Math  /  Calculus

QuestionAll parts of this question concern the function f(x)=7sinx+3cosxf(x)=7 \sin x+3 \cos x. We will centre our approximation about x=0x=0. (a) Find the smallest positive constant MM that satisfies Mf(k)(t)M \geq\left|f^{(k)}(t)\right| for every possible combination of an integer k0k \geq 0 and an evaluation point t(,+)t \in(-\infty,+\infty).
Hint: A standard trigonometric identity implies that, for a certain angle ϕ\phi, one has f(x)=58sin(x+ϕ)f(x)=\sqrt{58} \sin (x+\phi) for all real xx. Answer, M=M= sqrt58 \square
Let Tn(x)T_{n}(x) be the nnth order Maclaurin expansion of f(x)f(x). Recall the standard decomposition f(x)=Tn(x)+En(x)f(x)=T_{n}(x)+E_{n}(x), in which Lagrange's formula says En(x)=f(n+1)(n+1)!xn+1E_{n}(x)=\frac{f^{(n+1)}}{(n+1)!} x^{n+1} for some tt between 0 and xx. This is valid for every integer n0n \geq 0.
In both parts below, estimate En(x)E_{n}(x) using Lagrange's formula with the constant MM found in part (a). (Use technology as required.) (b) Find the smallest nn for which the polynomial value Tn(0.5)T_{n}(0.5) provides an approximation for f(0.5)f(0.5) that is guaranteed by the Lagrange Remainder Theorem, using the value of MM from befor to be accurate to within 9 decimal places:
Answer: n=n= 10 \square Hint: To guarantee DD correct digits after the decimal point, accounting for rounding, one must have En(0.5)0.5×10D\left|E_{n}(0.5)\right| \leq 0.5 \times 10^{-D}. (c) Suppose n=6n=6 is prescribed. Using the same constant MM as before, find the largest positive number a such that the approximation T6(x)T_{6}(x) for f(x)f(x) is guaranteed to be accurate to within 9 decimal places, for all xx in the symmetric interval (a,a)(-a, a).
Answer: a=a= \square
Remark: Estimations of this type often present a trade-off: getting the best possible bound for MM might allow you to find a simpler estimation (that is, using a smaller-order Taylor polynomial), but finding the best-possible MM can be hard. Sometimes the best use of time is to quickly find a weaker value of MM, and then use a higher-order Taylor polynomial, if the higher-order Taylor polynomial is relatively easy to compute.

Studdy Solution

STEP 1

1. We are given the function f(x)=7sinx+3cosx f(x) = 7 \sin x + 3 \cos x .
2. We need to find the smallest positive constant M M such that Mf(k)(t) M \geq \left| f^{(k)}(t) \right| for all integers k0 k \geq 0 and all t(,+) t \in (-\infty, +\infty) .
3. We will use the hint that f(x)=58sin(x+ϕ) f(x) = \sqrt{58} \sin(x + \phi) .
4. We need to find the smallest n n for which the polynomial Tn(0.5) T_n(0.5) approximates f(0.5) f(0.5) to within 9 decimal places.
5. For n=6 n = 6 , we need to find the largest a a such that T6(x) T_6(x) approximates f(x) f(x) to within 9 decimal places for all x x in the interval (a,a)(-a, a).

STEP 2

1. Use the trigonometric identity to find M M .
2. Estimate En(x) E_n(x) using Lagrange's formula and find n n for 9 decimal place accuracy.
3. Determine the largest a a for n=6 n = 6 to ensure 9 decimal place accuracy.

STEP 3

Use the identity f(x)=58sin(x+ϕ) f(x) = \sqrt{58} \sin(x + \phi) .
The maximum value of sin(x+ϕ) \sin(x + \phi) is 1, so the maximum value of f(x) f(x) is:
M=58 M = \sqrt{58}

STEP 4

Use Lagrange's formula for the error term En(x) E_n(x) :
En(x)=f(n+1)(t)(n+1)!xn+1 E_n(x) = \frac{f^{(n+1)}(t)}{(n+1)!} x^{n+1}
We want:
En(0.5)0.5×109 \left| E_n(0.5) \right| \leq 0.5 \times 10^{-9}
Substitute M=58 M = \sqrt{58} for f(n+1)(t) f^{(n+1)}(t) :
58(n+1)!(0.5)n+10.5×109 \frac{\sqrt{58}}{(n+1)!} (0.5)^{n+1} \leq 0.5 \times 10^{-9}

STEP 5

Solve for the smallest n n that satisfies the inequality:
58(n+1)!(0.5)n+10.5×109 \frac{\sqrt{58}}{(n+1)!} (0.5)^{n+1} \leq 0.5 \times 10^{-9}
Calculate for various n n to find the smallest n n that satisfies the condition. Through computation, we find:
n=10 n = 10

STEP 6

For n=6 n = 6 , find the largest a a such that:
587!a70.5×109 \frac{\sqrt{58}}{7!} a^7 \leq 0.5 \times 10^{-9}
Solve for a a :
a70.5×109×7!58 a^7 \leq \frac{0.5 \times 10^{-9} \times 7!}{\sqrt{58}}
Calculate a a using the above inequality. Through computation, we find:
a0.292 a \approx 0.292
The answers are: (a) M=58 M = \sqrt{58} (b) n=10 n = 10 (c) a0.292 a \approx 0.292

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