Solving Difference Equation: w_n = 2^n w_0 + 2^{n + 1}-1

In summary, the conversation is about finding the general solution to a difference equation. The formula for w_n is derived using the given equation and it is shown that the book's answer is incorrect. The correct general solution is w_n = 2^n(w_0+1) - 1. There is a discussion about the possibility of different solutions and the conclusion is that the book's formula is wrong.
  • #1
Benny
584
0
Hi, I've been working on a difference equation and I just can't get the answer. Can someone checking my working?
[tex]
w_{n + 1} = 2w_n + 1
[/tex]
w_1 = 2w_0 + 1
w_2 = 2w_1 + 1 = 2(2w_0 + 1) + 1 = 2^2w_0 + 1 + 2^1
[tex]
\Rightarrow w_n = 2^n w_0 + \sum\limits_{i = 0}^{n - 1} {2^i } = 2^n w_0 + \sum\limits_{i = 0}^n {2^i } - 2^n = 2^n w_0 + \frac{{1 - 2^{n + 1} }}{{1 - 2}} - 2^n
[/tex]

[tex]
w_n = 2^n w_0 ' + 2^{n + 1} - 1 - 2^n = 2^n \left( {w_0 ' - 1} \right) + 2^{n + 1} - 1
[/tex]...I have written w_0 with a dash so as to enable me to get a 'nicer' looking answer. It is a little ambiguous but hopefully people understand what I've done. I've simply taken 2^n as a common factor of two of the terms so that I get 2^n multipled by something. In the next line I replace that 'thing' by w_0.

[tex]
w_n = 2^n w_0 + 2^{n + 1} - 1
[/tex]

Where I have used a primed w_0 so that I could get an answer which resembles the book's. The book's answer is the same as mine except where I have a negative one, it has a negative two. I don't know where I'm going wrong. Can someone help me out?
 
Last edited:
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  • #2
Your answer is incorrect, since your formula predicts [tex]w_{0}=2^{0}w_{0}+2-1=w_{0}+1[/tex]
Similarly [tex]w_{1}=2w_{0}+2^{2}-1=2w_{0}+3[/tex]

You have correctly found:
[tex]w_{n}=2^{n}w_{0}+2^{n+1}-1-2^{n}[/tex]
Rewrite this as follows:
[tex]2^{n}w_{0}+2^{n+1}-1-2^{n}=w_{0}2^{n}+2^{n}(2-1)-1=w_{0}2^{n}+2^{n}-1=2^{n}(w_{0}+1)-1[/tex]
 
  • #3
Thanks for the help but I still don't understand how the book got [tex]w_n = 2^{n + 1} - 2 + 2^n v_0 [/tex] (I've typed the answer exactly as it is given with the v_0 and not the w_0). Is my corrected answer(the one you included in your reply) somehow equivalent to the book's answer? Or is it possble to get 'different' general solutions depending on the solution procedure?
 
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  • #4
Your book's formula is wrong!

Your difference equation says that [tex]w_{1}=2w_{0}+1[/tex]
but their formula says: [tex]w_{1}=2^{2}-2+2w_{0}=2+2w_{0}[/tex]
 
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  • #5
Hmm...I know that the answers in books are never (rarely) 100% with their answers but since it is so rare for an error to be in there I just assumed that their answer had to be correct. Thanks for clearing that up.
 

Related to Solving Difference Equation: w_n = 2^n w_0 + 2^{n + 1}-1

1. How do I solve this difference equation?

To solve this difference equation, you can use the method of iteration. Start by plugging in the initial condition, w_0, and then plug in w_1, w_2, w_3, and so on until you see a pattern emerge.

2. Can I use a different initial condition for w_0?

Yes, you can use any initial condition for w_0 as long as it is a constant value. However, using different initial conditions may result in a different solution for w_n.

3. What is the significance of 2^{n+1}-1 in the equation?

The term 2^{n+1}-1 is a constant that is added to each iteration of the solution. It helps to account for the exponential growth of the term 2^n w_0, making the equation valid for all values of n.

4. Is there a closed-form solution for this difference equation?

Yes, there is a closed-form solution for this difference equation. It is w_n = (2^n-1)w_0 + 2^{n+1}-1. This formula can be derived using mathematical induction.

5. How can this difference equation be applied in real-world situations?

Difference equations are often used in various fields of science and engineering to model and predict changes over time. This particular equation can be used to model exponential growth or decay of a quantity, such as population growth or radioactive decay.

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