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Subsitution - integrals I am attempting to solve this integral using substituion
$\int (x^2 +1) (x^3 +3x)^4dx$ I make $u=x^3+3x$ and then made $dx=du/(3x^2 + 3)$
I then got $1/3 \int (x^3+3x)^4$ I have no idea what to do now.
| Your substitution $u=x^3+3x$ is a good idea as it yields $\frac{du}{dx}=3x^2 + 3$
Substituting in you get:
$\int (x^2 +1) (x^3 +3x)^4dx$
$= \frac{1}{3} \int (3x^2 + 3)(x^3 + 3x)^4 dx$
$= \frac{1}{3} \int u^4 du$ (this is where you seemed to have gone wrong by converting to du yet leaving the integral in terms of x)
$= \frac{1}{15} u^5 + c$
$= \frac{1}{15} (x^3 + 3x)^5 + c$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there an easy way to determine when this fractional expression is an integer? For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer?
$$z=\frac{(1-x)-(1+x)y}{(1+x)+(1-x)y}$$
The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several factoring tricks without luck. The best I could do was find three solutions such that $0<x\le y\le z$. They are: $(2,5,8)$, $(2,4,13)$ and $(3,3,7)$.
The expression seems to converge pretty quickly to some non-integer between 1 and 2.
| We write out a solution, to check whether something like it can be "borrowed" for a problem set.
The numerator is $1-x-y-xy$, which can be written as $-[(x+1)(y+1)-2]$. The denominator is $-[(x-1)(y-1)-2]$. So our fraction is
$$\frac{(x+1)(y+1)-2}{(x-1)(y-1)-2}.$$
We can take $x=1$ or $y=1$. The ratio is then an integer, albeit negative. This gives infinitely many trivial solutions. We will list the non-trivial solutions.
By symmetry we can assume that $x \le y$. Let $x=2$. Then we want $(-1-3y)/(3-y)$ to be an integer. But
$$\frac{3y+1}{y-3}=3+\frac{10}{y-3}.$$
So $y-3$ must divide $10$, giving (since $y \ge x$), the solutions $y=2$, $4$, $5$, $8$, and $13$.
Next we deal with $x=3$. A calculation similar to the previous one (but shorter) gives that the only $y\ge 3$ are given by $y=3$ and $y=7$.
Next we deal with $x=4$. We need $\dfrac{5y+3}{3y-5}$ to be an integer. Any common divisor of these two numbers must divide $3(5y+3)-5(3y-5)$, which is $34$. The only divisor of $34$ which is of the form $3y-5$, where $y \ge 4$, is given by $y=13$.
Similarly, if $x=5$, we get the solution $y=8$.
For the rest, we use essentially the analysis of @Ragib Zaman. One needs to verify that $2x+2y< xy-x-y-1$, or equivalently that $(x-3)(y-3)>10$. This is true if $x\ge 6$ and $y\ge 7$. (It fails at $x=y=6$, but that is not a solution.)
| {
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Solving $A(x) = 2A(x/2) + x^2$ Using Generating Functions Suppose I have the recurrence:
$$A(x) = 2A(x/2) + x^2$$ with $A(1) = 1$.
Is it possible to derive a function using Generating Functions? I know in Generatingfunctionology they shows show to solve for recurrences like $A(x) = 2A(x-1) + x$. But is it possible to solve for the above recurrence as well?
| As per Qiaochu's comment on the answer so far, consider
$$a_n = 2 a_{\lfloor n/2 \rfloor} + n^2$$
with $a_1 = 1$ and $a_n = 1, 6, 11, 28, 37, 58, 71, 120,\dots$ for $n = 1,2,\dots$. Then
$$a_{2n} = 2 a_n + 4n^2 \quad\quad\text{and}\quad\quad a_{2n+1} = 2 a_n + 4n^2 + 4n + 1$$
where both recurrences are valid for $n\ge 1$. Working with each recurrence we can use generating functions to obtain a system of functional equations. Let
$$f(z) = \sum_{n=1}^{\infty}a_n z^n$$
be the generating function for the sequence of $a_n$'s. Working with the first equation we multiply by $z^{2n}$, sum over all $n\ge 1$
$$\sum_{n=1}^{\infty}a_{2n}z^{2n} = 2\sum_{n=1}^{\infty}a_n(z^2)^n + \sum_{n=1}^{\infty}4n^2z^{2n}$$
and obtain
$$ \frac{f(z) + f(-z)}{2} = 2f(z^2) + \frac{4 z^2 \left(1+z^2\right)}{\left(1-z^2\right)^3}$$
Working with the second equation, we multiply by $z^{2n+1}$, sum over all $n\ge 1$
$$\sum_{n=1}^{\infty}a_{2n+1}z^{2n+1} = 2z\sum_{n=1}^{\infty}a_n(z^2)^n + \sum_{n=1}^{\infty}(4n^2+4n+1)z^{2n+1}$$
and obtain
$$\frac{f(z)-f(-z)}{2} -z = 2zf(z^2)+\frac{z^3 \left(z^4-2 z^2+9\right)}{\left(1-z^2\right)^3}$$
We can obtain a solution by solving these functional equations.
EDIT: Adding the two equations together and simplifying we obtain
$$f(z) = \frac{z+z^2}{(1-z)^3} + 2(1+z)f(z^2)$$
We can iterate this equation to obtain better and better approximation of $f(z)$. I believe that if we iterate enough times to have $f(z^{2^t})$ as part of the approximation, then the approximation will be exact for the first $2^t-1$ coefficients.
| {
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Trick to find multiples mentally We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.
Example:
$21553$
Cut digits into pairs:
$2, 15, 53$
Multiply $53$ by $1, 15$ by $2, 2$ by $4$:
$8, 30, 53$
Add:
$8+30+53=91$
As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.
This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$
I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.
Exemple with $91$:
$91$
$9, 1$
$9\cdot3, 1\cdot1$
$27, 1$
$28$
My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?
| I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$ z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$
Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,
$$ 10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$
The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$ r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$
To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1)
+ \cdots $$
Since all this numbers $ (10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.
Using this approach we can derive divisibility of any integer.
| {
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How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed.
Can anyone help me solve it?
| It's kind of fun to "unveil" these mysterious formulas using only trigonometry, by which they all appear to be very simple angle relationships.
For $x>1$ consider the Figure below.
$\hskip1.5in$
Start with a right-angled triangle with hypothenuse $\overline{AC} = x+1$ and side $\overline{BC}=x-1$, so that
$$\alpha = \arcsin\left(\frac{x-1}{x+1}\right),$$
and, by Pythagorean Theorem,
$$\overline{AB} = 2\sqrt x.$$
Extend $BC$ to a segment $\overline{CD} = x+1.$ Then $\overline{BD} = 2$ and
$$\beta = \arctan \sqrt x.$$
Now use the fact that $\triangle ACD$ is isosceles and $\triangle ABD$ is right-angled to write
$$ \beta + (\beta - \alpha) = \frac{\pi}{2},$$
i.e.
$$2\arctan\sqrt x = \arcsin\left(\frac{x-1}{x+1}\right) + \frac{\pi}{2}.$$
For $0<x<1$ use the Figure below.
$\hskip1.5in$
Here $\overline{AC} = x+1$ and $\overline{BC} = 1-x$, so that
$$\alpha = -\arcsin\left(\frac{x-1}{x+1}\right).$$
Again we have $\overline{AB}=2\sqrt x$.
Extend $BC$ to a segment $\overline{BD} = 2$, so that $\overline{AC} = \overline{CD} = x+1$ and
$$\beta = \arctan\sqrt{x}.$$
Since $\triangle ACD$ is isosceles and $\triangle ABD$ is right-angled, we have, this time
$$(\alpha + \beta) + \beta = \frac{\pi}{2}.$$
Once the replacemente is done, this yields again the desired relationship, which therefore is valid for $x>0$. $\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$ Facing difficulty finding limit
$$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$
For starters I have trouble simplifying it
Which method would help in finding this limit?
| If you know that
$$\lim_{x\to\infty}\left(1 + \frac{a}{x}\right)^x = e^{a},$$
so that
$$\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x = e^{-1},$$
then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product,
$$\begin{align*}
\left(\frac{x}{x-1}\right)^{2x+1} &= \left(\left(\frac{x}{x-1}\right)^x\right)^2\left(\frac{x}{x-1}\right)\\
&= \left(\frac{1}{\left(\frac{x-1}{x}\right)^x}\right)^2\left(\frac{x}{x-1}\right)\\
&= \left(\frac{1}{\left(1 - \frac{1}{x}\right)^x}\right)^2\left(\frac{x}{x-1}\right).
\end{align*}$$
Then use limit laws to compute it.
| {
"language": "en",
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Simplifying trig expression I was working through some trig exercises when I stumbled upon the following problem:
Prove that: $ \cos(A+B) \cdot \cos(A-B)=\cos^2A- \sin^2B$.
I started out by expanding it such that
$$ \cos(A+B) \cdot \cos(A-B)=(\cos A \cos B-\sin A \sin B) \cdot (\cos A \cos B+ \sin A \sin B),$$
which simplifies to:
$$ \cos^2 A \cos^2 B- \sin^2 A \sin^2 B .$$
However, I don't know how to proceed from here. Does anyone have any suggestions on how to continue.
| Here is a detailed answer.Let's rock!
$$
\require{cancel}\begin{align}
\cos\left(A-B\right)\cdot\cos\left(A+B\right)&=\left(\cos A\cos B-\sin A\sin B\right)\left(\cos A\cos B+\sin A\sin B\right)\\
&=\cos^2A\cos^2B-\sin^2A\sin^2B\\
&=\cos^2A\left(1-\sin^2B\right)-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cos^2A-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cancelto{1}{\left(\cos^2A+\sin^2A\right)}\\
&=\cos^2A-\sin^2B
\end{align}
$$
I hope this helps.
| {
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How many ordered triple $ (p,a,b) $ is possible such that $p^a=b^4+4$? If we have a prime number $p$ and two natural numbers $a$ and $b$ such that $p^a=b^4+4$,
then how many such ordered triplets $(p,a,b)$ exist?
What should be the strategy to solve this one? The only I can see is $(5,1,1)$, is this the only one? if yes, how could we prove that?
| This is an old contest problem, I wish I could remember where I first saw it. Anyway, André's comment that $$b^4+4=(b^2-2b+2)(b^2+2b+2)$$ is the key to a solution.
Looking modulo $16$, we see that $b^4+4$ cannot be a power of $2$. For $b>1 $, both factors will be strictly greater then $1$, so that if $p^k|(b^4+4)$ then $p$ must divide both $b^2-2b+2$, and $b^2+2b+2$. Since $\gcd(b^2-2b+2,b^2+2b+2)$ must divide $4b$, and $p$ divides both terms, we see that $p|b$. This then implies that $p$ divides $4$ which is impossible.
If $b=1$, then we get the one special case.
| {
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Solving quadratic equation $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
Rearranging it I get: $1-x^2=x-x^2$, and so $x=1$. But the question Im doing says to find 2 solutions. How would I find the 2nd solution?
Thanks.
| $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
$$\frac{1}{x^2} = \frac{1}{x} $$
$$x^2-x=0$$
$$x(x-1)=0$$
The solutions are $x=1$ or $x=0$.
| {
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Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$
Is there any formula that tells this or why is it like that?
| \begin{align}
\sin(3x) &= \sin(x+2x) \tag{1} \\
\sin(\alpha+\beta) &= \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \tag{2} \\
\sin 2\alpha &= 2\cdot \sin \alpha \cdot \cos \alpha \tag{3} \\
\cos 2\alpha &= \cos^2 \alpha - \sin^2 \alpha \tag{4} \\
1 &= \sin^2 \alpha + \cos^2 \alpha \tag{5} %%
\end{align}
If you apply all these formulas you should get:
$$ \sin(3x)=3\cdot \sin x -4\cdot \sin^3 x $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solutions of some Diophantine equations Respected Mathematicians,
The Diophantine equation
$$2^x + 5^y = z^2$$
has solutions $$x = 3, y = 0, z = 3$$ and $$x = 2, y = 1, z = 3$$ I got these solutions by trial and error method. To be honest, these solutions are below the number $5$. So, I easily verified them by trial and error method. I would like to know the method which will give the solutions of the above equation, as well as the solutions of equations below.
a)
$$4^x + 7^y = z^2$$
b)
$$4^x + 11^y = z^2$$
Looking forward to your solution and support.
baba
| I'll do a piece of it to show you some methods you can try on the other pieces.
$2^x+5^y=z^2$. Let's do the case where $y=2s$ is even.
$2^x=z^2-(5^s)^2=(z+5^s)(z-5^s)$, so $z+5^s=2^m$ and $z-5^s=2^n$ with $m+n=x$. Eliminating $z$, $2\times5^s=2^m-2^n$, so $5^s=2^{m-1}-2^{n-1}$. The left side is odd, so the right side is odd, so $n=1$, and $5^s=2^{m-1}-1$. Left side is 1 modulo 4, so right side is 1 modulo 4, so we must have $m=2$. So if there's a solution with $y$ even, then $x=3$, $y=0$, $z=3$.
| {
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Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
| Edited Here is a much simpler version of the previous answer.
$$\int_0^1 \frac{x^4+1}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \int_0^1 \frac{x^2}{x^6+1}dx$$
After canceling the first fraction, and subbing $y=x^3$ in the second we get:
$$\int_0^1 \frac{x^4+1}{x^6+1}dx =\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0^1 \frac{1}{y^2+1}dy = \frac{\pi}{4}+\frac{\pi}{12}=\frac{\pi}{3} \,.$$
P.S. Thanks to Zarrax for pointing the stupid mistakes I did...
| {
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Find $DF$ in a triangle $DEF$ Consider we have a triangle $ABC$ where there are three points $D$, $E$ & $F$ such as point $D$ lies on the segment $AE$, point $E$ lies on $BF$, point $F$ lies on $CD$. We also know that center of a circle over ABC is also a center of a circle inside $DEF$. $DFE$ angle is $90^\circ$, $DE/EF = 5/3$, radius of circle around $ABC$ is $14$ and $S$ (area of $ABC$), K (area of DEF), $S/K=9.8$. I need to find $DF$. Help me please, I'd be very grateful if you could do it as fast as you can. Sorry for inconvenience.
| I refer to the diagram of Victor Liu's answer. This is an analytical
verification that $DF=8$ (which means that $u=2$) but omits some details. Using the equation of the circle centered at $(0,0)$ with radius $14$, the equation of $AE$ (tangent to the small circle at the point $(x,y)=(-8/5,6/5)$)
$$
y=\frac{4}{3}\left( x+\frac{8}{5}\right) +\frac{6}{5}
$$
and the equations of $BF$ ($y=-2)$ and $CD$ ($x=2$), we get the coordinates
of the vertices of triangle $ABC$:
$$
A\left( -\frac{8}{5}+\frac{24}{5}\sqrt{3},\frac{32}{5}\sqrt{3}+\frac{6}{5}
\right) ,\qquad B(-8\sqrt{3},-2),\qquad C(2,-8\sqrt{3}).
$$
The coordinates of the vertices of the right triangle's $DEF$ are
$$
D(2,8),\qquad E(-4,-2),\qquad F(2,-2).
$$
The lengths of the sides of $ABC$ computed by the distance formula are
$$
a =BC=14\sqrt{2}, \qquad
b =AC=\frac{42}{5}\sqrt{10},
\qquad
c =AB=\frac{56}{5}
\sqrt{5}.
$$
The semi-perimeter $p$ of $ABC$ is thus
$$
p=\frac{a+b+c}{2}=7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}.
$$
By Heron's formula the area of $ABC$ is
$$
S=S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)}.
$$
Since
$$
\begin{eqnarray*}
&&p(p-a)(p-b)(p-c) \\
&=&\left( 7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right) \left(
-7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right) \\
&&\times \left( 7\sqrt{2}-\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right)
\left( 7\sqrt{2}+\frac{21}{5}\sqrt{10}-\frac{28}{5}\sqrt{5}\right) \\
&=&\frac{1382976}{25},
\end{eqnarray*}
$$
we get
$$
S=\sqrt{\frac{1382976}{25}}=\frac{1176}{5}.
$$
The area of $DEF$ is $$K=S_{DEF}=\frac{EF\times DF}{2}=\frac{6\times 8}{2}=24$$
and the ratio $$\frac{S}{K}=\frac{1176/5}{24}=\frac{49}{5}=9.8,$$ as given.
Added: The scale is uniform throughout the following diagram drawn with the calculated equations
| {
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How to analyze (sum and convergence) step-by-step this series: $\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$? I have this series:
$$\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$$
and $$\sum_{n=1}^\infty\frac{4}{(n+1)(n+3)}$$
and I would like to know how to analyze, very step-by-step.
thanks
| One way to show convergence of the first series, is to first note that
$$ \frac{4}{(n+1)(n+2)} < \frac{4}{(n+1)^2}$$ for all $n$. Then write
$$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)^2} = 4\sum_{n=2}^{\infty} \frac{1}{n^2}. $$ Presumably, you already know that the latter series converges, and so you can apply the comparison test, to show that
$$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)(n+2)} $$ converges. The other one can be treated in a similar way.
To find the sum, write
$$ \frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2},$$ and solve for $A$ and $B$ (this is called a partial fractions decomposition). Then you should be able to determine the $n$'th partial sum of this series explicitly, and easily evaluate the sum (these types of series will be called telescopic in your textbook, for obvious reasons). The other one can be treated in a similar way, but the computations will be a bit more complicated.
Note: Just realized that the first part of my answer is made redundant by the second, but I'll leave it - it can't hurt to know more than one way of accomplishing something.
Edit: After seeing your comment above, I'll add some more details on the next step too:
$$ s_k = \sum_{n = 1}^{k} \frac{4}{(n+1)(n+2)} = 4\sum_{n = 1}^{k} \left( \frac{1}{n+1} - \frac{1}{n+2}\right).$$
Let us look more closely at the second sum
$$\sum_{n = 1}^{k} \left( \frac{1}{n+1} - \frac{1}{n+2}\right) = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots = \frac{1}{2} - \frac{1}{k+2}.$$
So $s_k = 4(1/2 - 1/(k+2))$ and hence the sum of the series is $s = \lim_{k \to \infty} s_k = 4/2 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ $n = 4k + 3 $
We start by letting $a \not\equiv 0\pmod n$ $\Rightarrow$ $a \equiv k\pmod n$ .
$\Rightarrow$ $a^{4k+2} \equiv 1\pmod n$
Now, I know that the contradiction will arrive from the fact that if we can show $a^2 \equiv 1 \pmod n $ then we can say $b^2 \equiv -1 \pmod n $ then it is not possible since solution exists only for $n=4k_2+1 $ so $a \equiv b\equiv 0 \pmod n $
So from the fact that $a^{2^{2k+1}} \equiv 1 \pmod n$ I have to conclude something.
| Assume that $a$ and $b$ are coprime. If $a$ and $b$ are odd, replace
$a$ and $ b$ by $(a-b)/2$ and $(a+b)/2$. Then $a^2 + b^2 = pm$, and by reducing $a$ and $b$
modulo $p$ you can make sure that $m < p$. Since $p = 4n+3$ and $a^2 + b^2 = 4k+1$,
we must have $m = 4j+3$. But then $m$ is divisible by a prime number $q = 4r+3$ strictly
smaller than $p$. Repeat until you find that $3$ is a sum of two squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/105034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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roots of complex polynomial - tricks What tricks are there for calculating the roots of complex polynomials like
$$p(t) = (t+1)^6 - (t-1)^6$$
$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$$\left( \frac{t+1}{t-1} \right)^6 = 1$$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which brings us to
$$\omega_k = e^{i \cdot k \cdot \frac{2 \pi}{6}}$$
So now we need to get the values from t for $k = 0,...5$.
How to get the values of t from the following identity then?
$$
\begin{align}
\frac{t+1}{t-1} &= e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\
(t+1) &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\
1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - t \\
1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t \cdot (e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1) \\
\end{align}
$$
And now?
$$
t = \frac{1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}}}{e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1}
$$
So I've got six roots for $k = 0,...5$ as follows
$$
t = \frac{1+e^{i \cdot k \cdot \frac{2 \pi}{6}}}{e^{i \cdot k \cdot \frac{2 \pi}{6}}-1}
$$
Is this right? But how can it be that the bottom equals $0$ for $k=0$?
I don't exactly know how to simplify this:
$$\frac{ \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} } + 1 }{ 1 - \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} }}$$
| Note that
$$(t+1)^6 - (t-1)^6=((t+1)^3-(t-1)^3)((t+1)^3+(t-1)^3)$$
(difference of squares).
When you simplify the first term in the product on the right, there is no $t^3$ term and no $t$ term! The second term in the product simplifies to $2t^3+6t$.
Remark: The solution by Arhabhata is the right one, it works if we replace $6$ by $n$. And when we set $\frac{t-1}{t+1}=e^{2\pi i k/n}$, where $k=1,2,\dots,n-1$, and solve for $t$, we get $-i$ times cotangents.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to integrate $1-\tan^4(x)$ without using secant? How to integrate such function:
$\int_{-\pi/3}^{\pi/3}1-\tan^4(x)$
I already found a solution using the trigonometric function secant. It looks like this:
$$u=\tan(x),\quad \frac{du}{dx}=\sec^2(x)$$
$$
\begin{align}
\int 1-\tan^4(x) & = \int 1 \;dx - \int\tan^4(x) \; dx \\
& = x-\int\tan^4(x) \; dx \\
& = x-\int\tan^2(x) \tan^2(x) \; dx \\
& = x-\int\tan^2(x) (\sec^2(x) - 1) \; dx \\
& = x-\int\tan^2(x) \sec^2(x) \; dx - \int\tan^2(x) \; dx \\
& = x-\int u^2 \; du - (\tan(x)-x) \\
& = x-\frac{1}{3}u^3 - (\tan(x)-x) \\
& = 2x-\frac{1}{3}\tan^3(x)-\tan(x)
\end{align}
$$
The only problem now is, that I have to find a solution without using secant. Can someone say me how to solve this without secant?
| Sasha's solution is very elegant. However, if you can't see a trick like this, you can always mechanically integrate any rational function of sine and cosine using the Weierstraß substitution.
With
$$\tan x=\frac{2t}{1-t^2}\;,$$
$$\frac{\mathrm dx}{\mathrm dt}=\frac2{1+t^2}\;,$$
$$t=\tan\frac x2\;,$$
the integral of $\tan^4 x$ becomes
$$
\begin{eqnarray}
&&\int\tan^4x\mathrm dx
\\
&=&
\int\left(\frac{2t}{1-t^2}\right)^4\frac2{1+t^2}\mathrm dt
\\
&=&
\int\frac{32t^4}{(1-t^2)^4(1+t^2)}\mathrm dt
\\
&=&
\int\left(\frac2{t^2+1}-\frac1{(t-1)^2}-\frac1{(t+1)^2}+\frac1{(t-1)^3}-\frac1{(t+1)^3}+\frac1{(t-1)^4}+\frac1{(t+1)^4}\right)\mathrm dt
\\
&=&
2\arctan t+\frac1{t-1}+\frac1{t+1}-\frac1{2(t-1)^2}+\frac1{2(t+1)^2}-\frac1{3(t-1)^3}-\frac1{3(t+1)^3}+ \color{gray}{\text{const}}
\\
&=&
2\arctan t+\frac{2t}{t^2-1}-\frac{8t^3}{3(t^2-1)^3}+ \color{gray}{\text{const}}
\\
&=&
x-\tan x+\frac13\tan^3 x+ \color{gray}{\text{const}}\;,
\end{eqnarray}
$$
where I used Wolfram|Alpha for pulling apart the fractions and putting them back together again.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Confused by Calc II question regarding derivative of rational integrals So here's the question:
If $f$ is a quadratic function such that $f(0) = 1$ and $\int \frac{f(x)}{x^2(x+1)^3}\,dx$ is a rational function, find the value of $f’(0)$.
What I've done so far is try to solve the integral using partial fractions i.e.
$\frac{f(x)}{x^2(x+1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2} + \frac{E}{(x+1)^3}$
Multiply out the denominator from the LHS to get:
$f(x) = Ax(x+1)^3 + B(x+1)^3 + Cx^2(x+1)^2 + Dx^2(x+1) + Ex^2$
when $x = 0$ I get that $B=1$.
At this point I'm stuck. I tried solving for the other variables but it gets insanely complicated. Wondering if anyone has a better strategy to solving the problem.
Thank you.
| You have
$f(x)=ax^2+bx +c$. That $f(0)=1$ gives you $c=1$. We have $f'(x)=2ax+b$; and so $f'(0)=b$.
The integrand can be written as
$$
{ax^2+ f'(0)x+1\over x^2(x+1)^3} = {A\over x\vphantom{ )^2}}+{B\over x^2\vphantom{ )^2}}+ {C\over (x+1)\vphantom{ )^2}}+{D\over (x+1)^2}+{E\over (x+1)^3}.
$$
Here's the important observation:
If the antiderivative of the above is a rational function, then $A=C=0$ (otherwise, it will contain logarithms).
Thus,
$$
{ax^2+ f'(0)x+1\over x^2(x+1)^3} = {B\over\vphantom{(^2} x^2}+ {D\over (x+1)^2}+{E\over (x+1)^3};
$$
or,
$$
{ax^2+ f'(0)x+1 } = {B }(x+1)^3+ {D }x^2(x+1)+{E }x^2.
$$
Setting $x=0$ in the above gives you $B=1 $.
Setting $x=-1$ in the above gives you $E=a-f'(0)+1$.
Also, comparing the $x^3$ terms, $B=-D$.
So:
$$
\eqalign{
& { ax^2+ f'(0)x+1 }\ =\ (x+1)^3- x^2(x+1)+{ (a-f'(0)+1)}x^2\cr
\iff& \color{maroon}{ax^2}+ f'(0)x+1\ =\ (\color{darkgreen}{x^3}+\color{darkblue}{3x^2}+3x+1) \color{darkgreen}{-x^3}\color{darkblue}{-x^2}+ \color{maroon}{ax^2} +({1-f'(0))}x^2 \cr
\iff&{ \hphantom{ax^2+} f'(0)x+1 }\ =\ 2x^2+3x+1+ \bigl (1 -f'(0)\bigr)x^2 \cr
\iff&
{ \hphantom{ax^2+} f'(0)x+1 }\ =\ \bigl(3 -f'(0)\bigr)x^2 +3x+1;
}
$$
whence, $f'(0)=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere.
Can anybody provide a hint as to how I should be looking at this problem?
| As you note, $\gcd(n^3+1,n^2+2) = \gcd(1-2n,n^2+2)$.
Now, continuing in that manner,
$$\begin{align*}
\gcd(1-2n, n^2+2) &= \gcd(2n-1,n^2+2)\\
&= \gcd(2n-1, n^2+2+2n-1)\\
&= \gcd(2n-1,n^2+2n+1)\\
&= \gcd(2n-1,(n+1)^2).
\end{align*}$$
Consider now $\gcd(2n-1,n+1)$. We have:
$$\begin{align*}
\gcd(2n-1,n+1) &= \gcd(n-2,n+1) \\
&= \gcd(n-2,n+1-(n-2))\\
&=\gcd(n-2,3)\\
&= 1\text{ or }3.
\end{align*}$$
Therefore, the gcd of $2n-1$ and $(n+1)^2$ is either $1$, $3$, or $9$. Hence the same is true of the original gcd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
Simple algebra. How can you factor out a common factor if it is $\frac{1}{\text{factor}}$ in one of the cases? I'm sure this is simple. I want to pull out a factor as follows...
I have the expression
$$\frac{a(\sqrt{x}) - (b + c)(\frac{1}{\sqrt{x}})c}{x}.$$
It would be useful for me to pull out the $\sqrt{x}$ from the numerator and try to simplify to remove the denominator, but how can I pull out the $\sqrt{x}$ from the right-most statement $\frac{1}{\sqrt{x}}$.
Thanks for your help!
| In general, if you have an expression $a+\frac{1}{c}b$ and you want to factor out the $\frac{1}{c}$, multiply the expression by $\frac{c}{c}$ like this:
$$
a+\frac{1}{c}b \\
\frac{c}{c}\left(a+\frac{1}{c}b\right) \\
\frac{1}{c}\left(c\cdot a + c\cdot \frac{1}{c}b\right) \\
\frac{1}{c}\left( ca+b \right)
$$
I find multiplying the whole expression by $\frac{c}{c}$ the clearest way to pull out these fractions, especially if the expression is part of a larger fraction, like yours. Then multiplying by $\frac{c}{c}$ is the same as just multiplying the numerator and denominator by $c$. For example,
$$
\frac{a+\frac{1}{c}b}{d+e} \\
\frac{c}{c}\frac{a+\frac{1}{c}b}{d+e} \\
\frac{c\left(a+\frac{1}{c}b\right)}{c\left(d+e\right)} \\
\frac{c\cdot a+c\cdot \frac{1}{c}b}{c\cdot d+c\cdot e} \\
\frac{ca+b}{cd+ce}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A minimization problem for a function involving maximum Let $a,b,c,d$ be constants in the interval $[-1,1]$. Define $$f(x,y)=\max\{|y-a|,1-b\}+\max\{1-x,1-y\}+\max\{|x-c|,1-d\}$$ for $ -1\le x\le 1, -1\le y\le 1.$
Prove, or disprove, that the minimum value of $f$ is $$\max\{2-b-c,2-a-d, 2-b-d\}.$$
Numerical evidence seems to show that this is true.
| That is true. Let $A=2-b-c, B=2-a-d, C=2-b-d$. Note that $f(x,y)\ge \max\{A,B,C\}$ for all $(x,y)$ in the domain. To prove that there exist $(x,y)$ in the domain such that $f(x,y)= \max\{A,B,C\}$ consider six cases: $A\le B\le C, A\le C\le B,\cdots$. For example in the case $C\le B\le A$, we have $$a\le b, c-d\le a-b.$$
If $1+c-d\le a+b-1$, we choose $x,y$ such that $$1+c-d\le x\le y \mbox{ and } a+b-1\le y\le 1+a-b,$$ and if $a+b-1\le 1+c-d$, we choose $x,y$ such that
$$1+c-d\le x\le y\le 1+a-b.$$
Direct checking shows that $f(x,y)= \max\{A,B,C\}=A.$ Other cases are similar; in some cases $x=y=1$ suffices.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get from $a\sqrt{1 + \frac{b^2}{a^2}}$ to $\sqrt{a^2 + b^2}$ I have the following expression: $a\sqrt{1 + \frac{b^2}{a^2}}$. If I plug this into Wolfram Alpha, it tells me that, if $a, b$ are positive, this equals $\sqrt{a^2 + b^2}$.
How do I get that result? I can't see how that could be done. Thanks
| If $a\ge0$,
$$\begin{align}
a\sqrt{1 + \frac{b^2}{a^2}}
&=\sqrt{a^2}\sqrt{1 + \frac{b^2}{a^2}}
\\
&=\sqrt{a^2\left(1 + \frac{b^2}{a^2}\right)}
\\
&=\sqrt{a^2 + b^2}.
\end{align}$$
($\sqrt{a^2}=|a|$ for all $a\in\mathbb{R}$ and $|a|=a$ when $a\ge0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Notation of indexers with multiples in a series If $\sigma _{n}=1+\dfrac {1} {2}+\dfrac {1} {3}+\ldots +\dfrac {1} {n}$ what series is given by $\sigma _{2n}$ ? Does that mean we only take the even terms now or does it mean every term is multiplied by 2 ?
| Since
$$\sigma_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}$$
is the sum of the reciprocals of $1$ up to $n$, we have that $\sigma_{2n}$ is
$$\sigma_{2n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}+\frac{1}{2n}$$
That is, we sum up to $2n$.
If we want to sum only even numbers, we'd have to change our notation and maybe write
$$\omega_n=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n-2}+\frac{1}{2n}=\frac{\sigma_{n}}{2}$$ and for odd numbers, put,
$$\kappa_n=\sigma_{2n}-\frac{\sigma_{n}}{2}=1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-3}+\frac{1}{2n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $ $$ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $$
I tried squaring and/or using $1-\cos x=2\sin^2{\frac{x}2}$, but no luck.
| If $t = \cos(x)$, we have $\sqrt{2-2t} + \sqrt{10-6t} = \sqrt{16-12 t^2}$. Square both sides, isolate the term with square roots, square again, and factor. The result should be equivalent to $(t+1)(t-1)(3t-2)^2=0$. $t=-1$ does not work, but the other factors do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Negative fractions - what's the difference? What's the difference between the following fractions:
$ \frac{-4}{-5}$
$ \frac{4}{-5}$
$ \frac{-4}{5}$
$ - \frac{4}{5}$
| Another way to think about this would be in terms of equivalence classes. If you are not familiar with this, it is pretty much how mathematicians say in the rationals that
$$\frac{1}{2} = \frac{2}{4} = \frac{4}{8}.$$
In fact one says that given two fractions
$$x = \frac{m}{n} \hspace{2mm} \text{and} \hspace{2mm} y= \frac{a}{b},$$
they are equal iff $mb - na = 0$. So in your case for example $\frac{-4}{5}$ and $-\frac{4}{5}$ are equal because
$$\frac{-4}{5} - \bigg((-1)\frac{4}{5}\bigg) = \frac{-4}{5} + \frac{4}{5} = \frac{-4 + 4}{5} = 0$$
recalling that $-\frac{4}{5} = (-1)\frac{4}{5}$. You can go on like this for all of them, e.g. for example the first and second are not equal because
$$\frac{-4}{-5} - \frac{4}{-5} = \frac{-4 - 4}{5} \neq 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Numbers are too large to show $65^{64}+64^{65}$ is not a prime I tried to find cycles of powers, but they are too big. Also $65^{n} \equiv 1(\text{mod}64)$, so I dont know how to use that.
| $$64^{65}+65^{64} = 6^{65}+7^{64} \pmod{29}$$
$65=2 \times 28+9, 64 = 2 \times 28 +8$, and also gcd$(29,36)$ = gcd$(29,49) = 1$
Therefore by Fermat's Little Theorem
If gcd$(a,p)= 1$, and $p$ is a prime then $a^{(p-1)} \hspace{3pt}\equiv \hspace{3pt}1 \pmod{p}$
$36^{29-1} \equiv 1 \pmod{29}, \hspace{5pt}49^{29-1} \equiv 1 \pmod{29} \hspace{3pt} \implies \hspace{3pt}
(6^{2})^{28} \equiv 1 \pmod{29}, \hspace{5pt} (7^{2})^{28} \equiv 1 \pmod{29}$
Therefore $6^{65} = 6^{(56+9)} \equiv 6^9 \pmod{29}, \hspace{5pt} 7^{64} = 7^{(56+8)} \equiv 7^8 \pmod{29}$
$$64^{65}+65^{64} \equiv 6^9+7^8 \pmod{29} \hspace{5pt} \equiv 22+7 \pmod{29} \equiv 0 \pmod{29}$$
Which shows that $$29 | (65^{64}+64^{65})$$
Therefore $65^{64}+64^{65}$ is not a prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "15",
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Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.
| Here is a simple "direct proof":
\begin{align*}
10^n+3 \times 4^n + 5&=10^n-1 +3 \times 2^{2n}+6 =9999..9+6 \times [2^{2n-1}+1] \\
&=9999..9+6 \times (2+1)(2^{2n-2}-2^{2n-3}+\cdots-2+1) \\
&= 9 \times [1111...1+2 \times (2^{2n-2}-2^{2n-3}+\cdots-2+1)]
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Compute the Unit and Class Number of a pure cubic field $\mathbb{Q}(\sqrt[3]{6})$ Find a unit in $\mathbb{Q}(\sqrt[3]{6})$ and show that this field has class number $h=1$.
I am done with the first part which is relatively simple:
Suppose that $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$. Then we have $\varepsilon=c+b\sqrt[3]{6}+a\sqrt[3]{6^2}$, since the integral base of $\mathbb{Q}(\sqrt[3]{6})$ can be written as $\{1,\sqrt[3]{6},\sqrt[3]{6^2}\}$. Thus,
$$\varepsilon=c+b\sqrt[3]{6}+a\sqrt[3]{6^2},$$
$$\sqrt[3]{6}\varepsilon=6a+c\sqrt[3]{6}+b\sqrt[3]{6^2},$$ $$\sqrt[3]{6^2}\varepsilon=6b+6a\sqrt[3]{6}+c\sqrt[3]{6^2}.$$
As we see it, the system of equations with variable $\varepsilon$ has only zero solution, since $\{1,\sqrt[3]{6},\sqrt[3]{6^2}\}$ is a base. Then $$\det\left(
\begin{array}{ccc}
c-\varepsilon & b & a \\
6a & c-\varepsilon & b \\
6b & 6a & c-\varepsilon \\
\end{array}
\right)
$$
is the minimal polynomial of $\varepsilon$. Since $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$ if and only if $N(\varepsilon)=\pm1$, we take $\varepsilon=0$ in the above polynomial, and $$\det\left(
\begin{array}{ccc}
c & b & a \\
6a & c & b \\
6b & 6a & c \\
\end{array}
\right)=\pm1.
$$
Compute the determinant we find that $a=33,~b=60,~c=109$ is one of the solutions. Hence a unit in $\mathbb{Q}(\sqrt[3]{6})$ is of the form $\varepsilon=109+60\sqrt[3]{6}+33\sqrt[3]{6^2}$.
For the second part of the problem, I have no idea how to show that $\mathbb{Q}(\sqrt[3]{6})$ is a principal ideal domain.
Any comment will be appreciated!
| The first idea for computing units in such fields is finding a generator of a purely ramified prime. Here $2 - \sqrt[3]{6}$ has norm $2$, hence
$$ (2 - \sqrt[3]{6})^3 = 2(1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2) $$
is $2$ times a unit.
Finding an element generating the prime ideal above $3$ is more difficult, but it turns out that $\beta = 3 + 2\sqrt[3]{6} + \sqrt[3]{6}^2$ is such an element with norm $3$. As above you now get
$$ (3 + 2\sqrt[3]{6} + \sqrt[3]{6}^2)^3
= 3(109+60\sqrt[3]{6} +33\sqrt[3]{6}^2), $$
and you get the unit you mentioned in your question. Actually we have
$$ \frac1{1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2} =
109+60\sqrt[3]{6} +33\sqrt[3]{6}^2 . $$
Finding elements of norms $5$ and $7$ is rather easy, which then shows that the ring ${\mathbb Z}[\sqrt[3]{6}]$ is principal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Show $ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$ Show $$ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$$
| If we make the standard "Weierstrass" $t=\tan(x/2)$ substitution, we get $\cos t=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2\,dt}{1+t^2}$. We end up quickly with
$$\int_0^\infty \frac{1+t^2}{1+t^4}\,dt.$$
But $1+t^4=(1-\sqrt{2}t+t^2)(1+\sqrt{2}t +t^2)$, so by partial fractions our integrand is
$$\frac{1}{2-2\sqrt{2}t+2t^2} +\frac{1}{2+2\sqrt{2}t+2t^2}.$$
Completing the squares, we end up with the integrand
$$\frac{1}{1+(\sqrt{2}t-1)^2}+\frac{1}{1+(\sqrt{2}t+1)^2}.$$
The substitutions $u=\sqrt{2} t-1$ and $u=\sqrt{2}t+1$ give
$$\int_{-1}^\infty \frac{1}{\sqrt{2}}\frac{du}{1+u^2}+\int_{1}^\infty \frac{1}{\sqrt{2}}\frac{du}{1+u^2}.$$
The first integral is $(1/\sqrt{2})(3\pi/4)$ and the second is $(1/\sqrt{2})(\pi/4)$. Add. We get $\pi/\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving an asymptotic lower bound for the integral $\int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$ This is a follow up to the great answer posted to https://math.stackexchange.com/a/125991/7980
Let $ 0 < r < \infty, 0 < s < \infty$ , fix $x > 1$ and consider the integral
$$ I_{1}(x) = \int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$$
Fix a constant $c^* = r^{\frac{1}{2r+2}} $ and let $x^* = x^{\frac{1}{1+r}}$.
Write $f(y) = \frac{x^2}{2y^{2r}} + \frac{y^2}{2}$ and note $c^* x^*$ is a local minimum of $f(y)$ so that it is a global max for $-f(y)$ on $[0, \infty)$.
We are trying to determine if there exist upper and lower bounds of the same order for large x. The coefficients in our bounds can be composed of rational functions in x or even more complicated as long as they do not have exponential growth. The Laplace expansion presented in the answer to the question cited above gives upper bounds.
In particular can we prove a specific lower bound:
Does there exist a positive constant $c_1(r,s)$ and such that for x>1 we have
$$I_1 (x) > \frac{c_1(r,s)}{x} \exp( - f(c^* x^*))$$ (it is ok in the answer if the function $\frac{1}{x}$ in the upper bound is replaced by any rational function or power of $x$)
| Let
$$
\phi_{r,x}(y)=-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\tag{1}
$$
Taking the first and second derivatives of $\phi_{r,x}(y)$ yields
$$
\phi_{r,x}^\prime(y)=r\frac{x^2}{y^{2r+1}}-y\tag{2}
$$
and
$$
\phi_{r,x}^{\prime\prime}(y)=-(2r+1)r\frac{x^2}{y^{2r+2}}-1\tag{3}
$$
Using $(2)$, $\phi_{r,x}(y)$ reaches a maximum at $y_0=(rx^2)^{\frac{1}{2r+2}}$. At that point,
$$
\phi_{r,x}(y_0)=-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\tag{4}
$$
Furthermore, $(3)$ gives that
$$
\frac12\phi_{r,x}^{\prime\prime}(y_0)=-(r+1)\tag{5}
$$
Standard stationary phase methods yield
$$
\begin{align}
&\int_0^\infty\exp\left(-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\right)\frac{\mathrm{d}y}{y^s}\\
&\sim\exp\left(\phi_{r,x}(y_0)\right)\int_0^\infty\exp\left(-(r+1)(y-y_0)^2\right)\frac{\mathrm{d}y}{y^s}\\
&\sim y_0^{-s}\exp\left(\phi_{r,x}(y_0)\right)\int_{-\infty}^\infty\exp\left(-(r+1)y^2\right)\mathrm{d}y\\
&=(rx^2)^{\frac{-s}{2r+2}}\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\sqrt{\frac{\pi}{r+1}}\tag{6}
\end{align}
$$
Where $f(x)\sim g(x)$ means that $\lim\limits_{x\to\infty}f(x)/g(x)=1$.
Estimate $(5)$ says that the kind of estimate sought above can be achieved only when $s\le r+1$.
Taking the derivative of $(2)$ yields
$$
\phi_{r,x}^{\prime\prime\prime}(y)=(2r+1)(2r+2)r\frac{x^2}{y^{2r+4}}\tag{7}
$$
which says that the second derivative of the exponent increases monotonically, whereas the second derivative of the quadratic approximation is constant. Since $\phi_{r,x}$ and its first and second derivatives match the quadratic approximation at $y_0$, we get that for $y\ge y_0$,
$$
\phi_{r,x}(y)\ge\phi_{r,x}(y_0)-(r+1)(y-y_0)^2\tag{8}
$$
Furthermore, since $(1+t)^{-s}\ge1-st$ for $t\ge0$, we get
$$
\begin{align}
&\int_0^\infty\exp\left(-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\right)\frac{\mathrm{d}y}{y^s}\\
&\ge\int_{y_0}^\infty\exp\left(\phi_{r,x}(y_0)-(r+1)(y-y_0)^2\right)y_0^{-s}\left(1-s\frac{y-y_0}{y_0}\right)\mathrm{d}y\\
&=y_0^{-s}\exp(\phi_{r,x}(y_0))\int_0^\infty\exp\left(-(r+1)t^2\right)\left(1-\frac{st}{y_0}\right)\mathrm{d}t\\
&=y_0^{-s}\exp(\phi_{r,x}(y_0))\left(\frac12\sqrt{\frac{\pi}{r+1}}-\frac{s}{2y_0\sqrt{r+1}}\right)\\
&=(rx^2)^{\frac{-s}{2r+2}}\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\left(\frac12\sqrt{\frac{\pi}{r+1}}-\frac{s}{2y_0\sqrt{r+1}}\right)\tag{9}
\end{align}
$$
For $x\ge x_0$, we get $(9)$ with $y_0=(rx_0^2)^\frac{1}{2r+2}$. This is the bound required as long as $s\le r+1$.
For example, if we set $\displaystyle x_0=\max\left(\frac{s^{r+1}}{\sqrt{r}},1\right)$, for $x\ge x_0$, we get
$$
\begin{align}
&\int_0^\infty\exp\left(-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\right)\frac{\mathrm{d}y}{y^s}\\
&\ge\left(\frac{\sqrt{\pi}-1}{2\sqrt{r+1}}r^{\frac{-s}{2r+2}}\right)x^{\frac{-s}{r+1}}\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\\
&\ge\left(\frac{\sqrt{\pi}-1}{2\sqrt{r+1}}r^{\frac{-s}{2r+2}}\right)\frac1x\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\tag{10}
\end{align}
$$
as long as $s\le r+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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inequality $(a+c)(a+b+c)<0$, prove $(b-c)^2>4a(a+b+c)$
If $(a+c)(a+b+c)<0,$
prove $$(b-c)^2>4a(a+b+c)$$
I will use the constructor method that want to know can not directly prove it?
| Because
$$(b-c)^2-4a(a+b+c)=(b-c)^2+(4a+8c)(a+b+c)-8(a+c)(a+b+c)=$$
$$=(2a+b+3c)^2-8(a+c)(a+b+c)>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
On solvable quintics and septics Here is a nice sufficient (but not necessary) condition on whether a quintic is solvable in radicals or not. Given,
$x^5+10cx^3+10dx^2+5ex+f = 0\tag{1}$
If there is an ordering of its roots such that,
$x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0\tag{2}$
or alternatively, its coefficients are related by the quadratic in f,
$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2
\tag{3}$
then (1) is solvable. This also implies that if $c\neq0$, then it has a solvable twin,
$x^5+10cx^3+10dx^2+5ex+f' = 0\tag{4}$
where $f'$ is the other root of (3). The Lagrange resolvent are the roots of,
$z^4+fz^3+(2c^5-5c^3e-4d^2e+ce^2+2cdf)z^2-c^5fz+c^{10} = 0\tag{5}$
so,
$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\tag{6}$
Two questions though:
I. Does the septic (7th deg) analogue,
$x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 – (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0\tag{7}$
imply such a septic is solvable?
II. The septic has a $5! = 120$-deg resolvent. While this is next to impossible to explicitly construct, is it feasible to construct just the constant term? Equating it to zero would then imply a family of solvable septics, just like (3) above.
More details and examples for (2) like the Emma Lehmer quintic in my blog.
| It turns out there is an infinite number of such septics, such as the Hashimoto-Hoshi septic,
$$\small x^7 - (a^3 + a^2 + 5a + 6)x^6 +
3(3a^3 + 3a^2 + 8a + 4)x^5 + (a^7 + a^6 + 9a^5 - 5a^4 - 15a^3 - 22a^2 -
36a - 8)x^4 - a(a^7 + 5a^6 + 12a^5 + 24a^4 - 6a^3 + 2a^2 - 20a - 16)x^3 + a^2(2a^6 + 7a^5 + 19a^4 + 14a^3 + 2a^2 + 8a - 8)x^2 - a^4(a^4 + 4a^3 + 8a^2 + 4)x + a^7=0$$
For example, let $a=1$ so,
$$1 - 17 x + 44 x^2 - 2 x^3 - 75 x^4 + 54 x^5 - 13 x^6 + x^7=0$$
which is the equation involved in $\cos\frac{\pi k}{43}$ and order its roots as,
$$x_1,\,x_2,\,x_3,\,x_4,\,x_5,\,x_6,\,x_7 =\\
r_1,\,r_2,\,r_5,\,r_6,\,r_3,\,r_7,\,r_4 =
\\ -0.752399,\; 0.0721331,\; 2.63744,\; 3.62599,\; 0.480671,\; 6.29991,\; 0.636246$$
where the $r_i$ is the root numbering in Mathematica. Then,
$$
x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 - (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/129655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
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Critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$ I would like some help finding the critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$. I tried solving $f_x=0, f_y=0$ (where $f_x, f_y$ are the partial derivatives) but the resulting equation is very complex. The exercise has a hint: think of $f_x-f_y$ and $f_x+f_y$. However, I can't see where to use it.
Thanks!
| Here $f_{x}=2x+y-\frac{1}{x^{2}}$ and $f_{y}=2y+x-\frac{1}{y^{2}}$
For critical point
$$f_{x}=0,\ f_{y}=0$$
$$2x+y-\frac{1}{x^{2}}=0\ ,\ 2y+x-\frac{1}{y^{2}}=0$$
$$2x^{3}+x^{2}y-1=0\ ,\ 2y^{3}+xy^{2}-1=0$$
Substracting this two equations we get,
$$ 2(x^{3}-y^{3})+xy(x-y)=0$$
$$ 2(x-y)(x^{2}+xy+y^{2})+xy(x-y)=0$$
$$(x-y)(2x^{2}+2xy+2y^{2}+xy)=0$$
$$x-y=0\ \Rightarrow x=y$$
Put $x=y$ in $2x^{2}+2xy+2y^{2}+xy=0$,we get,
$$ 2x^{2}+2x^{2}+2x^{2}+x^{2}=0\ \Rightarrow x=0 $$
Also $y=0$ \
$\therefore (0,0)$ is a critical point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
rotating a matrix Given a rectangular matrix $A$, what is the general form to rotate the matrix about the center term, e.g. such that
$$\pmatrix{a_{0,0} & a_{0,1} & a_{0,2} \\ a_{1,0} & a_{1,1} & a_{1,2} \\ a_{2,0} & a_{2,1} & a_{2,2}}\longrightarrow\pmatrix{a_{0,2} & a_{1,2} & a_{2,2} \\ a_{0,1} & a_{1,1} & a_{2,1} \\ a_{0,0} & a_{1,0} & a_{2,0}} $$
and possibly the reverse case as well.
| $$\pmatrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0}A^T = \pmatrix{a_{0,2} & a_{1,2} & a_{2,2} \\ a_{0,1} & a_{1,1} & a_{2,1} \\ a_{0,0} & a_{1,0} & a_{2,0}} =A_r $$
Edit: reverse is,
$$A_r^T \pmatrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0} = \pmatrix{a_{0,0} & a_{0,1} & a_{0,2} \\ a_{1,0} & a_{1,1} & a_{1,2} \\ a_{2,0} & a_{2,1} & a_{2,2}} = A $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
| Jordan, The derivative of your function is $\frac{1}{2} + \frac{\cos 2x}{2}$. Now note that $\cos 2x = \cos^2 x -\sin ^2 = \cos^2 x -1 -\cos^2 x =2\cos^2x -1$. Rearranging, you get $$\cos^2 x =\frac{\cos 2x}{2} + \frac{1}{2}.$$
$$
\begin{align*}
\cos 2x = \cos(x+x) & =\cos x \cos x -\sin x \sin x \\
& = \cos^2x -\sin^2x\\
& = \cos^2x -(1-\cos^2x)\qquad\text{because}~\cos^2x + \sin^2 x =1.\\
& = \cos^2x-1+ \cos^2x\\
& = 2\cos^2x-1
\end{align*}
$$
So, you have $\cos2x = 2\cos^2x -1$, which is the same as $\cos 2x + 1 = 2\cos^2x$. Divide both sides by 2 to get what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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} |
integration of fractions i am trying to integrate following equation
$$ \int\frac 1{(x^2-1)\cdot (x+2)}\,dx$$
i can represent $(x^2-1)=(x-1)(x+1)$ so,it would be converted in the following form
$$\int\frac1{(x^2-1)(x+2)}\,dx=\int \frac1{(x-1)(x+1)(x+2)}\,dx$$
or it is equal $$\int \frac1{(x-1)(x^2+3x+2)}\,dx$$ last one we can decompose into form
$$ \frac1{(x-1)(x^2+3x+2)}=\frac A{x-1}+\frac{Cx+D}{x^2+3x+2}$$ am i right?or did i miss some term?
| I think you can decompose it like this:
$$
\frac{1}{(x^2-1)\cdot(x+2)}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{c}{x+2}
$$
Thus we can solve the following equations:
$$
a+b+c=0\\3a+b=0\\2a-2b-c=1
$$
getting $a=1/6,b=-1/2,c=1/3$.
Therefore,
$$
\int\frac{dx}{(x^2-1)\cdot(x+2)}\\=\int\frac{1}{6}\cdot\frac{dx}{x-1}-\int\frac{1}{2}\cdot\frac{dx}{x+1}+\int\frac{1}{3}\cdot\frac{dx}{x+2}\\=\frac{1}{6}\cdot \log(x-1)-\frac{1}{2}\cdot \log(x+1)+\frac{1}{3}\cdot \log(x+2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/135155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What is so interesting about the zeroes of the Riemann $\zeta$ function? The Riemann $\zeta$ function plays a significant role in number theory and is defined by $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \qquad \text{ for } \sigma > 1 \text{ and } s= \sigma + it$$
The Riemann hypothesis asserts that all the non-trivial zeroes of the $\zeta$ function lie on the line $\text{Re}(s) = \frac{1}{2}$.
My question is:
Why are we interested in the zeroes of the $\zeta$ function? Does it give any information about something?
What is the use of writing $$\zeta(s) = \prod_{p} \biggl(1-\frac{1}{p^s}\biggr)^{-1}$$
| the HIlbert Polya operator which would prove Riemann Hypothesis is the Wu-Sprung genralzed model with potential
$$ f^{-1} (x)=\frac{4}{\sqrt{4x+1} } +\frac{1}{4\pi } \int\nolimits_{-\sqrt{x} }^{\sqrt{x}}\frac{dr}{\sqrt{x-r^2} } \left( \frac{\Gamma '}{\Gamma } \left( \frac{1}{4} +\frac{ir}{2} \right) -\ln \pi \right) -\sum\limits_{n=1}^\infty \frac{\Lambda (n)}{\sqrt{n} } J_0 \left( \sqrt{x} \ln n\right) $$
with boundary conditions $$ y(0)=0=y(\infty) $$ and $ H= -\frac{d^{2}}{dx^{2}}y(x)+f(x)y(x)=E_{n}$...................$E_{n}=\gamma_{n}^{2}$·
however mathematician do not like it, if we take the half derivative then we find the distributional Riemann-Weil formula for the zeros
$$ \begin{array}{l}
\sum\limits_{n=0}^{\infty }\delta \left( x-\gamma _{n} \right) + \sum\limits_{n=0}^{\infty }\delta \left( x+\gamma _{n} \right) =\frac{1}{2\pi } \frac{\zeta }{\zeta } \left( \frac{1}{2} +ix\right) +\frac{1}{2\pi } \frac{\zeta '}{\zeta } \left( \frac{1}{2} -ix\right) -\frac{\ln \pi }{2\pi } \\[10pt]
{} +\frac{\Gamma '}{\Gamma } \left( \frac{1}{4} +i\frac{x}{2} \right) \frac{1}{4\pi } +\frac{\Gamma '}{\Gamma } \left( \frac{1}{4} -i\frac{x}{2} \right) \frac{1}{4\pi } +\frac{1}{\pi } \delta \left( x-\frac{i}{2} \right) +\frac{1}{\pi } \delta \left( x+\frac{i}{2} \right) \end{array} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "72",
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Show that $ a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q \implies \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q $ Assume that $a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational，prove $\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier way
$\sqrt a + \sqrt b + \sqrt c = p \in \mathbb Q$,
$\sqrt a + \sqrt b = p- \sqrt c$,
$a+b+2\sqrt a \sqrt b = p^2+c-2p\sqrt c$,
$2\sqrt a\sqrt b=p^2+c-a-b-2p\sqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)\sqrt c$,
$\sqrt c=\frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}\in\mathbb Q$.
| Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$ and $n = [K: \mathbb{Q}]$. Then since $Tr_{K/\mathbb{Q}}(\sqrt{a}) = Tr_{\mathbb{Q}(\sqrt{a})/\mathbb{Q}} \circ Tr_{K/\mathbb{Q}(\sqrt{a})} (\sqrt{a})$, we have
$$ Tr_{K/\mathbb{Q}}(\sqrt{a}) = \begin{cases} 0,& \text{if } \sqrt{a} \notin \mathbb{Q} \\ n\sqrt{a}, &\text{if } \sqrt{a} \in \mathbb{Q}, \end{cases}$$
and same for $\sqrt{b}$ and $\sqrt{c}$.
By hypothesis $\sqrt{a} + \sqrt{b} +\sqrt{c} \in \mathbb{Q}$, so
$$ Tr_{K/\mathbb{Q}}(\sqrt{a}) + Tr_{K/\mathbb{Q}}(\sqrt{b}) + Tr_{K/\mathbb{Q}}(\sqrt{c}) = n\sqrt{a} + n \sqrt{b} + n \sqrt{c}.$$
It is easy to conclude that $\sqrt{a},\sqrt{b},\sqrt{c} \in \mathbb{Q}$.
| {
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Showing a series is a solution to a differential equation I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$
So i have:
$$y=\sum_{n=0}^{\infty} a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$
$$y''=\sum_{n=0}^{\infty}a_{n}n(n-1)x^{n-2}$$
then substituting these into the differential equation I get:
$$(1-2x+x^2)\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
$$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
relabeling the indexes:
$$\sum_{n=-2}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=-1}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
and then cancelling the $n=-2$ and $n=-1$ terms:
$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
but this doesn't give me what I want (I don't think) as I have $n^2$ terms as I would need
$(n^2+3n+2)a_{n+2}-(2n^2+n)a_{n+1}+(n^2-n-2)a_{n}=0$
I'm not sure where I have gone wrong?
Thanks very much for any help
| You are right till the last step.
You have $$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
which gives us
$$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$
and not
$$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - (2n^2+n) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$
as you have written.
Hence, setting the coefficients of $x^n$ to zero, we get that
$$\left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)=0$$
Factorizing $n+1$ out, we get what you need i.e.
$$\left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)= (n+1) \left((n+2) a_{n+2} - 2 n a_{n+1} + (n-2)a_n \right)$$
Hence, we get that $$(n+2) a_{n+2} - 2 n a_{n+1} + (n-2)a_n = 0$$
| {
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Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$? Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$?
I tried factoring individual components out, but it seemed really tedious.
| Tabulating the expression with respect to low primes:
$\bmod 2: 27^5 + 84^5 + 110^5 + 133^5 \equiv 1^5 + 0^5 + 0^5 + 1^5 \equiv 0 \implies n\equiv 0$
$\bmod 3: 27^5 + 84^5 + 110^5 + 133^5 \equiv 0^5 + 0^5 + -1^5 + 1^5 \equiv 0 \implies n\equiv 0$
$\bmod 5: 27^5 + 84^5 + 110^5 + 133^5 \equiv 2^5 + (-1)^5 + 0^5 + (-2)^5 \equiv -1 \implies n^5 \equiv n\equiv -1$
Collecting these gives $n\equiv 24\bmod 30$, which already points at $144$
$\bmod 7: 27^5 + 84^5 + 110^5 + 133^5 \equiv (-1)^5 + 0^5 + (-2)^5 + 0^5 \equiv -1+3 \equiv 2 \equiv n^5\equiv n^{-1} \implies n \equiv 4$
This confirms that $144$ is the only possible solution in range; the next modular equivalence solution would be far out of range at $144+210 = 354$.
| {
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Determinant of symmetric Matrix with non negative integer element Let \begin{equation*}
M=%
\begin{bmatrix}
0 & 1 & \cdots & n-1 & n \\
1 & 0 & \cdots & n-2 & n-1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-2 & \cdots & 0 & 1 \\
n& n-1 & \cdots & 1 & 0%
\end{bmatrix}%
\end{equation*}
How can you prove that $\det(M)=(-1)^n\cdot n \cdot 2^{n-1}$?
I just guess the formula in the right hand side by observing the calculation for small n but I can't prove for arbitrary n. Thanks everyone.
| Let's take a $4\times 4$ matrix (I don't want to type much).
$$\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 0 & 1 & 2 \\
2 & 1 & 0 & 1 \\
3 & 2 & 1 & 0
\end{vmatrix} $$
Since adding a row into another does not change determinant values. Add $-i'th$ row into $i+1$'th row.
$$\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 \\
1 & 1 & 1 & -1
\end{vmatrix} $$
Repeat the process with columns.
$$\begin{vmatrix}
0 & 1 & 1 & 1 \\
1 & -2 & 0 & 0 \\
1 & 0 & -2 & 0 \\
1 & 0 & 0 & -2
\end{vmatrix} = \frac{1}{2}\begin{vmatrix}
0 & 1 & 1 & 1 \\
2 & -2 & 0 & 0 \\
2 & 0 & -2 & 0 \\
2 & 0 & 0 & -2
\end{vmatrix} = \frac{1}{2}\begin{vmatrix}
3 & 1 & 1 & 1 \\
0 & -2 & 0 & 0 \\
0 & 0 & -2 & 0 \\
0 & 0 & 0 & -2
\end{vmatrix}$$
Now what you can say about its determinant?
| {
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"source": "stackexchange",
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Approximate a series with finite number of terms How it is possible to approximate:
$$\sum_{i=1}^{NR}{i\cdot \left( \dfrac{1}{1-p} \right)^i} $$
| Let $x = \dfrac1{1-p}$ and let $n = NR$. Then we are interested in the sum $\displaystyle \sum_{i=1}^{n} i x^i$.
\begin{align}
\sum_{i=1}^{n} i x^i & = x \left(\sum_{i=1}^{n} i x^{i-1} \right)\\
& = x \left( \sum_{i=1}^{n} \frac{d x^i}{dx} \right)\\
& = x \frac{d}{dx} \left( \sum_{i=1}^{n} x^i\right)\\
& = x \frac{d}{dx} \left( x\left(\frac{x^n - 1}{x-1} \right) \right)\\
& = x \left( \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \right)
\end{align}
Replacing $x$ by $\dfrac1{1-p}$ and $n$ by $NR$, we get that the solution is $$\left( \frac{NR}{p} - \frac1{p^2} + \frac1p \right) \left( \dfrac1{1-p}\right)^{NR} + \frac1{p^2} - \frac1p$$
| {
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Surprising approximation of weighted sum of binomial coefficients The following sum appeared in connection to the problem addition of angular momentum in physics:
$$
\frac{1}{2^{n+3}}\sum_{k=0}^n \left(\frac{n-2k-1}{\sqrt{k+1}}+\frac{n-2k+1}{\sqrt{n-k+1}}\right)^2 {n\choose k}
$$
The intriguing thing about this sum is extremely well approximated by $1-\frac{c}{n}$, where $c$ is some constant when $n\to\infty$. I wonder if there is a simple way to see this, or to get this lower bound by carefully approximating this rather unappealing sum.
| Notice that the sum can be given a probabilistic interpretation, since $\frac{1}{2^n} \binom{n}{k}$ is the point mass function of a symmetric Binomial random variable.
Let $X \sim \operatorname{Binom}\left(n, \frac{1}{2}\right)$. Then the sum in question equals
$$
S = \frac{1}{8} \mathbb{E} \left( \left( \frac{n-2X -1}{\sqrt{X+1}} + \frac{n-2X +1}{\sqrt{n-X+1}} \right)^2 \right)
$$
In the large $n$ limit, the de Moivre-Laplace theorem can be used, and $X$ can be approximated in distribution as $\frac{n}{2} + \frac{\sqrt{n}}{2} Z$, where $Z$ is the standard normal variable.
The sum $S$ then approximately equals:
$$
S \approx \frac{1}{8} \mathbb{E}\left( \left( \sqrt{2}\frac{1+\sqrt{n} Z}{\sqrt{2+n + \sqrt{n} Z}} + \sqrt{2}\frac{-1+\sqrt{n} Z}{\sqrt{2+n - \sqrt{n} Z}} \right)^2 \right) =
\mathbb{E}\left( Z^2 + \frac{3}{4n} Z^2 \left( Z^2-4\right) + \mathcal{o}\left(\frac{1}{n}\right) \right) = 1 - \frac{3}{4 n} + \mathcal{o}\left(\frac{1}{n}\right)
$$
where $\mathbb{E}(Z^{2k}) =(2k-1)!!$ was used, in particular $\mathbb{E}(Z^{2}) =1$ and $\mathbb{E}(Z^{4}) = 3$.
Added: The series expansion is derived using simple algebra:
$$\begin{eqnarray}
\frac{\pm 1 + \sqrt{n} Z}{\sqrt{2+n \pm \sqrt{n} Z}} &=& \frac{Z \pm \frac{1}{\sqrt{n}}}{\sqrt{1 + \frac{2}{n} \pm \frac{Z}{\sqrt{n}}}} \\ &=& \left( Z \pm \frac{1}{\sqrt{n}} \right) \left( 1 - \frac{1}{2} \left(\frac{2}{n} \pm \frac{Z}{\sqrt{n}} \right) + \frac{3}{8} \left(\frac{2}{n} \pm \frac{Z}{\sqrt{n}} \right)^2 + \mathcal{o}\left(\frac{1}{n} \right) \right) \\
&=& \left( Z \pm \frac{1}{\sqrt{n}} \right) \left( 1 \mp \frac{Z}{2 \sqrt{n}} + \frac{3 Z^2}{8 n} + \mathcal{o}\left(\frac{1}{n} \right) \right) \\
&=& Z \mp \frac{Z^2-2}{2 \sqrt{n}} + \frac{3 Z (Z^2-4)}{8n} + \mathcal{o}\left(\frac{1}{n} \right)
\end{eqnarray}
$$
| {
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Something is wrong with this proof, limit $\lim\limits_{(x,y) \to (0,0)} \frac{xy^3}{x^4 + 3y^4}$ Could someone please tell me what is wrong with this proof?
Show that $\lim\limits_{(x,y) \to (0,0)} \dfrac{xy^3}{x^4 + 3y^4}$ does not have a limit or show that it does and find the limit.
I know it is wrong because the limit doesn't exist, but this proof is contradicting me
Proof
Case 1
Assume for $x,y > 0$, then $x^4 + 3y^4 > x^4 > x > 0$
$$\begin{align*}
0 < x < x^4 + 3y^4 &\iff 0 < \dfrac{x}{x^4 + 3y^4} < 1 \\
& \iff 0 < \dfrac{x|y^3|}{x^4 + 3y^4} < |y^3| \\
& \iff \lim\limits_{(x,y) \to (0,0)} 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < \lim\limits_{(x,y) \to (0,0)}|y^3|\\
&\iff 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < 0
\end{align*}$$
Case 2. WLOG Assume $x,y <0$ and combine both cases.
What's wrong the ppoof? I don't find the flaw
| The statement $x^4>x$ is clearly wrong especially when $x$ is in the neighborhood of $0$.
Also, you need to look at cases $x>0,y<0$ and $x<0,y>0$. But the main error is the statement $x^4 > x$.
EDIT
The answer is that the function is not continuous at the origin. This can be seen as follows. Remember that in two dimensions there are infinitely many different directions from which you can approach origin unlike in one dimension where you need to look only at two different cases $x>0$ and $x<0$. Hence, in two dimensions it is not sufficient/advantageous to split it into just four cases $x,y>0$, $x,y<0$, $x>0,y<0$, $x<0,y>0$ and analyze.
Approach the origin along the straight line $y=mx$. Then we have that $$\lim_{\overset{x \rightarrow 0}{y \rightarrow 0}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{\overset{x \rightarrow 0}{y = mx}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{x \rightarrow 0} \dfrac{x(mx)^3}{x^4 + 3(mx)^4} \\ = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{x^4 + 3m^4 x^4} = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{(1 + 3m^4) x^4} = \lim_{x \rightarrow 0} \dfrac{m^3}{(1 + 3m^4)} = \dfrac{m^3}{1+3m^4}$$
Hence, approaching the origin along different straight lines give different answers.
For instance, if we approach the origin along the line $y = x$, the limit is $\dfrac14$.
If we approach the origin along the line $y = -x$, the limit is $-\dfrac14$.
If we approach the origin along the line $y = 0$, the limit is $0$.
This proves that the function is not continuous at the origin.
| {
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Calculate the volume between $z=x^2+y^2$ and $z=2ax+2by$ I'm trying to calculate the volume between the surfaces $z=x^2+y^2$ and $z=2ax+2by$ where $a>0,b>0$. Here's what I've tried:
First I noticed the projection of the volume to the xy plane is a circle: $(x-a)^2+(y-b)^2\leq a^2+b^2$. Using this I simplified the calculation of the integral for the volume a little. Marking $B$ as the circle we get that the volume is:
$$\iint_{}^{B} (2ax+2ay-x^2-y^2) = \iint_{}^{B} (a^2+b^2)-\iint_{}^{B} ((x-a)^2+(y-b)^2) $$
Using the symmetry of the circle we get:
$$\iint_{}^{B} ((x-a)^2+(y-b)^2) = 2\iint_{}^{B} ((x-a)^2)$$
And we can also use the formula for the area of a circle to get:
$$\iint_{}^{B} (a^2+b^2) = \pi (a^2+b^2)^2$$
So all I have left to do is calculate $\iint_{}^{B} ((x-a)^2)$, but this is where I get stuck. Trying to do it using iterated integrals becomes too complex (we have only covered Cartesian coordinates, so I can't use something like polar coordinates here).
I know the result is supposed to be $(1/2)\pi (a^2+b^2)^2$.
Assistance would be appreciated. Thanks!
| Integrate in $y$ first; this gives the additional factor of $2\sqrt{(a^2+b^2)-(x-a)^2}$, so that the question reduces to
$$\int_{a-\sqrt{a^2+b^2}}^{a+\sqrt{a^2+b^2}} (x-a)^2 \sqrt{(a^2+b^2)-(x-a)^2}\,dx$$
which may look scary, but is in fact a typical trigonometric substitution problem. Namely, $x=a+\sqrt{a^2+b^2}\sin \theta$ turns the integral into a multiple of $$\int_{-\pi/2}^{\pi/2} \sin^2\theta\, \cos^2\theta \,d\theta$$
Of course, with polar integrals this would have been much easier.
| {
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Need help with the integral $\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$ I'm having a problem resolving the following integral, spent almost all day trying.
Any help would be appreciated.
$$\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$$
| If you convert everything to sines and cosines, you get $$\frac{2\tan x+3}{5\sin^2x+4}=\frac{2\sin x+3\cos x}{5\sin^2x\cos x+4\cos x}\;,\tag{1}$$ which probably doesn’t look very promising. However, you can rewrite it as $$\frac{2\sin x+3\cos x}{\cos x(9-5\cos^2x)}=\frac{2\sin x}{\cos x(9-5\cos^2x)}+\frac3{9-5\cos^2x}\;.$$ The first term of this is nice: apart from a factor of $-1$, $\sin x$ is the derivative of $\cos x$, so it can be integrated by substituting $u=\cos x$ and using partial fractions. The second term still requires a bit of work. When sines and cosines don’t do the job, try secants and tangents:
$$\frac3{9-5\cos^2x}=\frac3{9-\frac5{\sec^2x}}=\frac{3\sec^2x}{9\sec^2x-5}=\frac{3\sec^2x}{9(\tan^2x+1)-5}=\frac{3\sec^2x}{9\tan^2x+4}\;,$$ which can be integrated by substituting $u=\tan x$.
If you don’t see any way forward from $(1)$, you can always jump directly to the stage of trying to get secants and tangents.
$$\frac{2\tan x+3}{5\sin^2x+4}=\frac{2\tan x+3}{5\tan^2x\,\cos^2x+4}=\frac{2\tan x+3}{5\frac{\tan^2x}{\sec^2x}+4}=\frac{2\sec^2x\tan x+3\sec^2x}{5\tan^2x+4\sec^2x}\;,$$
and after converting the $4\sec^2x$ in the denominator to $4\tan^2x-4$, we have
$$\frac{(2\tan x+3)\sec^2x}{9\tan^2x+4}\;;$$
$\sec^2x$ being the derivative of $\tan x$, the substitution $u=\tan x$ will turn this into a nice rational function of $u$. This is exactly what Chandrasekhar achieved directly in his solution by multiplying the fraction by $\dfrac{\sec^2x}{\sec^2x}$. That’s a nicer, more efficient way to go. My purpose in writing out this more roundabout route is to show that you often don’t have to see the really clever tricks if you can put together enough more routine manipulations.
| {
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Prove that $4^{2n} + 10n -1$ is a multiple of 25 Prove that if $n$ is a positive integer then $4^{2n} + 10n - 1$ is a multiple of $25$
I see that proof by induction would be the logical thing here so I start with trying $n=1$ and it is fine. Then assume statement is true and substitute $n$ by $n+1$ so I have the following:
$4^{2(n+1)} + 10(n+1) - 1$
And I have to prove that the above is a multiple of 25. I tried simplifying it but I can't seem to get it right. Any ideas? Thanks.
| Another solution, via congruences mod. $25$. First note $16\bmod 25$ generates a cyclic group of order $5$:
$$16^2\equiv 6,\quad16^3\equiv 6\cdot 16\equiv-4,\quad16^4\equiv6^2\equiv 11, \quad 16^5\equiv6\cdot-4\equiv1\mod25.$$
So let's examine each case:
*
*If $n\equiv 0\mod 5$, $\;16^n+10n-1\equiv1+0-1=0$.
*If $n\equiv 1\mod 5$, $\;16^n+10n-1\equiv16+10-1=25\equiv 0$.
*If $n\equiv 2\mod 5$, $\;16^n+10n-1\equiv 6+20-1=25\equiv 0$.
*If $n\equiv 3\mod 5$, $\;16^n+10n-1\equiv -4+30-1=25\equiv 0$.
*If $n\equiv 4\mod 5$, $\;16^n+10n-1\equiv 11+40-1=50\equiv 0$.
Thus in each case, $\;16^n+10n-1$ is divisible by $25$.
| {
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How do I show that this function is always $> 0$
Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +
\frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$
I can show that the first 3 terms are $> 0$ for all $x$:
$(x+1)^2 + 1 > 0$
But, I'm having trouble with the last two terms. I tried to show that the following was true:
$\frac{x^3}{3!} \leq \frac{x^4}{4!}$
$4x^3 \leq x^4$
$4 \leq x$
which is not true for all $x$.
I tried taking the derivative and all that I could ascertain was that the the function became more and more increasing as $x \rightarrow \infty$ and became more and more decreasing as $x \rightarrow -\infty$, but I couldn't seem to prove that there were no roots to go with this property.
| You have had some good ideas so far. You tried to see when this was true: $$\frac{x^3}{3!} \leq \frac{x^4}{4!}.$$
You rearranged this to $4x^3\leq x^4$ but you made an incorrect conclusion when you divided by $x^3$ (if $x<0$ then the inequality sign should flip). Instead, lets divide by $x^2$ to get $4x \leq x^2$ or $x(x-4)\geq 0.$ This is true when $x\leq 0$ or $x\geq 4$ so the desired inequality is true in that range.
For $0< x < 4$ we don't have $\frac{x^3}{3!} \leq \frac{x^4}{4!}$ but lets see if the other terms can save us. To do this, we need to see exactly how large $g(x) = x^3/3! - x^4/4!$ can be in $(0,4).$ We calculate that $g'(x) = -(x-3)x^2/6$ so $g$ increases when $0\leq x\leq 3$, the maximum occurs at $g(3)=9/8$, and then it decreases after that.
This is good, because the $1+x+x^2/2$ terms obviously give at least $1$ from $x=0$, and will give us more as $x$ gets bigger. So we solve $1+x+x^2/2=9/8$ and we take the positive solution which is $\frac{\sqrt{5}-2}{2} \approx 0.118.$ So the inequality is definitely true for $x\geq 0.12$ because $g$ is at most $9/8$ and $1+x+x^2/2$ accounts for that amount in that range.
Remember that $g$ was increasing between $x=0$ to $x=3$, so the largest $g$ can be in the remaining range is $g(0.12) = 873/315000 <1$, which is less than the amount $1+x+x^2/2$ gives us. So the inequality is also true for $0\leq x\leq 0.12$, so overall, for all $x.$
So all in all, the only trouble was for $x$ in $(0,4)$ and the contribution from the other terms was always enough to account for $x^3/3!$ when $x^4/4!$ wasn't enough.
| {
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Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is wrong.
I must be setting up the $A, B, C$ thing wrong but I do not know why.
|
My book is telling me that I have to complete the square
$$I=\begin{eqnarray*}
\int \frac{dx}{x^{2}-2x} &=&\int \frac{dx}{\left( x-1\right) ^{2}-1}\overset{
u=x-1}{=}\int \frac{1}{u^{2}-1}\,du=-\text{arctanh }u+C
\end{eqnarray*},$$
$$\tag{1}$$
where I have used the substitution $u=x-1$ and the standard derivative $$\frac{d}{du}\text {arctanh}=\frac{1}{1-u^{2}}\tag{2}$$
You just need to substitute $u=x-1$ to write $\text{arctanh }u$ in terms of $x$.
Added 2: Remark. If we use the logarithmic representation of the inverse hyperbolic function $\text{arctanh }u$
$$\begin{equation*}
\text{arctanh }u=\frac{1}{2}\ln \left( u+1\right) -\frac{1}{2}\ln \left(
1-u\right),\qquad (\text{real for }|u|<1)\tag{3}
\end{equation*}$$
we get for $u=x−1 $
$$\begin{eqnarray*}
I &=&-\text{arctanh }u+C=-\text{arctanh }\left( x-1\right) +C \\
&=&-\frac{1}{2}\ln x+\frac{1}{2}\ln \left( 2-x\right) +C \\
&=&\frac{1}{2}\left( \ln \frac{2-x}{x}\right) +C\qquad (0<x<2).
\end{eqnarray*}\tag{4}$$
Added. If your book does require using partial fractions then you can proceed as follows
$$\begin{equation*}
\int \frac{1}{u^{2}-1}\,du=\int \frac{1}{\left( u-1\right) \left( u+1\right)
}\,du=\int \frac{1}{2\left( u-1\right) }-\frac{1}{2\left( u+1\right) }du.
\end{equation*}$$
$$\tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/156566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Combining a radical and simplifying? How would I combine and simplify the following radical:
$$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}$$
| $\sqrt{\frac{A^2}{2}} - \sqrt{\frac{A^2}{8}} = \frac{A}{\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{2A}{2\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{A}{2\sqrt{2}}$
Assuming $A \geq 0$. If $A < 0$, you can replace with $|A|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recursive method.)
| Let $S = 2^0 + 2^1 + 2^2 + \cdots + 2^{n}$.
Then $2S = 2^1 + 2^2 + 2^3 + \cdots + 2^{n} + 2^{n+1}$.
Then
$$\begin{align*}
S = 2S - S &= & & 2^1 &+& 2^2 & + & 2^3 & + & 2^4 &+&\cdots &+& 2^{n} &+& 2^{n+1}\\
&& -2^0 -& 2^1 & - & 2^2 & - & 2^3 & - & 2^4 & - & \cdots & - & 2^n
\end{align*}$$
How much is that?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Proof about $z\cot z=1-2\sum_{k\ge1}z^2/(k^2\pi^2-z^2)$ In Concrete Mathematics, it is said that
$$z\cot z=1-2\sum_{k\ge1}\frac{z^2}{k^2\pi^2-z^2}\tag1$$
and proved in EXERCISE 6.73
$$z\cot z=\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}+\sum_{k=1}^{2^{n-1}-1}\frac z{2^n}\left(\cot\frac{z+k\pi}{2^n}+\cot\frac{z-k\pi}{2^n}\right)$$
The trigonmetric identity is not hard, but I cannot understand the rest:
It can be shown that term-by-term passage to the limit is justified, hence equation (1) is valid.
How can we conclude that? Thanks for help!
| This identity is also proven in this answer, but the limit of the trigonometric identity is a cute trick, too.
Concrete Mathematics claim:
For the limit claimed in Concrete Mathematics, we need a few things.
First, by inspecting the graph of $\frac{1-x\cot(x)}{x^2}$ for $-\frac{3\pi}{4}\le x\le\frac{3\pi}{4}$, we have
$$
\left|\frac1x-\cot(x)\right|\le|x|\tag{1}
$$
Next, the Mean Value Theorem says
$$
\begin{align}
|\cot(\delta+x)+\cot(\delta-x)|
&=|\cot(x+\delta)-\cot(x-\delta)|\\
&\le2\delta\sup_{[x-\delta,x+\delta]}\csc^2(\xi)\\
&\le\color{#C00000}{8\delta\,\csc^2(x)}\\
&\le\color{#C00000}{2\pi^2\delta/x^2}\tag{2}
\end{align}
$$
if $\color{#C00000}{2\delta\le|x|\le\frac{\pi}{2}}$.
Finally, note that since $0\le k< 2^{n-1}$, $0\le\frac{k\pi}{2^n}<\frac{\pi}{2}$
Using $(1)$, we get
$$
\begin{align}
&\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)-\left(\frac{z}{z+k\pi}+\frac{z}{z-k\pi}\right)\right|\\
&\le2\left|\frac{z}{2^n}\right|\frac{|z|+k\pi}{2^n}\tag{3}
\end{align}
$$
Using $(2)$, we get, for $2z\le k\pi$,
$$
\begin{align}
\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)\right|
&\le2\pi^2\left|\frac{z^2}{2^{2n}}\right|\left(\frac{2^n}{k\pi}\right)^2\\
&\le2\pi^2\left(\frac{z}{k\pi}\right)^2\tag{4}
\end{align}
$$
Estimate $(3)$ is used to control the difference between the series for small $k$, and $(4)$ to control the remainder in the sum of the cotangents for large $k$.
Pick an $\epsilon>0$, and find $m$ large enough so that $2z\le m\pi$ and
$$
\sum_{k=m}^\infty\frac{1}{k^2}\le\epsilon\tag{5}
$$
Then we have the following estimate for the tail of the sum
$$
\sum_{k=m}^\infty\frac{z^2}{k^2\pi^2-z^2}\le\frac43z^2\epsilon\tag{6}
$$
Combining $(4)$ and $(5)$ yields
$$
\sum_{k=m}^{2^{n-1}-1}\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)\right|\le2z^2\epsilon\tag{7}
$$
Summing $(3)$ gives
$$
\begin{align}
&\sum_{k=1}^{m-1}\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)-\left(\frac{z}{z+k\pi}+\frac{z}{z-k\pi}\right)\right|\\
&\le2\left|\frac{z}{2^n}\right|\frac{m|z|+m^2\pi/2}{2^n}\tag{8}
\end{align}
$$
Just choose $n$ big enough so that $(8)$ and $\displaystyle\left|\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}-1\right|$ are each less than $\epsilon$ and we get that the term-by-term absolute difference is less than
$$
\left(\frac{10}{3}z^2+2\right)\epsilon\tag{9}
$$
| {
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"url": "https://math.stackexchange.com/questions/159752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| There are two (three) ways to go. One, assume
$$x^4+1=(x^2+ax+1)(x^2-ax+1)$$
You'll get that
$${x^4} + 1 = {x^4} + \left( {2 - {a^2}} \right){x^2} + 1$$
Then $a=\sqrt 2$ (or the other, by symmetry)
$${x^4} + 1 = {x^4} + 1 = \left( {{x^2} + \sqrt 2 x + 1} \right)\left( {{x^2} - \sqrt 2 x + 1} \right)$$
The other ${x^2} = \tan \theta $, but it might get messy, unless you know how to use the Weierstrass substitution for example.
$$\int {\frac{{dx}}{{{x^4} + 1}}} = \int {\frac{{\left( {{{\tan }^2}\theta + 1} \right)d\theta }}{{{{\tan }^2}\theta + 1}}} \frac{1}{{2\sqrt {\tan \theta } }} = \int {\sqrt {\frac{{\cos\theta }}{{\sin\theta }}} \frac{{d\theta }}{2}} $$
$$\int {\sqrt {\frac{{\frac{{1 - {u^2}}}{{1 + {u^2}}}}}{{\frac{{2u}}{{1 + {u^2}}}}}} \frac{{du}}{{1 + {u^2}}}} = \int {\sqrt {\frac{{1 - {u^2}}}{{2u}}} \frac{{du}}{{1 + {u^2}}}} $$
However, Chandrasekar's is the best way to go, if you can figure it out.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 13
} |
What transforms under SU(2) as a matrix under SO(3)? A vector $\boldsymbol{r}$ in $\mathbb{R}^3$ transforms under rotation $\boldsymbol{A}$ to $\boldsymbol{r}'=\boldsymbol{Ar}$. It is equivalent to an SU(2) "rotation" as
$$\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdot\boldsymbol{\sigma} \right) \boldsymbol{h}^{-1},$$
where $\boldsymbol{h}$ is the counterpart of $\boldsymbol{A}$ in SU(2) given by the homomorphism between these two groups.
Now the question is, what would be the equivalent transformation in SU(2) of the rotation of a matrix in $\mathbb{R}^3$? In other words, what is the equivalent in SU(2) of $\boldsymbol{M}'=\boldsymbol{A}\boldsymbol{M}\boldsymbol{A}^{-1}$.
| Firstly, we need to map $\mathbb{R}^3$ to the representation space $V$ for $\mathrm{SU}(2)$. One possible map is given by the following formula:
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto
x \mathbf{I} +
y \mathbf{J} +
z \mathbf{K}$$
where
\begin{align}
\mathbf{I} & = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} &
\mathbf{J} & = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} &
\mathbf{K} & = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}
\end{align}
$\mathrm{SU}(2)$ acts on $V$ by conjugation: so for each $X$ in $V$ and each $A$ in $\mathrm{SU}(2)$, the ordinary matrix product $A X A^{-1}$ is in $V$. This is linear in $X$ and is indeed a linear representation of $\mathrm{SU}(2)$. Indeed, if
$$A = \begin{pmatrix} r e^{i \theta} & s e^{-i \phi} \\ -s e^{i \phi} & r e^{-i \theta} \end{pmatrix}$$
where $r, s, \theta, \phi$ are real numbers and $r^2 + s^2 = 1$, then $A \in \mathrm{SU}(2)$, and
\begin{align}
A \mathbf{I} A^{-1} & = (r^2 - s^2) \mathbf{I} + 2 r s \sin (\theta - \phi) \mathbf{J} - 2 r s \cos (\theta - \phi) \mathbf{K} \\
A \mathbf{J} A^{-1} & = 2 r s \sin (\theta + \phi) \mathbf{I} + (r^2 \cos 2 \theta + s^2 \cos 2 \phi) \mathbf{J} + (r^2 \sin 2 \theta - s^2 \sin 2 \phi) \mathbf{K} \\
A \mathbf{K} A^{-1} & = 2 r s \cos (\theta + \phi) \mathbf{I} - (r^2 \sin 2 \theta + s^2 \sin 2 \phi) \mathbf{J} + (r^2 \cos 2 \theta - s^2 \cos 2 \phi) \mathbf{K}
\end{align}
Thus, the induced action of $\mathrm{SU}(2)$ on $\mathbb{R}^3$ is given by the group homomorphism below,
$$\begin{pmatrix} r e^{i \theta} & s e^{-i \phi} \\ -s e^{i \phi} & r e^{-i \theta} \end{pmatrix} \mapsto
\begin{pmatrix}
r^2 - s^2 & 2 r s \sin (\theta + \phi) & 2 r s \cos (\theta + \phi) \\
2 r s \sin (\theta - \phi) & r^2 \cos 2 \theta + s^2 \cos 2 \phi & -r^2 \sin 2 \theta - s^2 \sin 2 \phi \\
-2 r s \cos (\theta - \phi) & r^2 \sin 2 \theta - s^2 \sin 2 \phi & r^2 \cos 2 \theta - s^2 \cos 2 \phi
\end{pmatrix}$$
and one may verify that the RHS is a matrix in $\mathrm{SO}(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of equations of 3rd degree I need help with the following system of equations:
$
2y^3 +2x^2+3x+3=0 $
$
2z^3 + 2y^2 + 3y + 3= 0 $
$2x^3 + 2z^2 + 3z + 3 = 0$
| The only real solution is $x = y = z = -1$.
Claim 1: $x,y,z \ge -1$.
Proof. Suppose that $x < -1$. Then $0 = 2y^3 + 2x^3 + 3x + 3 > 2y^3 + 2$, so that $y < -1$ also. Similarly it follows that $z < -1$. Hence if one of $x,y,z$ is smaller than $-1$, all of them are. But then if for example $x<z$, we have
$$0 = 2x^3 + 2z^2 + 3z + 3 < 2z^3 + 2z^2 + 3z + 3 = (z+1)(2z^2 + 3) < 0,$$
and we see that necessarily $x=y=z$, which implies that $x=y=z=-1$, contradiction.
Claim 2: $x,y,z \le -1$.
Proof. Suppose that $x > -1$ is the largest of $x,y,z$. So $z \le x$ and
$$0 = 2x^3 + 2z^2 + 3z + 3 \ge 2z^3 + 2z^2 + 3z + 3 = (z+1)(2z^2 + 3),$$
which implies that $z \le - 1$. By Claim 1. $z = -1$ and hence also $x = -1$ and $y = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Series Expansion An old problem from Whittaker and Watson I'm having issues with. Any guidance would be appreciated.
Show that the function
$$
f(x)=\int_0^\infty \left\{ \log u +\log\left(\frac{1}{1-e^{-u}} \right) \right\}\frac{du}{u}e^{-xu}
$$
has the asymptotic expansion
$$
f(x)=\frac{1}{2x}-\frac{B_1}{2^2x^2}+\frac{B_3}{4^2x^4}-\frac{B_5}{6^2x^6}+\;... \;,
$$
where
$$B_1, B_3, ...$$
are Bernoulli's numbers.
Show also that f(x) can be developed as an absolutely convergent series of the form
$$
f(x)=\sum_{k=1}^\infty\frac{c_k}{(x+1)(x+2)...(x+k)}
$$
| Note that:
$$\frac{d}{du} \left\{ \ln u + \ln \left( \frac{1}{1- e^{-u}}\right) \right\} = \frac{1}{u} - \frac{1}{e^u - 1} = -\sum_{n = 1}^{+\infty} \frac{B_n u^{n-1}}{n!}$$
then:
$$\ln u + \ln \left( \frac{1}{1- e^{-u}}\right) = -\sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} u^n $$
so we can rewrite the integral as:
$$-\int_0^{+\infty} \sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} u^n \cdot \frac{e^{-xu}}{u} \, du = - \sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} \cdot \frac{(n-1)!}{x^n} = - \sum_{n=1}^{+\infty} \frac{B_n}{n^2 x^n}$$
In other words it's equal to:
$$f(x) = \frac{1}{2x} - \frac{B_2}{2^2 \cdot x^2} + \frac{B_4}{4^2 \cdot x^4} - \ldots$$
I've used the same notation for Bernoulli numbers as it is at MathWorld.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the limit at x=0 Find the limit of $f(x)=\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}}$ (at x=0) so that $f(x)$ becomes continuous for all $x$. My answer is $2\sqrt{a}$. Am I right?
Sorry to state the question incorrectly. We have to define $f(x)$ at $x=0$ such that $f(x)$ is continuous for all $x$. In that case my answer is $2\sqrt{a}$.
Let f(x)=$[x]$+$[-x]$ be a function where [.] stands for the greatest integer not greater than x. For any integer $m$, what can we say about $lim_{x \to m}$. Is $f(x)$ contiuous at $x=m$. Sorry for asking such vague question but I am forgetting the exact wording and the options given. This was a question asked in a class test.
| EDIT::
$$ \frac{\sqrt{a^2}-\sqrt{a^2}}{\sqrt a - \sqrt a} = \frac{0}{0}\neq 2\sqrt a $$
You have to multiply by conjugate of both terms (in numerator and denominator) and get the following.
$$ \frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}} \\
= \frac{(a^2-ax+x^2)-(a^2+ax+x^2)}{(a-x)-(a+x)} \times \frac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \\
= \frac{-2ax}{-2x}\times \frac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}
\\
\text{ taking limit x } \rightarrow 0 \text{ we get}
=a \frac{2\sqrt a}{2 a} = \sqrt a$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding solutions to $(4x^2+1)(4y^2+1) = (4z^2+1)$ Consider the following equation with integral, nonzero $x,y,z$
$$(4x^2+1)(4y^2+1) = (4z^2+1)$$
What are some general strategies to find solutions to this Diophantine?
If it helps, this can also be rewritten as $z^2 = x^2(4y^2+1) + y^2$
I've already looked at On the equation $(a^2+1)(b^2+1)=c^2+1$
| Let $a$ be a positive integer.
Then
\begin{align}
(4a^2+1)(4((2a)^2)^2+1) &= 256a^6 + 64a^2 + 4a^2 + 1 \\
& = 4(64a^6 + 16a^4 + a^2) + 1 \\
&= 4(a^2(8a^2+1)^2)+1 \\
&= 4((8a^2+1)a)^2+1
\end{align}
so $(a, (2a)^2, (8a^2+1)a)$ is always a solution.
There are others as well.
| {
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"source": "stackexchange",
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Find all ordered pair of integers $(x,y)$
Obtain all ordered pair of integers $(x,y)$ such that $$x(x + 1) = y(y + 1)(y + 2)(y + 3)$$
I'm getting 8,
(0, 0), (0, -1), (0, -2), (0, -3)
(-1, 0), (-1, -1), (-1, -2), (-1, -3)
Please confirm my answer.
| Hint: It is easily proved that the product of four consecutive integers, plus $1$, is a perfect square. But $x(x+1)+1$ is hardly ever a perfect square!
Added: To prove that $y(y+1)(y+2)(y+3)+1$ is a perfect square, note that
$$y(y+1)(y+2)(y+3)=y(y+3)(y+1)(y+2)=(y^2+3y)(y^2+3y+2)=z(z+2),$$
where $z=y^2+3y$. And clearly $z(z+2)+1=(z+1)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$ I want to expand and test this $\{(n^3+1)^{1/3} - n\}$ for convergence/divergence.
The edited version is: Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$
| By direct inspection, for every pair of real numbers $A$ and $B$,
$$
A^3 - B^3 = (A-B) (A^2+AB+B^2).
$$
Choose now $A=\sqrt[3]{n^3+1}$ and $B=n$. Then
$$
(n^3+1)^{1/3} - n = \frac{n^3+1-n^3}{(n^3+1)^{2/3} + n (n^3+1)^{1/3}+n^2} \sim \frac{1}{n^2}
$$
as $n \to +\infty$. The limit is therefore zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine whether $\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$ is convergent or divergent. If convergent, find the sum. $$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$
It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = \frac{(-3)^{x-1}}{4^x}$.
$$\lim\limits_{x \to \infty}\frac{(-3)^{x-1}}{4^x}$$
Is this how I handle this exercise? I still cannot seem to get the answer $\frac{1}{7}$
| Let $q = \frac{-3}{4}$, $a_n = \frac{(-3)^{n-1}}{4^n}$, $b_0 = 0$, $b_n = b_{n-1} + a_n$.
$a_n = -\frac{1}{3} (\frac{-3}{4})^n = -\frac{1}{3} q^n$, hence $b_n = -\frac{1}{3} c_n$, where $c_0 = 0$, $c_n = c_{n-1} + q^n$.
The $c_n$ limit is equal to $q + q^2 + q^3 + \ldots = \frac{q}{1-q} = \frac{\frac{-3}{4}}{1-\frac{-3}{4}} = \frac{-3}{7}$, thus $b_n$ limit is equal to $\frac{1}{7}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$$\ \frac{ab}{a^{2}+3b^{2}} \leq \frac{1}{2 \sqrt{3}}=\frac{\sqrt{3}}{6} $$
$$\ \sum_{cyc} \frac{ab}{a^{2}+3b^{2}} \leq \frac{\sqrt{3}}{2} $$
But this is obviously is not working .
| By AM-Gm $$\sum_{cyc}\frac{ab}{a^2+3b^2}=\sum_{cyc}\frac{ab}{a^2+b^2+2b^2}\leq\sum_{cyc}\frac{ab}{2\sqrt{2b^2(a^2+b^2)}}=\frac{1}{2\sqrt2}\sum_{cyc}\sqrt{\frac{a^2}{a^2+b^2}}.$$
Thus, it remains to prove that $\sum\limits_{cyc}\sqrt{\frac{x}{x+y}}\leq\frac{3}{\sqrt2}$, which followos from C-S.
Indeed, $$\left(\sum\limits_{cyc}\sqrt{\frac{x}{x+y}}\right)^2\leq\sum_{cyc}\frac{x}{(x+y)(x+z)}\sum_{cyc}(x+z)=\frac{4(xy+xz+yz)(x+y+z)}{\prod\limits_{cyc}(x+y)}\leq\frac{9}{2},$$
where the last inequality it's just $\sum\limits_{cyc}z(x-y)^2\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Polynomial-related manipulation My question is:
Factorize: $$x^{11} + x^{10} + x^9 + \cdots + x + 1$$
Any help to solve this question would be greatly appreciated.
| $$
\begin{align}
& {}\quad (x^{11} + x^{10}) + (x^9 + x^8)+(x^7+x^6)+(x^5+x^4)+(x^3+x^2 )+( x + 1)\\[8pt]
& =x^{10}(x+1)+x^8(x+1)+x^6(x+1)+x^4(x+1)+x^2(x+1)+(x+1)\\[8pt]
& =(x+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\[8pt]
& =(x+1)(x^8(x^2+1)+x^4(x^2+1)+x^2+1)\\[8pt]
& =(x+1)((x^2+1)(x^8+x^4+1))\\[8pt]
& =(x+1)(x^2+1)(x^4+1-x^2)(x^4+1+x^2)\\[8pt]
& =(x+1)(x^2+1)(x^4+1-x^2)(x^2+1-x)(x^2+1+x)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$(x^n-x^m)a=(ax^m-4)y^2$ in positive integers How do I find all positive integers $(a,x,y,n,m)$ that satisfy $ a(x^{n}-x^{m}) = (ax^{m}-4) y^{2} $ and $ m\equiv n\pmod{2} $, with $ax$ odd?
| Some partial results.
Case 1) $n<m$.
Then $ax^m<4$.
The solutions of this inequality are:
$$
a=1, m=1, x=1,2,3,\\
m\geq2, x=1,
$$
and
$$
a=2,3, x=1, m\geq 1.
$$
From these we obtain two candidates (i) $x=1$ and (ii) $x=3$.
(i) Substituting into the original equation we get
$$
0=(a-4)y^2.
$$
Since $y>0$ and $ax^m=a<4$ there is no solution.
(ii) Substituting into the original equation we get
$$
3^n-3=(3-4)y^2.
$$
Since $n<m$ and $n\equiv m \pmod{2}$, thus $n=1$ which gives $y=0$ that is no solution.
Case 2) $n=m$.
Substituting into the original equation we have
$$
(ax^m-4)y^2=0.
$$
Since $y>0$ we get
$$
ax^m=4.
$$
Since $ax$ is odd there is no solution.
Thus we have justified that $n>m$.
So we can write
$$
x^m(x^{n-m}-1)a=(ax^m-4)y^2.
$$
Since $a$ is odd and $(a,ax^m-4)=1$, therefore $a|y^2$, and similarly $x^m|y^2$.
Furthermore $n=m+2k$ where $k$ is positive integer. Thus we obtain
$$
ax^m(x^{2k}-1)=(ax^m-4)y^2.
$$
Introducing the new variables $u:=ax^m$, $z:=x^k$ we obtain
$$
u(z^2-1)=(u-4)y^2,
$$
where $u,z$ are odd and $z>1$. Obviously $8|z^2-1$, thus $4|y$. Since
$$
u(z^2-1)=(u-4)y^2<uy^2,
$$
it follows $z<y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
how to solve system of linear equations of XOR operation? how can i solve this set of equations ? to get values of $x,y,z,w$ ?
$$\begin{aligned} 1=x \oplus y \oplus z \end{aligned}$$
$$\begin{aligned}1=x \oplus y \oplus w \end{aligned}$$
$$\begin{aligned}0=x \oplus w \oplus z \end{aligned}$$
$$\begin{aligned}1=w \oplus y \oplus z \end{aligned}$$
this is not a real example, the variables don't have to make sense, i just want to know the method.
| As I wrote in my comment, you can just use any method you know for solving linear systems, I will use Gauss:
$$
\begin{array}{cccc|c||l} \hline
x & y & z & w &\ & \\ \hline\hline
1 & 1 & 1 & 0 & 1 & \\
1 & 1 & 0 & 1 & 1 & \text{$+$ I}\\
1 & 0 & 1 & 1 & 0 & \text{$+$ I}\\
0 & 1 & 1 & 1 & 1 & \\ \hline
1 & 1 & 1 & 0 & 1 & \\
0 & 0 & 1 & 1 & 0 & \text{III}\\
0 & 1 & 0 & 1 & 1 & \text{II}\\
0 & 1 & 1 & 1 & 1 & \text{$+$ III}\\ \hline
1 & 1 & 1 & 0 & 1 & \\
0 & 1 & 0 & 1 & 1 & \\
0 & 0 & 1 & 1 & 0 & \\
0 & 0 & 1 & 0 & 0 & \text{$+$ III}\\ \hline
1 & 1 & 1 & 0 & 1 & \\
0 & 1 & 0 & 1 & 1 & \\
0 & 0 & 1 & 1 & 0 & \\
0 & 0 & 0 & 1 & 0 & \\\hline
\end{array}
$$
Now we can conclude $w = 0$ from line 4, which gives $z = 0$ from 3 and $y = 1$ from 2, and finally $x = 0$. So $(x,y,z,w) = (0,1,0,0)$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
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} |
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
| Here I give a detailed proof. Though steps could have been jumped to keep it short.
Without loss of generality we can assume that $a\ge b \ge c$
Let $$(-a+b+c)(a-b+c)(a+b-c)=S$$
$$\Rightarrow S=(-a+b+c)\{a-(b-c)\}\{a+(b-c)\}$$ $$\Rightarrow S=(-a+b+c)\{a^2-(b-c)^2\} $$ $$\Rightarrow S= (-a+b+c)\{a^2-b^2-c^2+2bc\}$$ $$\Rightarrow S=-(a^3+b^3+c^3)-2abc+b^2c+bc^2+ab^2+a^2b+ac^2+a^2c $$ $$\Rightarrow abc-S=(a^3+b^3+c^3)+3abc-(b^2c+bc^2+ab^2+a^2b+ac^2+a^2c) $$ $$\Rightarrow abc-S=(a^3-a^2b)+(b^3-b^2c)+(c^3-c^2a)+(abc-bc^2)+(abc-ab^2)+(abc-a^2c) $$$$\Rightarrow abc-S=a^2(a-b)+b^2(b-c)+c^2(c-a)+bc(a-c)+ab(c-b)+ac(b-a) $$ $$\Rightarrow abc-S=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)\{a(a-c)-b(b-c)\}+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)^2\{a^2-b^2+c(b-a)\}+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)^2\{a+b-c\}+c(c-a)(c-b)$$ Now $(c-a) \le 0$ and $(c-b) \le 0$ $$\Rightarrow c(c-a)(c-b)\ge 0 $$ and $$ (a-b)^2(a+b-c) \ge 0$$ This shows $$ abc-S \ge 0$$ $$\Rightarrow abc\ge S$$ $$\Rightarrow abc\ge (-a+b+c)(a-b+c)(a+b-c) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 0
} |
Trigonometric Identities: $\frac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$ $\dfrac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$
Right Side:
$1-\cos\theta$ either stays the same, or can be $1-\dfrac{1}{\sec\theta}$
Left Side:
$$\begin{align*}
&= \dfrac{\sin^2\theta}{1+\cos\theta}\\
&= \dfrac{1-\cos^2\theta}{1+\cos\theta}
&= \dfrac{(1-\cos\theta)(1+\cos\theta)}{1+cos\theta}
&= 1-\cos\theta
\end{align*}$$
Is this correct?
| Perhaps slightly simpler and shorter (FYI, what you did is correct):
$$\frac{\sin^2x}{1+\cos x}=1-\cos x\Longleftrightarrow \sin^2x=(1-\cos x)(1+\cos x)\Longleftrightarrow \sin^2x=1-\cos^2x$$
And since the last equality is just the trigonometric Pytahgoras Theorem we're done.
| {
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"url": "https://math.stackexchange.com/questions/170947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x+1=0$$
Edit:
$$2\cos^2x=\sin x+1$$
$$2(1-\sin^2x)=\sin x+1$$
$$2\sin^2x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$a=-1,\dfrac{1}{2}$$
$$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$
$$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$
$$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$
$$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$
I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$
| This is more a comment than an answer, but it is an important comment, since it affects the list of solutions. We seem to have solutions that come from $\sin x=\frac{1}{2}$, and solutions that come from $\sin x=-1$.
No problem with the $\sin x=\frac{1}{2}$ stuff. For the record, this gives the solutions $x=\frac{\pi}{6}+2k\pi$ and $x=\frac{5\pi}{6}+2k\pi$, where $k$ ranges over the integers, positive, negative, and $0$.
However, there is an issue with $\sin x=-1$. For then $\cos x=0$, and neither $\tan x$ nor $\sec x$ is defined. When we multiply through by $\cos x$, we must remember that if $\cos x=0$ we are multiplying both sides by $0$, so we are not getting an equivalent equation. Thus the "solutions" $-\frac{\pi}{2}+2k\pi$ must be discarded.
Remark: If we look at the function $\frac{\sin x+1}{\cos x}$, it turns out that it has a removable singularity at $x=-\frac{\pi}{2}$, and can be made continuous there by defining it to be $0$. In that sense, $x=-\frac{\pi}{2}$ is a solution. However, I doubt it would be accepted as a solution in the context the OP is in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
multiple choice summation problem
Let
$$X = \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001}.$$
Then
(A) $X < 1$
(B) $X > 3/2$
(C) $1 < X < 3/2$
(D) none of the above holds.
I assume that the answer is the third choice $1<X<3/2$. I integrate out $1/x$ in the interval $(1001, 3001)$ and get a result that satisfies only the choice C. Is this a Riemann sum? Please help.
| With respect to your Riemann sum approach: the idea is that for positive, decreasing functions, the Riemann sum and the integral carefully approximate each other, more or less as in the proof of the integral test of convergence.
If you'd like another sort of approach, we could approach it naively. Separate the sum into $250$ element blocks, $\frac{1}{1001}$ to $\frac{1}{1250}$ in the first block $B_1$, $\frac{1}{1251}$ to $\frac{1}{1500}$ in the second block $B_2$, and so on. We'll have $12$ blocks.
Note that $\frac{1}{5} = \frac{250}{1250} \leq B_1 \leq \frac{250}{1000} = \frac{1}{4}$. Similarly, we get that $\frac{1}{i + 4} \leq B_i \leq \frac{1}{i+3}$ for all of our blocks.
This means that our sum has upper and lower bounds:
$$ 1 < \frac{1}{5} + \frac{1}{6} + \dots \frac{1}{16} \leq B_1 + \dots + B_{12} \leq \frac{1}{4} + \dots + \frac{1}{15}< \frac{3}{2}$$
And this gives the desired inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?
I know how to prove
$ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $
in this way:
$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$
$\Longleftrightarrow \sin{20^0}.\cos{50^0}.\cos{10^0} = \sqrt{3}.\sin{50^0}.\sin{10^0}.\cos{20^0}$
$\Longleftrightarrow \frac{1}{2}\sin{20^0}(\cos{60^0}+\cos{40^0}) = \frac{\sqrt{3}}{2}(\cos{40^0}-\cos{60^0}).\cos{20^0}$
$\Longleftrightarrow \frac{1}{4}\sin{20^0}+\frac{1}{2}\sin{20^0}.\cos{40^0} = \frac{\sqrt{3}}{2}\cos{40^0}.\cos{20^0}-\frac{\sqrt{3}}{4}.\cos{20^0}$
$\Longleftrightarrow \frac{1}{4}\sin{20^0}-\frac{1}{4}\sin{20^0}+\frac{1}{4}\sin{60^0} = \frac{\sqrt{3}}{4}\cos{60^0}+\frac{\sqrt{3}}{4}\cos{20^0}-\frac{\sqrt{3}}{4}\cos{20^0}$
$\Longleftrightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8}$
Could this help to prove the first one and how ?Do i just need to know that $ \frac{1}{\tan\theta}=\tan(90^{\circ}-\theta) $ ?
| Due to $\tan 40^\circ*\tan 20^\circ$
$$=\frac{\sin 40^\circ*\sin 20^\circ}{\cos 40^\circ*\cos 20^\circ}$$
$$=\frac{4\sin 40^\circ*(\sin 20^\circ)^2}{4\cos 40^\circ*\cos 20^\circ*\sin 20^\circ}$$
$$=\frac{2\sin 40^\circ*(1-\cos 40^\circ)}{2\cos 40^\circ*\sin 40^\circ}$$
$$=\frac{2\sin 40^\circ-2\sin 40^\circ*\cos 40^\circ}{\sin 80^\circ}$$
$$=\frac{2\sin 40^\circ-\sin 80^\circ}{\cos10^\circ}$$
$$=\frac{\sin 40^\circ-(\sin 80^\circ-\sin 40^\circ)}{\cos10}$$
$$=\frac{\sin 40^\circ-2\cos 60^\circ*\sin 20^\circ}{\cos10}$$
$$=\frac{\sin 40^\circ-\sin 20^\circ}{\cos10}$$
$$=\frac{2\cos 30^\circ*\sin 10^\circ}{\cos10}$$
$$=\cot 30^\circ*\tan 10^\circ$$
Hence,
$$\tan 20^\circ*\tan 30^\circ*\tan 40^\circ$$
$$=\tan 30^\circ*\cot 30^\circ*\tan 10^\circ$$
$$=\tan 10^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $ (r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$ How would I verify the following trig identity?
$$(r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$$
My work thus far is
$$(r^2\cos^2A\sin^2A)+(r^2\sin^2A\sin^2A)+(r^2\cos^2A)$$
But how would I continue? My math skills fail me.
| Oh I didn't read robjohn's answer carefully before making this colourful answer.. I will leave it here anyways.
To continue on what you have:
$$ (\color{red}{r^2}\cos^2A\sin^2A)+(\color{red}{r^2}\sin^2A\sin^2A)+(\color{red}{r^2}\cos^2A) \\
= \color{red}{r^2}( \cos^2A\color{blue}{\sin^2A}+\sin^2A\color{blue}{\sin^2A}+\cos^2A) \\
= r^2( (\color{red}{\cos^2A+\sin^2A})\color{blue}{\sin^2A}+\cos^2A) \\
= r^2( (\color{red}{1})\sin^2A+\cos^2A) \\
= r^2( \color{red}{\cos^2A+\sin^2A}) \\
= r^2( \color{red}{1} ) \\
= r^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$ Recently, I ran across a product that seems interesting.
Does anyone know how to get to the closed form:
$$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$
I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something.
This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even.
So, I tried taking that into account, but must have approached it wrong.
How can this be shown? Thanks everyone.
| If $n$ is even, then the term with $k=n/2$ makes the product on the left $0$ and $\sin\left(\frac{n}{2}\pi\right)=0$. So assume that $n$ is odd.
$$
\begin{align}
\prod_{k=1}^n\cos\left(\frac{k\pi}{n}\right)
&=-\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)\tag{1}\\
&=-\prod_{k=1}^{n-1}\frac{\sin\left(\frac{2k\pi}{n}\right)}{2\sin\left(\frac{k\pi}{n}\right)}\tag{2}\\
&=\frac{-1}{2^{n-1}}\frac{\prod\limits_{k=\frac{n+1}{2}}^{n-1}\sin\left(\frac{2k\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{3}\\
&=\frac{(-1)^{\frac{n+1}{2}}}{2^{n-1}}\frac{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{4}\\
&=-\frac{\sin\left(n\frac\pi2\right)}{2^{n-1}}\tag{5}
\end{align}
$$
$(1)$: $\cos(\pi)=-1$
$(2)$: $\sin(2x)=2\sin(x)\cos(x)$
$(3)$: cancel $\sin\left(\frac{j\pi}{n}\right)$ in the numerator and denominator for even $j$ from $2$ to $n-1$
$(4)$: in the numerator, change variable $k\mapsto k+\frac{n-1}{2}$ and use $\sin(x+\pi)=-\sin(x)$
$(5)$: for odd $n$, $\sin\left(n\frac\pi2\right)=(-1)^{\frac{n-1}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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Trigonometry proof involving sum difference and product formula How would I solve the following trig problem.
$$\cos^5x = \frac{1}{16} \left( 10 \cos x + 5 \cos 3x + \cos 5x \right)$$
I am not sure what to really I know it involves the sum and difference identity but I know not what to do.
| Applying the same approach as that of this,
let $A\cos5x+B\cos3x+C\cos x=\cos^5x$
As $\cos 3x = 4\cos^3x-3\cos x$ and
$\cos 5x = 16\cos^5 x-20\cos^3 x+5\cos x$
$A(16\cos^5 x-20\cos^3 x+5\cos x) + B( 4\cos^3x-3\cos x)+ C\cos x=\cos^5x$
Comparing the coefficients of different powers of cosx,
5th power=>16A=1=>$A=\frac{1}{16}$ ,
3rd power=>-20A+4B=0=>B=5A=$\frac{5}{16}$ and
1st power=>5A-3B+C=0=>C=3B-5A$=3\cdot\frac{5}{16}-5\cdot\frac{1}{16}=\frac{5}{8}$
Alternatively, observe that the 3rd power of $\cos x$ is absent in the given expression.
But $A\cos5x+B\cos3x=A(16\cos^5 x-20\cos^3 x+5\cos x) + B( 4\cos^3x-3\cos x)$
$=16A\cdot\cos^5 x+ \cos^3x\cdot4(B-5A)+\cos x(5A-3B)$
So, B must be 5A to eliminate $cos^3x$
$A\cos5x+B\cos3x=A(\cos5x+5\cdot\cos3x)=A(16\cdot\cos^5 x - 10\cdot\cos x)$
Putting A=1, $\cos5x+5\cdot\cos3x=16\cdot\cos^5 x - 10\cdot\cos x$
So, we just need a little rearrangement.
| {
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Evaluation of $\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)} \, d\theta$ For a test I had to evaluate $\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta$. First I used substitution, with $z=e^{i\theta}$ and $d\theta=\frac{1}{iz}dz$, as shown:
$$\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta=\int_{|z|=1}\frac{1}{1+\frac{3}{4}(z+z^{-1})(z+z^{-1})}\frac{1}{iz}dz=\frac{4}{i}\int_{|z|=1}\frac{z}{3z^4+10z^2+3}dz$$
Then I used the residue theorem:
$$\int_{|z|=1}\underbrace{\frac{z}{3z^4+10z^2+3}}_{f(z)}dz=\int_{|z|=1}\frac{z}{(z^2+3)(z^2+\frac{1}{3})}dz=\int_{|z|=1}\frac{z}{(z-i\sqrt{3})(z+i\sqrt{3})(z-i\frac{i\sqrt{3}}{3})(z+\frac{i\sqrt{3}}{3})}dz$$
As only $\frac{i\sqrt{3}}{3}$ and $-\frac{i\sqrt{3}}{3}$ lie in the integration domain, I don't need to compute the other residues:
$$\operatorname{Res}(f;\frac{i\sqrt{3}}{3})=\frac{3}{16}$$ and
$$\operatorname{Res}(f;-\frac{i\sqrt{3}}{3})=\frac{3}{16}$$
Finally:
$$\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta=\frac{4}{i}2\pi i \frac{6}{16}=3\pi $$
However MATLAB doesn't agree with that:
EDU>> f=@(z) 1./(1+3*cos(z).^2);
quad(f,0,2*pi)
ans =
3.1416
Where did I go wrong ? I don't see what's wrong in using theorems this way and I did the computations 3 times. Ask for more steps in calculation if needed.
| Right after the first appearance of $f(z)$, when you factored the denominator, you forgot a factor $3$.
Note: Complex analysis might not be the simplest approach here, since the change of variables $t=\tan(θ)$ yields directly that, for every $a\gt-1$,
$$
\int_0^{2\pi}\frac{\mathrm d\theta}{1+a\cos^2(\theta)}=\frac{2\pi}{\sqrt{a+1}}.
$$
| {
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"url": "https://math.stackexchange.com/questions/186025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Infinite descent This Wikipedia article of Infinite Descent says:
We have
$ 3 \mid a_1^2+b_1^2 \,$.
This is only true if both $a_1$ and $b_1$ are divisible by $3$.
But how can this be proved?
| Well, we have two possibilities, first if neither $a_1$ nor $b_1$ is divisible by $3$ then that means that they have the form $a_1 = 3k \pm 1$ and $b_1 = 3t \pm 1$. Then $$a_1^2 + b_1^2 = (3k \pm 1)^2 + (3t \pm 1)^2 = 9k^2 \pm 6k + 1 + 9t^2 \pm 6t + 1 = 3A + 2$$
where $A = 3k^2 \pm 2k + 3t^2 \pm 2t$ and thus since $a_1^2 + b_1^2 = 3A + 2$ we can conclude that $3 \nmid a_1^2 + b_1^2$.
The other possibility is that exactly one of $a_1$ or $b_1$ is divisible by $3$. Let's assume that $a_1$ is divisible by $3$ but $b_1$ isn't. Then they have the form $a_1 = 3k$ and $b_1 = 3t \pm 1$. Thus
$$
a_1^2 + b_1^2 = (3k)^2 + (3t \pm 1)^2 = 9k^2 + 9t^2 \pm 6t + 1 = 3B + 1
$$
where $B = 3k^2 + 3t^2 \pm 2t$ so again since $a_1^2 + b_1^2 = 3B + 1$ then $3 \nmid a_1^2 + b_1^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Let,
$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$
$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$
$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$
| If you can find or sketch some 3D blocks, there is a fun geometric proof.
Fix some $n$. If you are doing this with real blocks, $n=3$ or $4$ should be convincing. Let's take $n=4$ for now.
Make a $4\times 4\times 1$ base, laid flat, which of course has volume $4^2$.
Now make a $3\times 3 \times 1$ brick and place it, laid flat, above the base with some corners aligned. Continue in this way up to the top, making smaller squares and always aligning with the same corner. You now have a 3D corner of stairs whose volume is $1^2+\cdots +n^2$.
Now the fun part. Make 5 more of these "stair corners", for a total of 6. These six toys can be turned sideways and upside down, and then pieced together to make an $n\times(n+1)\times(2n+1)$ rectangular solid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how to evaluate the integral $\frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx$ How to evaluate the 59 integral, possibly using real method?
$$\frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx$$
| Start by evaluating, for $-1<\Re(s)<4$, the following integral:
$$
I(s) = \int_0^\infty \frac{x^s}{(1+x^2)(1+x^4)} \mathrm{d}x = \frac{1}{2} \int_0^\infty x^s \left( \frac{1}{1+x^2} + \frac{1-x^2}{1+x^4} \right) \mathrm{d}x
$$
When $-1<\Re(s)<1$ we can rewrite this as a sum of integrals:
$$
I(s) = \frac{\pi}{4} \frac{1}{\cos\left(\frac{\pi}{2} s\right)} + \frac{\pi}{8} \frac{1}{\cos\left( \frac{\pi}{4} (s+1)\right)} - \frac{\pi}{8} \frac{1}{\cos\left( \frac{\pi}{4} (s+3)\right)}
$$
Where the following result was used:
$$
\int_0^\infty \frac{x^s}{1+x^n} \mathrm{d} x = \frac{\pi}{n} \frac{1}{\cos\left( \frac{\pi}{n} (s+1)\right)}
$$
The expression given above remains true for all $-1<Re(s)<4$ by the principle of analytic continuation. Now, we recover the original integral as:
$$
\frac{64}{\pi^3} \left( 15 I^{\prime\prime}(0) - 2 I^{\prime\prime}(1) \right)
$$
which gives the result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.
Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$
(b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that:
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$
| Following above motivations and applying AM-GM three times:
\begin{align}
&\frac13\left(\frac ab+\frac ab+\frac bc\right)+\frac13\left(\frac bc+\frac bc+\frac ca\right)+\frac13\left(\frac ca+\frac ca+\frac ab\right)+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&\ge \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&=\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&\ge4\left(\left(\frac{a+b+c}{3\sqrt[3]{abc}}\right)^2\right)^\frac14\\
&\ge4.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
Plane intersection by a mapping and different cases In $\mathbb{R}^3$ say we have the 2 planes $A=\{z=1\}$ and $B=\{x=1\}$. A line through 0 meeting $A$ at $(x,y,1)$ meets $B$ at $(1,y/x,1/x).$ Consider the map $\phi: A \rightarrow B$ defined by $(x,y) \mapsto (y' = y/x, z' = 1/x)$.
I'm trying to figure out the image under $\phi$ of
1) the line $ax = y + b$; the pencil of parallel lines $ax = y + b$ with fixed $a$ and variable $b$;
2) circles $(x-1)^2 + y^2 = c$ for variable $c,$ distinguishing the 3 cases $c>1, c = 1,$ and $c< 1$.
and to imagine the above as a perspective drawing by an artist sitting at $(0,0,0)$ and drawing figures from the plane $A$ on the plane $B$.
What happens to the points of the 2 planes where $\phi$ and $\phi^{-1}$ are undefined? Thanks!
| To answer your last question first, notice that a line through $0$ meeting $A$ at $(0,y,1)$ does not meet $B$ at all. This explains why $\phi$ is undefined in such cases. Correspondingly, pick any point on $B$ with $z = 0$ and any line through the origin and that point is wholly within the $xz$-plane, so will never hit $x = 1$, so is not the projection of any point on $A$, so $\phi^{-1}$ is undefined.
To understand how lines on $A$ work, think of lines as the intersection of planes. More specifically, for each line $\lambda$ in $A$ there is a unique plane $C$ through the origin such that $\lambda$ is the intersection of $A$ with $C$. Then the image under $\phi$ must be the intersection of $C$ with $B$ (since any "projection ray" from the origin through $\lambda$ lies in the plane $C$). Now, this intersection will be a line in $B$ (assuming the line was not $\{x = 0, z = 1\}$, in which case there is no intersection). So lines project to lines. Once we have that fact, it's easy to compute which line it is: just project any two points of $\lambda$, and join them up. If you really need an explicit formula, just ask.
Circles are a little trickier. Substitute $x=1/z\prime$ and $y=y\prime/z\prime$ into the equation, and get: \[\frac{1}{z^2}(y^2 + (1-z)^2)=c\]. What does this actually mean? Well, let's rearrange a little: \[\begin{align}
\frac{1}{z^2}(y^2 + 1 - 2z + z^2) &= c \\
y^2 + 1 - 2z + z^2 &= cz^2 \\
y^2 - 2z + (1-c)z^2 &= -1
\end{align}\]. At this point I want to divide by $1-c$ to complete the square, so I'm going to have to distinguish the $c=1$ case. In that case, we get \[\frac{1}{2}(y^2 + 1)=z\], which is a parabola. Otherwise: \[\begin{align}
y^2 + (1-c)(z^2 - \textstyle{\frac{2}{1-c}}z) &= -1 \\
y^2 + (1-c)((z-\textstyle{\frac{1}{1-c}})^2 - \textstyle{\frac{1}{(1-c)^2}}) &= -1 \\
y^2 + (1-c)(z-\textstyle{\frac{1}{1-c}})^2 &= \textstyle{\frac{1}{1-c}} - 1 \\
y^2 + (1-c)(z-\textstyle{\frac{1}{1-c}})^2 &= \textstyle{\frac{c}{1-c}}
\end{align}\]. For $c < 1$, this is an ellipse, while for $c > 1$, it is a hyperbola.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Complex numbers Here's the question:
z is a complex, and if $z^5 + z^4 + z^3 + z^2 + z + 1 = 0$ then $z^6=1$.
use this fact to calculate how many answers is there for:
$$z^5 + z^4 + z^3 + z^2 + z + 1 = 0$$
Thanks.
| $z^6 = 1$ if and only if $z^6 - 1 = (z-1)(z^5+z^4+z^3+z^2+z+1) = 0$. So the roots of $z^5+z^4+z^3+z^2+z+1$ consist of the roots of $z^6 = 1$ excluding the root $z = 1$, which leaves $5$ roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving differential equation for an expanding bubble I need to solve the equation
\begin{eqnarray}
R^3 \frac{d } {dt} \left [ \frac{4}{3} \rho_{\rm ext} \left ( \frac{dR}{dt} \right )^2 \right ]+ 4 p R^2 \frac{d R} {dt} =\frac{F_E}{4\pi}
\end{eqnarray}
Could you please help in this regard?
| Let $X=\dfrac{dR}{dt}$ ,
Then $R^3\dfrac{d}{dt}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]+4pR^2X=\dfrac{F_E}{4\pi}$
$R^3\dfrac{d}{dR}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]\dfrac{dR}{dt}+4pR^2X=\dfrac{F_E}{4\pi}$
$\dfrac{8\rho_{\rm ext}R^3X^2}{3}\dfrac{dX}{dR}=\dfrac{F_E}{4\pi}-4pR^2X$
Let $Y=\dfrac{1}{R^2}$ ,
Then $\dfrac{dX}{dR}=\dfrac{dX}{dY}\dfrac{dY}{dR}=-\dfrac{2}{R^3}\dfrac{dX}{dY}$
$\therefore-\dfrac{16\rho_{\rm ext}X^2}{3}\dfrac{dX}{dY}=\dfrac{F_E}{4\pi}-\dfrac{4pX}{Y}$
$-\dfrac{64\pi\rho_{\rm ext}X^2}{3}\dfrac{dX}{dY}=\dfrac{F_EY-16\pi pX}{Y}$
$(F_EY-16\pi pX)\dfrac{dY}{dX}=-\dfrac{64\pi\rho_{\rm ext}X^2Y}{3}$
This belongs to an Abel equation of the second kind.
Let $U=Y-\dfrac{16\pi pX}{F_E}$ ,
Then $Y=U+\dfrac{16\pi pX}{F_E}$
$\dfrac{dY}{dX}=\dfrac{dU}{dX}+\dfrac{16\pi p}{F_E}$
$\therefore F_EU\left(\dfrac{dU}{dX}+\dfrac{16\pi p}{F_E}\right)=-\dfrac{64\pi\rho_{\rm ext}X^2}{3}\left(U+\dfrac{16\pi pX}{F_E}\right)$
$F_EU\dfrac{dU}{dX}+16\pi pU=-\dfrac{64\pi\rho_{\rm ext}X^2U}{3}-\dfrac{1024\pi^2p\rho_{\rm ext}X^3}{3F_E}$
$F_EU\dfrac{dU}{dX}=-\left(\dfrac{64\pi\rho_{\rm ext}X^2}{3}+16\pi p\right)U-\dfrac{1024\pi^2p\rho_{\rm ext}X^3}{3F_E}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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How may I prove this inequality? Let $a, b, c$ be positive real, $abc = 1$. Prove that:
$$\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} \le \frac{1}{2+a} + \frac{1}{2+b}+\frac{1}{2+c}$$
I thought of Cauchy and AM-GM, but I don't see how to successfully use them to prove the inequality. Any hint, suggestion will be welcome. Thanks.
| $\frac{2}{1+a+b}-(\frac{1}{2+a}+\frac{1}{2+b})$
$=\frac{1}{1+a+b}-\frac{1}{2+a}+\frac{1}{1+a+b}-\frac{1}{2+b}$
$=\frac{1}{1+a+b}(\frac{1-b}{2+a}+\frac{1-a}{2+b})$
$=\frac{1}{(1+a+b)(2+a)(2+b)}((1-b)(2+b)+(1-a)(2+a))$
$≤\frac{1}{1\cdot 2\cdot 2}(4-(a+b+a^2+b^2))$ as $a,b>0,2+a>2$ and $a+b+1>1$
$\sum(\frac{2}{1+a+b}-(\frac{1}{2+a}+\frac{1}{2+b}))≤\frac{1}{4}(3\cdot 4-2(a+b+c)-2(a^2+b^2+c^2))≤0$ as $a^n+b^n+c^n≥3(abc)^{\frac{n}{3}}=3$ for any positive number $n$.
$\implies \sum2(\frac{1}{1+a+b})≤ \sum2(\frac{1}{2+a})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Sum of the sequence What is the sum of the following sequence
$$\begin{align*}
(2^1 - 1) &+ \Big((2^1 - 1) + (2^2 - 1)\Big)\\
&+ \Big((2^1 - 1) + (2^2 - 1) + (2^3 - 1) \Big)+\ldots\\
&+\Big( (2^1 - 1)+(2^2 - 1)+(2^3 - 1)+\ldots+(2^n - 1)\Big)
\end{align*}$$
I tried to solve this. I reduced the equation into the following equation
$$n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots$$
but im not able to solve it further. Can any one help me solve this equation out. and btw its not a Home work problem. This equation is derived from some puzzle.
Thanks in advance
| Others have given the correct answer; here’s how you could have simplified your incorrect expression.
$$\begin{align*}
n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots&=\sum_{k=1}^n(n-k+1)2^k\\
&=(n+1)\sum_{k=1}^n2^k-\sum_{k=1}^nk2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{k=1}^n\sum_{i=1}^k2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{i=1}^n\sum_{k=i}^n2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{i=1}^n\left(2^{n+1}-2^i\right)\\
&=(n+1)\left(2^{n+1}-2\right)-n2^{n+1}+\sum_{i=1}^n2^i\\
&=(n+1)\left(2^{n+1}-2\right)-n2^{n+1}+2^{n+1}-2\\
&=2\cdot2^{n+1}-2n-4\\
&=2^{n+2}-2n-4
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| $$x^3=-1$$
$$x^3+1=0$$
$$(x+1)(x^2+1-x)=0$$
$$x=-1 \quad\text{or}\quad x^2-x+1=0$$
When $$x^2-x+1=0, x= \frac{-1(\pm\sqrt{-3})}{2}$$
$$x= \frac{-1+\sqrt{3}i}{2} \quad\text{and}\quad \frac{-1-\sqrt{3}i}{2}$$
Which is equal to $e^{−i2π/3}$ and $e^{i2π/3}$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 3
} |
Solve sum for theta Is there any way to solve the following sum of trigonometric functions for theta without using a solver?
$$25\sin(\theta)-1.5\cos(\theta)=20$$
| $$
25\sin\theta-1.5\cos\theta = \sqrt{25^2+1.5^2}\left( \frac{25}{\sqrt{25^2+1.5^2}}\sin\theta - \frac{1.5}{\sqrt{25^2+1.5^2}}\cos\theta \right)
$$
$$
= \sqrt{25^2+1.5^2}(\cos\varphi\sin\theta-\sin\varphi\cos\theta) = \sqrt{25^2+1.5^2} \sin(\varphi-\theta).
$$
So you want
$$
\sin(\varphi-\theta)=\frac{20}{\sqrt{25^2+1.5^2}}.
$$
Take arcsines.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability of throwing multiple dice of at least a given face with a set of dice I know how to calculate the probability of throwing at least one die of a given face with a set of dice, but can someone tell me how to calculate more than one (e.g., at least two)?
For example, I know that the probability of throwing at least one 4 with two 6-sided dice is 27/216, or 1 - (3/6 x 3/6 x 3/6). How do I calculate throwing at least two 4s with four 6-sided dice?
| The probability of no 4 with four 6-sided dice$(p_1)=(\frac{5}{6})^4$
The probability of exactly one 4 with four 6-sided dice$(p_2)$
$=4\frac{1}{6}(\frac{5}{6})^3$ as here the combinations are $4XXX$ or $X4XX$ or $XX4X$ or $XXX4$ where $X$ is some other face$≠4$
So, the probability of at least two 4s with four 6-sided dice$=1-p_1-p_2$
$=1-((\frac{5}{6})^4+4\frac{1}{6}(\frac{5}{6})^3)$
$=1-(\frac{5}{6})^3(\frac{5}{6}+\frac{4}{6})=1-\frac{125}{144}=\frac{19}{144}$
The probability of throwing at least 4 by one 6-sided dice
$=\frac{3}{6}=\frac{1}{2}$
The possible combinations are $XXYY$, $XYXY$, $XYYX$, $YXXY$, $YXYX$, $YYXX$
where $1≤Y≤3,4≤X≤6$
So, the required probability of throwing exactly two occurrences of at least 4 is $^4C_2\frac{1}{2}\frac{1}{2}(1-\frac{1}{2})(1-\frac{1}{2})=\frac{3}{8}$ using Binomial Distribution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| This inequality can also be rewritten as
$$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \,,$$
which is just the AM-HM inequality.
A more direct proof would be to simply multiply:
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+ (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{c}{b}+\frac{b}{c}) \geq 3+2+2+2 =9$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
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Inequality. $\sum{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$ Let $a,b,c$ be the side-lengths of a triangle. Prove that:
I.
$$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
What I have tried:
\begin{eqnarray}
a-b+c&=&x\\
b-c+a&=&y\\
c-a+b&=&z.
\end{eqnarray}
So $a+b+c=x+y+z$ and $2a=x+y$, $2b=y+z$, $2c=x+z$ and our inequality become:
$$\sum_{cyc}{\frac{\sqrt{x}\cdot(x+2y+z)\cdot(x+y+2)}{4}} \geq 4\cdot(x+y+z)\cdot\sqrt{xyz}. $$ Or if we make one more notation : $S=x+y+z$ we obtain that:
$$\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} \Leftrightarrow$$
$$S^2(\sqrt{x}+\sqrt{y}+\sqrt{z})+S(y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z})+xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y} \geq 16S\sqrt{xyz}.$$ To complete the proof we have to prove that:
$$y\sqrt{x}+z\sqrt{x}+x\sqrt{y}+z\sqrt{y}+x\sqrt{z}+y\sqrt{z} \geq 16\sqrt{xyz}. $$
Is this last inequality true ?
II.
Knowing that: $$p=\frac{a+b+c}{2}$$ we can rewrite the inequality:
$$\sum_{cyc}{(2p-c)(2p-a)\sqrt{2(p-b)}} \geq 8p \sqrt{2^3 \cdot (p-a)(p-b)(p-c)} \Leftrightarrow$$
$$\sum_{cyc}{(2p-c)(2p-a)\sqrt{(p-b)}} \geq 16p \sqrt{(p-a)(p-b)(p-c)}$$
This help me ?
Thanks :)
| I couldn't solve this with elementary means.
Take $\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} $ and note that it is enough to prove it for $S=1$ because it is homogeneous. Substituting $x\to x^2$, etc. consider the function $f(x,y,z)=(x^2+1) (y^2+1) z+(x^2+1) y (z^2+1)+x (y^2+1) (z^2+1)-16 x y z$. It is enough to prove it is non-negative if $\sum_{cyc}{x^2}=1$.
Let's denote $p[i]=x^i+y^i+z^i$.
Now notice that the following decomposition is valid: $$12f(x,y,z)=12 p[1] - 32 p[1]^3 + p[1]^5 + 108 p[1] p[2] - 4 p[1]^3 p[2] + 3 p[1] p[2]^2 - 76 p[3] + 2 p[1]^2 p[3] - 2 p[2] p[3]=$$
Using our assumption $p[2]=1$ this simplifies to:$f(x,y,z)=123 p[1] - 36 p[1]^3 + p[1]^5 - 78 p[3] + 2 p[1]^2 p[3]$.
Lets put $x=p[1]$ and $a=p[3]$.
Because $p[2]=1$ we ca deduce the ranges for $a$ and $x$ as $1 \ge p[3] \ge 1/\sqrt{3}$ and $ \sqrt{3}\ge p[1] \ge 1$. Also we need to keep in mind that from the power means inequality $p[1]/3 \le \sqrt[3]{p[3]/3}$.
We need to prove that the function:$g(x)=123 x - 36 x^3 + x^5 - 78 a + 2 a x^2 $ is non-negative in our domain. It turns out that this problem is numerically tractable, and this function is decreasing within our domain and becomes zero only when $a=\sqrt{3}/3$, and $x=\sqrt{3}$. The calculations however are quite tedious.
I am sure that a better way to solve this exists.
| {
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"url": "https://math.stackexchange.com/questions/195185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$ using AM-GM I need to proof this inequality by AM-GM method.
Any ideas how to do it?
$$\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$$
| We may assume without loss of generality that $abc = k^3$, which enables us to make the substitution $\displaystyle a = \frac{kq}{p}, b = \frac{kr}{q}, c = \frac{kp}{r}$.
Now, $\displaystyle a(1+b) = \frac{kq}{p} \left(1+ \frac{kr}{q} \right) = \frac{k(q+kr)}{p}$.
Thus, the inequality reduces to proving (after cancelling $k$ from the denominator on both sides) $$\frac{p}{q+kr} +\frac{q}{r+kp} +\frac{r}{p+kq} \ge \frac{3}{1+k} $$
By Cauchy Schwarz, we get $$\left(\frac{p}{q+kr} +\frac{q}{r+kp} +\frac{r}{p+kq} \right) ( p(q+kr) + q(r+kp) + r(p+kq)) \ge (p+q+r)^2$$
Thus, we have that $$\frac{p}{q+kr} +\frac{q}{r+kp} +\frac{r}{p+kq} \ge \frac{(p+q+r)^2}{(1+k)(pq+qr+rp)} \ge \frac 3{1+k}$$ due to the well known $(p+q+r)^2 \ge 3(pq+qr+rp)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Generating integral triangles with two equal sides How can I generate all triangles which have integral sides and area, and exactly two of its three sides are equal?
For example, a triangle with sides ${5,5,6}$ satisfies these terms.
| Heron's formula says the area of a triangle with sides $a, b, c$ is
$$Area = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semiperimeter, $s = \frac{a+b+c}{2}$.
Now, your assumption is two sides are equal, so $a, b, b$. The area is now
$$Area = \sqrt{s(s-a)(s-b)^2} = (s-b)\sqrt{s(s-a)}$$
Now $s-b$ might not be an integer, if $s$ is not, but it will at worst be a fraction of the form $\frac{z}{2}$. I will deal more with this later.
For now, we want $\sqrt{s(s-a)}$ to be an integer, so $s(s-a)$ must be a perfect square. Now, $s = \frac{a + 2b}{2} = \frac{a}{2} + b$ so this simplifies to wanting
$$s(s-a) = (\frac{a}{2} + b)(\frac{a}{2} + b - a) = (b + \frac{a}{2})(b - \frac{a}{2}) = b^2 - \frac{a^2}{4}$$
a perfect square, so let's say it equals $c^2$. Thus, we want all integer solutions to
$$c^2 + \left(\frac{a}{2}\right)^2 = b^2$$
If we happen to get a solution such that $(s-b)\sqrt{s(s-a)}$ is not an integer, then multiply all side lengths by 2 to get a similar triangle with $s$ and thus $s-b$ an integer.
Other than that, the problem is reduced to finding all solutions to this equation, which is a well known problem with a well known solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve a simultaneous equation. How do we solve $|b-y|=b+y-2\;and\;|b+y|=b+2$? I have tried to square them and factorize them but got confused by and and or conditions.
| $b+2=|b+y|$ which is real, so is $b$
$y+b-2=|b-y|$ which is real, so is $y+b-2$ and $y$
(1)If $b \ge y, b-y=b+y-2\implies y=1 \implies |b+1|=b+2$ and $b \ge y=1$
So, $b+1 >0\implies |b+1|=b+1=b+2$ which has no finite solution.
(2) If $b<y, y-b=b+y-2\implies b=1, y>b=1$
So, $|1+y|=3\implies y+1=3\implies y=2$
The only solution $b=1,y=2$.
| {
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"url": "https://math.stackexchange.com/questions/202791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| $\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3} = \sqrt 6$ ?
$1 + i \sqrt 3 = 2 \exp \left( \dfrac {\pi}{3}i + 2 \pi n i \right)
\quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} = \sqrt 2 \exp \left( \dfrac {\pi}{6}i + \pi n i \right)
\quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} =
\pm \left( \dfrac{\sqrt 6}{2} + \dfrac{\sqrt 2}{2} i \right)$
Similarly
$\sqrt{1 - i \sqrt 3} =
\pm \left( \dfrac{\sqrt 6}{2} - \dfrac{\sqrt 2}{2} i \right)$
So there are four possible values of $\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3}$
One of then is $\sqrt 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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} |
difficulty working out equation $\int_{-a}^{a}2(\sqrt{a^2-x^2})dx $ =$\left.2[\frac{1}{2}\sqrt{a^2-x^2}+a^2\arcsin(\frac{x}{a})]\right|_{-a}^{a}$ I encountered the below formula in my text.
I know the author is using integration by substitution and double angle formula
but for some reason, every working on paper that i did is different from the answer below:
$\int_{-a}^{a}2(\sqrt{a^2-x^2})dx $
=$\left.2[\frac{1}{2}(\sqrt{a^2-x^2}+a^2\arcsin(\frac{x}{a}))]\right|_{-a}^{a}$
Can anyone help me to expand the above equation.
I can't wrap my head on how above answer is reproduced.
Thanks.
| This can be done with a standard $\sin$ substitution: $(1)\ x=a\sin(\theta)$ followed by a change of variables: $(2)\ \phi=2\theta$.
$$
\begin{align}
\int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\int_{-\pi/2}^{\pi/2}2a\cos(\theta)\,\mathrm{d}a\sin(\theta)\tag{1}\\
&=2a^2\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\theta\\
&=2a^2\int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,\mathrm{d}\theta\\
&=2a^2\int_{-\pi}^\pi\frac{1+\cos(\phi)}{4}\,\mathrm{d}\phi\tag{2}\\
&=2a^2\left[\frac{\phi+\sin(\phi)}{4}\right]_{\phi=-\pi}^{\phi=\pi}\\
&=\pi a^2
\end{align}
$$
However, the answer you show in the question looks like the answer that comes from an integration by parts: $u=\sqrt{a^2-x^2}$ and $\mathrm{d}v=\mathrm{d}x$ so that $v=x$ and $\mathrm{d}u=-\frac{x\,\mathrm{d}x}{\sqrt{a^2-x^2}}$
$$
\begin{align}
\int2\sqrt{a^2-x^2}\,\mathrm{d}x
&=2x\sqrt{a^2-x^2}+\int\frac{2x^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\
&=2x\sqrt{a^2-x^2}-\color{#C00000}{\int\frac{2(a^2-x^2)}{\sqrt{a^2-x^2}}\,\mathrm{d}x}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{3}
\end{align}
$$
Adding the integral in red to both sides of $(3)$ and dividing by $2$ yields
$$
\begin{align}
\int2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\frac12\left[2x\sqrt{a^2-x^2}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\right]\\
&=\frac12\left[2x\sqrt{a^2-x^2}+2a^2\,\sin^{-1}(x/a)\right]+C\tag{4}
\end{align}
$$
Evaluating $(4)$ at the limits of integration yields
$$
\begin{align}
\int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\left[x\sqrt{a^2-x^2}+a^2\,\sin^{-1}(x/a)\right]_{-a}^a\\
&=\pi a^2
\end{align}
$$
Not quite what you got, but of the same form.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Unsure about applying series comparison test Does this converge or diverge?
$$
\sum\limits_{n=1}^\infty (a_{n} = \frac{1}{2\sqrt{n} + \sqrt[3]{n}})
$$
The answer is: diverges by limit comparison to $\sum (b_{n} = \frac{1}{\sqrt{n}})$
If I look at $\lim_{n \to \infty}\frac{a_{n}}{b_{n}}$ I get
$$
\frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{n^\frac{1}{2}} {2n^{\frac{1}{2}} + n^{\frac{1}{2}}n^\frac{2}{3}} = \frac{1}{2 + n^\frac{2}{3}} = \frac{0}{0 + 1} = 0
$$ as $n\to\infty$
I would expect it to be some $c > 0$ or $\infty$
Also, because $a_{n} < b_{n}$ I don't think I can use the comparison test. I'm pretty sure I am missing something but not sure what.
Thanks
Update:
My algebra was wrong, how is this for finding the limit?
$$
\frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{1} {2 + \frac{1}{n^\frac{1}{6}}} = \frac{1} {\frac{2\sqrt[6]{n} + 1}{\sqrt[6]{n}}} = \frac{\sqrt[6]{n}}{2\sqrt[6]{n} + 1} = \frac{1}{2 + 0} = \frac{1}{2}
$$ as $n \to\infty$
And since $\frac{1}{2} > 0$ and $b_{n}$ diverges, $a_{n}$ diverges
| There seems to be a mistake in the algebra -- namely, you took $\sqrt[3]{n} = n^{1/2} n^{2/3}$. But the right-hand side of this is equal to $n^{1/2 + 2/3} = n^{3/6 + 4/6} = n^{7/6} \ne \sqrt[3]{n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.
| You can notice that many terms of
$$(b+5a-4)^2=b^2+10ab-8b-40a+25a^2+16$$
appear in the first equation. Similarly, in the first one, you can notice $(b+2a)^2$.
By algebraic manipulation you get that the original equations are equivalent to
$$
\begin{align}
(b+5a-4)^2&=4(a^2+1)\\
(b+2a)^2&=9(a^2+1)
\end{align}
$$
which implies $4(b+2a)^2=9(b+5a-4)^2$ and $2(b+2a)=\pm 3(b+5a-4)$. This should simplify things a little.
(In each of the two possibilities you can express $b$ using $a$ as a linear expression.
Then you will get a quadratic equation in $a$. Or you can start by eliminating $a$.)
| {
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"url": "https://math.stackexchange.com/questions/210454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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How to find the eigenvalues and eigenvector without computation? The given matrix is
$$
\begin{pmatrix}
2 & 2 & 2 \\
2 & 2 & 2 \\
2 & 2 & 2 \\
\end{pmatrix}
$$
so, how could i find the eigenvalues and eigenvector without computation?
Thank you
| Try to solve the following matrix equation, taking into account that your matrix is singular:
$$\begin{pmatrix}2&2&2\\2&2&2\\2&2&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2(x+y+z)\\2(x+y+z)\\2(x+y+z)\end{pmatrix}=2\begin{pmatrix}x+y+z\\x+y+z\\x+y+z\end{pmatrix}=\lambda\begin{pmatrix}x\\y\\z\end{pmatrix}$$
| {
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"url": "https://math.stackexchange.com/questions/211865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?
And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$
| This is little known, but you can solve the equation without much trigonometry.
WLOG, we can assume that $a^2+b^2=1$ (the coefficients can be normalized). Write
$$a\,S+b\,C=c,\\b\,C=c-aS,\\b^2(1-S^2)=(c-a\,S)^2,\\S^2-2ac\,S+c^2-b^2=0.$$
The solution of the quadratic equation is
$$S=ac\pm bd$$ where $d=\sqrt{1-c^2}$.
By symmetry,
$$C=bc\mp ad.$$
If you have enough with the values of the sine and the cosine, you can stop here. Otherwise
$$\theta=\arctan\frac SC.$$
For unnormalized $a,b$, the solution is
$$S=\frac{ac\pm bd}{a^2+b^2},\\
C=\frac{bc\mp ad}{a^2+b^2}$$ and
$$\color{green}{\theta=\arctan\frac{ac\pm bd}{bc\mp ad}}$$
where $d=\sqrt{a^2+b^2-c^2}$. It does not exist when $a^2+b^2<c^2$.
From the above, one can observe that the solution is also given by
$$\theta=\arctan\frac ab\pm\arctan\frac dc$$ but this takes two (costly) arc tangents instead of one.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$ prove the $$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$ for all n greater or equal to 2.
$\pi$ should be a big pi from $i=2$ to $n$ for $(1-1/i^2)$. I'm really confused about the $\prod$ function.
UPDATE:
Suppose$$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$
then $$\prod_{i=2}^{n+1} (1-1/i^2) = {n+2\over 2(n+1)}$$
so $$\prod_{i=2}^{n+1} (1-1/i^2) \times \left(1-{1\over (n+1)} \right)$$
then $${n+1\over 2n} \left(1-{1\over (n+1)} \right)$$
but this equals $1/2$?
| $$\displaystyle \prod_{k=2}^n f(k)$$ is a short hand notation for $$f(2) \times f(3) \times f(4) \times \cdots \times f(n-1) \times f(n)$$
For instance, $$\displaystyle \prod_{k=2}^{4} \left(1 - \dfrac1{k^2} \right) = \left(1 - \dfrac1{2^2} \right) \times \left(1 - \dfrac1{3^2} \right) \times \left(1 - \dfrac1{4^2} \right)$$
HINT
To solve the problem, note that $$\left(1 - \dfrac1{k^2} \right) = \dfrac{k-1}{k} \times \dfrac{k+1}k$$ Now write out the first few terms and see the cancellation.
| {
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How to get rid of a cubic root in a logarithmic limit? The formula:
$$\lim_{n \to \infty}\dfrac{\log (1 - n + n^2)}{\log (1 + n + n^{10})^{1/3}}.$$
Thanks for any advice!
| Hint:
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( 1-n+n^{2}\right) }{\frac{1}{3}
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right) } \\
&=&\frac{\log n^{2}+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\log
n^{10}+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) }.
\end{eqnarray*}
ADDED. Now that there is already a full answer I complete mine. Using the
rule $\log a^{r}=r\log a$, as commented by David Mitra, and $\log ab=\log
a+\log b$, manipulate algebrically the fraction and rewrite it as
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}%
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}%
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right) } \\
&=&\frac{\log n^{2}+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\log
n^{10}+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) } \\
&=&\frac{2\log n+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\times 10\log
n+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) } \\
&=&\frac{2+\frac{\log \left( 1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+%
\frac{1}{3}\frac{\log \left( 1/n^{10}+1/n^{9}+1\right) }{\log n}}.
\end{eqnarray*}
We thus have
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\frac{\log \left( 1-n+n^{2}\right) }{\log \left(
1+n+n^{10}\right) ^{1/3}} &=&\lim_{n\rightarrow \infty }\frac{2+\frac{\log
\left( 1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+\frac{1}{3}\frac{\log
\left( 1/n^{10}+1/n^{9}+1\right) }{\log n}} \\
&=&\frac{2+\displaystyle\lim_{n\rightarrow \infty }\frac{\log \left(
1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+\frac{1}{3}\displaystyle%
\lim_{n\rightarrow \infty }\frac{\log \left( 1/n^{10}+1/n^{9}+1\right) }{
\log n}} \\
&=&\frac{2+\frac{\log 1}{\displaystyle\lim_{n\rightarrow \infty }\log n}}{%
\frac{10}{3}+\frac{1}{3}\frac{\log 1}{\displaystyle\lim_{n\rightarrow \infty
}\log n}} \\
&=&\frac{2+0}{\frac{10}{3}+\frac{1}{3}\times 0}=\frac{3}{5}.
\end{eqnarray*}
| {
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Continued Fractions Approximation I have come across continued fractions approximation but I am unsure what the steps are.
For example How would you express the following rational function in continued-fraction form:
$${x^2+3x+2 \over x^2-x+1}$$
| Start out writing
$$\frac{x^2+3x+2}{x^2-x+1}=\frac{1}{\frac{(x^2+3x+2)-(4x+1)}{x^2+3x+2}}=\frac{1}{1-\frac{4x+1}{x^2+3x+2}}$$
and then iterate doing the same with the fraction in the denominator.
EDIT: Complete solution:
By polynomial long division we have $\frac{x^2+3x+2}{4x+1}=\frac{x}{4}+\frac{11}{16}+\frac{21}{16}\frac{1}{4x+1}$. Hence the above expression equals
$$\huge
\frac{1}{1-\frac{1}{x/4+11/16+\frac{1}{40x/21+10/21}}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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