(x+1)^(2012) = x^(2012)+1

[kinu]

How can i find Real Roots of \((x+1)^{2012} = x^{2012}+1\)

[João P. Ferreira]

Just a hint:

Remember the binomial theorem

\({\left(x+y\right)}^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^k\,\!\)

So:

\({\left(x+1\right)}^{2012}=\sum_{k=0}^{2012}{2012 \choose k}x^{2012-k}\times1^k\,\!\)

\({\left(x+1\right)}^{2012}=\sum_{k=1}^{2011}{2012 \choose k}x^{2012-k}+x^{2012}+1\)

So, you just have to solve

\(\sum_{k=1}^{2011}{2012 \choose k}x^{2012-k}=0\)

\(2012x^{2011}+2023066x^{2010}+...+2012x=0\)

It is a polynomial of degree 2011 (odd number). It should have (I suppose) several roots

[João P. Ferreira]

\(x=0\) is a evident solution...

[João P. Ferreira]

A simpler way out

\(x^{2012}\) is a curve (some how similar to parabola) with a minimum at \(x=0\)

\(x^{2012}+1\) is exactly the same curve but topped up 1 unit on y axis

\((x+1)^{2012}\) is exactly the same curve but shifted left 1 unit on x axis...

So I suppose they only "touch" each other in one point at \(x=0\)

Take care

[josesousa]

taking the derivatives of both sides

\(\frac{d}{dx}((x+1)^{2012})=2012(x+1)^{2011}\)
and
\(\frac{d}{dx}(x^{2012}+1)=2012.x^{2011}\)

function 1 is monotonous from \(]-\infty, -1[\) and from \(]-1, \infty\)
function 2 is monotonous from \(]-\infty, 0[\) and from \(]0, \infty\)

they both take the value 1 at x=0. !!1st solution, as Joao wrote!!

at the right of x=0, function 1 grows much faster than function 2 : \(x+1>x\),
immediatly at the left of x=0, function 2 has a negative derivative (thus is decreasing), while function 1 grows from x=-1 to x=0.
There there is no solution. However, at x=-1, func1 = 0, func 2 = 2

but, as x tends to be more negative, func2 has always a more negative derivative than func 1, so the zero at x=0 is unique.