Taylor’s Theorem: Error Analysis for Series Tacoma Narrows Bridge: November 7, 1940

Taylor series are used to estimate the value of functions (at least theoretically - now days we can usually use the calculator or computer to calculate directly.) An estimate is only useful if we have an idea of how accurate the estimate is. When we use part of a Taylor series to estimate the value of a function, the end of the series that we do not use is called the remainder. If we know the size of the remainder, then we know how close our estimate is.

For a geometric series, this is easy: ex. 2: Use to approximate over . Since the truncated part of the series is: , the truncation error is , which is . When you “truncate” a number, you drop off the end. Of course this is also trivial, because we have a formula that allows us to calculate the sum of a geometric series directly.

Taylor’s Theorem with Remainder If f has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I: Lagrange Form of the Remainder Remainder after partial sum Sn where c is between a and x.

Lagrange Form of the Remainder Remainder after partial sum Sn where c is between a and x. This is also called the remainder of order n or the error term. Note that this looks just like the next term in the series, but “a” has been replaced by the number “c” in . This seems kind of vague, since we don’t know the value of c, but we can sometimes find a maximum value for .

Lagrange Form of the Remainder If M is the maximum value of on the interval between a and x, then: Remainder Estimation Theorem We call this the Remainder Estimation Theorem.

Prove that , which is the Taylor series for sin x, converges for all real x. Since the maximum value of sin x or any of it’s derivatives is 1, for all real x, M = 1. Remainder Estimation Theorem so the series converges.

Alternating Series Estimation Theorem Since each term of a convergent alternating series moves the partial sum a little closer to the limit: Alternating Series Estimation Theorem For a convergent alternating series, the truncation error is less than the first missing term[, and is the same sign as that term]. A quick review of an error bound formula that we already know so we can see how it fits in with LaGrange… How far off is this from the actual sum? This is a good tool to remember, because it is easier than the LaGrange Error Bound.

Alternating Series Estimation Theorem Since each term of a convergent alternating series moves the partial sum a little closer to the limit: Alternating Series Estimation Theorem For a convergent alternating series, the truncation error is less than the first missing term[, and is the same sign as that term]. One more thing about this theorem: Are there any Taylor or MacClaurin Series for which we can use this instead of the LaGrange Error Bound? Think about it… sin x and cos x…but why? Think about it…

Recall that the Taylor Series for ln x centered at x = 1 is given by… Find the maximum error bound for each approximation. Because the series is alternating, we can start with… Wait! How is the actual error bigger than the error bound for ln 0.5?

Wait! How is the actual error bigger than the error bound for ln 0.5? Plugging in ½ for x makes each term of the series positive and therefore it is no longer an alternating series. So we need to use the La Grange Error Bound. The third derivative of ln x at x = c What value of c will give us the maximum error? Normally, we wouldn’t care about the actual value of c but in this case, we can graph 2c–3 and use the graph to find where the maximum value is.

The third derivative of ln x at x = c The question is what value of c between x and a will give us the maximum error? So we are looking for a number for c between 0.5 and 1. Let’s look at a graph of 2x–3. Since the function is decreasing between 0.5 and 1, it’s maximum is at 0.5 and therefore… c = 0.5

Wait! How is the actual error bigger than the error bound for ln 0.5? Which is larger than the actual error! Wait! How is the actual error bigger than the error bound for ln 0.5? Plugging in ½ for x makes each term of the series positive and therefore it is no longer an alternating series. So we need to use the La Grange Error Bound. The third derivative of ln x at x = c What value of c will give us the maximum error? Since we’ve established that c = 0.5, we now know that…

Find the Lagrange Error Bound when is used to approximate and . ex. 5: On the interval , decreases, so its maximum value occurs at the left end-point. Remainder after 2nd order term

p Find the Lagrange Error Bound when is used to approximate and . ex. 5: Remainder Estimation Theorem On the interval , decreases, so its maximum value occurs at the left end-point. error Error is less than error bound. Lagrange Error Bound p