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<Text-field style="Heading 1" layout="Heading 1">Introduction:</Text-field> restart; Handling infinite sequences and series in Maple is not always easy, but it does offer some tools. If we have the explicit formula for a sequence, testing convergenge is easy by taking a limit. Here, we define a function for the LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEibkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw==th coefficient, then use limit. a := n-> (n^2+1)/n^2; limit(a(n), n = infinity); Creating a finite list of the first few terms of a sequence is easy if we have an explicit formula, thanks to the seq command. l := seq(a(n), n = 1 .. 15); Looking at the floating-point form of the sequence members can sometimes give a strong clue as to what the sequence is converging to. evalf(%); To visualize the sequence, we can create a list of points... pointlist := [ seq([n,a(n)], n = 1..10) ]; ...then use a special syntax of plot: plot(pointlist,style=point); This graph produces visual confirmation of the limiting value. Maple sometimes gives an unexpected result for the limit of a divergent sequence. Here is an example. b := n -> (-1)^n*(n+1)/n; pointlist := [ seq([n, b(n)], n = 1..15) ]; plot(pointlist, style=point); As the graph suggests, this sequence diverges. But when we tell Maple to calculate the limit, we get the following: limit(b(n), n=infinity); Maple is trying to tell us that the sequence eventually takes values that are arbitrarily close to the interval [-1,1], which is a broader notion of convergence than we use in the course. A useful feature of Maple is the ability to solve certain recurrence relations. For example, the Fibonacci sequence is defined by two starting values and a recurrence equation. f:= n -> f(n-1)+f(n-2); f(0):= 1; f(1):= 1; seq( f(n), n=0..15 ); Maple can "solve" the recurrence, which gives us an explicit formula for the nth term, using rsolve: rsolve( c(n) = c(n-1)+c(n-2), c(n)); With just the defining recurrence equation, the result depends on the two starting values LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUkjbW5HRiQ2JFEiMEYnL0YzUSdub3JtYWxGJ0Y+ and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUkjbW5HRiQ2JFEiMUYnL0YzUSdub3JtYWxGJ0Y+. We can get more specific results by subs or eval, or by giving rsolve a complete set of defining equations. recur:= { c(n) = c(n-1)+c(n-2), c(0)=1, c(1)=1 }; rsolve( recur, c(n) ); We can use this output to define a function for the nth term of a sequence: fibo := unapply(%, n); evalf( seq(fibo(n),n=0..15) ); Whenever you see results like these in floating point, you should immediately wonder whether the results are supposed to be exactly integers that got rounded off a bit incorrectly--as is the case here. One advantage of the formula is that we can now take the limit. limit( fibo(n), n = infinity ); We can also see that the sequence consists of an exponentially decaying part and and exponentially growing part. evalf( fibo(n) ); Graphically, this becomes a straight line if we take the log of the terms. fibolist := [ seq( [n,ln(fibo(n))], n=0..15 ) ]: plot( fibolist, style = point ); Sometimes, rsolve gives us solutions that are so complex that it is hard to interpret them just by looking at them. For example: rsolve( {c(n) = 7*c(n-1)-2*c(n-2)+n, c(0) = 2, c(1) = 3}, c(n) ); f := unapply(%, n): We can still use evalf to get a feel for the numbers of the sequence. evalf( seq(f(n),n=1..15) ); evalf( f(n) ); Hence, this sequence also has exponential divergence to infinity. It's worth noting that Maple gets this wrong! limit( f(n), n=infinity ); This is further confirmation that Maple is a supplement, not a replacement, for your brain. print();
<Text-field style="Heading 1" layout="Heading 1">Exercises:</Text-field> restart; Instructions: The following exercises are all working with the Fibonacci recurrence relation, c(n) = c(n-1)+c(n-2); Ly1JImNHNiI2I0kibkdGJSwmLUYkNiMsJkYnIiIiISIiRixGLC1GJDYjLCZGJ0YsISIjRixGLA== Question 1. For the starting values c(0) = 3; c(1) = -2; Ly1JImNHNiI2IyIiISIiJA== Ly1JImNHNiI2IyIiIiEiIw== do the following: (1) Use rsolve to solve the recurrence with the given starting values. (2) Calculate and plot the first 10 values of the sequence. Give both exact and floating-point expressions for the nth term. (3) Determine the convergence/divergence of the sequence. Question 2. Find a pair of starting values LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUkjbW5HRiQ2JFEiMEYnL0YzUSdub3JtYWxGJ0Y+, LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUkjbW5HRiQ2JFEiMUYnL0YzUSdub3JtYWxGJ0Y+ such that the Fibonacci sequence converges to zero. (Hint: What term in equation is responsible for the exponential divergence? Make it go away.)
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