Alan

ok.  So plot the graph of cos(x) where x is is in degrees.  This time we have to home in on a small portion of the graph to see it clearly:

You can see they are still the same.

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(-{\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

However, your original post stated that the correct answer to the integral was -3.25.  This is only true if x is in radians.

I assume r is a scalar and I is the identity matrix.

It is normal to assume the angle is in radians unless it is accompanied by an explicit degree sign.

To see that cos(-3) is the same as cos(3) look at the graph below:

You can see that cos(-3) has exactly the same value as cos(3).  They are both ≈ -0.99.  

Do you mean (1/3)⁸ = (-1/3)ⁿ - 1 or  (1/3)⁸ = (-1/3)^n-1? 

If you mean (1/3)⁸ = (-1/3)^n-1 then n = 9 is an integer solution.

For other non-integer solutions you need to involve complex numbers. 

Look very carefully at your denominator Stu.  here is a copy of your last posted attempt:

The sin term should be multiplied by 4.005. You have added 4.005 to it.

Do you mean the probability of exactly 1 success in 4 trials given a probability of 0.1 of an individual success? If so, it is normal to put the 1 before the 4.

prob(k;n,p) = nCr(n,k)*p^k(1-p)^n-k    n = number of trials (4); k = number of successes (1); p = probability of individual success (0.1).

$${\mathtt{prob}} = {\left({\frac{{\mathtt{4}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.1}}}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.1}}\right)}^{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} \Rightarrow {\mathtt{prob}} = {\mathtt{0.291\: \!6}}$$

You were almost right Stu; just one small (but vital) mistake in the last line:  cos(-3) is +cos(3) not -cos(3).

The iterative method Bertie mentioned, to find the cube root of "a", say, is:

$$$$x_{n+1}=\frac{1}{3}(\frac{a}{x_n^2}+2x_n)$$

This is derived using Newton-Raphson.

However, in general, I wouldn't want to use this without a calculator of some sort! 

Rom's method is correct; he just used a value for r1 of 1 instead of the requested 4. 

1/R = 1/4 + 1/3  = 3/12 + 4/12 = 7/12

Hence R = 12/7 = 1.714 ohms

i.e. the same as DavidQD

1.  The point of putting 5.166 radians into the function was to show that the result was not zero.  In other words, 5.166 is not a root of the function (a "root" is a value that makes the function equal to zero).

2. The Newton-Raphson method requires iteration.  That is, you take the result that appears from your initial guess for x and you put it back into the formula to get another guess.  You then take the result of that and keep repeating the process until the output x is the same as the input x.  You have then converged on a solution.  Perhaps the image below might help.

 You can see that 5.166 is just the result of the first iteration when you use 4.005 as an initial guess.

You have used 4.005° inside the sin and cos functions, but you must have 4.005 radians.  Also you should have a 4.005 multiplying the sin term in the denominator (for the first iteration only, of course.  It should be the latest estimate of x for the other iterations).