How do I produce a function from a series?

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A simple example is 1+x+x^2=x^3+...=1/(1-x)

I know that is equals 1/(1-x), but how do you arrive at that? Thanks.
 
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There are lots of ways. The analysis route would be to show that 1+x+...+x^n=(1-x^{n+1})/(1-x). If |x|<1, then x^{n+1} -> 0 so the limit becomes just 1/(1-x).
 
well 1+x+x^2=x^3+... is like a+ar+ar^2+...
so that a=1 and r=x
and well if they stated that |x|<1 so this would imply that |r|<1

so then the sum to infinity(which is the sum of the infinite series1+x+x^2+x^3+...)

would be S_\infty=\frac{a}{1-r}

which is \frac{1}{1-x}
 
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sort of a generating function approach.

define a function f by the series:

f(x) = 1 + x + x^2 + x^3 + ...

now take a look at x*f(x):

xf(x) = x + x^2 + x^3 + ...

add them together:

f(x) - xf(x) = (1 + x + x^2 + x^3 + ... ) - (x + x^2 + x^3 + ... ) = 1.
notice the terms cancel out. so f(x) - xf(x) = 1.

factor out f(x) from the terms in the left-hand-side:

f(x)(1-x) = 1.

dividing both sides by (1-x) yields the result:

f(x) = 1/(1-x).
 
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the simplest way is do use polynomial division =]
 
antiemptyv said:
sort of a generating function approach.

define a function f by the series:

f(x) = 1 + x + x^2 + x^3 + ...

now take a look at x*f(x):

xf(x) = x + x^2 + x^3 + ...

add them together:

f(x) - xf(x) = (1 + x + x^2 + x^3 + ... ) - (x + x^2 + x^3 + ... ) = 1.
notice the terms cancel out. so f(x) - xf(x) = 1.

factor out f(x) from the terms in the left-hand-side:

f(x)(1-x) = 1.

dividing both sides by (1-x) yields the result:

f(x) = 1/(1-x).

This is doesn't tell you anything about the radius of convergence, which may or may not be relevant.
 
ZioX said:
This is doesn't tell you anything about the radius of convergence, which may or may not be relevant.

true. i just think it's a nice example of being able to play with a series to find an explicit formula, though this isn't the the most telling of its nature.
 
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