Why is this wrong?

[AvgStudent]

[math]\underbrace{x+x+x+...+x}_\text{x times}=x*x=x^2[/math]Differentiate both sides:
[math]\underbrace{\frac{dx}{dx}+\frac{dx}{dx}+\frac{dx}{dx}+...+\frac{dx}{dx}}_{\text{x times}}=2x[/math][math]\underbrace{1+1+1+...+1}_{\text{x times}}=2x[/math][math]x=2x[/math][math]1=2[/math]

 

[topsquark]

What if x is not an integer?

-Dan

 

[Steven G]

I just have a few question for you.

What do you get if you add 1 to itself 7 1/2 times?
What do you get if you add 1 to itself 2.92 times?
What do you get if you add 1 to itself sqrt(2) times?
What do you get if you add 1 to itself -5 times?
and my favorite one--What do you get if you add 1 to itself pi times?

 

[AvgStudent]

and my favorite one--What do you get if you add 1 to itself pi times?

I try to avoid [imath]\pi[/imath]. It'll go on and on forever

 

[Steven G]

I try to avoid [imath]\pi[/imath]. It'll go on and on forever

Why will it go on forever? How about sqrt(2), will it go on forever? How about 1/3?
If I asked you to add 1 to itself pi times, then it'll go on and on forever.
The only 'it' I can think of is the adding of 1's.
So if you add 1 to itself pi times, then you get 1+1 + 1 + ... = infinity?
That is strange as pi is less than 4 and if I add 1 to itself 4 times, 1+1+1+1, I get 4.
The point topsquark and I are trying to make is that you can't add 1 to itself other than a positive integer amount of times.

 

[BigBeachBanana]

Why will it go on forever? How about sqrt(2), will it go on forever? How about 1/3?
If I asked you to add 1 to itself pi times, then it'll go on and on forever.
The only 'it' I can think of is the adding of 1's.
So if you add 1 to itself pi times, then you get 1+1 + 1 + ... = infinity?
That is strange as pi is less than 4 and if I add 1 to itself 4 times, 1+1+1+1, I get 4.
The point topsquark and I are trying to make is that you can't add 1 to itself other than a positive integer amount of times.

I think he/she got the point. It was a pi joke.

 

[Harry_the_cat]

What if x is not an integer?

-Dan

But what if x IS an integer? What's wrong with the argument then?

 

[Cubist]

I think the main problem here is that the "x times" notation is really an obfuscated, hidden, operator that is being missed out in the differentiation process.

--

If this is changed to a constant "c times" then it works...

[math]\underbrace{x+x+x+...+x}_\text{c times}=c*x[/math]
[math]\underbrace{\frac{dx}{dx}+\frac{dx}{dx}+\frac{dx}{dx}+...+\frac{dx}{dx}}_{\text{c times}}=c[/math]
[math]\underbrace{1+1+1+...+1}_{\text{c times}}=c[/math]
[math]c=c[/math]
--

If we make the operator more explicit then it works...

[math]\underbrace{x+x+x+...+x}_\text{x times}=x^2[/math]
[math]x*x=x^2[/math]
Differentiate (product rule on the LHS):

[math]x*\frac{dx}{dx}+\frac{dx}{dx}*x=2x[/math]
[math]2x=2x[/math]