A maths puzzle (calculus)

[pokerfan]

A Calculus Conundrum

d/dx (x²) = is 2x.

So far, so good. High school stuff. But watch what happens when we write x² as the sum of x number of x's:

Let f(x) = x + x + ... + x     (x times)

Then:

f'(x) = d/dx (x + x + ... + x)     (x times)
= dx/dx + dx/dx + ... + dx/dx     (x times)
= 1 + 1 + ... + 1     (x times)
= x

d/dx (x²) = x?!

Obviously a fallacy. But where is it?

[Dr. Strange]

1. Obviously a fallacy?

Says who?

:evilgrin:

[pokerfan]

5. the derivative can't be both x and 2x.

"Two men say they're Jesus; one of them must be wrong..."

[Dr. Strange]

6. I challenge you and your law of the excluded middle!

I think I know the answer, but I'll let others chime in first.
Then I'll prove that 1 = 2 using Picard's Theorem.

[pokerfan]

8. THERE. ARE. TWO. X'S!

[TZ]

16. LOCKING

No X posts please!

[pokerfan]

18. THERE. ARE. TWO. WHOOPIES!

(Newlywed Game euphemisms.)

[skygazer]

2. Calculus, by its very definition, is a conundrum

Elementary algebra makes my brain hurt - I don't know what I was thinking clicking on this thread. :(

[TZ]

3. same reason I did no doubt.

Both masochists...:crazy:

[LSdemocrat]

4. because you can't write "x times" into the right hand side of the equation

It may make sense in a verbal and/or textual sense, but you can't just and a side note, "x times" into a mathematical equation.

[pokerfan]

7. pretend it's a summation then

[Dr. Strange]

9. That sort of assumes that you're summing x an integral number of times.

Whereas, you should allow for something like
3.5² = 3.5 + 3.5 + 3.5 + (0.5)(3.5).

[pokerfan]

10. a chicken and a half lays an egg and a half in a day and a half

but I see what you're saying.

But even limiting x to integers, where does the "logic" fail?

[Dr. Strange]

11. If you limit x to the integers...

then one place the logic fails is in the definition of the derivative. You're taking a limit which involves a continuous quantity, but the integers are discrete. To make it more interesting, you have to allow for x to be any real number.

[pokerfan]

12. that's fixable

Another potential objection is that, since the function is defined only when x is an integer, it is not continuous, and therefore not differentiable.

Consider, though, what happens if we extend the additive notation to cover positive real x.

For example, if x = 2.4, we write f(x) = x + x + 0.4x. If x = 2.5, we write f(x) = x + x + 0.5x, and so on.

Now, having restored continuity, we can again pose the question: why is the derivative of this function at x = 2.4 not equal to 1 + 1 + 0.4?

[Dr. Strange]

14. Then I say it's more "chain rule-y."

When taking the derivative of x + x + ... + x (x times) to get 1 + 1 + ... + 1 (x times), we've taken into account the rate of change of each x--but we haven't taken into account the fact that the number of xs is also changing with respect to x, and should thus figure into the derivative.

[pokerfan]

15. yeah, that's pretty much it

Here's the solution: http://www.qbyte.org/puzzles/p032s.html

The derivative at x = a is defined as the limit, as h tends to zero, of / h.
The fallacy lies in writing f(a+h) as a· (a+h). (From which f'(a) = a.)
The correct formulation is: f(a+h) = (a+h)· (a+h). Expanding, we find f'(a) = 2a, as expected!

[Lethe]

13. the definition of derivatives comes from limits

(x+deltax)^2 - x^^2 / deltax

=deltax(2x + deltax^2) / deltax

= 2x + deltax

which the limit of deltax->0 = 0

therefore it = 2x

problem solved

[billyoc]

17. Nice derivative, Hitler.

:evilgrin: