In the book " Number Theory IV Transcendental Numbers" written by Parsin and Shafarevich (book, page 104) it is asserted that to explicit a transcendence measure of a complex number $w$, it is sufficient to prove that there is an infinite sequence of polynomials $P_m(x)\in\mathbb Z[x]$ of fixed degree such that $$0<H(P_m)\le c^m\qquad e^{-\lambda_1m}\le|P_m(w)[\le e^{-\lambda_2m}$$ where $c$, $\lambda_1$, and $\lambda_2$ are constants, such that $c > 1, \lambda_l > \lambda_2 > 0$ with $H(P_m)=\max\{|\text{coefficients of $P_m$}|\}$. I do not know why. Can anyone give the argument behind this assertion or a reference? This result is used to give a transcendence measure of $\ln(r)$ ($r$ is a rational)

Thanks in advance

  • 2
    $\begingroup$ @fedor Why did you remove your post. It was very interesting, even it was not an answer to my question. $\endgroup$
    – joaopa
    Feb 18, 2023 at 20:49
  • $\begingroup$ Did you ask essentially the same question before? Just use polynomials $P_m(z)^k$. By the way, that book is written by N. I. Fel'dman and Yu. V. Nesterenko, and not by Parshin and Shafarevich (who are editors of the series). $\endgroup$ Feb 19, 2023 at 2:15
  • $\begingroup$ A detail answer would be welcome $\endgroup$
    – joaopa
    Feb 19, 2023 at 2:28
  • $\begingroup$ It's not a research question. It's an exercise for students, who just saw a definition of transcendence measure. By the way, the correct claim is that you can get a bound for a transcendence measure (not the actual measure, that is not known). $\endgroup$ Feb 19, 2023 at 2:40

1 Answer 1


Warning: this is for irrationaity measure, not transcendence measure.

Let $a/b$ be an approximation of $w$ such that $|w-a/b|=b^{-\kappa}$. Then $$P_m(a/b)=P_m(w)+(w-a/b)P_m'(\theta)$$ for certain $\theta$ between $a/b$ and $w$. Note that $P_m(a/b)$ is either 0 or at least $b^{-d}$ in absolute value, where $\deg P_m\leqslant d$. Choose $m$ such that $b^{-d}\geqslant 2e^{-\lambda_2 m}$, say, $m=\lceil \frac{\log 2+d\log b}{\lambda_2}\rceil$.

Then, if $|P_m(a/b)|\geqslant b^{-d}$, we get $$C(d,w)\cdot c^m b^{-\kappa}\geqslant |b^{-\kappa}P_m'(\theta)|=|(w-a/b)P_m'(\theta)|=|P_m(a/b)-P_m(w)|\geqslant e^{-\lambda_2 m},$$ thus $\kappa\log b\leqslant m(\log c+\lambda_2)+O(1)$ and $\kappa\leqslant d(1+\frac{\log c}{\lambda_2})+o(1)$.

If $P_m(a/b)=0$, then analogously $$C(d,w)\cdot c^m b^{-\kappa}\geqslant |b^{-\kappa}P_m'(\theta)|=|(w-a/b)P_m'(\theta)|=|P_m(w)|\geqslant e^{-\lambda_1 m},$$ so $\kappa\log b\leqslant m(\log c+\lambda_1)+O(1)$, and $\kappa\leqslant d\cdot \frac{\log c+\lambda_1}{\lambda_2}+o(1)$.

So, in both cases we may conclude that the irrationality measure of $w$ does not exceed $$d\cdot \frac{\log c+\lambda_1}{\lambda_2}.$$

  • $\begingroup$ Thanks for your comment. But you give an irrationality measure, not a transcendence measure. $\endgroup$
    – joaopa
    Feb 18, 2023 at 20:06
  • $\begingroup$ Thanks to your post, I manage to produce a transcendence measure. Soon, I will write a proof. This could help some people new to this subject. $\endgroup$
    – joaopa
    Mar 13, 2023 at 2:34

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