10.17-bo'lim : Seriya ilovalari
3. \(x = 0\) haqidagi \(f\left( x \right) = \ln \left( {3 + 4x} \right)\) ning Teylor seriyasi uchun \({T_3}\left( x \right)\), \({T_4}\left( x \right)\) va \({T_5}\left( x \right)\) ni yozing. \(\left[ { - \frac{1}{2},2} \right]\) oralig'i uchun bir xil grafikda Teylor ko'phadlari va \(f\left( x \right)\)ning uchtasini chizing.
Yechim
Bu erda qilishimiz kerak bo'lgan birinchi narsa - \(x = 0\) haqida \(f\left( x \right) = \ln \left( {3 + 4x} \right)\) uchun Teylor seriyasini olish. Yaxshiyamki, biz buni oldingi bo'limning 4-muammosida qildik. Mana biz ushbu muammoda olingan Teylor seriyasi.
\[ln\left( {3 + 4x} \right) = \ln \left( 3 \right) + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{{\left( {\frac{4}{3}} \right)}^n}}}{n}{x^n}} \]2-qadam
Bu masala uchun zarur bo'lgan uchta Teylor ko'phadlari.
\[\begin{align*}{T_3}\left( x \right) & = \ln \left( 3 \right) + \frac{4}{3}x - \frac{8}{9}{x^2} + \frac{{64}}{{81}}{x^3}\\ {T_4}\left( x \right) & = \ln \left( 3 \right) + \frac{4}{3}x - \frac{8}{9}{x^2} + \frac{{64}}{{81}}{x^3} - \frac{{64}}{{81}}{x^4}\\ {T_5}\left( x \right) & = \ln \left( 3 \right) + \frac{4}{3}x - \frac{8}{9}{x^2} + \frac{{64}}{{81}}{x^3} - \frac{{64}}{{81}}{x^4} + \frac{{1024}}{{1215}}{x^5}\end{align*}\]3-qadam
Mana bu muammo uchun grafik.
![The graphs of \(f\left( x \right) = \ln \left( {3 + 4x} \right)\) and the Taylor polynomials \({T_3}\left( x \right)\), \({T_4}\left( x \right)\) and \({T_5}\left( x \right)\) on the interval \(\left[ { - \frac{1}{2},2} \right]\). All four graphs run together near \(x = 0\), but as \(x\) moves away from zero the polynomial graphs diverge from the logarithm graph, some curving sharply upward and others turning steeply downward.](/Solutions/CalcII/TaylorSeriesApps//Solutions/CalcII/TaylorSeriesApps/TaylorSeriesApps_Files/image002.webp)
Biz \(x = 0\) “yaqinida” qolsak, ko'phadning grafiklari eksponensial funktsiya grafigiga juda yaqin ekanligini ko'rishimiz mumkin. Ammo, agar biz juda uzoqqa borsak, grafiklar haqiqatan ham eksponensial funktsiya grafigidan uzoqlasha boshlaydi.