Taylor series to estimate sums

  • Thread starter C.E
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  • #1
C.E
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1. Use Taylor's expansion about zero to find approximations as follows. You need
not compute explicitly the finite sums.

(a) sin(1) to within 10^-12; (b) e to within 10^-18:


3. I know that the taylor expansion for e is e=[tex]\sum_{n=1}^{\infty}\frac{1}[/tex]x[tex]^{n}[/tex]/n! and I aslo know that sine has a similar expansion my problem is with how to determine when the sum is in specific tolerence range, any ideas?
 

Answers and Replies

  • #2
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Does the 1 in sin(1) denote 1 degree? If so, you're going to have to use a value of pi/180.

Do you know the Maclaurin series for sin(x)? That's the Taylor's series for sin(x) about 0. If you don't know it, I suggest looking it up.

The Maclaurin series for sin(x) is an alternating series, and it is known that for a convergent alternating series, the error in approximating by the Maclaurin polynomial of a given degree is less than the absolute value of the first unused term.
 
  • #3
C.E
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The question does not, say (lets assume it is in radians). What about for e? that is not an altenating series.
 
  • #4
Dick
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If it's not alternating then use a form for a taylor series remainder term.
 
  • #5
36,442
8,413
The question does not, say (lets assume it is in radians). What about for e? that is not an altenating series.
You should check with the instructor. Although it would be ordinarily be reasonable to assume that 1 shown without any explicit dimensions meant radians, 1 radian is much farther away from 0 than is 1 degree, and this will definitely affect how many terms you need so that your answer is accurate to 11 or 12 decimal places. My best guess is that the problem is really about the sine of 1 degree.
 

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