I need help writing equations please...

[Alpha6]

So I had to solve the following problem...

X_n+1 - 5X_n + 4X_n-1 = 0 X₁ = 9 and X₂ = 33

So I set X = 1 and got:

X₁₊₁ - 5X₁ + 4X_1-1 = 0

(33) - 5(9) + 4X₀ = 0

33-45 + 4X₀ = 0

-12 + 4X₀ = 0

4X₀ = 12

X₀ = 3

So we know that X₀= 3, X₁=9, and X₂=33

Knowing this, how do I write the equation?? The equation is X_n = 1 + 2 * 4ⁿ However, I have no idea how to arrive at that equation if the book did not tell me. Any help please? There are many problems like these, I can always solve for X but I can't find the general equation.

 

[Brainwave]

solution

You would have been thought difference equation for this type of question. You can see see my solution in the attachment below

 

[Alpha6]

Thanks for replying.

I see why C₁=2 and C₂=1 but I wouldn't know how to work that out.

Do you use different numbers until one works out? A trial and error approach?

 

[JeffM]

So I had to solve the following problem...

X_n+1 - 5X_n + 4X_n-1 = 0 X₁ = 9 and X₂ = 33

So I set X = 1 and got:

X₁₊₁ - 5X₁ + 4X_1-1 = 0

(33) - 5(9) + 4X₀ = 0

33-45 + 4X₀ = 0

-12 + 4X₀ = 0

4X₀ = 12

X₀ = 3

So we know that X₀= 3, X₁=9, and X₂=33

Knowing this, how do I write the equation?? The equation is X_n = 1 + 2 * 4ⁿ However, I have no idea how to arrive at that equation if the book did not tell me. Any help please? There are many problems like these, I can always solve for X but I can't find the general equation.

You can virtually "see" this one if you already know the powers of 2: 3 is 1 greater than 2, 9 is 1 greater than 8, and 33 is one greater than 32, and 2, 8, and 32 are the successive odd powers of 2.

So \(\displaystyle x_n = 1 + 2^{(2n - 1)} = 1 + 2 * 2^{2n} = 1 + 2 * 4^n. \)

But perhaps you do not "see" that the powers of 2 are involved. What next?

If successive terms differ by a constant factor, you can discover the common factor as follows: \(\displaystyle x_{n + 1} = a * x^n \ne 0 \implies \dfrac{x_{n + 1}}{x_n} = a.\)

That does not work here because 9 / 3 = 3 and 33 / 9 > 3.

If successive terms differ by a constant, you can find the constant as follows: \(\displaystyle x_{n + 1} = x_n + a \implies a = x_{n + 1} - x_n.\)

That does not work here because 9 - 3 = 6 and 33 - 9 = 24. But maybe there is a ratio between these differences. 24 / 6 = 4.

We need more examples to see if there is a pattern.

\(\displaystyle x_3 - 5x_2 + 4x_1 = 0 \implies x_3 - 5 * 33 + 4 * 9 = 0 \implies x_3 = 165 - 36 = 129.\)

\(\displaystyle x_4 - 5x_3 + 4x_2 = 0 \implies x_3 - 5 * 129 + 4 * 33 = 0 \implies x_4 = 645 - 132 = 513.\)

So now the successive differences are 6, 24, 129 - 33 = 96, and 513 - 129 = 384.

But what are the ratios of the successive differences 24 / 6 = 4, 96 / 24 = 4, 384 / 96 = 4.

OK, it looks as though 4 is involved as follows:

\(\displaystyle \dfrac{x_{n + 1} - x_n}{x_n - x_{n - 1}} = 4 \implies x_{n + 1} - x_n = 4x_n - 4x_{n - 1} \implies x_{n + 1} - 5x_n + 4x_{n - 1} = 0.\) We are definitely on the right track.

In fact, 24 / 6 = 4, 96 / 6 = 16, 384 / 6 = 64. It looks as though powers of 4 are involved.

\(\displaystyle x_1 = 9 = 2 * 4 + 1 = 1 + 2 * 4^1.\)

\(\displaystyle x_2 = 33 = 2 * 16 + 1 = 1 + 2 * 4^2.\)

\(\displaystyle x_3 = 129 = 2 * 64 + 1 = 1 + 2 * 4^3.\)

\(\displaystyle x_4 = 513 = 2 * 256 + 1 = 1 + 2 * 4^4.\)

This looks very promising. If this pattern works \(\displaystyle 1 + 2 * 4^0 = 1 + 2 * 1 = 1 + 2 = 3 = x_0.\)

Now lets test.

\(\displaystyle x_n = 1 + 2 * 4^n \implies x_{n + 1} - 5x_n + 4x_{n - 1} = 1 + 2 * 4^{(n+1)} - 5\left(1 + 2 * 4^n\right) + 4(1 + 2 * 4^{n-1)}) =\)

\(\displaystyle 1 + 2 * 4^{(n + 1)} - 5 - 10 * 4^n + 4 + 8 * 4^{(n - 1)} =\)

\(\displaystyle 2 * 4^{(n + 1)} - 10 * 4^n + 8 * 4^{(n - 1)} =\)

\(\displaystyle 2 * 4^2 * 4^{(n - 1)} - 10 * 4^1 * 4^{(n-1)} + 8 * 4^{(n - 1)} =\)

\(\displaystyle 4^{(n - 1)}(2 * 4^2 - 10 * 4^1 + 8) = 4^{(n - 1)}(2 * 16 - 10 * 4 + 8) = 4^{(n - 1)}(32 - 40 + 8) = 4^{(n - 1)} * 0 = 0.\)

I do not know a general method for finding a closed form for a recursively defined function. I generally work out a number of terms of the recursion and then start taking differences, ratios, ratios of differences until I find some consistent element and then start trying to use that consistent element as part of a pattern. If you take this experimental approach, you must then test to make sure that the pattern is not spurious.

 

[Brainwave]

Try to check textbooks for reference. But how I got C1 and C2 is below.

 

[HallsofIvy]

So I had to solve the following problem...

X_n+1 - 5X_n + 4X_n-1 = 0 X₁ = 9 and X₂ = 33

So I set X = 1 and got:

What? There is no "X", without a subscript in the problem!
Oh,I see, you mean you set n equal to 1.

X₁₊₁ - 5X₁ + 4X_1-1 = 0

(33) - 5(9) + 4X₀ = 0

Did this sequence start with \(\displaystyle X_0\) rather than \(\displaystyle X_1\)? That's possible but if so, odd that the gave \(\displaystyle X_1\) and \(\displaystyle X_2\) as the initial values rather than \(\displaystyle X_0\) and \(\displaystyle X_1\).

33-45 + 4X₀ = 0

-12 + 4X₀ = 0

4X₀ = 12

X₀ = 3

So we know that X₀= 3, X₁=9, and X₂=33

Knowing this, how do I write the equation?? The equation is X_n = 1 + 2 * 4ⁿ However, I have no idea how to arrive at that equation if the book did not tell me. Any help please? There are many problems like these, I can always solve for X but I can't find the general equation.

If you are asking "What rule can I follow or what calculation can I do to get this without having to think", it just doesn't work that way. Instead, calculate some more values and look for a pattern! Knowing that \(\displaystyle X_1= 9\) and \(\displaystyle X_2= 33\),
\(\displaystyle X_{n+1}- 5X_n+ 4X_{n-1}= 0\)
With n= 2 \(\displaystyle X_3- 5X_2+ 4X1= X_3- 5(33)+ 4(9)= X_3- 129= 0\) so \(\displaystyle X_3= 129\).
With n= 3 \(\displaystyle X_4- 5X_3+ 4X_2= X_4- 5(129)+ 4(33)= X_4- 513\) so \(\displaystyle X_4= 513\).
With n= 4 \(\displaystyle X_5- 5X_4+ 4X_2= X_5- 5(513)+ 4(129)= X_5- 2049\) so \(\displaystyle X_5= 2049\).

Continue if you like but eventually you should recognize that 9= 1+ 8, 33= 1+ 32, 129= 1+ 128, 513= 1+ 512, and 2049= 1+ 2048.

That is, every number is 1 plus a power of 2: \(\displaystyle 8= 2^3\), \(\displaystyle 32= 2^5\), \(\displaystyle 128= 2^{7}\), \(\displaystyle 512= 2^9\), and \(\displaystyle 2048= 2^11\(\displaystyle . Noting further that those are all odd powers of 2, I can write them as \(\displaystyle 2^3= 2^1(2^2)= 2(4)\), \(\displaystyle 2^5= 2^1(2^2)(2^2)= 2(4^2)\), \(\displaystyle 2^7= 2^1(2^2)(2^2)(2^2)= 2(4^3)\), \(\displaystyle 2^9= 2^1(2^2)(2^2)(2^2)(2^2)= 2(4^4)\), and \(\displaystyle 2^{11}= 2^1(2^2)(2^2)(2^2)(2^2)(2^2)= 2(4^5)\). Do you see I now?\)\)