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Thread: Some Computation questions I dont undertstand

  1. #1
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    Some Computation questions I dont undertstand

    1. Let f: R --> R be defined by f(x) = x^2 - 2. Find f([-3,3]) and f^-1([-3,3]).

    2. For the infinite series sigma k=1 to infinity (1/3)^k find s1,s2, and s3.

    3. Find an example of a sequence (if possible) that contains subsequences converging to each number in the set {1,1/2,1/4,1/8,...}

    4. Suppose that {sn} = {2/n^2} is the sequence of partial sums ofr the series sigma k=1 to infinity ak. Find a1,a2,a3

    Can someone please explain to me and steer me in the right direction

    thank you.
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    Re: Some Computation questions I dont undertstand

    Quote Originally Posted by aris27 View Post
    1. Let f: R --> R be defined by f(x) = x^2 - 2. Find f([-3,3]) and f^-1([-3,3]).

    2. For the infinite series sigma k=1 to infinity (1/3)^k find s1,s2, and s3.

    3. Find an example of a sequence (if possible) that contains subsequences converging to each number in the set {1,1/2,1/4,1/8,...}

    4. Suppose that {sn} = {2/n^2} is the sequence of partial sums ofr the series sigma k=1 to infinity ak. Find a1,a2,a3

    Can someone please explain to me and steer me in the right direction
    Can you explain why $f[-3,3])=[-2,7]~?$
    If you want more help, show us some work.
    Last edited by Plato; Oct 3rd 2017 at 03:31 PM.
    Thanks from topsquark
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    Re: Some Computation questions I dont undertstand

    Quote Originally Posted by Plato View Post
    Can you explain why $f[-3,3])=[-2,7]~?$
    If you want more help, show us some work.
    P.S. $f^{-1}([-3,3])=[-\sqrt5,\sqrt5]$
    Last edited by Plato; Oct 3rd 2017 at 04:40 PM.
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    Re: Some Computation questions I dont undertstand

    The graph of y= x^2- 2 is a parabola opening upward with vertex at (0, -2) and x-intercepts (\sqrt{2}, 0) and (-\sqrt{2}, 0). Look at all the points in the interval [-3, 3] on the x-axis. What points on the y-axis are those mapped into? You can think of this as "from each point on the x-axis, go straight up (or down) from the x-axis to the corresponding point on the graph then straight left or right to the y-axis. Of course, both -3 and 3 are mapped to 9- 2= 7 but the points between are mapped to points from the minimum y value, -2, up to that 7.

    For the inverse, look at the points on the interval [-3, 3] on the y-axis. What values on the x-axis are they mapped to? Now, you go from a point on the y-axis straight left or right to the corresponding point on the graph, then up or down to the x-axis. Of course, there is no graph lower than y= -2 so the points from y= -2 to y= -3 correspond to NO points on the graph and so no points on the x-axis. For the points from -2 up to 3 you can go both left and right the "right-most" (and left-most) points coming when y= x^2- 2= 3 so x^2= 5 giving Plato's -\sqrt{5}, \sqrt{5}.

    For the second problem, I think that " s_1", " s_2", and " s_3", refer to the "partial sums". Surely that was given in the problem? If that is correct then s_1 is just the value for k= 1, s_2 is the sum of the values for k= 1 and k= 2, and s_3 is the sum for k= 1, 2, and 3. Once you know the definitions, that is easy arithmetic.

    For the fourth problem, can you think of a sequence that converges to 1? What about a sequence that converges to 1/2? Find a sequence that converges to each of 1, 1/2, 1/4, ... and put them together to make a single sequenc
    The fifth problem is the "opposite" of the second problem. For the series \sigma a_k, s_1= a_1, s_2= a_1+ a_2, and s_3= a_1+ a_2+ a_3. You are told that s_k= \frac{2}{n^2} so s_1= \frac{2}{1^2}= 2, s_2= \frac{2}{2^2}= \frac{1}{2}, and s_3= \frac{2}{3^2}= \frac{2}{9}. So a_1= s_1= 2. Since s_2= a_1+ a_2 a_2= s_2- a_1 and, since s_3= a_1+ a_2+ a_3, a_3= s_3- a_1- a_2= s_3- (a_1+ a_2)= s_3- s_2.
    Last edited by HallsofIvy; Oct 4th 2017 at 05:15 AM.
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