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I'm not strong in math and my academic advisor signed me up for discrete math. Need help with my homework.

a) Find the General Solution of the recurrence equation $$S_n = 3S_{n-1} - 10$$ for $ n \in \mathbb {N} $.

b) Determine the particular solution where $ S_0 = 15 $.

c) Use the formula in B to evaluate $S_6$ and check your answer using the recurrence equation itself.

Ok, so far, for the general solution, I'm going by the formula given in the book, Sn+1 = aSn +c

if a does not equal 1, the formula is Sn = a^nA + c/1-a

so far I have Sn = 3^n[ ] + 5

Am I on the right track and how do I find what goes into the brackets? Thanks

(sorry I have no idea how to format)

Antonio Montana
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  • Can you share what you've tried, and what you know about recurrence relations? For example, can you compute the first few terms given $S_0 = 15$? –  Nov 02 '13 at 00:13
  • Math.SE is not a place where people will do your homework for you. However, most users will be more than happy to help if you make an effort to solve the problem. – tylerc0816 Nov 02 '13 at 00:14
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    So = 15, S1 = 35, S2 = 95, S3 = 275.... I don't know if I'm right.. I'm not here for you to do my HW, just here to learn how to do it myself (correctly). It looks really stupid if I get it correct on my HW but do bad on exams, of course I don't want you to do my HW for me... Just here to learn correct ways and how to understand these as this is a brand new subject for me. Thanks – Antonio Montana Nov 02 '13 at 00:18

2 Answers2

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$a$) Consider the homogeneous recurrence relation $S_n=3S_{n-1}$ where $n\geq1$. The characteristic equation is $x-3=0$, where $x=3$ is the characteristic root. Thus the general solution is $S_n=a3^n$ where $a$ is a constant.

$b$) Let $S_n=b$ where $b$ is a constant, we do this because $b_n=-10$ is a constant. So the proper guess is a constant. This implies that $b=3b-10$ and so $b=5$. Thus $S_n=5$ is the particular solution. Combining the general solution and the particular solution we have $S_n=a3^n+5$ and using the fact that $S_0=15$ we see that $a=10$. Hence $S_n=10\cdot 3^n+5$.

$c$) We let $n=6$ and see that $S_6=10\cdot 3^6+5=10\cdot 729+5=7295$. We can substitute the necessary values into the original recurrence relation $S_n=3S_{n-1}-10$ and check that we have the correct value.

1233dfv
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  • You're awesome!!! Although I have a question for answer a. Wouldn't the general solution be in the form of 3x^2 - x - 10? If not, could you explain how you got Sn = a3^n? – Antonio Montana Nov 03 '13 at 07:36
  • :) You're welcome! – 1233dfv Nov 03 '13 at 07:40
  • How did you get $3x^2-x-10$? – 1233dfv Nov 03 '13 at 07:48
  • LOL I dunno? I transferred everything to the LHS of equation, and 3Sn-1 became 3x, and n became x^2. Sorry, I meant x^2 - 3x + 10. I thought a general solution is a general form? Or am I wrong on both? Excuse my ignorance, but my professor races through his lectures and english is also his second language. Care to explain how and why you got your answer for a? And what does a general solution mean? Thanks – Antonio Montana Nov 03 '13 at 08:11
  • I do realize I am wrong now though. Could you kindly explain the process in how you got there? – Antonio Montana Nov 03 '13 at 08:26
  • Sure. The general solution to a linear recurrence relation is the solution to the homogeneous part of the linear recurrence relation. A homogeneous linear recurrence relation is of the form $h_n=a_1h_{n-1}+a_2h_{n-2}+\cdots +a_kh_{n-k}+b_n$ where $b_n=0$. So in part $a$) the homogeneous part of the linear recurrence relation is $S_n=3S_{n-1}$ which becomes $S_n-3S_{n-1}=0$. Now we set up the characteristic equation and find the general solution to the homogeneous part of our linear recurrence relation. – 1233dfv Nov 03 '13 at 08:28
  • Ok, that's where I get in trouble. I've been searching through my book/notes, I didn't see that relation. There is a general solution though in which if there is a first-order linear recurrence equation in the form of Sn+1 = aSn +c, the general solution is if a = 1, Sn = A + nc. If a does not equal 1, Sn = a^nA + c/(1-a)... Was I looking at the wrong formula too? – Antonio Montana Nov 03 '13 at 08:40
  • I'm not sure I understand. If you would like a good outline for how to approach problems like the one you posted search "Combinatorics 6th edition by Brualdi pdf". Check the worked out examples in $7.4$ and $7.5$ similar to the approach I took. I think it will be helpful. – 1233dfv Nov 03 '13 at 08:51
  • Will do! Thank you soooo much for all of your help! You help me more than my professor! SMH LOL – Antonio Montana Nov 03 '13 at 08:52
  • No problem, I'm happy to help. Post more Discrete Mathematics problems if you get stuck, feel free to ask any questions in the future. :) Take care. – 1233dfv Nov 03 '13 at 09:00
  • Ooops, sorry, need more clarification. In answer a, what happened to the -10 in the recurrence relation Sn=3Sn−1 where n≥1? Why'd you drop it? How do you get to that equation? – Antonio Montana Nov 04 '13 at 01:43
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    We solve these problems in pieces. We first find the homogeneous solution where $b_n=0$. Basically, ignore the $-10$ when solving for the homogeneous part. Set up the characteristic equation, once this has been done you will find the general solution which is of the form $ax^n$ where $a$ is a constant and $x$ is the characteristic root. Now we can move onto the particular solution where we include $b_n=-10$. We make a guess as to what $S_n$ should be. Since $b_n$ is a constant we would assume that $S_n=b$ where $b$ is a constant. – 1233dfv Nov 04 '13 at 02:20
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    Substitute the guess into the recurrence relation and you will find the particular solution. Take the general solution and the particular solution and combine them and you have the solution to your original recurrence relation. If we are given initial conditions we can solve for the constants too. – 1233dfv Nov 04 '13 at 02:23
  • Ok... Struggling as all of this is unfamiliar, but trying! – Antonio Montana Nov 04 '13 at 02:38
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    It is difficult. A lot of these ideas are used for solving second order linear differential equations. However, I know you haven't see this stuff before because of you said you haven't taken calculus. It takes time, no worries. Did you check out the textbook I recommended? – 1233dfv Nov 04 '13 at 02:53
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HINT: Let $$f(x)=\sum_{n=0}^{\infty}S_nx^n=15+\sum_{n=1}^{\infty}S_nx^n=15+\sum_{n=1}^{\infty}(3S_{n-1}-10)x^n=$$ $$=15+3\sum_{n=1}^{\infty}S_{n-1}x^n-10\sum_{n=1}^{\infty}x^n=15+3xf(x)-10\frac{x}{x-1}$$ then solve for $f(x)$ to find generating function

Adi Dani
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  • LOL! I'm lost... Anyway to do this in layman's terms? Keep in mind I have no Calculus background (if that's calculus) – Antonio Montana Nov 02 '13 at 00:28
  • Hmm, ok, skip that... LOL I have no clue what that is. Let me focus on what I kind of know. Was my sequence right earlier though? So=15, S1=35, S2 = 95... – Antonio Montana Nov 02 '13 at 00:35