Maths again Exponential expansion

[Peter Pedant]

the question is

find a and b in termns of n if the expansions of (1+x)n and exp ( ax/(1+bx)) are the same up to x2

(1+x)n is easyy 1 +nx + n(n-1)/2! x2 etc

but the exponent : 1 + ax/1+bx + a2x2/(1+bx)2/2!...

still doesnt give ascending powers of x
so I am not able to equate and compare coefficients

please help (a level 1968)

[jake-the-peg]

http://www.efunda.com...xp_log/series_exp.cfm

Should help you if you think about it a bit

[jake-the-peg]

sorry ignore that I misunderstood what you were asking

[jake-the-peg]

I think you need to equate the coefficients in what you have so ignoring 1

nx= ax/(1+bx)
giving n=a/(1+bx)
(1+bx)=a/n

and
n(n-1)/2!=(a²x²/(1+bx)²)/2!

and solve simultaneously

[vascop]

I think you should solve it this way:
Write ax/(1+bx) as ax(1+bx)^-1 and expand as
ax(1-bx+b^2x^2+terms in x^3 or above)
so exp(ax(1-bx+b^2x^2+terms in x^3 or above)) is:
1+ax(1-bx+b^2x^2+terms in x^3 or above)+a^2x^2(1-bx+b^2x^2+terms in x^3 or above)^2
=1+ax-abx^2+a^2x^2+terms in x^3 or above.
If the x terms and x^2 terms in this series must be the same as those of
(1+x)^n=1+nx+n(n-1)x^2+terms in x^3 or above, then
a=n and n^2-nb=n(n-1)/2, so b=(n+1)/2 as long as n is not equal to 0.
You can see that this works for n=1 giving a=1 and b=1
and for n=2 giving a=2 and b=3/2

[vascop]

Sorry, the third line from the bottom should say ...b=1/2 as long as..., NOT b=(n+1)/2.
So a=n and b=1/2. So b does not depend on n.
This checks out OK if you try n=1 and n=2.

[Peter Pedant]

many thanks all who tried

[vascop]

Are you happy with my solution?

[Peter Pedant]

I didnt want to say
No because the only expansion is exp ax
and the other bit 1-bx never gets expanded

I thought it might involve two infinite expansions multiplying and expressing it up to x2 and throwing the rest away

but thanks anyway

[vascop]

If you look again carefully you will see that it is correct. Here is a link to a pdf file which shows how it works very clearly:
https://docs.google.c...yMzctYjYxMTRmMDAwMzRh

[vascop]

Peterpedant wrote:
"I thought it might involve two infinite expansions multiplying and expressing it up to x2 and throwing the rest away"
That is exactly what my solution does. If you follow the link you will see examples of how it works.

[Peter Pedant]

excellent ! I made it onto google docs and your answer made sense AND gave a reason why they made it up all those years ago

ALSO - you know I anwered this q in 1968 and got it out but couldnt remember how I had done it

also in getting me to think about it -for which I thank you - I saw that the alternative is to expand exp (ax(1-bx) to exp (ax - abx2) as exp(ax)exp(-abx2) as
(1+ax +a2x2/2.......)(1-abx2/2......) which lead to the same exp when multiplied out viz....
1 + ax + (a2/2-ab)x2

I thank you very very much and can asterisk your answer as the BEST as the toggle seems to be dead - manuy thanks again