Power Series. We already know how to express one function as a series. Take a look at following equation: 1 1 r = r n. n=0

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1 Math -0 Calc Power, Taylor, and Maclaurin Series Survival Guide One of the harder concepts that we have to become comfortable with during this semester is that of sequences and series We are working with infinite sums of complicated functions and are answering questions about convergence There are a large number of functions that we can express as one of these series and we must start by looking at the one we go into this chapter already knowing Power Series We already know how to express one function as a series Take a look at following equation: r = r n We all know this as the geometric series and know that it converges for all r such that < r < Since the geometric series converges over this interval we may think about f(x) = x and conclude that this also converges to xn over < x < It is to our benefit, then, to start recognizing when an expression we wish to evaluate is already in this form or when it can be made into this form Let us look at the following few examples Example : f(x) = +x This function can be rewritten as ( x ) and thus we may substitute x for x in our original geometric series We therefore have +x = ( x ) = ( x ) n = ( ) n x n We are not quite done working with this We have to make sure the function we have placed into our original position for x is within the right interval for convergence We need x = x < This is true when x < The endpoints, and, make for the divergent sum ( )n, so our interval of convergence is < x < Example : g(x) = x 3x 3 This expression can be rewritten as ( x ) ( ) 3 x3 We re going to deal with the right hand side of this expression, 3 x3 Let s replace x in our original power series with 3 x3 ( 3 x3 ) = ( ) n 3 x3 = ( ) n 3 (x 3 ) n = ( ) n 3 x 3n

2 Remember that this equation only holds over the interval of convergence, so we need to now figure out what that interval is Keeping in mind that a n = ( 3 n ) x 3n, let us apply the ratio test ) n+ x 3(n+) ) x 3n+3 lim n a n+ a n = lim n ( 3 ( 3 ) n x 3n = lim n ( 3 x 3n = lim 3 n Note that 3 x3 < when x 3 < Thus the radius of convergence is 3 3 convergence is 3 < x < x3 = 3 3 x3 and the interval of But wait! We still need to check the endpoints Putting in 3 3 for x we get ( )n and putting in 3 for x we get 3 n both of which are divergent series Therefore our interval of convergence is 3 < x < We re not done, however! We still need to multiply our power series by x Doing this will give us our final result, over the interval of convergence: x 3x 3 = x Example 3: h(x) = ln ( + x ) ( ) n 3 x 3n = x 3 n x 3n n = 3 n x 3n+ n+ While h(x) does not immediately look like what we ve been working with, all hope is not lost Let us take the first derivative, h (x) = x Using the series from Example and our rules for +x multiplication in Example, we can say that over the correct interval of convergence h (x) = (x) +x = (x) ( ) n x n = ( ) n x n+ Try to remember that this is only the series for the derivative! We re not nearly done yet! We still need to integrate h (x) to get h(x) ( ) n x n+ = ( ) n x n+ = ( ) n x n+ = ( ) ( ) n xn+ + c n + We need to catch up and figure a few things out at this point First, let us calculate our value for c It s in our best interest to choose a value for x( that will be easy to take this infinite sum ) for, say x = 0 Then we have ln ( + 0 0n+ ) = ln = 0 = ( )n +c = 0+c and so c = 0 n+ We need to determine our radius and interval of convergence Observe that a n = ( )n x n+ lim a n+ n a n = lim ( ) n+ x (n+)+ (n + ) n ((n + ) + )( ) n x n+ = lim x (n + ) n n +4 = x n+

3 It is true that x < when x <, so our radius of convergence is We need to check our endpoints, and At x = we have ( ) n ( )n+ n + = ( ) 3n+ n + = ( ) n n + which is a convergent alternating series Likewise, at x = we have ( ) n ()n+ n + = ( ) n n + which is the same convergent alternating series Thus, our interval of convergence is x Taylor and Maclaurin Series Now we are pretty good at working with power series, however there are only a few types of functions that we can represent using the techniques that we ve developed Let s look at the following equation: e4 x There is no way that we ve seen to evaluate this integral If we could write the function e 4 x as a series (a polynomial in terms of x, moreover), then we could integrate it (like our power series) Likewise, we could come up with some rules to approximate these integrals within arbitrary degrees of accuracy Let us assume that a function f(x) which is infinitely differentiable (that is, the n th derivative exists for all n N) can be written as a polynomial in the following form: f(x) =c 0 + c (x a)+c (x a) + c 3 (x a) 3 + We wish to ask ourselves: What could possibly fill in the coefficients c i? If we evaluate f(x) at a then we have the following f(a) =c 0 + c (a a)+c (a a) + = c 0 + c 0+c 0 + = c 0 Therefore c 0 = f(a) Let us note this is equivalent to c 0 = f(a) 0! since 0! = by definition Then what are we to do about c? Taking the first derivative of f(x), f (x) =c +c (x a)+ 3c 3 (x a) + From this observe that when evaluate f (x) at a we get the following f (a) =c +c (a a)+3c 3 (a a) + = c +c 0+3c = c As before we note that c = f (a) = f (a) Well, f(a) can be written as f (0) (a), and f (a) can be! written as f () (a) This observation helps us obtain the following relation: c i = f (i) (a) Therefore i! we may write f(x) as follows: f (n) (a)(x a) n f(x) = 3

4 This series is known as the Taylor Series for f(x) centered at a We say that f(x) is centered at a because the series is exactly accurate at x = a, and the radius of convergence will be centered at a as well If the a we choose is 0, then we call this series the Maclaurin Series of f(x), a special enough case to earn a separate name It will benefit us to work with a function and compute the corresponding series Let us use f(x) =e x and center it around a = 0 This is the Maclaurin Series for e x Example 4: f(x) =e x, a =0 Our first goal is to try to find an explicit formula for f (n) (0) This is fairly easy in this case, since (e x ) = e x Thus f (n) (0) = e 0 =, and we find the Maclaurin Series (good over the interval of convergence) f(x) =e x x n = Our second goal is to try to find the radius of convergence, and the interval of convergence Let us apply the ratio test: lim x n+ (n)! n x n (n + )! = lim x n n + =0 Since our limit does not depend on our choice of x, we say that this series converges for all x and that the radius of convergence is Furthermore, the interval of convergence is < x < This is great news! Our Maclaurin Series for e x converges for all choices for x This means the infinite sum x n equals e x for all x Example 5: g(x) =e 4 x, a =0 We first want to try to get g(x) into a form we ve seen before, or to find an easy to use formula for g (n) (0) If we look at the first few derivatives it will become painfully obvious that an easy formula is not going to be possible Instead let s note that g(x) =e 4 e x If we put x in for x in f(x) from Example 4 we will get that e x = ( x ) n = ( ) n x n The interval of convergence for this sum is also < x <, so we can now obtain our final series g(x) =e 4 x = Example 6: h(x) = sin(x), centered at a = π ( ) n x n e 4 Let s list at the derivatives for sin(x), and evaluate them at π 4

5 h(x) sin(x) h(π) 0 h (x) cos(x) h (π) - h (x) sin(x) h (π) 0 h (x) cos(x) h (π) There is an obvious pattern to the derivatives, so finding an explicit formula for the n th derivative should be fairly easy This will be the route we ll go Hopefully this list leads us to realize that only the odd derivatives will give us any coefficients in our series A formula for the series can be obtained with this information, which will be true over the interval of convergence: ( ) n+ (x π) n+ h(x) = sin(x) = (n + )! Now we ll talk about convergence of the series, by again applying the ratio test: lim ( ) n+ (x π) (n+)+ (n + )! n ((n + ) + )!( ) n+ (x π) n+ = lim (x π) n (n + 3)(n + ) =0 Again, we have a series which converges for all x R The radius of convergence is, and the interval of convergence is < x < This is again excellent! We have an alternating series which converges for all x R This will become helpful when we talk about remainder, which is now Taylor and Maclaurin Series with Remainder There s one last thing we should talk about, that s the concept of remainder It s not always convenient for us to look at an infinite sum; sometimes we just want to get a number to work with A remainder is how much different the infinite sum is from a finite sum evaluated at a certain point Conceptually this is similar to how much is left when we divide a number into another The remainder in that process tells us how close to a multiple of our divisor the dividend is Let T i = i f (n) (a)(x a) n We will call this the i th partial Taylor polynomial It s a polynomial of degree at most i (since f (i) (a) could certainly be 0) Let s look at the first few partial Taylor polynomials T 0 is just f(a), which is certainly a good estimate for f(x) at a, but not likely to be a good estimate for the entire function Next, T = f(a)+f (a)(x a) This expression is equal to the tangent line at x = a This is also equal to f(a) at x = a, but even better, it is much closer to f(x) for values very near to a It should also be reminiscent of the mean value theorem The series continues, each partial polynomial giving us a better and better approximation over the interval of convergence for the entire series We want to be able to answer a few questions, however 5

6 Over our interval of convergence or a subinterval, how far off is a partial Taylor Polynomial from the value of the full Taylor Series? Over our interval of convergence, over a subinterval, or at a specific value near the center of our series, how many terms do we need to get within a specific range of the correct answer? These are both questions about error, or remainder If we are lucky enough to have an alternating series, then our questions are fairly easily answered If T i = i f n (a)(x a) n is an alternating series, then the error term is T i+ T i More explicitly, the error term is f i+ (a)(x a) i+ (i+)! Example 7: Remainder for h(x) = sin x Looking at Example 5 we have an alternating series for sin(x) Let s say we want to know the largest error possible over the interval π < x < 3π, when we are looking at the 5th degree partial Taylor polynomial The corresponding 5 th degree Taylor polynomial, centered at π, is (x π)+ (x π)3 (x π)5 3! 5! This actually is T, so the error term is T 3 T = (x π)7 7! We want to look at how this behaves over the entire interval π < x < 3π (x π) 7 7! is largest when x is furthest to the right from π, so the error is at most = ( π which is roughly ( 3π π)7 7! However, if we are not lucky enough to have an alternating series then we have to resort to other means The remainder on the i th partial Taylor polynomial is defined in the following way Over an interval I containing both a and the x we wish to evaluate, we can find an ˆx (called x-hat) between x and a that makes the following function exactly equal to the remainder for T i evaluated at x: R i = f (i+) (ˆx)(x a) i+ (i + )! Plainly, when we want to figure out f(x) for an x not equal to a, then we can estimate the value by using a partial Taylor polynomial, plus or minus the remainder We won t know exactly what the remainder will be (if we did we wouldn t have to estimate!), so we want to figure at worst how far off we ll be We do this by centering our series at an easy to evaluate value a near the x we want, then trying to bound R i using information about the interval (x, a) or (a, x) Example 8: Remainder for f(x) =e x Let s use Example 4 to take a look at the 3 rd partial Maclaurin polynomial for e x, over the interval <x< Further let s figure out how bad our 3 rd partial Maclaurin polynomial will )7 7! 6

7 estimate e x over the entire interval This polynomial, T 3, is written as T 3 = + x + x + x3 3! We know that there exists an ˆx (, ) so that R 3 = eˆx x 4 4! So to find the remainder we need to figure out how bad we can make R 3 using possible choices for x and ˆx in our interval We want to look at R 3 rather than R 3 because we want to look at the magnitude of our error Clearly R 3 is largest when x =ˆx =, making R 3 = So our remainder for anything over 4! this interval is at most for the 4 3rd Taylor polynomial for e x Example 9: Remainder for g(x) = ln(x) Let us try one problem which asks the second question from above If g(x) = ln(x), how many terms do we need to compute ln(3) to within 00 of the correct answer? It is beneficial for us to center our Taylor series for ln(x) around e so that our polynomial is fairly accurate at 3 Computing the n th derivative of ln(x): g (n) (x) = ( )n+ (n )! x n for n>0 Again, this is magnificent! We have another alternating series, so our error is given by R n = g (n+) (3)(3 e) n+ (n + )! = ( ) n+ (3 e) n+ 3 n+ (n + )! = (3 e) n+ 3 n+ (n + ) Remember that what we want is for R n < Simple trial and error will show that this 000 happens for the first time for n = where R n is approximately Conclusion Hopefully this guide has given you some insight into power, Taylor, and Maclaurin series as well as some strategies to approach problems you will see involving them Remember the following steps when it comes to a series problem: Determine if the expression you wish to evaluate is in the form f(x) If it is, then you g(x) can make a power series! The power series will be in the form f(x) (g(x))n and will converge when g(x) < If the expression isn t obviously in the form f(x), can it instead be integrated or differentiated to get into that form? If it can, take the appropriate integral or derivative to get g(x) it into that form Once in that form, find the power series for it Remember: this is the power series for the integral or the derivative! We have to undo the damage by differentiating or integrating 7

8 If we previously integrated, then we don t have to figure out the constant, since differentiating will cancel it out If however we are integrating, then we need to remember to evaluate the function to find the constant It is important to choose an x that will evaluate cleanly 3 If those steps fail, then we can almost always construct a Taylor or Maclaurin Series for the expression If our expression is h(x), try to find an explicit formula for h (n) (x) (the n th derivative of h(x)) It will allow you to work more abstractly with the sum, and to write out the infinite sum concretely Otherwise, you will be left to work with a partial Taylor polynomial 4 If the question is asking about remainder, then remember to check if this is an alternating series If it is then the remainder is given as being less than or equal to the absolute value of the next term evaluated at the specific x that the question asks about If it s not alternating, then remember we have to play with x and ˆx within our interval to figure out just how large our error can be Remember that x is what we wish to evaluate, that our series should be centered at a nearby a that s easy to evaluate, and that our ˆx lives between x and a Also remember to make the expressions in the numerator large, and the expressions in the denominator as close to 0 as possible to make the remainder large over our interval Some things to think about Calculate a power series for ln ( + x 4 ), its radius of convergence, and its interval of convergence Calculate a power series for ( x), its radius of convergence, and its interval of convergence 3 Using the Maclaurin series for e x, show that (e x ) = e x 4 Using the Maclaurin series for sin(x), show that (sin(x)) = cos(x) 5 Calculate sin(6) using a Taylor series centered around π Be accurate within 00 6 Over the interval 3 < x < 5, how poorly does the tangent line at x = 4 approximate 3+x? How about the quadratic Taylor polynomial? 7 Find an explicit formula for the n th derivative of x 8