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Calculus 2 : Lagrange Error

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Example Question #61 : Taylor Series

Let $P_{5}(x)$ $\displaystyle P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for $\displaystyle f(x) = sin(x)$ , centered at x = 0 $\displaystyle x = 0$ .

What is the Lagrange error of the polynomial approximation to sin(1) $\displaystyle sin(1)$ ?

Possible Answers:

$\frac{1}{7!}$ $\displaystyle \frac{1}{7!}$

$\displaystyle 1$

$\frac{1}{6!}$ $\displaystyle \frac{1}{6!}$

$\displaystyle 0$

$\frac{1}{5!}$ $\displaystyle \frac{1}{5!}$

Correct answer:

$\frac{1}{7!}$ $\displaystyle \frac{1}{7!}$

Explanation:

The fifth degree Taylor polynomial approximating sin(x) $\displaystyle sin(x)$ centered at x=0 $\displaystyle x=0$ is:

$sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$ $\displaystyle sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

The Lagrange error is the absolute value of the next term in the sequence, which is equal to $\left |\frac{x^7}{7!} \right |$ $\displaystyle \left |\frac{x^7}{7!} \right |$ .

We need only evaluate this at $\displaystyle x=1$ and thus we obtain $\frac{1^7}{7!} = \frac{1}{7!}$ $\displaystyle \frac{1^7}{7!} = \frac{1}{7!}$

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Example Question #1 : Alternating Series With Error Bound

Which of the following series does not converge?

Possible Answers:

$\sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}$ $\displaystyle \sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}$

$\sum_{i=0}^{\infty} \frac{(-1)^n}{n}$ $\displaystyle \sum_{i=0}^{\infty} \frac{(-1)^n}{n}$

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ $\displaystyle \sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

$\sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$ $\displaystyle \sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$

$\sum_{i=0}^{\infty} 0.9999999999999^n$ $\displaystyle \sum_{i=0}^{\infty} 0.9999999999999^n$

Correct answer:

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ $\displaystyle \sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

Explanation:

We can show that the series $\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ $\displaystyle \sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ diverges using the ratio test.

$L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$ $\displaystyle L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$

$= \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$ $\displaystyle = \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$

n^2 $\displaystyle n^2$ will dominate over (n+1) $\displaystyle (n+1)$ since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that $\displaystyle n!$ is much greater than n^2 $\displaystyle n^2$ , and thus having $\displaystyle n!$ in the numerator will make the series diverge by the $n^{th}$ $\displaystyle n^{th}$ limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

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Calculus 2 : Lagrange Error

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Taylor Series

Example Question #1 : Alternating Series With Error Bound

All Calculus 2 Resources

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