How do you use the Maclaurin series for #f(x) = x^3 sinx^2#?

1 Answer
Aug 15, 2017

# f(x) = x^5 - (x^9)/(6) + (x^13)/(120) - (x^17)/(5040) + ... #

Or,

# f(x) = sum_(k=0)^oo (-1)^k \ (x^(4k+5))/((2k+1)!) #

Explanation:

We have:

# f(x) = x^3sin(x^2) #

I presume that you seek to derive a Maclaurin series for #f(x)# using the known series for #sinx#

Starting with the Maclaurin Series for #sinx#,

# sinx = x - x^3/(3!) + x^5/(5!) - x^7/(7!) + x^9/(9!) -... \ \ x in RR #

Then the series for #sin(x^2)# is:

# sin(x^2) = (x^2) - (x^2)^3/(3!) + (x^2)^5/(5!) - (x^2)^7/(7!) + ...#
# " " = x^2 - (x^6)/(6) + (x^10)/(120) - (x^14)/(5040) + ...#

Or in sigma notation:

# sin(x^2) = sum_(k=0)^oo (-1)^k \ (x^(4k+2))/((2k+1)!) #

And so if we multiply by #x^3# we get the desired series:

# f(x) = x^3sinx^2 #
# " " = x^3{ x^2 - (x^6)/(6) + (x^10)/(120) - (x^14)/(5040) + ... } #
# " " = (x^3)x^2 - (x^3)(x^6)/(6) + (x^3)(x^10)/(120) - (x^3)(x^14)/(5040) + ... #
# " " = x^5 - (x^9)/(6) + (x^13)/(120) - (x^17)/(5040) + ... #

Or,

# f(x) = sum_(k=0)^oo (-1)^k \ (x^(4k+2+3))/((2k+1)!) #
# " " = sum_(k=0)^oo (-1)^k \ (x^(4k+5))/((2k+1)!) #