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Found an unexpected expression: $$e^x -1=xe^{\theta.x}$$ where $\theta \in \left(0,1\right)$. However, I cannot prove it (the $\theta$ part). Any ideas? Thank you.

Abhishek Choudhary
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user314849
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  • Try the mean value theorem on $f(x) = e^x$. $f(x) = f(0) + f'(\xi) x$. – copper.hat May 14 '18 at 04:17
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    Can not prove what? That isn't always true. the the expression can not be proven because it is not true. I think you are supposed to solve for $x$ for values where it is true for a constant $\theta$. – fleablood May 14 '18 at 04:25
  • What does it mean "prove the $\theta$ part"? Why your expression is "unexpected"? – Taroccoesbrocco May 14 '18 at 09:27

1 Answers1

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Thank you greatly to everybody for your help. Mean Value Theorem works:

Fix x and consider the function $f(y)=\exp \left(xy\right)$ on the interval $y \in \left(0, 1\right)$. The function $f\left(y\right)$ is differentiable on this interval with respect to $y$. Then, by the MVT, there exists $\theta \in \left(0, 1\right)$ such that $$x \exp \left(x \theta\right)=\exp \left(x\right)-1.$$ Done.

user314849
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    So then your question was something like: For all $x$ show there exists $\theta$ such that $e^x-1=xe^{\theta x}$? Cause that was not clear to me from the way you phrased this question. – Mason May 14 '18 at 04:43
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    To be honest, I think you have the question backwards. What you have shown is that for any fixed $x$ there will a value of $\theta$ where the equation holds for the fixed $x$. I think what the question is actually asking is to prove for a fixed $\theta$ that there exist an $x$ where the equation is true. (It's very similar) – fleablood May 14 '18 at 04:52