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

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Homework Help Overview

The discussion revolves around finding the Taylor series for the function f(x) = e^(2x) using Taylor's formula. Participants are exploring the application of the Taylor series expansion and the correct substitution of variables.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to express e^(2x) using a series expansion, while others question the correctness of their substitutions and the adherence to using Taylor's formula specifically.

Discussion Status

There is ongoing exploration of different approaches to derive the series. Some participants have provided guidance on the steps needed to apply Taylor's formula correctly, while others are clarifying the importance of proper substitutions in the series expansion.

Contextual Notes

Participants note that the original poster is new to series and may require additional foundational understanding. There is an emphasis on following the specific instructions of the homework prompt regarding the use of Taylor's formula.

dejet
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Homework Statement


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

Homework Equations


[tex]\sum \frac{f^{n}(a)}{n!} (x-a)^{n}[/tex]

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

Homework Statement


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

Homework Equations


[tex]\sum \frac{f^{n}(a)}{n!} (x-a)^{n}[/tex]

any help as to where i start would be great. new to series...
You missed out one important section:
Homework Help Template said:

The Attempt at a Solution

Even if you are new to series, you must have some idea of how to start.
 
e^2x= 1+2x+ [tex]\frac{2x^{2}}{2!} +\frac{2x^{3}}{3!} +\frac{2x^{4}}{4!}+...[/tex]

there is more to it i think.
 
dejet said:
e^2x= 1+2x+ [tex]\frac{2x^{2}}{2!} +\frac{2x^{3}}{3!} +\frac{2x^{4}}{4!}+...[/tex]

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?
 
dejet said:
e^2x= 1+2x+ [tex]\frac{2x^{2}}{2!} +\frac{2x^{3}}{3!} +\frac{2x^{4}}{4!}+...[/tex]

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
 
rootX said:
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 ex 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.
 
so is this right?

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

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