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Lets have two recursive equations:

\begin{align} f(0) &= 2 \\ f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\ g(0) &= -2 \\ g(n+1) &= 3 \cdot g(n) + 12 \end{align}

We want a explicit equation for f(x) - g (x).

I firstly tried to do in manually for first $n$ numbers

\begin{array}{|c|c|c|c|} \hline n & f(n) & g(n) & f(n) - g(n) \\ \hline 0 & 2 & -2 & 4 \\ \hline 1 & 6 & 6 & 0 \\ \hline 2 & 26 & 30 & -4 \\ \hline 3 & 94 & 102 & -8 \\ \hline 4 & 306 & 318 & -12 \\ \hline \end{array}

We can deduce that $f(n) - g(n) = 4 - 4n$

But now we have to prove it.

Lets extend recursive equation $f(n)$: $$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$ for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$

We can simply check this by induction but I will skip it, so the question won't be so long.

Now lets put it together:

$$ f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\ = 4 \cdot 3^n + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\ = 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\ = 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 $$

As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.

How to prove that:

$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$

All I could do is this:

\begin{align} &8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\ &= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\ &= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\ &= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2}) \end{align}

And I made sum function out of it: $\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$

What to do next? Did I go the wrong direction anywhere?

Thank you for your fast responses.

trivial math is difficult
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Patrik Bašo
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5 Answers5

4

I think you can just use an induction proof, since you already have an intuition of the result, it is much easier to check :

$$\text{Let : } H_n :"w_n=f(n)-g(n)=4-4n" $$

First, for $n=0$ : $$w_0=4=4-4\times 0 $$ Hence, $H_0$ is true.
Let $n\in\mathbb{N} $, such that $H_n$ is true, let us show $H_{n+1}$ $$w_{n+1}=3w_n+8n-12=3(4-4n)+8n-12\\ = 12-12n+8n-12=-4n=4-4(n+1) $$

Hence, $H_{n+1} $ is true, so we can conclude that :

$$\forall n \in \mathbb{N}, f(n)-g(n)=4-4n $$

NHL
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  • Well i see I overengineered a little bit... But thank you :) – Patrik Bašo Jan 04 '21 at 14:16
  • For this kind of questions, it can be very tedious. However, computing the first values of the sequence is definately a good thing, it can give you some intuition on the result, and sometimes (like here), it lets you use proof by induction – NHL Jan 04 '21 at 14:18
  • Well you gave me answer, but still, im really want to know how to prove that Sum :D :) – Patrik Bašo Jan 04 '21 at 14:21
3

$$d(n+1)=3d(n)+8n-12,\\d(0)=4.$$

The homogeneous solution is

$$d_h(n)=3^nd_h(0).$$

Plugging the initial condition,

$$3^0d_h(0)=3\cdot4+8\cdot0-12=0.$$

Then with the ansatz $d_a(n)=an+b,$

$$an+a+b=3an+3b+8n-12$$ or by identification,$$d_a(n)=4-4n=d(n).$$

2

There is no need for induction, use a straight proof.

The equation is

$$f(n+1)-g(n+1)=3(f(n)-g(n))+8n-12$$ with $$f(0)-g(0)=4$$ and it does verify the solution

$$f(n)-g(n)=4-4n$$ as is shown by substitution,

$$4-4(n+1)=3(4-4n)+8n-12,$$ equivalent to $$-4n=-4n.$$

Furthermore, $$4=4-4\cdot 0.$$

  • I like the clarity of the algebra, but it seems like it would be clearer to just write $f(n+1)-g(n+1)$ as the left hand side of the fourth equation and then simplifying the right hand side as the next step (giving, well, an inductive proof). As it stands, there's a lemma hiding in here that says linear recurrences have a unique solution given boundary conditions - but, at least to my sense of style, that seems odd given that the elementary inductive proof is just as clear and just as short. – Milo Brandt Jan 04 '21 at 22:49
  • @MiloBrandt: the question was to prove that $4-4n$ is a solution. Uniqueness needs not be addressed. To check the solution of an equation, the straight method is to plug it. –  Jan 05 '21 at 08:08
1

So, denoting $w_n = f_n -g _n$, you can write down the equation \begin{align} w_0 = f_0-g_0 = 4, \qquad &w_{n+1} = f_{n+1}-g_{n+1} = 3(f_n-g_n)+8n -12, \quad\text{i.e.}\\ w_0 = 4, \qquad &w_{n+1} = 3 w_n +8n -12. \end{align}

This a liner difference equation for which you can have an explicit solution, in the same way that you could have computed explicitly $f_n$ and $g_n$ right from the beginning.

The general solution of the homogeneous equation $w_{n+1} = 3 w_n$ is $w_n^h = c 3^n$, and if you search for a particular solution of the form $w_n^* = a n + b$, you'll get $w_n^* = 4 - 4n$. So the general solution is $$ w_n = c 3^n +4 -4n. $$

Finally, using the initial condition $w_0=4$, we get $c=0$ and the solution is $w_n = 4 - 4n$.

trivial math is difficult
  • 691
PierreCarre
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1

This is an elegant problem. You can use induction on the difference and skip the brute force computation.

Proof by induction: Let S(n) be the claim that $f(n)-g(n)=4-4n$.

Base case: trivial.

Inductive case: Suppose S(n-1) is true. That is, suppose $f(n-1)-g(n-1)=4-4(n-1)$. Then $f(n)-g(n)=[3f(n-1)+8(n-1)]-[3g(n-1)+12]=3[f(n-1)-g(n-1)]+8n-20. $

By induction, that equals:$3[4-4(n-1)]+8n-20=12-12(n-1)+8n-20=12-12n+12+8n-20=24-4n-20=4-4n$

and you're done.

Hank Igoe
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