Upload
lyhanh
View
218
Download
2
Embed Size (px)
Citation preview
π is irrational
James A. Swenson
University of Wisconsin – Platteville
April 29, 2011MAA-WI Annual Meeting
James A. Swenson (UWP) π is irrational 4/29/11 1 / 23
Thanks for coming!
I hope you’ll enjoy the talk; please feel free to get involved!
James A. Swenson (UWP) π is irrational 4/29/11 2 / 23
Outline
1 History
2 Plan & Setup
3 Proof
James A. Swenson (UWP) π is irrational 4/29/11 3 / 23
Our topic
π is irrational
The first proof that π cannot be written in the form a/b, where a and bare whole numbers, was given in 1761 by Johann Lambert of Switzerland.
http://en.wikipedia.org/wiki/File:JHLambert.jpg
Some of Lambert’s other ideas
cosh and sinh in trig
area of hyperbolic triangles
map projections
the hygrometer
star systems
James A. Swenson (UWP) π is irrational 4/29/11 4 / 23
Lambert’s proof
I’m not going to show you Lambert’s proof: 58 dense pages.
James A. Swenson (UWP) π is irrational 4/29/11 5 / 23
Easier proofs
Mary Cartwright (Cambridge, 1945) and Ivan Niven (Oregon, 1947)produced proofs that don’t require any methods beyond Calc II.
http://www.lms.ac.uk/newsletter/339/339 12.html image: Konrad Jacobs
James A. Swenson (UWP) π is irrational 4/29/11 6 / 23
Source of this talk:
[1] Li Zhou and Lubomir Markov, Recurrent proofs of the irrationality of certaintrigonometric values, Amer. Math. Monthly 117 (2010), no. 4, 360–362.
Li Zhou, Polk St. College (FL)
http://search.intelius.com/Li-Zhou-MlEq61qz
Lubomir Markov, Barry Univ. (FL)
http://www.barry.edu/marc/faculty/Default.htm
James A. Swenson (UWP) π is irrational 4/29/11 7 / 23
Outline
1 History
2 Plan & Setup
3 Proof
James A. Swenson (UWP) π is irrational 4/29/11 8 / 23
Proving that something is impossible
Theorem
π is not a rational number.
Proof.
Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.
1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.
3 Prove: limn→∞
zn = 0.
Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!
James A. Swenson (UWP) π is irrational 4/29/11 9 / 23
Proving that something is impossible
Theorem
π is not a rational number.
Proof.
Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.
1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.
3 Prove: limn→∞
zn = 0.
Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!
James A. Swenson (UWP) π is irrational 4/29/11 9 / 23
Proving that something is impossible
Theorem
π is not a rational number.
Proof.
Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.
1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.
3 Prove: limn→∞
zn = 0.
Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!
James A. Swenson (UWP) π is irrational 4/29/11 9 / 23
Proving that something is impossible
Theorem
π is not a rational number.
Proof.
Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.
1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.
3 Prove: limn→∞
zn = 0.
Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!
James A. Swenson (UWP) π is irrational 4/29/11 9 / 23
Proving that something is impossible
Theorem
π is not a rational number.
Proof.
Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.
1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.
3 Prove: limn→∞
zn = 0.
Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!
James A. Swenson (UWP) π is irrational 4/29/11 9 / 23
Initial definitions: fn(x)
Definition
For any integer n ≥ 0, let fn(x) =
(πx − x2
)nn!
.
James A. Swenson (UWP) π is irrational 4/29/11 10 / 23
Initial definitions: fn(x)
Remark
On the interval [0, π], fn(x) ≥ 0.
James A. Swenson (UWP) π is irrational 4/29/11 10 / 23
Initial definitions: fn(x)
Remark
On the interval [0, π], fn(x) has its maximum value at x = π/2.
James A. Swenson (UWP) π is irrational 4/29/11 10 / 23
Initial definitions: InDefinition
For any integer n ≥ 0, let In =
∫ π
0fn(x) sin x dx.
James A. Swenson (UWP) π is irrational 4/29/11 11 / 23
What is In?
I0 = 2
I0 =
∫ π
0f0(x) sin(x) dx
=
∫ π
0
(πx−x2)0
0! sin x dx
=
∫ π
0sin x dx
= − cosπ + cos 0
= 1 + 1.
James A. Swenson (UWP) π is irrational 4/29/11 12 / 23
What is In?
I1 = 4
I1 =
∫ π
0
(πx−x2)1
1! sin x dx
IBP= (((((((((
(x2 − πx) cos x∣∣π0
+
∫ π
0(π − 2x) cos x dx
IBP= ((((((((
(π − 2x) sin x |π0 +
∫ π
02 sin x dx
= 2I0
= 4.
James A. Swenson (UWP) π is irrational 4/29/11 13 / 23
What is In?
I1 = 4
I1 =
∫ π
0
(πx−x2)1
1! sin x dx
IBP= (((((((((
(x2 − πx) cos x∣∣π0
+
∫ π
0(π − 2x) cos x dx
IBP= ((((((((
(π − 2x) sin x |π0 +
∫ π
02 sin x dx
= 2I0
= 4.
James A. Swenson (UWP) π is irrational 4/29/11 13 / 23
What is In?
I1 = 4
I1 =
∫ π
0
(πx−x2)1
1! sin x dx
IBP= (((((((((
(x2 − πx) cos x∣∣π0
+
∫ π
0(π − 2x) cos x dx
IBP= ((((((((
(π − 2x) sin x |π0 +
∫ π
02 sin x dx
= 2I0
= 4.
James A. Swenson (UWP) π is irrational 4/29/11 13 / 23
What is In?
I1 = 4
I1 =
∫ π
0
(πx−x2)1
1! sin x dx
IBP= (((((((((
(x2 − πx) cos x∣∣π0
+
∫ π
0(π − 2x) cos x dx
IBP= ((((((((
(π − 2x) sin x |π0 +
∫ π
02 sin x dx
= 2I0
= 4.
James A. Swenson (UWP) π is irrational 4/29/11 13 / 23
What is In?
In in terms of previous values
By the same method (using IBP twice), we find out that when n ≥ 2,
In = (4n − 2)In−1 − π2In−2.
For example:
I2 = (4 · 2− 2)I1 − π2I0 = 24− 2π2
I3 = (4 · 3− 2)I2 − π2I1 = 240− 24π2
I4 = (4 · 4− 2)I3 − π2I2 = 3360− 360π2 + 2π4
James A. Swenson (UWP) π is irrational 4/29/11 14 / 23
The form of In
Theorem
For each n ≥ 0: there is a polynomial gn(x), of degree ≤ n, having integercoefficients, such that In = gn(π). [Moreover, gn(x) is always even.]
What does this mean?
n In gn(x)
0 2 21 4 42 24− 2π2 24− 2x2
3 240− 24π2 240− 24x2
4 3360− 360π2 + 2π4 3360− 360x2 + 2x4
The theorem says that In will always look “like this” – the proof is routine,by strong induction, based on our recurrence:
In = (4n − 2)In−1 − π2In−2
James A. Swenson (UWP) π is irrational 4/29/11 15 / 23
Outline
1 History
2 Plan & Setup
3 Proof
James A. Swenson (UWP) π is irrational 4/29/11 16 / 23
π is irrational: step 1
What if π were rational?
Now suppose, for the sake of contradiction, that π is the rational numbera/b, where a and b are positive integers. Then, for example,
I4 = 3360− 360π2 + 2π4 =3360b4 − 360a2b2 + 2a4
b4
Thus b4I4 is the integer 3360b4 − 360a2b2 + 2a4.
Definition
For each n, let zn = bnIn.
James A. Swenson (UWP) π is irrational 4/29/11 17 / 23
π is irrational: step 1
What if π were rational?
Now suppose, for the sake of contradiction, that π is the rational numbera/b, where a and b are positive integers. Then, for example,
I4 = 3360− 360π2 + 2π4 =3360b4 − 360a2b2 + 2a4
b4
Thus b4I4 is the integer 3360b4 − 360a2b2 + 2a4.
Definition
For each n, let zn = bnIn.
James A. Swenson (UWP) π is irrational 4/29/11 17 / 23
π is irrational: step 2
Theorem
For each n, zn = bnIn is a positive integer.
Proof.
Recall that In =
∫ π
0fn(x) sin x dx , where fn(x) =
(x(π − x))n
n!. When
0 < x < π, both fn(x) and sin x are positive. Thus In > 0, and so zn > 0.
Next, recall that by our earlier theorem, In = gn(π), where gn is apolynomial with integer coefficients. We assumed that π = a/b; sincegn(x) has degree ≤ n, multiplying gn(π) by bn clears the denominators, sozn is an integer.
James A. Swenson (UWP) π is irrational 4/29/11 18 / 23
π is irrational: step 2
Theorem
For each n, zn = bnIn is a positive integer.
Proof.
Recall that In =
∫ π
0fn(x) sin x dx , where fn(x) =
(x(π − x))n
n!. When
0 < x < π, both fn(x) and sin x are positive. Thus In > 0, and so zn > 0.
Next, recall that by our earlier theorem, In = gn(π), where gn is apolynomial with integer coefficients. We assumed that π = a/b; sincegn(x) has degree ≤ n, multiplying gn(π) by bn clears the denominators, sozn is an integer.
James A. Swenson (UWP) π is irrational 4/29/11 18 / 23
π is irrational: step 2
Theorem
For each n, zn = bnIn is a positive integer.
Proof.
Recall that In =
∫ π
0fn(x) sin x dx , where fn(x) =
(x(π − x))n
n!. When
0 < x < π, both fn(x) and sin x are positive. Thus In > 0, and so zn > 0.
Next, recall that by our earlier theorem, In = gn(π), where gn is apolynomial with integer coefficients. We assumed that π = a/b; sincegn(x) has degree ≤ n, multiplying gn(π) by bn clears the denominators, sozn is an integer.
James A. Swenson (UWP) π is irrational 4/29/11 18 / 23
π is irrational: step 3
Theorem
limn→∞
zn = 0.
Proof by the Squeeze Theorem.
0 < zn = bn
∫ π
0
(πx − x2)n
n!sin x dx
≤ 1n!
∫ π
0bn(πx − x2)n dx
≤ 1n!
∫ π
0bn
(π · π
2−(π
2
)2)n
dx
(since the integrand is greatest when x = π/2)
= 1n!
∫ π
0bn
(π2
4
)n
dx
James A. Swenson (UWP) π is irrational 4/29/11 19 / 23
π is irrational: step 3
Theorem
limn→∞
zn = 0.
Proof by the Squeeze Theorem.
0 < zn = bn
∫ π
0
(πx − x2)n
n!sin x dx
≤ 1n!
∫ π
0bn(πx − x2)n dx
≤ 1n!
∫ π
0bn
(π · π
2−(π
2
)2)n
dx
(since the integrand is greatest when x = π/2)
= 1n!
∫ π
0bn
(π2
4
)n
dx
James A. Swenson (UWP) π is irrational 4/29/11 19 / 23
π is irrational: step 3
Theorem
limn→∞
zn = 0.
Proof by the Squeeze Theorem.
0 < zn = bn
∫ π
0
(πx − x2)n
n!sin x dx
≤ 1n!
∫ π
0bn(πx − x2)n dx
≤ 1n!
∫ π
0bn
(π · π
2−(π
2
)2)n
dx
(since the integrand is greatest when x = π/2)
= 1n!
∫ π
0bn
(π2
4
)n
dx
James A. Swenson (UWP) π is irrational 4/29/11 19 / 23
π is irrational: step 3
Theorem
limn→∞
zn = 0.
Proof by the Squeeze Theorem.
0 < zn = bn
∫ π
0
(πx − x2)n
n!sin x dx
≤ 1n!
∫ π
0bn(πx − x2)n dx
≤ 1n!
∫ π
0bn
(π · π
2−(π
2
)2)n
dx
(since the integrand is greatest when x = π/2)
= 1n!
∫ π
0bn
(π2
4
)n
dx
James A. Swenson (UWP) π is irrational 4/29/11 19 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
π is irrational: step 3
Proof (continued).
0 < zn ≤ 1n!
∫ π
0bn
(π2
4
)n
dx
= 1n!
∫ π
0
(π2b
4
)n
dx
= 1n! · π ·
(π2b4
)n= 1
n! ·ab ·(
a2
4b
)n=
a
b· (a2/4b)n
n!,
which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞
zn = 0.
James A. Swenson (UWP) π is irrational 4/29/11 20 / 23
Next steps?
Other things you can prove by this method
1 π2 is irrational. (Since gn(x) is even.)
2 e is irrational. (Use In =
∫ 1
0
(x − x2)n
n!ex dx .)
3 If r 6= 0 is rational, then er is irrational.
(Use In =
∫ r
0
(r x − x2)n
n!ex dx .)
4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)
James A. Swenson (UWP) π is irrational 4/29/11 21 / 23
Next steps?
Other things you can prove by this method
1 π2 is irrational. (Since gn(x) is even.)
2 e is irrational. (Use In =
∫ 1
0
(x − x2)n
n!ex dx .)
3 If r 6= 0 is rational, then er is irrational.
(Use In =
∫ r
0
(r x − x2)n
n!ex dx .)
4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)
James A. Swenson (UWP) π is irrational 4/29/11 21 / 23
Next steps?
Other things you can prove by this method
1 π2 is irrational. (Since gn(x) is even.)
2 e is irrational. (Use In =
∫ 1
0
(x − x2)n
n!ex dx .)
3 If r 6= 0 is rational, then er is irrational.
(Use In =
∫ r
0
(r x − x2)n
n!ex dx .)
4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)
James A. Swenson (UWP) π is irrational 4/29/11 21 / 23
Next steps?
Other things you can prove by this method
1 π2 is irrational. (Since gn(x) is even.)
2 e is irrational. (Use In =
∫ 1
0
(x − x2)n
n!ex dx .)
3 If r 6= 0 is rational, then er is irrational.
(Use In =
∫ r
0
(r x − x2)n
n!ex dx .)
4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)
James A. Swenson (UWP) π is irrational 4/29/11 21 / 23
Next steps?
Other things you can prove by this method
1 π2 is irrational. (Since gn(x) is even.)
2 e is irrational. (Use In =
∫ 1
0
(x − x2)n
n!ex dx .)
3 If r 6= 0 is rational, then er is irrational.
(Use In =
∫ r
0
(r x − x2)n
n!ex dx .)
4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)
James A. Swenson (UWP) π is irrational 4/29/11 21 / 23
Next steps?
One (true) thing you can’t prove by this method
π is transcendental. . .[1] F. Lindemann, Ueber die Zahl π, Math. Ann. 20 (1882), no. 2, 213–225
(German).
James A. Swenson (UWP) π is irrational 4/29/11 22 / 23
Thanks!
[1] Johann Heinrich Lambert, Memoire sur quelques proprietes remarquables desquantites transcendentes circulaires et logarithmiques, Histoire de l’Academie Royaledes Sciences et des Belles-Lettres de Berlin (1761), 265-322.
[2] Ivan Niven, A simple proof that π is irrational, Bull. Amer. Math. Soc. 53 (1947),509.
[3] Li Zhou and Lubomir Markov, Recurrent proofs of the irrationality of certaintrigonometric values, Amer. Math. Monthly 117 (2010), no. 4, 360–362.
Online resourcesBerlin-Brandenburgische Akademie der Wissenschaften: http://bibliothek.bbaw.de/
Spiral staircase: http://commons.wikimedia.org/wiki/File:Wfm mackintosh lighthouse.jpg
Ivan Niven: http://en.wikipedia.org/wiki/File:Ivan Niven.jpg
Mary Cartwright: http://en.wikipedia.org/wiki/File:Dame Mary Lucy Cartwright.jpg
Johann Lambert: http://en.wikipedia.org/wiki/Johann Heinrich Lambert
Ferdinand von Lindemann: http://en.wikipedia.org/wiki/File:Carl Louis Ferdinand von Lindemann.jpg
James A. Swenson (UWP) π is irrational 4/29/11 23 / 23