#03 (08/28/2023)

Taylor series

If all one wants is the formula for the Taylor series of f(x) about x = a, one can easily derive such a formula by assuming

f(x) = C0 +C1 (x − a) + C2 (x − a)2 + C3 (x − a)3 + …
(1)
where the coefficients can be determined by substituting x = a on the both sides and subsequent differentiation as
f(x) = f(a) + f′(a)(x − a) + f"(a)
2!
 (x − a)2 + f"′(a)
3!
 (x − a)3 + …
(2)
Example: Expand f(x) = exp (−1/x2) about x = 0.
One is tempted to use the Taylor series formula, i.e. f(x) = f(0) + f′(0) x + f"(0)/2! x2 + f"′(0)x3/3! + …. However, it turns out that all the coefficients are zero as shown in class, i.e. f(0) = f′(0) = f"(0) = …0 so the Taylor series for e−1/x2 is
exp⎛
⎝ 		− 1
x2
⎞
⎠ 		= 0 + 0 x + 0 x2
2!
 + …
(3)
This leads us to investigate under what condition the Taylor series is valid. In fact, this example is a rather pathetic one as f(z) = e−1/z2 in the complex plane has an essential singularity at z = 0 as shown in class.
We begin with an identity
⌠
⌡ 		x

a 
 f′(x) dx = f(x) − f(a),
(4)
or
f(x) = f(a) + ⌠
⌡ 		x

a 
 f′(x) dx.
(5)
By replacing f′(x) in Eq.(4) by f"(x), one obtains
⌠
⌡ 		x

a 
 f"(x) dx = f′(x) − f′(a),
(6)
or
f′(x) = f′(a) + ⌠
⌡ 		x

a 
 f"(x) dx,
(7)
which is substituted into Eq.(5) to yield
f(x) = f(a) + f′(a) (x − a) + ⌠
⌡ 		x

a 
 ⌠
⌡ 		x

a 
 f"(x) dx dx.
(8)
Using this relation recursively, one obtains
f(x)
=
f(a) + f′(a) (x − a) + f"(a)
2!
 (x − a)2 + f"′(a)
3!
 (x − a)3 + …+ ⌠
⌡ 		x

a 
 ⌠
⌡ 		x

a 
 …⌠
⌡ 		x

a 
 f(n)(x) dx dx …dx
=
f(a) + f′(a) (x − a) + f"(a)
2!
 (x − a)2 + f"′(a)
3!
 (x − a)3 + …+Rn,
(9)
where
Rn =

⌠
⌡ 		x

a 
 ⌠
⌡ 		x

a 
 …⌠
⌡ 		x

a 
n
 
 f(n)(x)

dx dx …dx
n 
 ,
(10)
is called the remainder of the Taylor series. Note that Eq.(9) is an identity which holds regardless of the condition of f(x) (no approximation).
If Rn → 0 as n→ ∞, we say that f(x) can be expanded by the Taylor series. In the example above for f(x) = e−1/x2, Rn does not go to 0 as n→ ∞.
The remainder of the Taylor series, Rn, can be assessed if one assumes that f(n)(x) is bounded as
m ≤ f(n)(x) ≤ M.
(11)
Substituting this into Eq.(10) yields
m (x − a)n
n!
 ≤ Rn ≤ M (x − a)n
n!
 .
(12)
From the inequality above, it follows that there must be a point ξ in (a, x) such that
Rn = f(n)(ξ) (x − a)n
n!
        a ≤ ξ ≤ x.
(13)
When n = 1, Eq.(13) is known as the mean value theory, i.e.
f(x) = f(a) + f′(ξ) (x − a),
(14)
or
f′(ξ) = f(x) − f(a)
x − a
 .
(15)
Exercise: Prove
1
x
 < ln x − ln a
x−a
 < 1
a
 .
(16)

Taylor series examples

Examples

  1. ln (1 + x)
    By integrating the both sides of
    1
    1 + x
    		 = 1 − x + x2 − x3 + x4 − x5 + …,
    		(17)
    one obtains
    ln (1 + x) = x − x2
    2
    		 + x3
    3
    		 − x4
    4
    		 + x5
    5
    		 − ….1
    		(18)
    Note that the convergence range of 1/(1 + x) is |x| < 1 so that the convergence region of ln x is also |x| < 1.
  2. arctan x
    By integrating
    1
    1+x2
    		 = 1 − x2 + x4 − x6 + …,
    		(19)
    one obtains
    arctan x = x − x3
    3
    		 + x5
    5
    		 − x7
    7
    		 +….
    		(20)
  3. Expand ln x about x = 2.

    ln x
    =
    ln ( (x−2)+ 2 )
    =
    ln ⎛
    ⎝     		2 (1+ x − 2
    2
    		 )⎞
    ⎠
    =
    ln 2 + ln ⎛
    ⎝     		1 + x−2
    2
    		⎞
    ⎠
    =
    ln 2 + ⎛
    ⎝     		x − 2
    2
    		⎞
    ⎠     		−
    ⎛
    ⎝     		x − 2
    2
    		⎞
    ⎠     		2

     
    2
    		 +
    ⎛
    ⎝     		x − 2
    2
    		⎞
    ⎠     		3

     
    3
    		 −….
  4. (1 + x)n (binomial expansion)

    (1+x)n = 1+ n x + n (n−1)
    2!
    		 x2 + n (n−1) (n−2)
    3!
    		 x3+….
    		(21)
    (Example):
    3
    √
     
    1003
     
    =
    (1000+3)1/3
    =
    10001/3 (1+0.003)1/3
     ∼ 
    10 ⎛
    ⎝     		1+0.003 × 1
    3
    		⎞
    ⎠
    =
    10.01.

Taylor series for multivariable functions

The Taylor series for a single variable function is rewritten as
f(x)
=
f(a) + f′(a) (x−a) + f"(a)
2!
 (x−a)2+ f"′(a)
3!
 (x−a)3 +…
=
f(a) + ⎛
⎝ 		(x−a) d
dx
⎞
⎠ 		f|a + 1
2!
 ⎛
⎝ 		(x−a) d
dx
⎞
⎠ 		2

 
 f|a + 1
3!
 ⎛
⎝ 		(x−a) d
dx
⎞
⎠ 		3

 
 f|a +…
(22)
When a function, f, has two variables, x and y, the Taylor series for f(x,y) can be rewritten by replacing ( (x−a)d/dx ) by ( (x−a)∂/∂x +(y−b) ∂/ ∂y) as
f(x,y)
=
f(a,b) + ⎛
⎝ 		(x−a) ∂
∂x
 +(y−b) ∂
∂y
⎞
⎠ 		f|(a,b)
+
1
2!
 ⎛
⎝ 		(x−a) ∂
∂x
 +(y−b) ∂
∂y
⎞
⎠ 		2

 
 f|(a,b)
+
1
3!
 ⎛
⎝ 		(x−a) ∂
∂x
 +(y−b) ∂
∂y
⎞
⎠ 		3

 
 f|(a,b)
+
…
(23)

Examples

As noted in class, the formula for the Taylor series can be bypassed in many cases:
  1. Expand sin (x + y) about (0, 0).
    Remembering that sin x = x − x3/3! + x5/5! − …,
    sin (x + y) = (x + y) − (x + y)3
    3!
    		 + (x + y)5
    5!
    		 −…
    		(24)
  2. Expand sin (x + y) about (π, π/2). In this case, the expansion must contain (x − π)(y − π/2) terms.

    sin (x + y)
    =
    sin ⎛
    ⎝     		(x − π) +π+ (y − π
    2
    		 )+ π
    2
    		⎞
    ⎠
    =
    sin ⎛
    ⎝     		(x − π) + (y − π
    2
    		 ) + 3π
    2
    		⎞
    ⎠
    =
    sin ⎛
    ⎝     		(x − π) + (y − π
    2
    		 )⎞
    ⎠     		cos 3π
    2
    		 +cos ⎛
    ⎝     		(x − π) +(y − π
    2
    		 )⎞
    ⎠     		sin 3π
    2
    =
    (−1) cos ⎛
    ⎝     		(x − π) + (y − π
    2
    		 )⎞
    ⎠
    =
    (−1)⎛
    ⎜
    ⎝     		1−
    ⎛
    ⎝     		(x − π) + (y − π
    2
    		 )⎞
    ⎠     		2

     
    2!
    		 +
    ⎛
    ⎝     		(x − π) + (y − π
    2
    		 )⎞
    ⎠     		4

     
    4!
    		 − …⎞
    ⎟
    ⎠     		,
    where sin (A + B) = sin A cos B + cos A sin B was used.
  3. Expand x2 y about (1, 2).

    x2 y = ( (x−1) + 1)2 ( (y − 2) + 2).
    		(25)
  4. Expand ln (x + y) about (1, 2).

    ln (x + y)
    =
    ln ⎛
    ⎝     		(x − 1) + 1 + (y − 2) + 2⎞
    ⎠
    =
    ln ⎛
    ⎝     		3 +(x − 1) + (y − 2)⎞
    ⎠
    =
    ln 3 ⎛
    ⎝     		1 + (x − 1) + (y − 2)
    3
    		⎞
    ⎠
    =
    ln 3 + ln⎛
    ⎝     		1+ (x − 1) + (y − 2)
    3
    		⎞
    ⎠
    =
    ln 3 + ⎛
    ⎝     		(x − 1) + (y − 2)
    3
    		⎞
    ⎠     		−
    ⎛
    ⎝     		(x − 1) + (y − 2)
    3
    		⎞
    ⎠     		2

     
    2
    		 +
    ⎛
    ⎝     		(x − 1) + (y − 2)
    3
    		⎞
    ⎠     		3

     
    3
    		 − …
Example of two variable functions
f(x, y) = ⎧
⎪
⎨
⎪
⎩
x y
x2 + y2
(x, y) ≠ (0, 0)
0
(x,y) = (0,0).
(26)
To find out what is lim(x, y)→ (0, 0) f(x,y), let (x, y) approach to (0, 0) along the line y = x. By setting x = t and y = t, f(x, y) = t2/(t2 + t2) = 1/2. However, if one approaches to (0, 0) along the x axis, f(x, y) = x ×0/(02 + x2) = 0 and they don't match. In short, f(x, y) does not have the limit as (x, y)→ (0,0).
taylor2var.jpg

Chain Differentiation

When f(x, y) is a composite function (i.e. x and y are also functions of other variables), the derivative of f with respect to the implicit variable can be achieved by using the chain differentiation rule, i.e.
df
dt
 = ∂f
∂x
 dx
dt
 + ∂f
∂y
 dy
dt
 .
(27)
An example was shown in class, i.e. this is a variation of derivative of composite functions.

Footnotes:

1 If one substitutes x=1 on the both sides, one obtains
ln 2 = 1 − 1
2
 + 1
3
 − 1
4
 + 1
5
 − …
(28)
Note that
1 + 1
2
 + 1
3
 + 1
4
 + 1
5
 + …,
(29)
diverges.

File translated from TEX by TTH, version 4.03.
On 24 Aug 2023, 13:15.