Math Calculus I Part 8 Power series, Taylor series

Series First.. a review of what we have done so far: 1. We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series. 2. We developed tests for convergence of series of constants. 3. We considered power series, derived formulas and other tricks for finding them, and know them for a few functions. 4. We used the ratio tests to determine intervals on which power series converge, and use the other tests to check convergence at the endpoints of the intervals.

Geometric series a/(1-r) = a + ar + ar 2 + ar provided |r|< 1. We often use partial fractions to detect telescoping series, for which we can calculate explicitly the partial sums s n

Tests for convergence for series of constants Fundamental divergence test (nth term must go to zero for convergence to be possible) Integral test Comparison and limit comparison tests Ratio test Root test Alternating series test

Power series f(x) = a 0 + a 1 x + a 2 x 2 + a 3 x where a n = f (n) (0)/n! We know the series for e x, sin(x), cos(x), 1/(1-x), and a few related functions.

Convergence of power series Before we get too excited about finding series, let's make sure that, at the very least, the series converge. Next week, we'll deal with the question of whether they converge to the function we expect. But for now, we'll assume that if they converge, they converge to the function they "came from". (Strictly speaking, this is not always true -- but it is true for a large class of functions, which includes nearly all the ones encountered in basic science and mathematics. This fact was not fully appreciated until the early part of the twentieth century.) Fortunately, most of the question of whether power series converge is answered fairly directly by the ratio test.

Recall that... for a series of constants, we have that the series converges (absolutely) if the the limit of the absolute value of is less than one, diverges if the limit is greater than one, and the test is indeterminate if the limit equals one. To use the ratio test on power series, just leave the x there and calculate the limit for each value of x. This will give an inequality that x must satisfy in order for the series to converge.

For the series for the exponential function...

Your turn... Calculate the series for the function sin(x) and determine for which x the series converges.

Here’s a more interesting example

What remains...

Final Conclusion

OK, your turn... A. -1 < x < 1 B. -2 < x < 2 C. 1/2 < x < 1/2 D. -2 < x < 2 E. -1/2 < x < 1/2

One more... A. -1 < x < 1 B. -1 < x < 1 C. 1 < x < 1 D. -1 < x < 1 E. 0 < x < 1

From these examples,...it should be apparent that power series converge for values of x in an interval that is centered at zero, i.e., an interval of the form [-a, a], (-a, a], [-a, a) or (-a, a) (where a might be either zero or infinity). The interval is called the interval of convergence and the number a is called the radius of convergence.

Let’s go back To finding series of functions:

The other way

Try this... Take the derivative of the series for sin(x) to get

Integrate both sides of the geometric series from 0 to x to get:

Negate both sides and replace x by (-x) everywhere to get:

Start from the geometric series again... And substitute x for x everywhere it appears to get 2

A challenge to think about... How to get the other one from previously

1. Limits: Series give a good idea of the behavior of functions in the neighborhood of 0: We know for other reasons that We could do this by series: Application of Series

This can be used on complicated limits... Calculate the limit: A. 0 B. 1/6 C. 1 D. 1/12 E. does not exist

Application of series (continued) 2. Approximate evaluation of integrals: Many integrals that cannot be evaluated in closed form (i.e., for which no elementary anti-derivative exists) can be approximated using series (and we can even estimate how far off the approximations are). Example: Calculate to the nearest

We begin by...

According to Maple... The last series is an alternating series with decreasing terms. We need to find the first one that is less than to ensure that the error will be less than According to Maple: evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5))); evalf(1/(13*factorial(6))); , ,

Keep going... So it's enough to go out to the 5! term. We do this as follows: Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1) *factorial(n)),n=0..5); evalf(%); =

and finally... So we get that to the nearest thousandth. Again, according to Maple, the actual answer (to 10 places) isevalf(int(exp(-x^2),x=0..1));

Try this... Sum the first four nonzero terms to approximate A B C D E