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Use the first three nonzero terms of Taylor’s formula for $\sin x$ to find an approximate value for the integral $\int_0^1 \frac{\sin x}{x}$ and give an estimate for the error.(It is understood that the quotient of $\frac{\sin x}{x}$ is equal to $1$ when $x=0$).

My problem:By Taylor's formula I have found the integral value $0.944611111$ But I do not know how to determine the error estimate.Please help me in finding the error estimate.

Thanks.

Flip
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2 Answers2

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Hint: Remember that if you are expanding $$f(x) \approx f(a) + f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n,$$ then the error is estimated by $$r_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},$$ for some $c$ between $x$ and $a$. Use that $|\sin c| \leq 1$ and $|\cos c| \leq 1$, and try to give a upper bound for $|r_n(x)|$.

Ivo Terek
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    Look at Jack's answer.. The idea is to find bounds for the function that you want to integrate, then integrate everything. I don't have the exact numbers here. – Ivo Terek Jan 21 '15 at 23:04
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Over $[0,1]$, by the Taylor expansion with Lagrange remainder, we have: $$ x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}\leq \sin x \leq x-\frac{x^3}{6}+\frac{x^5}{120},\tag{1}$$ so: $$ 1-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{5040}\leq \frac{\sin x}{x} \leq 1-\frac{x^2}{6}+\frac{x^4}{120},\tag{2}$$ and integrating both sides over $[0,1]$ we get: $$ 1-\frac{1}{18}+\frac{1}{600}-\frac{1}{35280}\leq I \leq 1-\frac{1}{18}+\frac{1}{600}\tag{3}$$ so the integral equals $\frac{1703}{1800}-\varepsilon$ and $\varepsilon$ is between zero and $\frac{3}{10^5}$, giving $I=\color{red}{0.946\ldots}$.

Jack D'Aurizio
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  • @Flip, I just used the Taylor series for the sine function in a neighbourhood of zero, with the Lagrange remainder, just as I wrote. – Jack D'Aurizio Jan 21 '15 at 23:08
  • In fact, we just have: $$ I=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\cdot (2n+1)!} $$ so the matter is just to truncate a fast-converging alternating series. – Jack D'Aurizio Jan 21 '15 at 23:22