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Taylor's Theorem watch

1. Find an estimate for Sin(31) using a Taylor expansion of 3rd order about x=pi/6 and justify the precision of your answer.

I've got my estimate for Sin(31) and I expect that they want us to use Taylor's theorem (because we truncated the Taylor series to 4 terms) to justify the precision by commenting on the upper bound of the error

Rn=mod([f^(n)(zeta)x^n]/n!) (apologise for lack of latex)

So my trouble is finding an expression for the nth derivative of Sin(x) at zeta (0<=zeta<=x). Because Rn is the mod of that expression, could you take max(d/dnSin(x)) by assuming that +/-Sin(x) and +/-Cos(x) both oscillate between -1 and 1 and thus just removing that term (taking the max for the upper bound) and stating that Rn<=mod(x^n/n!)? and since n! grows faster than x^n for all x,n then the series converges unconditionally for all x. My question is can we 'ignore' the nth derivative of sin(x) by taking the maximal value Sin(x)/cos(x) can take if we are considering an upper bound on the error.
2. (Original post by Protoxylic)
Find an estimate for Sin(31) using a Taylor expansion of 3rd order about x=pi/6 and justify the precision of your answer.

I've got my estimate for Sin(31) and I expect that they want us to use Taylor's theorem (because we truncated the Taylor series to 4 terms) to justify the precision by commenting on the upper bound of the error

Rn=mod([f^(n)(zeta)x^n]/n!) (apologise for lack of latex)

So my trouble is finding an expression for the nth derivative of Sin(x) at zeta (0<=zeta<=x). Because Rn is the mod of that expression, could you take max(d/dnSin(x)) by assuming that +/-Sin(x) and +/-Cos(x) both oscillate between -1 and 1 and thus just removing that term (taking the max for the upper bound) and stating that Rn<=mod(x^n/n!)? and since n! grows faster than x^n for all x,n then the series converges unconditionally for all x. My question is can we 'ignore' the nth derivative of sin(x) by taking the maximal value Sin(x)/cos(x) can take if we are considering an upper bound on the error.
You can remove the sine/cos term from the remainder to get an upper bound like you said. For your approximation you'll have to choose a suitable value of n, and the error will be less than (pi/180)^n / n! for the value of n that you chose. Make sure you understand the statement of Taylor's theorem (https://proofwiki.org/wiki/Taylor&#39;s_...m/One_Variable).

Notice in particular the 'there exists a η such that Rn=1(n+1)!(x−ξ)^(n+1)f(n+1)( ·)' part.
3. (Original post by metaltron)
You can remove the sine/cos term from the remainder to get an upper bound like you said. For your approximation you'll have to choose a suitable value of n, and the error will be less than (pi/180)^n / n! for the value of n that you chose. Make sure you understand the statement of Taylor's theorem (https://proofwiki.org/wiki/Taylor&#39;s_...m/One_Variable).

Notice in particular the 'there exists a Î· such that Rn=1(n+1)!(xâˆ’Î¾)^(n+1)f(n+1)(Î ·)' part.
Yeah I got the rest, I was just unsure about ignoring the sin/cos term, but now it seems blatantly obvious, cheers

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