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Challenging System of Equations (4 Variables)

  1. Feb 8, 2013 #1
    I am having a bit of trouble solving the following system of equations. I know what numbers solve the system, but I can only do so using a computer, such as WolframAlpha. The equations , for those interested, are the four partial derivatives for a Lagrange Multiplier. The four equations are as follows:

    It should be noted that a, b, and c are all positive numbers, and, therefore, must all be less than one. The solution is a=b=c=1/3, d=21/2, I just don't know how to get that.
    I greatly appreciate any help you can give to me for this problem. Thank you for your time.

    P.S. I hope this is the right section for this type of question, but I am very new here, so please forgive me if I was wrong.
  2. jcsd
  3. Feb 8, 2013 #2

    Simon Bridge

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    Welcome to PF;
    There's four equations and four unknowns.
    You solve it the same as you always do simultaneous equations.

    i.e. pick an equation, solve for one of the variables in that equation, substitute into the other three ... now you have three equations and three unknowns. Repeat until you are down to one.

    Can you identify where you keep getting stuck?
  4. Feb 9, 2013 #3


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    Hint: use symmetry. The fourth equation is symmetric, but what about the other three? What does that tell you about your solution set?
  5. Feb 9, 2013 #4
    I know how to solve a system of equations in general, but I always end up with obscene equations for each variable. I'll add to the first post what I tried, and why I haven't gotten a solution from it.
    I'm sorry, but I don't know what you mean by symmetric, or what it tells me. Does it mean that Fa=Fb=Fc?

    How do I edit the first post? The edit button doesn't show up.
    In the meantime, here is what I did, and how I got stuck:
    Subtracting the first and second equations, I got 2/(a-1)^2+9b=2/(b-1)^2+9a, which means b =(9a^2+/-√(-18a^3+54a^2-54a+19)-18a+8)/(9(a^2-2a+1)). I could substitute this equation in for b in any of the equations I have, but it is much to complicated for me to simplify without a computer more advanced than myself, unless, of course, I am simply not seeing a simple way of reducing this equation to something more manageable.
    Last edited: Feb 9, 2013
  6. Feb 9, 2013 #5

    Simon Bridge

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    You'll just have to add corrections as a reply. There's no shame in it.

    I can think of several different approaches, so I need to know how you are thinking about the problem in order to know which to suggest and how to do the suggesting.

    What I suggested could be thought of as the "brute force" approach - attacking the system blindly. With experience you earn to use clues, like symmetry, to help you to a shorter cut that may even produce a non-obnoxious form of equation when you get down to one variable. Even so - if you have one equation and one unknown, from a system like above, you should be able to find the solution.

    Note. that last equation is a+b+c=1 ... what class of geometric objects is that?
  7. Feb 9, 2013 #6
    I know that solving one variable in one equation is always theoretically possible, bit I can't solve the fifth degree equations that result from the method I outlined above (would you like me to go into more detail about what I did?). That is why I feel as though the "brute force" method, the one I tried, won't work.

    a+b+c=1 is a plane... How does that help?
  8. Feb 10, 2013 #7

    Simon Bridge

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    You can say more than that: which plane is it?
    What about the others?

    The way to approach math is by recognizing general classes of function/geometric object and understanding what their intersections can give rise to. This provides clues to possible shortcuts but you have to be prepared to explore.
    yes please.

    You ended up with a quintic function - have you put it in standard form and checked to see if it has an analytic solution?
    Last edited: Feb 10, 2013
  9. Feb 10, 2013 #8
    The plane has a normal vector of (1,1,1)...
    I don't know what you mean about the others, though.

    What do you mean by analytic solution? The rest of what I tried is shown below:
    This was just too massive for me to deal with. Is there an easier way of doing things?
  10. Feb 10, 2013 #9
    Have you tried the following substitution?

    [itex]d=d [/itex]

    And then substitute [itex]a,b,c,d[/itex] in terms of [itex]x,y,z,d[/itex], with [itex]x+1=a [/itex], [itex]y+1=b [/itex], [itex]z+1=c [/itex] and [itex]d=d [/itex].

    By doing so, your system of equations would become much easier to solve, wouldn't it?
  11. Feb 10, 2013 #10

    Simon Bridge

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    ... your main trouble, now I've seen you write it out, is that you have awkward square-roots and so on in there. You need to be able to factor things out better if you are to put the equation in standard form.

    I meant: what sort of geometric object do the other equations represent?
    Effectively, you are looking for the intersection of 4D surfaces with a 2D plane that is tilted in 3D (a,b,c) coordinate system provided.

    h6ss (above) had my next suggestion :)

    It amounts to choosing a more convenient coordinate system to do the math in.

    In a simpler problem - the intersection of a 3D surface and a plane will be either a curve or a point. It is easier to find the intersection if you do it if the plane has normal (0,0,1) isn't it? In general, you use the underlying symmetry of the problem to help with your choice of coordinate system.

    There's the shifted-rectangular one suggested; you could also rotate it (say, so that the x and y axis are in the plane a+b+c=1 and the z-axis points along (1,1,1)) if you liked but I'd try the shifted one first.
  12. Feb 10, 2013 #11
    I can't believe I never thought of that! From there the solution becomes fairly trivial:
    x=y or (9x^2y^2+2x+2y)=0
    Quadratic Formula (reject negative solution):
    y=x or y=(-1+√(1-18x^3))/(9x^2)
    if y=(-1+√(1-18x^3))/(9x^2) and y is positive and x is positive, then 9x^2 is positive and...
    x is negative, which is a contradiction, so x=y=z, and a=b=c
    Thank you so much!

    Even though I've solved it, I think what you say could help me with other problems where there isn't a simple substitution.

    Firstly, how would you factor things without the awkward radicals? Were you referring to the suggestion made by h6ss, or is there a different way?

    I am clueless as to what geometric object the other three equations represent. I'm not very fluent in naming 3-D objects by their equation, much less 4-D, and I'm not totally sure what to do with that information...

    I have always been curious as to how I could tilt an equation and move the axis as you suggested (I remember one problem in particular that would've been much easier if I could do that, and I had to use an exceptionally long way of solving it). Also, all I know about a normal vector is how to determine one and that they relate to slope (my 10th grade class is learning logarithms currently). Should I learn more about them to help me with this problem.

    If you're worried about giving away answers, this is just for practice and learning new concepts; I took the problem from a test whose due date is passed.
    Last edited: Feb 10, 2013
  13. Feb 10, 2013 #12

    Simon Bridge

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    It is a general strategy - try to find a geometry which is simple - solve the problem in that geometry - then transform the solution back. It sometimes takes several goes ... but it can be a very powerful method.

    That's right - the (a-1) etc was where the awkward radical came from. It may have been possible to factorize as you went using the fact that a,c,b,d > 0 but I don't know because I wouldn't have bothered with that. I couldn't tell what you'd tried because you didn't tell me and I wanted you to explore the system of equations more. As you marshal the facts about the system in front of you, ideas start to form.
    It's something you get used to with experience just like you learn to recognize 3-4-5 triangles when they show up.
    It may have been that the problem was part of a course in 4D surfaces so I could not be sure which approach to tell you about.

    The problem you had described a set of surfaces in 4D using (a,b,c,d) as axis. The surfaces intersected at only one point where everything was positive. The first three were basically the same surface rotated 90deg in different directions. In 2D they would be something like ##Ax^2y=B## and the 3D they'd be ##Ax^2(y+z)=B##.

    You are in the 10th grade or you teach 10th grade?
    One of the things you didn't say earlier was what level to approach the solution from.

    By 10th grade there should have been some work on the basic transformation, at least in 2D. You have translation, rotation, and reflection, within a coordinate system, and you can transform to a whole new system like polar, cylindrical, and spherical... and even some strange ones like those that crop up in relativity. But I don't think you have been encouraged to think of moving an actual set of axis - iirc that level has students thinking of the coordinates as fixed.

    If I had a function f(a,b,c,d) and moved the whole thing 1 unit in the +a direction, then the new function is f(a-1,b,c,d). It is the same as moving the entire coordinate axis one unit in the -a direction.

    The coordinate transform is the math way of looking at a problem from another POV.

    That's not so much of the problem as that you learn more if you do the discovery process yourself. It is also more likely that you will be learning at an appropriate level.

    Hopefully the discussion and the somewhat frustrating trail of having to nut through my earlier questions and hints gives you a deeper appreciation of math ;)
  14. Feb 11, 2013 #13
    I am in 10th grade. The reason I didn't mention that originally was because I know a few things that some other 10th graders may not know about math.

    I can easily translate a graph (or the axis, of course), but I don't know how to rotate anything other than a simple line in two dimensions (using trigonometry). How could I rotate the first three graphs in order to map them onto the a+b+c=1 plane, like you briefly mentioned? The other math problem I did where I wished I could rotate it was ##2(x^2+y^2)+x+y=5xy##, which I wished I could rotate 45 degrees in order to make it a hyperbola on which I'm more familiar with doing math, such as finding the "a" and "b" values.

    I don't really understand what you mean in the following section:
    How can you write a 4-D function in fewer dimensions?

    I greatly appreciate all of the help you've given me on this problem; you are a truly great person. Thank you! I apologize for taking so long to respond, but I was busy at school all day, up until now.

    P.S. Is there a time limit for editing posts? I can only edit this one now, and I just learned how to make the math look pleasant. If there is anyone with higher status that can edit existing posts, then I would appreciate it if you could add the "##" before and after each of my math statements. Sorry for my ignorance, and thanks to anyone who can do it.
  15. Feb 11, 2013 #14
    We know that three-dimensional objects have two-dimensional projected images. We call 3D projection the method of mapping three-dimensional points to a two-dimensional plane.

    For example, one can use contour lines (or contour map) to illustrate in a two-dimensional graph the steepness of slopes of a three-dimensional mountain.

    Similarly, a hypothetical four-dimensional object would have three-dimensional projections. Geometric objects in the space [itex]ℝ^{4}[/itex] can be projected first into the space [itex]ℝ^{3}[/itex] and then into the plane [itex]ℝ^{2}[/itex]. As a matter of fact, some very interesting shapes in [itex]ℝ^{3}[/itex] and [itex]ℝ^{2}[/itex] can be obtained by projecting curves of [itex]ℝ^{4}[/itex].
  16. Feb 11, 2013 #15

    Simon Bridge

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    In addition to what h6ss said above.:

    It can help you understand symmetries in higher dimensions to consider similarly formulated structured in fewer dimensions. i.e. something like ##w^2+x^2+y^2+z^2=R^2## in 4D has a lot in common with a sphere and a circle (3D and 2d)?

    ##x-y^2=1## would be a parabola in 2D ... how about ##z+x-y^2=1## ?

    On to your question about rotations:
    ##y-x^2=1## is the above parabola rotated 90##^\circ## (=+##\frac{\pi}{2}## radians - get used to radians)

    If I rotate the axis ##\frac{\pi}{2}##rad I'll get the first equation again.
    So a clockwise rotation of the object is the same as an anticlockwise rotation of the axis.

    Each point on your object can be considered a displacement from the origin - displacement is a vector, vectors are easily rotated by a matrix operation.

    i.e. rotating your object 45deg is a ##\frac{\pi}{4}##rad rotation ... that is the angle you get in a "1-1-√2" triangle so ##\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{ \sqrt{2}}## the (a,b) coordinate corresponding to a particular (x,y) coordinate would be given by:
    $$\left [ \begin{array}{c} a\\b\end{array}\right ]
    = \frac{1}{\sqrt{2}}\left [ \begin{array}{cc} 1 & -1\\ 1 & 1\end{array}\right ]
    \left [ \begin{array}{c} x\\y \end{array}\right ]
    =\frac{1}{\sqrt{2}}\left [ \begin{array}{c} x+y\\x-y \end{array}\right ]$$ ... which is to say: $$a=\frac{x+y}{\sqrt{2}}\; ; \;
    b=\frac{x-y}{\sqrt{2}}$$ ... solve for x and y in terms of a and b, substitute, and you do get a nicer hyperbola. You'll want to do a translation as well.

    In general (2D and 3D):

    Rotations in 4D are a bit trickier...

    Yes. When you start helping other people you'll soon see that this is a good idea.
    Last edited: Feb 11, 2013
  17. Feb 12, 2013 #16
    A somewhat similar problem with symmetry is

    maximize x_1 * x_2 * ... * x_n subject to sum(x_i) = a, x_i >= 0

    with the solution (obtained from the symmetric non-linear equations that you get from setting the partial derivatives = 0) x_i = a/n.
  18. Feb 12, 2013 #17


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    Subtract the first equation from the second to get [itex]2/(a-1)^2- 2/(b-1)^2= 0[/itex] or [itex](a- 1)^2= (b- 1)^2[/itex], so that a-1= b- 1, giving a= b, or a- 1= -(b-1)= -b+ 1, giving a= 2- b.

    You can do the same with other pairs of equations.

  19. Feb 12, 2013 #18

    Simon Bridge

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    2/(a-1)^2+9b+9c-d=0 ... (1)
    2/(b-1)^2+9a+9c-d=0 ... (2)
    ... doing (2)-(1) I get:
    \big (& 2/(b-1)^2 & +& 9a&+&9c&-&d&=&0& \big )\\
    - \big (& 2/(a-1)^2 & + & 9b&+&9c&-&d&=&0 &\big ) \\ \hline
    = \bigg (& \frac{2}{(a-1)^2} - \frac{2}{(b-1)^2}& +&9(a-b)& & & & & =&0& \bigg )

    How did you get rid of the 9(a-b) term?
    Doesn't have a big effect on your argument - you still get f(a)=f(b) => a=b ?
    Do the same with the other two and you get: a=b=c but d is different.
    No need for shifting the axis about :)

    But sometimes people have to go through the more painful ways first ;)
  20. Feb 12, 2013 #19
    So does that mean that if ##f(a)=f(b)=f(c)## then ##a=b=c## for every system of equations?
  21. Feb 12, 2013 #20

    Simon Bridge

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    No - consider
    ##\text{if }f(x)=x^2,\text{ then } f(a)=f(b) \Rightarrow a=\pm b##​
    ... so there are two possible values of a that satisfy the relation for a given value of b.
    In this case, all the unknowns have to be positive.
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