examples on how to find Taylor series from other known series


In this section we present numerous examples that provide a number of useful procedures to find new Taylor seriesMathworldPlanetmath from Taylor series that we already know. However, we are only worried about “computing” and we don’t worry (for now) about the convergence of the series we find.

We know “by heart” the following series:

ex = 1+x+x22+x33!+=n=0xnn!
cosx = 1-x22!+x44!-x66!+=n=0(-1)nx2n(2n)!
sinx = x-x33!+x55!-x77!+=n=0(-1)nx2n+1(2n+1)!
(1+x)p = 1+px+p(p-1)2!x2+p(p-1)(p-2)3!x3++p(p-1)(p-(n-1))n!xn+
ln(1+x) = x-x22+x33-x44+=n=1(-1)n+1xnn
Remark 1.

Recall that the first three have radius of convergenceMathworldPlanetmath R= but for the last two R=1.

Example 1.

Find the Taylor series about x=0 for sin(x2). If we try to take derivatives then we soon realize that consecutive derivatives get extremely hard to compute. However, one can do a simple trick. Since we know the Taylor series for sin(x) we can evaluate it at x2:

sinx=x-x33!+x55!-x77!+=n=0(-1)nx2n+1(2n+1)!
sin(x2)=(x2)-(x2)33!+(x2)55!-=x2-x63!+x105!-

One can also use the Σ notation:

sin(x2)=n=0(-1)n(x2)2n+1(2n+1)!=n=0(-1)nx2(2n+1)(2n+1)!.
Example 2.

Find the Taylor series about x=0 for e-x2 (this is a very important function, for example in probability theory). Again, we use the simple Taylor series of ex:

ex=1+x+x22+x33!+=n=0xnn!
e-x2=1+(-x2)+(-x2)22+(-x2)33!+=1-x2+x42-x63!+

Using the sigma notation we obtain:

e-x2=n=0(-x2)nn!=n=0(-1)nx2nn!.
Example 3.

Series can also be multiplied by x. For example, we find the Taylor series for xex:

xex=x(1+x+x22+x33!+)=x+x2+x32+x43!+)

or

xex=x(n=0xnn!)=n=0xn+1n!
Example 4.

Series can also be divided by x provided that the result has only non-negative exponentsPlanetmathPlanetmath. For example, we find the Taylor series for ln(1+x)x:

ln(1+x)x=x-x22+x33-x44+x=1-x2+x23-x34+

or

ln(1+x)x=n=1(-1)n+1xnnx=n=1(-1)n+1xn-1n
Example 5.

As well, we can multiply two Taylor series (term by term). Suppose we want to find the Taylor polynomial of degree 3 about x=0 of excosx. Then we can multiply the respective Taylor polynomials of degree 3 of ex and cosx and disregard any term higher than 3:

(1+x+x22+x36)(1-x22)
=(1-x22)+(x-x32)+(x22)+(x36)+

Since every other term in the product is of degree higher than 3 we disregard them. Thus:

T3(x)=(1-x22)+(x-x32)+(x22)+(x36)=1+x-x32+x36=1+x-x33.
Example 6.

Find the first three terms of the Taylor series for 1+2sinx. Since 1+x=(1+x)1/2 we know that:

1+x=1+x2-x28+

Thus:

1+2sinx=1+2sinx2-(2sinx)28+

Moreover, sinx=x-x33!+:

1+2sinx=1+2(x-x33!+)2-(2(x-x33!+))28+

By disregarding other than the first term in sinx we obtain the first three terms of the series are:

1+2sinx=1+2x2-(2x)28+=1+x-x22+
Example 7 (Differentiation).

Notice that we can deduce the series of sinx from the series for cosx by differentiating. Indeed ddxcosx=-sinx and we differentiate (term by term) the Taylor series of cosx we obtain the Taylor series of sinx (DO IT!).

Another example. Let us deduce the Taylor series of 1(1-x)2. Notice that ddx1(1-x)=1(1-x)2. Since:

11-x=1+x+x2+x3+

by deriving both sides we obtain:

1(1-x)2=0+1+2x+3x2+=n=0(n+1)xn

and if we derive again we obtain:

2(1-x)3=2+6x+12x2

Thus,

1(1-x)3=1+3x+6x2+10x3+=n=0(n+2)(n+1)2xn.
Example 8 (Integration).

Finally, we will deduce the Taylor series for arctanx using integration (term by term). Notice that:

11+x2𝑑x=arctanx+C

Moreover, since 1/(1-x)=1+x+x2+x3+ by substituting x by -x2 we obtain:

11+x2=1-x2+x4-x6+=n=0(-1)nx2n

Thus, the Taylor series of arctanx can be constructed integrating the previous one:

arctanx=11+x2𝑑x=(1-x2+x4-x6+)𝑑x=x-x33+x55-x77+

In Σ notation:

arctanx=11+x2𝑑x=n=0(-1)nx2ndx=n=0(-1)nx2n𝑑x=n=0(-1)nx2n+12n+1
Example 9.

As an application of the previous example, we compute π. Indeed:

arctanx=n=0(-1)nx2n+12n+1

converges between -1x1 and in particular

arctan1=π4

by the definition of tanx and arctanx. Thus:

π4=n=0(-1)n12n+12n+1=n=0(-1)n2n+1

and

π=4(n=0(-1)n2n+1)=4(1-13+15-17+19-)

For example, if one adds up to the 1/9 term, one obtains the approximation π3.33. Unfortunately, the convergence is very slow. If you want to have about m correct digits then you have to add about 10m/2 terms. For example, if you add 103/2=500 terms we get 3.143588.

Title examples on how to find Taylor series from other known series
Canonical name ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries
Date of creation 2013-03-22 15:05:47
Last modified on 2013-03-22 15:05:47
Owner alozano (2414)
Last modified by alozano (2414)
Numerical id 8
Author alozano (2414)
Entry type Example
Classification msc 41A58
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