Chapter 1.5, Problem 42E

Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Could you explain this using the formula I attached and polar coorindates

Could you explain this using the formula I attached and polar coordinates