Pratice exam confusion Showing the sumation is equal to the other wee fun

In summary, Ivey says that the differential equation has a recurrence relation, which is demonstrated by solving for x using the sumation method. He says that the equation can be simplified if the indices are changed, and that it is simpler to follow if the indices are changed to n-c and n=j+c.
  • #1
mr_coffee
1,629
1
Hello everyone, I'm studying for my exam. He gave us anice little pratice exam. Am i thinking too much about this one? Part (a) seems to easy.
http://img235.imageshack.us/img235/1554/lastscan9dx.jpg [Broken]I think I'm trying to make x^(n-c) be x^(n)
well to do this, i would add c to (n-c), but if u do this, you will have to subtract c from the indicies (the n under the sumation). So that would change, n = b, to n=b-c.

Is that all i had to do here?
Also in part b, I'm lost now! The directions say: show that a2 = 0, a3 = 0, and that the recurrance relation for this differential equation is: a_(n+4) = -a_n/((n+3)(n+4))
I've been using this as a referance to help me:http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp [Broken]
http://img126.imageshack.us/img126/2601/lastscan29ir.jpg [Broken]

THanks!
 
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  • #2
I think it is simpler to follow if you change indices- don't use "n" for both.
Let j= n-c. Then n= j+ c so an= aj+c, f(n)= f(j+ c) and xn-c= xj. Also, when n= b, j= b-c. Of course, when n= [itex]\infty[/itex], j= [itex]\infty[/itex]. That is:
[tex]\Sigma_{n=b}^\infty a_n f(n) x^{n-c}= \Sigma{j= b-c}^{\infty}a_{j-c}f(j-c}x^j[/tex].
Now, just change the index in the second sum from j to n- it doesn't change the value.

If [itex]y= \Sum_{n=0}^\infty a_n x^n[/itex] then
[tex]y"= \Sum_{n=2}^\infty n(n-1)a_n x^{n-2}[/tex]
(the sum starts at n= 2 because if n= 0 or n= 1 the term n(n-1) is 0). Putting that into the equation y"- x2y= 0, we get:
[tex]\Sigma_{n=2}^\infty n(n-1)a_n x^{n-2}- \Sigma_{n=0}^\infty a_n x^{n+2}[/tex]

In the first sum, let j= n-2 so n= j+ 2. The sum becomes
[tex]\Sigma_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j[/tex]
In the second sum, let j= n+2 so n= j- 2. The sum becomes
[tex]\Sigma_{j= 2}^\infty a_{j-2}x^j[/tex]
The point of that is, of course, to have the same exponent on x in both sums so we can "combine like terms". Since the second sum doesn't start until j= 2, we can write j=0 and j= 1 terms in the first sum separately:
[tex]2a_2+ 6a_3x+ \Sigma_{j=2}^{\infty}((j+2)(j+1)a_{j+2}- a{j-2})x^j= 0[/tex]
From that, it should be obvious that 2a2= 0 and 6a3= 0.
 
  • #3
Thanks again Ivey! it makes perfect sense, sorry about the delayed responce!
 

What is a "practice exam"?

A practice exam is a test or assessment that is designed to simulate a real exam or test-taking experience. It is often used as a study tool to help students prepare for an upcoming exam.

What does "confusion" refer to in this context?

In this context, "confusion" refers to a state of being uncertain or unclear about something, in this case, the calculation or problem being presented in the practice exam.

What is meant by "showing the summation is equal to the other wee fun"?

This phrase is likely referring to a mathematical equation or problem in which the summation, or total, of a series of numbers is equal to another value or expression. The phrase "wee fun" may be a playful way of describing the equation or problem.

Why is it important to show that the summation is equal to the other expression?

Showing that the summation is equal to the other expression is important because it demonstrates the accuracy and validity of the calculation or problem being presented. It also allows for better understanding and confirmation of the solution.

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To overcome confusion in a practice exam, it is important to thoroughly review and understand the material being tested. This may involve seeking clarification from a teacher or tutor, practicing similar problems, and actively engaging in the learning process.

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