Series Expansion for exponentials

[MrFahrenheit]

Hi there, this is my first post, I'm not an english speaker so I'm sorry if i misspel some words.
I don't know how to write summations here so I'm going to write sum(a_n) wich represents the sum from n=1 to n=infinity, and to make it easier to read I'm going to use some square brackets [] instead of normal brackets ().
I was learning about series expansions (modifying geometric series, Taylor series, etc) and then trying to find a series for sqrt(2) I found something more intresting.

Let's start with the series for e^x and ln(1+x):
e^x= sum( (x^n) / n! ) +1 (the +1 is because the series starts in 1)
ln(x+1)= sum( [ (-1)^(n+1) ] * (x^n) /n )
Then Let b and c be two real numers where 0 < b <= 1
Then we have: b^c = exp(ln(b^c)) = exp(c*ln(b))

Then write c*ln(b) as a series:
c*ln(b)=c * sum( [ (-1)^(n+1) ] [(b-1)^n] /n )

Then we write exp(c*ln(b)) using a series expansion with x=c*ln(b)
b^c = exp(c*ln(b)) = sum ( [ c * sum( [ (-1)^(n+1) ] [(b-1)^n] /n ) ]^n /n! ) +1

Bringing the c out of the exponent, breaking down (-1)^(n+1) into -(-1)^n
b^c = sum ( [ - sum( [ (-1)^n ] [(b-1)^n] /n ) ]^n * [ ( (c)^n ) /n!] ) +1

Finally, bringing the - in front of the second sum out of the exponent and merging it with the c^n to get (-c)^n and merging the (-1)^n with the (b-1)^n to get ( -(b-1) )^n = (1-b)^n
b^c = sum ( [ sum( [(1-b)^n] /n ) ]^n [ ( (-c)^n ) /n!] ) +1

About convergence, the series for e^x converges for all x and the series for ln(x+1) converges for |x|<1, so it does for |b-1|<1 which means -2 < b < 2 , but i'm going to only let b > 0 (if b=0 it seems to approach 0 but I'm not sure) and then I'm going to let b <= 1, so if you need to use b>1 you simply change the sign of c.

This series might be way too complicated (and you'd be right) but just for fun I tried using the Desmos calculator aproximation sqrt(2) (b= 1/2, c= -1/2) using 10 term for each series and i got 3 decimal places. I also tried 0^0 and it approached 1.

Here's a graph of the partial series (k determines the number of terms on both sums) for c = 1/2 vs the graph of sqrt(x). The graph is centered at x=1.

Desmos | Graphing Calculator

www.desmos.com

I would like to know your opinion and help if I made a mistake. I would appreciate any idea on how to use this, information if someone discovered this before and computational help graphing with a better graphing system.

Thanks for reading!

 

[HallsofIvy]

It's amusing that the only word you misspelled was "misspel"!

You have "e^x= sum( (x^n) / n! ) + 1" but don't say what values of n the sum is over. I presume the sum is from n=1 to infinity because if it were the more usual "0 to infinity" you wouldn't need the +1"! x^0= 1 and 0!= 1 so x^0/0!= 1. "e^x= sum (from 0 to infinity) x^n/n!" without the "+1".
And, unfortunately, your "Then we write exp(c*ln(b)) using a series expansion with x=c*ln(b)b^c = exp(c*ln(b)) = sum ( [ c * sum( [ (-1)^(n+1) ] [(b-1)^n] /n ) ]^n /n! ) +1"
simply makes no sense because you have used the same index, n, for both series.

 

[MrFahrenheit]

It's amusing that the only word you misspelled was "misspel"!

You have "e^x= sum( (x^n) / n! ) + 1" but don't say what values of n the sum is over. I presume the sum is from n=1 to infinity because if it were the more usual "0 to infinity" you wouldn't need the +1"! x^0= 1 and 0!= 1 so x^0/0!= 1. "e^x= sum (from 0 to infinity) x^n/n!" without the "+1".
And, unfortunately, your "Then we write exp(c*ln(b)) using a series expansion with x=c*ln(b)b^c = exp(c*ln(b)) = sum ( [ c * sum( [ (-1)^(n+1) ] [(b-1)^n] /n ) ]^n /n! ) +1"
simply makes no sense because you have used the same index, n, for both series.

Hi, thanks for replaying!
Yes, I ment the sum from 1 to infinity for the e^x series. About using the same index this is how it would look:
b^c = sum (from n=0 to n=infinity) ( [ sum (from i=1 to i=infinity) ( [(1-b)^i] /i ) ]^n [ ( (-c)^n ) /n!] )