Finding the general term for the sequence $a_n = \frac{3}{4}a_{n-1} +4e$

Question

How do I find the general term for the sequence $$a_n = \frac{3}{4}a_{n-1} +4e$$ using a generating function? If there is an easier way to do it without using a generating function, please tell me. Thanks in advance.

Answer 1

I'd write it as $$ a_{n+1}=\frac{3}{4}a_n+4e $$ so we can start counting from $0$. We also have $$ a_{n+2}=\frac{3}{4}a_{n+1}+4e $$ so $$ a_{n+2}=\frac{3}{4}a_{n+1}+a_{n+1}-\frac{3}{4}a_n $$ or $$ 4a_{n+2}-7a_{n+1}+3a_n=0 $$ which has, as characteristic polynomial $4X^2-7X+3$. The roots are $1$ and $3/4$, so the general solution is $$ a_{n}=\alpha 1^n+\beta\left(\frac{3}{4}\right)^n $$ with $a_1=3(a_0+16e)/4$ and $a_0$ arbitrary. Thus we get $$ \begin{cases} a_0=\alpha+\beta\\[2ex] \dfrac{3}{4}a_0+4e=\alpha+\dfrac{3}{4}\beta \end{cases} $$

Answer 2

As Did notes in comments above, getting a generating function is a bit of overkill here. That said, I figured I'd show how it works in this case for both exponential and ordinary generating functions.

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Exponential GF

We introduce $A_e (x)=\sum\limits_{n=0}^\infty \dfrac{a_n}{n!}x^n$ as the exponential generating function of $\{a_n\}$. Then

\begin{align} A_e'(x) = \sum_{n=1}^\infty \frac{a_n}{(n-1)!}x^{n-1} &= \frac34 \sum_{n=1}^\infty \frac{a_{n-1}}{(n-1)!}x^{n-1} +e\sum_{n=1}^\infty \frac{1}{(n-1)!}x^{n-1}\\ &= \frac34 \sum_{n=0}^\infty \frac{a_{n}}{(n)!}x^{n} +e\sum_{n=0}^\infty \frac{1}{(n)!}x^{n}\\ &=\frac34 A_e(x)+e^{x+1} \end{align} So now we have an inhomogeneous ODE for $A_e(x)$. If we solve this, then we can identify $a_n=A_e^{(n)}(0)$ and obtain the sequence.

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Ordinary GF

We introduce $A_o(x)=\sum\limits_{n=0}^\infty a_n x^n$ as the ordinary generating function of $\{a_n\}$. Then

\begin{align} A_o(x) &=a_0+\sum_{n=1}^\infty a_n x^n\\ &=a_0+\frac34\sum_{n=1}^\infty a_{n-1} x^n+e\sum_{n=1}^\infty x^n\\ &=a_0+\frac34 x A_o(x)+\frac{ex}{1-x} \end{align} We can solve this equation for $A_o(x)$ and expand in a power series to obtain $a_n$.

Answer 3

Generating functions are a good method, but for this straightforward problem there's another one, since you can get rid of the constant: $$ a_ n = \frac{3}{4} a_{n-1} + 4e\\ a_{n-1} = \frac{3}{4}a_{n-2} +4e\\ a_n - a_{n-1} = \frac{3}{4}(a_{n-1} -a_{n-2}) $$ Now set $\Delta a_n = a_n - a_{n-1}$ $$ \Delta a_n = \frac{3}{4} \Delta a_{n-1} = \bigg(\frac{3}{4}\bigg)^2 \Delta a_{n-2} = \ldots =\bigg(\frac{3}{4}\bigg)^{n-1} \Delta a_1 $$ This get s you a telescoping sum on the LHS and a geometric sum over $n$ on the RHS.