Very Easy Taylor Series Approximation Help

  • Thread starter Farzan
  • Start date
  • #1
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Homework Statement



Approximate f by a Taylor polynomial with degree n at the number a.

f(x) = x^(1/2)

a=4
n=2

4<x<4.2
(This information may not be needed for this, there are two parts but I only need help on the first)

Homework Equations



Summation [tex]f^(i) (a) * (x-a)^i / i![/tex]

The Attempt at a Solution



Derivatives...

[tex]x^(1/2)
0.5 * x^(-1/2)
-0.25 * x^(-3/2)[/tex]

So the Taylor series of order n=2 should be...

2 + 0.25(x-4) - (1/64)(x-4)(x-4)

Now, my question...

To find the approximation of square root of x, do I just plug in 4 to that?

That would make most of the terms zero, leaving me with 2

Does this mean that whenever the Taylor series is centered at a number, the first term is the approximation?


Ugh, sorry for the failure using tex tags :( It's my first post here
 

Answers and Replies

  • #2
Dick
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Yes. Exactly. If x=a the first term IS the approximation. The usual game is to approximate something like 4.1 using the series.
 
Last edited:
  • #3
HallsofIvy
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I think Dick is interpreting the question differently than I am.

What you have: 2+ (1/4)(x-4)- (1/64)(x- 4)2 alread is the quadratic approximation to [itex]\sqrt{x}[/itex] around x= 4.

Putting x= 4 would just give you the obvious [itex]\sqrt{4}= 2[/itex].
 

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