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asnaseer

  • 4 years ago

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?

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  1. Mr.Math
    • 4 years ago
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    In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.

  2. asnaseer
    • 4 years ago
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    it is differentiable and is continuous - just like \(y=x^2\)

  3. Mr.Math
    • 4 years ago
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    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
    • 4 years ago
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    your mistake is that you didn't differentiate the sum of all the x's properly

  5. agdgdgdgwngo
    • 4 years ago
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    but I forgot the exact mistake :-P

  6. asnaseer
    • 4 years ago
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    you can take one and half lots of something

  7. Mr.Math
    • 4 years ago
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    If we put f(x)=y=x+x+..+x (x times), could you tell me what f(0.5) for instance?

  8. asnaseer
    • 4 years ago
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    f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25

  9. TuringTest
    • 4 years ago
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    how can that be?

  10. Mr.Math
    • 4 years ago
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    Yeah, how can that be?

  11. asnaseer
    • 4 years ago
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    Mr.Math - you are probably correct in the strictest sense, but agdgdgdgwngo is closer to where the mistake is.

  12. Akshay_Budhkar
    • 4 years ago
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    how is f(0.5)= 0.25??

  13. asnaseer
    • 4 years ago
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    f(x)=x^2=x+x+... (taken x times)

  14. agdgdgdgwngo
    • 4 years ago
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    or \[\sum_{i=1}^{x}x\]

  15. asnaseer
    • 4 years ago
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    I didn't what to use that because it implies x must be an integer

  16. agdgdgdgwngo
    • 4 years ago
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    oh

  17. ktklown
    • 4 years ago
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    What does "taken x times" mean then if not a sum?

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

  19. Mr.Math
    • 4 years ago
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    Just quickly 2x=x means x=0, and doesn't mean 2=1.

  20. asnaseer
    • 4 years ago
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    yes Mr.Math - the real question is, how can I get two different results by differentiating the /same/ function?

  21. Akshay_Budhkar
    • 4 years ago
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    x cannot be zero?

  22. asnaseer
    • 4 years ago
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    at ostensibly the same function

  23. asnaseer
    • 4 years ago
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    *at least

  24. agdgdgdgwngo
    • 4 years ago
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    think of the product rule

  25. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    Mr.Math has hit the nail on the head!

  28. TuringTest
    • 4 years ago
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    but which rule exactly are we breaking and how?

  29. Akshay_Budhkar
    • 4 years ago
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    the product rule?

  30. Mr.Math
    • 4 years ago
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    then df/dx=(1+1+1..+1)+(1+1+..+1) (x times) for each brackets I think.

  31. agdgdgdgwngo
    • 4 years ago
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    right

  32. asnaseer
    • 4 years ago
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    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
    • 4 years ago
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    @TuringTest - the being broken is to treat "x" as a constant when it is not

  34. asnaseer
    • 4 years ago
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    *the rule

  35. Mr.Math
    • 4 years ago
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    Or the chain rule, let x+x+..+x (x times)=u, then we can apply the chain rule I think.

  36. TuringTest
    • 4 years ago
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    that's what I said, a variable number of terms. I just want to make it concrete in my mind.

  37. asnaseer
    • 4 years ago
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    I guess I've also learnt a thing or do from your replies on what a continuous function is - thx all

  38. asnaseer
    • 4 years ago
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    *or two

  39. Akshay_Budhkar
    • 4 years ago
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    but at the first place is x+x+x+... xtimes a continuous function to be differentiated?

  40. asnaseer
    • 4 years ago
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    @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
    • 4 years ago
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    btw asnaseer who is the guy in your profile pic? I can't stop wondering! :(

  42. asnaseer
    • 4 years ago
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    that highlights the fact that you cannot treat the x in "x times" as a constant

  43. asnaseer
    • 4 years ago
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    @agdgdgdgwngo - thats me in the pic - why? do I look odd?

  44. Akshay_Budhkar
    • 4 years ago
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    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
    • 4 years ago
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    yes - Mr.Math pointed that out earlier - I guess I used a bad example of trying to illustrate a specific point :-(

  46. Mr.Math
    • 4 years ago
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    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
    • 4 years ago
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    \(h\to 0\)*

  48. asnaseer
    • 4 years ago
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    no Mr.Math - the first set of "x+h" terms will appear "x+h" times

  49. FoolForMath
    • 4 years ago
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    This is a old problem.

  50. Akshay_Budhkar
    • 4 years ago
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    you came late ffm that is why it is old :P

  51. asnaseer
    • 4 years ago
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    yes FFM - I came across it several years ago and just recalled it :-)

  52. FoolForMath
    • 4 years ago
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    There has been a discussion on this one, let me do some digging.

  53. asnaseer
    • 4 years ago
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    you mean here on OS?

  54. Mr.Math
    • 4 years ago
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    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
    • 4 years ago
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    thats it Mr.Math - well done!

  56. Akshay_Budhkar
    • 4 years ago
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    So what is the conclusion of this question ( neglecting the fact that the function is discontinuous) ?

  57. asnaseer
    • 4 years ago
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    the mistake was in treating the "x" in "x times" as a "constant" when differentiating - this is wrong

  58. Akshay_Budhkar
    • 4 years ago
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    yea get it :D

  59. FoolForMath
    • 4 years ago
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    Here is it http://math.stackexchange.com/questions/1096/

  60. asnaseer
    • 4 years ago
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    I wasn't aware of that FFM - I ran into this many many years ago at uni

  61. TuringTest
    • 4 years ago
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    they mention the integer issue as well...

  62. FoolForMath
    • 4 years ago
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    It's okay, I can't imagine why you mean many many and many :D

  63. asnaseer
    • 4 years ago
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    ;-D

  64. Akshay_Budhkar
    • 4 years ago
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    yea @turing

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