Approximating $\sum_i \frac{a_i}{\lambda - \lambda_i} = \lambda$

Question

I am trying to solve the equation

$$ \sum_i^n \frac{a_i}{\lambda - \lambda_i} = \lambda $$

for some real constants $a_i, \lambda_i$, with $\lambda_1 > \lambda_2 > \lambda_3 > ... $ I have the hunch that I can approximate one of its solutions by taking only the first term and solving

$$ \frac{a_1}{\lambda - \lambda_1} = \lambda $$

but I'm having trouble formalizing this step. Under what conditions can I do so and how can I justify this formally? For example, if $\lambda_1 \gg \lambda_2 > \lambda_3 > ...$ then, whenever $\lambda \approx \lambda_1$, the original equation is approximately

$$ \frac{a_1}{\lambda - \lambda_1} + \frac{\sum_{i>1} a_i}{\lambda_1} = \lambda $$

Also, can I find other solutions to the original equation by focusing on other terms? Concretely, what can the solution to the following tell me?

$$ \frac{a_2}{\lambda - \lambda_2} = \lambda. $$

EDIT: I'm especially interested in the case where solving $a_1/(\lambda - \lambda_1) $ is a "good enough" approximation, as computing every $\lambda_1$ might be expensive.

Answer 1

Let $f(\lambda) = \sum_i a_i/(\lambda - \lambda_i)$. Suppose $a_1 > 0$. Then $\lim_{\lambda \to \lambda_1+} f(\lambda) = +\infty$, so $f(\lambda) > \lambda$ for $\lambda$ near $\lambda_1$ and slightly greater. On the other hand, $f(\lambda) \to 0$ as $\lambda \to +\infty$, so $f(\lambda) < \lambda$ when $\lambda$ is sufficiently large. By the Intermediate Value Theorem, there exists a solution to $f(\lambda) = \lambda$ somewhere in the interval $(\lambda_1, +\infty)$.

EDIT: You may be able to use some estimates to show that there is a solution near a solution to $a_1/(\lambda-\lambda_1) = \lambda$. Write $f(\lambda) = \frac{a_1}{\lambda - \lambda_1} + g(\lambda)$ where $$ g(\lambda) = \sum_{i=2}^n \frac{a_i}{\lambda - \lambda_i}$$ Suppose $p$ is a solution to $a_1/(\lambda - \lambda_1) = \lambda$, and in some interval $[p-\delta, p+\delta]$ around $p$ we have $f'(\lambda) - 1 < -m < 0$ and $|g(\lambda)| < \epsilon$. The $f(p+\delta) - (p+\delta) < -m \delta + \epsilon$ while $f(p-\delta) - (p-\delta) > m \delta - \epsilon$. If $m \delta - \epsilon > 0$, there must be a solution to $f(\lambda) = \lambda$ in the interval $(p - \delta, p+\delta)$. For example, this will work if $a_2,\ldots,a_n$ are sufficiently small.

Answer 2

Something I've found to be helpful. We may re-express \begin{align*} \sum_{i=1}^{n}\left\{\frac{a_i}{\lambda - \lambda_i} - \frac{\lambda}{n}\right\} = 0 \end{align*} Solving each summand individually (for positive solutions of $\lambda$): \begin{align*} f_i(\lambda) \overset{\text{def}}{=}\frac{a_i}{\lambda - \lambda_i} - \frac{\lambda}{n} = 0 \implies \lambda_i^* = \frac{1}{2}\left(\sqrt{4a_in + \lambda_i^2} + \lambda_i\right) \end{align*} Next, approximate $f_i(\lambda)$ with a Taylor series around $\lambda = \lambda_i^*$, \begin{align*} f_i(\lambda) &\approx a_i\left\{\frac{1}{\lambda_i^* - \lambda_i} - \frac{\lambda - \lambda_i^*}{\lambda_i^*-\lambda_i} + \frac{(\lambda - \lambda_i^*)^2}{(\lambda_i^*-\lambda_i)^3}\right\} \\ &=A_i\lambda^2 + B_i\lambda + C_i \end{align*} for constants $A_i, B_i, C_i$ after collecting the terms. Finally, you then solve \begin{align*} 0 = \sum_{i=1}^{n}f_i(\lambda) \approx \sum_{i=1}^{n}\{A_i \lambda^2 + B_i \lambda + C_i\} = A\lambda^2 + B\lambda + C \end{align*} In reality, Newton-Raphson could get a very close approximation within 4-5 iterations, but with this, you can get a closed-form approximation.

Answer 3

Just a few ideas.

For the case where the $a_i$'s are positive and $\lambda_1 \gg \lambda_2 >\cdots > \lambda_n$, the solution would be just above $\lambda_1$ and a possible first approximation would be given by the solution of $$\frac{a_1}{\lambda - \lambda_1}-(\lambda - \lambda_1)+\left(\sum_{i=2}^n \frac{a_i}{\lambda_1 - \lambda_i}-\lambda_1\right)=0\tag 1$$ which is quadratic in $(\lambda-\lambda_1)$ and we select the positive solution of it. Another possiblity would be to expand the equation around $\lambda_1$ using Taylor series to get $$\frac{a_i}{\lambda - \lambda_i}=\frac{a_i}{\lambda_i-\lambda_i }-\frac{a_i }{(\lambda_1 -\lambda_i)^2}(\lambda-\lambda_1)+O\left((\lambda-\lambda_1)^2\right)$$ making the equation to be $$\frac{a_1}{\lambda - \lambda_1}-(\lambda - \lambda_1)\left(1+\sum_{i=2}^n \frac{a_i}{(\lambda_1 - \lambda_i)^2} \right)+\left(\sum_{i=2}^n \frac{a_i}{\lambda_1 - \lambda_i}-\lambda_1\right)=0\tag 2$$

Just to test, for a few values of $n$, I used $a_i=1 \,\, \forall i$, $\lambda_1=2p_{n}$ and $\lambda_i=p_{n-i+1}$. Below are given some results

$$\left( \begin{array}{ccc} n & \text{using } (1)& \text{using } (2) & \text{exact} \\ 2 & 6.1622777 & 6.1622777 & 6.1686686 \\ 3 & 10.100248 & 10.098702 & 10.101653 \\ 4 & 14.071954 & 14.070869 & 14.072521 \\ 5 & 22.045695 & 22.045297 & 22.045833 \\ 6 & 26.038680 & 26.038346 & 26.038779 \\ 7 & 34.029541 & 34.029351 & 34.029583 \\ 8 & 38.026434 & 38.026262 & 38.026467 \\ 9 & 46.021819 & 46.021705 & 46.021836 \end{array} \right)$$ AT least for these specific cases, it seems that $(1)$ is more than sufficient.