Is the Infinite Derivative of a Function Equal to Zero?

  • Thread starter Thread starter Dauden
  • Start date Start date
  • Tags Tags
    Derivative
AI Thread Summary
The discussion centers on the concept of infinite derivatives and whether the derivative of infinity can be defined meaningfully. Participants debate the validity of expressions like d^inf/dx^inf and explore interpretations of infinite derivatives, with some arguing that taking derivatives of infinity is nonsensical since infinity is not a function. The conversation also touches on the outcomes of applying infinite derivatives to various functions, highlighting that while some results yield zero, others do not. Ultimately, the dialogue reveals a mix of humor and frustration over the complexities and ambiguities surrounding the mathematical treatment of infinity. The topic remains unresolved, emphasizing the challenges in defining and understanding infinite derivatives.
Dauden
Messages
42
Reaction score
0
I was just thinking about this earlier today...

Is the derivative of infinity equal to zero?

Then the more interesting question, is there such thing as an infinite derivative? (I battled with Latex for 20 minutes and can't figure out how to work it so bare with me)

d^inf/dx^inf

And assuming it does exist and that (d^inf/fx^inf) of any x^p is zero where p is a positive integer. Would the infinite derivative of x^inf equal zero?

Has anyone else come across anything like this or am I just assuming too much about the derivative itself?
 
Mathematics news on Phys.org
Derivatives are rates of change. Taking the derivative of infinity is simply non-sense since infinity is a limit, not a function.
 
Your language is confused. Are you taking the derivative of infinity (in which case we need a definition of "infinity" as a function), or are you taking the derivative an infinite number of times as suggested by your notation d^inf/fx^inf?

For the latter, I can interpret it thus: define \frac{d^\infty}{dx^\infty}f(x)=\lim_{n\to\infty}\frac{d^n}{dx^n}f(x). With that definition, \frac{d^\infty}{dx^\infty}x^7=0 but \frac{d^\infty}{dx^\infty}e^x=e^x.

For the former, if you define \infty(x)=\infty (on, say, the Riemann sphere) then \infty'(x)=0.

As to your mention of x^\infty, I don't know what that means. Formally, I could write \frac{d^\infty}{dx^\infty}x^\infty=\infty! but that doesn't have meaning in any structure I can think of.
 
CRGreathouse said:
As to your mention of x^\infty, I don't know what that means. Formally, I could write \frac{d^\infty}{dx^\infty}x^\infty=\infty! but that doesn't have meaning in any structure I can think of.

I disagree with your meaningless expression, CRGreathouse.

To me, the proper meaningless expression should be:
\frac{d^\infty}{dx^\infty}x^\infty=\infty^{\infty}x^{\infty}
:smile:
 
I still find CRGreathouse's meaningless expression more mathematically sound :biggrin:

Dauden, it's like asking what \infty -\infty is equal to. It could be zero, it could be finite, or it could be \pm \infty. In other words, since we don't know the answer to this, we can't tell you what the answer to the infinite derivative of a polynomial of infinite degree is.
 
arildno said:
I disagree with your meaningless expression, CRGreathouse.

To me, the proper meaningless expression should be:
\frac{d^\infty}{dx^\infty}x^\infty=\infty^{\infty}x^{\infty}
:smile:

That interpretation, like 0, is every bit as meaningful as my result. :rolleyes: :biggrin:
 
Mentallic said:
I still find CRGreathouse's meaningless expression more mathematically sound :biggrin:

:smile:

Yes. the sweet, sickly odour of a decomposing, dead corpse is still clinging to his expression; mine is utterly desiccated, and hence more pleasant to hang around with...
 
Back
Top