Compare the value of the function at x=0, with that of the series.
From the series find a general formula of the derivatives (any order) at x=0 .
Compare it with the first derivative obtained by direct differentiation of the function.
Check if the convergence interval of the series corresponds to the expected one.
Reminder
(1.3.5.4)
The expression of (1.3.5.4) is also called the Taylor polynomial.
The D'alambert's test of the series with terms an states:
(1.3.5.22)
... the power (Maclaurin) series with the domain of convergence :
(1.3.5.25)
... the multiplication of all the terms by a constant number does not affect the convergence of the series. If we use a power series of a function, the multiplication can be done by any expression of x. In such a case there are occasions where special care should be taken:
If at a point the series diverges to infinity, and this expression is zero, and if their product has a finite limit , then the new series converges there.
If at a point the expression diverges to infinity and the series vanishes and if the limit of their product diverges, then the new series does not converge there.
Otherwise, the convergence of the new series will not be affected.
Introduction (Part 0)
The relation (1.3.5.25) could be rewritten in a more convenient way for solving the exercise. This is not necessary and the exercise can be solved without that. However for the sake of convenience this will be done and used.
In this particular case the powers of x take all the integer values starting from 0, but on the other hand the starting value of the index is 1. It is more convenient to rewrite the expression with the index starting at zero also:
Of course this could be done directly from the first to the third expression, but that way is easier to understand.
Finally we are rewriting (1.3.5.25):
(1.3.5.25)
Parts 1-2
Solution of question 1
From the series (1.3.5.25) the term of x at power zero (1.3.5.4) yields:
On the other hand
in agreement with part 1.
Part 3-4
Solution of question 2
From the comparison of the terms of (1.3.5.4)
with (1.3.5.25), one obtains:
and therefore
Parts 5-6
Solution of question 3
From part 4 for n=1 one obtains
By direct differentiation of the function (1.3.5.25) and by use of the L'Hôpital rule (twice) one obtains
in agreement with part 5.
Parts 7-8
Solution of question 4
The D'Alambert's test gives for (1.3.5.25)
which yields the covergence interval:
exactly as the series of ln(1+x) with uncertain results about
For x=1 one obtains exactly the same result as in case of the series of ln(1+x): the same harmonic series with alternating sign corresponding to ln2.
For x=−1 the series is
which is the non-converging harmonic series.
Consequently the convergence interval is exactly that of the ln(1+x) series, as expected:
Score
Questions 1,2 give credit of 2 points each.
Questions 3,4 give credit of 3 points each.
By parts:
Parts 1,2,3,4,5,7 are worth 1 point each.
Parts 6,8 are worth 2 points each.