Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
John Wallis wikipedia , lookup
List of important publications in mathematics wikipedia , lookup
Approximations of π wikipedia , lookup
Vincent's theorem wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Positional notation wikipedia , lookup
Continued Fractions and Approximations Open Day 24th March, 2010 Dr. James Montaldi University of Manchester Continued Fractions – p. 1 Introduction Leap Years Continued Fractions – p. 2 How many days in a year? The Earth goes round the sun in 365.24219878 days. Just less than 365 and a quarter. Continued Fractions – p. 3 How many days in a year? The Earth goes round the sun in 365.24219878 days. Just less than 365 and a quarter. So, basically we need 365 days per year, Continued Fractions – p. 3 How many days in a year? The Earth goes round the sun in 365.24219878 days. Just less than 365 and a quarter. So, basically we need 365 days per year, and then study the fractional part 0.24219878 to see how to correct it. Continued Fractions – p. 3 How many days in a year? The Earth goes round the sun in 365.24219878 days. Just less than 365 and a quarter. So, basically we need 365 days per year, and then study the fractional part 0.24219878 to see how to correct it. The Julian Calendar (dating back to the Roman Empire) adds 1 extra day every four years (leap year). This averages out to 365.25 days per year. Continued Fractions – p. 3 How many days in a year? The Earth goes round the sun in 365.24219878 days. Just less than 365 and a quarter. So, basically we need 365 days per year, and then study the fractional part 0.24219878 to see how to correct it. The Julian Calendar (dating back to the Roman Empire) adds 1 extra day every four years (leap year). This averages out to 365.25 days per year. But this is too long by about 1 day each century. Continued Fractions – p. 3 Gregorian Calendar The Gregorian Calendar (introduced in Europe in 1582, and in England in 1752): (=0.25) +1 day every 4 years: 14 24 (=0.24) -1 day every 100 years: 100 97 +1 day every 400 years: 400 (=0.2425) ( compare: 0.24219878) Continued Fractions – p. 4 Gregorian Calendar The Gregorian Calendar (introduced in Europe in 1582, and in England in 1752): (=0.25) +1 day every 4 years: 14 24 (=0.24) -1 day every 100 years: 100 97 +1 day every 400 years: 400 (=0.2425) ( compare: 0.24219878) Error is now 1 day in 3300 years. But is there a better approximation? Continued Fractions – p. 4 “Gregorian Reformation" In 1752, 1st January in England, was 12th January in Europe!!! Continued Fractions – p. 5 “Gregorian Reformation" In 1752, 1st January in England, was 12th January in Europe!!! September S M Tu W 1 2 17 18 19 20 24 25 26 27 Th 14 21 28 1752 F 15 22 29 S 16 23 30 Continued Fractions – p. 5 CONTENTS Continued Fractions – p. 6 CONTENTS I Continued fractions Continued Fractions – p. 6 CONTENTS I Continued fractions II How do we work them out? Continued Fractions – p. 6 CONTENTS I Continued fractions II How do we work them out? III Approximations Continued Fractions – p. 6 What are continued fractions? . . . as if fractions weren’t bad enough!! Continued Fractions – p. 7 What are continued fractions? . . . as if fractions weren’t bad enough!! Here’s one: 81 1 = 1 + 1 56 2 + 4+ Shorthand: 81 56 1 6 = [1, 2, 4, 6] Continued Fractions – p. 7 What are continued fractions? . . . as if fractions weren’t bad enough!! And here’s another: 190 1 = 2 + 1 81 2 + 1+ Shorthand: 190 81 1 8+1 3 = [2, 2, 1, 8, 3] Continued Fractions – p. 7 And for irrational numbers? There are also continued fractions for irrational numbers Continued Fractions – p. 8 And for irrational numbers? There are also continued fractions for irrational numbers For example √ 2 = 1 + 1 2 + 1 2+ 1 1 2+ 2+... √ Shorthand: 2 = [1, 2, 2, 2, 2, . . .] Continued Fractions – p. 8 Part II How do we work these out? Continued Fractions – p. 9 The calculation Start with an example: Let’s take x = 2 · 75 = 2 34 Continued Fractions – p. 10 The calculation Start with an example: Let’s take x = 2 · 75 = 2 34 Integer part of 2.75 is 2 Continued Fractions – p. 10 The calculation Start with an example: Let’s take x = 2 · 75 = 2 34 Integer part of 2.75 is 2 So x = 2 + 34 Continued Fractions – p. 10 The calculation Start with an example: Let’s take x = 2 · 75 = 2 34 Integer part of 2.75 is 2 So x = 2 + 34 x=2+ 1 4/3 Continued Fractions – p. 10 The calculation Start with an example: Let’s take x = 2 · 75 = 2 34 Integer part of 2.75 is 2 So x = 2 + 34 x=2+ 1 4/3 x=2+ 1 1+ 13 Continued Fractions – p. 10 The calculation Start with an example: Let’s take x = 2 · 75 = 2 34 Integer part of 2.75 is 2 So x = 2 + 34 x=2+ 1 4/3 x=2+ 1 1+ 13 Shorthand: 11 4 = [2, 1, 3]. Continued Fractions – p. 10 Continued Fractions Take another fraction, say 43/30 Continued Fractions – p. 11 Continued Fractions Take another fraction, say 43/30 Now, 43 30 has an integer part and a fractional part: 43 13 =1+ 30 30 Continued Fractions – p. 11 Continued Fractions Take another fraction, say 43/30 Now, 43 30 has an integer part and a fractional part: 43 13 1 =1+ . =1+ 30 30 30/13 Continued Fractions – p. 11 Continued Fractions Take another fraction, say 43/30 Now, 43 30 has an integer part and a fractional part: 43 13 1 =1+ . =1+ 30 30 30/13 And 30/13 = 2 + 4/13, so Continued Fractions – p. 11 Continued Fractions Take another fraction, say 43/30 Now, 43 30 has an integer part and a fractional part: 43 13 1 =1+ . =1+ 30 30 30/13 And 30/13 = 2 + 4/13, so 1 1 43 . =1+ =1+ 30 30/13 2 + 4/13 Continued Fractions – p. 11 Continued Fractions . . . (from previous slide) 1 43 =1+ 30 2 + 4/13 Continued Fractions – p. 12 Continued Fractions . . . (from previous slide) 1 1 43 =1+ =1+ . 1 30 2 + 4/13 2 + 13/4 Continued Fractions – p. 12 Continued Fractions . . . (from previous slide) 1 1 43 =1+ =1+ . 1 30 2 + 4/13 2 + 13/4 Since 13/4 = 3 + 14 we have 1 43 =1+ 30 2 + 3+1 1 4 Continued Fractions – p. 12 Continued Fractions . . . (from previous slide) 1 1 43 =1+ =1+ . 1 30 2 + 4/13 2 + 13/4 Since 13/4 = 3 + 14 we have 1 43 =1+ 30 2 + 3+1 1 4 43 = [1, 2, 3, 4]. That is, 30 Continued Fractions – p. 12 CFs & Irrational Numbers NB: one can find the continued fraction of any number, even an irrational one. Continued Fractions – p. 13 CFs & Irrational Numbers NB: one can find the continued fraction of any number, even an irrational one. However, the important difference is that the procedure never stops. Continued Fractions – p. 13 CFs & Irrational Numbers NB: one can find the continued fraction of any number, even an irrational one. However, the important difference is that the procedure never stops. For example, √ 2=1+ 1 2+ 1 2+ 1 1 2+ 2+ 1 2+··· √ That is, 2 = [1, 2, 2, 2, 2, 2, . . .] Continued Fractions – p. 13 CFs & Irrational Numbers Calculation √ 2 = 1.414213562373095 Continued Fractions – p. 14 CFs & Irrational Numbers Calculation √ 2 = 1.414213562373095 = 1 + 0.414213562373095 Continued Fractions – p. 14 CFs & Irrational Numbers Calculation √ 2 = 1.414213562373095 = 1 + 0.414213562373095 1 = 1+ 2.414213562373095 Continued Fractions – p. 14 CFs & Irrational Numbers Calculation √ 2 = 1.414213562373095 = 1 + 0.414213562373095 1 = 1+ 2.414213562373095 1 = 1+ 2 + 0.414213562373095 Continued Fractions – p. 14 CFs & Irrational Numbers Calculation √ 2 = 1.414213562373095 = 1 + 0.414213562373095 1 = 1+ 2.414213562373095 1 = 1+ 2 + 0.414213562373095 = ... Continued Fractions – p. 14 CFs & Irrational Numbers Calculation √ 2 = 1.414213562373095 = 1 + 0.414213562373095 1 = 1+ 2.414213562373095 1 = 1+ 2 + 0.414213562373095 = ... √ Other examples are 3 = [1, 1, 2, 1, 2, 1, 2, 1, 2, . . .]. and √ 7 = [2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, . . .]. Continued Fractions – p. 14 CF of π Calculation π = 3.141592654 Continued Fractions – p. 15 CF of π Calculation π = 3.141592654 = 3 + 0.141592654 Continued Fractions – p. 15 CF of π Calculation π = 3.141592654 = 3 + 0.141592654 1 = 3+ 7.062513285 Continued Fractions – p. 15 CF of π Calculation π = 3.141592654 = 3 + 0.141592654 1 = 3+ 7.062513285 1 = 3+ 7 + 0.062513285 Continued Fractions – p. 15 CF of π Calculation π = 3.141592654 = 3 + 0.141592654 1 = 3+ 7.062513285 1 = 3+ 7 + 0.062513285 1 = 3+ 1 7 + 15+0.99659976 Continued Fractions – p. 15 CF of π Calculation π = 3.141592654 = 3 + 0.141592654 1 = 3+ 7.062513285 1 = 3+ 7 + 0.062513285 1 = 3+ 1 7 + 15+0.99659976 Continuing (after a lot of work) you get π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, . . .] Continued Fractions – p. 15 Continued Fractions Basic Theorem Continued Fractions – p. 16 Continued Fractions Basic Theorem If x is a rational number (fraction) then its CF terminates; Continued Fractions – p. 16 Continued Fractions Basic Theorem If x is a rational number (fraction) then its CF terminates; It x is irrational then its CF is infinite. Continued Fractions – p. 16 Continued Fractions Basic Theorem If x is a rational number (fraction) then its CF terminates; It x is irrational then its CF is infinite. If x is quadratic then the CF is repeating. Continued Fractions – p. 16 Continued Fractions Basic Theorem If x is a rational number (fraction) then its CF terminates; It x is irrational then its CF is infinite. If x is quadratic then the CF is repeating. There are other CFs with interesting patterns though: e − 1 = [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, . . .] Continued Fractions – p. 16 Continued Fractions Basic Theorem If x is a rational number (fraction) then its CF terminates; It x is irrational then its CF is infinite. If x is quadratic then the CF is repeating. There are other CFs with interesting patterns though: e − 1 = [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, . . .] And some with none, like π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] Continued Fractions – p. 16 Continued Fractions Basic Theorem If x is a rational number (fraction) then its CF terminates; It x is irrational then its CF is infinite. If x is quadratic then the CF is repeating. There are other CFs with interesting patterns though: e − 1 = [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, . . .] And some with none, like π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] (at least, no-one has found a pattern!) Continued Fractions – p. 16 Part III Approximations Continued Fractions – p. 17 A use for Continued Fractions Approximations of irrational numbers √ Find a fraction that’s a good approximation to 2. Continued Fractions – p. 18 A use for Continued Fractions Approximations of irrational numbers √ Find a fraction that’s a good approximation to 2. √ Use the continued fraction: 2 = [1, 2, 2, 2, 2, 2, . . .] Continued Fractions – p. 18 A use for Continued Fractions Approximations of irrational numbers √ Find a fraction that’s a good approximation to 2. √ Use the continued fraction: 2 = [1, 2, 2, 2, 2, 2, . . .] x0 = [1] = 1, x1 = [1, 2] = 1 + 1 2 = 32 . Continued Fractions – p. 18 A use for Continued Fractions Approximations of irrational numbers √ Find a fraction that’s a good approximation to 2. √ Use the continued fraction: 2 = [1, 2, 2, 2, 2, 2, . . .] x0 = [1] = 1, x1 = [1, 2] = 1 + 21 = 32 . x2 = [1, 2, 2] = 1 + 1/(2 + 21 ) = 7 5 Continued Fractions – p. 18 A use for Continued Fractions Approximations of irrational numbers √ Find a fraction that’s a good approximation to 2. √ Use the continued fraction: 2 = [1, 2, 2, 2, 2, 2, . . .] x0 x1 x2 x3 = [1] = 1, = [1, 2] = 1 + 21 = 32 . = [1, 2, 2] = 1 + 1/(2 + 21 ) = 75 √ 17 = [1, 2, 2, 2] = 12 . . . getting closer and closer to 2. Continued Fractions – p. 18 A use for Continued Fractions Approximations of irrational numbers √ Find a fraction that’s a good approximation to 2. √ Use the continued fraction: 2 = [1, 2, 2, 2, 2, 2, . . .] x0 x1 x2 x3 = [1] = 1, = [1, 2] = 1 + 21 = 32 . = [1, 2, 2] = 1 + 1/(2 + 21 ) = 75 √ 17 = [1, 2, 2, 2] = 12 . . . getting closer and closer to 2. They are called the first, second, third, . . . nth convergent, xn = abnn . Continued Fractions – p. 18 Accuracy: example √ How good are the approximations to 2 ? Continued Fractions – p. 19 Accuracy: example √ How good are the approximations to 2 ? Recall: √ 2 = 1.414213562 = [1, 2, 2, 2, 2, 2, 2, . . .] Continued Fractions – p. 19 Accuracy: example √ How good are the approximations to 2 ? Recall: √ 2 = 1.414213562 = [1, 2, 2, 2, 2, 2, 2, . . .] [1,2] = 3 2 = 1.5 Continued Fractions – p. 19 Accuracy: example √ How good are the approximations to 2 ? Recall: √ 2 = 1.414213562 = [1, 2, 2, 2, 2, 2, 2, . . .] [1,2] = 3 2 = 1.5 [1,2,2] = 7 5 = 1.4 Continued Fractions – p. 19 Accuracy: example √ How good are the approximations to 2 ? Recall: √ 2 = 1.414213562 = [1, 2, 2, 2, 2, 2, 2, . . .] [1,2] = 3 2 = 1.5 [1,2,2] = 7 5 = 1.4 [1,2,2,2] = 17 12 = 1.416667 Continued Fractions – p. 19 Accuracy: example √ How good are the approximations to 2 ? Recall: √ 2 = 1.414213562 = [1, 2, 2, 2, 2, 2, 2, . . .] [1,2] = 3 2 = 1.5 [1,2,2] = 7 5 = 1.4 [1,2,2,2] = 17 12 = 1.416667 [1,2,2,2,2] = 41 29 = 1.413793 Continued Fractions – p. 19 Accuracy: example √ How good are the approximations to 2 ? Recall: √ 2 = 1.414213562 = [1, 2, 2, 2, 2, 2, 2, . . .] [1,2] = 3 2 = 1.5 [1,2,2] = 7 5 = 1.4 [1,2,2,2] = 17 12 = 1.416667 [1,2,2,2,2] = 41 29 = 1.413793 [1,2,2,2,2,2] = 99 70 = 1.414285 Continued Fractions – p. 19 Accuracy When using approximations, it’s important to know how accurate they are. Continued Fractions – p. 20 Accuracy When using approximations, it’s important to know how accurate they are. Theorem If a b is one of the convergents for x, then a x − < b Continued Fractions – p. 20 Accuracy When using approximations, it’s important to know how accurate they are. Theorem If a b is one of the convergents for x, then a 1 x − < 2 . b b Continued Fractions – p. 20 Accuracy When using approximations, it’s important to know how accurate they are. Theorem If a b is one of the convergents for x, then a 1 x − < 2 . b b And you can say more: among all fractions with denominator no bigger than b, this convergent is the most accurate. Continued Fractions – p. 20 Accuracy: another example How good are the approximations to π ? Continued Fractions – p. 21 Accuracy: another example How good are the approximations to π ? Recall: π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] π = 3.14159265359 Continued Fractions – p. 21 Accuracy: another example How good are the approximations to π ? Recall: π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] [3, 7] = π = 3.14159265359 22 7 = 3.14285714286 Continued Fractions – p. 21 Accuracy: another example How good are the approximations to π ? Recall: π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] π = 3.14159265359 [3, 7] = 22 7 = 3.14285714286 [3, 7, 15] = 333 106 = 3.14150943396 Continued Fractions – p. 21 Accuracy: another example How good are the approximations to π ? Recall: π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] π = 3.14159265359 [3, 7] = 22 7 = 3.14285714286 [3, 7, 15] = 333 106 = 3.14150943396 [3, 7, 15, 1] = 355 113 = 3.14159292035 Continued Fractions – p. 21 Accuracy: another example How good are the approximations to π ? Recall: π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, . . .] π = 3.14159265359 [3, 7] = 22 7 = 3.14285714286 [3, 7, 15] = 333 106 = 3.14150943396 [3, 7, 15, 1] = 355 113 = 3.14159292035 103993 33102 = 3.14159265301 [3, 7, 15, 1, 292] = Continued Fractions – p. 21 Back to the Calendar The Earth goes round the sun in 365.24219878 days. Continued Fractions – p. 22 Back to the Calendar The Earth goes round the sun in 365.24219878 days. The Continued Fraction of 0.24219878 is [0, 4, 7, 1, 3, 5, 6, . . . ] Continued Fractions – p. 22 Back to the Calendar The Earth goes round the sun in 365.24219878 days. The Continued Fraction of 0.24219878 is [0, 4, 7, 1, 3, 5, 6, . . . ] First convergent for 0.24219878 is 14 . Continued Fractions – p. 22 Back to the Calendar The Earth goes round the sun in 365.24219878 days. The Continued Fraction of 0.24219878 is [0, 4, 7, 1, 3, 5, 6, . . . ] First convergent for 0.24219878 is 14 . So we should add an extra 1 day every 4 years (Leap Year). This forms the old Julian Calendar. But, Continued Fractions – p. 22 Back to the Calendar The Earth goes round the sun in 365.24219878 days. The Continued Fraction of 0.24219878 is [0, 4, 7, 1, 3, 5, 6, . . . ] First convergent for 0.24219878 is 14 . So we should add an extra 1 day every 4 years (Leap Year). This forms the old Julian Calendar. But, it’s not accurate enough . . . Continued Fractions – p. 22 The Calendar The second convergent for 0.24219878 is 7 29 . 8 33 . 31 . is 128 The third is The fourth Continued Fractions – p. 23 The Calendar The second convergent for 0.24219878 is 7 29 . 8 33 . 31 . is 128 The third is The fourth Gregorian calendar uses 97 400 Continued Fractions – p. 23 The Calendar The second convergent for 0.24219878 is 7 29 . 8 33 . 31 . is 128 The third is The fourth Gregorian calendar uses 97 400 Errors: Gregorian calendar: 1 day in 3300 years Continued Fractions – p. 23 The Calendar The second convergent for 0.24219878 is 7 29 . 8 33 . 31 . is 128 The third is The fourth Gregorian calendar uses 97 400 Errors: Gregorian calendar: 1 day in 3300 years 2rd convergent: 1 day in 1300 years Continued Fractions – p. 23 The Calendar The second convergent for 0.24219878 is 7 29 . 8 33 . 31 . is 128 The third is The fourth Gregorian calendar uses 97 400 Errors: Gregorian calendar: 1 day in 3300 years 2rd convergent: 1 day in 1300 years 3rd convergent: 1 day in 4400 years Continued Fractions – p. 23 The Calendar The second convergent for 0.24219878 is 7 29 . 8 33 . 31 . is 128 The third is The fourth Gregorian calendar uses 97 400 Errors: Gregorian calendar: 1 day in 3300 years 2rd convergent: 1 day in 1300 years 3rd convergent: 1 day in 4400 years 4th convergent: 1 day in 88000 years Continued Fractions – p. 23 A New Calendar? Add one day every fourth year, as usual, Continued Fractions – p. 24 A New Calendar? Add one day every fourth year, as usual, and then every 128th year, don’t have the extra day. Continued Fractions – p. 24 A New Calendar? Add one day every fourth year, as usual, and then every 128th year, don’t have the extra day. This gives 31 128 (the 4th convergent). Continued Fractions – p. 24 A New Calendar? Add one day every fourth year, as usual, and then every 128th year, don’t have the extra day. This gives 31 128 (the 4th convergent). It’s more accurate than the Gregorian calendar, and simpler. Continued Fractions – p. 24 A New Calendar? Add one day every fourth year, as usual, and then every 128th year, don’t have the extra day. This gives 31 128 (the 4th convergent). It’s more accurate than the Gregorian calendar, and simpler. (but they won’t do it!) Continued Fractions – p. 24 History: a coincidence The Gregorian calendar was introduced by Pope Gregory in 1582 in Catholic countries, and followed later in 1752 in England, Ireland, (most of) US. Continued Fractions – p. 25 History: a coincidence The Gregorian calendar was introduced by Pope Gregory in 1582 in Catholic countries, and followed later in 1752 in England, Ireland, (most of) US. Continued Fractions were first used by Bombelli to find an √ approximation to 13, also in the 16th century, although systematic understanding had to wait until Leonhard Euler in the 18th century. Continued Fractions – p. 25 History: a coincidence The Gregorian calendar was introduced by Pope Gregory in 1582 in Catholic countries, and followed later in 1752 in England, Ireland, (most of) US. Continued Fractions were first used by Bombelli to find an √ approximation to 13, also in the 16th century, although systematic understanding had to wait until Leonhard Euler in the 18th century. The End Continued Fractions – p. 25 Further information Much more information on this beautiful subject can be found by searching the internet. Eg, www.cut-the-knot.org A copy of this talk can is available online at www.maths.manchester.ac.uk/jm/CF.pdf Continued Fractions – p. 26