ADDITIONAL THEOREMS IN THE DIFFERENTIAL AND \\ INTEGRAL CALCULUS

Exercise LXIII, problem 10a, p. 289

Prove that $\ds\int_{0}^{\pi} \cos mx \cos nx\, dx$ and $\ds\int_{0}^{\pi} \sin mx \sin nx\, dx$ are each equal to zero except when $m = n$, when each is equal to $\frac{1}{2}\pi$; and that _0^ mx nx  dx = 2nn^2 - m^2,0pt minus 3pt_0^ mx nx  dx = 0, according as $n - m$ is odd or even.

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Exercise LXIII, problem 10b, p. 289

Prove that $\ds\int_{0}^{\pi} \cos mx \cos nx\, dx$ and $\ds\int_{0}^{\pi} \sin mx \sin nx\, dx$ are each equal to zero except when $m = n$, when each is equal to $\frac{1}{2}\pi$; and that _0^ mx nx  dx = 2nn^2 - m^2,0pt minus 3pt_0^ mx nx  dx = 0, according as $n - m$ is odd or even.

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Exercise LXIII, problem 10c, p. 289

Prove that $\ds\int_{0}^{\pi} \cos mx \cos nx\, dx$ and $\ds\int_{0}^{\pi} \sin mx \sin nx\, dx$ are each equal to zero except when $m = n$, when each is equal to $\frac{1}{2}\pi$; and that _0^ mx nx  dx = 2nn^2 - m^2,0pt minus 3pt_0^ mx nx  dx = 0, according as $n - m$ is odd or even.

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Exercise LXIII, problem 10d, p. 289

Prove that $\ds\int_{0}^{\pi} \cos mx \cos nx\, dx$ and $\ds\int_{0}^{\pi} \sin mx \sin nx\, dx$ are each equal to zero except when $m = n$, when each is equal to $\frac{1}{2}\pi$; and that _0^ mx nx  dx = 2nn^2 - m^2,0pt minus 3pt_0^ mx nx  dx = 0, according as $n - m$ is odd or even.

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Exercise LXIII, problem 1a, p. 289

Show that _a^b x^n  dx = b^n+1 - a^n+1n + 1, and in particular that _0^1 x^n  dx = 1n + 1.

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Exercise LXIII, problem 1b, p. 289

Show that _a^b x^n  dx = b^n+1 - a^n+1n + 1, and in particular that _0^1 x^n  dx = 1n + 1.

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Exercise LXIII, problem 2a, p. 289

$\ds\int_{a}^{b} \cos mx\, dx = \frac{\sin mb - \sin ma}{m}$, $\ds\int_{a}^{b} \sin mx\, dx = \frac{\cos ma - \cos mb}{m}$.

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Exercise LXIII, problem 2b, p. 289

$\ds\int_{a}^{b} \cos mx\, dx = \frac{\sin mb - \sin ma}{m}$, $\ds\int_{a}^{b} \sin mx\, dx = \frac{\cos ma - \cos mb}{m}$.

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Exercise LXIII, problem 3a, p. 289

$\ds\int_{a}^{b}\frac{dx}{1 + x^{2}} = \arctan b - \arctan a$, $\ds\int_{0}^{1}\frac{dx}{1 + x^{2}} = \tfrac{1}{4}\pi$. [There is an apparent difficulty here owing to the fact that $\arctan x$ is a many valued function. The difficulty may be avoided by observing that, in the equation _0^x dt1 + t^2 = x, $\arctan x$ must denote an angle lying between $-\frac{1}{2}\pi$ and $\frac{1}{2}\pi$. For the integral vanishes when $x = 0$ and increases steadily and continuously as $x$ increases. Thus the same is true of $\arctan x$, which therefore tends to $\tfrac{1}{2}\pi$ as $x \to \infty$. In the same way we can show that $\arctan x \to -\frac{1}{2}\pi$ as $x \to -\infty$. Similarly, in the equation _0^x dt1 - t^2 = x, where $-1 < x < 1$, $\arcsin x$ denotes an angle lying between $-\frac{1}{2}\pi$ and $\frac{1}{2}\pi$. Thus, if $a$ and $b$ are both numerically less than unity, we have _a^b dx1 - x^2 = b - a.]

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Exercise LXIII, problem 3b, p. 289

$\ds\int_{a}^{b}\frac{dx}{1 + x^{2}} = \arctan b - \arctan a$, $\ds\int_{0}^{1}\frac{dx}{1 + x^{2}} = \tfrac{1}{4}\pi$. [There is an apparent difficulty here owing to the fact that $\arctan x$ is a many valued function. The difficulty may be avoided by observing that, in the equation _0^x dt1 + t^2 = x, $\arctan x$ must denote an angle lying between $-\frac{1}{2}\pi$ and $\frac{1}{2}\pi$. For the integral vanishes when $x = 0$ and increases steadily and continuously as $x$ increases. Thus the same is true of $\arctan x$, which therefore tends to $\tfrac{1}{2}\pi$ as $x \to \infty$. In the same way we can show that $\arctan x \to -\frac{1}{2}\pi$ as $x \to -\infty$. Similarly, in the equation _0^x dt1 - t^2 = x, where $-1 < x < 1$, $\arcsin x$ denotes an angle lying between $-\frac{1}{2}\pi$ and $\frac{1}{2}\pi$. Thus, if $a$ and $b$ are both numerically less than unity, we have _a^b dx1 - x^2 = b - a.]

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Exercise LXIII, problem 4a, p. 289

$\ds\int_{0}^{1} \frac{dx}{1 - x + x^{2}} = \frac{2\pi}{3\sqrt3}$, $\ds\int_{0}^{1} \frac{dx}{1 + x + x^{2}} = \frac{\pi}{3\sqrt3}$

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Exercise LXIII, problem 4b, p. 289

$\ds\int_{0}^{1} \frac{dx}{1 - x + x^{2}} = \frac{2\pi}{3\sqrt3}$, $\ds\int_{0}^{1} \frac{dx}{1 + x + x^{2}} = \frac{\pi}{3\sqrt3}$

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Exercise LXIII, problem 5, p. 289

$\ds\int_{0}^{1} \frac{dx}{1 + 2x\cos\alpha + x^{2}} = \frac{\alpha}{2\sin\alpha}$ if $-\pi < \alpha < \pi$, except when $\alpha = 0$, when the value of the integral is $\frac{1}{2}$, which is the limit of $\frac{1}{2}\alpha\cosec\alpha$ as $\alpha \to 0$.

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Exercise LXIII, problem 6a, p. 289

$\ds\int_{0}^{\DPtypo{}{1}} \sqrtp{1 - x^{2}}\, dx = \tfrac{1}{4}\pi$, $\ds\int_{0}^{a} \sqrtp{a^{2} - x^{2}}\, dx = \tfrac{1}{4}\pi a^{2}$0pt minus 3pt $(a > 0)$.

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Exercise LXIII, problem 6b, p. 289

$\ds\int_{0}^{\DPtypo{}{1}} \sqrtp{1 - x^{2}}\, dx = \tfrac{1}{4}\pi$, $\ds\int_{0}^{a} \sqrtp{a^{2} - x^{2}}\, dx = \tfrac{1}{4}\pi a^{2}$0pt minus 3pt $(a > 0)$.

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Exercise LXIII, problem 7, p. 289

$\ds\int_{0}^{\pi} \frac{dx}{a + b\cos x} = \frac{\pi}{\sqrt{a^{2} - b^{2}}}$, if $a > |b|$. [For the form of the indefinite integral see liii. 3, 4. If $|a| < |b|$ then the subject of integration has an infinity between $0$ and $\pi$. What is the value of the integral when $a$ is negative and $-a > |b|$?]

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Exercise LXIII, problem 8, p. 289

$\ds\int_{0}^{\frac{1}{2}\pi} \frac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x} = \frac{\pi}{2ab}$, if $a$ and $b$ are positive. What is the value of the integral when $a$ and $b$ have opposite signs, or when both are negative?

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Exercise LXIII, problem 9a, p. 289

**’s integrals.** Prove that if $m$ and $n$ are positive integers then _0^2 mx nx  dx is always equal to zero, and _0^2 mx nx  dx,0pt minus 3pt_0^2 mx nx  dx are equal to zero unless $m = n$, when each is equal to $\pi$.

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Exercise LXIII, problem 9b, p. 289

**’s integrals.** Prove that if $m$ and $n$ are positive integers then _0^2 mx nx  dx is always equal to zero, and _0^2 mx nx  dx,0pt minus 3pt_0^2 mx nx  dx are equal to zero unless $m = n$, when each is equal to $\pi$.

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Exercise LXIII, problem 9c, p. 289

**’s integrals.** Prove that if $m$ and $n$ are positive integers then _0^2 mx nx  dx is always equal to zero, and _0^2 mx nx  dx,0pt minus 3pt_0^2 mx nx  dx are equal to zero unless $m = n$, when each is equal to $\pi$.

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Exercise LXIV, problem 1, p. 290

Evaluate $\ds\int_{a}^{b} x\, dx$ by dividing $\DPmod{(a, b)}{[a, b]}$ into $n$ equal parts by the points of division $a = x_{0}$, $x_{1}$, $x_{2}$, …, $x_{n} = b$, and calculating the limit as $n \to \infty$ of (x_1 - x_0)f(x_0) + (x_2 - x_1)f(x_1) + …+ (x_n - x_n-1)f(x_n-1).

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Exercise LXIV, problem 2, p. 290

Calculate $\ds\int_{a}^{b} x^{2}\, dx$ in the same way.

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Exercise LXIV, problem 3a, p. 290

Calculate $\ds\int_{a}^{b} x\, dx$, where $0 < a < b$, by dividing $\DPmod{(a, b)}{[a, b]}$ into $n$ parts by the points of division $a$, $ar$, $ar^{2}$, … $ar^{n-1}$, $ar^{n}$, where $r^{n} = b/a$. Apply the same method to the more general integral $\ds\int_{a}^{b} x^{m}\, dx$.

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Exercise LXIV, problem 3b, p. 290

Calculate $\ds\int_{a}^{b} x\, dx$, where $0 < a < b$, by dividing $\DPmod{(a, b)}{[a, b]}$ into $n$ parts by the points of division $a$, $ar$, $ar^{2}$, … $ar^{n-1}$, $ar^{n}$, where $r^{n} = b/a$. Apply the same method to the more general integral $\ds\int_{a}^{b} x^{m}\, dx$.

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Exercise LXIV, problem 4a, p. 290

Calculate $\ds\int_{a}^{b}\cos mx\, dx$ and $\ds\int_{a}^{b}\sin mx\, dx$ by the method of Ex. 1.

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Exercise LXIV, problem 4b, p. 290

Calculate $\ds\int_{a}^{b}\cos mx\, dx$ and $\ds\int_{a}^{b}\sin mx\, dx$ by the method of Ex. 1.

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Exercise LXIV, problem 5, p. 290

Prove that $n\sum\limits_{r=0}^{n-1} \dfrac{1}{n^{2} + r^{2}} \to \tfrac{1}{4}\pi$ as $n \to \infty$.

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Exercise LXIV, problem 6, p. 290

Prove that $\dfrac{1}{n^{2}} \sum\limits_{r=0}^{n-1} \sqrtp{n^{2} - r^{2}} \to \tfrac{1}{4}\pi$.

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Exercise LXV, problem 10, p. 293

Prove that [ tfrac12 < int_0^1 fracdxsqrtp4 - x^2 + x^3 < tfrac16pi. ]

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Exercise LXV, problem 11, p. 293

Prove that $(3x + 8)/16 < 1/\\sqrtp{4 - 3x + x^{3}} < 1/\\sqrtp{4 - 3x}$ if $0 < x < 1$, and hence that [ tfrac1932 < int_0^1 fracdxsqrtp4 - 3x + x^3 < tfrac23. ]

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Exercise LXV, problem 12, p. 293

Prove that [ .573 < int_1^2 fracdxsqrtp4 - 3x + x^3 < .595. ]

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Exercise LXV, problem 13, p. 293

If $\alpha$ and $\phi$ are positive acute angles then < _0^ dx1 - ^2^2 x < 1 - ^2^2. If $\alpha = \phi = \frac{1}{6}\pi$, then the integral lies between $.523$ and $.541$.

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Exercise LXV, problem 14, p. 293

Prove that [ left|int_a^b f(x), dxright| leq int_a^b|f(x)|, dx. ]

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Exercise LXV, problem 15, p. 293

If $|f(x)| \\leq M$, then [ left|int_a^b f(x)phi(x), dxright| leq Mint_a^b|phi(x)|, dx. ]

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Exercise LXV, problem 1a, p. 293

Show, by means of the direct definition of the definite integral, and equations Eq(1)--Eq(5) above, that CenterLineItemp(i)$\\ds\\int_{-a}^{a} \\phi(x^{2})\\, dx = 2\\int_{0}^{a} \\phi(x^{2})\\, dx$,quad $\\ds\\int_{-a}^{a} x\\phi(x^{2})\\, dx = 0$; CenterLineItemp(ii)$\\ds\\int_{0}^{\\frac{1}{2}\\pi} \\phi(\\cos x)\\, dx = \\int_{0}^{\\frac{1}{2} \\pi} \\phi(\\sin x)\\, dx = \\tfrac{1}{2} \\int_{0}^{\\pi} \\phi(\\sin x)\\, dx$; CenterLineItemp(iii)$\\ds\\int_{0}^{m\\pi} \\phi(\\cos^{2} x)\\, dx = m\\int_{0}^{\\pi} \\phi(\\cos^{2} x)\\, dx$, $m$ being an integer.

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Exercise LXV, problem 1b, p. 293

Show, by means of the direct definition of the definite integral, and equations Eq(1)--Eq(5) above, that CenterLineItemp(i)$\\ds\\int_{-a}^{a} \\phi(x^{2})\\, dx = 2\\int_{0}^{a} \\phi(x^{2})\\, dx$,quad $\\ds\\int_{-a}^{a} x\\phi(x^{2})\\, dx = 0$; CenterLineItemp(ii)$\\ds\\int_{0}^{\\frac{1}{2}\\pi} \\phi(\\cos x)\\, dx = \\int_{0}^{\\frac{1}{2} \\pi} \\phi(\\sin x)\\, dx = \\tfrac{1}{2} \\int_{0}^{\\pi} \\phi(\\sin x)\\, dx$; CenterLineItemp(iii)$\\ds\\int_{0}^{m\\pi} \\phi(\\cos^{2} x)\\, dx = m\\int_{0}^{\\pi} \\phi(\\cos^{2} x)\\, dx$, $m$ being an integer.

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Exercise LXV, problem 1c, p. 293

Show, by means of the direct definition of the definite integral, and equations Eq(1)--Eq(5) above, that CenterLineItemp(i)$\\ds\\int_{-a}^{a} \\phi(x^{2})\\, dx = 2\\int_{0}^{a} \\phi(x^{2})\\, dx$,quad $\\ds\\int_{-a}^{a} x\\phi(x^{2})\\, dx = 0$; CenterLineItemp(ii)$\\ds\\int_{0}^{\\frac{1}{2}\\pi} \\phi(\\cos x)\\, dx = \\int_{0}^{\\frac{1}{2} \\pi} \\phi(\\sin x)\\, dx = \\tfrac{1}{2} \\int_{0}^{\\pi} \\phi(\\sin x)\\, dx$; CenterLineItemp(iii)$\\ds\\int_{0}^{m\\pi} \\phi(\\cos^{2} x)\\, dx = m\\int_{0}^{\\pi} \\phi(\\cos^{2} x)\\, dx$, $m$ being an integer.

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Exercise LXV, problem 2, p. 293

Prove that $\ds\int_{0}^{\pi} \frac{\sin nx}{\sin x}\, dx$ is equal to $\pi$ or to $0$ according as $n$ is odd or or even.

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Exercise LXV, problem 3, p. 293

Prove that $\ds\int_{0}^{\pi} \sin nx \cot x\, dx$ is equal to $0$ or to $\pi$ according as $n$ is odd or even.

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Exercise LXV, problem 4, p. 293

If $\phi(x) = a_{0} + a_{1}\cos x + b_{1}\sin x + a_{2}\cos 2x + \dots + a_{n}\cos nx + b_{n}\sin nx$, and $k$ is a positive integer not greater than $n$, then _0^2 (x)  dx = 2a_0,0pt minus 3pt_0^2 kx (x)  dx = a_k,0pt minus 3pt_0^2 kx (x)  dx = b_k. If $k > n$ then the value of each of the last two integrals is zero.

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Exercise LXV, problem 5, p. 293

If $\phi(x) = a_{0} + a_{1} \cos x + a_{2}\cos 2x + \dots + a_{n}\cos nx$, and $k$ is a positive integer not greater than $n$, then _0^ (x)  dx = a_0,0pt minus 3pt_0^ kx (x)  dx = 12a_k. If $k > n$ then the value of the last integral is zero.

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Exercise LXV, problem 6, p. 293

Prove that if $a$ and $b$ are positive then [ int_0^2pi fracdxa^2cos^2 x + b^2sin^2 x = frac2piab. ]

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Exercise LXV, problem 7, p. 293

If $f(x) \leq \phi(x)$ when $a \leq x \leq b$, then $\ds\int_{a}^{b} f\, dx \leq \int_{a}^{b}\phi\, dx$.

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Exercise LXV, problem 8, p. 293

Prove that beginalignat*2 0 &< int_0^frac12pi sin^n+1x, dx &&< int_0^frac12pi sin^nx, dx, 0 &< int_0^frac14pi tan^n+1x, dx &&< int_0^frac14pi tan^nx, dx. endalignat*

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Exercise LXV, problem 9, p. 293

If $n > 1$ then [ .5 < int_0^frac12 fracdxsqrtp1 - x^2n < .524. ]

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Exercise LXVI, problem 1, p. 295

Prove that _a^b x f”(x)  dx = bf’(b) - f(b) - af’(a) - f(a).

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Exercise LXVI, problem 10, p. 295

Deduce that $u_{n}$ is equal to 2·4·6 …(n - 1)3·5·7 …n,0pt minus 3pt121·3·5 …(n - 1)2·4·6 …n, according as $n$ is odd or even.

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Exercise LXVI, problem 11, p. 295

**Second Mean Value Theorem.** If $f(x)$ is a function of $x$ which has a differential coefficient of constant sign for all values of $x$ from $x = a$ to $x = b$, then there is a number $\xi$ between $a$ and $b$ such that _a^b f(x)(x)  dx = f(a) _a^ (x)  dx + f(b) _^b (x)  dx.

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Exercise LXVI, problem 12a, p. 295

**’s form of the Second Mean Value Theorem.** If $f'(x)$ is of constant sign, and $f(b)$ and $f(a) - f(b)$ have the same sign, then _a^b f(x)(x)  dx = f(a) _a^X (x)  dx, where $X$ lies between $a$ and $b$.

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Exercise LXVI, problem 12b, p. 295

Prove similarly that if $f(a)$ and $f(b) - f(a)$ have the same sign, then _a^b f(x)(x)  dx = f(b) _X^b (x)  dx, where $X$ lies between $a$ and $b$.

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Exercise LXVI, problem 13, p. 295

Prove that |_X^X’ xx  dx| < 2X if $X' > X > 0$.

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Exercise LXVI, problem 14, p. 295

Establish the results of % [examples:lxv]Ex. lxv%. 1 by means of the rule for substitution.

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Exercise LXVI, problem 15, p. 295

Prove that _a^b F(x)  dx = _a^b F(a + b - x)  dx.

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Exercise LXVI, problem 16, p. 295

Prove that _0^12 ^m x^m x  dx = 2^-m _0^12 ^m x  dx.

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Exercise LXVI, problem 17, p. 295

Prove that _0^ x(x)  dx = 12_0^ (x)  dx.

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Exercise LXVI, problem 18, p. 295

Prove that _0^ xx1 + ^2 x  dx = 14^2.

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Exercise LXVI, problem 19, p. 295

Show by means of the transformation $x = a\cos^{2}\theta + b\sin^{2}\theta$ that _a^b (x - a)(b - x)  dx = 18(b - a)^2.

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Exercise LXVI, problem 2, p. 295

More generally, _a^b x^m f^(m+1)(x)  dx = F(b) - F(a), where multline* F(x) = x^m f^(m)(x) - mx^m-1 f^(m-1)x + m(m - 1)x^m-2 f^(m-2)x - … + (-1)^m m!  f(x). multline*

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Exercise LXVI, problem 20, p. 295

Show by means of the substitution $(a + b\cos x) (a - b\cos y) = a^{2} - b^{2}$ that _0^ (a + bx)^-n  dx = (a^2 - b^2)^-(n - 12) _0^ (a - by)^n-1  dy, when $n$ is a positive integer and $a > |b|$, and evaluate the integral when $n = 1$, $2$, $3$.

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Exercise LXVI, problem 21, p. 295

If $m$ and $n$ are positive integers then _a^b (x - a)^m (b - x)^n  dx = (b - a)^m+n+1 m!  n!(m + n + 1)!.

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Exercise LXVI, problem 3a, p. 295

Prove that _0^1 x  dx = 12- 1,0pt minus 3pt_0^1xx  dx = 14- 12.

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Exercise LXVI, problem 3b, p. 295

Prove that _0^1 x  dx = 12- 1,0pt minus 3pt_0^1xx  dx = 14- 12.

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Exercise LXVI, problem 4, p. 295

Prove that if $a$ and $b$ are positive then _0^12 xxx  dx(a^2^2x + b^2^2x)^2 = 4ab^2(a + b).

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Exercise LXVI, problem 5, p. 295

If f_1(x) = _0^xf(t)  dt,0pt minus 3ptf_2(x) = _0^xf_1(t)  dt, …,0pt minus 3ptf_k(x) = _0^x f_k-1(t)  dt, then f_k(x) = 1(k - 1)! _0^x f(t)(x - t)^k-1  dt.

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Exercise LXVI, problem 6, p. 295

Prove by integration by parts that if u_m, n = _0^1 x^m (1 - x)^n  dx, where $m$ and $n$ are positive integers, then $(m + n + 1) u_{m, n} = nu_{m, n-1}$, and deduce that u_m, n = m!  n!(m + n + 1)!.

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Exercise LXVI, problem 7, p. 295

Prove that if u_n = _0^14 ^nx  dx then $u_{n} + u_{n-2} = 1/(n - 1)$. Hence evaluate the integral for all positive integral values of $n$.

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Exercise LXVI, problem 8, p. 295

Deduce from the last example that $u_{n}$ lies between $1/\{2(n - 1)\}$ and $1/\{2(n + 1)\}$.

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Exercise LXVI, problem 9, p. 295

Prove that if u_n = _0^12 ^n x  dx then $u_{n} = \{(n - 1)/n\} u_{n-2}$. [Write $\sin^{n-1}x\sin x$ for $\sin^{n}x$ and integrate by parts.]

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Exercise Misc-VII, problem None, p. 300

The record holds no text for this problem.

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Exercise LX, problem 1, p. 275

0.375em plus 0.75em minus 0.25emProve that if $x = r\cos\theta$, $y = r\sin\theta$, so that $r = \sqrtp{x^{2} + y^{2}}$, $\theta = \arctan(y/x)$, then align* rx &= xx^2 + y^2, &ry &= yx^2 + y^2, &x &= -yx^2 + y^2, &y &= xx^2 + y^2, % xr &= , &yr &= , &x &= -r, &y &= r. align*

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Exercise LX, problem 2, p. 275

Account for the fact that $\dfrac{\dd r}{\dd x}\neq 1\bigg/\biggl(\dfrac{\dd x}{\dd r}\biggr)$ and $\dfrac{\dd \theta}{\dd x}\neq 1\bigg/\biggl(\dfrac{\dd x}{\dd \theta}\biggr)$. [When we were considering a function $y$ of one variable $x$ it followed from the definitions that $dy/dx$ and $dx/dy$ were reciprocals. This is no longer the [pg]276 case when we are dealing with functions of two variables. Let $P$ ([fig:46]Fig. 46) be the point $(x, y)$ or $(r, \theta)$. To find $\dd r/\dd x$ we must increase $x$, say by an increment $MM_{1} = \delta x$, while keeping $y$ constant. This brings $P$ to $P_{1}$. If along $OP_{1}$ we take $OP' = OP$, the increment of $r$ is $P'P_{1} = \delta r$, say; and $\dd r/\dd x = \lim(\delta r/\delta x)$. If on the other hand we want to calculate $\dd x/\dd r$, $x$ and $y$ %[Illustration: Fig. 46.] [2.25in]46p276 being now regarded as functions of $r$ and $\theta$, we must increase $r$ by $\Delta r$, say, keeping $\theta$ constant. This brings $P$ to $P_{2}$, where $PP_{2} = \Delta r$: the corresponding increment of $x$ is $MM_{1} = \Delta x$, say; and x/r = (x/r). Now $\Delta x = \delta x$: Of course the fact that $\Delta x = \delta x$ is due merely to the particular value of $\Delta r$ that we have chosen (viz. $PP_{2}$). Any other choice would give us values of $\Delta x$, $\Delta r$ proportional to those used here. but $\Delta r \neq \delta r$. Indeed it is easy to see from the figure that (r/x) = (P’P_1/PP_1) = , but (r/x) = (PP_2/PP_1) = , so that (r/r) = ^2. The fact is of course that *$\dd x/\dd r$ and $\dd r/\dd x$ are not formed upon the same hypothesis as to the variation of $P$.*]

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Exercise LX, problem 3, p. 275

Prove that if $z = f(ax + by)$ then $b(\dd z/\dd x) = a(\dd z/\dd y)$.

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Exercise LX, problem 4, p. 275

Find $\dd X/\dd x$, $\dd X/\dd y$, … when $X + Y = x$, $Y = xy$. Express $x$, $y$ as functions of $X$, $Y$ and find $\dd x/\dd X$, $\dd x/\dd Y$, ….

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Exercise LX, problem 5, p. 275

Find $\dd X/\dd x$, … when $X + Y + Z = x$, $Y + Z = xy$, $Z = xyz$; express $x$, $y$, $z$ in terms of $X$, $Y$, $Z$ and find $\dd x/\dd X$, ….

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Exercise LXI, problem 1, p. 277

Suppose $\phi(t) = (1 - t^{2})/(1 + t^{2})$, $\psi(t) = 2t/(1 + t^{2})$, so that the locus of $(x, y)$ is the circle $x^{2} + y^{2} = 1$. Then align* ’(t) &= -4t/(1 + t^2)^2,0pt minus 3pt’(t) = 2(1 - t^2)/(1 + t^2)^2, F’(t) &= -4t/(1 + t^2)^2f_x’ + 2(1 - t^2)/(1 + t^2)^2f_y’, align* where $x$ and $y$ are to be put equal to $(1 - t^{2})/(1 + t^{2})$ and $2t/(1 + t^{2})$ after carrying out the differentiations. [pg]278 0.375em plus 0.75em minus 0.25emWe can easily verify this formula in particular cases. Suppose, *e.g.*, that $f(x, y) = x^{2} + y^{2}$. Then $f_{x}' = 2x$, $f_{y}' = 2y$, and it is easily verified that $F'(t) = 2x\phi'(t) + 2y\psi'(t) = 0$, which is obviously correct, since $F(t) = 1$.

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Exercise LXI, problem 2a, p. 277

Verify the theorem in the same way when (*a*) $x = t^{m}$, $y = 1 - t^{m}$, $f(x, y) = x + y$; (*b*) $x = a\cos t$, $y = a\sin t$, $f(x, y) = x^{2} + y^{2}$.

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Exercise LXI, problem 2b, p. 277

Verify the theorem in the same way when (*a*) $x = t^{m}$, $y = 1 - t^{m}$, $f(x, y) = x + y$; (*b*) $x = a\cos t$, $y = a\sin t$, $f(x, y) = x^{2} + y^{2}$.

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Exercise LXI, problem 3, p. 277

One of the most important cases is that in which $t$ is $x$ itself. We then obtain D_xfx, (x) = D_xf(x, y) + D_yf(x, y)’(x). where $y$ is to be replaced by $\psi(x)$ after differentiation. It was this case which led to the introduction of the notation $\dd f/\dd x$, $\dd f/\dd y$. For it would seem natural to use the notation $df/dx$ for *either* of the functions $D_{x}f\{x, \psi(x)\}$ and $D_{x}f(x, y)$, in one of which $y$ is put equal to $\psi(x)$ before and in the other after differentiation. Suppose for example that $y = 1 - x$ and $f(x, y) = x + y$. Then $D_{x}f(x, 1 - x) = D_{x}1 = 0$, but $D_{x}f(x, y) = 1$. The distinction between the two functions is adequately shown by denoting the first by $df/dx$ and the second by $\dd f/\dd x$, in which case the theorem takes the form dfdx = fx + fy  dydx; though this notation is also open to objection, in that it is a little misleading to denote the functions $f\{x, \psi(x)\}$ and $f(x, y)$, whose forms as functions of $x$ are quite different from one another, by the same letter $f$ in $df/dx$ and $\dd f/\dd x$.

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Exercise LXI, problem 4, p. 277

If the result of eliminating $t$ between $x = \phi(t)$, $y = \psi(t)$ is $f(x, y) = 0$, then fx  dxdt + fy  dydt = 0.

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Exercise LXI, problem 5, p. 277

If $x$ and $y$ are functions of $t$, and $r$ and $\theta$ are the polar coordinates of $(x, y)$, then $r' = (xx' + yy')/r$, $\theta' = (xy' - yx')/r^{2}$, dashes denoting differentiations with respect to $t$.

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Exercise LXII, problem 1, p. 281

The area of an ellipse is given by $A = \pi ab$, where $a$, $b$ are the semiaxes. Prove that dAA = daa + dbb, and state the corresponding approximate equation connecting the increments of the axes and the area.

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Exercise LXII, problem 2, p. 281

Express $\Delta$, the area of a triangle $ABC$, as a function of (i) $a$, $B$, $C$, (ii) $A$, $b$, $c$, and (iii) $a$, $b$, $c$, and establish the formulae gather* d = 2daa + c  dBaB + b  dCaC,0pt minus 3ptd = A  dA + dbb + dcc, d= R(A  da + B  db + C  dc), gather* %[** TN: Sole instance of circumcircle, not hyphenated in the original] where $R$ is the radius of the circumcircle.

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Exercise LXII, problem 3, p. 281

The sides of a triangle vary in such a way that the area remains constant, so that $a$ may be regarded as a function of $b$ and $c$. Prove that ab = -BA,0pt minus 3ptac = -CA. [This follows from the equations da = ab  db + ac  dc,0pt minus 3ptA  da + B  db + C  dc = 0.

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Exercise LXII, problem 4, p. 281

If $a$, $b$, $c$ vary so that $R$ remains constant, then daA + dbB + dcC = 0, and so ab = -AB,0pt minus 3ptac = -AC. [Use the formulae $a = 2R\sin A$, …, and the facts that $R$ and $A + B + C$ are constant.]

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Exercise LXII, problem 5, p. 281

If $z$ is a function of $u$ and $v$, which are functions of $x$ and $y$, then zx = zu  ux + zv  vx,0pt minus 3ptzy = zu  uy + zv  vy. [We have dz = zu  du + zv  dv,0pt minus 3ptdu = ux  dx + uy  dy,0pt minus 3ptdv = vx  dx + vy  dy. Substitute for $du$ and $dv$ in the first equation and compare the result with the equation dz = zx  dx + zy  dy.]

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Exercise LXII, problem 6, p. 281

Let $z$ be a function of $x$ and $y$, and let $X$, $Y$, $Z$ be defined by the equations x = a_1 X + b_1 Y + c_1 Z,0pt minus 3pty = a_2 X + b_2 Y + c_2 Z,0pt minus 3ptz = a_3 X + b_3 Y + c_3 Z. Then $Z$ may be expressed as a function of $X$ and $Y$. Express $\dd Z/\dd X$, $\dd Z/\dd Y$ in terms of $\dd z/\dd x$, $\dd z/\dd y$. [Let these differential coefficients be denoted by $P$, $Q$ and $p$, $q$. Then $dz - p\, dx - q\, dy = 0$, or (c_1 p + c_2 q - c_3)  dZ + (a_1 p + a_2 q - a_3)  dX + (b_1 p + b_2 q - b_3)  dY = 0. [pg]283 Comparing this equation with $dZ - P\, dX - Q\, dY = 0$ we see that P = -a_1p + a_2q - a_3c_1p + c_2q - c_3,0pt minus 3ptQ = -b_1p + b_2q - b_3c_1p + c_2q - c_3.]

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Exercise LXII, problem 7, p. 281

If (a_1 x + b_1 y + c_1 z)p + (a_2 x + b_2 y + c_2 z)q = a_3 x + b_3 y + c_3 z, then (a_1 X + b_1 Y + c_1 Z) P + (a_2 X + b_2 Y + c_2 Z) Q = a_3 X + b_3 Y + c_3 Z. % [0]% (*Math. Trip.* 1899.)% [1]%

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Exercise LXII, problem 8, p. 281

**of implicit functions.** Suppose that $f(x, y)$ and its derivative $f_{y}'(x, y)$ are continuous in the neighbourhood of the point $(a, b)$, and that f(a, b) = 0,0pt minus 3ptf_b’(a, b) 0. Then we can find a neighbourhood of $(a, b)$ throughout which $f_{y}'(x, y)$ has always the same sign. Let us suppose, for example, that $f_{y}'(x, y)$ is positive near $(a, b)$. Then $f(x, y)$ is, for any value of $x$ sufficiently near to $a$, and for values of $y$ sufficiently near to $b$, an increasing function of $y$ in the stricter sense of [§]95. It follows, by the theorem of [§]108, that there is a unique continuous function $y$ which is equal to $b$ when $x = a$ and which satisfies the equation $f(x, y) = 0$ for all values of $x$ sufficiently near to $a$. Let us now suppose that $f(x, y)$ possesses a derivative $f_{x}'(x, y)$ which is also continuous near $(a, b)$. If $f(x, y) = 0$, $x = a + h$, $y = b + k$, we have 0 = f(x, y) - f(a, b) = (f_a’ + ) h + (f_b’ + ) k, where $\DPtypo{}{\epsilon}$ and $\eta$ tend to zero with $h$ and $k$. Thus kh = -f_a’ + f_b’ + -f_a’f_b’, or dydx = -f_a’f_b’.

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Exercise LXII, problem 9, p. 281

The equation of the tangent to the curve $f(x, y) = 0$, at the point $x_{0}$, $y_{0}$, is (x - x_0) f_x_0’(x_0, y_0) + (y - y_0) f_y_0’(x_0, y_0) = 0.

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Exercise LV, problem 1, p. 264

Suppose that $f(x)$ is a polynomial of degree $r$. Then $f^{(n)}(x)$ is identically zero when $n > r$, and the theorem leads to the algebraical identity f(a + h) = f(a) + hf’(a) + h^22! f”(a) + … + h^rr! f^(r)(a).

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Exercise LV, problem 10, p. 264

Show that the error in taking the root to be $\xi - (f/f') - \frac{1}{2}(f^{2}f''/f'^{3})$, where $\xi$ is the argument of every function, is in general of the third order.

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Exercise LV, problem 11, p. 264

The equation $\sin x = \alpha x$, where $\alpha$ is small, has a root nearly equal to $\pi$. Show that $(1 - \alpha)\pi$ is a better approximation, and $(1 - \alpha + \alpha^{2})\pi$ a better still. [The method of Exs. 7--10 does not depend on $f(x) = 0$ being an algebraical equation, so long as $f'$ and $f''$ are continuous.]

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Exercise LV, problem 12, p. 264

Show that the limit when $h \to 0$ of the number $\theta_{n}$ which occurs in the general Mean Value Theorem is $1/(n + 1)$, provided that $f^{(n+1)}(x)$ is continuous. [For $f(x + h)$ is equal to each of f(x) + …+ h^nn! f^(n)(x + _nh),0pt minus 3ptf(x) + …+ h^nn! f^(n)(x) + h^n+1(n + 1)! f^(n+1)(x + _n+1h), where $\theta_{n+1}$ as well as $\theta_{n}$ lies between $0$ and $1$. Hence f^(n)(x + _nh) = f^(n)(x) + hf^(n+1)(x + _n+1h)n + 1 But if we apply the original Mean Value Theorem to the function $f^{(n)}(x)$, taking $\theta_{n}h$ in place of $h$, we find f^(n)(x + _nh) = f^(n)(x) + _nhf^(n+1)(x + _nh), [pg]266 where $\theta$ also lies between $0$ and $1$. Hence _n f^(n+1)(x + _n h) = f^(n+1)(x + _n+1 h)n + 1, from which the result follows, since $f^{(n+1)}(x + \theta\theta_{n} h)$ and $f^{(n+1)}(x + \theta_{n+1} h)$ tend to the same limit $f^{(n+1)}(x)$ as $h \to 0$.]

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Exercise LV, problem 13, p. 264

Prove that $\{f(x + 2h) - 2f(x + h) + f(x)\}/h^{2} \to f''(x)$ as $h \to 0$, provided that $f''(x)$ is continuous. [Use equation (2) of [§]147.]

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Exercise LV, problem 14, p. 264

Show that, if the $f^{(n)}(x)$ is continuous for $x = 0$, then f(x) = a_0 + a_1x + a_2x^2 + …+ (a_n + _x) x^n, where $a_{r} = f^{(r)}(0)/r!$ and $\epsilon_{x} \to 0$ as $x \to 0$. It is in fact sufficient to suppose that *$f^{(n)}(0)$ exists*. See R. H. Fowler, “The elementary differential geometry of plane curves” (*Cambridge Tracts in Mathematics*, No. 20, p. 104).266

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Exercise LV, problem 15, p. 264

Show that if a_0 + a_1x + a_2x^2 + …+ (a_n + _x) x^n = b_0 + b_1x + b_2x^2 + …+ (b_n + _x) x^n, where $\epsilon_{x}$ and $\eta_{x}$ tend to zero as $x \to 0$, then $a_{0} = b_{0}$, $a_{1} = b_{1}$, …, $a_{n} = b_{n}$. [Making $x \to 0$ we see that $a_{0} = b_{0}$. Now divide by $x$ and afterwards make $x \to 0$. We thus obtain $a_{1} = b_{1}$; and this process may be repeated as often as is necessary. It follows that if $f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \dots + (a_{n} + \epsilon_{x}) x^{n}$, and the first $n$ derivatives of $f(x)$ are continuous, then $a_{r} = f^{(r)}(0)/r!$.]

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Exercise LV, problem 2, p. 264

By applying the theorem to $f(x) = 1/x$, and supposing $x$ and $x + h$ positive, obtain the result 1x + h = 1x - hx^2 + h^2x^3 - … + (-1)^n-1 h^n-1x^n + (-1)^n h^n(x + _n h)^n+1. [Since 1x + h = 1x - hx^2 + h^2x^3 - … + (-1)^n-1 h^n-1x^n + (-1)^n h^nx^n(x + h),0pt minus 3pt%[** TN: Quick spacing hack] we can verify the result by showing that $x^{n}(x + h)$ can be put in the form $(x + \theta_{n}h)^{n+1}$, or that $x^{n+1} < x^{n}(x + h) < (x + h)^{n+1}$, as is evidently the case.]

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Exercise LV, problem 3, p. 264

Obtain the formula multline* (x + h) = x + hx - h^22!x - h^33!x + … + (-1)^n-1h^2n-1(2n - 1)!x + (-1)^n h^2n2n!(x + _2n h), multline* the corresponding formula for $\cos(x + h)$, and similar formulae involving powers of $h$ extending up to $h^{2n+1}$.

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Exercise LV, problem 4, p. 264

Show that if $m$ is a positive integer, and $n$ a positive integer not greater than $m$, then (x + h)^m = x^m + m1x^m-1 h + … + mn - 1x^m-n+1 h^n-1 + mn(x + _n h)^m-n h^n. Show also that, if the interval $\DPmod{(x, x + h)}{[x, x + h]}$ does not include $x = 0$, the formula holds for all real values of $m$ and all positive integral values of $n$; and that, even if $x < 0 < x + h$ or $x + h < 0 < x$, the formula still holds if $m - n$ is positive.

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Exercise LV, problem 5, p. 264

The formula $f(x + h) = f(x) + hf'(x + \theta_{1}h)$ is not true if $f(x) = 1/x$ and $x < 0 < x + h$. [For $f(x + h) - f(x) > 0$ and $hf'(x + \theta_{1} h) = -h/(x + \theta_{1} h)^{2} < 0$; it is evident that the conditions for the truth of the Mean Value Theorem are not satisfied.]

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Exercise LV, problem 6, p. 264

If $x = -a$, $h = 2a$, $f(x) = x^{1/3}$, then the equation f(x + h) = f(x) + hf’(x + _1 h) is satisfied by $\theta_{1} = \frac{1}{2} ± \frac{1}{18}\sqrt{3}$. [This example shows that the result of the theorem may hold even if the conditions under which it was proved are not satisfied.]

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Exercise LV, problem 7, p. 264

**’s method of approximation to the roots of equations.** Let $\xi$ be an approximation to a root of an algebraical equation $f(x) = 0$, the actual root being $\xi + h$. Then 0 = f(+ h) = f() + hf’() + 12 h^2f”(+ _2h), so that h = -f()f’() - 12 h^2 f”(+ _2h)f’(). It follows that in general a better approximation than $x = \xi$ is x = - f()f’(). If the root is a simple root, so that $f'(\xi + h) \neq 0$, we can, when $h$ is small enough, find a positive constant $K$ such that $|f'(x)| > K$ for all the values of $x$ which we are considering, and then, if $h$ is regarded as of the first order of smallness, $f(\xi)$ is of the first order of smallness, and the error in taking $\xi - \{f(\xi)/f'(\xi)\}$ as the root is of the second order.

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Exercise LV, problem 8, p. 264

Apply this process to the equation $x^{2} = 2$, taking $\xi = 3/2$ as the first approximation. [We find $h = -1/12$, $\xi + h = 17/12 = 1.417\dots$, which is quite a good approximation, in spite of the roughness of the first. If now we repeat the process, taking $\xi = 17/12$, we obtain $\xi + h = 577/408 = 1.414\MS215\dots$, which is correct to $5$ places of decimals.

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Exercise LV, problem 9, p. 264

By considering in this way the equation $x^{2} - 1 - y = 0$, where $y$ is small, show that $\sqrtp{1 + y} = 1 + \frac{1}{2} y - \{\frac{1}{4}y^{2}/(2 + y)\}$ approximately, the error being of the fourth order.

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Exercise LVI, problem 1, p. 267

Let $f(x) = \sin x$. Then all the derivatives of $f(x)$ are continuous for all values of $x$. Also $|f^{n}(x)| \leq 1$ for all values of $x$ and $n$. Hence in this case $|R_{n}| \leq h^{n}/n!$, which tends to zero as $n \to \infty$ (% [examples:xxvii]Ex. xxvii%. 12) whatever value $h$ may have. It follows that (x + h) = x + hx - h^22!x - h^33!x + h^44!x + …, for all values of $x$ and $h$. In particular h = h - h^33! + h^55! - …, for all values of $h$. Similarly we can prove that (x + h) = x - hx - h^22!x + h^33! x + …,0pt minus 3pth = 1 - h^22! + h^44! - ….

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Exercise LVI, problem 2, p. 267

**Binomial Series.** Let $f(x) = (1 + x)^{m}$, where $m$ is any rational number, positive or negative. Then $f^{(n)}(x) = m(m - 1) \dots (m - n + 1) (1 + x)^{m-n}$ and Maclaurin’s Series takes the form (1 + x)^m = 1 + m1x + m2x^2 + …. When $m$ is a positive integer the series terminates, and we obtain the ordinary formula for the Binomial Theorem with a positive integral exponent. In the general case R_n = x^nn! f^(n)(_nx) = mnx^n(1 + _nx)^m-n, and in order to show that Maclaurin’s Series really represents $(1 + x)^{m}$ for any range of values of $x$ when $m$ is not a positive integer, we must show that $R_{n} \to 0$ for every value of $x$ in that range. This is so in fact if $-1 < x < 1$, and may be proved, when $0\leq x < 1$, by means of the expression given above for $R_{n}$, since $(1 + \theta_{n}x)^{m-n} < 1$ if $n > m$, and $\dbinom{m}{n} x^{n} \to 0$ as $n \to \infty$ (% [examples:xxvii]Ex. xxvii%. 13). But a difficulty arises if $-1 < x < 0$, since $1 + \theta_{n}x < 1$ and $(1 + \theta_{n}x)^{m-n} > 1$ if $n > m$; knowing only that $0 < \theta_{n} < 1$, we cannot be assured that $1 + \theta_{n}x$ is not quite small and $(1 + \theta _{n}x)^{m-n}$ quite large. In fact, in order to prove the Binomial Theorem by means of Taylor’s Theorem, we need some different form for $R_{n}$, such as will be given later ([§]162).

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Exercise LVII, problem 1, p. 268

Verify the result when $\phi(x) = (x - a)^{m}$, $m$ being a positive integer, and $\xi = a$.

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Exercise LVII, problem 2, p. 268

Test the function $(x - a)^{m} (x - b)^{n}$, where $m$ and $n$ are positive integers, for maxima and minima at the points $x = a$, $x = b$. Draw graphs of the different possible forms of the curve $y = (x - a)^{m} (x - b)^{n}$.

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Exercise LVII, problem 3, p. 268

Test the functions $\sin x - x$, $\sin x - x + \dfrac{x^{3}}{6}$, $\sin x - x + \dfrac{x^{3}}{6} - \dfrac{x^{5}}{120}$, …, $\cos x - 1$, $\cos x - 1 + \dfrac{x^{2}}{2}$, $\cos x - 1 + \dfrac{x^{2}}{2} - \dfrac{x^{4}}{24}$, … for maxima or minima at $x = 0$.

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Exercise LVIII, problem 1, p. 270

Find the limit of x - (n + 1)x^n+1 + nx^n+2/(1 - x)^2, as $x \to 1$. [Here the functions and their first derivatives vanish for $x = 1$, and $f''(1) = n(n + 1)$, $\phi''(1) = 2$.]

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Exercise LVIII, problem 2a, p. 270

Find the limits as $x \to 0$ of (x - x)/(x - x),0pt minus 3pt(nx - nx)/(nx - nx).

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Exercise LVIII, problem 2b, p. 270

Find the limits as $x \to 0$ of (x - x)/(x - x),0pt minus 3pt(nx - nx)/(nx - nx).

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Exercise LVIII, problem 3, p. 270

Find the limit of $x\{\sqrtp{x^{2} + a^{2}} - x\}$ as $x \to \infty$. [Put $x = 1/y$.]

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Exercise LVIII, problem 4a, p. 270

Prove that _x n (x - n)x= (-1)^n,0pt minus 3pt_x n 1x - n x- (-1)^n(x - n) = (-1)^n6, $n$ being any integer; and evaluate the corresponding limits involving $\cot x\pi$.

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Exercise LVIII, problem 4b, p. 270

Prove that _x n (x - n)x= (-1)^n,0pt minus 3pt_x n 1x - n x- (-1)^n(x - n) = (-1)^n6, $n$ being any integer; and evaluate the corresponding limits involving $\cot x\pi$.

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Exercise LVIII, problem 4c, p. 270

Prove that _x n (x - n)x= (-1)^n,0pt minus 3pt_x n 1x - n x- (-1)^n(x - n) = (-1)^n6, $n$ being any integer; and evaluate the corresponding limits involving $\cot x\pi$.

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Exercise LVIII, problem 5a, p. 270

Find the limits as $x \to 0$ of 1x^3(x - 1x - x6),0pt minus 3pt1x^3(x - 1x + x3).

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Exercise LVIII, problem 5b, p. 270

Find the limits as $x \to 0$ of 1x^3(x - 1x - x6),0pt minus 3pt1x^3(x - 1x + x3).

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Exercise LVIII, problem 6a, p. 270

$(\sin x\arcsin x - x^{2})/x^{6} \to \frac{1}{18}$, $(\tan x\arctan x - x^{2})/x^{6} \to \frac{2}{9}$, as $x \to 0$.

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Exercise LVIII, problem 6b, p. 270

$(\sin x\arcsin x - x^{2})/x^{6} \to \frac{1}{18}$, $(\tan x\arctan x - x^{2})/x^{6} \to \frac{2}{9}$, as $x \to 0$.

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Exercise LIX, problem 1, p. 272

Let $\phi(x) = ax + b$, so that $y = \phi(x)$ is a straight line. The conditions for contact at the point for which $x = \xi$ are $f(\xi) = a\xi + b$, $f'(\xi) = a$. If we determine $a$ and $b$ so as to satisfy these equations we find $a = f'(\xi)$, $b = f(\xi) - \xi f'(\xi)$, and the equation of the tangent to $y = f(x)$ at the point $x = \xi$ is y = xf’() + f() - f’(), or $y - f(\xi) = (x - \xi)f'(\xi)$. Cf. % [examples:xxxix]Ex. xxxix%. 5.

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Exercise LIX, problem 10, p. 272

Verify that the curvature of a circle is constant and equal to the reciprocal of the radius; and show that the circle is the only curve whose curvature is constant.

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Exercise LIX, problem 11, p. 272

0.375em plus 0.75em minus 0.25emFind the centre and radius of curvature at any point of the conics $y^{2} = 4ax$, $(x/a)^{2} + (y/b)^{2} = 1$.

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Exercise LIX, problem 12, p. 272

In an ellipse the radius of curvature at $P$ is $CD^{3}/ab$, where $CD$ is the semi-diameter conjugate to $CP$.

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Exercise LIX, problem 13, p. 272

Show that in general a conic can be drawn to have contact of the fourth order with the curve $y = f(x)$ at a given point $P$. [Take the general equation of a conic, viz. ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, and differentiate four times with respect to $x$. Using suffixes to denote differentiation we obtain align* ax + hy + g + (hx + by + f) y_1 &= 0, a + 2hy_1 + by_1^2 + (hx + by + f) y_2 &= 0, 3(h + by_1) y_2 + (hx + by + f) y_3 &= 0, 4(h + by_1) y_3 +3by_2^2 + (hx + by + f) y_4 &= 0. align* If the conic has contact of the fourth order, then these five equations must be satisfied by writing $\xi$, $\eta$, $\eta_{1}$, $\eta_{2}$, $\eta_{3}$, $\eta_{4}$, for $x$, $y$, $y_{1}$, $y_{2}$, $y_{3}$, $y_{4}$. We have thus just enough equations to determine the ratios $a : b : c : f : g : h$.]

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Exercise LIX, problem 14, p. 272

An infinity of conics can be drawn having contact of the third order with the curve at $P$. Show that their centres all lie on a straight line. [Take the tangent and normal as axes. Then the equation of the conic is of the form $2y = ax^{2} + 2hxy + by^{2}$, and when $x$ is small one value of $y$ may be expressed (Ch.V, [misc:V]Misc. Ex. 22) in the form y = 12ax^2 + (12ah + _x) x^3, where $\epsilon_{x} \to 0$ with $x$. But this expression must be the same as y = 12f”(0) x^2 + 16f”’(0) + ’_x x^3, where $\epsilon'_{x} \to 0$ with $x$, and so $a = f''(0)$, $h = f'''(0)/3f''(0)$, in virtue of the result of % [examples:lv]Ex. lv%. 15. But the centre lies on the line $ax + hy = 0$.]

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Exercise LIX, problem 15, p. 272

Determine a parabola which has contact of the third order with the ellipse $(x/a)^{2} + (y/b)^{2} = 1$ at the extremity of the major axis.

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Exercise LIX, problem 16, p. 272

The locus of the centres of conics which have contact of the third order with the ellipse $(x/a)^{2} + (y/b)^{2} = 1$ at the point $(a\cos\alpha, b\sin\alpha)$ is the diameter $x/(a\cos\alpha) = y/(b\sin\alpha)$. [For the ellipse itself is one such conic.]

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Exercise LIX, problem 2, p. 272

The fact that the line is to have simple contact with the curve completely determines the line. In order that the tangent should have *contact of the second order* with the curve we must have $f''(\xi) = \phi''(\xi)$, *i.e.* $f''(\xi) = 0$. A point at which the tangent to a curve has contact of the second order is called a **of inflexion**. %[** TN: Differs from the modern definition]

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Exercise LIX, problem 3, p. 272

Find the points of inflexion on the graphs of the functions $3x^{4} - 6x^{3} + 1$, $2x/(1 + x^{2})$, $\sin x$, $a\cos^{2}x + b\sin^{2}x$, $\tan x$, $\arctan x$.

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Exercise LIX, problem 4, p. 272

Show that the conic $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$ cannot have a point of inflexion. [Here $ax + hy + g + (hx + by + f)y_{1} = 0$ and a + 2hy_1 + by_1^2 + (hx + by + f)y_2 = 0, suffixes denoting differentiations. Thus at a point of inflexion a + 2hy_1 + by_1^2 = 0, or a(hx + by + f)^2 - 2h(ax + hy + g)(hx + by + f) + b(ax + hy + g)^2 = 0, or (ab - h^2)ax^2 + 2hxy + by^2 + 2gx + 2fy + af^2 - 2fgh + bg^2 = 0. But this is inconsistent with the equation of the conic unless af^2 - 2fgh + bg^2 = c(ab - h^2) or $abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0$; and this is the condition that the conic should degenerate into two straight lines.]

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Exercise LIX, problem 5, p. 272

The curve $y = (ax^{2} + 2bx + c)/(\alpha x^{2} + 2\beta x + \gamma)$ has one or three points of inflexion according as the roots of $\alpha x^{2} + 2\beta x + \gamma = 0$ are real or complex. [The equation of the curve can, by a change of origin (cf. % [examples:xlvi]Ex. xlvi%. 15), be reduced to the form = /(A^2 + 2B+ C) = /A(- p)(- q), where $p$, $q$ are real or conjugate. The condition for a point of inflexion will be found to be $\xi^{3} - 3pq\xi + pq(p + q) = 0$, which has one or three real roots according as $\DPtypo{\{pq(p - q)\}}{\{pq(p - q)\}^{2}}$ is positive or negative, *i.e.* according as $p$ and $q$ are real or conjugate.] [pg]273

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Exercise LIX, problem 6, p. 272

Discuss in particular the curves $y = (1 - x)/(1 + x^{2})$, $y = (1 - x^{2})/(1 + x^{2})$, $y = (1 + x^{2})/(1 - x^{2})$.

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Exercise LIX, problem 7, p. 272

Show that when the curve of Ex. 5 has three points of inflexion, they lie on a straight line. [The equation $\xi^{3} - 3pq\xi + pq(p + q) = 0$ can be put in the form $(\xi - p)(\xi - q)(\xi + p + q) + (p - q)^{2}\xi = 0$, so that the points of inflexion lie on the line $\xi + A(p - q)^{2}\eta + p + q = 0$ or $A\xi - 4(AC - B^{2})\eta = 2B$.]

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Exercise LIX, problem 8, p. 272

Show that the curves $y = x\sin x$, $y = (\sin x)/x$ have each infinitely many points of inflexion.

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Exercise LIX, problem 9, p. 272

**of a circle with a curve. Curvature. A much fuller discussion of the theory of curvature will be found in Mr Fowler’s %[** TN: Reference on page 272 of orig. points to page 266.] tract referred to on p.272266.** The general equation of a circle, viz. (x - a)^2 + (y - b)^2 = r^2, (1) contains three arbitrary constants. Let us attempt to determine them so that the circle has contact of as high an order as possible with the curve $y = f(x)$ at the point $(\xi, \eta)$, where $\eta = f(\xi)$. We write $\eta_{1}$, $\eta_{2}$ for $f'(\xi)$, $f''(\xi)$. Differentiating the equation of the circle twice we obtain align (x - a) + (y - b)y_1 &= 0, (2) 1 + y_1^2 + (y - b)y_2 &= 0. (3) align If the circle touches the curve then the equations (1) and (2) are satisfied when $x = \xi$, $y = \eta$, $y_{1} = \eta_{1}$. This gives $(\xi - a)/\eta_{1} = -(\eta - b) = r/\sqrtp{1 + \eta_{1}^{2}}$. If the contact is of the second order then the equation (3) must also be satisfied when $y_{2} = \eta_{2}$. Thus $b = \eta + \{(1 + \eta_{1}^{2})/\eta_{2}\}$; and hence we find a = - _1(1 + _1^2)_2,0pt minus 3ptb = + 1 + _1^2_2,0pt minus 3ptr = (1 + _1^2)^3/2_2. The circle which has contact of the second order with the curve at the point $(\xi, \eta)$ is called the **of curvature**, and its radius the **of curvature**. The **of curvature** (or simply the *curvature*) is the reciprocal of the radius: thus the measure of curvature is $f''(\xi)/\{1 + [f'(\xi)]^{2}\}^{3/2}$, or d^2d^2 / 1 + (dd)^2^3/2.

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