Finding P(n) for a Challenging Sum Series: A Hint for Simplifying the Expression

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In summary, the conversation is about simplifying the expression for P(n) where n is a non-negative integer. The solution involves using trigonometric formulas and substituting variables, leading to the simplified form P(n) = (n-1) * pi / 2 - sum of arctan(sqrt(i)) from i=2 to n. Another way to write this is P(n) = (n-1) * pi / 2 - (n-1) * arctan(sqrt(2)/2).
  • #1
FermatPell
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Homework Statement


Let

[tex]P(x) = \displaystyle\sum\limits_{k=1}^\infty arctan (\frac{x-1}{(k+x+1)sqrt(k+1) + (k+2)sqrt(k+x)})[/tex] (infite sum from K=1 to infinity)

a) Simplify the expression for P(n), where n is a non-negative integer

Homework Equations



[tex]tg(a-b)=\frac{tg(a)-tg(b)}{1+tg(a)tg(b)}[/tex]

The Attempt at a Solution



Im looking for a hint. My first idea was to somehow use trig formulae, but i can't factorize the denominator or split it into partial fractions. Now I am stuck, any hints?
 
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  • #2
I think I sort of see how to do this...Let's give it a go. I'll post more later if something else comes to me:

First, there's that pesky arctangent in there. So, let's use the relation you have to get rid of it. That is, we need to propose two variables [tex]a[/tex] and [tex]b[/tex] that give us

[tex]\Sigma\arctan[\tan(a-b)]=\Sigma(a-b)[/tex]

Now let's find those coefficients...
 
  • #3
I've tried that I the result was really ugly. Something like this:

[tex]a = \frac{\frac{x-1}{(k+x+1)sqrt(k+1)}+sqrt(C)}{2}[/tex]

where C is a very complicated fraction. I don't think that's the way to go

EDIT: I've succeeded. :D :D but followed a different path..

MOD's please delete this post, i don't want to double post
 
Last edited:
  • #4
Is there a simpler form of this?

[tex]$\sum\limits_{i=2}^n \arctan(\sqrt{i})$[/tex]
 
  • #5
Care to post the solution?
 
  • #6
Substitute u=sqrt(k+1), v=sqrt(k+x). After some algebra, the final solution is:

[tex]P(n) = (n-1) \frac{\pi}{2} - \sum\limits_{i=2}^n \arctan(\sqrt{i})[/tex]

Now i need another way to write that
 
  • #7
That's neat! The square root of the imaginary term can be rewritten as

[tex]i=e^{i\tfrac{\pi}{2}}[/tex]
[tex]\sqrt{i}=e^{i\tfrac{\pi}{4}}[/tex]
[tex]\sqrt{i}=\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}[/tex]
[tex]\sqrt{i}=\tfrac{\sqrt{2}}{2}\left(1+i\right)[/tex]
 
  • #8
No, you misunderstood it. i is not an imaginary term, its a variable (integer) that goes from 2 to n
 
  • #9
Ah.

Seems like you've got this one. I'm curious, could you post the solution when you get it? I'd be much obliged.
 

1. What is a challenging sum series?

A challenging sum series is a mathematical sequence in which the terms are added together to form a sum. The series is considered challenging because it may involve complex patterns, require advanced techniques to solve, or have an infinite number of terms.

2. How do you solve a challenging sum series?

To solve a challenging sum series, you need to first identify the pattern or rule governing the series. You can then use mathematical techniques, such as algebra or calculus, to find a formula for the series. Alternatively, you can use numerical methods, such as summation or approximation, to find the sum of the series.

3. What makes a sum series challenging?

A sum series can be considered challenging due to its complexity, difficulty in finding a closed form solution, or infinite number of terms. Some series are challenging because they require advanced mathematical knowledge or techniques to solve.

4. What are some real-world applications of challenging sum series?

Challenging sum series have many applications in various fields, including physics, engineering, and finance. For example, they can be used to model the behavior of complex systems, calculate the total cost or revenue of a project, or analyze the convergence of algorithms.

5. How can I improve my skills in solving challenging sum series?

To improve your skills in solving challenging sum series, you can practice solving different types of series, study advanced mathematical concepts, and learn various techniques for finding the sum of a series. You can also seek guidance from experienced mathematicians or attend workshops and seminars on sum series.

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