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How do you find the Taylor series of #f(x)=e^x# ?

Answer 1

Aurora Alvarado

Taylor series at #x=0# (also called Maclaurin series) for #f(x)# is

#f(x)=sum_{n=0}^infty{f^{(n)}(0)}/{n!}x^n#.

Since if #f(x)=e^x#, then

#f(x)=f'(x)=f''(x)=cdots=f^{(n)}(x)=e^x#,

so,

#f(0)=f'(0)=f''(0)=cdots=f^{(n)}(0)=e^0=1#

Hence, the Maclaurin series is

#f(x)=sum_{n=0}^infty 1/{n!}x^n=sum_{n=0}^infty x^n/{n!}#

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Answer 2

Christopher Allers

To find the Taylor series of ( f(x) = e^x ), you can use the formula for the Taylor series expansion of a function about a point ( a ) as follows:

[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \ldots ]

For the function ( f(x) = e^x ), you can choose ( a = 0 ) (since it simplifies calculations), and then calculate the derivatives of ( e^x ) at ( x = 0 ) to find the coefficients. The derivatives of ( e^x ) are simply ( e^x ) itself. Therefore, at ( x = 0 ), all derivatives equal 1.

So, the Taylor series expansion of ( e^x ) about ( x = 0 ) (also known as the Maclaurin series) is:

[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots ]

Or, more succinctly:

[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.