Thread: inverse tan x power series

For the power series of inverse tan (x) the geometric series is:
1 - t ^2 + t^ 4 - t^ 6 +...+ t^( 4n)

find the sum and hence prove that for 0 < t < x:
1/(1+t2) < 1 - t^2 + t^ 4 - t^ 6 +...+ t^(4n)


I havent got a clue how to approach this.
thanks
jacs

Originally Posted by jacs

For the power series of inverse tan (x) the geometric series is:
1 - t ^2 + t^ 4 - t^ 6 +...+ t^( 4n)

I'm a bit puzzled by the reference to the series for the inverse tan
function here. It seems unrelated to the problem.

Just a mo - it seems that this series is the derivative (term-by-term)
of the Gregory series:



which may be what it wants you to use.

RonL

I am more than a bit puzzled myself, in fact, have no idea. But that is what the question said. I have got only as far as:
a = 1 and n = 2n + 1



but i am not sure about my n = 2n + 1 and even if that is right, i have no idea how to proceed from there to get the proof.

Originally Posted by CaptainBlack

I'm a bit puzzled by the reference to the series for the inverse tan
function here. It seems unrelated to the problem.

Just a mo - it seems that this series is the derivative (term-by-term)
of the Gregory series:



which may be what it wants you to use.

RonL

Start with:



then differentiate to get:

.

Now if we truncate the series after a positive term the remainder is again
a geometric series like what we had originally but multipled by -1 times the
last term included in the sum, and the sum of the series comprising the
remainder is negative. Which will prove the result (after a bit of jiggery-pokery).


(or use the remainder form for taylor series, that should also give the
result)

We also need the restriction that .


RonL

There is a much simpler way. If you used a finite geometric expansion and then take the limit it will hold for but you will have a remainder term. It can be shown very easily that the remainder term approaches zero because,

Since the dominating series converges to zero so too the remainder term.

Originally Posted by ThePerfectHacker

There is a much simpler way. If you used a finite geometric expansion and then take the limit it will hold for but you will have a remainder term. It can be shown very easily that the remainder term approaches zero because,

Since the dominating series converges to zero so too the remainder term.

Except the question apparently wants the student to use the series for
arctan.

RonL