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LIMIT SPEED 40 LIMIT SPEED 80 LIMIT SPEED 100 LIMIT SPEED 20 LIMIT SPEED 60 Dessert: Hilbert’s 13th Problem, in Full Colour (Page 2) How go least blur? Step 1 Step 4 Step 2. There exists φ : [0, 1] → [0, 1] so that for every ǫ> 0 and every f : [0, 1]×[0, 1] → R there exists a g : [0, 1+λ] → R so that |f (x, y ) − g (φ(x)+ λφ(y ))| <ǫ on a set of area at least 1 − ǫ in [0, 1] × [0, 1]. The key. “Iterated pooriﬁcation”. Step 3. There exist φ i : [0, 1] → [0, 1] (1 ≤ i ≤ 5) so that for every ǫ> 0 and every f : [0, 1] × [0, 1] → R there exists a g : [0, 1+ λ] → R so that |f (x, y ) − 5 i=1 g (φ i (x)+ λφ i (y ))| < 2 3 + ǫ f ∞ for every x, y ∈ [0, 1]. The key. “Shift the chocolates”. . . Exercise 1. Do the m-dimensional case. Exercise 2. Do R m instead of just I m . Step 4. We are done. The key. Learn from the artillery! Set Tg := ∑ 5 i=1 g(φi(x)+ λφi(y)), f1 := f , M := f , and iterate “shooting and adjusting”. Find g1 with g1 ≤ M and f2 := f1 − Tg1≤ 3 4 M. Find g2 with g2≤ 3 4 M and f3 := f2 − Tg2≤ ( 3 4 ) 2 M. Find g3 with g3≤ ( 3 4 ) 2 M and f4 := f3 − Tg3≤ ( 3 4 ) 3 M. Continue to eternity. When done, set g = ∑ g k and note that f = Tg as required. Propaganda. I love handouts! • I have nothing to hide and you can take what you want, forwards, backwards, here and at home. • What doesn’t ﬁt on one sheet can’t be done in one hour. • It takes learning and many hours and a few pennies. The audience’s worth it! • There’s real math in the handout viewer! . . . then iterate. g 2 φ(x)+ λφ(y ) g 2 -coloured φ(x)+ λφ(y ) φ 0,1,2 g 1 g 1 -coloured φ(x)+ λφ(y ) Video and more at http://www.math.toronto.edu/ ~ drorbn/Talks/Fields-1411/ 89

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LIMIT

SPEED40

LIMIT

SPEED80

LIMIT

SPEED100

LIMIT

SPEED20

LIMIT

SPEED60

Dessert: Hilbert’s 13th Problem, in Full Colour (Page 2)

How goleast blur?

Step 1 Step 4

Step 2. There exists φ : [0, 1] → [0, 1] so that for every ǫ > 0and every f : [0, 1]×[0, 1] → R there exists a g : [0, 1+λ] → Rso that |f(x, y) − g(φ(x) + λφ(y))| < ǫ on a set of area atleast 1 − ǫ in [0, 1] × [0, 1].The key. “Iterated poorification”.

Step 3. There exist φi : [0, 1] → [0, 1] (1 ≤ i ≤ 5) sothat for every ǫ > 0 and every f : [0, 1] × [0, 1] → Rthere exists a g : [0, 1 + λ] → R so that

|f(x, y) −5∑

i=1

g(φi(x) + λφi(y))| <

3+ ǫ

)‖f‖∞

for every x, y ∈ [0, 1].The key. “Shift the chocolates”. . .

Exercise 1. Do the m-dimensional case.Exercise 2. Do Rm instead of just Im.

Step 4. We are done.The key. Learn from the artillery!Set Tg :=

∑5i=1 g(φi(x) + λφi(y)), f1 := f , M := ‖f‖, and

iterate “shooting and adjusting”. Find g1 with ‖g1‖ ≤ Mand ‖f2 := f1 − Tg1‖ ≤ 3

4M . Find g2 with ‖g2‖ ≤ 3

4M and

‖f3 := f2 − Tg2‖ ≤ ( 34)2M . Find g3 with ‖g3‖ ≤ ( 3

4)2M and

‖f4 := f3 − Tg3‖ ≤ ( 34)3M . Continue to eternity. When done,

set g =∑

gk and note that f = Tg as required.

Propaganda. I love handouts! • I have nothing to hide and youcan take what you want, forwards, backwards, here and at home.• What doesn’t fit on one sheet can’t be done in one hour. • Ittakes learning and many hours and a few pennies. The audience’sworth it! • There’s real math in the handout viewer!

. . . then iterate.

φ(x) + λφ(y)

g2-coloured φ(x) + λφ(y)

φ0,1,2

g1-coloured φ(x) + λφ(y)

Video and more at http://www.math.toronto.edu/~drorbn/Talks/Fields-1411/

http://www.math.toronto.edu/~drorbn/Talks/Fields-1411/

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