Odd function effect in desmos

Question

Graph in desmos

(Zoom in on its negative side starting at x=-30)
Why is it choppy? Why does it stop at x=-35.63?

Desmos:

Symbolab:

Is it just that the graphers can't handle it?

Thanks in advance.

Answer 1

The problem lies in the second term of your function, being $ln(\sqrt{e^x+1}-x)$. What happens is, that if $x$ becomes negative, that e-power rapidly goes to zero. What happens then, is that you are asking the program to take the ln of a number so incredibly close to zero, that the arithmetic part can't handle it. When I type in $e^{-35}$ in my TI, and add $1$ and take square root, it flatly gives me $1$, which is of course not right either. So DESMOS is not the only program struggling here. Best thing is to combine the ln terms and simplify (to plainly $x$)and regraph. DEMSOS has no issues then. Technology is great but has limitations...

Answer 2

You will see a helpful clue if you instead plot $\log_2\mathopen{}\left(\sqrt{e^x+1}-1\right)\mathclose{}$, which maybe you need to enter in Desmos as $\ln\mathopen{}\left(\sqrt{e^x+1}-1\right)\mathclose{}/\ln(2)$.

The output values are then $\{-52,-51,-50.4\ldots,-50,\ldots\}$.

I would deduce that Desmos treats real numbers as having 52 bits past the decimal and no more. So for large enough negative $x$ (Apparently close to $-35$) you have in binary that $$\sqrt{e^x+1}-1\approx0.\overbrace{0\ldots01}^{52\text{ bits}}$$ and the $\log_2$ gives $-52$.

Then as far as Desmos is concerned, the next largest real number is $0.\overbrace{0\ldots010}^{52\text{ bits}}$, and $\log_2$ returns $-51$. And then the next largest real number is $0.\overbrace{0\ldots011}^{52\text{ bits}}$, and $\log_2$ returns $-50.4\ldots$.

So you are seeing Desmos have to round to discrete values on the order of $2^{-52}$. Furthermore, once $x$ is just a bit beyond $-35$, then Desmos has to round $$\sqrt{e^x+1}-1\approx0.\overbrace{0\ldots00}^{52\text{ bits}}$$ and now the logarithm will give no output at all.

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Once you understand this issue, you can use expressions that are algebraically equivalent but will not ask Desmos to compute at that order of magnitude. Here, you can replace $\ln\mathopen{}\left(\sqrt{e^x+1}-1\right)\mathclose{}$ with the equivalent $x-\ln\mathopen{}\left(\sqrt{e^x+1}+1\right)\mathclose{}$. When I do so, now I have a smooth-looking graph in the negative direction forever. (It cuts off quickly in the positive direction, but some kind of if/then definition of the function, using one expression for positive $x$ and another for negative $x$ should work in both direction.)