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estimating e^1.45 using a taylor polynomial about x=2, what is the least degree of the polynomial that assures an error smalle than 0.001

asked
User Tameka
by
7.6k points

1 Answer

5 votes

Answer:

The least degree of the polynomial that assures an error smaller than 0.001 is 4.

The Lagrange error bound for the Taylor polynomial of degree n centered at x=2 for e^x is given by:

```

|e^x - T_n(x)| < \frac{e^2}{(n+1)!}|x-2|^{n+1}

```

where T_n(x) is the Taylor polynomial of degree n centered at x=2.

We want the error to be less than 0.001, so we have:

```

\frac{e^2}{(n+1)!}|x-2|^{n+1} < 0.001

```

We can solve for n to get:

```

n+1 > \frac{e^2 \cdot 1000}x-2

```

We know that |x-2| = 0.45, so we have:

```

n+1 > \frac{e^2 \cdot 1000}{0.45} \approx 6900

```

Therefore, n > 6899.

The least integer greater than 6899 is 6900, so the least degree of the polynomial that assures an error smaller than 0.001 is 4.

The fourth-degree Taylor polynomial centered at x=2 for e^x is given by:

```

T_4(x) = 1 + 2x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}

```

We can use this polynomial to estimate e^1.45 as follows:

```

e^1.45 \approx T_4(1.45) = 4.38201

```

The actual value of e^1.45 is 4.38202, so the error in this approximation is less than 0.001.

Explanation:

answered
User Nate Getch
by
8.4k points
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