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I Reverse Series Problem

  1. Oct 18, 2016 #1
    So all these college classes are really a growing experience for my teenage boys (home schooled). Last night my older son kept us up late persevering on a Calculus problem. Now, I remember a lot about sequences and series from my own days in Calculus and from teaching Calc 1, 2, and 3 at the Air Force Academy.

    But my son's prof turned the tables with this homework question. Most Calc 2 questions on sequences and series provide the summation and want to know whether it converges and what it is equal to. This problem provided what the series summed to and wanted an expression for the terms in the sum:

    Find An if the sum of An from 1 to n is (-6n + 15)/(2n + 8).

    One big difference between high school and college math is having to go down a few blind alleys and back out if not headed to a solution. It took us a while to finally make a table of n values, the sum values, and then the difference between the last two terms in the sum column, recognizing these were the numerical values of the An. We're pretty good at recognizing patterns, but this one was hard.

    Eventually, our google-fu failed us (no online help to be found), and I retrieved the Stewart Calculus book that had been holding up our big screen TV. Finding the section covering the same topic, I perused it and realized it would have similar problems and hints for a solution. I handed it to my son, and the light went on as he read. We needed to express the difference column in our table as the ratio of two integers (rather than decimals) and look for the patterns in the numerator and denominator separately. We also needed to focus on the An for n > 1, because A1 would not fit the pattern.

    He quickly realized the numerator was simply a linear function of n, while I typed the denominator values into a graphing program, fit to a quadratic, and realized it was a perfect fit. So, the An was so challenging to find, because it was a rational function of n (a linear function divided by a quadratic function).

    It's a growing experience for students to leave the high school cocoon where the teacher demonstrates every problem type for the more mature problem solving expectations of college where there is often a significant gap between the examples covered in class and the more challenging homework problems. At times during the process, my son realized he would be safe leaving this problem until later (when he could ask his prof for help) and he recognized none of the other students would likely solve it either. But it niggled at him. He couldn't go to bed until he had solved it.

    I think "reverse problems" are common challenges in college Calculus, physics, and chemistry courses. It's not like early grades when they teach you multiplication and division separately. Here, they teach you multiplication, and leave it to you to figure out division. They teach you to compute where a projectile will land from the initial conditions, and then expect you to compute the initial conditions from where it lands.
     
  2. jcsd
  3. Oct 18, 2016 #2
    For those dying to know:

    An = -13(n - 1)/(11n^2-19.8n+15.8) for n > 1
     
  4. Oct 18, 2016 #3

    PeroK

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    If you have a formula, ##S_n##, for sum to ##n## terms, then ##A_n = S_n - S_{n-1}##.
     
  5. Oct 18, 2016 #4
    Yes, indeed.

    We worked it out the hard way.
     
  6. Oct 18, 2016 #5

    PeroK

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    Are you sure the answer isn't:

    ##A_n = -\frac{39}{2(n+3)(n+4)} \ \ (n > 1)## and ##A_1 = \frac{9}{10}##
     
  7. Oct 18, 2016 #6
    No, not sure. I was trusting the student with the details and focusing on the process.

    Have to check that. Thanks.
     
  8. Oct 18, 2016 #7
    Yep, the student confirms that he kept working after I went to bed last night, and the answer that I awoke to on the white board was just an intermediate step in the process. Thanks for clarifying this.
     
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