From using the Taylor Polynomial on several functions, we ...



Section 2C:

CONCLUSION (Taylor polynomials):

In our experiments with Taylor polynomials, we observed that by fixing the basepoint and then increasing the degree, our approximation of the graph of f(x) was better over a wider interval. This was true for the functions Sin(3x), Exp(x), 1+5x2+17x3-3x+5, Sqrt(x), and (x+1)/(x-1). When we studied Sqrt(x), we made an additional observation: by increasing the basepoint and leaving the degree the same, the graph of the Taylor polynomial was a better approximation to Sqrt(x) over a wider interval (see figure below). For example, consider n = 2, and change the base from a = 10 to a = 20 (i.e., from Taypol[2,10] to Taypol[2,20]). We notice on the graph of Taypol[2,10] that we have a "good" interval [5,15]. (On this interval we have close visual agreement of the Mathematica graphs of Sqrt(x) and Taypol[2,10].) When we shift to basepoint

a = 20, we see that Taypol[2,20] has a "good" interval [14,32]. We made similar observations comparing Taypol[3,5] and Taypol[3,15]. On the graph of Taypol[3,5], we had a "good" interval [2,10], and for the graph of Taypol[3,15], we had a "good"

interval [5,30].

[pic]

n = 3, a =5, and a( =15

Figure: The effect of shifting the basepoint to a "flat" region

We see in these examples that as the Sqrt(x) becomes flatter, the good interval becomes wider. This is due to the fact that for large values of x, the successive derivatives f(1)(x), f(2)(x), f(3)(x), f(4)(x), … of the square root function f(x) = Sqrt(x) converge to 0.

To explain this further, we consider the formula for the Taylor polynomial for f of degree n based at x = a:

T(n,a)(x) ( T(x) = f(a) + (1 ( i ( n f(i)(a)(x-a)i/i!

We want to see how this formula explains the widening of the good interval as we change the base point to the right. For example, consider changing from a = 5 to a = 15. For

Y = Sqrt(x), let’s tabulate some derivatives at these points:

| |A=5 |a( = 15 |

|Y(0) = x(1/2) = Sqrt(x) |2.223 |3.872 |

|Y(1) = 1/2x-(1/2) = 1/2x-(1/2) |.2236 |.12909 |

|Y(2) = 1/4x(-3/2) = -1/4x(3/2) |-.0481 |-.0043 |

|Y(3 ) = 3/8x(-5/2 ) = 3/8x(5/2) |.0067 |.00043 |

|Y(4) = -15/16x(-7/2) = -15/16x(7/2) |-.0033 |-.0000717 |

By looking at this table, we can see that the values of the derivatives at base point

a( = 15 are approaching zero faster than at a = 5. We are now going to estimate T(3,5)(x) and T(3,15)(x() at points x and x( equidistant from their respective basepoints a = 5 and

a( = 15:

[pic]

For example, at x = 10 (5 units from a = 5) and at x( = 20 (5 units from a( = 15) we have:

| |a = 5, x = 10 | a = 10, x = 15 |

| f(1) (a)(x-a) |-10 |-.12345 |

|f(2) (a)((x-a)2/(2!)) |50 |.0685 |

| f(3) (a)((x-a)3/(3!)) |-250 |-.038125 |

|f(4) (a)((x-a)4/(4!) |1250 |.02117 |

Now, using the values from the above table, we can compare the absolute errors using the third Taylor polynomial T3:

|T(3,5)(10)-Sqrt(10)| = |4.09498-Sqrt(10)| = .93262

|T(3,15)(20)-Sqrt(20)| = |.48113-Sqrt(20)| = .10899

.93262 > .10899

As we expected, the error at x = 10 is much larger than the error at x = 20.

Now, using the values from the above table, we can compare the absolute error using T4:

|T(4,5)(10)-Sqrt(10)| = 1.185

|T(4,15)(20)-Sqrt(20)| = .11057

1.185 > .11057

Again, the error at x = 10 is much large than the error at x = 20.

Our observation was correct: the approximating sum using successive derivatives at basepoint 15 was smaller than the sum using derivatives at base point 5. Therefore, we have a wider “good interval” when we use a = 15 as compared to a = 5.

We will now experiment with: f(x) = (x+1)/(x-1)

[pic]

The above graph shows the regions between a = 2 and x = 7 (interval length x-a = 5), and between a = 10 and x = 15 (interval length x-a = 5). The graph of f(x) is flatter as the second interval, so we expect a better Taylor approximation on the second interval.

From the following table, we can say that the values of the derivatives at base point

a( = 10 are approaching zero, but the values of the derivatives at base a = 2 are increasing rapidly.

| | a = 2 |a( = 10 |

|Y(0) = (a+1)/(a-1) |3 |1.222 |

|Y(1) = -(2)/(a-1)2 |-2 |-.02469 |

|Y(2) = (4)/(a-1)3 |4 |.00548 |

|Y(3) = -(12)/(a-1)4 |-12 |-.00183 |

|Y(4) = (48)/(a-1)5 |48 |.000813 |

We are going to use the same method as in the last example in order to calculate the values of T3 from the following table. The idea here is to show again that we get a better approximation if the basepoint is in a flatter region. We work at basepoint a = 2 and

a( = 10 and points x = 7 (5 units from a) and x( = 15 (5 units from a().

| |a = 2, x = 7 |a = 10, x = 15 |

|f(1) (a)(x-a) |-10 |-.12345 |

|f (2) (a)((x-a)2/(2!)) |50 |.0685 |

|f(3) (a)((x-a)3/(3!)) |-250 |-.038125 |

|f(4) (a)((x-a)4/(4!) |1250 |.02117 |

Now, using the values from the above table, we can compare the absolute errors |T(3,2)(7)-f(7)| and |T(3,10)(15)-f(15)| :

|T(3,2)(7)-f(7)| = 208.3333

|T(3,10)(15)-f(15)| = .0139

208.3333 > .0139

The error using a = 2 is much larger than the error using a = 10 in the flatter region.

We can make the same test for T4:

|T(4,2)(7)-f(7)| = 1041.6666

|T(4,10)(15)-f(15)| = .0072

1041.6666 > .0072

Again, the error using a = 2 is much larger than the error using a = 10 in the flatter region.

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