Thread: What is the name of this function?

1. What is the name of this function?

y=(ax+b)^r
where "a", "b", and "r" are real constants and "r>0".
It is clear that for "r=1", this equation is general form of a linear function; and also for positive integer values of "r" (i.e. Natural numbers), this equation is a polynomial function. But I don't know whether there is a special function name for positive real values of "r" or not. Any reply is appreciated.

2. Originally Posted by mottaghi
y=(ax+b)^r
where "a", "b", and "r" are real constants and "r>0".
It is clear that for "r=1", this equation is general form of a linear function; and also for positive integer values of "r" (i.e. Natural numbers), this equation is a polynomial function. But I don't know whether there is a special function name for positive real values of "r" or not. Any reply is appreciated.
When r is a positive integer you have a polynomial functions.

When r is not, then I do not know what you have.

3. Originally Posted by mottaghi
y=(ax+b)^r
where "a", "b", and "r" are real constants and "r>0".
It is clear that for "r=1", this equation is general form of a linear function; and also for positive integer values of "r" (i.e. Natural numbers), this equation is a polynomial function. But I don't know whether there is a special function name for positive real values of "r" or not. Any reply is appreciated.
It is a binomial function. See Binomial expansion somewhere.

4. Originally Posted by ticbol
It is a binomial function. See Binomial expansion somewhere.

Thank you very much for your contribution.
But I think binomial is an expresion of the form (a+bx) or (1+x) and so on which have two nomials, and not of the form (a+bx)^r. For any power of "r", the binomial " (a+bx) " can be expanded using binomial series...

5. Originally Posted by mottaghi
Thank you very much for your contribution.
But I think binomial is an expresion of the form (a+bx) or (1+x) and so on which have two nomials, and not of the form (a+bx)^r. For any power of "r", the binomial " (a+bx) " can be expanded using binomial series...
(x+y)^r
(33x -0.2y)^r
They are binomial functions too.
Any two terms inside the parentheses, be they variables or constants, can be called a binomial "function".

(2 +3)^4 = 2^4 +4[2^3 *3] +6[2^2 *3^2] +4[2 *3^3] +3^4
= 16 + 4[24] +6[36] +4[54] +81
= 625

(2 +3)^4 = (5)^4 = 625

6. Originally Posted by ticbol
(x+y)^r
(33x -0.2y)^r
They are binomial functions too.
Any two terms inside the parentheses, be they variables or constants, can be called a binomial "function".

(2 +3)^4 = 2^4 +4[2^3 *3] +6[2^2 *3^2] +4[2 *3^3] +3^4
= 16 + 4[24] +6[36] +4[54] +81
= 625

(2 +3)^4 = (5)^4 = 625

Dear "ticbol"
Thank you again. Knowing that:
(x+y)^2 = x^2 + 2xy + y^2 ,
do you mean the collected form is a binomial, while the expanded form is a polynomial (because has more than two terms)? On the other words, do you mean that polynomial function is a sub_set of binomial function?

7. Originally Posted by mottaghi
Dear "ticbol"
Thank you again. Knowing that:
(x+y)^2 = x^2 + 2xy + y^2 ,
do you mean the collected form is a binomial, while the expanded form is a polynomial (because has more than two terms)? On the other words, do you mean that polynomial function is a sub_set of binomial function?
I don't know about the polynomial function is a subset of binomial function. For sure (ax +by +cz)^r will produce polynomial functions too eventhough (ax +by +cz)^r is not a binomial function.

The (x+y)^2 is a binomial function.
It's expansion x^2 +2xy +y^2 is a polynomial function.
Is the polynomial function a subset of the original binomial function? I don't know.
What I'm saying, your y = (ax +b)^r can be called a binomial function in that it obeys a binomial expansion.

To confuse you more, , say,
y = (x^2 -5x +6) / (x -3)
That is a rational or "fractional" function.
Simplify or "expand" that,
y = [(x-3)(x-2)]/(x-3)
y = x-2
The simplified form now is not "fractional" anymore.
The original is not the same as the "expanded" or simplified form.
So, back to your (x+y)^2, a binomial, .....blah, blah....

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Ah, the (ax +by +cz)^r can also be binomial if it is first treated as [ax +(by +cz)]^r. That's only one of the three possible forms. [(ax +by) +cz]^r . [(ax +cz) +by]^r. Play with the binomial expansions of the 3 forms and you should get the same result in the end.