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B Can you reason out a differential equation using logic?

  1. Jul 27, 2016 #1
    Let us suppose that we are modeling a particle traversing some distance in space & that f(x)=x². Let the x-axis be time (hrs) & the y-axis be position in space (miles). What is the instantaneous velocity of said particle at t=4 hours?

    Let's assign variables to each coordinate in our x, y table. So A(2,4), B(3,9), C(4,16), etc. & see what we can figure out. From A to B, the particle traversed a distance of 5 miles, hence 9-4=5. From B to C, the particle traversed a distance of 7 miles. C to D, 9 miles, & so on. What's the particle doing? The particle is traversing a larger & larger distance over each time interval. It's accelerating. Well at what rate? At a rate of 2mph/hr. In terms of variables, the y variable is accelerating twice as fast with respect to the x variable, hence f'(x)=2x. Using this argument, I can solve the velocity (the slope) at any point on the curve x², including x=4, which is 8mph.

    The wonderful thing here is that you could reason this problem out without using any calculus. No need for limits here. You can also use this argument to find the total distance traversed over any time interval. My younger sibling told me I wasn't smart enough to reason out a problem involving a differential equation & so farshe's right! I can find h(t) of a falling body using f=ma, but I can't reason it out like I can do for the differential & the integral calculus. I'm sure there are many individuals on this forum that are much brighter than I am. Maybe give it a go? & don't feel limited to falling bodies either...

    Thanks!



    *I am aware that acceleration is typically used to describe the 2nd derivative, especially in physics problems like the one I listed. I have used acceleration to describe the 1st derivative ...even though that's exactly what acceleration is ...the rate at which something is changing with respect to the other. Anyways!
     
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  3. Jul 28, 2016 #2
    Let's ignore the fact that you've given the function as a closed form and just focus on the points you've chosen for now. The way you've solved it, it is not immediately obvious that this formula holds for non-integer inputs. For instance, I see no reason that the function you're referring to isn't just a sequence of straight lines whose slope doubles every time you hit another integer.

    To accommodate those numbers in between, you'd in essence have to, for every point (assuming you know all of the points on the graph) calculate the small change in output as you vary the input by a small amount, to make sure all those middle values are accounted for. But that's just a derivative, isn't it?*


    *Of course, derivatives are generally defined rigorously in terms of limits, but the explanation can still stand.
     
  4. Jul 28, 2016 #3
    I don't understand what "closed form" means.

    I'll choose some new points for you. A(.5,.25), B(1,1), C(1.5, 2.25), D(2,4), E(2.5,6.25), F(3,9), G(3.5,12.25), H(4,16), I(4.5,20.25). From A to C the particle traversed 2 miles, from B to C the particle traversed 4 miles, from C to E 6 miles, & so on. Half the x inputs I've included are non-integers & the rate that I've already defined, 2mph/hr, still holds true ...between the whole integer inputs.

    You could come back at me again with the same argument you had before about the sequence of straight lines & all, but I could cut my x inputs in half again, you know, just to make sure that rate of change is the same between the [integer & non-integer] x inputs I've already listed, & you & I both know it'd still be 2mph/hr. Heck, we could do this indefinitely. It's kind of like that delta epsilon definition that I think you're referring to where for any number...


    So yeah, I can do this forever & I'm going to say that the y's are changing twice as fast with respect to the x's EVERYWHERE along the curve x².

    -I do have a learning disability, but I know ^ is correct.




    Anyone interested in taking a stab at reasoning out a differential equation like I've done with the differential calculus?!
     
  5. Jul 28, 2016 #4
    My point is that you haven't solved this problem without using calculus, or by "logic alone," because in order to be sure that you can find the velocity everywhere, you need the derivative, which you get by applying your method to ever smaller intervals.

    Now, the closest thing I can think of to what you want is that many differential equations are solved simply by guessing the solution.
     
  6. Jul 28, 2016 #5
    Allow me to elaborate on "guessing the solution" with a simple example. Suppose we have the differential equation $$y + y' = 0$$
    We could solve this by separation of variables: rewrite as $$y + \frac{dy}{dx} = 0$$ or $$y = -\frac{dy}{dx} \Rightarrow \int dx = -\int \frac{dy}{y}$$ Finally, $$x = -\ln(y) + C_0 \Rightarrow y = Ce^{-x}$$

    Alternately, we could start with $$y = -y'$$ and ask ourselves "what function is equal to its own derivative?" Well, that's ##e^{x}##, but we have to accommodate that minus sign, so we make it ##e^{-x}##, and finally, we note that this equality also holds if the function is multiplied by some constant, so our solution is ##y = C e^{-x}##. So, we can go through the usual process to solve this ODE (separation of variables), but we can also deduce the answer by guessing a solution and modifying it a bit to meet our needs.
     
  7. Jul 28, 2016 #6
    You're both right & wrong. I was wrong in saying that could find the velocity everywhere. Let me restate it, f'(x)=2x, but only at the points that I've found this argument to be true. In my original example, this would satisfy x=1, x=2, x=3, x=4... The question was what was the particle's velocity at t=4. x=4 is a point that I've found my argument to be true, so I can indeed say that the particle's velocity at t=4 is 8mph. Who cares what the particles velocity was at t=3.9. That wasn't what I had asked & plus that's not at a point which I've found my argument to be true, so I can't apply f'(x)=2x anyways. I guess this rules out trying to integrate, because I can't prove to you what the particle was doing at t=3.9.

    I don't know how to express that constraint mathematically, but I can find this relationship (the derivative) using ANY two x points so I don't know how that isn't saying that it'd apply to ALL x inputs, but maybe you could clear that up for me. And if you disagree with me, then give me an x input & I'll make sure that I've found my argument to be true at that point. And then I'll be able to say that f'(x)=2x.



    Now I will ask again. You sure you can't reason out a differential equation? ...even if you'd be limited to certain points?
     
    Last edited: Jul 28, 2016
  8. Jul 29, 2016 #7

    Mark44

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    You might be able to come up with an educated guess, as @axmls was describing, but this works in only the simplest of cases.
     
  9. Jul 29, 2016 #8

    vela

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    What if the function were ##f(x) = x^2 + \sin \pi x##? Your values for A, B, and C would be the same, so your logic would lead again to ##f'(x) = 2x## and f'(4) = 8 mph, but neither of those conclusions is correct.
     
  10. Jul 30, 2016 #9
    Why is this not legitimate proof that f'(x)=2x of f(x)=x²? Because even though I've found this relationship (the derivative) to be true at all the points I've chosen so far, I can't prove to you it applies to all x inputs all at once? You can't prove to me that f'(x)=2x of f(x)=x² either. In order to prove that, hypothetically, h would have to be zero. But if it were zero, you'd get indeterminate form, so the best thing you can do is make h approach zero indefinitely & say "well so far, it looks like it's probably 2x." So why is it okay for you to make the jump & say it's 2x (even though you really can't prove it), but I can't say it when I'm making my intervals smaller? Because of the delta epsilon def'n? I can play that argument too. Read the second part of my statement over again > give me an x input & I'll make sure that I've found my argument to be true at that point. And then I'll be able to say that f'(x)=2x.

    I never said I could find the derivative of that particular function, but I guess if I say that I can apply my method to one function (f(x)=x²), then I should be able to apply it to other functions as well ...unless I can prove otherwise.

    And I would have approached that problem differently. Consider f(x)=x² again with points A(0,0), B(1,1), C(2,4), D(3,9), & E(4,16). If f'(x)=2x, then that would imply that the average velocity at each 1 hour time interval from t=0 to t=4 would be 4 mph, hence f'(0)=0, f'(4)=8. 8+0=8. 8/2=4. That quantity 4, multiplied by the time interval 4 would be 16. This would yield the total distance traversed. Looking at our f(x)=x² table, which is our position (in miles) VS time function, I can see that the particle indeed had traveled a total of 16 miles at the 4 hour time period.

    Now let us consider f(x)=x²+2x with points A(0,0), B(1,3), C(2,8), D(3,15), & E(4,24). Using my method, I would have argued that f'(x)=2x just like I had found in f(x)=x². Now let's find the total distance traversed at t=4 using f'(x)=2x. I would have found it to be 16 again, but if you look at the x y table of f(x)=x²+2x at x=4, you'll see it's 24, not 16. That's a difference of 8. This tells me there is more to the derivative. We jumped 4 times between 5 time intervals. I know the derivative needs to be the same at each time interval, so I would have argued the remaining value missing from the derivative would have been 2, since 8/4=2. Sometimes, at least from what I have found, there are parts of the derivative that can be hidden in the functional outputs of f(x).

    Unfortunately, I would have found that f'(x)=2x with your particular function. I would have also found that the total distance traversed at t=4 using f'(x)=2x would have been 16 too, like it is in the y output of the function you gave me. But, can't say that's true for intervals starting at x=.5 so I wouldn't have argued it in the first place:)
     
  11. Aug 13, 2016 #10

    chiro

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    Hey INTP_ty.

    If you are asking how to solve for a differential equation by trying to look at the DE form and come up with some "intuitive" guess of what it should look like then I guess you will have to link the DE with a geometric basis in the same way that you do so for functions.

    I'll give you an example.

    In harmonic analysis you can represent functions as a set of orthogonal basis vectors and an example of this is a Hermite function basis. The idea is you use the Gram-Schmidt "process" to obtain a set of independent ("orthogonal") functions and you write your function as a linear combination of those.

    Derivatives are technically functions and provided you have the right basis, you can see the different derivatives (which should be independent unless they are constant in general) as being part of a geometric "basis" in the same way a vector is or an orthogonal function is.

    If you get that intuition, then I'm sure it will be useful for "you".
     
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