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I know so far for a derivative to exist.

-The point should not exist as a discontinuity

-It should not have a vertical tangent

-There should be no sharp corner/ cusp at the point

$$(-2)^x=\begin{cases} 2^x & s=\left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\left(2^x\right) & s=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\ \\ \text{undefined} & s=\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{even integer}} \end{cases} $$

I learned these kinds of functions "disconnected" meaning there are extremely close defined points such as at $x=3/100001$ and $5/100001$ but also contains extremely close undefined points such as at $1/50000$ and $2/50000$.

I was not sure if this function was continous but @epimorphic told me these disconnected functions are indeed continuous. For example with $\left|{x}^{\frac{1}{2x}}\right|$.

"Your function $x↦\left|{x}^{\frac{1}{2x}}\right|$ is actually continuous, because continuity is determined at and only at points in the domain of the function."

But then my professor told me, "You can even turn disconnected functions and take the derivative of the complex plane and convert it to a real function".

So then is it possible for the whole domain of $(-2)^{x}$ to have a derivative and a tangent line at defined points (such as $\left(3,-8\right)$)?

Is there infromation of examples or online sources so I can dwell deeper into this kind of approach in real and complex analysis?

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  • $\begingroup$ I made a suggestion that can help with an answer. $\endgroup$ – Arbuja Nov 7 '15 at 23:46
  • $\begingroup$ Can someone explain why this should be closed? $\endgroup$ – Arbuja Nov 17 '15 at 13:53
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    $\begingroup$ Let's assume you are talking about real functions, as studied early in calculus. You need to have a function $f$ be defined in an open interval containing point $x_0$ in order to define continuity of $f$ at $x_0$. There are weaker notions (left continuous, right continuous) that only require $f$ to be defined on half-open intervals terminating at $x_0$. $\endgroup$ – hardmath Nov 17 '15 at 13:57
  • $\begingroup$ @hardmath Thank you, I will clarify my question since it is unclear what I am asking? $\endgroup$ – Arbuja Nov 17 '15 at 14:02
  • $\begingroup$ Yes, it is unclear. What do you mean by "disconnected function" as mentioned in title, and does this form part of your question? You quote a statement about a function that doesn't stand on its own (context missing) from epimorphic, but elsewhere in the question you seem to be asking about differentiability. $\endgroup$ – hardmath Nov 17 '15 at 14:05
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This is not an answer but this could lead to an answer

Taking the definition of a limit.... $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

I found that there can be a derivative of $(-2)^x$ but only if $h\to0$ in a way that the numerator is even and the denominator is odd (even/odd). Such as $h\to\frac{2}{5}\to\frac{2}{101}\to\frac{2}{1000001}\to0$.

This is because using odd numerators and denominators (odd/odd) will result in the derivative diverging to $\infty$.

This is because for $(-2)^x$ the output is negative when x is (odd/odd) or positive when x is (even/odd).

For example $(-2)^{-1/15}=-.954841$ and $(-2)^{12/71}=1.124289$, which means. $$(-2)^x=\begin{cases} 2^x & s=\left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\left(2^x\right) & s=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\ \\ \text{undefined} & s=\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{even integer}} \end{cases} $$

So then when we apply the formal definition of $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ where $f(x)=\left({-2}\right)^{x}$ $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}$$

First if $h=\text{odd}/\text{even}$ such as $1/2$ and we take $x$ as $odd/odd$ or $x=1/3$ then $f(x)$ is already undefined. This means the entire limit is undefined. Now to get $f(x+h)$ we must get $x+h$ which is $5/6$ which means $f(x+h)$ is also undefined whenver $h$=odd/even. Thus...

$$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}-2^x}{h}$$ $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{\text{undefined}-\text{undefined}}{h}$$ $$\text{Limit does not exist}$$

Now if $h=\text{odd}/\text{odd}$ for example $h=1/3$ and let $x$ be (even/odd) fraction like $x=2/3$. So $f(x)=\left({2^x}\right)$; however $x+h=1$. This shows under these conditions $x+h$ will always be (odd/odd) and thus $f(x+h)=-\left(2^{x+h}\right)$

So then when x is odd/odd the limit will be $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}-2^x}{h}$$
$$\lim_{h\to0}\frac{\left(\left({-{2}^{h}-1}\right)*{{2}^{x}}\right)}{h}\approx$$ $$\lim_{h\to0}\frac{-2\left(2\right)^{x}}{h}=$$ $$\lim_{h\to{0}}\frac{1}{h}*{-2\left(2\right)^{x}}$$ Now since$-2({2})^x$ is always postive we have to analyze $\lim_{h\to0}\frac{1}{h}$. Since that part of the limit does not exist there is no derivative.

$$\text{Limit Does Not Exist}$$

If $h=\text{odd}/\text{odd}$ for example $h=1/3$ but $x$ is (odd/odd) fraction like $x=1/5$. So $f(x)=-\left(2^x\right)$; however, $x+h=\frac{8}{15}$. This shows that under these conditions $x+h$ is always (even/odd) so $f(x+h)=2^{x+h}$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{2^{x+h}+2^x}{h}$$ $$\lim_{h\to0}\frac{\left(2^{h}+1\right)\left({2^x}\right)}{h}\approx$$ $$\lim_{h\to0}\frac{2\left({2}^{x}\right)}{h}=$$ $$\lim_{h\to0}\frac{1}{h}*2\left({2}^{x}\right)$$

Since $2\left({2}\right)^{x}$ is always positive but also has $\lim_{h\to0}\frac{1}{h}$ with a limit that does not exist there no derivative.

$$\text{Limit Does Not Exist}$$

But if $h=\text{even}/\text{odd}=2/3$ and $x=\text{odd}/{\text{odd}}=1/3$ then $f(x)=-\left(2^x\right)$. Since $x+h=3/3$, under these conditions, $x+h$ is always (odd/odd) and so $f(x+h)=-\left(2^{x+h}\right)$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}+2^{x}}{h}=-\ln{(2)}{2^{x}}\to\text{For x=odd/odd}$$

And if $h=\text{even}/\text{odd}=2/3$ and $x=\text{even}/{\text{odd}}=2/3$ then $f(x)=2^x$. Since $x+h=4/3$, under these conditions, $x+h$ is always (even/odd) and thus $f(x+h)=2^{x+h}$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{2^{x+h}-2^x}{h}=\ln{(2)}{2^x}\to\text{For x=even/odd}$$

$$\frac{d}{dx}(-2)^x=\begin{cases} \ln(2)2^x & s=\left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\ln(2)2^x & s=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\end{cases}=\text{limit does not exist}$$

According to what I have heard from mathematicians it is possible to define a derivative at cluster points. However for the defined cluster points of each group must have the same limit.

Take $\lim_{x\to\infty}\left({-1}^{x}\right)$ for example. It ossilates between $-1,1,-1,1$ and cannot exist. If I choose (even/odd) for $x\to\infty$ ($2/3\to2000/3\to200000/3$) then the only get $1$ but if I choose (odd/odd) for $x\to\infty$ ($1/3\to10/3\to10000/3$) then I only get $-1$. Thus all sequences for defined intervals of values that approaches some value must have the same limit.

For the sake of derivatives use replace $x$ with $h$. All intervals of $h$ from the domain of $x$ must have the same derivative.

Thus for $\left({-2}\right)^{x}$ the derivative must not exist. However if we instead took $|\left({-2}\right)^{x}|$ then the derivative would be $\ln{\left(2\right)}|\left(-2\right)^{x}|$. Infact when you put the absolute value the derivative satisfies the mean value theorem and rolles theorem.

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    $\begingroup$ this answer makes no sense to me. First thing, in the definition of derivative, you need $\lim_{h\to0}$, not $\lim_{x\to0}$. Next I do not see how you get that $\infty$ or $-\infty$ as the values of the limit. Don't know if this could make an "official" answer, being posted it is official, I would say, but it does not seem clear or correct. $\endgroup$ – Mirko Nov 18 '15 at 1:51
  • $\begingroup$ Working on this I am editing. $\endgroup$ – Arbuja Nov 18 '15 at 2:33
  • $\begingroup$ I think I have done everything correctly. $\endgroup$ – Arbuja Jan 5 '16 at 14:01

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