User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

To do
I'll need this theorem in order to solidify some results with the definite integrals above.

Let $f$ be a differentiable real function. Then:


 * $\dfrac {f(x+h) - f(x-h)} {2h} \to f\,'(x)$ as $h \to 0$ --GFauxPas (talk) 22:28, 19 December 2012 (UTC)


 * Is the "2" in the denominator extraneous? --GFauxPas (talk) 22:31, 19 December 2012 (UTC)


 * Not unless $f'(x) = 0$. Try multiplying both sides with $2$. --Lord_Farin (talk) 23:30, 19 December 2012 (UTC)


 * Oh, okay, right. So to prove this I'm going to use that $f\,'_+ = f\,'_-$, from Limit iff Limits from Left and Right, right? And then what's the next step, can I have a hint?


 * It may be worthwhile to add definitions for "derivative from the left" and "derivative from the right". This also allows for neater structuring of Definition:Differentiable/Real Function/Interval.


 * Furthermore, it appears that the following equivalent formulation of differentiability is not covered yet (it's rather trivial that they're equivalent but needs proof nonetheless):


 * Let $x \in \R$ be a real number.


 * Let $f$ be a real function whose domain is a neighborhood of $x$.


 * Then $f$ is said to be differentiable at $x$ iff the following limit exists:


 * $f' \left({x}\right) := \displaystyle \lim_{h \to 0} \frac {f \left({x + h}\right) - f \left({x}\right)} {h}$


 * If this is the case, $f' \left({x}\right)$ is called the derivative of $f$ at $x$.


 * With this definition, we observe that your expression is (one half times) $\displaystyle \lim_{h \to 0} \frac {f \left({x + h}\right) - f \left({x}\right)} {h} + \frac {f \left({x - h}\right) - f \left({x}\right)} {-h} = 2 f' \left({x}\right)$, using that $-h \to 0$ iff $h \to 0$. --Lord_Farin (talk) 08:16, 20 December 2012 (UTC)


 * Equivalence of Definitions of the Derivative, Definition:Right-Hand Derivative, Definition:Left-Hand Derivative --GFauxPas (talk) 12:10, 20 December 2012 (UTC)

Is there a such thing as a one-sided (real valued) definite integral? --GFauxPas (talk) 13:54, 20 December 2012 (UTC)


 * The closest thing I can think of is Definition:Stieltjes Function of Measure on Real Numbers. --Lord_Farin (talk) 14:24, 20 December 2012 (UTC)

Continuity of Inverse Functions
Larson, appendix, A12

Theorem
Let $f: \mathbb I \to \mathbb J$ be a real function, where $\mathbb I$ is some real interval.

Let $f$ admit an inverse.

Then, if $f$ is continuous on its domain, then $f^{-1}$ is continuous on its domain.

Proof
Let $f$ be invertible on $\mathbb I$.

Part 1:


 * $f$ is a bijection from Bijection iff Inverse is Bijection.


 * $f$ is then strictly monotone from Continuous Injection of Interval is Strictly Monotone.

Part 2:

Because $f$ is a bijection, $\mathbb J = f\left({\mathbb I}\right)$.

From Image of Interval by Continuous Function is Interval, $\mathbb J$ is an interval.


 * $\downarrow$ PROVE THIS $\downarrow$

Let $a$ be an interior point of $\mathbb J$. Then $f^{-1} \left({a}\right)$ is an interior point of $\mathbb I$.


 * $\uparrow$ PROVE THIS $\uparrow$

Let $\epsilon > 0$. From Interval Defined by Betweenness there exists some $0 < \epsilon_1 < \epsilon$ such that:


 * $\mathbb I\,' = \left ({f^{-1}\left({a}\right) - \epsilon_1 \,.\,.\, f^{-1}\left({a}\right) + \epsilon_1} \right)$

is a subinterval of $\mathbb I$.

From Image of Subset is Subset of Image, $f\left({\mathbb I\,'}\right) \subseteq \mathbb J$.

From Interval Defined by Betweenness there exists some $\delta > 0$ such that:


 * $\left ({a - \delta \,.\,.\, a + \delta} \right)$

is a subinterval of $f\left({\mathbb I\,'}\right)$.

Because $ \left({\R, \vert \cdot \vert}\right)$ is a metric space, the above interval can be considered as an open ball:


 * $x \in \left ({a - \delta \,.\,.\, a + \delta} \right) \iff \left \vert {x - a} \right \vert < \delta$.

Similarly:


 * $\left \vert {x - a} \right \vert < \delta \implies \left \vert {f^{-1}\left({y}\right) - f^{-1}\left({a}\right)} \right \vert < \epsilon_1 < \epsilon$.

This proves that $f^{-1}$ is continuous at $a$, from the definition of continuity.

As $a$ was an arbitrary interior point, all that remains to be proven is to address the endpoints of $\mathbb J$, if $\mathbb J$ is half-open or closed.

--GFauxPas (talk) 03:24, 1 February 2013 (UTC)


 * Let me tend to the two indicated steps only. Second point simply follows by Image of Subset is Subset of Image. First point is more intricate. An attempt:
 * We have that $f^{-1}$ is strictly monotone by Inverse of Strictly Monotone Function.
 * Let $a$ be an interior point of $\mathbb J$; we find $b, c \in \mathbb J$ such that $b < a < c$.
 * By monotonicity of $f^{-1}$, $f^{-1} \left({b}\right) < f^{-1} \left({a}\right) < f^{-1} \left({c}\right)$.
 * By definition of $\mathbb J$, $f^{-1} (b), f^{-1} (c) \in \mathbb I$.
 * Thus $f^{-1} (a)$ is not an end point of $\mathbb I$.
 * Hence by Interior Point of Interval, $f^{-1} (a)$ is an interior point of $\mathbb I$.
 * Here, Interior Point of Interval is a reference to a result that a non-end point of an interval is an interior point. I hope that suffices. --Lord_Farin (talk) 13:55, 4 February 2013 (UTC)


 * That's very helpful, thanks! What about continuity at endpoints, do I go to the definition of one-sided limit? --GFauxPas (talk) 14:05, 4 February 2013 (UTC)


 * Yes, at least that's what arises from going into topology and back. --Lord_Farin (talk) 16:46, 4 February 2013 (UTC)


 * Wouldn't Interior Point of Interval follow directly from the definition of an interval? Is it as trivial as it seems, or am I missing something subtle? --GFauxPas (talk) 01:17, 5 February 2013 (UTC)


 * It's very intuitive and trivial, but since "interior point" is topological language, there is merit in having such a page. --Lord_Farin (talk) 14:11, 5 February 2013 (UTC)

Proof 4
Same methodology as Derivative of Exponential at Zero/Proof 3:

--GFauxPas (talk) 05:04, 4 June 2013 (UTC)