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(noting that the lower limit is n now, instead of 2)

it should be \sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{n}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} right?

So it's right?

How many terms of the series \sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} would you need to add to find its sum to within 0.01? Here's what i got: let f(n) = \frac{1}{n(ln\;n)^{2}} . Since f(n) is continuous, positive and decreasing for all n over the interval [2,\infty] , we can use...

I just want to confirm these two questions. Thanks in advance. (1) Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix. \left(\begin{array}{uvwxyz}1 & 5 & 2 & -6 & 9 & 0 \\0 & 0 & 1 & -7 & 4 & -8\\0 & 0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0...

thank you cronxeh, you've been very helpful!

ah, the question was written down wrong in the matrix...it's suppose to be \left(\begin{array}{xyzk}3 & i & 2+i & 3i\\-i & 1 & 1 & 1\\1 & 1 & 2+i & i\end{array}\right) (notice the last element is an 'i' not a 1) Thanks for everyones reply...i'm going to work with this, and see where it...

here's what i got... \left(\begin{array}{xyzk}3 & i & 2+i & 3i\\-i & 1 & 1 & 1\\1 & 1 & 2+i & 1\end{array}\right) R2 = R2 * i \left(\begin{array}{xyzk}3 & i & 2+i & 3i\\1 & i & i & i\\1 & 1 & 2+i & 1\end{array}\right) R1 = R1 - R2 \left(\begin{array}{xyzk}2 & 0 & 2 & 2i\\1 & i & i...

wow! thanks for the detailed solution! I need to practice these complex systems. Thanks!

I guess i didn't get the last one completely because I've been having a hard time with this one. Solve 3x + iy + (2+i)z = 3i -ix + y + z = 1 x + y + (2+i)z = i i've tried dividing through by the leading coefficent, and re-arranging it...but can't seem to get the right answer...

thanks guys...i think i got it...

Find the solution in C to the following linear system of equations. (a) (1-i)z + 4w = 2 + 8i (b) 3z + (1+i)w = 1 + 5i I tried expanding but that didn't get me anywhere. Then i put it in a matrix, but i didn't know how to go from there. Any suggestions? Thanks.

sorry, the lagrange remainder is suppose to be f^{n+1}(p) * (x-a)^{n+1} ----------- (n+1)!

sorry, T_2(x) is suppose to be 1 + (2/3)x - (1/9)x^{2} i believe the lagrange remainder is: f^{n+1}(p) * (x-a)^{n+1} ----------- (n+1)! where p is between (a,n). The problem...

1. Let f(x) = (1+x)^{2/3} (a) find the taylor polynomial T_2(x) of f expanded about a = 0. i got 1 + (1/3)x - (1/9)x^{2} For the rest, i have no idea how to do...any help would be greatly appreciated. (b) For the givven f write the lagrange remainder formula for the error term...