Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
З цієї книги
Результати 1-5 із 43
Сторінка 25
... Euler ) regard- ing homogeneous functions of any number of variables , which from the frequent applications made of it ought to be noticed in this place . If u be a homogeneous algebraic function of n dimensions SUCCESSIVE ...
... Euler ) regard- ing homogeneous functions of any number of variables , which from the frequent applications made of it ought to be noticed in this place . If u be a homogeneous algebraic function of n dimensions SUCCESSIVE ...
Сторінка 26
... Euler , Calc . Diff . p . 188 . In applying this theorem to transcendental functions of algebraical functions , it is to be observed that it is not suffi- cient that these last should be homogeneous , it is also necessary that they ...
... Euler , Calc . Diff . p . 188 . In applying this theorem to transcendental functions of algebraical functions , it is to be observed that it is not suffi- cient that these last should be homogeneous , it is also necessary that they ...
Сторінка 54
... , therefore - - tan - 1 ( x + h ) = tan - 1x + sin y sin y h h sin 2y ( sin y ) 2 1 2 h3 h1 + sin 3y ( sin y ) 3 - sin 4y ( sin y ) + & c . From this development Euler * has deduced many re- markable 3 4 54 DEVELOPMENT OF FUNCTIONS .
... , therefore - - tan - 1 ( x + h ) = tan - 1x + sin y sin y h h sin 2y ( sin y ) 2 1 2 h3 h1 + sin 3y ( sin y ) 3 - sin 4y ( sin y ) + & c . From this development Euler * has deduced many re- markable 3 4 54 DEVELOPMENT OF FUNCTIONS .
Сторінка 55
Duncan Farquharson Gregory William Walton. From this development Euler * has deduced many re- markable theorems , some of which are subjoined . In the preceding example let h = -x , then tan - 1 ( x + h ) = tan - 10 = 0 ; therefore tan ...
Duncan Farquharson Gregory William Walton. From this development Euler * has deduced many re- markable theorems , some of which are subjoined . In the preceding example let h = -x , then tan - 1 ( x + h ) = tan - 10 = 0 ; therefore tan ...
Сторінка 72
... Euler , Calc . Diff . p . 519 . m - 2 ao ( 5 ) Let U = εаo + a1 x + Aq x2 + As before , assume u = A。 + Â ̧x + Â1⁄2∞2 + & c . + А „ x ” + & c . By taking the logarithmic differentials we find ( n + 1 ) An + 1 = a1 An + 2a1⁄2 An − 1 + ...
... Euler , Calc . Diff . p . 519 . m - 2 ao ( 5 ) Let U = εаo + a1 x + Aq x2 + As before , assume u = A。 + Â ̧x + Â1⁄2∞2 + & c . + А „ x ” + & c . By taking the logarithmic differentials we find ( n + 1 ) An + 1 = a1 An + 2a1⁄2 An − 1 + ...
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a² b2 a²x² angle arbitrary constant asymptote becomes C₁ c²x² Cambridge circle co-ordinates condition Crelle's Journal curvature curve cycloid determine differential coefficients differential equation dx dx dx dy dx dy dx dx² dy dx dy dy dy dy dz dz dz eliminate ellipse equal Euler factor formula fraction function Geometry gives Hence hypocycloid infinite intersection John Bernoulli Let the equation lines of curvature locus logarithmic logarithmic spiral Multiply negative origin parabola perpendicular plane of reference radius SECT singular solution spiral Substituting subtangent surface tangent plane theorem triangle vanish whence x²)³