Maclaurin Polynomial

Jun 2008
58
0
Sanger, CA
Q: Find the Maclaurin polynomial of order 3 for the function f(x) = e^(-2x).

I know that the Maclaurin polynomial for e^x = 1 + x + x^2/2! + x^3/3! + ...

So my question is if I would have to plug in (-2x) in the Maclaurin polynomial of e^x so I get

e^(-2x) = 1 - 2x + 2x^2 - 4x^3/3 + ...

Is this right or wrong?
 
Jul 2010
456
138
to get the Maclaurin use the formula :

\(\displaystyle f_{(x)}= \displaystyle \sum_{n=0} ^{\infty} \displaystyle \frac {f_{(0)}^{(n)}}{n!} x^n\)

where is \(\displaystyle f_{(0)}^{(n)}\) nth derivative...


so it goes :D

\(\displaystyle f_{(0)}=e^{-2\cdot0}=1\)

\(\displaystyle f'_{x}=(e^{-2x})'=\displaystyle -\frac {2}{e^{2x}} \mid _{x=0} = -2\)

\(\displaystyle f''_{x}=(e^{-2x})''=\displaystyle \frac {4}{e^{2x}} \mid _{x=0} = 4\)

\(\displaystyle f'''_{x}=(e^{-2x})''' =\displaystyle -\frac {8}{e^{2x}} \mid _{x=0} =-8\)

\(\displaystyle f^{(4)}_{x}=(e^{-2x})^{(4)}=\displaystyle \frac {16}{e^{2x}} \mid _{x=0} = 16\)


and so on .... and u have :


\(\displaystyle f_{(x)}= \displaystyle \sum _{n=0} ^{\infty} \frac {(-1)^{n}}{n!} \cdot ...... \)

i think u can do that :D
 
Last edited:
Dec 2009
1,506
434
Russia
Q: Find the Maclaurin polynomial of order 3 for the function f(x) = e^(-2x).

I know that the Maclaurin polynomial for e^x = 1 + x + x^2/2! + x^3/3! + ...

So my question is if I would have to plug in (-2x) in the Maclaurin polynomial of e^x so I get

e^(-2x) = 1 - 2x + 2x^2 - 4x^3/3 + ...

Is this right or wrong?

It is right! (But why?)
 
Jun 2008
58
0
Sanger, CA
I know the formula I was just looking to see some work to see if I did the problem correctly. In another problem that I had in an exam consisted of the Maclaurin polynomial of 1/(1-x) = 1 + x + x^2 + X^3 + ...

and the professor wanted us to find the power series representation of 1/(1-x^3) and to figure that out you needed to put an x^3 wherever there was an x in the original given equation.

So I thought since I know the Maclaurin polynomial of e^x I thought that you were able to plug in (-2^x) wherever there was an x in that original polynomial.
 
Jul 2010
456
138
I know the formula I was just looking to see some work to see if I did the problem correctly. In another problem that I had in an exam consisted of the Maclaurin polynomial of 1/(1-x) = 1 + x + x^2 + X^3 + ...

and the professor wanted us to find the power series representation of 1/(1-x^3) and to figure that out you needed to put an x^3 wherever there was an x in the original given equation.

So I thought since I know the Maclaurin polynomial of e^x I thought that you were able to plug in (-2^x) wherever there was an x in that original polynomial.
yes but i think yours 3rd member is wrong :D should be \(\displaystyle 4x^2\)


just to note that when u mentioned that one \(\displaystyle \displaystyle \frac {1}{1-x} \) that is from \(\displaystyle \displaystyle \sum _{n=0} ^{\infty} x^n \) when \(\displaystyle |x|<1\) we had have a lot of times to find \(\displaystyle \displaystyle \frac {1}{(1-x)^2} \) or \(\displaystyle \displaystyle \frac {1}{(1-x)^3} \) which are just a derivate of \(\displaystyle \displaystyle \frac {1}{1-x} \) and actually using that Maclouren \(\displaystyle \displaystyle \frac {1}{1-x} = \displaystyle \sum _{n=0} ^{\infty} x^n\) u can have a lot of another to do based on that one, you just have to tune it so it looks like that one :D

there is like 5 - 10 depending on school to school that u should know by heart and everything goes around them (at least here ) :D:D:D



Edit:just remembered something to note :D
 
Last edited:

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
yes but i think yours 3rd member is wrong :D should be \(\displaystyle 4x^2\)
No, the third term for \(\displaystyle e^x\) is \(\displaystyle \frac{1}{2}x^2\). Replacing x by -2x gives \(\displaystyle \frac{(-2x)^2}{2}= \frac{4x^2}{2}= 2x^2\) as katchat64 has.

katchat64, yes, your method is completely correct.
 
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Jul 2010
456
138
No, the third term for \(\displaystyle e^x\) is \(\displaystyle \frac{1}{2}x^2\). Replacing x by -2x gives \(\displaystyle \frac{(-2x)^2}{2}= \frac{4x^2}{2}= 2x^2\) as katchat64 has.

katchat64, yes, your method is completely correct.
thank you :D for some reason i leave factorials as they are :D because in my mind i always thinking of it as sum there and then just looking for how does changes... leaving down that factorial :D thanks again :D