( ) f a, b n f(b) = f(a) + f (a)(b a) + + f (n 1) (a) (n 1)! (b a)n 1 + R n, R n = b a f (n) (b t)n 1 (t) (n 1)! dt. : R n = b a f (n) (b t
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1 ) f, b n fb) = f) + f )b ) + + f n 1) ) n 1)! b )n 1 + R n, R n = f n) b t)n 1 t) n 1)! dt. : R n = f n) b t)n 1 t) n 1)! dt ] b b b t)n 1 + n 1)! = f n 1) b )n 1 ) + R n 1. n 1)! R n = [f n 1) t) f n 1) b t)n t) n )! dt R n R n 1 = f n 1) ) R 1 = b )n 1. n 1)! f t)dt = fb) f). R n = R n R n 1 ) + R n 1 R n ) + + R R 1 ) + R 1 = f n 1) b )n 1 ) n 1)! f n ) b )n ) f b ) ) f) + fb), n )! 1! fb) 1
2 .1 ) f, b n φt) n φ n) )fb) f)) = b ){φ n 1) 1)f b) φ n 1) )f )} b ) {φ n ) 1)f ) b) φ n ) )f ) )} + + 1) n 1 b ) n {φ1)f n) b) φ)f n) )} + R n+1 R n+1 = 1) n b ) n+1 φt)f n+1) + tb ))dt : R n+1 R n+1 = 1) n b ) n+1 { [ φt) f n) + tb )) b = 1) n b ) n φ1)f n) b) φ)f n) )) + S n ] 1 S n = 1) n 1 b ) n φ t) f n) + tb )) dt b = 1) n 1 b ) [ n 1 φ t)f n 1) + tb )) ] 1 } φ t) f n) + tb )) dt b + 1) n b ) n 1 φ t)f n 1) + tb ))dt = 1) n 1 b ) n 1 φ 1)f n 1) b) φ )f n 1) )) + S n 1 S n 1 S n S 1 S k = 1) k 1 b ) k 1 φ n+1 k) 1)f k 1) b) φ n+1 k) )f k 1) )) + S k 1 S 1 S 1 = b ) φ n) t)f + tb ))dt φ n φ n) φ n) ) = φ n) 1)) S 1 = φ n) )fb) f)). k = 1,,, n + 1) R n+1 =R n+1 S n ) + S n S n 1 ) + + S S 1 ) + S 1 = 1) n b ) n φ1)f n) b) φ)f n) )) + 1) n 1 b ) n 1 φ 1)f n 1) b) φ )f n 1) )) + + φ n) )fb) f))
3 3 ) u R n+1 = 1) n b ) n φ f n+1) u)du 3.1) b u = + tb ) du = b )dt u b t 1 3.1) ) + tb ) 1) n b ) n φ f n+1) + tb ))b )dt b = 1) n b ) n+1 φt)f n+1) + tb ))dt = R n+1 φt) = t 1) n.1 φ k) t) = n!fb) f)) = b ) 3.1) n! n k)! t 1)n k k n) { f b) n! ) } { f ) b ) f b) n! } 1!! f ) + + 1) n 1 b ) n { f n) b) 1) n f n) ) + R n+1 } fb) = f) + b ) f ) 1! + + b ) n f n) ) n! R n+1 = 1) n b ) n = b t) n f n+1) t)dt + b ) f )! + R n+1 3.) n! ) n t b 1 f n+1) t)dt n! R n+1 n! = b t) n f n+1) t)dt n! 3.) 3
4 4 4.1 ) fx), gx) [, b] < b) gx) > gx) = fx)gx)dx = fc) gx)dx < c < b gx) = 1 fx)dx = fc)b ) : fx) [, b] M, m gx) [, b] m gx) > m fx) M mgx) fx)gx) Mgx) gx)dx fx)gx)dx M gx)dx > m fx)gx)dx gx)dx fx)gx)dx gx)dx gx)dx gx)dx M 4.1) = fc) c < c < b) fx)gx)dx = fc) gx)dx 4.1 4
5 R n = f n) b t)n 1 t) n 1)! dt = b c)n 1 b ) f n c) n 1)! < c < b) 5 R n+1 = 1) n b ) n+1 φt)f n+1) + tb ))dt = 1) n b ) n+1 φc)f n+1) + cb )) < c < 1) φt) = t m t 1) m n = m) φ n) )fb) f)) = b ){φ n 1) 1)f b) φ n 1) )f )} b ) {φ n ) 1)f ) b) φ n ) )f ) )} + + 1) n k 1 b ) n k {φ k) 1)f n k) b) φ k) )f n k) )} + + 1) n 1 b ) n {φ1)f n) b) φ)f n) )} + R n+1 φ k) ) φ k) 1) ) k fg) k) = f k) g + kf k 1) g + + f k j) g j) + + fg k) j { ft) = t m gt) = t 1) m f k) t) = { m! k = m) k m) g k) t) = mm 1) m k + 1)t 1) m k) 1) m k m t = ) k φ k) ) = fg) k) ) = f m) )g k m) ) m ) k m!) = 1) m k m m k)! t = 1 φ k) 1) = fg) k) 1) = ) k m!) m m k)! 5
6 ) m k m fg) k) ) = fg) k) 1) = fb) = f) + m k=1 φ m) )fb) f)) = m)!fb) f)) ) φ 1) k 1 b ) k m k 1) φ m ) f k) b) φm k ) φ m ) f k) ) k=1 + R m+1 m)! m k m φ m k) ) = φ m k) = m ) m k)! k fb) = f) + f k) ) 1) k f k) b))b ) k + R m+1 m)! m m)! R m+1 m)! = 1)m b )m+1 m)! t m t 1) m f m+1) + tb ))dt { gt) = t m t 1) m F t) = f m+1) + tb )) 4.1 { m gt) m gt) R m+1 m)! = 1 m)! b )m+1 f m+1) + cb )) t m t 1) m dt m [ t t m t 1) m m+1 dt = t 1)m m + 1 ] 1 t m t 1) m dt m m + 1 tm+1 t 1) m 1 dt = m t m+1 t 1) m 1 dt m + 1 mm 1) 1 = t m+ t 1) m dt m + 1)m + ). = 1) m 1 mm 1)m ) 3 m + 1)m + ) m 1)m = 1) m mm 1)m ) 3 m + 1)m + ) m)m + 1) t m dt 6
7 m! R m+1 = 1) m m!) m + 1)! m)! = 1)m b ) m+1 f m+1) + cb )) < c < 1) m)!m + 1)! 5.1 f, b n fb) = f) + m k=1 m k)! m)! ) k f k) ) 1) k f k) b))b ) k + R m+1 m m)! R m+1 = 1)m b ) m+1 f m+1) + cb ))dt < c < 1) m + 1)! 6, 3 φt) = t m t 1) m n = m) m = 1 φt) = tt 1) 5.1 R 3 fb) f) = 1 b )f b) + f )) + R 3 fb) = f) + 1 b )f b) + f )) + R 3 6.1) R 3 = b ) 3 t t)f 3) + tb ))dt 6.1) f) = p+1 fb) = b p+1 p) b p p b p+1 = p+1 + b )p + 1)bp + p ) ) p + 1)bp p + 1)bp = + R 3 + p+1 1 p + 1 b p+1 p 1) + p + 1)b p = p + 1)b p + p 1) p+1 R 3 ) + R 3 b = p + 1)bp + p 1) p+1 R 3 + p 1)b p + p + 1) p p 1)b p + p + 1) 6.) p 7
8 1) b =, p = 6.) = R R 3 n+1 = 6 n + n n 1 = 1 { n } 61) + 1)3 = = ) 5 = /5) + 7/5)3 3 = = /5) 4 = = ) 3 1) b = 3, p = 6.) 3 = R R 3 n+1 = 9 n + n n 1 = 1 { n } 3 91) + 1)3 = = ) 6 = /6) + 1/6)3 3 = = /6) 4 = =
9 7 e e e lim n = e.71) n n) : n = ) n nk ) n! = n n k)!k! n = 1 + n 1 n nn 1) 1 ) nn 1)n ) 1 ) ! n 3! n nn 1) 1 1 n + n! n) = ) 1 n! ) 1 1 n n) 3! ) 1 ) 1 n 1 ) 1 n n n n! n+1 = n + 1 n + 1 ) 1! ) 1 n + 1 ) ) 1 1 n n + 1 n + 1 n + 1 ) ) 1 3! + 1 n + 1)! 7.1) 1 k! n+1 n+1 1 n + 1)! n < n+1 { n } 7.1) 1 1 ), 1 ),, 1 k ) n n n n < ! + 1 3! n! 1! = 1 1 3! = 1 1 ) 3 <, 1 n! = 1 1 ) n 1 3 n < n < ) ) n 1 + = = 3 9
10 n < 3 { n } { n } n < 3 n 1 + n) 1 n = e < 3 lim n φt) = t m t 1) m n = m) e 1) m = φt) = t t 1) 5.1 fb) = f) {b )1f b) + 1f )) b ) f b) f ))} + R ) R 5 R 5 = b )5 t t 1) f 5) + tb ))dt 7.) fx) = e x = e b b = 1 e b = e {b1eb + 1e ) b e b e )} + R 5 4 e b = 1 + 6b + b 1 6b + b + R 5 4 1b + b e = R = R 5 14 = R 5 14 R 5 = t t 1) e t dt < t < 1 e t < e R 5 < e t t 1) dt = e 3 e < 3 R 5 < 1 1 R 5 14 < < 1 1
11 ) m = 3 φt) = t 3 t 1) fb) = f) {b )36f b) + 36f )) b ) 7f b) 7f )) +b ) 3 6f b) + 6f ))} + R ) R 7 R 7 = b )7 t 3 t 1) 3 f 7) + tb ))dt 7.3) fx) = e x = e b = e {b36eb + 36e ) b 7e b 7e ) + b 3 6e b + 6e )} + R 7 7 e b b = 1 e b = b + 7b + 6b b + 7b 6b 3 + R b + 7b 6b e = R = R 7 46 = R 7 46 R 7 = t 3 t 1) 3 e t dt < t < 1 e t < e R 7 < e t 3 t 1) 3 dt = e 1 14 = e 14 e < 3 R 7 < 3 14 R 7 46 < <
12 [1] E. W. Cheney nd T. H. Southrd, A survey of methods for rtionl pproximtion, with prticulr reference to new method bsed on formul of Drboux, SIAM Rev ) [] P. M. Hummel nd C. L. Seebeck, A generliztion of Tylor s expnsion, Amer. Mth. Monthly ) [3] [4] H. S. Wll, A modifiction of Newton s method, Amer. Mth. Monthly ), [5] E. T. Whittker nd G. N. Wtson, A Course of Modern Anlysis: 4th edition, Cmbridge University Press,
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A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
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December 28, 2018
e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................
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7-12.dvi
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C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1
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Chap9.dvi
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di-problem.dvi
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IS-LM (interest) 100 (net rate of interest) (rate of interest) ( ) = 100 (2.1) (gross rate of interest) ( ) = 100 (2.2)
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