## asnaseer Group Title How good is your understanding of differentiation? $$y=x^2$$ can be written as $$y=x*x$$ which can also be written as $$y=x+x+x+...+x$$ (i.e. "x" lots of x). using this we get:\begin{align} y&=x^2\\ \therefore \frac{dy}{dx}&=2x\\ &\text{but we can also write this as:}\\ y&=x+x+x+...+x\quad\text{i.e. "x" lots of x}\\ \therefore \frac{dy}{dx}&=1+1+1+…+1\quad\text{i.e. "x" lots of 1's}\\ &=x\\ \therefore 2x&=x\\ \therefore 2&=1 \end{align}where is the mistake in this? 2 years ago 2 years ago

1. Mr.Math Group Title

In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.

2. asnaseer Group Title

it is differentiable and is continuous - just like $$y=x^2$$

3. Mr.Math Group Title

Not really. x here can take only integer values, so it's defined from N to R. While y=x^2 is defined from R to R. So they are tow different functions.

4. agdgdgdgwngo Group Title

your mistake is that you didn't differentiate the sum of all the x's properly

5. agdgdgdgwngo Group Title

but I forgot the exact mistake :-P

6. asnaseer Group Title

you can take one and half lots of something

7. Mr.Math Group Title

If we put f(x)=y=x+x+..+x (x times), could you tell me what f(0.5) for instance?

8. asnaseer Group Title

f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25

9. TuringTest Group Title

how can that be?

10. Mr.Math Group Title

Yeah, how can that be?

11. asnaseer Group Title

Mr.Math - you are probably correct in the strictest sense, but agdgdgdgwngo is closer to where the mistake is.

12. Akshay_Budhkar Group Title

how is f(0.5)= 0.25??

13. asnaseer Group Title

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

14. agdgdgdgwngo Group Title

or $\sum_{i=1}^{x}x$

15. asnaseer Group Title

I didn't what to use that because it implies x must be an integer

16. agdgdgdgwngo Group Title

oh

17. ktklown Group Title

What does "taken x times" mean then if not a sum?

18. asnaseer Group Title

but I guess Mr.Math is right - we cannot say x^2=x+x+x+... (x times) for non-integers?

19. Mr.Math Group Title

Just quickly 2x=x means x=0, and doesn't mean 2=1.

20. asnaseer Group Title

yes Mr.Math - the real question is, how can I get two different results by differentiating the /same/ function?

21. Akshay_Budhkar Group Title

x cannot be zero?

22. asnaseer Group Title

at ostensibly the same function

23. asnaseer Group Title

*at least

24. agdgdgdgwngo Group Title

think of the product rule

25. TuringTest Group Title

I don't think you can differentiate a variable number of terms like that, but I can't seem to apply the right rule.

26. Mr.Math Group Title

f(x)=x+x+..+x (x times). The key is "x" times, since we're differentiating with respect to x. If it's f(x)=x+x+..+x ("n" times), what we did would be correct.

27. asnaseer Group Title

Mr.Math has hit the nail on the head!

28. TuringTest Group Title

but which rule exactly are we breaking and how?

29. Akshay_Budhkar Group Title

the product rule?

30. Mr.Math Group Title

then df/dx=(1+1+1..+1)+(1+1+..+1) (x times) for each brackets I think.

31. agdgdgdgwngo Group Title

right

32. asnaseer Group Title

if you re-did this using the definition of differentiation then you would get the right result. i.e.$f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$

33. asnaseer Group Title

@TuringTest - the being broken is to treat "x" as a constant when it is not

34. asnaseer Group Title

*the rule

35. Mr.Math Group Title

Or the chain rule, let x+x+..+x (x times)=u, then we can apply the chain rule I think.

36. TuringTest Group Title

that's what I said, a variable number of terms. I just want to make it concrete in my mind.

37. asnaseer Group Title

I guess I've also learnt a thing or do from your replies on what a continuous function is - thx all

38. asnaseer Group Title

*or two

39. Akshay_Budhkar Group Title

but at the first place is x+x+x+... xtimes a continuous function to be differentiated?

40. asnaseer Group Title

@Akshay Budhkar - if you use this to evaluate the differential of y = x+x+...+x (x times), then you will get the right answer:$f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$

41. agdgdgdgwngo Group Title

btw asnaseer who is the guy in your profile pic? I can't stop wondering! :(

42. asnaseer Group Title

that highlights the fact that you cannot treat the x in "x times" as a constant

43. asnaseer Group Title

@agdgdgdgwngo - thats me in the pic - why? do I look odd?

44. Akshay_Budhkar Group Title

No i was just wondering if the function was continuous, i know that formula, but if a function is not continuous we cannot differentiate it, and i feel this is not a continuous function

45. asnaseer Group Title

yes - Mr.Math pointed that out earlier - I guess I used a bad example of trying to illustrate a specific point :-(

46. Mr.Math Group Title

Wait!! $f'(x)=\lim_{h\to 0}{x+h+x+h+\dots+x+h-x-x-\dots-x \over h}$ $=\lim_{x\to 0}{xh \over h}=x$

47. Mr.Math Group Title

$$h\to 0$$*

48. asnaseer Group Title

no Mr.Math - the first set of "x+h" terms will appear "x+h" times

49. FoolForMath Group Title

This is a old problem.

50. Akshay_Budhkar Group Title

you came late ffm that is why it is old :P

51. asnaseer Group Title

yes FFM - I came across it several years ago and just recalled it :-)

52. FoolForMath Group Title

There has been a discussion on this one, let me do some digging.

53. asnaseer Group Title

you mean here on OS?

54. Mr.Math Group Title

Right! $\lim_{h \to 0}{x+h+x+h+..+x+h-x-x-\dots -x \over h}$ $=\lim_{h \to 0}{hx+(x+h)h \over h}=2x$

55. asnaseer Group Title

thats it Mr.Math - well done!

56. Akshay_Budhkar Group Title

So what is the conclusion of this question ( neglecting the fact that the function is discontinuous) ?

57. asnaseer Group Title

the mistake was in treating the "x" in "x times" as a "constant" when differentiating - this is wrong

58. Akshay_Budhkar Group Title

yea get it :D

59. FoolForMath Group Title
60. asnaseer Group Title

I wasn't aware of that FFM - I ran into this many many years ago at uni

61. TuringTest Group Title

they mention the integer issue as well...

62. FoolForMath Group Title

It's okay, I can't imagine why you mean many many and many :D

63. asnaseer Group Title

;-D

64. Akshay_Budhkar Group Title

yea @turing