Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like e, 7 or ln(2)? After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let f(x) = e. Part (a) The tangent line to f = e at x = 0 is L(x) = The formula L(a) 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 function to make it better fit the graph of e* for a close to 0. With the linearization, we had both f(x) and f'(x) share the same value as the linearization at a = 0. We will now use a quadratic approximation P₂(x) to f(x) = e centered at x = 0 which has the property that P₂(0) = f(0), P₂(0) = f'(0), and P(0) = f'(0). (i) Let P₂(x) = 1 + x + 2. Compute the following: P₂(0) = P₂(0) = P" (0) = , this should equal f(0) = , this should equal f'(0) = , this should equal f"(0) = = f(0) = f'(0) = 8. ,f"(0) Now that you have shown the equalities above, use P₂(a) 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 a around 0. For example, let P3(x) = 1+x+²+. Show that P3 (0) = f(0), Pg(0) = f'(0), Pg(0) = f'(0), and P" (0) = f(0) by completing the following: P3 (0) = P3(0) P" (0) = Now use P3(x) to approximate e in a way similar to how you did so with P₂(x) above. In particular, P3(1) = ; 1: 9. 9. ; 9. ; 9.
Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like e, 7 or ln(2)? After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let f(x) = e. Part (a) The tangent line to f = e at x = 0 is L(x) = The formula L(a) 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 function to make it better fit the graph of e* for a close to 0. With the linearization, we had both f(x) and f'(x) share the same value as the linearization at a = 0. We will now use a quadratic approximation P₂(x) to f(x) = e centered at x = 0 which has the property that P₂(0) = f(0), P₂(0) = f'(0), and P(0) = f'(0). (i) Let P₂(x) = 1 + x + 2. Compute the following: P₂(0) = P₂(0) = P" (0) = , this should equal f(0) = , this should equal f'(0) = , this should equal f"(0) = = f(0) = f'(0) = 8. ,f"(0) Now that you have shown the equalities above, use P₂(a) 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 a around 0. For example, let P3(x) = 1+x+²+. Show that P3 (0) = f(0), Pg(0) = f'(0), Pg(0) = f'(0), and P" (0) = f(0) by completing the following: P3 (0) = P3(0) P" (0) = Now use P3(x) to approximate e in a way similar to how you did so with P₂(x) above. In particular, P3(1) = ; 1: 9. 9. ; 9. ; 9.
Chapter6: Exponential And Logarithmic Functions
Section6.1: Exponential Functions
Problem 57SE: Repeat the previous exercise to find the formula forthe APY of an account that compounds daily....
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Transcribed Image Text:Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like e, π or ln(2)? After all, the only operations a calculator can really
perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This activity provides the first steps in understanding how this process
works. Throughout the activity, let f(x) = e.
Part (a)
The tangent line to f = eª at x = 0 is L(x) =
The formula L(x) can be used to appriximate e since L(1) ≈ f(1) = e. In particular, L(1)
Part (b)
X
The linearization of eª 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 eª for a close to 0. With the linearization, we had both f(x)
and f'(x) share the same value as the linearization at î : 0. We will now use a quadratic approximation P₂(x) to f(x) = eª centered at x = 0 which has the property that
P₂(0) = f(0), P₂(0) = f'(0), and P″(0) = ƒ"(0).
=
(i) Let P₂(x) = 1 + x + 2. Compute the following:
P₂(0)
P₂(0) =
P" (0) =
=
P3 (0)
=
this should equal ƒ(0)
this should equal ƒ'(0) :
=
this should equal fƒ"(0) =
=
Now that
you have shown the equalities above, use P₂(x) to approximate e by observing that P₂(1) ≈ ƒ(1). In particular P₂(1)
P₂(0) =
P" (0) =
f(0)
‚ ƒ'(0) =
ƒ" (0) =
=
2
=
(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 around 0. For example, let P3(x) = 1 +x+22² +2³. Show that P3(0) = ƒ(0), Pģ(0) = ƒ'(0), Pg'(0) = ƒ”(0), and P!"(0) = f'(0) by
completing the following:
x3
6
=
▶
=
;
;
;
=
Now use P3(x) to approximate e in a way similar to how you did so with P₂(x) above. In particular, P3(1) =
-
▼
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