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 function to make it better fit the graph of e 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 P(z) to f(z)= e centered at z=0 which has the property that P₂(0)=f(0), P,(0) = f'(0), and P(0) = f"(0). (i) Let Pa(z)=1+z+. Compute the following: P₂(0) = P/(0) = Pr(0) = , this should equal f(0) = ✓, this should equal f'(0) = , this should equal f"(0) = 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 Ps(z)=1+z+5 +6. Show that Ps(0) = f(0), P(0) = f'(0), Pr(0) = f"(0), and P(0) = f(0) by completing the following: P₂(0) = f(0) = P(0) = , f'(0) = Pr(0) = f"(0)= Now use Ps(z) to approximate e in a way similar to how you did so with P₂(z) above. In particular, P₂(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 function to make it better fit the graph of e 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 P(z) to f(z)= e centered at z=0 which has the property that P₂(0)=f(0), P,(0) = f'(0), and P(0) = f"(0). (i) Let Pa(z)=1+z+. Compute the following: P₂(0) = P/(0) = Pr(0) = , this should equal f(0) = ✓, this should equal f'(0) = , this should equal f"(0) = 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 Ps(z)=1+z+5 +6. Show that Ps(0) = f(0), P(0) = f'(0), Pr(0) = f"(0), and P(0) = f(0) by completing the following: P₂(0) = f(0) = P(0) = , f'(0) = Pr(0) = f"(0)= Now use Ps(z) to approximate e in a way similar to how you did so with P₂(z) above. In particular, P₂(1) -
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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q13
Transcribed Image Text: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 function
to 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 that
P₂(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)
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