$$a=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}\\b=\sum_{n=1}^\infty\frac{x^{3n-2}}{(3n-2)!}\\c=\sum_{n=1}^\infty\frac{x^{3n-1}}{(3n-1)!}$$Find $a^3+b^3+c^3-3abc$: $(a)\ 1$ $(b)\ 0$ $(c)-1$ $(d)-2$  Please help me solve this question. I added $a,b$ and $c$. It gives me the expansion of $e^x$. But i dont know how to use it.
How do I prove that $$ a^3+b^3+c^3−3abc = 1 $$ if   $ a = 1+\frac{x^3}{3!}+\frac{x^6}{6!}......$ , $b=x+\frac{x^4}{4!}+\frac{x^7}{7!}....$ , $c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}......$ ?  I have tried the following :-  I discovered that $a+b+c=e^x$ and that $a=\frac{d(b)}{dx}$, $b=\frac{d(c)}{dx}$, $c=\frac{d(a)}{dx}$ . But I am unable to proceed further with these information.
Surface of genus $g$ does not retract to circle (Hatcher exercise) I'm trying exercise 9 on page 53 in  but I need some help with it. The exercise is:  In the surface $M_g$ of genus $g$, let $C$ be a circle that separates $M_g$ into two compact subsurfaces $M_h^\prime $ and $M_k^\prime$ obtained from the closed surfaces $M_h$ and $M_k$ by deleting an open disk from each. Show that $M_h^\prime$ does not retract onto its boundary circle $C$, and hence $M_g$ does not retract onto $C$. [Hint: abelianize $\pi_1$.] But show that $M_g$ does retract onto the nonseparating circle $C^\prime$ in the figure.    My first question is: assume there wasn't the hint, how would I think of abelianising? What does it mean exactly?  I thought I could do this by contradiction: if it retracts the induced map $i_\ast : \pi_1(C) \rightarrow \pi_1(M_h^\prime) $ is injective. I know $\pi_1(C) \cong \mathbb{Z}$, then I computed $\pi_1(M_h^\prime) \cong \mathbb{Z}$ and then I think I'm stuck. Right? Do you agree with $\pi_1(M_h^\prime) \cong \mathbb{Z}$ and being stuck after that?  What do I need to know to make progress? Many thanks for your help! Rescuing/Compiling old TeX file starting with “\input amstex” I have come across an ancient (1995) LaTeX file which was (surely) once compliant, but now does not compile at all. Is there any way to rescue it?  My actual question is much more precise. I want a copy of the follow paper on the arXiv:    Downloading the pdf source produces something which looks like a document on a4 paper printed on US letter (the bottoms of some pages are cut off). However, if you download the TeX source from the arXiv, then it doesn't compile (even adding .tex, naturally). I'm using an up to date version of TeXshop on a mac. How to initialize private static members in C++? What is the best way to initialize a private, static data member in C++?  I tried this in my header file, but it gives me weird linker errors:  class foo {     private:         static int i; };  int foo::i = 0;   I'm guessing this is because I can't initialize a private member from outside the class.  So what's the best way to do this? I followed the official step-by-step "Visual Studio Code and Unity" guide, but it seems I messed up something, because Intellisense doesn't see any Unity related stuff.  Like the MonoBehaviour class and its members (like .transform), or the Unity callbacks, like Start or Update. What is the difference between "Redirection" and "Pipe"? This question may sound a bit stupid, but I can not really see the difference between redirection and pipes.  Redirection is used to redirect the stdout/stdin/stderr, e.g. ls &gt; log.txt.  Pipes are used to give the output of a command as input to another command, e.g. ls | grep file.txt.  But why are there two operators for the same thing?  Why not just write ls &gt; grep to pass the output through, isn't this just a kind of redirection also? What I am missing? Is the unit sphere in $\Bbb R^4$ path connected? I have a computer (Asus T200TA) with 2G RAM and 32-bit UEFI [Yep, what a wonderful idea to put 32-bit UEFI!]. I have two options:   Install a 64-bit system, boot manually and install grub-efi-ia32. Drawback: 2G is not much, systems with such low RAM are said to run better on 32-bit systems. Install a 32-bit version. (but from my state of knowledge it's not trivial to boot)   Which option would be better? If one were to take a similar path as  in shooting candid street photography today, should they be concerned about laws or other realistic norms of today? I am wondering specifically in the US speaking broadly of all states. I am not looking for legal advice, but anything that one photographer to another would caution against as potentially being grey or obviously illegal.  Similar previous question: Is there a difference between them? If so, how and when are they used?  For example:     I fixed a little/small typo. Why does the question's closed status disappear when clicking to load a new edit? Hide some fields in the NewForm.aspx for a custom list Is there a string math evaluator in .NET? If I have a string with a valid math expression such as:  String s = "1 + 2 * 7";   Is there a built in library/function in .NET that will parse and evaluate that expression for me and return the result? In this case 15. How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? How many blocks is my database How can I write a query using Oracle SQL Developer to find how many blocks my database is using? (tables &amp; indexes) What is the difference between a URI, a URL and a URN? People talk about URLs, URIs, and URNs as if they're different things, but they look the same to the naked eye.  What are the distinguishable differences between them? Distribution of orbital velocities in a disk galaxy for N-body simulation? How to remove session entries from LightDM? $ which echo echo: shell built-in command. $ which ls /bin/ls $ which cat /bin/cat   Why is echo not an independent utility like ls, ps, cat etc? Why is it shell specific? Any good reasons? Callback animation to begin only after ALL children have finished animating Did the Nazgûl retain their Rings of Power? In The Fellowship of the Rings, Frodo sees the true appearance of the Nazgûl:      He was able to see beneath their black wrappings. There were five tall figures: two standing on the lip of the dell, three advancing. In their white faces burned keen and merciless eyes; under their mantles were long grey robes; upon their grey hairs were helms of silver; in their haggard hands were swords of steel. Their eyes fell on him and pierced him, as they rushed towards him.   The Fellowship of the Ring - J.R.R. Tolkien - chapter 11, A Knife In the Dark   There's no mention of the Nazgûl wearing the rings of power that Sauron gave them, although their hands are referenced. Did the Nazgûl retain their rings of power once they became wraiths? Or did something else happen to the rings once the Nazgûl emerged?&#185;  &#185; As an FYI, I am currently re-reading The Lord of the Rings trilogy and am just pages from being done with The Fellowship of the Ring. I'm very familiar with the Peter Jackson movies. So that's where I'm coming from as far as canon knowledge goes.