Taylor's Series

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Taylor's theorem states that, for any reasonable function f(x), the value of f at a point (x+[tex]\delta[/tex]) can be expressed as an infinite series involving f and its derivatives at the point of x:

f(x+[tex]\delta[/tex]) = f(x) + f '(x)[tex]\delta[/tex] + [tex]\frac{1}{2!}[/tex]f ''(x)[tex]\delta[/tex][tex]^{2}[/tex] + ....

where the primes denote successive derivatives of f(x). (Depending on the function this series may converge for any increment [tex]\delta[/tex] or only for values of [tex]\delta[/tex] less then some nonzero 'radius of convergence'.) This theorem is enormously useful, especially for small values of [tex]\delta[/tex], when the first one or two terms of the series are often an excellent approximation. Find the taylor's series for ln(1+[tex]\delta[/tex]). Do the same for cos [tex]\delta[/tex].


We just started Taylor's theorems now, and it seems like there are many of them. Why exactly are they good and what purpose do they serve?

Is finding the Taylor's series for ln (1+[tex]\delta[/tex]) a matter of substituting it into the above equation or is there more to it?

Thanks again!
 
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Answers and Replies

  • #2
learningphysics
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In lots of problems, you may have to find f(x+delta) - f(x)... where delta is very small... your calculator might only give you a rounded figure of 0 because the numbers are too close together...

However if you take the taylor series of f(x+delta)... then subtract f(x)... you get: f '(x)delta + f ''(x)delta^2. This will give a good approximation of f(x+delta) - f(x).

That's just one example of why taylor series are useful. I'm sure there are many more.

Is finding the Taylor's series for ln (1+ delta) a matter of substituting it into the above equation or is there more to it?

Yeah, just substitute into the equation, using x = 1. That should work.
 
  • #3
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Effectively solving for delta?
 
  • #4
learningphysics
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Effectively solving for delta?

I don't understand what you mean.
 
  • #5
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When everything is subbed in, am I solving for delta?
 
  • #6
learningphysics
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When everything is subbed in, am I solving for delta?

No... you'll just be getting a polynomial formula for ln(1+delta)... in terms of delta...
 
  • #7
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If I'm subbing x=1, my ln (1+delta) goes where in the formula?
 
  • #8
learningphysics
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If I'm subbing x=1, my ln (1+delta) goes where in the formula?

ln (x + delta) is the left hand side... right hand side is f(x) + f'(x)delta... where f(x) = ln(x)

ln(x + delta)

ln(x+delta) = ln(x) + (1/x)delta + ....

So ln(1+delta) = ln(1) + (1/1)delta + ....
 
  • #9
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So ln(1+delta) = ln(1) + (1/1)delta + ....

is the bolded delta in the denominator?
 
  • #10
learningphysics
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So ln(1+delta) = ln(1) + (1/1)delta + ....

is the bolded delta in the denominator?

No. I'm using the formula from your original post exactly... delta doesn't go in the denominator...

I got the 1/1.... because the deriative of lnx = 1/x... so f'(1) = 1/1
 
  • #11
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Ok, just making sure I was reading it right.

Is that as far as you can go? Since the 2nd derivative would be 0?
 
  • #12
learningphysics
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Ok, just making sure I was reading it right.

Is that as far as you can go? Since the 2nd derivative would be 0?

How do you get 0?

f(x) = ln(x)

f'(x) = 1/x

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

So f''(1) = -1 not zero...
 
  • #13
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Oh, sub afterwards.
I was deriving 1/1.
 
  • #14
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I assume I don't need to do all that many though?
 
  • #15
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Is this far enough to do the problem?

ln (x+[tex]\delta[/tex]) = ln(x) + [tex]\frac{1}{x}[/tex][tex]\delta[/tex] + ([tex]\frac{1}{2!}[/tex])[tex]\frac{1}{x^{2}}[/tex]([tex]\delta^{2}[/tex])
 
  • #16
learningphysics
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Is this far enough to do the problem?

ln (x+[tex]\delta[/tex]) = ln(x) + [tex]\frac{1}{x}[/tex][tex]\delta[/tex] + ([tex]\frac{1}{2!}[/tex])[tex]\frac{1}{x^{2}}[/tex]([tex]\delta^{2}[/tex])

Actually I think the question wants you to keep going... and find a general formula for the entire series... do you see a pattern to the terms of ln(1+delta) ?
 
  • #17
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denominator and delta are increasing by powers of 1. along with that 1/2!, and its denominator increasing..
 
  • #18
learningphysics
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Is this far enough to do the problem?

ln (x+[tex]\delta[/tex]) = ln(x) + [tex]\frac{1}{x}[/tex][tex]\delta[/tex] + ([tex]\frac{1}{2!}[/tex])[tex]\frac{1}{x^{2}}[/tex]([tex]\delta^{2}[/tex])

careful... it needs to be (-1/x^2) for the third term...
 
  • #19
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... i cant say i see a pattern
 
  • #20
learningphysics
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denominator and delta are increasing by powers of 1. along with that 1/2!, and its denominator increasing..

yes, and also, you should have alternating signs...

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

f'''(x) = 2/x^3

I recommend finding the first 4 or 5 terms of the series... then you should see the pattern...
 
  • #21
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except for the first 2 though? (alternating signs)
and the pattern will boil down to some formula?
 
  • #22
learningphysics
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except for the first 2 though? (alternating signs)
and the pattern will boil down to some formula?

yeah, do 5 terms altogether... get the formula for ln(x+delta) for 5 terms.
 
  • #23
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I looked up the higher derivatives for ln x and only found to 4...

ln(x+[tex]\delta[/tex]) = ln (x) + [tex]\frac{1}{x}[/tex][tex]\delta[/tex] + [tex]\frac{1}{2!}[/tex] [tex]\frac{-1}{x^{2}}[/tex][tex]\delta^{2}[/tex] + [tex]\frac{1}{3!}[/tex][tex]\frac{2}{x^{3}}[/tex][tex]\delta^{3}[/tex] + [tex]\frac{1}{4!}[/tex] [tex]\frac{-4}{x^{6}}[/tex][tex]\delta[/tex][tex]^{4}[/tex] + ...
 
  • #24
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The question asked for the taylor's series, I need to do more you think?
 
  • #25
learningphysics
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I looked up the higher derivatives for ln x and only found to 4...

ln(x+[tex]\delta[/tex]) = ln (x) + [tex]\frac{1}{x}[/tex][tex]\delta[/tex] + [tex]\frac{1}{2!}[/tex] [tex]\frac{-1}{x^{2}}[/tex][tex]\delta^{2}[/tex] + [tex]\frac{1}{3!}[/tex][tex]\frac{2}{x^{3}}[/tex][tex]\delta^{3}[/tex] + [tex]\frac{1}{4!}[/tex] [tex]\frac{-4}{x^{6}}[/tex][tex]\delta[/tex][tex]^{4}[/tex] + ...

The last term is wrong... it should be -6/x^4... yeah, that's enough terms... now simplify the numbers... ie get rid of the factorials...
 
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