Suppose linear interpolation is used to approximate the function f(x) = 2 sin (2 z) over...

 

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Suppose linear interpolation is used to approximate the function f(x) = 2 sin (2 z) over the interval 0<x</4. Find the linear interpolation polynomial that passes through f(x) at 0 = 0 and 2₁ = π/4. How accurately does this polynomial approximate the underlying function f(z) over the interval 0 <I</4? (a) The values of the function at the given points are fo=f(zo) = 0 (b) Enter the linear interpolation polynomial p₁ (2) that passes through the given points. Use the "Plot" link to check that your answer passes through the given points. The blue curve is the underlying function f(x), while the red curve is your response. Note that the MapleTA syntax for π is Pi. P₁(z) = 8*x/Pi (c) The difference between the underlying function f(x) and the linear interpolation polynomial p₁ (z) is the error €₁ (2) = f(x) - P₁ (z). Use the polynomial interpolation error theorem to write down an expression for the error. Assume that t is a variable in the smallest interval that contains {0, π/4, z). The error is €1(x) = and f1 = f(1) = 2 (d) Normally, we do not know t (which depends on z), hence we are unable to precisely evaluate the error. Nevertheless, if we have some information about the underlying function f(x) and its derivatives, we can use the polynomial error theorem to bound the absolute value of the error by writing €₁(z)| ≤U, where U is an expression that is independent of t (it might depend on though). This is a guarantee that the absolute value of the error never exceeds U. Clearly, we want to find the smallest possible value of U that we can (an error bound of the form <1 (2)| ≤oo is not particularly useful!). (i) In this case, we know that | sin (2 t) | is 1 The number C= maximum value for 0 ≤t≤ π/4. (ii) Hence, |4₁ (1)| ≤ C¹ (1-1) | for 0 ≤ ≤ π/4. Enter the number C Use the "Plot" link to check that the actual error is less than the bound. The blue curve is the actual error |₁ (2)| = |f(z) - P₁ (2)| while the red curve is Cz (2-1) ADP Suppose linear interpolation is used to approximate the function f(x) = 2 sin (2 z) over the interval 0<x</4. Find the linear interpolation polynomial that passes through f(x) at 0 = 0 and 2₁ = π/4. How accurately does this polynomial approximate the underlying function f(z) over the interval 0 <I</4? (a) The values of the function at the given points are fo=f(zo) = 0 (b) Enter the linear interpolation polynomial p₁ (2) that passes through the given points. Use the "Plot" link to check that your answer passes through the given points. The blue curve is the underlying function f(x), while the red curve is your response. Note that the MapleTA syntax for π is Pi. P₁(z) = 8*x/Pi (c) The difference between the underlying function f(x) and the linear interpolation polynomial p₁ (z) is the error €₁ (2) = f(x) - P₁ (z). Use the polynomial interpolation error theorem to write down an expression for the error. Assume that t is a variable in the smallest interval that contains {0, π/4, z). The error is €1(x) = and f1 = f(1) = 2 (d) Normally, we do not know t (which depends on z), hence we are unable to precisely evaluate the error. Nevertheless, if we have some information about the underlying function f(x) and its derivatives, we can use the polynomial error theorem to bound the absolute value of the error by writing €₁(z)| ≤U, where U is an expression that is independent of t (it might depend on though). This is a guarantee that the absolute value of the error never exceeds U. Clearly, we want to find the smallest possible value of U that we can (an error bound of the form <1 (2)| ≤oo is not particularly useful!). (i) In this case, we know that | sin (2 t) | is 1 The number C= maximum value for 0 ≤t≤ π/4. (ii) Hence, |4₁ (1)| ≤ C¹ (1-1) | for 0 ≤ ≤ π/4. Enter the number C Use the "Plot" link to check that the actual error is less than the bound. The blue curve is the actual error |₁ (2)| = |f(z) - P₁ (2)| while the red curve is Cz (2-1) ADP

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