Math gurus UNITE!

Our calculus professor gave us an extra credit assignment the other day, and I just can't seem to figgure it out for the life of me. Any help on it would be greatly appreciated.

Its basically a proof- each step is listed below. Obviously, the proof doesn't hold true, but I need to know where it falls apart.

When it say X times, it simply means that there are X values in the set. Meaning 1 + 1 + ... + 1 (X Times) Means there are X number of 1's in the equation.

Also, x^2 is my notation for x squared (x*x). In case thats a bit vague and all.

And D(x) means to take the derivative of whatevers in the ()'s.

---

For x > 0:

x = 1 + 1 + ... + 1 (x times)

x^2 = x + x + ... + x (x times)

D(x^2) = D(x + x + ... + x (x times) )

D(x^2) = D(x) + D(x) + ... + D(x) (x times)

2x = 1 + 1 + ... + 1 (x times)

2x = x

2 = 1

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Any ideas? Sorry if this is vague, but I really want to know what is going on here. Especially since the prof. is evil about stuff like this and probably won't tell us the answer.

Thanks in advance.

The error actually occurs at the beginning:

For x > 0,

x = 1 + 1 + 1 + 1 .... (x times)

That's just not true in general. It's only true for integers (like -5, 0, 1, 10). For example, what does the expression mean if you set x = 0.5? Or even worse, what about x = pi?

Why doesn't the proof work if x is an integer? That would still "prove" that 2 = 1.

The rule,

D(x^2) = 2x

...is not a rule for a particular number x. It is a rule for the function x^2, which is a continuous (and hence differentiable) function.

What the proof is trying to say is something like this:

Hey Monqui, you know that 2x is the derivative of x^2, right? Now that's true at every point x, so in general the derivative of a real number must be twice its square root, and in particular this must be true of an integer.

But that's certainly not true. Let's take k = 16. Then D(k) = 8, right? But now express k as part of the function 2x. D(2x) = 2 for all x, so D(k) = D(8) = 2. But then 2 = 8.

The point is that there is no such thing as the derivative of a number. Only functions have derivatives.

Yes, in saying that x^2 = x + x + .. + x it's not continous and you can't differentiate.

you can tell who got to go to a decent english school cant ya :/

yeah yeah, I knew that and my school was bad enough that one of the uni's I applied for gave me a special concession where I needed less points to get in becasue of my school :razz:

I knew it would be something obvious like that... Thanks a ton guys!

When he introduced the problem, he said that the less you know about derivitives the better. He normally gives the assignment out just before spring break so that the students can take the problem to their old teachers from high-school. The normal reaction by them is supposedly "wtf."

Ugh. I feel so stupid now :/

:eek:

It's topics like these that remind me why I come here :smile:

Interesting point, I heard that Isaac Newton invented Calculas while a student at Cambridge university, however decided not to tell anyone for about 27 years.

I heard that Newton and Leibniz could never agree on who had first invented calculus, and that the consensus now is "they both did".

Yeah, Gollum is right. And I believe that if you go to German schools (in Germany), they get really offended if you tell them Newton invented it. Although it really isn't fair to say that one "invented it," since they really did come up with many of the same concepts around the same time.

Plus it depends on whether you 'invent' or 'discover' maths... :razz: (I know half the crap they teach us in Algebra is blatantly made up though!)

You lost me at 'x ='

:biggrin:

Yeah, I suppose inventing math would involve something like "x := 2 + 5 => x = 4" or something.

edit

I need to stop doing Ada...

I knew it would be something obvious like that

...



heh, you lost me at Our calculus professor