Extension of BinomialTheorem

1. May 10, 2012

eddybob123

Thank you in advance, I need help proving or disproving this. In the binomial theorem, with a power (a+b)^n, I need to prove that a^n + b^n is greater than the rest, or in other words, (a+b)^n - (a^n + b^n).

2. May 10, 2012

DonAntonio

Uh? I think you forgot to add some info and/or to write some symbols, as it seems to the question doesn't make sense as it is.

DonAntonio

3. May 10, 2012

Matt Benesi

It depends on what you use for a,b, and n. In some cases, a^n + b^n will be greater than "the rest", in others, it won't.

4. May 11, 2012

HallsofIvy

I think you mean that you want to prove that $(a+ b)^n- (a^n+ b^n)$ is positive. As Matt Benesi said, that depends upon what a and b are.

5. May 11, 2012

dodo

If the statement is suspicious, the first thing is to try a few numerical examples, to try to find a counterexample. Here is a hint: 1+1=2. :)

P.S.: My understanding is that the OP tries to prove or disprove$$a^n + b^n > (a+b)^n - (a^n + b^n)$$

6. May 11, 2012

DonAntonio

Too many assumptions: shall we let the OP to tell us what he meant, please?

DonAntonio

7. May 11, 2012

eddybob123

That is correct. It can also be assumed that a and b do not equal 1.

8. May 11, 2012

DonAntonio

Since the OP already wrote a post saying this is correct, this is the same as $$2(a^n+b^n)>(a+b)^n$$ which is greatly false, for example: for $\,\,a=1\,,\,b=2\,,\,n=3\,\,,\,\,or\,\,a=2\,,\,b=3\,,\,n=4\,\,$ , and infinite counterexamples more.

DonAntonio

9. May 11, 2012

Matt Benesi

Question: can a and b be less than 1? Can they be negative?

Just some interesting cases: in all of these cases, x^n+y^n = "all the rest".

n=2 and a=b.

For n=3 we have $x=-y$, $x=y\times\left[2-\sqrt{3}\right]$, $x=y\times\left[2+\sqrt{3}\right]$.

Getting more complicated for n=4, just one example (out of 4):
$$x=\sqrt{2\,\sqrt{3}+3}\,y+\sqrt{3}\,y+y$$
or to preserve the format used above:
$$x=y\times\left[\sqrt{2\,\sqrt{3}+3}\,+\sqrt{3}\,+1\right]$$

Last edited: May 11, 2012
10. May 12, 2012

eddybob123

a and b both must be integral and greater than 1. (a^n + b^n) must be greater than (a+b)^n - (a^n + b^n) by obvious reasoning. First of all, the two terms, a^n and b^n, contains the highest power of the binomial expansion. The next highest power,n-1, should be greater than a^n or b^n only when n is greater than the following coefficient

11. May 12, 2012

DonAntonio

Either you don't understand mathematically what is going on here or else you're misunderstanding and/or misreading big time

the inequality you want/must prove.

You wrote above "(a^n + b^n) must be greater than (a+b)^n - (a^n + b^n) by obvious reasoning", which means Gauss knows what, but

this inequality is $\,\,a^n+b^n>(a+b)^n-(a^n+b^n)\Longleftrightarrow 2(a^n+b^n)>(a+b)^n\,\,$ , which already was show to be

false for lots and lots of options...Please do read and write carefully what you exactly want to achieve.

DonAntonio

12. May 12, 2012

micromass

Please state exactly what you want to prove.