{ "id": "4909912d4e61b603dbec3fd68a4faa342dccef7e", "text": "XXII. Consideration of various Points of Analysis. By John F. W. Herschel, Esq. F. R. S.\n\nRead May 19, 1814.\n\nOn the Calculus of Generating Functions.\n\nIn whatever point of view we consider the theory of generating functions, whether as the fertile source of new discoveries, or as a medium for exhibiting, in the most comprehensive and uniform point of view, results already known; we shall find fresh cause to admire the profound and original genius of its author. To the latter of these objects it is, however, more peculiarly adapted, and perhaps, in the present state of analytics, this may even be considered as the more precious advantage of the two. Such has been the indefatigable activity of those illustrious men, who have devoted themselves to the pursuit of mathematical science, that analysis must be considered as already adequate to every purpose to which we can reasonably hope to see it applied. The attention of the scientific observer must now be directed to those elevated stations, from which distinct and extended views of its arrangement as a whole can be obtained. The calculus of generating functions affords such a station, and commands a wider and more magnificent prospect than any which has yet been opened to the view of the speculative philosopher. It becomes interesting then to extend its application as far as possible in this line, and to introduce it on every occasion where there seems any\nprobability of its coming successfully into play. Such have been in part the considerations which determined me to adopt it as a vehicle in laying before the Society some results of a singular and interesting nature, derived indeed originally from other principles, but which, like all the rest, flow with the utmost facility from the first elements of this calculus.\n\nIn the following pages I have uniformly made use of the functional or characteristic notation; together with the method of separating (where it could conveniently be done) the symbols of operation from those of quantity. This method I have, perhaps, extended and carried somewhat farther than has hitherto been customary; but, I trust, without losing sight of its grand and ultimate object, the union of extreme generality with conciseness of expression. To avoid the necessity of continual explanation, I shall here set down the leading points of the system.\n\nI. The signs : \\( \\times () \\) are used to separate the symbol of operation from that of the quantity operated upon, thus:\n\n\\[\nf(x), \\phi : \\log x, \\left\\{ \\frac{d}{dx} - 1 \\right\\}^n \\times \\phi(x).\n\\]\n\nII. 1. The combination of two operations is represented by placing their symbols together in their proper order. Thus, \\( \\phi(\\psi(x)) \\) is simply written \\( \\phi\\psi(x) \\). For example, if \\( \\phi(x) = 1 + x \\), and \\( \\psi(x) = x^2 \\), then \\( \\phi\\psi(x) = 1 + x^2 \\), and \\( \\psi\\phi(x) = (1 + x)^2 \\).\n\n2. When several operations combined are considered as one, their characteristics are inclosed in parentheses (). Thus\n\n\\[\nf \\log \\phi(x) = (f \\log \\phi) : x.\n\\]\n\n3. The repetition of the same operation \\( f \\) being denoted (by the first rule of this article) by \\( ff, fff, \\&c. \\) may be more con-\ncisely represented thus, $f^2, f^3, \\ldots, f^n$; and this furnishes us with a general and very simple notation for the reverse operation of that denoted by $f$. For since $f^m f^n(x) = f^{m+n}(x)$, if we make $m = -1$, $n = +1$, we find $f^{-1} f(x) = f^0(x) = x$ with the operation $f$ no times performed, $= x$. Thus $f^{-1}$ is the characteristic of that operation which must be performed on $f(x)$ to reduce it to $x$: that is, of the reverse operation. This is surely a simpler and more expressive method than that of inverting the characteristic,* accentuating it on the left side,† or below;‡ or other similar contrivances. For instance,\n\n$$\\log^{-1} x = e^x = 1 + \\frac{x}{1} + \\frac{x^2}{1.2} + \\&c.$$ \n$$\\tan^{-1} x = \\frac{x}{1} - \\frac{x^3}{3} + \\frac{x^5}{5} - \\&c.$$ \n\n4. If a combination of operations, as $\\phi \\psi \\&c.$ be $n$ times repeated in their order, thus; $\\phi \\psi \\phi \\psi \\ldots (x)$, it will, by the second and third rules of this article be denoted by $(\\phi \\psi)^n : x$. It must be observed, however, that $(\\phi \\psi)^n$ is not the same as $\\phi^n \\psi^n$, except in some particular cases.\n\nIII. 1. If any number of functions of a symbol $x$ be added, subtracted, or otherwise combined, the resulting function is expressed by the same combination of their characteristics, observing the following conditions.\n\n2. The actual multiplication of two functions $\\phi(x)$ and $\\psi(x)$ is expressed by inserting a full point between their characteristics, thus, $\\phi \\cdot \\psi(x) = \\phi(x) \\cdot \\psi(x)$.\n\n3. The actual elevation of a function to any power is thus\n\n---\n\n* Laplace. Journal de l'Ecole Polytechnique. No. 15.\n† Monge. Savans Etrangers. 1773.\n‡ Knight. Philosophical Transactions, 1811. Part I.\nexpressed, \\( \\{ f(x) \\}^n = \\{ f \\}^n : x \\), to distinguish it from \\( f^n(x) \\), whose signification has already been explained.\n\n4. If \\( F(x) \\) be developable in any series of the form\n\n\\[\nax^\\alpha + bx^\\beta + cx^\\gamma + \\&c.\n\\]\n\nthe following abbreviations are used\n\n\\[\n(F : \\phi) : x = a \\cdot \\phi^\\alpha(x) + b \\cdot \\phi^\\beta(x) + c \\cdot \\phi^\\gamma(x) + \\&c.\n\\]\n\nand, \\((F \\{ \\phi \\}) : x = a \\cdot \\{ \\phi(x) \\}^\\alpha + b \\cdot \\{ \\phi(x) \\}^\\beta + \\&c.\\)\n\nThus (for example)\n\n\\[\n\\frac{1}{1 - \\frac{d}{dx}} \\times y = y + \\frac{dy}{dx} + \\frac{d^2y}{dx^2} + \\&c.\n\\]\n\n\\[\n\\frac{1}{1 - \\{ \\log_{-1} \\}} \\times x = x + e^x + e^{2x} + \\&c.\n\\]\n\nIV. 1. D is used as the sign of derivation. It is, properly speaking, the sign of an operation performed, not on quantity, but on the characteristic which it immediately precedes; by which the operation denoted by that characteristic is altered. For instance: \\( D \\sin. = \\cos.; D \\cos. = -\\sin. \\). But it must be observed that \\( D \\log_{-1} = \\log_{-1} \\).\n\n2. The sign D affects only the characteristic next following it, thus, \\( D \\phi f(x) = (D \\phi) : f(x) \\). If it be intended to affect a combination of operations, the rule II, 2 must be observed. Thus, \\( D (\\phi f) : x, D^n (\\psi \\log_{-1}) : \\log_{-1} x \\).\n\nV. Every functional characteristic is affected by all the characteristics preceding it, in the same manner as if it were a symbol of quantity.\n\nVI. Every characteristic of operation performed on quantity affects all which follows it, as if it were one symbol. Thus if \\( f(x) = ax^\\alpha + bx^\\beta + \\&c.; \\) we shall have\n\\[ fD\\phi(x) = a \\cdot \\{ D\\phi(x) \\}^\\alpha + b \\cdot \\{ D\\phi(x) \\}^\\beta + \\&c. \\]\n\nThis rule does not extend to the signs \\( D, d, \\Delta, \\delta, J, \\Sigma \\), according to the remark in IV. 1.\n\nThese rules will suffice to explain whatever may appear obscure or capricious in the following sheets. We shall now proceed to their practical application.\n\nIf \\( \\phi(t) \\) be a function of \\( t \\), developable in a series\n\n\\[ A_{-\\infty} t^{-\\infty} + \\ldots + A_0 + A_1 t + \\ldots A_x t^x + \\ldots A_\\infty t^\\infty \\]\n\n\\( \\phi(t) \\) is said to be the generating function of \\( A_x \\), and it may be said to be taken with respect to \\( t \\). To this we shall appropriate a peculiar symbol \\( G_t \\), as follows:\n\n\\[ \\phi(t) = G_t \\{ A_x \\} \\]\n\nWhen only one symbol \\( t \\) is used, the index below the \\( G \\) may be understood, and our equation will be\n\n\\[ \\phi(t) = G \\{ A_x \\}. \\]\n\nIf \\( \\phi(t, t') \\) be a function of \\( t, t' \\), developable in a double series\n\n\\[ A_{-\\infty,-\\infty} t^{-\\infty} t'^{-\\infty} + \\ldots A_{-\\infty,y} t^{-\\infty} t'^y + \\ldots A_{-\\infty,\\infty} t^{-\\infty} t'^\\infty \\]\n\\[ + A_{x,-\\infty} t^x t'^{-\\infty} + \\ldots A_{x,y} t^x t'^y + \\ldots A_{x,\\infty} t^x t'^\\infty \\]\n\\[ + A_{\\infty,-\\infty} t^\\infty t'^{-\\infty} + \\ldots A_{\\infty,y} t^\\infty t'^y + \\ldots A_{\\infty,\\infty} t^\\infty t'^\\infty. \\]\n\n\\( \\phi(t, t') \\) is said to be the generating function of \\( A_{x,y} \\) with respect to \\( t, t' \\), and may be thus expressed;\n\n\\[ \\phi(t, t') = G_{t,t'} \\{ A_{x,y} \\} \\]\n\nand so on, if there be any number of symbols \\( t, t', t'', \\&c \\) To\ndenote the sum of all the terms of a series, we shall use the sign $S$, thus,\n\n$$\\phi(t) = S \\left\\{ A_x t^x \\right\\}, \\text{ and in like manner } \\phi(t, t') = S^2_{x,y} \\left\\{ A_{x,y} t^x \\cdot t'^y \\right\\}. $$\n\nFor $t$ let $ht$ be written, and we obtain\n\n$$\\phi(ht) = S \\left\\{ A_x (ht)^x \\right\\} = S \\left\\{ A_x h^x \\cdot t^x \\right\\} = G \\left\\{ A_x h^x \\right\\}. $$\n\nThus, if the generating function of $A_x$ be $\\phi(t)$, that of $A_x h^x$ will be $\\phi(ht)$.\n\nLet this equation be multiplied by $t^{-r}$, and we get\n\n$$t^{-r} \\phi(ht) = S \\left\\{ A_x h^x \\cdot t^{x-r} \\right\\} = S \\left\\{ A_{x+r} h^{x+r} \\cdot t^x \\right\\} = G \\left\\{ A_{x+r} h^{x+r} \\right\\}. $$\n\nIf then the generating function of $A_x$ be $\\phi(t)$, that of $A_{x+r} h^{x+r}$ will be $t^{-r} \\cdot \\phi(ht)$.\n\nAgain, it is easy to see that\n\n$$a \\cdot G \\left\\{ A_x \\right\\} + b \\cdot G \\left\\{ B_x \\right\\} + \\&c. = G \\left\\{ aA_x + bB_x + \\&c. \\right\\} $$\n\nand thus we have\n\n$$(at^{-\\alpha} + bt^{-\\beta} + ct^{-\\gamma} + \\&c.) \\cdot \\phi(ht) = G \\left\\{ aA_{x+\\alpha} h^{x+\\alpha} + bA_{x+\\beta} h^{x+\\beta} + \\&c. \\right\\} $$\n\nand, if $h = 1$\n\n$$(at^{-\\alpha} + bt^{-\\beta} + \\&c.) \\cdot \\phi(t) = G \\left\\{ aA_{x+\\alpha} + bA_{x+\\beta} + \\&c. \\right\\};(1). $$\n\nLet us express the function $aA_{x+\\alpha} + bA_{x+\\beta} + \\&c$ by the symbol $\\nabla A_x$ and let\n\n$$at^\\alpha + bt^\\beta + ct^\\gamma + \\&c. = f(t) $$\n\n$\\nabla A_x$ then will be the same as $f(A_x)$, provided that in the\ndevelopment of $f(A_x)$, $A_{x+i}$ be everywhere written for $\\{A_x\\}^i$,\nand to derive the generating function of $\\nabla A_x$ from that of $A_x$, we\nhave only to multiply the latter by $f(\\frac{1}{t})$, thus\n\n$$G \\{\\nabla A_x\\} = \\phi(t) \\cdot f\\left(\\frac{1}{t}\\right); \\ldots (2)$$\n\nLet now $\\nabla^2 A_x$, $\\nabla^3 A_x$, &c. denote the respective values of the\nexpression\n\n$$aA_{x+\\alpha} + bA_{x+\\beta} + \\&c.$$ \n\nWhen instead of $A_x$ we write $\\nabla A_x$, $\\nabla^2 A_x$, &c. that is,\n\n$$\\nabla^{i+1} A_x = a\\nabla^i A_{x+\\alpha} + b\\nabla^i A_{x+\\beta} + \\&c.$$ \n\nand it is easy to see that we shall have\n\n$$G \\{\\nabla^i A_x\\} = \\{f\\left(\\frac{1}{t}\\right)\\}^i \\cdot \\phi(t); \\ldots . (3).$$\n\nLet us now denote by $'f\\left(\\frac{1}{t}\\right)$ the expression\n\n$$'a \\cdot t^{-\\alpha} + 'b \\cdot t^{-\\beta} + \\&c.$$ \n\nand by $'\\nabla^i A_x$ the function formed from $'f\\left(\\frac{1}{t}\\right)$ in the same\nmanner as we formed $\\nabla^i A_x$ from $f\\left(\\frac{1}{t}\\right)$. The equation (3)\nthen may be thus written\n\n$$G \\{\\nabla^i A_x\\} = \\{(f'f^{-1}) : 'f\\left(\\frac{1}{t}\\right)\\}^i \\cdot \\phi(t)$$\n\nIf then $\\{f'f^{-1}(t)\\}^i$ be developed by the ordinary methods\ninto a series of the form\n\n$$a_{-\\infty} t^{-\\infty} + \\ldots \\ldots a_x t^x + \\ldots \\ldots a_\\infty t^\\infty = S_z \\{a_x t^x\\}$$\n\nwe shall obtain\n\n$$G \\{\\nabla^i A_x\\} = S_z \\{a_x \\{f'\\left(\\frac{1}{t}\\right)\\}^x\\} \\cdot \\phi(t) = S_z \\{a_x \\cdot G \\{\\nabla^x A_x\\}\\},$$\nand of course\n\n\\[ \\nabla^i A_x = a_{-\\infty} \\nabla^{-\\infty} A_x + \\ldots \\ldots a_0 A_x + a_1 \\nabla A_x + \\ldots \\ldots a_\\infty \\nabla^\\infty A_x; \\ldots \\ldots (4). \\]\n\nThus we may always develope \\( \\nabla^i A_x \\) in a series containing only the successive orders of \\( \\nabla A_x \\), such as \\( \\nabla^n A_x \\), &c.\n\nIf the developement of \\( \\{ f'f^{-1}(t) \\}^i \\) contain no negative powers of \\( t \\), we have\n\n\\[ a_z = \\frac{D^\\infty \\{ f'f^{-1} \\}^i : o}{1.2 \\ldots \\ldots z} \\]\n\nand consequently\n\n\\[ \\nabla^i A_x = \\{ f'f^{-1} \\}^i : o \\cdot A_x + \\frac{D \\{ f'f^{-1} \\}^i : o}{1} \\cdot \\nabla A_x + \\frac{D^2 \\{ f'f^{-1} \\}^i : o}{1.2} \\cdot \\nabla^2 A_x + \\&c. \\ldots \\ldots (5). \\]\n\nLet \\( \\nabla A_x = A_{x+1} - A_x \\), and we have \\( f'(1/t) = \\frac{1}{t} - 1 \\),\n\nand \\( f^{-1}(t) = 1 + t \\), whence we obtain\n\n\\[ \\nabla^i A_x = \\{ f(1) \\}^i \\cdot A_x + \\frac{D \\{ f \\}^i : i}{1} \\Delta A_x + \\frac{D^2 \\{ f \\}^i : i}{1.2} \\Delta^2 A_x + \\&c.; \\ldots \\ldots (6) \\]\n\nfor it is evident that when \\( t = 0 \\), \\( D^\\infty \\{ f(1+t) \\}^i \\) becomes \\( D^\\infty \\{ f \\}^i : 1 \\).\n\nTo take a particular case, let \\( \\nabla A_x = A_{x+1} \\) and\n\n\\[ \\nabla^i A_x = A_{x+i} f\\left(\\frac{1}{t}\\right) = \\frac{1}{t}, \\text{ and } D^\\infty \\{ f \\}^i : 1 = i(i-1) \\ldots \\ldots (i-z+1), \\text{ whence } \\]\n\n\\[ A_{x+i} = A_x + \\frac{i}{1} \\cdot \\Delta A_x + \\frac{i(i-1)}{1.2} \\cdot \\Delta^2 A_x + \\&c. (7). \\]\n\nAgain, if we suppose \\( \\nabla A_x = \\Delta A_x = A_{x+1} - A_x \\), and \\( \\nabla A_x = A_{x+1} \\), we shall obtain from (5)\n\\[ \\Delta^i A_x = A_{x+i} - \\frac{i}{1} \\cdot A_{x+i-1} + \\frac{i(i-1)}{1 \\cdot 2} \\cdot A_{x+i-2} - \\&c; \\ldots (8) \\]\n\nBut to proceed. We have,\n\n\\[ G \\left\\{ x^i A_x \\right\\} = S \\left\\{ A_x x^i \\cdot t^x \\right\\} = \\frac{t}{dt} d \\frac{t}{dt} d \\ldots \\ldots \\frac{t}{dt} \\cdot d\\phi(t) \\]\n\nor which is the same thing,\n\n\\[ G \\left\\{ x^i A_x \\right\\} = \\frac{1}{d \\log t} d \\frac{1}{d \\log t} d \\ldots \\ldots d \\left\\{ \\phi \\log^{-1} : \\log t \\right\\} \\]\n\nNow,\n\n\\[ \\frac{d \\left\\{ \\phi \\log^{-1} : \\log t \\right\\}}{d \\log t} = D \\left( \\phi \\log^{-1} \\right) : \\log t \\times \\frac{d \\log t}{d \\log t} = (\\phi \\log^{-1}) : \\log t \\]\n\n\\[ \\frac{d \\left\\{ D \\left( \\phi \\log^{-1} \\right) : \\log t \\right\\}}{d \\log t} = D^2 \\left( \\phi \\log^{-1} \\right) : \\log t \\]\n\nand so on. Thus our equation becomes\n\n\\[ G \\left\\{ x^i A_x \\right\\} = D^i \\left( \\phi \\log^{-1} \\right) : \\log t \\]\n\nand, if \\( f(x) = ax^\\alpha + bx^\\beta + \\&c \\), we see that\n\n\\[ G \\left\\{ A_x \\cdot f(x) \\right\\} = (aD^\\alpha + bD^\\beta + \\&c) \\left( \\phi \\log^{-1} \\right) : \\log t \\]\n\n\\[ = (f : D) \\left( \\phi \\log^{-1} \\right) : \\log t ; \\ldots \\ldots (9) \\]\n\nIf \\( f(x) \\) be a rational integral function of \\( x \\), the second member of this equation will require only the ordinary rules of the differential calculus for its formation, and of course the first may be rigorously obtained.\n\nConceive \\( f(t) \\) and \\( F(t') \\) to be developed into the two series\n\n\\[ S_x \\left\\{ a_x t^x \\right\\} \\text{ and } S_y \\left\\{ A_y t^y \\right\\}, \\text{ and let us consider the double series} \\]\n\n\\[ \\sigma = S_{x,y}^2 \\left\\{ a_x A_y \\cdot h^{xy} \\cdot t^x \\cdot t^y \\right\\} \\]\n\nFirst, \\( \\sigma \\) may be expressed as follows,\n\n\\[ \\sigma = S_x \\left\\{ a_x t^x \\cdot S_y \\left\\{ A_y \\cdot (h^* t')^y \\right\\} \\right\\} = S_x \\left\\{ a_x t^x \\cdot F(t' h^*) \\right\\} \\]\n\nBut by a similar mode of reasoning, we should also find\nMr. Herschel on various points of Analysis.\n\n\\[ \\sigma = S_y \\left\\{ A_y t'^y \\cdot f(th'^y) \\right\\} \\]\n\nor,\n\n\\[ \\sigma = S_x \\left\\{ A_x t'^x \\cdot f(th'^x) \\right\\} \\]\n\nfor since \\( x \\) and \\( y \\) vary through all their values, these two sums are identical. Equating then the values of \\( \\sigma \\)\n\n\\[ S \\left\\{ A_x t'^x \\cdot f(th'^x) \\right\\} = G_t \\left\\{ a_x \\cdot F(t'h'^x) \\right\\} \\]\n\n\\[ = (F : t'h'^D) (f \\log^{-1}) : \\log t \\]\n\nLet \\( t' = 1 \\), and for \\( h \\) writing \\( h^{-1} \\), and adding or subtracting\n\n\\[ S \\left\\{ A_x \\cdot \\frac{f(th'^x) \\pm f(th'^{-x})}{2} \\right\\} = \\frac{F(b^D) \\pm F(b^{-D})}{2} (f \\log^{-1}) : \\log t; (10) \\]\n\nII. On Logarithmic Transcendents.\n\nThe equation we have just arrived at affords us a method of exhibiting, in a finite form, the sum of the series in its first member, provided we possess the means of obtaining the second; and it appears, by what we have before remarked, that this can be performed, whenever \\( F(h^D) \\pm F(h^{-D}) \\) is a rational integral function of \\( D \\). This includes among the forms of \\( F \\) those remarkable functions denominated by Mr. Spence, \"Logarithmic Transcendents,\" and we shall now proceed, by the help of the general property demonstrated by that author in his \"Essay &c.\" to derive from these principles the summation of one of the most extensive classes of series which has yet received discussion.\n\nAdopting Mr. Spence's notation, we will represent the series\n\n\\[ \\frac{x^n}{1^n} - \\frac{x^a}{2^n} + \\frac{x^3}{3^n} - \\&c. \\]\n\nby the symbol \\( nL(1 + x) \\). The property then alluded to is as follows:\ncontinued as far as it will go without involving negative powers of log. \\( x \\). Supposing then \\( F(t) = nL(1 + t) \\), and writing \\( e^\\theta \\) for \\( h \\) we obtain\n\n\\[\nS \\left\\{ \\frac{(-1)^x + 1}{x^n} \\cdot f(t^{e^\\theta}) + \\frac{(-1)^n \\cdot f(t^{-e^\\theta})}{2} \\right\\} =\n\\]\n\n\\[\n= \\left( \\frac{nL(2)}{1,2 \\ldots n} D^n + \\frac{n-2L(2)}{1,2 \\ldots (n-2)} D^{n-2} + \\&c. \\right) (f \\log^{-1}) : \\log t; \\quad (11)\n\\]\n\nA very remarkable case of this equation is when \\( t = 1 \\), or \\( \\log t = 0 \\), for then \\( \\frac{D'(f \\log^{-1}) : \\log t}{1,2 \\ldots i} \\) is the coefficient of \\( t^i \\) in the development of \\( f \\log^{-1}(t) \\) or of \\( f(e^t) \\). If then we suppose\n\n\\[\nf(e^t) = a_0 + a_1 t + a_2 t^2 + \\ldots a_n t^n + \\&c.\n\\]\n\nwe shall have the following equation:\n\n\\[\nS \\left\\{ \\frac{(-1)^x + 1}{x^n} \\cdot f(t^{e^\\theta}) + \\frac{(-1)^n \\cdot f(t^{-e^\\theta})}{2} \\right\\} = nL(2) \\cdot a_n \\theta^n + n-2L(2) \\cdot a_{n-2} \\theta^{n-2} + \\&c.; \\quad (12)\n\\]\n\nThe second member being continued so long as it does not involve negative powers of \\( \\theta \\).\n\nWith regard to the functions \\( nL(2) \\), \\( nL(2) \\), &c. we have, as is well known\n\n\\[\nnL(2) = 1 - 1 + 1 - 1 + \\&c. = \\frac{1}{2}, \\text{ and}\n\\]\n\n\\[\n2xL(2) = \\frac{(2x-1)(-1)^{2x}}{1,2 \\ldots (2x)} \\cdot B_{2x-1}\n\\]\n\n\\( B_{2x-1} \\) being the \\( x \\)th number of BERNOULLI.\n\nThe equation (12) by assigning specific forms to \\( f(t) \\) affords an indefinite variety of interesting results, of which we shall only notice a few, the most remarkable.\n\n1. Let \\( f(t) = \\frac{-1}{\\sqrt{-1} + t} \\), and for \\( n \\) write \\( 2n - 1 \\), and \\( \\theta \\sqrt{-1} \\) for \\( \\theta \\).\nThe usual exponential expression for tan. $\\theta$, reduced into two fractions, gives\n\n$$\\tan. \\theta = \\frac{-1}{\\sqrt{-1} + i\\sqrt{-1}} - \\frac{-1}{\\sqrt{-1} + i\\sqrt{-1}}; \\ldots \\ldots \\ldots \\ldots (13)$$\n\nThus, the first member of (12) becomes\n\n$$\\frac{1}{2} S \\left[ (-1)^x + 1 \\cdot \\frac{\\tan. x\\theta}{x^{2n-1}} \\right] \\text{ or, } \\frac{1}{2} \\left\\{ \\frac{\\tan. \\theta}{1^{2n-1}} - \\frac{\\tan. 2\\theta}{2^{2n-1}} + \\frac{\\tan. 3\\theta}{3^{2n-1}} - \\&c. \\right\\}$$\n\nIn order to obtain the coefficients $a_1, a_3, \\ldots, a_{2n-1}$, in the second, we have,\n\n$$f(t) = \\frac{-1}{\\sqrt{-1} + t} = a_0 + a_1 \\cdot t + a_2 \\cdot t^2 + \\&c.$$ \n\nNow $f(t)$ may also be thrown into the form\n\n$$\\frac{1}{2} \\left\\{ \\frac{-1}{\\sqrt{-1} + t} - \\frac{-1}{\\sqrt{-1} + i-t} \\right\\} + \\frac{1}{2} \\left\\{ \\frac{-1}{\\sqrt{-1} + t} + \\frac{-1}{\\sqrt{-1} + i-t} \\right\\}$$\n\nwhich is the same as\n\n$$\\frac{1}{2} \\tan. \\left( \\frac{t}{\\sqrt{-1}} \\right) + \\frac{1}{2} \\sqrt{-1} - \\frac{1}{2} \\sec. \\left( \\frac{t}{\\sqrt{-1}} \\right)$$\n\nThus the even values of $a_x$ are given by the development of sec. $\\left( \\frac{t}{\\sqrt{-1}} \\right)$ and the odd by that of tan. $\\left( \\frac{t}{\\sqrt{-1}} \\right)$ and hence it is easy to see that\n\n$$a_{2x-1} = \\frac{(-1)^x + 1}{\\sqrt{-1}} \\cdot \\frac{2^{2x-1} \\cdot (2^{2x-1})}{1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot (2x)} \\cdot B_{2x-1}; \\quad (14)$$\n\nand\n\n$$a_{2x} = \\frac{-1}{1 \\cdot 2 \\cdot \\ldots \\cdot (2x) \\cdot 2^{2x+1}} \\left\\{ 1^{2x} - \\left\\{ 3^{2x} - \\frac{2x+1}{1} \\cdot 1^{2x} \\right\\} + \\left\\{ 5^{2x} - \\&c. \\right\\} - \\&c. \\right\\}; \\quad (15)$$\n\nbut a general value of $a_x$ may readily be obtained by the immediate consideration of the function $f(t)$ itself, as follows:\n\nWe know that\n\\[ a_x = \\frac{1}{1,2,\\ldots,x} \\cdot D^x \\frac{-1}{\\sqrt{-1 + \\{\\log^{-1}\\}}} : 0 \\]\n\nNow we have\n\n\\[ D(f \\log^{-1}) : t = \\frac{-t}{(\\sqrt{-1 + t})^2} \\]\n\n\\[ D^2 (f \\log^{-1}) : t = \\frac{t^2 - \\sqrt{-1} \\cdot t}{(\\sqrt{-1 + t})^3} \\]\n\n\\[ D^x (f \\log^{-1}) : t = \\frac{0A \\cdot e^{xt} + 1A \\cdot e^{(x-1)t} + \\ldots x^{-1}A \\cdot e^t}{(\\sqrt{-1 + t})^x + 1} \\]\n\nIn order to determine the numerator of this fraction, we shall adopt the elegant artifice used by Laplace * on a similar occasion.\n\n\\[ 0A \\cdot e^{xt} + \\ldots x^{-1}A \\cdot e^t = (\\sqrt{-1 + t})^{x+1} \\cdot D^x \\left\\{ \\frac{-1}{t + \\sqrt{-1}} \\right\\} \\]\n\n\\[ = -(\\sqrt{-1 + t})^{x+1} \\cdot D^x \\left\\{ e^{-t} - \\sqrt{-1} \\cdot e^{-2t} - e^{-3t} + \\sqrt{-1} \\cdot \\&c. \\right\\} \\]\n\n\\[ = (-1)^{x+1} \\cdot (\\sqrt{-1 + t})^{x+1} \\cdot \\left\\{ 1 \\cdot e^{-t} - \\sqrt{-1} \\cdot 2^x \\cdot e^{-2t} - 3^x \\cdot e^{-3t} + \\&c. \\right\\} \\]\n\nNow, as this equation is rigorous, and the first member contains only positive powers of \\( e^t \\), the negative powers in the second must destroy each other, and may therefore be neglected.\n\nExpanding then \\( (\\sqrt{-1 + t})^{x+1} \\) in powers of \\( e^t \\), multiplying together the two series, and retaining only positive powers of \\( e^t \\), we find\n\n* Laplace. Mém. de l’Acad. 1779. Sur l’usage du calc. des diff. partielles dans la théorie des suites.\n\\[ A \\cdot e^{xt} + \\ldots x^{-1} A \\cdot s^t = \\]\n\n\\[ = (-1)^{x+1} \\left\\{ \\begin{array}{c}\n1^x \\cdot e^{xt} - \\left\\{ 2^x - \\frac{x+1}{1} \\cdot 1^x \\right\\} \\sqrt{-1} \\cdot e^{(x-1)t} \\\\\n- \\left\\{ 3^x - \\frac{x+1}{1} \\cdot 2^x + \\frac{(x+1) \\cdot x}{1.2} \\cdot 1^x \\right\\} \\cdot e^{(x-2)t} + \\&c.\n\\end{array} \\right\\} \\]\n\nAnd after the substitution of \\( o \\) for \\( t \\), or \\( 1 \\) for \\( e^t \\) and its powers we obtain,\n\n\\[ a_x = \\frac{(-1)^{x+1}}{1.2 \\ldots \\ldots x(1 + \\sqrt{-1})^{x+1}} \\times \\]\n\n\\[ \\left\\{ \\begin{array}{c}\n1^x - \\left\\{ 2^x - \\frac{x+1}{1} \\cdot 1^x \\right\\} \\sqrt{-1} \\\\\n- \\left\\{ 3^x - \\frac{x+1}{1} \\cdot 2^x + \\frac{(x+1) \\cdot x}{1.2} \\cdot 1^x \\right\\} + \\&c.\n\\end{array} \\right\\}; \\ldots (16) \\]\n\nThe equation (12) will thus take the following form,\n\n\\[ S \\left\\{ (-1)^{x+1} \\cdot \\tan x \\theta \\right\\} = Y_1 \\theta + Y_3 \\theta^3 + \\ldots Y_{2n-1} \\theta^{2n-1}; (17) \\]\n\nwhere\n\n\\[ Y_{2x-1} = \\frac{2^x (2^x - 1) \\cdot (2^{2n-2x-1}-1)}{1.2 \\ldots \\ldots (2x) \\times 1.2 \\ldots \\ldots (2n-2x)} \\cdot B_{2x-1} \\cdot B_{2n-2x-1}; (18) \\]\n\n2. Retaining the same form of the function \\( f \\), and of course the same value of \\( a_x \\), for \\( n \\) write \\( 2n \\), and for \\( \\theta, \\theta \\cdot \\sqrt{-1} \\), and the first member of (12) becomes\n\n\\[ \\frac{1}{2} \\left\\{ \\sqrt{-1} \\cdot 2^n L(2) - S \\left\\{ \\frac{(-1)^{x+1}}{x^{2n}, \\cos x \\theta} \\right\\} \\right\\} \\]\n\nAnd the second,\n\n\\[ a_o \\cdot 2^n L(2) - a_2 \\cdot 2^{n-2} L(2) \\cdot \\theta^2 + \\ldots + (-1)^n \\cdot 0 L(2) \\cdot \\theta^{2n} \\cdot a_{2n} \\]\n\nNow, since \\( a_o = \\frac{-1}{1 + \\sqrt{-1}} \\), collecting into one the coefficients of \\( 2^n L(2) \\), viz.: \\( \\sqrt{-1} - 2a_o \\), we find their sum equal to 1, and of course,\n\nMDCCCXIV.\n\\[ S \\left\\{ \\frac{(-1)^x + 1}{x^{2n} \\cdot \\cos x\\theta} \\right\\} = 2^n L(2) + Y_2 \\theta^2 + Y_4 \\theta^4 + \\ldots Y_{2n} \\theta^{2n}; \\quad (19) \\]\n\nwhere\n\n\\[ Y_{2x} = \\frac{(-1)^x (2^{2n-2x-1}-1) \\cdot \\pi^{2n-2x}}{1 \\cdot 2 \\cdot \\ldots (2x) \\times 1 \\cdot 2 \\cdot \\ldots (2n-2x) \\cdot 2^{2x}} B_{2n-2x-1}. \\]\n\n\\[ \\left\\{ 1^{2x} - \\left\\{ 3^{2x} - \\frac{2x+1}{1} \\cdot 1^{2x} \\right\\} \\right\\} \\]\n\n\\[ + \\left\\{ 5^{2x} - \\&c \\right\\} - \\&c \\quad (20) \\]\n\nIf in this equation for \\( \\theta \\) we write \\( \\pi + \\theta \\), we shall obtain the sum of the series\n\n\\[ \\frac{1}{1^{2n} \\cdot \\cos \\theta} + \\frac{1}{2^{2n} \\cdot \\cos 2\\theta} + \\frac{1}{3^{2n} \\cdot \\cos 3\\theta} + \\&c; \\quad \\ldots \\ldots \\quad (21) \\]\n\nAnd by addition, of\n\n\\[ \\frac{1}{1^{2n} \\cdot \\cos \\theta} + \\frac{1}{3^{2n} \\cdot \\cos 3\\theta} + \\frac{1}{5^{2n} \\cdot \\cos 5\\theta} + \\&c; \\quad \\ldots \\ldots \\quad (22) \\]\n\n3. Let \\( f(t) = \\frac{\\sqrt{-1} - t}{\\sqrt{-1} + t} \\), and let \\( 2n \\) be written for \\( n \\) and \\( \\theta \\sqrt{-1} \\) for \\( \\theta \\); and the first member of (12) becomes\n\n\\[ \\sqrt{-1} \\cdot S \\left\\{ \\frac{(-1)^x + 1}{x^{2n} \\cdot \\cos x\\theta} \\right\\} \\]\n\nNow the development of \\( \\frac{-1}{\\sqrt{-1} + t} \\) being \\( a_0 + a_1 t + \\&c \\), that of \\( f(t) \\) will be\n\n\\[ (\\sqrt{-1} - t) \\left\\{ a_0 + a_1 t + \\&c \\right\\} = \\]\n\n\\[ = \\left\\{ (1 - \\sqrt{-1}) + \\frac{t}{1} + \\frac{t^2}{1 \\cdot 2} + \\&c \\right\\} \\cdot \\left\\{ a_0 + a_1 t + \\&c \\right\\} \\]\n\nand the coefficient of \\( t^x \\) will be found equal to\n\n\\[ (1 - \\sqrt{-1}) \\cdot a_x + \\frac{a_{x-1}}{1} + \\frac{a_{x-2}}{1 \\cdot 2} + \\ldots \\ldots \\frac{a_0}{1 \\cdot 2 \\cdot \\ldots \\cdot x} \\]\n\nThus the application of (12) gives the following equation\n\\[ S \\left\\{ \\frac{(-1)^x + 1}{x^{2n} \\cdot \\cos x \\theta} \\right\\} = 2^n L(2) + Y_2 \\theta^2 + Y_4 \\theta^4 + \\ldots Y_{2n} \\theta^{2n} \\]\n\nwhere\n\n\\[ Y_{2x} = \\frac{(-1)^x \\cdot (2^{2n-2x-1}-1)}{\\sqrt{-1} \\cdot 1.2 \\ldots (2n-2x)} B_{2n-2x-1} \\left\\{ (1 - \\sqrt{-1}) \\right\\} \\]\n\n\\[ a_{2x} + \\frac{a_{2x-1}}{1} + \\ldots \\frac{a_0}{1.2 \\ldots (2x)} \\]\n\nWhich compared with the value of \\( Y_{2x} \\) before found (20)\n\ngives the following singular equations,\n\n\\[ 0 = a_{2x} + \\frac{a_{2x-2}}{1.2} + \\frac{a_{2x-4}}{1.2.3.4} + \\ldots \\frac{a_0}{1.2 \\ldots (2x)}; \\ldots \\ldots (23) \\]\n\n\\[ a_{2x} = \\sqrt{-1} \\cdot \\left\\{ \\frac{a_{2x-1}}{1} + \\frac{a_{2x-3}}{1.2.3} + \\ldots \\frac{a_1}{1.2 \\ldots (2x-1)} \\right\\} - \\frac{1}{2} \\cdot \\frac{1}{1.2 \\ldots (2x)}. \\quad (24) \\]\n\nThe latter of these two equations affords the even values of \\( a_x \\)\n\nin terms of the odd, and hence we are enabled to express the\n\nsum of the series \\( \\frac{1}{1^{2x+1}} - \\frac{1}{3^{2x+1}} + \\frac{1}{5^{2x+1}} - \\&c = 2^{2x+1} C(1)* \\)\n\nby means of the numbers of Bernouilli, which Euler appears\n\nto have considered as impracticable.† We need hardly re-\n\n* See \"Essay on Logarithmic Transcendents,\" page 51. I should not omit to observe that the equations in p. 69 of that work, expressing the value of the function \\( nC(x) + (-1)^n \\cdot nC(x^{-1}) \\) when combined with our equation (10) by making\n\n\\[ F(t) = nC(t) = \\frac{t}{1n} - \\frac{t^3}{3n} + \\&c, \\] afford a series of results, highly interesting, but\n\nwhich the necessary limits of this paper forbid me at present to dilate on.\n\n† \"Per hos autem numeros Bernouillianos secans exprimi non potest, sed requirit\n\n\"alios numeros qui in summas potestatum reciprocarum imparium ingrediuntur, si\n\n\"enim ponatur,\n\n\\[ 1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\&c = \\alpha \\cdot \\frac{\\pi}{2^2} \\]\n\n\\[ 1 - \\frac{1}{3^3} + \\frac{1}{5^3} - \\frac{1}{7^3} + \\&c = \\beta \\cdot \\frac{\\pi^3}{1.2 \\cdot 2^4}, \\&c. \\]\n\n\\[ 3N2 \\]\nmark that the imaginary form here assumed by \\(a_{zx}\\) is merely apparent.\n\nTo proceed. Let the equation (17), multiplied by \\(d\\theta\\), be integrated between the limits 0 and \\(\\theta\\), and we find after all reductions.\n\n\\[\n(\\sec \\theta)^{\\frac{1}{2}} \\cdot (\\cos 2\\theta)^{\\frac{1}{2}} \\cdot (\\sec 3\\theta)^{\\frac{1}{3}} \\cdot \\&c. = \\log^{-1}\n\\]\n\n\\[\n\\left\\{ o + Y_1 \\cdot \\frac{6^2}{2} + \\ldots \\ldots Y_{2n-1} \\cdot \\frac{6^{2n}}{2n} \\right\\}; \\ldots (25).\n\\]\n\nThe value of \\(Y_{2n-1}\\) being given by the equation (18)\n\nAgain, if we suppose, for the sake of brevity\n\n\\[\n2^n L(2) + Y_2 \\cdot \\theta^2 + \\ldots Y_{2n} \\cdot \\theta^{2n} = U(\\theta)\n\\]\n\nand \\(2^n L(2) \\cdot \\frac{0}{1} + Y_2 \\cdot \\frac{6^3}{3} + \\ldots Y_{2n} \\cdot \\frac{6^{2n+1}}{2n+1} = D^{-1} U(\\theta)\\),\n\nwe shall obtain, by operating in the same manner on (19) and the equations derived from it, expressing the values of the series (21) and (22).*\n\n\\[\n\\text{erit } \\alpha = 1, \\beta = 1, \\gamma = 5, \\&c. \\text{ ex hisque valoribus obtinebitur }\n\\]\n\n\\[\n\\sec x = \\alpha + \\frac{\\beta}{1.2} x^2 + \\frac{\\gamma}{1.2.3.4} x^4 + \\&c.”\n\\]\n\nEuler. Inst. Calc. Diff. Pars posterior. Cap. VIII. p. 542.\n\nThe general value of \\(2x+1C(1)\\) as deduced from our equation (24) in terms of the numbers of Bernouilli is as follows;\n\n\\[\n2x+1C(1) = \\left( \\frac{\\pi}{2} \\right)^{2x+1} \\left\\{ \\frac{2^{2x-1}(2^{2x}-1)}{1.2.3 \\ldots (2x)} \\cdot \\frac{B_{2x-1}}{1} - \\frac{2^{2x-3}(2^{2x-2}-1)}{1.2 \\ldots (2x-2)} \\cdot \\frac{B_{2x-3}}{1.2.3} + \\ldots \\ldots \\ldots + (-1)^{x-1} \\cdot \\frac{2 \\cdot (2^2-1)}{1.2} \\cdot \\frac{B_1}{1.2 \\ldots (2x-1)} \\right\\}\n\\]\n\n\\[\n+ (-1)^x \\cdot \\frac{1}{2} \\cdot \\frac{1}{1.2 \\ldots (2x)}\n\\]\n\n* The constant added to complete these integrals is determined by making \\(\\theta = 0\\) in which case since \\(\\cot \\left( \\frac{\\pi}{4} - x\\theta \\right) = 1\\), their first members vanish when \\(n\\) is greater\nMr. Herschel on various points of Analysis.\n\n\\[ S \\left\\{ \\frac{(-1)^{x+1}}{x^{2n+1}} \\log \\cot \\left( \\frac{\\pi}{4} - \\frac{x\\theta}{z} \\right) \\right\\} = D^{-1} U(\\theta); \\ldots (26) \\]\n\n\\[ S \\left\\{ \\frac{1}{x^{2n+1}} \\cdot \\log \\cot \\left( \\frac{\\pi}{4} - \\frac{x\\theta}{z} \\right) \\right\\} = D^{-1} U(\\pi) - D^{-1} U(\\pi + \\theta); (27) \\]\n\n\\[ S \\left\\{ \\frac{1}{(2x-1)^{2n+1}} \\log \\cot \\left( \\frac{\\pi}{4} - \\frac{(2x-1)\\theta}{z} \\right) \\right\\} = \\]\n\n\\[ D^{-1} \\times \\left\\{ U(\\pi) + U(\\theta) - U(\\pi + \\theta) \\right\\}; \\ldots \\ldots \\ldots (28). \\]\n\nIn the last of these equations, if we write \\( -\\left( \\frac{\\pi}{2} + 2\\theta \\right) \\) for \\( \\theta \\), we obtain the following equation, corresponding to (25)\n\n\\[ \\left( \\frac{1}{1} \\right)^{2n+1} \\left( \\frac{1}{3} \\right)^{2n+1} \\left( \\frac{1}{5} \\right)^{2n+1} \\]\n\n\\[ (\\tan \\theta) \\cdot (\\cot 3\\theta) \\cdot (\\tan 5\\theta) \\&c = \\]\n\n\\[ \\log^{-1} \\left\\{ \\frac{D^{-1} U(\\pi) + D^{-1} U(-\\frac{\\pi}{2} - 2\\theta) - D^{-1} U(\\frac{\\pi}{2} - 2\\theta)}{2} \\right\\}; \\ldots (29). \\]\n\nThese are but a few of the very singular results which may be deduced from our equation (12); but I shall forbear to extend this paper to an unnecessary length by any farther applications of it.\n\nthan unity; but when \\( n = 0 \\), the series \\( S \\left\\{ \\frac{1}{x^{2n+1}} \\right\\} \\) and \\( S \\left\\{ \\frac{1}{(2x-1)^{2n+1}} \\right\\} \\) becoming infinite, they take the forms \\( \\infty \\times \\log (1) \\) which is altogether vague and inconclusive. Our equations (27) and (28), (29), then, are defective in this case, and we can only conclude that the function\n\n\\[ (\\tan \\theta)^{\\frac{1}{3}} \\cdot (\\cot 3\\theta)^{\\frac{1}{3}} \\&c. \\]\n\nis independent on \\( \\theta \\), or constant. There seems reason to conclude from other principles however that this constant is \\( \\pm \\frac{\\pi^2}{8} \\), or more generally, \\( \\pm \\frac{\\pi^2}{8} \\) \\( i \\) being any integer, positive or negative.\nIII. On Functional Equations.\n\nThe determination of functions from given conditions is a point of such importance, not only in the partial differential calculus, but also in a variety of other branches, that it has occupied the attention of the most eminent Analysts, and it must be confessed, not without considerable success. Their researches, however, have hitherto extended no farther than to such conditions as involve only the unknown function, $\\phi$ without any of its superior or inferior orders, $\\phi^2$, $\\phi^3$, ... &c, $\\phi^{-1}$, &c. It is to equations of this latter kind, therefore, that we now propose to direct our attention.\n\nThe successive orders of any function $f(x)$ may be produced, either by actually writing $f(x)$ for $x$ in the expression of $f(x)$, in which case the general value of $f^z(x)$ must be concluded from induction; or more elegantly by the following method.\n\nAssume $f^z(x) = u_z$ and we have $f^{z+1}(x) = u_{z+1}$\n\n\\[ o = u_{z+1} - f(u_z) \\]\n\nan equation of differences whose integral will be of the form\n\n\\[ u_z = F(z, C) \\]\n\n$C$ being an arbitrary quantity independent on $z$. Let $z = 0$, and we have\n\n\\[ F(0, C) = u_0 = f^0(x) = x \\]\n\nan equation which gives $C$ in functions of $x$.\n\nFor example; let $f(x) = 2x^2 - 1$, and we have\n\n\\[ o = u_{z+1} - 2u_z^2 + 1 \\]\n\nand integrating,\n\\[ u_x = \\frac{1}{2} \\left\\{ C^{2x} + C^{-2x} \\right\\} \\]\n\nNow, if we make \\( u_0 = x \\), we get\n\n\\[ C = x + \\sqrt{x^2 - 1} \\]\n\nand consequently,\n\n\\[ f(x) = \\frac{1}{2} \\left\\{ (x + \\sqrt{x^2 - 1})^{2x} + (x - \\sqrt{x^2 - 1})^{2x} \\right\\}. \\]\n\nIf we suppose \\( \\phi(x, y) \\) to denote any function of \\( x \\) and \\( y \\),\n\nand conceive this expression substituted for \\( x \\), as follows;\n\n\\[ \\phi \\left\\{ \\phi(x, y), y \\right\\}, \\]\n\nwe shall have the second partial function, taken with respect to \\( x \\), which we may denote thus, \\( \\phi^{2,1}(x, y) \\). If we repeat, or conceive repeated, this operation \\( m \\) times, we shall have the \\( m \\)th partial function with respect to \\( x \\):\n\n\\[ \\phi^{m,1}(x, y) = \\phi \\left\\{ \\phi^{m-1,1}(x, y), y \\right\\}. \\]\n\nIf the \\( m \\)th partial function with respect to \\( x \\) be in like manner successively substituted \\( n \\) times for \\( y \\) in the expression \\( \\phi(x, y) \\),\n\nwe shall obtain a result,\n\n\\[ \\phi^{m,n}(x, y) = \\phi \\left\\{ x, \\phi^{m,n-1}(x, y) \\right\\}, \\]\n\nand so on for more variables, \\( z, w, \\&c. \\) — An equation containing any number of the successive orders \\( \\phi^0(x) = x, \\phi(x), \\ldots, \\phi^n(x) \\), of a function \\( \\phi \\), and from which \\( \\phi \\) is to be determined, is called a functional equation of the \\( n \\)th order, in \\( \\phi \\).\n\nThus the equation\n\n\\[ o = \\phi^2(x) - (1 + b) \\cdot \\phi(x) + bx \\]\n\nis a functional equation of the second order, and is satisfied by the following\n\n\\[ \\phi(x) = a + bx. \\]\nAn equation between any number of the partial functions $\\phi^m, n, \\&c$ ($x, y, z, \\&c$) for determining the form of $\\phi(x, y, z, \\&c)$ is called an equation of partial functions, and its order may be denoted in the same manner. Thus\n\n$$o = \\phi^{2,1}(x, y) + \\phi^{1,2}(x, y) - (a + b + 1) \\cdot \\phi(x, y) + c$$\n\nis an equation of partial functions of the second order with two variables, and is satisfied by the equation\n\n$$\\phi(x, y) = ax + by + c$$\n\nLet $\\phi(x)$ be a function of $x$, and a certain number of constants $a, b, c, \\ldots$. And from this expression conceive $\\phi^\\alpha(x), \\phi^\\beta(x) \\&c$ to be successively formed which will be functions also of $a, b, \\&c$. If the number of these constants be $n$, we may thus produce $n + 1$ such functions of them, which will be respectively equal to the several orders of $\\phi$ which they represent. Thus we have $n + 1$ equations involving the $n$ quantities $a, b, c, \\&c$ which may therefore be eliminated, and the resulting equation between $x, \\phi^\\alpha(x), \\phi^\\beta(x) \\&c$ will therefore be independent on them. As far then as regards this equation they are arbitrary, so that in reascending from a functional equation which contains $n + 1$ different orders of $\\phi$ (not including $\\phi^0(x)$ or $x$) $n$ arbitrary constants must be introduced. The reasoning here made use of is sufficiently plausible, and in fact, no other than has been adduced in demonstration of well known and important truths. The conclusion too, to the extent of its literal meaning, is correct.\n\nBut we have here to notice a paradox of a very singular nature, viz.: that even in the simplest cases imaginable (such as $\\phi^2(x) = x$) the general expression for $\\phi(x)$ may contain, not one or two, but an unlimited number of arbitrary constants, nay\neven one or more arbitrary functions. A nearer attention to every step of the above reasoning will explain this paradox. But what has been said will serve to make us cautious in trusting implicitly to all its other applications.\n\n**Problem I.** To determine $\\phi(x)$ from the equation $\\phi^2(x) = x$. Assume $z$ a function of $x$, and $u$ a functional characteristic, which shall satisfy the following conditions\n\n$$x = u_z, \\phi(x) = u_{z+1}.$$ \n\nFrom these, we obtain\n\n$$\\phi(x), \\text{that is, } \\phi(u_z) \\text{ or } (\\phi u)_z = u_{z+1}; \\ldots \\ldots (a)$$\n\nand $\\phi^2(x)$ or $\\phi \\{ \\phi(x) \\} = (\\phi u)_{z+1} = x = u_z; \\ldots \\ldots (b)$\n\nand, subtracting, $(\\phi u)_{z+1} - (\\phi u)_z = -(u_{z+1} - u_z)$\n\nthat is, $\\Delta \\{ (\\phi u)_z + u_z \\} = 0; \\ldots \\ldots \\ldots \\ldots (c)$\n\nand integrating,\n\n$$0 = (\\phi u)_z + u_z + C.$$ \n\nNow by cross-multiplication of the equations $(a)$ and $(b)$ we find,\n\n$$u_{z+1} \\cdot (\\phi u)_{z+1} = u_z \\cdot (\\phi u)_z.$$ \n\nThus the function $u_z \\cdot (\\phi u)_z$ does not vary when $z$ changes to $z + 1$, and of course must be considered as constant in the integration of $(c)$. $C$ therefore may be any function of $u_z \\cdot (\\phi u)_z$, and thus our equation becomes\n\n$$0 = u_z + (\\phi u)_z + f \\{ u_z \\cdot \\phi(u_z) \\}$$\n\nor $$0 = x + \\phi(x) + f \\{ x \\cdot \\phi(x) \\}$$\n\nan equation from which $\\phi(x)$ may be obtained for any assigned form of the function $f$. Thus if $f(x) = a + bx,$\n\\[ o = a + x + (1 + bx) \\cdot \\phi(x) \\]\n\nand \\( \\phi(x) = -\\frac{a+x}{1+bx} \\)\n\nwhich satisfies the condition proposed; and by giving \\( f \\) other forms, we should obtain other values of \\( \\phi(x) \\).\n\nThe subsidiary function \\( z \\), and the characteristic \\( u \\) are not then necessary to be known but as a matter of curiosity. They may however be found when \\( \\phi \\) is determined, by the resolution of the equation of differences \\( \\phi(u_z) = u_{z+1} \\) which gives the form of the function \\( u_z \\) in \\( z \\), and \\( z \\) is given by the equation \\( x = u_z \\), or \\( z = u^{-1}(x) \\).\n\nAliter. Assume as before, \\( x = u_z \\), \\( \\phi(x) = u_{z+1} \\)\n\nthen we have\n\n\\[ \\phi(u_{z+1}) = u_z; \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots (d). \\]\n\nNow, \\( x = u_z \\) therefore \\( \\phi(x) = \\phi(u_z) \\), that is, \\( u_{z+1} = \\phi(u_z) \\)\n\nand for \\( z \\) writing \\( z-1 \\), \\( u_z = \\phi u_{z-1} \\) which being substituted in (d) gives\n\n\\[ \\phi(u_{z+1}) = \\phi(u_{z-1}). \\]\n\nNow this is a perfect function \\( \\phi \\) on both sides, and of course, taking the inverse function \\( \\phi^{-1} \\) on both sides\n\n\\[ u_{z+1} = u_{z-1} \\]\n\nwhence,\n\n\\[ u_z = C \\{ \\cos 2\\pi z \\} + (-1)^z \\cdot C' \\{ \\cos 2\\pi z \\} \\]\n\n\\( C \\) and \\( C' \\) being two arbitrary functional characteristics. Now \\( u_z = x \\), and consequently\n\n\\[ x = C \\{ \\cos 2\\pi z \\} + (-1)^z \\cdot C' \\{ \\cos 2\\pi z \\}. \\]\n\nFrom this conceive \\( z \\) found in functions of \\( x \\), and call it \\( Z(x) \\) then,\n\\[ u_{z+1} = \\phi(x) = C \\{ \\cos 2\\pi Z(x) \\} - (-1)^{Z(x)} \\cdot C' \\{ \\cos 2\\pi Z(x) \\}. \\]\n\nThis method applies also to the more general equation \\( \\phi^n(x) = f(x) \\), by the substitutions \\( f(x) = u_z, \\phi(x) = u_{z+1} \\), but, owing to the transcendental equations it introduces, must be regarded as totally ineffectual and useless.\n\n**Prob. II.** Given \\( \\phi^n(x) = f(x) \\). Required at least one satisfactory value of \\( \\phi(x) \\).\n\nLet the general expression of \\( f^n(x) \\) in functions of \\( z \\) and \\( x \\) found according to the method above explained be \\( F \\{ z, x \\} \\):\n\nwe have then \\( \\phi^n = f \\), and \\( \\phi = f^{1/n} \\), that is\n\n\\[ \\phi(x) = f^{1/n}(x) = F \\left\\{ \\frac{1}{n}, x \\right\\}. \\]\n\n**Ex. 1.** Let \\( f(x) = 2x^2 - 1 \\), or \\( \\phi^2(x) = 2x^2 - 1 \\), and we find\n\n\\[ f^n(x) = \\frac{1}{2} \\left\\{ (x + \\sqrt{x^2 - 1})^{2^n} + (x - \\sqrt{x^2 - 1})^{2^n} \\right\\} \\]\n\nand of course\n\n\\[ f^{1/n}(x) = \\phi(x) = \\frac{1}{2} \\left\\{ (x + \\sqrt{x^2 - 1})^{n\\sqrt{2}} + (x - \\sqrt{x^2 - 1})^{n\\sqrt{2}} \\right\\}. \\]\n\nWe may here observe that any one of the \\( n \\) values of \\( n\\sqrt{2} \\) will equally afford a satisfactory value of \\( \\phi(x) \\).\n\n**Ex. 2.** Let \\( \\phi^n(x) = \\frac{\\alpha + \\beta x}{\\gamma + \\delta x} = f(x) \\)\n\nAssume \\( \\omega = \\frac{\\beta - \\gamma}{2}, \\lambda = \\frac{\\beta + \\gamma}{2}, \\mu = \\left\\{ \\omega^2 + \\alpha \\delta \\right\\}^{1/2}, \\nu = \\frac{\\lambda - \\mu}{\\lambda + \\mu} \\)\n\nand we shall find\n\n\\[ f^{1/n}(x) \\text{ or } \\phi(x) = \\frac{\\left\\{ \\alpha + (\\omega - \\mu)x \\right\\}^{1/n} - \\left\\{ \\alpha + (\\omega + \\mu)x \\right\\}^{1/n}}{\\left\\{ \\delta x - (\\omega + \\mu) \\right\\}^{1/n} - \\left\\{ \\delta x + (\\omega - \\mu) \\right\\}^{1/n}} \\]\n\nwhere any of the \\( n \\) values of \\( \\nu^{1/n} \\) may be taken, and thus as\nmany different values of $\\phi(x)$ be obtained. This example depends on the integration of an equation of differences of the form\n\n$$o = u_{z+1} \\cdot u_z + A \\cdot u_{z+1} + B \\cdot u_z + C$$\n\na particular case of which had been previously integrated by Laplace in the Journal de l’Ecole Polytechnique.\n\nEx. 3. To take an instance of the application of these equations to geometrical problems, let $AM$ be an hyperbola whose axis is $CP$ and centre $C$, and let it be required to find a curve $am$ such that drawing the ordinate $PMm$, making $Cd = Pm$ and again erecting $dl$ parallel to $PM$, if this be repeated $n$ times the last ordinate $fk$ shall be equal to $PM$. Let $y = \\phi(x)$ be the equation of $am$ and $y^2 = (1 - e^2)(a^2 - x^2)$ that of $AM$, then $dl = \\phi(Cd) = \\phi(Pm) = \\phi(x)$ and in like manner, $fk = \\phi^n(x) = PM$, that is, $\\phi^n(x) = f(x) = \\sqrt{(1 - e^2)(a^2 - x^2)}$; consequently,\n\n$$f^n(x) = \\phi(x) = \\left\\{ (e^2 - 1)^{\\frac{1}{n}} \\cdot x^2 - \\frac{e^2 - 1}{e^2 - 2} \\left( (e^2 - 1)^{\\frac{1}{n}} - 1 \\right) \\cdot a^2 \\right\\}^{\\frac{1}{2}}.$$\n\nThus we see, that $am$ is also an hyperbola, whose centre is $C$, and calling $a'$ and $e'$ its semiaxis and excentricity, we have\n\n$$e' = \\sqrt{(e^2 - 1)^{\\frac{1}{n}} + 1}, \\text{ and } a' = a \\cdot \\left\\{ \\frac{e^2 - 1}{e^2 - 2} \\right\\}^{\\frac{1}{2}}.$$\n\nIf $AM$ be a right angled hyperbola, or $e = \\sqrt{2}$, we shall have $e' = \\sqrt{2}$ and $a' = \\frac{a}{\\sqrt{n}}$; that is, $am$ is also a right angled hyperbola, having its axis $\\frac{1}{\\sqrt{n}}$ part of that of $AM$. If $e$\nbe < 1, and \\( n \\) odd, we shall have \\( e' \\) also < 1. Thus if the curve \\( AM \\) be an ellipse, \\( am \\) is also a concentric ellipse. The equations of Ex. 2. geometrically expressed afford a property of the hyperbola something similar.\n\nIV. On differential Equations of the first degree.\n\nAny equation of this species may be reduced to the form\n\n\\[\no = u + {}^1A \\cdot Du + {}^2A \\cdot D^2u + \\ldots {}^nA \\cdot D^n u + X; \\ldots (1)\n\\]\n\n\\( u \\) being the unknown, and \\( {}^1A, \\ldots {}^nA, X \\), known functions of \\( x \\). To integrate it, assume the following equations\n\n\\[\n\\begin{align*}\nu + {}^1\\alpha Du &= u^{(1)} \\\\\nu^{(1)} + {}^2\\alpha Du^{(1)} &= u^{(2)} \\\\\n&\\vdots \\\\\nu^{(n-2)} + {}^{n-1}\\alpha Du^{(n-2)} &= u^{(n-1)} \\\\\nu^{(n-1)} + {}^n\\alpha Du^{(n-1)} + X &= u^{(n)} = o\n\\end{align*}\n\\]\n\nFrom these, eliminating successively \\( u^{(1)}, \\ldots u^{(n-1)} \\), we obtain\n\n\\[\n\\begin{align*}\nu^{(1)} &= u + {}^1\\alpha \\cdot Du \\\\\nu^{(2)} &= u + \\left( {}^1\\alpha + {}^2\\alpha + {}^2\\alpha D^1\\alpha \\right) \\cdot Du + {}^1\\alpha \\cdot {}^2\\alpha \\cdot D^2u \\\\\n&\\vdots \\\\\no = u^{(n)} &= u + \\left\\{ {}^1\\alpha + {}^2\\alpha \\left( 1 + D^1\\alpha \\right) + \\&c \\right\\} \\cdot Du + \\ldots {}^1\\alpha \\cdot {}^2\\alpha \\ldots {}^n\\alpha \\cdot D^n u + X; \\ldots (3)\n\\end{align*}\n\\]\n\nThe comparison of the coefficients of this equation with \\( {}^1A, \\ldots {}^nA \\), gives \\( n \\) equations for determining \\( {}^1\\alpha, \\ldots {}^n\\alpha \\), into which \\( X \\) does not enter. Consequently these functions are independent on \\( X \\), and therefore, the same as if \\( X = o \\). Now the successive integration of (3) gives\n\\[ u = \\left\\{ \\log^{-1} \\int \\frac{dx}{x^\\alpha} \\right\\} \\int \\log^{-1} \\int \\left( \\frac{dx}{x^\\alpha} - \\frac{dx}{x^\\alpha} \\right) \\int \\ldots \\ldots \\ldots \\]\n\n\\[ \\int \\log^{-1} \\int \\left( \\frac{dx}{n-1} - \\frac{dx}{n^\\alpha} \\right) \\int -X dx^n \\int \\frac{dx}{n^\\alpha} \\log^{-1} \\int \\frac{dx}{n^\\alpha}; \\ldots \\ldots (4) \\]\n\nwhich by writing \\( o \\) for \\( X \\), and adding a constant at each integration becomes,\n\n\\[ u = ^1C \\cdot \\log^{-1} \\int \\frac{dx}{x^\\alpha} + ^2C \\cdot \\left\\{ \\log^{-1} \\int \\frac{dx}{x^\\alpha} \\right\\} \\int \\frac{\\log^{-1} \\int \\left( \\frac{dx}{x^\\alpha} - \\frac{dx}{x^\\alpha} \\right)}{x^\\alpha} \\cdot dx + \\&c. (5) \\]\n\nNow, if \\( ^{(1)}u, ^{(2)}u, \\ldots ^{(n)}u \\) be the particular integrals of\n\n\\[ o = u + ^1A \\cdot Du + \\ldots ^nA \\cdot D^n u; \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots (6) \\]\n\nwe shall have, when \\( X = o \\),\n\n\\[ u = ^1C \\cdot ^{(1)}u + ^2C \\cdot ^{(2)}u + ^3C \\cdot ^{(3)}u + \\&c. \\]\n\nAnd comparing this with the expression (5), we find\n\n\\[ \\frac{-1}{x^\\alpha} = D \\log ^{(1)}u; \\]\n\n\\[ \\frac{-1}{x^\\alpha} = D \\log \\left\\{ ^1a \\cdot ^{(1)}u \\cdot D \\frac{^{(2)}u}{^{(1)}u} \\right\\}; \\]\n\n\\[ \\frac{-1}{x^\\alpha} = D \\log \\left\\{ ^1a \\cdot ^2a \\cdot ^{(1)}u \\cdot D \\frac{^{(2)}u}{^{(1)}u} \\cdot D \\frac{D \\frac{^{(3)}u}{^{(1)}u}}{D \\frac{^{(2)}u}{^{(1)}u}} \\right\\}; \\]\n\n\\[ \\&c = \\&c. \\]\n\nSuppose now that by any means whatever we can discover \\( n-1 \\) particular integrals \\( ^{(1)}u, \\ldots ^{(n-1)}u \\), of (6), the original equation deprived of its last term: then by the help of the first \\( n-1 \\) of these equations, the values of \\( ^1a, \\ldots ^{n-1}a \\), are given,\nand from these, \\( n \\alpha \\) may be derived by considering that the comparison of the equations (1) and (3) gives\n\n\\[\n1 \\alpha \\cdot 2 \\alpha \\ldots n-1 \\alpha \\cdot n \\alpha = nA, \\text{ or } n \\alpha = nA \\cdot (1 \\alpha \\ldots n-1 \\alpha)^{-1}\n\\]\n\nHaving obtained \\( 1 \\alpha, \\ldots, n \\alpha \\), nothing more is requisite for obtaining a complete integral of (1), than to substitute their values in equation (4).\n\nThe method here delivered of obtaining the known theorems respecting the equation\n\n\\[\no = u + 1ADu + \\ldots nA \\cdot D^n u + X\n\\]\n\nappears to have the advantage in point of conciseness over any I have hitherto met with; a sufficient apology for the revival of a subject whose theory, and whose difficulties have been so long and completely understood.\n\nIn the case when \\( X = o \\) and \\( 1A, \\ldots nA \\) are constant, the method of separating the symbols of operation from those of quantity, may be introduced with great elegance.\n\nLet \\( p, q, r, \\&c \\) be the roots of\n\n\\[\no = D^n + \\frac{n-1A}{nA} D^{n-1} + \\ldots + \\frac{1}{nA}\n\\]\n\nand the equation (1) becomes\n\n\\[\no = (D - p)(D - q) \\ldots \\&c : u\n\\]\n\nwhich is satisfied by either of the equations\n\n\\[\no = (D - p) : u, \\quad o = (D - q) : u, \\&c. \\text{ or, }\nDu = pu, \\quad Du = qu, \\&c.\n\\]\n\nNow these equations integrated give the following\n\n\\[\nu = e^{px}, \\quad u = e^{qx}, \\quad u = e^{rx}, \\&c.\n\\]\n\nwhich are the particular integrals of the proposed, and of course its complete integral will be\n\n\\[\nu = C_1 e^{px} + C_2 e^{qx} + C_3 e^{rx} + \\&c.\n\\]\nIf there be \\( m \\) roots equal to \\( p \\), we have\n\n\\[\n(D - p)^m : u = 0, \\quad \\text{or} \\quad (D - p)^m : u \\times e^{-px} = 0.\n\\]\n\nNow, \\((D - p)^m : u \\cdot e^{-px} = D^m \\{ u \\cdot e^{-px} \\} = 0.\\)\n\nTherefore, integrating \\( m \\) times\n\n\\[\nu \\cdot e^{-px} = ^0C + ^1C \\cdot x + \\ldots + ^{m-1}C \\cdot x^{m-1}\n\\]\n\nand\n\n\\[\nu = \\{ ^0C + ^1Cx + \\ldots + ^{m-1}C \\cdot x^{m-1} \\} e^{px}\n\\]\n\nwhich is the part of the integral arising from the equal roots \\( p \\).\n\nJOHN F. W. 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W. Herschel", "year": 1814, "volume": "104", "journal": "Philosophical Transactions of the Royal Society of London", "page_count": 30, "jstor_url": "https://www.jstor.org/stable/107442" } }