# Various problems - help,suggestion?

1. Aug 22, 2008

### R A V E N

Problem 1:Prove that the following system of linear equations has unique solutions(sorry,I dont know precise english term for this kind of system) and if it has,solve it:

$$x_1+x_2+x_3+x_4=4$$

$$x_1+2x_2+3x_3+x_4=7$$

$$x_2+x_3+2x_4=4$$

$$x_1-x_2+x_3-2x_4=-1$$

$$2x_1-2x_2+x_3-x_4=0$$

So, four variables,five equations.I dont see how any method could be applied here:Gaussian elimination,Cramers rule,matrix method.Any hint just to get stared would be enough.

Last edited: Aug 22, 2008
2. Aug 22, 2008

### dirk_mec1

Look for dependent equations and see if there are four independent equations then calculate the determinant.

3. Aug 22, 2008

### R A V E N

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.

*started

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

### HallsofIvy

Staff Emeritus
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.

5. Aug 22, 2008

### R A V E N

Yeah,choosing random four equations and solving system by using Cramers rule works.Thanks,HallsofIvy.

6. Aug 22, 2008

### R A V E N

Problem 2:Solve this equation($$z$$ - complex number):

$$(z^2-2)^6+z^6=0$$

Any hints?

7. Aug 22, 2008

### Dick

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.

8. Aug 23, 2008

### R A V E N

Thanks,Dick.

Problem 3:
Calculate:

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

Even hint how to start would be appreciated.

9. Aug 23, 2008

### R A V E N

Actually,correct form is:

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

By using trigonometric identities,I transformed above into:

$$\lim_{x\to 0}\left(\tan\frac{1}{\cos{2x}}\right)^{\frac{\cos{2x}}{\sin{2x}}}$$

Last edited: Aug 23, 2008
10. Aug 23, 2008

### Dick

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.

11. Aug 23, 2008

### R A V E N

Yeah,instead of attempting anything and ramble in various faulty courses,its better to wait advice for correct approach.

On what basis did you concluded that it is $$1^\infty$$ type of a limit?

Last edited: Aug 23, 2008
12. Aug 23, 2008

### Dick

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.

13. Aug 23, 2008

### R A V E N

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 dont 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.

14. Aug 23, 2008

### Dick

What I gave just shows you what KIND of limit it is. It doesn't solve the limit. Did you figure out the value?

15. Aug 23, 2008

### R A V E N

I will try. I know that solution is $$e$$.

Last edited: Aug 23, 2008