Problem of the Month (August 2004)

This month's problem is to approximate famous mathematical constants using only the first n digits (each used exactly once) and the mathematical symbols + – * / ( ) . and ^. For each n, what is the best approximation you can get to the following constants?

What other constants can you approximate well in this manner?


ANSWERS

Richard Sabey, Joseph DeVincentis, Jeremy Galvagni, Bill Clagett, Bertrand Winter, Berend Jan van der Zwaag, Ambrus Zsbán, and Gerrit de Blaauw sent answers this month, with Richard Sabey and Gerrit de Blaauw furnishing the vast majority of results.

Approximations of γ using 1–n
n expression value error author
1 1 1. .42 EF
2 .1.2 .6 .053 BC
3 3–1/2 .5773 .00013 EF
4 (4–.3+.1)2 .57722 1.04E–5 RS
5 (4–3+.1.5)–2 .5772153 –2.72E–7 RS
6 (.6((.1+.4)–.3)–.2).5 .577215656 –8.38E–9 GB
7 3^(–.5–2^(–4–.7^(.1–6))) .577215666 1.21E–9 RS
8 (.4^((6+3*.8)^–7)+.1^.5)^–2 .5772156648 –8.10E–12 RS
9 .8^(.2674^9–1/3)–.5 .5772156648 –4.25E–12 RS

Approximations of e using 1–n
n expression value error author
1 1 1. –1.7 EF
2 1+2 3. .28 EF
3 3–.1–.2 2.70 –.018 EF
4 2(.1+.3)–.4 2.71829 9.16E–6 RS
5 3.5+4–(.12) 2.718284 1.68E–6 GB
6 ((.4+6.2)*5–1)/3 2.71828180 –2.66E–8 BC
7 2^((.3+.1)^–.4–5^–7.6) 2.718281812 –1.57E–8 RS
8 (1+2^–76)^(4^38+.5) 2.718281828 3.96E–47 RS
9 (1+9–47*6)3285 2.718281828 –2.01E–18457734525360901453873570 RS

Approximations of π using 1–n
n expression value error author
1 1 1. –2.1 EF
2 1+2 3.0 –.14 EF
3 (1/.3)–.2 3.13 –.0082 BC
4 .4–(2.1).3 3.14156 –3.26E–5 GB
5 3+.2/(.4.1+.5) 3.141598 5.92E–6 BC
6 (4+6/(3(.2)(.1))).5 3.14159263 –1.38E–8 BC
7 3+1/(7+4(6–5–2)) 3.14159269 4.33E–8 RS
8 2^(5^.4)–.6–.1^(38/7) 3.141592654 5.15E–10 RS
9 2^5^.4–.6–(.3^9/7)^(.8^.1) 3.14159265359 6.60E–13 RS

Approximations of Feigenbaum's constant 4.66920160910299 using 1–n
n expression value error author
1 1 1. –3.6 RS
2 1/.2 5. .33 BC
3 (.1)–2/3 4.64 –.027 RS
4 4*1+2/3 4.666 –.0025 RS
5 (.3–.1+.4)2/.5 4.66923 2.94E–5 RS
6 (.3–(.61.5)–.4)2 4.6692017 1.27E–7 GB
7 .2 + (4 + .5(–13/6)).7 4.66920161 8.59E–9 GB
8 4+.8^(3*.6)–.2^(7+.1^.5) 4.66920161 1.01E–8 RS
9 4+.8^(2*.9)–(7*6)^–3.15 4.669201608 –3.04E–10 RS

Corey Plover suggested the following variant: use the first n decimal digits of a constant instead of the digits 1 through n. He asks, for a given constant, what is the smallest n that gives a closer approximation than the trivial decimal expansion. Here are some answers.

constant digits expression value error author
γ .57721 (1+(7/.7)^–.5)^–2 .5772153 –2.72E–7 RS
e 2.71 (2–.1)/.7 2.714 –3.99E–3 RS
π 3.141 (.4^.1)/.3+.1 3.1414 –1.14E–4 RS

This got me thinking. What is the best approximation to these constants using ANY n digits? Here's the best known:

Approximations of γ using any digits
expression value error author
1 .6 .6 .023 AZ
2 3–.5 .5773 .00013 EF
3 3–1/2 .5773 .00013 EF
4 (1+.1^.5)^–2 .5772154 –2.72E–7 GB
5 5*.3^(.6^–(.8^–.6)) .5772154 –2.57E–9 GB
6 ((6–(.1^(–.7)))/((.1^(–.4))–.8)) .5772156642 –6.27E–10 AZ

Approximations of e using any digits
expression value error author
1 3 3. .28 EF
2 2.7 2.70 –.018 EF
3 2.4–.4 2.71829 9.16E–6 RS
4 (.2+6^.2)/.6 2.71828180 –2.66E–8 GB
5 (1+9–9)99 2.718281825 –3.50E–9 RS
6 (1+9–9)99+.5 2.718281828 1.50E–18 RS

Approximations of π using any digits
expression value error author
1 3 3. –.14 EF
2 .1–.5 3.16 .026 RS
3 .8–4+.7 3.1414 –1.86E–4 RS
4 25.4–.6 3.141596 3.72E–6 RS
5 .8^–(.1^–.1+5)–.9 3.14159268 3.36E–8 GB
6 ((7–(.3^(–.7)))^(.7^(.8^.8))) 3.141592656 –3.14E–9 AZ

Approximations of Feigenbaum's constant 4.66920160910299 using any digits
expression value error author
1 5 5. .33 GB
2 9^.7 4.656 –.014 GB
3 6^.86 4.668 3.50E–4 GB
4 9–6^(.8^.9) 4.669198 –3.33E–6 GB
5 (8^.7)^(.6^–.6–.3) 4.669201600 9.50E–10 GB
6 8^((.6^(.1–.7)–.3)*.7) 4.669201600 9.50E–10 GB

And this got Richard Sabey thinking. What is the best approximation to these constants using the digits 1 through n in that order? Here's the best known:

Approximations of γ using 1–n in order
n expression value error author
1 1 1. .42 RS
2 .1^.2 .6 .053 RS
3 1.2^–3 .578 .0014 RS
4 (1–2^.3)/–.4 .5778 6.45E–4 RS
5 (1+.2^.3/4)*.5 .5771 –8.64E–5 RS
6 1/(2–.3+.4^.5–.6) .5772153 –2.72E–7 RS
7 (.1–(23^–.45–.6))^.7 .5772155 –7.13E–8 GB
8 ((.1+(2^(3^.4))*5*.6)^.7)/8 .57721567 1.34E–8 GB
9 .1^–(.2/(.3–4^–(.5^–((6*7)^–.8))–.9)) .5772156644 –5.01E–10 GB

Approximations of e using 1–n in order
n expression value error author
1 1 1. –1.7 RS
2 1+2 3. .28 RS
3 1+2–.3 2.70 –.018 RS
4 .1+2+.3^.4 2.717 –4.80E–4 RS
5 –.1+(–.2+.3)^–.45 2.7183 1.01E–4 RS
6 (1+2)^(.3+.4^5+.6) 2.718284 2.30E–6 RS
7 –((.1–2)/.3^.4–.5+6/7) 2.71828185 2.17E–8 GB
8 ((.1^–.2)–((3–.4)^(.5–.6))/.7)^–.8 2.718281825 –2.77E–9 GB
9 (1+2^(–3*(4+5)))^(.6*.7+8^9) 2.718281826 –1.62E–9 RS

Approximations of π using 1–n in order
n expression value error author
1 1 1. –2.1 RS
2 1+2 3.0 –.14 RS
3 .1^(–.2–.3) 3.16 .020 RS
4 (.1+2^.3)^4 3.13 –.0018 RS
5 (.1/.2^.3^4)^–.5 3.1417 1.39E–4 RS
6 .1^(–2/3)+4.5–6 3.14158 –3.81E–6 RS
7 (1+2^–.3)^.4^(–.5/.6^.7) 3.1415924 –1.88E–7 RS
8 .1^(–2/3)+(4/.5)^–6–.7–.8) 3.14159264 –5.28E–9 GB
9 (.1+2^–(3–(4^–((.5+6^–7)^8))))*9 3.14159262 2.43E–8 GB

Approximations of Feigenbaum's constant 4.66920160910299 using 1–n in order
n expression value error author
1 1 1. –3.6 RS
2 1/.2 5. .33 RS
3 (.1)–2/3 4.64 –.027 RS
4 1*2/3+4 4.666 –.0025 RS
5 .1^.2*(.3–4)/–.5 4.6690 –1.17E–4 RS
6 1/(.2–3^–4)–.5^.6 4.6691 –8.19E–6 RS
7 .12*(.3+(4+.5^–.6)*7) 4.6692019 3.07E–7 GB
8 .1^–(2/((3/4.5)^(6–.7–8))) 4.669201630 2.04E–8 GB
9 1+2^((3–.4*(.5–.6^7)–.8)^.9) 4.669201613 4.02E–9 GB

Richard Sabey also sent results for expressions without using the decimal point, both in any order and in ascending order. Joseph DeVincentis sent results that use the digits 0–n. And thanks to Berend Jan van der Zwaag for finding a bunch of typos.

In 2008, I extended these results by considering the best approximations to various constants using n copies of a given digit. Here are the best results.

Best Approximations to π with n Copies of the Digit k
n \ k 2 3 4 5 6 7
1 1 + 1
= 2
1 + 1 + 1
= 3
(. 1)–1/(1+1)
= 3.162
1 + (1 + 1)1.1
= 3.143
.1^-(.1+(.1^.1)/(1+1))
= 3.1416 (GB)
1.1^(11+(1-.1)^-.1)
= 3.141598 (GB)
2 2 + 2
= 4
2 + 2(. 2)
= 3.149
22(. 2)(. 2)
= 3.144
2 + 2 – 2–(. 22)
= 3.1414
2*(.2+.2+2.2^.2)
= 3.1416 (GB)
2^(2-.2^-(.2^.2-.2^-.2))
= 3.14158 (GB)
3 3.3
= 3.3
3 + (. 3)/3
= 3.1
333(. 3)
= 3.140
3 × [(3 – . 3)(. 3) – . 3]
= 3.1413
.3^-((3-.3)^-((3-.3)^-3))
= 3.14157 (GB)
3/(.33/(.3-3^-3)-.3)
= 3.1415929 (GB)
4 4 – (. 4)
= 3.6
4 – (. 4) – (. 4)
= 3.2
(. 4)–4(. 4)×(. 4)
= 3.139
4 – 4–(. 44)/4
= 3.1414
((.4+.4)^-(.4^-(.4^4)))/.4
= 3.141594 (GB)
(4-4/(.4-.4*4^.4))^.4
= 3.141593 (GB)
5 5 × (. 5)
= 2.5
(5 + 5)(. 5)
= 3.162
5 / [5 × (. 5)](. 5)
= 3.162
5 – (. 5)–[5–(. 5)/(. 5)]
= 3.1411
.5+(5^.5)^(.5+.5^.5)
= 3.141591 (GB)
.5+.5+.5^-((.5+.5^.5)^.5)
= 3.14159268 (GB)
6 6(. 6)
= 2.930
6.6(. 6)
= 3.103
[6 + (. 6)(. 6)](. 6)
= 3.140
(. 6)[6 ×(. 6) –(. 6)–(. 6)]
= 3.142
6*6*.666^6
= 3.1415 (GB)
(6^.6)/(.6^-(6^-(.6/6))-.6)
= 3.141593 (GB)
7 7(. 7)
= 3.905
7 – 7(. 7)
= 3.095
[(. 7)–7 – 7](. 7)
= 3.147
77/[7+7×(. 7)]
= 3.1413
(7*7)^((7*.7)^-.77)
= 3.14158 (GB)
7^-(.7-7^(.7^(7-.7^-.7)))
= 3.1415927 (GB)
8 (. 8) + (. 8)
= 1.6
(. 8)–8(. 8)
= 3.247
(. 8)–8×(. 8)×(. 8)
= 3.135
8(. 8) / [(. 8) + (. 88)]
= 3.1416
(8/((.8+.8)/.8^.8))^.8
= 3.1416 (GB)
(8^-((.8+.8)^-(8^(.8^.8))))^-8
= 3.1415929 (GB)
9 (. 9)–9
= 2.581
(. 9)–9/(. 9)
= 2.868
9 × (. 9) × (. 9)9
= 3.138
(. 9) × [(. 9)(. 9) + (. 9)–9]
= 3.1416
(9-.99*.9)*.9^9
= 3.1415927 (GB)
(.9-.9*(.99-9))*.9^9
= 3.1415927 (GB)

Best Approximations to e with n Copies of the Digit k
n \ k 2 3 4 5 6 7
1 1 + 1
= 2
1 + 1 + 1
= 3
1 + 1 + (. 1)(. 1)
= 2.794
(1 + 1) × [(. 1) + (. 1)–(. 1)]
= 2.717
1-.1+(.1+.1)/.11
= 2.7181 (GB)
(1+(.1+.1)^-.1)/(1-.1-.1)
= 2.71827 (GB)
2 2 + (. 2)
= 2.2
2 + (. 2)(. 2)
= 2.725
(. 2)[(.2)–(. 2)–2]
= 2.713
2[2×(. 2)]–2×(. 2)
= 2.71829
2^(((2^.2)*.2^.2)^-2)
= 2.71829 (GB)
2+(2*(2+2^-2))^-.22
= 2.71827 (GB)
3 3 – (. 3)
= 2.7
3 × 3 × (. 3)
= 2.7
3 × 3–(. 3)×(. 3)
= 2.717
[(. 3)–3 – 3 × 3](. 3)
= 2.7184
(3-.3+.3^-3)^.3-.3
= 2.71827 (GB)
3-.3^(.333+3^-.3)
= 2.7182817 (GB)
4 4 – (. 4)
= 3.6
4 × (. 4)(. 4)
= 2.772
(. 4)–[(. 4)+(. 4)(. 4)]
= 2.722
[(4 + 4) / 4](. 4)–(. 4)
= 2.71829
.4+4^(4^-((.4^-.4)/4))
= 2.718287 (GB)
4^((4-.4/4)^-(.4-.4*.4))
= 2.7182815 (GB)
5 5 × (. 5)
= 2.5
(. 5) + 5(. 5)
= 2.736
55(. 5)×(. 5)
= 2.723
5 × (. 55) – (. 5)5
= 2.7187
.5^-((5*.5)^(.5-.5/5))
= 2.71829 (GB)
(.5+.5+5^-5)^(.5+5^5)
= 2.71828185 (GB)
6 6(. 6)
= 2.930
[6 – (. 6)](. 6)
= 2.751
(. 6)–(. 6) + (. 6)–(. 6)
= 2.717
(. 6)–[(. 6)×6(. 66)]
= 2.7183
(6/6+6^-6)^(6^6)
= 2.71825 (GB)
(.6+.6+6^(.6+.6))/(6*.6)
= 2.71828180 (GB)
7 7(. 7)
= 3.905
(. 7) / 7–(. 7)
= 2.733
7.7(. 7)×(. 7)
= 2.7188
7[(. 77)–7–(. 7)]
= 2.71822
.7^-((7+7*7)^(7^-.7))
= 2.7183 (GB)
(.7^-((7+7)^(7^-.77)))/.7
= 2.7182817 (GB)
8 (. 8) + (. 8)
= 1.6
8 – 8(. 8)
= 2.722
(. 8)[(. 8)–8(. 8)]
= 2.716
(. 8) + 8 × (. 8)8×(. 8)
= 2.7180
.8*(.8+.8^-8)^(.8*.8)
= 2.7181 (GB)
(8/(8-8^-8))^(8/8^-8)
= 2.71828183 (GB)
9 (. 9)–9
= 2.581
(. 9) + (. 9) + (. 9)
= 2.7
9–(. 9) + (. 9)–9
= 2.719
(. 9) + (. 9) + (. 9)(. 9)×(. 9)
= 2.7181
9*(.9+.9)-(9+9)^.9
= 2.7182815 (GB)
(.9/(.9+9^-9))^-(.9/9^-9)
= 2.718281824 (GB)

Best Approximations to γ with n Copies of the Digit k
n \ k 2 3 4 5 6 7
1 (. 1)(. 1)
= .794
[(. 1) × (. 1)](. 1)
= .630
(1 + 1)–(. 1)(. 1)
= .576
(. 1) / (. 11)(. 1)(. 1)
= .5773
.1*((.1+.1)^-1.1-.1)
= .5773 (GB)
(1+.1^(1/(1+1)))^-(1+1)
= .57723 (GB)
2 (. 2)(. 2)
= .724
2(. 2) / 2
= .574
2 – (. 2)–(. 22)
= .575
[(. 2) / (2 – (. 2))]2–2
= .5773
.2+.2^(.2^(.2^(.2^.2)))
= .577218 (GB)
2^-((.2^-.2-.2^-(.2^2))^.2)
= .5772151 (GB)
3 (. 3) + (. 3)
= .6
(3 + 3)–(. 3)
= .584
3–3/[3+3]
= .5773
(. 3)–(. 3) – [(. 3) + (. 3)](. 3)
= .5771
3*(.3^(.3^.3))-3^-.3
= .57722 (GB)
.3/.3^((3+(.3+.3)^-3)^-.3)
= .5772157 (GB)
4 4–(. 4)
= .574
(. 4) / (. 4)(. 4)
= .5770
(. 4)4×(. 4) / (. 4)
= .5770
[(. 4)(. 4) – (. 44)](. 4)
= .57722
.4*(.4+(4/(4-.4))^.4)
= .577217 (GB)
(44+4)^-((4-.4)^-4)-.4
= .5772154 (GB)
5 (. 55)
= .55
(. 555)
= .555
[5 × (. 5) + (. 5)]&ndash(. 5)
= .5773
(. 5) / (. 5)[(. 5)(. 5)–(. 5)]
= .5771
.5^(.5+(.5^-.55)/5)
= .5772158 (GB)
.5/(.5+5^-((5^(.5^.5))/5))
= .5772158 (GB)
6 (. 66)
= .66
6–(. 6) / (. 6)
= .569
[6 / 6 – (. 6)](. 6)
= .5770
[66–6 – (. 6)](. 6)
= .5771
.6-(6^.6)*((.6^6)/6)
= .5772151 (GB)
(.6+(6-(.6^-.6)/6)^.6)/6
= .5772158 (GB)
7 (. 7) × (. 7)
= .49
7 × 7(. 7)
= .576
[(. 7) × (. 7)](. 77)
= .5773
(. 7) – [(. 7) / (7 + 7)](. 7)
= .5771
(.7^(.7*(.7+.7)))^-.7-.7
= .577211 (GB)
.7-((7-.7)^-(7^-(.7^-.7)))/7
= .5772158 (GB)
8 (. 8) × (. 8)
= .64
(. 8) – 8–(. 8)
= .610
8–8–(. 8)×(. 8)
= .57723
[(. 8) × (. 8)]8(. 8)/8
= .57727
(.8*8^(.8^8.8))^-8
= .577216 (GB)
((8^-((.8^8)*(.8^.8)))/.8)^8
= .577216 (GB)
9 (. 9)9
= .387
[(. 9) + (. 9)]–(. 9)
= .589
[(. 9) × (. 9)](. 9)–9
= .580
[(. 9) + 9–(. 9)] / [(. 9) + (. 9)]
= .576
(9^(.9-.9^(9+9)))/9
= .57723 (GB)
(9-.9/9)/(9+9-.9^-9)
= .577216 (GB)

Best Approximations to Feigenbaum's Constant with n Copies of the Digit k
n \ k 2 3 4 5 6 7
1 1 + 1
= 2
1 / [(. 1) + (. 1)]
= 5
1 / [(. 11) + (. 1)]
= 4.762
(1 + 1)–1.1 / (. 1)
= 4.665
.1^-((.1^.1-.1)^1.1)
= 4.67 (GB)
.1+.1+1/(1-.1^.11)
= 4.66921 (GB)
2 2 + 2
= 4
22.2
= 4.594
2–(. 2)/2 / (. 2)
= 4.665
2 + 2 + (. 2)2–2
= 4.668
.2+(2*2^.2)^(2-.2)
= 4.6691 (GB)
.2^-(.2^(2^-(2+.2^-(.2^.2))))
= 4.669203 (GB)
3 3 + 3
= 6
3(. 3) / (. 3)
= 4.634
(. 33)–3(. 3)
= 4.671
3[3 + (. 3) × (. 3)](. 3)
= 4.6697
3*(.3+.3+.3^(3^-3))
= 4.6691 (GB)
.3^-((.3+.3)^(.3^3))+3^.3
= 4.6692014 (GB)
4 4 + (. 4)
= 4.4
4 + (. 4)(. 4)
= 4.693
4 + (. 4)(. 44)
= 4.668
4 / [(. 4) + (. 4)](. 4)(. 4)
= 4.6690
4+.4^(.4/(.4^(.4/4)))
= 4.6691 (GB)
.44-(4/.4^.4-4/.4)
= 4.669200 (GB)
5 5 – (. 5)
= 4.5
(. 5)–5(. 5)
= 4.711
5 × (. 5)(. 5)/5
= 4.665
[5 – (. 5)–(. 5)–(. 5)] / (. 5)
= 4.6697
5-(5^-(.5*5^.5))/.5
= 4.669205 (GB)
.5^5+5.5^((5-.5)/5)
= 4.669203 (GB)
6 6 – (. 6)
= 5.4
6 – (. 6)–(. 6)
= 4.641
6 – (. 6) – (. 6)(. 6)
= 4.663
6 – 6(. 6)6×(.6)
= 4.670
6^(6/(6+.6^(.6^6)))
= 4.6691 (GB)
6*(.6^((6-.6)/66))^6
= 4.66921 (GB)
7 7 × (. 7)
= 4.9
(. 7) + 7(. 7)
= 4.604
(. 77) + 7(. 7)
= 4.674
(. 7) × [7 – (7×(. 7))–(. 7)]
= 4.6698
7^(.7^(.7^(.7+.7*.7)))
= 4.6691 (GB)
((.77+.7^.7)/(7+7))^-.7
= 4.6692019 (GB)
8 8(. 8)
= 5.278
(. 8) / (. 8)8
= 4.768
(. 88) / 8–(. 8)
= 4.644
[8 / (. 8)](. 8)×(. 8)(. 8)
= 4.668
.8*(8/(.8+.8)+.8^.8)
= 4.669209 (GB)
8/(8^.8-.8^-(8.8^.8))
= 4.6691 (GB)
9 (. 9)–9
= 2.581
9 / [(. 9) + (. 9)]
= 5
[(. 9) + (. 9)] / 9(. 9)
= 4.646
(. 9) + (. 9) + (. 9)–9 / (. 9)
= 4.667
9^.9-.99*.9^-9
= 4.6693 (GB)
.9/(.99+.9-(.9+.9)^.9)
= 4.669207 (GB)

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 5/15/08.