Solving Complex Math Problems: Recurrence Relations and Limits Explained

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In summary: You found a clever way to simplify the product into a power series and then use the definition of e to solve the limit. As for the related question, yes, \lim\limits_{n\rightarrow\infty}(1+\frac{x}{n})^n=e^x. This can be proven using the binomial theorem and the definition of e as the limit of (1+\frac{1}{n})^n as n approaches infinity. Keep up the good work with your self-teaching! In summary, the conversation discusses two limit problems in mathematics. The first problem involves a recurrence relation and the solution can be found by using the derivative of a geometric series. The second problem involves a product series and can be solved by
  • #1
hamsterman
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I'm trying to teach myself maths, and I have a couple of problems I can't figure out.

1. [itex]\lim\sum\limits_{k=1}^{n}\frac{2k-1}{2^k}[/itex]. I see that this is [itex]=\lim\sum\limits_{k=1}^{n}\frac{k}{2^{k-1}}-1[/itex] but I have no idea what to do with that. I tried writing it as [itex]\lim\frac{\sum\limits_{k=1}^{n}2^{n-k}k}{2^{n-1}}-1[/itex] and then writing the top as [itex]x_n[/itex] where [itex]x_1=1, x_k=2x_{k-1}+k[/itex], but I failed. I hardly have any experience with recurrence relations.. And it seems overly complex.

2. [itex]\lim\prod\limits_{j=1}^{n}(1+\frac{aj}{n^2}), a\in\mathbb{R}[/itex]. (It might be that a can only be positive, I don't remember..). I have no ideas about this at all. The answer is [itex]e^{a/2}[/itex], I think.
A related question, is [itex]\lim\limits_{n\rightarrow\infty}(1+\frac{x}{n})^n=e^x[/itex] ? I don't see how that works..

Thanks for your time
 
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  • #2
To part 1: You apparently know the geometric series, right? If you express it in terms of
[tex] \sum_{n=0}^\infty r^n [/tex] and then look at the derivative with respect to r, what do you get?
 
  • #3
To part 2: If I were you, I would expand the product into a power series in a and see if you can find the right coefficients. It might be a tedious task though.
 
  • #4
Thanks for the first. It seems that I've solved the second:
[itex]\lim\prod\limits_{j=1}^{n}(1+\frac{aj}{n^2})=\lim\\sqrt[n^2]{\prod\limits_{j=1}^{n}(1+\frac{aj}{n^2})^{n^2}}=\lim\sqrt[n^2]{\prod\limits_{j=1}^{n}e^{aj}}=\lim e^{\frac{a}{n^2}\sum\limits_{j=1}^{n}j}=\lim e^{\frac{a}{n^2}\frac{n(n+1)}{2}}=e^{a/2}[/itex]
 
  • #5
Wow, that's what I call elegant :-o
Well done!
 

1. What are recurrence relations in mathematics?

Recurrence relations are mathematical equations that define a sequence of numbers in terms of one or more previous terms. They are used to represent patterns and relationships in various mathematical and scientific contexts.

2. How are recurrence relations used to solve complex math problems?

Recurrence relations are used to break down complex problems into smaller, more manageable parts. By defining a sequence of numbers, we can use recurrence relations to find a general formula or pattern that can be applied to solve the problem.

3. What is the importance of understanding limits in solving recurrence relations?

Limits are essential in solving recurrence relations because they help us determine the behavior of a sequence as it approaches infinity. They also provide a way to check the correctness of our solutions and identify any potential errors.

4. What are some common techniques for finding limits in recurrence relations?

Some common techniques for finding limits in recurrence relations include substitution, algebraic manipulations, and using L'Hopital's rule. It is also important to understand the properties of limits and how they can be applied in different situations.

5. Can recurrence relations be used in real-life applications?

Yes, recurrence relations are used in various real-life applications, such as in computer science, engineering, economics, and physics. They are particularly useful in modeling and predicting the behavior of complex systems and processes.

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