I get a close error/remainder to this series, but it's wrong why?

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  • #1
s3a
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Homework Statement


The Taylor series for f(x) = ln(sec(x)) at a = 0 is ##Σ_{n=0}^{∞} c_n x^n##.

(a) Find the first few coefficients. (I don't need help for this part.)

(b) Find the exact error in approximating ln(sec(-0.1)) by its fourth-degree Taylor polynomial at a = 0.

Homework Equations


E(x) = f(x) - ##P_n (x)##, or in this case, E(-0.1) = f(-0.1) - ##P_4 (-0.1)##


The Attempt at a Solution


I found the cofficients of each term of the polynomial, but I don't feel it's necessary to show you how I did that, since you can just confirm that I'm right by looking here.:
http://www.wolframalpha.com/input/?i=ln(sec(x))+maclaurin+polynomial

If I'm correct, for this problem, the error is the difference between the function and the 4th degree polynomial.:
E(-0.1) = f(-0.1) - ##P_4 (-0.1)## = ln(sec(-0.1)) – (1/2*0.1^2+1/12*0.1^4) = 2.22899019757457996644E-8 ≠ 2.22899018157481E-8

Notice how my answer (which is the one on the left-hand side) differs slightly from the correct answer (which is the one on the right-hand side) of the problem.

Where is this small disagreement in error/remainder values coming from?

Any input would be greatly appreciated!
 

Answers and Replies

  • #2
Dick
Science Advisor
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Homework Statement


The Taylor series for f(x) = ln(sec(x)) at a = 0 is ##Σ_{n=0}^{∞} c_n x^n##.

(a) Find the first few coefficients. (I don't need help for this part.)

(b) Find the exact error in approximating ln(sec(-0.1)) by its fourth-degree Taylor polynomial at a = 0.

Homework Equations


E(x) = f(x) - ##P_n (x)##, or in this case, E(-0.1) = f(-0.1) - ##P_4 (-0.1)##


The Attempt at a Solution


I found the cofficients of each term of the polynomial, but I don't feel it's necessary to show you how I did that, since you can just confirm that I'm right by looking here.:
http://www.wolframalpha.com/input/?i=ln(sec(x))+maclaurin+polynomial

If I'm correct, for this problem, the error is the difference between the function and the 4th degree polynomial.:
E(-0.1) = f(-0.1) - ##P_4 (-0.1)## = ln(sec(-0.1)) – (1/2*0.1^2+1/12*0.1^4) = 2.22899019757457996644E-8 ≠ 2.22899018157481E-8

Notice how my answer (which is the one on the left-hand side) differs slightly from the correct answer (which is the one on the right-hand side) of the problem.

Where is this small disagreement in error/remainder values coming from?

Any input would be greatly appreciated!
Who ever generated that 'correct' answer probably used a floating point calculator that was only accurate to 16 decimal places. There's no real reason to list that many decimal places. Especially when the last 12 or so aren't even correct.
 
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  • #3
s3a
800
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Who ever generated that 'correct' answer probably used a floating point calculator that was only accurate to 16 decimal places. There's no real reason to list that many decimal places. Especially when the last 12 or so aren't even correct.
Okay, so ln(sec(-0.1)) – (1/2*0.1^2+1/12*0.1^4) is the correct answer, right?

Also, I have to consider only up to, and including, n=4 and not n=5, right? I ask since the n=5 case yields 0, so I just want to make sure that I understand the procedure correctly too, and it's not just a coincidence that I get the correct answer (because of the n=5 case being 0, in this case).

Assuming everything is fine concerning what I said above, could you please tell me why one would use Lagrange's remainder formula instead of just computing ##E(x) = f(x) – P_n(x)## (which seems easier and yields and accurate answer)?
 
  • #4
lurflurf
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We have
0.1->2.22899019757457996644E-8
0.0999999998804874->2.22899018157481E-8

The second answer results from a less accurate calculation.
To see exactly how that happens we would need to know more about how each calculation was performed.
 
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  • #5
lurflurf
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...why one would use Lagrange's remainder formula instead of just computing ##E(x) = f(x) – P_n(x)## (which seems easier and yields and accurate answer)?
Lagrange's remainder formula can be used when we cannot calculate f(x).
We use the Maclaurin polynomial because f is expensive or impossible to compute. If we compute f(x) to high accuracy we might not even need to use the Maclaurin polynomial.
 
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  • #6
s3a
800
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Thanks guys! :)
 

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