[Math] Solve the recurrence relation by taking the logarithm of both sides and making the substitution $b_n = \lg a_n$

discrete mathematicslogarithmsrecurrence-relations

Solve this recurrence relation:

$$a_n = \left(\frac{a_{n-2}}{a_{n-1}}\right)^{\frac{1}{2}}$$

by taking the logarithm of both sides and making the substitution

$$b_n = \lg a_n$$

A couple years ago I took precalc and a couple years before that I took College Algebra to brush up since before then I'd been out of school for a couple of years already. So I'm extremely rusty and this first step, getting the logarithm of both sides, has me confused enough… Like how would I even get the logarithm of the left side?? and then what exactly is that "substitution" supposed to be substituting?

So log'ing both sides I get

$$\ln(a_{n}) = \frac{1}{2}\ln(a_{n-2}) – \frac{1}{2}\ln(a_{n-1})$$

Thank you Alex below but I still don't get how that translates to the substitution equation given… or how to solve for b..? Sorry… Please help me understand better…

Would b_n =

$$b_{n} = \frac{1}{2}[\ln(a_{n-2})-\ln(a_{n-1})]$$

? Then still, how to solve that…

Best Answer

Remember the properties of the log function: $a \log b = \log b^a$ and $\log \frac{a}{b} = \log a - \log b$. Use it to obtain the recurrence $b_n = \frac{1}{2} b_{n-2} - \frac{1}{2}b_{n-1}$. Then solve it using generating functions or difference equations.

Related Solutions

Trying to solve a recurrence relation by using generating functions: $a_n=3a_{n-1} + a_{n-2}$

I'm going to rewrite the problem as $$a_{n+2} = 3a_{n+1} + a_n; \; a_0 = 2,\; a_1=1.$$ Multiply by $x^n$, sum, and let $A(x) = \sum_{n\ge 0}a_nx^n$. So, $$\sum_{n\ge 0}a_{n+2}x^n = 3\sum_{n\ge 0}a_{n+1}x^n + \sum_{n\ge 0}a_n x^n,$$ and we can see that $\sum_{n\ge 0}a_{n+2}x^n = a_2 + a_3x + \cdots = (1/x^2)(A(x)-a_0-a_1x)$ and similarly $\sum_{n\ge 0}a_{n+1}x^n = a_1 + a_2x + \cdots = (1/x)(A(x)-a_0)$, so we obtain $$\frac{1}{x^2}(A(x) - a_0 - a_1x) = \frac{3}{x} (A(x)-a_0 ) + A(x).$$

Substituting for $a_0$ and $a_1$ and solving for $A(x)$ yields your $F(x)$, ie $$A(x) = \frac{2-5x}{1-3x-x^2}.$$

Now the for the partial fraction decomposition, which is not so ugly if we keep our heads straight. Note that $1-3x-x^2$ has roots $\alpha_1 = -\frac{3}{2} - \frac{\sqrt{13}}{2}$ and $\alpha_2 = -\frac{3}{2} + \frac{\sqrt{13}}{2}$ and we want $$\frac{1}{1-3x-x^2} = \frac{1}{(x-\alpha_1)(x-\alpha_2)} = \frac{A}{(x-\alpha_1)} + \frac{B}{(x-\alpha_2)}.$$ Using this equation and solving for $A$ and $B$ yields \begin{align} \frac{1}{1-3x-x^2} &= \frac{1}{(\alpha_1-\alpha_2)(x-\alpha_1)} + \frac{1}{(\alpha_2-\alpha_1)(x-\alpha_2)}\\ &=\frac{1}{-\sqrt{13}(x-\alpha_1)} + \frac{1}{\sqrt{13}(x-\alpha_2)} \end{align}

So we have

\begin{align} A(x) &= (2-5x)\left(\frac{1}{-\sqrt{13}(x-\alpha_1)} + \frac{1}{\sqrt{13}(x-\alpha_2)} \right)\\ &=(2-5x) \left( \frac{1}{\alpha_1\sqrt{13}} \cdot \frac{1}{1-(x/\alpha_1)} + \frac{1}{-\alpha_2\sqrt{13}} \cdot \frac{1}{1-(x/\alpha_2)} \right)\\ &= (2-5x) \left( \frac{1}{\alpha_1\sqrt{13}} \sum_{n\ge 0} \left(\frac{1}{\alpha_1} \right)^n x^n + \frac{1}{-\alpha_2\sqrt{13}} \sum_{n\ge 0} \left(\frac{1}{\alpha_2} \right)^n x^n \right)\\ &= \frac{(2-5x)}{\sqrt{13}} \left(\sum_{n\ge 0} \left[ \left(\frac{1}{\alpha_1} \right)^{n+1} - \left(\frac{1}{\alpha_2} \right)^{n+1}\right] x^n \right) \end{align}

Can you take it from here?

Solve the recurrence relation $a_{n+1}=\frac{2a_n^2}{1-2a_n^2}$

This answer to your question may not have a closed form, but it is not that complicated. Define the real constants $$ w :=\! \sqrt{3}, \; p :=\! (-1\!+\!w)/2, \; q :=\! (-1\!-\!w)/2. \tag{1} $$ Define the recurrence function $$ f(x) := 2x^2/(1-2x^2). \tag{2} $$ Notice that $\,p\,$ is a repelling while $\,q\,$ is an attracting fixed point of $\,f(x).\,$

Define the convergent power series $$ s(x) \!:=\! x \!+\!\! \left(\frac12\!-\!\frac56w\right)\!x^2 \!+\!\! \left(\frac73\!-\!\frac23w\right)\!x^3 \!\!+\! O(x^4).\! \tag{3} $$ Then the following functional equation holds $$ f(q+s(x)) = q+s(-2px) \tag{4} $$ where the the coefficients of $\,s(x)\,$ are uniquely determined so that the functional equation holds. Thus the convergent ($\;|\!-\!2p|<1\;$) sequence for any $\,x_0\,$ defined by $$ a_n := q + s((-2p)^n x_0) \tag{5} $$ satisfies $\,a_n \to q\,$ as $\,n\to\infty\,$ and the recursion $$ a_{n+1} = f(a_n). \tag{6} $$

For your information here is a PARI/GP code to calculate the power series for $\,s(x)\,$

w = quadgen(12); p = (-1+w)/2; q = (-1-w)/2;
f(x) = 2*x^2/(1-2*x^2); 
nxt(n) = {my(s, c='c);
  s = truncate(stx + O(x)^n) + c*x^n*(1 + O(x));
  s = truncate(f(q+s) - (q+subst(s, x, -2*p*x)))/x^n;
  sx -= x^n * polcoeff(s, 0, c)/polcoeff(s, 1, c)};
sx = x ; for(n=2, M=9, nxt(n)); print(sx + x*O(x)^M);

Best Answer

Related Solutions

Trying to solve a recurrence relation by using generating functions: $a_n=3a_{n-1} + a_{n-2}$

Solve the recurrence relation $a_{n+1}=\frac{2a_n^2}{1-2a_n^2}$

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