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Maclaurin series

So the question is to determine the Maclaurin series up to term in x^3. I'm stuck on one particluar one. I keep feeling this has clicked then get set back.

(1 + x)*e^-x

So I have worked out the differentials
(1+x)*e^-x
-xe^-x
(x-1)*e^-x
-(x-2)*e^-x

Subbed in for when x = 0
1
-e^-1
e^-2
e^-1

And then put it into the formula for Maclaurin series but I am going wrong somewhere because it isn't the right answer. Any help is appreciated thanks. I don't know if i'm starting off wrong on this on or halfway....
Reply 1
Avatar for math42
math42
20
Original post by RHCPfan
So the question is to determine the Maclaurin series up to term in x^3. I'm stuck on one particluar one. I keep feeling this has clicked then get set back.

(1 + x)*e^-x

So I have worked out the differentials
(1+x)*e^-x
-xe^-x
(x-1)*e^-x
-(x-2)*e^-x

Subbed in for when x = 0
1
-e^-1
e^-2
e^-1

And then put it into the formula for Maclaurin series but I am going wrong somewhere because it isn't the right answer. Any help is appreciated thanks. I don't know if i'm starting off wrong on this on or halfway....


Huh? Subbing in x = 0 you should get
1
0
-1
2
Original post by RHCPfan
So the question is to determine the Maclaurin series up to term in x^3. I'm stuck on one particluar one. I keep feeling this has clicked then get set back.

(1 + x)*e^-x

So I have worked out the differentials
(1+x)*e^-x
-xe^-x
(x-1)*e^-x
-(x-2)*e^-x

Subbed in for when x = 0
1
-e^-1
e^-2
e^-1

And then put it into the formula for Maclaurin series but I am going wrong somewhere because it isn't the right answer. Any help is appreciated thanks. I don't know if i'm starting off wrong on this on or halfway....


Your differentials are correct.

f(x)=(1+x)⋅e−x=e−x+xe−xf(x)=(1+x)\cdot e^{-x}=e^{-x}+xe^{-x}
f′(x)=−e−x+e−x−xe−x=−xe−xf'(x)=-e^{-x}+e^{-x}-xe^{-x}=-xe^{-x}
f′′(x)=−e−x+xe−xf''(x)=-e^{-x}+xe^{-x}
f′′′(x)=e−x+e−x−xe−xf'''(x)=e^{-x}+e^{-x}-xe^{-x}

but when you sub in x=0, there shouldn't be any ee terms because e0=1e^0=1
Reply 3
Avatar for RHCPfan
RHCPfan
OP
12
Original post by 1 8 13 20 42
Huh? Subbing in x = 0 you should get
1
0
-1
2


Original post by RDKGames
Your differentials are correct.

f(x)=(1+x)⋅e−x=e−x+xe−xf(x)=(1+x)\cdot e^{-x}=e^{-x}+xe^{-x}
f′(x)=−e−x+e−x−xe−x=−xe−xf'(x)=-e^{-x}+e^{-x}-xe^{-x}=-xe^{-x}
f′′(x)=−e−x+xe−xf''(x)=-e^{-x}+xe^{-x}
f′′′(x)=e−x+e−x−xe−xf'''(x)=e^{-x}+e^{-x}-xe^{-x}

but when you sub in x=0, there shouldn't be any ee terms because e0=1e^0=1

Oh man I've just seen where I have gone wrong! Thank you both!! Saying I was subbing x=0 when I was doing 0,1,2. It's been a long day haha.
Reply 4
Avatar for RHCPfan
RHCPfan
OP
12
Original post by 1 8 13 20 42
Huh? Subbing in x = 0 you should get
1
0
-1
2


Original post by RDKGames
Your differentials are correct.

f(x)=(1+x)⋅e−x=e−x+xe−xf(x)=(1+x)\cdot e^{-x}=e^{-x}+xe^{-x}
f′(x)=−e−x+e−x−xe−x=−xe−xf'(x)=-e^{-x}+e^{-x}-xe^{-x}=-xe^{-x}
f′′(x)=−e−x+xe−xf''(x)=-e^{-x}+xe^{-x}
f′′′(x)=e−x+e−x−xe−xf'''(x)=e^{-x}+e^{-x}-xe^{-x}

but when you sub in x=0, there shouldn't be any ee terms because e0=1e^0=1


Oh man I've just seen where I have gone wrong! Thank you both!! Saying I was subbing x=0 when I was doing 0,1,2. It's been a long day haha.

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