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CHAPTER XI

EXPANSIONS
ROLLE'S THEOREM    I75
ential coefficients of the function. The first of these theorems is Rolle's Theorem, which states that
If f (x) and f' (x) (i.e. x f (x)) are continuous over the range
x = a to x = b, and if f (x) = o when x = a and when x = b, then for at least one value of x between a and b, f' (x) will be zero.
Y

C
Fig. 23.
The proof is as follows :
Since f (a) = f (b) = o, f (x) cannot always be increasing or decreasing. Hence for at least one value between x = a and x = b there will be a change from an increase to a decrease or vice versa. For the particular value of x for which this is so, f' (x) must be zero, which proves the proposition.
That the theorem is self-evident may be seen from the above diagram (Fig. 23).
If the curve represents the continuous function y = f (x) and if y = f (x) = o for the values x = a and x = b (i.e. at the points A and B), then at the points C, D, E the Y    C.
tangents to the curve are parallel to the
x-axis. That is, at these points f' (x) is zero.
It should be noted that f (x) must be 0    X
continuous within the given range. If there    A    B
be a discontinuity such as at the points C    Fig. 24.
and C' in Fig. 24, there is no unique differential coefficient; conthe theorem does not apply and f' (x) is not necessarily zero in the range.
Since difficulties may arise in dealing with multiple-valued functions, it is advisable to restrict the above proof to single-valued functions of x.

17b    DIFFERENTIAL CALCULUS
3. Mean Value Theorem.
As before, let f (x) and f' (x) be continuous in the range x = a
to x = b and let m = f     (bb      —    f (a) , so that a
f(b)—f(a)—m(b—a)=o.
Replace b by x in the left-hand side of this expression and let ~(x)=f(x)—f(a)—m(x—a).
Then obviously (a) = o and we have shown that (b) = o. Therefore Rolle's Theorem holds, since f (x) and f' (x) are continuous in the given range.
Hence 0' (x) will be zero for at least one value of x (x1 say) between a and b.
But    q (x) = f (x) — f (a) — m (x — a).
Therefore 0' (x) = f' (x) — m on differentiating.
Hence since 9' (xi) = o, then f' (x1) = m.
Therefore    f     (bb — a   (a) =f (x1).


This is the Mean Value Theorem, and may be stated thus :
If f (x) and f' (x) are continuous in the range x = a to x = b, then there is at least one value of x (xt say) between x = a and x = b
such that f _('_b   —    f (a) = f' (xi). Y a

This is equivalent to saying that if y = f (x) is a continuous curve between the values A (x = a) and B (x = b), then there is at least one value of x, (x1), where a < x < b,
for which the tangent to the curve o A'
is parallel to the chord AB. That    Fig. 25.
is, if the tangent at this point C make an angle with the x-axis,
B'N _ BK — f (b) — f (a)
tanv=A'N AK— b—a
A more convenient form may be obtained for the result of this theorem. Since x1 lies between a and b we may write x1=a+ 01 (b — a),
where 01 is a positive proper fraction.

MEAN VALUE THEOREM    177
The mean value theorem becomes therefore
.f(b)—.f(a)= f,[a+ei(b—a)], b—a
or if b —a= h, so that b= a+h, then
f (a + h) — f (a)=hf'(a+01h);
i.e.    f (a + h) = f (a) + hf' (a + 01h).


Then cb (a) = o and (b) = o ; and ck (x) satisfies the conditions of Rolle's Theorem.
Therefore for a value xi say between x = a and x = b, (x1) = o.
But    c6' (x) =f' (x) —f' (a) — (x — a) p,
and this vanishes for the values a and xt .
Therefore 0" (x) = o for some value of x (x2 say) between a and x1, i.e. between a and b.
But    (x) = f„ (x) — p.
Hence since 0" (x2) = o, then p = f" (x2).
f(b)—.f(a)— (b—a)f'(a)= (b — a)2 f" (x,).
If as before we put x2=a+ 02 (b — a) andb — a=h,we have
f (b) = f (a) + hf' (a) + 2h2f" (a + 02h).
I78    DIFFERENTIAL CALCULUS
Consider the general case, where all the derivatives are con:
Let
f (b) —[f (a) + (b — a) f' (a) + (b 2 )a)zf„ (a) + ... + ((n —a1) i    n-1) (a)]
4     (b — a)n
and let
0 (x) = f (b) – f (x) – (b – x) f' (x) – b2)x)zf„ (x)—...
_ ((n xI)_li f (n-1) (x) – (b - x)n q.
As before 0 (a), 0 (b) are both zero. Since 0 (x), 0' (x) are continuous, 0' (x) is zero for a value x1 in the given range. But by differentiation
n 1
0' (x) _ – b (n - - x)) ~ f (n) (x) + n (b – x)n-1 q,
all the remaining terms in the expression for 0 (x) vanishing on differentiation.
(b—x1)n f(n)
(n – I)!    (x1)+n (b-x1)n-1q=o.
q = fcnn      (xi) since b + x1. Ifx1=a+B(b–a)andb–a=hwesee that
2
.f (b) = f (a) + hf ' (a) + 2 f„ (a)
+ (nl~n 1)_I fcn-1) (a) + n I f(n) (a + Oh).
An expansion for f (x + h) in terms of f (x) and ascending powers of h is at once evident. Replace b by x + h and write x for a sothatb – a= (x+h)–x=h as before.
Then
2! f(x+h)=f(x)+hf'(x)+f„ (x)+31 f,,,(x)+...
+     I) ! f(n-1) (x) + ~ I f(n) (x + Oh).
This is Taylor's Theorem.
TAYLOR'S THEOREM
If in the above expansion we put x = o we have f(h)=f(o)+hf'(o)+2`~ f„(o)+3i f" (o)+...
n 1    n
+ (n Z 1) I {(n—1) (o) + _.I f(n
or putting x for h
f(x)=f(o)+xf'(a)+2i f„ (o)+3i f(a)+...
+ (nxn 1) 1 f ("_I) (a) + y I f(n) (Ox).
In this form the expansion is known as Stirling's or Maclaurin's Theorem.
6. It will be noticed that the first n terms in Taylor's Theorem are r
of the form 1 f (r) (x). The (n + 1)th term is of the same form but
involves a different value of the variable. This term is called the "remainder" term after n terms and is denoted by Rn (x). If
Lt Rn (x) is zero, then f (x + h) can be expanded as an infinite n—>oo
and will be convergent. We may state therefore that if f (x) and its successive differential coefficients are continuous within the given range, then
f (x) + hf , (x) + 2 ~ f" (x) + ... + ~nhn 1)      f (n-1) (x) + ...
converges to the limit f (x + h), provided that Lt Rn (x) is zero. n —.
 Igo    DIFFERENTIAL CALCULUS
This is Schlomilch's form. The Lagrange form follows immediately by putting p = n. If we put p = I we obtain another form,
Rn (x) _ jln(n1 - O)i_I f(n> (x + Oh),
due to Cauchy.
8. Examples on the above theorems.
In obtaining expansions for various functions of x it is more convenient to use Maclaurin's form than to use Taylor's. More-over, since the condition for a convergent series applies equally to both forms, it is strictly necessary to prove that Lt Rn (x), i.e.
n n
Lt x f (n) (Ox), is zero before assuming that an infinite series can
n-~T71.
represent the function. For example, on expanding (I + x)n by Maclaurin's Theorem different conditions arise according to the values of x and n, and a complete investigation of the convergency of the various series involves further mathematical analysis. In the examples that follow it will be assumed that Lt Rn (x) is zero and
that the function in question is the sum of an infinite convergent series.
Example 1.
Expand log (I + x + x2) as far as the term involving x3.
f (x) = log (1 + x + x2), and f (o) = o,
fr(x)_ I+2.t'
{ /    I -y-x+x12'
J (x) 11+2C+x2)2 (I+x+x2)
+ x 3 + x
(    2)2r    f"(°)=1,
ft// (x) =I (I     ) +x)2+(I+(x-+x2
(        )3'    frrr(o)__4.
By Maclaurin's Theorem :
.f (x) =.f (o) + x f' (o) + 2 i f" (o) + 3 i f"' (o) + ....
2    3 log(1+x+x2)=x+x -2x +....
2    3
f' (o) = 1,

ILLUSTRATIVE EXAMPLES    181
Notes on this example :
 J (T)    2'x2 (I + Bx)2
But for all positive values of x this is positive, since B is a positive proper fraction.
(x + 2) log (I + x) — 2x is positive when x is positive, i.e.    (x + 2) log (I + x) > 2x.
9. Formation of a differential equation.
This method can be employed with advantage for the expansion of certain functions without the use of the above series. It must be assumed that the given function f (x) can be expanded in the form

182    DIFFERENTIAL CALCULUS
a,, + alx + a2x2 + ... + arxr + ..., and if on differentiating f (x) a simple relation between the coefficients is evident, we can obtain the required expansion. It should be noted that the first one or two terms of the expansion may have to be found by a different method, such as by substitution of numerical values on both sides of the identity.
Example 3.
Expand log (I + x + x2) in ascending powers of x. (Cf. Ex. 1, p. 180.)
Let
log (I+x+x2)=ao+alx+a2x2+a3x3+.. +arx'+.... Then by differentiating,
I + 2X
I+x+x2=a1+2a2x+3a3x2+...+rarxr-1+...,
or I + 2x = (I+ x + x2) (a1 + 2a2x + 3aIx2 + ... + rarxr-I + ...). Equating coefficients of powers of x,
    al = I,    al = I,
    2a2 + al = 2,    a2 = ,
    3a3+2a2+a1= 3,    a3= — 3,
    4a4 + 3a3 + 2a2 = o,    a4 = +,

    5a5 + 4a4 + 3a3 = 0,    a5 = ,
and so on, the law of formation of the coefficients being rar + (r — I) ar-1 + (r — 2) ar-2 = 0,
except for the first two terms.
2    2    5
log(I+x+x2)=x+- - -+ +x...,
    2    3    4 5
since ao is obviously zero.
Example 4.
If y = log (I + sin x) obtain a relation between the first three difcoefficients of y with respect to x, and hence expand y in an ascending series of powers of x.
f (x) = y = log (1 + sin x),    f' (x) = cos x 1+sinx
„    I    cos x
f” (x) - - I + sin x'    f (x) = (1 + sin x)2'

.'. f,,, (x) + f' (x) f" (x) = 0.

Let    f(x) = as+a1x+a2x2+a3x3+a4x4+a5 x' +....

183
THE SERIES xj(ex— i)
Then    f' (x) = a1 + 2a2x + 3a3x2 + 4a4x3 + 5(20x4 + ..., f" (x) = 2a2 + 6a3x + 12a4x2 + 2oa5x3 + ..., f" (x) = 6a3 + 24a4X + 6oa5 x2 +
Now    f (o) = o,    or    ao = o,
f' (o) = I,    giving a1 = I,
f" " (0) _ — I,    ,,    a2 = 11— ,
f"' (o) = I,    ,,    a3 = g•
Since f"' (x) + f' (x) f " (x) = o, we have, by multiplying together the expansions for f' (x) and f" (x) and equating coefficients of x, 4(222 + 6(21(23 = — 24a4, so that a4 = — 11; .
Similarly, equating coefficients of x2,
6a2a3 + I2a2a3 + 12a1a4 = — 6oa5 and
and so on.
x2 x3 x4 x' ,, log(I+sin x)=x — 2+6 12+24
10. The series ex x 1.
The coefficients in the expansion of ex x I are of greatimportance in the higher branches of mathematics, and the method of obtaining the series in ascending powers of x is an excellent example of the application of the above principles.
x    x_x_ ex+ I
ex—I+2 2 ex—I
Let -.    =a +a x+a x2+a x3+a x4+....
2 ex— I— o    I    z    3    a
Change the sign of x: the left-hand side becomes
xe-x+i    x ex+1
- - '    or    '
2e-x— i    2ex— I
x ex+ I 2 ex— I
Add: then on dividing both sides by 2, we have x ex+ I =ao +azx2+a4x4+...
2.ex—1
Consider

184    DIFFERENTIAL CALCULUS
which shows that no odd powers of x occur in the expansion of x ex + I
2'ex— I'
x
Let    f (x) =    x + x_ x e+ I
ex— I 2 2 ex— I


Then    exf (x) = f (x) + zx + xex. Differentiate with respect to x:


ex[f(x)+f'(x)]=f'(x)+2+2ex(I+x). Similarly, on successive differentiation,
ex [f " (x) + 2f' (x) + f (1)] = f" (x) + 2ex (2 + x),
ex Lf,,, (x) + 13f" (x) + 3f' (x) +f (x)J =f"' (x) + 2ex (3 + x),
and the law of formation is evident.
Moreover, if we expand f (x) by Maclaurin's Theorem, we have f (x) = f (o) + xf' (o) + z , f„ (o) + 3 i f" (o)
and since there are no odd powers of x in the expansion off (x), it
follows that
f, (o) = f,,, (o) = f v (o) = ... = o.


Now ex [f (x) + f' (x)] = f' (x) + 2 + 2e' (I + x),
so that, when x = o,
f (o) = + = I since f' (o) = o.


Similarly, by substitution of x = o in the next equation but one, we obtain
if,,, (o) + 3f" (o) + 3f' (o) +f (o)) = f,,, (o) + ze° (3 + o),
and, remembering that f' (o) and f"' (o) are zero, we find that f" (o)=6.
Similarly, f I° (o) _ —f vI (o) = a'z , f veil (o) = — zs, ...
x_ x_ x ex + I    I x2    I x4    I xs I x8
ex—I 2 2' ex—I—IT6'2130.41+42.6—30.81
x    x I x2    I x4    I x6    I x8
or    ex-i=I—2+6.2! 30.4! 42.61-30.81+....

DIFFERENTIATION OF A KNOWN SERIES    185
The coefficients obtained above are denoted by B1, B,, B3, B4 ... ; these are called Bernouilli's Numbers. We have, therefore, that
x2    x4    6    x8
x-x-I = — 2x + B1 2 I — B2 x~ + B3 — B4 g I + .... 4•


The expansion of x-- - may be obtained otherwise by assuming that
ex —
i.e. that
/    x2 x3
x= (x+—,+ i+...)(ao+alx+a2x2+...+a,.xT+...),
3
and equating coefficients of powers of x on both sides of the identity.
11. Differentiation of a known series.
It can be shown that if an infinite series converges to a value f (x) within a given range, then the series formed by differentiating each term of the original series is f' (x), provided that the second series is convergent for all values of the variable within the given range.
It sometimes happens that the function whose expansion is required is the differential coefficient of a function whose exis known. By the application of the simple process of differentiating each term of the known expansion the required result is easily obtained.
Example 5.

If    log (I — x — x2) = — u1x — zzi2x2 — 3u3x3 — ...,
prove that    un = u,_1 + un-2•
d    I
dx log (I — x — x2) = —     x 2xx2
But dxlog(1 —x—x2)=—u1—u2x—u3x2—u4x3—...,


I +2x
— I — x — x2 — u1 —u2x — u3 x2 — u4 x3 — ... ,
i.e. - I    + 2x 2 is the generating function of the series I — x — x
u1 + u2x + u3x2 + ... + un_2xn_3 + un_1xn_2 + unxn-1 + .... .. un — un-1 — 11n_2 = 0, 0r un = un_1 + un-2, which proves the proposition.

x
=as+alx+a2x2+...+a,x''+...,
ex— I

186    DIFFERENTIAL CALCULUS 12. Trigonometrical series.


Let    f (x) = sin x.
Then    f' (x) = cos x = sin (x + 2) ,
f" (x) = cos (x +    = sin (x + 2 - , 2
and the nth differential coefficient is sin (x + n 2) .
f (o) = o ; f' (o) = sin 2 = I ; f" (o) = sin 22 = o; f"' (o) = sin 32 = — I ; f Iv (o) = sin z? = 0,


and so on.
Even derivatives will obviously be zero, and odd derivatives + I and — I alternately.
Therefore by Maclaurin's Theorem,
x3 x5 x' sinx=x — 3+     +...,
5    71


and it is easy to show that the series is convergent for all values of x. x2
Similarly    cos x = – l + – 6c~ + ... .
24!
This result can be obtained by replacing nO by a in the expansion for cos nO in Example 11, p. I7, and by then considering the limit of the expansion as n-). oo . It will be seen, however, that by finding the limit n->- co in the series
cos a = (cos a/n)n -112 (cos a/n)n-2 (sin a/n)2 + n4 (cos a/n)"-4 (sin a/n)4 — ... a double limit is involved. It is therefore simpler to obtain the cosine and sine series by Maclaurin's Theorem as above.
The series for tan x does not take a simple form: the first few terms can be obtained by division of the sine series by the cosine series, or by Maclaurin's Theorem.
The series for the inverse functions sin-1 x, cos-1 x, tan-' x are easily evaluated by the integration of (I — x2)-i and (I + x2)-I

RELATION BETWEEN A AND D    187

13. If ux be a rational integral function of x, the expansion for ux in terms of advancing differences is
ux=u0+x;Au0+x2A2u0+...+x,,Anu0+..., or, in factorial notation,
x(l)    x'2)    x(n)
ur=u0+ I~ Au0+ a! A2u0+...+ n Anuo+....
We may compare this with the Maclaurin expansion for ux:

x    x2    xn
ux=u0+I~Du0+2!D2u0+...+nDnu+....
Separating the symbols,
2    3    n
ux=(I+IiD+Z~D2+x, D3+...+ Z)Di ! ...Ju0
    3    /
Now whatever the degree of the polynomial in x represented byux we may treat the expression in brackets as an infinite series, for higher derivatives than, say, the nth will vanish.
Consequently, we can express ux shortly as exD uo a convenient form for relating finite differences to differential calculus.

(I +A)xu0=ux=exDIto, or symbolically    (1 + A)x exD.

This result is analogous to the important relation in the Theory of Compound Interest, (1 + i)n = ens, where 3 is the force of interest corresponding to a rate of interest i.
EXAMPLES II I. Prove that ex+h = ex + hex + ex + ... .
 188    DIFFERENTIAL CALCULUS
 EXAMPLES    189
2 Show that ex eosx — 1 + x + z — 3 • • . , and find the coefficient
18. Expand 1~1 +x in ascending powers of x by first forming a — x)
differential equation.
19. If esln-'x = ao + a1x + a2x2 + ... + arx'' + •.. , prove that (n + 2) an+2 __    )Zan
n    n + I
x>Iog(I+x)>x—2x2.