Answer to Question #305134 in Calculus for Talal

Question #305134

Write down 𝑇3(𝑥), 𝑇4(𝑥), 𝑎𝑛𝑑 𝑇5(𝑥) for the Taylor series of 𝑓(𝑥) = ln (3 + 4𝑥) about 𝑥 = 0


1
Expert's answer
2022-03-03T13:50:37-0500
f(x)=ln(3+4x)f(x)=\ln(3+4x)


f(0)=ln(3+4(0))=ln3f(0)=\ln(3+4(0))=\ln3

f(x)=43+4xf'(x)=\dfrac{4}{3+4x}

f(0)=43+4(0)=43f'(0)=\dfrac{4}{3+4(0)}=\dfrac{4}{3}


f(x)=16(3+4x)2f''(x)=-\dfrac{16}{(3+4x)^2}

f(0)=16(3+4(0)2=169f''(0)=-\dfrac{16}{(3+4(0)^2}=-\dfrac{16}{9}

f(x)=128(3+4x)3f'''(x)=\dfrac{128}{(3+4x)^3}

f(0)=128(3+4(0))3=12827f'''(0)=\dfrac{128}{(3+4(0))^3}=\dfrac{128}{27}

f(4)(x)=1536(3+4x)4f^{(4)}(x)=-\dfrac{1536}{(3+4x)^4}

f(4)(0)=1536(3+4(0))4=51227f^{(4)}(0)=-\dfrac{1536}{(3+4(0))^4}=-\dfrac{512}{27}

f(5)(x)=24576(3+4x)5f^{(5)}(x)=\dfrac{24576}{(3+4x)^5}

f(5)(0)=24576(3+4(0))5=819281f^{(5)}(0)=\dfrac{24576}{(3+4(0))^5}=\dfrac{8192}{81}

T3(x)=ln3+43x89x2+6481x3T_3(x)=\ln3+\dfrac{4}{3}x-\dfrac{8}{9}x^2+\dfrac{64}{81}x^3



T4(x)=ln3+43x89x2+6481x36481x4T_4(x)=\ln3+\dfrac{4}{3}x-\dfrac{8}{9}x^2+\dfrac{64}{81}x^3-\dfrac{64}{81}x^4



T5(x)=ln3+43x89x2+6481x36481x4T_5(x)=\ln3+\dfrac{4}{3}x-\dfrac{8}{9}x^2+\dfrac{64}{81}x^3-\dfrac{64}{81}x^4

+10241215x5+\dfrac{1024}{1215}x^5


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