WorkbenchPress x=y=0, y=−x and y=1 in turn, and read what each substitution tells you. Then drag the coefficients on the right until every light shows ✓.
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What this move does
A functional equation holds for allx and y, so you may plug in whatever values suit you best. The usual steps:
Plug in x=y=0 (or x=y=1) to find special values such as f(0) and f(1).
Plug in y=−x or x=0 to find the parity; plug in y=x to relate f(2x) and f(x).
Plug in y=1 to get a recurrence between f(x+1) and f(x); then you know the values at all the integers.
Guess f, and substitute it back into the original equation to check.
The right side of the workbench is a “candidate tester”: every substitution you pick is used to test the current candidate f. One substitution rules out only some candidates; several together leave just one.
Predict first
f(x+y)+f(x−y)=2f(x)+2f(y) for all real numbers, and f(1)=1. What is f(2)?
Choose an answer first; then you can see the result.
✓You predicted it!✗First find f(0), then put x=y=1.
x=y=0: 2f(0)=4f(0), so f(0)=0. x=y=1: f(2)+f(0)=4f(1), so f(2)=4.
More generally, y=x gives f(2x)=4f(x): doubling the input multiplies the value by 4, very much like x2. In the candidate tester below, try f(x)=x2, f(x)=3x2 and f(x)=x2+x.
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Every constant multiple of x2 passes all the checks. Add a linear term x, and the “even function” line is the first to show ✗.
Work through one
Example: f(x + y) = f(x) + f(y) + 2xy, f(1) = 3
Step 1 / 5
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First find f(0)
Put x=y=0: f(0)=2f(0), so f(0)=0. The candidate f≡0 on the right passes, of course. It passes this test, yet it fails f(1)=3.
f(0)=2f(0)⇒f(0)=0
y = −x
y = 1: a recurrence
Guess, and check
What is still missing
When it fails
Break it
Counterexample lab: remove one condition
Conclusion: the only solution of f(xy)=f(x)f(y)−f(x+y)+1 is f(x)=x+1← still true?
With f(1)=2: y=1 gives f(x+1)=f(x)+1, x=1,y=−1 gives f(−1)=0, … The candidate x+1 passes every test.
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Remove the condition f(1)=2. Look at the candidate f(x)≡1 (a=b=0,c=1): all36 spot checks of the equation pass. In fact 1=1⋅1−1+1 always holds. Only the “Condition” meter shows ✗.
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So the extra condition is not decoration: without it, the equation has two solutions. When you plug in values, notice which step uses the condition. x=1,y=−1 gives f(−1)(1−f(1))=0, and that is exactly where the two paths split.
Practice
Exercise 1Basic
f(x+y)=f(x)+f(y)+xy for all real numbers, and f(1)=1. Find f(4).
Hint · direction
Put y=x.
Hint · method
f(2x)=2f(x)+x2. Get f(2) from f(1), then f(4).
Hint · first step
f(2)=2f(1)+1=3.
Full solution
f(4)=2f(2)+4=10. (In fact f(x)=2x2+x satisfies the equation, and f(4)=10.)
Exercise 2Basic
f(xy)=f(x)+f(y) for all positive x,y, and f(2)=1. Find f(41).
Hint · direction
First find f(1).
Hint · method
x=y=1 gives f(1)=0; then put y=x1.
Hint · first step
f(2)+f(21)=f(1)=0, so f(21)=−1.
Full solution
f(41)=f(21)+f(21)=−2.
Exercise 3Advanced
f(xy)=f(x)f(y)−f(x+y)+1 for all real numbers, and f(1)=2. Find f(−2).
Hint · direction
Use the substitutions from the workbench.
Hint · method
x=y=0 gives f(0)=1; y=1 gives f(x+1)=f(x)+1.
Hint · first step
f(x)=f(x+1)−1, so f(−1)=f(0)−1=0.
Full solution
f(−2)=f(−1)−1=−1.
Exercise 4Challenge
Find all functions f:R→R with f(x)f(y)−f(xy)=x+y for all real x,y. (Enter the formula for f(x).)
Hint · direction
First find f(0).
Hint · method
x=y=0: f(0)2−f(0)=0, so f(0)=0 or 1. In each case, put y=0 and see.
Hint · first step
If f(0)=0, put y=0: 0−0=x for every x, a contradiction. So f(0)=1.
Full solution
y=0: f(x)⋅1−1=x, so f(x)=x+1. Check: (x+1)(y+1)−(xy+1)=x+y ✓. The only solution is f(x)=x+1.
The wrong step: f(x+y)+f(x−y)=2f(x)+2f(y). x=y=0 gives f(0)=0, x=0 shows f is even, so f(x)=x2.
What went wrong? Think first, then open
Diagnosisf(0)=0 and “even” are only necessary conditions. Plenty of functions satisfy them (for example f(x)=x4), and x4 does not satisfy the original equation; on the other hand f(x)=3x2 does satisfy it too. Plugging in values by itself cannot give “f(x)=x2”.
✓
The right step: Plugging in gives properties and special values. Either keep going until you derive a formula for f (with suitable extra conditions), or guess a solution, substitute it back into the original equation to check, and explain why there are no others.
“Plugging in gives clues, not the answer”
Functional Equations: Plugging in Values ← → step · F full screen · Esc exit