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# infinite division function dealie?

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is there any way to make thispattern repeat infinitely but without me typing it forever? like is there a function for this

TheJonyMyster  Jun 23, 2015

#1
+26971
+10

I don't know of a universally accepted, concise representation for this, but I guess you could use something like the list form used for continued fractions. If we call the whole thing g(f(x)) then here this might look like:

g(f(x)) = [f(x); f(x+1), f(x+2), ... , f(x+n)]

Notice that the first term is followed by a semicolon, the others by a comma.

Or you could use

$$g(f(x))=f(x)+\frac{f(x)}{f(x+1)+}\frac{f(x+1)}{f(x+2)+}...\frac{f(x+n-1)}{f(x+n)}$$

a notation also used for continued fractions.

If you want an infinite repeat, you could just end each of the above with ... instead of adding the terms with n.

Have a look at https://en.wikipedia.org/wiki/Continued_fraction for other possibilities.

.

Alan  Jun 23, 2015
#1
+26971
+10

I don't know of a universally accepted, concise representation for this, but I guess you could use something like the list form used for continued fractions. If we call the whole thing g(f(x)) then here this might look like:

g(f(x)) = [f(x); f(x+1), f(x+2), ... , f(x+n)]

Notice that the first term is followed by a semicolon, the others by a comma.

Or you could use

$$g(f(x))=f(x)+\frac{f(x)}{f(x+1)+}\frac{f(x+1)}{f(x+2)+}...\frac{f(x+n-1)}{f(x+n)}$$

a notation also used for continued fractions.

If you want an infinite repeat, you could just end each of the above with ... instead of adding the terms with n.

Have a look at https://en.wikipedia.org/wiki/Continued_fraction for other possibilities.

.

Alan  Jun 23, 2015