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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Riemannian Trigonometric
Series
H. Vic Dannon
[email protected]
May, 2011
Abstract:
Riemann derived necessary conditions for the
equality of a function to its Fourier Series, and claimed that
these conditions are sufficient. We observe that they are not.
Riemann claimed that a Trigonometric Series is the convolution
of its second primitive with the Dirichlet Kernel,
cos(x − t ) + cos 2(x − t ) + cos 3(x − t ) + ... .
But that infinite Series diverges to infinity at x = t , and cannot
be integrated.
Riemann claimed that a function with infinitely many maxima or
minima on any interval has diverging Fourier Coefficients. But
his proof is incomplete, and the claim is unproven
Riemann presents examples of Fourier Series expansions of
pathologically constructed functions. These examples support
Fourier’s claim that any function equals its Fourier Series.
Keywords: Trigonometric Series, Fourier Series, Riemann (x)
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Function, Riemann (x) Series, Integral, Integration, Dirichlet
Kernel,
2000 mathematics subject Classification: 26A42, 26A30,
26A15, 42A16, 42A20, 42B05, 43A50,
Contents
Introduction
1. Riemann’s 1st and 2nd Theorems
2. Riemann’s 3rd Theorem
3. Unproven equality of f (x ) to its Trigonometric Series
4. Unproven equality of Fourier Series to a convolution of F (t ) with a
Dirichlet Kernel
5. Unproven divergence of Fourier Coefficients
6. Fourier Series of Riemann’s (x ) function
7. Fourier Series of Riemann’s (2x ) function
8. Fourier Series of Riemann’s (3x ) function
9. Fourier Series of Riemann (x ) Series
10. Other Non-integrable Fourier Series
11. Fourier Series with An g 0
Introduction
In his paper “On the Representation of a Function by a
Trigonometric Series”,
Riemann presents his integral for a
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
bounded function, and reviews the Dirichlet Conditions sufficient
for the equality of a function to its associated Fourier Series.
Riemann wrote
“The Integral
1
2π
sin ( (n + 12 )(x − α) )
∫ f (α) sin ( 1 (x − α) ) d α
α =−π
2
α=π
approaches the value
f (x )
infinitesimally closely when
n → ∞ ”.
However, at x = α , when n → ∞ ,
sin ( (n + 12 )(x − α) )
sin ( 12 (x − α) )
f (α)
sin ( (n + 12 )(x − α) )
→ ∞,
is unbounded function,
sin ( 12 (x − α) )
and the integral does not exist as a Riemann Integral.
This raises doubts regarding Riemann attempts there to
obtain necessary and sufficient conditions for a function to
equal its associated Fourier Series,
represent a Trigonometric Series as a convolution with the
Dirichlet Kernel
show that a function with infinitely many maxima and
minima, has divergent Fourier coefficients
We follow through the paper to find which of its results hold.
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
1.
Riemann’s 1st
2nd Theorems
and
1.1 Necessary Condition for Convergence of a
Trigonometric Series
If for any x , the Trigonometric Series
a1 sin x + a2 sin 2x + ... + 12 b0 + b1 cos x + b2 cos 2x + ... =
= 12 b0 + a1 sin x + b1 cos x + a2 sin 2x + b2 cos 2x + ... ,
A0
A1 (x )
A2 (x )
converges to a function
f (x ) .
Then for any x ,
An (x ) → 0 , as n → ∞
1.2
The first primitive of f (x ) is the series
C '+ A0x − a1 cos x + b1 sin x − 12 a2 cos 2x + 12 b2 sin 2x + ...
1.3
The second primitive of f (x ) is the series
C + C ' x + 12 A0x 2 −a1 sin x − b1 cos x − 12 a2 sin 2x −
2
−A1 (x )
1
b
22 2
− 1 A2 (x )
4
4
cos 2x + ...
Gauge Institute Journal, Volume 7, No 3, August 2011
= C + C ' x + 12 A0x 2 − A1(x ) −
1.4
H. Vic Dannon
1
22
A2 (x ) −
1
32
A3 (x )...
The second primitive of f (x ) converges for any x to a
continuous Integrable function
F (x ) .
Proof: Convergence
Since for any x , An (x ) → 0 , then for k = n + 1, n + 2,... ,
Ak (x ) < ε ,
and for the tail of the second primitive series we have
−
Since
An +1(x )
(n + 1)2
1
2
(n + 1)
bounded by
+
−
⎛
⎞
1
1
− ... < ε ⎜⎜⎜
+
+ ... ⎟⎟⎟ .
(n + 2)2
⎠⎟
⎝ (n + 1)2 (n + 2)2
An +2 (x )
1
2
(n + 2)
+ ... is the tail of 1 +
1
22
+
1
32
+ ... , it is
1
, and we have
n
<
ε
.
n
Therefore, the tail of the second primitive series can be made
arbitrarily small, and that series converges to F (x ) .
Continuity
For any x , and for δ > 0 ,
F (x + δ ) − F (x ) ≤ C ' δ + b0 (x + 12 δ )δ +
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Gauge Institute Journal, Volume 7, No 3, August 2011
+ A1(x + δ ) − A1(x ) + ... +
1
n2
An (x + δ ) − An (x ) +
H. Vic Dannon
1
(n +1)2
An +1(x + δ ) − An +1(x ) + ...
For k = 1...n ,
1
k2
Ak (x + δ ) − Ak (x ) = ak (sin(kx + δ ) − sin kx ) + bk (cos(kx + δ ) − cos kx )
≤ ak sin(kx + δ ) − sin(kx ) + bk cos(kx + δ ) − cos(kx ) ,
and applying the Intermediate Value Theorem to each term,
= ak cos ξk δ + bk sin ηk δ
≤ ak δ + bk δ .
For k = n + 1, n + 2,... ,
1
k2
Ak (x + δ ) − Ak (x ) ≤
1
k2
( Ak (x + δ )
<
2
k2
ε.
+ Ak (x ) ) ,
and by 1.1,
Therefore, F (x + δ ) − F (x ) is bounded by
{C '
+ b0 ( x + 12 δ ) + a1 + b1 + ... + an + bn
} δ + 2ε ( (n +11)
≤ { C ' + b0 ( x + 12 δ ) + a1 + b1 + ... + an + bn
2
+
1
(n + 2)2
} δ + 2ε n1
+ ...
)
≡ η.
Consequently, given x , and arbitrarily small η > 0 , there is n so
that η − 2ε n1 > 0 ,
and there is a quadratic equation solution,
δ > 0 , so that F (x + δ ) − F (x ) < η .
1.5 If the Trigonometric Series converges to f (x )
Then for infinitesimals α , and β
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
=
4αβ
= A0 + A1
sin α sin β
sin 2α sin 2β
sin 3α sin 3β
+ A2
+ A3
+ ...
α
β
2α
2β
3α
3β
Proof:
The terms C + C ' x in the series for F (x ) yield zero.
The term 12 b0x 2 yields
1b
2 0
(x + α + β )2 − (x + α − β )2 − (x − α + β )2 + (x − α − β )2
=
4αβ
= 12 b0
2αβ + 2αβ + 2αβ + 2αβ
= b0
4αβ
The term − 12 ak sin(kx ) yields
k
−
1
k
ak
2
=−
sin k(x + α + β ) − sin k(x + α − β ) − sin k (x − α + β ) + sin k(x − α − β )
4αβ
⎡ sin k (x + α + β ) + sin k (x − α − β ) ⎤ − ⎡ sin k(x + α − β ) + sin k (x − α + β ) ⎤
⎦ ⎣
⎦
ak ⎣
4αβ
k2
=−
=−
=−
1
⎤
⎡
⎤ ⎡
⎣ 2 sin(kx )cos k(α + β ) ⎦ − ⎣ 2 sin(kx )cos k(α − β ) ⎦
a
k
4αβ
k2
1
1
k
2
1
k
2
ak 2 sin(kx )
cos k (α + β ) − cos k (α − β )
4αβ
ak 2 sin(kx )
−2 sin(k α)sin(k β )
4αβ
= ak sin(kx )
sin(k α) sin(k β )
kα
kβ
The term − 12 bk cos(kx ) yields
k
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Gauge Institute Journal, Volume 7, No 3, August 2011
−
1
k
b
2 k
=−
H. Vic Dannon
cos k (x + α + β ) − cos k(x + α − β ) − cos k(x − α + β ) + cos k(x − α − β )
4αβ
⎡ cos k(x + α + β ) + cos k(x − α − β ) ⎤ − ⎡ cos k(x + α − β ) + cos k(x − α + β ) ⎤
⎦ ⎣
⎦
bk ⎣
2
4
αβ
k
=−
=−
=−
1
⎤
⎡
⎤ ⎡
⎣ 2 cos(kx )cos k(α + β ) ⎦ − ⎣ 2 cos(kx )cos k(α − β ) ⎦
b
k
4αβ
k2
1
1
k
b 2 cos(kx )
2 k
1
k
b 2 cos(kx )
2 k
= bk cos(kx )
cos k (α + β ) − cos k(α − β )
4αβ
−2 sin(k α)sin(k β )
4αβ
sin(k α) sin(k β )
kα
kβ
Therefore,
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
=
4αβ
= b0 + (a1 sin x + b1 cos x )
A0
A1
= A0 + A1
sin α sin β
sin 2α sin 2β
+ (a2 sin 2x + b2 cos 2x )
+ ...
2α
2β
α
β
A2
sin α sin β
sin 2α sin 2β
sin 3α sin 3β
+ A2
+ A3
+ ... .
α
β
2α
2β
3α
3β
1.6 If the Trigonometric Series converges to f (x )
Then for an infinitesimal α ,
F (x + 2α) − 2F (x ) + F (x − 2α)
4α2
⎛ sin α ⎞⎟2
⎛ sin 2α ⎞⎟2
= A0 + A1 ⎜⎜
⎟ + A2 ⎜⎜
⎟ + ...
⎜⎝ α ⎠⎟
⎝⎜ 2α ⎠⎟
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
1.7 Denote the negative tails of the Trigonometric Series by
ε1 = −A1 − A2 − A3 − A4 − ... = A0 − f (x )
ε2 =
ε3 =
− A2 − A3 − A4 − ... = A0 + A1 − f (x )
− A3 − A4 − ... = A0 + A1 + A3 − f (x )
..……………………………………………………………………
εn =
=A0 + A1 + A3 + ... + An −1 − f (x )
Then, A0 = f (x ) + ε1
A1 = ε2 − ε1
A2 = ε3 − ε2
A3 = ε4 − ε3
……………..
An = εn +1 − εn
1.8 If the Trigonometric Series converges to f (x )
Then
F (x + 2α) − 2F (x ) + F (x − 2α)
4α 2
− f (x ) =
2
2
⎧⎪
⎧⎪⎛
⎛
⎪
⎛ sin α ⎟⎞2 ⎪⎫⎪
⎛ sin 2α ⎞⎟2 ⎪⎫
⎛ sin 3α ⎞⎟2 ⎟⎞⎟
sin α ⎞⎟
⎜⎜ ⎛⎜ sin 2α ⎞⎟
⎪
⎪
⎜
⎜
⎜
= ε1 ⎨1 − ⎜
⎟ + ε2 ⎨⎜
⎟ −⎜
⎟ + ε3 ⎜ ⎜
⎟ − ⎜⎜
⎟ ⎟ + ...
⎜ α ⎠⎟ ⎬⎪
⎜ α ⎠⎟
⎜ 2α ⎠⎟ ⎬⎪
⎜ 2α ⎠⎟
⎜
⎪
⎪
⎝
⎝
⎝
⎝
⎝⎜ 3α ⎠⎟ ⎠⎟⎟
⎜
⎪
⎪
⎪
⎪
⎝
⎪
⎪
⎩
⎭⎪
⎩⎪
⎭
Proof: By 1.6, and 1.7,
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Gauge Institute Journal, Volume 7, No 3, August 2011
F (x + 2α) − 2F (x ) + F (x − 2α)
4α2
=
A0
f (x )+ ε1
H. Vic Dannon
=
⎛ sin α ⎟⎞2
⎛ sin 2α ⎞⎟2
⎛ sin 3α ⎞⎟2
⎜
⎜
+ A1 ⎜
⎟ + A2 ⎜
⎟ + A3 ⎜⎜
⎟ + ... .
⎜⎝ α ⎟⎠
⎜⎝ 2α ⎠⎟
⎝⎜ 3α ⎠⎟
ε3 −ε2
ε2 −ε1
ε4 −ε3
1.9 If the Trigonometric Series converges to f (x )
Then
F (x + 2α) − 2F (x ) + F (x − 2α)
4α 2
→ f (x ) , as α → 0 .
Proof:
By 1.8, we aim to show that
⎛ sin α ⎞⎟2
⎛ sin α ⎞⎟2 ⎛ sin 2α ⎞⎟2
⎛ sin 2α ⎞⎟2 ⎛ sin 3α ⎞⎟2
⎜
⎜
⎜
ε1 1 − ⎜
⎟ + ε2 ⎜
⎟ −⎜
⎟ + ε3 ⎜⎜
⎟ − ⎜⎜
⎟ + ...
⎜⎝ α ⎠⎟
⎝⎜ α ⎠⎟
⎝⎜ 2α ⎠⎟
⎝⎜ 2α ⎠⎟
⎝⎜ 3α ⎠⎟
vanishes as α → 0 .
Although each of the terms vanishes as α → 0 , there are
infinitely many terms, and ∞ × 0 is undefined. Therefore, we
have to separate the infinite series into parts.
Our first separation is based on the fact that the nth tail of the
Trigonometric Series, εn of 1.7, vanishes as n → ∞ . Hence, for
arbitrarily small δ > 0 , there is M , so that for any n > M ,
δ > εn .
As α → 0 , we have
sin(k α )
kα
→ 1 , and the partial sum till n = M ,
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
⎛ sin α ⎞⎟2
⎛ sin α ⎞⎟2 ⎛ sin 2α ⎞⎟2
⎜
ε1 1 − ⎜
⎟ + ε2 ⎜⎜
⎟ − ⎜⎜
⎟ + ... + εM
⎜⎝ α ⎠⎟
⎝⎜ α ⎠⎟
⎝⎜ 2α ⎠⎟
⎛ sin(M − 1)α ⎞⎟2 ⎛ sin M α ⎞⎟2
⎟
⎟ − ⎜⎜
⎜⎜
⎝⎜ (M − 1)α ⎠⎟
⎝⎜ M α ⎠⎟
vanishes as α → 0 .
We need to show that the tail with n > M , vanishes as α → 0 .
That requires another separation:
For small enough α > 0 , there is N > M , so that
(N − 1)α < π ,
and
Nα > π .
For n > M , we have δ > εn , and
⎛ sin M α ⎞⎟2 ⎛ sin(M + 1)α ⎞⎟2
εM +1 ⎜⎜
⎟ − ⎜⎜
⎟ + ... + εN
⎝⎜ M α ⎠⎟
⎝⎜ (M + 1)α ⎠⎟
⎪⎧⎪ ⎛ sin M α ⎞2 ⎛ sin(M + 1)α ⎞2
⎟⎟ − ⎜⎜
⎟⎟ + ... +
≤ δ ⎨ ⎜⎜
⎪⎪ ⎝⎜ M α ⎠⎟
⎝⎜ (M + 1)α ⎠⎟
⎩⎪
For t > 0 , the function
⎛ sin(N − 1)α ⎞⎟2 ⎛ sin N α ⎞⎟2
⎜⎜
⎟ − ⎜⎜
⎟
⎝⎜ (N − 1)α ⎠⎟
⎝⎜ N α ⎠⎟
⎛ sin(N − 1)α ⎞⎟2 ⎛ sin N α ⎞⎟2
⎜⎜
⎟ − ⎜⎜
⎟
⎝⎜ (N − 1)α ⎠⎟
⎝⎜ N α ⎠⎟
sin t
has the derivative
t
cos t
1
1⎛
sin t ⎞⎟
− sin t = ⎜⎜ cos t −
⎟⎟ ,
2
⎜
t
t
t
⎝
⎠
t
which is negative because for small enough arc-length t ,
11
⎪⎫⎪
⎬
⎪⎪
⎭⎪
,
Gauge Institute Journal, Volume 7, No 3, August 2011
t < tan(t ) =
Therefore, the function
H. Vic Dannon
sin(t )
cos(t )
sin t
is decreasing.
t
Thus, for α > 0 ,
sin k α sin(k + 1)α
−
> 0,
kα
(k + 1)α
and
2
2
⎛ sin k α ⎞⎟
⎛ sin(k + 1)α ⎞⎟
⎟ − ⎜⎜
⎟
⎜⎜⎜
⎟
⎝ kα ⎠
⎝⎜ (k + 1)α ⎠⎟
⎛
⎞⎟
⎜⎜
⎛ sin k α sin(k + 1)α ⎞⎟ ⎜ sin k α sin(k + 1)α ⎟⎟⎟
⎟> 0
= ⎜⎜
−
+
⎟⎜
(k + 1)α ⎠⎟ ⎜⎜⎜ k α
(k + 1)α ⎟⎟⎟
⎝⎜ k α
⎟
⎜⎝ >0
⎠⎟
>0
>0
Consequently,
⎪⎧⎪ ⎛ sin M α ⎞2 ⎛ sin(M + 1)α ⎞2
⎟⎟ − ⎜⎜
⎟⎟ + ... +
δ ⎨ ⎜⎜
⎪⎪ ⎝⎜ M α ⎠⎟
⎝⎜ (M + 1)α ⎠⎟
⎩⎪
⎛ sin(N − 1)α ⎞⎟2 ⎛ sin N α ⎞⎟2
⎜⎜
⎟ − ⎜⎜
⎟
⎝⎜ (N − 1)α ⎠⎟
⎝⎜ N α ⎠⎟
⎪⎫
⎪⎬
⎪⎪
⎭⎪
2
⎧⎪⎛
⎛ sin(M + 1)α ⎞⎟2
⎛ sin(N − 1)α ⎞⎟2 ⎛ sin N α ⎞⎟2 ⎫⎪⎪
⎪⎜ sin M α ⎞⎟
⎜
= δ ⎨⎜
⎟ −⎜
⎟ + ... + ⎜⎜
⎟ − ⎜⎜
⎟ ⎬
⎜⎝ M α ⎠⎟
⎜ (M + 1)α ⎠⎟
⎜ (N − 1)α ⎠⎟
⎪
⎝
⎝
⎝⎜ N α ⎠⎟ ⎪⎪
⎩⎪⎪
⎭⎪
⎪⎧⎪⎛ sin M α ⎞2 ⎛ sin N α ⎞2 ⎪⎫
⎟⎟ − ⎜⎜
⎟⎟ ⎪⎬
= δ ⎨⎜⎜
⎪⎪⎝⎜ M α ⎠⎟
⎝⎜ N α ⎠⎟ ⎪⎪
⎩⎪
⎭⎪
⎛ sin M α ⎞⎟2
< δ ⎜⎜
⎟
⎜⎝ M α ⎠⎟
Since
sin(M α )
Mα
→ 1 , as α → 0 , the sum is bounded by
We are left with the terms with n = N + 1, N + 2,...
Then, δ > εn , and
12
δ.
Gauge Institute Journal, Volume 7, No 3, August 2011
εN +1
⎛ sin N α ⎞⎟2 ⎛ sin(N + 1)α ⎞⎟2
⎜⎜
⎟ − ⎜⎜
⎟ + εN + 2
⎜⎝ N α ⎟⎠
⎝⎜ (N + 1)α ⎠⎟
H. Vic Dannon
⎛ sin(N + 1)α ⎞⎟2 ⎛ sin(N + 2)α ⎞⎟2
⎜⎜
⎟ + ... ,
⎟ − ⎜⎜
⎝⎜ (N + 1)α ⎠⎟
⎝⎜ (N + 2)α ⎠⎟
2
⎧⎪ ⎛
⎫⎪
⎛ sin(N + 1)α ⎞⎟2
⎛ sin(N + 1)α ⎞⎟2 ⎛ sin(N + 2)α ⎞⎟2
⎪ sin N α ⎞⎟
< δ ⎨ ⎜⎜
⎟⎟ − ⎜⎜
⎟⎟ + ⎜⎜
⎟⎟ − ⎜⎜
⎟⎟ + ... ⎪⎬
⎪⎪ ⎝⎜ N α ⎠
⎪⎪
⎝⎜ (N + 1)α ⎠
⎝⎜ (N + 1)α ⎠
⎝⎜ (N + 2)α ⎠
⎩⎪
⎭⎪
2
2
⎛ sin k α ⎞⎟
⎛ sin(k + 1)α ⎞⎟
Since ⎜⎜⎜
⎟⎟ > 0 , we can remove the absolute
⎟⎟ − ⎜⎜
⎜
⎝ kα ⎠
⎝ (k + 1)α ⎠
value sign, and we have
⎫⎪
⎪⎧⎛ sin2(N α) sin2[(N + 1)α] ⎞⎟ ⎛⎜ sin2[(N + 1)α] sin2[(N + 2)α ] ⎞⎟
⎟⎟ + ⎜
⎟⎟ + ...⎪
= δ ⎪⎨⎜⎜⎜
−
−
⎬.
⎜ [(N + 1)α]2
2
2 ⎟
2 ⎟
⎪
⎪
⎜
⎜
+
+
(
N
α
)
[(
N
1)
α
]
[(
N
2)
α
]
⎝
⎠
⎝
⎠
⎪
⎪
⎩
⎭
Nevertheless, the telescopic series sum,
δ
sin2(N α)
(N α)2
, has no limit for
N → ∞ , and α → 0 . Thus, we need another series separation:
Now, for n > N ,
sin2[(n − 1)α]
[(n − 1)α]2
−
sin2[n α]
[n α]2
=
⎧
⎪ sin2[(n − 1)α] sin2[(n − 1)α] ⎪⎫⎪ ⎧
⎪
sin2[(n − 1)α] sin2[n α] ⎫⎪⎪
⎪
⎪
=⎨
−
−
⎬+⎨
⎬
2
2
2
⎪
⎪
⎪
α
α
α
−
[(
n
1)
]
[
n
]
[
n
]
[n α]2 ⎪⎪⎭
⎪ ⎩
⎪
⎩⎪
⎭
sin2[(n − 1)α] ⎧⎪⎪ 1
1 ⎫⎪⎪
1
=
− ⎬+
sin2[(n − 1)α] − sin2[n α]}
⎨
{
2
2
2
2
2
⎪⎩⎪ [n − 1]
α
n ⎪⎪⎭ n α
The first term
=δ
⎪
⎪
sin2[(n − 1)α] ⎧
1
1 ⎫
⎪
− ⎪
⎨
⎬
2
2
2
⎪
α
n ⎪⎪
⎪ [n − 1]
⎩
⎭
generates the series
⎧ 1
⎫
⎪
sin2(N α) ⎪
1
1
1
⎪
+ ... ⎪
−
+
−
⎨
⎬
2
⎪
⎪
α2
[N + 1]2 [N + 1]2 [N + 2]2
⎪N
⎪
⎩
⎭
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Gauge Institute Journal, Volume 7, No 3, August 2011
=δ
≤δ
≤δ
H. Vic Dannon
sin2(N α)
N 2α2
1
2 2
N α
1
π2
The second term is
1
2 2
n α
{ sin2[(n − 1)α] − sin2[nα]} =
=
=
1
n 2α2
( sin[(n − 1)α] − sin[nα])( sin[(n − 1)α] + sin[nα])
1
n 2α2
=−
=−
( 2 sin
1
n 2α 2
(2n −1)α
2
cos α2
)( 2 cos
(2n −1)α
sin(− α2 )
2
)
sin[(2n − 1)α]sin α
sin α
1
sin[(2n − 1)α]
α
n 2α
Thus, for infinitesimal α , the second term is bounded by
and generates the series
⎫⎪
1 ⎧⎪⎪
1
1
+
+ ... ⎪⎬
δ ⎨
2
2
⎪⎭⎪
α ⎪⎩⎪ (N + 1)
(N + 2)
≤δ
1
1
α (N + 1)
14
1
2
n α
,
Gauge Institute Journal, Volume 7, No 3, August 2011
<δ
H. Vic Dannon
1
π
{
In summary, the Series is bounded by δ 1 +
1
π2
+
1
π
} ,and we let
δ ↓ 0.
1.10 Riemann’s 1st Theorem
If the Trigonometric Series converges to f (x )
Then
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
→ f (x ) ,
4αβ
for infinitesimals α , and β , so that
α
β
, and are finite.
α
β
Proof: By 1.9,
F (x + 2α1 ) − 2F (x ) + F (x − 2α1 )
(2α1 )2
F (x + 2α2 ) − 2F (x ) + F (x − 2α2 )
(2α2 )2
= f (x ) + δ1
= f (x ) + δ2
Hence,
F (x + 2α1 ) − 2F (x ) + F (x − 2α1 ) = (2α1 )2 f (x ) + (2α1 )2 δ1
F (x + 2α2 ) − 2F (x ) + F (x − 2α2 ) = (2α2 )2 f (x ) + (2α2 )2 δ1
Subtracting the second equation from the first,
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H. Vic Dannon
F (x + 2α1 ) − F (x + 2α2 ) − F (x − 2α2 ) + F (x − 2α1 )
(2α1 )2 − (2α2 )2
= f (x ) +
(2α1 )2
(2α1 )2 − (2α2 )2
δ1 +
=
(2α2 )2
(2α1 )2 − (2α2 )2
δ2
Denote
2α1 = α + β
2α2 = α − β
Then,
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
=
4αβ
= f (x ) +
(α + β )2
(α − β )2
δ1 +
δ2
4αβ
4αβ
1⎛α
β⎞
1⎛α
β⎞
= f (x ) + ⎜⎜ + 2 + ⎟⎟⎟ δ1 + ⎜⎜ − 2 + ⎟⎟⎟ δ2
4 ⎝⎜ β
α⎠
4 ⎝⎜ β
α⎠
For infinitesimal α , and β , so that
coefficients of
α
β
, and
are finite, the
α
β
δ1 , and δ2 are bounded, and we let δ1 ↓ 0 , and
δ2 ↓ 0 , to obtain the equality to f (x ) .
1.11 Riemann’s 2nd Theorem
If the Trigonometric Series converges to f (x )
Then For any x ,
F (x + 2α) − 2F (x ) + F (x − 2α)
→ 0,
2α
α → 0,
16
as
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Proof: By 1.6,
⎧
⎫
⎪
⎪
⎛ sin α ⎞⎟2
⎛ sin 2α ⎞⎟2
F (x + 2α) − 2F (x ) + F (x − 2α)
⎪
⎪
⎜
= 2α ⎨ A0 + A1 ⎜
⎟⎟ + A2 ⎜⎜
⎟⎟ + ... ⎬
⎪⎪
⎪
2α
⎝⎜ α ⎠
⎝⎜ 2α ⎠
⎪
⎪
⎩⎪
⎭
We need to show that the series in
{ } is bounded.
As in the proof of 1.9, we separate the series into three parts:
First, since the series converges to f (x ) , for ε > 0 , there is M , so
that for n > M , we have ε > An .
For infinitesimal α , the partial sum till M ,
⎛ sin α ⎞⎟2
⎛ sin(2α) ⎞⎟2
⎛ sin(M α) ⎞⎟2
⎜
⎜
A0 + A1 ⎜
⎟ ,
⎟ + A2 ⎜
⎟ + ... + AM ⎜⎜
⎝⎜ α ⎠⎟
⎝⎜ 2α ⎠⎟
⎝⎜ M α ⎠⎟
has a limit A0 + A1 + A2 + ... + AM , and is bounded by a number Q .
Second, for small enough α > 0 , there is N > M , so that
(N − 1)α < π ,
and
Nα > π .
For the part of the series from n = M + 1 , to n = N , we have
An < ε , and
⎛ sin[(M + 1)α] ⎞⎟2
⎛ sin[(M + 2)α] ⎞⎟2
⎛ sin(N α) ⎞⎟2
⎜
⎜
AM +1 ⎜
⎟ ,
⎟ + AM + 2 ⎜
⎟ + ... + AN ⎜⎜
⎜⎝ (M + 1)α ⎠⎟
⎝⎜ (M + 2)α ⎠⎟
⎝⎜ N α ⎠⎟
is bounded by
2
⎧⎪⎛
⎛ sin[(M + 2)α] ⎞⎟2
⎛ sin(N α) ⎞⎟2 ⎫⎪⎪
sin[(M + 1)α] ⎞⎟
⎪
⎜
⎜
ε ⎨⎜
⎟ +⎜
⎟ + ... + ⎜⎜
⎟ ⎬.
⎪⎪⎜⎝ (M + 1)α ⎠⎟
⎝⎜ (M + 2)α ⎠⎟
⎝⎜ N α ⎠⎟ ⎪⎪
⎩⎪
⎭⎪
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H. Vic Dannon
Since sin(t ) ≤ t , the series from n = M + 1 , to n = N is bounded
by
ε {1 + 1 + ... + 1 } < εN < ε
N −M
π
α
Third, the tail of series for n > N ,
⎛ sin[(N + 1)α] ⎞⎟2
⎛ sin[(N + 2)α] ⎞⎟2
⎜
AN +1 ⎜
⎟ + AN + 2 ⎜⎜
⎟ + ... ,
⎜⎝ (N + 1)α ⎠⎟
⎝⎜ (N + 2)α ⎠⎟
is bounded by
⎧⎪
⎫⎪
1
1
1
⎪⎬
ε ⎪⎨
+
+
...
+
⎪⎩⎪ (N + 1)2 α2 (N + 2)2 α2 (N + 3)2 α2
⎪⎭⎪
<ε
<
1
1
N + 1 α2
ε1
.
απ
Therefore,
⎧⎪
⎫⎪
⎛ sin α ⎞⎟2
⎛ sin 2α ⎞⎟2
⎪
⎪
⎜
⎜
2α ⎨ A0 + A1 ⎜
⎟⎟ + A2 ⎜
⎟⎟ + ... ⎬
⎪⎪
⎪⎪
⎝⎜ α ⎠
⎝⎜ 2α ⎠
⎩⎪
⎭⎪
is bounded by
⎧⎪
⎛
π
α ⎫⎪
1⎞
2α ⎪⎨Q + ε + ε ⎪⎬ = 2αQ + 2ε ⎜⎜ π + ⎟⎟⎟ .
⎪⎩⎪
α
π ⎪⎪⎭
π⎠
⎝⎜
We let α ↓ 0 , and ε ↓ 0 .
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H. Vic Dannon
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
2.
Riemann’s 3rd Theorem
2.1 If the Trigonometric Series converges to f (x )
c > b are arbitrary constants
λ(x ) is continuous on [b, c ]
x =c
Then μ
2
∫
F (x )cos(μ[x − a ])λ(x )dx =
x =b
x =c
= μ2
∫ (C + C ' x + 12 A0x 2 ) cos(μ[x − a ])λ(x )dx +
x =b
x =c
−μ2
∫ 12 ⎡⎣ {a1 sin a + b1 cos a } cos(μ + 1)(x − a ) + {a1 cos a + b1 sin a } sin(μ + 1)(x − a ) ⎤⎦ λ(x )dx
x =b
Bμ +1 (x )
x =c
−μ
2
∫ 12 ⎡⎣ {a1 sin a + b1 cos a } cos(μ − 1)(x − a ) − {a1 cos a + b1 sin a } sin(μ − 1)(x − a ) ⎤⎦ λ(x )dx
x =b
−
−
μ2
22
μ2
22
Bμ−1 (x )
x =c
∫
x =b
1
2
⎡ {a2 sin 2a + b2 cos 2a } cos(μ + 2)(x − a ) + {a2 cos 2a + b2 sin 2a } sin(μ + 2)(x − a ) ⎤ λ(x )dx
⎣
⎦
Bμ + 2 ( x )
x =c
∫
x =b
1
2
⎡ {a2 sin 2a + b2 cos 2a } cos(μ − 2)(x − a ) − {a2 cos na + b2 sin 2a } sin(μ − 2)(x − a ) ⎤ λ(x )dx
⎣
⎦
B μ − 2 (x )
− ………
{an sin nx + bn cos nx } cos μ(x − a ) = Bμ+n (x ) + Bμ−n (x )
An
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H. Vic Dannon
Bμ'' +n (x ) = −(μ + n )2 Bμ +n (x ) .
Bμ'' −n (x ) = −(μ − n )2 Bμ−n (x ) .
Bμ+n (x ) → 0 , as n → ∞ .
Bμ−n (x ) → 0 , as n → ∞ .
Proof:
An cos μ(x − a ) = an sin nx cos μ(x − a ) + bn cos nx cos μ(x − a )
= an sin(nx − na + na )cos μ(x − a ) + bn cos(nx − na + na )cos μ(x − a )
= an { sin n(x − a )cos na + cos n(x − a )sin na } cos μ(x − a )
+bn { cos n(x − a )cos na + sin n(x − a )sin na } cos μ(x − a )
= ⎣⎡ an cos na + bn sin na ⎤⎦ sin n(x − a )cos μ(x − a )
+ ⎡⎣ an sin na + bn cos na ⎤⎦ cos n(x − a )cos μ(x − a )
= ⎡⎣ an cos na + bn sin na ⎤⎦ 12 { sin(μ + n )(x − a ) − sin(μ − n )(x − a )}
+ ⎡⎣ an sin na + bn cos na ⎤⎦ 12 { cos(μ + n )(x − a ) + cos(μ − n )(x − a )}
=
1
2
{ ⎡⎣ an cos na + bn sin na ⎤⎦ sin(μ + n )(x − a ) + ⎡⎣ an sin na + bn cos na ⎤⎦ cos(μ + n )(x − a )}
Bμ +n (x )
+ 12 { ⎡⎣ an sin na + bn cos na ⎤⎦ cos(μ − n )(x − a ) − ⎡⎣ an cos na + bn sin na ⎤⎦ sin(μ − n )(x − a )}
Bμ −n (x )
Therefore,
An cos μ(x − a ) = Bμ +n (x ) + Bμ−n (x ) ,
Bμ'' +n (x ) = −(μ + n )2 Bμ +n (x )
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H. Vic Dannon
Bμ'' −n (x ) = −(μ − n )2 Bμ−n (x ) .
and since an → 0 , and bn → 0 ,
Bμ+n (x ) → 0 , as n → ∞ .
Bμ−n (x ) → 0 , as n → ∞ .
2.2 If the Trigonometric Series converges to f (x )
c > b are arbitrary constants
λ(x ) has a continuous derivative λ '(x ) on [b, c ]
λ(b) = λ(c ) = 0
λ '(b) = λ '(c ) = 0
λ ''(x ) has finitely many maxima and minima
x =c
Then μ 2
∫
F (x )cos(μ[x − a ])λ(x )dx =
x =b
x =c
=μ
2
∫ (C + C ' x + 12 A0x 2 ) cos(μ[x − a ])λ(x )dx +
x =b
+
+
μ2
x =c
∫
(μ + 1)2 x =b
μ2
Bμ +1(x )λ ''(x )dx +
x =c
∫
22 (μ + 2)2 x =b
Bμ +2 (x )λ ''(x )dx +
μ2
x =c
∫
(μ − 1)2 x =b
μ2
Bμ−1(x )λ ''(x )dx
x =c
∫
22 (μ − 2)2 x =b
Bμ−2 (x )λ ''(x )dx
……………………………………………………………………..
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H. Vic Dannon
x =c
∫
Bμ +n (x )λ ''(x )dx → 0 , as μ + n → ∞
x =b
x =c
∫
Bμ−n (x )λ ''(x )dx → 0 , as μ − n → ∞
x =b
μ2
−
Proof:
n2
=−
μ2
n2
x =c
∫
An cos(μ[x − a ])λ(x )dx =
x =b
x =c
∫
Bμ +n (x )λ(x )dx −
x =b
x =c
μ2
∫
n2
Bμ−n (x )λ(x )dx
x =b
Substituting
Bμ'' +n (x ) = −(μ + n )2 Bμ +n (x ) ,
Bμ'' −n (x ) = −(μ − n )2 Bμ−n (x ) ,
=
x =c
μ2
2
2
n (μ + n )
∫
Bμ'' +n (x )λ(x )dx
+
x =b
x =c
μ2
2
2
n (μ − n )
∫
Bμ'' −n (x )λ(x )dx
x =b
Integrating by parts,
x =c
∫
x =b
x =c
Bμ'' +n (x )λ(x )dx
=
∫
λ(x )dBμ' +n (x )
x =b
x =c
= ⎡⎢ λ(x )Bμ' +n (x ) ⎤⎥
−
⎣
⎦ x =b
0
23
x =c
∫
x =b
Bμ' +n (x )d λ(x )
λ '(x )dx
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
x =c
= − ∫ λ '(x ) Bμ' +n (x )dx
x =b
dBμ +n
⎫
x =c
⎪⎧⎪
⎪
⎪
x =c
⎪
= − ⎨ ⎡⎣ λ '(x )Bμ +n (x ) ⎤⎦
− ∫ Bμ +n (x )d λ '(x ) ⎪
⎬
x =b
⎪⎪
⎪
⎪
x =b
λ
''(
)
x
dx
⎪⎩
⎪
⎭
0
x =c
=
∫
Bμ +n (x )λ ''(x )dx .
x =b
Hence, −
=
μ
2 x =c
n2
An cos(μ[x − a ])λ(x )dx =
x =b
x =c
μ2
2
∫
2
n (μ + n )
∫
Bμ +n (x )λ ''(x )dx +
x =b
x =c
μ2
2
2
n (μ − n )
∫
Bμ−n (x )λ ''(x )dx .
x =b
Now, since λ ''(x ) has finitely many maxima and minima, then,
by [Riemann] or [Dan1], it is integrable on [b, c ]. Therefore, by the
Riemann-Lebesgue Theorem, if μ → ∞ , we have,
x =c
∫
λ ''(x )sin μxdx → 0 ,
x =b
and
x =c
∫
λ ''(x )cos μxdx → 0 .
x =b
Consequently, if μ → ∞ ,
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
x =c
∫
Bμ +n (x )λ ''(x )dx → 0 ,
x =b
and
x =c
∫
Bμ−n (x )λ ''(x )dx → 0 .
x =b
2.3 Riemann’s 3rd Theorem
If the Trigonometric Series converges to f (x )
c > b are arbitrary constants
λ(x ) has a continuous derivative λ '(x ) on [b, c ]
λ(b) = λ(c ) = 0
λ '(b) = λ '(c ) = 0
λ ''(x ) has finitely many maxima and minima
x =c
Then μ2
∫
F (x )cos(μ[x − a ])λ(x )dx → 0 , as μ → ∞ .
x =b
Proof: In three parts
Part 1:
x =c
The first term in the expansion of μ 2
∫
F (x )cos(μ[x − a ])λ(x )dx
x =b
in 2.2, that needs to vanish as μ → ∞ is
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H. Vic Dannon
x =c
μ
2
∫ (C + C ' x + 12 A0x 2 ) cos(μ[x − a ])λ(x )dx .
x =b
x =c
We confirm that μ2
∫
C ' x cos(μ[x − a ])λ(x )dx vanishes after two
x =b
consecutive integrations by parts.
x =c
μ 2C '
∫
x =b
x =c
x cos(μ[x − a ]) λ(x )dx = μC '
1
μ
d { sin(μ[x −a ])}
∫
x λ(x )d { sin(μ[x − a ])}
x =b
dx
⎧⎪
⎫⎪
x =c
⎪⎪
⎪
x =c
= μC ' ⎨ ⎡⎣ x λ(x )sin(μ[x − a ]) ⎤⎦ x =b − ∫ sin(μ[x − a ])d { x λ(x )} ⎪⎬
⎪⎪
⎪⎪
x
=
b
⎪⎩
⎭⎪
0
⎧
⎫
⎪
⎪
⎪
⎪
x =c
⎪
⎪
⎪
⎪
⎪
= −μC ' ⎨ ∫ sin(μ[x − a ]) { λ(x ) + x λ '(x )}dx ⎪
⎬
⎪
⎪
⎪
⎪
x =b 1 d { cos(μ[x −a ])}
⎪
⎪
−
⎪
⎪
⎪
⎪
dx
μ
⎪
⎪
⎩
⎭
⎧⎪ x =c
⎫⎪
⎪⎪
⎪
= C ' ⎨ ∫ { λ(x ) + x λ '(x )}d { cos(μ[x − a ])} ⎪⎬
⎪⎪
⎪⎪
⎩⎪ x =b
⎭⎪
x =c
⎧
⎫
⎪
⎪
⎪
⎪
x =c
⎪
= C ' ⎨[λ(x ) + x λ '(x )][cos(μ[x − a ])]x =b − ∫ cos(μ[x − a ])d { λ(x ) + x λ '(x )} ⎪
⎬
⎪
⎪
⎪
⎪
x
=
b
0
⎪
⎪
⎩
⎭
x =c
= −C '
∫
cos(μ[x − a ])[2λ '(x ) + λ ''(x )]dx .
x =b
Since 2λ '(x ) + λ ''(x ) is integrable on [b, c ], by the Riemann-
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Lebesgue Lemma, the last integral vanishes as μ → ∞ .
The rest of the first term vanishes similarly.
Part 2:
Next we show that as μ → ∞ , the series
μ2
x =c
∫
12 (μ + 1)2 x =b
Bμ +1(x )λ ''(x )dx +
μ2
x =c
∫
22 (μ + 2)2 x =b
Bμ +2 (x )λ ''(x )dx + ...
vanishes.
By 2.2,
x =c
∫
Bμ +n (x )λ ''(x )dx → 0 , as μ → ∞ .
x =b
Thus, for ε > 0 , and for all μ > Μ
x =c
∫
Bμ +n (x )λ ''(x )dx < ε .
x =b
Hence, the series is bounded by
⎧
⎫
⎪
⎪
⎧
⎫
⎪
⎪
⎪⎪
⎪
μ2
μ2
1
1
⎪
⎪
ε⎨
+
+ ... ⎬ = ε ⎨
+
+ ... ⎪⎬
2
2
⎪⎩
⎪⎭⎪
⎪⎪ (1 + 1 )2 22 (1 + 2 )2
⎪⎪
22 (μ + 2)2
⎪ 1 (μ + 1)
μ
μ
⎪
⎪
⎩
⎭
⎧1
⎫
1
π2
⎪
⎪
⎪
< ε⎪
+
+
...
=
ε
⎨ 2
⎬
⎪
⎪
6
22
⎪1
⎪
⎩
⎭
and we let ε ↓ 0 .
Part 3:
We show that as μ → ∞ , the series
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Gauge Institute Journal, Volume 7, No 3, August 2011
μ2
x =c
∫
12 (μ − 1)2 x =b
Bμ−1(x )λ ''(x )dx +
H. Vic Dannon
x =c
μ2
∫
22 (μ − 2)2 x =b
Bμ−2 (x )λ ''(x )dx + ...
vanishes.
By 2.2,
x =c
∫
Bμ−n (x )λ ''(x )dx → 0 , as μ → ∞ .
x =b
Thus, for ε > 0 , and for all μ > Μ
x =c
∫
Bμ−n (x )λ ''(x )dx < ε .
x =b
Hence, the series is bounded by
1
1
⎧
⎫
⎪
⎫
⎪⎧⎪
⎪
⎪
μ2
μ2
1 ⎪⎪⎪
μ
μ
⎪
⎪
ε⎨
+
+ ... ⎬ = ε ⎨
+
+ ... ⎬
2
2
2
2
2
2
2
2
⎪⎩⎪ 1 (μ − 1)
⎪⎭⎪
⎪⎪
μ ⎪⎪ ( 1 ) (1 − 1 )
2 (μ − 2)
( μ2 ) (1 − μ2 )
μ
⎪ μ
⎪
⎩
⎭
The terms in { } sum up to the Lower Riemann Sum for the
function
1
x 2 (1 − x )2
,
over a partition with subintervals of length
Thus, the Series is bounded by
1
ε
μ
x =∞
∫
x =−∞
1
2
2
x (1 − x )
28
dx .
1
,
μ
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Decomposing the integrand,
1
2
2
x (1 − x )
=
A + Bx
x
2
+
C + D(1 − x )
(1 − x )2
.
1 = A(1 − x )2 + Bx (1 − x )2 + Cx 2 + D(1 − x )x 2
x =0⇒A=1
x = 1 ⇒C = 1
1 = (1 − x )2 + Bx (1 − x )2 + x 2 + D(1 − x )x 2
x =2⇒
1 = 1 + 2B + 4 − 4D ⇒
− 2B + 4D = 4
x = −1 ⇒
1 = 4 − 4B + 1 + 2D ⇒
4B − 2D = 4
B =2
D =2
1
x 2 (1 − x )2
=
1
x2
+
2
1
2
.
+
+
x (1 − x )2 1 − x
⎫⎪⎪
1
1
1 ⎧⎪⎪ 1
1
=
−
+
+
−
−
dx
x
x
2
log
2
log(1
)
⎨
⎬ + Const .
μ ∫ x 2 (1 − x )2
μ ⎪⎩⎪ x 1 − x
⎪⎪
⎭
=
1 ⎧⎪⎪ 1
1
x ⎫⎪⎪
−
+
+
2
log
⎨
⎬ + Const .
⎪⎭⎪
−
−
x
1
x
1
x
μ⎪
⎩⎪
x =μ
1
x =−
μ
⎧
⎫
1
1
1⎪
1
1
x ⎪
⎪
⎪
=
−
+
+
2
log
dx
⎨
⎬
∫ x 2(1 − x )2
⎪
⎪
1
1
μ x =−
μ
x
x
x
−
−
⎪
⎪x =−μ
⎩
⎭
μ
1
x =1−
μ
⎫
1⎧
1
1
x
⎪
⎪
⎪
+ ⎪
+ 2 log
⎨− +
⎬
μ⎪
1−x⎪
⎪ x 1−x
⎪x = 1
⎩
⎭
μ
29
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
x =μ
1⎧
1
x ⎫
⎪⎪ 1
⎪
⎪
+ ⎨− +
+ 2 log
⎬
μ ⎪⎩⎪ x 1 − x
1−x⎪
⎪x =1+ 1
⎭
μ
⎧⎪
⎪ 1
⎪⎪
1
2
1 1
2
−1 ⎫⎪⎪ ⎧
−μ ⎫
= ⎪⎨1 +
+ log
− log
⎬ + ⎪⎨ 2 −
⎬
⎪⎩⎪
⎪⎭
μ +1 μ
μ + 1 ⎪⎭⎪ ⎪⎩
μ
1
+
μ
μ
1
+
μ
μ
⎪
⎪
⎧⎪ 1
⎫⎪ ⎧⎪
1
1
1
1 ⎫⎪⎪
+ ⎪⎨
+ 1 + 2 log(μ − 1)⎪⎬ + ⎪⎨1 +
− 2 log
⎬
⎪⎩⎪ 1 − μ
⎪⎭⎪ ⎪⎩⎪
1−μ
μ
μ
μ − 1⎪
⎪⎭
⎧⎪ 1
⎧
⎫
1 1
1
μ ⎫⎪
⎪⎬ + ⎪
⎪⎨ 1 + 1 − 2 1 log[−(μ + 1)]⎪⎬⎪
+ ⎪⎨ −
+
+ 2 log
⎪⎭⎪
⎪⎩⎪ μ2 μ 1 − μ
μ
1 − μ ⎪⎭⎪ ⎪⎩⎪ μ + 1
μ
= 4+
2
2
1 1
1 1
+
−
+
μ +1 1−μ μ1+ μ μ1−μ
+
2
{ log(μ − 1) − 2 log(−1) − log(μ + 1)}
μ
Letting μ → ∞ ,
x =μ
1
1
ε
∫ x 2(1 − x )2 dx → 4ε .
μ x =−
μ
And we let ε ↓ 0 .
30
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
3.
Unproven equality of f (x ) to its
Trigonometric Series
Riemann claimed that the necessary conditions of his 1st and 3rd
Theorem, are also sufficient for a periodic f (x ) to equal its
Fourier Series.
His plausibility argument is fatally flawed, leaving his claim
unproven.
We state Riemann’s claim, and follow his attempted proof, till its
breakdown.
3.1 Unproven equality of f (x ) to its Trigonometric Series
Let f (x ) be periodic with period 2π .
Then,
f (x ) = 12 b0 + a1 sin x + b1 cos x + a2 sin 2x + b2 cos 2x + ...
A0
A1 (x )
A2 (x )
so that for each x , An → 0 , as n → ∞ .
⇔
There is a continuous F (x ) so that
31
Gauge Institute Journal, Volume 7, No 3, August 2011
(I)
H. Vic Dannon
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
→ f (x ) ,
4αβ
for infinitesimals α , and β , so that
α
β
, and are finite.
α
β
(II) for any c > b , there is λ(x ) with λ '(x ) continuous on (b, c )
λ(b) = λ(c ) = 0
λ '(b) = λ '(c ) = 0
λ ''(x ) bounded and has finitely many maxima and minima
x =c
and μ2
∫
F (x )cos(μ[x − a ])λ(x )dx → 0 , as μ → ∞ .
x =b
Riemann’s Non-Proof:
( ⇒ ) By the 1st Theorem, and the 3rd Theorem.
( ⇐ ) By (I), there is a continuous F (x ) so that
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
→ f (x ) ,
4αβ
which is periodic with period 2π .
Thus,
F "(x ) = f (x ) .
Take C ' , and A0 so that
Φ(x ) ≡ F (x ) − C ' x − 12 A0x 2
is periodic with period 2π , and form a trigonometric series
32
Gauge Institute Journal, Volume 7, No 3, August 2011
C − (a1 sin x + b1 cos x ) −
1
22
H. Vic Dannon
(a2 sin 2x + b2 cos 2x ) − ... ,
A1 (x )
A2 (x )
where
t =π
1
C =
∫ Φ(t )dt ,
2π t =−
π
and
t =π
1
1
− An (x ) =
∫ Φ(t )cos[n(x − t )]dt .
π t =−
n2
π
Integrating twice by parts,
=−
1
t =π
∫
n 2π t =−π
Φ "(t )cos[n(x − t )]dt
Now, for infinitesimals α , and β , so that
α
β
, and are finite,
α
β
F (x ) satisfies
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
→ f (x )
4αβ
C ' x satisfies
C'
{(x + α + β ) − (x + α − β ) − (x − α + β ) + (x − α − β )} = 0
4αβ
1
A x2
2 0
1A
2 0
4αβ
satisfies
{(x + α + β )2 − (x + α − β )2 − (x − α + β )2 + (x − α − β )2 } = 0
Hence, for infinitesimals α , and β , so that
33
α
β
, and
are finite,
α
β
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Φ(x + α + β ) − Φ(x + α − β ) − Φ(x − α + β ) + Φ(x − α − β )
→ f (x ) .
4αβ
Thus,
Φ "(x ) = f (x ) .
Substituting this in the integral above, we obtain
−
1
t =π
∫
n 2 π t =−π
f (t )cos[n(x − t )]dt .
Therefore,
1
An (x ) =
π
t =π
∫
f (t )cos[n(x − t )]dt .
t =−π
Hence,
A0 + A1(x ) + A2(x ) + ...
is the Trigonometric Series associated with the periodic function
f (x ) . So far, no hint about its equality to f (x ) .
⎧
⎪1, x ∈ (−π, π)
By (II), for [b, c ] = [−π, π ] , and for λ(x ) = ⎪
,
⎨
⎪
0,
otherwise
⎪
⎩
λ(x ) and λ '(x ) are continuous on (b, c ) ,
λ(b) = λ(c ) = 0 ,
λ '(b) = λ '(c ) = 0 ,
λ ''(x ) bounded and has finitely many maxima and
minima,
Hence,
34
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
x =π
n
2
∫
F (x )cos(n[x − a ])λ(x )dx → 0 , as n → ∞ .
x =−π
That is,
x =π
n2
∫
F (x )cos(n[x − a ])dx → 0 , as n → ∞ .
x =−π
By integration by parts, we confirm that
x =π
n2
∫
x =−π
(C ' x + 12 A0x 2 )cos(n[x − a ])dx → 0 , as n → ∞ .
Therefore,
x =π
n2
∫
x =−π
(F (x ) − C ' x − 12 A0x 2 )cos(n[x − a ])dx → 0 , as n → ∞ .
Φ(x )
Namely,
t =π
n
2
∫
Φ(t )cos(n[t − x ])dt → 0 , as n → ∞ .
t =−π
−An (x )
Thus, An (x ) → 0 , as n → ∞ , for each x .
Clearly, An (x ) → 0
is only a necessary condition for the
convergence of the Trigonometric Series.
To establish that f (x ) is equal to the Trigonometric Series,
Riemann needed to show that as n → ∞ , for each x ,
A0 + A1(x ) + ...An (x ) − f (x ) → 0 .
35
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Riemann’s concludes with
“It follows by Theorem 1 of the preceding section that the
series
A0 + A1 + A2 + ...
converges to the function f (x ) , wherever it converges”.
Clearly, Riemann’s 1st Theorem of 1.10 assumes that the function
equals its trigonometric series, to obtain condition (I).
Applying Theorem 1 here, means assuming that the function
equals its trigonometric series.
Thus, Riemann’s proof is based on assuming its result.
His claim that conditions (I) and (II) are sufficient remains
unproven.
36
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
4.
Unproven
equality
of
Fourier
Series to a Convolution of F (t )
with Dirichlet Kernel
Riemann attempted to represent the Trigonometric Series as a
convolution of F (t ) with the Dirichlet Kernel.
He failed because the Dirichlet Kernel is the infinite Series
cos(x − t ) + cos 2(x − t ) + cos 3(x − t ) + ...
that diverges to infinity at x = t .
4.1 Unproven equality of Fourier Series to a convolution of
F (t ) with the Dirichlet Kernel
If
f (x ) is periodic with period 2π ,
There is a continuous F (x ) so that
(I)
F (x + α + β ) − F (x + α − β ) − F (x − α + β ) + F (x − α − β )
→ f (x ) ,
4αβ
for infinitesimals α , and β , so that
37
α
β
, and are finite.
α
β
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
(II) for any c > b , there is λ(x ) with λ '(x ) continuous on
(b, c )
λ(b) = λ(c ) = 0
λ '(b) = λ '(c ) = 0
λ ''(x ) bounded and has finitely many maxima and minima
x =c
and μ
2
∫
F (x )cos(μ[x − a ])λ(x )dx → 0 , as μ → ∞ .
x =b
(III) ρ(t ) and ρ '(t ) are continuous on (b, c ) so that
ρ(b) = ρ(c ) = 0 ,
ρ '(b) = ρ '(c ) = 0 ,
ρ ''(t ) is bounded and has finitely many maxima and minima,
and at a fixed point b < x < c ,
ρ(x ) = 1 ,
ρ '(x ) = 1 ,
ρ "(x ) = 1 ,
ρ '''(x ) and ρ ''''(x ) are finite and continuous
Then, as n → ∞ ,
1
{ A0 + A1(x ) + ...An (x )} −
2π
⎧ sin (2n +1)(x −t ) ⎫
⎪
d2 ⎪
⎪
⎪
⎪
⎪
2
(
)
F
t
⎨
⎬ ρ(t )dt → 0 .
∫
−
2⎪
x
t
⎪
sin
dt
⎪
⎪
t =b
2
⎪
⎪
⎩
⎭
t =c
Hence, as n → ∞ ,
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
A0 + A1(x ) + A2 (x ) + ... converges to f (x )
⇔
1
2π
⎧ sin (2n +1)(x −t ) ⎪
⎫
d2 ⎪
⎪
⎪
⎪
2
∫ F (t ) dt 2 ⎨⎪ sin x −t ⎪⎬⎪ ρ(t )dt converges to f (x )
⎪
⎪
t =b
2
⎪
⎪
⎩
⎭
t =c
Riemann’s Non-Proof:
Keeping the notations of the unproven 3.1,
A1(x ) + A2 (x ) + ...An (x ) =
1
=
π
1
=
π
⎧
⎫
⎪
⎪
⎪
⎪
⎪
⎪ dt
2
2
2
1
(
(
)
'
)
1
cos(
)
...
cos
(
)
−
−
−
−
−
−
−
F
t
C
t
A
t
x
t
n
n
x
t
⎨
⎬
0
∫
2
⎪
⎪
⎪
2
2
t =−π
D
x
−
t
D
n
x
−
t
cos(
)
cos
(
)
⎪
⎪
}
} ⎪
Φ(t )
t {
⎪ t{
⎪
⎩
⎭
t =π
t =π
∫
Φ(t )Dt2 { cos(x − t ) + ... + cos n(x − t )}dt
t =−π
Since cos(x − t ) + ... + cos n(x − t ) =
(2n +1)(x −t )
2
x
t
2 sin −
2
sin
,
t =π
⎧ sin (2n +1)(x −t ) ⎫
⎪
1
d2 ⎪
⎪
⎪
⎪
⎪
2
(
)
A1(x ) + ...An (x ) =
t
Φ
⎨
⎬ dt .
∫
x
−
t
2
⎪
⎪
2π
sin
dt
⎪
⎪
t =−π
2
⎪
⎪
⎩
⎭
Therefore,
1
{ A1(x ) + ...An (x )} −
2π
⎧ sin (2n +1)(x −t ) ⎪
⎫
d2 ⎪
⎪
⎪
⎪
2
∫ Φ(t ) dt 2 ⎨⎪ sin x −t ⎪⎬⎪ ρ(t )dt =
⎪
⎪
t =b
2
⎪
⎪
⎩
⎭
t =c
39
Gauge Institute Journal, Volume 7, No 3, August 2011
1
=
2π
H. Vic Dannon
t =c
(2n +1)(x −t ) ⎫
⎧
⎧ sin (2n +1)(x −t ) ⎫
⎪
⎪
1
d 2 ⎪⎪⎪ sin
d2 ⎪
⎪
⎪
⎪
⎪
⎪
⎪
2
2
(
)
(
)
t
dt
t
Φ
−
Φ
⎨
⎬
⎨
⎬ ρ(t )dt .
∫
∫
x
−
t
x
−
t
2⎪
2
⎪
⎪
⎪
π
2
sin
sin
dt
dt
⎪⎩⎪
⎪
⎪
⎪
t =−π
t =b
2
2
⎪
⎪
⎪
⎭
⎩
⎭
t =π
⎧
t ∉ (b, c )
⎪ 1,
Denoting λ(t ) = ⎪⎨
, we have,
⎪
1 − ρ(t ), t ∈ (b, c )
⎪
⎩
1
=
2π
⎧ sin (2n +1)(x −t ) ⎫
⎪
d2 ⎪
⎪
⎪
⎪
⎪
2
Φ
(
)
t
⎨
⎬ λ(t )dt
∫
x
−
t
2⎪
⎪
sin
dt
⎪
⎪
t =−π (F (t )−C ' t − 1 A t 2 )
2
⎪
⎪
⎩
⎭
0
2
t =π
Now, λ(t ) , and λ '(t ) are continuous
λ "(t ) has finitely many maxima and minima
For t = x , λ(x ) = 1 − ρ(x ) = 0
λ '(x ) = −ρ '(x ) = 0
λ ''(x ) = −ρ ''(x ) = 0
λ '''(x ) , and λ ''''(x ) are finite and continuous
Riemann claims that by his 3rd Theorem, as n → ∞ ,
t =π
⎪⎪ sin (2n +1)(x −t ) ⎫
⎪
2 ⎧
1
d
⎪
2
⎪
⎪
2
1
(F (t ) − C ' t − 2 A0t )
⎨
⎬ λ(t )dt → 0 ,
∫
x
−
t
2
⎪
⎪
2π
sin
dt
⎪⎩⎪
⎪
t =−π
2
⎪
⎭
and claims further that by integration by parts
t =c
⎪ sin (2n +1)(x −t ) ⎫
⎪
2 ⎧
1
⎪
2 d ⎪
⎪
⎪
2
1
(
'
)
C
t
A
t
+
⎨
⎬ ρ(t )dt → A0 .
0
∫
2
x
−
t
2
⎪⎪ sin
⎪
2π
dt
⎪
t =b
2
⎪
⎩⎪
⎭
Thus, as n → ∞ ,
40
Gauge Institute Journal, Volume 7, No 3, August 2011
1
{ A0 + A1(x ) + ...An (x )} −
2π
H. Vic Dannon
⎧ sin (2n +1)(x −t ) ⎫
⎪
d2 ⎪
⎪
⎪
⎪
⎪
2
(
)
F
t
⎨
⎬ ρ(t )dt → 0 .
∫
x
−
t
2⎪
⎪
sin
dt
⎪
⎪
t =b
2
⎪
⎪
⎩
⎭
t =c
Rather then fill in the details, we proceed to Riemann’s purpose
of this derivation, the representation of f (x ) by the Dirichlet
Integral.
To start with, note that Riemann did not prove in 3.1 that as
n → ∞ , A0 + A1(x ) + A2 (x ) + ... equals f (x ) .
Therefore, if the convolution
1
2π
⎧ sin (2n +1)(x −t ) ⎫
⎪
d2 ⎪
⎪
⎪
⎪
⎪
2
(
)
F
t
⎨
⎬ ρ(t )dt
∫
2⎪
x
−
t
⎪
sin
dt
⎪
⎪
t =b
2
⎪
⎪
⎩
⎭
t =c
converges, its limit need not be f (x ) .
However, since
(2n +1)(x −t ) ⎫
⎧
⎪⎪
d 2 ⎪⎪⎪ sin
⎪⎬ = − cos(x − t ) − 22 cos 2(x − t ) − ... − n 2 cos n(x − t ) ,
2
⎨
x
−
t
2⎪
⎪⎪
dt ⎪ sin 2
⎩⎪
⎭⎪
then, as n → ∞ , we obtain the infinite series
− cos(x − t ) − 22 cos 2(x − t ) − 32 cos 3(x − t ) − ...
that diverges to infinity at t = x .
Therefore, the integral cannot be defined, and the question of its
convergence is mute.
Riemann attempted to represent f (x ) by an integral that does
not exist.
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Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
5.
Unproven Divergence of Fourier
Coefficients
Riemann attempted to show that a function with infinitely many
maxima or minima may have diverging Fourier Coefficients. But
his proof is incomplete, and the claim is not proven.
5.1 Unproven divergence of Fourier Coefficients
d
x ν cos x1 } ,
{
dx
0 ≤ x ≤ 2π , 0 < ν < 12 ,
is integrable,
but has infinitely many maxima and minima,
and diverging Fourier Coefficients
Riemann’s Non-Proof:
Since
d
x ν cos x1 } = νx ν −1 cos x1 + x ν (− sin x1 )(− 12 ) ,
{
x
dx
x =2π
∫
the Fourier Coefficient
x =0
d
x ν cos x1 } cos n(x − a )dx has the term
{
dx
x =2 π
∫
x =0
xν
1
x2
sin( x1 )cos n(x − a )dx .
42
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Since sin( x1 )cos n(x − a ) = 12 sin ⎡⎢⎣ x1 + n(x − a ) ⎤⎥⎦ + 12 sin ⎡⎢⎣ x1 − n(x − a ) ⎤⎥⎦ ,
the Fourier coefficient has the term
x =2 π
∫
x =0
x ν −2 sin ⎡⎢⎣ x1 + n(x − a ) ⎤⎥⎦ dx .
The slowest change of the sign of
sin ⎡⎢⎣ x1 + n(x − a ) ⎤⎥⎦ is in the
neighborhood of the point where
d ⎡1
dx ⎣⎢ x
+ n(x − a ) ⎤⎦⎥ = 0 ,
−
1
x2
+n = 0,
x =
1
n
.
Expanding
y(x ) =
1
x
+ n(x − a ) ,
in a second order Taylor Polynomial about x =
y(x ) ≈ y(
1
)
n
+ y '(
1
)(x
n
−
3
2n 2
3
= 2 n − na + n 2 (x −
3
dy
= 2n 2 (x −
dx
(x −
1 2
)
n
=
1
) + 12 y ''( 1 )(x
n
n
0
2 n −na
1 ),
n
y −(2 n −na )
3
n2
.
Therefore,
43
1
n
1 2
) ,
n
,
−
1 2
)
n
Gauge Institute Journal, Volume 7, No 3, August 2011
x >
x <
1
n
⇒ x−
1
n
⇒
1
n
y −(2 n −na )
=
3
,
and
n4
x−
1
n
H. Vic Dannon
3
dy
= 2n 4 y − (2 n − na )
dx
y −(2 n −na )
=−
3
n4
,
and
3
dy
= −2n 4 y − (2 n − na )
dx
x =2 π
∫
The contribution from the term
x =0
the interval [0,
2 ]
n
x ν −2 sin ⎡⎢⎣ x1 + n(x − a ) ⎤⎥⎦ dx in
is
x=
∫
2
n
x ν −2 sin ydx .
x =0
Substituting
x
ν −2
⎛ 1 ⎞⎟ν −2
∼ ⎜⎜
⎟ ,
⎜⎝ n ⎠⎟
the integral is
x=
≈n
1− ν
2
∫
2
n
sin ydx
x =0
x= 2
⎛ x = 1n
⎞⎟
n
⎜
⎟
1− ν ⎜
= n 2 ⎜⎜ ∫ sin ydx + ∫ sin ydx ⎟⎟⎟ .
⎜⎜
⎟⎟
x= 1
⎝ x =0
⎠
n
Changing the integration variable to y ,
x =0⇒
3
y ≈ 2 n − na + n 2 (0 −
= 3 n − na
44
1 2
)
n
Gauge Institute Journal, Volume 7, No 3, August 2011
x =
1
n
⇒
y = 2 n − na
x =
2
n
⇒
y ≈ 2 n − na + n 2 (
3
2
n
−
H. Vic Dannon
1 2
)
n
= 3 n − na
In [0,
1 ]
,
n
x <
1
n
,
dy
dx = −
2n
3
4
y − (2 n − na )
,
and the first integral transforms to
y = 3 n −na
⎛
⎞⎟
dy
sin y
⎜⎜
− 43 1
⎟
dy .
⎟⎟ = n
∫ sin y ⎜⎜⎜ − 43
∫
2
⎟
−
−
y
(2
n
na
)
⎝
−
−
⎠
2
n
y
(2
n
na
)
y = 3 n −na
y = 2 n −na
y = 2 n −na
In [
1
n
,
2
],
n
x >
1
n
,
dy
dx =
2n
3
4
y − (2 n − na )
,
and the second integral transforms to
y = 3 n −na
⎛
⎞⎟
3 1
dy
sin y
⎜⎜
−
dy .
⎟⎟⎟ = n 4
∫ sin y ⎜⎜⎜ 34
∫
2
⎟
y
(2
n
na
)
−
−
⎝ 2n y − (2 n − na ) ⎠
y = 2 n −na
y = 2 n −na
y = 3 n −na
Therefore,
x=
n
1− ν
2
∫
2
n
x =0
sin ydx = n
1 −ν
4 2
y = 3 n −na
∫
y =2 n −na
45
sin y
dy .
y − (2 n − na )
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
Changing variable to
ξ = y − 2 n − na
=n
1 −ν
4 2
ξ= n
∫
ξ =0
Since
1
2
> ν > 0 , we have,
1
4
sin(ξ + 2 n − na )
dξ
ξ
− ν2 > 0 , and as n → ∞ ,
1 −ν
2
n4
→ ∞.
But we don’t know what is
ξ= n
lim
n →∞
∫
ξ =0
sin(ξ + 2 n − na )
dξ .
ξ
Riemann wrote
“If
ξ =∞
∫
ξ =0
sin(ξ + β )
dξ ,
ξ
which equals
π sin(β + π4 )
is not zero…”
demonstrating oblivion to the dependence of the integrand on n .
Thus, leaving the proof hanging on an “if”, incomplete, and his
claim unproven.
46
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
6.
Fourier Series of Riemann’s (x )
Function
6.1 The Riemann (x ) Function
⎧
⎪ x − nearest integer, if x ≠ n +
(x ) = ⎪
⎨
⎪
0, if x = n + 21
⎪
⎩
1
2
6.2 (x ) is periodic with period 1 ,
is continuous in ..., − 23 < x < − 21 , − 21 < x < 21 ,
3
2
1
2
is discontinuous at….. − , − ,
47
1 3
, ,…
2 2
1
2
< x < 23 ,..
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
and has the Fourier series
(x ) =
⎞
1 ⎛⎜ sin 2πx sin 2 ⋅ 2πx
sin 3 ⋅ 2πx
− ... ⎟⎟⎟
−
+
⎜⎜
π⎝ 1
2
3
⎠
A detailed discussion of Riemann’s (x ) Function appears in
[Dan1].
48
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
7.
Fourier Series of Riemann’s (2x )
Function
7.1 (2x ) is periodic with period
1
,
2
is continuous in .., − 43 < x < − 14 , − 14 < x < 41 ,
1
4
< x < 43 ,….
3
4
1
4
is discontinuous at ….. − , − ,
has the graph
49
1 3
, ,…..
4 4
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
and has the Fourier series
(2x ) =
⎞⎟
1 ⎛⎜ sin 4πx sin 2 ⋅ 4πx sin 3 ⋅ 4πx
−
+
−
...
⎟
⎜
2
3
π ⎜⎝ 1
⎠⎟
50
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
8.
Fourier Series of Riemann’s (3x )
Function
8.1 (3x ) is periodic with period 1 ,
3
is continuous in …, − 63 < x < − 16 , − 16 < x < 61 ,
1
6
< x < 63 ,….
3
6
1
6
is discontinuous at ….. − , − ,
has the graph
51
1 3
, ,…..
6 6
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
and has the Fourier series
(3x ) =
⎞
1 ⎜⎛ sin 6πx sin 2 ⋅ 6πx
sin 3 ⋅ 6πx
− ... ⎟⎟⎟
−
+
⎜⎜
π⎝ 1
2
3
⎠
52
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
9.
Fourier Series of Riemann’s (x )
Series
9.1
Riemann’s (x) Series is
f (x ) =
9.2
(x ) (2x ) (3x )
+
+
+ ...
1
2
3
Riemann (x) Series is unbounded in any interval
Hence,
9.3
Riemann (x) Series is nowhere integrable
But
9.4
Riemann (x) Series converges at each rational x = q
And,
9.5
At each rational x = q , Riemann’s (q) Series has the
Fourier Series
53
Gauge Institute Journal, Volume 7, No 3, August 2011
f (q ) =
H. Vic Dannon
⎞
1 ⎛⎜ sin 2πq sin 2 ⋅ 2πq sin 3 ⋅ 2πq sin 4 ⋅ 2πq sin 5 ⋅ 2πq sin 6 ⋅ 2πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
π⎝ 1
2
3
4
5
6
⎠
+
⎞
1 1 ⎛⎜ sin 4πq sin 2 ⋅ 4πq sin 3 ⋅ 4πq sin 4 ⋅ 4πq sin 5 ⋅ 4πq sin 6 ⋅ 4πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
2 π⎝ 1
2
3
4
5
6
⎠
+
⎞
1 1 ⎛⎜ sin 6πq sin 2 ⋅ 6πq sin 3 ⋅ 6πq sin 4 ⋅ 6πq sin 5 ⋅ 6πq sin 6 ⋅ 6πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
3 π⎝ 1
2
3
4
5
6
⎠
+
⎞
1 1 ⎛⎜ sin 8πq sin 2 ⋅ 8πq sin 3 ⋅ 8πq sin 4 ⋅ 8πq sin 5 ⋅ 8πq sin 6 ⋅ 8πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
4 π⎝ 1
2
3
4
5
6
⎠
+
⎞
1 1 ⎛⎜ sin10πq sin 2 ⋅ 10πq
sin 3 ⋅ 10πq sin 4 ⋅ 10πq
sin 5 ⋅ 10πq sin 6 ⋅ 10πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
5 π⎝
1
2
3
4
5
6
⎠⎟
+
⎞
1 1 ⎛⎜ sin 12πq sin 2 ⋅ 12πq
sin 3 ⋅ 12πq sin 4 ⋅ 12πq
sin 5 ⋅ 12πq sin 6 ⋅ 12πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
6 π⎝
1
2
3
4
5
6
⎠⎟
+
⎞
1 1 ⎜⎛ sin 14πq sin 2 ⋅ 14πq
sin 3 ⋅ 14πq sin 4 ⋅ 14πq
sin 5 ⋅ 14πq sin 6 ⋅ 14πq
−
+
−
+
−
+ ... ⎟⎟⎟
⎜⎜
⎟⎠
7 π⎝
1
2
3
4
5
6
+ ……………………………………………………………………………………
=
1
sin 2πq
π
2
+ sin 6πq
3π
1
− sin 8πq
4π
2
+ sin 10πq
5π
+
2
sin 14πq
7π
− ……………………………..
Riemann gave the formula
54
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
9.6 f (q ) has the Fourier Series
⎧
⎪
⎨⎪
∑
⎪
⎪ θ =divisor
⎩
⎫⎪ sin(1 ⋅ 2πq ) ⎧⎪⎪
−(−1)θ ⎬⎪
+⎨
∑
⎪
⎪⎩⎪ θ =divisor
π
⎪
⎭
of 1
⎫⎪ sin(2 ⋅ 2πq )
−(−1)θ ⎬⎪
+ ...
⎪
2
π
⎭⎪
of 2
A detailed discussion of Riemann’s (x ) Series appears in [Dan1].
55
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
10.
Other
Non
Integrable
Fourier
Series
Riemann supplies other examples of Fourier series that converge
at infinitely many points, but are nowhere integrable:
10.1
If as n → ∞ , c0 > c1 > ... > cn ↓ 0
c0 + c1 + ... + cn → ∞
x
p
is a rational in its lowest terms.
2π
q
Then,
c0 + c1 cos x + c2 cos 22 x + c3 cos 32 x + ...
and
c1 sin x + c2 sin 22 x + c3 sin 32 x + ...
converge
⇔
c0 + c1 cos x + c2 cos 22 x + ... + cq −1 cos(q − 1)2 x = 0
and
c1 sin x + c2 sin 22 x + .... + cq −1 sin(q − 1)2 x = 0
56
Gauge Institute Journal, Volume 7, No 3, August 2011
10.2
H. Vic Dannon
⎧ (1 − e ix )
⎫
⎪
⎪
− log(1 − e ix ) (1 − e 2ix )
− log(1 − e 2ix )
⎪
⎪
Im
log
log
...
+
+
⎨
⎬
2
3
ix
3
2
ix
⎪
⎪
2
dx
e
e
⎪ 1
⎪
⎩
⎭
d2
is a Trigonometric Series that converges infinitely often
on any interval.
Its term by term integral
⎧
⎫
⎪ (1 − e ix )
⎪
− log(1 − e ix ) (1 − e 2ix )
− log(1 − e 2ix )
d
⎪
Im ⎪⎨
log
log
...
+
+
⎬
3
ix
3
2
ix
⎪
⎪
dx
1
2
e
e
⎪
⎪
⎩
⎭
diverges infinitely often on any interval.
Its second term by term integral
⎪⎧ (1 − e ix )
⎪⎫⎪
− log(1 − e ix ) (1 − e 2ix )
− log(1 − e 2ix )
Im ⎪⎨
log
log
...
+
+
⎬
⎪⎩⎪ 13
⎪⎭⎪
23
e ix
e 2ix
is a Trigonometric Series
57
Gauge Institute Journal, Volume 7, No 3, August 2011
H. Vic Dannon
11.
Fourier Series with An g 0
Riemann supplies an example of Fourier series that converges at
infinitely many points, although its coefficients do not vanish as
n → ∞.
sin(1! πx ) + sin(2! πx ) + sin(3! πx ) + ...
11.1
converges at each rational
x . (Then it is a finite
sum).
and at infinitely many irrationals such as
sin(1) , cos(1) ,
2m
(2m + 1)
(e − e1 ) , etc.
, (2m + 1)e ,
4
e
References
[Dan1] Dannon, H. Vic, “Riemannian Integration”, Gauge Institute Journal,
Volume 7, No. 2, May 2011.
[Dan2] Dannon, H. Vic, “Infinitesimals”, Gauge Institute Journal, Volume 7, No. 1,
February 2011.
[Riemann] Riemann, Bernhard, “On the Representation of a Function by a
Trigonometric Series”.
1. “Collected Papers, Bernhard Riemann”, translated from the 1892 edition by
Roger Baker, Charles Christenson, and Henry Orde, Paper XII, pages 219256, Kendrick press, 2004
2. “God Created the Integers” Edited by Stephen Hawking, pages 826-859,
Running Press, 2005.
58