Verified answer

Checking this on a computer, which has an accuracy of probably about 30 decimal places gives:

x^4 - 48x^3 - 12x^2 - 33x = -1613.0000000019244114817039141789

So when we all the final 1613 we get:

-1.9244114817039141788992672849282 * 10^-9

Which is very close, but not quite equal to zero. So pi is not a root of that equation, but it is very close. In fact you could use the exact root of that equation as an approximation to pi.

Try it yourself using the windows calculator.

If you set the calculator to scientific mode and copy this string in:

py4-48*py3-12*py2-33*p+1613

This will evaluate the equation with x = pi. p is the keyboard shortcut to the value of pi on the calculator, y is the shortcut to the power of function.

From the website quickmath.com you can solve the equation to obtain an exact answer and an approximate answer. The approximate answer to the equation is:

x = 3.141592653588443...

while pi is

3.141592653589792328...

So as you can see the first few decimal places are the same but after a while the two answers differ.

It is not a root of that polynomial:

http://www.google.com/search?q=pi%5E4+-+48pi%5E3+-...

However, for any ε>0, you can find an algebraic (non-transcendental) number α with its own characteristic polynomial P(x), where P(α) = 0 and |P(π)|<ε.

This is because the algebraic numbers are dense in the real line, and because polynomials are continuous.

http://en.wikipedia.org/wiki/Dense_set

If you forget to worry about round-offs, you might come to the conclusion that P(π)=0, but that will be untrue for any P. This error is the source of your confusion. There is nothing "supposed" about the transendentialism of π, and there is no gap or error in his proof.

The roots of your polynomial are (approximately):

-1.69506 ± 2.78715 i, 48.2485, and...

3.141592653588443

Compare:

3.141592653588443

3.141592653589793 = π

So you have (re)discovered the interesting fact that roots of polynomials can approximate π, but there is none that is exactly π. Sorry to disappoint you.

A Transcendental Number is any number that is not an Algebraic Number. (i.e., it is not a root of any polynomial equation with rational coefficients.)

Examples of transcendental numbers include π and e

Your polynomial does not have π as a root. It is close, but no cigar.

Pi is transcendental. It is not the root of that polynomial. If you plug it in you will not even get zero. (Do not use a computer, do it by hand and you will see this.)

I agree with my first colleague..pi is transcendental..the number which is the root of your polynomial is only approximate..

See

http://en.wikipedia.org/wiki/Transcendental_number...

second paragraph

Pi is not the root of that polynomial.