# Math Help - Taylor Series Question..

1. ## Taylor Series Question..

Hi, I'm new here. I have big exams for a scholarship in just over two weeks, which includes material beyond the scope of my university course and one of those exams is in Maths, so I've registered here to get some help on questions I'm not sure about.

I've attached a past exam question I'm having some difficulty with, which relates to Taylor Series.

I'm not sure about the first part, and if someone could explain to me or link me to a site explaining what exactly is to be proved when asked to prove Taylor's Theorem and how to go about it, that'd be great.

I can get the terms for the second part, but since it doesn't say how many terms to give, I'm guessing it's the general term they're after, and I'm unsure how to derive the general term - there doesn't seem to be any definite pattern arising.

And for the last part, I'm getting:
$f(x) = \frac{1\pm\sqrt{1-4x}}{2x}$
And I can't see how there is a solution where f(x)!=infinity when x=0....

2. Hi

$f(x) = \sqrt{1-4x}$

To get the Taylor's series of f(x) around x=0

$\sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!}\:x^n$

you need to get $f^{(n)}(0)$

If $f^{(n)}(x) = u_n\1-4x)^{\frac{1}{2}-n}" alt="f^{(n)}(x) = u_n\1-4x)^{\frac{1}{2}-n}" />

you can show that

$u_{n+1} = 2\:u_n\2n-1)" alt="u_{n+1} = 2\:u_n\2n-1)" />

which leads to $u_n = -2\:\frac{(2n-2)!}{(n-1)!}$

Then $f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!}\:x^n = \sum_{n=0}^{+\infty} \frac{u_n}{n!}\:x^n$

For the last part, you are getting:
$f(x) = \frac{1\pm\sqrt{1-4x}}{2x}$

But $f(x) = \frac{1+\sqrt{1-4x}}{2x}$ has an infinite limit for x=0

3. Does
$f(x) = \frac{1-\sqrt{1-4x}}{2x}$ not have an infinite limit also?

4. Originally Posted by Jimmeh
Does
$f(x) = \frac{1-\sqrt{1-4x}}{2x}$ not have an infinite limit also?
No
The limit is 1

5. Ah, I get it now:

$f(x) = \frac{1-\sqrt{1-4x}}{2x}\times\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}} = \frac{2}{1+\sqrt{1-4x}}$

Which is 1 when x=0.

Thank You!