This is exactly what I was intending.
Your geometric method does seem interesting, but I do not quite seem to understand it.
The above is what you had tried to show in your post for 3^2. Could you explain what you meant by "delta". The image seem to show 1+3+5=9. That is what the...
I have been working on representing the powers of numbers as a summation.
This is as far as I have gotten.
Power: 2
m^2 = \sum_{n=1}^m \left(2n -1\right)
Power: 3
m^3 = \sum_{n=1}^m \left(3n^2 -3n +1\right)
Power: 4
m^4 = \sum_{n=2}^m \left[6*(4n-6) * \left(\sum_{a=1}^{m-n+1}...
@HallsofIvy
I cannot believe I did not see that. I learned that in differential equations class and even used it often in adv engg math. Thanks a ton. This just makes solving the problem a whole lot easier (might I say, cleaner too). Kudos to you ! I will post the completed solution once I...
Thanks a ton micromass !
Your pointers were very helpful. Just one last question. If the equations were not homogeneous, which would be the simplest way to solve it, undetermined coefficients or variation of parameters ?
PS: I tried to follow the example on matrix differential equation on...
I am not sure what you meant by 'diagonizable'.
But, I tried solving for the eigenvectors as follows (wiki):
\begin{pmatrix} 1 & -1 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \lambda \begin{pmatrix} \alpha \\ \beta \end{pmatrix}
This gave me,
\hat{v_{1}} =...
Thanks for pointing that out. Never looked at the problem that way.
I think it is a eigenvalue problem.
\begin{vmatrix}\begin{bmatrix} 1 & -1 \\ -4 & 1 \end{bmatrix}- \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\end{vmatrix} = 0
And I solve for lambda and get \lambda_{1} = 3...