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hey guys ive attached a jpg with the question:
f(x)= sin3x
f'(x)= 3cos3x
f''(x) = -3.3sin3x
f'''(x) = 3.3.3cos3x
I took a=0 and began to input it into the derivations:
f(0) = 0
f'(0) = 3
f''(0) = 0
f''' (0) = 27
but when i put it into the maclaurin series i get 3x + 27x^3/3!
shudnt i get the original formula again? what have i done wrong?
Your question does not make sense. You are making a trigonometric function into an infinite polynomial. Why would you expect any visual similarity?
Besides the sign error on the 3rd derivative, you have it.
so the answer i have given is the answer that the question expects?
I would suggest an expansion about $\displaystyle x = 0$ of
$\displaystyle f(x) = f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+\frac{f'''(0)x^3} {3!}+\dots =$ $\displaystyle \sin(3\times 0)+3\cos(3\times 0)x-\frac{9\sin(3\times 0)x^2}{2!}-\frac{27\cos(3\times 0)x^3}{3!}+\dots$
Not quite. You need to do two more things:
1) Fix the error in the 3rd derivative.
2) Find the 5th derivative so that you can add the 3rd non-zero term. You've only two.
