L13-quadraticApprox.mws
Calculus I
Lesson 13: Quadratic Approximation
We want to approximate a given function f(x) at x=a with a second degree polynomial.
Express the second degree polynomial as P(x) = A + B (x - a) + C
We want:
P(a) = f (a)
P'(a) = f ' (a)
P''(a) = f '' (a)
Hence, we want:
A = f (a)
B = f ' (a)
2 C = f '' (a)
Thus,
P(x) = f (a) + f ' (a) (x-a) + [ f '' (a) / 2 ]
Conclusion: The quadratic approximation to f(x) near x = a is given by:
P(x) = f (a) + f ' (a) (x-a) + [ f '' (a) / 2 ]
Example
1
Find the quadratic approximation to f(x) = sec (x) for a = 0.
Plot both f(x) and P(x) on the same axes near 0.
>
restart:
>
f:= x -> sec(x);
>
f(0);
>
D(f)(0);
>
D(D(f))(0);
>
P:= x -> 1 + 0.5 * (x)^2;
>
plot({f(x),P(x)}, x = -1..1, color=[blue,brown]);
Example
2
Find the quadratic approximation to f x) = sin (x) for a =
.
Plot both f(x) and P(x) on the same axes near
.
Then plot both functions on a larger domain; is P(x) a good estimate
for f (x) away from
? .
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f:= x -> sin(x);
>
f(Pi/6);
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D(f)(Pi/6);
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evalf(%);
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D(D(f))(Pi/6);
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P:= x -> 0.5 + .8660254040 * (x - Pi/6) - 0.25 *( x - Pi/6)^2;
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evalf(Pi/6);
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plot({f(x),P(x)}, x = 0..1, color=[brown,blue]);
The graph below shows that P(x) is NOT a good approximation to f (x) for x far from
.
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plot({f(x),P(x)}, x = -2*Pi..2*Pi, color=[brown,blue]);
3) Find the quadratic approxiamtion for f (x) =
near a = 1.
Plot both f (x) and P (x) on the same axes near 1.
Plot both f (x) and P (x) on the same axes for a larger domain; is P(x) a good estimate away from 1?
Determine the values of x for which the quadratic approximation is accurate to within 0.01.
Can you use a plot to help?
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f:= x -> 1/(1+x)^2;
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f(1);
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D(f)(1);
>
D(D(f))(1);
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P:= x -> 0.25 - 0.25 * (x - 1) + (3/16) * ( x - 1)^2;
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plot({f(x),P(x)}, x = 0..2, color=[blue,brown]);
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plot({f(x),P(x)}, x = 0..10, color=[blue,brown]);
From the above graph, we see that P (x) is NOT a good estimate for f (x) for x not near 1.
We are to find x values for which | f(x) - P (x) | < 0.01, i.e.,
we want:
-0.01 < f (x) - P (x) < 0.01
OR
f (x) - 0.01 < P (x) < f (x) + 0.01.
>
g1:= x -> f(x) - 0.01;
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g2:= x -> f(x) + 0.01;
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plot({P(x), g1(x), g2(x)}, x = 0.5..3, color=[blue, brown, magenta]);
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solve({-0.01 < f (x) - P (x), x > 0}, x );
>
solve({f (x) - P (x) < 0.01, x > 0}, x);
From the plot and solve, we see that | P (x) - f (x) | < 0.01 for
x in (
,
). Our answer is a numerical estimate.