Finding the sum of inverse trigonometric expression

[Raghav Gupta]

Homework Statement

Find $\sum_1^{23} tan^{-1}(\frac{1}{n^2+n+1})$

Homework Equations

$tan^{-1}x + tan^{-1}y = tan^{-1}(\frac{x+y}{1-xy} )$

The Attempt at a Solution

I think we have to split the question in a form of relevant equation given above.
First thing what should I do?

 

[phyzguy]

In order to use the identity you have been given, you want to get the term \frac{x+y}{1-xy} into the form of \frac{1}{n^2+n+1}. This requires that you find x and y in terms of n. Do you see a way to do this?

 

[Raghav Gupta]

Yes but it is by trial and error method.
1/(n²+n+1) can be written as (n+1-n)/(1+n(n+1))
Now it is seperatable.

 

[Ray Vickson]

Homework Statement

Find $\sum_1^{23} tan^{-1}(\frac{1}{n^2+n+1})$

Homework Equations

$tan^{-1}x + tan^{-1}y = tan^{-1}(\frac{x+y}{1-xy} )$

The Attempt at a Solution

I think we have to split the question in a form of relevant equation given above.
First thing what should I do?

One possibility is to apply the equation in (2) recursively: If $t_k = 1/(k^2+k+1)$, let $T_n = \sum_{k=1}^n \arctan(t_k) \equiv \arctan(S_n)$. Then $S_1 = t_1$ and $$S_k = \frac{S_{k-1}+t_k}{1-S_{k-1} t_k}$$
for $k = 2, \ldots, 23$.

 

[phyzguy]

Yes but it is by trial and error method.
1/(n²+n+1) can be written as (n+1-n)/(1+n(n+1))
Now it is seperatable.

An easier way is to see that in order for the forms to be the same, we must have x+y=1. Then solve for y in terms of x and substitute into 1-xy = n^2+n+1. Then solve for x, then y.

 

[SammyS]

An easier way is to see that in order for the forms to be the same, we must have x+y=1. Then solve for y in terms of x and substitute into 1-xy = n^2+n+1. Then solve for x, then y.

@Raghav Gupta ,
Use this method suggested by phyzguy together with the difference identity:

$\displaystyle\ \tan^{-1}x - \tan^{-1}y = \tan^{-1}(\frac{x-y}{1+xy} )\$ .​

 

[Raghav Gupta]

It's getting quadratic.
I am getting $x = \frac{1 ± \sqrt{1+4n^2 + 4n}}{2}$
following both of you @phyzguy and @SammyS

 

[phyzguy]

It's getting quadratic.
I am getting $x = \frac{1 ± \sqrt{1+4n^2 + 4n}}{2}$
following both of you @phyzguy and @SammyS

OK, keep going. 1+4n^2+4n is a perfect square, so you can take the square root and simplify it further. There are two solutions for x and y, but you will see they are really the same solution, with x and y interchanged.

 

[Raghav Gupta]

Thanks, got it.
Sorry @Ray Vickson for not following your method, as I am not so acquainted with it.

 

[Ray Vickson]

Thanks, got it.
Sorry @Ray Vickson for not following your method, as I am not so acquainted with it.

No problem: whatever works for you is fine by me.

However, just to clarify (in case you did not understand): one way of computing $T_{23} = \sum_{k=1}^{23} \arctan(t_k)$ is to compute the successive sums. Let $T_2 = \arctan(t_1) + \arctan(t_2) = \arctan(S_2)$ (using the equation in (2) to find $S_2$), then $T_3 = \sum_1^3 \arctan(t_k) = T_2 + \arctan(t_3) = \arctan(S_2) + \arctan(t_3) = \arctan(S_3)$, where we find $S_3$ in terms of $S_2$ and $t_3$ from equation (2). If you keep going like that you can find $T_4, T_5, T_6, \ldots, T_{23}$.

 

[Raghav Gupta]

No problem: whatever works for you is fine by me.

However, just to clarify (in case you did not understand): one way of computing $T_{23} = \sum_{k=1}^{23} \arctan(t_k)$ is to compute the successive sums. Let $T_2 = \arctan(t_1) + \arctan(t_2) = \arctan(S_2)$ (using the equation in (2) to find $S_2$), then $T_3 = \sum_1^3 \arctan(t_k) = T_2 + \arctan(t_3) = \arctan(S_2) + \arctan(t_3) = \arctan(S_3)$, where we find $S_3$ in terms of $S_2$ and $t_3$ from equation (2). Continue like that until you are done.

The mood is not coming for solving by this method.
Will see it later if required. Although thanks.