Solving Tailor Series Question: My Attempt and Explanation

[transgalactic]

i added the question and how i tried to solve it in the link

http://img146.imageshack.us/my.php?image=img8217ix2.jpg

 

[Gib Z]

Ok first of all, since we are cube rooting, to get O(x^4) the series inside the cube root must be continued up to the 12th power. Also, you have the right series for the thing inside. Just put that into the Taylor Series for the cube root. It might not work though, because of troubles involved with the cube roots series.

 

[transgalactic]

i tried to build the series for x^1/3
in order to do that I've built the first derivative
the 2nd etc..
but when i put 0 in the derivative i get zeros for each member
so the series for the cube root is the cube root itself

so my other idea is the take this function as a whole and make a series
out of it
using the derivatives till the 4th power
which is very long .

is there any other way
because the first way is not working
?

 

[Rainbow Child]

Use the series expansion

$$(1+u)^\alpha=1+\alpha\, u+\dots+\frac{\alpha\,(\alpha-1)\dots (\alpha-n+1)}{n!}\, u^n+\dots$$

 

[transgalactic]

i tried to use this formula but i didnt understand how the "n" member
works
and where to stop devloping this formula??

i show in my link
how i tried to use it

http://img518.imageshack.us/my.php?image=img8221tm9.jpg

 

[Rainbow Child]

The series expansion of $\sin(2\,x)$ is

$$\sin(2\,x)=2\,x-\frac{2^3}{3!}\,x^3+\frac{2^4}{5!}\,x^5+\dots \quad (1)$$

and the series expansion of $(1+u)^{1/3}$ is

$$(1+u)^{1/3}=1+\frac{1}{3}\,u-\frac{1}{9}\,u^2+\frac{5}{81}\,u^3-\frac{10}{243}\,u^4+\dots$$

For the 4^th power is enough to plug for $u$ the first two terms of (1)

i tried to use this formula but i didnt understand how the "n" member
works

$$n=2\rightarrow \frac{\alpha\,(\alpha-1)}{2!}$$

$$n=3\rightarrow \frac{\alpha\,(\alpha-1)\,(\alpha-2)}{3!}$$

$$n=4\rightarrow \frac{\alpha\,(\alpha-1)\,(\alpha-2)\,(\alpha-3)}{4!}$$

$$\dots\dots\dots\dots\dots\dots$$

 

[transgalactic]

i got to the conclution that it doesn't matter
what what the length of each series as long as in the end we get
a series that on the 4th power
or on the 5th power in which case we delete the 5th power member

is that true??

 

[Gib Z]

Ok basically for now, yes, continue the series for an infinite number of terms, then truncate at the end. With time you will learn short cuts with the Big-Oh notation, but practice makes perfect.