Rewriting some Series / Factorials

[DivGradCurl]

Let

$$u = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + \frac{x^{9}}{9!} + \dotsb$$

$$v = x + \frac{x^{4}}{4!} + \frac{x^{7}}{7!} + \frac{x^{10}}{10!} + \dotsb$$

$$w = \frac{x^{2}}{2!} + \frac{x^{5}}{5!} + \frac{x^{8}}{8!} + \dotsb$$

Show that

$$u^3 + v^3 + w^3 - 3 u v w = 1$$

Well, here what I've done:

$$u = \sum _{n=0} ^{\infty} \frac{x^{3n}}{(3n)!}$$

$$v = \sum _{n=0} ^{\infty} \frac{x^{3n+1}}{(3n+1)!}$$

$$w = \sum _{n=0} ^{\infty} \frac{x^{3n+2}}{(3n+2)!}$$

Further work can be simplified if I rewrite those series in terms of familiar functions.

The first one is not too hard to obtain:

$$u = \sum _{n=0} ^{\infty} \frac{x^{3n}}{(3n)!} = \sum _{n=0} ^{\infty} \frac{x^{3n}}{n!3^n} = \sum _{n=0} ^{\infty} \frac{ \left( \frac{x^3}{3} \right) ^{n}}{n!} = e^{x^3 / 3}$$

Unfortunately, the others got me stuck. I can't find ways to rewrite the following factorials:

$$(3n+1)! \qquad \mbox{ and } \qquad (3n+2)!$$​

Is there any trick?

Thanks

 

[DivGradCurl]

I made a mistake:

$$\prod _{i=0} ^{n} (3i) = n! 3^n \neq (3n)! \Longrightarrow u \neq e^{x^3 / 3}$$

Sorry about that.

 

[arildno]

I am POSITIVE that there is a trick here; most probably, it has to do with forming combinations of your original power series into the familiar power series for sine and cosine.

 

[Hurkyl]

The method is related to sines and cosines, but doesn't use them at all.

You might be familiar with the sinh and cosh functions -- the taylor series for cosh z has all the even terms of e^z, and the taylor series for sinh z has all of the odd terms.

One can write sinh and cosh in terms of the two functions e^z and e^((-1)z) -- the idea is that the (-1) uniformly alters every odd term in the taylor series for e^z, but leaves the even terms alone (since (-1)^2 = 1). Once you have the taylor series for e^z and e^((-1)z), you can then solve for the other functions.


You want to do the same thing here, except you want a modification that has period 3... i.e. that t^3 = 1. The three possibilities, e^z, e^(tz), and e^(t^2 z) give you three series you can manipulate...

 

[arildno]

From what I can see, setting x=iy, we ought to have:
$$u+v+w=\cos(y)+i\sin(y)$$

I might be wrong, but I don't think so.

At any rate, even if I'm right about this, it doesn't necessarily follow that an approach along my line would be as easy as following Hurkyl's advice.

 

[Hurkyl]

That is true, but a simpler relation is that u(x) + v(x) + w(x) = e^x. (e^(ix) is, of course, cos x + i sin x)

 

[DivGradCurl]

Thank you so much for all the hints. I'm not used to working with hyperbolic functions, but it sounds like it might be useful to learn more about them now. Let me check with you whether or not I roughly understand what you mean. I've searched the web and found:

$$\sinh (x) = \frac{e^x - e^{-x}}{2} = \sum _{n=0} ^{\infty} \frac{x^{2n+1}}{(2n+1)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \dotsb$$

$$\cosh (x) = \frac{e^x + e^{-x}}{2} = \sum _{n=0} ^{\infty} \frac{x^{2n}}{(2n)!} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dotsb$$

They are directly related to

$$e^x = \sum _{n=0} ^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dotsb$$

$$e^{-x} = \sum _{n=0} ^{\infty} (-1)^n \frac{x^{n}}{n!} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dotsb$$

We also have:

$$e^z = \sum _{n=0} ^{\infty} \frac{z^{n}}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dotsb$$

$$e^{tz} = \sum _{n=0} ^{\infty} \frac{(tz)^{n}}{n!} = 1 + tz + \frac{t^2 z^2}{2!} + \frac{t^3 z^3}{3!} + \dotsb$$

$$e^{t^2 z} = \sum _{n=0} ^{\infty} \frac{(t^2 z)^{n}}{n!} = 1 + t^2 z + \frac{t^4 z^2}{2!} + \frac{t^6 z^3}{3!} + \dotsb$$

From those three, I can get to the following:

$$e^{ix} = \sum _{n=0} ^{\infty} \frac{(ix)^{n}}{n!} = 1 + ix - \frac{x^2}{2!} + \frac{ix^3}{3!} + \dotsb$$

$$e^{-ix} = \sum _{n=0} ^{\infty} (-1)^n \frac{(ix)^{n}}{n!} = 1 - ix + \frac{x^2}{2!} - \frac{ix^3}{3!} + \dotsb$$

And that has a close relationship to my problem, since

$$u(x) + v(x) + w(x) = e^{ix} = \cos x + i\sin x$$

What I initially had in mind was to individually manipulate the series representation of the given functions so that each is related to some familiar function. It is a bit confusing now that I try to figure out a way to rewrite them, since they appear together above. I'm not sure on know how to proceed. I am supposed to manipulate $$e^{ix}$$ and $$e^{-ix}$$ to go any further? Thanks.

 

[Hurkyl]

Here is your problem: you're manipulating things with a period of 2 or 4 (i.e. -1 or i), trying to make something with a period of 3...

 

[TenaliRaman]

If i get Hurkyl's Hint correctly,then think ::cube root of unity[/Color]::

-- AI

 

[DivGradCurl]

Is this what you mean?

$$(1)^{1/3} = i^{4n/3} = 1$$

 

[TenaliRaman]

You missed a pi i guess, but yes. You can work in terms of omega instead of i.
Note that $$\omega^3 = 1$$ and $$\omega^2 + \omega + 1 = 0$$.

-- AI

 

[DivGradCurl]

How about $$i^{(4\pi n) /3}$$ ? As a result, we may find:

$$n=0 \Longrightarrow i^{(4\pi n) /3} = 1$$

$$n=1 \Longrightarrow i^{(4\pi n) /3} = \cos \left( \frac{2\pi ^2}{3} \right) + i \sin \left( \frac{2\pi ^2}{3} \right) = (-1) ^{2\pi /3}$$

$$n=2 \Longrightarrow i^{(4\pi n) /3} = \cos \left( \frac{4\pi ^2}{3} \right) + i \sin \left( \frac{4\pi ^2}{3} \right) = (-1) ^{4\pi /3}$$

$$n=3 \Longrightarrow i^{(4\pi n) /3} = \cos \left( 2\pi ^2 \right) + i \sin \left( 2\pi ^2 \right) = 1$$

Does $$x = 2\pi ^2$$ ?

I'm not sure about what you mean by $$\omega$$. Does $$\omega = i^{(4\pi n) /3}$$ ?
Also, how does the relationship $$\omega^2 + \omega + 1 = 0$$ fit into the problem (I'm really lost)?

Any help is highly appreciated.

 

[Hurkyl]

It's customary to use $\omega$ or $\xi$ as the symbol for a (primitive) primitive root of unity.

Generally, when manipulating complex exponents, it's better to do everything to the base e, so the cube roots of 1 are $e^{2\pi n / 3}$.

By the way, your calculations in the last post are entirely wrong. For example, $\cos 2\pi^2 n$ is not 1.

 

[DivGradCurl]

Sorry about that.

I guess we might ultimately find

$$u(x) + v(x) + w(x) = e^{i\left( \frac{2\pi}{3}n \right)} \qquad n \in \mathbb{N}$$

My question is:

Is it now possible to individually manipulate the given functions? If so, how?

Thank you very much.

 

[krab]

Still don't know what you're getting at.
$$u(x) + v(x) + w(x) = e^x$$, directly. No imaginary numbers are needed for this. So your equation is only consistent if $x=i(2\pi n/3)$. How does that help? Why are you restricting the value of x? You are meant to prove it for all x.

 

[krab]

Explicit formulas exist for u,v, and w in terms of exponentials and cosines. I won't give them here, since this is a homework forum. But if you have access to a mathematics library, there is a very nice paper in the journal of the Society for Industrial and Apllied Mathematics Vol.15, p.618:
http://www.jstor.org/view/00361399/di974630/97p0414b/

 

[krab]

Also: You can prove the relation u^3+v^3+w^3-3uvw=1 by differentiating it and using the relations: u'=w, v'=u, w'=v.

 

[dextercioby]

Also: You can prove the relation u^3+v^3+w^3-3uvw=1 by differentiating it and using the relations: u'=w, v'=u, w'=v.

Yes,Krab,your method unfortunately involves the operation of differentiation which "erases" the constant in the RHS of the equality.So basically,it helps u prove that
$$u^{3}+v^{3}+w^{3}-3uvw=C$$
,where C is an arbitrary constant.U'll need to 'fix' it,and that is done through an initial condition:
Take the function
$$f(t)=:u^{3}(t)+v^{3}(t)+w^{3}(t)-3u(t)v(t)w(t)$$
It verifies $$f(0)=1$$
and at the same time,as u indicated
$$f'(t)=0$$
which means it is constant,no matter 't' and has the same value for all 't',implicitely for 't=0'.
So one finds $$f(t)=1$$

Quod erat demonstrandum.

Daniel.

PS.This kind of proof reminds me of the proof of the inverse of Stone's theorem.

 

[DivGradCurl]

The purpose of the homework help section at physicsforums.com is to help people get their homework done, not actually do it for them. I am aware of that. I'm not asking you to do anything other than give me some pointers. I'm grateful for them. So, let's get down to the nitty-gritty.

I shouldn't have restricted the value of x. What I intended to to was to apply the idea of using the the cube roots of 1 into this problem. As you've just showed me, that is not the right way to proceed. So, what happens to the $$e^{i\left( \frac{2\pi}{3}n \right)}$$ Hurkyl and TenaliRaman guided me to? I simply don't see how that becomes a $$e^x$$. Maybe, that goes in at a different stage of the solution. Also, I'd like to illustrate why I have difficulty dealing with $$u(x) + v(x) + w(x) = e^x$$:

$$\sum _{n=0} ^{\infty} \left[ \frac{x^{3n}}{(3n)!} \right] + \sum _{n=0} ^{\infty} \left[ \frac{x^{3n+1}}{(3n+1)!} \right] + \sum _{n=0} ^{\infty} \left[ \frac{x^{3n+2}}{(3n+2)!} \right] = \sum _{n=0} ^{\infty} \frac{x^{n}}{n!}$$

My first impression is: "then what?". It doesn't seem to follow from this relationship a way to write u, v, and w in a familiar way. It is different from something like:

$$k = \sum _{n=0} ^{\infty} \frac{x^{n}}{n!}$$

I would know that k = e^x. This doesn't seem to be same case:


$$u = e^x - \sum _{n=0} ^{\infty} \left[ \frac{x^{3n+1}}{(3n+1)!} \right] - \sum _{n=0} ^{\infty} \left[ \frac{x^{3n+2}}{(3n+2)!} \right]$$

Is it easy to find u^3 now? No. The reason why I insist on this point is that I think I first need to get to $$u^3 ; v^3 ; w^3 ; -3 u v w$$. From those pieces I would prove

$$u^3 + v^3 + w^3 - 3 u v w = 1$$

for all x.

Thank you

 

[dextercioby]

Me and Krab have suggested an elegant proof involving differential calculus.
If you want to struggle with getting the third powers of each series,please be my guest.

Daniel.

 

[DivGradCurl]

This is much simpler. I see what you mean.

Thank you

 

[TenaliRaman]

thiago,
i hope u are not averse to another solution.

The proof hurkyl was trying to hint at is absolute beauty! Elegance personified in mathematical symbol! and u know what u don't have to cube anything and its just 5 line proof! I am going to write the proof down here just to document it somewhere before i forget the method.

I will be referring to u,v,w as u(x),v(x) and w(x). I will be using the omega $$\omega$$ notation because i am used to it and keep in mind the two identities i gave in my last post.

now look at this magic,
$$u(x) + v(x) + w(x) = e^x$$
$$u(\omega x) + v(\omega x) + w(\omega{x}) = u(x) + \omega{v(x)} + \omega^2{w(x)} = e^{\omega x}$$
$$u(\omega^2 x) + v(\omega^2 x) + w(\omega^2 x) = u(x) + \omega^2 v(x) + \omega w(x) = e^{\omega^2 x}$$

therefore,
$$(u(x) + v(x) + w(x))(u(x) + \omega v(x) + \omega^2 w(x))(u(x) + \omega^2 v(x) + \omega w(x))$$
$$= (u(x) + v(x) + w(x))(u(\omega x) + v(\omega x) + w(\omega x))(u(\omega^2 x) + v(\omega^2 x) + w(\omega^2 x))$$
$$= e^x * e^{\omega x} * e^{\omega^2 x}$$
$$= e^{x(1+\omega+\omega^2)}$$
$$= 1$$

Now its a well known identity that,
$$(a+b+c)*(a+b\omega+c\omega^2)*(a+b\omega^2+c\omega) = a^3+b^3+c^3-3abc$$

et voila! QED!

-- AI

 

[krab]

Yes, and BTW, write the first 3 equations in matrix form, invert the matrix, and you have explicit expressions for u, v, w in terms of exponentials.

 

[TenaliRaman]

Ah! Indeed!

-- AI