# numerically solving

I am trying to solve for [Lambda] as a function of a (unknown) from this expression

202.[2.51521 + 1/(-1 + E^(202. λ))] +
802.[2.52457 + 1/(-1 + E^(802. λ))] +
1802.[2.52632 + 1/(-1 + E^(1802. λ))] +
3202.[2.52694 + 1/(-1 + E^(3202. λ))] +
5002.[2.52722 + 1/(-1 + E^(5002. λ))]=a


I tried with Reduce and FindRoot, and as neither of them helped I think closed form expression can not be obtained. Any suggestion how to get numerical solution? Thanks in advance.

• You can use FindRoot to solve once given specific values for a. – Daniel Lichtblau Apr 23 '15 at 17:20
• @DanielLichtblau Only after those square brackets are replaced by parenthesis – Dr. belisarius Apr 23 '15 at 17:21
• Anyway, the function is too steep and you'll probably need a few tricks – Dr. belisarius Apr 23 '15 at 17:22
• @DanielLichtblau Thanks. But I want to express λ as a function of a. Sorry about the square brackets. – reach2brb Apr 23 '15 at 17:41
• [I have nothing against square brackets.] – Daniel Lichtblau Apr 23 '15 at 19:23

f[x_] = 202. (2.51521 + 1/(-1 + E^(202. x))) + 802. (2.52457 + 1/(-1 + E^(802. x))) + 1802. (2.52632 + 1/(-1 + E^(1802. x))) + 3202. (2.52694 + 1/(-1 + E^(3202. x))) + 5002. (2.52722 + 1/(-1 + E^(5002. x)));

$-0.000231828 - 1.11206 \times 10^{-8} a - 5.50983 \times 10^{-13} a^2$
So we have three steps here. First I solve it for different a, then store them in data. And then I go for a polynomial fit. As you can see within the defined range of a [-100,100] it is almost linear. You can check with different a values. For higher a you have to take more terms in your polynomial.