How do you find the power series for #f(x)=int arctan(t^3)dt# from [0,x] and determine its radius of convergence?

Question

Luke Alvarez · Accepted Answer

#int_0^x arctant^3dt = sum_(n=0)^oo (-1)^n x^(6n+4)/((6n+4)(2n+1))#

with radius of convergence #R=1# Start from: #arctanx = int_0^x (dt)/(1+t^2)# Now the integrand function is the sum of a geometric series of ratio #-t^2#: #1/(1+t^2) = sum_(n=0)^oo (-1)^n(t^2)^n = sum_(n=0)^oo (-1)^nt^(2n)# so: #arctan x = int_0^x sum_(n=0)^oo (-1)^nt^(2n)# This series has radius of convergence #R=1#, so in the interval #x in (-1,1)# we can integrate term by term: #arctan x = sum_(n=0)^oo int_0^x (-1)^nt^(2n)dt = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)# and the resulting series has the same radius of convergence. Now substitute #x= t^3#. the radius of convergence does not change since #abs x < 1 => abs(t^3) < 1#: #arctan t^3 = sum_(n=0)^oo (-1)^n (t^3)^(2n+1)/(2n+1) = sum_(n=0)^oo (-1)^n t^(6n+3)/(2n+1)# and integrate again term by term: #int_0^x arctant^3dt = sum_(n=0)^oo (-1)^n int_0^x t^(6n+3)/(2n+1)dt = sum_(n=0)^oo (-1)^n x^(6n+4)/((6n+4)(2n+1))#

William Abdalla · Answer

undefined To find the power series for $ f(x) = \int_0^x \arctan(t^3) \, dt $ and determine its radius of convergence, we first find the indefinite integral of $ \arctan(t^3) $. Then, we express it as a power series using the Maclaurin series expansion. Finally, we determine the radius of convergence using the ratio test.

First, we integrate $ \arctan(t^3) $ with respect to $ t $:
$$ \int \arctan(t^3) \, dt $$

Now, we express $ \arctan(t^3) $ as a power series:
$$ \arctan(t^3) = t^3 - \frac{t^9}{3} + \frac{t^{15}}{5} - \frac{t^{21}}{7} + \cdots $$

Next, we integrate each term of the series to obtain the indefinite integral:
$$ \int \arctan(t^3) \, dt = \frac{1}{4}t^4 - \frac{1}{12}t^{10} + \frac{1}{20}t^{16} - \frac{1}{28}t^{22} + \cdots $$

Now, we need to determine the radius of convergence of this power series. We use the ratio test, which states that if $ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $, then the series converges absolutely if $ L < 1 $ and diverges if $ L > 1 $.

Applying the ratio test to the series, we have:
$$ L = \lim_{n \to \infty} \left| \frac{\frac{1}{4}t^{4(n+1)} - \frac{1}{12}t^{10(n+1)} + \frac{1}{20}t^{16(n+1)} - \frac{1}{28}t^{22(n+1)} + \cdots}{\frac{1}{4}t^{4n} - \frac{1}{12}t^{10n} + \frac{1}{20}t^{16n} - \frac{1}{28}t^{22n} + \cdots} \right| $$

Simplify and take the limit:
$$ L = \lim_{n \to \infty} \left| \frac{t^4}{4} \cdot \frac{1 - \frac{1}{3}t^6 + \frac{1}{5}t^{12} - \frac{1}{7}t^{18} + \cdots}{1 - \frac{1}{3}t^6 + \frac{1}{5}t^{12} - \frac{1}{7}t^{18} + \cdots} \right| $$

$$ L = \lim_{n \to \infty} \left| \frac{t^4}{4} \right| = \frac{|t|^4}{4} $$

The radius of convergence $ R $ is the reciprocal of the absolute value of $ L $:
$$ R = \frac{1}{\frac{|t|^4}{4}} = \frac{4}{|t|^4} $$

Thus, the radius of convergence of the power series for $ f(x) $ is $ R = \frac{4}{|t|^4} $.

Charles Arocha · Answer

undefined To find the power series for $ f(x) = \int_0^x \arctan(t^3) \, dt $ and determine its radius of convergence, follow these steps:

1. Differentiate $ \arctan(t^3) $ with respect to $ t $ to find its power series representation.
2. Integrate the resulting power series term by term to obtain the power series for $ f(x) $.
3. Determine the radius of convergence of the obtained power series using the ratio test or another appropriate convergence test.

Let's denote $ g(t) = \arctan(t^3) $. Then we differentiate $ g(t) $ with respect to $ t $ to find its power series representation:

$$ g'(t) = \frac{d}{dt} \arctan(t^3) $$

Using the chain rule, this becomes:

$$ g'(t) = \frac{1}{1 + (t^3)^2} \cdot (3t^2) $$

Simplify this expression to get:

$$ g'(t) = \frac{3t^2}{1 + t^6} $$

Now, integrate $ g'(t) $ term by term to find the power series representation of $ f(x) $:

$$ f(x) = \int_0^x \frac{3t^2}{1 + t^6} \, dt $$

The radius of convergence of this power series can be determined using the ratio test or another convergence test on the resulting series.