Manipulating Taylor Series
To find the Taylor series of complicated functions, it's often easiest to manipulate the Taylor series of simpler functions.
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Manipulating Taylor Series. In Justin Math: Calculus. https://justinmath.com/manipulating-taylor-series/
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To find the Taylor series of complicated functions, itβs often easiest to manipulate the Taylor series of simpler functions, such as those given below.
Multiplying by a Constant
For example, to compute the Taylor series of xex centered at x=0, we can take the elementary Taylor series ex=ββn=0xnn! and multiply it by x.
Though not strictly necessary, we can make the exponent on the match the index of summation by changing the index of summation to k=n+1.
In this case, since we are multiplying the series by a constant, the interval of convergence of the series will stay the same: ββ<x<β.
This is because a convergent series has a finite sum, and multiplying by a constant cannot cause a finite number to become infinite; whereas a divergent series has an infinite sum, and multiplying by a constant cannot cause an infinite number to become finite.
Multiplying Two Series
Similarly, to compute the Taylor series of ex1βx around x=0, we can multiply the two elementary Taylor series ex=ββn=0xnn! and 11βx=ββn=0xnn!.
Defining a new index of summation m=n+k, we can write the series in order of increasing powers of x.
The interval of convergence of a product of series is at least the intersection of the seriesβ individual intervals of convergence.
Here, recalling that the interval of convergence of the Taylor series of ex is (ββ,β) and the interval of convergence of the Taylor series of 11βx is (ββ,β), we determine that the interval of convergence for the Taylor series of ex1βx must be at least the intersection (ββ,β)β©(β1,1)=(β1,1).
If we go through the trouble of performing a test of convergence, itβs possible that we might find a larger interval of convergence β but just based on the intervals of convergence of the two series being multiplied, we can say with certainty that the product converges for at least β1<x<1, without needing to perform any tests of convergence.
Adding Two Series
Sometimes, we can take advantage of the fact that itβs easier to add or subtract series than to multiply series.
For example, to find the Taylor series of ex(1βex) around x=0, one option is to multiply the Taylor series of ex and 1βex.
However, an easier route is to simplify the expression to exβe2x, and then subtract the Taylor series of e2x from the Taylor series of ex. To compute the Taylor series of e2x, we can substitute 2x for x in the Taylor series of ex.
Then, we can proceed with subtracting the Taylor series.
Again, the interval of convergence of a sum or difference of series is at least the intersection of the seriesβ individual intervals of convergence.
The series for ex converges for ββ<x<β, so the series for e2x converges for ββ<2x<β, which simplifies to ββ<x<β. The intersection is given by (ββ,β)β©(ββ,β)=(ββ,β), so the interval of convergence of the series for exβe2x is at least ββ<x<β.
Note that this interval contains all real numbers, so the interval canβt get any bigger. Thus, the interval of convergence of the series for exβe2x is ββ<x<β.
Using Differentiation and Integration
We can also use differentiation and integration to simplify the process of finding Taylor series. For example, to find the Taylor series of sin2x, one option is to multiply the series of sinx by itself β but an easier option is to differentiate to yield a simpler result, then find the Taylor series of the simpler result, and then integrate the Taylor series to get back to the original function.
To solve for the constant of integration, we can substitute x=0.
Thus, we have
Though not strictly necessary, we can clean up the series a bit by changing the index of summation to k=n+1.
Neither differentiating nor integrating a Taylor series changes its interval of convergence, so the interval of convergence of the series for sin2x is the same as the interval of convergence of the series for sin2x, which is ββ<x<β.
Substitution
In the previous examples, we computed the series for sin2x and e2x by substituting 2x for x in the series for sinx and ex. We can extend this idea to more clever substitutions.
For example, to compute the series of the function 12+x5, we can substitute βx52 for x in the elementary series 11βx=ββn=0xn.
After substitution, the interval of convergence becomes β1<βx52<1, which simplifies to β5β2<x<5β2.
Exercises
Compute the Taylor series for the following functions, centered at x=0. (You can view the solution by clicking on the problem.)
1)f(x)=xln(1+x2)
Solution:
ββn=1(β1)n+1nx2n+1β1<xβ€1
2)f(x)=1x2arctan(2x3)
Solution:
ββn=0(β1)n22n+12n+1x6n+1β3β12β€xβ€3β12
3)f(x)=11+2x
Solution:
ββn=0(β2)nxnβ12<x<12
4)f(x)=cos(βΟx)
Solution:
ββn=0(β1)nΟn(2n)!xn0β€xβ€β
5)f(x)=ln(e+x)
Solution:
1+ββn=1(β1)n+1nenxnβe<xβ€e
6)f(x)=(cosx+sinx)(cosxβsinx)
Solution:
ββn=0(β1)n22n(2n)!x2nββ<x<β
7)f(x)=sin2(3x)
Solution:
ββn=1(β1)n+132n22nβ1(2n)!x2nββ<x<β
8)f(x)=cos2(Οx)
Solution:
1+ββn=1(β1)nΟ2n22nβ1(2n)!x2nββ<x<β
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Manipulating Taylor Series. In Justin Math: Calculus. https://justinmath.com/manipulating-taylor-series/
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