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I Why is this function constant in this interval?

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  1. Feb 18, 2017 #1
    This question has a little bit of physics in it, but it's mostly maths.

    If I have force, or any function f(z), I was told that I can assume it to be constant only in the interval dz.

    However, in this case, I had to calculate the work done by the spring force as a function of y

    img_20170218_162826_01-min-jpg.113388.jpg

    Over here, I assumed the spring force, which is a function of its elongation x (F = -kx) to be constant in the interval dy and integrated and this gave me the correct answer

    I want to know why the error vanished over here. Shouldn't spring force only be constant in the interval dx and not dy?

    I also want to know, in general, if I have a function, how to decide whether it is constant in some particular interval/in which cases the error will vanish as I take the limit and integrate. Or are forces/functions constant for any infinitesimal intervals such as Rdθ, dy/cosθ, dz etc etc.?
     
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  3. Feb 18, 2017 #2

    haruspex

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    It depends whether the derivative is bounded. If it is, we can approximate the function in the vicinity of a point using the Taylor expansion: f(x+dx)=f(x)+f'(x)dx+... So for small dx f(x+dx) is approximately f(x). But this will not work for x ln(x) in the vicinity of 0.
     
  4. Feb 18, 2017 #3
    I understand that I can assume f(x) to be constant in the interval [x, x+dx), but in my case, I have assumed it to be constant in an unrelated interval [y,y+dy) and still gotten the correct answer. I want to know why, and if I can, in general, assume any function to be constant in any infinitesimal interval such as Rdθ, dy/cosθ, dz etc
     
  5. Feb 18, 2017 #4

    haruspex

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    It is not unrelated. x and y are directly related, so you can recast f as a function of y.
     
  6. Feb 19, 2017 #5
    I don't understand how being able to recast f as a function of y is relevant.
    Also, can I, in general, assume a force or function to be constant in any infinitesimal interval such as Rdθ, dy/cosθ, dz etc
     
  7. Feb 19, 2017 #6

    Stephen Tashi

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    The existence of a definite integral of a function does not require that the function be constant in any particular interval. The notion that a function ##f(x)## must be "constant" is small interval of length ##dx## in order for ##\int f(x) dx## to exist isn't a correct mathematical statement. The correct statement is that if ##f(x)## is continuous in the closed interval ##[a,b]## then ##\int_a^b f(x) dx## exists and can be computed, in the usual way, by using an antiderrivative. (See "The Fundamental Theorem of Calculus".)

    A person trying to formulate an intuitive explanation of what "continuous" means to a child might resort to using the words that ##f(x)## is "constant" in each "infinitesimal" interval of length ##dx##, but that is a mangled description of the actual meaning of "continuous".
     
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