1991
DOI: 10.1007/bf03167680
|View full text |Cite
|
Sign up to set email alerts
|

A generalization of Hata-Yamaguti’s results on the Takagi function

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1

Citation Types

1
28
0

Year Published

1995
1995
2020
2020

Publication Types

Select...
6

Relationship

2
4

Authors

Journals

citations
Cited by 19 publications
(29 citation statements)
references
References 4 publications
1
28
0
Order By: Relevance
“…In this section, we prove Theorem 2 in a similar manner to the proof of Theorem 2.2 [18] which generalized Hata and Yamaguti's results connecting the Takagi function with Lebesgue's singular function [9]. We prepare some lemmas without proof.…”
Section: Proof Of Theoremmentioning
confidence: 99%
“…In this section, we prove Theorem 2 in a similar manner to the proof of Theorem 2.2 [18] which generalized Hata and Yamaguti's results connecting the Takagi function with Lebesgue's singular function [9]. We prepare some lemmas without proof.…”
Section: Proof Of Theoremmentioning
confidence: 99%
“…Hata and Yamaguti showed that T is a unique continuous solution of the system (2). More generally, they obtained the following: In [15] we showed that the condition oo is a one Y]n=o [cn[ < oc necessary and gave a concrete representation of the solution. oo a unique continuous solution ifand on y if~n=o Ic~l < c~.…”
mentioning
confidence: 94%
“…In our previous paper [9], noticing the close connection between s(n) and the binomial measure, and using the result obtained in Sekiguchi and Shiota [13], we have obtained ah explicit formula of Sk(N). In this paper, by use of a method similar to that of [9], we study the exponential sum F(~, N).…”
Section: Introductionmentioning
confidence: 99%
“…This method is direct and does not require solving the inductive formula (see [9]). We notice that the higher order derivatives of L(r, x) with respect to r play ah important part in the explicit formula of Sk(N) and that 10L(r,x) 5 of ~=1/2 is the Takagi function (see Hata and Yamaguti [7], [13]). …”
Section: Introductionmentioning
confidence: 99%
See 1 more Smart Citation