Using Taylor's Theorem and big O

1. Sep 25, 2012

SMA_01

1. The problem statement, all variables and given/known data

Use Taylor's Theorem to show that

√(x+1)=1+(1/2)x+O(x2)

for x sufficiently small.

Here's what I did:

f(x)= √x+1

f'(x)= (1/2)(x+1)(-1/2)

Then using x0=0,

f(0)= 1, f'(0)=1/2.

√x+1=1+(1/2)x-(1/8)x2(cx+1)(-3/2)

So, then using h as a parameter:

l√(h+1) -1-(1/2)h l ≤ 1/8(h2)

Finally,

√(h+1) = 1+(1/2)h+O(h2)

Is this correct?

I' m having difficulty understanding the meaning of O, can someone please explain in simple terms?
Thank you.

Last edited: Sep 25, 2012
2. Sep 25, 2012

LCKurtz

If you expand $f(x)$ in a Taylor series about $x=0$ you get$$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$You need to write out the first few terms of that for your function. The notation $O(x^2)$ means the other terms are bounded by a constant times $x^2$. So you group all the higher order terms together and call them $O(x^2)$.

3. Sep 25, 2012

SMA_01

LCKurtz- Thank you. I just edited my original question, I think I solved it right, but am not completely sure. How do I know when to stop expanding the series? Is it once I get x^2?

4. Sep 25, 2012

LCKurtz

Yes, the terms from $x^2$ and higher are all $O(x^2)$. I like to think of $O(x^2)$, very informally mind you, as "Other terms containing $x$ to at least second power". Not very precise but easy to remember.

Last edited: Sep 25, 2012
5. Sep 25, 2012

SMA_01

Much easier to understand, thank you!