Examples of the Processes of the Differential and Integral Calculus |
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Page 15
... multiply h ' , we find that the term in ... 1.2 ( u + u'h ) " is n ( n - 1 ) ( n in ( u + u'h ) ” - 1 h2 is -1 - 1 ) ... multiplying by 1.2 ... r , we obtain for the th differential coefficient of u " ď ( u " ) dx = n ( n − 1 ) ... ( n ...
... multiply h ' , we find that the term in ... 1.2 ( u + u'h ) " is n ( n - 1 ) ( n in ( u + u'h ) ” - 1 h2 is -1 - 1 ) ... multiplying by 1.2 ... r , we obtain for the th differential coefficient of u " ď ( u " ) dx = n ( n − 1 ) ... ( n ...
Page 16
... multiplied by 1.2 ... r . Now expanding each term by the binomial theorem , we have for the coefficient of u ' ' 1 ... multiplying by 1.2 ... r , we find d ' ( u " ) ' u ' = 2n ( 2n − 1 ) ... ( 2n − r + 1 ) un -T nr ( r − 1 ) e2 + dx ...
... multiplied by 1.2 ... r . Now expanding each term by the binomial theorem , we have for the coefficient of u ' ' 1 ... multiplying by 1.2 ... r , we find d ' ( u " ) ' u ' = 2n ( 2n − 1 ) ... ( 2n − r + 1 ) un -T nr ( r − 1 ) e2 + dx ...
Page 19
... Multiplying these together , taking only the coefficient of h ' and multiplying it by 1.2 ... r , we find d'u = dx + € ¤¤2 { c ' ( 2x ) ' + r ( r − 1 ) c ' − 1 ( 2 x ) ' −2 r ( r 1 ) ... ( r− 3 ) 1.2 - c - 2 ( 2x ) ' ' + & c ...
... Multiplying these together , taking only the coefficient of h ' and multiplying it by 1.2 ... r , we find d'u = dx + € ¤¤2 { c ' ( 2x ) ' + r ( r − 1 ) c ' − 1 ( 2 x ) ' −2 r ( r 1 ) ... ( r− 3 ) 1.2 - c - 2 ( 2x ) ' ' + & c ...
Page 21
... multiplying by ( e2 + 1 ) ' + 1 we must have ɑ ‚ € TM a + ɑ , -16 ( r − 1 ) * + & c . + a1e * ( 1 ) . ( ε * + 1 ) ' + 1 d u dx ' = -2x -3- Now as u = € -E + € - & c . d'u dx 2x − ( ~ = ( − ) ' { l'e ̄ ′ — 2 ′ e ̃2 * + 3 ′ € ...
... multiplying by ( e2 + 1 ) ' + 1 we must have ɑ ‚ € TM a + ɑ , -16 ( r − 1 ) * + & c . + a1e * ( 1 ) . ( ε * + 1 ) ' + 1 d u dx ' = -2x -3- Now as u = € -E + € - & c . d'u dx 2x − ( ~ = ( − ) ' { l'e ̄ ′ — 2 ′ e ̃2 * + 3 ′ € ...
Page 32
... multiplying by ay , we have ( a + y ) * d2 u dy2 du + ( a + y ) dy = d'u dx ? differentiating a third time and again multiplying by a + y , du we see that ď u d'u ( a + y ) 3 + 3 ( a + y ) 2 + ( a + y ) dy3 dy dy and therefore d3 u + bu ...
... multiplying by ay , we have ( a + y ) * d2 u dy2 du + ( a + y ) dy = d'u dx ? differentiating a third time and again multiplying by a + y , du we see that ď u d'u ( a + y ) 3 + 3 ( a + y ) 2 + ( a + y ) dy3 dy dy and therefore d3 u + bu ...
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