1. Verify the terms given of the following Taylor’s Series: \[\begin{aligned} &(1) & \tan x &= x + \tfrac{1}{3} x^{3} + \tfrac{2}{15} x^{5} + \dots, \\ &(2) & \sec x &= 1 + \tfrac{1}{2} x^{2} + \tfrac{5}{24} x^{4} + \dots, \\ &(3)\quad & x\csc x &= 1 + \tfrac{1}{6} x^{2} + \tfrac{7}{360} x^{4} + \dots, \\ &(4) & x\cot x &= 1 – \tfrac{1}{3} x^{2} – \tfrac{1}{45} x^{4} – \dots.\end{aligned}\]

 

2. Show that if \(f(x)\) and its first \(n + 2\) derivatives are continuous, and \(f^{(n+1)}(0) \neq 0\), and \(\theta_{n}\) is the value of \(\theta\) which occurs in Lagrange’s form of the remainder after \(n\) terms of Taylor’s Series, then \[\theta_{n} = \frac{1}{n + 1} + \frac{n}{2(n + 1)^{2}(n + 2)} \left\{\frac{f^{(n+2)}(0)}{f^{(n+1)}(0)} + \epsilon_{x}\right\}x,\] where \(\epsilon_{x} \to 0\) as \(x \to 0\). [Follow the method of Ex. LV. 12.]

 

3. Verify the last result when \(f(x) = 1/(1+ x)\). [Here \((1 + \theta_{n}x)^{n+1} = 1 + x\).]

 

4. Show that if \(f(x)\) has derivatives of the first three orders then \[f(b) = f(a) + \tfrac{1}{2}(b – a) \{f'(a) + f'(b)\} – \tfrac{1}{12}(b – a)^{3} f”'(\alpha),\] where \(a < \alpha < b\). [Apply to the function \[\begin{gathered} f(x) – f(a) – \tfrac{1}{2}(x – a) \{f'(a) + f'(x)\}\\ – \left(\frac{x – a}{b – a}\right)^{3} [f(b) – f(a) – \tfrac{1}{2}(b – a) \{f'(a) + f'(b)\}]\end{gathered}\] arguments similar to those of § 147.]

 

5. Show that under the same conditions \[f(b) = f(a) + (b – a) f’\{\tfrac{1}{2}(a + b)\} + \tfrac{1}{24}(b – a)^{3}f”'(\alpha).\]

 

6. Show that if \(f(x)\) has derivatives of the first five orders then \[f(b) = f(a) + \tfrac{1}{6}(b – a) [f'(a) + f'(b) + 4f’\{\tfrac{1}{2}(a + b)\}] – \tfrac{1}{2880}(b – a)^{5} f^{(5)}({\alpha}).\]

 

7. Show that under the same conditions \[f(b) = f(a) + \tfrac{1}{2}(b – a) \{f'(a) + f'(b)\} – \tfrac{1}{12}(b – a)^{2} \{f”(b) – f”(a)\} + \tfrac{1}{720}(b – a)^{5} f^{(5)}(\alpha).\]

 

8. Establish the formulae

\[(i) \quad \begin{vmatrix} f(a) & f(b)\\ g(a) & g(b) \end{vmatrix} = (b – a) \begin{vmatrix} f(a) & f'(\beta)\\ g(a) & g'(\beta) \end{vmatrix} \]

where \(\beta\) lies between \(a\) and \(b\), and

\[(ii) \quad\begin{vmatrix} f(a) & f(b) & f(c)\\ g(a) & g(b) & g(c)\\ h(a) & h(b) & h(c) \end{vmatrix} = \tfrac{1}{2} (b – c)(c – a)(a – b) \begin{vmatrix} f(a) & f'(\beta) & f”(\gamma)\\ g(a) & g'(\beta) & g”(\gamma)\\ h(a) & h'(\beta) & h”(\gamma) \end{vmatrix} \]

where \(\beta\) and \(\gamma\) lie between the least and greatest of \(a\), \(b\), \(c\). [To prove (ii) consider the function \[\phi(x) = \begin{vmatrix} f(a) & f(b) & f(x)\\ g(a) & g(b) & g(x)\\ h(a) & h(b) & h(x) \end{vmatrix} – \frac{(x – a)(x – b)}{(c – a)(c – b)} \begin{vmatrix} f(a) & f(b) & f(c)\\ g(a) & g(b) & g(c)\\ h(a) & h(b) & h(c) \end{vmatrix},\] which vanishes when \(x = a\), \(x = b\), and \(x = c\). Its first derivative, by Theorem B of § 121, must vanish for two distinct values of \(x\) lying between the least and greatest of \(a\), \(b\), \(c\); and its second derivative must therefore vanish for a value \(\gamma\) of \(x\) satisfying the same condition. We thus obtain the formula \[\begin{vmatrix} f(a) & f(b) & f(c)\\ g(a) & g(b) & g(c)\\ h(a) & h(b) & h(c) \end{vmatrix} = \tfrac{1}{2}(c – a)(c – b) \begin{vmatrix} f(a) & f(b) & f”(\gamma)\\ g(a) & g(b) & g”(\gamma)\\ h(a) & h(b) & h”(\gamma) \end{vmatrix}.\] The reader will now complete the proof without difficulty.]

 

9. If \(F(x)\) is a function which has continuous derivatives of the first \(n\) orders, of which the first \(n – 1\) vanish when \(x = 0\), and \(A \leq F^{(n)}(x) \leq B\) when \(0 \leq x \leq h\), then \(A(x^{n}/n!) \leq F(x) \leq B(x^{n}/n!)\) when \(0 \leq x \leq h\).

Apply this result to \[f(x) – f(0) – xf'(0) – \dots – \frac{x^{n-1}}{(n – 1)!} f^{(n-1)}(0),\] and deduce Taylor’s Theorem.

 

10. If \(\Delta_{h}\phi(x) = \phi(x) – \phi(x + h)\), \(\Delta_{h}^{2}\phi(x) = \Delta_{h}\{\Delta_{h}\phi(x)\}\), and so on, and \(\phi(x)\) has derivatives of the first \(n\) orders, then \[\Delta_{h}^{n}\phi(x) = \sum_{r=0}^{n}(-1)^{r} \binom{n}{r} \phi(x + rh) = (-h)^{n} \phi^{(n)}(\xi),\] where \(\xi\) lies between \(x\) and \(x + nh\). Deduce that if \(\phi^{(n)}(x)\) is continuous then \(\{\Delta_{h}^{n}\phi(x)\}/h^{n} \to (-1)^{n}\phi^{(n)}(x)\) as \(h \to 0\). [This result has been stated already when \(n = 2\), in Ex. LV. 13.]

 

11. Deduce from Ex. 10 that \(x^{n-m}\, \Delta_{h}^{n} x^{m} \to m(m – 1) \dots (m – n + 1)h^{n}\) as \(x \to \infty\), \(m\) being any rational number and \(n\) any positive integer. In particular prove that \[x\sqrt{x} \{\sqrt{x} – 2\sqrt{x + 1} + \sqrt{x + 2}\} \to -\tfrac{1}{4}.\]

 

12. Suppose that \(y = \phi(x)\) is a function of \(x\) with continuous derivatives of at least the first four orders, and that \(\phi(0) = 0\), \(\phi'(0) = 1\), so that \[y = \phi(x) = x + a_{2}x^{2} + a_{3}x^{3} + (a_{4} + \epsilon_{x})x^{4},\] where \(\epsilon_{x} \to 0\) as \(x \to 0\). Establish the formula \[x = \psi(y) = y – a_{2}y^{2} + (2a_{2}^{2} – a_{3})y^{3} – (5a_{2}^{3} – 5a_{2}a_{3} + a_{4} + \epsilon_{y})y^{4},\] where \(\epsilon_{y} \to 0\) as \(y \to 0\), for that value of \(x\) which vanishes with \(y\); and prove that \[\frac{\phi(x)\psi(x) – x^{2}}{x^{4}} \to a_{2}^{2}\] as \(x \to 0\).

 

13. The coordinates \((\xi, \eta)\) of the centre of curvature of the curve \(x = f(t)\), \(y = F(t)\), at the point \((x, y)\), are given by \[-(\xi – x)/y’ = (\eta – y)/x’ = (x’^{2} + y’^{2})/(x’y” – x”y’);\] and the radius of curvature of the curve is \[(x’^{2} + y’^{2})^{3/2}/(x’y” – x”y’),\] dashes denoting differentiations with respect to \(t\).

 

14. The coordinates \((\xi, \eta)\) of the centre of curvature of the curve \(27ay^{2} = 4x^{3}\), at the point \((x, y)\), are given by \[3a(\xi + x) + 2x^{2} = 0, \quad \eta = 4y + (9ay)/x.\quad\]

 

15. Prove that the circle of curvature at a point \((x, y)\) will have contact of the third order with the curve if \((1 + y_{1}^{2})y_{3} = 3y_{1}y_{2}^{2}\) at that point. Prove also that the circle is the only curve which possesses this property at every point; and that the only points on a conic which possess the property are the extremities of the axes. [Cf. Ch. VI, Misc. Ex. 10 (iv).]

 

16. The conic of closest contact with the curve \(y = ax^{2} + bx^{3} + cx^{4} + \dots + kx^{n}\), at the origin, is \(a^{3}y = a^{4}x^{2} + a^{2}bxy + (ac – b^{2})y^{2}\). Deduce that the conic of closest contact at the point \((\xi, \eta)\) of the curve \(y = f(x)\) is \[18\eta_{2}^{3}T = 9\eta_{2}^{4}(x – \xi)^{2} + 6\eta_{2}^{2}\eta_{3}(x – \xi)T + (3\eta_{2}\eta_{4} – 4\eta_{3}^{2})T^{2},\] where \(T = (y – \eta) – \eta_{1}(x – \xi)\).

 

17. Homogeneous functions.1 If \(u = x^{n} f(y/x, z/x, \dots)\) then \(u\) is unaltered, save for a factor \(\lambda^{n}\), when \(x\), \(y\), \(z\), … are all increased in the ratio \(\lambda : 1\). In these circumstances \(u\) is called a homogeneous function of degree \(n\) in the variables \(x\), \(y\), \(z\), …. Prove that if \(u\) is homogeneous and of degree \(n\) then \[x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} + \dots = nu.\] This result is known as Euler’s Theorem on homogeneous functions.

 

18. If \(u\) is homogeneous and of degree \(n\) then \(\partial u/\partial x\), \(\partial u/\partial y\), … are homogeneous and of degree \(n – 1\).

 

19. Let \(f(x, y) = 0\) be an equation in \(x\) and \(y\) (e.g. \(x^{n} + y^{n} – x = 0\)), and let \(F(x, y, z) = 0\) be the form it assumes when made homogeneous by the introduction of a third variable \(z\) in place of unity ( \(x^{n} + y^{n} – xz^{n-1} = 0\)). Show that the equation of the tangent at the point \((\xi, \eta)\) of the curve \(f(x, y) = 0\) is \[xF_{\xi} + yF_{\eta} + zF_{\zeta} = 0,\] where \(F_{\xi}\), \(F_{\eta}\), \(F_{\zeta}\) denote the values of \(F_{x}\), \(F_{y}\), \(F_{z}\) when \(x = \xi\), \(y = \eta\), \(z = \zeta = 1\).

 

20. Dependent and independent functions. Jacobians or functional determinants. Suppose that \(u\) and \(v\) are functions of \(x\) and \(y\) connected by an identical relation \[\begin{equation*} \phi(u, v) = 0. \tag{1}\end{equation*}\]

Differentiating (1) with respect to \(x\) and \(y\), we obtain \[\begin{equation*} \frac{\partial \phi}{\partial u}\, \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v}\, \frac{\partial v}{\partial x} = 0,\quad \frac{\partial \phi}{\partial u}\, \frac{\partial u}{\partial y} + \frac{\partial \phi}{\partial v}\, \frac{\partial v}{\partial y} = 0, \tag{2} \end{equation*}\] and, eliminating the derivatives of \(\phi\), \[\begin{equation*} J = \begin{vmatrix} u_{x} & u_{y}\\ v_{x} & v_{y} \end{vmatrix} = u_{x}v_{y} – u_{y}v_{x} = 0, \tag{3} \end{equation*}\] where \(u_{x}\), \(u_{y}\), \(v_{x}\), \(v_{y}\) are the derivatives of \(u\) and \(v\) with respect to \(x\) and \(y\). This condition is therefore necessary for the existence of a relation such as (1). It can be proved that the condition is also sufficient; for this we must refer to Goursat’s Cours d’ Analyse, vol. i, pp. 125 et seq.

Two functions \(u\) and \(v\) are said to be dependent or independent according as they are or are not connected by such a relation as (1). It is usual to call \(J\) the Jacobian or functional determinant of \(u\) and \(v\) with respect to \(x\) and \(y\), and to write \[J = \frac{\partial(u, v)}{\partial(x, y)}.\]

Similar results hold for functions of any number of variables. Thus three functions \(u\), \(v\), \(w\) of three variables \(x\), \(y\), \(z\) are or are not connected by a relation \(\phi(u, v, w) = 0\) according as \[J = \begin{vmatrix} u_{x} & u_{y} & u_{z}\\ v_{x} & v_{y} & v_{z}\\ w_{x} & w_{y} & w_{z} \end{vmatrix} = \frac{\partial(u, v, w)}{\partial(x, y, z)}\] does or does not vanish for all values of \(x\), \(y\), \(z\).

 

21. Show that \(ax^{2} + 2hxy + by_{2}\) and \(Ax^{2} + 2Hxy + By^{2}\) are independent unless \(a/A = h/H = b/B\).

 

22. Show that \(ax^{2} + by^{2} + cz^{2} + 2fyz + 2gzx + 2hxy\) can be expressed as a product of two linear functions of \(x\), \(y\), and \(z\) if and only if \[abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0.\]

[Write down the condition that \(px + qy + rz\) and \(p’x + q’y + r’z\) should be connected with the given function by a functional relation.]

 

23. If \(u\) and \(v\) are functions of \(\xi\) and \(\eta\), which are themselves functions of \(x\) and \(y\), then \[\frac{\partial(u, v)}{\partial(x, y)} = \frac{\partial(u, v)}{\partial(\xi, \eta)}\, \frac{\partial(\xi, \eta)}{\partial(x, y)}.\] Extend the result to any number of variables.

 

24. Let \(f(x)\) be a function of \(x\) whose derivative is \(1/x\) and which vanishes when \(x = 1\). Show that if \(u = f(x) + f(y)\), \(v = xy\), then \(u_{x}v_{y} – u_{y}v_{x} = 0\), and hence that \(u\) and \(v\) are connected by a functional relation. By putting \(y = 1\), show that this relation must be \(f(x) + f(y) = f(xy)\). Prove in a similar manner that if the derivative of \(f(x)\) is \(1/(1 + x^{2})\), and \(f(0) = 0\), then \(f(x)\) must satisfy the equation \[f(x) + f(y) = f\left(\frac{x + y}{1 – xy}\right).\]

 

25. Prove that if \(f(x) = \int_{0}^{x} \frac{dt}{\sqrt{1 – t^{4}}}\) then \[f(x) + f(y) = f\left\{ \frac{x\sqrt{1 – y^{4}} + y\sqrt{1 – x^{4}}}{1 + x^{2}y^{2}} \right\}.\]

 

26. Show that if a functional relation exists between \[u = f(x) + f(y) + f(z),\quad v = f(y)f(z) + f(z)f(x) + f(x)f(y),\quad w = f(x)f(y)f(z),\] then \(f\) must be a constant. [The condition for a functional relation will be found to be \[f'(x)f'(y)f'(z) \{f(y) – f(z)\} \{f(z) – f(x)\} \{f(x) – f(y)\} = 0.]\]

 

27. If \(f(y, z)\), \(f(z, x)\), and \(f(x, y)\) are connected by a functional relation then \(f(x, x)\) is independent of \(x\).

If \(u = 0\), \(v = 0\), \(w = 0\) are the equations of three circles, rendered homogeneous as in Ex. 19, then the equation \[\frac{\partial(u, v, w)}{\partial(x, y, z)} = 0\] represents the circle which cuts them all orthogonally.

 

29. If \(A\), \(B\), \(C\) are three functions of \(x\) such that \[\begin{vmatrix} A & A’ & A”\\ B & B’ & B”\\ C & C’ & C” \end{vmatrix}\] vanishes identically, then we can find constants \(\lambda\), \(\mu\), \(\nu\) such that \(\lambda A + \mu B + \nu C\) vanishes identically; and conversely. [The converse is almost obvious. To prove the direct theorem let \(\alpha = BC’ – B’C\), …. Then \(\alpha’ = BC” – B”C\), …, and it follows from the vanishing of the determinant that \(\beta\gamma’ – \beta’\gamma = 0\), …; and so that the ratios \(\alpha : \beta : \gamma\) are constant. But \(\alpha A + \beta B + \gamma C = 0\).]

 

30. Suppose that three variables \(x\), \(y\), \(z\) are connected by a relation in virtue of which (i) \(z\) is a function of \(x\) and \(y\), with derivatives \(z_{x}\), \(z_{y}\), and (ii) \(x\) is a function of \(y\) and \(z\), with derivatives \(x_{y}\), \(x_{z}\). Prove that \[x_{y} = – z_{y}/z_{x},\quad x_{z} = 1/z_{x}.\]

[We have \[dz = z_{x}\, dx + z_{y}\, dy,\quad dx = x_{y}\, dy + x_{z}\, dz.\] The result of substituting for \(dx\) in the first equation is \[dz = (z_{x} x_{y} + z_{y})\, dy + z_{x}x_{z}\, dz,\] which can be true only if \(z_{x} x_{y} + z_{y} = 0\), \(z_{x} x_{z} = 1\).]

 

31. Four variables \(x\), \(y\), \(z\), \(u\) are connected by two relations in virtue of which any two can be expressed as functions of the others. Show that \[y_{z}^{u}z_{x}^{u}x_{y}^{u} = -y_{z}^{x}z_{x}^{y}x_{y}^{z} = 1,\quad x_{z}^{u}z_{x}^{y} + y_{z}^{u}z_{y}^{x} = 1,\] where \(y_{z}^{u}\) denotes the derivative of \(y\), when expressed as a function of \(z\) and \(u\), with respect to \(z\).

 

32. Find \(A\), \(B\), \(C\), \(\lambda\) so that the first four derivatives of \[\int_{a}^{a+x} f(t)\, dt – x[Af(a) + Bf(a + \lambda x) + Cf(a + x)]\] vanish when \(x = 0\); and \(A\), \(B\), \(C\), \(D\), \(\lambda\), \(\mu\) so that the first six derivatives of \[\int_{a}^{a+x} f(t)\, dt – x[Af(a) + Bf(a + \lambda x) + Cf(a + \mu x) + Df(a + x)]\] vanish when \(x = 0\).

 

33. If \(a > 0\), \(ac – b^{2} > 0\), and \(x_{1} > x_{0}\), then \[\int_{x_{0}}^{x_{1}} \frac{dx}{ax^{2} + 2bx + c} = \frac{1}{\sqrt{ac – b^{2}}} \arctan\left\{ \frac{(x_{1} – x_{0}) \sqrt{ac – b^{2}}} {ax_{1}x_{0} + b(x_{1} + x_{0}) + c} \right\},\] the inverse tangent lying between \(0\) and \(\pi\).2

 

34. Evaluate the integral \(\int_{-1}^{1} \frac{\sin\alpha\, dx}{1 – 2x\cos\alpha + x^{2}}\). For what values of \(\alpha\) is the integral a discontinuous function of \(\alpha\)?

[The value of the integral is \(\frac{1}{2}\pi\) if \(2n\pi < \alpha < (2n + 1)\pi\), and \(-\frac{1}{2}\pi\) if \((2n – 1)\pi < \alpha < 2n\pi\), \(n\) being any integer; and \(0\) if \(\alpha\) is a multiple of \(\pi\).]

 

35. If \(ax^{2} + 2bx + c > 0\) when \(x_{0} \leq x \leq x_{1}\), \(f(x) = \sqrt{ax^{2} + 2bx + c}\), and \[y = f(x),\quad y_{0} = f(x_{0}),\quad y_{1} = f(x_{1}),\quad X = (x_{1} – x_{0})/(y_{1} + y_{0}),\] then \[\int_{x_{0}}^{x_{1}} \frac{dx}{y} = \frac{1}{\sqrt{a}} \log \frac{1 + X\sqrt{a}}{1 – X\sqrt{a}},\quad \frac{-2}{\sqrt{-a}} \arctan\{X\sqrt{-a}\},\] according as \(a\) is positive or negative. In the latter case the inverse tangent lies between \(0\) and \(\frac{1}{2}\pi\). [It will be found that the substitution \(t = \dfrac{x – x_{0}}{y + y_{0}}\) reduces the integral to the form \( 2\int_{0}^{X} \frac{dt}{1 – at^{2}}\).]

 

36. Prove that \[\int_{0}^{a} \frac{dx}{x + \sqrt{a^{2} – x^{2}}} = \tfrac{1}{4}\pi.\]

 

37. If \(a > 1\) then \[\int_{-1}^{1} \frac{\sqrt{1 – x^{2}}}{a – x}\, dx = \pi\{a – \sqrt{a^{2} – 1}\}.\]

 

38. If \(p > 1\), \(0 < q < 1\), then \[\int_{0}^{1} \frac{dx}{\sqrt{\{1 + (p^{2} – 1)x\}\{1 – (1 – q^{2}) x\}}} = \frac{2\omega}{(p + q)\sin\omega},\] where \(\omega\) is the positive acute angle whose cosine is \((1 + pq)/(p + q)\).

 

39. If \(a > b > 0\), then \[\int_{0}^{2\pi} \frac{\sin^{2}\theta\, d\theta}{a – b\cos\theta} = \frac{2\pi}{b^{2}} \{a – \sqrt{a^{2} – b^{2}}\}.\]

 

40. Prove that if \(a > \sqrt{b^{2} + c^{2}}\) then \[\int_{0}^{\pi} \frac{d\theta}{a + b\cos\theta + c\sin\theta} = \frac{2}{\sqrt{a^{2} – b^{2} – c^{2}}} \arctan \left\{\frac{\sqrt{a^{2} – b^{2} – c^{2}}}{c}\right\},\] the inverse tangent lying between \(0\) and \(\pi\).

 

41. If \(f(x)\) is continuous and never negative, and \(\int_{a}^{b} f(x)\, dx = 0\), then \(f(x) = 0\) for all values of \(x\) between \(a\) and \(b\). [If \(f(x)\) were equal to a positive number \(k\) when \(x = \xi\), say, then we could, in virtue of the continuity of \(f(x)\), find an interval \({[\xi – \delta, \xi + \delta]}\) throughout which \(f(x) > \frac{1}{2}k\); and then the value of the integral would be greater than \(\delta k\).]

 

42. Schwarz’s inequality for integrals. Prove that \[\left(\int_{a}^{b} \phi\psi\, dx\right)^{2} \leq \int_{a}^{b} \phi^{2}\, dx \int_{a}^{b} \psi^{2}\, dx.\]

[Use the definitions of §§ 156 and 157, and the inequality \[\left(\sum\phi_{\nu}\psi_{\nu}\, \delta_{\nu}\right)^{2} \leq \sum\phi_{\nu}^{2}\, \delta_{\nu} \sum\psi_{\nu}^{2}\, \delta_{\nu}\] (Ch. I, Misc. Ex. 10).]

 

43. If \[P_{n}(x) = \frac{1}{(\beta – \alpha)^{n} n!} \left(\frac{d}{dx}\right)^{n} \{(x – \alpha)(\beta – x)\}^{n},\] then \(P_{n}(x)\) is a polynomial of degree \(n\), which possesses the property that \[\int_{\alpha}^{\beta} P_{n}(x)\theta(x)\, dx = 0\] if \(\theta(x)\) is any polynomial of degree less than \(n\). [Integrate by parts \(m + 1\) times, where \(m\) is the degree of \(\theta(x)\), and observe that \(\theta^{(m+1)}(x) = 0\).]

 

44. Prove that \[\int_{\alpha}^{\beta} P_{m}(x) P_{n}(x)\, dx = 0\] if \(m \neq n\), but that if \(m = n\) then the value of the integral is \((\beta – \alpha)/(2n + 1)\).

 

45. If \(Q_{n}(x)\) is a polynomial of degree \(n\), which possesses the property that \[\int_{\alpha}^{\beta} Q_{n}(x)\theta(x)\, dx = 0\] if \(\theta(x)\) is any polynomial of degree less than \(n\), then \(Q_{n}(x)\) is a constant multiple of \(P_{n}(x)\).

[We can choose \(\kappa\) so that \(Q_{n} – \kappa P_{n}\) is of degree \(n – 1\): then \[\int_{\alpha}^{\beta} Q_{n}(Q_{n} – \kappa P_{n})\, dx = 0,\quad \int_{\alpha}^{\beta} P_{n}(Q_{n} – \kappa P_{n})\, dx = 0,\] and so \[\int_{\alpha}^{\beta} (Q_{n} – \kappa P_{n})^{2}\, dx = 0.\] Now apply Ex. 41.]

 

46. Approximate Values of definite integrals. Show that the error in taking \(\tfrac{1}{2}(b – a) \{\phi(a) + \phi(b)\}\) as the value of the integral \(\int_{a}^{b} \phi(x)\, dx\) is less than \(\tfrac{1}{12}M(b – a)^{3}\), where \(M\) is the maximum of \(|\phi”(x)|\) in the interval \({[a, b]}\); and that the error in taking \((b – a)\phi\{\tfrac{1}{2}(a + b)\}\) is less than \(\tfrac{1}{24}M(b – a)^{3}\). [Write \(f'(x)= \phi(x)\) in Exs. 4 and 5.] Show that the error in taking \[\tfrac{1}{6}(b – a)[\phi(a) + \phi(b) + 4\phi\{\tfrac{1}{2}(a + b)\}]\] as the value is less than \(\tfrac{1}{2880}M(b – a)^{5}\), where \(M\) is the maximum of \(\phi^{(4)}(x)\). [Use Ex. 6. This rule, which gives a very good approximation, is known as Simpson’s Rule. It amounts to taking one-third of the first approximation given above and two-thirds of the second.]

Show that the approximation assigned by Simpson’s Rule is the area bounded by the lines \(x = a\), \(x = b\), \(y = 0\), and a parabola with its axis parallel to \(OY\) and passing through the three points on the curve \(y = \phi(x)\) whose abscissae are \(a\), \(\tfrac{1}{2}(a + b)\), \(b\).

It should be observed that if \(\phi(x)\) is any cubic polynomial then \(\phi^{(4)}(x) = 0\), and Simpson’s Rule is exact. That is to say, given three points whose abscissae are \(a\), \(\tfrac{1}{2}(a + b)\), \(b\), we can draw through them an infinity of curves of the type \(y = \alpha + \beta x + \gamma x^{2} + \delta x^{3}\); and all such curves give the same area. For one curve \(\delta = 0\), and this curve is a parabola.

 

47. If \(\phi(x)\) is a polynomial of the fifth degree, then \[\int_{0}^{1} \phi(x)\, dx = \tfrac{1}{18}\{5\phi(\alpha) + 8\phi(\tfrac{1}{2}) + 5\phi(\beta)\},\] \(\alpha\) and \(\beta\) being the roots of the equation \(x^{2} – x + \frac{1}{10} = 0\).

 

48. Apply Simpson’s Rule to the calculation of \(\pi\) from the formula \(\tfrac{1}{4}\pi = \int_{0}^{1} \frac{dx}{1 + x^{2}}\). [The result is \(.7833\dots\). If we divide the integral into two, from \(0\) to \(\tfrac{1}{2}\) and \(\tfrac{1}{2}\) to \(1\), and apply Simpson’s Rule to the two integrals separately, we obtain \(.785\ 391\ 6\dots\). The correct value is \(.785\ 398\ 1\dots\).]

 

49. Show that \[8.9 < \int_{3}^{5} \sqrt{4 + x^{2}}\, dx < 9.\]

 

50. Calculate the integrals \[\int_{0}^{1} \frac{dx}{1 + x},\quad \int_{0}^{1} \frac{dx}{\sqrt{1 + x^{4}}},\quad \int_{0}^{\pi} \sqrt{\sin x}\, dx,\quad \int_{0}^{\pi} \frac{\sin x}{x}\, dx,\] to two places of decimals. [In the last integral the subject of integration is not defined when \(x = 0\): but if we assign to it, when \(x = 0\), the value \(1\), it becomes continuous throughout the range of integration.]


  1. In this and the following examples the reader is to assume the continuity of all the derivatives which occur.↩︎
  2. In connection with Exs. 33–35, 38, and 40 see a paper by Dr Bromwich in vol. xxxv of the Messenger of Mathematics.↩︎

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