Is this form possible?

The trouble is that you clearly don't know the meaning of the words you are using, so it is senseless to try to answer. In any case, it would contribute nothing to your stated goal, so I suppose the answer is no.
 
Dr.Peterson said:
The trouble is that you clearly don't know the meaning of the words you are using, so it is senseless to try to answer. In any case, it would contribute nothing to your stated goal, so I suppose the answer is no.
Click to expand...

This was clarified - the meaning was eventually understood.
Polynomial with no variables, only pi, coefficients, radicals
and a single ± whose variability reflect the two solutions
1 and phi^3.
 
nothing said:
Please do not scapegoat your own nature onto others - I asked a question. Would a radical expression whose only two solutions are 1 and 4.23606797... count? It's a yes/no question.
Click to expand...
Given that you are having trouble with some very basic terminology and even admit that you don't know much Math I think it's reasonable that we will be skeptical.

I am still trying to figure out what this means
I am interested in polynomials whose roots are both real and irrational.

Take [math]ax^2+bx \pm c = 1, 4.23606797...[/math]
wherein only by changing the +/- (not the value) of c, 1 becomes [math]\phi ^3[/math] and vice versa.
I am wondering if such a thing exists.
Click to expand...
It looks like your question is if [math]ax^2 + bx \pm c = 0[/math] then the roots are 1 and [math]\phi ^3[/math] for both. Let's take a look:
[math]x = \dfrac{-b \pm \sqrt{ b^2 - 4ac} }{2a}[/math]Since c only appears under the radical then you are asking if there is ever a case where
[math]b^2 - 4ac = b^2 + 4ac[/math][math]-4ac = 4ac[/math]Since a cannot be 0 by definition c = 0.

This means your original quadratic must be [math]ax^2 + bx[/math]. But this has zero at x = 0, which is neither 1 nor [math]\phi ^3[/math].

So the answer is no.

Did I get the question right?

-Dan
 
topsquark said:
Given that you are having trouble with some very basic terminology and even admit that you don't know much Math I think it's reasonable that we will be skeptical.

I am still trying to figure out what this means

It looks like your question is if [math]ax^2 + bx \pm c = 0[/math] then the zeros are 1 and [math]\phi ^3[/math] for both. Let's take a look:
[math]x = \dfrac{-b \pm \sqrt{ b^2 - 4ac} }{2a}[/math]Since c only appears under the radical then you are asking if there is ever a case where
[math]b^2 - 4ac = b^2 + 4ac[/math][math]-4ac = 4ac[/math]Since a cannot be 0 by definition c = 0.

This means your original quadratic must be [math]ax^2 + bx[/math]. But this has zero at x = 0, which is neither 1 nor [math]\phi ^3[/math].

So the answer is no.

Did I get the question right?

-Dan
Click to expand...

No, not close. Best to start clean:

Polynomial radical expression with no variables, less a single ±
which produces the 1, 4.236... when flipped back and forth.
There is no x or 0.
 
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nothing said:
No, not close. Best to start clean:

Polynomial radical expression with no variables, less a single ±
which produces the 1, 4.236... when flipped back and forth.
There is no x or 0.
Click to expand...
But polynomials are defined as having an unknown! They take the form [math]ax^n + bx^{n-1} + \text{ ... } + cx + d[/math].

I'm going to out on another limb. Are you looking for an expression like (and I'm just making this up) [math]( \sqrt{6} )^3 + 2 \pi ^2 - 3[/math]? It's not a polynomial but it contains the list of exponents and no variables. Or maybe [math]3(5)^2 + 2(5) - 6[/math]?

-Dan
 
nothing said:
Polynomial radical expression with no variables, less a single ±
which produces the 1, 4.236... when flipped back and forth.
Click to expand...
Cubist said:
Perhaps OP is looking for \(\displaystyle \phi^2 \pm \phi \) since

\(\displaystyle \phi^2+\phi=\phi^3 \)
\(\displaystyle \phi^2-\phi=1 \)
Click to expand...

I think you've got it. Though we can't be sure how to interpret "polynomial radical expression", it seems that the goal is an expression [MATH]A\pm B[/MATH] such that [MATH]A+B = \phi^3[/MATH] and [MATH]A-B = 1[/MATH]. And the solution to that system of equations is [MATH]A = \phi^2[/MATH], [MATH]B = \phi[/MATH].

Or, to put it in terms of radicals, [MATH]A = \frac{3+\sqrt{5}}{2}[/MATH], [MATH]B = \frac{1+\sqrt{5}}{2}[/MATH]. So the desired expression is [MATH]\frac{3+\sqrt{5}}{2}\pm\frac{1+\sqrt{5}}{2}[/MATH].
 
nothing said:
Please do not scapegoat your own nature onto others - I asked a question. Would a radical expression whose only two solutions are 1 and 4.23606797... count? It's a yes/no question.
Click to expand...
I have no reason as yet to believe you are a troll, but the question you are asking has been quite incoherent.

Are you asking whether it is possible to have a quadratic function with real coefficients that has two real roots, one of which is rational and one irrational? Virtually every noun and adjective in that question have very specific meanings. You cannot expect to get helpful answers if you do not bother to learn what those specific meanings are.
 
topsquark said:
I don't see any reason that they couldn't be. As Romsek says, the quadratic would be [math](x - 1)(x - \phi ^3 )[/math]. On the other hand, when you expand this out to the standard [math]ax^2 + bx + c[/math] form none of a, b, and c are going to be integers in general.

-Dan
Click to expand...
Are you sure about that? I suspect that a=1 for what you wrote.
 
If I have guessed what your specific question is, then here we go.

[MATH]\text {Let } p \text { be an ARBITRARY irrational number, which} \implies p \ne 0.[/MATH]
[MATH]\text {Let } q \text { be an ARBITRARY rational number such that } q \ne 0.[/MATH]
[MATH]\text {Let } r \text { be an ARBITRARY real number such that } r \ne 0.[/MATH]
[MATH]\therefore f(x) = r(x - p)(x - q) = r\{x^2 - x(p + q) + pq = rx^2 - x(pr + qr) + pqr.[/MATH]
[MATH]\text {Let } a = r,\ b = -\ (pr + qr), \text { and } c = pqr.[/MATH]
[MATH]\therefore f(x) = ax^2 + bx + c \implies[/MATH]
[MATH]x = \dfrac{- \ \{ - \ (pr + qr)\} \pm \sqrt{\{-\ (pr + qr)\}^2\} - 4r(pqr)}}{2r} = \dfrac{r(p + q) \pm \sqrt{p^2r^2 + 2pqr^2 + q^2r^2 - 4pqr^2}}{2r} =[/MATH]
[MATH]\dfrac{r(p + q) \pm r\sqrt{p^2 + 2pq + q^2 - 4pq}}{2r} = \dfrac{p + q \pm \sqrt{p^2 - 2pq + q^2}}{2} = \dfrac{p + q \pm \sqrt{(p - q)^2}}{2}.[/MATH]
[MATH]\therefore x = \dfrac{p + q + p - q}{2} = p \text { or } x = \dfrac{p + q - p + q}{2} = q.[/MATH]
Far from being rare, there are an infinite number of quadratic functions that have a non-zero rational root and an irrational root.
 
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topsquark said:
But polynomials are defined as having an unknown! They take the form [math]ax^n + bx^{n-1} + \text{ ... } + cx + d[/math].

I'm going to out on another limb. Are you looking for an expression like (and I'm just making this up) [math]( \sqrt{6} )^3 + 2 \pi ^2 - 3[/math]? It's not a polynomial but it contains the list of exponents and no variables. Or maybe [math]3(5)^2 + 2(5) - 6[/math]?

-Dan
Click to expand...

Really? Okay, then what I am after is a quadratic expression. It is in the form Jeff posted.

Cubist said:
Perhaps OP is looking for \(\displaystyle \phi^2 \pm \phi \) since

\(\displaystyle \phi^2+\phi=\phi^3 \)
\(\displaystyle \phi^2-\phi=1 \)
Click to expand...

Dr.Peterson said:
I think you've got it. Though we can't be sure how to interpret "polynomial radical expression", it seems that the goal is an expression [MATH]A\pm B[/MATH] such that [MATH]A+B = \phi^3[/MATH] and [MATH]A-B = 1[/MATH]. And the solution to that system of equations is [MATH]A = \phi^2[/MATH], [MATH]B = \phi[/MATH].

Or, to put it in terms of radicals, [MATH]A = \frac{3+\sqrt{5}}{2}[/MATH], [MATH]B = \frac{1+\sqrt{5}}{2}[/MATH]. So the desired expression is [MATH]\frac{3+\sqrt{5}}{2}\pm\frac{1+\sqrt{5}}{2}[/MATH].
Click to expand...

Thank you both - this helps immensely. Can I ask what is the proper name for such an expression?
Just an expression in terms of radicals?
Also, why has this not been used to prove the Reimann Hypothesis as true?
 
nothing said:
Just an expression in terms of radicals?
Also, why has this not been used to prove the Reimann Hypothesis as true?
Click to expand...
Polynomial functions have a finite number of addends, each of which is a known number times a variable raised to a power that is a known non-negative integer. The degree of a polynomial is the highest power where the multiplying number is not zero.

[MATH]f(x) = - 3x + 4[/MATH] is a polynomial of degree 1.

[MATH]f(x) = 2x^4 - 3x + 4[/MATH] is a polynomial of degree 4.

The zeroes of f(x) are the values of x such that f(x) = 0. Polynomials of degree higher than 4 may not have zeroes expressible solely in radicals. Even the meaning of the term "a solution by radicals" is a bit comples, but polynomials of degree 5 and higher are not generally solvable by radicals.


The Riemann Hypothesis is not about a polynomial function at all, but about a much more complex function. It involves an infinite number of addends, each of which is a known number raised to the power of a variable, which variable need not even be a real number. We have gone from the finite to the infinite and are no longer talking about known non-negative integer powers.

I have a suggestion for anyone not a mathematician (such as me) when confronted with a problem that professional mathematicians have not been able to solve in decades. If you think you have a simple solution, perhaps you have misunderstood the problem because the mathematicians would almost certainly have stumbled on a simple solution at some time over the last 150 years.
 
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JeffM said:
Polynomial functions have a finite number of addends, each of which is a known number times a variable raised to a power that is a known non-negative integer. The degree of a polynomial is the highest power where the multiplying number is not zero.

[MATH]f(x) = - 3x + 4[/MATH] is a polynomial of degree 1.

[MATH]f(x) = 2x^4 - 3x + 4[/MATH] is a polynomial of degree 4.

The zeroes of f(x) are the values of x such that f(x) = 0. Polynomials of degree higher than 4 may not have zeroes expressible solely in radicals. Even the meaning of the term "a solution by radicals" is a bit comples, but polynomials of degree 5 and higher are not generally solvable by radicals.


The Riemann Hypothesis is not about a polynomial function at all, but about a much more complex function. It involves an infinite number of addends, each of which is a known number raised to the power of a variable, which variable need not even be a real number. We have gone from the finite to the infinite and are no longer talking about known non-negative integer powers.

I have a suggestion for anyone not a mathematician (such as me) when confronted with a problem that professional mathematicians have not been able to solve in decades. If you think you have a simple solution, perhaps you have misunderstood the problem because the mathematicians would almost certainly have stumbled on a simple solution at some time over the last 150 years.
Click to expand...

Thanks Jeff.

I believe what I have is a (or, the?) radical expression of a "null" golden quadratic.
By plotting this single quadratic into mathway, along with a sin-of-theta I got this (colors inverted):

f5b.jpg

zooming in:

0.5.jpg

g2b.jpg
 
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