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 \[\left(\int_{a}^{b} \phi\psi\, dx\right)^{2} \leq \int_{a}^{b} \phi^{2}\, dx \int_{a}^{b} \psi^{2}\, dx.\]

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.]

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