find the maclarin series expansion of e^-2x
Answers
The Maclaurin series of f(x)=e−2x is
f(x)=1+(−2x)+(−2x)22!+(−2x)33!+...
First Solution Method: The Maclaurin Series of y=ez is
y=1+z+z22!+z33!+z44!+...
Let z=−2x.
Then f(x)=e−2x=ez and f(x) has the same Maclaurin series as the one above except we set z=−2x and get
f(x)=1+(−2x)+(−2x)22!+(−2x)33!+...
I used the well known Maclaurin series for y=ez to get the answer. If this series has not been discussed in class, you should use the general definition of a Maclaurin series to get the answer.
The Maclaurin series of f(x) is
f(x)=f(x=0) +f'(x=0)1!x
+f''(x=0)2!x2
+f'''(x=0)3!x3+...
^Couldn't put all terms on same line, sorry for poor formatting.
Anyways, the first term is f(x=0). Here, f(x=0)=e−2(0)=1.
The second term is f'(x=0)1!x=−2e−2(0)1x=−2x
The third term is f''(x=0)2!x2=(−2)2e−2(0)2!x2=(−2x)22!
These are the same terms as in the Maclaurin series I wrote above.
By observing a pattern, the nth term of the series is (−2x)nn!
Using a summation sign, the Maclaurin series of f(x) can be written instead as
f(x)=Σn=∞n=0[(−2x)nn!]