# Are there systemic method to producing transcendental numbers?

#### theName()

the title is the question.

#### matt grime

Homework Helper
Yes. pi, pi+1, pi+2, pi+any rational number....

#### theName()

Yes. pi, pi+1, pi+2, pi+any rational number....
what about transendential numbers that are not the multiple, addition, substraction.... of some other transendential numbers?

#### matt grime

Homework Helper
Every number sum of two transcendental numbers.

Sorry to appear obfuscatory, but your question is so open as to not have a meaningful answer. The first number known to be transcendental was shown to be so by demonstrating some property of rate of convergence about it truncated decimal expansions. That might be a method you consider good enough to meet your criteria.

In general it is *very* hard to even decide whether a number is transcendental.

#### Gib Z

Homework Helper
Well, The Tangent, Sine or Cosine of Any Rational Value, when In Radians, Is Irrational. eg $$\sin 1$$ is irrational, Or the Natural Log Of any positive rational number that is not equal to 1.

The Louisville Constant is a good example of a Louisville Number, which is a general class of numbers which are all transcendental.

These numbers all fulfill the property: $$0<|x-p/q|<1/q^n$$ for integers p and q, q>1 and n is any positive integer. One property of the Louisville numbers is that they can be reasonably well approximated by rational numbers. As algebraic irrational numbers do not have this property, we can prove Louisville numbers are transcendental.

First we prove they are irrational by letting x = a Liouville number, x is irrational. Assume otherwise; then there exists integers c, d with x = c/d. Let n be a positive integer such that 2n−1 > d. Then if p and q are any integers such that q > 1 and p/q ≠ c/d, then

$$|x-p/q|=|c/d-p/q| >= 1/dq > 1/(2^{n-1} q) >= 1/q^n$$ (where >= means more or equal to)

which contradicts the definition of Liouville number. Therefore They are not rational. Then we use the property that irrational algebraic numbers cannot be approximated by rational numbers very well. Since Louisville numbers do not have this property, they can not be algebraic and therefore irrational.

Another Way of finding transcendental numbers is by using the Gelfond-Schnieder theorem. It states that if a and b are algebraic numbers, where a is not equal to zero or one, and b is not a real rational number, then a^b is a transcendental number. I believe perhaps this one may be the most helpful to you. This can generate many transcendental numbers, just let a equal some integer or so, not equal to 0 or 1, and b any irrational number, and we have a^b being transcendental. Eg 2^pi, 2^(root3) , or (sqrt2)^(sqrt2).

I have seen this be used to prove the transcendence of pi or e, cant remember, by Euler's Identity: $$e^{i\pi}=-1$$. I can't be bothered to write the relativity simple proof, its a little exercise for you :)

From what i've written here you should be able to recognise that ln(3), cos 3, and 2^root 2 all fall into the set of transcendental numbers, I hope i helped.

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#### gabee

The first thing I thought of when I read your question is the use of a power series or Taylor polynomial to describe or approximate a transcendental function. For example, the transcendental function $$sin\,x$$ can be represented by:

$$sin\,x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n-1)!}$$

Wikipedia has a good article on Taylor series here: http://en.wikipedia.org/wiki/Taylor_series

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#### Gib Z

Homework Helper
O, I just read over my post and realised I said something that was correct, but not intended. The sin, tan, or cos of any rational radian value is transcendental, as well as irrational. same goes for the natural log. And same goes for the Louisville numbers, they are not algebraic and therefore transcendental.

#### d_leet

O, I just read over my post and realised I said something that was correct, but not intended. The sin, tan, or cos of any rational radian value is transcendental, as well as irrational.
Any rational radian value other than zero.

#### Gib Z

Homework Helper
O yes, forgot about zero :) ty

#### agentredlum

Yes. pi, pi+1, pi+2, pi+any rational number....

which is .1 + .001 + .000001 + .0000000001 + .000000000000001 etc isn't this transcendental? The number of 0's between 1's keep increasing by one and the digit 1 appears in the triangular number position. All other digit positions 0

Explanation: .1 position 1 triangular
.101 positions 1 and 3 1,3 triangular
.101001 positions 1,3,6 all triangular
.1010010001 positions 1,3,6,10 all triangular etc.

#### HallsofIvy

Homework Helper
I believe you mean Liouville numbers, not "Louisville numbers". Named for Joseph Liouville who proved the existance of transcendental numbers in 1844.

#### SteveL27

the title is the question.
There could never be a systematic method for generating all transcendentals, because there are uncountably many transcendentals; but only countably many "systems," if by system you mean an algorithm or procedure expressible as a finite string of symbols from a countable alphabet.

Familiar transcendentals such as pi and e are expressible by simple descriptions or algorithms; but most transcendentals don't have that property.