Problem 19 in §8.7
January 25, 2013
Two smart students brought this problem to my attention, so I assume most of you will have difficulties on it.
However, I didn’t have enough time to finish 19 of §8.7. In fact, I changed my mind in the middle to give you more
time to think about problem 19. I am changing the weight of this problems to 2, and it will definitely be graded.
Pn
x2k+1
Let f (x) = sin(πx), Tn (x) = k=0 (−1)k π 2k+1 (2k+1)!
. Below are some hints:
(i) a = 0 for Maclaurin series, here we have to show ∀x ∈ R, the series converges to f (x).
(ii) In order to show limn→∞ Tn (x) = f (x), we need to show limn→∞ Rn = limn→∞ (f (x) − Tn (x)) = 0.
(iii) Taylor inequality is only one way to bound Rn . In this case, it will not work because as n → ∞, there does
not exist a constant M such that ∀n ∈ N+ , f (n+1) (x) ≤ M . Hence the condition of Taylor inequality does not
hold.
(iv) When all the conditions of some theorem does not hold, what we usually do is to try some other theorem or
go to a theorem that is more general than the one we tried. What theorems do we have? Which theorem is
more general than Taylor inequality? think, think, think, · · ·
What is the other important theorem that showed up in in-class quiz 2? Ha, Taylor theorem! The importance
of Taylor theorem has been emphasized again and again, it would be weird to not have any practice on it.
(v) What does Taylor thoerem tell us? It says that the remainder exactly equals the value of the last term of the
(n + 1)th Taylor polynomial at some y. Since for all y, | sin(πy)| ≤ 1, can you derive bn as an upper bound of
|Rn |? In other words, which bn you can derive such that ∀n, |Rn | ≤ bn ?
(vi) The squeeze theorem say that if bn > 0, limn→∞ bn = 0, then limn→∞ |Rn | = 0.
(vii) How can you show limn→∞ bn = 0?
P∞
(viii) Theorem 6 on page 570 says that if n=1 bn converges, then limn→∞ bn = 0.
P∞
(ix) How can you show n=1 bn converges?
(x) You should be able to figure out the rest (compare this to the first example on the homework 2 handout).
Note: the solution given in the solution manual is wrong, so if you have access to the solution manual, don’t
listen to it.
I encourage you to think through the hints above. If somehow you just don’t get it, turn over the page. If you
have worked out your solution, congratulations! turn over the page to check your answer and read the important
notes.
1
checking the condition of the Taylor Theorem ...,
(n+1)
(y) n+1
x
, since ∀y ∈ R, |f (n+1) (y)| ≤ π n+1 ,
∃y, s.t. Rn (x) = f (n+1)!
bn =
π n+1 |x|n+1
≥ |Rn |.
(n + 1)!
Note the existence of y P
is necessary here to construct bn .
Consider the series
bn , use the ratio test to show that it converges:
bn+1 = lim πx = 0.
lim n→∞
n→∞ n + 2 bn
Show lim |Rn | = 0, hence show lim Rn = 0 (by what?)
Important Notes
In the context of ratio test or other convergence test, we think of x
being fixed as a finite number. Why? because sticking a variable x
into the expression of a series is a syntax sugar for us to study a whole
bunch of different series, these series differs because different value’s
of x can be plugged in. When judging their convergence, we are not
interested in the case of x going to infinity: when x = ∞, the Taylor
series is useless in the sense that it does not approximate anything.
Therefore in the context of convergence tests, we always think of x
as a finite number.
On the other hand, when we are using a convergent Taylor series
to approximate a function or its value, we thinks of the series as a
function of x. It is definitely not a function of n because n is just the
“dummy index.”
The above notes illustrates an importance principle of resolving ambiguity. If you are not sure how to identify the independent variable
of a function, ask this question go back to the question: what is
the context where this mathematical entity is used?
2
(1)
(2)