enter image description hereHello, I’m new to math and I consider myself a hobbyist. Could someone explain the type of proof this is and how it works? I can see that it appeared to be a proof by contradiction but I don’t understand where the integration is coming from. The theorem relates to the two cycle solutions of difference equations. Picture is uploaded.
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2Since $f'$ is continuous in the (I assume compact) interval $I$, it is integrable, so for the fundamental theorem of calculus it is $f(b)-f(a)=\int_a^b f'(x) \text{d}x$. Of course it is $\int_a^b 1 \text{d}x=b-a$. For the future questions, please don't upload pictures for such short statements and please show us your efforts to solve the problem. Thank you. – Bernkastel Apr 21 '22 at 05:55
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I believe, proof is clear. $1+f' 0$ and $f'$ is continuous $\Rightarrow 1+f'$ is always positive or always negative in $I$, then integral of $1+f'$ is not zero. Then use fundamental theorem of calculus to find integral value in case $x_1\neq x_0$. – Ivan Kaznacheyeu Apr 21 '22 at 08:15
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Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Martin R Nov 28 '22 at 08:05
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The proof does not need to use integration. In fact, I would find the following proof more natural.
Suppose there is a 2-cycle, i.e. $x,y\in I$ such that $x\ne y$ and $f(x)=y, f(y)=x$. Per Mean Value Theorem, there is a $\xi$ between $x$ and $y$ such that:
$$f'(\xi)=\frac{f(y)-f(x)}{y-x}=\frac{x-y}{y-x}=-1$$
... but this is a contradiction, because it implies $f'(\xi)+1=0$.
I believe the original proof is a variant of this (there is such a thing as a Mean Value Theorem for integrals), but it obscures the basic idea a bit. Which (the basic idea) is: if the points $(x,y)$ and $(y,x)$ are swapped, the "chord" on the function graph joining those points has the slope (gradient) $-1$ and so there must be a point in between with a tangent which also has gradient $-1$.