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Using Taylor's Theorem and big O

  • Thread starter SMA_01
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  • #1
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Homework Statement



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:

Answers and Replies

  • #2
LCKurtz
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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)##.
 
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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?
 
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LCKurtz
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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?
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.
 
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  • #5
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Much easier to understand, thank you!
 

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