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If $$f(x) =\sum_{n=0}^{\infty}x^ n$$ Then Determine the function $f(x)$. Discuss the domain of $f(x)$. Discuss the domain of the derivative of $f(x)$. Thanks!

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What is $x_n$? Do you mean $x^n$? –  Robert Israel Feb 7 '13 at 7:23
    
@RobertIsrael: The OP wrote x n. –  Babak S. Feb 7 '13 at 7:35
    
Do you know about geometric series? –  Daenerys Naharis Feb 7 '13 at 7:48

1 Answer 1

If $ x = 1 $, then the infinite series diverges to $ \infty $. Hence, suppose that $ x \in \mathbb{R} \setminus \{ 1 \} $. Then for all $ N \in \mathbb{N} $, we have $$ \sum_{n=0}^{N} x^{n} = \frac{1 - x^{N+1}}{1 - x}. $$ The sum on the left-hand side of this identity converges as $ N \to \infty $ if and only if the sequence $ (x^{N+1})_{N \in \mathbb{N}} $ converges if and only if $ |x| < 1 $. Therefore, $ \text{Dom}(f) = (-1,1) $, and $$ \forall x \in (-1,1): \quad f(x) = \lim_{N \to \infty} \frac{1 - x^{N+1}}{1 - x} = \frac{1}{1 - x}. $$ Finally, observe that by either the Quotient Rule or the Chain Rule, $ f $ is differentiable on $ (-1,1) $.

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Range is upto ∞ not upto N so please tell me accordingly –  Haider Feb 7 '13 at 9:36
    
I’m not sure what you mean, so the best that I can say is that $ f(x) $ is undefined if $ x \leq -1 $, is $ \dfrac{1}{1 - x} $ if $ -1 < x < 1 $ and is $ + \infty $ if $ 1 \leq x $. –  Haskell Curry Feb 7 '13 at 9:52
    
It is necessary to first consider the finite sum $ \displaystyle \sum_{n=0}^{N} x^{n} $ because $ \displaystyle \sum_{n=0}^{\infty} x^{n} \stackrel{\text{def}}{=} \lim_{N \to \infty} \sum_{n=0}^{N} x^{n} $. –  Haskell Curry Feb 7 '13 at 9:56
    
thnx a lot ............ –  Haider Feb 7 '13 at 10:09

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