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Maths - proof

Shinjimonkey

I need to prove that (1 + x/n)^n tends to e^x as n tends to infinity. could someone give me a step by step proof?

Reply 1

RDKGames

Study Forum Helper

Original post by Shinjimonkey

I need to prove that (1 + x/n)^n tends to e^x as n tends to infinity. could someone give me a step by step proof?

Do you know where to start? Have you tried anything?

Reply 2

Shinjimonkey

Original post by RDKGames

Do you know where to start? Have you tried anything?

I am trying to:

ln(1+x/n)^n
= n ln (1+x/n)

but I do not know whether this is right or not

Reply 3

RDKGames

Study Forum Helper

Original post by Shinjimonkey

I need to prove that (1 + x/n)^n tends to e^x as n tends to infinity. could someone give me a step by step proof?

We know that:

\displaystyle (1+\frac{x}{n})^n=e^{n \ln(1+\frac{x}{n})}

\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^r}{r!}+...

\displaystyle \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^r\frac{x^r}{r!}+...

Can you work from there??

Reply 4

Shinjimonkey

Original post by RDKGames

We know that:

\displaystyle (1+\frac{x}{n})^n=e^{n \ln(1+\frac{x}{n})}

\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^r}{r!}+...

\displaystyle \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^r\frac{x^r}{r!}+...

Can you work from there??

how do we know the first rule that you have given?

Reply 5

RDKGames

Study Forum Helper

Original post by Shinjimonkey

how do we know the first rule that you have given?

Because

\displaystyle e^{n \ln(1+\frac{x}{n})}= e^{\ln[(1+\frac{x}{n})^n]}=(1+ \frac{x}{n})^n

Simple logarithm manipulation whereby

e^{\ln(b)}=b

(edited 7 years ago)

Reply 6

Shinjimonkey

Original post by RDKGames

Because

\displaystyle e^{n \ln(1+\frac{x}{n})}= e^{\ln[(1+\frac{x}{n})^n]}=(1+ \frac{x}{n})^n

Simple logarithm manipulation whereby

e^{\ln(b)}=b

I see... as n tends to infinity, x/n tends to 0, so ln(1+x/n) tends to 0, therefore
n ln(1+x/n) = 0. then wouldn't (1+x/n)^n tend to 1?

Reply 7

RDKGames

Study Forum Helper

Original post by Shinjimonkey

I see... as n tends to infinity, x/n tends to 0, so ln(1+x/n) tends to 0

Correct.

Original post by Shinjimonkey

therefore n ln(1+x/n) = 0. then wouldn't (1+x/n)^n tend to 1?

Not quite. Using the expansion for

\ln(1+x)

you would find that:

\displaystyle n \ln(1+\frac{x}{n}) = n[\frac{x}{n}-\frac{x^2}{2n^2}+\frac{x^3}{3n^3}-...]

\displaystyle \Rightarrow n \ln(1+\frac{x}{n}) = x-\frac{x^2}{2n}+\frac{x^3}{3n^2}-...

Therefore from the RHS you can see that as n goes to infinity, you are left with

x

and not a 0.

Reply 8

Shinjimonkey

Original post by RDKGames

Correct.

Not quite. Using the expansion for

\ln(1+x)

you would find that:

\displaystyle n \ln(1+\frac{x}{n}) = n[\frac{x}{n}-\frac{x^2}{2n^2}+\frac{x^3}{3n^3}-...]

\displaystyle \Rightarrow n \ln(1+\frac{x}{n}) = x-\frac{x^2}{2n}+\frac{x^3}{3n^2}-...

Therefore from the RHS you can see that as n goes to infinity, you are left with

x

and not a 0.

I don't quite understand why ln(1+x/n) expands to that?

Reply 9

RDKGames

Study Forum Helper

Original post by Shinjimonkey

I don't quite understand why ln(1+x/n) expands to that?

I hope you understand where the expansion for

\displaystyle \ln(1+x)

comes from.

If you're okay with that, the transition from

\displaystyle \ln(1+x)

\displaystyle n \ln(1+\frac{x}{n})

is quite straight forward as

\displaystyle x \mapsto \frac{x}{n}

so you replace every

x

with

\displaystyle \frac{x}{n}

. Then of course, the whole series is simply multiplied by

n

(edited 7 years ago)

Reply 10

Shinjimonkey

Original post by RDKGames

I hope you understand where the expansion for

\displaystyle \ln(1+x)

comes from.

If you're okay with that, the transition from

\displaystyle \ln(1+x)

\displaystyle n \ln(1+\frac{x}{n})

is quite straight forward as

\displaystyle x \mapsto \frac{x}{n}

so you replace every

x

with

\displaystyle \frac{x}{n}

. Then of course, the whole series is simply multiplied by

n

I do not quite understand why the expansion of ln(1+x) come from? Could you explain please? thank you for the help!

Reply 11

Shinjimonkey

Original post by RDKGames

I hope you understand where the expansion for

\displaystyle \ln(1+x)

comes from.

If you're okay with that, the transition from

\displaystyle \ln(1+x)

\displaystyle n \ln(1+\frac{x}{n})

is quite straight forward as

\displaystyle x \mapsto \frac{x}{n}

so you replace every

x

with

\displaystyle \frac{x}{n}

. Then of course, the whole series is simply multiplied by

n

I understand now. I just realised that ln(!+x) = integral of 1/(1+t) to the limits x to 0. Thanks for the help!

Reply 12

RDKGames

Study Forum Helper

Original post by Shinjimonkey

I do not quite understand why the expansion of ln(1+x) come from? Could you explain please? thank you for the help!

It follows from Maclaurin's Series whereby any function

f(x)

can be rewritten as a sum of derivatives with increasing powers of x.

\displaystyle f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...+\frac{f^r(0)}{r!}x^r+...

So let

\displaystyle f(x)=\ln(1+x)

and perform this. You will get out with an infinite summation.

https://www.youtube.com/watch?v=ibGZapvnB-M

(edited 7 years ago)

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