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Maclaurin question

Just to clear this up, if you have say e^(-x^2) then can you just use the expansion for e^x = 1+x+x^2/2+... and sub in (-x^2) instead of x?
Reply 1
Avatar for yusufu
yusufu
16
this is a bit hazy in my memory but i think you can. :smile:
Reply 2
Avatar for BovineBeast
BovineBeast
C4>O7
Just to clear this up, if you have say e^(-x^2) then can you just use the expansion for e^x = 1+x+x^2/2+... and sub in (-x^2) instead of x?


No, because the derivative of e^(-x^2) is different to that of e^x. The derivative of e^(-x^2) is -2xe^(-x^2).

This means, by the way, that it can't be expanded as a Maclaurin series, since f'(0), f''(0), etc. will all be zero, so you get the expansion e^(-x^2)=1, which only works for x=0. I think you might be able to do a Taylor series, though.
Reply 3
Avatar for C4>O7
C4>O7
OP
5
In the heinemann FP3 book they just come up with the e^(-x^2) series by saying "the expansion for e^x is.....so taking -x^2 in place of x.."
So it certainly would appear to work for that.
But Im having a hard time accepting that it works like that for all functions.

Does it work for a composite function fg(x) provided g(0)=0 or some other condition like that??
btw I have no idea if the answer is really simple or really complex :s-smilie:
BovineBeast
No, because the derivative of e^(-x^2) is different to that of e^x. The derivative of e^(-x^2) is -2xe^(-x^2).

This means, by the way, that it can't be expanded as a Maclaurin series, since f'(0), f''(0), etc. will all be zero, so you get the expansion e^(-x^2)=1, which only works for x=0. I think you might be able to do a Taylor series, though.
Work out f''(0)

f'(x) = -2xe-x2
f''(x) = -2e-x2 + 4x2e-x2

f''(0) = -2

The odd derivatives (1st, 3rd, 5th etc) are going to be zero, because they are the coefficents of x2k+1 and the expansion for e-x2 isn't going to have any odd powers of x is it? The even coefficents will be just right to give you the same exponential power series but with -x2 instead of x.
Reply 5
Avatar for dvs
dvs
10
C4>O7
In the heinemann FP3 book they just come up with the e^(-x^2) series by saying "the expansion for e^x is.....so taking -x^2 in place of x.."
So it certainly would appear to work for that.
But Im having a hard time accepting that it works like that for all functions.

Does it work for a composite function fg(x) provided g(0)=0 or some other condition like that??
btw I have no idea if the answer is really simple or really complex :s-smilie:

Suppose you have a function f(x) that you expressed as a power series, then:
f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n + ...

Now:
f(g(x)) = a_0 + a_1 g(x) + a_2 (g(x))^2 + ... + a_n (g(x))^n + ...

However you have to be very careful here. For instance, g(x) may not be defined for some values, so that f(g(x)) doesn't make sense for those values. Another thing you have to be careful about is convergence. In the case of e^x the Taylor series converges for all x, but for other functions this is usually not true. So, for example, we can have the power series for f(x) converge for some x_0, while the power series for f(g(x)) diverges when x=x_0.

Edit:
So to answer your original question: yes, it's valid to replace x by -x^2 in the expansion of e^x to get the expansion of e^(-x^2). This new expansion also converges for all x.
Reply 6
Avatar for BovineBeast
BovineBeast
AlphaNumeric

The odd derivatives (1st, 3rd, 5th etc) are going to be zero, because they are the coefficents of x2k+1 and the expansion for e-x2 isn't going to have any odd powers of x is it? The even coefficents will be just right to give you the same exponential power series but with -x2 instead of x.


Oh, fair point. I wasn't really paying attention.

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