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It is, in particular, in my book "Combinatorial algebra: syntax and semantics". The length of each "petal" is at most the length of the word (times 4) plus the Dehn function multiplied by a constant (the max of lengths of the relators times 4). The book contains a proof. A longer proof with a slightly worse estimate is in the introduction to my joint paper with Birget and Rips. It is where we prove that the word problem in a group with Dehn function $f(n)$ can be solved by a nondeterministic Turing machine in time $\sim f(n)$. And of course the original paper by Madlener and Otto contains a (nonlinear) estimate of the lengths of conjugates.

It is, in particular, in my book "Combinatorial algebra: syntax and semantics" which can be found on my Web site. The length of each "petal" is at most the length of the word plus the Dehn function multiplied by a constant. The book contains a proof (in fact a stronger result is proved). It is essentially a copy of our proof (with A.Yu. Olshanskii) from the paper https://arxiv.org/pdf/math/9811107.pdf of 1998. A longer proof with a slightly worse estimate is in the introduction to my joint paper with Birget and Rips (there is only one such paper in the arXiv). It is where we prove that the word problem in a group with Dehn function $f(n)$ can be solved by a nondeterministic Turing machine in time $\sim f(n)$. And of course the original paper by Madlener and Otto contains a (nonlinear) estimate of the lengths of conjugates. Related statements about a double exponential relationship between Dehn function and isodiametric function of a f.p. group was proved by Gersten.

It is, in particular, in my book "Combinatorial algeba: syntax and semamtics". The length of each "petal" is at most the lngth of the word (times 4) plus the Dehn function multiplied by a constant (the max of lengths of the relators times 4). The book contains a proof. A longer proof with a slightly worse estimate is in the introduction to my joint paper with Birget and Rips. It is where we prove that the word problem in a group with Dehn function $f(n)$ can be solved by a nondeterministic Turing machine in time $\sim f(n)$. And of course the original paper by Madlener and Otto contains a (nonlnear) estimate of the lengths of conjugates.

It is, in particular, in my book "Combinatorial algebra: syntax and semantics". The length of each "petal" is at most the length of the word (times 4) plus the Dehn function multiplied by a constant (the max of lengths of the relators times 4). The book contains a proof. A longer proof with a slightly worse estimate is in the introduction to my joint paper with Birget and Rips. It is where we prove that the word problem in a group with Dehn function $f(n)$ can be solved by a nondeterministic Turing machine in time $\sim f(n)$. And of course the original paper by Madlener and Otto contains a (nonlinear) estimate of the lengths of conjugates.

It is, in particular, in my book "Combinatorial algeba: syntax and semamtics". The length of each "petal" is at most the lngth of the word (times 4) plus the Dehn function multiplied by a constant (the max of lengths of the relators times 4). The book contains a proof.

It is, in particular, in my book "Combinatorial algeba: syntax and semamtics". The length of each "petal" is at most the lngth of the word (times 4) plus the Dehn function multiplied by a constant (the max of lengths of the relators times 4). The book contains a proof. A longer proof with a slightly worse estimate is in the introduction to my joint paper with Birget and Rips. It is where we prove that the word problem in a group with Dehn function $f(n)$ can be solved by a nondeterministic Turing machine in time $\sim f(n)$. And of course the original paper by Madlener and Otto contains a (nonlnear) estimate of the lengths of conjugates.