Math / Calculus

QuestionUNIVERSITY OF CALGARY
Department of Mathematics and Statistics University of Calgary MATH 249 Introductory Calculus Fall 2024 Written Lab Task \#2 Due Date: Friday, November 22 at 11:59 PM Mountain Time Upload your solution to Gradescope [Video on how to submit]
Taylor's Remainder Theorem (see CLP-1 section 3.4.9) Suppose that $f$ is an $n+1$ -times differentiable function on an interval $I$ containing the point $x=a$ . There exists a point $c$ strictly between $x$ and $a$ so that $f(x)=T_{n}(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ where $T_{n}(x)$ is the $n$ -th degree Taylor polynomial of $f$ centred at $x=a$ . Moreover, if $M$ is a constant and $\left|f^{(n+1)}(c)\right| \leq M$ , then we can estimate the error in the approximation of $f$ by $T_{n}$ as follows: $\left|f(x)-T_{n}(x)\right| \leq \frac{M}{(n+1)!}|x-a|^{n+1}$
In practice, we take $M$ to be the maximum of $\left|f^{(n+1)}(x)\right|$ on the interval $I$ . Question: (a) Approximate $\sin (1)$ using the 5-th degree Taylor polynomial of $f(x)=\sin (x)$ centred at $x=0$ . (b) Use Taylor's Remainder Theorem to approximate the error between $\sin (1)$ and the 5 -th degree Taylor polynomial approximation on the interval $I=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Hint: You can use the closed interval method to find " $M$ " in this context. (c) Let $T_{n}(x)$ be the $n$ -th degree Taylor polynomial for $f(x)=\sin (x)$ centred at $x=0$ . Using Taylor's Remainder Theorem, find the smallest integer $n$ so that $\left|f(x)-T_{n}(x)\right| \leq 0.0001$ for all $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
Hint: See the Fall 2021 Written Question Exam Solution / Key and the Written Lab Task \#1 solutions posted to D2L for a sample of what good mathematical writing looks like. Reminder: Follow the checklist on the Written Lab Task Rubric. (C) University of Calgary

Studdy Solution

STEP 1

What is this asking? We need to approximate $\sin(1)$ using a Taylor polynomial, estimate the error of this approximation, and find how many terms we need for a super accurate approximation across a wider interval. Watch out! Radians vs degrees!
Remember, Taylor series for trigonometric functions are defined using radians.
Also, don't mix up the degree of the Taylor polynomial with the degree symbol for angles.

STEP 2

1. Approximate $\sin(1)$
2. Estimate the error
3. Find the smallest n for desired accuracy

STEP 3

The Taylor series for $\sin(x)$ centered at $x = 0$ is given by: $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

STEP 4

The 5th-degree Taylor polynomial, $T_5(x)$ , is: $T_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}$

STEP 5

Substituting $x = 1$ , we get: $T_5(1) = 1 - \frac{1^3}{3!} + \frac{1^5}{5!} = 1 - \frac{1}{6} + \frac{1}{120} = \frac{120 - 20 + 1}{120} = \frac{101}{120}$ So, $\sin(1) \approx \frac{\mathbf{101}}{\mathbf{120}}$ .

STEP 6

Since we used the 5th-degree Taylor polynomial ( $n=5$ ), we need the 6th derivative of $\sin(x)$ . The derivatives of $\sin(x)$ cycle through $\cos(x)$ , $-\sin(x)$ , $-\cos(x)$ , and back to $\sin(x)$ .
The 6th derivative is $-\sin(x)$ .

STEP 7

We need to find the maximum value of $|-\sin(x)|$ on the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
Since $|\sin(x)|$ is at most 1, and $\sin(\frac{\pi}{2}) = 1$ , we have $M = |-\sin(\frac{\pi}{2})| = |-1| = \mathbf{1}$ .

STEP 8

The error is given by: $|f(x) - T_n(x)| \leq \frac{M}{(n+1)!} |x-a|^{n+1}$ In our case, $x = 1$ , $a = 0$ , $n = 5$ , and $M = 1$ .
Thus, the error is at most: $\frac{1}{(5+1)!} |1-0|^{5+1} = \frac{1}{6!} \cdot 1^6 = \frac{1}{720}$

STEP 9

We want to find the smallest integer $n$ such that: $\frac{M}{(n+1)!} |x-0|^{n+1} \leq 0.0001$ for all $x$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
Since $M=1$ and the maximum value of $|x|$ on this interval is $\frac{\pi}{2}$ , we have: $\frac{1}{(n+1)!} \left(\frac{\pi}{2}\right)^{n+1} \leq 0.0001$

STEP 10

We can test values of $n$ starting from $n=0$ and increasing until the inequality is satisfied.
For $n=7$ : $\frac{1}{8!} \left(\frac{\pi}{2}\right)^8 \approx 0.00016 > 0.0001$ For $n=8$ : $\frac{1}{9!} \left(\frac{\pi}{2}\right)^9 \approx 0.000025 < 0.0001$

STEP 11

Thus, the smallest integer $n$ that satisfies the condition is $\mathbf{8}$ .

STEP 12

(a) $\sin(1) \approx \frac{101}{120}$ (b) The error is at most $\frac{1}{720}$ . (c) The smallest integer $n$ is 8.

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Studdy Solution

STEP 1

STEP 2

1. Approximate $\sin(1)$
2. Estimate the error
3. Find the smallest n for desired accuracy

STEP 3

The Taylor series for $\sin(x)$ centered at $x = 0$ is given by: $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

STEP 4

The 5th-degree Taylor polynomial, $T_5(x)$ , is: $T_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}$

STEP 5

Substituting $x = 1$ , we get: $T_5(1) = 1 - \frac{1^3}{3!} + \frac{1^5}{5!} = 1 - \frac{1}{6} + \frac{1}{120} = \frac{120 - 20 + 1}{120} = \frac{101}{120}$ So, $\sin(1) \approx \frac{\mathbf{101}}{\mathbf{120}}$ .

STEP 6

Since we used the 5th-degree Taylor polynomial ( $n=5$ ), we need the 6th derivative of $\sin(x)$ . The derivatives of $\sin(x)$ cycle through $\cos(x)$ , $-\sin(x)$ , $-\cos(x)$ , and back to $\sin(x)$ .
The 6th derivative is $-\sin(x)$ .

STEP 7

We need to find the maximum value of $|-\sin(x)|$ on the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
Since $|\sin(x)|$ is at most 1, and $\sin(\frac{\pi}{2}) = 1$ , we have $M = |-\sin(\frac{\pi}{2})| = |-1| = \mathbf{1}$ .

STEP 8

The error is given by: $|f(x) - T_n(x)| \leq \frac{M}{(n+1)!} |x-a|^{n+1}$ In our case, $x = 1$ , $a = 0$ , $n = 5$ , and $M = 1$ .
Thus, the error is at most: $\frac{1}{(5+1)!} |1-0|^{5+1} = \frac{1}{6!} \cdot 1^6 = \frac{1}{720}$

STEP 9

STEP 10

We can test values of $n$ starting from $n=0$ and increasing until the inequality is satisfied.
For $n=7$ : $\frac{1}{8!} \left(\frac{\pi}{2}\right)^8 \approx 0.00016 > 0.0001$ For $n=8$ : $\frac{1}{9!} \left(\frac{\pi}{2}\right)^9 \approx 0.000025 < 0.0001$

STEP 11

Thus, the smallest integer $n$ that satisfies the condition is $\mathbf{8}$ .

STEP 12

(a) $\sin(1) \approx \frac{101}{120}$ (b) The error is at most $\frac{1}{720}$ . (c) The smallest integer $n$ is 8.

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Studdy Solution

STEP 1

STEP 2

1. Approximate sin⁡(1)\sin(1)sin(1)2. Estimate the error3. Find the smallest n for desired accuracy

STEP 3

STEP 4

The 5th-degree Taylor polynomial, T5(x)T_5(x)T5​(x), is: T5(x)=x−x33!+x55! T_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} T5​(x)=x−3!x3​+5!x5​

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

STEP 10

STEP 11

Thus, the smallest integer nnn that satisfies the condition is 8\mathbf{8}8.

STEP 12

(a) sin⁡(1)≈101120\sin(1) \approx \frac{101}{120}sin(1)≈120101​ (b) The error is at most 1720\frac{1}{720}7201​. (c) The smallest integer nnn is 8.

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1. Approximate $\sin(1)$
2. Estimate the error
3. Find the smallest n for desired accuracy

The 5th-degree Taylor polynomial, $T_5(x)$ , is: $T_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}$

Thus, the smallest integer $n$ that satisfies the condition is $\mathbf{8}$ .

(a) $\sin(1) \approx \frac{101}{120}$ (b) The error is at most $\frac{1}{720}$ . (c) The smallest integer $n$ is 8.