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| (x+1)^(2012) = x^(2012)+1 https://forumdematematica.org/viewtopic.php?f=15&t=186 |
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| Autor: | kinu [ 06 fev 2012, 17:51 ] |
| Título da Pergunta: | (x+1)^(2012) = x^(2012)+1 |
How can i find Real Roots of \((x+1)^{2012} = x^{2012}+1\) |
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| Autor: | João P. Ferreira [ 07 fev 2012, 18:08 ] |
| Título da Pergunta: | Re: (x+1)^(2012) = x^(2012)+1 |
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 |
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| Autor: | João P. Ferreira [ 08 fev 2012, 10:19 ] |
| Título da Pergunta: | Re: (x+1)^(2012) = x^(2012)+1 |
\(x=0\) is a evident solution... |
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| Autor: | João P. Ferreira [ 08 fev 2012, 10:29 ] |
| Título da Pergunta: | Re: (x+1)^(2012) = x^(2012)+1 |
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 |
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| Autor: | josesousa [ 08 fev 2012, 22:26 ] |
| Título da Pergunta: | Re: (x+1)^(2012) = x^(2012)+1 |
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. |
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