An analytical exercise

5 Pages·1776·0.053 MB·English

by  Leonhard Euler

An analytical exercise∗
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0
0
Leonhard Euler†
2

c
e
D

8 1. In considering the infinite product for expressing cosine in terms of

any angle, which is

1
v
π 1 1 1 1
4 cos = (1− )(1− )(1− )(1− ) etc.,
2n nn 9nn 25nn 49nn
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1
it comes to mind to investigate a method which itself works by the nature
2
1 of the value of this product, which we in fact already know to be equal to
4
cos π . Indeed, it is possible, though with many labors which were brought
0 2n
/ about by artifice, for a geometric demonstration of this to be given which I
O
do not consider at all unpleasant.
H
.
h 2. Therefore I set,
t
a
m 1 1 1
S = (1− )(1− )(1− ) etc.,
: nn 9nn 25nn
v
i
X and then the logarithms produced by me are summed:
r
a 1 1 1
lS = l(1− )+l(1− )+l(1− )+ etc.,
nn 9nn 25nn
∗Delivered October 3, 1776. Originally published as Exercitatio analytica, Nova Acta
Academiae Scientarum Imperialis Petropolitinae 8 (1794), 69-72, and republished in
Leonhard Euler, Opera Omnia, Series 1: Opera mathematica, Volume 16, Birkh¨auser,
1992. A copy of the original text is available electronically at the Euler Archive, at
www.eulerarchive.org. This paper is E664 in the Enestr¨om index.
†Date of translation: December 8, 2004. Translated from the Latin by Jordan Bell,
2nd year undergraduate in Honours Mathematics, School of Mathematics and Statistics,
Carleton University, Ottawa, Ontario, Canada. Email: [email protected]. Part
of this translation was written during an NSERC USRA supervised by Dr. B. Stevens.
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and since
1 1 1 1 1
l(1− ) = − − − − −etc.,
x x 2xx 3x3 4x4
the successive series set out with their signs changed are:
1 1 1 1
−lS = + + + +etc.
nn 2n4 3n6 4n8
1 1 1 1
+ + + + +etc.
9nn 2·92n4 3·93n6 4·94n8
1 1 1 1
+ + + + +etc.
25nn 2·252n4 3·253n6 4·254n8
1 1 1 1
+ + + + +etc.
49nn 2·492n4 3·493n6 4·494n8
etc.
3. But now if we arrange each of the rows vertically in columns, we will
obtain the following series for −lS:
1 1 1 1 1
−lS = (1+ + + + +etc.)
nn 32 52 72 92
1 1 1 1 1
+ (1+ + + + +etc.)
2n4 34 54 74 94
1 1 1 1 1
+ (1+ + + + +etc.
3n6 36 56 76 96
1 1 1 1 1
+ (1+ + + + +etc.
4n8 38 58 78 98
etc.
Thus in this way, the investigation is led to the summation of series of even
powers of the harmonic progression, 1, 1, 1, 1, etc.
3 5 7
4. By putting π = ρ for the sake of brevity, it can then be seen that the
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sums of powers can be represented in the following manner:
1 1 1
1+ + + +etc. = Aρ2,
32 52 72
1 1 1
1+ + + +etc. = Bρ4,
34 54 74
1 1 1
1+ + + +etc. = Cρ6,
36 56 76
etc.
2
with the first one taken as A = 1, and then the other letters determined by
2
the preceding ones in the following way:
2 2 2
B = A2,C = ·2AB,D = (2AC +BB),
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2 2
E = (2AD +2BC),F = (2AE +2BD+CC),etc.
9 11
the truth of which will soon shine forth from a most beautiful conjunction of
analysis.
5. With the substituted values, this series is obtained:
aρ2 1 Bρ4 1 Cρ6 1 Dρ8
−lS = + · + · + · +etc.
nn 2 n4 3 n6 4 n8
If we put ρ = x, so that x = π , then this series is obtained:
n 2n
1 1 1
−lS = Axx+ Bx4 + Cx6 + Dx8 +etc.
2 3 4
So that we can get rid of the fractions 1, 1, 1, etc., we differentiate, and then
2 3 4
after dividing the result by 2∂x it follows,
∂S
− = Ax+Bx3 +Cx5 +Dx7 +etc.
2S∂x
6. Then for the sake of brevity we set − ∂S = t, so that we have
2S∂x
t = Ax+Bx3 +Cx5 +Dx7 +etc.,
for which the square is seen to be this series:
tt =A2xx +2ABx4 +2ACx6 +2ADx8 +2AEx10 +etc.
+BB +2BC +2BD +etc.
+CC +etc.
and in this, each of the powers of x are the very same as those found in the
formulas which express the determination of the letters A,B,C,D: the only
thing missing are the coefficients 2, 2, 2, etc.
3 5 7
3
7. But we can indeed introduce these coefficients by integrating after we
have multiplied through by 2∂x. So it will be obtained that
2 2 2
2 tt∂x = A2x3 + ·2ABx5 + (2AC +BB)x7+
Z 3 3 7
2 9
(2AD+2BC)x9 + (2AE +2BD+CC)x11 +etc.
9 11
Since at present,
2 2 2
A2 = B, ·2AB = C, (2AC +BB) = D,etc.,
3 3 7
with these values substituted in, we then attain this series:
2 tt∂x = Bx3 +Cx5 +Dx7 +Ex9 +etc.
Z
8. Therefore with the series we had from before:
t = Ax+Bx3 +Cx5 +Dx7 +etc.,
this equation clearly follows:
t = Ax+2 tt∂x,
Z
which when differentiated yields
1 1
∂t = A∂x+2tt∂x = ∂x+2tt∂x, because A = .
2 2
From this we therefore have 2∂t = ∂x(1 +4tt), from which it is ∂x = 2∂t ,
1+4tt
whose integral is at hand, namely x = A tang.2t,1 for which the addition of
a constant is not desired, since by setting x = 0 then t should simultaneously
vanish. Hence with this equation that has been discovered, if the quantity x
isseen asananglethenitwillbe2t = tang.x. Sinceitwasalreadyt = − ∂S ,
2S∂x
from this are deduced these equations:
∂S ∂S ∂xsinx
− = tang.x, and − = .
S∂x S cosx
1Translator: Euler uses here A tang. 2t to express the arctangent of 2t, and later uses
tang. x to stand for the tangent of x.
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9. Therefore, since it is ∂xsinx = −∂cosx, it will be ∂S = ∂cosx, and
S cosx
thenby integrating, lS = lcosx+C, which tobeconsistent shouldbedefined
such that setting x = 0 makes lS = 0. Thus it will be C = 0, so lS = lcosx,
and then so that the numbers can come forth, S = cosx.
10. Moreover, since we already had that x = π , from this it is clear
2n
that the value being sought out for S is revealed to be S = cos π , which
2n
agrees totally with what was known from before. Therefore, analysis most
excellently confirms the former relation between the letters A,B,C,D, which
I introduced in earlier calculations.
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