I am trying to find critical points of a function, and to do that, i have to equate the partial derivative of the function to zero, which is:
$\frac{\partial T_o}{\partial \gamma}=\frac{C_1}{\gamma^k}-\frac{C_4}{\gamma^{k-1}}-C_3$
Now, this results in a k-order polynomial that I require the roots to. $C_1,C_4$ and $C_3$ are constants. The constraints that I have to narrow down my result to a minimum number of critical points are as follows:
1.$\gamma\in R^+$ & $0\leq\gamma\leq1$, where $R^+$ is the set of all positive real numbers.
2.$k\in Z^+$ and $k>1$, where $Z^+$ is the set of all positive integers.
3.If a general solution is not obtained by just using the above constraints, an additional cap on $k$ can be used as $2\leq k\leq20$
I require a general term or formula for finding out the roots that satisfy the above condition. I have had no success in finding out ways to solve this equation. Please help me. Thanks.
You're trying to solve $$C_1 - C_4 \gamma - C_3 \gamma^k = 0 $$ You won't get solutions in radicals if $k \ge 5$, and the solutions for $k = 3$ and $k=4$ are quite unpleasant. Numerical methods can be used. Also, there is a series solution that should converge if $\left|C_1^{k-1} C_3/C_4^k\right|$ is sufficiently small:
$$ \eqalign{\gamma &= \sum_{m=1}^\infty (-1)^m \frac{C_1^{mk-m+1} C_3^m}{C_4^{mk+1} m!} \prod_{j=0}^{m-2} (mk-j) \cr &= \frac{C_1}{C_4} - \frac{C_1^k C_3}{C_4^{k+1}} + \frac{2k\; C_1^{2k-1} C_3^2}{2!\; C_4^{2k+1}} - \frac{3k(3k-1)\; C_1^{3k-2} C_3^{3}}{3!\; C_4^{3k+1}} + \frac{4k(4k-1)(4k-2)\; C_1^{4k-3} C_3^4}{4!\; C_4^{4k+1}} + \ldots }$$
If I'm not mistaken, what you need for convergence is $$ \left| \frac{C_1^{k-1} C_3}{C_4^k} \right| < \frac{(k-1)^{k-1}}{k^k}$$