Find the Maclaurin series expansion: F(x) = x*sin(x)?

Answer 1

f(0) = 0
f'(0) = sin(0) + 0*cos(0) = 0
f''(0) = cos(0) + cos(0) - 0*sin(0) = 2
f'''(0) = -sin(0) + -sin(0) - sin(0) - 0*cos(0) = 0
4th derviative = -cos(0) + -cos (0) - cos(0) - cos(0) + 0*sin(0) = -4
5th = sin(0) + sin(0) + sin(0) + sin(0) + sin(0) + 0*cos(0) = 0
6th = cos(0) + cos(0) ... = 6

P(x) = 0 + 0(x) + 2x^2/2 + 0x^3/3! - 4x^4/4! + 0x^5/5! + 6x^6/6!
= x^2 - x^4/3! + x^6/5! + ...

summation from n=1 to infinity of (-1)^(n-1) * x^(2n) / (2n-1)!

Answer 2

with a view to compute a maclaurin sequence, you may evaluate the expression and all of its derivatives at x=0. in case you are able to no longer do this at once, then limits will suffice -- compute the shrink as x->0 of x/sin(x). L'Hopital's rule: lim x->0 = one million/cos(x), which comes out to one million. you're actually not completed, you in ordinary terms have the 1st term in the sequence. you nonetheless could differentiate the expression (it is hassle-free sufficient, you employ the product rule with x*csc(x) and that provides you with what you like), and in case you are able to no longer at once evaluate later expressions, then use limits lower back. If any of the bounds do no longer exist, then the expression has no MacLaurin sequence. this time-honored you might have a MacLaurin sequence with a radius of convergence of pi, if I remember properly. the 2d does not have a MacLaurin sequence because of the fact the shrink as x->0 of two/sin(x) does not exist. even nevertheless, the 2d does have a Taylor sequence approximately 0

Answer 3

The general nth term of a Maclaurin series is given by:
f(x) = [d^n/dx^n f(0)][x^n]/n!
Let F(x) = f(x)
F'(x) = sin x + sin x(cos x) = sin x(1+cos x)
F"(x) = cos x(1+cosx) + sin x(0-sin x) = cos x(1+cos x) - sin^2 x
.
. all the way, as many terms as you like until you
. get to the nth term
.
F^n(x)

Answer 4

Since sin(x), x in radians, is x-x^3/3!+x^5/5!...
The expansion of xsin(x) is x^2-x^4/3!+ x^6/5!...