Hi, could someone please help me with this question? Thanks
Calculate the Maclaurin series of f(x)=4x^3*sin(x^4). Explicitly use the Maclaurin series for sine. Using this series for f(x), verify that the integral of 4x^3*sin(x^4)dx=-cos(x^4)+C.
we know that
sin(u) = u - u^3/3! + u^5/5! - ...
so, sin(x^4) = x^4 - x^12/3! + x^20/5! - ...
so, f(x) = 4x^7 - 4x^15/3! + 4x^23/5! - ...
it is certainly obvious that d/dx -cos(x^4) = 4x^3 sin(x^4). Using series,
-cos(x^4) = -1 + x^8/2! - x^16/4! + x^24/6! + ...
so, its derivative is
8x^7/2! - 16x^15/4! + 24x^23/6! - ...
= 4x^7 - 4x^15/3! + 4x^23/5! - ...
and they are the same
Sure! To find the Maclaurin series of a function, we need to use the formula for the Maclaurin series of the individual terms and then multiply them together.
First, let's find the Maclaurin series of sin(x). The Maclaurin series for sin(x) is:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
Now, we need to find the Maclaurin series for the term x^3 in f(x)=4x^3*sin(x^4). To do this, we can use the formula for the Maclaurin series of x^n, which is simply the term itself:
x^3 = x^3
Putting it all together, we can write the Maclaurin series for f(x) as:
f(x) = (4x^3)*(x^3 - (x^3)^3/3! + (x^3)^5/5! - (x^3)^7/7! + ...)
Simplifying this expression, we get:
f(x) = 4x^3*(x^3 - (x^9/3!) + (x^15/5!) - (x^21/7!) + ...)
Now, let's distribute the 4x^3 term into each term inside the parentheses:
f(x) = 4x^6 - 4x^12/3! + 4x^18/5! - 4x^24/7! + ...
To verify the integral of f(x) = 4x^3*sin(x^4)dx = -cos(x^4) + C, we need to find the derivative of -cos(x^4) and see if it matches f(x).
The derivative of -cos(x^4) with respect to x is:
d/dx (-cos(x^4)) = -4x^3*sin(x^4)
As we can see, the derivative matches f(x), so we can conclude that the integral of 4x^3*sin(x^4)dx is indeed equal to -cos(x^4) + C.
Hope this explanation helps clarify the process of finding the Maclaurin series and verifying the integral!