Question

using Maclaurin series create a polynimial replacement for e^(2x) times sin(2x). four terms needed

Answer 1

You have to just calculate the first,second and third derivative of this function, find their value for x = 0 and then use the MacLaurin polynomial approximation.\[f(x)\approx f(0)+xf'(0)/2! + x^2f''(0)/3! + x^3f'''(0)/4!\]

Answer 2

sorry it's \[f(x) \approx f(0)/0! + xf'(0)/1! + x^2f''(0)/2! + x^3f'''(0)/3! \] the polynomial approximation of f(x) around zero.

Answer 3

This power-series expression is only convergent in a circle centred at some point P in the complex plane; I don't know why one talks of an approximation, there is no such qualification required if you see the proof. Also the Taylor series expansion could be summarized with sigma notation. This would require a generalisation of the given function's derivative (assuming quite validly infinite differentiability); one that's provided by the Cauchy formula. I guess the more 'simpler' approach would (as suggested by the alex..) be to find explicitly the first three derivatives: an application of the chain and product rules for differentiation.
I hope I've been of some help.

Answer 4

The Maclaurin series for each is given by \[e ^{2x}=\sum_{n=0}^{\infty}((2x) ^{n}/n!)=1+2x+2x ^{2}+(4/3)x ^{3}+...\]

\[\sin(2x)=\sum_{n=0}^{\infty}((2x)^{2n+1}/(2n+1)!)=(2x)+(2x)^{3}/3! +...\]

To obtain the first four nonzero terms, just multiply these. Add the (2x) term through the x^4 terms. This will be your fourth degree Maclaurin approximation.