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Two days ago, I tried to create an infinite series that might be able to generate a transcendental number, and when I checked the proper definition, it was mentioned that, it is a number that cannot be expressed as a root of any non zero polynomial with integral coefficients and that it is irrational. (Please tell me if I am right)

And I came up with this series:

$$\gamma = \sum_{r=0}^\infty\frac{1}{(5^r)(10^{f(r)})},f(r)=4^r$$ The basic idea that I had was to generate a number, such that its decimals will be randomised by the digits of the numbers of geometric progression of 2.

$$\frac{1}{5^1}=0.2$$ $$\frac{1}{5^2}=0.04$$ $$\frac{1}{5^3}=0.008$$ $$\frac{1}{5^4}=0.0016$$ $$\frac{1}{5^5}=0.00032$$ By means of trial and error, I found $10^{4^r}$ to be a perfect factor to addup zeroes in the decimal And I believe that the function would produce something like this:

$0.10002.........$

My questions are:

  • Is the definition of transcendental number I mentioned legit?
  • Does the series produce a transcendental number?
  • If it does not, can I by chance define a $f(r)$ such that it gives a trascendental number?

I am a complete novice to this, so feel free to leave your advice and opinions here! Thank you!

J. W. Tanner
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Teflon
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    Check out Roth's theorem on algebraic numbers. I think you'll find that the partial sums of your series give rational approximations to $\gamma$ that are too good to be consistent with Roth's theorem if $\gamma$ were algebraic. – Greg Martin Apr 18 '24 at 15:22
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    Most numbers are transcendental because there are only countably many algebraic numbers. It is easy to create an infinite sum that evaluates to a transcendental, but often hard to prove it is transcendental. – Ross Millikan Apr 18 '24 at 15:39

1 Answers1

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It looks as if the zeroes in the decimal expansion of your number may occur in long enough chunks to make it a Liouville number and thus transcendental.

If not, you can probably modify your construction to make the argument work.

Ethan Bolker
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