# The Linear Map and Its Seven Cases On [[MATH310F26(Day 27) - One step at a time, no dice|Day 27]] we cut the first-order map $y_{n+1} = f(n, y_n)$ down to the simplest one we could, the first-order, linear, autonomous, homogeneous map $y_{n+1} = a\,y_n, \qquad a \in \mathbb{R}.$ Recursing back to the start gives the closed form $y_n = a^n y_0, \qquad n = 0, 1, 2, \dots$ The value of $a$ alone decides what the map does. The starting value $y_0$ only sets the scale, and its sign mirrors the picture. The applet below runs the map for two starting values at once so you can check that claim for yourself. <div class="applet" data-applet="linear-map"></div> ## Reading the applet - Each stem is one click of $n$. The orange stems start from the first $y_0$ and the green stems from the second. A faint line joins the tips so the zigzag is easy to follow. The green stems sit a little to the right so the two never hide each other. - The dashed curves are the envelope, drawn for each starting value. When $a > 0$ every $y_n$ has the sign of $y_0$, so there is one curve, $y_0 a^t$, and it passes through every stem tip. When $a < 0$ the sign flips every click, so there are two curves, $\pm|y_0||a|^t$, and the tips alternate between them. At $a = 0$ there is no envelope. - The case buttons set $a$ to the values used in the Day 27 figures, $1,\ 1.15,\ 0.8,\ -1,\ -1.15,\ -0.8,\ 0$. The slider covers everything from $-1.5$ to $1.5$. - Drag either dot at $n = 0$ to move a starting value, or use its slider. - The vertical axis is fixed from $-2.5$ to $2.5$. When a value leaves that window its stem ends in a triangle at the edge. Turn on **autoscale** to fit the axis to the data instead. - **run the map** reveals the clicks one at a time, three per second. **one click** adds a single click. **show all** puts every click back. - Hover over a click to read off $y_n = a^n y_0$ for both starting values. The applet computes each value by iterating $y_{n+1} = a y_n$, the way the map itself works, and prints the closed form beside it. The two agree to within about $4 \times 10^{-16}$ (relative) for every $n \le 40$, which is rounding in the last digit. ## The seven cases | case | $a$ | what $y_n$ does | |---|---|---| | 1 | $a = 1$ | equilibrium, $y_n = y_0$ | | 2 | $a > 1$ | growth, $\lvert y_n \rvert \to \infty$ | | 3 | $0 < a < 1$ | decay, $\lvert y_n \rvert \to 0$ | | 4 | $a = -1$ | sign oscillation, $y_n = (-1)^n y_0$ | | 5 | $a < -1$ | growth with sign alternation | | 6 | $-1 < a < 0$ | decay with sign alternation | | 7 | $a = 0$ | trivial, $y_n = 0$ for $n \ge 1$ | ## Worked examples **Growth, $a = 1.15$, $y_0 = 1$.** Each click multiplies by $1.15$, so $y_{15} = 1.15^{15} \approx 8.137$. The first value past the top of the fixed axis is $y_7 = 1.15^7 \approx 2.660$, because $y_6 \approx 2.313$ still fits. Starting from $y_0 = -0.5$ instead, the stems leave the bottom of the window later, at $n = 12$, since they start half as far from zero. With both starting values and 15 clicks, 13 values are off the fixed axis. **Decay, $a = 0.8$, $y_0 = 1$.** Now $y_{15} = 0.8^{15} \approx 0.0352$. The value drops below half between $n = 3$ ($0.8^3 = 0.512$) and $n = 4$ ($0.8^4 \approx 0.410$). **Alternation, $a = -1.15$.** The sizes are exactly those of $a = 1.15$, with the sign flipping every click, so $y_{15} = (-1.15)^{15} \approx -8.137$ for $y_0 = 1$ and $\approx 4.069$ for $y_0 = -0.5$. ## Try these > [!question] 1. Same size, different sign > Press case 2 and then case 5, and compare the heights of the stems. What changes and what doesn't? Explain it with the closed form. > [!success]- Answer > The heights $|y_n|$ are identical. Only the signs differ. Since $(-a)^n = (-1)^n a^n$, the map with $-a$ is the map with $a$ with every odd click flipped. > [!question] 2. Two starting values > With the first $y_0 = 1$ and the second $y_0 = -0.5$, pick any $a$ and any click $n$. What is the first $y_n$ divided by the second $y_n$? Does it depend on $a$ or $n$? > [!success]- Answer > It is always $-2$, whatever $a$ and $n$ are, because $a^n \cdot 1 \,/\, \big(a^n \cdot (-0.5)\big) = -2$. The second sequence is the first one scaled by $-\tfrac12$. This is linearity at work. Scale the start and the whole sequence scales with it, so $y_0$ cannot change which case you are in. > [!question] 3. Doubling time > Set $a = 1.15$ and the first $y_0 = 1$. Using **one click**, find the first $n$ at which $y_n \ge 2$. Then check your answer with logarithms. > [!success]- Answer > At $n = 5$, since $1.15^5 \approx 2.011$ while $1.15^4 \approx 1.749$. Solving $1.15^n = 2$ gives $n = \ln 2 / \ln 1.15 \approx 4.96$, just under 5 clicks. > [!question] 4. Start at zero > Drag one starting value to $y_0 = 0$ and move $a$ through all seven cases. What happens, and why? > [!success]- Answer > It stays at zero for every $a$. Since $a \cdot 0 = 0$, the value $0$ is a fixed point of the map no matter what $a$ is. In the readout this is the line that says $0$ is a fixed point for every $a$. > [!question] 5. Period three? > At $a = -1$ the sequence repeats every two clicks. Is there a real $a$ for which a nonzero $y_0$ comes back after exactly three clicks and not sooner? > [!success]- Answer > No. Coming back after three clicks means $y_3 = a^3 y_0 = y_0$, so $a^3 = 1$, and the only real cube root of 1 is $a = 1$. But at $a = 1$ the sequence is already back after one click. The other two cube roots of 1 are complex, so a real linear map can repeat with period 1 ($a = 1$) or period 2 ($a = -1$) and nothing longer. Keep this in mind for when the maps become nonlinear. > [!question] 6. What autoscale hides > Set $a = 1.5$, show 40 clicks, and turn on autoscale. Most of the stems look like they sit at zero. Are they zero? How big is $y_{20}$ compared with $y_{40}$? > [!success]- Answer > None of them is zero. $y_{40} = 1.5^{40} \approx 1.106 \times 10^7$, and $y_{20}/y_{40} = 1.5^{-20} \approx 3.0 \times 10^{-4}$, so the stem for $y_{20}$ is about $3 \times 10^{-4}$ as tall as the stem for $y_{40}$, well under a pixel. A linear axis can only show a range of sizes this wide by squashing the small values flat. That is why the fixed axis is the default, since it shows how fast growth leaves the window. > [!question] 7. Back to continuous time > On Day 27 we saw that $y = y_0 e^{kt}$, sampled at $t = 0, 1, 2, \dots$, is the map with $a = e^{k}$. Which of the seven cases can a first-order ODE reproduce this way? > [!success]- Answer > Only cases 1, 2 and 3. Since $e^k > 0$ for every real $k$, sampling the ODE always gives $a > 0$. Case 2 at $a = 1.15$ is $k = \ln 1.15 \approx 0.140$ and case 3 at $a = 0.8$ is $k = \ln 0.8 \approx -0.223$. Cases 4, 5 and 6 need $a < 0$, and case 7 needs $a = 0$, which $e^k$ never reaches. ## Credits Built for MATH310 Fall 2026 to match the Day 27 figures (the stems, the two starting values $1$ and $-0.5$, and the dashed envelope).