Recursive Sequences & Series - IB Mathematics Analysis and Approaches

Consider the sequence given by $u_1=1$ and $u_{n+1} = u_n + \frac{1}{u_n}$ for $n \ge 1$ . Which statement best describes the behavior of this sequence?
Answer: It diverges to infinity. Since $u_1=1$ is positive, and the recurrence relation adds a positive term $1/u_n$ at each step, all terms $u_n$ will be positive. This means $u_{n+1} - u_n = 1/u_n > 0$ , so the sequence is strictly increasing. If the sequence converged to a finite limit $L$ , we would have $L = L + \frac{1}{L}$ , which implies $0 = \frac{1}{L}$ . This equation has no solution. Since the sequence is increasing and does not converge to a finite limit, it must diverge to infinity.
Consider the sequence defined by the recurrence relation $u_{n+1} = u_n^2 - 4u_n + 6$ with $u_1 = 2.5$ . Determine the behavior of the sequence as $n \to \infty$ .
Answer: It converges to 2. To find the possible limits, we solve for the fixed points of the function $f(x) = x^2 - 4x + 6$ by setting $f(L)=L$ . This gives $L = L^2 - 4L + 6$ , which simplifies to $L^2 - 5L + 6 = 0$ , or $(L-2)(L-3) = 0$ . The possible limits are $L=2$ and $L=3$ . To determine which limit the sequence converges to, we check the stability of these fixed points. The derivative is $f'(x) = 2x - 4$ . At $L=2$ , $|f'(2)| = |2(2)-4| = 0 < 1$ , so this is a stable (attracting) fixed point. At $L=3$ , $|f'(3)| = |2(3)-4| = 2 > 1$ , so this is an unstable (repelling) fixed point. Since the starting value $u_1 = 2.5$ is closer to 2 than 3, and 2 is a stable fixed point, the sequence converges to 2.
A sequence is defined by $u_{n+1} = \sqrt{4 - u_n}$ with $u_1 = 1$ . Find the limit of the sequence, if it exists.
Answer: $\frac{-1 + \sqrt{17}}{2}$ When you encounter a recursive sequence where each term depends on the previous one, you need to find what value the sequence approaches as n gets very large. This involves finding the fixed point of the recurrence relation. To find the limit L of this sequence, assume the sequence converges. If $u_n \to L$ as $n \to \infty$ , then $u_{n+1} \to L$ as well. Substituting into the recurrence relation: $L = \sqrt{4 - L}$ Squaring both sides gives us $L^2 = 4 - L$ , which rearranges to $L^2 + L - 4 = 0$ . Using the quadratic formula: $L = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$ This gives two potential limits: $\frac{-1 + \sqrt{17}}{2} \approx 1.56$ and $\frac{-1 - \sqrt{17}}{2} \approx -2.56$ . Since we need $L = \sqrt{4 - L} \geq 0$ , we must have $L \geq 0$ . Only the positive solution satisfies this condition. Choice A gives the correct positive solution. Choice B represents the negative solution from our quadratic equation, but this is impossible since the sequence terms are all non-negative (being defined by square roots). Choice C (the value 2) would give $\sqrt{4-2} = \sqrt{2} \neq 2$ , so it's not a fixed point. Choice D is incorrect because we can verify convergence by checking that the sequence is bounded and monotonic. Study tip: For recursive sequences, always check that your proposed limit actually satisfies the recurrence relation and respects any domain constraints (like non-negativity for square roots).
The population $P_n$ of a species in a protected area after $n$ years is modelled by the recurrence relation $P_{n+1} = P_n + 0.1 P_n \left(1 - \frac{P_n}{1000}\right)$ . If the initial population $P_1$ is 200, what is the long-term stable population?
Answer: 1000 The long-term stable population is a stable fixed point of the recurrence relation. Let the limit be $L$ . Then $L = L + 0.1L(1 - L/1000)$ . This simplifies to $0 = 0.1L(1 - L/1000)$ . The solutions (equilibrium populations) are $L=0$ and $1 - L/1000 = 0$ , which gives $L=1000$ . To check stability, let $f(P) = P + 0.1P(1-P/1000) = 1.1P - 0.0001P^2$ . Then $f'(P) = 1.1 - 0.0002P$ . At $L=0$ , $|f'(0)| = 1.1 > 1$ (unstable). At $L=1000$ , $|f'(1000)| = |1.1 - 0.0002(1000)| = |1.1 - 0.2| = 0.9 < 1$ (stable). Therefore, the population will converge to the stable equilibrium of 1000.
A sequence is defined by $u_1=0$ and $u_{n+1} = \frac{1}{2-u_n}$ . The sequence converges to a limit $L$ . Find the value of the derivative of the function $f(x) = \frac{1}{2-x}$ at $x=L$ .
Answer: 1 First, we find the limit $L$ by solving the fixed-point equation $L = \frac{1}{2-L}$ . This gives $L(2-L) = 1$ , which simplifies to $2L - L^2 = 1$ , or $L^2 - 2L + 1 = 0$ . This factors as $(L-1)^2 = 0$ , so the only fixed point and thus the limit is $L=1$ . Next, we find the derivative of $f(x) = (2-x)^{-1}$ . Using the chain rule, $f'(x) = -1(2-x)^{-2}(-1) = \frac{1}{(2-x)^2}$ . Finally, we evaluate the derivative at the limit $L=1$ : $f'(1) = \frac{1}{(2-1)^2} = \frac{1}{1^2} = 1$ .
A sequence is defined by the recurrence relation $u_{n+1} = \frac{1}{2}\left(u_n + \frac{9}{u_n}\right)$ for $n \ge 1$ , with $u_1 = 1$ . Find the limit of this sequence as $n \to \infty$ .
Answer: 3 Let the limit of the sequence be $L$ . As $n \to \infty$ , $u_n \to L$ and $u_{n+1} \to L$ . The recurrence relation becomes $L = \frac{1}{2}\left(L + \frac{9}{L}\right)$ . Multiplying by $2L$ gives $2L^2 = L^2 + 9$ , which simplifies to $L^2 = 9$ . The possible limits are $L = 3$ and $L = -3$ . Since the first term $u_1 = 1$ is positive, and the recurrence relation involves averaging positive numbers, all subsequent terms $u_n$ will be positive. Therefore, the limit must be positive. The limit is $L=3$ .
For a sequence defined by $u_{n+1} = \frac{4u_n - 2}{u_n + 1}$ with $u_1 = 3$ , find its limit as $n \to \infty$ .
Answer: 2 Let the limit be $L$ . Then $L = \frac{4L - 2}{L + 1}$ . This leads to $L(L+1) = 4L-2$ , which simplifies to $L^2+L=4L-2$ , or $L^2-3L+2=0$ . Factoring gives $(L-1)(L-2)=0$ , so the possible limits are $L=1$ and $L=2$ . To check stability, let $f(x) = \frac{4x-2}{x+1}$ . The derivative is $f'(x) = \frac{4(x+1) - 1(4x-2)}{(x+1)^2} = \frac{6}{(x+1)^2}$ . At $L=1$ , $|f'(1)| = \frac{6}{(1+1)^2} = \frac{6}{4} = 1.5 > 1$ (unstable). At $L=2$ , $|f'(2)| = \frac{6}{(2+1)^2} = \frac{6}{9} = \frac{2}{3} < 1$ (stable). The sequence will converge to the stable fixed point, which is 2.
Let a sequence be defined by $u_{n+1} = u_n^3 - 3u_n$ . The fixed points of the associated function $f(x) = x^3 - 3x$ are $0, 2, -2$ . Which of these fixed points is/are stable (attracting)?
Answer: None of them are stable To determine the stability of the fixed points, we examine the derivative of $f(x) = x^3 - 3x$ , which is $f'(x) = 3x^2 - 3$ . A fixed point $L$ is stable if $|f'(L)| < 1$ . We check this condition for each fixed point: At $L=0$ , $|f'(0)| = |3(0)^2 - 3| = |-3| = 3$ . Since $3 > 1$ , this fixed point is unstable. At $L=2$ , $|f'(2)| = |3(2)^2 - 3| = |12 - 3| = 9$ . Since $9 > 1$ , this fixed point is unstable. At $L=-2$ , $|f'(-2)| = |3(-2)^2 - 3| = |12 - 3| = 9$ . Since $9 > 1$ , this fixed point is also unstable. Therefore, none of the fixed points are stable.
Let a sequence be defined by $u_{n+1} = e^{u_n} - 1$ with $u_1 = 0.1$ . Which statement best describes the long-term behavior of the sequence?
Answer: It diverges to infinity. Let $f(x) = e^x - 1$ . The fixed points are solutions to $L = e^L - 1$ . One solution is clearly $L=0$ . To analyze stability, we find the derivative $f'(x) = e^x$ . At the fixed point $L=0$ , $f'(0) = e^0 = 1$ . Since $|f'(0)|=1$ , the stability test is inconclusive. However, for any $x > 0$ , $e^x > x+1$ , which means $e^x - 1 > x$ . So, if $u_n > 0$ , then $u_{n+1} = e^{u_n} - 1 > u_n$ . Since $u_1 = 0.1 > 0$ , the sequence is strictly increasing. An increasing sequence either converges to a limit or diverges to infinity. Since it is moving away from the only fixed point $L=0$ , it cannot converge. Therefore, the sequence must diverge to infinity.
A sequence is defined by $u_{n+1} = k u_n (1-u_n)$ , where $k$ is a positive constant and the initial value $u_1$ is in the interval $(0, 1)$ . Given that the sequence converges to a limit $L > 0$ , find the range of values for $k$ .
Answer: $1 < k < 3$ The limit $L$ must be a fixed point of $f(x) = kx(1-x)$ . We solve $L = kL(1-L)$ . One solution is $L=0$ . Since we are given $L>0$ , we can divide by $L$ to get $1 = k(1-L)$ , which gives $L = 1 - 1/k$ . For $L>0$ , we must have $1-1/k > 0$ , which means $k>1$ . For the sequence to converge to this limit, the fixed point must be stable. The derivative is $f'(x) = k - 2kx$ . The condition for stability is $|f'(L)| < 1$ . Substituting $L = 1-1/k$ , we get $f'(L) = k - 2k(1-1/k) = k - (2k-2) = 2-k$ . The stability condition is $|2-k| < 1$ , which means $-1 < 2-k < 1$ . This simplifies to $-3 < -k < -1$ , or $1 < k < 3$ . This range also satisfies the earlier condition $k>1$ .
The recurrence relation $u_{n+1} = \frac{1}{2}\left(u_n + \frac{5}{u_n}\right)$ with $u_1 = 2$ is an application of Newton's method to find a root of a function $g(x)=0$ . What is the exact value of the limit of this sequence?
Answer: $\sqrt{5}$ The limit $L$ of the sequence must satisfy $L = \frac{1}{2}\left(L + \frac{5}{L}\right)$ . This simplifies to $2L^2 = L^2 + 5$ , which gives $L^2=5$ , so $L = \pm\sqrt{5}$ . Since $u_1=2$ is positive, all subsequent terms will be positive, so the limit must be $\sqrt{5}$ . This recurrence relation is generated by applying Newton's method, $x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}$ , to find the roots of the function $g(x) = x^2 - 5$ . Therefore, the limit is a root of $x^2-5=0$ .
A sequence is defined by $u_{n+1} = 1.5u_n(1-u_n)$ . This sequence is known to converge to its non-zero fixed point, $L=1/3$ . What is the largest open interval of starting values $u_1$ for which this convergence occurs?
Answer: $(0, 1)$ When analyzing recursive sequences with fixed points, you need to determine the basin of attraction - the set of initial values that lead to convergence at the fixed point. For the sequence $u_{n+1} = 1.5u_n(1-u_n)$ , let's find all fixed points by solving $L = 1.5L(1-L)$ . This gives us $L = 1.5L - 1.5L^2$ , which simplifies to $1.5L^2 - 0.5L = 0$ , so $L(1.5L - 0.5) = 0$ . The fixed points are $L = 0$ and $L = \frac{1}{3}$ . To find the basin of attraction for $L = \frac{1}{3}$ , we need to determine where the sequence behavior changes. The critical points occur when the derivative $f'(u) = 1.5(1-2u) = 0$ , giving us $u = \frac{1}{2}$ . At this point, $f(\frac{1}{2}) = 1.5 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{8}$ . For this type of logistic map, convergence to the non-zero fixed point occurs when $0 < u_1 < 1$ . Values outside this interval either diverge or behave chaotically. Choice A) $(0,1)$ correctly identifies this interval. Choice B) $(-\infty, \infty)$ is wrong because negative values and values greater than 1 don't converge to $\frac{1}{3}$ . Choice C) $(0, \frac{1}{3})$ is too restrictive - initial values between $\frac{1}{3}$ and 1 also converge. Choice D) $(\frac{1}{3}, 1)$ misses the convergent values below $\frac{1}{3}$ . Study tip: For logistic maps $u_{n+1} = ru_n(1-u_n)$ , the basin of attraction for the non-zero fixed point typically spans the interval $(0,1)$ when the parameter allows convergence.
A sequence is defined by the recurrence relation $u_{n+1} = f(u_n)$ . The sequence is known to converge to a limit $L$ , where $f(L) = L$ . Which of the following conditions on the derivative of $f$ at $L$ is sufficient to ensure convergence for a starting value sufficiently close to $L$ ?
Answer: $|f'(L)| < 1$ For a sequence defined by $u_{n+1} = f(u_n)$ to converge to a fixed point $L$ , the fixed point must be stable or 'attracting'. The general condition for a stable fixed point is that the absolute value of the derivative of the function at the fixed point is less than 1, i.e., $|f'(L)| < 1$ . If $|f'(L)| > 1$ , the point is unstable (repelling). If $|f'(L)| = 1$ , the test is inconclusive. While $f'(L)=0$ is a condition for very rapid (quadratic) convergence and implies stability, $|f'(L)| < 1$ is the general sufficient condition.
A sequence is defined by $u_{n+1} = \sqrt{a + u_n}$ , where $a > 0$ and $u_1 = 1$ . The sequence converges to a limit of 5. Find the value of $a$ .
Answer: 20 If the sequence converges to a limit $L=5$ , then this limit must be a fixed point of the function $f(x) = \sqrt{a+x}$ . Therefore, $L = \sqrt{a+L}$ . Substituting $L=5$ , we get $5 = \sqrt{a+5}$ . Squaring both sides gives $25 = a+5$ , which yields $a = 20$ . Distractor '24' comes from the error $L^2 = a + u_1 \implies 25 = a+1$ . Distractor '4' comes from the error $L = a+u_1 \implies 5 = a+1$ .
The behavior of a sequence $u_{n+1} = f(u_n)$ can be analyzed using a cobweb diagram, which involves the graphs of $y=f(x)$ and $y=x$ . If the sequence converges to a limit $L$ , what does this correspond to on the diagram?
Answer: The x-coordinate of a point of intersection of $y=f(x)$ and $y=x$ . A limit $L$ of a sequence defined by $u_{n+1}=f(u_n)$ must be a fixed point of the function $f$ . This means $L$ must satisfy the equation $L = f(L)$ . Graphically, the solutions to this equation are the x-coordinates of the points where the graph of $y=f(x)$ intersects the graph of $y=x$ . A cobweb diagram visually shows the sequence of values $u_n$ converging towards such an intersection point if the fixed point is stable.
The sequence $u_{n+1} = 2u_n$ has a fixed point at $L=0$ . This fixed point is unstable. If a sequence starts with a value $u_1$ very close to 0, but not equal to 0, how will the sequence behave?
Answer: The terms will move away from 0. An unstable fixed point, also known as a repelling fixed point, is one where nearby values are pushed away by the iteration. For $u_{n+1} = 2u_n$ , each term is twice the previous term. So if $u_1 = 0.01$ , the sequence is $0.01, 0.02, 0.04, ...$ , whose terms are moving away from 0. This is confirmed by the stability condition: $f(x)=2x$ , so $f'(x)=2$ . At $L=0$ , $|f'(0)|=2 > 1$ , confirming the point is unstable.
A sequence is defined by $u_1=1$ and $u_{n+1} = \frac{u_n}{u_n+1}$ for $n \ge 1$ . Find the value of $\sum_{n=1}^{4} u_n$ .
Answer: $\frac{25}{12}$ When you encounter a recursively defined sequence, your first step is to calculate several terms using the given recurrence relation to identify any patterns before attempting to find the sum. Given $u_1 = 1$ and $u_{n+1} = \frac{u_n}{u_n + 1}$ , let's find the first four terms: $u_1 = 1$ $u_2 = \frac{u_1}{u_1 + 1} = \frac{1}{1 + 1} = \frac{1}{2}$ $u_3 = \frac{u_2}{u_2 + 1} = \frac{\frac{1}{2}}{\frac{1}{2} + 1} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}$ $u_4 = \frac{u_3}{u_3 + 1} = \frac{\frac{1}{3}}{\frac{1}{3} + 1} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4}$ Notice the pattern: $u_n = \frac{1}{n}$ . Now we can find the sum: $\sum_{n=1}^{4} u_n = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{25}{12}$ This confirms answer A is correct. Answer B ( $\frac{13}{12}$ ) would result from incorrectly calculating just $u_1 + u_2 + u_3$ . Answer C ( $\frac{35}{12}$ ) might come from computational errors in the fraction arithmetic. Answer D (4) suggests someone mistakenly thought the sum equals the number of terms. Strategy tip: With recursive sequences, always compute the first few terms explicitly to spot patterns. Many IB questions reward recognizing that complex recursive definitions often simplify to familiar sequences like $\frac{1}{n}$ or geometric progressions.
The iteration $x_{n+1} = x_n - (x_n^2 - a)$ is used to find a solution to $x^2=a$ , where $a$ is a positive constant. For which values of $a$ will this iteration converge to $\sqrt{a}$ if started sufficiently close to the root?
Answer: $0 < a < 1$ The iteration is given by the function $f(x) = x - x^2 + a$ . A fixed point $L$ satisfies $L=L-L^2+a$ , which simplifies to $L^2=a$ , so $L = \pm\sqrt{a}$ . For the iteration to converge to $L=\sqrt{a}$ , this fixed point must be stable. We check the derivative: $f'(x) = 1 - 2x$ . The stability condition is $|f'(\sqrt{a})| < 1$ . This means $|1 - 2\sqrt{a}| < 1$ , which is equivalent to $-1 < 1 - 2\sqrt{a} < 1$ . Subtracting 1 gives $-2 < -2\sqrt{a} < 0$ . Dividing by -2 and reversing the inequalities gives $0 < \sqrt{a} < 1$ . Squaring gives $0 < a < 1$ .
Consider the sequence defined by $u_{n+1} = 3 - u_n$ and $u_1 = 1$ . Which statement accurately describes the sequence?
Answer: It is periodic with period 2. Let's compute the first few terms of the sequence: $u_1 = 1$ , $u_2 = 3 - u_1 = 3 - 1 = 2$ , $u_3 = 3 - u_2 = 3 - 2 = 1$ , $u_4 = 3 - u_3 = 3 - 1 = 2$ . The sequence of terms is 1, 2, 1, 2, ... . The sequence oscillates between 1 and 2, so it is periodic with period 2. It does not converge. The value 1.5 is the fixed point of the function $f(x) = 3-x$ , since $L=3-L \implies 2L=3 \implies L=1.5$ , but the sequence does not approach this value.

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