Thread: inverse tan x power series

1. 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

2. 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:

$
\arctan(x)=x-x^3/3+x^5/5- .. +(-1)^n x^{2n+1}/(2n+1) ..
$

which may be what it wants you to use.

RonL

3. 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
$
r = -t^2
$

$
Sn = ((-t^2)^{2n+1} - 1)/(-t^2 - 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.

4. 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:

$
\arctan(x)=x-x^3/3+x^5/5- .. +(-1)^n x^{2n+1}/(2n+1) ..
$

which may be what it wants you to use.

RonL

$
\arctan(x)=x-x^3/3+x^5/5- .. +(-1)^n x^{2n+1}/(2n+1) ..
$

then differentiate to get:

$
\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}=-x^2+x^4- .. +(-1)^n x^{2n} ..
$
.

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 $|x|<1$.

RonL

5. There is a much simpler way. If you used a finite geometric expansion and then take the limit it will hold for $x=\pm 1$ but you will have a remainder term. It can be shown very easily that the remainder term approaches zero because,
$\int_0^t \frac{x^n}{1+x^2} dx<\int_0^t \frac{x}{1+x^2}$
Since the dominating series converges to zero so too the remainder term.

6. 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 $x=\pm 1$ but you will have a remainder term. It can be shown very easily that the remainder term approaches zero because,
$\int_0^t \frac{x^n}{1+x^2} dx<\int_0^t \frac{x}{1+x^2}$
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