# Mathematica Equation Solving

Suppose I wanted to solve the following for $a[1], a[2], a[3]$ in the Mathematica code:

(a[1]+a[3])*t+(a[1]+2*a[2])*t^2=0


with $a[1] = 1$ where $t$ is just a variable and the equation above is identically 0. How would I automate this in Mathematica to get $a[3] = -1, a[2] = -1/2$?

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In general, you can do it this way:

a[1] = 1;
Solve[# == 0] & /@
CoefficientList[(a[1] + a[3])*t + (a[1] + 2*a[2])*t^2, t]

Out[1] = {{{}}, {{a[3] -> -1}}, {{a[2] -> -(1/2)}}}


What the above does is it first collects the coefficients of the polynomial using CoefficientList and passes them to Solve which solves each of them.

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a[1] = 1; Solve[{a[1] + a[3] == 0, a[1] + 2*a[2] == 0}, {a[2], a[3]}]

Yes, so let's say I have a differential equation $f''+f=0$ whose solution I'm going to write as $f(t)=\sum_{n = 1}^{100}a_{n}t^{n-1}$. It would be nice to compute the values of $a_{n}$. –  Shayla Jun 8 '11 at 4:14