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

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.

• Do you know the behavior of $a_i$? If $a_i$ was very small (went to zero) after one or two terms then you can neglect them. But since the $\lambda_i$ are just getting smaller, the component terms of the sum are simply approaching $\dfrac{a_i}{\lambda}$. – JacobCheverie Apr 15 '19 at 15:04
• Can we assume $a_i, \lambda_i$ are positive? – Tom Chen Apr 15 '19 at 15:06
• @TomChen Yes, although I'm interested in the general case too. – Leo Apr 15 '19 at 15:09
• @JacobCheverie this is my intuition too though I'm not sure how large the difference $\lambda_1 - \lambda_2$ must be for me to make that approximation. And in that case, what's the size of the error term. – Leo Apr 15 '19 at 15:12

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.

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.

• Excellent! Any ideas on how to approximate that solution using the solution to $a_1/(\lambda - \lambda_1) = \lambda$? Or any other methods? – Leo Apr 15 '19 at 15:22

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.