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Taylor Series to x places of decimals

by *Alice*
Tags: decimals, places, series, taylor
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*Alice*
#1
Jan4-06, 01:00 AM
P: 26
Hi all,

here's the problem:

given: [tex]tan^(-1)= x - x^3/3 + x^5/5[/tex]

using the result [tex]tan^(-1) (1)= pi/4[/tex]

how many terms of the series are needed to calculate pi to ten places of decimals?


note: this is supposed to say tan^(-1) and tan^(-1)[1] respectively

Does anyone know whether there's a formula to calculate the accuracy/number of decimals of the Taylor expansion? In my books I found a formula to calculate the approximation of error from the remainder term - not sure, how that would help, though.

The above seems to be one of those standard questions that one should definitely be able to solve in order to pass an exam. So, if anyone knows - help's much appreciated!
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Tide
#2
Jan4-06, 01:06 AM
Sci Advisor
HW Helper
P: 3,147
It's an alternating convergent series so the error is bounded by the term following the last term you keep.
*Alice*
#3
Jan4-06, 01:25 AM
P: 26
OK...so that's the error formula I found:

[tex]Rn(x) = (x-a)^n / n! * f^n * (k)[/tex]

where Rn(x) is the remainder term and represents the error in approximating f(x) by the above (n-1)th order power series.

k is not known.

So now I would calculate k by finding the upper limit of the error by differentiating Rn(x) with respect to k and finding the maximum.

then I would put that in the above equation and substitute 10^(-10) for Rn(x) and x=1, a=0

Does that make any sense ?

dicerandom
#4
Jan4-06, 01:45 AM
P: 308
Taylor Series to x places of decimals

Quote Quote by *Alice*
note: this is supposed to say tan^(-1) and tan^(-1)[1] respectively
You can do that like this:

\tan^{-1}(1)
[tex]\tan^{-1}(1)[/tex]
Tide
#5
Jan4-06, 03:20 AM
Sci Advisor
HW Helper
P: 3,147
Alice,

You know that the [itex]n^{th}[/itex] term in the series is [itex](-1)^{n-1}x^{2n-1}/(2n-1)[/itex] so you should be able to find the upper bound on the error.
*Alice*
#6
Jan4-06, 03:42 AM
P: 26
Thanks Tide - so that would go like this:

[tex](-1)^{n-1} * x^{n-1} / (2n-1) = 10^{-10}[/tex]

so, now let x = 1

[tex] 1^n / (2n-1) = 1 / (2n-1) = 10^{-10}[/tex]

solve for n:
n = 5 * 10^9

which is what the solutions manual says as well.

Thank you so much!!!!
Tide
#7
Jan4-06, 01:17 PM
Sci Advisor
HW Helper
P: 3,147
You are very welcome! :)


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