q12 and q13 Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like e, or ln(2)? After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations thatmake up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let f(z) = e².Part (a)The tangent line to f = e² at z = 0 is L(z)The formula L(z) can be used to appriximate e since L(1) ≈ f(1) = e. In particular, L(1) = Part (b)The linearization of e* does not provide a good approximation to e since 1 is not very close to 0. To obtain a better approximation, we alter our approach a bit. Instead of using a straight line to approximate e, we put an appropriate bend in our estimating functionto make it better fit the graph of e² for z close to 0. With the linearization, we had both f(z) and f'(z) share the same value as the linearization at z = 0. We will now use a quadratic approximation P₂(z) to f(z) = e centered at z=0 which has the property thatP₂(0) = f(0), P₂(0) = f'(0), and P(0) = f"(0).(i) Let P₂(z)=1+z+. Compute the following:1.P₂(0) =P'(0) =P" (0)this should equal f(0) =this should equal f'(0) =, this should equal f"(0):1.1.1.1.;Now that you have shown the equalities above, use P₂(z) to approximate e by observing that P₂(1) ≈ f(1). In particular P₂(1) =(ii) We can continue approximating e with polynomials of larger degree whose higher derivatives agree with those of f at 0. This turns out to make the polynomials fit the graph of f better for more values of z around 0. For example, let P3(z)=1+2+2 +3.Show that P3(0) = f(0), P'(0) = f'(0), P″(0) = ƒ"(0), and P!" (0) = f" (0) by completing the following:P3(0) =P'(0) =4, f(0) =&, f'(0) =f" (0)P" (0) =Now use P3(z) to approximate e in a way similar to how you did so with P₂(z) above. In particular, P3(1)

Question

q12  and  q13 Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like e, or ln(2)? After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations thatmake up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let f(z) = e².Part (a)The tangent line to f = e² at z = 0 is L(z)The formula L(z) can be used to appriximate e since L(1) ≈ f(1) = e. In particular, L(1) = Part (b)The linearization of e* does not provide a good approximation to e since 1 is not very close to 0. To obtain a better approximation, we alter our approach a bit. Instead of using a straight line to approximate e, we put an appropriate bend in our estimating functionto make it better fit the graph of e² for z close to 0. With the linearization, we had both f(z) and f'(z) share the same value as the linearization at z = 0. We will now use a quadratic approximation P₂(z) to f(z) = e centered at z=0 which has the property thatP₂(0) = f(0), P₂(0) = f'(0), and P(0) = f&#34;(0).(i) Let P₂(z)=1+z+. Compute the following:1.P₂(0) =P'(0) =P&#34; (0)this should equal f(0) =this should equal f'(0) =, this should equal f&#34;(0):1.1.1.1.;Now that you have shown the equalities above, use P₂(z) to approximate e by observing that P₂(1) ≈ f(1). In particular P₂(1) =(ii) We can continue approximating e with polynomials of larger degree whose higher derivatives agree with those of f at 0. This turns out to make the polynomials fit the graph of f better for more values of z around 0. For example, let P3(z)=1+2+2 +3.Show that P3(0) = f(0), P'(0) = f'(0), P″(0) = ƒ&#34;(0), and P!&#34; (0) = f&#34; (0) by completing the following:P3(0) =P'(0) =4, f(0) =&, f'(0) =f&#34; (0)P&#34; (0) =Now use P3(z) to approximate e in a way similar to how you did so with P₂(z) above. In particular, P3(1)

Accepted Answer

Please check step 2 for solution.!