Week 9 - Maple explanation PDF

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MATH1231 Maple TA Explanations for Week 9, SEM 2 Gerald Huang September 21, 2018

Contents 1 Taylor polynomial of a polynomial

3

1.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

1.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

2 Taylor polynomial of a function

7

2.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

2.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

2.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

2.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

2.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

3 Lagrange formula for the remainder 3.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

10 10

3.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

3.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

3.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

3.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

4 Classifying stationary points

14

4.1

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

4.2

Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

4.3

Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15

4.4

Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15

4.5

Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

2

1 1.1

Taylor polynomial of a polynomial Question 1

A Taylor polynomial is a polynomial approximation of a given function or polynomial around a particular point. To construct a Taylor polynomial, we find respective derivatives around a particular point. For our approximation to be accurate, our Taylor polynomial should have the same f (3) value. So if we need to construct our polynomial, we should get f (3) = 9 as the ”constant” term. Next, we need our first derivative to match at x = 3. So, taking the derivative, we find that f ′ (3) = 2(3) = 6. We can then construct our Taylor polynomial to be: y = 9 + 6(x − 3). Note that (x − 3) comes from the fact that we are approximating our Taylor polynomial near x = 3. A Taylor polynomial around x = 0 is called a maclaurin series.

3

1.2

Question 2

A different way in looking at it is by first constructing some polynomial. This polynomial should be of degree 2 if we’re looking at the first 2 derivatives. So, let’s suppose our Taylor series is of the form: p(x) = a(x − 2)2 + b(x − 2) + c Remember that we’re making this approximation near x = 2. Now, if we want our approximation to be somewhat accurate, the first place to start is by having our 0th derivative to match at x = 2. So, we want p(2) = f (2). p(2) simply gives us c, and f (2) gives us 8. So we conclude that c = 8. Next, we want p′ (2) = f ′ (2). p′ (x) = 2a(x − 2) + b p′ (2) = b f ′ (x) = 3x2 f ′ (2) = 12 ⇒ b = 12 And again, we want p′′ (2) = f ′′ (2). p′′ (x) = 2a p′′ (2) = 2a f ′′ (x) = 6x f ′′ (2) = 12 ⇒ a = 6 4

So, our Taylor polynomial is: y = 6(x − 2)2 + 12(x − 2) + 8.

1.3

Question 3

Again, it’s the same idea with the previous sections.

1.4

Question 4

Reminder: Rewrite

5

1 as x−1 . x

1.5

Question 5

Here, a and b can be anything you want, except for 0. Keep the rest as it is, and it should work!!

6

2

Taylor polynomial of a function

2.1

Question 1

2.2

Question 2

From Question 1, we found the first 3 derivatives evaluated at x = that the Taylor series is of the form:

pn (x) =

n X f (k) (x) . k! k=0

7

π . 2

Note

Then we are simply finding the Taylor series for n = 5.

2.3

ln

Question 3

our Taylor polynomial is an approximation, then we can approximate  Since  7 with our Taylor series near x = 7/6 and near x = 1/6 for the sin function. 6

2.4

Question 4

8

2.5

Question 5

9

3

Lagrange formula for the remainder

3.1

Question 1

When we take the fourth derivative of ex , we are left with ex . So evaluating it at x = c gives us ec . Since y = ex is an increasing function, then when we restrict c to be between -0.01 and 0, then our maximum value will occur when c = 0, which turns out to be 1. So, our remainder must be no greater than 1 · x4 /24 = x4 /24.

10

3.2

Question 2

Notice that the third derivative is 3! which is 6. So, at any point in the interval [1, x], we end up with: R3 (x) =

3.3

3!(x − 1)3 = (x − 1)3 . 3!

Question 3

11

3.4

Question 4

Hint: |cos(x)| = 1 and |sin(x)| = 1.

12

3.5

Question 5

13

4

Classifying stationary points

4.1

Question 1

4.2

Question 2

14

4.3

Question 3

4.4

Question 4

15

4.5

Question 5

So, recalling the Taylor theorem, we get: f (x) = p2 (x) + R3 (x). We eventually want to find an expression for f (x), so we’ll need to find two things: p2 (x) is the Taylor approximation around x = 0 with degree 2. We’re given the derivatives around x = 0, so we conclude that:

p2 (x) = f (0) + f ′ (0)(x − 0) +

f ′′ (0)(x − 0)2 2!

= 0 + 3x − x2 . We then need to find an expression for the remainder, R3 (x) at x = 0. By Lagrange’s formula, we get: R3 (x) =

f (3)(x) (x − 0)3 . 3! 16

It’s important to note that this works for relatively close approximations for x around x = 0. So given the interval [0, x], we need to find an upper bound to find the maximal approximation. Let’s suppose we restrict x to be between 0 and 1, so that:    (3)  f (x)  ≤ 12. Then we’ve essentially contained R3 (x) to be no greater than we conclude that:

12 6

= 2. Thus,

R3 (x) ≤ 2x3 . So, we deduce that f (x) ≤ 3x − x2 + 2x3 for x = 0+ . Then, we deduce that the limit is:

lim

x→0+

3x − (3x − x2 + 2x3 ) 3x − f (x) = lim 2 + x x→0 x2 2 3 x − 2x = lim x2 x→0+ 2 x (1 − 2x) = lim + x2 x→0 = lim+ (1 − 2x) x→0

= 1.

– END OF MAPLE –

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