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Power Series - Exponential

I have been asked to find the first three non-zero terms of

e^{(u-1)^2}

where u is very small.

I started using the expansion of e^x

1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...

But in the answer, it gives

e(1-2u+3u^2)

Where does that come out of? :/

Reply 1

Farhan.Hanif93

Original post by String.

I have been asked to find the first three non-zero terms of

e^{(u-1)^2}

where u is very small.

I started using the expansion of e^x

1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...

But in the answer, it gives

e(1-2u+3u^2)

Where does that come out of? :/

Notice that, in order to use the expansion of

e^x

with

x=(u-1)^2

, you need to consider the contribution from every term in the expansion since they each give some

\text{const}, u, u^2

terms.

Hence you need to consider the constant,

u

and

u^2

terms (and hope those are all non-zero!*) that arise from

\dfrac{((u-1)^2)^n}{n!}

for each natural n and then add the result together.

*[For, if not, you'll have to look at higher order terms i.e.

u^3, u^4

etc. - fortunately, however, the question gives you the result so you know where to look]

Observe that,

\dfrac{(1-u)^{2n}}{n!} = \dfrac{1}{n!} [1 + ^{2n}\mathrm{C}_1 (-u) + ^{2n}\mathrm{C}_2 (-u)^2 +…]

by binomial expansion, where the "…" produces terms of higher order (that are irrelevant to us here). So the first three non-zero terms of the expansion of

e^{(u-1)^2}

arise from:

\displaystyle\sum_{n=0} ^{\infty} \dfrac{1}{n!} [1 + ^{2n}\mathrm{C}_1 (-u) + ^{2n}\mathrm{C}_2 (-u)^2]

(be wary of the values of n where this simplifies slightly [or doesn't make sense and therefore needs to neglect certain terms])

Think about how to evaluate these sums using the fact that

e=\displaystyle\sum_{n=0}^ {\infty} \dfrac{1}{n!}

(edited 10 years ago)

Reply 2

ztibor

Original post by Farhan.Hanif93

\displaystyle\sum_{n=0} ^{\infty} \dfrac{1}{n!} [1 + ^{2n}\mathrm{C}_1 (-u) + ^{2n}\mathrm{C}_2 (-u)^2]

Think about how to evaluate these sums using the fact that

e=\displaystyle\sum_{n=0}^ {\infty} \dfrac{u^n}{n!}

Do you think to

\displaystyle e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!}

(edited 10 years ago)

Reply 3

Farhan.Hanif93

Original post by ztibor

So You think to

\displaystyle e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!}

Jesus that was bad. Typos all over the shop in that post. Thanks.

Reply 4

davros

Original post by String.

I have been asked to find the first three non-zero terms of

e^{(u-1)^2}

where u is very small.

I started using the expansion of e^x

1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...

But in the answer, it gives

e(1-2u+3u^2)

Where does that come out of? :/

Probably the simplest way to see it is to expand (u-1)^2 first and then note that you can take out a factor e leaving something else which requires you to use the power series expansion.

Reply 5

DFranklin

Original post by davros

Probably the simplest way to see it is to expand (u-1)^2 first and then note that you can take out a factor e leaving something else which requires you to use the power series expansion.