Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$
| Use the fact that $ \displaystyle \frac{1}{m+1} = \int^1_0 x^m dx$ to get $$ \sum_{n=1}^{\infty} \int^1_0 \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$
$$ = \int^1_0 \sum_{n=1}^{\infty} \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$
$$= \int^1_0 \frac{ x+1-2x^2}{1-x^3} dx= \int^1_0 \frac{2x+1}{x^2+x+1} dx = \log (x^2+x+1) \biggr|^1_0=\log 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/197595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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finding the sum of the absolute values for the roots How to find the sum of the absolute values for the roots of this equation:
$$x^4-4x^3-4x^2+16x-8=0$$
| Hint $\ $ Suppose that $\rm\ g(x\!+\!1)\, =\, f(x^2),\:$ and that $\rm\:f(x)\:$ has roots $\rm\:0< s < 1 < r.\:$
Then $\rm\:g\:$ has roots $\rm\:1\!-\!\sqrt{r}\, <\, 0\, <\, 1+\sqrt{r},\: 1\pm\sqrt{s},\:$ with absolute sum $\rm\ 2 + 2\,\sqrt{r}.$
In your case $\rm\:f(x) =\, x^2 - 10\,x + 1\:$ has roots $\rm\: 0 < 5 -2 \sqrt{6} < 1 < 5 + 2\sqrt{6}\:$ therefore, by above, we deduce that $\rm\:g\:$ has absolute root sum $= 2 + 2\sqrt{5+2\sqrt{6}}\, =\, 2\,(1 + \sqrt{2} + \sqrt{3}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200298",
"timestamp": "2023-03-29T00:00:00",
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What is the coefficient of the $x^3$ term in the expansion of $(x^2+x-5)^7$ (See details)? I fail to see a simple way to answer this.
As such, this is my long winded approach:
Using the multinomial theorem,
$$(x_1 + x_2 + \cdots + x_m)^n
= \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
\prod_{1\le t\le m}x_{t}^{k_{t}},$$
we have the specific parameters $m=3$, $n=7$, $x_1=x^2$, $x_2=x$, and $x_3=-5$.
Via the theorem,
$$(x^2+x-5)^7=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}\prod_{1 \le t \le 3}x_{t}^{k_t}=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}x^{2k_1}x^{k_2}(-5)^{k_3}.$$
The coefficient of the $x^3$ term is the summation of the multinomial coefficient multiplied by the $(-5)^{k_3}$ factor evaluated at all the solutions of the equation $2k_1+k_2=3$ where $0\le k_1\le 7$ and $0 \le k_2 \le 7$.
Those values are $(k_1,k_2)=\{(1,1),(0,3)\}.$ Given that $k_1+k_2+k_3=7$, $k_3$ are respectively $5$ and $4$.
Hence, the coefficient of the $x^3$ is
$${7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4.$$
Given that the definition of the multinomial coefficient is
$${n \choose k_1,k_2,\ldots ,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!},$$
$$
\begin{align}
{7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4&=\frac{7!(-5)^5}{1!1!5!}+\frac{7!(-5)^4}{0!3!4!}\\
&=7\cdot 6(-5)^5+\frac{7\cdot 6\cdot 5(-5)^4}{3!}\\
&=-109375
\end{align}$$
This could be atrociously wrong. Either way, I am desperate for a much simpler process. This is ridiculous to do in a timed testing environment without the formulas given.
I would like to see a very simple but also very general way of arriving at the correct answer (preferably without college methods, but I am open to any methods).
What says you, Math.SE?
| This is pretty much the same as joriki's approach, but the emphasis here is on a slightly different idea. Write
$$
[x^2+x+(-5)]^7=\sum_{k=0}^7{7\choose k}(x^2+x)^k(-5)^{7-k}\,.
$$
Now note that for $k\ge 4$, the power $(x^2+x)^k$ starts with $x^4$ or higher power, and for $k<2$, it ends with $x^2$ or lower power. Thus only 2 terms ($k=2$ and $k=3$) have $x^3$ in them. Looking at what the corresponding coefficients are, we get
$$
{7\choose 3}(-5)^4+{7\choose 2}\cdot 2\cdot (-5)^5
$$
The rest of the computation is the same as in joriki's answer.
| {
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I don't understand $\sqrt{-9i}$. I try to visualise it on a graph, where x is real numbers and y is the imaginary numbers.
$\sqrt{9} = (3,0)$ and $(-3,0)$.
$\sqrt{-9} = \sqrt{-1} \times \sqrt{9} = (0,3) $ and $(0,-3)$.
$\sqrt{9i}$ =
$\sqrt{-9i}$ =
Basically, I have some trouble representing the numbers visually on the graph.
Thanks.
| First: one can (hear) talk about a square root. We might say that a number $a$ is a square root of $b$ is $a^2 = b$. In this sense both $3$ and $-3$ are square roots of $9$.
Second: Most of the time (IMO) when one comes across the radical sign $\sqrt{}$, then one is thinking about the square root also known as the principal square root. For the non-negative real numbers, the square root of $b\geq 0$ is then defined to the the unique positive number $a$ such that $a^2 = b$. Hence we say that the square root of $9$ is equal to $3$ and we write $\sqrt{9} = 3$. (Granted, one might consider the radical sign as denoting the set consisting of all the square roots of a number). Note that for this setup we think og $\sqrt{}$ as a function from $[0,\infty) \to [0,\infty)$.
For complex numbers we also can talk about a square root or the (principal) square root. For the square root of a complex number $z = re^{i\theta}$, with $r\geq 0$ and $-\pi < \theta \leq \pi$ one usually defined the square root as: $\sqrt{re^{i\theta}} = \sqrt{r}e^{i\theta/2}$. So with this definition we have $$\begin{align}
\sqrt{i} &= \sqrt{e^{i\pi/2}} = e^{i\pi/4} = \frac{1}{\sqrt{2}}(1+i) = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \\
\sqrt{-i} &= \sqrt{e^{i(-\pi/2)}} = e^{-i\pi/4} = -\frac{1}{\sqrt{2}}(1+i).
\end{align}
$$
And you would then get for example $\sqrt{9i} = \frac{3}{\sqrt{2}}(1+i)$.
Graphically you would then represent $\sqrt{9i}$ as the point $(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}})$
Note that with this definition certain familiar rules don't hold. You for example do not have that $\sqrt{ab} = \sqrt{w}\sqrt{z}$ for all complex numbers $w$ and $z$. If you did, then you would have
$$
\begin{align}
1 &= \sqrt{1} \\
&= \sqrt{(-1)(-1)} \\
&= \sqrt{-1}\sqrt{-1}\\
&= i\cdot i\\
&= -1.
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim\limits_{n \to\infty} a_n =\lim\limits_{n \to\infty} \frac{2n-1}{3n+2} = \frac{2}{3}$ Prove: $\lim\limits_{n \to\infty} a_n =\lim\limits_{n \to\infty} \dfrac{2n-1}{3n+2} = \dfrac{2}{3}$ using the definition of the limit.
This is what I have so far:
*
*Let $\epsilon > 0$ and take $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$. This is my reasoning:
Solve: $\left|\dfrac{2n-1}{3n+2} - \dfrac{2}{3}\right| < \epsilon$
We get: $\left|\dfrac{-7}{9n+6}\right| < \epsilon$ $\iff$ $\left|\dfrac{-7}{9n+6}\right| < \left|\dfrac{7}{9n}\right| < \epsilon$
Take $n > 1$ so we can drop the absolute value sign and Solve for $n$:
$\dfrac{7}{9\epsilon} < n$
So where would I go from here? Also, does taking $n > 1$ mean we have to do $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$?
| $\left|\dfrac{2n-3}{3n+2} - \dfrac{2}{3}\right| < \epsilon$
$\implies\left|\dfrac{-13}{9n+6}\right| < \epsilon$ $ \text{ and } $$\left|\dfrac{-13}{9n+6}\right| < \left|\dfrac{13}{9n}\right| < \epsilon$
Thus,
Given $\epsilon\gt 0,$
For $\forall n\gt \dfrac{13}{9\epsilon}, \left|\dfrac{2n-3}{3n+2} - \dfrac{2}{3}\right| < \epsilon $
Hence the limit of $\dfrac{2n-3}{3n+2} $ is $\dfrac{2}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove/disprove $\sum x^2y^2 \ge \sum x^3y$ Prove $x^2y^2+y^2z^2+z^2x^2 \ge$ or $\le x^3y+y^3z+z^3x$ where $x,y,z$ are real numbers.
Actually, I have reached here from this problem:
Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
| If you set x = 0 and y = 1, then $x^2y^2+y^2z^2+z^2x^2$ becomes $z^2$ and $x^3y+y^3z+z^3x$ becomes $z$. Clearly the relation between them depends on the value of $z$. As for the original inequality, you can prove that $x^4 + y^4 + z^4 +x^2y^2+y^2z^2+z^2x^2\ge 2(x^3y+y^3z+z^3x)$ and $x^4 + y^4 + z^4 \ge x^3y+y^3z+z^3x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of digits and product of digits is equal (3 digit number) My child got a question in school (grade) that is:
Find biggest and smallest 3 digits number, which has sum of it's digits equal to product of those digits.
Help please since I cannot explain my child this question.
Example would be:
$1+2+3=1\cdot 2\cdot 3$
$3+2+1=3\cdot 2\cdot 1$
Numbers 123 and 321
But this is just example of this, how to solve it as a problem.
| Clearly $0$ cannot be one of the digits: if it were, the product would be $0$, and the sum would not. We can’t have two $1$’s: if the remaining digit is $c$, the product is $c$, but the sum is $c+2$. This means that the smallest two digits must be at least $1$ and $2$.
It’s easy to check that the digits cannot all be the same: the only solutions of the equation $a^3=3a$ are $a=0$ and $=\pm\sqrt3$, none of which is possible in this problem. Thus, if the digits are $a,b$, and $c$, with $a\le b\le c$, we must have $a<c$.
We already know that $b\ge 2$; suppose that $b\ge 3$; then $abc\ge bc\ge 3c>a+b+c$, since we know that $a<c$. Thus, there is no solution with $b\ge 3$. There is also no solution with $a=b=2$: that would require $4c=4+c$, $3c=4$ and $c=4/3$, which is impossible. Thus, every solution must have $a=1$ and $b=2$. This means that $2c=3+c$, whose only solution is $c=3$.
Thus, the three digits must be $1,2$, and $3$. The smallest number that can be formed from them is $123$, and the largest is $321$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/202679",
"timestamp": "2023-03-29T00:00:00",
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Trigonometric proof If $4 \tan(\alpha - \beta) = 3 \tan \alpha $, then prove that
$$\tan \beta = \frac{\sin(2 \alpha)}{7 + \cos(2 \alpha)}$$
This is not homework and I've tried everything so I would just like a straight answer thank you in advance.
| Recall that $$\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$$
Hence $$\tan\alpha-4\tan\beta=3\tan^2\alpha\tan\beta \hspace{8pt}\Rightarrow\hspace{8pt}\tan\alpha=(3\tan^2\alpha+4)\tan\beta\hspace{8pt}\Rightarrow$$
$$\begin{align*}\tan\beta=&\frac{\tan\alpha}{3\tan^2\alpha+4}=\frac{\frac{\sin\alpha}{\cos\alpha}}{3\frac{\sin^2\alpha}{\cos^2\alpha}+4}=\frac{\sin\alpha\cos\alpha}{3\sin^2\alpha+4\cos^2\alpha}\\
=& \frac{\frac12\sin(2\alpha)}{3+\cos^2\alpha}=\frac{\frac12\sin(2\alpha)}{3+\frac{\cos(2\alpha)+1}{2}}=\frac{\sin(2\alpha)}{7+\cos(2\alpha)}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/206421",
"timestamp": "2023-03-29T00:00:00",
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How can I calculate $\int_{0}^{2\pi}\sqrt{2-\sin(2t) }dt$? I'm trying to calculate the perimeter of a curve $C$ in ${\mathbb R}^3$ where $C$ is given by
$$
\begin{cases}
x^2+y^2=1\\
x+y+z=1
\end{cases}
$$
Things boil down to calculating $$
\int_{0}^{2\pi}\sqrt{2-\sin(2t) }dt$$
using $\vec{r}(t)=(\cos t,\sin t,1-\sin t-\cos t)$. Is this an elliptic integral so that one can not find its value? Is there any other way to find the perimeter?
| Using symmetry:
$$
\int_0^{2\pi} \sqrt{2 - \sin(2t)} \mathrm{d}t = 2 \int_0^{\pi} \sqrt{2 - \sin(2t)} \mathrm{d}t = \int_0^{2 \pi} \sqrt{2 - \sin(t)} \mathrm{d}t
$$
Now we use $\sin(t) = 1 - 2 \sin^2\left(\frac{\pi}{4} - \frac{t}{2} \right)$:
$$
\int_0^{2 \pi} \sqrt{2 - \sin(t)} \mathrm{d}t = \int_0^{2\pi} \sqrt{1+2 \sin^2\left(\frac{\pi}{4} - \frac{t}{2} \right)} \mathrm{d} t = 2\int_{-\tfrac{\pi}{4}}^{\tfrac{3 \pi}{4}} \sqrt{1+2 \sin^2(u)} \mathrm{d} u
$$
The anti-derivative is not elementary (uses elliptic integral of the second kind):
$$
\int \sqrt{1 - m \sin^2 \phi} \, \mathrm{d} \phi = E\left(\phi|m\right) + C
$$
Thus the parameter of interest equals
$$
2 \left( E\left(\left.\frac{3 \pi}{4}\right|-2\right) - E\left(\left.-\frac{\pi}{4}\right|-2\right)\right) = 2 \left( E\left(\left.\frac{3 \pi}{4}\right|-2\right) + E\left(\left.\frac{\pi}{4}\right|-2\right)\right)
$$
Inspired by Mhenni's result:
$$
2\int_{-\tfrac{\pi}{4}}^{\tfrac{3 \pi}{4}} \sqrt{1+2 \sin^2(u)} \mathrm{d} u = 2\int_{0}^{\pi} \sqrt{1+2 \sin^2(u)} \mathrm{d} u = 4 \int_0^{\frac{\pi}{2}} \sqrt{1+2 \sin^2(u)} \mathrm{d} u = 4 E(-2)
$$
where $E(m)$ is the complete elliptic integral of the second kind.
Numerical verification in Mathematica:
In[67]:= NIntegrate[Sqrt[2 - Sin[2 t]], {t, 0, 2 Pi},
WorkingPrecision -> 20]
Out[67]= 8.7377525709848047416
In[68]:= N[4 (EllipticE[-2]), 20]
Out[68]= 8.7377525709848047416
| {
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"timestamp": "2023-03-29T00:00:00",
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Sides of triangle and an altitude Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Let $h$ be the altitude drawn on the side of length $a$ Then is $a^2 + 4h^2 - (b+c)^2$ always negative ?
| Let's suppose the triangle is acute
We have
$$
a_1^2 + h^2 = b^2\\
a_2^2 + h^2 = c^2
$$
and by Cauchy-Schwarz inequality
$$
a_1\cdot a_2 + h\cdot h \leq \sqrt{a_1^2 + h^2}\sqrt{a_2^2 + h^2} = bc
$$
Summing the above relations we get
$$
a^2 + 4h^2 = a_1^2 + a_2^2 + 2a_1 a_2 + 4h^2 \leq b^2 + c^2 +2bc = (b + c)^2
$$
A similar reasoning shows the inequality is true even for obtuse triangles.
| {
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Find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$ I'm trying to find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$, as part of a basic number systems course.
I'm rather lost with this question. For $10!$ I tried writing the terms out and just extracting powers of $2$ manually, getting $2^8$ as the highest powers of $2$, with $10! = (2^8)(14175)$ as the result.
I'm fairly confident that the answer is correct (although I'm not sure, so confirmation of that would be great!), but this method is rather crude for larger numbers, so I suspect that it isn't the right way to do it.
If anyone can point me in the right direction I would be very grateful.
| One way to do it is to count the ammount of numbers divisible by $2,4,8,...$ between $1$ and $n$. For example, look at $40!$:
There are $\left\lfloor\frac{40}{32}\right\rfloor=1$ numbers divisible by $32$ (between $1$ and $40$) and each contributes $2^5$.
There are $\left\lfloor\frac{40}{16}\right\rfloor=2$ numbers divisible by $16$ and each contributes $2^4$.
There are $\left\lfloor\frac{40}{8}\right\rfloor=5$ numbers divisible by $8$ and each contributes $2^3$.
There are $\left\lfloor\frac{40}{4}\right\rfloor=10$ numbers divisible by $4$ and each contributes $2^2$.
There are $\left\lfloor\frac{40}{2}\right\rfloor=20$ numbers divisible by $2$ and each contributes $2^1$.
Hence the biggest power that will divide $40!$ is $1+2+5+10+20=38$ (i.e. $(2^5)^1(2^4)^2(2^3)^5(2^2)^{10}(2^1)^{20}=2^{38}$), since that way a number divisible by $32$ is counted $5$ times, a number divisible by $16$ is counted $4$ times and so on.
Applying the same method to $10!$ you get: $1$ number divisible by $8$, $2$ numbers divisible by $4$, $5$ numbers divisible by $2$. So total of $1+2+5=8$. Hence $2^8$ is the highest power.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a+b=c \times d$ and $a\times b = c + d$ There is a 'nice' relationship between the integers (1,5) and (2,3) as
$$1+5=2 \times 3;$$
$$1\times 5 = 2 + 3.$$
So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$
$$a\times b = c + d.$$
To find this, $a, b, c, d$ must satisfy $$(a+1)(b+1)=(c+1)(d+1).$$ However, this condition is only necessary but not sufficient.
Any idea?
| The following approach is based on the idea that, for positive integers $x,y$, the product $xy$ typically exceeds the sum $x+y$. We can apply this to show that in the problem of the question, either $a=b=c=d=2$ or else at least one of $a,b,c,d$ is 1; in this case it's easy to arrive at the only remaining solution $1,5;2,3$.
Suppose for two positive integers $x$ and $y$, that $x,y \ge k$ for some fixed $k \ge 2$. Then from $(x-k)(y-k) \ge 0$ we have, after expanding and rearranging, that $xy \ge k(x+y-k)$.
Now assume that each of $a,b,c,d$ is at least $k$. Of course we also use the assumptions $ab=c+d,cd=a+b$ of the question. We then have $ab \ge k(a+b-k)=k(cd-k)$, and also $cd \ge k(c+d-k)=k(ab-k)$. Putting these together we have
$ ab \ge k(k(ab-k)-k)=k^2ab-(k^3+k^2)$.
This inequality simplified is
$$ab \le \frac{k^2}{k-1}.$$
Now if $k=2$ here we obtain $ab \le 4$ which with the assumptions $a,b \ge 2$ leads to $a=b=2$, and (by symmetry or by the initial relations) also $c=d=2$.
If $k=3$ then we obtain $ab \le 9/2 = 4.5$, but this cannot hold since we're assuming $a,b \ge 3$ so that in fact $ab \ge 9$.
So by considering how products are typically larger than sums, we have shown that, except for the solution $a=b=c=d=2$, one of the values $a,b,c,d$ must be 1. Putting say $a=1$ into the two equations, one easily gets $a=1,b=5$ and that $c,d$ are $2,3$ insome order.
| {
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Line integral over ellipse in first quadrant
Evaluate $ \int_{C} xy\,ds $ where C is the arc of the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ in the first quadrant.
Let $x = a\cos t$ and $ y= b\sin t$ and use a substitution of $ u = a^2 \sin^2t + b^2\cos^2t $ to simply the expression under the sqrt root when dealing with $ ds$.
I eventually get to $$\frac{ab}{2(a^2-b^2)} [\frac{2}{3} \sqrt{(a^2\sin^2t + b^2 \cos^2t)^3}]$$evaluated between $\pi/2$ and $0$.
Should I take $\sqrt{a^6} = a^3$ here? (my thoughts being that $ a>0 $ in first quadrant)
| The correct setup is
$$\int_{0}^{2 \pi} ab\cos(t) \sin(t) \sqrt{a^2 \cos^2t+b^2\sin^2t}dt$$
As you hypothesized. Let $u = a^2 \cos^2t+b^2\sin^2t$ and $du = 2b^2\cos(t)\sin(t) - 2a^2\sin(t) \cos(t) dt$
Factor out $\cos(t) \sin(t)$, this will cancel out the same factor upstairs and in the end you should end up with
$$\int \dfrac{ab \sqrt{u} du}{2b^2 - 2a^2}$$
Factor out the constants and integrate. Then do the definite integral (which is very similar).
I leave the rest to you
| {
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} |
limit of convergent series What is the limit of $U_{n+1} = \dfrac{2U_n + 3}{U_n + 2}$ and $U_0 = 1$?
I need the detail, and another way than using the solution of $f(x)=x$, as $f(x) = \frac{2x+3}{x+2}$ because I can't show that $f(I) \subseteq I$ as $I = ]-\infty; -2[~\cup~ ]-2; +\infty[$.
| First note that
$$
\begin{align}
U_{n+1}
&=\frac{2U_n+3}{U_n+2}\\
&=2-\frac1{U_n+2}\tag{1}
\end{align}
$$
Therefore, if $U_n\ge1$, then $\frac53\le U_{n+1}\lt2$.
$$
\begin{align}
U_{n+1}-U_n
&=\frac1{U_{n-1}+2}-\frac1{U_n+2}\\
&=\frac{U_n-U_{n-1}}{(U_n+2)(U_{n-1}+2)}\tag{2}
\end{align}
$$
Since $(U_n+2)(U_{n-1}+2)\ge11$, $(2)$ says that the sequence is monotonic. Thus, because the sequence is monotonic and bounded above and below, it converges to some limit, $U$.
$(1)$ says that
$$
U=\frac{2U+3}{U+2}\Rightarrow U^2=3\tag{3}
$$
Since $U\ge0$, we get that $U=\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/227723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How to get the center and the axes of an ellipse Get the center and the semimajor/semiminor axes of the following ellipses:
$$x^2-6x+4y^2=16$$
$$2x^2 - 4x+3y^2+6y=7$$
How would one get these? I have no clue. I have a problem with merely rewriting these in the traditional ellipse equation.
| $$x^2-6x+4y^2=16$$
$$(x-3)^2-9+4y^2=16$$
$$(x-3)^2+4y^2=25$$
$$\frac{(x-3)^2}{25}+\frac{4y^2}{25}=1$$
$$\frac{(x-3)^2}{5^2}+\frac{y^2}{(\frac {5}{2})^2}=1$$
center is $O(3,0)$ AND axes are $a=5,b=5/2$, for second ellipse you can proceed similarly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/227869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing that $\frac{1+nx^2}{(1+x^2)^n}$ converges to $0$ I aim to show that $\{a_n\} = \frac{1+nx^2}{(1+x^2)^n}$ converges to $0$. The following two facts seem obvious:
(1) $\forall n \in \mathbb{N}, a_n \ge 0$ (since each $a_n$ is the product of two positive numbers).
(2) $\forall n \in \mathbb{N}, a_n \ge a_{n+1}$.
Yet this is not quite enough to show that $a_n \rightarrow 0$ since the sequence could very well converge to some other $c > 0$.
| The claim is not entirely true; it is valid if and only if $ x \in \mathbb{R} \setminus \{ 0 \} $. Here is a solution that uses neither transcendental functions nor calculus techniques such as l'Hôpital's Rule.
Firstly, observe that the limit is $ 1 $ if $ x = 0 $. Next, let $ n \in \mathbb{N} $. By the Binomial Theorem, we have
$$
\forall x \in \mathbb{R}: \quad (1 + x^{2})^{n} \geq 1 + n x^{2} + \frac{n(n - 1)}{2} x^{4}.
$$
Hence,
\begin{align}
\forall x \in \mathbb{R}: \quad \frac{(1 + x^{2})^{n}}{1 + n x^{2}}
&\geq \frac{1 + n x^{2} + \frac{n(n - 1)}{2} x^{4}}{1 + n x^{2}} \\\\\\
&= \frac{1 + n x^{2}}{1 + n x^{2}} + \frac{\frac{n(n - 1)}{2} x^{4}}{1 + n x^{2}} \\\\\\
&= 1 + \frac{\frac{n(n - 1)}{2} x^{4}}{1 + n x^{2}} \\\\\\
&= 1 + \frac{\frac{1}{n} \left[ \frac{n(n - 1)}{2} x^{4} \right]}{\frac{1}{n} \left( 1 + n x^{2} \right)} \\\\\\
&= 1 + \frac{\frac{n - 1}{2} x^{4}}{\frac{1}{n} + x^{2}} \\\\\\
&= 1 + \frac{n - 1}{2} \cdot \frac{x^{4}}{\frac{1}{n} + x^{2}} \\\\\\
&\geq 1 + \frac{n - 1}{2} \cdot \frac{x^{4}}{1 + x^{2}}.
\end{align}
If $ x \neq 0 $, then the expression in the last line diverges to $ \infty $ as $ n \rightarrow \infty $. It follows that
$$
\forall x \in \mathbb{R} \setminus \{ 0 \}: \quad \lim_{n \rightarrow \infty} \frac{(1 + x^{2})^{n}}{1 + n x^{2}} = \infty.
$$
The original limit must therefore be $ 0 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Quadratic congruence There is question
For each of prime $p$, show that the congruence
$x^2 \equiv1 \pmod {p^a}$
has precisely two solutions.
Continue and show that the congruence
$x^2 \equiv 1 \pmod {2^a} $
has one solution if $a=1$, two solutions if $a=2$, and four solutions if $a \ge 3$.
I don't know how to do. Help please?
| If $a\ge 3,$ and $2^a\mid (x-1)(x+1), x$ must be odd,
So, $$2^{a-2}\mid \left(\frac{x-1}2\right)\left(\frac{x+1}2\right)$$
Now, $$\frac{x+1}2-\frac{x-1}2=1$$ So, $$\left(\frac{x-1}2,\frac{x+1}2\right)=1$$
Now, either $$2^{a-2}\mid \left(\frac{x-1}2\right)$$ or $$2^{a-2}\mid \left(\frac{x+1}2\right)$$
If $$2^{a-2}\mid \left(\frac{x-1}2\right)\implies 2^{a-1}\mid(x-1)\implies x\equiv 1\pmod {2^{a-1}}\equiv 1,1+2^{a-1}\pmod {2^a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Problem of finding the limit of reductive sequence Let $\lbrace u_n \rbrace$ be a sequence defined by : $u_1=1$, $u_{n+1}=u_{n}+\dfrac{u_{n}^{2}}{a}$ for a be a positive real constant.
Find : $\text{lim}_{n\rightarrow \infty}\left(\dfrac{u_1}{u_2}+...+\dfrac{u_n}{u_{n+1}}\right)$
If I denote $S_n$ by $\dfrac{u_1}{u_2}+...+\dfrac{u_n}{u_{n+1}}$ then $S_n$ is increasing sequence, but I could not prove that it is bounded above. May be $\text{lim}_{n\rightarrow \infty} S_n =\infty$ ?
Please help me to solve it. Thanks
| Shortly after finishing the proof, I realized that it is sufficient to observe the following identity
$$ \sum_{k=1}^{n} \frac{u_k}{u_{k+1}} + \frac{a}{u_{n+1}} = a $$
and the fact that $u_{n} \to \infty$ as $n\to\infty$. Indeed,
$$ \frac{a}{u_n} = \frac{u_n + a}{u_n + \frac{u_n^2}{a}} = \frac{u_n + a}{u_{n+1}} = \frac{u_n}{u_{n+1}} + \frac{a}{u_{n+1}}$$
and since $u_1 = 1$, the identity follows. Then by noting that
$$ u_{n} \geq u_{n-1}^2 \geq \cdots \geq (u_2)^{2^{n-2}} \to \infty \quad \text{as} \quad n\to\infty, $$
the conclusion follows.
I leave my old and sophisticated proof, which served a motivation.
Finally I succeeded in proving my observation that
$$ \sum_{n=1}^{\infty} \frac{u_n}{u_{n+1}} = a.$$
Step 0. Notation
By the substitution $u_n = \dfrac{a}{x_n}$, we have
$$ x_1 = a, \qquad x_{n+1} = \frac{x_n^2}{1+x_n}, \qquad \frac{u_n}{u_{n+1}} = \frac{x_{n+1}}{x_n}.$$
We denote $x_n = x_n(a)$ whenever the dependence of the sequence $(x_n)$ on the variable $a$ is required. Then we denote the limit as
$$s(a) := \sum_{n=1}^{\infty} \frac{u_n}{u_{n+1}} = \sum_{n=1}^{\infty} \frac{x_{n+1}(a)}{x_{n}(a)} \tag{1}$$
Since each summand is non-negative, it either converges absolutely or diverges to $+\infty$.
Step 1. Convergence and simple estimation of $s(a)$
Let $g(x)$ and $h(x)$ be functions defined on $x > 0$ by
$$ g(x) = \frac{x^2}{1+x} \quad \text{and} \quad h(x) = \frac{g(x)}{x} = \frac{x}{1+x}.$$
One the one hand, since $0 < h(x) < 1$,
$$ x_{n+1} = h(x_n) x_n < x_n$$
and $(x_n)$ decreases and converges to $0$ as $n \to \infty$. Then from the observation that $h(x)$ is an increasing function, we have
$$ x_{n+1} = h(x_n) x_n \leq h(x_1) x_n.$$
Inductively applying this inequality we obtain
$$ x_n \leq h(x_1)^{n-1} x_1 = \left(\frac{a}{1+a}\right)^{n-1} a. $$
On the other hand,
$$ x_{n+1} \leq x_{n}^2 \leq \left(\frac{a}{1+a}\right)^{n-1} a x_n.$$
Therefore we have
$$ s(a) \leq \sum_{n=1}^{\infty} \left(\frac{a}{1+a}\right)^{n-1} a = a + a^2.$$
In particular, the defining summation $(1)$ converges absolutely. Since
$$ s(a) \geq \frac{x_2(a)}{x_1(a)} = \frac{a}{1+a} \geq a - a^2, $$
we obtain the following estimate:
$$ \left| s(a) - a \right| \leq a^2. \tag{2}$$
Step 2. Further estimation
Since $x_{n+1} = g(x_n)$, we have $x_{n+1}(a) = x_{n}(g(a))$. This gives the following identity:
$$ s(a) = \frac{x_{2}(a)}{x_{1}(a)} + \sum_{n=1}^{\infty} \frac{x_{n+1}(g(a))}{x_{n}(g(a))} = \frac{a}{1+a} + s\left(\frac{a^2}{1+a}\right). \tag{3}$$
Now we claim the following estimation:
$$ \left| s(a) - a \right| \leq a^{2^n} \quad \text{for all} \quad a > 0 \text{ and } n \geq 1. \tag{4} $$
The case $n = 1$ reduces to $(2)$, hence is true for the initial case $n = 1$. Thus assume that $(4)$ holds for $n$. By $(3)$,
$$\left| s(a) - a \right|
= \left| \frac{a}{1+a} + s\left(\frac{a^2}{1+a}\right) - a \right|
= \left| s\left(\frac{a^2}{1+a}\right) - \frac{a^2}{1+a} \right|
\leq \left(\frac{a^2}{1+a}\right)^{2^n}
\leq a^{2^{n+1}}.$$
Therefore $(4)$ holds for $n+1$, and consequently for all $n$ by mathematical induction.
Step 3. Proof of the main claim
Now we claim that $s(a) = a$ for all $a > 0$. Let us first consider the case $a \in (0, 1)$. Then taking $n \to \infty$ to the estimation $(4)$, we have $s(a) = a$.
Now we consider the case $a \geq 1$. Then note that $(3)$ can be written as
$$ s(a) = \frac{x_2(a)}{x_1(a)} + s(x_2(a)). $$
Thus inductively applying this identity, we obtain
$$ s(a) = \sum_{k=1}^{m-1} \frac{x_{k+1}(a)}{x_k(a)} + s(x_m(a)).$$
But we know that $x_m(a) < 1$ for large $m$. Thus with such $m$, the previous result implies
$$ s(a) = \sum_{k=1}^{m-1} \frac{x_{k+1}(a)}{x_k(a)} + x_m(a).$$
But since
$$ \frac{x_{k+1}}{x_k} + x_{k+1} = h(x_k) + g(x_k) = x_k,$$
the above sum reduces to $s(a) = a$ as desired, completing the proof. ////
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/231204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
two limit questions related to $\sin$ I stuck with these 2 limits, can you help me please?
$1.\displaystyle\quad \lim_{n \to \infty }\frac{\sin1+2\sin\frac{1}{2}+3\sin\frac{1}{3}+\cdots+n\sin\frac{1}{n}}{n}$
$2.\displaystyle\quad \lim_{n \to \infty }\frac{n}{\frac{1}{\sin1}+\frac{1/2}{\sin1/2}+\frac{1/3}{\sin1/3}+\cdots+\frac{1/n}{\sin1/n}}
$
Thanks in advance.
| We will use unnecessarily explicit inequalities to prove the result.
In the first limit, the general term on top can be rewritten as $\dfrac{\sin(1/k)}{1/k}$. This reminds us of the $\frac{\sin x}{x}$ whose limit as $x\to 0$ we needed in beginning calculus.
Note that for $0\lt x\le 1$, the power series
$$x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\cdots$$
for $\sin x$ is an alternating series. It follows that for $0\lt x\le 1$,
$$x-\frac{x^3}{6}\lt \sin x\lt x.$$
and therefore
$$1-\frac{x^2}{6}\lt \frac{\sin x}{x}\lt 1.$$
Put $x=1/k$. We get
$$1-\frac{1}{6k^2}\lt \frac{\sin(1/k)}{1/k} \lt 1.\tag{$1$}$$
Add up, $k=1$ to $k=n$, and divide by $n$ Recall that
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots =\frac{\pi^2}{6}.$$
We find that
$$1-\frac{\pi^2}{36n}\lt \frac{\sin1+2\sin\frac{1}{2}+3\sin\frac{1}{3}+\cdots+n\sin\frac{1}{n}}{n}\lt 1.$$
From this, it follows immediately that our limit is $1$.
A very similar argument works for the second limit that was asked about. It is convenient to consider instead the reciprocal, and calculate
$$\lim_{n \to \infty }\frac{\frac{1}{\sin1}+\frac{1/2}{\sin1/2}+\frac{1/3}{\sin1/3}+\cdots+\frac{1/n}{\sin1/n}}{n}.$$
We can then use the inequality
$$1\lt \frac{1/k}{\sin(1/k)} \lt \frac{1}{1-\frac{1}{6k^2}},$$
which is simple to obtain from the Inequalities $(1)$.
Having the $1-\frac{1}{6k^2}$ in the denominator is inconvenient, so we can for example use the inequality $\dfrac{1}{1-\frac{1}{6k^2}}\lt 1+\dfrac{1}{k^2}$ to push through almost the same proof as the first one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/231256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Definite integrals using u substitution (verification needed) Would someone mind verifying this?
$
\int_{0}^{\ln(\pi + 1)}e^x \sin(e^x - 1) \space dx
$
$
u = e^x - 1
\Rightarrow \frac{du}{dx} = e^x
\Rightarrow du = e^x \space dx
\Rightarrow dx = \frac{1}{e^x} \space du
$
$
\int_{0}^{\ln(\pi + 1)} e^x \sin(u) \frac{1}{e^x} \space du
= \int_{0}^{\ln(\pi + 1)} \sin(u) \space du
= -\cos(u) \space |_{0}^{\ln(\pi + 1)}
= -\cos(e^x - 1) \space |_{0}^{\ln(\pi + 1)}
= [-\cos(e^{\ln(\pi + 1)} - 1)] - [-\cos(e^{0} - 1)]
= [-\cos(\pi + 1-1)] - [-\cos(1 - 1)]
= [-\cos(\pi)] - [-\cos(0)]
= -\cos(\pi) + 1
$
and a second one
$\int_{\frac{\pi^2}{4}}^{\pi^2}\frac{\cos(\sqrt{x})}{\sqrt(x)} \space dx \\
$
$
u = \sqrt{x}
\Rightarrow \frac{du}{dx} = \frac{1}{2 \sqrt{x}}
\Rightarrow du = \frac{1}{2 \sqrt{x}} \space dx
\Rightarrow dx = \frac{du}{\frac{1}{2 \sqrt{x}}} = 2\sqrt{x} \space du
$
$
\int_{\frac{\pi^2}{4}}^{\pi^2} = \frac{\cos(u)}{\sqrt(x)} 2\sqrt{x} \space du
= 2 \int_{\frac{\pi^2}{4}}^{\pi^2} \cos(u) \space du
= 2\sin(\sqrt{x}) \space |_{\frac{\pi^2}{4}}^{\pi^2}
$
$
= 2\sin(\sqrt{x}) \space |_{\frac{\pi^2}{4}}^{\pi^2}
= [2\sin(\sqrt{\pi^2})] - [2\sin(\sqrt{\frac{\pi^2}{4}})]
= [2\sin(\pi)]-[2\sin\frac{\pi}{2}]
= [2(0)]-[2(1)]
= 0 - 2
= -2
$
(Thanks. :))
| When you were doing the u subsitution, why did you bring the e^x to the du side? When subsituting, you want to eliminate all functions of x and change them into a function in terms of u. Also you will get a new interval when making u-substitutions!
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$
Determine
$$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}.$$
Multiplying and dividing by $\sqrt{x}$ yields
$$\frac{(x+7)^2\sqrt{x^2+2x}}{7x^3-2x^2}$$
where I would like to approximate the squareroot for sufficiently large $x$ with
$$\frac{(x+7)^2\sqrt{x^2+2x+1}}{7x^3-2x^2}=\frac{(x+7)^2(x+1)}{7x^3-2x^2}=\frac{x^3(49/x^3+63/x^2+15/x+1)}{x^3(7-2/x)}\longrightarrow 1/7.$$
Can anyone confirm that my approximation is valid and does anyone know how to solve this in a more "usual" way like I did?
| As $x \to +\infty$ we have $(x + 7)^2 \sim x^2$ (i.e. $\lim_{x \to \infty} \dfrac{(x+7)^2}{x^2} = 1$), $\sqrt{x+2} \sim \sqrt{x}$, and in the denominator
$7 x^2 \sqrt{x} - 2 x \sqrt{x} \sim 7 x^2 \sqrt{x}$, so
$$ \dfrac{(x+7)^2 \sqrt{x+2}}{7 x^2 \sqrt{x} - 2 x \sqrt{x}} \sim \dfrac{x^2 \sqrt{x}}{7 x^2 \sqrt{x}} = \frac{1}{7} $$
Thus the answer is $1/7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/235866",
"timestamp": "2023-03-29T00:00:00",
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Let a, b, and c be the sides of a triangle. Prove that $(a+b-c)(b+c-a)(c+a-b)\leq abc$
Possible Duplicate:
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers
The title says it all. I've been trying to prove this for hours! Help me!
| What I did was wrong.
Use the substitution $a=x+y$, $b=y+z$ and $c=z+x$ in the initial inequality to have $$8xyz\le(x+y)(y+z)(z+x)$$ and that follows directly from $AM-GM$ since
$$\begin{array}{ll}x+y\ge 2\sqrt{xy}\\ y+z\ge 2\sqrt{yz}\\ z+x\ge 2\sqrt{zx}\\(x+y)(y+z)(z+x)\ge 8\sqrt{x^2y^2z^2}=8xyz\\ \end{array}$$
NB: $AM-GM$ applies only because the sides of a triangle are positive, thus $a,b,c>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/236725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Implicit Differentiation of $x^2y+y^5\sec(x)=5$. Problem
I have began teaching myself Calc I and I've come across the following problem:
Find $\frac{dy}{dx}$ for the following:
$$x^2y+y^5\sec(x)=5.$$
I automatically presumed this was an implicit differentiation. However, since I'm somewhat new to implicit differentiation, my solution looks messy. It is below, and my question is: Am I doing this right and are there ways I could improve (whether it be in terms of my notation, method, or something else)?
Solution
\begin{align}
x^2y+y^5\sec(x)&=5\\
\frac{d}{dx}\left(x^2y+y^5\sec(x)\right)&=\frac{d}{dx}5\\
\frac{d}{dx}x^2y+\frac{d}{dx}y^5\sec(x)&=\frac{d}{dx}5\\
2xy+x^2\frac{d}{dx}(y)+\frac{d}{dx}(y^5)\sec(x)+y^5\sec(x)\tan(x)&=0\\
2xy+x^2\frac{d}{dx}(y)+5y^4\frac{d}{dx}(y)\sec(x)+y^5\sec(x)\tan(x)&=0\\
x^2\frac{d}{dx}(y)+5y^4\frac{d}{dx}(y)\sec(x)&=-2xy-y^5\sec(x)\tan(x)\\
\frac{d}{dx}(y)\left(x^2+5y^4\sec(x)\right)&=-2xy-y^5\sec(x)\tan(x)\\
\frac{dy}{dx}&=\frac{-2xy-y^5\sec(x)\tan(x)}{x^2+5y^4\sec(x)}.
\end{align}
Side question
How could I put this into Wolfram Alpha?
| $$x^2y+y^5\sec(x)=5.$$
$$2xy+x^2y'+5y^4y'\sec(x)+y^5\sec'x=0$$
$$y'(x^2+5y^4\sec x)=-2xy-y^5\frac{\sin x}{\cos^2x }$$
$$y'=\frac{dy}{dx}=\frac{-2xy-y^5\frac{\sin x}{\cos^2x }}{x^2+5y^4\sec x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/242154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Prove $\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \csc^2 (2x)$ I'm having massive issues proving this identity:
$$\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \csc^2 (2x)$$
How is this proven?
| We know that $\cos^2(\theta) + \sin^2(\theta) = 1$, for any $\theta$ including $\theta = 2x$
Hence the left hand side of the equation is $\cot^2(2x) + 1$
Look at our first identity, let's divide it by $\sin^2(\theta)$ yielding
$\displaystyle (\cos^2(\theta))/(\sin^2(\theta)) + 1 = \frac{1}{\sin^2(\theta)}$
And gives us
$\displaystyle\cot^2(\theta) +1 = \frac{1}{\sin^2(\theta)}$
we let $\theta = 2x$ and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a row-reduced echelon matrix. $A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\ 1& -2& 1& 1\end{pmatrix}$.
I should find a row-reduced echelon matrix $R$ which is row equivalent to $A$ and an invertible $3 \times 3$ matrix $P$ such that $R = PA$.
I know that if a matrix is row equivalent to another that means that we can obtain such a matrix by using elementary row operations. But how should i apply this to such question?
| This another approach, however; Don's way explains it to you in a brief solid form.
$A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\ 1& -2& 1& 1\end{pmatrix}\xrightarrow{R_1+R_2\mapsto R_2}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\ 1& -2& 1& 1\end{pmatrix}
\xrightarrow{-R_1+R_3\mapsto R_3}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& -4& 0& 1\end{pmatrix}\\\xrightarrow{-R_2+R_1\mapsto R_1}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\ 0& -4& 0& 1\end{pmatrix}\xrightarrow{2R_2+R_3\mapsto R_3}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\ 0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{-1}{2}R_3+R_2\mapsto R_2}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &0 &-0.5\\ 0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{3}{8}R_3+R_1\mapsto R_1}\begin{pmatrix} 1& 0& 0& \frac{-13}{8}\\ 0 &2 &0 &-0.5\\ 0& 0& 8& 11\end{pmatrix}$
and the last matrix is the row-reduced echelon form of $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/243141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why ${ \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent , but ${ \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent? I don't understand why ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent, but ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent and its limit is equal to ${ \displaystyle\frac{\pi^2}{6} }$. In both cases, ${n^{th}}$ term tends to zero, so what makes these series different?
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + }$ ...
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} =1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + }$ ...
| Summation is closely related to integration. For a non-increasing function $f$ we have
$$\sum_{n=t+1}^\infty f(n) \leq \int_t^\infty f(x)\mathrm{d}x \leq \sum_{n=t}^\infty f(n)$$
or expressed differently
$$-f(t) + \sum_{n=t}^\infty f(n) \leq \int_t^\infty f(x)\mathrm{d}x \leq \sum_{n=t}^\infty f(n)$$
or equivalently
$$\int_t^\infty f(x)\mathrm{d}x \leq \sum_{n=t}^\infty f(n) \leq f(t) + \int_t^\infty f(x)\mathrm{d}x$$
So if the integral exists, the sum diverges iff the integral is $\infty$ - the sum bounds the integral and vice versa.
Let's consider $p>1$:
\begin{align}
\int_1^{\infty} \frac{1}{x^p} \mathrm{d}x &= \int_1^{\infty} x^{-p} \mathrm{d}x \\
&= \frac{1}{1-p}(\lim_{x\rightarrow\infty} x^{1-p} - 1^{1-p}) \\
&= \frac{1}{p-1}
\end{align}
which means that the corresponding sum converges. We even get some bounds, in particular
$$ \frac{1}{p-1} \leq \sum_{n=1}^\infty n^{-p} \leq \frac{1}{p-1} + 1$$
If $p = 1$ then
\begin{align}
\int_1^{\infty} \frac{1}{x} \mathrm{d}x &= \lim_{x\rightarrow\infty} (\ln x) - \ln 1 \\
&= \infty
\end{align}
which means that the corresponding sum diverges.
So we can conclude that the sum $\sum_{n=1}^\infty n^{-p}$ converges iff $p>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/243482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 3
} |
Integral - how would you compute this expression? $$\int\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^4-1}}dx$$
I tried many types of substitutions, but I got nowhere.
| Hint:
Notice that $x^4-1 = (x^2-1)(x^2+1)$, and then split into 2 fractions:
$$\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^4-1}} = \frac{\sqrt{x^2+1}}{\sqrt{(x^2+1)(x^2-1)}} - \frac{\sqrt{x^2-1}}{\sqrt{(x^2+1)(x^2-1)}}$$
Then simplify and you get 2 easier integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the sum of a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ I was thinking about this problem:
Given a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ ,how can i show that its sum is a/an rational/irrational number,given that the series converges?
Could someone point me in the right direction? Thanks everyone in advance.
| Find the closed form and take the limit in the case of infinite sums. That is, you find the closed form of the sum $\sum_{4 \le k \le m}\frac{1}{k^2-1}$ and evaluate $\lim_{m \to \infty}(\text{closed form of the sum})$.
In this case, you apply partial fraction decomposition to $\frac{1}{k^2-1}$ and arrive at $$\frac{1}{k^2-1}=\frac{-\frac{1}{2}}{k+1}+\frac{\frac{1}{2}}{k-1}=\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right).$$
Taking the partial sum $$\sum_{4 \le k \le m}\frac{1}{k^2-1}=\frac{1}{2}\sum_{4 \le k \le m}\frac{1}{k-1}-\frac{1}{k+1}.$$
The sum telescopes, so we see that $$\sum_{4 \le k \le m}\frac{1}{k+1}-\frac{1}{k-1}=\color{red}{\frac{1}{3}}-\color{green}{\frac{1}{5}}+\color{red}{\frac{1}{4}}-\frac{1}{6}+\color{green}{\frac{1}{5}}-\frac{1}{7}+\dots+\color{green}{\frac{1}{m-1}}-\frac{1}{m-3}+\color{red}{\frac{1}{m}}-\frac{1}{m-2}+\color{red}{\frac{1}{m+1}}-\color{green}{\frac{1}{m-1}}=\frac{1}{3}+\frac{1}{4}+\frac{1}{m}+\frac{1}{m+1}.$$
Taking the limit, we have $$\lim_{m \to \infty}\frac{1}{3}+\frac{1}{4}+\frac{1}{m}+\frac{1}{m+1}=\frac{1}{3}+\frac{1}{4}+\lim_{m \to \infty}\frac{1}{m}+\frac{1}{m+1}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}.$$
Recalling the factor of $\frac{1}{2}$, we arrive at $\sum_{4 \le k \le \infty}\frac{1}{k^2-1}=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
can we have a triangle with sides $1, x$ and $x^2$? Can we have a triangle with sides $1$, $x$ and $x^2$? And what could $x$ be?
I try to approach this question by making 3 inequalities.
$1+x>x^2$,
$1+x^2>x$,
$x^2+x>1$
and they come with different quadratic inequalities
$x^2-x-1<0$ (solution is $(1+\sqrt 5)/2 > x > (1-\sqrt 5)/2$ ) ;
$x^2-x+1>0$ (solution is $x > (-1+\sqrt 5)/2$ or $x < (-1-\sqrt 5)/2$ )
$x^2+x-1>0$ (no real solution)
Then I start to struggle with the next step.... Thank you
| So you already have the inequalities:
$$1+x>x^2\Longleftrightarrow x^2-x-1<0\Longleftrightarrow \frac{1-\sqrt 5}{2}<x<\frac{1+\sqrt 5}{2}$$
$$1+x^2>x\Longleftrightarrow x^2-x+1>0\Longrightarrow\,\text{always true}$$
$$x^2+x>1\Longleftrightarrow x^2+x-1>0\Longleftrightarrow x<\frac{-1-\sqrt 5}{2}\,\,\vee\,\,x>\frac{-1+\sqrt 5}{2}$$
The common interval to all the above inequalities is
$$\frac{-1+\sqrt 5}{2}<x<\frac{1+\sqrt 5}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/250362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
What is wrong with my polynomial division? $x^3-2x^2-5x+6$ $$x^3-2x^2-5x+6$$
I want to get the solutions of this. I did a polynomial division.
First, I know that $(x+1)$ is a factor since $1^3-2\cdot1^2-5\cdot1+6 = 0$
So my division goes like this:
$$..........x^2-3x-2$$
$$(x+1)|\overline{x^3-2x^2-5x+6}$$
$$x^3+x^2$$
$$......\overline{0-3x^2}-5x$$
$$........-3x^2-3x$$
$$...............\overline{0-2x}+6$$
$$....................2x-2$$
$$....................\overline{0 + 8}$$
(Sorry for the improvised formatting. Ignore any dots you see there.)
So I get, at the end, the quadratic $x^2-3x-2+8=x^2-3x+6$
However, $\triangle = (-3)^2-4(1)(6)=9-4(6)=9-24=-15$
Therefore, there are no solutions since $\triangle$ is negative.
But I definitely did something wrong, since I do know that the solutions are $1,-2,3$
What did I do wrong?
| $(x+1)$ isn't a factor; you discovered that $1$ is a zero of your polynomial, hence $(x-1)$ is a factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/252915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Zeta function and fractions of sums of natural numbers From my work on the Goldbach conjecture I have formulated the following series
$ζ[ς]=\sum\limits_{f=2}^∞ \frac{1}{f^ς}-\frac{2}{(f+1)f-2}$ where $ς$ a natural number. If $ς=2$ we have the series
$\frac{1}{2^2}-\frac{1}{2}+\frac{1}{3^2}-\frac{1}{5}+\frac{1}{4^2}-\frac{1}{9}+\frac{1}{5^2}-\frac{1}{14}$....
Does anyone know for which $ς$ these series are convergent?
| Your sum can be rewritten as
$$
\sum_{f=2}^{\infty} \left( \frac{1}{f^s} - \frac{2}{(f+1)f-2} \right) = \sum_{f=2}^{\infty} \frac{1}{f^s} - \sum_{g=2}^{\infty} \frac{2}{(g+1)g-2}.
$$
Now, the right-hand sum can be evaluated explicitly since it telescopes. Indeed,
$$
\begin{align*}
\sum_{g=2}^{\infty} \frac{2}{(g+1)g-2} &= 2\sum_{g=2}^{\infty} \frac{1}{(g+1)g-2} \\
&= 2\sum_{g=2}^{\infty} \frac{1}{(g-1)(g+2)} \\
&= \frac{2}{3} \sum_{g=2}^{\infty} \left(\frac{1}{g-1} - \frac{1}{g+2}\right) \\
&= \frac{2}{3} \left(1 \color{red}{- \frac{1}{4}} + \frac{1}{2} \color{blue}{- \frac{1}{5}} + \frac{1}{3} \color{violet}{- \frac{1}{6}} \color{red}{+ \frac{1}{4}} \color{violet}{- \frac{1}{7}} \color{blue}{+ \frac{1}{5}} \color{violet}{- \frac{1}{8}} + \cdots\right) \\
&= \frac{2}{3}\left(1 + \frac{1}{2} + \frac{1}{3}\right) \\
&= \frac{11}{9},
\end{align*}
$$
where the terms in $\color{red}{\text{red}}$ cancel with each other, the terms in $\color{blue}{\text{blue}}$ cancel with each other, and the terms in $\color{violet}{\text{violet}}$ cancel with something that isn't shown.
Then, since
$$
\sum_{f=2}^{\infty} \frac{1}{f^s} = \sum_{f=1}^{\infty} \frac{1}{f^s} - 1 = \zeta(s) - 1,
$$
where $\zeta$ is the Riemann zeta function, we see that your sum is equal to
$$
\sum_{f=2}^{\infty} \left( \frac{1}{f^s} - \frac{2}{(f+1)f-2} \right) = \zeta(s) - 1 - \frac{11}{9} = \zeta(s) - \frac{20}{9}.
$$
The Riemann zeta function converges for all $s$ with $\Re(s) > 1$, so the same is true for your sum.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\lim\limits_{x\rightarrow+\infty}\left[(x^3+x^2+1)^{1/3}- (x^2+x+1)^{0.5}\right] = -1/6$? Every method I use seem to get me to something to the extent of $0/0$, stuff I can't work with. Wolfram Alpha claims the answer to this is $-1/6$ but they offer no step by step solution.
Would appreciate any tips and help.
| $$ (x^3+x^2+1)^{1/3} - (x^2+x+1)^{0.5} = x \cdot (1+x^{-1}+x^{-3})^{1/3} - x \cdot (1+x^{-1}+x^{-2})^{0.5} $$
then I suggest Taylor series for both terms.
$$ x \cdot \left [ 1+\frac{1}{3} \cdot x^{-1}-\frac{1}{9} \cdot x^{-2} + ... - 1-\frac{1}{2}\cdot x^{-1}-\frac{3}{8} \cdot x^{-2} - ...\right] = x \cdot [ -\frac{1}{6} x^{-1} + ... ] = -\frac{1}{6} + o(x^{-1})$$
which has limit of $-\frac{1}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution for $\log_7x+\log_{\frac17}x^2=\log_{49}x-3$ What is the right solution for $\log_7x+\log_{\frac17}x^2=\log_{49}x-3$. What logarithm identities used?
| $$\log_7 x + \log_{1/7} x^2 =\log_{7^2} (x) - 3 $$
$$\log_7 x + \frac{\log_7(1/7)}{\log_7 x^2} =\frac{\log_77^2}{\log_7 (x) }-3 $$
$$\log_7 x + \frac{-1}{\log_7 x^2} =\frac{2}{\log_7 (x) }-3 $$
$\log_7 x=t$ then we have
$$t-\frac{1}{t}=\frac{2}{t}-3$$
$$t^2+3t-3=0$$
$$\log_7x_1=\frac{-3+\sqrt{21}}{2},x_1=7^{\frac{-3+\sqrt{21}}{2}}$$
$$\log_7x_2=\frac{-3-\sqrt{21}}{2},x_2=7^{\frac{-3-\sqrt{21}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve recurrence equation $f(n) = f(n-5) + f(n-10)$? How to solve recurrence equation $f(n) = f(n-5) + f(n-10)$? For something like fibonacci sequence $f(n+1) = f(n) + f(n-1)$ I can solve for the quadratic equation $x^2-x-1=0$ then $f(n) = A x_1 + A^\prime x_2$. But what should I do for this one?
| Although a topic in Discrete math, I attempted this question using the tools I learned in linear algebra.
Rewrite it as $f(n+10) = f(n+5) +f(n)$, this can be represented as
\[ \left( \begin{array}{c}
f_{n+10} \\
f_{n+5} \end{array} \right) = \left( \begin{array}{cc}
1 &1 \\
1& 0 \end{array} \right) \left( \begin{array} {c}
f_{n+5} \\
f_n \end{array} \right)
\]
\[ \left( \begin{array}{c}
f_{n+10} \\
f_{n+5} \end{array} \right) = \left( \begin{array}{cc}
1 &1 \\
1& 0 \end{array} \right)^{n+5} \left( \begin{array} {c}
f_{1} \\
f_{-4} \end{array} \right)
\]
Then diagonalize $\left( \begin{array}{cc}
1 &1 \\
1& 0 \end{array} \right)$ to find eigenvalues, which is the same as the ones for fibonacci, $1/2 (1-\sqrt5), 1/2 (1+\sqrt5)$
then $f_{n+5} = \alpha (1/2 (1-\sqrt5))^{n+5} +\beta (1/2 (1+\sqrt5))^{n+5} $
Is this correct? If not please advise.
| {
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"url": "https://math.stackexchange.com/questions/255083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $(a,b)=1$ then prove $(a+b, ab)=1$.
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.
$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of $a$.
Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.
So $a(k-l)=b, \ a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].
But $(a,b)=1$ so $a$ cannot divide $b$.
We have a similar argument for $b$.
So $a+b$ is not divisible by any factors of $ab$.
Therefore, $(a+b, ab)=1$.
Would this be correct? Am I missing anything?
| If $d$ divides $a+b$ and $d$ divides $ab$, then $d$ divides $a(a+b)-ab = a^2$. Similarly, $d$ divides $b^2$. Thus $d$ divides $(a^2,b^2$). But $(a^2,b^2) = (a,b)^2 = 1$, so $d=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 8
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How to solve indeterminations of the type $0/0$ I am unable to find these limits:
1)
$$
\lim_{x \to 1} \frac{3(1 - x^2) - 2(1 - x^3)}{(1 - x^3)(1 - x^2)}
$$
2)
$$
\lim_{x \to 0} \frac{\sqrt{1 - 2x - x^2} - (x + 1)}{x}
$$
3)
$$
\lim_{x \to 0} \frac{\sqrt{x + 2} + \sqrt{x + 6} - \sqrt{6} - \sqrt{2}}{x}
$$
4)
$$
\lim_{x \to 0} \frac{1 - \sqrt[3]{1 - x}}{1 + \sqrt[3]{3x - 1}}
$$
My interest is not in the answers, but in the algebraic manipulations i can use to eliminate the indeterminations of the type $0/0$.
My english skills are not so good, i'm sorry for this.
| 1) Factorise both the numerator and denominator so that the terms $(1-x)^n$ can be simplified
2) Use $$\alpha-\beta=\frac{\alpha^2-\beta^2}{\alpha+\beta}$$
3) Write
$$
\sqrt{x + 2} + \sqrt{x + 6} - \sqrt{6} - \sqrt{2}=(\sqrt{x + 2} - \sqrt{2})+ (\sqrt{x + 6} - \sqrt{6})
$$
and use what we did in $2$.
4) Use $$\alpha-\beta=\frac{\alpha^3-\beta^3}{\alpha^2+\alpha\beta+\beta^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/258780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you construct a lattice from its basis or its Gram Matrix? I'm really having trouble trying to understand this. A few weeks back, I got pretty interested in sphere packing and I'm trying to grasp the idea of using a matrix to represent the basis of a lattice. I've been using lattices like this one $\left(\begin{array}\12 & 0 & 0\\ 1 & \sqrt3 & 0\\1 & \frac{1}{\sqrt3} &\frac{2\sqrt6}{3}\end{array}\right)$ for the fcc lattice, because this represents the three vectors $(2,0,0)$, $(1,\sqrt3,0)$, and $(1,\frac{1}{\sqrt3},\frac{2\sqrt6}{3})$ that make the fundamental parallelipipid for the lattice. I think this lattice is also represented by $A_3$. It says here (archive) that the Gram Matrix of $A_3$ is $\left(\begin{array}\12 & -1 & 0\\-1 & 2 & -1\\0 & -1 & 2\end{array}\right)$. How is the Gram Matrix related to the basis I've been using? What about the basis they give? Is there a fast way to construct a lattice and find its density from a Gram Matrix? Everything I've found so far is over my head.
| What they actually do on the website, which saves a bunch of square root signs, is to realize the lattice in $\mathbb R^4$ as
$$
\left( \begin{array}{rrrr}
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0\\
0 & 0 & 1 & -1
\end{array}
\right) \cdot
\left( \begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0\\
0 & -1 & 1 \\
0 & 0 & -1
\end{array}
\right) =
\left( \begin{array}{rrr}
2 & -1 & 0 \\
-1 & 2 & -1\\
0 & -1 & 2
\end{array}
\right).
$$
Here you are given basis vectors as the three rows of the rectangular matrix on the left, or the columns of the matrix on the right, which is its transpose. That is the fundamental relation, call it
$$ B \; B^T = G. $$
For the version you want, squre matrices over the integers are possible:
$$
\left( \begin{array}{rrr}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right) \cdot
\left( \begin{array}{rrr}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right) =
\left( \begin{array}{rrr}
2 & 1 & 1 \\
1 & 2 & 1\\
1 & 1 & 2
\end{array}
\right).
$$
If you really want, you can write this in lower/upper triangular square matrices. This can be done with CHOLESKY, which can be done by hand here. You should have been using
$$
\left( \begin{array}{rrr}
\sqrt 2 & 0 & 0 \\
\sqrt{ \frac{1}{2} } & \sqrt{ \frac{3}{2} } & 0 \\
\sqrt{ \frac{1}{2} } & \sqrt{ \frac{1}{6} } & \sqrt{ \frac{4}{3} }
\end{array}
\right) \cdot
\left( \begin{array}{rrr}
\sqrt 2 & \sqrt{ \frac{1}{2} } & \sqrt{ \frac{1}{2} } \\
0 & \sqrt{ \frac{3}{2} } & \sqrt{ \frac{1}{6} } \\
0 & 0 & \sqrt{ \frac{4}{3} }
\end{array}
\right) =
\left( \begin{array}{rrr}
2 & 1 & 1 \\
1 & 2 & 1\\
1 & 1 & 2
\end{array}
\right).
$$
Finally, the relationship required for distinct gram matrices to represent the same lattice is called "equivalence." If you have gram matrices $G,H,$ equivalence means there is an integer matrix $P$ with $\det P = 1$ such that
$$ P G P^T = H. $$
Naturally, if you take $Q = P^{-1}$ you get
$$ Q H Q^T = G $$ as well.
Gaze in awe:
$$
\left( \begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0 \\
0 & -1 & 1
\end{array}
\right) \cdot
\left( \begin{array}{rrr}
2 & 1 & 1 \\
1 & 2 & 1\\
1 & 1 & 2
\end{array}
\right) \cdot
\left( \begin{array}{rrr}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{array}
\right) =
\left( \begin{array}{rrr}
2 & -1 & 0 \\
-1 & 2 & -1\\
0 & -1 & 2
\end{array}
\right).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with infinite sum Can you guys give me a hint on evaluating $$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)}?$$
I have tried partial fractions but the series is not telescopic (at least I cannot see it)...
| Hint:
$$\frac{1}{n(n+2)(n+4)}=\frac{1}{4}\frac{n+4-n}{n(n+2)(n+4)}=\frac{1}{4}[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}]$$
In general:
$$\frac{1}{n(n+d)...(n+kd)}=\frac{1}{kd}\left[\frac{1}{n(n+d)...(n+(k-1)d)}-\frac{1}{(n+d)(n+2d)...(n+kd)}\right]$$
Now if you let $a_n=\frac{-1}{kd}(\frac{1}{n(n+d)...(n+(k-1)d)})$, we find that:
$$\frac{1}{n(n+d)...(n+kd)}=a_{n+d}-a_n$$
Thus, the sum $\sum_{n=1}^{\infty}\frac{1}{n(n+d)...(n+kd)}$ is a telescoping sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Estimating Residues in McLauren's Polynomial Can someone please help me figure out why it is enough to take $n\geq 9 $ in order to get that:
$$\displaystyle \frac{ 9 \cdot 2^{n+1} } {(n+1)!} < \frac{1}{100}\;\;?$$
Thanks in advance
| First note that $a_n = \dfrac{2^{n+1} \cdot 9}{(n+1)!}$ is a monotone decreasing sequence for $n \geq 1$. Hence, if $a_n < \epsilon$, then $a_m < \epsilon$ for all $m \geq n$.
$$a_n = \dfrac{2^{n+1} \cdot 9}{(n+1)!} = \dfrac{3 \cdot 2^{n-2} \cdot 4!}{(n+1)!}$$
$$a_8 = \dfrac{3 \cdot 2^6 \cdot 4!}{9!} = \dfrac{3 \cdot 2^6}{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5} = \dfrac{2^2}{9 \cdot 7 \cdot 5} = \dfrac4{315} > \dfrac4{400} = \dfrac1{100}$$
$$a_9 = \dfrac{3 \cdot 2^7 \cdot 4!}{10!} = \dfrac{3 \cdot 2^7}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5} = \dfrac{2^2}{5 \cdot 9 \cdot 7 \cdot 5} = \dfrac4{1575} < \dfrac4{400} = \dfrac1{100}$$
Since $a_9 < \dfrac1{100}$ and $a_n$ is a decreasing sequence, we have that $a_n < \dfrac1{100}$ for all $n \geq 9$.
| {
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Getting basic step for calculating the limit of e I have the following limit:
$$
\lim_{x\to \infty}{\bigg( \frac{x^2 + 3x - 1}{x^2 + 3} \bigg) }^\frac{x -2}{2}
$$
We had no explanations for calculating such limits i looked over few poor textbook examples and i understand the result of this limit will be $ e^X $ where X will be what i get from expanding the limits $ \frac {x-2}{2} $.
And that first step is to get the following form:
$$
\lim_{x\to\infty}{ \bigg ( 1 + \frac{1}{x+1} \bigg )^{x+1} } = e
$$
Are there any simpler steps to solving such a limit ?
| $$\lim_{x\to \infty}{\left( \frac{x^2 + 3x - 1}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( \frac{x^2 + 3-3+3x - 1}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( \frac{x^2 + 3+3x - 4}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( 1+\frac{3x - 4}{x^2 + 3} \right) }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( 1+\frac{3x - 4}{x^2 + 3} \right) }^{\frac{x^2+3}{3x-4}\cdot \frac{3x-4}{x^2+3}\cdot\frac{x -2}{2}}\overset{def}{=}L$$
Since exists $\lim\limits_{x\to \infty}{\left( \dfrac{3x-4}{x^2+3}\cdot\dfrac{x -2}{2} \right) }=\dfrac{3}{2}$, then $L=e^\frac{3}{2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help in proving $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum\limits_{k=0}^{n} \frac{1}{k!} < 3 $. I am trying to prove this statement for all $ n \geq 1 $ using induction:
$$
\left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3.
$$
I said:
*
*Base case $ n = 1 $:
$$
\left( 1 + \frac{1}{1} \right)^{1} \leq \sum_{k=0}^{1} \frac{1}{k!} < 3,
$$
which is okay.
*Induction step: Suppose that $ \displaystyle \left( 1 + \frac{1}{n}
\right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3 $ for a given $ n \in
\mathbb{N} $.
Transition from $ n \to n + 1 $:
$$
\displaystyle \left( 1 + \frac{1}{n + 1} \right)^{n+1}
= \left( 1 + \frac{1}{n + 1} \right)^{n} \left( 1 + \frac{1}{n + 1} \right)
= \ldots \text{Help} \ldots
< 3.
$$
I need some guidance for proof-writing (-thinking) in orders.
| To show that $\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!} \lt 3$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots +\frac{1}{n!}$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=1+1+\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{1}{3*4}+\frac{1}{3*4*5}+\dots+\frac{1}{3*4*5*\dots*n}\right)$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}\lt1+1+1$
| {
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Linear algebra about cyclic group and torsion factors Express the commutative group $Z^{3}/(f_{1}.f_{2},f_{3})$ as a direct sum of cyclic group where
$f_{1}=(4,6,9), f_{2}=(2,4,12), f_{3}=(4,8,16)$
my answer is $Z[x]/(-11x^{2}+22x-128)$.
I wonder if my calculation is wrong, for this is not a direct sum of cyclic group, hope someone can give me a more clear answer.
How to find the number of non-isomorphic commutative group of a certain order, like $100$ and $p^{3}$ where $p$ is a prime.
I know that I can use Kronecker Decomposition Theorem, finite abelian group theorem.
| The smith normal form for
$$\begin{pmatrix} 4 & 6 &9\\ 2 & 4 & 12 \\ 4 & 8 & 16\end{pmatrix}$$ can be found by several steps. We first have by substracting two times middle row from top and bottom row:
$$\begin{pmatrix} 0 & -2 &-15\\ 2 & 4 & 12 \\ 0 & 0 & -8\end{pmatrix}$$ which is the same as
$$\begin{pmatrix} 2 & 4 & 12 \\0 & -2 &-15\\ 0 & 0 & -8\end{pmatrix}$$
Then we substracting the first column from second and third column, and do the same for second column:
$$\begin{pmatrix} 2 & 0 & 0 \\0 & -2 &-1\\ 0 & 0 & -8\end{pmatrix}$$
and then we have:
$$\begin{pmatrix} 2 & 0 & 0 \\0 & 0 &-1\\ 0 & 16 & -8\end{pmatrix}$$
changing order we have:
$$\begin{pmatrix} 2 & 0 & 0 \\0 & 16 &0\\ 0 & 0 & -1\end{pmatrix}$$
| {
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Find all roots of $\,(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$ The question is to find all complex roots of
$$(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$$
and it is meant to be solved by hand.
Is there any quick way to solve this using some trick that I'm not aware of? The solution given by Wolfram used a lot of multiplying out and factoring that would be very difficult to think of or write. Also, just by observation $x=0$ and $x=-6$ are solutions.
| This is another approach but with a lot more computation required than that of Marvis. It could still be done by hand as stated in the original post with some patience. Expanding the factors we get $$x^6 + 18x^5 + 130x^4 + 480x^3 + 949x^2 + 942x + 360=360$$ Then subtract $360$ from each side to get $$x^6 + 18x^5 + 130x^4 + 480x^3 + 949x^2 + 942x=0$$ Because it was noted by observation that $0$ and $-6$ are solutions let’s divide the polynomial by $x+6$. So
$$\frac{ x^6 + 18x^5 + 130x^4 + 480x^3 + 949x^2 + 942x}{x+6}=x^5+12x^4+58x^3+132x^2+157x$$ Because $0$ is also a solution we can factor out an $x$ for free to get $$x^4+12x^3+58x^2+132x+157=0$$ This is a quartic which we can convert to a depressed quartic by setting $$x=u-\frac{12}{4}$$ and expanding the following as seen in Quartic function
$$\left(u-\frac{12}{4}\right)^4+\frac{12}{4}\left(u-\frac{12}{4}\right)^3+ \frac{58}{4}\left(u-\frac{12}{4}\right)^2+ \frac{132}{4}\left(u-\frac{12}{4}\right)+\frac{157}{4}=0$$
This expands to
$$u^4 + 4u^2 + 40=0$$ which is a biquadratic equation. Solving this quadratic for $u^2$ we get solutions
$$u=\pm \sqrt{-2\pm6i}$$ so all the complex solutions to the original quartic and also the original polynomial are
$$x=-3\pm \sqrt{-2\pm6i}$$ getting the $-3$ above from the $-\frac{12}{4}$ term used to depress the original quartic.
| {
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About implicit differential equation? $\displaystyle \begin{align*}
& 0<x<1\wedge f\left( x \right)=\int_x^1 \frac{\left( 1-t \right)^2}{t^2} \text{d}t \\
& \text{Prove}:\ \ f\left( x \right)\ge \frac{2\left( 1-x \right)^3}{3x\left( 1+x \right)}
\end{align*}$
| Let $g(x) = \frac {2(1-x)^3}{3x(1+x)}$. We want to show that $f(x) \geq g(x)$. Since we have $f(1) = 0, g(1) = 0$, it suffices to show that $f'(x) \leq g'(x)$ for $x\in [0,1]$.
From definition of $f(x)$, we have $f'(x) = - \frac {(1-x)^2}{x^2} $. By Wolfram Alpha (or do this yourself), we have $g'(x) = -\frac {2 (-1+x)^2 (1+4 x+x^2)}{3 x^2 (1+x)^2}$. Hence, it suffices to show that $ - \frac {(1-x)^2}{x^2} \leq -\frac {2 (-1+x)^2(1+4x + x^2}{3x^2(1+x)^2}$, or equivalently that $3(1+x)^2 \geq 2(1+4x+x^2)$ (Why is the inequality sign switched?), which reduces to $(1-x)^2 \geq 0$.
| {
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Find value of $k$ for which the equation has real roots What can be the value of $k$ for which the equation $9x^2+2kx-1=0$ has real roots?
Thing should be known
*
*When the quadratic equation has real roots, then $d=b^2-4ac \ge 0$ .
Where $a$, $b$ and $c$ are the constant terms of a quadratic equation $ax^2+bx+c=0$.
| Besides to @Rankeya's answer, you can see that: $$9x^2+2kx=1\Leftrightarrow 9x^2+2kx+\frac{k^2}{9}=1+\frac{k^2}{9}\\\Leftrightarrow \left(3x+\frac{k}{3}\right)^2=\frac{k^2}{9}+1\Leftrightarrow 3x+\frac{k}{3}=\pm\sqrt{\frac{k^2}{9}+1}$$ or $$3x=\pm\sqrt{\frac{k^2}{9}+1}-\frac{k}{3}$$ which shows that you always have two distinct solutions for $x$ for all $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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calculate $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}$ $$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(2x)}{\cos(x)-\sin(x)}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{2\cos^{2}(x)-1}{\cos(x)-\sqrt{1-\cos^{2}(x)}}$$
$$t=\cos(x)$$
$$\lim_{x\rightarrow\frac{\pi}{4}}\frac{2t^{2}-1}{t-\sqrt{1-t^{2}}}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{(2t^{2}-1)(t+\sqrt{t^{2}-1})}{2t^{2}-1}$$
$$\lim_{x\rightarrow\frac{\pi}{4}}(2t^{2}-1)(t-\sqrt{t^{2}-1})=3(\sqrt{2}-1)$$
Please help me to find an error. Correct answer is $\sqrt{2}$. Thanks
| The third line is in error: you forgot to cancel the factor of $2 t^2 - 1$ and you switched the sign of $\sqrt{1-t^2}$. If you correct, you'll see that
$$ \lim_{x \rightarrow \frac{\pi}{4}} \frac{\cos{2 x}}{\cos{x} - \sin{x}} = \lim_{t \rightarrow \frac{1}{\sqrt{2}}} \left [ t + \sqrt{1-t^2} \right ] = \sqrt{2} $$
| {
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Inequality. $\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$ Prove that :
$$\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
I try to apply Cauchy-Buniakowski and I obtaine the followin:
$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$
So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
Another trying :
$$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that:
$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:
$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or
$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or
$$x^2+y^2+z^2+xy+yz+zx \leq 18$$
$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$
Yes, it is ok :)
thanks :)
| I found an easier proof: Let $a=\sqrt{x^2+xy+3}$ and $b,c$ so on. Since $xy+yz+xz\leq x^2+y^2+z^2\leq 9$ we know that $a^2+b^2+c^2\leq 27$. Then $3(abc)^{\frac{2}{3}}\leq a^2+b^2+c^2\leq 27$, i.e. $(abc)^{\frac{1}{3}}\leq 3$. Then
\begin{align}
LHS&=\sum_{cyc}\frac{1}{a}\\
&=\frac{ab+bc+ac}{abc}\\
& \geq\frac{3(abc)^{\frac{2}{3}}}{abc}\\
&= 3(abc)^{-\frac{1}{3}}\\
&\geq 1.
\end{align}
| {
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Find the area of the pentagon formed in the plane with the fifth roots of unity as its vertices Find the area of the pentagon formed in the plane with the fifth roots of
unity as its vertices.
is there any formula to solve this type of problem?
| Consider the lines joining the vertices of the pentagon to the center. Each of these triangles are isosceles, with side length 1 and vertex angle $\frac {2 \pi}{5}$.
Hence, the area of the pentagon is $\frac {5}{2} \sin \frac {2\pi}{5}$, which we can evaluate to be $ \frac {5}{4} \sqrt{ \frac {5+ \sqrt{5}}{2} } $.
In general, the area of the n-gon is $\frac {n}{2} \sin \frac {2\pi}{n}$.
If you have to use complex numbers to approach this question, then since the cross product uses $\sin \theta$, hence the area of one of these triangles will be $\frac {1}{2} \left \| 1 \times \omega \right \| = \frac {1}{2} \sin \frac {2 \pi}{5}$.
| {
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How to evaluate the following integral using hypergeometric function? May I know how this integral was evaluated using hypergeometric function?
$$\int \sin^n x\ dx$$
Wolframalpha showed this result but with no steps
Thanks in advance.
| Assuming $n$ is a non-negative integer, you could use binomial theorem:
$$\begin{eqnarray}
\sin^n(x) &=& \left( \frac{\exp(i x) - \exp(-i x)}{2i}\right)^n = \frac{1}{2^n i^n} \sum_{m=0}^n \binom{n}{m} (-1)^m \exp\left( i \left(n-2m\right)x \right) \\ &=&
\frac{1}{2^n i^n} \sum_{m=0}^n \binom{n}{m} (-1)^m \left(\cos\left(\left(n-2m\right)x \right) + i \sin\left( \left(n-2m\right)x \right) \right)
\end{eqnarray}
$$
Since the left-hand-side is real we only keep cosines for even $n$:
$$\begin{eqnarray}
\sin^{2n}(x) &=& \frac{1}{2^{2n}} \sum_{m=0}^{2n} \binom{2n}{m}\left(-1\right)^{n-m} \cos\left(\left(2n-2m\right)x\right) \\ &\stackrel{\text{symmetry}}{=}& \frac{1}{2^{2n}} \binom{2n}{n} + \frac{1}{2^{2n-1}} \sum_{m=0}^{n-1} \binom{2n}{m}\left(-1\right)^{n-m} \cos\left(2 \left(n-m\right)x\right) \\
&\stackrel{m\to n-m} =& \frac{1}{2^{2n}} \binom{2n}{n} + \frac{1}{2^{2n-1}} \sum_{m=1}^n \binom{2n}{n+m} (-1)^n \cos(2 m x) \tag{1}
\end{eqnarray}$$
and, likewise, only sines for odd $n$:
$$
\sin^{2n+1}(x) = \frac{1}{2^{2n}} \sum_{m=0}^n \binom{2n+1}{n+1+m} (-1)^m \sin\left((2m+1)x\right) \tag{2}
$$
We can now integrate element-wise:
$$
\int \sin^{2n}(x) \, \mathrm{d}x = \frac{1}{2^{2n}} \binom{2n}{n} x + \frac{1}{2^{2n-1}} \sum_{m=1}^n \binom{2n}{n+m} (-1)^n \frac{\sin(2 m x)}{2m} + \text{const.}
$$
$$
\int \sin^{2n+1}(x) \, \mathrm{d}x = -\frac{1}{2^{2n}} \sum_{m=0}^n \binom{2n+1}{n+1+m} (-1)^m \frac{\cos\left((2m+1)x\right)}{2m+1} + \text{const.}
$$
To obtain a hypergeometric function, let $u = \sin(x)$. Then
$$
\int \sin^n(x)\, \mathrm{d}x = \int \frac{u^n}{\sqrt{1-u^2}} \mathrm{d}u
$$
Now see this answer of mine on how to find the anti-derivative of $\int u^a (1-u)^b \mathrm{d} u$. Applying the same principles, we find:
$$
\int \frac{u^n}{\sqrt{1-u^2}} \mathrm{d}u =\int u^n \cdot {}_1F_0\left(\left.\begin{array}{c} \frac{1}{2} \\ - \end{array} \right| u^2 \right) \mathrm{d} u = \int \frac{\mathrm{d}}{\mathrm{d}u} \left( \frac{u^{n+1}}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| u^2 \right) \right) \mathrm{d} u
$$
Thus, we have:
$$
\int \sin^n(x) \, \mathrm{d}x = \frac{\sin^{n+1}(x)}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| \sin^2(x) \right) + \text{const.} \tag{3}
$$
This works where $u = \sin(x)$ is invertible. To extend validity of the answer, differentiate it. We would get
$$
\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\sin^{n+1}(x)}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| \sin^2(x) \right) \right) = \frac{\sqrt{\cos^2(x)}}{\cos(x)} \sin^{n} (x)
$$
and since the pre-factor $\frac{\sqrt{\cos^2(x)}}{\cos(x)}$ is a differential constant, i.e. its derivative is zero, we arrive at:
$$
\int \sin^n(x) \, \mathrm{d}x = \frac{\cos(x)}{\sqrt{\cos^2(x)}} \frac{\sin^{n+1}(x)}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| \sin^2(x) \right) + \text{const.} \tag{4}
$$
This can be related to the answer provided by Wolfram|Alpha, and thus by Mathematica, using Kummer's relations.
| {
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Exponent questions algebra How would I solve the following two exponent questions?
(1) The first question is
$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}$$
I got $\quad \displaystyle \frac{-xy^{-2}}{x^2+y},\;\;$but this does not seem to be correct.
(2) My second question is
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}$$
I got $\quad\displaystyle \frac{A^6B^4}{1/9},\quad$ but my book's answer is $\quad\displaystyle \frac{1}{9A^6B^4}$
| We are not solving, we are simplifying: For the first, note that
$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}\;=\; \frac{xy^2}{x^{-2} + y^{-1}}\;=\;\frac{xy^2}{\large\frac{1}{x^2} + \frac{1}{y}}$$
Try now to multiply numerator and denominator by $x^2y$:
$$\frac{xy^2}{\left(\large\frac{1}{x^2} + \frac{1}{y}\right)}\cdot \frac{(x^2y)}{(x^2y)} \quad = \quad\frac {x\cdot x^2 \cdot y^2 \cdot y}{\left(\large\frac{x^2y}{x^2} + \frac{x^2y}{y}\right)}\quad =\quad \frac{x^3y^3}{y + x^2} $$
For the second, again, we are simplifying:
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2} \quad = \quad \frac{3^{-2}}{A^{(-3)(-2)}B^{(-2)(-2)}}\quad = \quad \frac{1/9}{A^6B^4}\quad =\quad \frac{1}{9A^6B^4}$$
(as per my now deleted comment below) Alternatively, for the second problem, we proceed as follows:
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}\quad = \quad\left(\frac{A^{-3}B^{-2}}{3}\right)^2 \quad=\quad \frac{A^{-6}B^{-4}}{9} \quad = \quad \frac{1}{9A^6B^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/279637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Primes congruent to 1 mod 6 I came across a claim that I found interesting, but can't seem to prove for some reason. I have the feeling it should be easy
a prime $p$ can be written in the form $p = a^2 -ab +b^2$ for some $a,b\in\mathbb{Z}$ if and only if $p\equiv 1\bmod{6}$
| If $(a,b)=d,d^2\mid (a^2-ab+b^2)$
If $d>1,a^2-ab+b^2$ can not be prime $\implies d=1$
Now, if prime $p=a^2-ab+b^2\implies p\mid (a+b)(a^2-ab+b^2)\implies p\mid (a^3+b^3)$
So, $$a^3\equiv(-b)^3\pmod p\implies \left(-\frac ab\right)^3\equiv 1\pmod p\implies ord_p \left(-\frac ab\right)\mid 3$$
If $ord_p \left(-\frac ab\right)=1,p(=a^2-ab+b^2)\mid (a+b)$
If $(a^2-ab+b^2)\mid(a+b),(a^2-ab+b^2)\mid(a+b)^2$
$\implies (a^2-ab+b^2)\mid 3ab$
$\implies (a^2-ab+b^2)\mid 3$ as $(a^2-ab+b^2,a)=(b^2,a)=1$ as $(a,b)=1$
But, $a^2-ab+b^2>3,$ for $a,b>2$
$$\implies ord_p \left(-\frac ab\right)= 3\implies 3\mid \phi(p)\implies p\equiv1\pmod 3\equiv1,4\pmod 6$$
Hence, $p\equiv1\pmod 6$ as $p\equiv4\pmod 6$ is even and $p>2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/280551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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If $X$ is uniform on the interval $[0, 2]$ , find the PDF of $X^2-2X$ If $X$ is uniform on the interval $[0, 2]$ find the PDF of $Y =
X^2-2X$.
I solved for $X$ and got $x=1 + \sqrt{1+y}$ or $x=1 - \sqrt{1+y}$. Not sure what to do from here.
| Note that $Y=(X-1)^2-1$. Now let $W=X-1$. It is nearly obvious that $W$ is uniform on $[-1,1]$.
We find the distribution of $W^2$. Since $Y=W^2-1$, it is then easy to find the distribution of $Y$.
Let $Z=W^2$. Then (for $0\le z\le 1$),
$$\Pr(Z\le z)=\Pr(W^2\le z)=\Pr(|W|\le \sqrt{z})=\frac{2\sqrt{z}}{2}=\sqrt{z}.$$
Remark: The problem can also be solved using a variant of your calculation. We have $Y\le y$ if and only if $X^2-2X\le y$. This inequality holds if and only if
$$1-\sqrt{1+y}\le X\le 1+\sqrt{1+y}.$$
If $y\ge 0$, the above probability is $1$. So the cdf of $Y$ is $1$ for $y \ge 0$.
If $y\lt -1$, the above condition is an impossible one, so the probability is $0$. Thus the cdf of $Y$ is $0$ when $y\lt -1$. So the only interesting $y$ are between $-1$ and $0$.
For $-1\le y\le 0$, the length of the interval $[1-\sqrt{1+y},1+\sqrt{1+y}]$ is $2\sqrt{1+y}$, and the interval lies within $[0,2]$. So the probability we land in that interval is $\frac{2\sqrt{1+y}}{2}$, that is, $\sqrt{1+y}$.
Now that we have the cdf of $Y$, we can differentiate to get the density.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/280799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve this equation for R/r? How do I solve this expression for $\frac{R}{r}$?
$\frac{1}{(r-R)^3}-\frac{y}{R^2(r-R)}=\frac{1}{r^3}$
| Rearrange the equation to get
$$1 - y \left ( 1 - \frac{r}{R} \right )^2 = \left ( 1 - \frac{r}{R} \right )^3 $$
Set $z = 1-r/R$ and get the equation:
$z^3+y z^2-1=0$
Get a root at $z_0(y)$ numerically or otherwise. Then $R/r$ is then
$$\frac{R}{r} = \frac{1}{1-z_0(y)} $$
NB There is an analytical expression for $z_0$:
$$ z_0(y) = \frac{1}{3} \left(\frac{\sqrt[3]{-2 y^3+3 \sqrt{3} \sqrt{27-4
y^3}+27}}{\sqrt[3]{2}}+\frac{\sqrt[3]{2} y^2}{\sqrt[3]{-2 y^3+3 \sqrt{3}
\sqrt{27-4 y^3}+27}}-y\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the limit of $\frac{e^{6x}-2e^{3x} + 1}{x^2}$, as $x \rightarrow 0$? I am to calculate $\frac{e^{6x}-2e^{3x} + 1}{x^2}$ when $x$ goes towards $0$.
I find that
$$\frac{e^{6x}-2e^{3x} + 1}{x^2} = \frac{(e^{3x}-1)^2}{x^2} = \left(\frac{e^{3x}-1}{x}\right)^2$$
$$\left(\frac{e^{3x}-1}{x}\right)^2 \rightarrow 1^2$$
but according to the answer in the book I am incorrect. It agrees with me halfway through, but ends with
$$\frac{(e^{3x}-1)^2}{x^2} = 9\left(\frac{e^{3x}-1}{3x}\right)^2 \rightarrow 9 \times 1^2$$
While this is correct mathematically, why would it be $3$ and $9$ instead of for example $4$ and $16$ or, as in my case, $1$ and $1$? I don't see the relevance of adding the $3$ and $3^2$.
| I realised my problem while typing in the question.
The common limit I was thinking of is $\frac{e^{x}-1}{x} \rightarrow 1$, and not $\frac{e^{nx}-1}{x} \rightarrow 1$, as is the case of $\frac{e^{3x}-1}{x}$.
Thus, I have to change the denominator into $3x$, and the only way of doing so is to add $3$ to the denominator and the numerator.
The only way of doing so without messing up the beautiful numerator is to add $3^2 = 9$ to the outside, forming $9(\frac{e^{3x}-1}{3x})^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $a$ and $b$ in the given cubic polynomial Find $a$ and $b$ such that $x+1$ and $x+2$ are factors of the polynomials $x^3+ax^2-bx+10$.
Here I am not sure that how can I obtain the value of $a$ and $b$, I tried to multiply $x+1$ and $x+2$ to obtain a quadratic equation by which I divided the obtained quadratic polynomial with the above given cubic polynomial, But it didn't worked. How can I overcome my answer.
| If $x+1$ is a factor of $p(x) = x^3+ax^2-bx+10$, Then Sub. $x=-1$ in $p(x) = 0$
we Get $p(-1)=0\Leftrightarrow -1+a+b+10=0\Leftrightarrow a+b=-9$
Similarly If $x+2$ is a factor of $p(x) = x^3+ax^2-bx+10$, Then Sub. $x=-2$ in $p(x) = 0$
we Get $P(-2)=0\Leftrightarrow -8+4a+2b+10=0\Leftrightarrow 2a+b=-1$
After Solving these two equations, We Get $a=8$ and $b=-17$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/284364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $x_1^6 +x_2^6$ of this quadratic equation without solving it I got this question for homework and I've never seen anything similar to it.
Solve for $x_1^6+x_2^6$ for the following quadratic equation where $x_1$ and $x_2$ are the two real roots and $x_1 > x_2$, without solving the equation.
$25x^2-5\sqrt{76}x+15=0$
I tried factoring it and I got $(-5x+\sqrt{19})^2-4=0$
What can I do afterwards that does not constitute as solving the equation? Thanks.
| \begin{align}
x_1^6+x_2^6 &= (x_1^2+x_2^2)^3-3x_1^4x_2^2-3x_1^2x_2^4 \\
&= (x_1^2+x_2^2)^3-3(x_1x_2)^2(x_1^2+x_2^2).
\end{align}
Since $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$, therefore:
\begin{align}
x_1^6+x_2^6 &= ((x_1+x_2)^2-2x_1x_2)^3-3(x_1x_2)^2((x_1+x_2)^2-2x_1x_2) \\
&= \left( \left(\frac{5\sqrt{76}}{25}\right)^2 -2\left(\frac{15}{25}\right) \right)^3 -3\left(\frac{15}{25}\right)^2 \left( \left(\frac{5\sqrt{76}}{25}\right)^2 -2\left(\frac{15}{25}\right) \right).
\end{align}
The values of $x_1x_2$ and $x_1+x_2$ come from the following argument:
\begin{align}
25(x-x_1)(x-x_2) &= 25x^2-25(x_1+x_2)x+25x_1x_2 \\
&= 25x^2-5\sqrt{76}+15.
\end{align}
Now equate the cofficents of both polynomials to get the values of $x_1x_2$ and $x_1+x_2$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Simple Summation Proof with identities Using some of the identities, determine the value of
$\sum_0^5$ ${12 \choose i}$
. You may use the
substitution ${12 \choose 6}$
= 924, but you may not evaluate the individual chooses.
Proofs of summations are a hard topic for me, hate to show no work, but i am unsure of how to approach this question. Is there any tips as well that you can give when approaching summation problems with proofs? Thankyou.
| $$\begin{array}?
&& 2^{12} \\ &=& (1+1)^{12} \\&=& \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6} + \binom{12}{7} + \binom{12}{8} + \binom{12}{9} + \binom{12}{10} + \binom{12}{11} + \binom{12}{12} \\
&=& \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6} + \binom{12}{5} + \binom{12}{4} + \binom{12}{3} + \binom{12}{2} + \binom{12}{1} + \binom{12}{0} \\
&=& 2\left(\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}\right) + \binom{12}{6}
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differential Equation $y' = \frac{ty(4-y)}{(1+y)}$ $y' = \frac{ty(4-y)}{(1+y)}$
given $ y_0 = 2 $
At what time $t$ when the solution will first be 3.9
I tried solving this but it didnt work out so well.
What I did was:
*
*separate dy/dx and move all $y$ terms to one side with $dy$
*Some how I don't know how to deal with the $dx$, so I couldn't come up with the solution , $y$.
| Separate the $y$ and $t$ terms on either side of the equal side and get
$$ t \, dt = \frac{1+y}{y (4-y)} dy $$
The right-hand side may be broken up as
$$\begin{align} \frac{1+y}{y (4-y)} &= \frac{1}{y (4-y)} + \frac{1}{4-y} \\ &=\frac{1}{4} \left ( \frac{1}{y} + \frac{1}{4-y} \right ) + \frac{1}{4-y} \\ &= \frac{1}{4 y} + \frac{5}{4} \frac{1}{4-y} \end{align}$$
Now integrate both sides:
$$ \frac{1}{2} t^2 + C = \frac{1}{4} [\log{y} - 5 \log{(4-y)}] $$
or
$$ 2 t^2 + C = \log{\left [ \frac{y}{(4-y)^5} \right ]}$$
Using $y(0)=2$ implies that $C=-\log{16}$. Then
$$t^2 = \frac{1}{2} \log{\left [ \frac{16 y}{(4-y)^5} \right ]}$$
When $y=3.9$, $t$ will be
$$t^2 = \frac{1}{2} \log{\left [ \frac{62.4}{(0.1)^5} \right ]}$$
This gives
$$t \approx 2.797$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three inflection points are on a line Given the function $f(x)=\frac{x+1}{x^2+1}$. I am asked to show that $f$ has three inflection points which are lie on a same line.
I know what to find those points by taking $f'$ and then $f''$, but how can I show that there is a line contaning these points?
| Unless, I miscalculated, the numerator of $f''$ becomes
$$2x^3+6x^2-6x-2.$$
Next we find the roots of $x^3+3x^2-3x-1$. Luckily $x_1:=1$ is a root and the polynomial factors as $(x-1)(x^2+4x+1)$. The other two roots hence are $x_{2,3}:=-2\pm\sqrt3$.
Their square: ${x_{2,3}}^2 = 7\mp4\sqrt3 $. And we get the three points:
$$\begin{align} x_1&=1 & y_1:=f(x_1)&=1 \\
x_{2,3}&=-2\pm\sqrt3 & y_{2,3}:=f(x_{2,3})&= \frac{-1\pm\sqrt3}{8\mp 4\sqrt3}=\ldots=\frac{1\pm\sqrt3}4.
\end{align}$$
Then the slopes $\displaystyle\frac{y_2-y_1}{x_2-x_1}$ and $\displaystyle\frac{y_3-y_1}{x_3-x_1}$ become
$$\frac{y_{2,3}-y_1}{x_{2,3}-x_1} = \frac{\frac{1\pm\sqrt3}4-1}{-3\pm\sqrt3} =
\frac14\cdot\frac{-3\pm\sqrt3}{-3\pm\sqrt3} = \frac14. $$
$$ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286541",
"timestamp": "2023-03-29T00:00:00",
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How to prove $|f(x)|\leq \frac{3}{2}$ for all $x\in [-1, 1]$
Let $f(x) = ax^2 + bx + c$ where $a, b, c$ are real numbers. Suppose $f(-1),f(0), f(1) \in[-1, 1]$. Prove that $|f(x)|\leq \frac{3}{2}$ for all $x\in [-1, 1]$.
Here I need to show $|f(x)|\leq \frac{3}{2}$. It means $f(x)$ lies between $-3/2$ and $3/2$. But how can I show that. If I equate $f'(x)$ to $0$ I get $x=-b/2a$.Then stuck. Please help.
| Let $I = [-1, 1]$. We know that:
$f(0) = c \in I$, $f(1) = a + b + c \in I$ and $f(-1) = a - b + c \in I$.
If we subtract the third from the second, we have that:
$f(1) - f(-1) = 2b \Rightarrow b = \frac{f(1)-f(-1)}{2} \in I$
Now, assume that $a>0$. Let $x_m$ and $x_M$ be the points where we have, respectively, the minimum and maximum of the parabola in $I$.
The maximum point candidates are $-1$ and $1$ (borders of $I$). In both cases we will have that $f(x_M) \leq 1 < \frac{3}{2}$.
The candidates to be the minimum are $-1, 1$ (borders of $I$) and $-\frac{b}{2a}$ if this is in $I$. If the minimum is attained at $-1$ or at $1$, then we are ok since in both cases we will have that $-\frac{3}{2} < -1 \leq f(x_m)$.
If $x_m = -\frac{b}{2a} \in I$, then the minimum is $f(x_m) = c - \frac{b^2}{4a} = c + x_m \frac{b}{2}$.
We know that $b, x_m, c \in I$ and these fact yeld to $f(x_m) = c + x_m \frac{b}{2}\geq -1 + x_m \frac{b}{2} \geq -1 + (-1)\frac{1}{2} = -\frac{3}{2}$.
This means that, in all the possible cases ($x_m = -1$ or $x_m= 1$ or $x_m = -\frac{b}{2a} \in I$), we have $f(x_m) \geq -\frac{3}{2}$.
So we have that $|f(x)| \leq \frac{3}{2} ~ \forall x \in I \wedge a > 0$
Similar arguments holds assuming that $a < 0$.
Finally, if $a = 0$, then we have a line and minimum and maximum in $I$ are attained on the border points $-1$ and $+1$, so again we are bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!
| Consider
$$ I_{n+2} = \int_0^{\pi/4} \tan(x)^{n+2} \mathrm dx = \int_0^{\pi/4} \tan(x)^{n}(\sec(x)^2-1) \mathrm dx $$
$$ = -\int_0^{\pi/4} \tan(x)^{n} \mathrm dx + \int_0^{\pi/4} \tan(x)^{n}\sec(x)^2 \mathrm dx $$
$$ \implies I_{n+2}+I_n=\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n^3}+\frac{1}{n^5}-\dots$$
$$\implies 2I_n\sim \frac{1}{n} \implies I_n \sim \frac{1}{2n}, $$
as $ n \to \infty. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
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$\int \frac{x^{2} \arctan x}{1+x^{2}}dx$ So I have the integral
$$\int \frac{x^{2} \arctan x}{1+x^{2}}dx$$
how can I solve this integral without substituting $u=\arctan x$?
Because I think that if I do this, lets suppose that in the end I'll have as a solution $\tan u + \cos u$ when I replace $u$, it will look strange.
| Since
$$\int\frac{x^2}{1+x^2}dx=\int\left(1-\frac{1}{1+x^2}\right)dx=x-\arctan x+C$$
$$\int\frac{\arctan x}{1+x^2}dx=\int\arctan x\,\,d(\arctan x)=\frac{\arctan^2x}{2}+K$$
Thus, integrating by parts:
$$\begin{align*}u=\arctan x&;\;\;u'=\frac{1}{1+x^2}\\{}\\v'=\frac{x^2}{1+x^2}&;\;\;v=x-\arctan x\end{align*}$$
so
$$\int\frac{x^2\arctan^2 x}{1+x^2}dx=x\arctan x-\arctan^2x-\int \frac{x}{x^2+1}dx+\int\frac{\arctan x}{1+x^2}dx\,dx=$$
$$=x\arctan x-\frac{\arctan^2x}{2}-\frac{1}{2}\log(1+x^2)+C$$
| {
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"url": "https://math.stackexchange.com/questions/291510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}$ I'm being asked to evaluate the limit of:
$$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}$$
There aren't any solutions, so I'm just wondering if I'm going about this properly, I first factored the numerator and denominator to:
$$\frac{(x+4)(x+1)}{(x+4)(x-1)}\;.$$
then cancelled both of the $(x+4)$ and plugged in $4$ for the $x$'s to get an answer of $5/3$.
Am I doing this right?
| $$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}=\lim_{x\to 4}\frac{x^2+x+4x+4}{x^2-x+4x-4}=$$
$$=\lim_{x\to 4}\frac{x(x+1)+4(x+1}{x(x-1)+4(x-1)}=\lim_{x\to 4}\frac{(x+1)(x+4)}{(x-1)(x+4)}=\lim_{x\to 4}\frac{(x+1)}{(x-1)}=\frac{5}{3}$$
Directly
$$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}=\frac{40}{24}=\frac{5}{3}$$
and
$$\lim_{x\to -4}\frac{x^2+5x+4}{x^2+3x-4}=\lim_{x\to -4}\frac{(x+1)}{(x-1)}=\frac{3}{5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Definite integral, quotient of logarithm and polynomial: $I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$ I was thinking this integral : $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$$
What I do is use a Reciprocal subsitution, easy to show that:
$$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{\lambda ^2x^2+\lambda x+1}\text{d}x
=\frac{1}{\lambda}\int_0^{\infty}\frac{(\ln x-\ln \lambda)^2}{x^2+x+1}\text{d}x \\
=\frac{\ln ^2\lambda}{\lambda}\int_0^{\infty}\frac{1}{x^2+x+1}\text{d}x+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2x}{x^2+x+1}\text{d}x \\
=\frac{\ln ^2\lambda}{\lambda}\frac{2\pi}{3\sqrt{3}}+\frac{2}{\lambda}\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x$$
But I hav no idea on process the remaining integral:(
Anyone knows how to solve it?
THX guys! I came with another method might work for this:
Recall the handy GF $$\frac{1}{x^2+x+1}=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$
Then we have :
$$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\frac{2}{\left(k+1\right)^3}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$
With $$\sum_{k=1}^{\infty}\frac{\sin (kx)}{k^3}=\frac{\pi ^2}{6}x-\frac{\pi}{4}x^2+\frac{x^3}{12}$$
What can we arrived?
| $$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} = \int_0^1 \dfrac{1-x}{1-x^3} \ln^2(x) dx = \int_0^1 (1-x)\ln^2(x) \left(\sum_{k=0}^{\infty} x^{3k}\right) dx$$
Now note that
$$\underbrace{\int_0^1 x^n \ln^2(x)dx = \int_{-\infty}^0 e^{nt} t^2 e^t dt}_{x \mapsto e^t} = \int_0^{\infty} t^2 e^{-(n+1)t}dt = \dfrac2{(1+n)^3}$$
Hence,$$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} dx = \sum_{k=0}^{\infty} \left(\dfrac2{(1+3k)^3} - \dfrac2{(2+3k)^3} \right) = \dfrac{8 \pi^3}{81 \sqrt3}$$
Let us call
$$\sum_{k=0}^{\infty} \dfrac1{(1+3k)^3} =f$$
and
$$\sum_{k=0}^{\infty} \dfrac1{(2+3k)^3} =g$$
We are interested in $f-g$.
We have
$$\text{Li}_3(\omega) = \sum_{k=1}^{\infty} \dfrac{\omega^k}{k^3} = \omega f + \omega^2 g + \dfrac{\zeta(3)}{27}$$
$$\text{Li}_3(\omega^2) = \sum_{k=1}^{\infty} \dfrac{\omega^{2k}}{k^3} = \omega^2 f + \omega g + \dfrac{\zeta(3)}{27}$$
where $\text{Li}_s(x)$ is the polylogarithm function defined as
$$\text{Li}_s(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k^s}$$
Polylgarithm function satisfies a nice identity namely
$$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ where $B_n(x)$ are Bernoulli polynomials. Take $n=3$ and $x = 1/3$ to get that
$$\text{Li}_3(\omega) - \text{Li}_3(\omega^2) = - \dfrac{(2\pi i)^3}{3!}B_3(1/3) = - \dfrac{(2\pi i)^3}{3!} \dfrac1{27} = \dfrac{8 \pi^3}{6 \times 27}i = \dfrac{4 \pi^3}{81}i$$
We also have that $$\text{Li}_3(\omega) - \text{Li}_3(\omega^2) = (\omega-\omega^2)(f-g) = \sqrt{3}i(f-g)$$
Hence, we get that
$$f-g = \dfrac{4 \pi^3}{81 \sqrt3}$$
We can even get the values of $f$ and $g$ in terms of $\zeta(3)$. Note that $$f+g + \dfrac{\zeta(3)}{27} = \zeta(3) = \text{Li}_3(1)$$
Hence,
$$f = \dfrac{13}{27} \zeta(3) + \dfrac{2 \pi^3}{81 \sqrt3}; \,\,\,\,\,\,\,\, g = \dfrac{13}{27} \zeta(3) - \dfrac{2 \pi^3}{81 \sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Find a basis for a solution set of a linear system I'm trying hard with this exercise, but it is breaking my back.
Find a basis for the solution set of the given homogeneous linear system
$3x_1+x_2+x_3=0$
$6x_1+2x_2+2x_3=0$
$-9x_1-3x_2-3x_3=0$
I do what I know I need to do. First I get the solution set of the system by reducing like this:
$\begin{pmatrix}
3 & 1 & 1 \\
6 & 2 & 2 \\
-9 & -3 & -3 \end{pmatrix} \leadsto \begin{pmatrix}
3 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \leadsto\begin{pmatrix}
1 & 1/3 & 1/3 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}$
So I know $\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix}$
That being the general solution.
Now, giving the values for $r$ and $s$ according to the standard vectors $i$, $j$
$\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix} = r \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix} + s\begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix}$
From my results, the basis will be:
$ ( \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix}, \begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix})$
But instead, the book answer (I'm self-studying )is:
$ ( \begin{bmatrix}
-1\\
3\\
0\end{bmatrix}, \begin{bmatrix}
-1\\
0\\
3\end{bmatrix})$
Any idea on what I'm doing wrong? Thank you :)
| I don't follow this step.
$\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix} = r \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix} + s\begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix}$
Wouldn't you want to write this as:
$\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix} = r \begin{bmatrix}
\frac{-1}{3}\\
1\\
0\end{bmatrix} + s\begin{bmatrix}
\frac{-1}{3}\\
0\\
1\end{bmatrix}$
If you write out the eigenvalues, you get $\lambda = 1, 0 , 0$
The corresponding eigenvectors are:
$(1, 0, 0), (-1/3, 0, 1), (-1/3, 1, 0)$ and these should look familiar to your solution.
Regards
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Factor out into ellipse complex numbers Factor into equation of ellipse the following sets:
$$|z+2i|+ |z-2i| =6$$
$$|z-3|-|z+3|=4$$
I got to the part of taking one of the square roots and bringing it to the other side and then squaring both sides but can't seem to simplify it after that.
Detailed steps would be kindly appreciated, thanks!
| $$|x+yi+2i|+|x+yi-2i|=6$$
$$\sqrt{(x+yi+2i)(x-yi-2i)}+\sqrt{(x+yi-2i)(x-yi+2i)}=6$$
$$\sqrt{x^2+y^2+4y+4}+\sqrt{x^2+y^2-4y+4}=6$$
Square both sides:
$$8+2x^2+2y^2+2\sqrt{x^2+y^2+4y+4}\sqrt{x^2+y^2-4y+4}=36$$
Move $8+2x^2+2y^2$ to the RHS and square again:
$$4(16 + 8 x^2 + x^4 - 8 y^2 + 2 x^2 y^2 + y^4)=784 - 112 x^2 + 4 x^4 - 112 y^2 + 8 x^2 y^2 + 4 y^4$$
Leading to:
$$9x^2+5y^2=45$$
Or:
$$\frac{x^2}{5}+\frac{y^2}{9}=1$$
The second problem is similar.
If you wanted to "cheat" or check your work, you could use the fact that in an ellipse, $r_1 + r_2 = 2a$, in your case $r_1 + r_2 = 6 \to a = 3$, and also the distance between focii is $4 = 2c$. Thus you could have found the coefficient of $y^2$ as $1/a^2=1/9$, and that of $x^2$ as $1/(a^2-c^2)=1/5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/296654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Geometrical meaning of $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$ If $\displaystyle \arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$. Then locus of complex no. $z$ is:
My try: Let $z = x+iy\;, $ We get $\displaystyle \displaystyle\arg\left(\frac{x+iy-1}{x+iy+1}\right)=\frac{\pi }{3}$
$$\displaystyle \arg\left(\frac{(x^2+y^2-1)+2ixy}{(x+1)^2+y^2}\right) = \frac{\pi}{3}$$
$$\displaystyle x^2+y^2-\frac{2}{\sqrt{3}}y-1=0$$
So Locus of $z$ is a Circle.
My question is: by using geometry how can I interpret the given equation $$\displaystyle \arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$$
| Suppose you have a circle $\mathscr{C}$ centered at $O$ and $A$, $B$ are two points on $\mathscr{C}$. $A$ and $B$ split $\mathscr{C}$ into two arcs. If you pick an arbitrary point $C$ from one of its arc, the angle $\measuredangle{ACB}$ will be independent of your choice of $C$ and equal to $\frac12\measuredangle{AOB}$.
Let's identify the complex plane $\mathbb{C}$ with the Eucliean plane $\mathbb{R}^2$
and take $A = 1+0i$, $B = -1+0i$ and $C = z$. Remember,
$\arg(z - 1)$ is the angle between the line $AC$ and the real axis.
$\arg(z + 1)$ is the angle between $BC$ and the real axis.
This means $ \arg(\frac{z-1}{z+1}) = \arg(z-1) - \arg(z+1)$ is nothing but the angle $\measuredangle{ACB}$.
Since the corresponding angle $\measuredangle{AOB} = 2 \measuredangle{ACB} = \frac{2\pi}{3}$ is one third of $360^\circ$, the locus for $\arg(\frac{z-1}{z+1}) = \frac{\pi}{3}$ is an arc on the circumcircle for some isosceles triangle $\triangle ABD$ which has $A$, $B$ as two of its vertices.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\}$ Evaluate
$$\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\}$$
where $\{x\}$ is the fractional part of $x$. Some suggestions here? Thanks!
| For $k \ge 1$, we have
$$\{\sqrt{k^2+2k+2}\} = \sqrt{(k+1)^2+1)} - (k+1) = \frac{1}{2k}( 1 + m_k )$$
for some bounded sequence $m_k$ which $\to 0$ as $k \to \infty$.
Let $M$ be an upper bound for $|m_k|$. Rewrite the expression in the limit as:
$$\begin{align}
&\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\}\\
=&\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\frac{1}{2k}( 1 + m_k )\\
=&\frac{1}{2^{n+1}}\sum_{k=1}^{n}\dbinom{n}{k}( 1 + m_k )\\
=&\frac{2^n-1}{2^{n+1}} + \frac{1}{2^{n+1}}\sum_{k=1}^{n}\dbinom{n}{k} m_k
\end{align}
$$
For any $\epsilon > 0$, pick a $N_1$ large enough such that $|m_k| < \epsilon$ for $k > N_1$. For any $n > N_1$, we have:
$$\begin{align}
\left|\frac{1}{2^{n+1}}\sum_{k=1}^{n}\dbinom{n}{k} m_k\right|
&\le \frac{M}{2^{n+1}}\sum_{k=1}^{N_1}\dbinom{n}{k} + \frac{\epsilon}{2^{n+1}}\sum_{k=N_1+1}^{n}\dbinom{n}{k}\\
&\le \frac{M}{2^{n+1}}\sum_{k=1}^{N_1}\dbinom{n}{k} + \frac{\epsilon}{2}
\end{align}
$$
Notice for fixed $N_1$, $\frac{M}{2^{n+1}}\sum_{k=1}^{N_1}\dbinom{n}{k} \to 0$ as $n \to \infty$. We can pick another $N_2 > N_1$ such that this part is $< \frac{\epsilon}{2}$ whenever $n > N_2$. For $n > N_2$, we then have:
$$
\left|\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\} - \frac{1}{2}\right| \le \epsilon + \frac{1}{2^{n+1}}
$$
Since $\epsilon$ can be arbitrary small, we conclude:
$$\lim_{n\to\infty} \frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\} = \frac12$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$
| $\frac{1}{1\times 2}+\frac{1}{2\times 3}+\ldots+\frac{1}{n(n+1)}$
$=(1-\frac12)+(\frac12-\frac13)+\ldots+(\frac{1}{n}-\frac{1}{n+1})$
$=1-\frac{1}{n+1}$
$=\frac{n}{n+1}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluation by methods of complex analysis $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm{dx}$ How would we evaluate the below integral by methods of complex analysis?
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm{dx}$$
I asked the question a while ago, but at that time I didn't specify this requirement.
Now, I'm only interested in a complex way. Thanks !!!
Sis.
| We have
$$
\int_0^1 \frac{\ln(1+x)}{x^2+1}dx=
\left[\begin{array}{l}
t=\frac{1}{x} \\
dt=-\frac{1}{x^2}dx
\end{array}\right]=\int_1^\infty \frac{\ln(t+1)-\ln{t}}{1+t^2}dt=
\left[\begin{array}{l}
u=t-1 \\
du=dt
\end{array}\right] =
\int_0^\infty \frac{\ln(u+2)-\ln(u+1)}{1+(u+1)^2}du.
$$
I am quite confident that this integral can be handled with a keyhole contour in much a similar way as this integral. If you want me to do more details, just comment, and I will update when I have the time.
EDIT: Continued the argument. Please notify me about misprints or other errors.
I will show how to compute
$$\int_0^\infty \frac{\ln(x+2)}{1+(x+1)^2}dx$$
(the other one is solved analogously).
We now consider the integral
$$\int_C \frac{\log^2(z+2)}{1+(z+1)^2}dz,$$
where $C$ is the keyhole contour above. Note the choice of logarithm! It is imperative that we take $0<\arg{z}<2\pi$ for the logarithms (or $2\pi n<\arg{z}<2\pi (n+1)$ for some $n\in\mathbb{Z}$).
The integrand has poles at $z=-1\pm i$, and by Cauchy's integral formula, we get
$$
\int_C \frac{\log^2(z+2)}{1+(z+1)^2}dz=
\int_C \frac{\log^2(z+2)}{(z+1+i)(z+1-i)}dz=
2\pi i\left(\frac{\log^2(1-i)}{-2i}+
\frac{\log^2(1+i)}{2i}\right)=
\pi(\log^2(1+i)-\log^2(1-i))=
\pi\left(\left(\ln\sqrt{2}+\frac{i\pi}{4}\right)^2-\left(\ln\sqrt{2}+\frac{7i\pi}{4}\right)^2\right)=
\frac{6\pi^3-3i\pi^2\ln{2}}{2}
.$$
The integral along $\gamma$ goes to zero as the inner radius goes to zero, and the integral along $\Gamma$ goes to zeros as $R\rightarrow\infty$ (this can be shown using the ML-inequality).
Therefore, in the limit we only get contributions from the straight parts, so
$$
\int_0^R\frac{\log^2(x+i\varepsilon+1)}{(1+x+i\varepsilon)^2+1}dx+
\int_R^0\frac{\log^2(x-i\varepsilon+1)}{(1+x-i\varepsilon)^2+1}dx\rightarrow \frac{6\pi^3-3i\pi^2\ln{2}}{2}.
$$
But we also have
$$
\int_0^R\frac{\log^2(x+i\varepsilon+2)}{(1+x+i\varepsilon)^2+1}dx+
\int_R^0\frac{\log^2(x-i\varepsilon+2)}{(1+x-i\varepsilon)^2+1}dx\rightarrow
\int_0^\infty\frac{\ln^2(x+2)-\left(\ln(x+2)+2i\pi\right)^2}{(1+x)^2+1}dx=
\int_0^\infty\frac{4\pi^2-4i\pi\ln(x+2)}{(1+x)^2+1}dx,
$$
so
$$\int_0^\infty\frac{4\pi^2-4i\pi\ln(x+2)}{(1+x)^2+1}dx=\frac{6\pi^3-3i\pi^2\ln{2}}{2}.$$
By considering the imaginary part, we get that
$$\int_0^\infty \frac{\ln(x+2)}{(x+1)^2+1}dx=\frac{3\pi\ln{2}}{8}.$$
It is possible to compute
$$\int_0^\infty \frac{\ln(x+1)}{(x+1)^2+1}dx$$
analogously, and then take the difference in order to find the value of the original integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Verifying Trigonometric Identities: $2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$ Verify that $$ 2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$$
| $$ \begin{align*}
2 \cos^2(x)-1&=2 \cos^2(x)-\sin^2(x)-\cos^2(x)\\
&= \cos^2(x)-\sin^2(x)\\
&=(\cos(x)-\sin(x))\cdot (\cos(x)+\sin(x))\\
&= \cos^2(x) (1-\tan(x)) \cdot(1+\tan(x))\\
&=\cos^2(x) (1-\tan^2(x))\\
&=\frac{\cos^2(x)}{\sin^2(x)+\cos^2(x)} \cdot (1-\tan^2(x))\\
&=\frac{1-\tan^2(x)}{1+\tan^2(x)}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Formula for the sequence formed by the digits of the natural numbers
Consider the following sequence:
$$1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 2, 0, 2, 1, 2, 2, 2, 3, \ldots$$
which is formed by extracting the digits of the natural numbers. Is there any formula for the general term of this sequence?
All I can think of is an algorithm to obtain its terms:
"Set $n = 1$. While $n$ is not too big {extract digits from $n$ and insert them into the sequence, then set $n = n + 1$}"
| Given a number $n$, the total number of digits to represent $1$ to $n$ can be obtained as follows.
Let $k = \left \lfloor \log_{10}(n) \right \rfloor$. Then the number of digits from $1$ to $10^k-1$ is given by
$$\sum_{l=1}^k l \cdot 9 \cdot 10^{l-1} = k \cdot 10^k - \dfrac{10^k-1}9$$
Hence, the number of digits needed to list $1$ to $n$ is
$$f(n) = \underbrace{k \cdot 10^k - \dfrac{10^k-1}9}_{\text{Number of digits from $1$ to $10^k-1$}} + \overbrace{(k+1)\left(n+1-10^k\right)}^{\text{Number of digits from $10^k$ to $n$}} = (n+1)(k+1) - \dfrac{10^{k+1}-1}9$$
where $k = \left \lfloor \log_{10}(n) \right \rfloor$. The function you are after is the inverse of $f(n)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is
(a) 0 (b) 2
(c) 4 (d) none of these
Solution: put $y = x + (3+5)/2 = x+4$
the equation becomes
=> $(y-1)^4 + (y+1)^4 = 16$ ---- (i)
=> $2{y^4 + 6(y)^2 + 1 } = 16$ --------(ii)
My question is how was (i) converted to (ii)? I just couldn't get it. Please help?
| $$
\begin{align}
(y−1)^4 +(y+1)^4
&= \left(y^4 − \binom{4}{1}y^3 + \binom{4}{2}y^2 − \binom{4}{3}y + 1\right) \\
&+ \left(y^4 + \binom{4}{1}y^3 + \binom{4}{2}y^2 + \binom{4}{3}y + 1\right) \\
&=2(y^4 + 6y^2 + 1)
\end{align}
$$
| {
"language": "en",
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"answer_count": 4,
"answer_id": 2
} |
Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$
The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.
$$x = 2\tan\theta$$
$$dx = 2\sec^2\theta d\theta$$
$$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{\sec^2\theta} 2\sec^2\theta \,d\theta}$$
$$ = \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$
$$ = 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$
We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$.
$$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) $$
$$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$
But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$
There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.
| Since
$$
\frac52 \ln (\frac{x^2+4}{4})=\frac52 \ln(x^2+4)-\frac52 \ln 4,
$$
the difference between the two answers is an additive constant, which can be absorbed into the constant of integration $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Method to find $\sin (2\pi/7)$ I just thought a way to find $\sin\frac{2π}{7}$.
Considering the equation $x^7=1$
$⇒(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0$
$⇒(x-1)[(x+\frac1 x)^3+(x+\frac1 x)^2-2(x+\frac1 x)-1]=0$
We can then get the 7 solutions of x, but the steps will be very complicated, especially when solving cubic equation, and expressing x as a+bi. The imaginary part of the second root of x will be $\sin\frac{2π}{7}$.
Besides this troublesome way, are any other approach? Thank you.
| Just for laughs, we can at least in principle compute $\cos{(\pi/7)}$ by observing that
$$\sin{\frac{3 \pi}{7}} = \sin{\frac{4 \pi}{7}}$$
Using a combination of double-angle forumlae, we end up with a cubic equation for $\cos{(\pi/7)}$:
$$8 \cos^3{\frac{\pi}{7}} - 4 \cos^2{\frac{\pi}{7}} - 4 \cos{\frac{\pi}{7}}+1=0$$
This equation has one real solution which is $\cos{(\pi/7)}$. The bad news is that the expression is unwieldy at best:
$$\cos{\frac{\pi}{7}}=\frac{1}{6} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 i\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}\right)$$
The imaginary part of this expression is of course zero. The real part, however, ends up being expressed in terms of a sine and cosine of another angle, and I think the point of an exercise like this is to not do that. Anyway, I hope this adds to the discussion above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate: $\int_{0}^\infty e^{-x^2} \cos^n(x) dx $ How to evaluate:
$$ \int_0^\infty e^{-x^2} \cos^n(x) dx$$
Someone has posted this question on fb. I hope it's not duplicate.
| Using the power reduction formula for $\cos$ for odd powers:
\begin{align}
I &= \int_0^\infty e^{-x^2} \cos^n(x) \, dx \\
&= \int_0^\infty e^{-x^2} \left( \frac{2}{2^n} \sum_{k=0}^{(n-1)/2} \binom{n}{k} \cos{((n-2k)x)} \right) \, dx \\
&= \frac{2}{2^n} \sum_{k=0}^{(n-1)/2} \binom{n}{k} \int_0^\infty e^{-x^2} \cos{((n-2k)x)} \, dx \\
\end{align}
The inner integral is a generalization of the Gaussian integral and can be evaluated using differentiation under the integral sign:
$$
\int_0^\infty e^{-x^2} \cos{((n-2k)x)} \, dx = \frac{1}{2} \sqrt{\pi} e^{-\frac{1}{4}(n-2 k)^2}
$$
Therefore, we have:
$$
I = \frac{\sqrt{\pi}}{2^n} \sum_{k=0}^{(n-1)/2} \binom{n}{k} e^{-\frac{1}{4}(n-2 k)^2} \qquad n \text{ odd}
$$
I don't think there is a nice closed form for this sum.
As for even powers, the same method yields:
\begin{align}
I &= \int_0^\infty e^{-x^2} \cos^n(x) \, dx \\
&= \int_0^\infty e^{-x^2} \left( \frac{1}{2^n} \binom{n}{n/2} + \frac{2}{2^n} \sum_{k=0}^{n/2-1} \binom{n}{k} \cos{((n-2k)x)} \right) \, dx \\
&= \frac{\sqrt{\pi}}{2^{n+1}} \binom{n}{n/2} + \frac{2}{2^n} \sum_{k=0}^{n/2-1} \binom{n}{k} \int_0^\infty e^{-x^2} \cos{((n-2k)x)} \, dx
\end{align}
Therefore:
$$
I = \frac{\sqrt{\pi}}{2^{n+1}} \binom{n}{n/2} + \frac{\sqrt{\pi}}{2^n} \sum_{k=0}^{n/2-1} \binom{n}{k} e^{-\frac{1}{4}(n-2 k)^2} \qquad n \text{ even}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A two variable Inequality Show that $(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2$
Can we go like this : $(a^2+b^2)=(a+b)^2-2ab \& (a^4+b^4) = (a^2+b^2)^2-2a^2b^2$ and right hand side of the inequality : $[(a+b)^3-3ab(a+b)]^2$
| Are there any restrictions on $a,b$? RHS is $a^3+b^3 = (a+b)(a^2+b^2-ab)$. Clearly for all $|a|,|b|>1 \ \ a^2+b^2 >a^2+b^2 -ab$ and $ a^4+ b^4> a+b$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^2+x+7 \equiv 0\pmod{81}$ Solve $x^2+x+7\equiv 0 \pmod{81}$
My work:
Prime factorization $81 = 9^2 = 3^4$
Test the value $x\equiv0,1,2$ for $x^2+x+7\equiv0\mod{3}$
we have $x\equiv1\mod{3}$ works.
Now life this to $\mod{3^2} = \mod{9}$
Let $x=1+3k$ for some integer$k$
$(1+3k)^2 + (1+3k) +7 \equiv0\mod{9}$
$1+6k+0+1+3k+7\equiv0\mod{9}$
$9+9k\equiv0\mod9$
$1+k\equiv0\mod9$
$k\equiv-1\mod9$
$k=-1+9m$ for some integer m
Lift again to $\mod81$
$(-1+9m)^2 + (-1+9m)+7\equiv0\mod81$
$-9m+7\equiv0\mod81$
I cant continue... how can I make it work???
thanks!!
| Let's argue following your approach: As you say, $x$ must be of the form $3k+1$. We then have that $$x^2+x+7=9k^2+9k+9=9(k^2+k+1).$$ This is a multiple of $81=9\times 9$ iff $9$ divides $k^2+k+1$. Again, arguing as you did, we see that $k$ must have the form $3t+1$ for some integer $t$, and then $$k^2+k+1=9t^2+9t+3=9(t^2+t)+3,$$ which never is a multiple of $9$, so there are no solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\int \frac{1}{x^{7} - x} ~ d{x} $. How do I evaluate the following indefinite integral?
$$
\int \frac{1}{x^{7} - x} ~ d{x}.
$$
Could someone give me some advice as to what method I should use or the steps that I should take?
Note: The OP originally requested for help in evaluating $ \displaystyle \int \left( \frac{1}{x^{7}} - x \right) ~ d{x} $, which may not have been his/her actual intention.
| $$
\begin{aligned}
\int \frac{1}{x^{7}-x} d x &=\int \frac{1}{x\left(x^{6}-1\right)} d x \\
&=\int\left(\frac{x^{5}}{x^{6}-1}-\frac{1}{x}\right) d x \\
&=\int \frac{x^{5} d x}{x^{6}-1}-\ln |x| \\
&=\frac{1}{6} \ln \left|x^{6}-1\right|-\ln |x|+C \\
&=\frac{1}{6} \ln \left|1-\frac{1}{x^{6}}\right|+C .
\end{aligned}
$$
Replacing 6 by $n$ and changing the sign yields the general integral
$$\int \frac{1}{x\left(x^{n}\pm1\right)} d x=\mp \frac{1}{n} \ln \left|1\pm\frac{1}{x^{n}}\right|+C\tag*{}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you evaluate this limit? $\lim\limits_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$ $$\lim_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$$
This is what I have tried so far, $$\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10}+x^2)^{\frac{1}{2}}}$$
$$\begin{align}
&\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10})^{\frac{1}{2}}+(x)^{\frac{1}{2}}}\\
&\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{5})+(x)}\\
&\lim_{x\to-\infty}\frac{x^2}{x}\\
&\lim_{x\to-\infty}x = -\infty\\
\end{align}$$
Is this the correct approach to finding the limit?
| Observe that for $x<0$, we have $$\begin{align}\sqrt{2x^{10}+x^2} &= \left(\sqrt{x^2}\right)^5\sqrt{2+\frac1{x^8}}\\ &= |x|^5\sqrt{2+\frac1{x^8}}\\ &= -x^5\sqrt{2+\frac1{x^8}}.\end{align}$$ Does that get you started?
Let me approach this two other ways, rigorously, and loosely. Note that the function is undefined for $x=0$, so we will only consider $x\neq 0$ in the following. To study the end behavior of this function, we will ultimately be interested only in the terms of highest degree on top and on bottom. Ideally, we'd like to rewrite the quotient in an equivalent way so that at least one of the highest degree terms is constant. Once we've done that, we'll proceed to the limit. I will call the function $f(x).$
Rigorous:
You expressed a desire to divide top and bottom by $x^5,$ so I will do it that way. The key fact that I will be using several times is that for any real $\alpha,$ we have $$|\alpha|=\sqrt{\alpha^2}.$$ The following manipulations hold for any $x\neq 0:$ $$\begin{align}f(x) &= \frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}\\ &= \frac{x^{-5}}{x^{-5}}\cdot\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}\\ &= \frac{x^{-5}(2x^5+x^2)}{x^{-5}\sqrt{2x^{10}+x^2}}\\ &= \frac{2+x^{-3}}{x^{-5}\sqrt{x^2(2x^8+1)}}\\ &= \frac{2+x^{-3}}{x^{-5}\sqrt{x^2}\sqrt{2x^8+1}}\\ &= \frac{2+x^{-3}}{x^{-5}|x|\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{x^{-4}\sqrt{2x^8+1}}.\end{align}$$ Observe that $x^{-4}$ is positive for all $x\neq 0,$ so in particular, $$x^{-4}=\left|x^{-4}\right|=\sqrt{(x^{-4})^2}=\sqrt{x^{-8}}.$$
Hence, we have $$\begin{align}f(x) &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{x^{-4}\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{x^{-8}}\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{x^{-8}\left(2x^8+1\right)}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}\end{align}$$ for all $x\neq 0$. Now, note that $$\lim_{x\to-\infty}\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}=\sqrt{2},$$ as I believe you've already calculated. Also note that $$\frac{x}{|x|}=\begin{cases}1 & \text{if }x>0,\\-1 & \text{if }x<0,\end{cases}$$ and so $$\lim_{x\to-\infty}\frac{x}{|x|}=-1.$$ Therefore, $$\lim_{x\to-\infty}f(x)=\left[\lim_{x\to-\infty}\frac{x}{|x|}\right]\cdot\left[\lim_{x\to-\infty}\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}\right]=-1\cdot\sqrt{2}=-\sqrt{2}.$$
Loose:
This gets back to what Babak S. mentions in the comment below. When dealing with end-behavior of polynomials, only the highest-degree term ultimately matters. To that end, we can (roughly speaking) "drop" all the terms in numerator and denominator except those of highest degree, and then find the limit that way. That is, $$\begin{align}\lim_{x\to-\infty}f(x) &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2x^{10}}}\\ &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2}\sqrt{x^{10}}}\\ &= \frac{2}{\sqrt{2}}\cdot\lim_{x\to-\infty}\frac{x^5}{\sqrt{(x^5)^2}}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}\frac{x^5}{|x^5|}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}-1\\ &= -\sqrt{2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315471",
"timestamp": "2023-03-29T00:00:00",
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Finding $x+y$ given that $(2^{28}-1)$ is divisible by $x,y$ and between $120 ,130$ The number $(2^{28}-1)$ is divisible by $x,y$.If each of $x,y$ is between $120 ,130$ .How to find $x+y$
| Using the decomposition $a^2-b^2=(a+b)(a-b)$ you can factorize $(2^{28}-1)$ as
$$
2^{28}-1
=
(2^{14}+1)(2^{14}-1)
=
(2^{14}+1)(2^7+1)(2^7-1)
$$
Now, $2^7+1=129$ and $2^7-1=127$...
| {
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Prove: If $n=2^k-1$, then $\binom{n}{i}$ is odd for $0\leq i\leq n$ Kinda stuck on this one. Help is appreciated. I'm going for either a direct or contrapositive proof.
Prove: If $n=2^k-1$, for $k\in\mathbb{N}$, then every entry in Row $n$ of Pascal's Triangle is odd.
I've been considering entry $i$ in row $n$ of Pascal's Triangle, so for $0\leq i\leq n$, we have:
$$
\binom{n}{i}
=\frac{n!}{i!(n-i)!}
=\frac{(2^k-1)!}{i!(2^k-1-i)!}
$$
I've tried manipulating this in a bunch of ways, including using the fact that $\binom{n}{i}=\binom{n-1}{i-1}+\binom{n-1}{i}$, but nothing's panned out.
| If $n=2^k-1$, for $k \in N$, then every entry in Row $n$ of Pascal's Triangle is odd.
$Proof.$ (Direct) Suppose $n=2^k-1$, for $k \in N$. Then $n+1 = 2^k$. Each entry in Row $n+1$ of Pascal's Triangle is given by $\binom{n+1}{i}$, where $0\le i \le n+1$. Clearly, $\binom{n+1}{0}=\binom{n+1}{n+1}=1$, so let's consider two cases for $0<i<n+1$: one where $i$ is odd, and the other where $i$ is even.
$Case 1.$ Let $i$ be odd. If $n+1=2^k$, then $n+1$ is even. The meaning of $\binom{n+1}{i}$ is the total number of subsets of size $i$ from a set, $A$, with $n+1$ elements. Since $n+1$ is even, we can arbitrarily pair elements from $A$ so that any given element is related to exactly one other element in the set which we could call its “complement.” Then for any given subset of $A$ of size $i$, there will be exactly one other subset of size $i$ which contains the unpaired complements of the first subset. Since there will always exist such pairings for all odd size subsets of sets with even cardinalities, $\binom{n}{k}$ is divisible by 2 for all even $n>0$ and odd $k$ where $0<k<n$. Then $\binom{n+1}{i}$ is even for all odd $i$ in $0<i<n+1$.
$Case 2$. Let $i$ be even. Note $\binom{n+1}{i}=\binom{ 2^{k}}{i}=$ ${ 2^{k}! } \over {i! (2^k - i)!}$ $=$ ${ (2^k)(2^k - 1)(2^k - 2)...( 2^k - i + 1 ) } \over { (1)(2)(3)...(i) }$.
Both the numerator and denominator contain $i$ factors. Because $\binom{n+1}{i}$ is an integer, we know that the number of $2$'s in the prime factorization of the numerator must be greater than or equal to that of the denominator. If there are more $2$'s in the numerator, $\binom{n+1}{i}$ is even, and else the $2$'s cancel completely and $\binom{n+1}{i}$ is odd. Then let us only consider the even factors of $\binom{n+1}{i}$. From ${ (2^k)(2^k - 1)(2^k - 2)...( 2^k - i + 1 ) } \over { (1)(2)(3)...(i) }$ we consider only the even factors: ${ (2^k)(2^k - 2)...( 2^k - i + 2 ) } \over { (2)(4)...(i) }$ which has ${i}\over{2}$ factors in both the numerator and denominator. Since both the numerator and denominator contain the same number of factors, we can multiply every factor in both by ${1}\over{2}$ and maintain the equality:
${ (2^k)(2^k - 2)...( 2^k - i + 2 ) } \over { (2)(4)...(i) }$ $=$ $ \frac{1} {2}(2^k) \frac{1}{2}(2^k - 2)... \frac{1}{2}( 2^k - i + 2 ) \over \frac{1}{2}(2)\frac{1}{2}(4)... \frac{1}{2}(i) $ $=$ ${ (2^{k-1})(2^{k-1} - 1)...( 2^{k-1} - \frac{i}{2} + 1 ) } \over { (1)(2)...( \frac{i}{2}) }$.
But this last equality is nothing more than $\binom{2^{k-1}}{{i}/{2}}$. This means if $\binom{2^{k-1}}{{i}/{2}}$ is even then $\binom{n+1}{i}$ is even as well. As we already know from $Case 1$, since $2^{k-1}$ is even, if $\frac{i}{2}$ is odd, $\binom{2^{k-1}}{{i}/{2}}$ is even. On the other hand, if $\frac{i}{2}$ is even, the process can be repeated until reaching some binomial of the form $ \binom{2^{k-j}}{{i}/{2^j}} $ where $\frac{i}{2^j}$ is odd for $0<j<k$, as $i<2^k$. Thus $\binom{n+1}{i}$ is even for all even $i$ where $0<i<n+1$.
Finally, because $\binom{n+1}{i}$ is even for all $0<i<n+1$, and because $\binom{n+1}{i} = \binom{n}{i} + \binom{n}{i-1}$, the parity for all $\binom{n}{i}$ must be the same.
Therefore, since $\binom{n}{0} = 1$, $\binom{n}{i}$ is odd for all $0 \le i \le n$, and every entry in Row $n$ of Pascal's Triangle is odd. $\blacksquare$
| {
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Finding the maximum value If $a^2+b^2+c^2+d^2=4$ are real numbers, then how to find the maximum value for:$$a^3+b^3+c^3+d^3$$
| $x_1=a,x_2=b,x_3=c,x_4=d,n=4,v=\sum\limits_{k=1}^nx_i^2=4$
$\sum\limits_{k=1}^nx_i^3\le\sum\limits_{k=1}^n\left|x_i^3\right|=\sum\limits_{k=1}^n\left|x_i\right|^3$ so we can assume $\boxed{\forall i\in\{1,\dots,n\}, x_i \ge 0}$
Since $v=\sum\limits_{k=1}^nx_i^2$, $\forall i\in\{1,\dots,n\},x_i^2\le v$ that is $x_i \le \sqrt{v}$
So $\sum\limits_{k=1}^nx_i^3\le \sum\limits_{k=1}^n\sqrt{v}x_i^2=\sqrt{v}\sum\limits_{k=1}^nx_i^2\le\sqrt{v}v=v^\frac{3}{2}$
$\boxed{\sum\limits_{k=1}^nx_i^3\le v^\frac{3}{2}}$
If we take take $x_1=\sqrt{v}$ and $\forall i \in\{2,\dots,n\}, x_i = 0$
$\sum\limits_{k=1}^nx_i^2=\left(\sqrt{v}\right)^2=v$
$\sum\limits_{k=1}^nx_i^3=\left(\sqrt{v}\right)^3=v^{\frac{3}{2}}$
So $\boxed{\exists (x_1,\dots x_n)\in\mathbb{R}^n, \sum\limits_{k=1}^nx_i^2=v \land \sum\limits_{k=1}^nx_i^3=v^{\frac{3}{2}}}$
So since you know the sum is less than $v^{\frac{3}{2}}$ and this value is attained for some $ (x_1,\dots x_n)$, you can conclude that $v^{\frac{3}{2}}$ is the maximum of $\sum\limits_{k=1}^nx_i^3$.
And as you can see, the number of variables and the sum of the squares do not influence that result.
| {
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How many integer solutions to the equation (discrete math) How many integer solutions to the equation $x_1 + x_2 + x_3 = 15$ so that $x_1 \ge 3, x_2 \ge 2$ and $x_3 \ge 0$ ? I honestly don't know where to start on this
Thanks
| Let's consider the number of solutions of $x_1 + x_2 + x_3 = n$, where $x_1 \in \{3, 4, ...\}$, $x_2 \in \{2, 3, ...\}$ and $x_3 \in \mathbb{N} \cup \{0\}$.
Let $a_n$ be the number of solutions of that equation. You're looking for $a_{15}$.
We can easily find the generating function of $a_n$:
$$g(x) = (x^3 + x^4 + ...)(x^2 + x^3 + ...)(1+x+x^2+...)$$
So that the coefficient of $x^n$ is $a_n$:
$$g(x) = \displaystyle{\sum_{n=0}^\infty a_n x^n}$$
Therefore, for $|x|<1$, we conclude that:
$$g(x) = \dfrac{x^3}{1-x} \cdot \dfrac{x^2}{1-x} \cdot \dfrac{1}{1-x} = x^5 \cdot (1-x)^{-3}$$
Using the Generalized Binomial Theorem, and since $\binom{-\alpha}{k} = \binom{\alpha + k - 1}{k}(-1)^k$:
$$g(x) = x^5 \displaystyle{\sum_{k=0}^\infty \binom{-3}{k} (-1)^k x^k} = x^5 \displaystyle{\sum_{k=0}^\infty \binom{2+k}{k}x^5} = \displaystyle{\sum_{k=0}^\infty \dfrac{(k+1)(k+2)}{2}x^{k+5}}$$
Therefore:
$$
a_n = \left \{
\begin{matrix}
0 & n < 5 \\
\dfrac{(k+1)(k+2)}{2} & n \geq 5 \end{matrix}
\right .
$$
| {
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"answer_id": 2
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$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
| If for $c>0, xy=c^2\implies y=\frac{c^2}x\implies x+y=x+\frac{c^2}x=z$(say)
So, $x^2-zx+c^2=0$
As this is a quadratic equation in $x$ and $x$ is real,
the discriminant $(-z)^2-4\cdot1\cdot c^2\ge 0\implies z^2\ge4c^2\implies $ either $z\ge2c$ or $z\le -2c$
As $x,y>0,x+y>0$
So, $x+y=x+\frac1x=z\ge2c$
| {
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"url": "https://math.stackexchange.com/questions/317831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 10
} |
Differential equation $y'=\frac{1-xy}{y-x^2}$ I tried many thing but I could not find a method to solve the differential equation $$y'=\frac{1-xy}{y-x^2}$$
$$y'=-x+\frac{1-x^3}{y-x^2}$$
$Z=y-x^2$
$$Z'+2x=-x+\frac{1-x^3}{Z}$$
$$Z(Z'+3x)=1-x^3$$
Could you please give me hint which method can be used for such first order differential equations?
Thanks a lot for advice and answers
| $y'=\dfrac{1-xy}{y-x^2}$
$(y-x^2)y'=1-xy$
Let $u=y-x^2$ ,
Then $y=u+x^2$
$y'=u'+2x$
$\therefore u(u'+2x)=1-x(u+x^2)$
$uu'+2xu=1-xu-x^3$
$uu'=-3xu-x^3+1$
Try to solve this ODE by Wolfram Alpha, you will discover that the general solution is implicitly expressed by $x$ and $x+\dfrac{x^3-1}{u}$ .
Since Wolfram Alpha discover that the substitution $v=x+\dfrac{x^3-1}{u}$ leads the ODE becomes a separable ODE:
Let $v=x+\dfrac{x^3-1}{u}$ ,
Then $u=\dfrac{x^3-1}{v-x}$
$u'=\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}$
$\therefore\dfrac{x^3-1}{v-x}\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}=-3x\dfrac{x^3-1}{v-x}-x^3+1$
$\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}=-3x-(v-x)$
$3x^2(v-x)-(x^3-1)(v'-1)=-3x(v-x)^2-(v-x)^3$
$(x^3-1)(v'-1)=3x^2(v-x)+3x(v-x)^2+(v-x)^3$
$(x^3-1)v'-x^3+1=3x^2(v-x)+3x(v-x)^2+(v-x)^3$
$(x^3-1)v'=x^3+3x^2(v-x)+3x(v-x)^2+(v-x)^3-1$
$(x^3-1)v'=(x+v-x)^3-1$
$(x^3-1)v'=v^3-1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving for $b$ in $25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$ What are the steps to get from:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
to:
$$b = \frac{\sqrt{5}}{5} + \frac{2\sqrt{5} - \sqrt{10}}{10}$$
Thanks.
| Divide by $5$, then subtract the term $$5\left(\frac{\sqrt{10} - 2 \sqrt 5}{50}\right)$$ from the both sides of the equation:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
$$\iff 5 \left(\frac{\sqrt{10} - 2 \sqrt{5}}{50}\right) + b = \frac {\sqrt 5}{5}\tag{divide by 5}$$
$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf -} 5 \left(\frac{\color{blue}{\bf \sqrt{10} - 2 \sqrt{5}}}{50}\right)\tag{subtract term to left of b}$$
$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf +} \color{red}{\bf 5} \left( \frac{\color{blue}{\bf 2 \sqrt{5} - \sqrt{10}}}{\color{red}{\bf 50}}\right) \tag{"reversal of sign"}$$
$$\iff b = \frac{\sqrt 5}{5} + \left(\frac{2 \sqrt 5 - \sqrt{10}}{10}\right)\tag{cancel common factor 5}$$
| {
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"url": "https://math.stackexchange.com/questions/320864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find last items of order to get its sum Find the sum of order:
$$\sum_{n=1}^{∞}\left(\frac{\frac{3}{2}}{2n+3}-\frac{\frac{3}{2}}{2n-1}\right)$$
There is how they count it in book:
$$s_{n} = \left(\frac{3}{10}-\frac{3}{2}\right)+\left(\frac{3}{14}-\frac{1}{2}\right)+\left(\frac{1}{6}-\frac{3}{10}\right)+\left(\frac{3}{22}-\frac{3}{14}\right)+...+\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$
$$s_{n} = \frac{-3}{2}+\frac{-1}{2}+\frac{3}{4n+2}+\frac{3}{4n+6}$$
$$s = \lim_{n->∞}s_{n} = \lim_{n->∞}\left [\frac{-3}2 - \frac12 + \frac3{4n+2} + \frac3{4n+6}\right ] = -2$$
I understand how to solve the lim, I just dont understand, how to get those last items from order, I mean this items:
$$\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$
UPDATE
Now I understand how to get those "last" items. But I'm confused now, why are they even in the sum inside of lim of $s_n$? If I would keep counting next items, they would get canceled. For example, there is $\frac3{4n+2}$ in $s_n$, if I count n+1 item, this would get canceled. So why do we count them in $s_n$ if only first two fractals {$\frac{-3}2, \frac12 $} couldn't be canceled (if not thinking of negative n).
Could anyone explain please?
| For the update: If you look at the denominators, the first fraction goes $5,7,9,11,\ldots$ The second fraction goes $1,3,5,7,9\ldots$ As they enter with opposite signs, all the matching ones can be canceled with one from the other set. We are left with just $1$ and $3$. Because these are in the deominators, that is what becomes $\frac {-\frac 32}1 + \frac {-\frac 32}3=-2$. This is commonly true for telescoping sums. The surviving terms are the first few, which are not canceled. All the rest vanish through cancellation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trouble with factorising a polynomial I'm supposed to show that:
$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$
and the required answers are: $$ P=5, Q=4, R=-4 $$
I tried to solve this with partial fractions like so:
$$5(x-1)(x+2) = A(x+3) + B(x-2)$$
$\implies$ $A$=4, $B$=-4
$\implies$ $Q$=4, R=-4
But where does $P$=5 come from?
Or should I have first multiplied out the numerator and denominator and then used long division to solve?
| You are given the wrong expression. After some calculation I figure out that in order for these two expressions to be equivalent,
$$\frac{5(x-1)(x-2)}{(x-2)(x+3)} = 5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$
instead of having $\frac{5(x-1)(x-2)}{(x-2)(x+3)}$, it should be
$$\frac{5(x-1)(x+2)}{(x-2)(x+3)}$$
Now solve it again it would work well.
My approach to solving it is
$$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = \frac{5x^2+5-10}{(x-2)(x+3)} =\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)}$$
$$\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)} = \frac{5x+15+\frac{40}{(x-2)}}{(x+3)}$$
$$\frac{5x+15+\frac{40}{(x-2)}}{(x+3)} = 5+\frac{40}{(x-2)(x+3)}$$
$$5+\frac{40}{(x-2)(x+3)}=5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The use of master theorem appriopriately I have a recurrence relation and trying to use master theorem to solve it. The recurrene relation is:
$$T(n) = 3T\left(\tfrac n5\right) + \sqrt n$$
Can i use the master theorem in that relation? If so, can i say that $T(n)=\Theta(\sqrt n)$?
Thanks
| This recurrence has an explicit solution when $T(0) = 0$ the same way as was done here.
Let $$n = \sum_{k=0}^{\lfloor \log_5 n \rfloor} d_k 5^k$$ be the base $5$ digit representation of $n.$ We assume that $T(0) = 0$ and that
$$ T(n) = T(3 n/5) + \lfloor \sqrt n \rfloor.$$
It is not difficult to see that we have the following exact formula for all $n:$
$$ T(n) = \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j
\Bigg\lfloor\sqrt{\sum_{k=j}^{\lfloor \log_5 n \rfloor} d_k 5^{k-j}}\Bigg\rfloor.$$
Now to get an upper bound on this consider the case where all digits are equal to four.
$$ T(n) \le \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j
\Big\lfloor\sqrt{5^{\lfloor \log_5 n \rfloor -j +1}-1} \Big\rfloor
< \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j \sqrt{5}^{\lfloor \log_5 n \rfloor -j +1} =
\sqrt{5}^{\lfloor \log_5 n \rfloor +1}
\sum_{j=0}^{\lfloor \log_5 n \rfloor} \left(\frac{3}{\sqrt 5}\right)^j \\
= \sqrt{5}^{\lfloor \log_5 n \rfloor +1}
\frac{\left(\frac{3}{\sqrt 5}\right)^{\lfloor \log_5 n \rfloor +1}-1}{\frac{3}{\sqrt 5}-1} = \frac{3^{\lfloor \log_5 n \rfloor +1}-\sqrt{5}^{\lfloor \log_5 n \rfloor +1}}{\frac{3}{\sqrt 5}-1}.$$
For a lower bound, suppose that the leading digit is one and the rest are zero, giving
$$ T(n) \ge \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j
\lfloor\sqrt{5^{\lfloor \log_5 n \rfloor-j}}\rfloor
> \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j
\left(\sqrt{5^{\lfloor \log_5 n \rfloor-j}} - 1\right) \\
= \sqrt{5}^{\lfloor \log_5 n \rfloor}
\sum_{j=0}^{\lfloor \log_5 n \rfloor} \left(\frac{3}{\sqrt 5}\right)^j
- \frac{1}{2} \left( 3^{\lfloor \log_5 n \rfloor+1} - 1\right) \\=
\sqrt{5}^{\lfloor \log_5 n \rfloor}
\frac{\left(\frac{3}{\sqrt 5}\right)^{\lfloor \log_5 n \rfloor+1}-1}{\frac{3}{\sqrt 5}-1}
- \frac{1}{2} \left( 3^{\lfloor \log_5 n \rfloor+1} - 1\right)\\=
\frac{1}{5}
\frac{3^{\lfloor \log_5 n \rfloor+1}-\sqrt{5}^{\lfloor \log_5 n \rfloor +1}}
{\frac{3}{\sqrt 5}-1}
- \frac{1}{2} \left( 3^{\lfloor \log_5 n \rfloor+1} - 1\right)
$$
Taking the leading terms of the two bounds together we have shown that
$$ T(n) \in \Theta\left(3^{\lfloor \log_5 n \rfloor}\right) =
\Theta\left(3^{\log_5 n}\right) =
\Theta\left(5^{\log_5 3 \log_5 n}\right) =
\Theta\left(n^{\log_5 3}\right).$$
Here we have used the fact that $1/5\, \left( 3/5\,\sqrt {5}-1 \right) ^{-1}-1/2>0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Whether $\sum \limits_{n=1}^{\infty} \dfrac{\left(-1\right)^n n^2}{n^2+5} =\sum \limits_{n=1}^{\infty} \left(-1\right)^n -\dfrac{5}{n^2+5} $ $$\sum \limits_{n=1}^{\infty} \dfrac{\left(-1\right)^n n^2}{n^2+5} =\sum \limits_{n=1}^{\infty} \left(-1\right)^n -\dfrac{5}{n^2+5} $$
I don't trust my solution, mostly because I don't have much experience with using $\left(-1\right)^n$.
Because both summations have the same starting point ($1$) and ending point ($\infty$). Thus $a_n = b_n$.
$$\dfrac{\left(-1\right)^n n^2}{n^2+5} = \left(-1\right)^n - \dfrac{5}{n^2+5} $$
$$\dfrac{\left(-1\right)^n n^2}{n^2+5} + \dfrac{5}{n^2+5} = \left(-1\right)^n $$
$$ \left(-1\right)^n\dfrac{n^2+5}{n^2+5} = \left(-1\right)^n$$
$$ \dfrac{n^2+5}{n^2+5} = \dfrac{\left(-1\right)^n}{\left(-1\right)^n} $$
$$ 1 = 1 $$
| $$ \dfrac{\left(-1\right)^n n^2}{n^2+5} = \dfrac{\left(-1\right)^n ((n^2+5)-5)}{n^2+5} = \dfrac{\left(-1\right)^n (n^2+5)}{(n^2+5)} - \dfrac{\left(-1\right)^n 5}{n^2+5} $$ $$=(-1)^n- \dfrac{\left(-1\right)^n 5}{n^2+5} $$
You missed $(-1)^n$ on R.H.S. with $5$ in numerator.
| {
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"timestamp": "2023-03-29T00:00:00",
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With $xy+yz+zx=-1$, proving: $x^2+2y^2+2z^2 .....$ Assuming $xy+yz+zx=-1$, prove that :
$$x^2+2y^2+2z^2 \geq \frac{1+\sqrt{17}}{2}$$
| You can do this by suitable coefficients and a variant of Cauchy-Schwarz inequality.
First, note that $\dfrac{a_1^2}{b_1} + \dfrac{a_2^2}{b_2} + \dfrac{a_3^2}{b_3} \ge \dfrac{(a_1 + a_2 + a_3)^2}{b_1 + b_2+b_3}$. This readily follows from Cauchy Schwarz, for real numbers $a_i$ and $b_i$.
So, $\dfrac{x^2}{a} + \dfrac{y^2}{b} + \dfrac{z^2}{b} \ge \dfrac{(x + y + z)^2}{a + 2b} = \dfrac{x^2 + y^2 + z^2 - 2}{a + 2b}$
$$ \implies \left( \frac{1}{a} - \frac{1}{a+2b} \right)x^2 + \left( \frac{1}{b} - \frac{1}{a+2b} \right)\left(y^2+z^2\right) \ge \frac{-2}{a+2b}$$
Comparing with the LHS we want in our equality, we have two equations to solve to get the right $a, b$ :
$$ \frac{1}{a} - \frac{1}{a+2b} = 1 \text{ and } \frac{1}{b} - \frac{1}{a+2b} = 2$$
Solving these, you have two possible solutions, choose the one which gives you the desired RHS.
In this case, $b = \dfrac{7+\sqrt{17}}{8} $ and $a = -\dfrac{3+\sqrt{17}}{2} $ do the trick and give RHS of $\dfrac{1+\sqrt{17}}{2} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Continued fractions for $\sqrt{x} $ and beyond, valid formula? For $x > 0$, is this trick valid?
I use $$ ( \sqrt{x}-1)(\sqrt{x}+1)=x-1 $$
then $$ \sqrt{x}+1 = \frac{x-1}{\sqrt{x}+1-2} $$
so I can use iterations to get the rational approximant
$$ \sqrt{x} \approx \frac{a(0)+a(1)x+...a(n)x^{n}}{b(0)+b(1)x+..+b(m)x^{m}} $$
By repeating this procedure I can use this method to get a rational approximant to $ x^{2^{-n}} $ for integer $n$ valid for positive $x$.
| It is possible to use ratio of polynomials from continues fraction expansion:
$$\sqrt x = 1 + \cfrac{x-1}{2+ \cfrac{x-1}{2
+ \cfrac{x-1}{...}}}
$$
so $f_{k}=(2+(x-1)\cdot f_{k-1})^{-1}$ and $y_k=1+(x-1)\cdot f_k$ with initial condition $f_0=1$ and $k\in\mathbb{N}$. For example:
$$
y_{6}=
{x \cdot {{x^3+21 \cdot x^2+35 \cdot x+7}\over{7 \cdot x^3+35 \cdot x^2+21 \cdot x+1}}}$$ $$
y_{10}=
{x \cdot {{x^5+55 \cdot x^4+330 \cdot x^3+462 \cdot x^2+165 \cdot x+11}\over{11 \cdot x^5+165 \cdot x^4+462 \cdot x^3+330 \cdot x^2+55 \cdot x+1}}}
$$
The main property is exact zero value at $x=0$ and ${d\over dx }y_k(x)|_{x=0}=k+1$ so for infinitely large $k$ the derivative trends to infinity.
Note that different properties of $y_k$ for even and odd $k$ is obtained: for even $k$ and $x\to\infty$ function $y_k$ trends to slant asymptote and for odd $k$ trends to constant.
Square root approximation
Green lines is for even $k$ and blue for odd, red line is for $\sqrt x$.
Open question is closed form for polynomial coefficients. As can be noted the coefficients at the $y_k$ function's numerator and denominator are in reverse order. This function can be used for $|x|$ approximation via substitution $x=t^2$ with important features $y_k(t)|_{t=0}=0$ and $y_k(t)|_{t=1}=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Range of $\sin^2x-\sin x +1$ How can we find the range of $f(x) =\sin^2x-\sin x +1$?
The function can be written as $(\sin x-\frac{1}{2})^2+\frac{3}{4}$.
Range of $\sin x$ function is given by : $-1 \leq \sin x \leq 1$. Please guide how to get the result.
| So, $-3/2 \le \sin(x) -1/2 \le 1/2$. Squaring, $0\le (\sin(x)-1/2)^2 \le 9/4$. Now add $3/4$ to see that $3/4 \le (\sin(x) - 1/2)^2 + 3/4 \le 3$. So we know that the range lies in the interval $[3/4, 3]$. Now $sin(3\pi/2) = -1$ so
$f(-3\pi/2) = 9/4$. We have $\sin(\pi/3) = 1/2$ so $f(\pi/3) = 3/4$. We now know that $3$ and $3/4$ lie in the range of $f$. By the intermediate value theorem, the range of $f$ is the whole interval.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding integers $a,b$ and $c$ such that $a^3+b^3 = c^3$ We are learning about the the Pythagorean Theorem in class. It says that $a^2+b^2 = c^2$. My homework problem says the following:
Find integers $a,b$ and $c$ such that $a^3+b^3 = c^3$.
How do I solve this equation?
I've been starting with $(3,4,5), (4,4,5)$ etc. Basically I am starting Pythogrean triples.
| If $a=0,b^2=c^2\implies b=\pm c$ and $b^3=c^3\implies b=c$
$\implies b=c$
So, $(0,b,b)$ is a solution
Similarly, $(b,0,b)$ is a solution
If $c\ne0, \left(\frac ac\right)^2+\left(\frac bc\right)^2=1$
$\implies \frac ac<1\implies \left(\frac ac\right)^2>\left(\frac ac\right)^3$
Similarly, $\left(\frac bc\right)^2>\left(\frac bc\right)^3$
$\implies \left(\frac ac\right)^3+\left(\frac bc\right)^3<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330363",
"timestamp": "2023-03-29T00:00:00",
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Triple Integration The problem is as follows: Compute the intergal
$$I= \iiint_B \frac{x^4+2y^4}{x^4 +4y^4 +z^4 }\text{d}x\,\text{d}y\,\text{d}z $$
where $B$ is the unit ball defined by $B=$ {$(x,y,z)∣x^ 2 +y^ 2 +z^ 2 \leq 1$} .
The official solution is tricky: The change of variable $(x,y,z) \mapsto (z,y,x)$ transforms the integral into
$$ I= \iiint_B \frac{z^4+2y^4}{x^4 +4y^4 +z^4 }\text{d}x\,\text{d}y\,\text{d}z \qquad(1)$$
hence $2I= \iiint_B 1\, \text{d}x\,\text{d}y\,\text{d}z = 4\pi/3$. Therefore, $I=2\pi/3$.
My question is: what is meant by $(x,y,z)\mapsto (z,y,x)$ , isn't it ambiguous? Somebody have another solution saying that by symmetry, we have $I$ equals $(1)$ and adding gives the answer. I wanna ask if symmetric here means that the permutation of the variables preserves the domain $D$? Why if it is symmetric, then the two integrals are the same?
The next question is to compute the integral
$$J=\int_0^1 \int_0^1 \int_0^1 \cos^2(\frac{\pi}{6}(x+y+z)) \text{d}x\,\text{d}y\,\text{d}z$$
The solution uses similar technique in above: substitutes $x=1-u. y=1-v, z=1-w$, then we will get $$J=\int_1^0 \int_1^0\int_1^0 \cos^2(\frac{\pi}{2}-\frac{\pi}{6}(u+v+w)) \text{d}u\,\text{d}v\,\text{d}w$$ but why it also equals to
$$\int_0^1 \int_0^1\int_0^1 \cos^2(\frac{\pi}{2}-\frac{\pi}{6}(u+v+w)) \text{d}u\,\text{d}v\,\text{d}w$$
did I do something wrong??
Thank you for answering such a dumb question!
| For your first question, the solution is not so mysterious. You are integrating over a region symmetric in $x$ and $z$. Therefore you can switch $x$ for $z$ and get the same answer. Specifically:
$$x^2+y^2+z^2=1 \implies z^2+y^2+x^2=1$$
$$\begin{align}I=\iiint_B dx\,dy\,dz \frac{x^4+2 y^4}{x^4+4 y^4+z^4} &= \iiint_B dz\,dy\,dx \frac{z^4+2 y^4}{z^4+4 y^4+x^4}\\ &= \iiint_B dx\,dy\,dz \frac{z^4+2 y^4}{z^4+4 y^4+x^4}\end{align}$$
Note that the volume element is the same after substitution. You may then add the two equal quantities as you mentioned above:
$$\begin{align}2I &= \iiint_B dx\,dy\,dz \frac{x^4+2 y^4+z^4+2 y^4}{x^4+4 y^4+z^4} \\ &= \iiint_B dx\,dy\,dz\\ &= \frac{4 \pi}{3}\\\implies I&=\frac{2 \pi}{3} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding an inflection point Can anyone help me find an inflection point for the following function without using graphing calculator.
Determine the intervals of concavity and points of inflection for the function
$f(x)=x\sqrt{25-x^2}$
for my first derivative I did
$f'(x)=1\sqrt{25-x^2}+(x)\frac{1}{2\sqrt{25-x^2}}(-2x)$
for my common denominator I got
$\frac{50-4x^2}{2\sqrt{25-x^2}}$
for my second derivative I got
$f''(x)=\frac{(-4x)\sqrt{25-x^2}-(25-2x^2)(1/2)\sqrt{25-x^2}(-2x)}{\sqrt{(25-x^2)^2}}$
But I am having trobule finding the inflection point.
| First, simplify your $f'(x)$ before taking the second derivative.
$$f'(x)= \dfrac{(\sqrt{25-x^2})^2 - (x^2)}{\sqrt{25-x^2}} $$
$$ = \dfrac{25 - 2x^2}{\sqrt{25-x^2}}$$
Now use the quotient rule to compute $f''(x)$.
(You need to correct your $f''(x)$ and simplify, to find the common denominator), then determine when $f''(x) = 0$. That's where an inflection point would be.
Of course, we also need to check $x = 5, x = -5$ because that is where the denominator will be zero, and the derivatives (first and second) undefined. So just check what is happening there $f(0)$, however, will be zero as well.
$$f''(x) = \frac{(-4x)\sqrt{25-x^2}-(25-2x^2)\large\frac{-x}{\sqrt{25-x^2}}}{25-x^2} $$
$$ = \frac{(-4x)(25-x^2)-(25-2x^2)(-x)}{(25-x^2)\sqrt{25-x^2}} $$
$$ = \frac{-75x+2x^3}{(25-x^2)^{\large\frac32}} $$
$$ = \frac{x(2x^2 - 75)}{(25-x^2)^{\large\frac32}} $$
$$ = \frac{2x(x^2 - \frac{75}{2})}{(25 - x^2)^{\large \frac 32}}$$
Now, when is $f''(x) = 0$? Be careful with $x = 5, x= -5$: you want to check the point, but note that neither the first nor the second derivative is defined there.
When $2x = 0,\; f''(x) = 0$, as is $f'$ and $f$.
When $x^2 - 75/2 = 0, f''(x) = 0 \implies x \pm 5\sqrt{\frac 32} = 0 \implies $
But to confirm that that there is an inflection point at $x=0$, you'll want to check whether $f''(x)$ changes signs on there, from negative to positive, or positive to negative. It does, indeed change signs at $x = 0$: as $x \to 0$ from the left, $f''(x) > 0$, and when $x \to 0$ from the right, $f''(x) <0$. So there is, indeed, an inflection point at $(0, 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Partial Fraction Decomposition Problem... half answered... $$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$
I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:
$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{Cx+D}{x^2+3}$$
Multiplying by the LCD : $(x+3)^2(x^2+3)$
I am left with :
$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2$$
By setting $x=-3:B=-4$
Now is where I am running into trouble. Now that I can substitute B into the original decomposition equation, There is no value of x that will leave only one variable to solve for. Please lend me a hand you guys(and girls). Thanks!
| $$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2=$$
$$=(A+C)x^3+(3A+B+6C)x^2+(3A+6D+9C)x+(9A+3B+9D)$$
equating the coefficients next to the same power of $x$ we get following system
$$A+C=5$$
$$3A+B+6C=19$$
$$3A+6D+9C=27$$
$$9A+3B+9D=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$ Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?
| Let $a=\sqrt{1-\sqrt{x}}, b=\sqrt{1+\sqrt{x}}$, and $b>a>0$ for all $x\in\left(0,\frac{1}{2}\right)$. One could get that $ab=\sqrt{1-x}$, $b^2-a^2=2\sqrt{x}$, $1-a^2b^2=\sqrt{x}$, and $b^2+a^2=2$. Then, one could obtain: $b+a=\sqrt{2}\cdot\sqrt{1+ab}$ and $b-a=\sqrt{2}\sqrt{1-ab}$.
Now, when one would prove that the value of the above function is equal to $\sqrt{2}+1$ for all $x\in\left(0,\frac{1}{2}\right)$. Given $y=f(x)=\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}=\frac{b+\sqrt{1+ab}}{a+\sqrt{1-ab}}\equiv \sqrt{2}+1$. So, it only required to prove that $y^2-2y-1=0$. The proof is the same as to prove that $\left(\frac{b+\sqrt{1+ab}}{a+\sqrt{1-ab}}\right)^2-2\left(\frac{b+\sqrt{1+ab}}{a+\sqrt{1-ab}}\right)-\left(\frac{a+\sqrt{1-ab}}{a+\sqrt{1-ab}}\right)$ shall be equal to zero for all $x\in\left(0,\frac{1}{2}\right)$. In the case of a non-trivial solution being existed in, the portion of denominator shall not be equal to zero, and the numerator shall be zero. After expansion, one got the portion of numerator as follows: $(b+\sqrt{1+ab})^2-2(b+\sqrt{1+ab})(a+\sqrt{1-ab})-(a+\sqrt{1-ab})^2=b^2-a^2+2ab-2ab+2(b-a)\sqrt{1+ab}-2(b+a)\sqrt{1-ab}-2\sqrt{1-a^2b^2}\equiv 2\sqrt{x}+2\sqrt{2}\sqrt{1-ab}\sqrt{1+ab}-2\sqrt{2}\sqrt{1+ab}\sqrt{1-ab}-2\sqrt{1-a^2b^2}=2\sqrt{x}+2\sqrt{2}\sqrt{1-a^2b^2}-2\sqrt{2}\sqrt{1-a^2b^2}-2\sqrt{x}=0$. So, the proof was done, and we show that $y=f(x)=\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}\equiv\sqrt{2}+1$, $\forall x\in\left(0,\frac{1}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$
We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.
Any idea? Or any other method to establish the convergence?
| I figured I'd take a different approach to it, by providing upper and lower bounds.
For an upper bound, we'll assume that all values in the positive sequences are the largest of them, and all values in the negative sequences are the smallest. That is, for instance...
$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} < \frac{3}{4}$
And
$\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}>\frac{4}{10} = \frac{2}{5}$
For the positive terms, we have that the sums are bounded above by $\frac{2n+1}{n(2n+1)+1} < \frac{1}{n}$, while the negative terms' sums are bounded below by $\frac{2}{2n+1} > \frac{1}{n+1}$. As such, we can say easily that the series cannot be larger than
$$
1+\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = 2
$$
Similarly, the positive sums are bounded below by $\frac{1}{n+1}$ and the negative sums are bounded above by $\frac{2n}{n(2n-1)+1} < \frac{1}{n-1}$. And so, the series cannot be smaller than
$$
\frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n+1} - \frac{1}{n-1}\right) = -1
$$
As such, we know that the series must be less than 2 and greater than -1, so we know that the series does not diverge, and thus we only need to show that it does not fail to converge due to oscillation. As both the positive and negative sum terms behave as $1/n$ in the limit as $n \to \infty$ (thereby preventing cyclic behaviour), we know that the series must converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 5
} |
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