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18. Taylor series 201
Now we show that the function f is differentiable at the point 0 and its
derivative f0is continuous at the given point.
To find the derivative of the function f at the point x 6= 0, it suffices to
use the theorem on the derivative of the superposition and formulas for the
derivatives of the corresponding elementary functions:
f0(x) =e
−
1
x
2
0
= e
−
1
x
2
·
2
x
3
, x 6= 0. (6)
To find f0(0), we need to use the definition of the derivative:
f0(0) = lim
x→0
f(x) −f (0)
x
= lim
x→0
e
−
1
x
2
x
= lim
t→∞
e
−t
2
1
t
= lim
t→∞
t
e
t
2
= 0.
The limit is 0, since the exponential function e
t
2
grows at infinity faster
than any power function including the linear function t.
We proved that the function f is differentiable at the point 0 and its
derivative at this point is 0. It remains to show that the function f0is
continuous at the point 0. To do this, we find the limit f0(x) as x → 0 using
the previously obtained formula (6):
lim
x→0
f0(x) = lim
x→0
2e
−
1
x
2
x
3
= lim
t→∞
2t
3
e
t
2
= 0.
In this case, we again used the fact that the exponential function grows at
infinity faster than any power function.
So, we have proved that the function is continuously differentiable at the
point 0 and f0(0) = 0.
To study the second derivative of the function f, we need to investigate
the first derivative of the function f0, for which we already have the following
representation:
f0(x) =
e
−
1
x
2
·
2
x
3
, x 6= 0,
0, x = 0.
At all points except the point 0, the second derivative can be found by
differentiation formulas:
f00(x) =e
−
1
x
2
·
2
x
3
0
= e
−
1
x
2
·
2
x
3
2
+e
−
1
x
2
·−
6
x
4
= e
−
1
x
2
4
x
6
−
6
x
4
.
Thus, the derivative f00(x) can be represented as e
−
1
x
2
P
6
1
x
, where P6(t)
is a polynomial of degree 6. It follows from this representation that the limit
f00(x) as x → 0 is 0:
lim
x→0
f00(x) = lim
x→0
e
−
1
x
2
P
6
1
x
= lim
t→∞
P6(t)
e
t
2
= 0.
202 M. E. Abramyan. Lectures on integral calculus and series theory
At the point 0, the second derivative also exists and is equal to zero:
f00(0) = lim
x→0
f0(x) −f0(0)
x
= lim
x→0
2e
−
1
x
2
x
3
x
= lim
x→0
2e
−
1
x
2
x
4
= lim
t→∞
2t
4
e
t
2
= 0.
Using the method of mathematical induction, we can prove for the func-
tion f that its derivative of any order k is representable for x 6= 0 in the form
f
(k)
(x) = P (
1
x
)e
−
1
x
2
, where P(t) is some polynomial. Using this representation and finding limits similar to those given above, we can prove that, at the
point 0, there exist derivatives of f of any order that are equal to zero.
So, the function f is infinitely differentiable on the set R, and all its
derivatives at the point 0 are equal to zero. If the function f were expanded
into a Taylor series in a certain interval (−R, R) with a center at the point
x0= 0, then this would mean that the function f is identically equal to zero
on this interval:
f(x) =
∞
X
k=0
f
(k)
(0)
k!
xk=
∞
X
k=0
0
k!
xk= 0.
But from the definition of the function f, it follows that it is nonzero
at any point x 6= 0. The obtained contradiction means that the function f
does not expand into a Taylor series in a neighborhood of zero, i. e., it is
not a real analytic function in this neighborhood. Note that the function f
expands into a Taylor series in a neighborhood of any point x 6= 0.
We have shown that there exist infinitely differentiable functions that can-
not be expanded into a Taylor series.
Additional remarks on the properties
of the considered function 3.16A/15:20 (03:39)
The reason for the “bad” behavior of the function e
−
1
x
2
in a neighborhood
of the point 0 becomes more clear if we consider this function not on the real
axis but on a complex plane consisting of numbers of the form z = x + iy,
where x, y ∈ R and i is the imaginary unit for which the relation i2= −1
holds. We do not even need to specify how the function ezbehaves for complex
numbers. It is enough to note that if we consider the numbers z = iy, where
y ∈ R, and find the limit as y → 0, then, for the function f (z), we get
lim
y→0
e
−
1
(iy)
2
= lim
y→0
e
1
y
2
= lim
t→∞
e
t
2
= +∞.
Therefore, the point 0 on the complex plane is a singular point for the
function e
−
1
z
2
. This circumstance is the reason that in a neighborhood of zero
18. Taylor series 203
this function does not expand into a Taylor series neither on the complex
plane, nor on the real axis.
Sufficient condition for the existence of a Taylor series.
Expansions of exponent, sine, and cosine
into a Taylor series
Relationship between the existence of a Taylor series
and the behavior of the remainder term
of Taylor’s formula 3.16A/18:59 (06:28)
Suppose that the function f is infinitely differentiable in a neighborhood of
the point x0. Then, for the function f , Taylor’s formula holds for any n ∈ N
(see [1, Ch. 22]):
f(x) =
n
X
k=0
f
(k)
(x0)
k!
(x − x0)k+ rn(x0, x).
Here rn(x0, x) is the remainder term of Taylor’s formula. Recall the repre-
sentation of the remainder term in the Lagrange form. In this representation,
the point ξ appears; this is some point located between x0and x:
rn(x0, x) =
f
(n+1)
(ξ)
(n + 1)!
(x − x0)
n+1
. (7)
We rewrite Taylor’s formula in the following form:
rn(x0, x) = f (x) −
n
X
k=0
f
(k)
(x0)
k!
(x − x0)k.
If the remainder term vanishes as n → ∞ for all x from some neighborhood
of the point x0, then the limit of the right-hand side of the equality also exists
and is equal to 0, i. e., there exists a limit of partial sums
P
n
k=0
f
(k)
(x0)
k!
(x−x0)
k
as n → ∞ and this limit is f (x).
But the limit of partial sums, by definition, is the sum of the power series
P
∞
k=0
f
(k)
(x0)
k!
(x−x0)k. Thus, in this case, the function f expands into a Taylor
series in some neighborhood of the point x0:
f(x) =
∞
X
k=0
f
(k)
(x0)
k!
(x − x0)k.
Therefore, if we succeed in formulating the condition under which the
remainder term of Taylor’s formula vanishes as n → ∞ on some interval,
then this condition will be sufficient for the function to expand into a Taylor
series on this interval.
204 M. E. Abramyan. Lectures on integral calculus and series theory
Sufficient condition for the existence
of a Taylor series 3.16A/25:27 (08:54)
Theorem (on a sufficient condition for the existence of
a Taylor series).
Let the function f be infinitely differentiable on the interval (x0−R, x0+R)
and let the condition for uniform boundedness of the derivatives of the function f of all orders on this interval be fulfilled:
∃M > 0 ∀k ∈ N ∀x ∈ (x0− R, x0+ R) |f
(k)
(x)| ≤ M . (8)
Then the function f expands into a Taylor series on the interval
(x0− R, x0+ R).
Remark.
It follows from this theorem that the function e
−
1
x
2
considered in the previous example does not satisfy condition (8) in a neighborhood of the point 0,
since if it satisfies this condition, then it would expand into a Taylor series
in this neighborhood, but we proved that this is not true. Therefore, it can
be stated that, for any interval (−R, R), the set of all derivatives of a given
function is not bounded, i. e., for any value M > 0, there exists a point
x ∈ (−R, R) and an order of derivative k such that the value |f
(k)
(x)| is
greater than M .
Proof.
We show that, under condition (8), the remainder term of the Taylor
formula rn(x0, x) vanishes as n → ∞ for all x ∈ (x0− R, x0+ R). As noted
above, this ensures the existence of a Taylor series for the function f on
a given interval.
If x ∈ (x0− R, x0+ R), then using the representation of the remainder
term in the Lagrange form (7) and condition (8), we obtain the following
estimate for rn(x0, x):
|rn(x0, x)| =
|f
(n+1)
(ξ)|
(n + 1)!
|x − x0|
n+1
<
MR
n+1
(n + 1)!
. (9)
In this estimate, we took into account that the point ξ is located between x
0
and x and therefore also belongs to the interval (x0− R, x0+ R).
The right-hand side of estimate (9) does not depend on x and vanishes
as n → ∞, since the factorial (n + 1)! grows faster than any exponential
function R
n+1
(see the theorem on the convergence of the sequence
q
n
n!
in [1, Ch. 5]). Thus, we have proved that lim
n→∞rn(x0
, x) = 0 for any
x ∈ (x0− R, x0+ R); therefore, the function f expands into a Taylor series
on the interval (x0− R, x0+ R).
18. Taylor series 205
Taylor series expansion
of exponent, sine, and cosine 3.16A/34:21 (09:28)
1. Consider the function exin some neighborhood of the point 0:
x ∈ (−R, R), R > 0. This is an infinitely differentiable function, and its
derivatives of any order coincide with the initial function:
(ex)
(k)
= ex, k ∈ N.
Therefore, all derivatives are uniformly estimated by the value eRon the
interval (−R, R):
|(ex)
(k)
| = |ex| < eR.
Thus, all the conditions of the previous theorem are satisfied; therefore, the
function exexpands into a Taylor series on the interval (−R, R). Given Taylor’s formula centered at point 0 for the function ex, we obtain the following
expansion of this function into a Taylor series:
ex=
∞
X
k=0
x
k
k!
. (10)
Since the value R > 0 can be chosen arbitrarily, the expansion (10) is
valid for any point x ∈ R. Consequently, the radius of convergence of this
power series is +∞ (note that this result can also be obtained by the Cauchy–
Hadamard formula).
2. Consider the functions sin x and cosx in some neighborhood of the
point 0: x ∈ (−R, R), R > 0. These are infinitely differentiable functions,
and the following formulas are valid for their derivatives of any order:
(sin x)
(k)
= sinx +
kπ
2
, (cos x)
(k)
= cosx +
kπ
2
, k ∈ N.
All derivatives of these functions are estimated by 1 for any x ∈ R:
|(sin x)
(k)
| =
sinx +
kπ
2
≤ 1, |(cos x)
(k)
| =
cosx +
kπ
2
≤ 1.
By the previous theorem, the functions sin x and cosx expand in the
Taylor series on the interval (−R, R) for any R > 0. Given Taylor’s formulas
centered at point 0 for the functions sin x and cos x, we obtain the following
expansions:
sin x =
∞
X
k=0
(−1)kx
2k+1
(2k + 1)!
, cos x =
∞
X
k=0
(−1)kx
2k
(2k)!
.
These expansions are valid for any point x ∈ R, the radius of convergence
of the obtained power series is +∞.
206 M. E. Abramyan. Lectures on integral calculus and series theory
Taylor series expansion
of a power function 3.16B/00:00 (04:25)
When considering Taylor’s formula for a power function of the form
(1 + x)α, α ∈ R, we noted that Taylor’s formula for this function is de-
fined only for |x| < 1. It is natural to expect that a similar restriction will
occur for the expansion of this function into a Taylor series. Terms of the
series can be obtained from the corresponding Taylor’s formula:
1 + αx +
α(α − 1)
2
x2+
α(α − 1)(α − 2)
3!
x3+ ··· =
= 1 +
∞
X
k=1
α . . . (α − k + 1)
k!
xk. (11)
Theorem (on the expansion of a power function into a Tay-
lor series).
The power function (1 + x)αexpands into the Taylor series (11) on the
interval (−1, 1):
(1 + x)α= 1 +
∞
X
k=1
α . . . (α − k + 1)
k!
xk. (12)
Proof5.
First, we investigate the convergence of series (11) for a fixed value of x.
To do this, we use the limit D’Alembert test. Denote the common term
of the series by ak=
α...(α−k+1)
k!
xkand find the limit lim
k→∞
a
k+1
a
k
under
the assumption that ak6= 0 for all k ∈ N (note that the situation ak= 0 is
possible only for α ∈ N; in this case, the Taylor series turns into a polynomial
of finite degree α):
lim
k→∞
a
k+1
a
k
= lim
k→∞
α . . . (α − k)x
k+1
· k!
(k + 1)! ·α . . . (α − k + 1)x
k
=
= lim
k→∞
|α − k|
k + 1
· |x| = |x|.
Thus, according to the limit d’Alembert test, series (11) converges for
|x| < 1 and diverges for |x| > 1.
To prove that the series (11) converges to the function (1 + x)αfor |x| < 1,
it suffices to show that, for these values of x, we have the limit relation
lim
n→∞rn(x0
, x) = 0, where rn(x0, x) is the remainder term in the corre-
sponding Taylor’s formula with n terms.
5
There is no proof of this theorem in video lectures.
18. Taylor series 207
We use the representation of the remainder term in the Cauchy form:
rn(x0, x) =
f
(n+1)
x0+ θ(x − x0)(1 −θ)
n
n!
(x − x0)
n+1
.
Here θ is some value lying in the range from 0 to 1.
For the function f (x) = (1 + x)α, in the case x0= 0, we get
rn(0, x) =
α . . . (α − n)(1 + θx)
α−n−1
(1 −θ)
n
n!
x
n+1
=
=
(α − 1) . . .α − 1 −(n − 1)
n!
x
n
×
×αx(1 + θx)
α−1
·
(1 −θ)
n
(1 + θx)
n
. (13)
The expression bn(x) =
(α−1)...(α−1−(n−1))
n!
xnis a common term of the se-
ries (11) corresponding to the exponent (α−1). We have already proved that
this series converges for |x| < 1, therefore, due to the necessary convergence
condition, bn(x) → 0 as n → ∞ for all |x| < 1.
Thus, in order to prove that rn(0, x) vanishes as n → ∞, it suffices to
establish the boundedness of the two remaining factors on the right-hand
side of equality (13).
The first of these factors is αx(1 + θx)
α−1
. It does not depend on n and
is bounded for any |x| < 1.
The second of the factors is
(1−θ)
n
(1+θx)
n
. Note that the following estimate holds
for any |x| < 1:
1 + θx ≥ 1 −θ|x| > 1 − θ.
Using this estimate, we obtain
(1 −θ)
n
(1 + θx)
n
<
1 −θ
1 −θ
n
= 1.
So, the first factor on the right-hand side of (13) vanishes, the other two
are bounded, therefore rn(0, x0) also vanishes as n → ∞ for any |x| < 1. This
means that the series (11) converges to the function (1 + x)αfor |x| < 1.
208 M. E. Abramyan. Lectures on integral calculus and series theory
Taylor series expansions of the logarithm and arcsine
Taylor series expansion of the logarithm 3.16B/04:25 (10:55)
Having the expansion of a power function into a Taylor series (12), we can
easily obtain expansions for other elementary functions using theorems on the
integration of a power series.
Theorem (on the expansion of the logarithm into a Taylor
series).
The function ln(1 +x) expands in the following Taylor series on the interval (−1, 1):
ln(1 + x) = x −
x
2
2
+
x
3
3
− ··· =
∞
X
k=1
(−1)
k+1xk
k
. (14)
Proof.
Find the derivative of the function ln(1 + x):
ln(1 + x)
0
=
1
1 + x
.
The derivative is a power function of the form (1 + x)αfor α = −1. This
function expands into the Taylor series (12) on the interval (−1, 1). Note that
the required expansion can be obtained in a simpler way using the formula of
the sum of the terms of an infinite geometric progression with the ratio (−x):
1
1 + x
=
∞
X
k=0
(−x)k=
∞
X
k=0
(−1)kxk.
We integrate both sides of this equality from 1 to x assuming that |x| < 1:
Z
x
0
1
1 + t
dt =
Z
x
0
∞
X
k=0
(−1)ktkdt.
The left-hand side of the resulting equality is
Z
x
0
1
1 + t
dt = ln |1 + t|
x
0
= ln |1 + x| = ln(1 + x). (15)
On the right-hand side, we can use the theorem on the integration of
a power series and move the sign of the integral under the sign of the infinite
sum:
Z
x
0
∞
X
k=0
(−1)ktkdt =
∞
X
k=0
Z
x
0
(−1)ktkdt =
∞
X
k=0
(−1)
k
t
k+1
k + 1
x
0
=
18. Taylor series 209
=
∞
X
k=0
(−1)
k
x
k+1
k + 1
=
∞
X
k=1
(−1)
k+1xk
k
. (16)
By virtue of the theorem on the integration of a power series, the radius
of convergence of the series obtained on the right-hand side of equality (16)
coincides with the radius of convergence of the initial series.
Equating the expressions in the right-hand sides of equalities (15) and (16),
we obtain the proved equality (14).
Taylor series expansion of the arcsine 3.16B/15:20 (13:03)
Now consider the function arcsin x. This function is differentiable on the
interval (−1, 1). Find its derivative:
(arcsin x)0=
1
√
1 −x
2
= (1 −x2)
−
1
2
.
This derivative can be represented as (1 −t)α, where t = x2and α = −
1
2
.
Given the expansion of the power function in the Taylor series (12), we
obtain
1 + (−t)
−
1
2
= 1 +−
1
2
(−t) +−
1
2
−
3
2
1
2!
(−t)2+
+−
1
2
−
3
2
−
5
2
1
3!
(−t)3+ ··· =
= 1 +
t
2
+
1 ·3
22· 2!
t2+
1 ·3 ·5
23· 3!
t3+ ··· =
= 1 +
∞
X
k=1
1 ·3 · ··(2k − 1)
2k· k!
tk.
The resulting expression can be simplified by using the double factorial
function, which is denoted by n!! and is equal to the product of all numbers
from 1 to n of the same parity as the number n: n!! = n(n − 2)(n − 4) . . .
For example, 5!! = 5 ·3 · 1 = 15, 6!! = 6 · 4 ·2 = 48.
In our case, 1 · 3 · ··(2k − 1) = (2k − 1)!! and, in addition,
2k· k! = (2 · 1)(2 · 2) · ··(2 · k) = (2k)!!. Using double factorials and taking
into account that t = x2, we obtain the Taylor series for the function
1
√
1−x
2
for |x| < 1:
1
√
1 −x
2
= 1 +
∞
X
k=1
(2k − 1)!!
(2k)!!
x2k. (17)
Now we integrate both sides of equality (17) from 0 to x for |x| < 1:
210 M. E. Abramyan. Lectures on integral calculus and series theory
Z
x
0
(1 −t2)
−
1
2
dt = x +
Z
x
0
∞
X
k=1
(2k − 1)!!
(2k)!!
t2kdt.
On the left-hand side, we get arcsinx, on the right-hand side, we apply
the theorem on the integration of a power series and move the sign of the
integral under the sign of the infinite sum:
x +
Z
x
0
∞
X
k=1
(2k − 1)!!
(2k)!!
t2kdt = x +
∞
X
k=1
Z
x
0
(2k − 1)!!
(2k)!!
t2kdt =
= x +
∞
X
k=1
(2k − 1)!!
(2k)!!
·
x
2k+1
2k + 1
.
The resulting power series converges for |x| < 1 and is equal to the function
arcsin x. Thus, we have proved the following theorem.
Theorem (on the expansion of the arcsine into a Taylor series).
The function arcsin x expands into a Taylor series on the interval (−1, 1):
arcsin x = x +
∞
X
k=1
(2k − 1)!!
(2k)!!
·
x
2k+1
2k + 1
.
Additional remarks 3.16B/28:23 (06:25)
1. Acting in a similar way, we can obtain the Taylor series expansion for
the function arctan x on the interval (−1, 1). To do this, it is sufficient to find
the derivative (arctan x)0=
1
1+x
2
= (1 + x2)−1, expand the power function
(1 + x2)−1into a Taylor series, and then integrate both sides of the resulting
equality.
2. If we have some Taylor series and it is required to determine a function that expands into this series, then sometimes it is possible to solve this
problem by performing term-by-term integration of the initial series. If, as
a result of integration, a series arises corresponding to some known function f (x), then this means, by virtue of the theorem on the differentiation
of a power series, that the initial series corresponds to the function f0(x).
Another way to solve this problem is the term-by-term differentiation of the
initial series and finding the function g(x) corresponding to the differentiated
series. In this case, the initial series will correspond to the function obtained
by integrating the function g(x). All considered series will have the same
interval of convergence.
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