# Product Series

Sikz
$$0 = \prod_{n=a}^b (x-n)$$

I've not taken any calculus classes yet (I'm learning this stuff on my own because I need it for things that I'm working on), so this may be an elementary question... But is it possible to solve for a variable "within" a product series like this?

In the above example, is it possible to solve for x? If so, how, and if not, why?

Homework Helper
In this case, every 'n' (so going from a to b) will be a solution of this equation.

Homework Helper
Assuming:

$$a, b \in \mathbb{N}$$

Then all that tells us is that:

$$a \leq x \leq b \quad \text{and} \quad x \in \mathbb{N}$$

Sikz
I understand that, but I need to use every value of n in another equation. I have an equation similar to the one I just showed you and another... And the other requires an input of every integer between a and b. I figured I could do this with a product series... But then I realized that I can't solve for x, so I can't substitute anything into the other equation for where the variable x is now.

That's why I need to solve for x, or figure out that it can't be done.

Homework Helper
o.k, think about it this way:

0 = (x - [a])(x - [a+1])(x - [a+2])...(x - [b-1])(x - )

Which means one of those brackets is equal to 0, let's say it is (x - [t]) for $a \leq t \leq b$ and $t \in \mathbb{N}$ therefore x - t = 0 and x = t.

Sikz
Zurtex said:
o.k, think about it this way:

0 = (x - [a])(x - [a+1])(x - [a+2])...(x - [b-1])(x - )

Which means one of those brackets is equal to 0, let's say it is (x - [t]) for $a \leq t \leq b$ and $t \in \mathbb{N}$ therefore x - t = 0 and x = t.

I understand that. I didn't get this equation from the internet or something; I constructed it delibrately as an equation that would give me x = all intergers such that a<=x<=b.

To simplify it to the basics, I need x to be defined by

$$0 = \prod_{n=a}^b (x-n)$$

And then solve for values of y:

$$0 = \prod_{k=c}^d (y-x(2+k))$$

As you can see, I need to rearrange the first equation, if possible, into an x= format; that way I can substitute whatever is equal to x into the second equation in place of x.

I understand that x is all integers greater than or equal to a and less than or equal to b. I made the equation specifically for that purpose; but I need x on its own, so I can substitute.

If this is impossible, I have an alternate (trig-based) form of the equations that act to the same purpose; I have the same problem with them, the inability to isolate x. And while the problem would be easier to overcome with those equations, these are preferrable. So, unless what I want to do is impossible, I'd prefer to use these if anyone can help me ><.

Thanks to you guys who have replied so far ^_^.

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daveb
I'm assuming that the index for the second was supposed to be k, not n.

The easiest way is to use set notation, so that with a, b, c, d, n, and k integers:

From the first equation, you know that x is an element of {n | a < n < b }

The using the second equation, y is an element of {n*(2+k) | a < n < b}, c < k < d}.

Sikz
daveb said:
I'm assuming that the index for the second was supposed to be k, not n.

The easiest way is to use set notation, so that with a, b, c, d, n, and k integers:

From the first equation, you know that x is an element of {n | a < n < b }

The using the second equation, y is an element of {n*(2+k) | a < n < b}, c < k < d}.

Yes, it was supposed to be k! I'm sorry, that's what I get for copying and pasting instead of rewriting the code from scratch. I've edited it to what it should be.

And the set notation is a good idea... It would work excellently, but is there no way to put all of those values into a single variable (isolate x in the first equation or use some other method?)? It would be most convenient to have everything in one simple equation instead of having extra definitions and sets and things floating around.

If it is not possible to isolate the x in the first equation, perhaps someone could explain the (presumably) more simple problem of simplifying an equation that contains sine? I've got an equation for a two waves that intersect with the x-axis and each other at every integer between a and b... But I've been unable to perform the simple process of solving for those intersections... When I've tried (assuming I'm setting up such an elementary problem correctly ><), I always run into the sine... And if I cancel it out with sine$$^{-1}$$, I don't get all of the values that the actual equation gives-- because sine modulates whatever input I give it, and I can't account for that in my solution...

I'd prefer to isolate x in the equation I've posted above... But if that isn't possible, the intersection of the two wave functions and the x-axis would be sufficient.

Dr Avalanchez
If I understand your problem correctly, this is just an application of the implicit function theorem? f(x,n) is your function, P=(a,b) is a point satisfying f(P)=0. If df/dx<>0 in P, then there exists a unique function g so that f(g(n),n)=f(x,n) (so g(n)=x) in an open interval round x.

Sikz
If I understand your problem correctly, this is just an application of the implicit function theorem? f(x,n) is your function, P=(a,b) is a point satisfying f(P)=0. If df/dx<>0 in P, then there exists a unique function g so that f(g(n),n)=f(x,n) (so g(n)=x) in an open interval round x.

I'm not exactly sure of what you mean, having not learned this level of mathematics... But, from what I can tell and from what I've researched, I'm pretty sure that you're right. That is what I'm trying to do, although I don't know how to use differential equations or the implicit function theorem-- I only about them and what they are in basic terms ><.

Dr Avalanchez
It only means that your function is solvable (and unique) in an open interval round x. It does not tell what or how you should solve it. It only tells that your quest for such a function will not be fruitless.

In most cases, it is sufficient just to know that such a function exists, without having to search explicitly for one. You can also find things like first deriviative in a point, without having to know the function.

Do you want such a function explicitly? (I take it we assume a<=b, positive integers)

Sikz
Dr Avalanchez said:
It only means that your function is solvable (and unique) in an open interval round x. It does not tell what or how you should solve it. It only tells that your quest for such a function will not be fruitless.

In most cases, it is sufficient just to know that such a function exists, without having to search explicitly for one. You can also find things like first deriviative in a point, without having to know the function.

Do you want such a function explicitly? (I take it we assume a<=b, positive integers)

Thank you for the explanation.

Yes, I want such a function explicitly. And yes again, a<=b, positive integers.

Sikz
Thank you all for your help, but the answer randomly came to me. I've found the function, so this thread is no longer necessary.

Thank you all very much ^_^