Solving Power Series Limits: Find ak Value for S 4^n z^(3n)

[sweetvirgogirl]

so there is a power series
S 4^n z^(3n)
and upper limit being infinity and lower limit being 0. (S means sigma)
then my book says, ak = 4^ (k/3) if k = 0, 3, 6...
and ak equals 0 otherwise.
i don't get how they came up for the value of ak

i might be missing something silly ...
any clues?
thanks!

 

[dextercioby]

Well, the way you asked it

S=\sum_{n=0}^{\infty} \left(4z^{3}\right)^{n}

exists iff |4z^{3}|<1.

Daniel.

 

[Galileo]

They are comparing it to the general form of a power series:
\sum_{n=0}^\infty a_n z^n=a_0+a_1z+a_2z^2+...

You have
\sum_{n=0}^\infty 4^nz^{3n}=1+4z^3+4^2z^6+...

Now what is the coefficient of z^k in the given power series? (It's clearly zero if k is not a multiple of 3).

 

[sweetvirgogirl]

They are comparing it to the general form of a power series:
\sum_{n=0}^\infty a_n z^n=a_0+a_1z+a_2z^2+...

You have
\sum_{n=0}^\infty 4^nz^{3n}=1+4z^3+4^2z^6+...

Now what is the coefficient of z^k in the given power series? (It's clearly zero if k is not a multiple of 3).

since you seem to have a good understanding on power series and such ...
could you also tell me how they expand rational functions?
like i know 1/(1-x) = 1 +x +x^2/2! ...
but how did they come up with that?

and any other interesting rules about power series expansion of rational functions?

 

[roger]

since you seem to have a good understanding on power series and such ...
could you also tell me how they expand rational functions?
like i know 1/(1-x) = 1 +x +x^2/2! ...
but how did they come up with that?

and any other interesting rules about power series expansion of rational functions?

you meant 1+x+x^2...

by binomial theorem ?

 

[sweetvirgogirl]

you meant 1+x+x^2...

by binomial theorem ?

oops yeah ... there is no division by n factorial involved ...

man i forgot all that power series and stuff ...

thanks!

 

[Galileo]

It's called a geometric series