Given a Polynomial of Degree Bound 8 Find 8 Distinct Points to - - PowerPoint PPT Presentation

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Given a Polynomial of Degree Bound 8 Find 8 Distinct Points to - - PowerPoint PPT Presentation

Sep 02, 2023 338 likes •594 views

Given a Polynomial of Degree Bound 8 Find 8 Distinct Points to Efficiently Evaluate it at

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SLIDE 1
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 4

slide-2
SLIDE 2
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 4

slide-3
SLIDE 3
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 4

slide-4
SLIDE 4
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 4

slide-5
SLIDE 5
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 4

slide-6
SLIDE 6
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 4

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SLIDE 7
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

We save roughly half the work. STRATEGY: Set for 0 4

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SLIDE 8
  • Given a Polynomial of Degree Bound 2

Find 2 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 1

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SLIDE 9
  • Given a Polynomial of Degree Bound 2

Find 2 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 1

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SLIDE 10
  • Given a Polynomial of Degree Bound 2

Find 2 Distinct Points to Efficiently Evaluate it at

1 1

STRATEGY: We will evaluate any polynomial of degree bound 2 at 1 1

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SLIDE 11
  • Given a Polynomial of Degree Bound 4

Find 4 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 2

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SLIDE 12
  • Given a Polynomial of Degree Bound 4

Find 4 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 2

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SLIDE 13
  • Given a Polynomial of Degree Bound 4

Find 4 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 2

slide-14
SLIDE 14
  • Given a Polynomial of Degree Bound 4

Find 4 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 2

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SLIDE 15
  • Given a Polynomial of Degree Bound 4

Find 4 Distinct Points to Efficiently Evaluate it at

STRATEGY: Set for 0 2

slide-16
SLIDE 16

Given a Polynomial of Degree Bound 4 Find 4 Distinct Points to Efficiently Evaluate it at

  • STRATEGY: Set for 0 2
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SLIDE 17

Given a Polynomial of Degree Bound 4 Find 4 Distinct Points to Efficiently Evaluate it at

  • STRATEGY: Set for 0 2
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SLIDE 18

Given a Polynomial of Degree Bound 4 Find 4 Distinct Points to Efficiently Evaluate it at

  • Observe that we evaluate both and at

and .

But we decided to always evaluate polynomials of degree bound 2 at 1 and 1. So,

1 ⇒ 1 and 1 ⇒

1 .

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SLIDE 19

Given a Polynomial of Degree Bound 4 Find 4 Distinct Points to Efficiently Evaluate it at

  • So, we evaluate any polynomial of degree bound 4 at

1, and 1,

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SLIDE 20
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

Observe that we evaluate both and at

, , and .

But we decided to always evaluate polynomials of degree bound 4 at 1, , 1 and . So,

1 ⇒ 1, ⇒

  • ,

1 ⇒ , and ⇒ ! .

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SLIDE 21
  • Given a Polynomial of Degree Bound 8

Find 8 Distinct Points to Efficiently Evaluate it at

So, we evaluate any polynomial of degree bound 8 at 1,

  • , ,

!

  • and

1,

  • , ,

!

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SLIDE 22

Given a Polynomial of Degree Bound " #$ Find " #$ Distinct Points to Efficiently Evaluate it at

degree bound how did we find the points to evaluate the polynomial at? the points point property 2 ⋯ ⋯ ⋯

1, 1

all 2nd roots of unity 2 take positive and negative square roots of points used for degree bound 2 which are already the 2nd roots of unity

1, , 1,

  • all 4th roots of unity

2 take positive and negative square roots of points used for degree bound 2 which are already the 4th roots of unity

1, 1 2 , 1, 1 2 , , 1 2 , , 1 2

all 8th roots of unity 2 take positive and negative square roots of points used for degree bound 2 which are already the 8th roots of unity

1, 2 2 2 2 2 2 , 1,

  • 2

2 2 2 2 2 , ⋯ , ⋯ , ⋯ , ⋯

all 16th roots of unity ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

⋯ ⋯ ⋯

⋯ ⋯ ⋯ 2'! take positive and negative square roots of points used for degree bound 2(! which are already the 2(!th roots of unity

⋯ ⋯ ⋯

all 2'!th roots of unity ) 2' take positive and negative square roots of points used for degree bound 2(! which are already the 2(!th roots of unity

⋯ ⋯ ⋯

all 2'th roots of unity (i.e., )th roots of unity)

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SLIDE 23

How to Find all "th Roots of Unity

Euler’s Formula: For any real number *, cos * sin * 01 The )th roots of unity are: 1, 2, 2 , 2 , … … … , 2 !, where 2 cos

4 sin 4

567 8 is known as the primitive )th roots of unity.

The result above can be derived using Euler’s Formula. Euler’s formula follows very easily from the following three power series each of which holds for ∞ * ∞: cos * 1

15 ! 1; ! 1< ! 1= >! ⋯

sin * *

1? ! 1@ ! 1A ! 1B C! ⋯

01 1 *

15 ! 1? ! 1; ! 1@ ! 1< ! 1A ! 1= >! ⋯

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SLIDE 24

How to Find all "th Roots of Unity

Observe that for (any) real numbers * and D, cos * sin * E 01 E 0 E1 cos D* sin D* Also observe that for any integer F, cos F G 2Π sin F G 2Π 1 G 0 1 Then the )th root of 1 (unity) is 1

I 8 cos F G 2Π sin F G 2Π I 8 cos F G

4 J

sin F G

4

  • Observe that cos F G

4 J

sin F G

4

  • takes ) distinct values for 0 F ), and then simply repeats

those values for F 0 and F K ). When F 1, we have cos F G

4 J

sin F G

4

  • cos

4 J

sin

4

  • 2 primitive )th root of 1.

Clearly, for any F, cos F G

4 J

sin F G

4

  • cos

4 J

sin

4

  • '

2 ' Hence, 1

I 8 cos F G

4 J

sin F G

4

  • 2 ', for F 0, 1, 2, … , ) 1.

In other words, the )th roots of 1 (unity) are: 1, 2, 2 , 2 , … … … , 2 !

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SLIDE 25

Coefficient Form ⇒ ⇒ ⇒ ⇒ Point-Value Form

L ) M Θ 1 , N ) 1, 2L ) 2 Θ ) , OPQ0RST0. Θ ) log )

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