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Various problems - help,suggestion?

  1. Aug 22, 2008 #1
    Problem 1:Prove that the following system of linear equations has unique solutions(sorry,I don`t know precise english term for this kind of system) and if it has,solve it:






    So, four variables,five equations.I don`t see how any method could be applied here:Gaussian elimination,Cramer`s rule,matrix method.Any hint just to get stared would be enough.
    Last edited: Aug 22, 2008
  2. jcsd
  3. Aug 22, 2008 #2
    Look for dependent equations and see if there are four independent equations then calculate the determinant.
  4. Aug 22, 2008 #3
    What is dependant equation?If I am right,it is one equation from the system multiplied by some constant to produce additional equation.But I do not see any.

    Last edited: Aug 22, 2008
  5. Aug 22, 2008 #4


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    Well, try to solve it! Since you have 4 variables, you should be able to solve any 4 of the equations for those variables. Once you have done that, do they satisfy the fifth equation?
    A slightly more formal way, though not any better, would be to set up the 5 by 4 matrix of coefficients and row-reduce it. If the last row turns out to be all 0's then there is a solution, otherwise, there is not.
  6. Aug 22, 2008 #5
    Yeah,choosing random four equations and solving system by using Cramer`s rule works.Thanks,HallsofIvy.
  7. Aug 22, 2008 #6
    Problem 2:Solve this equation([tex]z[/tex] - complex number):


    Any hints?
  8. Aug 22, 2008 #7


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    z is not a root. So it's safe to divide both sides by z^6. That gives you ((z^2-2)/z)^6=(z-2/z)^6=(-1). So z-2/z is equal to one of the six sixth roots of (-1). Solve z-2/z=k and set k to be each of the six sixth roots. You'll get a total of twelve roots (as you should, it's a twelfth degree equation). That's pretty tedious. But you can do it.
  9. Aug 23, 2008 #8

    Problem 3:


    Even hint how to start would be appreciated.
  10. Aug 23, 2008 #9
    Actually,correct form is:

    [tex]\lim_{x\to 0}\left[\tan\left(\frac{\pi}{4}+x\right)\right]^{\cot{2x}}[/tex]

    By using trigonometric identities,I transformed above into:

    [tex]\lim_{x\to 0}\left(\tan\frac{1}{\cos{2x}}\right)^{\frac{\cos{2x}}{\sin{2x}}}[/tex]
    Last edited: Aug 23, 2008
  11. Aug 23, 2008 #10


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    I think you slipped up on the trig identities, but you don't really need them anyway. You have a 1^(infinity) type limit. Take the log to change it to an infinity*0 limit and rearrange it to apply l'Hopital's rule.
  12. Aug 23, 2008 #11
    Yeah,instead of attempting anything and ramble in various faulty courses,it`s better to wait advice for correct approach.

    On what basis did you concluded that it is [tex]1^\infty[/tex] type of a limit?
    Last edited: Aug 23, 2008
  13. Aug 23, 2008 #12


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    No, it's better for you to ramble in faulty courses, otherwise I won't help at all. :) x-> 0, x+pi/4 -> pi/4. tan(pi/4)=1. 2x -> 0, cos(2x) -> 1, sin(2x) -> 0, cos(2x)/sin(2x)=cot(2x)->infinity. Hence 1^infinity.
  14. Aug 23, 2008 #13
    And it was that simple?(By the way,everything is simple if you have enough time and energy to invest into studying,thinking and looking for advice on various places).

    I just wonder why those 800+ pages mathematical books which have zillion things in it don`t have any example which would for the moment escape rigorous mathematical world and demonstrate things like that and other in more human-friendly fashion?

    For the sake of more successful and contented both teachers and students.
  15. Aug 23, 2008 #14


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    What I gave just shows you what KIND of limit it is. It doesn't solve the limit. Did you figure out the value?
  16. Aug 23, 2008 #15
    I will try. :smile: I know that solution is [tex]e[/tex].
    Last edited: Aug 23, 2008
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