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Basic calculus question

  1. Dec 8, 2004 #1
    Hi

    I have a basic question on calculus ....please can someone explain to me :


    Let f(x) = x^2

    y + delta y = f ( x + delta x )

    The book says that a small increase in x will cause a small increase in y.

    But if I put lets say 3 and 2 for x and dx y is 5^2 which is 25.

    But this isn't the same as f(3) + f(2) which equals 13 ?

    What have I done wrong ?




    Also, can f ( a+b+c ) be treated as f(a) + f(b) + f(c) ?


    Thanks a lot for any help


    Roger
     
  2. jcsd
  3. Dec 8, 2004 #2
    As I've always been told, dx is an infinitesimal change in x, so really really really small. Also, f(a + b + c) can't be treated as f(a) + f(b) + f(c), I don't think. f(a + b + c) can't be treated like that in your example anyway:

    f(x) = x^2
    f(a + b + c) = (a + b + c)^2
    f(a) + f(b) + f(c) = a^2 + b^2 + c^2
     
  4. Dec 8, 2004 #3
    f(a+b+c) doesn't wqual to f(a)+f(b)+f(c)

    now suppose that f(x)=X+1

    and let a=1 & b=2 & c=3

    so a+b+c=6

    f(6)=6+1=7

    but

    f(1)=1+1=2
    f(2)=2+1=3
    f(3)=3+1=4

    f(1)+f(2)+f(3)=2+3+4=9

    that is f(a+b+c) does not equal to f(a)+f(b)+f(c)
     
  5. Dec 8, 2004 #4

    HallsofIvy

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    In general f(x+ y) is NOT the same as f(x)+ f(y).

    Except for some extremely weird function, the only functions for which
    f(x+ y)+ f(x)+ f(y) are f(x)= Cx- that is, a constant time x.
     
  6. Dec 8, 2004 #5
    Sorry I don't quite understand the last bit ?
    please can you explain further ?




    But my main question is the first section I originally wrote :

    Let f(x) = x^2

    y + delta y = f ( x + delta x )

    The book says that a small increase in x will cause a small increase in y.

    But if I put lets say 3 and 2 for x and dx y is 5^2 which is 25.

    But this isn't the same as f(3) + f(2) which equals 13 ?



    I understand dx is infinitesimall but I just plugged in a larger value for ease of calculation.

    In the example above, why doesn't change in x give a change in y ?

    The book gave the statement I outlined in red above.......

    But isn't it strictly speaking inaccurate because a small change in x gives a small change in y but the value of that small change in y is still bigger than x
    because the function is applied to x ( in this case f(x)=x^2 ) ?


    please can someone explain ?
    thanks

    roger
     
    Last edited: Dec 8, 2004
  7. Dec 8, 2004 #6

    matt grime

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    Do the math, as they say:

    (x+dx)^2 = x^2+2xdx+(dx)^2

    where on earth do you even get that f(2)+f(3) should be 25? What are you doing there?

    Are you saying you think dy = f(dx)? Cos that's how it appears.

    dy = f'(x)dx +o((dx)^2) is what's going on really,.
     
    Last edited: Dec 8, 2004
  8. Dec 8, 2004 #7

    Thanks

    Roger
     
  9. Dec 8, 2004 #8

    matt grime

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    Obviously f(x+dx) doesn't equal f(x)+f(dx), and no one says it does.

    A small change of dx changes y by approximately dx.f'(x) at x. that is what the derivative does.
     
  10. Dec 8, 2004 #9
    Is it a small change in x or dx ?

    Why is it only approximate ?
     
  11. Dec 8, 2004 #10

    arildno

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    roger:
    You should ALWAYS think of [tex]\bigtriangleup{y}[/tex] as (when y=f(x)):
    [tex]\bigtriangleup{y}=f(x+\bigtriangleup{x})-f(x)[/tex]
    Hence, in your example, we have:
    [tex]\bigtriangleup{y}=2x\bigtriangleup{x}+(\bigtriangleup{x})^{2}[/tex]
    When [tex]\bigtriangleup{x}[/tex] is TINY, the second term on the right-hand side is much less in magnitude than the first term (provided x is non-zero).
    Hence, we may then write:
    [tex]\bigtriangleup{y}\approx{2x}\bigtriangleup{x}[/tex]
    which you should recognize as:
    [tex]\bigtriangleup{y}\approx{f}'(x)\bigtriangleup{x}[/tex]
     
  12. Dec 9, 2004 #11

    matt grime

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    To repeat myself:

    (x+dx)^2 = ......
     
  13. Dec 9, 2004 #12
    thanks.

    roger.
     
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