CHAPTER XIII
DIFFERENTIAL CALCULUS. EXPANSIONS.
TAYLOR'S AND MACLAURIN'S THEOREMS
1. It is often necessary to expand f (x) in a series of ascending powers of x. This has been done by ordinary algebraic or trigonometrical methods in such cases as (x + a)", ex, loge (1 + x), sin x, etc.
The various methods which can be employed may be summarised as follows:
- I.By algebraic or trigonometrical methods.
By the use of Taylor's or Maclaurin's Theorem.By the use of a differential equation.By differentiating or integrating a known series. [For integration, see Chapters XVI—XX.]2. I. An example will make this method sufficiently clear.Example 1.et-1 =1+— t++++...3t 2+12 720+ by actual division.3. II. Taylor's and Maclaurin's Theorems.Assume that f (x + h) can be expanded in a series of positive integral powers of h.Letf(x+h)=a+bh+ch2+dh3+.... Differentiating with regard to h,f (x + h) = b + 2ch + 3dh2 + ...,f"(x+h)=2c +3. 2dh+...,and so on.Put h= 0, then we havea =f (x), b =f' (x), c =f /2 ix) , etc.Hencef(x+h)=f(x)+hf'(x)+2' f„(x)+3 f," (x)+...+nf(n)(x)+.... This result is known as Taylor's Theorem.
78 DIFFERENTIAL CALCULUS
- 4.If in the above result we put x = 0 we obtain f(h)=f(0)+hf'(0)+2-,f"(0)+...+n f(n)(0)+..., which, by altering the notation, may be written
f(x) =f(0) + xf (0) + 2 !f „ (0) + ... + n 1 n> (0) + ...
This result is known as Stirling's or Maclaurin's Theorem.The student should verify the common algebraic and trigonoexpansions by means of the above theorems.Example 2.f (x) = log (1 + ex),f (0) = log 2,xex1f' (0) _f ( ) —— 1 — 1+ex1+ex'f” (x) (1+6x)2,f" (0)=f,,,(x)-(160— +e)$'f" (0) = 0,f" (x)_ex—4e2x+e3x(0),(1 + ex)°etc.Whence, by using Maclaurin's Theorem,
log (1+ex)=f(x)=log2+-+-- 92""
- The above results are not universal in their application. Thus it can be shown that the method fails if f (x) or one of its derivatives becomes infinite or discontinuous between the specified range of values of x. Further, the series obtained must be a convergent series. Lagrange has shown that the remainder after the first n terms have
itn been taken from Taylor's series can be expressed as —fn) (x + Oh), nwhere 0 is a positive proper fraction. The corresponding value of the remainder in the case of Maclaurin's Theorem is e f (n) (0x) Unless therefore those expressions tend to vanish when n becomes infinite, the series is divergent and the method fails.
EXPANSIONS 79
6. III. The use of a differential equation.
The following are examples of the use of this method.
Example 3.
y=(1 +x)n=ao+a,x+a2x2+a,x3+.... Then y, = n (l + x)"-' or (1 + x) y, = ny.
Differentiating the first equation,
y,=a,+2a2x+3a,x2+..., Therefore
(1 +x)(a,+ 2azx+3a,x2+...) =n(a,+a,x+a,x2+...). Whence, by comparing coefficients,
a,=nao,
2a2+a,=na„
3a, + 2a2 = nag,
etc.
Putting x = 0 in the original equation we have a,=1,
and successively
a,=na, =n,
- (n—1)n(n—1)
a2 = --- 2 a, =2 i,
- (n—2)n(n—1)(n—2)
a,= a3,=3!etc.Hence(1 + x)" = 1 + nx + (2) x2 + (3) x° + .... Example 4. y=sin 'x=a,+a,x+a2x2+..., 1yl 1~1 — x2and(1 — x2) W = 1.
Differentiating (1 — x2) 2y, y, — 2xy,2 = 0
or(1 — x2) y2 = xy,.
Nowsin-' x = — sin-' (— x). Thereforea,+a,x+a,x°+a, x3+... —(a,—a,x+a2x2—+...). Whenceao=a,=a,=... =0.
80 DIFFERENTIAL CALCULUS Thus we may write
y=a,x+ asx3+...+a21+,x2''+,+...,
a, + a3x2+...++..., x
y,=a, + 3a3x2+... + 2n + 1 aa„+,xan+...,
y2= 3.2a3x+...+2n+1.2n.a.n}1x2ri''+.... Therefore
(1-x2)(3.2.a3x+... +2n+1.2n.asn+,x"'-'+...) x(a,+3a3x'+... + 2n + 1 a.,,,+,x3n + ...). Whence a, = 3.2as,
3as=5.4a,-3.2as,
(2n + 1) a+,=(2n+3)(2n+ 2)aa-(2n+1)2n.a2n+, or (2n+1)2a2„+1 =(2n+ 3) (2n+2)a2n+a.
But in the limit when x -.0, = 1. x
Therefore a, =1,
and accordingly as = 213'
1.3
ab_2.4.5' etc.
1 x3 1.3 a°
and sin-1x=x+2'3+2.4'5+ .
7. IV. Differentiation or integration of a known series.
The method is sufficiently indicated by the following examples. Example 5. sinx=x - i+5! -
..,
d sin xcosx=l-2 +4 -....
We have therefore obtained the expansion of cos x in terms of a from that of sin x.
EXPANSIONS 81
Example 6. Let
y=loge (1—x)=ao+a,x+a2x'+...,
dy 1
=— — x =a,+2a2x+3asx2+....
dx 1
But 11x=1+x+x2+x3+.... (x<1)
Therefore, comparing coefficients,
a,=—1, a,=--, a3=— 1, etc.
Also putting x = 0 in the original equation, we find ao = 0.
Hence log, (1— x) = — x — 2 — 3 — .... (x < 1)
Thus our knowledge of the expansion of 1 x in terms of x enables.
us to find the expansion of loge (1 — x).
H. T.11. L 6