Answer to Question #305134 in Calculus for Talal

Question

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

Answer 1

$$f(x)=\ln(3+4x)$$

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

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

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

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

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

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

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

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

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

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

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

$$T_3(x)=\ln3+\dfrac{4}{3}x-\dfrac{8}{9}x^2+\dfrac{64}{81}x^3$$

$$T_4(x)=\ln3+\dfrac{4}{3}x-\dfrac{8}{9}x^2+\dfrac{64}{81}x^3-\dfrac{64}{81}x^4$$

$$T_5(x)=\ln3+\dfrac{4}{3}x-\dfrac{8}{9}x^2+\dfrac{64}{81}x^3-\dfrac{64}{81}x^4$$

$$+\dfrac{1024}{1215}x^5$$