The Volterra Function

whose derivative is Riemann Non-integrable

In[]:=

ClearAll["Global`*"];

The Smith-Volterra-Cantor Set

In this section, we construct several iterations of Smith-Volterra-Cantor set by marking its pair of points indicating the interval being removed.

In[]:=

ClearAll[SVC,SVCG];SVC[0]={0,1};SVC[n_]:=Ifn>0,Sort@FlattenJoinTableIfOddQ[k]&&k<Length[SVC[n-1]],

SVC[n-1][[k]]+SVC[n-1][[k+1]]

2

-

1

2

n

4

,

SVC[n-1][[k]]+SVC[n-1][[k+1]]

2

+

1

2

n

4

,Nothing,{k,1,Length[SVC[n-1]]},SVC[n-1],SVC[0];

SVC[4]

Out[]=

0,

9

128

,

11

128

,

5

32

,

7

32

,

37

128

,

39

128

,

3

8

,

5

8

,

89

128

,

91

128

,

25

32

,

27

32

,

117

128

,

119

128

,1

In[]:=

SVCG=SVC[4];Graphics[Table[If[OddQ[k],Line[{{SVCG[[k]],0},{SVCG[[k+1]],0}}]],{k,1,Length[SVCG]}],ImageSizeLarge,PlotRange{{-0.2,1.2},{-0.2,0.2}}]

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The Volterra Function

The Volterra function is constructed recursively, by using

2

x

sin

1

x

.

1

.

Consider subinterval

0,

1

8



, find the maximum zero

z

1

for the derivative of

2

x

sin

1

x

in this interval, and make a piecewise defined function

f

1

such that

f

1

(x)

2

x

sin

1

x

on

[0,

z

1

]

but

f

1

(x)

2

z

1

sin

1

z

1

on



z

1

,

1

8



. Define the function

f

1

(x)

symmetrically on



1

8

,

1

4



, set

f

1

(x)0

elsewhere, and shift the graph of

f

1

(x)

to the right by the amount

3

8

, such that the

supp

f

1

⊂

3

8

,

5

8



.

2

.

Consider subinterval

0,

1

2·

2

4



, find the maximum zero

z

2

for the derivative of

2

x

sin

1

x

inside of this interval, and make a piecewise defined function

f

2

such that

f

2

(x)

2

x

sin

1

x

on

[0,

z

2

]

but

f

2

(x)

2

z

2

sin

1

z

2

on



z

2

,

1

2·

2

4



. Define the function

f

2

(x)

symmetrically on



1

2·

2

4

,

1

2

4



, set

f

2

(x)0

elsewhere, and shift the graph of

f

2

(x)

to the right by the amount

5

32

and

25

32

respectively.

3

.

Do the same thing again and again.

4

.

The Volterra function is defined by adding altogether all such functions.

In[]:=

Chi[a_,b_,x_]:=Piecewise[{{1,a<x<b}},0];V[a_,b_,x_]:=Module{z},z=Maxu/.NSolve2uSin

1

u

-Cos

1

u

==0&&u>

b-a

100

&&u<

b-a

2

,u;

2

(x-a)

Sin

1

x-a

Chi[a,a+z,x]+

2

(b-x)

Sin

1

b-x

Chi[b-z,b,x]+

2

z

Sin

1

z

Chi[a+z,b-z,x];Vd[a_,b_,x_]:=Module{z},z=Maxu/.NSolve2uSin

1

u

-Cos

1

u

==0&&u>

b-a

100

&&u<

b-a

2

,u;2(x-a)Sin

1

x-a

-Cos

1

x-a

Chi[a,a+z,x]+2(b-x)Sin

1

b-x

-Cos

1

b-x

Chi[b-z,b,x];

In[]:=

Volterra[x_]:=Sum[If[OddQ[k],0,V[SVCG[[k]],SVCG[[k+1]],x]],{k,1,Length[SVCG]-1}];DVolterra[x_]:=Sum[If[OddQ[k],0,Vd[SVCG[[k]],SVCG[[k+1]],x]],{k,1,Length[SVCG]-1}];Plot[Evaluate[Volterra[x]],{x,0,1},PlotRange->All]//AbsoluteTimingPlot[Evaluate[DVolterra[x]],{x,0,1},PlotRange->All]//AbsoluteTiming

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45.7245,



Out[]=

46.0275,

