1 = 2
...and:
2^n+1 – 2 = 2^1+1 – 2 = 2^2 – 2 = 4 – 2 = 2
So (*) works for n = 1.
Assume, for n = k, that (*) holds; that is, that
2 + 2^2 + 2^3 + 2^4 + ... + 2^k = 2^k+1 – 2
Let n = k + 1.
I understand this part. This is where I get lost. (I understand that first we set n=k and then n=k+1 but the equations are sloppy and I don't follow how both sides are "equal" as the original statement asks me to prove.
2 + 2^2 + 2^3 + 2^4 + ... + 2^k + 2^k+1 | (I assume this is the left side)
= [2 + 2^2 + 2^3 + 2^4 + ... + 2^k] + 2^k+1| (still on the left side)
= [2^k+1 – 2] + 2^k+1 | (now we are on the right side??)
= 2×2^k+1 – 2 | (right side simplified)
= 2^1×2^k+1 – 2
= 2^k+1+1 – 2
= 2^(k+1)+1 – 2
Then (*) works for n = k + 1.
Ok, so what is the final conclusion? Looks to me the sides are not equal but it's not evident what the final answer is. Seems like the statement "Then (*) works for n = k + 1." suggests that the statement holds true but I fail to see why. Thanks!
| The goal of the induction step is to show that (*) holds for $n = k+1$, in particular, we want to show $$2 + 2^2 + \cdots + 2^{k+1} = 2^{(k+1)+1}-2$$
The induction hypothesis says $2 + 2^2 + \cdots + 2^{k} = 2^{k+1}-2$. So we start with the LHS of what we want to show, and transform it to the RHS, using things we know are true. In particular
$2 + 2^2 + \cdots + 2^{k+1} = (2 + 2^2 + \cdots 2^k)+ 2^{k+1}$
$= (2^{k+1}-2) + 2^{k+1}$ by the induction hypothesis.
$= 2 \cdot 2^{k+1} - 2$ re-arranging
$= 2^{(k+1)+1} - 2$ (re-arranging some more)
And so, we have showed the LHS of what we wanted equals the RHS of what we wanted, which completes the induction.
| {
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Is there a mathematical way to solve this problem? The question goes something like this:
How many different ways can you add up 2, 3, and 5 to get a sum of 12. Numbers can be repeated, and all three numbers do not have to be used for each solution. For instance, 5 + 2 + 2 is an answer, and so is 2 + 5 + 2.
The answer is 5 different ways:
*
*5 + 5 + 2
*2 + 2 + 2 + 2 + 2 + 2
*3 + 3 + 3 + 3
*3 + 3 + 2 + 2 + 2
*5 + 3 + 2 + 2
Is there some equation or formula that can be used to solve this problem, or does it have to be done manually?
| Here's a way to enumerate all sums of $5$, $3$, and $2$ that add up to $n$,
if we do not consider the order of the terms in the sums to matter.
Let's consider just sums of $2$ and $3$ at first.
If $n > 1$, there is at least one such sum that adds up to $n$.
To find a specific sum, take the remainder of $n$ after dividing by $3$,
and treat each remainder as a separate case:
Case 0: $n = 3m$. Then the sum is simply
$$
n = 3m = \overbrace{3+3+\cdots+3}^{m\ \text{times}}
$$
Case 1: $n = 3m + 1$, where $m \geq 1$. Then the sum is
$$
n = 3(m-1) + 2\cdot2
= \overbrace{3+3+\cdots+3}^{m - 1\ \text{times}} + 2 + 2
$$
Case 2: $n = 3m + 2$. Then the sum is
$$
n = 3m + 2 = \overbrace{3+3+\cdots+3}^{m\ \text{times}} + 2
$$
These are the maximum numbers of $3$s that can appear in the sum,
depending on which case applies.
Once we have found such a sum in which the term $3$ appears $k$ times,
we can substitute $2+2+2$ for $3+3$ until the number of $3$s remaining
is less than $2$. There are therefore $\lfloor k/2 \rfloor + 1$
distinct sums consisting only of $2$s and $3$s without regard to order.
If we now consider sums using $2$, $3$, and $5$ as terms,
we can convert any such sum to a sum of just $2$s and $3$s by
substituting $5=3+2$ until no $5$s remain.
Working backwards, any sum using $2$, $3$, and $5$ can be obtained
by starting with a sum consisting only of $2$s and $3$s
of the form $n = 3a + 2b$ and then replacing $3+2$ with $5$
(rearranging the order of terms as needed) up to $\min\{a,b\}$ times.
In summary to generate all possible sums of $2$, $3$, and $5$
adding up to $n$, we find an integer $k$ such that either
$n = 3k$, $n=3k+2$, or $n=3k+4$
(only one of these will be possible).
We use the substitution $3+3=2+2+2$ to generate a list of
$\lfloor k/2 \rfloor + 1$ sums
of the form $n = 3a + 2b$ (including the original sum).
From each of the resulting sums,
using the substitution $3+2=5$, we generate a
list of $\min\{a,b\} + 1$ sums of $2$, $3$, and $5$.
If we do not actually do the final step above, but simply
count $\min\{a,b\} + 1$ for each sum of the form $n = 3a + 2b$,
it is possible in this way to count the number of sums without actually listing all of them.
Moreover, we can subdivide the list of sums of $2$s and $3$s into
two sublists: all the sums that have more $2$s than $3$s,
and the sums that do not.
In the first sublist, if the number of $3$s in the sums has a minimum $p$
and maximum $p+2q$, then the total number of sums generated from this
sublist of sums is $(p+q+1)(q + 1)$.
In the second sublist, if the number of $2$s in the sums has a minimum $r$
and maximum $r+3s$, then the total number of sums generated from this
sublist of sums is $\frac12(2r+3s+2)(s + 1)$.
Hence all we actually need to do is to find the minimum and maximum values
of $a$ for which $3a+2b=n$ and $0 \leq a < b$,
and the minimum and maximum values of $b$ for which
$3a+2b=n$ and $0\leq b \leq a$, determine the values $p$ and $q$
or $r$ and $s$ to plug into each of the formulas above,
and add the two results together.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove Lucas numbers and Fibonacci numbers relation $F_n = \frac{L_{n - 1} + L_{n + 1}}{5}$ $F_n$, is the $n$th term of the Fibonacci sequence.
$L_n$ is the $n$th Lucas number.
I want to prove that $F_n = \dfrac{L_{n-1}+L_{n+1}}5 $.
Things I know:
*
*$L_n$+$L_{n+1}=L_{n+2}$
*$F_{n-1} + F_{n+1}=L_n$.
The base case is easy to prove. I will omit proving the base case it is simple.
My attempt:
$F_{n+1} = \dfrac{L_{n}+L_{n+2}}5 $.
Attempt one :using and manipulating $#2$ make all $L_n$ in terms of $F_n$.
Attempt two:
Using $#2$ two make $F_{n+1}$ in terms of $L_n$.
| The Fibonacci numbers are defined recursively by
$$
F_n =
\begin{cases}
1 & \text{if $n = 1$}\\
1 & \text{if $n = 2$}\\
F_{n - 1} + F_{n - 2} & \text{if $n > 2$}
\end{cases}
$$
and the Lucas numbers are defined recursively by
$$
L_n =
\begin{cases}
2 & \text{if $n = 0$}\\
1 & \text{if $n = 1$}\\
L_{n - 1} + L_{n - 2} & \text{if $n > 1$}
\end{cases}
$$
Proof. Let $P(n)$ the statement
$$F_n = \frac{L_{n + 1} + L_{n - 1}}{5}$$
We will prove $P(n)$ holds for all positive integers $n$ using strong induction.
Let $n = 1$. Observe that $L_2 = L_1 + L_0 = 1 + 2 = 3$. Thus,
$$\frac{L_{1 + 1} + L_{1 - 1}}{5} = \frac{L_2 + L_0}{5} = \frac{3 + 2}{5} = \frac{5}{5} = 1 = F_1$$
so $P(1)$ holds.
Let $n = 2$. Observe that $L_3 = L_2 + L_1 = 3 + 1 = 4$. Thus,
$$\frac{L_{2 + 1} + L_{2 - 1}}{5} = \frac{L_3 + L_1}{5} = \frac{4 + 1}{5} = \frac{5}{5} = 1 = F_2$$
so $P(2)$ also holds.
Assume $P(n)$ holds for each positive integer $n \leq m$, where $m \geq 2$. Then
$$F_m = \frac{L_{m + 1} + L_{m - 1}}{5}$$
and
$$F_{m - 1} = \frac{L_{(m - 1) + 1} + L_{(m - 1) - 1}}{5} = \frac{L_m + L_{m - 2}}{5}$$
Let $n = m + 1$. Then
\begin{align*}
\frac{L_{(m + 1) + 1} + L_{(m + 1) - 1}}{5} & = \frac{L_{m + 2} + L_{m}}{5}\\
& = \frac{L_{m + 1} + L_m + L_{m - 1} + L_{m - 2}}{5}\\
& = \frac{L_{m + 1} + L_{m - 1} + L_m + L_{m - 2}}{5}\\
& = \frac{L_{m + 1} + L_{m - 1}}{5} + \frac{L_m + L_{m - 2}}{5}\\
& = F_m + F_{m - 1}\\
& = F_{m + 1}
\end{align*}
Thus, $P(n)$ holds for each positive integer $n$.$\blacksquare$
| {
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"question_score": "2",
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Finding formula for sum $S_n = \sum_{i=1}^n i(i-1)$ How do I find the formula for this sum
$S_n = \sum_{i=1}^n i(i-1)$
Wolfram alpha gives me the correct formula which is $ \frac{1}{3} (n-1)n(n+1)$, but I'm interested in how I get to that and how I can approach this kind of problem in general.
| This is much simpler. Let's use the notation $x^{\underline{k}} = x (x - 1) \dotsm (x - k + 1)$ for arbitrary $x$ and $k \in \mathbb{N}_0$.
Prove by induction on $n$ that:
$\begin{align}
\sum_{0 \le k < n} k^{\underline{m}} = \frac{n^{\underline{m + 1}}}{m + 1}
\end{align}$
Base: For $n = 0$, this is true, the left hand side is an empty sum and the right hand side is zero.
Induction: Asume it is true for $n$, check $n + 1$:
$\begin{align}
\sum_{0 \le k < n + 1} k^{\underline{m}}
&= \sum_{0 \le k < n} k^{\underline{m}} + n^{\underline{m}} \\
&= \frac{n^{\underline{m + 1}}}{m + 1} + n^{\underline{m}} \\
&= \frac{n^{\underline{m}} \cdot (n - m)
+ n^{\underline{m}} \cdot (m + 1)}
{m + 1} \\
&= \frac{n^{\underline{m}} (n + 1)}{m + 1} \\
&= \frac{(n + 1)^{\underline{m + 1}}}{m + 1}
\end{align}$
For your specific case:
$\begin{align}
\sum_{1 \le i \le n} i (i - 1)
&= \sum_{0 \le i < n + 1} i^{\underline{2}} \\
&= \frac{(n + 1)^{\underline{3}}}{3} \\
&= \frac{(n + 1) n (n - 1)}{3} \\
&= \frac{n^3 - n}{3}
\end{align}$
| {
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How do I solve the system of congruences $x \equiv 1 \pmod{3}$, $x \equiv 2 \pmod{4}$, $x \equiv 4 \pmod{5}$, $x \equiv 2 \pmod{7}$? Sorry for the weird title, but I don't know how else to phrase this.
Say I have a set of numbers, and the remainders each get by dividing them by a certain number. For instance:
$x \equiv 1 \pmod{3}$
$x \equiv 2 \pmod{4}$
$x \equiv 4 \pmod{5}$
$x \equiv 2 \pmod{7}$
I know the math is similar to finding the least common multiple, but I can't grok it. How do I find $x$ in the above equations, assuming they all use the same $x$?
Additionally, after that's found, how do I find further multiples that fit these restrictions?
| Since $x\equiv2\pmod{4}$ and $x\equiv2\pmod{7}$, we must have that $x\equiv2\pmod{28}$.
We can use the Extended Euclidean Algorithm as implemented in this answer to solve the following equivalences.
Since $140=5\cdot28$, solve
$$
\begin{align}
x&\equiv1\pmod{3}\\
x&\equiv0\pmod{140}
\end{align}
$$
getting $x\equiv280\pmod{420}$.
Since $84=3\cdot28$, solve
$$
\begin{align}
x&\equiv1\pmod{5}\\
x&\equiv0\pmod{84}
\end{align}
$$
getting $x\equiv336\pmod{420}$.
Since $15=3\cdot5$, solve
$$
\begin{align}
x&\equiv1\pmod{28}\\
x&\equiv0\pmod{15}
\end{align}
$$
getting $x\equiv225\pmod{420}$.
We can get the solution to the original equivalence as
$$
x\equiv\overbrace{1\cdot280}^{\substack{1\pmod{3}\\0\pmod{4}\\0\pmod{5}\\0\pmod{7}}}+\overbrace{4\cdot336}^{\substack{0\pmod{3}\\0\pmod{4}\\4\pmod{5}\\0\pmod{7}}}+\overbrace{2\cdot225}^{\substack{0\pmod{3}\\2\pmod{4}\\0\pmod{5}\\2\pmod{7}}}\equiv\overbrace{\ 394\ }^{\substack{1\pmod{3}\\2\pmod{4}\\4\pmod{5}\\2\pmod{7}}}\pmod{420}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \times 1 + 8 \times 2) +1, (8 \times 1 + 8 \times 2 + 8 \times 3 + 8 \times 4) +1 ....$
equals to
$1^2, 3^2, 5^2, 7^2 ...$
and my brain stuck here and cannot find out the relationship between two series.
Can someone give me a hint why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$
| As the terms grow linearly, their average is also the average of the extreme values. Hence the sum for the $n$ first terms
$$S_n=n\cdot\frac{0+8(n-1)}2+1=4n^2-4n+1=(2n-1)^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A hemisphere is inscribed in a cube Finding the largest cube inscribed in a hemisphere has been considered here previously. So let's consider the reverse relationship:
A hemisphere is inscribed in a cube with an edge of $1m$. What is the maximum radius of the hemisphere ?
Obviously a whole sphere of radius $\frac{1}{2}$ can be inscribed in the cube, but could a hemisphere of larger radius fit in there somehow?
| I forward and edit this proof from a middle school math teacher
Note that a sphere inscribe to a cubiod is tangent to all the six faces of the cuboid, then a hemisphere inscribe to a cuboid should be tangent to at least three cuboid-faces adjacent to each other.
This means, the (sphere) center of the hemisphere locates at a diagonal line of the cuboid.
Since a hemisphere is tangent to at least three cuboid faces adjacent-to-each other, and the three cuboid faces are perpendicular to each other and they have a common intersection point, let it be $O$, we can set up a 3D Descartes coordiate with $O$ as origin, and the three intersection lines of the three tangent faces as $x,y,z$-axes.
Suppose such a hemisphere is half of a sphere with radius $r$ centered at $A$. Then the coordinate of $A$ is $(r,r,r)$; the implicit equation of the sphere/hemisphere is $(x-r)^2+(y-r)^2+(z-r)^2=r^2$.
Suppose a plane $ax+by+cz=1$ (then $ar+br+cr=1$, $a>0,b>0,c>0$)passes through $A$ and separates sphere $A$ into two halfs, one of which is what we expect to find.
**Here the $W$ is the key, but I don't really understand **
why $W$ and $a\le b, c$ can be assumed together. from the picture at the bottom, $W$ has three possibilities, for each one, we have to assume different inequalities to prove that $3-\sqrt{6}$ is the maximum $r$.
Without loss of generality, we can assume $a\le b, c$, then $a^2+b^2+c^2\le \dfrac{3}{2}(b^2+c^2)$
and it is easy to verify that point $W(x_w,y_w,z_w)=$
$\left(\dfrac{r \sqrt{b^2+c^2}}{\sqrt{a^2+b^2+c^2}}+r,\;r-\dfrac{a b r}{\sqrt{b^2+c^2}
\sqrt{a^2+b^2+c^2}},\;r-\dfrac{a c r}{\sqrt{b^2+c^2} \sqrt{a^2+b^2+c^2}}\right)$
is on both the plane $ax+by+cz=1$ and the hemisphere.
Easy to know that:
$$\left(1+\sqrt{\dfrac{2}{3}}\right)r\le x_w=\dfrac{r \sqrt{b^2+c^2}}{\sqrt{a^2+b^2+c^2}}+r\le1$$
which means:
$$r\le \dfrac{1}{1+\sqrt{\dfrac{2}{3}}}=3-\sqrt{6}$$
$r$ maximizes when $a=b=c=\dfrac{1}{3r}=\dfrac{1}{3}+\dfrac{\sqrt{6}}{9}$
The results can be visualized as:
| {
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} |
Prove that $8640$ divides $n^9 - 6n^7 + 9n^5 - 4n^3$. I found this problem in a book, I can't solve it unfortunately.
Prove that for all integer values $n$, $n^9 - 6n^7 + 9n^5 - 4n^3$ is divisible by $8640.$
So far I've noticed that $8460 = 6! \times 12$, also I've tried to simplify that expression and I've found that it's equal to this $n^3(n^3-3n-2)(n^3-3n+2)$, but I can't move on after that.
| We can further note that $-1$ is a root of $n^3-3n-2$, and that $1$ is a root of $n^3-3n+2$.
You can find these roots by the Rational Root theorem:
Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.
Other techniques can also be found in this question and its answers.
Then we simplify it to \begin{align*} n^3(n+1)(n^2-n-2)(n-1)(n^2+n-2) &= n^3(n+1)(n-2)(n+1)(n-1)(n+2)(n-1)\\ &=n^3(n+1)^2(n-1)^2(n-2)(n+2) \end{align*}
There are 6 factors two in 8640.
*
*If $n$ is even, $n, n-2$ and $n+2$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2.
*If $n$ is odd, $n-1$ and $n+1$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2.
There are 3 factors three in 8640.
*
*If $n \equiv 0 \mod 3$, then $n$ is divisible by 3, and then $27 \mid n^3$.
*If $n \equiv 1 \mod 3$, then $n-1$ and $n+2$ are divisible by 3.
*If $n \equiv 2 \mod 3$, then $n+1$ and $n-2$ are divisible by 3.
The factor 5 can be taken care of since the product is a multiple of five consecutive integers, namely a product of $(n-2)(n-1)n(n+1)(n+2)$.
| {
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An alternative way to compute $a^n+b^n+c^n+d^n$ I was doing the following problem:
given $a, b, c, d \in \mathbb{R}$ and
$a+b+c+d=1$, $a^2+b^2+c^2+d^2=2$, $a^3+b^3+c^3+d^3=3$ and $a^4+b^4+c^4+d^4=4$.
Find $a^n+b^n+c^n+d^n$ (I am looking for a closed kind of formula)
Now, I can already do it in one way. We can find the elementary symmetric polynomials using newton's identity and then solve the quartic to find the values of $a, b, c, d$.
But I realise that my method is computationally heavy. And more importantly once the number of variables is increased there is no guarantee that we can solve for the roots. Also, the problem does not really need the values of $a, b, c, d$ explicitly.
The above observations motivate me to find an alternative way to find a an alternative technique to solve problems like this(may be in a more general setting).
So, can someone help me?
| If we take $a,b,c,d$ as the roots of a monic polynomial
$$ p(x)=x^4-e_1 x^3+e_2 x^2-e_3 x+e_4 $$
Newton's identities give:
$$ \left\{\begin{array}{rcl}e_1 &=& 1\\ 2e_2 &=& e_1-2 \\ 3e_3 &=& e_2 - 2e_1 +3 \\ 4e_4 &=& e_3-2e_2+3e_1-4 \end{array}\right. $$
hence it follows that:
$$ e_1=1,\quad e_2 = -\frac{1}{2},\quad e_3 = \frac{1}{6},\quad e_4 = \frac{1}{24}, $$
$$ p(x) = x^4-x^3-\frac{x^2}{2}-\frac{x}{6}+\frac{1}{24} $$
and:
$$ a^n+b^n+c^n+d^n = \text{Tr}\,\begin{pmatrix} 0 & 0 & 0 & -\frac{1}{24} \\
1 & 0 & 0 & \frac{1}{6} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1\end{pmatrix}^n .$$
Since the dominant root of $p(x)$ is about $1.42016$ and the other ones are quite close to zero, $a^n+b^n+c^n+d^n$ almost behaves like $2^{n/2}$ for large $n$s.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ax^2 + (b+c)x + d+ e=0$ has solutions on $[1,\infty)$, so that $ax^4+bx^3+cx^2+dx+e=0$ has solutions on $[1,\infty)$
Let $ax^2 + (b+c)x + d+ e=0, a, b, c, d, e \in \mathbb R$ has solutions on the interval $[1, +\infty)$. Prove that the polynomial $f(x) = ax^4 + bx^3 + cx^2+ dx + e =0$ has solutions on $[1, +\infty)$.
I used the intermediate value theorem, but not found $x_0, x_1 \in [1, +\infty) $ such that $f(x_0)f(x_1) < 0.$
Thank for helps.
| Nobody can prove that, since it is false. Suppose that $a=b=1$, that $c=-5$, that $d=4$, and that $e=0$. Then\begin{align}ax^2+(b+c)x+d+e=0&\iff x^2-4x+4=0\\&\iff(x-2)^2=0\\&\iff x=2.\end{align}So, yes, $ax^2+(b+c)x+d+e=0$ has solutions on $[1,+\infty)$ (actually, all of its solutions belong to that interval). However, \begin{align}ax^4+bx^3+cx^2+dx+e=0&\iff x^4+x^3-5x^2+4x=0\\&\iff x=0\vee x^3+x^2-5x+4=0\end{align}and the equation $x^3+x^2-5x+4=0$ has one and only one real root, which is approximately $-3.06$. Therefore, the equation $ax^4+bx^3+cx^2+dx+e=0$ has no real root on $[1,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680361",
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"source": "stackexchange",
"question_score": "1",
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Good idea to find below sum this is the question.
Purpose is to find sum of $$f\left(\frac{1}{14}\right)+f\left(\frac{2}{14}\right)+f\left(\frac{3}{14}\right)+...+f\left(\frac{13}{14}\right)$$for function $$f(x)=\frac{4^x}{4^x+2}$$Is there a nice method to find $f\left(\frac{1}{14}\right)+f\left(\frac{2}{14}\right)+f\left(\frac{3}{14}\right)+...+f\left(\frac{13}{14}\right)$ ?
| Yup, a nice thing to notice is the value of
$$
\begin{align}
f(x)+f(1-x)
&=\frac{4^x}{4^x+2}+\frac{4}{4+2\cdot4^x}\\
&=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}\\[3pt]
&=\cdots\,?
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Vectors that form basis in $\mathbb{R}^3$ and transition matrix
Consider the following set of vectors in $\mathbb{R}^3:$
$u_0 = (1,2,0),~ u_1 = (1,2,1), ~u_2 = (2,3,0), ~u_3 = (4,6,1)$
Explain why each of the two subsets $B_0 = \left\{u_0, u_2,
u_3\right\}$ and $B_1 = \left\{u_1, u_2, u_3\right\}$ forms a basis of
$\mathbb{R}^3$. If we write $[\mathbf{x}]_0$ and $[\mathbf{x}]_1$ for the coordinates of the vector $\mathbf{x}$ in terms of these two basis, find the precise transition matrix which inter-relates these two sets of coordinates. If $\mathbf{x} = 4\mathbf{e}_1+4\mathbf{e}_3$, what are $\mathbf [\mathbf{x}]_0$ and $[\mathbf{x}]_1$?
Writing the vectors of the subset $B_0$ as the columns of a matrix we have:
$\mathbf{A}_0: = \begin{pmatrix}
1&2&4 \\
2&3&6 \\
0&0 &1
\end{pmatrix} \to \begin{pmatrix}
1&2&4 \\
0&-1&-4 \\
0&0 &1
\end{pmatrix} \implies \det(\mathbf{A}_0) =(1)(-1)(1) = -1 $
As the columns of a $3 \times 3$ matrix whose determinant is non-zero, these vectors form a basis for $\mathbb{R}^3.$
Similarly, writing the vectors of the subset $B_1$ as the columns of a matrix we have:
$\mathbf{A}_1: = \begin{pmatrix}
1&1&2 \\
3&2&6 \\
0&1 &1
\end{pmatrix} \to \begin{pmatrix}
1&1&2 \\
0&-1&0 \\
0&0 &1
\end{pmatrix} \implies \det(\mathbf{A}_1) =(1)(-1)(1) = -1 $
As the columns of a $3 \times 3$ matrix whose determinant is non-zero, these vectors form a basis for $\mathbb{R}^3.$
I don't know what it wants me to do beyond this point however. Could someone please explain what a transition matrix is?
| The given vector $x$
$$
x = 4 e_1+ 4 e_3
$$
can be represented according the first or the second basis too.
\begin{align}
x
&= x_1^{(0)} b_1^{(0)} + x_2^{(0)} b_2^{(0)} + x_3^{(0)} b_3^{(0)}
= (x_1^{(0)},x_2^{(0)},x_3^{(0)})^T = [x]_0 \\
&= x_1^{(1)} b_1^{(1)} + x_2^{(1)} b_2^{(1)} + x_3^{(1)} b_3^{(1)}
= (x_1^{(1)},x_2^{(1)},x_3^{(1)})^T = [x]_1
\end{align}
This gives a set $[x]_i$ of coordinates, a coordinate vector, for each basis. You are asked to give the matrix $T$, which would transform the first set into the second set of coordinates.
$$
[x]_1 = T [x]_0
$$
Solution:
From $B_0$ to standard basis $e_i$ we get via the matrix
$$
A_0 =
(u_0, u_2, u_3) =
\begin{pmatrix}
1 & 2 & 4 \\
2 & 3 & 6 \\
0 & 0 & 1
\end{pmatrix}
$$
From $B_1$ to standard basis we get via the matrix
$$
A_1 =
(u_1, u_2, u_3) =
\begin{pmatrix}
1 & 2 & 4 \\
2 & 3 & 6 \\
1 & 0 & 1
\end{pmatrix}
$$
From $B_0$ to $B_1$ we get by going from $B_0$ to standard basis via $A_0$ and from standard basis to $B_1$ via $A_1^{-1}$:
\begin{align}
T &= A_1^{-1} A_0 \\
&=
\begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & 0 \\
-1 & 0 & 1
\end{pmatrix}
\end{align}
We have
\begin{align}
[x]_0 &= A_0^{-1} [x] \\
&=
\left(
\begin{array}{rrr}
-3 & 2 & 0 \\
2 & -1 & -2 \\
0 & 0 & 1
\end{array}
\right)
\begin{pmatrix}
4 \\
0 \\
4
\end{pmatrix} \\
&=
\left(
\begin{array}{rrr}
-12 \\
0 \\
4
\end{array}
\right)
\end{align}
and
\begin{align}
[x]_1 &= A_1^{-1} [x] \\
&=
\left(
\begin{array}{rrr}
-3 & 2 & 0 \\
-4 & 3 & -2 \\
3 & -2 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
4 \\
0 \\
4
\end{array}
\right)
\\
&=
\left(
\begin{array}{rrr}
-12 \\
-24 \\
16
\end{array}
\right)
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1682893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you integrate $(x+2)\ln(x-3)$? I got $$\left(\frac {x^2}{2} +2x\right)\ln(x-3)-\left(\frac {x^2}{4}-\frac {7x}{2} -\frac {21}{2}\right)\ln(2x-6)$$ as my answer... Not sure If I got it right. Please correct me, thank you!
| Hint. You may use an integration by parts
$$
\begin{align}
\int(x+2)\ln(x-3)\:dx&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+2)^2}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+7)(x-3)+25}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\left(x+7+\frac{25}{x-3}\right)\:dx.
\end{align}
$$ Then conclude easily.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 0
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Solving linear recursive equation $a_n = a_{n-1} + 2 a_{n-2} + 2^n$. I wish to solve the linear recursive equation:
$a_n = a_{n-1} + 2a_{n-2} + 2^n$, where $a_0 = 2$, $a_1 = 1$.
I have tried using the Ansatz method and the generating function method in the following way:
Ansatz method
First, for the homogenous part, $a_n = a_{n-1} + 2a_{n-2}$, I guess $a_n = \lambda^n$ as the solution, and substituting and solving for the quadratic, I get $\lambda = -1, 2$. So, $a_n = \alpha (-1)^n + \beta 2^n$. Then, for the inhomogenous part, I guess $a_n = \gamma 2^n$, to get $\gamma 2^n = \gamma 2^{n-1} + 2\gamma 2^{n-2} + 2^n$, whence $2^n=0$, which means, I suppose, that this guess is not valid. These are the kind of guesses that usually work, so I don't know why it fails in this particular case, and what to do otherwise, so I tried the generating function method.
Generating function method
Let
$$
A(z) = \sum_{i=0}^{\infty} a_k z^k
$$
be the generating function for the sequence $\{ a_n \}_{n \in \mathbb{N} \cup {0}}$. Then, I try to write down the recursive relation in terms of $A(z)$:
$$
A(z) = zA(z) + 2z^2 A(z) + \frac{1}{1-2z} + (1 - 2z),
$$
where the last term in the brackets arises because of the given initial conditions. Then, solving for $A(z)$,
$$
\begin{align}
A(z) &= \frac{1}{(1+z)(1-2z)^2} + \frac{1}{1+z}\\
&= \frac{2}{9}\frac{1}{1-2z} + \frac{2}{3}\frac{1}{(1-2z)^2} + \frac{10}{9}\frac{1}{1+z}\\
&=\frac{2}{9} \sum_{k=0}^{\infty} 2^k z^k + \frac{2}{3} \sum_{k=0}^{\infty} (k+1)2^k z^k + \frac{10}{9} \sum_{k=0}^{\infty} (-1)^k z^k\\
&= \sum_{k=0}^\infty \frac{(3k+4)2^{k+1} + (-1)^k 10}{9} z^k.
\end{align}
$$
So,
$$
a_k = \frac{(3k+4)2^{k+1} + (-1)^k 10}{9}.
$$
But then, $a_1 = 2$, whereas we started out with $a_1 = 1$.
At first, I thought that maybe the generating function method did not work because some of the series on the right hand side were not converging, but they all look like they're converging for $|z| < 1/2$. I rechecked my calculations several times, so I don't think there is any simple mistake like that. It would be great if someone could explain to me what exactly is going wrong here.
| The $1-2z$ in your implicit equation for $A(z)$ is not correct, it should be $1-3z$:
$$\begin{eqnarray*}A(z)&=& a_0 + a_1 z + \sum_{n \geq 2} a_n z^n \\ &=& 2+ z + \sum_{n \geq 2} a_{n-1} z^n + \sum_{n \geq 2} 2 a_{n-2} z^n + \sum_{n \geq 2} 2^n z^n \\
&=&2+ z + \sum_{n \geq 1} a_n z^{z+1} + \sum_{n \geq 0} 2 a_n z^{n+2} + \sum_{n \geq 2} (2z)^n \\
&=&2+ z + \bigl(z A(z)-2 z\bigr) + 2 z^2 A(z) + \left(\frac{1}{1-2z}-1-2z\right)\\
&=&1-\color{red}{3} z + z A(z) + 2z^2 A(z) + \frac{1}{1-2z}\end{eqnarray*}$$
Now you can use your method to compute the coefficients of $A(z)$. This is done in detail in Brian M. Scott's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix representation of Heisenberg group Equiped with the law $(a,b,c)\circ (a',b',c') = (a + a', b + b', c + c' + ab'),$ the matrix representation of the Heisenberg group $H^3$ is given by
$$
\begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}; \quad
a,b,c\in \mathbb{R}.
$$
My question, what the matrix representation of $H^3$, if the law in $H^3$ is given by
$$(x,y,z) . (x',y',z') = (x + x', y + y', z + z' + xy'-yx').$$
thank you in advance
| I think the answer is as follows, we have:
*
*$$ \begin{pmatrix}
1 & a & c\\
0 & 1 & b\\
0 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
1 & a' & c'\\
0 & 1 & b'\\
0 & 0 & 1\\
\end{pmatrix}=
\begin{pmatrix}
1 & a+a' & c+c'+ab'\\
0 & 1 & b+b'\\
0 & 0 & 1\\
\end{pmatrix} $$
is corresponds to $(a,b,c)\circ (a',b',c') = (a + a', b + b', c + c' + ab')$.
. And $$ \begin{pmatrix}
1 & 0 & 0 & y \\
x & 1 & b & z \\
0 & 0 & 1 & - x\\
0 & 0 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & y' \\
x' & 1 & y' & z' \\
0 & 0 & 1 & - x'\\
0 & 0 & 0 & 1 \\
\end{pmatrix}
=\begin{pmatrix}
1 & 0 & 0 & y+y' \\
x+x' & 1 & y+y' & z + z' + xy'-yx' \\
0 & 0 & 1 & - x-x'\\
0 & 0 & 0 & 1 \\
\end{pmatrix}$$
is corresponds to $(x,y,z).(x',y',z')=(x + x', y + y', z + z' + xy'-yx')$.
It follows that, the matrix representation of $H^3$, if the law in $H^3$ is $(x,y,z) . (x',y',z') = (x + x', y + y', z + z' + xy'-yx')$, is given by
$$\begin{pmatrix}
1 & 0 & 0 & y \\
x & 1 & y & z \\
0 & 0 & 1 & - x\\
0 & 0 & 0 & 1 \\
\end{pmatrix}, \quad x,y,z \in \mathbb R$$
thank you for any remark or comment.
| {
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} |
Find sum of $1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$ Find the sum of the following series :
$$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots$$ and
$$1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$$
For the first series , we have , $$(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}+\cdots$$
$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots=\frac{1}{2}\log 2$$
But I am unable to set the second series to find its sum..Please help.
| The first is $\sum_{i=0}^{\infty}[\frac 1{2i+1} - \frac 1{2(2i+1)} - \frac 1{2(2i+1) + 2}]= \sum_{i=0}^{\infty}[\frac 1{2i+1} - \frac 1{2(2i+1)} - \frac 1{4(i+1)}]$
which is $\sum_{i=0}^{\infty}[\frac {4(i+1)}{4(2i+1)(i+1)} - \frac {2(i+1)}{4(2i+1)(i+1)} - \frac {2i+1}{4(2i+1)(i+1)}]=\sum_{i=0}^{\infty}\frac {1}{4(2i+1)(i+1)}$ which is ... I don't know. Somehow you seem to believe $\sum \frac 1{2(2i+1)(i+1)} = \log 2$ which... I won't deny although I don't know how you got that.
The second is $\sum_{i=0}^{\infty}[\frac{1}{6i + 1} + \frac{1}{6i+3} + \frac{1}{6i+5} - \frac{1}{6i + 2} - \frac{1}{6i+4} - \frac{1}{6i + 6}] = \sum_{i=0}^{\infty}[\frac{1}{2(3i + 1) - 1} - \frac{1}{2(3i + 1)}$$ + \frac{1}{2(3i + 2) - 1} - \frac{1}{2(3i + 2)} + \frac{1}{2(3i + 3) - 1} - \frac{1}{2(3i + 3)}]$
$=\sum_{j= 1}^{\infty}(\frac{1}{2j-1} - \frac{1}{2j}) = \sum_{j=1}^{\infty}\frac 1{2j(2j-1)} = \frac 1 2 \sum_{i=0}^{\infty}\frac 1{(i+1)(2i+1)}$ which is... I don't know. But again, I won't deny it is $\log 2$ although I don't know how you got that.
But the point to note, the second series is exactly twice the first series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2} $ . Using AM-HM inequality we get $t(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2) \geq 2^{2}.$ let $ m =(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2)$ and using AM-GM inequality we get $m \leq 5/3$.
But from this point my inequality signs are getting mixed up. Am I on right track?
| Hint: $f(x,y) = f(x) = \dfrac{4}{4-x^2} +\dfrac{9x^2}{9x^2-1}$. Let $u = x^2$, then $0 \leq u < 4$, and you get a relatively nice rational function in $u$ that you can use derivative to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Problem with indefinite integration I introduce shift $t^5=2x -3$,but it doesn't help $$\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}dx$$
| Starting from
$$I=\int \frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\text dx$$
One substitution approach is $u=\sqrt[5]{(2x-3)^4},\,\text du=\frac 8{5\sqrt[5]{2x-3}}\text dx,$ which yields
$$I=\frac 58\int\frac{2u^\frac 14-1}{u^\frac 54+1}\text du$$
If $v^4 = u,4v^3\text dv=\text du$, then we get
$$I=\frac 52\int\frac{4v^3(2v-1)}{v^5+1}\text dv$$
With some partial fraction decomposition, we get
$$\frac 1{v^5+1}=\frac 1{v(v+1)}\left(\frac 1{v^2-\varphi v+1}+\frac 1{v^2+\frac v\varphi+1}\right)$$
where $\varphi^2-\varphi-1=0$, and thus
$$I=\frac 52\int\frac{4v^2(2v-1)}{v+1}\left(\frac 1{v^2-\varphi v+1}+\frac 1{v^2+\frac v\varphi+1}\right)\text dv$$
Now we are in the realm of relatively simple integral terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$A, B$ and $C$ can do a piece of work
$A, B$ and $C$ can do a piece of work in $18$, $24$ and $30$ days respectively. $A$ and $B$ work together for a certain number of days. Then $B$ leaves and $C$ joins. They work together for half the number of days for which $A$ and $B$ had worked together. After that $A$ goes away and $B$ rejoins the work. If $B$ and $C$ together complete the remaining work in one third of the number of days for which $A$ and $B$ had previously worked together, for how many days does each of them work?
My attempt:
In $18$ days, $A$ can do $1$ work
In $1$ day, $A$ can do $\frac{1}{18}$ work
In $1$ day, $B$ can do $\frac{1}{24}$ work
In $1$ day, $C$ can do $\frac{1}{30}$ work
| Let:
$x$ the number of days that $A$ and $B$ worked together,
then $A$ and $C$ worked together $x/2$ days,
and $B$ and $C$ worked $x/3$ days together.
So, we have
\begin{align}
x\left(\frac{1}{18}+\frac{1}{24}\right)+\frac{x}{2}\left(\frac{1}{18}+\frac{1}{30}\right)+\frac{x}{3}\left(\frac{1}{24}+\frac{1}{30}\right)&=1
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequalities Using the AM-GM inequality, the Cauchy-Schwarz Inequality, and brute force Prove that if $x,y,z$ are positive real numbers then the following inequality holds
$$ \frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z . $$
I have tried everything to listed in the title like the AM-GM inequality and the Cauchy-Schwarz Inequality but it seems to be getting me nowhere. I also tried brute force but that really just led me to going in circles.
| We have $a^2+b^2\ge 2ab$ for all $a,b\in\mathbb R$, because this is equivalent to $(a-b)^2\ge 0$, which is true.
Since $x,y,z>0$, we get:
$$\frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2}$$
$$\leq \frac{x+y}{2xy}+\frac{y+z}{2yz}+\frac{z+x}{2zx}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
Equality holds if and only if $x=y=z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simple method to solve a geometry question for junior high school student Rencently, my sister asked me a geometry question that came from her mock examination, please see the following graph.
Here,
*
*$\angle DOE=45°$
*the length of $DE$ is constant, and $DE=1$. Namely, $OD,OE$ are changeable.
*$\triangle DEF$ is equilateral triangle.
Q: What is the maximum length of $OF$?
My solution
Denote $OD,OE,\angle ODE$ as $x,y,\theta$, respectively.
Via sine theorem
$$
\begin{cases}
ED^{2} = OE^{2} + OD^{2} - 2OE \times OD\cos \angle EOD \\[6pt]
\cos \theta = \dfrac{EO^{2} + ED^{2} - OD^{2}}{2 EO \times ED}
\end{cases}
$$
$$
\begin{align}
1^{2} &= x^{2} + y^{2} - 2xy\cos 45^{\circ} \\
&= x^{2} + y^{2} - \sqrt{2} xy
\end{align}
$$
$$
\implies
\begin{cases}
\color{red}{xy} = \dfrac{x^{2} + y^{2} - 1}{\sqrt{2}} \color{red}{\leq}
\dfrac{x^{2} + y^{2}}{2} \implies x^{2} + y^{2} \color{red}{\leq}
2 + \sqrt{2} \\[6pt]
\cos \theta = \dfrac{x^{2} + 1 - y^{2}}{2x}
\end{cases}
$$
Via cosine theorem
$$
\frac{y}{\sin \theta} = \frac{DE}{\sin \angle EOD} = \frac{1}{\sin
45^{\circ}} \implies \sin \theta = \frac{y}{\sqrt{2}}
$$
$$\begin{align}
OF^{2} &= EO^{2} + EF^{2} - 2EO \times EF\cos \angle OEF \\
&= x^{2} + 1^{2} - 2x\cos(\theta + 60^{\circ}) \\
&= x^{2} + 1 - 2x(\cos \theta \cos 60^{\circ} - \sin \theta \sin
60^{\circ}) \\
&= x^{2} + 1 - 2x\left(\frac{x^{2} + 1 - y^{2}}{2x} \frac{1}{2} -
\frac{y}{\sqrt{2}} \frac{\sqrt{3}}{2}\right) \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3} xy}{\sqrt{2}} \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3}}{\sqrt{2}}
\frac{x^{2} + y^{2} - 1}{\sqrt{2}} \\
&= \frac{(\sqrt{3} + 1)(x^{2} + y^{2})}{2} + \frac{1 - \sqrt{3}}{2}
\\
&\color{red}{\leq} \frac{(\sqrt{3} + 1)(2 + \sqrt{2})}{2} + \frac{1 -
\sqrt{3}}{2} = \frac{1}{2}(3 + \sqrt{3} + \sqrt{2} + \sqrt{6})
\end{align}
$$
However, for junior high school student, she doesn't learn the following formulae:
*
*sine theorem
*cosine therem
*$\cos(x+y)=\cos x \cos y-\sin x \sin y$
*fundamental inequality $x y\leq \frac{x^2+y^2}{2}$
Question
*
*Is there other simple/elegant method to solve this geometry question?
Update
Thanks for MXYMXY's hint
Here, the line $O'F$ pass the center of the circle. Namely, $O'D=OF$
In Rt $\triangle O'OF$, the inequality $O'F>OF$ holds.
| The hardest step is the first one: guessing the correct answer. By symmetry, it's plausible that $OF$ bisecting $\angle DOE$ will give either a maximum or minimum length. Another likely guess is when $O$, $D$, and $E$ are collinear, but for the given angles of $45^\circ$ and $60^\circ$ that would make $F$ lie outside the angle $DOE$, so "common sense" would eliminate that possibility.
So, how to prove the symmetrical diagram does give the longest line? Imagine the triangle $DEF$ is fixed in space. (Note, the diagram given in the question, including the angle $\theta$, is perhaps designed to mislead the student into thinking about $O$ being fixed, and $D$ and $E$ moving).
The locus of points $O'$ such that $\angle DO'E$ is constant is the arc of a circle with chord $DE$ and center $C$ as in @Ameet Sharma's diagram. On the other hand the locus of points $O'$ such that $O'F$ is constant is a circle center $F$. At high school geometry level (and even in Eulid!) is it "obvious" that one circle lies outside the other one, and they have a common tangent at point $O$.
Now that you know the symmetrical figure is the solution, calculating the length is simple trigonometry.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Simple epsilon-delta question Is it true that for every $ε > 0$, there is $δ > 0$, such that $0 < |x−2| < δ ⇒ |(x^2 −x)−2| < ε$?
Now I know that $|(x^2 −x)−2|$ is same as $|(x-2)(x+1)|$, but I am not sure how to link that with the first bit of info given. In general epsilon-delta proofs confuse me.
So I start by saying that there is an epsilon s.t $|(x^2 −x)−2| < ε$. And if this is true then there is a delta s.t $0 < |x−2| < δ$. Or is it the other way around?
Now, if $|(x^2 −x)−2| < ε$ then $|(x-2)(x+1)| < ε$ and $|x-2||x+1| < ε$ and
$$|x-2|<\frac{ε}{|x+1|}$$ But since epsilon is always positive and so is $|x+1|$ then a delta always exists.
Is my proof correct or totally wrong? I feel as though all I have done is rearranged the equation, and not really proved anything.
| Let the $\epsilon = \epsilon_0$ satisfying $|x^2-x-2| < \epsilon_0$. Initially choose $\delta$ to be $1$. We will refine this delta.
$-\epsilon_0 < x^2-x-2 < \epsilon_0$
$\implies -\epsilon_0+\frac{9}{4} < x^2-x+\frac{1}{4} < \epsilon_0+\frac{9}{4}$
$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 < \epsilon_0+\frac{9}{4}$
$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 $
Now, if $\epsilon_0 < \frac{9}{4}$
$\implies \sqrt{-\epsilon_0+\frac{9}{4}} < x-\frac{1}{2} $
$\implies \sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2} < x+1 $
Remember we had $|x^2-x-2| < \epsilon_0$. Then
$-\epsilon_0 < (x-2)(x+1) < \epsilon_0$
$\implies -\frac{\epsilon_0}{x+1} < (x-2) < \frac{\epsilon_0}{x+1}$
$\implies |x-2| < \frac{\epsilon_0}{x+1} < \frac{\epsilon_0}{\sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2}}$.
Therefore choose $\delta = min( 1 , \frac{\epsilon_0}{\sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2}} )$.
(note that for $\epsilon_0 > \frac{9}{4}$, choose $\delta$ as if $\epsilon_0 = \frac{9}{4}$, this will eventually satisfy the condition)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\int\frac{dx}{(x^4-1)^{3/2}}$ Can $$\int\frac{dx}{(x^4-1)^{3/2}}$$ be represented in terms of elementary integrals?
I have tried multiple substitutions, such as $x^2=\sec(t)$ or $\sqrt{x^4-1}=t$, but nothing seems to be working. Please provide some insight.
| Case $1$: $|x^4|\leq1$
Then $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfrac{i}{(1-x^4)^\frac{3}{2}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{i(2n+1)!x^{4n}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{i(2n+1)!x^{4n+1}}{4^n(n!)^2(4n+1)}+C$
Case $2$: $|x^4|\geq1$
Then $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfrac{dx}{x^6\left(1-\dfrac{1}{x^4}\right)^\frac{3}{2}}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-6}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-5}}{4^n(n!)^2(-4n-5)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{(2n+1)!}{4^n(n!)^2(4n+5)x^{4n+5}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\cos(x)\cos(2x)\cos(4x)=1/8$ I have to solve $\cos(x)\cos(2x)\cos(4x)=1/8$.
I can express it for $x$ only with $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\cos(4x)=\cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to proceed after... Any suggestions?
| For $x\ne n\pi$ (it's true, because $ \cos{x}\cdot\cos{2x}\cdot\cos{4x}\,\Big|_{x=n\pi} \ne\frac{1}{8}$)
$$ \cos{x}\cdot\cos{2x}\cdot\cos{4x}=\frac{\sin{x}\cdot\cos{x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}} = \\
=\frac{1}{2}\cdot \frac{\sin{2x}\cdot\cos{2x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{4}\cdot \frac{\sin{4x}\cdot\cos{4x}}{\sin{x}}=\frac{1}{8}\cdot \frac{\sin{8x}}{\sin{x}},$$
therefore,
$$\sin{8x}=\sin{x}.$$
| {
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"url": "https://math.stackexchange.com/questions/1697505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3] 2}{2}$$
contradiction, but I'm not at all sure about that. How should I proceed?
| This is an elementary but very good question. We have $$k = 2\sqrt{3} + 3\sqrt[3]{2} - 1$$ so that $$\sqrt[3]{2} = \frac{k + 1 - 2\sqrt{3}}{3}$$ and if $k$ is rational then it means that that $\sqrt[3]{2}$ is a quadratic irrationality i.e. it is a root of a quadratic equation with rational coefficients.
Hardy proves in "A Course of Pure Mathematics" that $\sqrt[3]{2}$ is not the root of such a quadratic equation. I show his method here. Let us suppose on the contrary that $\sqrt[3]{2}$ is a root of the equation $$ax^{2} + bx + c = 0\tag{1}$$ where $a \neq 0 \neq c$ and $a, b, c$ are integers. Now $\sqrt[3]{2}$ is also a root of $$x^{3} - 2 = 0\tag{2}$$ Multiplying $(1)$ by $2$ we get $$2ax^{2} + 2bx + 2c = 0$$and using $(2)$ we get $$2ax^{2} + 2bx + cx^{3} = 0$$ or $$cx^{2} + 2ax + 2b = 0\tag{3}$$ Multiply $(1)$ by $c$ and $(3)$ by $a$ and subtract to get $$(bc - 2a^{2})x + c^{2} - 2ab = 0$$ Now $x = \sqrt[3]{2}$ is irrational and hence the above equation is possible only when $c^{2} - 2ab = 0$ and $bc - 2a^{2} = 0$. This means that $c^{2} = 2ab$ and $bc = 2a^{2}$ so that $b^{2}c^{2} = 4a^{4}$ or $2ab^{3} = 4a^{4}$ or $b^{3} = 2a^{3}$. This is not possible unless $a = 0, b = 0$ because $\sqrt[3]{2}$ is irrational. We have reached a contradiction and hence $k$ is irrational.
Note: In terms of abstract algebra the above is a very elementary proof that polynomial $x^{3} - 2$ is irreducible over $\mathbb{Q}$ and $[\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}] = 3$. A proof more in line with techniques of abstract algebra is done via the concept of GCD. The GCD of polynomials in $(1)$ and $(2)$ also has $\sqrt[3]{2}$ as its root and since it is irrational it follows that the GCD has degree $2$ so that the polynomial in $(1)$ is the GCD. Hence $ax^{2} + bx + c$ is a factor of $x^{3} - 2$ and thus $x^{3} - 2$ has a linear factor so that $(2)$ has a rational root and thus the contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with Square roots and Condition Given $a+b+c=1$, prove that $\sqrt{a+\frac{(b-c)^2}4}+\sqrt{b}+\sqrt{c}\le \sqrt{3}$.
So far, I have tried to apply cauchy schwarz somehow because this works well with square roots and the inequality signs match up. However, this nonhomogeneity is tripping me up, so I would like to know how I could solve this inequality. Thanks!
| By Cauchy-Schwarz inequality
$$\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2
\le \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)(1+2)
$$
$$\Longleftrightarrow \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)\le 1$$
since $1-a=b+c$
$$\Longleftrightarrow \dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\le b+c$$
$$\Longleftrightarrow \dfrac{(b-c)^2}{4}\le\dfrac{(\sqrt{b}-\sqrt{c})^2}{2}$$
$$\Longleftrightarrow (\sqrt{b}+\sqrt{c})^2\le 2$$
and
$$b+c\le 1\Longrightarrow (\sqrt{b}+\sqrt{c})^2\le[1+1](b+c)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$ I've came across this problem some hours ago and, although it looks (and possibly is) just some algebra calculus, I can't get on the right track.
Find $x$, $y$ integers such that
$$ (x+1)^2 + (y+1)^2 + xy(x+y+3)=2 $$
Some basic algebra will lead to:
$$
(x+y)^2 + x(xy+2) + y(xy+2) + xy = 0$$
$$(x+y)(xy+x+y+2)=-xy$$
Now, some divisibility properties should end the problem.
Well, maybe my approach isn't the right one, but I'm just stuck at this point.
A piece of advice or a hint would be apreciated.
| The origin $(0,0)$, $(0,-2)$ and $(-2,0)$ are the only diophantine (integer) solutions. Simplifying the equation and taking note of the symmetry about $y=x$ ($x$ & $y$ are interchangeable, so we will eventually only see symmetric terms such as xy and x+y), we see that
$$\begin{align}
0 &= (x+1)^2 + (y+1)^2 + xy(x+y+3) - 2
\\&= x^2+2x + y^2+2y + xy(x+y+3)
\\&= x^2+y^2 + 2(x+y) + xy(x+y+3)
\\&= (x+y)^2 + 2(x+y) + xy(x+y+1)
\\&= (x+y+1)^2 + xy(x+y+1) - 1
\\&= u^2 + 2uv - 1
\\&= \left(u+v\right)^2 - \left(1+v^2\right)
\end{align}$$
for $u=x+y+1$ & $v=xy/2$, a quadratic equation with solution $u=-v\pm\sqrt{1+v^2}$, or $2(x+y+1)=-xy\pm\sqrt{4+(xy)^2}$. Next, note that $x$ & $y$ can't both be odd, because if they were, then we would have $u$ odd $\implies 2v=1-u^2$ even $\implies v\in\mathbb{Z}\implies xy$ even, leading to a contradiction. But then $2|xy$ so that $v\in\mathbb{Z}$. However, then the radical $\sqrt{1+v^2}$ can only be an integer if $v^2$ is both a perfect square and one less than a perfect square. This can only happen when $v=xy=0$, i.e., when at least one of $x,y$ is zero as noted above.
Another approach:
$$\begin{align}0
&=(x+y)^2+x(xy+2)+y(xy+2)+xy\\
&=(x+y)^2+(x+y)(xy+2)+xy\\
&=u^2+ut+t-2
\end{align}$$
for (integers) $$\begin{align}t&=xy+2\\u&=x+y\end{align}$$ so that
$$t(1+u)=2-u^2$$ or $$t=\frac{2-u^2}{1+u}=\frac{1-u}{1+u}+\frac1{1+u}.$$
For $t$ to be an integer, we must have $u+1=\pm1$, or $u=x+y\in\{0,-2\}$.
In both cases, $t=2\implies xy=0\implies x$ or $y$ is zero.
This only admits the three solutions $(x,y)=(0,0),(0,-2),(-2,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find Nth formula of recursive formula $a_n=a_{n-1}+n(n-1)a_{n-2}$ $$a_n=a_{n-1}+n(n-1)a_{n-2}$$
$$a_0=1, a_1=-\frac{1}{2}$$
Is it possible to find explicit formula for $a_n$ just by using $a_0$ and $a_1$?
I know how to solve this problem if $a_n=Aa_{n-1}+Ba_{n-2}$ where $A$ and $B$ is some real number(constant), but in this problem that is not a case.
| For the second question, note that
$$\begin{align*}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}&=\begin{bmatrix}Aa_{n-1}+Ba_{n-2}\\ a_{n-1}\end{bmatrix}=\begin{bmatrix}A &B\\ 1 & 0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}A&B\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\ a_{n-3}\end{bmatrix}\\&=\cdots=\begin{bmatrix}A&B\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_1\\a_0\end{bmatrix}\end{align*}$$
With a similar argument, we would get that
$$\begin{align*}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}&=\begin{bmatrix}a_{n-1}+n(n-1)a_{n-2}\\ a_{n-1}\end{bmatrix}=\begin{bmatrix}1 &n(n-1)\\ 1 & 0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}\\&=\begin{bmatrix}1&n(n-1)\\1&0\end{bmatrix}\begin{bmatrix}1 &(n-1)(n-2)\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\ a_{n-3}\end{bmatrix}\\&=\cdots \\&=\left(\prod_{k=0}^{n-2} \begin{bmatrix}1&(n-k)(n-k-1)\\1&0\end{bmatrix}\right)\begin{bmatrix}a_1\\a_0\end{bmatrix}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Prove $\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$ Im trying to prove the following equation:
$$\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$$
I have tried to do it several times now but get stuck at different places. I don't know if it's a calculation error or the question is defective. It would be really helpful if someone could point out the method.
we are to use $$z - \frac1{z} = 2i\sin x$$
$$z^n + (\frac1{z})^n = 2\cos nx$$
| Raising to the $6^{th}$ power we get
$$
\eqalign{
64i^6\sin^6 x &= \bigl(z-{1\over z}\bigr)^6 \cr
&= z^6 - 6z^5{1\over z} + 15z^4{1\over z^2} -20z^3{1\over z^3} + 15z^2{1\over z^4} - 6z{1\over z^5} + {1\over z^6} \cr
&= z^6 - 6z^4 + 15z^2 -20 + 15{1\over z^2} - 6{1\over z^4} + {1\over z^6} \cr
&= z^6 + {1\over z^6} - 6\bigl(z^4+{1\over z^4}\bigr) + 15\bigl(z^2+{1\over z^2}\bigr) -20 \cr
&= 2\cos6x -6\cdot 2\cos4x +15\cdot 2\cos2x -20 \ \ \ so \ \ \ that \cr
-32\sin^6x &= \cos6x -6\cos4x +15\cos2x -10 \cr
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Integrate the expression using any method I have an expression to integrate, which I don't know how to integrate, tried to factorize and solve, but got really ugly expression at the end. Would appreciate any help:
$$\int \frac{dx}{(x-1)\sqrt{x^2-x+1}}$$
| That edit makes it easier. Let $x-\frac12=\frac{\sqrt3}2\tan\theta$. Then
$$\begin{align}\int\frac{dx}{(x-1)\sqrt{x^2-x+1}} & =\int\frac{d\theta}{\frac{\sqrt3}2\sin\theta-\frac12\cos\theta}=\int\frac{d\theta}{\sin(\theta-\pi/6)} \\
& =-\ln(\csc(\theta-\pi/6)+\cot(\theta-\pi/6))+C_1 \\
& =\ln\left(\frac{\sin(\theta-\pi/6)}{1+\cos(\theta-\pi/6)}\right)+C_1 \\
& = \ln\left(\frac{\frac{\sqrt3}2\sin\theta-\frac12\cos\theta}{1+\frac{\sqrt3}2\cos\theta+\frac12\sin\theta}\right)+C_1\end{align}$$
From the Pythagorean theorem,
$$\sin\theta=\frac{x-\frac12}{\sqrt{x^2-x+1}},\,\,\cos\theta=\frac{\frac{\sqrt3}2}{\sqrt{x^2-x+1}}$$
$$\begin{align}\int\frac{dx}{(x-1)\sqrt{x^2-x+1}} & =\ln\left(\frac{\frac{\sqrt3}2(x-\frac12)-\frac12\frac{\sqrt3}2}{\sqrt{x^2-x+1}+\frac{\sqrt3}2\frac{\sqrt3}2+\frac12(x-\frac12)}\right)+C_1 \\
& =\ln\left(\frac{x-1}{\sqrt{x^2-x+1}+\frac12x+\frac12}\right)+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerator). The given limit has a $\frac{0}{0}$ form. I tried using taylor series but the it made the problem more complicated.
| $\sin(1+x)=x+\mathcal O(x^3)\;$, and likewise for the logarithmic function, so:
$$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\cong$$
$$\frac{\left(\sin^3x\cos^2x\right)\cot\left(x^3\right)\left(x^4\right)}{\left(\sqrt{x^2+2}-\sqrt2\right)\left(x^2\right)}=\cos x^3\;\frac{x^3}{\sin x^3}\;\cos^2x\;\frac{\sin^3x}{x^3}\left(\sqrt{x^2+2}+\sqrt2\right)\longrightarrow$$
$$\xrightarrow[x\to0]{}1\cdot1\cdot1\cdot1\cdot2\sqrt2=2\sqrt2=\sqrt8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that the two powers are equal
Prove that: $$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1} = \dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)^{180} - \left(1-\dfrac{\sqrt{3}}{2}\right)^{180}}{\sqrt{3}}$$ where: $$a = \dfrac{1+\sqrt{3}}{\sqrt{2}}.$$
This seemed like a very challenging question but the fact that we have a telescoping binomial sum in the numerator of the LHS helps. I think if we can simplify the LHS sufficiently we might be able to prove it by just equating both sides of the equation.
| HINT:
Observe that $a^2=2+\sqrt3$
$\implies1+\dfrac{\sqrt3}2=\dfrac{a^2}2$
and $1-\dfrac{\sqrt3}2=\dfrac1{2a^2}$
Use Componendo and Dividendo to find $\dfrac{a^2-1}{a^2+1}=?$
Now
$$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1}=\left[\left(\dfrac{a^2}2\right)^{180}-\left(\dfrac1{2a^2}\right)^{180}\right]\cdot\dfrac{a^2-1}{a^2+1}$$
Can you reach home from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let's denote them by $I_1$ and $I_2$ respectively.
$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$.
If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?
| Assume $a\in\mathbb{R}^+$
$$\int_{0}^{a}\frac{\sqrt{a+x}}{\sqrt{a-x}}\space\text{d}x=$$
Substitute $u=\sqrt{a+x}$ and $\text{d}u=\frac{1}{2\sqrt{a+x}}\space\text{d}x$.
THis gives a new lower bound $u=\sqrt{a+0}=\sqrt{a}$ and upper bound $u=\sqrt{a+a}=\sqrt{2a}$:
$$\int_{\sqrt{a}}^{\sqrt{2a}}\frac{2u^2}{\sqrt{2a-u^2}}\space\text{d}u=2\int_{\sqrt{a}}^{\sqrt{2a}}\frac{u^2}{\sqrt{2a-u^2}}\space\text{d}u=$$
Substitute $u=\sqrt{2a}\sin(s)$ and $\text{d}u=\sqrt{2a}\cos(s)\space\text{d}s$.
Then $\sqrt{2a-u^2}=\sqrt{2a-2a\sin^2(s)}=\sqrt{2a}\cos(s)$ and $s=\arcsin\left(\frac{u}{\sqrt{2a}}\right)$.
THis gives a new lower bound $s=\arcsin\left(\frac{\sqrt{a}}{\sqrt{2a}}\right)=\frac{\pi}{4}$ and upper bound $s=\arcsin\left(\frac{\sqrt{2a}}{\sqrt{2a}}\right)=\frac{\pi}{2}$:
$$4a\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2(s)\space\text{d}s=4a\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{1}{2}-\frac{\cos(2s)}{2}\right]\space\text{d}s=$$
$$4a\left[\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1\space\text{d}s-\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos(2s)\space\text{d}s\right]=\frac{a(2+\pi)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding dy/dx with two elements? I don't understand the problem. How can I find $\frac{dy}{dx}$, given that $y = \frac{u^2 - 1}{u^2 + 1}$ and $u = \sqrt[3]{x^2 + 2}$?
For the next step, the book says:
$\frac{dy}{du} = \frac{4u}{(u^2+1)^2}$.
Am I supposed to place the value of $u$ into the $y$ fraction?
If I put it into the numerator, it won't solve for 4u. Is the book wrong?
Also, where does du come from? It says dx before.
| You have two functions:
$$
y=f(u)=\frac{u^2-1}{u^2+1} \qquad u=g(x)=\sqrt[3]{x^2+1}
$$
so $y$ is a composite function:
$$
y=f(g(x))=\frac{\left(\sqrt[3]{x^2+1}\right)^2-1}{\left(\sqrt[3]{x^2+1}\right)^2+1}
$$
Maybe that, in this form, you know how find the derivative using the chain rules:
$$
\frac{dy}{dx}=f'(g(x))g'(x)
$$
Anyway, your book gives you the derivative
$$\frac{dy}{du}=\frac{df}{du}=f'(u)=f'(g(x))=\frac{4u}{(u^2+1)^2}=\frac{4\sqrt[3]{x^2+1}}{\left(\left(\sqrt[3]{x^2+1}\right)^2+1\right)^2}$$
so you have to calculate
$$
g'(x)=\frac{d}{dx}\sqrt[3]{x^2+1}=\frac{2x}{3\sqrt[3]{(x^2+1)^2}}
$$
and multiply to complete the chain rule.
$$
\frac{dy}{dx}=\frac{4\sqrt[3]{x^2+1}}{\left(\left(\sqrt[3]{x^2+1}\right)^2+1\right)^2}\frac{2x}{3\sqrt[3]{(x^2+1)^2}}=
\frac{8x}{3\sqrt[3]{x^2+1}\left(\left(\sqrt[3]{x^2+1}\right)^2+1\right)^2}
$$
Deriving directly the function $y=f(g(x))$, using the quotient and the chain rules you can find:
$$
y'=\frac{8}{3}x\frac{1}{\sqrt[3]{x^2+1}}\frac{1}{\left(\left(\sqrt[3]{x^2+1} \right)^2+1\right)^2}
$$
note that:
$$
\frac{8}{3}x\frac{1}{\sqrt[3]{x^2+1}}=4\sqrt[3]{x^2+1}\left(\frac{2}{3}x\frac{1}{\sqrt[3]{(x^2+1)^2}}\right)=4u(x)u'(x)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find the integral $\int \frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$ $$\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}=\frac{2\sqrt[5]{2x-3}-1}{2\sqrt[5]{2x-3}(x-1)}=\frac{1}{x-1}-\frac{1}{2\sqrt[5]{2x-3}(x-1)}$$
$$\int \frac{1}{x-1}\mathrm dx=\ln|x-1|+c$$
$$\int \frac{1}{2\sqrt[5]{2x-3}(x-1)}\mathrm dx=\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx$$
Substitution $u=2x-3,du=2dx$ gives
$$\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$$
How to evaluate $\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du$?
| Hint. Let $u^{1/5}=v$ so $$\int\frac{du}{u^{6/5}+u^{1/5}}=5\int\frac{v^4}{v^6+v}dv=5\int\frac{v^3}{v^5+1}dv$$
and factorize $v^5+1$ as $$(1 + v)(v^4 - v^3 + v^2 - v + 1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the integral $\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$ $$\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$$$=2\int \frac{x^3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx+3\int \frac{x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
For these two integrals, I have tried Euler substitutions and various types of substitutions of irrational functions, but it seems that those substitutions don't simplify the integral.
What substitution is useful for this type of integrals?
| $$\int\frac{2x^3+3x^2}{\left(2x^2+x-3\right)\sqrt{x^2+2x-3}}\space\text{d}x=$$
Using the Euler subsitution:
$x=\frac{u^2+3}{2u+2}$ and $\text{d}x=\left(\frac{2u}{2u+2}-\frac{2\left(u^2+3\right)}{\left(2u+2\right)^2}\right)\space\text{d}u$.
And when we substitute back we get $u=x+\sqrt{x^2+2x-3}$:
$$\int\frac{\left(u^2+3\right)^2}{2\left(u^2-1\right)^2}\space\text{d}u=\frac{1}{2}\int\frac{\left(u^2+3\right)^2}{\left(u^2-1\right)^2}\space\text{d}u=$$
$$\frac{1}{2}\int\left[\frac{4}{(u+1)^2}+\frac{4}{(u-1)^2}+1\right]\space\text{d}u=$$
$$\frac{1}{2}\left[4\int\frac{1}{(u+1)^2}\space\text{d}u+4\int\frac{1}{(u-1)^2}\space\text{d}u+\int1\space\text{d}u\right]=$$
$$\frac{1}{2}\left[4\int\frac{1}{(u+1)^2}\space\text{d}u+4\int\frac{1}{(u-1)^2}\space\text{d}u+u\right]=$$
Substitute $s=u+1$ and $\text{d}s=\text{d}u$.
And Substitute $p=u-1$ and $\text{d}p=\text{d}u$:
$$\frac{1}{2}\left[4\int\frac{1}{s^2}\space\text{d}s+4\int\frac{1}{p^2}\space\text{d}p+u\right]=$$
$$\frac{1}{2}\left[u-\frac{4}{s}-\frac{4}{p}\right]+\text{C}=$$
$$\frac{1}{2}\left[u-\frac{4}{u+1}-\frac{4}{u-1}\right]+\text{C}=$$
$$\frac{1}{2}\left[x+\sqrt{x^2+2x-3}-\frac{4}{x+\sqrt{x^2+2x-3}+1}-\frac{4}{x+\sqrt{x^2+2x-3}-1}\right]+\text{C}=$$
$$\frac{1}{2}\left[\frac{(3+x)(2x-3)}{\sqrt{(3+x)(x-1)}}\right]+\text{C}=\frac{(3+x)(2x-3)}{2\sqrt{(3+x)(x-1)}}+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of product of two normal distribution densities I want to compute the integral:
$\displaystyle \int^{\infty} _{-\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-x)^2}{2}} \frac{1}{\sqrt{2\pi}ab} e^{-\frac{x^2}{2(ab)^2}} dx$
Maybe we can use that for a normal distribution with mean $\mu$ and variance $\sigma^2$ we have
$\displaystyle \int^{\infty} _{-\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx = 1$
In an effort to write the integral in this form, I tried to take the exponents together. This gives:
$\displaystyle -\frac{(y-x)^2}{2} - \frac{x^2}{2(ab)^2} = \frac{-[(ab)^2 (y-x)^2 + x^2]}{2(ab)^2} = \frac{-[(ab)^2 (y^2 -2xy +x^2) + x^2]}{2(ab)^2}$
But this leads to nowhere. Any suggestions?
| Actually, you're on the right track. Just keep on going with
$$\begin{align}(ab)^2(y^2-2xy+x^2)+x^2 & =(a^2b^2+1)x^2-2a^2b^2yx+a^2b^2y^2 \\
& =(a^2b^2+1)\left[x^2-2\frac{a^2b^2y}{a^2b^2+1}x\right]+a^2b^2y^2 \\
& =(a^2b^2+1)\left[\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2-\frac{a^4b^4y^2}{(a^2b^2+1)^2}\right]+a^2b^2y^2 \\
& =(a^2b^2+1)\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2+\left[a^2b^2-\frac{a^4b^4}{a^2b^2+1}\right]y^2 \\
& =(a^2b^2+1)\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2+\left(\frac{a^2b^2}{a^2b^2+1}\right)y^2 \\
\end{align}$$
Then we know that
$$\int_{-\infty}^{\infty}e^{\frac{-(a^2b^2+1)\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2}{2a^2b^2}}dx=\int_{-\infty}^{\infty}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\sqrt{2\pi}\sigma$$
Where
$$\mu=\frac{a^2b^2y}{a^2b^2+1}$$
$$\sigma=\frac{ab}{\sqrt{a^2b^2+1}}$$
So now you have
$$\begin{align}\int_{-\infty}^{\infty}\frac1{\sqrt{2\pi}}e^{-\frac{(y-x)^2}{2}}\frac1{\sqrt{2\pi}ab}e^{-\frac{x^2}{2(ab)^2}}dx & =\frac1{\sqrt{2\pi}}\frac1{\sqrt{2\pi}ab}\frac{\sqrt{2\pi}ab}{\sqrt{a^2b^2+1}}e^{-\frac{y^2}{2\left(\sqrt{a^2b^2+1}\right)^2}} \\
& = \frac1{\sqrt{2\pi}\sqrt{a^2b^2+1}}e^{-\frac{y^2}{2\left(\sqrt{a^2b^2+1}\right)^2}}
\end{align}$$
So the means add, $0+0=0$, as do the variances, $1+a^2b^2=\left(\sqrt{a^2b^2+1}\right)^2$, just like they are supposed to.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Elliptic integrals Knowing that
$$\int^{2\pi}_{0}\sqrt{a^2\cos^{2}t+b^2\sin^{2}t}dt＝\int_{0}^{2\pi}\sqrt[]{1+\cos^{2}t}dt$$
how to find the value of $a$ and $b$?
Elliptic integral? Thank you!
| $$\int^{2\pi}_{0}\sqrt{a^2\cos^{2}(t)+b^2\sin^{2}(t)}dt = |a|\int^{2\pi}_{0}\sqrt{1-\left(1-\frac{b^2}{a^2}\right)\sin^{2}(t)}dt =4|a|\text{E}\left(1-\frac{b^2}{a^2}\right)$$
E$(X)$ is the complete elliptic integral of second kind.
$$\int^{2\pi}_{0}\sqrt{1+\cos^{2}(t)}dt =\sqrt{2}\int^{2\pi}_{0}\sqrt{1-\frac{1}{2}\sin^{2}(t)}dt = 4\sqrt{2}\text{ E}\left(\frac{1}{2}\right)$$
The equation to be solved is :
$$|a|\text{E}\left(1-\frac{b^2}{a^2}\right) = \sqrt{2}\text{ E}\left(\frac{1}{2}\right)$$
Let $X=1-\frac{b^2}{a^2}\quad$ Hense $\quad X\leq 1\quad$ and E$(X)$ is real.
$a=\pm\frac{b}{\sqrt{1-X}}\quad$ This transforms the equation to :
$$\text{E}(X) = \sqrt{2}\text{ E}\left(\frac{1}{2}\right)\frac{\sqrt{1-X} }{|b|} \qquad (1)$$
On $-\infty<X\leq 1$ the functions E$(X)$ and $\sqrt{1-X}$ are both smooth functions. Both are smooth and decreassing.
Given a value of $b$, the curves E$(X)$ and $\sqrt{1-X}$ intersect each other on one point only.
For example, given $b=1\quad\to\quad X=\frac{1}{2}\quad\to\quad a=\pm \sqrt{2}$
So, they are an infinity of solutions : To each arbitrary value of $b$, the transcendantal equation (1) has a root to which corresponds a value of $X$ and a value of $|a|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral through $u$-substitution v. multiplying out I have the integral:
$ \int (x^2)(x^3-1)^2 \, dx $ Through $u$-substitution, I can write this is equal to $ \frac {1}{3} \int (3x^2)(x^3-1)^2 dx $ which equals $ \frac{1}{3}\cdot\frac{(x^3-1)^3}{3}$.
However, if rather than using $u$-substitution, I multiply within $ \int (x^2)(x^3-1)^2 \, dx $ from the beginning, I don't get the same answer. For example: $ \int (x^2)(x^3-1)^2 \, dx $ = $ \int (x^2)(x^6 -2x^3 + 1) \, dx = \int (x^8 -2x^5 + x^2) \, dx$. The anti derivative of this becomes $ \frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} $.
However, these anti-derivatives are not equal. Through u-substitution, when $x=1$, the anti derivative evaluates to zero. In the second route, when I plug in $1$, I get $1/9$.
Where did I go wrong? From what I see, the $u$-substitution version is correct.
| Ahhh, the question is what is the function at the point (1,0). And the constant will change if I use the first equation v. the second. So the first equation will have a constant of 0, which has a solution function of $ \frac {1}{3} * \frac {(x^3-1)^3}{3}$ with its constant = 0.
And the second way $ \frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} - \frac{1}{9} $ where 1/9 is my constant. I assume now the two functions are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
partial fractions
$$\frac{1}{1-x^2}$$
$$\frac{1}{1-x^2}=\frac{a}{1-x}+\frac{b}{1+x}$$
$$1=a+ax+b-bx$$
$$1=a+b+x(a-b)$$
$a+b=1$ and $x(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$
$$2a=1\Rightarrow a=\frac{1}{2}$$
$b=\frac{1}{2}$
$$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
Where I went wrong?
|
Notice:
*
*$$\frac{1}{1-x^2}=\frac{1}{-(x-1)(x+1)}=-\frac{1}{(x-1)(x+1)}$$
So, we get:
$$-\frac{1}{(x-1)(x+1)}=-\frac{a}{x-1}-\frac{b}{x+1}$$
And, now we can see that:
$$a(x+1)+b(x-1)=1\Longleftrightarrow a-b+(a+b)x=1$$
So, we get:
*
*$$a-b=1\Longleftrightarrow a=b+1\Longleftrightarrow a=-\frac{1}{2}+1=\frac{1}{2}$$
*$$a+b=0\Longleftrightarrow b+1+b=0\Longleftrightarrow 2b=-1\Longleftrightarrow b=-\frac{1}{2}$$
So, now we have the partial fractions:
$$\frac{1}{1-x^2}=-\frac{\frac{1}{2}}{x-1}-\frac{-\frac{1}{2}}{x+1}=\frac{1}{2(x+1)}-\frac{1}{2(x-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3 Prove that if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3
| $$
(3k)^2 = 3k' \; , (3k+1)^2 = 3k'+1 \; and \; (3k+2)^2 = 3k'+4=3k''+1
$$
(so there's no square of the form $3k + 2$)
If $a = 3k$ or $b=3k$ we are done, assume the opposite:
$a\mod3 \neq b \mod 3 \neq 0$
$$
(3m+1)^2+(3n+2)^2 = 3p+1+4=3p'+2 = c^2
$$
(not possible)
$a\mod3 = b \mod 3 \neq 0 $
$$
(3m+1)^2+(3n+1)^2 = 3p+1+1=3p'+2 = c^2
$$
$$
(3m+2)^2+(3n+2)^2 = 3p+4+4=3p'+2 = c^2
$$
(not possible)
So at least one must be divisible by 3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral $\int_0^{2\pi} \frac{d \theta}{5-3 \cos \theta}$ $$\int_0^{2\pi} \frac{d \theta}{5-3 \cos \theta}$$
My attempt:
Let $z=e^{i\theta}$ which gives $d\theta = \frac{dz}{iz}$
Thus,
$$\oint_C \frac{1}{5-3(\frac{z+z^{-1}}{2})}\frac{dz}{iz}$$
$$=\frac{1}{i}\oint_C \frac{dz}{(3z-1)(z-3)}$$
We can ignore the singularity at $z=3$ because it lies outside the unit circle and thus we don't need to account for it when computing the residues.
$$=\frac{1}{i}\oint_C f(z) \, dz = 2\pi [\operatorname{Res}(f(z),\frac{1}{3})]$$
$$\operatorname{Res}(f(z),\frac{1}{3})=\frac{\lim_{z\to1/3} (3z-1)\cdot \frac{1}{(3z-1)(z-3)}}{0!}$$
$$=-\frac{9}{8}$$
Which yields:
$$=\frac{1}{i}\oint_C f(z) \, dz = 2\pi [-\frac{9}{8}]$$
$$=-\frac{9\pi}{4}$$
But how can I have a negative answer for an integration??? Did I make a mistake somewhere??
| It is true that in order to find the roots, you can multiply both sides of
$$
5z-\frac{3z^2} 2 - \frac 3 2 = 0
$$
by $-2$, getting
$$
-10z + 3z^2 + 3 = 0,
$$
and then using your favorite method of solving quadratic equations, concluding that the roots are $3$ and $1/3$.
But that does not mean that $-10z+3z^2 + 3z$ is equal to $(-1/2)(10z + 3z^2 + 3)$ when the value of $z$ is such that the value of $-10z+3z^2 + 3z$ is not $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Jordan form exercise What am I doing wrong?
I've been learning how to put matrices into Jordan canonical form and it was going fine until I encountered this $4 \times 4$ matrix:
$A=\begin{bmatrix}
2 & 2 & 0 & -1 \\
0 & 0 & 0 & 1 \\
1 & 5 & 2 & -1 \\
0 & -4 & 0 & 4 \\
\end{bmatrix}
$
Which has as only eigenvalue $\lambda_1=\lambda_2=\lambda_3=\lambda_4=2$ with 2 corresponding eigenvectors, which I will for now call $v_1$ and $v_2$:
$v_1 = \pmatrix{0\\0\\1\\0}, v_2=\pmatrix{-3 \\ 1 \\ 0 \\ 2}
$
2 eigenvectors means 2 Jordan blocks so I have 2 possibilities:
$J= \pmatrix{2 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2} $ or $ J= \pmatrix{2 & 1 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 0 & 2}
$
I consider the first possibility. This gives me the relations:
$Ax_1=2x_1 \\
Ax_2=x_1+2x_2 \\
Ax_3=2x_3+x_2 \\
Ax_4=2x_4 \\
$
where $x_1$ and $x_4$ should be $v_1$ and $v_2$. From the second relation $(A-2I)x_2=x_1$ I see
$\pmatrix{0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2} \pmatrix{a \\ b \\ c \\ d} =\pmatrix{0 \\ 0 \\ 1 \\ 0} $
( $v_2= \pmatrix{ -3 \\ 1 \\ 0 \\ 2}
$ will give an inconsistent system)
Now I get that $x_2 = \pmatrix{-2 \\ 1 \\ 0 \\ 2}
$
From the third relation $(A-2I)x_3=x_2$:
$\pmatrix{0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2} \pmatrix{e \\ f \\ g \\ h} =\pmatrix{-2 \\ 1 \\ 0 \\ 2} $
But this system is inconsistent as well! No matter which vectors I try in which places, when I try to generalize eigenvectors I seem to always end up with some inconsistency.
Is there something staring me in the face that I am overlooking? Or am I doing it completely wrong (even though this method worked fine for me before)?
Sorry for the lengthiness and thank you in advance.
| Take a look at
$$
(A-2I)^2 = \begin{bmatrix}
0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2 \\
\end{bmatrix}^2
$$
The $a_{1,2}$ entry is non-zero. What does that tell you?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Sum of the n nth roots of unity 1) If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$
2) Then apparently: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}=\frac{0}{1-z}=0$
Could someone please explain why: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}$?
| Let $z^0+z^1+z^2...z^{n-1}= p(z)$.
Then $z \ p(z)=z^1+z^2+z^3...z^{n}$.
Hence $p(z)-zp(z)=1-z^n$.
$p(z)(1-z)=1-z^n$.
$p(z)=\frac{1-z^n}{1-z}$
If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. Then there is a formula (that is easy to prove) that states for any $i \in \mathbb{R}$, $z^i=\cos\frac{2i\pi}{n}+i\sin\frac{2i\pi}{n}$. Hence $z^n=\cos\frac{2n\pi}{n}+i\sin\frac{2n\pi}{n}=1$. Substitute that in the last expression of $p(z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$
| If $\sin4x=\sin3x,$
$4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$ $$4\cos x\cos2x=3-4\sin^2x\iff4\cos x(2\cos^2x-1)=3-4(1-\cos^2x)$$
$$\iff8\cos^3x-4\cos^2x-4\cos x+1=0$$
Again if $\sin4x=\sin3x,$
$4x=n\pi+(-1)^n3x$ where $n$ is any integer
For even $n=2m$(say), $x=2m\pi\implies \sin x=?$ where $m$ is any integer
For odd $n=2m+1$(say), $7x=(2m+1)\pi$ where $m$ is any integer
$\implies x=\dfrac{(2m+1)\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$
But for $m\equiv3\pmod7,\sin x=0$
So, the roots of $$8t^3-4t^2-4t+1=0$$ are $\cos\dfrac{(2m+1)\pi}7$ where $m\equiv0,1,2\pmod7$
as $\cos(-y)=\cos y,\cos\dfrac\pi7=\cos\dfrac{\pi\{2(-1)+1\}}7$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Contour integral of $f(z) = \frac{1}{z^2+iz+6} $ Need help evaluating a certain contour integral.
$f(z) = \frac{1}{z^2+iz+6} $
Steps so far:
Poles: $ z^2+iz+6 \rightarrow \frac{-i \pm \sqrt{-1-24}}{2}=0 \rightarrow z_0 = +2i, -3i $
Residues: $ a_{-1} = \lim{z \to z_{i}} [ \frac{d^{n-1}}{dz^{n-1}} (z-z_{0})^n f(z)] $
Plug all this in I get:
$\frac{1}{z+3i} +\frac{1}{z-2i} = a_{-1} $
$ \therefore \oint \frac{1}{z^2+iz+6}dz = -2\pi i[\frac{1}{z+3i}+\frac{1}{z-2i} ]$
Have I done something wrong?
| Let $f(z)$ be defined as
$$\begin{align}
f(z)&=\frac{1}{z^2+iz+6}\\\\
&=\frac{1}{(z+i3)(z-i2)}\\\\
&=\frac{1}{i5}\left(\frac{1}{z-i2}-\frac{1}{z+i3}\right)
\end{align}$$
Therefore, $f(z)$ has poles at $z=-i3$ and $z=i2$. Integration of $f(z)$ around a closed rectifiable contour $C$ yields
$$\oint_C f(z)\,dz=\begin{cases}
0&,\text{if}\,\,C\,\,\text{does not encircle either pole}\\\\
\frac{2\pi}{5} &,\text{if}\,\,C\,\,\text{encircles only the pole at}\,\,z=i2\\\\
-\frac{2\pi}{5}&,\text{if}\,\,C\,\,\text{encircles only the pole at}\,\,z=-i3\\\\
0&,\text{if}\,\,C\,\,\text{encircles both poles}\\\\
\end{cases}$$
NOTE:
To use the limit definition for evaluating residues, we have for the residue at $z=i2$
$$\begin{align}\text{Res}\left(\frac{1}{z^2+iz+6},z=i2\right)&=\lim_{z\to i2}\frac{z-i2}{z^2+iz+6}\\\\
&=\lim_{z\to i2}\frac{1}{z+i3}\\\\
&=\frac{1}{i5}
\end{align}$$
and for the residue at $z=-i3$
$$\begin{align}\text{Res}\left(\frac{1}{z^2+iz+6},z=-i3\right)&=\lim_{z\to -i3}\frac{z+i3}{z^2+iz+6}\\\\
&=\lim_{z\to -i3}\frac{1}{z-i2}\\\\
&=-\frac{1}{i5}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Irreducible factorisation of polynomial over quotient field Let $F=\mathbb{Z}_3[x]/<x^2+1>$.
Factor $x^4+2$ into irreducibles in $F[x]$.
I know that $F$ is a field since $x^2+1$ is irreducible.
The usual way to find out that a polynomial is irreducible is that it has no roots. But how to do it if the element of F is now of the form $g(x)+<x^2+1>$ where $g(x)\in\mathbb{Z}_3[x]$ and the zero element of $F$ is $<x^2+1>$. Any help is appreciated :)
| we find that $F = \mathbb{Z}_3[x] / (x^2+1)$ (which is the field with $9$ elements) has a cyclic multiplicative group with $8$ elements, whose generator is $x+1$
(proof : $(x+1)^2 = 2x$, $(2x)^2 = 2$, $2^2 = 1$, hence $(x+1)^8 = 1$ but $(x+1)^4 \ne 1$)
hence in $F$ the equation $y^4 = 1$ has $4$ solutions for $y$ : $$(x+1)^2 = 2x, \qquad(x+1)^4 = 2, \qquad (x+1)^6 = x, \qquad (x+1)^8 = 1$$ and that's cool because it means that in $F[y]$, the polynomial $y^4+2$ has $4$ distinct roots, hence :
$$y^4+2 = (y-2x)(y-2)(y-x)(y-1) = (y+x)(y+1)(y+2x)(y+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For large $n$, show that $\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $ nearly equals $\frac{1}{2}$. For large $n$, show that $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $$ nearly equals $\frac{1}{2}$.
Integrating by parts we get $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2}=\Bigg(\frac{x^n}{1+x^2}\Bigg)^{1}_{0} - \int\limits_{0}^1 \frac{2x^{n+1}}{(1+x^2)^2}dx$$
The integral $$\int\limits_{0}^1 \frac{2x^{n+1}}{(1+x^2)^2}dx\leq\int\limits_{0}^1 {2x^{n+1}}dx =\frac{2}{n+2}$$
which $\rightarrow 0$ as $n\rightarrow \infty$
Hence the value of the main integral is $1$. Where am I wrong ?
| As commented:
$$\left.\left(\frac{x^n}{1+x^2}\right)\right|_0^1=\frac1{1+1}-\frac0{1+0}=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloor$.
So we want to prove $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$ and $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
Since $\lfloor x\rfloor\le x$, we can just start from here and prove
$\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
But for $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$, I have no idea how to start.
| One way to do it is to write $x=kn+r+\epsilon$, where $0\le r\le n-1$ is an integer and $0\le\epsilon\lt1$, so that $0\le r+\epsilon\lt n$. Then
$$\left\lfloor{\lfloor x\rfloor\over n}\right\rfloor=\left\lfloor{kn+r\over n}\right\rfloor=\left\lfloor k+{r\over n}\right\rfloor=k=\left\lfloor k+{r+\epsilon\over n}\right\rfloor=\left\lfloor {kn+r+\epsilon\over n} \right\rfloor=\left\lfloor {x\over n} \right\rfloor$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
What is the indefinite integral of $\sqrt[4] {\tan \left( x\right) }$ $$\int\sqrt[4] {\tan \left( x\right) } dx$$
I'm really stuck right now with this integral, so any kind of advice would be appreciated.
Perhaps a pretty nifty substitution that can save me from partial fraction decomposition.
| $$ \small \left( x^2 + \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 + \sqrt{2 - \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 - \sqrt 2} \; x + 1 \right) = x^8 + 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$ How do we integrate
$$\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$$
Could someone give me some hint for this question?
| Let $$I = \int\frac{x^2+20}{(x\sin x+5 \cos x)^2}dx$$
Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $x^{2\times 5-2} = x^8$
So $$I = \int\frac{x^8(x^2+20)}{x^8(x\sin x+5\cos x)^2}dx = \int\frac{x^8(x^2+20)}{(x^5\sin x+5x^4\cos x)^2}dx$$
Now put $x^5\sin x+5x^4 \cos x=t\;,$ Then $x^3\cos x(x^2+20)dx=dt$
So $$I = \int x^5 \sec x \cdot \frac{x^2+20}{(x\sin x+5\cos x)^2}dx$$ (Using Integration by parts )
So $$I = -\frac{x^5\sec x}{x^5\sin x+5x^4\cos x}+\int \frac{x^5 \sec x\tan x+5x^4\sec x}{x^5\sin x+5x^4\cos x}dx$$
So $$I = -\frac{x^5\sec x}{x^5\sin x+5x^4\cos x}+\int \sec^2 x \cdot \frac{x^5\sin x+5x^4\cos x}{x^5\sin x+5x^4\cos x}dx$$
So $$I = -\frac{x^5\sec x}{x^5\sin x+5x^4\cos x}+\tan x+\mathcal{C}=\frac{5\sin x-x\cos x}{x\sin x+5\cos x}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the infinite sum $$S = {\frac{2}{5} \\+ \frac{2}{5}\cdot\frac{3}{7} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19}\cdot\frac{17}{23} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19}\cdot\frac{17}{23}\cdot\frac{21}{25} \\+ ...}$$ ?
I tried to figure out the general form of the terms, but this is unfamiliar to me. How should I proceed?
| If we group the series in units of pair, the first few pairs are
$$\require{cancel}
\begin{array}{rll}
p_0 :& \frac25 \left(1+\frac37\right) &= \frac25 \frac{10}{7} = \frac47\\
p_1 :& \frac{\color{red}{2}}{\cancel{5}}\frac37\frac{\cancel{5}}{\color{red}{11}}\left(1+\frac{9}{13}\right)
&= \color{red}{\frac{2}{11}}\frac{3}{7}\frac{22}{13} = \frac47\frac{3}{13}\\
p_2 :& \frac{\color{red}{2}}{\cancel{5}}\frac37\frac{\cancel{5}}{\cancel{11}}\frac{9}{13}\frac{\cancel{11}}{\color{red}{17}}
\left(1+\frac{15}{19}\right) &= \color{red}{\frac{2}{17}}\frac{3}{7}\frac{9}{13}\frac{34}{19} = \frac{4}{7}\frac{3}{13}\frac{9}{19}
\end{array}
$$
From this, we see the general pattern for pair $p_n$ is
$\displaystyle\;\frac{4}{7}\prod_{k=0}^{n-1}\frac{6k+3}{6k+13}$.
The series at hand becomes
$$\sum_{n=0}^\infty p_n
= \frac47\sum_{n=0}^\infty \prod_{k=0}^{n-1} \frac{6k+3}{6k+13}
= \frac47\sum_{n=0}^\infty \frac{\left(\frac36\right)_n}{\left(\frac{13}{6}\right)_n}
= \frac47 {}_2F_1\left(1,\frac36; \frac{13}{6}; 1 \right)
$$
where $(\alpha)_n = \prod\limits_{k=0}^{n-1}(\alpha+k)$ is the Pochhammer symbol and ${}_2F_1(a,b;c;z)$ is the Hypergeometric function.
WA knows how to evaluate RHS exactly. The result is a surprisely simple number $1$. This suggest we can prove it ourselves. We can simplify above expression using
Gamma function and Beta function.
For any $a, b > 0$, we have
$$\frac{(a)_n}{(b)_n} = \frac{\Gamma(b)}{\Gamma(a)}\frac{\Gamma(a+n)}{\Gamma(b+n)}
= \frac{\Gamma(b)}{\Gamma(a)\Gamma(b-a)}\frac{\Gamma(a+n)\Gamma(b-a)}{\Gamma(b+n)}
= \frac{B(a+n,b-a)}{B(a,b-a)}$$
Using the integral representation of beta function
$$B(a,b) = \int_0^1 (1-s)^{a-1} s^{b-1} ds$$
We find
$$\begin{align}
\sum_{n=0}^\infty \frac{(a)_n}{(b)_n}
&= \frac{1}{B(a,b-a)}\int_0^1 (1-s)^{b-a-1} \sum_{n=0}^\infty s^{a+n-1} ds\\
&= \frac{1}{B(a,b-a)}\int_0^1 (1-s)^{b-a-2} s^{a-1}ds
= \frac{B(a,b-a-1)}{B(a,b-a)}\\
&= \frac{\cancel{\Gamma(a)}\Gamma(b-a-1)}{\Gamma(b-1)}
\frac{\Gamma(b)}{\cancel{\Gamma(a)}\Gamma(b-a)} = \frac{\Gamma(b-a-1)}{\Gamma(b-a)}\frac{\Gamma(b)}{\Gamma(b-1)}\\
&= \frac{b-1}{b-a-1}
\end{align}
$$
Using this, we find the series at hand equals to
$$\sum_{n=0}^\infty p_n = \frac{4}{7}\left(\frac{\frac{13}{6}-1}{\frac{13}{6}-\frac{3}{6}-1}\right) = 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 2,
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} |
Minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$
Find the minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$
$\alpha=\sqrt{3-2\sqrt{2}} \implies -\alpha^2-3=2\sqrt{2} \implies (\frac{-\alpha^2-3}{2})^2-2=0$
$\implies (\frac{-\alpha^2-3}{2})^2-2=\frac{\alpha^4+6\alpha^2+9}{4}-2=\alpha^4+6\alpha^2+1=0$
So I believe $P=\alpha^4+6\alpha^2+1$ is a candidate for a minimal polynomial
Let $x=\alpha^2 \implies P=x^2+6x+1=(x+3)^2-8$
Could P be the minimal polynomial? It clearly has $\alpha$ as root, but I am not sure if there is another polynomial of lower degree also having $\alpha$ as a root
Would appreciate your guidance on this
| Hint: Note that $(1-\sqrt 2)^2=3-2\sqrt{2}$.
Therefore, $\alpha = \pm(1-\sqrt 2)$. Since $\alpha>0$, we must have $\alpha = -1+\sqrt2$.
Then $\alpha^2 = 3-2\sqrt{2} = -2\alpha +1$ and so the minimal polynomial of $\alpha$ is $x^2+2x-1$.
Clearly, $\mathbb Q(\alpha)=\mathbb Q(\sqrt 2)$, because $\alpha = -1+\sqrt2 \in \mathbb Q(\sqrt 2)$ and $\sqrt 2 = \alpha+1 \in \mathbb Q(\alpha)$.
| {
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Show that $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $ Show that if $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $
$\gcd(a,b)=d\Rightarrow\ d\mid a,b\Rightarrow\ \ d^2\mid a^2,b^2\Rightarrow\ d^2\mid\gcd(a^2,b^2)$.
But to complete the proof we must show that: $\gcd(a^2,b^2)\mid d^2$
How can I achieve this?
| By dividing $d$ on both sides and using $\gcd(ac,bc)=c\gcd(a,b)$, we may assume that $\gcd(a,b)=d=1$, so you only need to show that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$. Suppose that $p$ is a prime that divides $a^2$ and $b^2$, then $p$ divides $a$ and $b$, this contradict to the assumption $\gcd(a,b)=1$, so there is no prime divides $a^2$ and $b^2$, it follows that $\gcd(a^2,b^2)=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Doubt in solving $\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$ Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of
$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ Letting
$$\theta=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ taking $\sin$ both sides and using $sin(A+B)=sinAcosB+cosAsinB$ we get
$$\sin\theta=\sin\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\cos\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)+\cos\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\sin\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)$$ so
$$\sin\theta=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}$$
Now my doubt is will $\theta=\frac{\pi}{4}$ or $\theta=\frac{3\pi}{4}$ ?
My book has taken $\theta=\frac{3\pi}{4}$
| See $\sec^{-1}(\sqrt{5})+\csc^{-1}(\frac{\sqrt{10}}{3})=135$ so $x=45$ if we take $\theta=\pi/4$ then sum would be $\pi/2$ and not $\pi$ thus book has taken $\theta=3\pi/4$
| {
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"source": "stackexchange",
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Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| Hint. One may observe that
$$
6+2\sqrt{5}=(\sqrt{5})^2+2\sqrt{5}+1=(\sqrt{5}+1)^2.
$$
| {
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} |
Evaluation of $\int \frac{1-\sin x}{(1+\sin x)\cos x}dx$ Evaluate $$I=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}$$
I tried in the following way:
$$1-\sin x=1-\cos\left(\frac{\pi}{2}-x\right)=2 \sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
Similarly $$1+\sin x=2 \cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$
So
$$I=\int \tan^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx=\int \sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx-\int \sec x \:dx$$
If $$J=\int \sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx$$ Applying parts for $J$ we get
$$J=-2\sec x \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)+2\int \sec x \tan x \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)dx $$
But i am clueless from here
| Multiply both numerator and denominator by $\;1+\sin x\;$ , so you get ( observe that $\;\cos x=(1+\sin x)'\;$):
$$\int\frac{\cos x}{(1+\sin x)^2}dx=-\frac1{1+\sin x}+K$$
| {
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Let $a_{n}=(1-\frac{1}{\sqrt{2}})\ldots(1-\frac{1}{\sqrt{n+1}}), n \geq 1 $. Then $\lim_\limits{n \rightarrow \infty} a_{n} $ Options are (A) equals 1 (B) does not exist (C) equals $\frac{1}{\sqrt{\pi}}$ (D) 0.
I have multiplied each numerator with $\sqrt{n+1}+1$, $\forall n = 1$ to $\infty$ and then got $a_{n}=\frac{n!}{(2+\sqrt{2})(3+\sqrt{3})\ldots((n+1)+\sqrt{n+1})}$. So this must converge to 0 right since the fraction comes as $\frac{n!}{(n+1)!+\mathrm{something}}$ ?
| We can write
$$a_n = \frac{\sqrt{2}-1}{\sqrt{2}}\frac{\sqrt{3}-1}{\sqrt{3}}\cdots \frac{\sqrt{n+1}-1}{\sqrt{n+1}} = \frac{1}{\sqrt{2}(\sqrt{2}+1)}\frac{2}{\sqrt{3}(\sqrt{3}+1)}\cdots \frac{n}{\sqrt{n+1}(\sqrt{n+1}+1)}$$
For every positive integer $k$, $\sqrt{k}(\sqrt{k}+1) > \sqrt{k}\cdot\sqrt{k} = k$; therefore
$$a_n < \frac{1}{2}\frac{2}{3}\cdots \frac{n}{n+1}=\frac{1}{n+1}$$
Since $\lim \frac{1}{n+1} = 0$ and $a_n$ is positive, by the squeeze theorem we must have $\lim a_n = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Catalan numbers formula derivation I'm trying to follow a proof of the Catalan numbers being equal to $\frac{1}{n+1} {2n \choose n}$ from the recurrence relation $C_n = C_0C_{n-1}+C_1C_{n-2}+...+C_{n-2}C_{1}+C_{n-1}C_0$
Now it's seen that the generating function satisfies $xf^2-f+1=0$ so $f=\frac{1-\sqrt{1-4x}}{2x}$ since the other root has a pole at 0 but I'm struggling to see how $f=\frac{1-\sqrt{1-4x}}{2x}$ can be expanded to obtain the necessary power series
I've seen Wikipedia's proof but I don't see how ${ \frac{1}{2} \choose n} = \frac{(-1)^{n+1}}{4^n(2n-1)} {2n \choose n}$ nor how this gets lost into ${2n \choose n}$ via plugging y=-4x and putting it into the expression $f=\frac{1-\sqrt{1-4x}}{2x}$
| Hint: By binomial theorem, we have $$(1 + z)^\alpha = \sum\limits_{k \geq 0} \binom{\alpha}{k} z^k.$$
Use this for $\alpha = \frac{1}{2}$ on the square root term, and equate coefficients.
Edit: Here, we define $$\binom{\alpha}{k} = \frac{\alpha(\alpha - 1)\cdots(\alpha - (k-1))}{k!}.$$
Note that this agrees with our usual definition when $\alpha$ is an integer.
Edit 2: We have \begin{align}
\frac{1 + \sum_{n \geq 0} \binom{2n}{n} \frac{x^n}{2n - 1}}{2x} &= \frac{1 + (-1) + \sum_{n \geq 1} \binom{2n}{n} \frac{x^n}{2n - 1}}{2x} \\
&= \sum\limits_{n \geq 1}\binom{2n}{n} \frac{x^{n-1}}{(2n-1)2} \\
&= \sum\limits_{n \geq 0} \binom{2n+2}{n+1} \frac{x^n}{2(2n + 1)} \\
&= \sum\limits_{n \geq 0} \frac{(2n + 2)!}{(n+1)!(n+1)!} \frac{x^n}{2(2n + 1)} \\
&=\sum\limits_{n \geq 0} \frac{(2n + 2)(2n+1)(2n)!}{(n+1)(n+1)(n!)(n!)} \frac{x^n}{2(2n + 1)}\\
&= \sum\limits_{n \geq 0} \binom{2n}{n}\frac{x^n}{n + 1}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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Find the remainder when $169\times144^{25}$ is divided by $13^{4}$
Find the remainder when $169\times144^{25}$ is divided by $13^{4}$
Meanwhile I reduced it to
$\dfrac{13^{2}\times 144^{25}}{13^{4}} \\
=\dfrac{144^{25}}{13^{2}} \\
=\dfrac{12^{50}}{13^{2}} \\
$
and Euler function of $13^{2}$ is coming to be $156$
I look for a short and simple way.
| The problem boils down to computing $144^{25}\pmod{13^2}$, or $25^{25}\pmod{13^2}$, or $25^{26}\pmod{13^2}$, or $25^{13}\pmod{13^2}$. By the binomial theorem:
$$ 25^{13} = (26-1)^{13} = \sum_{k=0}^{13}\binom{13}{k}(-1)^{13-k}(26)^k $$
but for every $k\geq 2$ we have $26^k\equiv 0\pmod{13^2}$ and $\binom{13}{1}(26)^1\equiv 0\pmod{13^2}$ too, so
$$ 25^{13}\equiv -1\pmod{13^2},\qquad 25^{26}\equiv 1\pmod{13^2},$$
$$ 144^{25}\equiv -25^{25}\equiv 27\pmod{13^2} $$
and:
$$ 13^2\cdot 144^{25} \equiv \color{red}{4563}\pmod{13^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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order $a$ = 5, $a^3b = ba^3$. show that that $ab = ba$. Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.
Here is what I think:
We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. Now,
$a^3b = ba^3$ then $a^{-2}b = ba^{-2}$ then $ba^2 = a^2b$ then $ba^{-4} = a^{-4}b$ then $a^{-1}b = ba^{-1}$ then multiply on the left and right by $a$ to get $ba=ab$. Is this correct or is there a simpler approach?
| The point is that since the exponent $3$ is coprime to the order $5$ of $a$, the orders of $a$ and $a^{3}$ are the same, and thus so are the subgroups generated by each one. Therefore the centralizer of $b$, which contains $a^{3}$, also contains $a$.
| {
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"source": "stackexchange",
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Show that $U_{14}\cong U_{18}$? Because $|U_{14}|=|U_{18}|$ and both of them is cyclic and commutative, so i just need define a function $f : U_{14}\to U_{14}$ that f is homomorphism and bijective.
$f(1)=1, f(3)=5, f(5)=7, f(9)=11, f(11)=13, f(13)=17$
Is that function homomorphism and bijective? If not, what is the function?
| Since
\begin{align}
\varphi(14)
&=14\left(1-\frac{1}{2}\right)\left(1-\frac{1}{7}\right)=6 \\[6px]
\varphi(18)
&=18\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=6
\end{align}
we know the two groups can be isomorphic.
If we consider $3\in U_{14}$ and its powers, we get
$$
1=3^0,\quad
3=3^1,\quad
9=3^2,\quad
13=-1=3^3
$$
so the order of $3$ is six.
Similarly, if we consider $5$ in $U_{18}$, we get
$$
1=5^0,\quad
5=5^1,\quad
7=5^2,\quad
17=-1=5^3
$$
so the order of $5$ is six.
Since the two groups are cyclic of the same order, they're isomorphic. You find all isomorphisms by sending one generator of $U_{14}$ (for instance $3$) to any generator of $U_{18}$; the number is $\varphi(6)=2$, so the generators are $5$ and $5^{-1}=11$. The isomorphisms are
$$
f(1)=1, f(3)=5, f(9)=7, f(13)=17, f(11)=13, f(5)=11
$$
and
$$
g(1)=1, g(3)=11, g(9)=13, g(13)=17, g(11)=7, g(5)=5
$$
where $3^k$ is sent to $5^k$ or to $11^k$ respectively.
| {
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"source": "stackexchange",
"question_score": "2",
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Prove if $A$ is positive definite, then $A_{11}$,$A_{22}$ are positive definite. Here ${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\
{{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}}
\end{array}} \right)$.
I can prove $\mathbf{A}_{11}$ is positive definite but need help with $\mathbf{A}_{22}$. By Sylvester’s criterion, $|\mathbf{A}_{11} |>0$ since $|\mathbf{A}_{11} |$ is a principle minor. Thus $\mathbf{A}_{11}$ is invertible and we have
${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\
{{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{\mathbf{O}} \\
{{{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}}&{\mathbf{I}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\
{\mathbf{O}}&{\mathbf{S}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}} \\
{\mathbf{O}}&{\mathbf{I}}
\end{array}} \right)$
where ${\mathbf{S}} = {{\mathbf{A}}_{22}} - {{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}$ is the Schur complement. Since
${\left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{\mathbf{O}} \\
{{{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}}&{\mathbf{I}}
\end{array}} \right)^{\text{T}}} = \left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{{{\left( {{\mathbf{A}}_{11}^{\text{T}}} \right)}^{ - 1}}{\mathbf{A}}_{21}^{\text{T}}} \\
{\mathbf{O}}&{\mathbf{I}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\mathbf{I}}&{{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}} \\
{\mathbf{O}}&{\mathbf{I}}
\end{array}} \right)$
we have $\mathbf{A}$ is congruent with $\left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\
{\mathbf{O}}&{\mathbf{S}}
\end{array}} \right)$. If $$ is positive definite, then $\left( {\begin{array}{*{20}{c}}
{{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\
{\mathbf{O}}&{\mathbf{S}}
\end{array}} \right)$ is positive definite, and $_{11}$,$\mathbf{S}$ are both positive definite, since a block diagonal matrix is positive definite iff each diagonal block is positive definite.
Anyone can help continue to prove $_{22}$ is also positive definite? Thank you!
| Positive definite implies that the scalar product of $x^TAx\geq 0$ for all $x$, then you just have to take $x=e_2$ and you get your result
| {
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prove $\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$ Today I found the identity : $$\sum_{k=1}^n(-1)^{k+1}\cos^{2n+1}\left(\dfrac{k\pi}{2n+1}\right)=\frac{1}{2}-\frac{2n+1}{2^{2n+1}}$$.
How to prove or disprove this?
Thank you.
| Now I can solve it.
Let $a=e^{\frac{\pi}{2n+1}i}$. We will get $$\sum_{k=1}^n(-1)^k\cos^{2n+1}\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^{2n+1}}\sum_{k=1}^n(-1)^k(a^k+a^{-k})^{2n+1}$$
Using the binomial theorem and switching the sums yield
$$\frac{1}{2^{2n+1}}\sum_{k=1}^n(-1)^k(a^k+a^{-k})^{2n+1}=\frac{1}{2^{2n+1}}\sum_{r=0}^{2n+1}\binom{2n+1}{r}\sum_{k=1}^na^{2rk}$$.
For $r=0$ and $r=2n+1$, we have $\sum_{k=1}^na^{2rk}=n$ because $a^{2r}=1$.
For $r\in\{1,2,...,2n\}$, we have $\sum_{k=1}^na^{2rk}=\frac{a^{2r}(a^{2nr}-1)}{a^{2r}-1}$.
Hence,
$$\sum_{r=0}^{2n+1}\binom{2n+1}{r}\sum_{k=1}^na^{2rk}=2n+K$$
,Where $K=\sum_{r=1}^{2n}\binom{2n+1}{r}\frac{a^{2r}(a^{2nr}-1)}{a^{2r}-1}$.
Let $A_r=\binom{2n+1}{r}\frac{a^{2r}(a^{2nr}-1)}{a^{2r}-1}$. We obtain $A_{2n+1-r}=\binom{2n+1}{2n+1-r}\frac{a^{2n+1-r}((-1)^{2n+1-r}-a^{2n+1-r})}{a^{2({2n+1-r})}-1}$.
Since $a^{2n+1}=-1$, we can simplify $A_r=\binom{2n+1}{r}\frac{(-a)^r-a^{2r}}{a^{2r}-1}$ and $A_{2n+1-r}=\binom{2n+1}{r}\frac{1-(-a)^r}{a^{2r}-1}$.
We see that $A_{r}+A_{2n+1-r}=-\binom{2n+1}{r}$.
Since $K =\sum_{r=1}^{2n}A_r=\sum_{r=1}^{2n}A_{2n+1-r}$,
we obtain $K=\frac{1}{2}\sum_{r=1}^{2n}(A_{r}+A_{2n+1-r})=-\frac{1}{2}\sum_{r=1}^{2n}\binom{2n+1}{r}=-\frac{1}{2}(2^{2n+1}-2)=1-4^n$
Hence, $$\sum_{k=1}^n(-1)^k\cos^{2n+1}\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^{2n+1}}(2n+K)=\frac{2n+1}{2^{2n+1}}-\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of recursive sequence $n^2q_n=1+(n-1)^2q_{n-1}+2(n-2)q_{n-2}$ When looking at this riddle, I came across the following sequence for the frequency of sampled integers between 1 and $n$ in a without replacement/without neighbour sampling:
$$q_1=1,\quad q_2=1/2,\quad q_n=\frac{1}{n^2}+\frac{(n-1)^2}{n^2}q_{n-1}+\frac{2(n-2)}{n^2}q_{n-2}\quad (n>2)$$
and I wonder if there is a generic mathematical approach to computing an analytic solution for $$\lim_{n\to\infty} q_n$$Assessing the numerical limit with an R code like
q=rep(1,1e7)
for (n in 3:1e7) q[n]=(1+2*q[n-2]+(n-1)*q[n-1])/n
q[1e7]/1e7
led me to $0.432332...$ And a probabilistic reasoning indicates that $(n>1)$ $$\frac{1}{3}\le q_n\le \frac{1}{2}$$
| Let $a_n = (n+1)q_{n+1}$. We have $a_0 = a_1 = 1$ and
$$(n+1)a_n = 1 + na_{n-1} + 2a_{n-2}\quad\text{ for }\quad n > 1\tag{*1}$$
Let $f(z) = \sum\limits_{n=0}^\infty a_n z^n$, multiply $(*1)$ by $z^n$ and start to sum from $n = 2$, we get
$$\begin{align}
& \left(z\frac{d}{dz} + 1 \right)(f(z) - 1 - z) = \frac{z^2}{1-z} + \left(z\frac{d}{dz}\right)(z(f(z)-1)) + 2z^2f(z)\\
\iff &
zf' + f - 1 -2z = \frac{z^2}{1-z} + z^2 f' + zf - z + 2z^2 f\\
\iff &
z(1-z)f' + (1-z-2z^2)f = \frac{1}{1-z}
\end{align}
$$
Solving the ODE give us
$$f(z) = \frac{1-e^{-2z}}{2z(1-z)^2} = \frac{A}{(1-z)^2} + \frac{B}{(1-z)} + g(z)\tag{*2}$$
where
$\displaystyle\;\begin{cases}
A &= \frac{1-e^{-2}}{2}\\
B &= \frac{1-3e^{-2}}{2}\\
\end{cases}\;
$
and $g(z) = \sum\limits_{n=0}^\infty g_n z^n$
is some function analytic over all $\mathbb{C}$.
Expanding $(*2)$ as a power series and compare coefficients of $z^n$ on both sides, we get
$$(n+1)q_{n+1} = a_n = (n+1)A + B + g_n$$
Since $g(z)$ is entire, its coefficients of taylor expansion $g_n$ is bounded.
As a result,
$$
\begin{align}
\lim_{n\to\infty} q_{n+1}
&= A + \lim_{n\to\infty} \frac{B + g_n}{n+1} = A = \frac{1 - e^{-2}}{2}\\
&\approx 0.432332358381693654053000252513757798296184227045212059265
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
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For each pair of events A and B, find P(A|B) and P(B|A). I've got a simple problem here, but I just want to ensure that I'm not losing a simple concept.
Relevant equations:
$$P(A|B) = \frac{P(A \cap B)}{P(B)},$$
$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
Let $(S,P)$ be a sample space with $S = \{1, 2, 3, 4, 5\}$, and,
$P(1) = P(2) = 0.1$$P(3) = P(4) = 0.2$
$P(5) = 0.4$
For each pair of events $A$ and $B$, find $P(A|B)$ and $P(B|A)$.
a. $A = \{1, 2, 3\}$; $B = \{2, 3, 4\}$
b. $A = \{1, 2, 3\}$; $B = \{4, 5\}$
c. $A = \emptyset$; $B = \{2, 3, 4\}$
d. $A = \{1, 2, 3, 4, 5\}; B = \{4, 5\}$
If I am asked to find the probability of one of these sets, do I just add the values together? Is there any instance I would multiply them; such as when I'm finding $P(A \cap B)$ vs. just $P(A)$ or $P(B)$? Do I need to use Inclusion-Exclusion?
Here's what I found.
a. $P(A|B) = \frac{0.3}{0.5} = \frac{3}{5}$, $P(B|A) = \frac{0.3}{0.4} = \frac{3}{4}$
b. $P(A|B) = 0$, $P(B|A) = 0$
c. $P(A|B) = 0$, $P(B|A) = undefined$
d. $P(A|B) = \frac{0.3}{0.3} = 1$, $P(B|A) = \frac{0.3}{1} = \frac{3}{5}$
| Your approach is correct, as are your answers. However, your solutions in part (d) are incorrect.
Since $A = \{1, 2, 3, 4, 5\}$ is the entire sample space, $P(A) = 1$. Since $P(4) = 0.2$ and $P(5) = 0.4$ and $B = \{4, 5\}$, $P(B) = P(4) + P(5) = 0.2 + 0.4 = 0.6$. Since $B \subseteq A$, $P(A \cap B) = P(B)$. Hence,
\begin{align*}
P(A \mid B) & = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = \frac{0.6}{0.6} = 1\\
P(B \mid A) & = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)} = \frac{0.6}{1} = 0.6
\end{align*}
| {
"language": "en",
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"question_score": "5",
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| $$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(a+c)$$
$$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(2-c)+bc(2-a)+ca(2-b)$$
$$12=8+2(ab+bc+ca)-3abc$$
Now,
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$2(ab+bc+ca)=4-6=-2$$
Thus,
$$12=8-2-3abc$$
$$abc=-2$$
Let
$$f(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
$$f(x)=x^3-2x^2-x+2$$
The values of $a,b,c$ are the roots of $f(x)$.
By trial and error, clearly $1$ is a root.
Then, $$x^3-2x^2-x+2=(x-1)(x^2-x-2)=(x-1)(x+1)(x-2)$$
The values of $a,b,c$ are thus $-1,1,2$ in any order.
| {
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"source": "stackexchange",
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How to show that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is strictly increasing for $x \geq 1$? I am trying to prove that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is a strictly increasing function for $x \geq 1$.
I try to do this by showing that $f'(x)>0$ for all $x \geq 1$.
We have $$f'(x)= \arctan(x \sin^{2} \frac{1}{x}) + \frac{x}{1+x^{2} \sin^{4} \frac{1}{x}} \left(\frac{-2\cos\frac{1}{x} \sin \frac {1}{x}}{x} + \sin^{2} \frac{1}{x}\right),$$
so it's not clear if this quantity is positive.
Now I can show that $x \arctan \left(\frac{1}{x}\right)$ is strictly increasing (by computing its derivative and using simple trig inequalities to show that it is positive), and that $$x \sin^{2} \left(\frac{1}{x}\right)- \frac{1}{x}>0$$ So my aim is somehow to compare $f(x)$ with $x\arctan \frac{1}{x}$ and use the above inequality to prove that $f'(x)>0$.
But now I am stuck. Any suggestions as to how to proceed? Or any other approach to the problem would be appreciated!
| I have proved it as follows. Taking the expression for $f'(x)$, multiplying through by $(1+x^{2} \sin^{4} \frac{1}{x})$, we need to prove the following inequality, call it (a):
$$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \mbox{(a)}$$
Observe that since $2\sin\frac{1}{x} \cos \frac{1}{x} < 2x\sin \frac{1}{x}$ for $x>1$, if we prove that the L.H.S. side of (a) is greater than $2x \sin \frac{1}{x}$, we have automatically proved (a) itself.
Let $y = x \sin^{2}\frac{1}{x}$ and consider
$$g(y)=(1+y^{2}) \arctan y + y - 2y = (1+y^{2}) \arctan y - y$$
Now $g'(y) = 2y \arctan y>0$ if $y>0$ and thus $g(y)$ is strictly increasing; now $g(0)=0$ and hence $g(y)>0$ for $y>0$.
Thus (a) is indeed true, and so $f'(x)>0$ for $x>1$. Proof is complete.
Any opinions on this?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without calculating them determine whether $36^2+1$ and $154^2+1$ are prime and find the prime factors if not prime I know that $36^2 + 1$ is prime, $154^2 + 1$ is not, both are equal to $1 \bmod 4$. The prime divisors of $154^2 + 1$ should also be of the form $1 \bmod 4$. Tried showing this by Wilson's theorem however I don't feel like I'm getting any where.
| The square roots of $-1 \pmod 5$ are $2,3 \pmod 5$ but $154 \equiv 4 \pmod 5.$
The square roots of $-1 \pmod {13}$ are $5,8 \pmod {13}$ but $154 \equiv 11 \pmod {13}.$
The square roots of $-1 \pmod {17}$ are $4,13 \pmod {17}$ but $154 \equiv 1 \pmod {17}.$
The square roots of $-1 \pmod {29}$ are $12,17 \pmod {29}$ but $154 \equiv 9 \pmod {29}.$
The square roots of $-1 \pmod {37}$ are $6,31 \pmod {37}.$ $$154 \equiv 6 \pmod {37}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is the sum of reciprocals of all products from $2$ to $n-1$ always $0.5n-1$? I was looking up riddles for my math classes to work on for the end of the year and found the following riddle. http://mathriddles.williams.edu/?p=129
I followed the advice and started working with examples of small numbers and stumbled upon a pattern that I wanted to generalize.
$$\frac{1}{2}(1)=0.5$$
$$\frac{1}{3}\left(1+\frac{1}{2}\right)=0.5$$
$$\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2\cdot 3}\right)=0.5$$
$$\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 4}+\frac{1}{2\cdot 3\cdot 4}\right)=0.5$$
$$\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{2\cdot 5}+\frac{1}{3\cdot 4}+\frac{1}{3\cdot 5}+\frac{1}{4\cdot 5}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{2\cdot 3\cdot 5}+\frac{1}{2\cdot 4\cdot 5}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{2\cdot 3\cdot 4\cdot 5}\right)=0.5$$
If my pattern doesn't make sense, I'm taking $\frac{1}{n}$ and multiplying it by the sum of the reciprocals of all unique products for $2$ to $n-1$ and it comes out to 0.5 each time up to $n=7$ (I have not tested any higher values). Equivalently, if you multiply both sides by $n$ then subtract $1$, you see that all the reciprocals sum to $0.5n-1$.
I don't know where to start with generalizing this pattern as I have never seen explicit formulas for such sums, so I wanted to see if anyone knew if this was the case for all $n$ and how one could prove or disprove it.
| If you factor it as $\frac{1}{n}(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{n-1})$ it telescopes nicely.
| {
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How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? The solution is supposed to be derived using Inclusion-Exclusion.
Here is my attempt at a solution:
Let $A_0$= sequences where there are two $0$'s that appear in the sequence.
...
$A_{9}$=sequences where there are two $9$'s that appear in the sequence.
I want the intersection of $A_0^{'}A_1^{'}...A_9^{'}$= $N-S_1+S_2$ because you can only have at most two digits who are used exactly two times each in a $5$ digit sequence.
$N=10^5$, $S_1=10\cdot \binom{5}{2}\cdot[9+9\cdot 8 \cdot 7]$, and $S_2=10 \cdot 9 \cdot \binom{5}{4} \cdot8$.
The $S_1$ term comes from selecting which of the ten digits to use twice, selecting which two places those two digits take, and then either having the same digit used three times for the other three places, or having different digits used for the other three digits.
The $S_2$ term comes from selecting which two digits are used twice, selecting where those four digits go, and then having eight choices for the remaining spot.
So my answer becomes $10^5 -10 \cdot \binom{5}{2} \cdot [9+9 \cdot 8 \cdot 7]+10 \cdot 9 \cdot \binom{5}{4} \cdot 8$.
Am I doing this correctly?
| The structure of the analysis is the same as yours. We count the bad strings, where some digit appears exactly twice.
We first count the strings where say the digit $0$ appears exactly twice. Where the $0$'s are can be chosen in $\binom{5}{2}$ ways. For each of these ways, the remaining $3$ places can be filled in $9^3$ ways.
So our first estimate of the number of bad strings is $(10)\binom{5}{2}(9^3)$.
But we have double-counted, for example, the strings where $0$ and $1$ appear exactly twice. There are $\binom{5}{2}$ ways to choose where the $0$'s go, and for each there are $\binom{3}{2}$ ways to choose where the $1$'s go. The remaining spot can be filled in $8$ ways. Multiply by $\binom{10}{2}$ for the number of ways to choose the two digits that appear twice.
Putting things together, we find that the number of bad strings is
$$(10)\binom{5}{2}(9^3)-\binom{10}{2}\binom{5}{2}\binom{3}{2}(8).\tag{1}$$
Finally, subtract (1) from $10^5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard summation involving binomial and quadratic What is $$\sum \frac{2r^2-98r+1}{(100-r)({100\choose r})}$$ Where $r\in [1,99]$I have reduced it to $$\frac{(2r^2-98r+1)}{(100){99\choose r}}$$ what to do further? Partial fractions don't seem to help.
| Just to note that, in general $\sum_{r=0}^n \frac{P(r)}{\binom nr}$ can be expressed in terms of $\sum_{r=0}^n \frac1{\binom nr}$ and $n$ only.
Let's denote $c_{np} = \sum_{r=0}^n \frac{r^p}{\binom nr}$. Note that
$$
c_{n1} = \sum_{r=0}^n\frac r{\binom nr}
= \sum_{r=0}^n\frac{n-r}{\binom n{n-r}}
= \sum_{r=0}^n\frac{n-r}{\binom nr}
= nc_{n0}-c_{n1}
\implies c_{n1} = \frac n2c_{n0}
\tag1
$$
similarly we have
$$ c_{n3} = \frac32n c_{n2} - \frac14n^3 c_{n0}. \tag3 $$
From this all odd terms can be written in terms of the even terms.
These can also be generated using the telescoping sum method from https://math.stackexchange.com/a/1767041, i.e. if we can have the sum
$$ \sum_{r=0}^n \frac{p(r) - q(r+1)}{\binom nr} = p(0) - q(n+1) $$
requiring $\binom n{r-1}p(r)=\binom nrq(r) \implies rp(r) = (n+1-r)q(r)$, i.e. the following equation is satisfied:
$$ \sum_{r=0}^n \frac{(n+1-r)s(r)-(r+1)s(r+1)}{\binom nr} = (n+1)\bigl(s(0) - s(n+1)\bigr) $$
OP's problem is recovered with $s(r)=r$ (and $n=99$), which proves
$$ -2c_{n2} + (n-1)c_{n1} - c_{n0} = -(n+1)^2
\iff c_{n2} = \frac12(n+1)^2-\frac12(n+1)(n-2)c_{n0} \tag2 $$
Similarly using $s(r)=r^{p-1}$ we could find the relation of $c_{np}$ in terms of the lower terms.
Besides telescoping sum, we could also prove (2) using mathematical induction though it is pretty tedious. We will rely on
\begin{align}
c_{k+1,0} &= \sum_{r=0}^{k+1} \frac1{\binom{k+1}r} \\
&= 1 + \sum_{r=1}^{k+1} \frac1{\binom{k+1}r} \\
&= 1 + \sum_{r=0}^{k} \frac1{\binom{k+1}{r+1}} \\
&= 1 + \sum_{r=0}^{k} \frac1{\frac{k+1}{r+1}\binom kr}\\
&= 1 + \frac{c_{k1} + c_{k0}}{k+1} \\
&= 1 + \frac{\frac k2c_{k0} + c_{k0}}{k+1} \\
\Longleftrightarrow c_{k0} &= \frac{k+1}{\frac12k + 1}(c_{k+1,0}-1), \tag{a}
\end{align}
then
\begin{align}
c_{k+1,2} &= \sum_{r=0}^{k+1} \frac{r^2}{\binom{k+1}r} \\
&= \sum_{r=1}^{k+1} \frac{r^2}{\binom{k+1}r} \\
&= \sum_{r=0}^{k} \frac{(r+1)^2}{\binom{k+1}{r+1}} \\
&= \sum_{r=0}^{k} \frac{(r+1)^2}{\frac{k+1}{r+1}\binom kr} \\
&= \frac1{k+1} \sum_{r=0}^{k} \frac{(r+1)^3}{\binom kr} \\
&= \frac{c_{k3} + 3c_{k2} + 3c_{k1} + c_{k0}}{k+1} \\
&= \frac{(3+\frac32k)c_{k2} + (-\frac14k^3 + \frac32k + 1)c_{k0}}{k+1} &(\text{apply (1), (3)}) \\
&= \frac{(k+2)\bigl(6(k+1)^2 + (k^2+k-2)c_{k0}\bigr)}{8(k+1)} &(\text{apply (2) for }n=k) \\
&= \frac{(k+2)\left(6(k+1)^2 + (k^2+k-2)\frac{k+1}{\frac12k+1}(c_{k+1,0}-1)\right)}{8(k+1)} &(\text{apply (a)}) \\
&= \frac12(k+2)^2 + \frac14(k-1)(k+2)c_{k+1,0} &(\text{simplify yourself ☹})
\end{align}
Thus proves (2) for all $n\ge 0$.
There is no closed form of $c_{n0}$, though from Calculate sums of inverses of binomial coefficients we could eliminate the binomial term:
$$ c_{n0} = \sum_{r=0}^n \frac1{\binom nr} = \frac{n+1}{2^n}\sum_{r=0}^n \frac{2^r}{r+1}.\tag0 $$
| {
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"question_score": "3",
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Prove that $\frac{\pi^3}{48} \le \int_0^{\pi/2}\frac{x^2}{2-\sin(x)}\,dx \le \frac{\pi^3}{24}$ Is it possible to prove that
$$\frac{\pi^3}{48} \le \int_0^{\pi/2}\frac{x^2}{2-\sin(x)}\,dx \le \frac{\pi^3}{24}$$
without evaluating the integral?
| Without evaluating this integral:
Note $1\le 2-\sin x\le 2$ on $\;[0,\frac\pi2]$, hence $\;\dfrac{x^2}2\le \dfrac{x^2}{2-\sin x}\le x^2 $, and by the positivity of the integral:
$$\frac{\pi^3}{48}=\int_0^\tfrac\pi2\dfrac{x^2}2\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2\dfrac{x^2}{2-\sin x}\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2 x^2\,\mathrm d\mkern1mu x=\frac{\pi^3}{24}.$$
| {
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$1+x^4\leq 2(y-z)^2$ and switching of $x,y,z$ Find all triples of real numbers $x,y,z$ such that $1+x^4\leq 2(y-z)^2$, $1+y^4\leq 2(z-x)^2$, and $1+z^4\leq 2(x-y)^2$.
Beside $(1,0,-1)$ and permutations, I can't find any others. We cannot have two equal numbers. Maybe the inequality $1+a^4\geq 2a^2$ helps? It yields $|x-y|\geq |z|$ and analogous.
| The only solutions are $(-1,0,1)$ and permutations.
Subtracting $2x^2,$ $2y^2,$ and $2z^2$ respectively from each inequality gives
\begin{align*}
(1-x^2)^2&=1+x^4-2x^2\leq 2(y-z)^2-2x^2&=&-2(x-y+z)(x+y-z)\\
(1-y^2)^2&=1+y^4-2y^2\leq 2(z-x)^2-2y^2&=&-2(x+y-z)(-x+y+z)\\
(1-z^2)^2&=1+z^4-2z^2\leq 2(x-y)^2-2z^2&=&-2(-x+y+z)(x-y+z).
\end{align*}
Since the left-hand-sides are non-negative, these inequalities can be multiplied:
$$0\leq (1-x^2)^2(1-y^2)^2(1-z^2)^2\leq -8(-x+y+z)^2(x-y+z)^2(x+y-z)^2\leq 0.$$
So one of $-x+y+z,$ $x-y+z,$ or $x+y-z$ is zero. After a permutation of variables we have $x+y-z=0,$ which forces $(1-x^2)^2=(1-y^2)^2=0,$ so $x,y\in\{-1,1\}.$ We cannot have $x=y$ because $1\leq 1+z^4\leq 2(x-y)^2.$ So $x=-y\in\{-1,1\},$ which gives $z=x+y=0.$
| {
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Can anyone show that $\int_0^1\left[\frac{B-1}{1-x}+\frac{B(x^B-x)}{(1-x)(1-x^B)}\right]\sum _{n=1}^{\infty}x^{B^n-1}dx=\gamma$ Euler's constant
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{k}-\ln(n)=\gamma$$
It numerical value is $\gamma=0.5772166...$
Let B be valid for all integers, where $B\ge 2$
Show that,
$$\int_0^1\left[\frac{B-1}{1-x}+\frac{B(x^B-x)}{(1-x)(1-x^B)}\right]\sum_{n=1}^{\infty}x^{B^n-1}dx=\gamma$$
I will be greatly appreciated if anyone can prove it for me.
| Let we set $f_B(x)=\sum_{n\geq 1}x^{B^n}$. We have to compute:
$$ I_B=\int_{0}^{1}\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right)\,f_B(x)\,\frac{dx}{x} $$
but since $f_B(x^B) = x^B+f_B(x)$, by setting $x=z^B$ it follows that:
$$ I_B = \int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)\cdot(z^B+f_B(x))\cdot\frac{dz}{z}$$
and by using the digamma function we also have:
$$ \int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)z^{B-1}\,dz = -\gamma-\psi\left(\frac{1}{B}\right) $$
as well as:
$$ \int_{0}^{1}\left(\frac{B^{n+1}}{1-z^{B^n}}-\frac{B^n}{1-z^n}\right)z^{B^n-1}\,dz = -\gamma-\psi\left(\frac{1}{B}\right) $$
hence:
$$\begin{eqnarray*} I_B &=& -\gamma-\psi\left(\frac{1}{B}\right)+\int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)f_B(z)\,\frac{dz}{z}\end{eqnarray*}$$
and the claim follows by using induction and the fact that, in a neighbourhood of zero:
$$ \psi(x)=-\frac{1}{x}-\gamma+\zeta(2)x+O(x^2). $$
| {
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"url": "https://math.stackexchange.com/questions/1769607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of three consecutive cubes equals a perfect square I have found this problem in an old German textbook:
Find all sets of three consecutive integers such that the sum of their cubes is a perfect square.
We can write
$$S = (x-1)^3 + x^3 + (x+1)^3 = (x-1+x+1)((x-1)^2 - (x-1)(x-1) + (x+1)^2) + x^3$$
which reduces to
$$S = 3x(x^2 + 2).$$
If we set $x^2 + 2 = 3x$, we get
$$x^2 - 3x + 2 = 0 \iff (x-1)(x-2) = 0$$
and we thus obtain the solutions $(0,1,2)$ and $(1,2,3)$.
At first I conjectured that these are the only solutions, but I couldn't prove this. However, I was wrong: $(23,24,25)$ also satisfies the relationship.
It is worth noting that these are the only solutions for $x \leq 100000$. Can anyboy help me prove that these are the only ones? Or otherwise, help me find all other triples?
| Here is an approachable (but not exhaustive) algebraic route to obtain the solution $23,24,25$.
OP obtains $S=3x(x^2+2)$. Plainly, $x^2+2$ cannot itself be a square (there are no two integer squares which differ by $2$), so OP makes the (limiting) assumption that to obtain a square, you must set $3x=x^2+2$. It is also possible, however, that $3x=a\cdot b^2$ and $x^2+2=a\cdot c^2$, yielding a product $a^2b^2c^2$. Furthermore, since we have one explicit factor of $3$ to account for, one of $x$ or $x^2+2$ must contain another factor of $3$.
Let's assume here that $x=3k$. Then $3x(x^2+2)=9k(9k^2+2)$; and if that is a square, then so is $k(9k^2+2)$. Now $gcd(k,(9k^2+2))=1,2$. If it is $1$, then $k=d^2$ and $(9k^2+2)=9d^4+2$. But this is impossible because $9d^4+2$ is $2$ greater than a square, and therefore cannot be a square.
So we conclude that $k=2n$ and $k(9k^2+2)=2n(36n^2+2)=4n(18n^2+1)$. Again, since this last expression is a square, so is $n(18n^2+1)$. Here, $gcd(n,(18n^2+1))=1$, so we arrive at $n=d^2$ and $18n^2+1=18d^4+1$.
$18d^4+1=m^2$ has the solution $d=2,m=17$, from which $n=4,k=8,x=24$, yielding the overall solution $23,24,25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Evaluate a double integral.
To find:
$$I =\int\int_Rx(1+y^2)^{\frac{-1}{2}}dA$$
R is the region in the first quadrant enclosed by $y=x^2$, $y=4$, and $x=0$
$$y=x^2, y=4,x=0, (x= y^\frac{1}{2})$$
$$R=((x,y), 0 \le y \le 4, x^2 \le y \le 4)$$
i.e.
$$R=((x,y), 0 \le x \le 2, x^2 \le y \le 4)$$
$$I=\int_0^2\int_{x^2}^4x(1+y^2)^{\frac{-1}{2}}\,dy\,dx$$
Where to from here?
| \begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
I&=& \int_0^4\left(\int_0^{\sqrt{y}}\frac{x}{\sqrt{1+y^2}}dx\right)dy\\
&=& \int_0^4\frac{1}{\sqrt{1+y^2}}\left(\int_0^{\sqrt{y}}xdx\right)dy \\
&=& \frac{1}{2} \int_0^4\frac{y}{\sqrt{1+y^2}} dy\\
&=&\frac{1}{2}\left[\sqrt{1+y^2}\right]^4_0\\
&=&\frac{1}{2}\left(\sqrt{17}-1\right)
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$ So, my teacher gave us this to compute yesterday, and I'm completly confused on how should I proceed :
$$\frac{1^4 + 2012^4 +2013^4}{1^2 + 2012^2 + 2013^2}$$
I've tried several ways, but most of them are very long, for example I've simplified both numbers :
$2012^2 = (2 * 10^3)^2 + 12^2 + 24 * 10^3$
$2013^2 = (2 * 10^3)^2 + 13^2 + 26 * 10^3$
I can't see how this could help me solve this problem . So how how should I go with this kind of problems in general ?
| Hint The given triple $(1, 2012, 2013)$ has the special feature that $1 + 2012 = 2013$. This motivates writing the quotient as
$$\frac{a^4 + b^4 + (a + b)^4}{a^2 + b^2 + (a + b)^2}$$
for $a = 1, b = 2012$.
Additional hint Expanding the numerator, we can see that it factors as $2(a^2+ab+b^2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Inequalities involving the Roots of unity Let $\epsilon \ne 1$ be a nth root of unity. Then prove that,
$\bullet \ |1-\epsilon|\ge\frac{2}{n-1}$
$\bullet |\sin\frac{k}{n}| \ge \frac{1}{n-1}$
Here are my solutions:
$\bullet$ To prove $\implies \ |1-\epsilon| \ge\frac{2}{n-1}$
We have $$\sum_{i=1}^{n-1} {\epsilon}^i = -1 $$
$$\implies \sum_{i=1}^{n-1} ({\epsilon}^i -1 ) = -n $$
$$\implies (\epsilon - 1) (\epsilon^{n-2} + 2\epsilon^{n-3} + \cdots + (n-2)\epsilon + n-1) = -n $$
Taking absolute value on both sides
$$|\epsilon -1||\epsilon^{n-2} + 2\epsilon^{n-3} + \cdots + (n-2)\epsilon + n-1| = n$$
$$\implies |1-\epsilon| (|\epsilon^{n-2}| + 2 |\epsilon^{n-3}| + \cdots (n-2)|\epsilon| + n-1) \ge n$$
$$\implies |1-\epsilon| \cdot \frac{n(n-1)}{2} \ge n $$
$$\implies \boxed{|1-\epsilon| \ge \frac{2}{n-1}}$$
$\bullet$ To prove $\implies |\sin\frac{k\pi}{n}| \ge \frac{1}{n-1}$
Using $|1-\epsilon| \ge \frac{2}{n-1}$
we have,
$$| 1- \cos\frac{2k\pi}{n} -i\sin\frac{2k\pi}{n}| \ge \frac{2}{n-1}$$
$$\implies |2\sin^2\frac{k\pi}{n} - 2i\sin\frac{k\pi}{n} \cdot \cos\frac{k\pi}{n}| \ge \frac{2}{n-1}$$
$$\implies |\sin\frac{k\pi}{n}||\sin\frac{k\pi}{n} -i\cos\frac{k\pi}{n}| \ge \frac{1}{n-1}$$
$$\boxed{|\sin\frac{k\pi}{n}| \ge \frac{1}{n-1}}$$
Do you have any other solution?!!
| HINT:
Let $f(x)=\sin\left(\frac{\pi}{x}\right)-\frac{1}{x-1}$ for $x\ge 2$. Note that $f(0)=0$ and $\lim_{x\to \infty}f(x)=0$. The second derivative is given by
$$f''(x)=-\frac{2}{(x-1)^3}+\frac{2\pi\cos\left(\frac{\pi}{x}\right)}{x^3}-\frac{\pi^2\sin\left(\frac{\pi}{x}\right)}{x^4}$$
Show that $f''(x)>0$ for $x\ge 2$ and conclude from the concavity that $f(x)> 0$ for $x>2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying the result formula for depressed Cubic After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?
P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.
| \begin{align*}
&\left(10+\sqrt{108}\right)^{1/3}-\left(-10+\sqrt{108}\right)^{1/3}\\
&=\left(10+6\sqrt{3}\right)^{1/3}-\left(-10+6\sqrt{3}\right)^{1/3}\\
&=\left((1+\sqrt{3})^3\right)^{1/3}-\left((\sqrt{3}-1)^3\right)^{1/3}\\
&=(1+\sqrt{3})-(\sqrt{3}-1)\\
&=2
\end{align*}
The formula:
$$(A+B\sqrt{3})^3=A^3+3\sqrt{3}A^2B+9B^2A+3\sqrt{3}B^3$$ is useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Since
$$
\begin{align}
\frac{x^2+3}{x^6(x^2+1)}
&=\frac1{x^6}+\frac2{x^6(x^2+1)}\\
&=\frac1{x^6}+\frac{2(x^6+1)}{x^6(x^2+1)}-\frac2{x^2+1}\\
&=\frac1{x^6}+\frac{2(x^4-x^2+1)}{x^6}-\frac2{x^2+1}\\
&=\frac3{x^6}-\frac2{x^4}+\frac2{x^2}-\frac2{x^2+1}
\end{align}
$$
we have
$$
\int\frac{x^2+3}{x^6(x^2+1)}\,\mathrm{d}x
=C-\frac3{5x^5}+\frac2{3x^3}-\frac2x-2\arctan(x)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
| I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5a^2}\geq \frac{a+b}{18}$$
Proof:
We have with $x=\frac{a}{b}$ :
$$\frac{x^3}{13x^2+5}+\frac{1}{13+5x^2}\geq \frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)\geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}+\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{9}$$
If we have $\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq\sum_{cyc}\frac{a^3}{13a^2+5c^2}$
We have :
$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq \frac{a+b+c}{18}$$
But also
$$\frac{(a-\epsilon)^3}{13(a-\epsilon)^2+5b^2}+\frac{(b)^3}{13(b)^2+5(c+\epsilon)^2}+\frac{(c+\epsilon)^3}{13(c+\epsilon)^2+5(a-\epsilon)^2}\geq \frac{a+b+c}{18}$$
If we put $a\geq c $ and $\epsilon=a-c$
We finally obtain :
$$\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{18}$$
So all the cases are here so it's proved !
| {
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"url": "https://math.stackexchange.com/questions/1777075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 5,
"answer_id": 4
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How to solve $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } }$ for $\cos(\alpha)$ using half-angle formula. I have $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } } $
How can I get $\cos(\alpha ) $ from this?
I know this identitiy.
$\cos(\frac{\alpha }{2} )=\sqrt{\frac{1+\cos(\alpha ) }{2} } $
But just cant figure out, how to do it.
| If we have that $a,b$ are given then we can use the above answers to say that
$$\sqrt{\frac{1+\cos(\alpha)}{2}}=\frac{a}{\sqrt{a^2+b^2}}$$
Then
$$\cos(\alpha) = \frac{2a^2}{a^2+b^2}-1$$
If instad you have $\sin(\alpha/2) = \frac{a}{\sqrt{a^2+b^2}}$ you can use that $\sin(\alpha/2) = \sqrt{1 - \cos^2(\alpha/2)} = \sqrt{1 - \frac{1+\cos(\alpha)}{2}} = \frac{a}{\sqrt{a^2+b^2}}$ and you solve again for $\cos(\alpha)$.
If you want to exchange $\cos(\alpha)$ for $\sin(\alpha)$ it is always possible using $\cos^2+\sin^2=1$ and the identities that came from that. I think that in the comments was that what you pointed out.
$\cos^2(2\alpha) = \cos^2(\alpha)-\sin^2(\alpha) = 2\cos^2(\alpha)-1$
$\sin^2(2\alpha) = 4\sin^2(\alpha)\cos^2(\alpha)=4\sin^2(\alpha)(1 - \sin^2(\alpha))$
So what possibly be other interpretation of your question is that given
$$\cos(\alpha/2) = \frac{a}{\sqrt{a^2+b^2}}$$
what is $\sin(\alpha)$? We know that, because of this we have that
$$\cos(\alpha/2) = \frac{b}{\sqrt{a^2+b^2}}$$
So that $\cos^2+\sin^2 = 1$ and then we can use that
$$\sin(\alpha) = 2\cos(\alpha/2)\sin(\alpha/2) = \frac{2ab}{a^2+b^2}$$
And the answers quoted respect the fundamental equation of trigonometry.
| {
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"url": "https://math.stackexchange.com/questions/1777306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.
I want to know if my attempt is correct.
First $a^{13} \equiv (a^3)^4 \cdot a \equiv a^4 \cdot a \equiv a^3 \cdot a^2 \equiv a \cdot a^2 \equiv a^3 \equiv a \pmod 3$.
Second, $(a^7)^2 \cdot a^{-1} \equiv a^2 \cdot a^{-1} \equiv a \pmod 7$.
Third, $a^{12} \equiv 1 \pmod{13} \implies a^{13} \equiv a \pmod{13}$.
So, $3, 7, 13 | a^{13} - a$. Since they are relatively prime in pairs, $$3 \cdot 7 \cdot 13 | a^{13} - a \implies a^{13} \equiv a \pmod{ 3 \cdot 7 \cdot 13}$$
Is it correct? Could you suggest an easier or a better proof?
Thanks.
| By the Chinese remainder theorem, we have that $ R = \mathbb{Z}/(3 \cdot 7 \cdot 13)\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z} $, so the equation $ X^{13} - X = 0 $ holds for $ R $ if it holds for each factor ring on the right. On the other hand, we know by Fermat's little theorem that this equation does hold for each ring on the right, and the result follows. (Note that $ 2, 6 | 12 $, which is why the equation holds.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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