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Recurrence Relations (need help)

Solve the following for \(a_n\) given that \(a_0=0\) and \(a_1=1\)

  1. \(a_n=a_{n-1}+2a_{n-2}+n\)

  2. \(a_n=a_{n-1}+2a_{n-2}+n^2\)

  3. \(a_n=2a_{n-1}-a_{n-2}+n^2\)

  4. \(a_n=a_{n-1}+2a_{n-2}+2^n\)

  5. \(a_n=a_{n-1}a_{n-2}\) (with initial conditions \(a_0=1\) and \(a_1=2\))

  6. \(a_n^2=a_{n-1}a_{n-2}\) (with initial conditions \(a_0=1\) and \(a_1=2\))

  7. \(a_n=na_{n-1}+a_{n-2}\)

  8. \(a_n=a_{n-1}^2+a_{n-2}\)

  9. \(a_n=a_{n-1}^2+a_{n-2}^2\)

Note by Aditya Raut
3 years, 11 months ago

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For numbers 1-3, the trick is to use the characteristic equation to solve. First, eliminate all terms that are not in the form of \(ia_j\):

\[\begin{align*}a_n&=a_{n-1}+2a_{n-2}+n\\a_{n-1}=a_{n-2}+2a_{n-3}+(n-1)\end{align*}\]

Subtracting the two, we get

\[a_n-a_{n-1}=a_{n-1}+a_{n-2}-2a_{n-3}+1\]

But there's still a \(1\) in there! Do it again!

\[\begin{align*}a_n-a_{n-1}=a_{n-1}+a_{n-2}-2a_{n-3}+1\\ a_{n-1}-a_{n-2}=a_{n-2}+a_{n-3}-2a_{n-4}+1\end{align*}\]

Subtract:

\[\begin{align*}a_n-2a_{n-1}+a_{n-2}=a_{n-1}-3a_{n-3}+2a_{n-4}\\a_n-3a_{n-1}+a_{n-2}+3a_{n-3}-2a_{n-4}=0\end{align*}\]

Finally, we have it in the form we want. Now, we'll use the characteristic equation, which is an equation that looks like this:

\[x^4-3x^3+x^2+3x-2=0\]

Basically, it takes its coefficients from the recurrence relation. Solving, we get \(x=-1,1,1,2\). From generating functions, we can now say that

\[a_n=c_1\cdot(-1)^n+(c_2+c_3n)\cdot1^n+c_4\cdot2^n\]

What we did here was take the roots of the equation and use their exponentials to develop the relation. For repeated roots, we have to create an entire polynomial with an ascending degree: if the root \(x_0\) is repeated \(k\) times, it'll be \((c_0+c_1n+c_2n^2+\dots+c_kn^{k-1})x_0^n\). Lastly, we solve for \(c_1,c_2,c_3,c_4\) using a system of equations. Note that since there are four constants, we need to know four terms of the sequence, so we must compute \(a_2=3\) and \(a_3=8\). Solve the system of equations

\[\begin{cases}0=c_1\cdot(-1)^0+(c_2+c_3\cdot0)\cdot1^0+c_4\cdot2^0\\1=c_1\cdot(-1)^1+(c_2+c_3\cdot1)\cdot1^1+c_4\cdot2^0\\3=c_1\cdot(-1)^2+(c_2+c_3\cdot2)\cdot1^2+c_4\cdot2^2\\8=c_1\cdot(-1)^3+(c_2+c_3\cdot3)\cdot1^3+c_4\cdot2^3\end{cases}\]

Eventually, we get \((c_1,c_2,c_3,c_4)=\left(-\frac1{12},-\frac54,-\frac12,\frac43\right)\), so

\[a_n=-\frac1{12}\cdot(-1)^n-\frac54-\frac{n}2+\frac43\cdot2^n\]

That technique takes care of 1-3. It also takes care of 5 as you can let \(a_n=2^{b_n}\) with \(b_0=0\) and \(b_1=1\) such that \(b_n=b_{n-1}+b_{n-2}\), the famous Fibonacci sequence. Similarly, 6 has \(a_n=2^{b_n}\), \(b_0=0\), \(b_1=1\), and \(2b_n=b_{n-1}+b_{n-2}\).

Cody Johnson - 3 years, 10 months ago

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Cody !!!! Even question 4 can be solved by your technique !!!! Just a little manipulation

Given thing \(a_n=a_{n-1}+2a_{n-2}+2^n\)

Next recurrence \(a_{n-1}=a_{n-2}+2a_{n-3}+2^{n-1}\)

JUST MULTIPLY THIS BY 2 ON BOTH SIDES

\(2a_{n-1}=2a_{n-2}+4a_{n-3}+2^n\)

Now subtract from given !!!!

You get \(a_n-3a_{n-1}+4a_{n-3}=0\)

comparable to \(x^3-3x^2+4=0\)

Has roots \(x=-1,2,2\)

And then we get \(a_n=c_1(-1)^n+(c_2+nc_3)2^n\)

Solve to get \(c_1=\frac{1}{9}\) , \(c_2=\frac{-1}{9}\) . \(c_3=\frac{2}{3}\)

And then \(a_n=\dfrac{1}{9}(-1)^n-\dfrac{2^n}{9}+\dfrac{2}{3} n2^n \)

Aditya Raut - 3 years, 10 months ago

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Wow, nice thought @Aditya Raut :)

Happy Melodies - 3 years, 7 months ago

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Very good!

Cody Johnson - 3 years, 10 months ago

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Thank you very very much Cody !!! Now please try for last three questions ...I found no way to solve them :(

Aditya Raut - 3 years, 10 months ago

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Wow, that's insane. :P

Finn Hulse - 3 years, 9 months ago

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There is a typo at line 7: should be \(a_n\) instead of \(a_{n-2}\).

Happy Melodies - 3 years, 10 months ago

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Try with generating functions.

Alessio Di Lorenzo - 3 years, 10 months ago

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Solve \(a_{n}\) in terms of function of n only with initial values \(a_{0}=0, a_{1}=1, a_{n}=a_{n-1}^2 + a_{n-2}^2 \)



Here we observe one inequality. As series is growing,

\(a_{n} => a_{n-1}\) It implies

\(a_{n-1}^2+ a_{n-2}^2 <= 2a_{n-1}^2 \)

i.e.\( 2a_{n+1}^2<=a_{n} <= 2a_{n-1}^2\)

Observe \(a_{n}=2a_{n-1}^2\) is a series whose solution is

\(a_{n}=2^2^n\)

Hence we conclude that our solution will be of the same type

\(a_{n}=f^2^n\)

where f is a real number less than 2 (and definitely greater than zero.) Now all our efforts are to find f which satisfies above equation. As I worked out for calculating f,I came to know that it is irrational number and its value depends upon value of

\(a_{0},a_{1},a_{2},a_{3}...\)

For exact expression for f see this ,

" www.fq.math.ca/Scanned/11-4/aho-a.pdf "

Value of f up to fourteen significant digit is 1.11148222558297....

Another thing to point here is that \(a_{n}\)=nearest integer to \(f^2^n\)

as value may be floating we have to take into account only integer values.

This is analytic solution but is correct for first 8th term.....For other terms more accurate value of f is needed otherwise it is perfect solution. Below is the verification of my solution...As i had said more significant digits are required for correctness.

\( n\) \( a_{n} \) \( f^2^n\)

 0                                  0                                             1
 1                                  1                                             1.2353
 2                                  1                                             1.5261
 3                                  2                                             2.3292
 4                                  5                                             5.4255
 5                                29                                            29.4361
 6                              866                                            866.485
 7                        750797                                           750797.4995
 8          563696885165                                            563696885249

Aamir Faisal Ansari - 3 years, 10 months ago

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Haa...one thing.....Not applicable for n=0 case

Aamir Faisal Ansari - 3 years, 10 months ago

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The questions 7 to 9 are extremely tricky ..... No method worked :(

Aditya Raut - 3 years, 10 months ago

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I don't think the method of characteristics equation will work here... Instead, multiply the RHS (\(a_n\)) and LHS by \(x^n\), then sum the terms with respect to \(n\). Then, look for power series and simplify the expression,

Happy Melodies - 3 years, 5 months ago

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Hmm Qn 7 is hard... (Refer to this link)[http://www.math.upenn.edu/~wilf/gfology2.pdf]. I think can still multiply by \(x^n\) on both sides then take sum. Differentiate the \(n(a_{n-1})\) term (with summation) then solve differential eqn.... Haven't worked that out but I think it might work. If you have no idea what I am talking about, the link above is a good guide to Generating Functions, which is very useful in solving recurrence type problems.

Happy Melodies - 3 years, 5 months ago

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@Happy Melodies I have worked that out the differential equation and it turns out to be unsolvable. The differential equaiton turns out to be \(x^{ 2 }f^{ ' }\left( x \right)+({ x }^{ 2 }+x-1)f(x)+x=0\) where \(\\ f(x)=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ x }^{ n } } \)

Ronak Agarwal - 3 years, 5 months ago

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@Ronak Agarwal Hmm i am not familiar with differential equation ... (Just started learning calculus) anyone can help?

Happy Melodies - 3 years, 5 months ago

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@Happy Melodies If you want to learn, I learned from Khan Academy. But if you want help in explaining this to Ronak, I'm afraid I can't help.

Finn Hulse - 3 years, 5 months ago

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@Finn Hulse Alright I will maybe after my exams

Happy Melodies - 3 years, 5 months ago

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@Happy Melodies what are the terms LBS and RHS?

Mardokay Mosazghi - 3 years, 5 months ago

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@Mardokay Mosazghi Right Hand Side (of equation) for RHS and Left Hand Side for LHS

Happy Melodies - 3 years, 5 months ago

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Anyone knows how to solve Q2? I'm stuck with the characteristic equation method as well as generating funcs method...

Happy Melodies - 3 years, 10 months ago

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Do as Cody did !!!

given thing is \(a_n=a_{n-1}+2a_{n-2}+n^2\)

Do for \(a_{n-1}=a_{n-2}+2a_{n-3}+(n-1)^2\)

Which is \(a_{n-1}=a_{n-2}+2a_{n-3}+n^2-2n+1\).....(i)

Subtract this from Given one......... you will get \(a_{n}=2a_{n-1}+a_{n-2}-2a_{n-3}+2n-1\).........(ii)

Now write the next recurrence............. \(a_{n-1}=2a_{n-2}+a_{n-3}-2a_{n-4}+2(n-1)-1\)

\(a_{n-1}=2a_{n-2}+a_{n-3}-2a_{n-4}+2n-3\)................(iii)

Subtract (iii) from (ii) to get

\(a_{n}=3a_{n-1}-a_{n-2}-3a_{n-3}+2a_{n-4}+2\)

But we got "2" here ...... Eliminate it from again one more recurrence !!!!!!!! \(a_{n-1}=3a_{n-2}-a_{n-3}-3a_{n-4}+2a_{n-5}+2\)

Subtracting, you get \(a_n=3a_{n-1}-4a_{n-2}-2a_{n-3}+5a_{n-4}-2a_{n-5}\)

And now we find characteristic equation

\(x^5-4x^4+4x^3+2x^2-5x+2=0\)

Has it's roots \(x=-1,1,1,1,2\)

Next things you can find in Cody's solution :)

Aditya Raut - 3 years, 10 months ago

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What's q2?

Alessio Di Lorenzo - 3 years, 10 months ago

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As in the above point 2? Solving for \(a_n\)

Happy Melodies - 3 years, 10 months ago

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See i guess that n cannot take values 0 & 1 but when you start with n=2 and go on till 3 or 4 then you will start getting values now the difference of 2 consecutive terms will start forming an A.P. or G.P. that can be solved by method of differences,see if this works!!!

Nitish Sharma - 3 years, 10 months ago

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