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Thread: Complicated equation with 3 unknowns - please help!

  1. #1
    wtw
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    Question Complicated equation with 3 unknowns - please help!

    Hi all,

    I'm doing my final year maths project, and it has appeared to be a bit beyond my limits... It would be great if you helped me somehow!

    So, I have 3 points from a graph and a general equation. With this data, I have to find a, b, and c of this equation.

    The equation is: Complicated equation with 3 unknowns - please help!-zrzut-ekranu-2018-03-05-o-20.04.48.png

    While the points I'm given are: (1,0) and (4.5, 0.5) and (7.15,0.7)

    As far as I know, I have to substitute x and y with given coordinates of points and solve a three unknowns system of equations. But to do so, I have to (probably) transform the equations to a form of 'a+b+c = x'.

    Can you please tell me if I'm correct and whether it can be done?

    Best wishes!
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  2. #2
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    Re: Complicated equation with 3 unknowns - please help!

    What does $*$ mean? Is that multiplication? If so, then you wind up with a logical contradiction.

    This is because $e^x > 0$ for all $x\in \mathbb{R}$. So, if $y=0$, then $c=0$. If $c=0$, then $y=0$ for all $x$. Are you missing a decimal?

    Edit: I will assume that the first point you list is incorrect. Instead, you have the following three points:
    $(1,y_0), (4.5,0.5),(7.15,0.7)$.

    This gives the following equations:

    Equation 1: $\dfrac{y_0}{c} = e^{-a(1-b)^2}$
    Equation 2: $\dfrac{0.5}{c} = e^{-a(4.5-b)^2}$
    Equation 3: $\dfrac{0.7}{c} = e^{-a(7.15-b)^2}$

    Take ln of each side:

    Equation 1a: $\ln y_0 - \ln c = -a(1-b)^2$
    Equation 2a: $\ln 0.5 - \ln c = -a(4.5-b)^2$
    Equation 3a: $\ln 0.7 - \ln c = -a(7.15-b)^2$
    Last edited by SlipEternal; Mar 5th 2018 at 10:41 AM.
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  3. #3
    wtw
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    Re: Complicated equation with 3 unknowns - please help!

    Thanks for such a quick answer!

    Yes, * is multiplication. Also, you're absolutely right - I've rewritten a wrong point. All the coordinates are: (4.5, 0.5) and (6.33, 0.9) and (7.15,0.7).
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    Re: Complicated equation with 3 unknowns - please help!

    Quote Originally Posted by wtw View Post
    Thanks for such a quick answer!

    Yes, * is multiplication. Also, you're absolutely right - I've rewritten a wrong point. All the coordinates are: (4.5, 0.5) and (6.33, 0.9) and (7.15,0.7).
    Then, you wind up with the equations:

    Eq 1: $0.5 = c\cdot e^{-a(4.5-b)^2}$
    Eq 2: $0.9 = c\cdot e^{-a(6.33-b)^2}$
    Eq 3: $0.7 = c\cdot e^{-a(7.15-b)^2}$

    We can rewrite these as:

    Eq 1a: $\ln 0.5 - \ln c = a(4.5-b)^2$
    Eq 2a: $\ln 0.9 - \ln c = a(6.33-b)^2$
    Eq 3a: $\ln 0.7 - \ln c = a(7.15-b)^2$
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  5. #5
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    Re: Complicated equation with 3 unknowns - please help!

    No, you don't want "a+ b+ c= x". For one thing a, b, and c are constants and x is a variable. Do exactly what you said, putting the numbers you are given into the general equation you get
    First point: x= 4.5, y= 0.5 so this becomes $0.5= ce^{-a(4.5- b)^2}$
    Second point: x= 6.33, y= 0.9 this becomes $0.9= ce^{-a(6.33- b)^2}$
    Third point: x= 7.16, y= 0.7 so this becomes $0.7= ce^{-a(7.16- b)^2}$

    Three equations to solve for three values, a, b, and c. Because of the exponentials I would start by taking the logarithm of both sides of each equation:
    $log(0.5) log(c)- a(4.5- b)^2$
    $log(0.9)= log(c)- a(6.33-b)^2$
    $log(0.7)= log(c)- a(7.16- b)^2$

    Subtracting the first equation from the second eliminates c:
    $log(0.9)- log(0.5)= a((4.5- b)^2- (6.33- b)^2= a(20.25- 9b+ b^2- 40.0689+ 12.66b- b^2= a(3.66b-19.8189)$

    Subtracting the third equation from the second also eliminates c:
    $log(0.9)- log(0.7)= a((7.16- b)^2- (6.33- b)^2)= a(51.2656- 14.32b+ b^2- 40.0689+ 12.66b- b^2)= a(11.1967- 1.66b)$

    Dividing the first of those two equations by the second eliminates a leaving an equation for b:
    $\frac{log(0.9)- log(0.5)}{log(0.9)- log(0.7)}= \frac{3.66b- 19.8189}{11.1967- 1.66b}$

    That should be easy to solve.
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    Re: Complicated equation with 3 unknowns - please help!

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