‫‪ -5‬ﺍﻟﻨﺸﺮ ﺍﶈﺪﻭﺩ‬
‫‪ – 1.5‬ﺗﻌﺎﺭﻳﻒ ‪:‬‬
‫‪ -1‬ﺍﻟﺪﺍﻟﺔ ‪ f‬ﻣﻌﺮﻓﺔ ﰲ ﺍﳉﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ ﺍﻟﺼﻔﺮ‪،‬ﺃﻭ ﺣﱴ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ﻟﻠﺼﻔﺮ‪،‬ﻧﻘﻮﻝ ﺇﻥ ﳍﺎ ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍ‪)،‬ﻣﻨﺘﻪ(‪،‬‬
‫ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪، n‬ﺇﺫﺍ ﻭﺟﺪ ﻛﺜﲑ ﺣﺪﻭﺩ )‪ Ρ(x‬ﺩﺭﺟﺘﻪ ﺃﻗﻞ ﺃﻭ ﺗﺴﺎﻭﻱ ‪ n‬ﲝﻴﺚ ﺗﻜﻮﻥ ‪:‬‬
‫∞→‪x‬‬
‫) (‬
‫‪, y→0‬‬
‫) (‬
‫‪f ( x) = Ρ ( x) + 0 x n ,‬‬
‫‪ -2‬ﺍﻟﺪﺍﻟﺔ )‪ f (x‬ﺍﳌﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ ‪، x0‬ﺃﻭ ﺣﱴ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ ‪ ، x0‬ﻧﻘﻮﻝ ﺇﻥ ﳍﺎ ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍ ﰲ‬
‫ﺟﻮﺍﺭ ‪ x0‬ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪، n‬ﺇﺫﺍ ﻛﺎﻥ ﻟﻠﺪﺍﻟﺔ )‪، g (x‬ﺣﻴﺚ )‪ g (x) = f (x0 + y‬ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪، n‬ﺃﻱ‬
‫ﻋﻨﺪﻫﺎ ﻧﻜﺘﺐ‪:‬‬
‫)‬
‫‪, x → x0‬‬
‫ﺫﻟﻚ ﻷﻥ ) ‪(x → x0 ) ⇔ ( y → 0‬‬
‫‪n‬‬
‫‪g ( y) = Ρ( y) + 0 y n‬‬
‫(‬
‫) ‪f ( x) = f ( x0 + y) = g ( y) = Ρ( x − x0 ) + 0 ( x − x0‬‬
‫‪n‬‬
‫‪x = y + x0 ,‬‬
‫) ‪Ρ( x − x0 ) = a 0 + a 1 ( x − x0 ) + ...... + a n ( x − x0‬‬
‫‪ -3‬ﺍﻟﺪﺍﻟﺔ )‪ f (x‬ﺍﳌﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎﻟﻠﻤﺎﻻ ‪‬ﺎﻳﺔ‪،‬ﻧﻘﻮﻝ ﺇﻥ ﳍﺎ ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍﹰ ﰲ ﻫﺬﺍ ﺍﳉﻮﺍﺭ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪ ، n‬ﺇﺫﺍ ﻛﺎﻧﺖ‬
‫ﻟﻠﺪﺍﻟﺔ )‪، g ( y‬ﺣﻴﺚ‬
‫‪1‬‬
‫‪f  ‬‬
‫‪ y‬‬
‫‪1‬‬
‫= )‪, g ( y‬‬
‫‪y‬‬
‫=‪x‬‬
‫ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪ ، n‬ﺃﻱ‬
‫‪, y→0‬‬
‫) ‪g ( y ) = Ρ ( y ) + 0( y‬‬
‫ﻋﻨﺪﻫﺎ ﻧﻜﺘﺐ‪:‬‬
‫‪n‬‬
‫‪1‬‬
‫‪ 1   1  ‬‬
‫‪f ( x) = f   = g ( y) = Ρ  + 0   ‬‬
‫‪ x    x  ‬‬
‫‪ y‬‬
‫∞→‪, x‬‬
‫ﺃﻱ ﺃﻥ‬
‫∞→‪, x‬‬
‫ﺃﻣﺜﻠﺔ ‪:‬‬
‫‪ -1‬ﺍﻟﺪﺍﻟﺔ‬
‫‪f‬‬
‫ﺣﻴﺚ‬
‫‪1‬‬
‫‪1− x‬‬
‫‪a‬‬
‫‪a1‬‬
‫‪ 1 ‬‬
‫‪+ ..... + nn + 0 n ‬‬
‫‪x‬‬
‫‪x‬‬
‫‪x ‬‬
‫‪f ( x) = a 0 +‬‬
‫= )‪ ، (x ≠ 1) ، f (x‬ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‬
‫ﻧﻘﺴﻢ ‪ 1‬ﻋﻠﻰ ‪ 1 − x‬ﻭﻓﻘﺎ ﻟﺘﺰﺍﻳﺪ ﺍﻷﺱ ﺣﱴ ﺍﻟﺪﺭﺟﺔ ‪ ، n‬ﳓﺼﻞ ﻋﻠﻰ‬
‫‪1‬‬
‫‪x n +1‬‬
‫‪= 1 + x + x 2 + ........ + x n +‬‬
‫‪1− x‬‬
‫‪1− x‬‬
‫ﲟﺎ ﺃﻥ‬
‫‪x n +1‬‬
‫‪x‬‬
‫‪lim n = lim‬‬
‫‪=0‬‬
‫‪x→ 0 x‬‬
‫‪x→ 0 1 − x‬‬
‫‪،‬ﻓﺈﻥ‬
‫‪1‬‬
‫‪x n +1‬‬
‫‪= 0 xn‬‬
‫‪1− x‬‬
‫) (‬
‫‪, x→0‬‬
‫ﻭﻣﻨﻪ ﻭﺟﺪ ﻛﺜﲑ ﺣﺪﻭﺩ )‪، Ρ(x‬ﺣﻴﺚ‬
‫‪Ρ( x) = 1 + x + x 2 + ....... + x n‬‬
‫ﻣﻦ ﺃﺟﻠﻪ ﻳﻜﻮﻥ‬
‫) (‬
‫‪f ( x ) = Ρ ( x) + 0 x n‬‬
‫‪, x→0‬‬
‫ﻭﺑﺎﻟﺘﺎﱄ )‪ f (x‬ﳍﺎ ﻧﺸﺮ ﳏﺪﻭﺩ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ‪.‬‬
‫‪ -2‬ﺍﻟﺪﺍﻟﺔ ‪ ، f (x) = log1 + 1 ‬ﰲ ﺟﻮﺍﺭ ) ∞ ‪. (+‬‬
‫‪x‬‬
‫‪‬‬
‫ﺣﺴﺐ ﺍﻟﺘﻌﺮﻳﻒ ‪، 3‬ﳒﻌﻞ‬
‫‪1‬‬
‫)‪g ( y) = f   = log(1 + y‬‬
‫‪ y‬‬
‫ﺍﻟﺪﺍﻟﺔ )‪ g ( y‬ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‬
‫ﺣﺴﺐ ﺍﳌﺜﺎﻝ ‪ 4‬ﻓﻘﺮﺓ ‪، 3.4.5‬ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻳﻜﻮﻥ‬
‫) (‬
‫‪n‬‬
‫‪y 2 y3 y 4‬‬
‫‪n −1 y‬‬
‫‪+‬‬
‫‪−‬‬
‫)‪+ ...... + (− 1‬‬
‫‪+ 0 yn‬‬
‫‪n‬‬
‫‪2‬‬
‫‪3‬‬
‫‪4‬‬
‫‪g ( y) = log(1 + y) = y −‬‬
‫ﺃﻱ ﺃﻥ ‪ g‬ﻗﺎﺑﻠﺔ ﻟﻠﻨﺸﺮ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‪،‬ﻭﺑﺎﻟﺘﺎﱄ ﺣﺴﺐ ﺍﻟﺘﻌﺮﻳﻒ ‪، 3‬ﺍﻟﺪﺍﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ ﻟﻠﻨﺸﺮ ﰲ ﺟﻮﺍﺭ ) ∞ ‪ ، (+‬ﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ‬
‫∞‪, x → +‬‬
‫‪1‬‬
‫‪1‬‬
‫‪ 1 ‬‬
‫‪n −1 1‬‬
‫)‪− 2 + ...... + (− 1‬‬
‫‪+ 0 n ‬‬
‫‪n‬‬
‫‪x 2x‬‬
‫‪nx‬‬
‫‪x ‬‬
‫= )‪f ( x‬‬
‫‪ -3‬ﺍﻟﺪﺍﻟﺔ ‪، f (x) = e x‬ﰲ ﺟﻮﺍﺭ ﺍﻟﻮﺍﺣﺪ‬
‫‪g ( y) = e ⋅ e y‬‬
‫ﻧﻀﻊ ‪ ، g ( y) = f ( y + 1) = e y+1 , y = x − 1‬ﺃﻱ‬
‫ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‪،‬ﺃﻱ ﻋﻨﺪﻣﺎ ‪، y → 0‬ﺣﺴﺐ ﺍﳌﺜﺎﻝ ‪ 1‬ﻓﻘﺮﺓ ‪، 3.4.5‬ﻳﻜﻮﻥ‬
‫‪, y→0‬‬
‫‪ y y2‬‬
‫‪‬‬
‫‪yn‬‬
‫‪g ( y) = e ⋅ e y = e 1 + +‬‬
‫‪+ ...... +‬‬
‫‪+ 0 yn ‬‬
‫!‪n‬‬
‫!‪ 1! 2‬‬
‫‪‬‬
‫) (‬
‫ﺃﻱ ﺃﻥ ‪ g‬ﻗﺎﺑﻠﺔ ﻟﻠﻨﺸﺮ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‪،‬ﻭﺑﺎﻟﺘﺎﱄ ‪ f‬ﻗﺎﺑﻠﺔ ﻟﻠﻨﺸﺮ ﰲ ﺟﻮﺍﺭ ﺍﻟﻮﺍﺣﺪ ‪ ،‬ﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ‬
‫‪2‬‬
‫‪n‬‬
‫‪‬‬
‫(‬
‫(‬
‫)‪x − 1‬‬
‫)‪x − 1‬‬
‫‪n ‬‬
‫‪, x →1‬‬
‫‪+ ....... +‬‬
‫) )‪+ 0(( x − 1‬‬
‫‪f ( x) = e 1 + ( x − 1) +‬‬
‫‪‬‬
‫‪‬‬
‫!‪n‬‬
‫!‪2‬‬
‫‪‬‬
‫‪‬‬
‫ﻣﻼﺣﻈﺔ ‪ :‬ﻛﺜﲑ ﺍﳊﺪﻭﺩ )‪ Ρ(x‬ﰲ ﺍﻟﺘﻌﺎﺭﻳﻒ ﺍﻟﺴﺎﺑﻘﺔ‪،‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ‪،‬ﻳﺴﻤﻰ ﺍﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻠﺪﺍﻟﺔ‬
‫‪x→0‬‬
‫‪x → x0 ,‬‬
‫ﻧﺘﺎﺋﺞ ‪:‬‬
‫‪x → ∞,‬‬
‫‪ -1‬ﺍﻟﻨﺸﺮ ﰲ ﺍﳌﺎﻻ ‪‬ﺎﻳﺔ ﺃﻭ ﰲ ﻧﻘﻄﺔ ‪، (x0 ≠ 0) ، x0‬ﻳﺆﻭﻝ ﺇﱃ ﺍﻟﻨﺸﺮ ﰲ ﺍﻟﺼﻔﺮ‪.‬‬
‫‪2‬‬
‫‪f‬‬
‫ﻋﻨﺪﻣﺎ‬
‫‪ -2‬ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ ﻟﻠﻨﺸﺮ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪، n‬ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻭﺟﺪ ﻛﺜﲑ ﺣﺪﻭﺩ )‪ Ρ(x‬ﻣﻦ ﺍﻟﺪﺭﺟﺔ ﺃﻗﻞ‬
‫‪، lim‬ﻭ‬
‫ﺃﻭ ﺗﺴﺎﻭﻱ ‪، n‬ﻭﺩﺍﻟﺔ )‪ α ( x‬ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﺃﻭ ﺟﻮﺍﺭ ﻣﺜﻘﻮﺏ ﻟﻠﺼﻔﺮ ﲢﻘﻖ ‪α ( x) = 0‬‬
‫‪x→ 0‬‬
‫)‪f ( x) = Ρ( x) + x nα ( x‬‬
‫‪ -3‬ﺇﺫﺍ ﻛﺎﻧﺖ ﻟﻠﺪﺍﻟﺔ ‪ f‬ﻧﺸﺮﺍ ﳏﺪﻭﺩ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪، n‬ﻓﺈﻧﻪ ﻭﺣﻴﺪ‪.‬‬
‫‪ -4‬ﺇﺫﺍ ﻛﺎﻧﺖ ‪ f‬ﲤﻠﻚ ﻧﺸﺮﺍ ﳏﺪﻭﺩ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪، n‬ﻓﺈ‪‬ﺎ ﲤﻠﻚ ﻧﺸﺮﺍ ﳏﺪﻭﺩ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ‬
‫‪ k‬ﻣﻦ ﺃﺟﻞ ‪. k ≤ n‬‬
‫‪ -5‬ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ ‪ f‬ﻓﺮﺩﻳﺔ‪)،‬ﺯﻭﺟﻴﺔ(‪،‬ﻭﲤﻠﻚ ﻧﺸﺮﺍ ﳏﺪﻭﺩ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‪،‬ﻓﺈﻥ ﺟﺰﺋﻬﺎ ﺍﻟﺮﺋﻴﺴﻲ ﻳﻜﻮﻥ ﻓﺮﺩﻳﺎﹰ‪)،‬ﺯﻭﺟﻴﺎﹰ(‪.‬‬
‫‪ f ( x) = Ρ( x) + 0(xn ) , x → 0 ) -6‬ﺣﻴﺚ ‪⇔ ( Ρ( x) = a 0 + a 1 x + ..... + a n x n‬‬
‫‪‬‬
‫‪‬‬
‫‪f ( x) − a 0 − a 1 x...... − a n −1 x n −1‬‬
‫‪f ( x) − a 0‬‬
‫‪ a n = lim‬‬
‫‪‬‬
‫(‬
‫)‬
‫‪a‬‬
‫‪,‬‬
‫‪lim‬‬
‫‪,‬‬
‫‪lim‬‬
‫‪a‬‬
‫‪f‬‬
‫‪x‬‬
‫=‬
‫=‬
‫‪0‬‬
‫‪1‬‬
‫‪n‬‬
‫‪‬‬
‫‪‬‬
‫‪x→ 0‬‬
‫‪x→ 0‬‬
‫‪x→ 0‬‬
‫ﺍﻟﱪﻫﺎﻥ ‪:‬‬
‫‪‬‬
‫‪x‬‬
‫‪‬‬
‫‪x‬‬
‫‪ -1‬ﻭﺍﺿﺢ ﻣﻦ ﺍﻟﺘﻌﺮﻳﻒ ‪. 1،2،3‬‬
‫‪ -2‬ﻭﺍﺿﺢ ﻣﻦ ﺗﻌﺮﻳﻒ ﺍﻟﺼﻔﺮ ﺍﻟﺼﻐﲑ ‪،‬ﻓﻘﺮﺓ ‪. 2.9.3‬‬
‫‪ -3‬ﻧﻔﺮﺽ ﺃﻥ ‪ f‬ﲤﻠﻚ ﻧﺸﺮﻳﻦ ﳏﺪﻭﺩﻳﻦ‪،‬ﺃﻱ ‪:‬‬
‫‪, x→0‬‬
‫‪, x→0‬‬
‫) (‬
‫) ‪+ ο (x‬‬
‫) (‬
‫‪f ( x) = g ( x) + ο (x ) = b‬‬
‫‪f ( x) = Ρ( x) + ο x n = a 0 + a 1 x + ........ + a n x n + ο x n‬‬
‫‪n‬‬
‫‪+ b1 x + ........ + bn x n‬‬
‫‪n‬‬
‫‪0‬‬
‫ﻧﱪﻫﻦ ﺃﻥ )‪، Ρ(x) = g (x‬ﺃﻱ ‪:‬‬
‫) ‪a 0 + a 1 x + ........ + a n x n + ο (x n ) = b0 + b1 x + ........ + bn x n + ο (x n‬‬
‫ﻋﻨﺪﻧﺎ‬
‫ﻭﻣﻨﻪ‬
‫) ‪(a 0 − b0 ) + (a 1 − b1 )x + ......(a n − bn )x n = ο (x n ) − ο (x n ) = ο (x n‬‬
‫‪∀i = 0,1........, n, a i = bi‬‬
‫ﻋﻨﺪﻣﺎ ‪ x → 0‬ﳓﺼﻞ ﻋﻠﻰ ‪، a 0 = b0‬ﻭﻳﺒﻘﻰ‬
‫‪x→0‬‬
‫‪,‬‬
‫) ‪(a 1 − b1 )x + ......(a n − bn )x n = ο (x n‬‬
‫)‬
‫‪n‬‬
‫‪(a 1 − b1 ) + ......(a n − bn )x n −1 = ο (x‬‬
‫)‪, (x ≠ 0‬‬
‫ﺃﻱ‬
‫‪x‬‬
‫‪ο xn‬‬
‫‪. lim‬‬
‫ﻋﻨﺪﻣﺎ ‪ x → 0‬ﳓﺼﻞ ﻋﻠﻰ ‪، a 1 = b1‬ﻷﻥ ‪= 0‬‬
‫‪x→ 0‬‬
‫‪x‬‬
‫‪n‬‬
‫ﻭﻫﻜﺬﺍ ﻧﻜﺮﺍﺭ ﺍﻟﻌﻤﻠﻴﺔ ﳓﺼﻞ ﻋﻠﻰ ‪، a n − bn = ο xn‬ﻭﻣﻨﻪ ‪، a n = bn‬‬
‫‪x‬‬
‫) (‬
‫) (‬
‫ﻭﺑﺎﻟﺘﺎﱄ ﻳﻜﻮﻥ )‪، Ρ(x) = g (x‬ﺃﻱ ﺍﻟﻨﺸﺮ ﻭﺣﻴﺪ‪.‬‬
‫‪ -4‬ﻋﻨﺪﻧﺎ‬
‫‪3‬‬
‫ﻷﻥ ‪) = 0‬‬
‫‪ο (x‬‬
‫‪. lim‬‬
‫‪n‬‬
‫‪x→ 0‬‬
‫‪n‬‬
‫‪x‬‬
‫) (‬
‫) (‬
‫‪f ( x) = Ρ( x) + ο x n = a 0 + a 1 x + ...... + a n x n + ο x n‬‬
‫ﲟﺎ ﺃﻥ ‪، k < n‬ﻓﺈﻥ‬
‫‪,‬‬
‫‪x→0‬‬
‫) (‬
‫‪f ( x) = a 0 + a 1 x + ...... + a k x k + a k +1 x k +1 + ....... + a n x n + ο x n‬‬
‫ﻻﺣﻆ ﺃﻥ‬
‫) (‬
‫‪a k +1 x k +1 + ....... + a n x n + ο x n‬‬
‫‪lim‬‬
‫‪=0‬‬
‫‪x→ 0‬‬
‫‪xk‬‬
‫ﻭﺑﺎﻟﺘﺎﱄ ﻳﻜﻮﻥ‬
‫‪x→0‬‬
‫) (‬
‫‪f ( x) = a 0 + a 1 x + ...... + a k x k + ο x k‬‬
‫‪,‬‬
‫ﺃﻱ ﺃﻥ ﻟﻠﺪﺍﻟﺔ ‪ f‬ﻧﺸﺮ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ‪. k‬‬
‫‪ -5‬ﻋﻨﺪﻧﺎ ‪ f‬ﺯﻭﺟﻴﺔ ﻭﻗﺎﺑﻠﺔ ﻟﻠﻨﺸﺮ‪ ،‬ﺃﻱ )‪، f (x) = f (− x‬ﻭ‬
‫‪x→0‬‬
‫ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ‬
‫) (‬
‫‪f ( x) = a 0 + a 1 x + ...... + a n x n + ο x n‬‬
‫‪,‬‬
‫) (‬
‫‪f (− x) = a 0 − a 1 x + a 2 x − a 3 x + ...... + (− 1) a n x n + ο x n‬‬
‫‪n‬‬
‫ﻭﻣﻨﻪ ﺣﺴﺐ ﻭﺣﺪﺍﻧﻴﺔ ﺍﻟﻨﺸﺮ‪،‬ﻧﺴﺘﻨﺘﺞ ﺃﻥ‬
‫‪a 0 = a 0 , a1 = −a1 , a 2 = a 2 , a 3 = −a 3 .......... a n = (− 1) a n‬‬
‫‪n‬‬
‫ﺃﻱ ﺃﻥ ‪، a n = 0‬ﻣﻦ ﺃﺟﻞ‬
‫ﻭﺑﺎﻟﺘﺎﱄ ﻳﻜﻮﻥ‬
‫‪k = 0,1......., n = 2k + 1‬‬
‫) (‬
‫‪f ( x) = a 0 + a 2 x 2 + ..........a 2 k x 2 k + ο x 2 k‬‬
‫‪, x→0‬‬
‫ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﰲ ﺣﺎﻟﺔ‬
‫‪f‬‬
‫ﻓﺮﺩﻳﺔ ﳓﺼﻞ ﻋﻠﻰ‬
‫‪, x→0‬‬
‫)‬
‫(‬
‫‪f ( x) = a 1 x + a 3 x 3 + ..........a 2 k +1 x 2 k +1 + ο x 2 k +1‬‬
‫ﻭﺑﺎﻟﺘﺎﱄ ﻧﻘﻮﻝ ‪:‬‬
‫ﺃ‪ -‬ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ‬
‫‪f‬‬
‫ﺯﻭﺟﻴﺔ ‪،‬ﻓﺈﻥ ﻧﺸﺮﻫﺎ ﺍﶈﺪﻭﺩ ﻣﻦ ﺍﻟﺪﺭﺟﺔ‬
‫) (‬
‫‪f ( x) = a 0 + a 2 x 2 + ..........a 2 n x 2 n + ο x 2 n‬‬
‫‪, x→0‬‬
‫ﺏ‪ -‬ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ‬
‫ﻓﺮﺩﻳﺔ‪ ،‬ﻓﺈﻥ ﻧﺸﺮﻫﺎ ﺍﶈﺪﻭﺩ ﻣﻦ ﺍﻟﺪﺭﺟﺔ‬
‫‪f‬‬
‫‪, x→0‬‬
‫‪ -6‬ﻭﺍﺿﺢ ﻷﻥ‬
‫) (‬
‫‪2n‬‬
‫ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻫﻮ‬
‫)‬
‫‪2n + 1‬‬
‫(‬
‫ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ﻫﻮ‬
‫‪f ( x) = a 1 x + a 3 x 3 + ..........a 2 n +1 x 2 n +1 + ο x 2 n +1‬‬
‫) (‬
‫‪ο xn‬‬
‫‪ο xn‬‬
‫‪= ........ = lim n = 0‬‬
‫‪x→ 0‬‬
‫‪x→ 0‬‬
‫‪x‬‬
‫‪x‬‬
‫) (‬
‫‪lim ο x n = lim‬‬
‫‪4‬‬
‫‪x→ 0‬‬
‫‪.‬‬
‫ﻗﻀﻴﺔ ‪ :‬ﺇﺫﺍ ﻛﺎﻥ ﺍﳌﺸﺘﻖ ﻣﻦ ﺍﻟﺪﺭﺟﺔ‬
‫‪n‬‬
‫ﻟﻠﺪﺍﻟﺔ‬
‫ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ ‪،‬ﺃﻱ‬
‫) (‬
‫‪f ( x) = a 0 + a 1 x + ..........a n x n + ο x n‬‬
‫‪, x→0‬‬
‫ﺍﻟﺮﻫﺎﻥ ‪:‬‬
‫) ‪f ( n ) (0‬‬
‫ﻭﻣﻨﻪ ﻳﻜﻔﻲ ﺃﺧﺬ‬
‫‪f‬‬
‫ﰲ ﺍﻟﺼﻔﺮ ﻣﻮﺟﻮﺩ‪ ،‬ﻓﺈﻥ ﺍﻟﺪﺍﻟﺔ‬
‫‪f‬‬
‫ﲤﻠﻚ ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍ ﻣﻦ ﺍﻟﺪﺭﺟﺔ‬
‫ﻣﻮﺟﻮﺩ ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﻓﻘﺮﺓ ‪، 2.4.5‬ﺍﻟﺪﺍﻟﺔ ‪ f‬ﺗﻜﺘﺐ ﻛﺎﻟﺘﺎﱄ ‪:‬‬
‫) ‪f ′(0‬‬
‫‪f (n ) (0 ) n‬‬
‫‪f ( x) = f (0 ) +‬‬
‫‪x + ......... +‬‬
‫) ‪x + ο (x n‬‬
‫‪, x→0‬‬
‫!‪n‬‬
‫!‪1‬‬
‫)‪f ′(0‬‬
‫) ‪f (n ) (0‬‬
‫= ‪............., a n‬‬
‫!‪n‬‬
‫!‪1‬‬
‫ﺃﻱ ﻧﺄﺧﺬ )‪، Ρ(x‬ﻫﻮ ﻛﺜﲑ ﺣﺪﻭﺩ ﺗﺎﻳﻠﻮﺭ )‪ ϕ (x,0‬ﻟﻠﺪﺍﻟﺔ‬
‫‪, x→0‬‬
‫ﺃﻣﺜﻠﺔ ‪ :‬ﺻﻴﻐﺔ ﻣﺎﻛﻠﻮﺭﺍﻥ ﺑﺒﺎﻗﻲ ﺑﻴﺎﻧﻮ ﻟﻠﺪﻭﺍﻝ‬
‫‪f‬‬
‫= ‪a1‬‬
‫‪a 0 = f (0) ,‬‬
‫ﻭﻧﻜﺘﺐ‬
‫) (‬
‫‪f ( x) = Ρ ( x) + ο x n‬‬
‫‪, ln(1 + x) , cos x , sin x , e x‬‬
‫ﺍ‪‬ﻮﺩﺓ ﰲ ﺍﻷﻣﺜﻠﺔ ‪ 5،4،3،2،1‬ﻓﻘﺮﺓ ‪ ،3.4.5‬ﲤﺜﻞ ﻧﺸﺮﺍ ﳏﺪﻭﺩﺍ ﳍﺎ ﰲ ﺟﻮﺍﺭ ﺍﻟﺼﻔﺮ‪.‬‬
‫‪5‬‬
‫‪(1 + x)α‬‬
‫‪α ∈ℜ‬‬
‫‪n‬‬