The students can use the following theorems to answer sequences questions.
Theorem : Let {𝑎𝑛 } be a sequence. Then lim |𝑎𝑛 | = 0 ⟺ lim 𝑎𝑛 = 0.
𝑛→∞
𝑛→∞
Proof is an exercise.
Theorem : For any 𝛼 ∈ 𝑅, lim
𝑛→∞
𝛼𝑛
( )=0
𝑛!
Proof is omitted.
Sandwich Theorem for Sequences : Let {𝑎𝑛 } , {𝑏𝑛 } and {𝑐𝑛 } be sequences with the property
𝑏𝑛 ≤ 𝑎𝑛 ≤ 𝑐𝑛 ∀𝑛 > 𝑛1 for some 𝑛1 ∈ 𝑍+ .
If lim 𝑏𝑛 = 𝐿 = lim 𝑐𝑛 , then lim 𝑎𝑛 = 𝐿 .
𝑛→∞
𝑛→∞
𝑛→∞
Proof is an exercise. The proof will be very similar to the same proof regarding functions.
Series II
Theorem 1
For a series ∑ 𝑎𝑛 , define
𝑠𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 ∀𝑛. Suppose {𝑠𝑛 } is a bounded sequence. Let {𝑏𝑛 } be a decreasing
sequence of positive terms such that lim 𝑏𝑛 = 0
𝑛→∞
Then the series ∑ 𝑎𝑛 𝑏𝑛 is convergent.
Proof is omitted.
Theorem 1 is known as Dirichlet’s test.
Example 1
∞
Consider ∑ 𝑎𝑛
𝑖=1
1
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
(𝑛 + 1)3𝑛
where 𝑎𝑛 =
5
𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
{(𝑛 + 2)3𝑛
1
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
𝑛
Define 𝑐𝑛 = { 3
5
𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
3𝑛
By an example in the previous lecture, ∑ 𝑐𝑛 is convergent.
∴ If 𝑠𝑛 = 𝑐1 + 𝑐2 + …… + 𝑐𝑛 , then ∃ 𝐿 ∈ 𝑅 such that lim 𝑠𝑛 = 𝐿.
𝑛→∞
It follows that {𝑠𝑛 } is bounded.
1
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
𝑛
+
1
Now define, 𝑏𝑛 = {
1
𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
𝑛+2
1
Case 1 𝑛 is even. Write 𝑛 = 2𝑚
1
1
𝑏𝑛+1 − 𝑏𝑛 =
−
= 0
2𝑚 + 1 + 1
2𝑚 + 2
Case 2 𝑛 is odd. Write 𝑛 = 2𝑚 + 1
1
1
𝑏𝑛+1 − 𝑏𝑛 =
−
<0
2𝑚 + 2 + 2
2𝑚 + 2
∴ 𝑏𝑛+1 − 𝑏𝑛 ≤ 0 ∀𝑛 ∈ 𝑍+
∴ {𝑏𝑛 } is a decreasing sequence such that lim 𝑏𝑛 = 0
𝑛→∞
∴ By Dirichlet’s test, ∑ 𝑐𝑛 𝑏𝑛 is convergent.
∞
∴ ∑ 𝑎𝑛 is convergent
𝑖=1
Example 2
∞
Consider ∑
𝑛=2
1
ln 𝑛
Let 𝑓(𝑥) = 𝑥 − ln 𝑥
1
Then 𝑓 ′ (𝑥) = 1 − > 0 ∀ 𝑥 > 1
𝑥
∴ 𝑓(𝑥) is increasing in (1, ∞).
𝑓(1) = 1 > 0
∴ 𝑓(𝑥) > 0 ∀ 𝑥 ∈ (1, ∞)
∴ 𝑛 > ln 𝑛 ∀ 𝑛 ∈ {2,3,4,…….}
1
1
∴0< <
∀𝑛 >1
𝑛 ln 𝑛
1
We know that, ∑ 𝑛 is divergent.
Hence, from the comparison test of the same chapter,
∞
1
∑
is divergent.
ln 𝑛
𝑛=2
Example 3
Consider ∑ 𝑛2 𝑒 −𝑛
2
2
If 𝑎𝑛 = 𝑛2 𝑒 −𝑛 , then
2
(𝑛 + 1)2 𝑒 −(𝑛+1)
𝑎𝑛+1
|
|=|
|
2
𝑎𝑛
𝑛2 𝑒 −𝑛
𝑛 + 1 2 𝑛2 −(𝑛2 +2𝑛+1)
1 2 −2𝑛−1
=(
) 𝑒
= (1 + ) 𝑒
𝑛
𝑛
2
𝑎𝑛+1
1
∴ lim |
| = lim (1 + ) lim 𝑒 −2𝑛−1
𝑛→∞ 𝑎𝑛
𝑛→∞
𝑛 𝑛→∞
=1∙0=0
2
∴ By the ratio test, ∑ 𝑛2 𝑒 −𝑛 is convergent.
2
Now, we will give a proposition which will be very useful when handling series involving 𝑙𝑛
function.
Proposition 1 Let {𝑎𝑛 } be a decreasing sequence of positive terms.
∞
∞
Then ∑ 𝑎𝑛 is convergent if and only if the series ∑ 2𝑛 𝑎2𝑛 is convergent.
𝑛=1
𝑛=1
Proof is omitted
Example 4
1
. We can assume that the
𝑛 (ln 𝑛) (ln (ln 𝑛))
summation begins at the minimum 𝑛 value for which 𝑎𝑛 is meaningful.
2𝑛
𝑛 𝑛
2 𝑎2 = 𝑛
2 (ln 2𝑛 ) ln (ln 2𝑛 )
Consider ∑ 𝑎𝑛 where 𝑎𝑛 =
=
Then
𝑛
2 𝑏
2𝑛
1
= 𝑏𝑛 (say)
𝑛 ln 2 ln(𝑛 ln 2)
2𝑛
= 𝑛
2 ln 2 ln (2𝑛 ln 2)
=
1
(ln 2)[ln 2𝑛 + ln(ln 2)]
=
1
(ln 2)[𝑛 ln 2 + ln(ln 2)]
=
1
𝛼𝑛+𝛽
where 𝛼, 𝛽 ∈ 𝑅
Here 𝛼 = (𝑙𝑛2)2 > 0. So one can easily prove the divergence of
∑
1
𝛼𝑛+𝛽
by applying the limit comparison test.
∴ ∑ 2𝑛 𝑏2𝑛 is divergent.
∴ By proposition 1, ∑ 𝑏𝑛 is divergent where 𝑏𝑛 = 2𝑛 𝑎2𝑛
∴ Again by proposition 1, ∑ 𝑎𝑛 is divergent.
3
Sequences of Functions
Let 𝐼 be an interval. If for each 𝑛 ∈ 𝑍+ , ∃ a function 𝑢𝑛 : 𝐼 → 𝑅, then we say that {𝑢𝑛 } is a
sequence of functions on 𝐼 to 𝑅.
Suppose {𝑢𝑛 } is such a sequence. Take any 𝑥 ∈ 𝐼 and fix it. Now, {𝑢𝑛 (𝑥)} is a sequence of real
numbers similar to the ones we have seen in chapter 5. It would be interesting to know for
which values of 𝑥, {𝑢𝑛 (𝑥)} is convergent. It may or may not be convergent for all 𝑥 in 𝐼.
Definition 1 𝐼 is an interval in 𝑅. Let {𝑢𝑛 } be a sequence of functions on 𝐼 to 𝑅.
Let 𝐴 ⊆ 𝐼. (𝐴 is not necessarily an interval.)
Suppose ∃ a function 𝑢: 𝐴 → 𝑅. We say that the sequence {𝑢𝑛 } converges on 𝐴 to 𝑢 if for each
𝑥 ∈ 𝐴, the sequence {𝑢𝑛 (𝑥)} is convergent to 𝑢(𝑥). When that happens, 𝑢 is called the limit of
the sequence {𝑢𝑛 } on 𝐴. We may state this fact by writing 𝑢𝑛 → 𝑢 with the dependence on 𝐴
being understood. We may even write one of the following two statements to explain this fact.
i) lim 𝑢𝑛 = 𝑢 on 𝐴
𝑛→∞
ii) lim 𝑢𝑛 (𝑥) = 𝑢(𝑥) ∀𝑥 ∈ 𝐴
n→∞
Example 5
For each 𝑛 ∈ 𝑍+ ,
sin 𝑛𝑥
∀𝑥 ∈ 𝑅
𝑛
Take any 𝑥 ∈ 𝑅. Let 𝜀 > 0.
By Archimedean property, ∃𝑛0 ∈ 𝑍+ such that
1
𝑛0 >
𝜀
| sin 𝑛𝑥|
1
|𝑢𝑛 (𝑥) − 0| =
≤
𝑛
𝑛
1
𝑛 > 𝑛0 ⟹ < 𝜀 ⟹ |𝑢𝑛 (𝑥) − 0| < 𝜀
𝑛
∴ ∀𝑥 ∈ 𝑅, 𝑢𝑛 (𝑥) → 0
Hence, if 𝑢(𝑥) = 0 ∀𝑥 ∈ 𝑅, then
lim 𝑢𝑛 = 𝑢 on 𝑅.
let 𝑢𝑛 (𝑥) =
𝑛→∞
It is worthwhile to check whether the choice of 𝑛0 in this example depends on the particular
point 𝑥. The reader will immediately see that it didn’t depend on 𝑥. That observation should
prepare us for our next definition.
Definition 2 Let 𝐼 be an interval with a sequence of functions {𝑢𝑛 } defined on 𝐼 and, let 𝐴
be a subset of 𝐼 such that a function 𝑢 is defined on 𝐴. If ∀ 𝜀 > 0, ∃𝑛0 ∈ 𝑍+ such that
∀ 𝑥 ∈ 𝐴, 𝑛 > 𝑛0 ⟹ |𝑢𝑛 (𝑥) − 𝑢(𝑥)| < 𝜀 where 𝑛0 doesn’t depend on 𝑥, then we say that
the sequence {𝑢𝑛 } is uniformly convergent on 𝐴 to the function 𝑢.
As we have seen, the sequence {𝑢𝑛 } in example 5 is uniformly convergent to 𝑢 on all of 𝑅.
4
Clearly, when a sequence {𝑢𝑛 } is uniformly convergent on 𝐴, it is convergent on 𝐴. But, the
converse of this is false. The following example will illustrate that.
Example 6
𝑥
𝑛
and 𝑢(𝑥) = 𝑥 ∀𝑥 ∈ 𝑅.
Take any 𝑥 ∈ 𝑅. Let 𝜀 > 0.
|𝑥|
|𝑢𝑛 (𝑥) − 𝑢(𝑥)| =
𝑛
|𝑥|
∃𝑛0 ∈ 𝑍+ such that 𝑛0 >
𝜀
|𝑥|
𝑛 > 𝑛0 ⟹
< 𝜀
𝑛
⟹ |𝑢𝑛 (𝑥) − 𝑢(𝑥)| < 𝜀
∴ The sequence {𝑢𝑛 } converges on 𝑅 to 𝑢.
But, 𝑛0 depends on 𝑥.
∴ It is not uniformly convergent on 𝑅.
Let 𝑢𝑛 (𝑥) = 𝑥 +
But, it can be proven that {𝑢𝑛 } is uniformly convergent to 𝑢 on any bounded interval 𝐼.
Let 𝐼 be a bounded interval. Then ∃𝑀 ∈ 𝑅 such that
−𝑀 < 𝑥 < 𝑀 ∀𝑥 ∈ 𝐼
∴ |𝑥| < 𝑀 ∀𝑥 ∈ 𝐼
𝑀
Take 𝑛0 ∈ 𝑍+ such that 𝑛0 >
𝜀
𝑀
Then 𝑛 > 𝑛0 ⟹
< 𝜀
𝑛
|𝑥|
⟹
< 𝜀 ∀𝑥 ∈ 𝐼
𝑛
⟹ |𝑢𝑛 (𝑥) − 𝑢(𝑥)| < 𝜀
∴ {𝑢𝑛 } is uniformly convergent to 𝑢 on 𝐼.
Do we expect the limit of a convergent sequence of continuous functions to be continuous?
The following proposition will show under what conditions we can have the assurance of it.
Proposition 2 𝐼 is an interval. Let {𝑢𝑛 } be a sequence of continuous functions defined on 𝐼.
Suppose {𝑢𝑛 } converges uniformly on 𝐼 to a function 𝑢 defined on 𝐼. Then 𝑢 is continuous on
𝐼.
Proof: Let 𝜀 > 0. By assumption, ∃𝑛0 ∈ 𝑍+ such that
𝜀
𝑛 > 𝑛0 ⟹ |𝑢𝑛 (𝑥) − 𝑢(𝑥)| <
∀𝑥 ∈ 𝐼
3
Let 𝑎 ∈ 𝐼. Take 𝑛1 ∈ 𝑍+ such that 𝑛1 > 𝑛0 . Now, 𝑢𝑛1 (𝑥) is continuous at 𝑥 = 𝑎.
∴ ∃ 𝛿 > 0 such that
𝜀
|𝑥 − 𝑎| < 𝛿 ⟹ |𝑢𝑛 1 (𝑥) − 𝑢𝑛 1 (𝑎)| <
3
5
Because 𝑛1 > 𝑛0 ,
𝜀
and
3
𝜀
|𝑢𝑛1 (𝑥) − 𝑢(𝑥)| <
∀𝑥 ∈ 𝐼
3
∴ |𝑥 − 𝑎| < 𝛿 ⟹
|𝑢𝑛1 (𝑥) − 𝑢(𝑥)| + |𝑢𝑛1 (𝑥) − 𝑢𝑛 1 (𝑎)| + |𝑢𝑛1 (𝑎) − 𝑢(𝑎)| < 𝜀 ⟹
|𝑢(𝑥) − 𝑢(𝑎)| < 𝜀 by triangle inequality
∴ 𝑢 is continuous at 𝑥 = 𝑎 for any 𝑎 ∈ 𝐼.
∴ 𝑢 is continuous on 𝐼.
|𝑢𝑛1 (𝑎) − 𝑢(𝑎)| <
The converse of proposition 2 is false. In example 6, 𝑢(𝑥) = 𝑥 is continuous on 𝑅.
But, the convergence 𝑢𝑛 → 𝑢 is not uniform.
Suppose we have a sequence of continuous functions {𝑢𝑛 (𝑥)}. If we prove that 𝑢𝑛 converges
to a function 𝑢 by using the definition, then by checking the proof, we would know whether
the convergence is uniform or not.
But, if we prove the convergence 𝑢𝑛 → 𝑢 by using a known theorem about sequences, then we
wouldn’t be able to say whether the convergence is uniform or not in that manner. In that type
of situation, we might be able to use proposition 2 to conclude that the convergence is not
uniform.
Let 0 < 𝑘 < ∞. For each 𝑛 ∈ 𝑍+ , define
𝑥𝑛
𝑢𝑛 (𝑥) =
𝑥 ∈ [0,1]
𝑘 + 𝑥𝑛
Clearly, 𝑢𝑛 (𝑥) ≥ 0 ∀𝑥.
Example 7
Case 1: 𝑥 ∈ [0,1)
1
1
𝑥𝑛
(𝑥)
≤
⟹
0
≤
𝑢
≤
𝑛
𝑘 + 𝑥𝑛
𝑘
𝑘
By a theorem on sequences,
𝑥𝑛
lim
=0
𝑛→∞ 𝑘
∴ By sandwich theorem for sequences,
lim 𝑢𝑛 (𝑥) = 0
𝑛→∞
Case 2: 𝑥 = 1. Then
1
𝑢𝑛 (𝑥) =
∀𝑛
1+𝑘
So, 𝑢𝑛 (𝑥) → 𝑢(𝑥) where
0
𝑖𝑓
𝑢(𝑥) = { 1
𝑖𝑓
1+𝑘
6
𝑥 ∈ [0,1)
𝑥=1
Hence, lim− 𝑢(𝑥) = 0 ≠
𝑥→1
1
1+𝑘
∴ lim− 𝑢(𝑥) ≠ 𝑢(1)
𝑥→1
∴ 𝑢 is not continuous on [0,1]. Hence, by proposition 2, the convergence 𝑢𝑛 → 𝑢 is not
uniform.
Power Series
Now, we will
∞
introduce
a
special
type
of
series.
A
series
of the form
∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛 where 𝑥 is a variable is called a power series about the
𝑛=0
point 𝑎. As with a regular series, it is acceptable to write this series as ∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛 .
The most common type of power series would be a series about the point 0. Such a series is
written as 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + …… or ∑ 𝑎𝑛 𝑥 𝑛 .
Regarding a general power series, we are interested in knowing about the set of convergence
given by
𝑆 = {𝑥|𝑥 ∈ 𝑅, ∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛 converges}
Let 𝑢 ∈ 𝑆. If we can prove that all the 𝑥 values which lie in between 𝑢 and 𝑎 also belong to 𝑆,
then we have proven that 𝑆 is an interval centered at 𝑎. That will be the first theorem regarding
power series.
Theorem 2 Consider the power series ∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛 . The set of values 𝑥 for which this series
converges is an interval centred at 𝑎.
Proof is omitted
Definition 3 Consider ∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛 . The interval we have discovered in the above theorem
is called the interval convergence of the series. Suppose this interval is written as
(𝑎 − 𝑅, 𝑎 + 𝑅) or [𝑎 − 𝑅, 𝑎 + 𝑅]. This 𝑅 is called the radius of convergence of the series. For
the rest of the chapter, we shall take (𝑎 − 𝑅, 𝑎 + 𝑅) as the interval of convergence while
keeping in mind that the inclusion of end points is possible.
Theorem 3
Consider ∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛 . If the conditions on 𝑎𝑛 ’s are such that the right hand
sides of the following formulas are well defined, then the radius of convergence 𝑅 is given by
1
i) 𝑅 =
𝑎
lim | 𝑛+1
|
𝑎
𝑛→∞
ii) 𝑅 =
𝑛
1
1
lim |𝑎𝑛 |𝑛
𝑛→∞
7
Proof: According to the ratio test, the series is convergent when
lim |
𝑛→∞
𝑎𝑛+1 (𝑥 − 𝑎)𝑛+1
|<1
𝑎𝑛 (𝑥 − 𝑎)𝑛
𝑎𝑛+1
i.e. when lim |
| |𝑥 − 𝑎| < 1
𝑛→∞ 𝑎𝑛
1
i.e. when |𝑥 − 𝑎| <
𝑎
lim | 𝑛+1
|
𝑎
𝑛→∞
∴𝑅=
𝑛
1
𝑎 +1
lim | 𝑛 |
𝑎𝑛
𝑛→∞
1
ii) According to the root test, the series is convergent when lim (|𝑎𝑛 (𝑥 − 𝑎)𝑛 |𝑛 ) < 1
𝑛→∞
1
𝑛
i.e. when lim (|𝑎𝑛 | |𝑥 − 𝑎|) < 1
𝑛→∞
1
i.e. when |𝑥 − 𝑎| <
1
lim (|𝑎𝑛 |𝑛 )
𝑛→∞
1
∴𝑅=
1
lim (|𝑎𝑛 |𝑛 )
𝑛→∞
Example 8 i) Consider ∑ 2𝑛 (𝑥 − 1)𝑛
By the above theorem, the radius of convergence 𝑅 is given by
1
1
𝑅=
=
lim (2)
2
𝑛→∞
Note that we will get the same value from the other formula. Hence, the
1 3
interval of convergence is ( , ) .
2 2
3
1
When 𝑥 = , the series become ∑ 2𝑛 ( 𝑛 ) = ∑ 1 which is divergent.
2
2
1
When 𝑥 =
too the series becomes divergent.
2
Hence, the inclusion of end points is impossible in this case.
ii) Consider ∑ 𝑛! (𝑥 − 1)𝑛
𝑅=
1
(𝑛+1)!
lim | 𝑛! |
𝑛→∞
=
1
=0
lim (𝑛+1)
𝑛→∞
This means the series converges only at 𝑥 = 1.
8
On the other hand, if we consider the series ∑
𝑅=
1
𝑛!
lim |(𝑛+1)!|
(𝑥 − 1)𝑛
, then
𝑛!
= lim (𝑛 + 1) = ∞
𝑛→∞
𝑛→∞
Now, this means that the interval of convergence is (−∞, ∞) = 𝑅
Taylor Series
Suppose 𝑓(𝑥) is 𝑛 times differentiable in an open interval 𝐼. Let 𝑎 ∈ 𝐼. From Taylor’s theorem,
we have seen that ∀𝑥 ∈ 𝐼,
𝑓 ′ (𝑎)
𝑓 (2) (𝑎)
𝑓 (𝑛−1) (𝑎)
2
(𝑥 − 𝑎) +
(𝑥 − 𝑎) + .… +
(𝑥 − 𝑎)𝑛−1
𝑓(𝑥) = 𝑓(𝑎) +
(𝑛 − 1)!
1!
2!
+
𝑓 (𝑛) (𝜉)
(𝑥 − 𝑎)𝑛
𝑛!
for some 𝜉 between 𝑥 and 𝑎.
Let us make the following two assumptions.
i) 𝑓 can be differentiated as many times as we wish.
𝑓 (𝑛) (𝜉)
(𝑥 − 𝑎)𝑛 = 0
ii) ∀𝑥, 𝜉 ∈ 𝐼, lim
𝑛→∞
𝑛!
Then the above expression can be written as
𝑓(𝑥) = 𝑓(𝑎) +
𝑓 ′ (𝑎)
𝑓 (2) (𝑎)
𝑓 (3) (𝑎)
(𝑥 − 𝑎) +
(𝑥 − 𝑎)2 +
(𝑥 − 𝑎)3 + .…
1!
2!
3!
This power series is called the Taylor series of 𝑓(𝑥) about the point 𝑥 = 𝑎. Note that the
function 𝑓(𝑥) is equal to the Taylor series if and only if the conditions i) and ii) above are
satisfied. A function satisfying those two conditions is said to be analytic at the point 𝑥 = 𝑎.
Example 9
Let 𝑓(𝑥) = sin 𝑥
Then 𝑓(𝑥) is infinitely many times differentiable.
∀𝑛 ∈ 𝑍+ and ∀𝜉 ∈ (−∞, ∞), 𝑓 𝑛 (𝜉) = ± sin 𝜉 or ± cos 𝜉
∴ |𝑓 𝑛 (𝜉)| = 1
𝑓 𝑛 (𝜉) 𝑛
𝑥𝑛
∴|
𝑥 | = | | ∀𝑥 ∈ 𝑅
𝑛!
𝑛!
𝑛
𝑥
𝑓 𝑛 (𝜉) 𝑛
Now lim
= 0 . ∴ lim |
𝑥 |=0
𝑛!
𝑛→∞ 𝑛!
𝑛→∞
Hence, lim
𝑓 𝑛 (𝜉)𝑥 𝑛
=0
𝑛!
∴ 𝑓(𝑥) is analytic at 𝑥 = 0.
𝑛→∞
𝑓 (1) (𝑥) = cos 𝑥
,
(3) (𝑥)
𝑓
= −cos 𝑥 ,
𝑓 (2) (𝑥) = −sin 𝑥
𝑓 (4) (𝑥) = sin 𝑥
9
∴ 𝑓 ′ (0) = 1 , 𝑓 (2) (0) = 0 ,
𝑓 (3) (0) = −1
𝑓 (4) (0) = 0 , 𝑓 (5) (0) = 1 etc.
Because 𝑓(𝑥) is analytic at 𝑥 = 0, we can write
𝑓 (1) (0)
𝑓 (2) (0) 2 𝑓 (3) (0) 3
𝑓(𝑥) = 𝑓(0) +
𝑥+
𝑥 +
𝑥 + ………
1!
2!
3!
𝑥3 𝑥5
∴ sin 𝑥 = 𝑥 −
+
− ……..
3!
5!
This is the Taylor series of sin 𝑥 about 𝑥 = 0.
Because of the usefulness of this result, this can be given as the part i) of our next theorem.
Theorem 4
The following Taylor series are valid in the indicated intervals.
If an exam question says “using the general formula” the student is not allowed to use this
theorem.
𝑥3 𝑥5 𝑥7
i) sin 𝑥 = 𝑥 −
+
− + ….. − ∞ < 𝑥 < ∞
3!
5! 7!
𝑥2 𝑥4 𝑥6
ii) cos 𝑥 = 1 −
+
− + ….. − ∞ < 𝑥 < ∞
2!
4! 6!
𝑥3 𝑥5 𝑥7
iii) tan−1 𝑥 = 𝑥 −
+
− + ….. − 1 ≤ 𝑥 ≤ 1
3
5
7
2
3
4
𝑥
𝑥
𝑥
iv) 𝑒 𝑥 = 1 + 𝑥 +
+
+ + ….. − ∞ < 𝑥 < ∞
2!
3! 4!
𝑥2 𝑥3 𝑥4
v) ln|1 + 𝑥| = 𝑥 −
+
− + ….. − 1 < 𝑥 ≤ 1
2
3
4
The proof is an exercise for the student.
10