# Taylor series with using geometric series

1. May 18, 2015

### Pietervv

• Member warned about not using the homework template
The question is:

Determine the Taylor series of f(x) at x=c(≠B) using geometric series

f(x)=A/(x-B)4

My attempt to the solution is:

4√f(x) = 4√A/((x-c)-B = (4√A/B) * 1/(((x-c)/B)-1) = (4√A/-B) * 1/(1-((x-c)/B))

using geometric series : 4√f(x) = (4√A/-B) Σ((x-c)/B)n

f(x)= A/B4 * Σ((x-c)/B)4n with abs((x-c)/B)<1

But i am not very sure if this is true.

2. May 18, 2015

### HallsofIvy

Staff Emeritus
f(x)= A/(x- B)4 is NOT the same as A/(x- c- B)^4 so $\sqrt[4]{f(x)}$ is NOT equal to $\sqrt[4]{A/((x- c)- B)^4}$. To put the "c" in you need
$f(x)= A/((x- c)- (B- c))^4$.

Last edited by a moderator: May 18, 2015
3. May 18, 2015

### Ray Vickson

As HallofIvy has pointed out, the manipulations you have done above are incorrect. After correcting them, you will be left with an an expression that involves, in part, an expansion of the form
$$\frac{1}{(1-u)^4} = c_0 + c_1 u + c_2 u^2 + \cdots .$$
That is a not a "geometric" series, because you have the 4th power in the denominator. I think the title of the problem is incorrect---you cannot do it using a geometric series (at least, not directly). However, you can start with a geometric series and do some calculus manipulations on it to obtain the series above.

Last edited: May 18, 2015
4. May 18, 2015

### Pietervv

Then the c disappears in the nominator and my f(x) is again f(x)= A/(x- B)4 after put the c in??

And is my tactic for solving this problem, with taking the 4√ the good tactic, or is it totaly the wrong way to solve it?

5. May 18, 2015

### Ray Vickson

No, it is totally wrong. In simplified notation, here is what you said:
$$\left(\sum_n a_n y^n\right)^4 = \sum_n a_n^4 \, y^{4n} \; \Longleftarrow \; \text{False!}$$

6. May 22, 2015

### Pietervv

For the people who are still interested, after watched the solution, the key was to first determine the series of g(x)=A/(x-B) using the geometric series (1) and then take the third derivative of both sides of the equation obtained by step 1

7. May 22, 2015

### Ray Vickson

Yes, that is exactly what I meant in Post #3 when I said to start with a geometric series and perform calculus manipulations on it.