$\pi^2=(1\ 2\  3\  4\  5),\pi \in S_{5} \\$  where $\pi$ is the cycle. I have calculated the following: $(1 2 3 4 5)(1 2 3 4 5)^2 = (1 3 5 2 4)(1 3 5 2 4),\\(1 2 3 4 5)(1 2 3 4 5)^3 = (1 4 2 5 3)(1 4 2 5 3)\\(1 2 3 4 5)(1 2 3 4 5)^4 = (1 5 4 3 2)(1 5 4 3 2)$ and after $(1 2 3 4 5)(1 2 3 4 5)^{10}$ I get the identical function. Well $S_{5}$ has $5!$ elements, how can I finish this exercise?
120648
$\pi^2=(1\ 2\  3\  4\  5),\pi \in S_{5} \\  \\ \pi^2=(1\ 2\  3\  4\  5\ 6),\pi \in S_{10} $  These are two different equations.Where $\pi$ is the cycle of the permutation, I know that the set $S_{5}$ has $5!$ elements and $\in S_{10}$ has $10!$ elements.
87989
Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$. Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$. Derive a formula for $F$. Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its inverse.  If we express $S$ in polar coordinates: $$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$ then, from simple trigonometric relations, the inverse function $\Phi$ is: $$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$ and its inverse $F$ is: $$F(t)=\frac\pi{2}-2\arctan t$$ After changing coordinates back to cartesian, $F$ will look like: $$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$ which can be further simplified to: $$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$ Is this the formula of $F$? Is it correct? And I know that to show that $\phi$ : $\mathbb R$ →  $S$ \ {$p$} is a bijective map, I have to show that $\phi$ is injective and surjective. But how do I do that? Can someone please help me show that $\phi$ is bijective? and how can I derive a formula for its inverse?
107592
Benoit Cloitre offered two 'mirror sequences', which allow to compute $\pi$ and $e$ in similar ways:  $$u_{n+2}=u_{n+1}+\frac{u_n}{n}$$  $$v_{n+2}=\frac{v_{n+1}}{n}+v_{n}$$    $$u_1=v_1=0$$  $$u_2=v_2=1$$    $$\lim_{n \to \infty} \frac{n}{u_n}=e$$  $$\lim_{n \to \infty} \frac{2n}{v_n^2}=\pi$$    The formulation and the proof can be seen .     What do you think - is it just a coincidence or is there some deeper meaning in this mirror algorithm about the connection of two constants?     By @EricStucky in the comment, the better question:     Is there any connection between $e$ and $π$ which is essentially different than Euler's formula?   Of course, I expect an answer related to my own question about this 'mirror sequence'  If, on the other hand, someone shows a clear relation between this sequence and Euler's formula, that's fine too
67116
I want to solve the following:     Let $p$ be a prime number and let $a$ be a p-cycle in $S_p$, and let $b$ be a transposition in $S_p$. Show $S_p$ is generated by $a$ and $b$.   My attempt  Write $a=(y_1 \space y_2 \space ...\space y_p)$ and $b=(z_1 \space z_2)$. WLOG, $y_1=z_1=1$. WLOG, we can also assume $a=(1 \space 2 \space ... \space p).$   Let $σ\in S_p$. It should be obvious that $σ$ is a product of transpositions, so to show $σ \in S_p$, it suffices to show every transposition can be written in terms of $(1 \space 2 \space ... \space p)$ and $(1 \space z_2)$. But how do I show this?  Duplicate? I think not.  Although another question (already answered on math.SE) is similar to mine, I do not believe that mine is a duplicate. The other question is equivalent to mine only for the special case when $z_2=2$, but not for general $z_2$.
35278
I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!
1302
What is wrong with this proof?    Is $\pi=4?$
102064
I have three functions:      $$f(x) = x+1 ,\; g(x) = x - 1 ,\; h(x) = 2x$$   I want to find $g\circ f$ such that $g(f(x))$, which equals $(x+1)-1=x$ but how?   I don't understand the steps.  Also, I want to find $h\circ f$ which equals $h(f(x)) = 2x-1$ but I dont know how to get that answer either.  Thanks
54137
I have to find an easy ring which this ring is isomorphic with.  I can prove that $5 = (2+i)(2-i) \in (2+i)$ but this doesn't help me further.  My book says the answer is $\mathbb {Z_5} $
63937
Let G be a topological group and let $s_1$ and $s_2$ be loops in G (both loops are based at the identity e of G). Is it true that the loop $s_1s_2$ (where the multiplication is the one of the group structure of G) is equal, in $\pi_1(G,e)$, to the loop $s_1*s_2$ where this product is given by first going around $s_1$ and then $s_2$ (i.e., do we have $[s_1s_2] = [s_1*s_2]$)? If yes, what is the proof for this?
10800
There is a 'nice' relationship between the integers (1,5) and (2,3) as  $$1+5=2 \times 3;$$ $$1\times 5 = 2 + 3.$$  So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$  $$a\times b = c + d.$$  To find this, $a, b, c, d$ must satisfy $$(a+1)(b+1)=(c+1)(d+1).$$ However, this condition is only necessary but not sufficient. Any idea?
48862
If $p$ is a prime and $(a,p^2)=p, (b,p^3)=p^2$,calculate $(a+b,p^4)$ $$$$ That's what I thought: $$$$ $$(a,p^2)=p \Rightarrow a=kp$$ $$(b,p^3)=p^2 \Rightarrow b=l p^2$$  So,$$(a+b,p)=(p+p^2)$$  Is it right or am I wrong?
119233
I have already tried with $S_3$, and indeed, the product is $(13)$, and $(13)^2=e$    But what about this: I define + in this way:$45=2$,$26=3$, $1$ is the identity. therefore, $123456=2326=233=3$, and as you can see from the picture, the inverse of $3$ is not itself, thus contradicts the statement.  My question: What is wrong with my argument?  And How do you prove the statement? it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied.
34735
The problem reads as follows:     Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of    $$\cos(36^\circ)$$   it says that you may find useful the following identities: $$\cos^2 t+\sin^2 t = 1,\\ \sin 2t = 2\sin t\cos t,\\ \sin 3t = 3\sin t - 4\sin^3 t. $$  Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?  Thanks in advance :)
34607
How is $3\pi-10$ element of $[-\pi/2,\pi/2]$? According to me this means that $530$ is an element of $[-90,90]$. i.e. Range of $\arcsin$ function.  (Note:- I was trying to find the value of $\arcsin(\sin 10))$)
85502
[question:]      Prove by induction that the i th Fibonacci number satisfies the equality $$F_i=\frac{\phi^i-\hat{\phi^i}}{\sqrt5}$$where $\phi$ is the golden ratio and $\hat{\phi}$ is its conjugate. [end]  I've had multiple attempts at this, the most fruitful being what follows, though it is incorrect, and I cannot figure out where I am going wrong:  [my answer:] What follows is an incorrect approach, after which comes the approach that I should have done  Inductive Hypothesis: $$F_i=\frac{\phi^i-\hat{\phi^i}}{\sqrt5}, i\in\mathbb{N}  $$   (I include 0) Base case: $$F_0=\frac{\phi^0-\hat{\phi^0}}{\sqrt5}=0$$  Proof: $$\begin{eqnarray*} F_{i+1}&amp;=&amp;\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ &amp;=&amp;\frac{\phi*\phi^i-\hat{\phi}*\hat{\phi^i}}{\sqrt5}\\ &amp;=&amp;\frac{\frac{1+\sqrt5}{2}*\phi^i-\frac{1-\sqrt5}{2}*\hat{\phi^i}}{\sqrt5}\\ &amp;=&amp;\frac{\frac{\phi^i+\sqrt5\phi^i}{2}-\frac{\hat{\phi^i}-\sqrt5\hat{\phi^i}}{2}}{\sqrt5}\\ &amp;=&amp;\frac{\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{2}}{\sqrt5}\\ &amp;=&amp;\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{2*\sqrt5}\\ &amp;=&amp;\frac{1}{2}\left(\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{\sqrt5}\right)\\ &amp;=&amp;\frac{1}{2}\left(\frac{\sqrt5\phi^i}{\sqrt5} + \frac{\sqrt5\hat{\phi^i}}{\sqrt5} +\frac{\phi^i-\hat{\phi^i}}{\sqrt5}\right)\\ &amp;=&amp;\frac{1}{2}\left(\phi^i+\hat{\phi^i}+F_{i}\right)\text{   by inductive hypothesis}\\ &amp;=&amp;\frac{1}{2}\left(\sqrt5*...\right)\\ \end{eqnarray*}$$ Actually I just saw my error in that line (I eventually multiplied by $\frac{\sqrt5}{\sqrt5}$ and  substituted for $F_i$, but I see that the conjugates are added there, not subtracted).  Correct approach: $$\begin{eqnarray*} F_{i+1}&amp;=&amp;F_{i} + F_{i-1}\\ &amp;=&amp;\frac{\phi^i-\hat{\phi^i}}{\sqrt5}+\frac{\phi^{i-1}-\hat{\phi^{i-1}}}{\sqrt5}\\ &amp;=&amp;\frac{\left(\phi+\hat{\phi}\right)\left(\phi^i-\hat{\phi^i}\right)-\phi\hat{\phi}\left(\phi^{i-1}-\hat{\phi^{i-1}}\right)}{\sqrt5}\text{ (see answer for why this works)}\\ &amp;=&amp;\frac{\phi^{i+1}-\phi\hat{\phi^i}+\hat{\phi}\phi^i-\hat{\phi^{i+1}}-\phi^i\hat{\phi}+\phi\hat{\phi^i}}{\sqrt5}\\ &amp;=&amp;\frac{\phi^{i+1}-\hat{\phi^{i+1}}-\phi\hat{\phi^i}+\phi\hat{\phi^i}-\hat{\phi}\phi^i+\hat{\phi}\phi^i}{\sqrt5}\\ &amp;=&amp;\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ \text{Q.E.D., punk problem.} \end{eqnarray*}$$  From what I searched on the web, there is no available source proving this of the Fibonacci sequence - all inductive hypothesis proofs prove the actual sequence with induction.   Bill, it's always easy to forget about the inherent properties of special numbers... I didn't even realize $\phi$ + $\hat{\phi}$ = 1 in your first hint, as well as the multiplication, until you pointed it out. Alex, I'm sure that going through the problem that way will produce the result, also, but with Bill's, it's much faster.    Thanks for the help.
119410
So I am asked to find all the isomorphisms between $G = (\mathbb Z_4,+)$ and $H = (U_5,*)$.  I solved it as follows: we will have two isomorphism corresponding to the two generators of $U_5$.  The first isomorphism : $\phi([x]_4)$ = $[2]_5^x$   The second isomorphism is: $\phi([x]_4)$ = $[3]_5^x$   Notice that [2] and [3] are the only generators of $U_5$ one can check it systematically or use the fact that $a^m$ is a generator iff gcd(m,4) = 1.  So I just wanted a verification of my work I showed that they are bijective, well defined, and is a homomorphism.
105100
For $n=3$, $\sigma(n)=4$, a perfect square. Calculating further was not yielding positive results.  I was wondering is there a way to find all such an $n$, like some algorithm?  We know that if $n=p_1^{k_1}p_2^{k_2}\dots p_r^{k_r}$ then $\sigma(n)=\frac {p_1^{k_1+1}-1}{p_1-1} \frac{p_2^{k_2+1}-1}{p_2-1}\dots \frac {p_r^{k_r+1}-1}{p_r-1}$. So we have to find $n$ such that $\sigma(n)=\frac {p_1^{k_1+1}-1}{p_1-1} \frac{p_2^{k_2+1}-1}{p_2-1}\dots \frac {p_r^{k_r+1}-1}{p_r-1}=t^2$ for some integer $t$. If each $\frac{p_i^{k_i+1}-1}{p_i-1}$ is a square we will get some such numbers (not all though), but let us consider $\frac{p_i^{k_i+1}-1}{p_i-1}=m^2$.   Working a example, If I take $p_i=17 $ , (or any prime such that $p_i-1$ is a perfect square), then if I can find a $k_i$ such that ${p_i^{k_i+1}-1}=16.m^2$ I will get an $n$, so now when can this happen, is something I have to try out.   But out of curiosity I am asking has this been worked out before. Problem looks broad to me, Is there a solution to this problem?