Is there an alternate nth power function without exponents?

[eddybob123]

Hi guys I am just wondering if there exists an alternate nth power function that doesn't involve any exponents. Thank you in advance.

 

[DonAntonio]

Hi guys I am just wondering if there exists an alternate nth power function that doesn't involve any exponents. Thank you in advance.

What is an "alternate n-th power function"?

DonAntonio

 

[Mark44]

"nth power" is another way of saying that the exponent is n.

 

[DonAntonio]

"nth power" is another way of saying that the exponent is n.

That I know, what is "aternate" such a function, though? Perhaps he meant "alternative"? But then he'd be asking whether there's an exponential function that doesn't involve exponents...!

DonAntonio

 

[dipole]

if n is an integer you can just express it as a sum or product.

 

[Mark44]

eddybob123,
If by "power function" you mean a function in which a variable is raised to a constant power (xⁿ), then no, there is no way to express it without an exponent.

If that's not what you mean by power function, then please clarify for us what you're asking.

if n is an integer you can just express it as a sum or product.

I don't see what this has to do with what the OP asked.

 

[eddybob123]

But there should obviously be one that exists. For example, if one lists out the squares of the integers and find their differences, one will achieve all the positive odd numbers. Likewise, if one lists out the cubes of the integers, and find their differences and the differnece of their difference, one will achieve all the multiples of 6. I was just wondering whether their was a general formula for all the nth powers.

 

[DonAntonio]

But there should obviously be one that exists. For example, if one lists out the squares of the integers and find their differences, one will achieve all the positive odd numbers. Likewise, if one lists out the cubes of the integers, and find their differences and the differnece of their difference, one will achieve all the multiples of 6. I was just wondering whether their was a general formula for all the nth powers.

What does this have to do at all with your original question??

DonAntonio

 

[chiro]

Hi guys I am just wondering if there exists an alternate nth power function that doesn't involve any exponents. Thank you in advance.

Hey eddybob123.

One way that you could do it is first decompose into integer and non-integer parts (i.e. x^(a+b) = x^a * x^b where a is an integer and b is the fractional part).

From this you can reduce the x^a to a summation (humungous but huge non-the-less) which then leaves how to resolve x^b.

For this part you can use a power-series expansion that is of the form of an 'infinite-polynomial'.

Technically you can't calculate this result in a finite-manner, but you can write the summation out in terms of a power series with integer powers and because they are in integer powers, you can use the ideas for summations from integer powers and do it this way.

So for example let's look at x^(2.5).

Take x^(2.5) = x^2 * x^(1/2). x^2 can be written out as a summation (you need to take into account the fractional part of x as well).

Now x^(1/2) has a power series expansion with integer powers. Again you will need to take into account the fractional parts and deal with these in a way that translates to summation.

Your answer most likely won't lend itself to finite-computation of integers or rational numbers, but it still can be expressed in terms of a summation.

When x is not an integer you have x = c + d where c is an integer and d is the fractional part. So now from this you have to consider (c+d)^f for some f.

The thing is though, that you may have to do many of these transformations repeatedly to get something in terms of rational numbers, but it can be done. Just don't expect the final transformation to be a simple one.

If x is an integer or a rational number, use the method presented first in this response. If however it is not rational, then you will need to use more general tools which will be a lot harder.

 

[haruspex]

if one lists out the cubes of the integers, and find their differences and the differnece of their difference, one will achieve all the multiples of 6.

Yes, you can achieve rⁿ (plus smaller terms) by doing n summations up to r of the constant n!:
1! S(1, r) = Ʃ 1 = r
2! S(2, r) = 2! Ʃ S(1, r) = r(r+1) = r² + r
3! S(3, r) = 3! Ʃ S(2, r) = r(r+1)(r+2) = r³ + 2r² + 2r
etc.
Is that what you had in mind?

 

[dipole]

eddybob123,
If by "power function" you mean a function in which a variable is raised to a constant power (xⁿ), then no, there is no way to express it without an exponent.

If that's not what you mean by power function, then please clarify for us what you're asking.


I don't see what this has to do with what the OP asked.

xⁿ = x*x*x*x... n times. This can also be written as a sum.

 

[eddybob123]

Yes, haruspex, except with just general powers x^n expressed in summation form [n{SUM}k=1]f(x,n)

 

[Mark44]

xⁿ = x*x*x*x... n times. This can also be written as a sum.

OK, I'm game - show me.

 

[haruspex]

3! S(3, r) = 3! Ʃ S(2, r) = r(r+1)(r+2) = r³ + 2r² + 2r

Er.. I mean = r³ + 3r² + 2r

 

[eddybob123]

I know that for all x^n, that the terms p_n are related to n, and that x^n=p_1 + p_2 + p_3 + ... + p_n. I just need to find a direct function relating all values of p.