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Reducing the function of a function to the ind. variable

  1. Mar 15, 2015 #1

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    Hey, PF:

    If I have a function ##f(x)## where ##x## is itself a function of another variable (say time), is the following then true? $$f(x)=f(x(t))=f(t)$$

    I ask this because if I have the following system of differential equations

    $$\frac{dA}{dt}=-Bb$$
    $$\frac{dB}{dt}=-Aa$$

    where litte ##a## and ##b## are constants, and I solve for ##B(t)## and ##A(t)## as functions of ##t## and solve for ##B(A)## and ##A(B)## as functions of each other, can I reduce ##A(B)## to ##A(t)##?

    In my mind, it makes sense that $$\begin{align} B(A)=&B(A(t)) \\ =&B(t)\end{align}$$ If this is in fact true, what is this process, or principle, called?

    Here's the link outlining the problem:
    http://courses.ncssm.edu/math/POW/POW07_08/Calculus Challenge #14 SOLUTION.pdf

    I'm specifically trying to understand how ##A(t)## on pg. 4 relates to ##A## (As a function of ##B##) on page 5.

    Please keep in mind that my background in math is limited to some multivariable calculus and some ordinary differential equations.

    Thank you,
     
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  3. Mar 15, 2015 #2

    andrewkirk

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    I'm afraid you can't do what you suggest. If you have B as a function of A and A as a function of t, then to express B as a function of t you need to do what is called 'function composition', and that will give a different result from what you have suggested.

    An easy way to approach (4) is to divide the first equation by the second, which gives:
    [itex]\frac{\frac{dA}{dt}}{\frac{dB}{dt}}={\frac{bB}{aA}}[/itex]
    Cross-multiplying we get
    [itex]aA\frac{dA}{dt}=bB\frac{dB}{dt}[/itex]
    Integrate both sides with respect to [itex]t[/itex] gives
    [itex]\frac{a}{2}A^2=\frac{b}{2}B^2+C[/itex] for some [itex]C[/itex] that is not a function of [itex]t[/itex].
    Take the square root, rearrange and you have [itex]A[/itex] as a function of [itex]B[/itex] and the unknown constant [itex]C[/itex]. The latter can then be determined by the initial conditions you are given, ie the initial sizes of the two fleets: [itex]A_0[/itex] and [itex]B_0[/itex].
     
  4. Mar 15, 2015 #3

    Fredrik

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    There seems to be some confusion here between functions, and variables that are functions of some other variable. If x and y are variables that represent real numbers, then a constraint like x+y=1 would make y a "function of" x in the sense that the value of y is completely determined by the value of x. But y is still not a function. It started out as a variable that represents a real number, and it can't be turned into something else by an equation that describes how its value is related to the value of another variable.

    However, we can define a function f by f(r)=1-r for all real numbers r. Then if the values of x and y are consistent with the constraint, we have y=f(x). It would be confusing to denote this function by y instead of f. This is the sort of thing you're doing.

    Consider e.g. the kinetic energy T at time t of a mass m dropped at time 0. If we denote its velocity at time t by v, the values of v,t and T will satisfy the equations v=gt and T=½mv2, and therefore also the equation T=½mg2t2. So we have ½mv2=T=½mg2t2. Now we can introduce functions T1 and T2, defined by ##T_1(r)=\frac 1 2mr^2## and ##T_2(r)=\frac 1 2 mg^2r^2## for all ##r\in\mathbb R##. If the values of v and t are consistent with the constraints, we have ##T_1(v)=T_2(t)##. But we certainly don't have ##T_1=T_2##, as you can see by verifying that ##T_1(1)=m/2## and ##T_2(1)=g^2/2##.
     
  5. Mar 15, 2015 #4

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    Thanks for your responses, Fredrik and Andrew (I'm still trying to understand your post, Fredrik).

    I ran into function composition and decomposition while trying to understand this problem, and if I recall correctly, it seems like most function composition consists of substitution (which I thought could be applied in this context). Looking back at the solutions, it seems to me that they might in fact be a pair of parametric equations as a function of the parameter, ##t##. If this is the case, can't we then apply function composition to the parametric solutions ##A(t)## and ##B(t)## to derive the solutions that relate ##A## as a function of ##B##?

    In the broader sense, I suppose I would like to know whether it is possible to work backwards and derive algebraically from a pair of non-differential parametric equations as a function of ##t## an implicit function of ##t## which relates both original functions to each other (without utilizing any calculus).

    Here's a resource which mentions function composition of parametric equations (see pg. 5,6)
    gmu.edu
     
  6. Mar 15, 2015 #5

    Stephen Tashi

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    It's not clear what process are describing by the above statement.

    The notation for functions is ambiguous. In mathematics, a function has a domain and a codomain. The abbreviation f(t) for a function indicates a element of the codomain will be denoted by f(t) and an element of the domain will be denoted by t. There is no requirement that a function f(t) be given by a simple formula or expression such as f(t) =3 + sin(t).

    However, in practical applications a function f(t) is often associated with a formula or algorithm. When that is the case, the notation f(t) also conveys the meaning of "apply the formula for the function to the value t".

    For example, if the energy expenditure of a machine is a function of the distance it moves and the distance it moves is a function of time then one may think of energy E as being "a function of time" and express it as E(t). However, if we are dealing with specific formulas like E(d) = 3d + 6 giving energy as a function of distance and d(t) = exp(2t) then the notation E(t) would be interpreted as E(t) = 3t + 6 which would not be the function E(d(t)) = 3 exp(2t) + 6, which is the one that is appropriate to this example.
     
  7. Mar 15, 2015 #6

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    Thanks for the example and clarification, Stephen!

    So I suppose I used the wrong notation to express my question. It still seems to me that ##A(t)## and ##B(t)## can be use to express either ##B## or ##A## as an implicit function of ##t##. This would be my ##A## as a function of ##B## or vice versa.
     
  8. Mar 15, 2015 #7

    Fredrik

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    I know that you know everything I'm about to say already, but the OP doesn't.

    f(t) isn't the notation for a function. It's the notation for a number, specifically the output produced by f when the input is t. Since the value of f(t) (i.e. the number that the notation represents) is completely determined by the value of t, one can still say that f(t) "is a function of" t.

    "t" is a variable that represents a number.
    "f" is a variable that represents a function.
    "f(t)" is an expression (=string of text) that represents a number.
    f(t) is a function of t in the sense that the value of (=number represented by) the expression "f(t)" is determined by the value of the variable "t".

    It's (unfortunately) fairly standard to say that f(t) is a function, even though the expression "f(t)" represents a number. This is an abuse of terminology. Most people seem to be OK with it. I have a strong dislike for it, but even I use it occasionally, because statements like
    "The derivative of the function f such that f(x)=x3 for all real numbers x is the function g such that g(x)=3x2 for all real numbers x."​
    are just annoying. So I say "the derivative of x3 is 3x2", and cringe a little every time.

    Let me know if there's anything I need to clarify or elaborate on.
     
  9. Mar 15, 2015 #8

    andrewkirk

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    It is not always possible in practice. To make it work you generally need to be able to make ##t## the subject of your equation that gives ##A## in terms of ##t##. If you can get that, you can substitute it for ##t## in your equation that gives ##B## in terms of ##t## and voila, you have an eqation for ##B## in terms of ##A## with no refs to ##t##.

    That is easy with simple equations like linear, quadratic, or basic log or trig, but in general it is not possible to get a formula for ##t## in terms of ##A##. So, given a value of ##B##, one has to use numerical techniques to find ##A##.

    The example you were given is simple enough that this technique can work.
     
  10. Mar 15, 2015 #9

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    Awesome explanation! That was exactly what I was trying to express, but you managed to say it more elegantly. I think I'm beginning to understand this problem at a deeper level now.

    By numerical techniques, I assume you mean we cannot apply symbolic techniques in those cases. I would also assume that no symbolic solution exists to relate ##A## to ##B## in those more complicated cases.

    A few quick follow-up questions:

    1. Would you happen to have an example of one of those cases where a numerical method must be used?
    2. Does this topic have a specific name or classification, or does it all fall under function composition. (I'm interested in reading up on it.)
     
  11. Mar 15, 2015 #10

    andrewkirk

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    Yes, 'numerical techniques' means using a computer to find the solution, because no symbolic formula can be written for it. Most polynomial equations of order 5 or more cannot be solved symbolically. So ##A=t^5+t^4+t^2+t+3## might be an example. There is no formula for the solutions but computer algorithms exist to search and approximate them. I'm afraid I don't know a specific name for this topic, but the areas of 'Theory of Equations' and 'Numerical Analysis' have considerable overlap with it.
     
  12. Mar 15, 2015 #11

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    Thanks for your enlightening responses, Andrew!
     
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