How do you find the power series for #f(x)=int tln(1-t)dt# from [0,x] and determine its radius of convergence?

Answer 1

#f(x) = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))# with radius of convergence #R=1#.

We have:

#f(x) = int_0^x tln(1-t)dt#

Focus on the function:

#ln(1-x)#

We know that:

# ln (1-x) = -int_0^x (dt)/(1-t)#

where the integrand function is the sum of the geometric series:

#sum_(n=0)^oo t^n = 1/(1-t)#

then we can integrate term by term, and we have:

#ln(1-x) = -int_0^x (sum_(n=0)^oo t^n)dt = -sum_(n=0)^oo int_0^x t^ndt =- sum_(n=0)^oo x^(n+1)/(n+1)#

and mutiplying by x term by term:

#xln(1-x) = - sum_(n=0)^oo x^(n+2)/(n+1)#

We can substitute this expression in the original integral and integrate again term by term:

#f(x) = int_0^x tln(1-t)dt = - int_0^x ( sum_(n=0)^oo t^(n+2)/(n+1))dt = -sum_(n=0)^oo int_0^x t^(n+2)/(n+1)dt = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))#

To determine the radius of convergence we can then use the ratio test:

#abs (a_(n+1)/a_n) = abs (frac (x^(n+4)/((n+4)(n+2))) (x^(n+3)/((n+3)(n+1)))) = abs (x) ((n+3)(n+1))/((n+4)(n+2))#
#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)#
and the series is absolutely convergent for #abs(x) <1# which means the radius of convergence is #R=1#
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Answer 2

To find the power series for ( f(x) = \int_0^x t \ln(1 - t) , dt ), we can integrate the given function term by term to obtain its power series representation. After integrating term by term, we'll then determine the radius of convergence by applying the ratio test to the resulting series.

First, let's find the antiderivative of ( t \ln(1 - t) ):

[ \int t \ln(1 - t) , dt ]

Using integration by parts with ( u = \ln(1 - t) ) and ( dv = t , dt ), we get:

[ \int t \ln(1 - t) , dt = -\frac{t^2}{2} \ln(1 - t) + \frac{t^2}{4} + C ]

Now, we have the antiderivative. To find the power series representation, we'll integrate this expression term by term:

[ f(x) = -\frac{x^2}{2} \ln(1 - x) + \frac{x^2}{4} + C ]

The power series representation of ( f(x) ) is:

[ f(x) = -\sum_{n=1}^{\infty} \frac{x^{n+2}}{2(n+1)} + \sum_{n=1}^{\infty} \frac{x^{n+2}}{4} + C ]

Now, let's determine the radius of convergence, ( R ), using the ratio test:

[ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| ]

Where ( a_{n} ) represents the coefficients of the power series. In this case, ( a_{n} ) is either ( \frac{1}{2(n+1)} ) or ( \frac{1}{4} ) depending on the series.

After finding ( R ), we can determine the interval of convergence for the power series by considering the convergence behavior at the endpoints of the interval.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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