n cannot be

Thanks Dr Johannes for pointing out the mistake. However, n is 0.

Are you asking for an analogy to a2 - b2 = (a -b)(a + b) or a3-b3 (n is integer)? I don't have an answer in this case. But if your problem stems from machine calculations, producing null for a not equal to b and n close to zero, then the working remedy might be: an - bn = an (1 - (b/a)n).

Thanks Dr Marek for your response, I, however, want to have an expression for an-bn which could some how simplify a term in an equation containing multiple parameters (non-linear eqn). I have tried solving the the equations by various software, but the outcomes are disobeying the experimental values in every instance. This, eventually led me to simplify the equations and which, finally, brings this problem.

The question ought to be specified: what kind of information do you expect?

The following inequality might be helpful for the question:

a^n-b^n=(a-b)nc^(n-1) where c is a number between a and b.

This is my trick: A^n= \sum_{X=1}^{A} (X^n-(X-1)^n) if A,n Integers. Where I call M_n=(X^n-(X-1)^n) the complicate integer modulus. For A =Rationals use x=X/K, then you can arrange the Sum as a Step Sum, with Rational Step (or index) x=1/K, 2/K etc... . K must divide A rational so A=P/Q, then K=Q or one of its multiple. Here the example for A= P/2 n=2, K=Q=2: A^2=(P/2)^2= \sum:{1/2}^{P/2}2x/2-1/2^2. For a generic n,k there is a M_{n,K} develope. for K going to infinite you've the integral. here why FLT works for n=2.

Dear Arpan Sarkar ,

I think you need the expansion of this expression aⁿ - bⁿ.

If n is a positive integer, then

aⁿ- bⁿ = (a-b)(aⁿ⁻¹+ aⁿ⁻²b+....+abⁿ⁻¹+bⁿ).

If n=(p/q) rational positive integer, then

ap/q-bp/q= (a1/q )p - (b1/q )p

=((a1/q ) - (b1/q ))((a1/q )p -1+ (a1/q )p -2(b1/q )+....

+(a1/q )(b1/q )p-2+(b1/q )p-1 ).

Similar argument for irrational exponents, uses the formula

eln(x) = x and expand.

Best regards

n=0.5 is a particular case.

Therefore,

the expression will be

a^n -b^n = (a^0.25 + b^0.25)(a^0.25 - b^0.25) is case when n = 0.5

But for general case

I agree with Issam Kaddoura.

Cheers

Pls see, and apply on what shown, for the develop of (X-1)^n with rational n: https://en.wikipedia.org/wiki/Binomial_theorem

For n0, you have the inequality:

a^n - b^n

Inequality a^n-b^n

It is exactly:

|a^n - b^n| 0, n

The series in powers of z of exp(zlog(a)) converges slow for small a. In fact one need to take more than m=e/zlog(a)/ of terms of the series to quarantee the remainder tend to zero (/x/ means the absolute value of x).

if a and b larger than zero and n is a fraction we can make the change: a^n = c, b^n = d and we have a^n - b^n = c - d = (c^(1/2) - d(1/2)) (c^(1/2) + d^(1/2)) = (a^(n/2) - b(n/2)) (a^(n/2) + b^(n/2)) by the same way if n is a Root of rank m

I would like to thank all the researchers and scholars for their answers. I really appreciate your time.