Question

Give the mclaurin series for f ( x ) = cos ( x 2 ) .

Answer 1

Sure, I can help you with that.

The Maclaurin series for f(x) = cos(x^2) is:

f(x) = 1 - x^4/2! + x^8/4! - x^12/6! + ...

This can be derived from the Maclaurin series of cos(x) using the chain rule and the fact that the derivative of x^2 is 2x.

Answer 2

The Maclaurin series for f(x) = cos(x^2) is found by replacing every x in the Maclaurin series for cos(x) with x^2, yielding an infinite series with even powers of x and alternating signs.

To find the Maclaurin series for f(x) = cos(x2), we need to express the function as a power series expansion around x=0. The Maclaurin series of cos(x) is given by:

cos(x) = 1 - x2/2! + x4/4! - x6/6! + ...

In the case of cos(x2), we just replace every x in the series for cos(x) with x2, to get:

cos(x2) = 1 - (x2)2/2! + (x2)4/4! - (x2)6/6! + ...

This simplifies to:

cos(x2) = 1 - x4/2! + x8/4! - x12/6! + ...

Thus, the Maclaurin series for f(x) = cos(x2) is an infinite series with terms of even powers of x, starting from 0, with alternating signs.