Finding a derivative at an imaginary point (say, Y = sqrt(2 - x^2) and x = 3)

[Eigendorf]

Hello again!

I have a question about how you would find the derivative at an imaginary point.

Lets say I had a function Y = sqrt(2 - x^2)

AND I wanted to find the derivative at the point x = 3.

Plugging in X = 3 to the function I get the Point P(3, sqrt(-7)) or P(3, sqrt(7)I) I for imaginary sqrt(-1)

I'm assuming I can just find the derivative in a normal fashion and then plug in X = 3 to get the imaginary derivative but this doesn't sound right.

If I took the derivative of sqrt(2-x^2) I'd get:

Y' = -x / sqrt(2 - x^2)

Then plugging in X = 3 I'd get

Y' = -3 / sqrt(2 - 9) or -3/[sqrt(7)*I]

Any thoughts on places to read about this and if I'm thinking about this in the correct manner?

 

[ksdhart2]

Everything you've done looks fine to me. Generally speaking, because the real numbers are a subset of the complex numbers, most functions that are defined over the reals are also defined on the complex plane. As long as the function is defined and continuous at whatever point, there's nothing that I know of that stops you from evaluating the derivative at a point that would make the function have a complex value. Similarly, there's nothing stopping you from evaluating a derivative at a point that itself is complex. For instance, plugging in the point x = 3 + 5i gives:

\(\displaystyle \dfrac{-(3+5i)}{\sqrt{2-(3+5i)^2}} = \dfrac{-(3+5i)}{\sqrt{2-(-16 + 30i)}} = -\dfrac{17 \sqrt{\frac{2}{3}}}{(3-5i)^{\frac{3}{2}}}\)

 

[Eigendorf]

Thank you for the explanation.

I don't know why they never talk about complex and imaginary numbers in math classes. They don't seem horrible but nobody seems to talk about them.

One more question if you don't mind.

How did you make that nicely formatted equation?

 

[tkhunny]

One step at a time. There are many discussion to be had about Complex Numbers. No need to have them when they would only confuse most students. If you wish to branch out on your own, feel free to do so.

Anyway, you should check your DEFINITION of a derivative. Is it DEFINED at imaginary points? If not, there is no sense calculating or substituting. Your definitions are more important than your opinions. It's not a matter of being more progressive or thoughtful or creative. It's just wrong.

 

[ksdhart2]

Thank you for the explanation.

I don't know why they never talk about complex and imaginary numbers in math classes. They don't seem horrible but nobody seems to talk about them.

I don't really understand it either. In my highschool algebra class, complex numbers were given a brief mention in a sort of "Oh, by the way, these are things that exist. We're not going to talk about them though." The first math class I remember really talking about them in depth was Differential Equations. I do know the complex numbers are very very useful in electrical engineering, though.

One more question if you don't mind.

How did you make that nicely formatted equation?

It's called LaTeX. There's a lot of neat stuff you can do with it, but it does have kind of a steep learning curve. But for the basic stuff, it shouldn't be too bad. The forum has a built-in parser for it, so you can type, for example:

[tex]\sqrt{\dfrac{3}{5}} + \dfrac{\sqrt{6}}{4^{\frac{7}{9}}}[/tex]

And it will render as

\(\displaystyle \sqrt{\dfrac{3}{5}} + \dfrac{\sqrt{6}}{4^{\frac{7}{9}}}\)

You may have noticed I used both \dfrac and \frac. \dfrac makes big fractions, which isn't always desirable, particularly in exponents. Here's a few examples of the comparison:

[tex]\dfrac{3}{5}[/tex] vs. [tex]\frac{3}{5}[/tex]

\(\displaystyle \dfrac{3}{5}\) vs. \(\displaystyle \frac{3}{5}\)

and

[tex]2^{\dfrac{3}{5}}[/tex] vs. [tex]2^{\frac{3}{5}}[/tex]

\(\displaystyle 2^{\dfrac{3}{5}}\) vs. \(\displaystyle 2^{\frac{3}{5}}\)

I generally like the bigger fractions as they're easier to discern which numbers are which (particularly, to my eyes, 2 and 3 look very very similar when they're small), but the big fractions in an exponent just don't look right to me.

 

[Eigendorf]

One step at a time. There are many discussion to be had about Complex Numbers. No need to have them when they would only confuse most students. If you wish to branch out on your own, feel free to do so.

Anyway, you should check your DEFINITION of a derivative. Is it DEFINED at imaginary points? If not, there is no sense calculating or substituting. Your definitions are more important than your opinions. It's not a matter of being more progressive or thoughtful or creative. It's just wrong.

Oh, I wasn't trying to be creative or anything. I just came across a question on a test I took recently that was similar to what I posted although a little more complicated. I just wrote a simplified version of the question here that struck at the issue I was having understanding the solution to the question.

I'm pretty sure it was just an oversight on the math teachers part to choose a value that was outside the Real plane. I haven't studied anything beyond basic imaginary number stuff you do in basic algebra and I wasn't sure if the basic calculus rules applied to complex numbers.

I'm assuming for complex numbers you'd just take the derivative of the "Real" part and then the derivative of the "imaginary" part but in this case the function is Real but the point X = 3 lies outside of its Real interval.

I'm going to buy a book on this just to read about it so I'm not so lost when I do have to take differential equations or whatever engineering class may require this.

 

[mmm4444bot]

Regarding the LaTex implementation here, you may right-click on any formatted expressions to see the code (in a pop-up window). Select Show Math As --> TeX Commands :cool: