# How do you find #f^8(0)# where #f(x)=cos(x^2)#?

# f^((8))(0) = 1680 #

It doesn't matter how we get the power series—it could come from the Binomial Theorem, a Taylor series, or a Maclaurin series—we know that the power series for a function is unique.

By definition:

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To find ( f^8(0) ), where ( f(x) = \cos(x^2) ), we need to repeatedly apply the function ( f(x) ) eight times to the initial value ( x = 0 ).

Starting with ( f(x) = \cos(x^2) ), we find:

[ f^2(x) = f(f(x)) = \cos\left((\cos(x^2))^2\right) ]

[ f^3(x) = f(f^2(x)) = \cos\left(\cos\left((\cos(x^2))^2\right)^2\right) ]

[ \vdots ]

[ f^8(x) = f(f^7(x)) = \cos\left(\cos\left(\cos\left(\cos\left(\cos\left(\cos\left(\cos\left(\cos(x^2))^2\right)\right)^2\right)\right)^2\right)\right) ]

Then, to find ( f^8(0) ), we substitute ( x = 0 ) into ( f^8(x) ) as follows:

[ f^8(0) = \cos\left(\cos\left(\cos\left(\cos\left(\cos\left(\cos\left(\cos\left(\cos(0)^2\right)\right)^2\right)\right)^2\right)\right)^2\right) ]

Evaluating this expression will give us the value of ( f^8(0) ), which represents the result of applying the function ( f(x) = \cos(x^2) ) eight times to ( x = 0 ).

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