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Math Help - Adv. Calc

  1. #1
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    Adv. Calc

    Let {Xn}, {yn} be real sequences and b,x,y be real numbers.

    Prove (using limit theorems)

    If Xn goes to X and Yn goes to Y as n goes to infinity, then
    i)lim (Xn + Yn) = X+Y
    ii)lim (bXn) = bX
    iii) lim (XnYn) = XY
    (all limits n going to infinity)

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  2. #2
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    Quote Originally Posted by taypez View Post
    Let {Xn}, {yn} be real sequences and b,x,y be real numbers.

    Prove (using limit theorems)

    If Xn goes to X and Yn goes to Y as n goes to infinity, then
    i)lim (Xn + Yn) = X+Y
    Let \lim \ x_n = x and \lim \ y_n = y.
    This means,
    |x_n - x|< \frac{\epsilon}{2} \mbox{ where }n>N_1
    |y_n-y|<\frac{\epsilon}{2} \mbox{ where }n>N_2
    Thus,
    |x_n - x|< \frac{\epsilon}{2} \mbox{ and }|y_n-y|< \frac{\epsilon}{2} \mbox{ where }n>N \mbox{ where }N=\max\{N_1,N_2\}.

    Thus,
    |x_n+y_n - x - y| \leq |x_n - x|+|y_n-y| < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon \mbox{ for }n>N.
    Q.E.D.
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    Quote Originally Posted by taypez View Post
    Let {Xn}, {yn} be real sequences and b,x,y be real numbers.

    Prove (using limit theorems)

    If Xn goes to X and Yn goes to Y as n goes to infinity, then
    i)lim (Xn + Yn) = X+Y
    ii)lim (bXn) = bX
    iii) lim (XnYn) = XY
    (all limits n going to infinity)

    Thanks
    Theorem: \lim (x + y) = \lim x + \lim y

    since \lim_{n \to \infty}x_n = X and \lim_{n \to \infty}y_n = Y

    we have, \lim_{n \to \infty} (x_n + y_n) = \lim_{n \to \infty}x_n + \lim_{n \to \infty}y_n = X + Y

    now look at the limit theorems and try the others
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    Quote Originally Posted by taypez View Post
    iii) lim (XnYn) = XY
    The second one is really easy try that one yourself. This one is the hardest of all of them.

    Lemma: Convergent sequences are bounded.

    Proof: Let \{ x_n \}\to x be a convergent sequence then for \epsilon = 1 we have |x_n-x|< 1 for x\geq N\in \mathbb{N}^+. Thus, |x_n|\leq |x_n-x|+|x| <1+|x|. This means x_N,x_{N+1},... are all bounded by |x|+1. So let A = \max\left\{ |x_1|,|x_2|,...,|x_N|,|x|+1 \right\} then it means |x_n|\leq A for all n=1,2,3,.... Q.E.D.

    Now we can prove your problem.

    We know that the sequence y_n is bounded by some positive number A.
    This means,
    |x_n-x| < \frac{\epsilon}{A+|x|} \mbox{ for }n>N_1
    |y_n-y| < \frac{\epsilon}{A+|x|} \mbox{ for }n>N_2.
    Thus,
    |x_n-x|<\frac{\epsilon}{A+|x|} \mbox{ and }|y_n-y|< \frac{\epsilon}{A+|x|} \mbox{ for }n>N \mbox{ where }N=\max \{ N_1,N_2 \}.

    This means,
    |x_ny_n-xy| \leq |x_ny_n-xy_n|+|xy_n-xy| = |y_n||x_n-x|+|x||y_n-y| < A\cdot \frac{\epsilon}{A+|x|} + |x|\cdot \frac{\epsilon}{A+|x|} = \epsilon \mbox{ for }n>N.
    Q.E.D.
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  5. #5
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    adv calc

    Doesn't that just show that
    lim (Xn + Yn) = lim Xn + lim Yn?
    How do I show that lim Xn goes to X?
    Also, since your proof is based on convergence, will I need to prove convergence?
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    Quote Originally Posted by taypez View Post
    How do I show that lim Xn goes to X?
    Also, since your proof is based on convergence, will I need to prove convergence?
    The problem says you can assume it. Meaning it says "If xn and yn are convergent sequences ... " this means we are accepting that these sequences converge no need to prove that.
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  7. #7
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    adv calc

    Doesn't that just show that
    lim (Xn + Yn) = lim Xn + lim Yn?
    How do I show that lim Xn goes to X?
    NEVERMIND. DUH!

    Also, since your proof is based on convergence, will I need to prove convergence.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taypez View Post
    lim (Xn + Yn) = lim Xn + lim Yn?
    How do I show that lim Xn goes to X?
    Also, since your proof is based on convergence, will I need to prove convergence?
    Quote Originally Posted by taypez View Post
    ... If Xn goes to X and Yn goes to Y as n goes to infinity...
    ..
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    Thanks!
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