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taylor series

i have

cos(z-z^2)

which i need to apply taylor series up to and including z^5.

i'm only on the second derivative;

f''(a) =

(-4z^2 + 4z - 1)cos(z^2 - z)-2sin(z^2 - z)

and it'll only get longer when i'm on the 5th derivative. surely i need to do something to the

cos(z-z^2)

before differentiating so many times? (unless i'm doing the derivatives completely wrong?)

are there any rules that could help me? i've looked at trig rules which doesn't help i don't think! thanks

Reply 1

Phugoid

In the Taylor series

f(a)

is the function at some constant point, a. You choose what it is. If you choose it to be zero, it's a Maclaurin series.

So, for example, for a = 0, your 2nd derivative simplifies largely:

f''(0) = (-4(0)^2 + 4(0) - 1) \cos(0^2 - 0) - 2 \sin(0^2 - 0) = (-1)\cos(0) - 2 \sin(0)

= -1

And hence, the third term of your series would be:

\frac{f''(a)}{2!} (z-a)^2 = \frac{f''(0)}{2} (z)^2 = \frac{(-1)}{2} z^2 = -\frac{1}{2}z^2

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