- #1
DivGradCurl
- 372
- 0
Let
[tex] u = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + \frac{x^{9}}{9!} + \dotsb [/tex]
[tex] v = x + \frac{x^{4}}{4!} + \frac{x^{7}}{7!} + \frac{x^{10}}{10!} + \dotsb [/tex]
[tex] w = \frac{x^{2}}{2!} + \frac{x^{5}}{5!} + \frac{x^{8}}{8!} + \dotsb [/tex]
Show that
[tex] u^3 + v^3 + w^3 - 3 u v w = 1 [/tex]
Well, here what I've done:
[tex] u = \sum _{n=0} ^{\infty} \frac{x^{3n}}{(3n)!} [/tex]
[tex] v = \sum _{n=0} ^{\infty} \frac{x^{3n+1}}{(3n+1)!} [/tex]
[tex] w = \sum _{n=0} ^{\infty} \frac{x^{3n+2}}{(3n+2)!} [/tex]
Further work can be simplified if I rewrite those series in terms of familiar functions.
The first one is not too hard to obtain:
[tex] 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} [/tex]
Unfortunately, the others got me stuck. I can't find ways to rewrite the following factorials:
Is there any trick?
Thanks
[tex] u = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + \frac{x^{9}}{9!} + \dotsb [/tex]
[tex] v = x + \frac{x^{4}}{4!} + \frac{x^{7}}{7!} + \frac{x^{10}}{10!} + \dotsb [/tex]
[tex] w = \frac{x^{2}}{2!} + \frac{x^{5}}{5!} + \frac{x^{8}}{8!} + \dotsb [/tex]
Show that
[tex] u^3 + v^3 + w^3 - 3 u v w = 1 [/tex]
Well, here what I've done:
[tex] u = \sum _{n=0} ^{\infty} \frac{x^{3n}}{(3n)!} [/tex]
[tex] v = \sum _{n=0} ^{\infty} \frac{x^{3n+1}}{(3n+1)!} [/tex]
[tex] w = \sum _{n=0} ^{\infty} \frac{x^{3n+2}}{(3n+2)!} [/tex]
Further work can be simplified if I rewrite those series in terms of familiar functions.
The first one is not too hard to obtain:
[tex] 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} [/tex]
Unfortunately, the others got me stuck. I can't find ways to rewrite the following factorials:
[tex] (3n+1)! \qquad \mbox{ and } \qquad (3n+2)! [/tex]
Is there any trick?
Thanks