Asked by KIKSY
Thankyou very much!
I'm having trouble with this question about taylor series.
how would i go about doing the following questoin?
thank you!
find the maclaurin series for the function
sin(x)-x+((x^3)/3!)
how do i do this?
thanks aagain
I'm having trouble with this question about taylor series.
how would i go about doing the following questoin?
thank you!
find the maclaurin series for the function
sin(x)-x+((x^3)/3!)
how do i do this?
thanks aagain
Answers
Answered by
MathMate
A Maclaurin series expansion is a power series expansion about 0.
Thus:
f(x)=f(0)+f'(0)*x + f"(0)*x^2/2! + f<sup>(3)</sup>(0)x^3/3! + ....
for sin(x),
all the even ordered derivatives (f(0),f"(0)...) are sin(0), so the terms left are:
sin(x)=cos(0)x-cos(0)*x^3/3!+cos(0)x^5/5!-...
=x-x^3/3!+x^5/5!-x^7/7!+....
So the given expression removes the first two terms of the expansion:
sin(x)-x+((x^3)/3!)
=x-x^3/3!+x^5/5!-x^7/7!+.... -x+x^3/3!
=x^5/5!-x^7/7!+....
Thus:
f(x)=f(0)+f'(0)*x + f"(0)*x^2/2! + f<sup>(3)</sup>(0)x^3/3! + ....
for sin(x),
all the even ordered derivatives (f(0),f"(0)...) are sin(0), so the terms left are:
sin(x)=cos(0)x-cos(0)*x^3/3!+cos(0)x^5/5!-...
=x-x^3/3!+x^5/5!-x^7/7!+....
So the given expression removes the first two terms of the expansion:
sin(x)-x+((x^3)/3!)
=x-x^3/3!+x^5/5!-x^7/7!+.... -x+x^3/3!
=x^5/5!-x^7/7!+....
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