a. Another Maclaurin series we examined is for sin x: a. Find the series for frac{sin(x^{2})}{x} . Express the result in the summation notation, not j

Ayaana Buck 2021-01-24 Answered
a. Another Maclaurin series we examined is for sin x: a. Find the series for sin(x2)x . Express the result in the summation notation, not just as a partial listing of the series terms.
b. Another Maclaurin series we examined is for sin x:
d. Find the series for integral sin(x2)x
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Expert Answer

Benedict
Answered 2021-01-25 Author has 108 answers

The maclurin series is the Taylor series expansion about point a = 0.
Find the value for f(x) at a = 0.
f(x)=sinx
f(a)=sina
f(0)=sin0
So, f(0)=0
Consider the formula for the Taylor series.
Find the value for the derivatives required to find the Taylors series.
f(x)=n=0fn(a)n!(xa)n
=f(a)0!(xa)0+f1(a)1!(xa)1+f2(a)2!(xa)2+...
=0+f1(a)1!(xa)1+f2(a)2!(xa)2+...
=f1(a)1!(xa)1+f2(a)2!(xa)2+...
Further, simplify.
Now, f(x)=ddx(sinx)
So, f1(x)=cosx
f1(0)=cos0
=1
f(x)=ddx(cosx)
So, f2(x)=sinx
f2(0)=sinx
=0
f(x)=ddx(sinx)
f3(x)=cosx
f3(0)=cos0
=1
Then, find the Taylor series centered at a = 0.
Thus, find the Maclurins series.
At a=0
sin(x2)x=x2xx66x+x10120x...
x16x5+1120x9...
x1!13!x5+15!x9...
n=0(1)nx4n+1(2n+1)!
Hence, sin(x2)x=n=0(1)nx4n+1(2n+1)!

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Jeffrey Jordon
Answered 2021-12-16 Author has 2064 answers

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