Using Taylor Formula to Find Series of f(x)=e^{2x}

[dejet]

Homework Statement

using the Taylor Formula, find the series for the function f(x)=e^{2x}

Homework Equations

\sum \frac{f^{n}(a)}{n!} (x-a)^{n}

any help as to where i start would be great. new to series...

 

[Hootenanny]

Homework Statement

using the Taylor Formula, find the series for the function f(x)=e^{2x}

Homework Equations

\sum \frac{f^{n}(a)}{n!} (x-a)^{n}

any help as to where i start would be great. new to series...

You missed out one important section:

The Attempt at a Solution

Even if you are new to series, you must have some idea of how to start.

 

[dejet]

e^2x= 1+2x+ \frac{2x^{2}}{2!} +\frac{2x^{3}}{3!} +\frac{2x^{4}}{4!}+...

there is more to it i think.

 

[Hootenanny]

e^2x= 1+2x+ \frac{2x^{2}}{2!} +\frac{2x^{3}}{3!} +\frac{2x^{4}}{4!}+...

there is more to it i think.

No, that isn't correct. The question states that you must use Taylor's formula, have you tried using it?

 

[rootX]

e^2x= 1+2x+ \frac{2x^{2}}{2!} +\frac{2x^{3}}{3!} +\frac{2x^{4}}{4!}+...

there is more to it i think.

e^x = 1 + x + x^2/2!+ x^3/3! + ...

You substituted 2x for x

so .. x^2 should be (2x)^2 not 2x^2

 

[Hootenanny]

e^x = 1 + x + x^2/2!+ x^3/3! + ...

You substituted 2x for x

so .. x^2 should be (2x)^2 not 2x^2

To the OP: Although, substituting 2x for x in the Taylor series for e^x as rootX has done is a totally valid (and preferable) method, the question specifically states that Taylor's Formula should be used. Try using it.

 

[dejet]

so is this right?

2^(n-1) f(x)=1+2x+4x^2/2!+8x^3/3!+...+2^(n-1)/(n!)...

 

[rootX]

so is this right?

2^(n-1) f(x)=1+2x+4x^2/2!+8x^3/3!+...+2^(n-1)/(n!)...

no!

Steps:
1. Find f' , f'', f''', ... for f(x) = e^(2x)
2. plug values in your taylor formula

No straight jumping to the answer.