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Find the least $\lambda$ for which the sequence $\{b_n\}$ defined by $b_1=1$, $b_2=\lambda-1$ and $b_{n+2}=\lambda(b_{n+1}-b_n)$ is always positive.

I guess $\lambda=4$, which yields $b_n=2^{n-1}(n+1)$. This indeed works. For $\lambda=3.9$, I used Python to disprove.

l=3.9
a=1
b=2.9
for i in range(20):
    c=l*(b-a)
    a=b
    b=c
print(a,b)

I also found that $\lambda=4$ is the least parameter for which the quadratic $x^2-\lambda(x-1)$ has real roots.

Actually I find the question here. But the solution seems too tricky. I want to find a more natural one.

youthdoo
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  • What have you tried? For your problem where the recurrence relation is an equation rather than an inequality given in the link, it's seems pretty straightforward. Can you solve the characteristic equation and find the closed form for $b_n$? – Calvin Lin Sep 27 '23 at 12:34
  • "Can you solve the characteristic equation and find the closed form for bn ?" Unfortunately no, since by computer we have$$\frac{2^{-n-1} \left(\sqrt{L-4} \left(\left(L-\sqrt{L-4} \sqrt{L}\right)^n+\left(L+\sqrt{L-4} \sqrt{L}\right)^n\right)+\sqrt{L} \left(\left(L+\sqrt{L-4} \sqrt{L}\right)^n-\left(L-\sqrt{L-4} \sqrt{L}\right)^n\right)\right)}{\sqrt{L-4} L},$$which is hopeless to do by hand. – youthdoo Sep 27 '23 at 13:56

3 Answers3

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I have previously described in detail a general solution for every sequence of the form $f_n=af_{n-1}+bf_{n-2}$ with arbitrary $f_0,f_1$. The solution can be expressed as

$$ f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta} $$

where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.

What is happening in your problem is that when $\lambda<4$, $\alpha$ and $\beta$ are complex, and moreover they are conjugates. Consider that $\alpha,\beta=re^{\pm i\theta}$, then

$$ \frac{\alpha^n-\beta^n}{\alpha-\beta}=\frac{r^n(e^{in\theta}-e^{-in\theta})}{r(e^{i\theta}-e^{-i\theta})}=\frac{r^n\sin n\theta}{r\sin \theta}\\ \frac{\alpha^n+\beta^n}{\alpha+\beta}=\frac{r^n(e^{in\theta}+e^{-in\theta})}{r(e^{i\theta}+e^{-i\theta})}=\frac{r^n\cos n\theta}{r\cos \theta} $$

Then the solution can be expressed as

$$ \begin{align} &f_n=(A\sin n\theta+B\cos n\theta)r^n\\ &B=f_0\\ &A=\frac{f_1/r-f_0\cos\theta}{\sin\theta} \end{align} $$

So the solution is seen to periodic with $\pm$-terms and that's why the outcome can only be positive when $\lambda\ge4$.

Cye Waldman
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1

The "natural" approach by just pushing our brute force way through.
For simplicity of typing, let $ \lambda = L$.

The solutions to the characteristic equation $x^2 - Lx + L = 0 $ are $ \alpha = \frac{L+ \sqrt{(L-4)L}}{2}$ and $ \beta = \frac{L- \sqrt{(L-4)L}}{2}$.
We have $ \alpha + \beta = L, \alpha\beta = L, \alpha - \beta = \sqrt{ (L-4)L}$.
The solutions to the recurrence relation is of the form $b_n = A\alpha^n + B\beta ^n$.
The initial conditions are $b_0 = \frac{1}{L}$ (Check this!), $b_1 = 1$.
So we want to solve for $ \frac{1}{L} = A + B , 1 = (A+B) \frac{L}{2} + (A-B) \frac{\sqrt{(L-4)L}}{2}$.
Solving this system gives $A+B = \frac{1}{L}$ and $ A-B = \frac{1}{\sqrt{(L-4)L}}$, or that $A = \frac{ \sqrt{L} + \sqrt{L-4} } {2 L \sqrt{L - 4} }$ and $B = \frac{ - \sqrt{L} + \sqrt{L-4} } {2 L \sqrt{L - 4} }$.
Hence $ b_n = \frac{1}{2L \sqrt{L-4}} [\sqrt{L-4} (\alpha^n + \beta^n) + \sqrt{L} (\alpha^n - \beta^n)]$, which OP stated in the comments.
This can be further simplified to

$$b_n = \frac{1}{2}[ \frac{(\alpha^n + \beta^n)}{\alpha+\beta} + \frac{(\alpha^n - \beta^n)}{\alpha-\beta}] = \frac{\alpha^{n+1} - \beta^{n+1}}{\alpha^2 - \beta^2} . $$

Observing that $b_{2^k-1} = (\alpha^{2^k} + \beta^{2^k}) b_{2^{k-1} -1} = \prod_{j=1}^k (\alpha^{2^j} + \beta^{2^j}) $, we choose to focus on $c_k = (\alpha^{2^k} + \beta^{2^k}) $ and show that this sequence isn't always positive, which would then yield the desired negative $b_n$.

Proof by contradiction. Suppose $c_k$ is always positive.
We have $c_{k+1} = c_k^2 - 2L^{2^k}$, which gives $c_{k+1} - 2L^{2^k} = ( c_k - 2L^{2^{k-1} } ) ( c_k + 2L^{2^{k-1} }) = (c_1 - 2L) \prod_{j=1}^k ( c_j + 2L^{2^{j-1} }) $.
Notice that $c_1 - 2L = L^2 - 4L < 0$.
Since $c_k$ is always positive, thus $\prod_{j=1}^k ( c_j + 2L^{2^{j-1} }) > \prod_{j=1}^k ( 2L^{2^{j-1} }) = 2^k L^{2^k - 1 } $.
Then, for a large enough $K$, $2^K (4L-L^2) > 2L$, which would give $c_{K+1} - 2L^{2^K} = (c_1 - 2L) \prod_{j=1}^k ( c_j + 2L^{2^{j-1} }) < - 2L^{2^K}$, which results in $c_{K+1} < 0 $ as desired.

Thus, we have the stronger statement that eventually one of $b_{2^k - 1}$ is negative.

Calvin Lin
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Claim: For $\lambda < 4$, the sequence (for any positive starting values) is eventually negative.

Suppose not. Let $ c_n = \frac{b_{n+1}}{b_{n} }$, which is always positive.
The condition transforms to $c_{n+1} + \frac{\lambda}{c_{n}} = \lambda $.
Thus we have $\lambda \geq 2\sqrt{ \lambda}\sqrt{ \frac{ c_{n+1}} { c_{n}}}$, or that $ 1 > \frac{\lambda}{2\sqrt{\lambda}} \geq \sqrt{ \frac{ c_{n+1}} { c_{n}}}$, so $c_n$ is a decreasing sequence.
We also have $c_n > 1$ from the condition.
Thus, $c_n$ is a decreasing sequence that is bounded below, so it has a limit, which we denote by $c^*$.
However, there are no real solutions to $c^* + \frac{\lambda}{c^*} = \lambda $, which is a contradiction.

Calvin Lin
  • 77,541