# Taylor series remainder

## Main Question or Discussion Point

How is the Taylor remainder of a series (with given Taylor expansion) expressed if you want to make a calculation with known error? e.g. if I want to calculate π to, say, 12 decimal places using the previously-derived result π=4*arctan(1) and the Taylor series for arctan(x), how will I work out how many terms I need (or, imagine that number of decimal places is high enough that trial and error is not efficient)?

The problem is that equating the error (e.g. 5*10^(-13) in my example above) to the integral form of the remainder leaves us an expression that can be numerically (but certainly not analytically) solved, whereas I am looking for a method that works (even approximately) by hand and handheld calculator.

mfb
Mentor
(or, imagine that number of decimal places is high enough that trial and error is not efficient)?
It is always possible as you have to calculate every term individually.

The problem is that equating the error (e.g. 5*10^(-13) in my example above) to the integral form of the remainder leaves us an expression that can be numerically (but certainly not analytically) solved
You can find an analytic upper limit I think.

FactChecker
Gold Member
The remainder after expanding n terms of a Taylor series around point a is equal to ((x-a)n/n!)f(n)(x') for some x' in (a,x). So you can bound the error at x if you can bound the n'th derivative.

The remainder after expanding n terms of a Taylor series around point a is equal to ((x-a)n/n!)f(n)(x') for some x' in (a,x). So you can bound the error at x if you can bound the n'th derivative.
How would this work, for, e.g. my example of π=4*arctan(1)? It seems I would have to take the nth derivative of arctan(x) - which doesn't exactly suggest itself easily to me - even then, I'm not sure what would have to be done with the remaining 3 variables (x, x', n) to find an upper bound for n?

mfb
Mentor
It seems I would have to take the nth derivative of arctan(x)
Yes. It might give ugly expressions, but no one said calculating the 12th decimal digit that way was easy without a computer.
even then, I'm not sure what would have to be done with the remaining 3 variables (x, x', n) to find an upper bound for n?
x is where you want to evaluate the series (at 1), a the point your taylor series is based on (0), x' can be anything between 0 and 1, you have to find an upper bound that is valid in the whole range. n is just the derivative you have.

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