Math  /  Calculus

QuestionTaylor's Remainder Theorem (see CLP-1 section 3.4.9) Suppose that ff is an n+1n+1-times differentiable function on an interval II containing the point x=ax=a. There exists a point cc strictly between xx and aa so that f(x)=Tn(x)+f(n+1)(c)(n+1)!(xa)n+1f(x)=T_{n}(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} where Tn(x)T_{n}(x) is the nn-th degree Taylor polynomial of ff centred at x=ax=a. Moreover, if MM is a constant and f(n+1)(c)M\left|f^{(n+1)}(c)\right| \leq M, then we can estimate the error in the approximation of ff by TnT_{n} as follows: f(x)Tn(x)M(n+1)!xan+1\left|f(x)-T_{n}(x)\right| \leq \frac{M}{(n+1)!}|x-a|^{n+1}
In practice, we take MM to be the maximum of f(n+1)(x)\left|f^{(n+1)}(x)\right| on the interval II.
Question: (a) Approximate ln(1.5)\ln (1.5) using the 3-rd degree Taylor polynomial of f(x)=ln(x)f(x)=\ln (x) centred at x=1x=1. (b) Use Taylor's Remainder Theorem to approximate the error between ln(1.5)\ln (1.5) and the 3-rd degree Taylor polynomial approximation on the interval I=[12,32]I=\left[\frac{1}{2}, \frac{3}{2}\right]. (c) Let Tn(x)T_{n}(x) be the nn-th degree Taylor polynomial for f(x)=ln(x)f(x)=\ln (x) centred at x=1x=1. Using -Taylor's Remainder Theorem, find the smallest integer nn so that f(x)Tn(x)0.0001\left|f(x)-T_{n}(x)\right| \leq 0.0001 for all x[12,32]x \in\left[\frac{1}{2}, \frac{3}{2}\right].

Studdy Solution

STEP 1

What is this asking? We need to approximate ln(1.5)\ln(1.5) using a Taylor polynomial, estimate the error of this approximation, and find how many terms we need for a super accurate approximation within a given interval. Watch out! Remember that the interval given affects the maximum value of the n+1n+1-th derivative, which is crucial for calculating the error bound.
Also, don't forget to center the Taylor polynomial at the specified point!

STEP 2

1. Approximate ln(1.5)\ln(1.5)
2. Estimate the error
3. Find the necessary degree for the desired accuracy

STEP 3

We want the 3rd-degree Taylor polynomial of f(x)=ln(x)f(x) = \ln(x) centered at x=1x = 1.
Remember, the formula for the nn-th degree Taylor polynomial is: Tn(x)=k=0nf(k)(a)k!(xa)k T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k Here, a=1a = 1 and n=3n = 3.
Let's find the derivatives of f(x)f(x) at x=1x = 1: f(1)=ln(1)=0 f(1) = \ln(1) = 0 f(x)=1x    f(1)=1 f'(x) = \frac{1}{x} \implies f'(1) = 1 f(x)=1x2    f(1)=1 f''(x) = -\frac{1}{x^2} \implies f''(1) = -1 f(x)=2x3    f(1)=2 f'''(x) = \frac{2}{x^3} \implies f'''(1) = 2 So, our Taylor polynomial is: T3(x)=0+11!(x1)12!(x1)2+23!(x1)3=(x1)12(x1)2+13(x1)3 T_3(x) = 0 + \frac{1}{1!}(x-1) - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3

STEP 4

Now, plug in x=1.5x = 1.5 into T3(x)T_3(x): T3(1.5)=(1.51)12(1.51)2+13(1.51)3=0.512(0.5)2+13(0.5)3=0.50.125+0.041666...0.4167 T_3(1.5) = (1.5-1) - \frac{1}{2}(1.5-1)^2 + \frac{1}{3}(1.5-1)^3 = 0.5 - \frac{1}{2}(0.5)^2 + \frac{1}{3}(0.5)^3 = 0.5 - 0.125 + 0.041666... \approx \mathbf{0.4167}

STEP 5

We need the **4th derivative** of f(x)f(x) for the error estimate: f(4)(x)=6x4 f^{(4)}(x) = -\frac{6}{x^4} On the interval [12,32][\frac{1}{2}, \frac{3}{2}], the absolute value f(4)(x)|f^{(4)}(x)| is largest when xx is smallest, so at x=12x = \frac{1}{2}: f(4)(12)=6(12)4=616=96 |f^{(4)}(\frac{1}{2})| = |-\frac{6}{(\frac{1}{2})^4}| = |-6 \cdot 16| = 96 So, we take M=96M = \mathbf{96}.

STEP 6

Using Taylor's Remainder Theorem, the error is bounded by: f(x)T3(x)M4!x14 |f(x) - T_3(x)| \leq \frac{M}{4!}|x-1|^4 For x=1.5x = 1.5 and M=96M = 96, we have: 96241.514=4(0.5)4=40.0625=0.25 \frac{96}{24}|1.5 - 1|^4 = 4 \cdot (0.5)^4 = 4 \cdot 0.0625 = \mathbf{0.25}

STEP 7

We want f(x)Tn(x)0.0001|f(x) - T_n(x)| \leq 0.0001 for all xx in [12,32][\frac{1}{2}, \frac{3}{2}].
The (n+1)(n+1)-th derivative of f(x)=ln(x)f(x) = \ln(x) is f(n+1)(x)=(1)nn!xn+1f^{(n+1)}(x) = (-1)^n \frac{n!}{x^{n+1}}.
The maximum of f(n+1)(x)|f^{(n+1)}(x)| on [12,32][\frac{1}{2}, \frac{3}{2}] occurs at x=12x = \frac{1}{2}, and is M=n!2n+1M = n! 2^{n+1}.
We want: n!2n+1(n+1)!x1n+10.0001 \frac{n! 2^{n+1}}{(n+1)!} |x-1|^{n+1} \leq 0.0001 2n+1n+1x1n+10.0001 \frac{2^{n+1}}{n+1} |x-1|^{n+1} \leq 0.0001

STEP 8

Since x112|x-1| \leq \frac{1}{2} for xx in [12,32][\frac{1}{2}, \frac{3}{2}], we have: 2n+1n+1(12)n+10.0001 \frac{2^{n+1}}{n+1} (\frac{1}{2})^{n+1} \leq 0.0001 1n+10.0001 \frac{1}{n+1} \leq 0.0001 n+110000 n+1 \geq 10000 n9999 n \geq 9999 So, the smallest integer nn is 9999\mathbf{9999}.

STEP 9

(a) ln(1.5)0.4167\ln(1.5) \approx 0.4167 (b) The error is at most 0.250.25. (c) The smallest integer nn is 99999999.

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