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Thread: Complex Analysis, Sequence

  1. #1
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    Complex Analysis, Sequence

    I assume that by sequence they mean $\displaystyle {\alpha\choose 1}$$\displaystyle +{\alpha\choose 2}+...+{\alpha\choose n}$
    I do not understand what should be $\displaystyle n$... Is it complex? Usual binomial coefficients have both $\displaystyle n $ and $\displaystyle \alpha$ integers and $\displaystyle n\leq\alpha$. In this problem can $\displaystyle n$ be more than $\displaystyle \alpha$??
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  2. #2
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    Notice that they have "redefined" the numerator of the binomial coefficient as $\displaystyle (\alpha)(\alpha- 1)..$ rather than $\displaystyle \alpha !$ because factorial, is only defined for positive integers. Because they do use the notation "n!" in the denominator, n must be a non-negative integer.

    You cannot even ask the question "can n be bigger than $\displaystyle \alpha$" because the complex number system is not an ordered field!

    However, "sequence" does NOT mean a sum. That is a "series". They mean just the numbers $\displaystyle \left(\begin{array}{c}0 \\ \alpha\end{array}\right)$, $\displaystyle \left(\begin{array}{c}1 \\ \alpha\end{array}\right)$, ... themselves. Basically, the first thing you are asked to prove is that there is some real number M such that $\displaystyle \left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$ for all n.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    N
    However, "sequence" does NOT mean a sum. That is a "series". They mean just the numbers $\displaystyle \left(\begin{array}{c}0 \\ \alpha\end{array}\right)$, $\displaystyle \left(\begin{array}{c}1 \\ \alpha\end{array}\right)$, ... themselves. Basically, the first thing you are asked to prove is that there is some real number M such that $\displaystyle \left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$ for all n.

    you meant $\displaystyle {\alpha\choose\ n }< M$. I ask this because you wrote $\displaystyle \left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$
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    Quote Originally Posted by andreas View Post
    you meant $\displaystyle {\alpha\choose\ n }< M$. I ask this because you wrote $\displaystyle \left|\left(\begin{array}{c} n \\ \alpha\end{array}\right)\right|< M$
    What HallsofIvy wrote is correct. When you want to show a sequence is bounded you have to show there is a positive number larger than the absolute value of each term. Therefore, you must include absolute values.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    What HallsofIvy wrote is correct. When you want to show a sequence is bounded you have to show there is a positive number larger than the absolute value of each term. Therefore, you must include absolute values.

    I agree with that but why should $\displaystyle \alpha$ and $\displaystyle n$ exchange their places?

    ______________

    I do not understand how to deal with inequalities where on one side is a complex number and on the other only real number... Can somebody explain intuitively what is going on?
    Last edited by andreas; Feb 13th 2009 at 11:12 AM.
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