# 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?

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
nexpressionvalueerrorauthor
111. .42EF
2.1.2 .6.053BC
33–1/2 .5773.00013EF
4(4–.3+.1)2 .577221.04E–5RS
5(4–3+.1.5)–2 .5772153–2.72E–7RS
6(.6((.1+.4)–.3)–.2).5.577215656–8.38E–9GB
73^(–.5–2^(–4–.7^(.1–6))) .5772156661.21E–9RS
8(.4^((6+3*.8)^–7)+.1^.5)^–2 .5772156648–8.10E–12RS
9.8^(.2674^9–1/3)–.5 .5772156648–4.25E–12RS

Approximations of e using 1–n
nexpressionvalueerrorauthor
11 1.–1.7EF
21+2 3..28EF
33–.1–.2 2.70–.018EF
42(.1+.3)–.4 2.718299.16E–6RS
53.5+4–(.12) 2.7182841.68E–6GB
6((.4+6.2)*5–1)/3 2.71828180–2.66E–8BC
72^((.3+.1)^–.4–5^–7.6) 2.718281812–1.57E–8RS
8(1+2^–76)^(4^38+.5) 2.7182818283.96E–47RS
9(1+9–47*6)3285 2.718281828–2.01E–18457734525360901453873570RS

Approximations of π using 1–n
nexpressionvalueerrorauthor
11 1.–2.1EF
21+2 3.0–.14EF
3(1/.3)–.2 3.13–.0082BC
4.4–(2.1).3 3.14156–3.26E–5GB
53+.2/(.4.1+.5)3.1415985.92E–6BC
6(4+6/(3(.2)(.1))).53.14159263–1.38E–8BC
73+1/(7+4(6–5–2))3.141592694.33E–8RS
82^(5^.4)–.6–.1^(38/7) 3.1415926545.15E–10RS
92^5^.4–.6–(.3^9/7)^(.8^.1) 3.141592653596.60E–13RS

Approximations of Feigenbaum's constant 4.66920160910299 using 1–n
nexpressionvalueerrorauthor
11 1.–3.6RS
21/.2 5..33BC
3(.1)–2/3 4.64–.027RS
44*1+2/3 4.666–.0025RS
5(.3–.1+.4)2/.5 4.669232.94E–5RS
6(.3–(.61.5)–.4)2 4.66920171.27E–7GB
7.2 + (4 + .5(–13/6)).7 4.669201618.59E–9GB
84+.8^(3*.6)–.2^(7+.1^.5) 4.669201611.01E–8RS
94+.8^(2*.9)–(7*6)^–3.15 4.669201608–3.04E–10RS

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.
constantdigitsexpressionvalueerrorauthor
γ.57721(1+(7/.7)^–.5)^–2.5772153–2.72E–7RS
e2.71 (2–.1)/.72.714–3.99E–3RS
π3.141 (.4^.1)/.3+.13.1414–1.14E–4RS

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
expressionvalueerrorauthor
1.6.6 .023AZ
23–.5 .5773.00013EF
33–1/2 .5773.00013EF
4(1+.1^.5)^–2 .5772154–2.72E–7GB
55*.3^(.6^–(.8^–.6)) .5772154–2.57E–9GB
6((6–(.1^(–.7)))/((.1^(–.4))–.8)) .5772156642–6.27E–10AZ

Approximations of e using any digits
expressionvalueerrorauthor
13 3..28EF
22.7 2.70–.018EF
32.4–.42.718299.16E–6RS
4(.2+6^.2)/.6 2.71828180–2.66E–8GB
5(1+9–9)99 2.718281825–3.50E–9RS
6(1+9–9)99+.5 2.7182818281.50E–18RS

Approximations of π using any digits
expressionvalueerrorauthor
13 3.–.14EF
2.1–.5 3.16.026RS
3.8–4+.7 3.1414–1.86E–4RS
425.4–.6 3.1415963.72E–6RS
5.8^–(.1^–.1+5)–.9 3.141592683.36E–8GB
6((7–(.3^(–.7)))^(.7^(.8^.8))) 3.141592656–3.14E–9AZ

Approximations of Feigenbaum's constant 4.66920160910299 using any digits
expressionvalueerrorauthor
15 5..33GB
29^.7 4.656–.014GB
36^.86 4.6683.50E–4GB
49–6^(.8^.9) 4.669198–3.33E–6GB
5(8^.7)^(.6^–.6–.3) 4.6692016009.50E–10GB
6 8^((.6^(.1–.7)–.3)*.7) 4.6692016009.50E–10GB

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
nexpressionvalueerrorauthor
111. .42RS
2.1^.2 .6.053RS
31.2^–3 .578.0014RS
4 (1–2^.3)/–.4 .57786.45E–4RS
5(1+.2^.3/4)*.5 .5771–8.64E–5RS
61/(2–.3+.4^.5–.6) .5772153–2.72E–7RS
7(.1–(23^–.45–.6))^.7 .5772155–7.13E–8GB
8((.1+(2^(3^.4))*5*.6)^.7)/8 .577215671.34E–8GB
9.1^–(.2/(.3–4^–(.5^–((6*7)^–.8))–.9)) .5772156644–5.01E–10GB

Approximations of e using 1–n in order
nexpressionvalueerrorauthor
11 1.–1.7RS
21+2 3..28RS
31+2–.3 2.70–.018RS
4.1+2+.3^.4 2.717–4.80E–4RS
5–.1+(–.2+.3)^–.45 2.71831.01E–4RS
6(1+2)^(.3+.4^5+.6) 2.7182842.30E–6RS
7–((.1–2)/.3^.4–.5+6/7) 2.718281852.17E–8GB
8((.1^–.2)–((3–.4)^(.5–.6))/.7)^–.8 2.718281825–2.77E–9GB
9(1+2^(–3*(4+5)))^(.6*.7+8^9) 2.718281826–1.62E–9RS

Approximations of π using 1–n in order
nexpressionvalueerrorauthor
11 1.–2.1RS
21+2 3.0–.14RS
3.1^(–.2–.3) 3.16.020RS
4(.1+2^.3)^4 3.13–.0018RS
5(.1/.2^.3^4)^–.5 3.14171.39E–4RS
6.1^(–2/3)+4.5–6 3.14158–3.81E–6RS
7(1+2^–.3)^.4^(–.5/.6^.7) 3.1415924–1.88E–7RS
8.1^(–2/3)+(4/.5)^–6–.7–.8) 3.14159264–5.28E–9GB
9(.1+2^–(3–(4^–((.5+6^–7)^8))))*9 3.141592622.43E–8GB

Approximations of Feigenbaum's constant 4.66920160910299 using 1–n in order
nexpressionvalueerrorauthor
11 1.–3.6RS
21/.2 5..33RS
3(.1)–2/3 4.64–.027RS
41*2/3+4 4.666–.0025RS
5.1^.2*(.3–4)/–.5 4.6690–1.17E–4RS
61/(.2–3^–4)–.5^.6 4.6691–8.19E–6RS
7.12*(.3+(4+.5^–.6)*7) 4.66920193.07E–7GB
8.1^–(2/((3/4.5)^(6–.7–8))) 4.6692016302.04E–8GB
91+2^((3–.4*(.5–.6^7)–.8)^.9) 4.6692016134.02E–9GB

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 through 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 \ k234567
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 \ k234567
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 \ k234567
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 \ k234567
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.