Hands first, words secondPress “Iterate once →” a few times and watch the red path bounce between the curve and the dashed line y=x. Then drag k to 1.5 and to −0.5, and try the other functions.
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Try this
Take f(x)=21x+1 (k=0.5), start at x0=0 and keep pressing “Iterate once”. Which number are the values in the orbit table closing in on? Which intersection point in the picture does it belong to?
Drag k to 1.5. Same starting point: what happens to the orbit now?
Drag k to −0.5. The red path is no longer a “staircase”. What shape does it wind into?
Switch to f(x)=1−x1 and start at x0=3. After how many iterations are you back where you started? Try other starting points.
Switch to f(x)=x2+k with k=−2. Put the starting point at −1 or at 2. Does the orbit still move?
Predict first
f(x)=1−x1. What is f(2025)(3)? (f(n) means applying fn times in a row.)
Choose an answer first; then you can see the result.
✓You predicted it!✗First compute f(3), f(f(3)) and f(f(f(3))) by hand, and see what happens.
f(3)=−21, f(−21)=32, f(32)=3: back after three steps. In fact f(3)(x)=x for every x=0,1, so every orbit has period 3. Since 2025=3×675, f(2025)(3)=3.
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The cobweb closes up into a triangle and then runs around it again and again. Drag the starting point: every starting point has its own “triangle”.
Why
Definition. Let f:D→D. Set f(0)(x)=x and f(n+1)(x)=f(f(n)(x)). Then f(n) is called the n-th iterate of f. The sequence x0,f(x0),f(2)(x0),… is called the orbit of x0.
To keep iterating forever, the values of f must land in the domain D again; that is why the definition asks for f:D→D. A point with f(x∗)=x∗ is called a fixed point.
How a cobweb diagram is drawn
Step 1 / 5
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Start
Take x0=0 on the x-axis.
x0=0
Go vertically to the curve
Go across to y = x
Keep going
The fixed point
The identity in the last step is already a “formula for the n-th iterate”. How to find f(n)(x) in general is the subject of the next stop.
Remove one condition
Break it
Counterexample lab: remove one condition
Conclusion: every orbit of f(x)=k(x−2)+2 tends to the fixed point 2← still true?
k=−0.5: at each step the distance to 2 is multiplied by 21 and the sign alternates, so the cobweb winds into a box that keeps getting tighter.
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k=1.5: f(n)(x)−2=1.5n(x−2), so the distance is multiplied by 1.5 at each step. The fixed point 2 is still there, but no orbit except its own ever reaches it. Press “Iterate once” a few times and see where the cobweb runs off to.
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Having a fixed point does not mean orbits go there. Whether a fixed point “attracts” or “repels” depends on whether the graph near it is flatter or steeper than y=x.
Practice
Exercise 1Basic
f(x)=2x+1. Find f(3)(1).
Hint · direction
Substitute one step at a time.
Hint · method
Compute f(1), feed the result back into f, three times in all.
Hint · first step
f(1)=3, f(3)=7.
Full solution
f(7)=15, so f(3)(1)=15.
Exercise 2Basic
f(x)=x2. Write down f(3)(x).
Hint · direction
It is not [f(x)]3.
Hint · method
f(2)(x)=f(f(x)): substitute the whole of x2 into f.
Hint · first step
f(2)(x)=(x2)2=x4.
Full solution
f(3)(x)=f(x4)=(x4)2=x8. In general f(n)(x)=x2n.
Exercise 3Advanced
f(x)=x+1x−1. Find f(2026)(2).
Hint · direction
Compute the first few iterates and look for a period.
Hint · method
Find a formula for f(2)(x). It is very simple.
Hint · first step
f(2)(x)=x+1x−1+1x+1x−1−1=2x−2=−x1, so f(4)(x)=−−1/x1=x: the period is 4.
Full solution
2026=4×506+2, so f(2026)(2)=f(2)(2)=−21.
Exercise 4Challenge
f(x)=x2−2. How many distinct real roots does the equation f(f(x))=x have?
Hint · direction
Every root of f(x)=x is also a root of f(f(x))=x.
Hint · method
Expand to get a quartic, then divide it by f(x)−x=x2−x−2.
x2−x−2=0 gives the fixed points −1,2; x2+x−1=0 gives 2−1±5. These two points map to each other: they form an orbit of period 2. That makes 4 real roots. In the cobweb diagram choose x2+k with k=−2 and put the starting point near 0.618.