Fn+2=Fn+1+Fn for n0 where F0=0 and F1=1. The resulting sequence 1,1,2,3,4,5,8,13,21, is a...

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image Fn+2=Fn+1+Fn for n0 where F0=0 and F1=1. The resulting sequence 1,1,2,3,4,5,8,13,21, is a very famous sequence in mathematics and is called the Fibonaci sequence. This sequence is thought to model many natural phenomena such as number of seeds in a sunflower and anything which grows in a spiral form. It is so famous in fact that it has a journal devoted entirely to it. As a note, while Fibonacci's work Liber Abaci introduced this sequence to the western world, it had been described earlier Sanskrit texts going back as early as the sixth century. Since Fn+2 is determined by previous values Fn+1 and Fn, the relation Fn+2=Fn+1+Fn is called a recurrence relation. Recursive formulas are nice because they are typical easier to use and to determine based on the situation. However, it is certainly tedious to use for computing outputs for large values of k and there is no procedural computation to determine a value of k to give you a particular output. It turns out that there is a fascinating formula that gives the kth term of the Fibonaci sequence directly, without using the recurrence relation. It is sufficient to answer the following questions and clearly solve the problems. While you need to show steps and potentially do a bit of explaining of what you're doing, you do not need to explain the "Why?" part of anything. The more explaining of the background, theory, and general processes the more evidence you'll have to support a higher grade. The prompts in blue are not required, but give you some ideas for where you could do this deeper dive into the material. (1) Complete the chart using the recursive formula given. (2) Consider the matrix A=[0111]. Compute the first five powers of A by hand. What do you notice? (3) Use the recurrence relation to explain why xn=[FnFn+1]=Axn1. (4) Explain why xn=Anx0 and use this relationship (and technology if you're not looking to practice this many matrix multiplications by hand) to verify this relationship and compare your results to the table above. (5) Determine the characteristic polynomial of A, and use it to show that the eigenvalues of A are 215. You may be tempted to use the decimal approximation of these values, don't do it. (6) Because we're going to do a lot of algebraic simplifying, it is worth noticing something nice about these eigenvalues now. Compute the product of the two eigenvalues and simplify as much as possible. (7) Determine an eigenvector associated to each eigenvalue. (For each eigenvalue, you'll have infinitely many possible eigenvectors. The choice of the free variable can make the particular vector you choose seem nicer to you. That is a good choice... You can avoid fractions altogether and avoid having radicals in both expressions, but something similar to one of the eigenvalues will show up somewhere.)

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