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Maclaurin and Taylor Series
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Maclaurin Taylor Series
KUS objectives BAT Adapt the Maclaurin series to find solutions of differential equations WB16: Given that π¦=π₯ π π₯ a) Show that π π π¦ π π₯ π = π+π₯ π π₯ b) Find the Taylor expansion of π¦=π₯ π π₯ in ascending powers of π₯+1 as far as the term in π₯+1 4
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π π₯ =π₯ π π₯ = 1 π β1+ 1 2 π₯+1 2 + 1 3 π₯+1 3 + 1 8 π₯+1 4 + β¦
WB16 solutions π π₯ =π₯ π π₯ π βπ =β π π π β² π₯ =π₯ π π₯ + π π₯ = 1+π₯ π π₯ π β² (βπ)=π π β²β² (βπ)= π π π β²β² π₯ = π π₯ + 1+π₯ π π₯ = 2+π₯ π π₯ β¦ β¦ π π (βπ)= πβπ π π π π₯ = π+π₯ π π₯ π π =π π + π β² π (πβπ)+ πβ²β²(π) π! (πβπ) π + πβ²β²β²(π) π! (πβπ) π β¦ π π₯ =π₯ π π₯ =β 1 π + π₯+1 Γ0+ π₯ ! Γ 1 π π₯ ! Γ 2 π + π₯ ! Γ 3 π +β¦ π π₯ =π₯ π π₯ = 1 π β π₯ π₯ π₯ β¦
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Using the alternative notation for evaluated derivatives :
Notes The Maclaurin series for f(x) can be adapted to find solutions of differential equations Maclaurin series: Using the alternative notation for evaluated derivatives : By substituting given values into a differential equation, and differentiating the DE to find higher order derivatives, the coefficients can be found and βplugged intoβ the standard form above In fact, differential equations that cannot be solved using the methods seen already for 1st and 2nd order DEs can be solved in this way. Also, higher order differential equations involving, say, can also be solved.
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given that when π₯=0, π¦=1 , and ππ¦ ππ₯ =2
WB use the Taylor method to find a series solution, in ascending powers of x, up to and including the term in x3, of π 2 π¦ π π₯ 2 =π¦β sin π₯ (1) given that when π₯=0, π¦=1 , and ππ¦ ππ₯ =2 Substitute to (1) π₯=0, π¦=1 , and ππ¦ ππ₯ =2 to get π 2 π¦ π π₯ 2 = 1 β sin 0 =1 Differentiate (1) to get π 3 π¦ π π₯ 3 = ππ¦ ππ₯ β cos π₯ (π) Substitute to (2) π₯=0, π¦=1 , and ππ¦ ππ₯ =2 to get π 3 π¦ π π₯ 3 = 2 β πππ 0 =1 π π₯ =π 0 + π β² 0 π₯+ πβ²β²(0) 2! π₯ 2 + πβ²β²β²(0) 3! π₯ 3 β¦ Substitute into the standard form π¦=1+2π₯+ π₯ 2 2 Γ1+ π₯ 3 6 Γ1+β¦ π¦=1+2π₯+ π₯ π₯ β¦
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a) Differentiating original equation
WB π₯ ππ¦ ππ₯ =π₯+4 π¦ (1) a) Show that 1+2π₯ π 2 π¦ π π₯ 2 = π¦β1 ππ¦ ππ₯ (2) b) Differentiate equation (1) with respect to x to obtain an equation involving π 2 π¦ π π₯ π 2 π¦ π π₯ 2 , π 2 π¦ π π₯ 2 , x and y given that when π₯=0, π¦= 1 2 c) find a series solution for y, in ascending powers of x, up to and including the term in x3 a) Differentiating original equation b) Differentiating again:
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b) Differentiating again:
WB 18 b) Differentiating again: (3) Sub in (1) Sub in (2) Sub in (3) π π₯ =π 0 + π β² 0 π₯+ πβ²β²(0) 2! π₯ 2 + πβ²β²β²(0) 3! π₯ 3 β¦ c) Substitute into the standard form π¦= 1 2 +π₯+ π₯ 2 2 Γ3+ π₯ 3 6 Γ8+β¦ π¦= 1 2 +π₯ π₯ π₯ 3 +β¦
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One thing to improve is β
KUS objectives BAT Adapt the Maclaurin series to find solutions of differential equations self-assess One thing learned is β One thing to improve is β
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