COSMO · STUDY + TEACHING DECK Algebra May 23, 2026
C. A question worth sharing Functional Equations Use substitutions, normalization, and a hidden additive rule to reconstruct an unknown function.
Think Compare Explain
Today’s route A question becomes an explanation. Try Enter through a problem. A partial attempt counts.
Connect Use substitution, normalization, additive to name what you notice.
Convince Build an argument a partner can check.
The goal is a method someone else can use.
The opener Start with a question. Opening problem A function f on the nonnegative integers satisfies f(x+y) = f(x) + f(y) - 2xy and f(1) = 5. Find f(10).
Think · 2 min Sketch, test a small case, or make a first guess.
Pair · 3 min Compare methods. Find one thing you agree on and one question.
Be ready to share your route, even if it is unfinished.
Compare before checking Put two attempts in conversation. Explain Tell a partner how you started. Let them finish before responding.
Question Which step feels certain? Which step still needs evidence?
Build Combine one useful idea from each attempt into a new route.
Can you explain your partner’s method in your own words?
Board conversation Notice before naming. Look for structure 01 Substitute 0 first: it often exposes the value at the identity. 02 Set one input equal to 1 to turn the equation into a controlled recurrence. 03 Ask whether a simple correction term makes the rule additive. Make a conjecture We think ___ because ___. A case that might break this is ___.
What changed? What stayed the same?
Useful language Now the language has a job. substitution choose inputs that reveal a special value or a simpler equation
normalization a known value such as f(0) or f(1) that fixes the family
additive a rule g(x+y) = g(x) + g(y) that is easy to build on integers
Use one of these words to explain something you already noticed.
The core idea Keep the idea. Explain the conditions. f ( x + y ) = f ( x ) + f ( y ) + x y ⟹ g ( n ) = f ( n ) − n ( n − 1 ) 2 is additive f(x+y)=f(x)+f(y)+xy\quad\Longrightarrow\quad g(n)=f(n)-\frac{n(n-1)}{2}\text{ is additive} f ( x + y ) = f ( x ) + f ( y ) + x y ⟹ g ( n ) = f ( n ) − 2 n ( n − 1 ) is additive The idea Use substitutions, normalization, and a hidden additive rule to reconstruct an unknown function.
What to remember A carefully chosen correction can turn a nonlinear-looking rule into a linear one.
What information does this method keep?
Worked example One route, one decision at a time The problem Suppose f(x+y) = f(x) + f(y) + xy for nonnegative integers and f(1) = 4. Find f(5).
Predict each next step 01 Let h(n) = n(n - 1)/2; then h(x+y) = h(x) + h(y) + xy. 02 Subtract h from f: g(n) = f(n) - h(n) satisfies g(x+y) = g(x) + g(y). 03 On nonnegative integers, g(n) = n g(1) = 4n because h(1) = 0. 04 Thus f(5) = 4(5) + 5(4)/2 = 20 + 10 = 30. Which step needs a reason, not just a calculation?
See the structure A correction reveals the straight line Read the representation The original rule bends by a predictable amount xy. Remove the triangular-number correction, and the remaining function grows one equal step at a time.
Look together Point to the part that explains the relationship. Change one input and predict what changes.
What can you see here that was harder to see in words?
Two approaches Compare the reasoning. Recursive build 01 Use f(n+1) to move one input at a time. 02 Good for a first table of values. 03 The correction term may stay hidden. Closed form 01 Find an additive correction once. 02 Good for a large input such as 10 or 100. 03 Requires checking the identity carefully. Which route can you justify? What would you test first?
Reasoning clinic Find the step that needs repair. Investigate these concerns 01 Find an additive correction once. 02 Good for a large input such as 10 or 100. 03 Requires checking the identity carefully. Repair together Name the missing condition. Try a small counterexample. Rewrite one step so it is justified.
A useful correction explains why the original step fails.
Work together Build a function table Your shared task Teams receive different correction terms and test whether the transformed values have constant first differences.
Explore 6 min - calculate f(0) through f(5)
Compare 5 min - propose a correction
Defend 4 min - defend the transformed rule
Everyone should be able to explain the group’s final claim.
Share → connect What did the room discover? Bring your evidence Share a conjecture, a useful example, or a route that changed.
Connections to investigate 01 The right substitution can replace a mysterious function with a familiar sequence. 02 On integers, an additive rule is determined by its value at 1. 03 A closed form is useful only after its domain and normalization have been checked. Which observation explains the mechanism?
Try → explain → check Your turn to build the route The problem A function on the nonnegative integers satisfies f(x+y) = f(x) + f(y) + 3xy and f(1) = 7. Find f(4).
Predict each next step 01 Set y = 0; the equation forces f(0) = 0. 02 For n to n+1, f(n+1) - f(n) = 7 + 3n. 03 Add the four increments: 7 + 10 + 13 + 16 = 46. 04 So f(4) = 46; equivalently f(n) = 7n + 3n(n - 1)/2. Which step needs a reason, not just a calculation?
Convince a skeptic Turn a pattern into a proof. The claim For f(x+y) = f(x) + f(y) + xy, subtracting n(n - 1)/2 creates an additive function.
Build the argument 01 Let h(n) = n(n - 1)/2, so h(x+y) - h(x) - h(y) = xy. 02 Set g = f - h. The +xy terms cancel when g(x+y) is expanded. 03 Therefore g(x+y) = g(x) + g(y), and repeated addition gives g(n) = n g(1). What does each step rely on? Where are the conditions used?
Stretch together Take the idea somewhere new. Transfer challenge A function on the nonnegative integers satisfies f(x+y) = f(x) + f(y) + xy and f(2) = 9. Find f(8).
Choose a route Name the objects, the conditions, and a relationship that might help. Start with one test case.
What transfers from the earlier example? What is different?
Peer review Check the argument together. Restate Explain the other person’s route before evaluating it.
Test Try a small case or boundary case. Does every step still make sense?
Revise Fix an assumption, add a reason, or find a second method.
Where did your thinking change?
Beyond this problem Where might this idea help? A connection The same move appears in recurrence solving, finite differences, and contest problems where a predictable quadratic drift hides a linear pattern.
Discuss the model 01 What objects or quantities are represented? 02 Which relationship from today is preserved? 03 What assumption would you check before trusting it? Invent one more situation with the same structure.
Exit ticket Leave with an explanation. Question 1 Why is y = 0 a useful first substitution in this family of equations?
Question 2 What correction turns a +xy rule into an additive rule?
Question 3 Why does additivity determine g(n) once g(1) is known on nonnegative integers?
Choose one. Write an answer with a reason, then compare with a partner.
Reflect together Make your thinking visible. I used to think… Name a claim or approach that changed.
Now I can explain… Give the reason, not only the result.
I still wonder… Write a question worth bringing to the next circle.
Share one sentence. Ask one follow-up question.
The idea to keep A carefully chosen correction can turn a nonlinear-looking rule into a linear one. Try changing +xy to -2xy and predict the correction before calculating any values.
A good question belongs to everyone in the room. ← 1 / 21 →
Go to slide 01 · COSMO · Math circle 02 · Today’s route 03 · The opener 04 · Compare before checking 05 · Board conversation 06 · Useful language 07 · The core idea 08 · Worked example 09 · See the structure 10 · Two approaches 11 · Reasoning clinic 12 · Work together 13 · Share → connect 14 · Try → explain → check 15 · Convince a skeptic 16 · Stretch together 17 · Peer review 18 · Beyond this problem 19 · Exit ticket 20 · Reflect together 21 · The idea to keep Show all steps Speaker notes Present ↗
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Functional Equations Welcome the room. Students can begin with examples, drawings, or questions. Let students calculate f(0) before introducing any named technique. Keep the correction h(n) visible beside f(n) so the cancellation can be checked. Ask which steps use the integer domain; do not silently extend the conclusion to all real functions. Require a quick substitution check after any proposed closed form. A question becomes an explanation. Introduce the learning arc without teaching the theorem in advance. Invite students to choose a personal goal: make a conjecture, ask a useful question, or explain someone else’s idea. Start with a question. Allow quiet attempts before inviting discussion. Opener answer: Set y = 0: f(0) = 0. Define g(n) = f(n) + n(n - 1). Then g(x+y) = g(x) + g(y), so on nonnegative integers g(n) = n g(1) = 5n. Therefore f(n) = 5n - n(n - 1), and f(10) = 50 - 90 = -40. Let students calculate f(0) before introducing any named technique. Keep the correction h(n) visible beside f(n) so the cancellation can be checked. Ask which steps use the integer domain; do not silently extend the conclusion to all real functions. Require a quick substitution check after any proposed closed form. Give everyone quiet thinking time. Ask a pair to compare approaches, then invite a different pair to restate the argument. Ask what would convince someone who disagrees. Put two attempts in conversation. Opener solution: Set y = 0: f(0) = 0. Define g(n) = f(n) + n(n - 1). Then g(x+y) = g(x) + g(y), so on nonnegative integers g(n) = n g(1) = 5n. Therefore f(n) = 5n - n(n - 1), and f(10) = 50 - 90 = -40. Select two contrasting methods from the room. Record their assumptions and compare what each preserves. If no one has a route yet, return to a smaller case rather than announcing the answer. Notice before naming. Record student language before supplying formal terms. A pattern from examples is a conjecture. Ask why it continues or where it could fail. Give everyone quiet thinking time. Ask a pair to compare approaches, then invite a different pair to restate the argument. Ask what would convince someone who disagrees. Now the language has a job. Connect each term to a student observation from the opener. Ask students to give an example and a non-example; check conditions rather than memorized wording. Keep the idea. Explain the conditions. Translate the key relationship into ordinary language, then read the notation together. Ask what each symbol represents and when the relationship is valid. One route, one decision at a time Show the problem first. Invite a proposed first step before revealing one step at a time. Let h(n) = n(n - 1)/2; then h(x+y) = h(x) + h(y) + xy. Subtract h from f: g(n) = f(n) - h(n) satisfies g(x+y) = g(x) + g(y). On nonnegative integers, g(n) = n g(1) = 4n because h(1) = 0. Thus f(5) = 4(5) + 5(4)/2 = 20 + 10 = 30. After the final step, ask a partner to reconstruct the method without looking. A correction reveals the straight line Ask students to describe the diagram or formula before interpreting it for them. Have a partner identify the condition that makes the representation accurate. Give everyone quiet thinking time. Ask a pair to compare approaches, then invite a different pair to restate the argument. Ask what would convince someone who disagrees. Compare the reasoning. Give each pair one route to defend and one to question. Ask when each approach is useful and which condition decides whether it works. Avoid treating experimentation as a mistake. Find the step that needs repair. These are concerns to investigate, not all claims of mathematical error. Invite a student to explain why a tempting route is incomplete before proposing a repair. Build a function table Groups of 2–3: rotate who records, who questions, and who explains. Offer a hint only after asking what the group has tried. Let students calculate f(0) before introducing any named technique. Keep the correction h(n) visible beside f(n) so the cancellation can be checked. Ask which steps use the integer domain; do not silently extend the conclusion to all real functions. Require a quick substitution check after any proposed closed form. Save two contrasting approaches for the debrief. What did the room discover? Hear group observations before revealing the connections. The right substitution can replace a mysterious function with a familiar sequence. On integers, an additive rule is determined by its value at 1. A closed form is useful only after its domain and normalization have been checked. Ask for a boundary case that tests the claim. Acknowledge unfinished work that produced a useful question. Your turn to build the route Show the problem first. Invite a proposed first step before revealing one step at a time. Set y = 0; the equation forces f(0) = 0. For n to n+1, f(n+1) - f(n) = 7 + 3n. Add the four increments: 7 + 10 + 13 + 16 = 46. So f(4) = 46; equivalently f(n) = 7n + 3n(n - 1)/2. After the final step, ask a partner to reconstruct the method without looking. Turn a pattern into a proof. Show the claim first. Ask the room to propose a reason. Let h(n) = n(n - 1)/2, so h(x+y) - h(x) - h(y) = xy. Set g = f - h. The +xy terms cancel when g(x+y) is expanded. Therefore g(x+y) = g(x) + g(y), and repeated addition gives g(n) = n g(1). Reveal one link at a time, checking that students can explain why it follows. Ask a partner to find any unstated assumption. Take the idea somewhere new. Challenge solution: Let h(n) = n(n - 1)/2 and g(n) = f(n) - h(n). Then g is additive. Since f(2) = 9 and h(2) = 1, g(2) = 8, so g(1) = 4. Hence f(8) = 8(4) + h(8) = 32 + 28 = 60. Allow groups to choose a representation. Ask for a plan before a calculation. Give a hint only when the group can say where it is stuck. Check the argument together. Challenge solution: Let h(n) = n(n - 1)/2 and g(n) = f(n) - h(n). Then g is additive. Since f(2) = 9 and h(2) = 1, g(2) = 8, so g(1) = 4. Hence f(8) = 8(4) + h(8) = 32 + 28 = 60. Invite two routes. Ask the room to check the logic before comparing answers. Offer the solution as a response to a specific student question, then invite a reconstruction. Where might this idea help? Start with an example students recognize. Keep the discussion about how the mathematical structure transfers, and where the model stops being faithful. Leave with an explanation. Answer 1: It forces f(0)=0 and removes an unknown special value before other substitutions. Answer 2: Subtract h(n)=n(n-1)/2, since h(x+y)-h(x)-h(y)=xy; the corrected function g=f-h is additive. Answer 3: Every nonnegative integer is a sum of copies of 1, so repeated additivity gives g(n)=n g(1), with g(0)=0. Make your thinking visible. Exit answer 1: It forces f(0)=0 and removes an unknown special value before other substitutions. Exit answer 2: Subtract h(n)=n(n-1)/2, since h(x+y)-h(x)-h(y)=xy; the corrected function g=f-h is additive. Exit answer 3: Every nonnegative integer is a sum of copies of 1, so repeated additivity gives g(n)=n g(1), with g(0)=0. Collect one unresolved question and one explanation from each group. Use the exit evidence to decide what to revisit next time. A carefully chosen correction can turn a nonlinear-looking rule into a linear one. End with a student explanation, not a final lecture. Invite a new example or a question the group could continue investigating. Enable JavaScript for slide navigation and reveals, or open the full PDF .