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Use Taylor's Theorem to estimate the error in approximating $\sinh 2x$ by $2x + 4/3x^3$ on the interval $[-0.5,0.5]$.

For this question, I use the Taylor's remainder formular, $$ R_n(x)= \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!}$$ and I get $R_4(0.5) = 0.012\,85$.

Is this correct?

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  • $\begingroup$ A related problem. $\endgroup$ Jun 11, 2013 at 6:32
  • $\begingroup$ @Karen: Please see my update, I left out that this is an odd function and we could have done better at bounding the error, so I agree with your answer - see details below. Sorry for the confusion! $\endgroup$
    – Amzoti
    Jun 11, 2013 at 14:06

1 Answer 1

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Generally, you want to use:

$$\displaystyle |R_{n}(x)| \le \frac{M_{n+1}}{(n+1)!}|x-a|^{n+1}$$

where, $\displaystyle R_{n}(x) = f(x) -T_n(x)$ is the remainder term and $T_n(x)$ is the Taylor polynomial of degree $n$ for $f(x)$, centered at $x = a$.

For this problem, we have an odd function, $f(x) = \sinh 2x$ on the interval $\displaystyle \left[-\frac{1}{2},\frac{1}{2}\right]$, so we can take:

$$\displaystyle T_4(x) = P_0(x) + P_1(x) + P_2(x) + P_3(x) + P_4(x) = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 + \frac{f^{(4)}(0)}{4!}(x-0)^3= 0 + 2x + 0 + \frac{8 x^3}{6} + 0 = 2x + \frac{4}{3} x^3$$

We also have: $f^{(5)}(x) = 32 \cosh x$, so

Max $\displaystyle |f^{(5)}(x)|$ on $\displaystyle -\frac{1}{2} \le x \le \frac{1}{2}$ occurs on the endpoints, that is $\displaystyle x = \pm \frac{1}{2}$, so:

$$\displaystyle M_5 = \max_{-\frac{1}{2} \le x \le \frac{1}{2}} \left|f^{(5)}(x)\right| = 32 \cosh(1) \approx 49.37858$$

So, the upper error bound is given by:

$$\displaystyle |f(x) -T_4(x)| = |R_4(x)| \le \frac{M_5}{5!}|x-0|^5 = \frac{49.37858}{5!}\left|\frac{1}{2}\right|^5 = 0.0128590052083333$$

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  • $\begingroup$ that's because R3(x) = R4(x), isn't it? are you using taylor's remainder formular? $\endgroup$
    – Karen
    Jun 11, 2013 at 6:38
  • $\begingroup$ Yes, do you know whether I should use R3(x) or R4(x)? $\endgroup$
    – Karen
    Jun 11, 2013 at 6:46
  • $\begingroup$ I would think it natural to use $R_4$, since the function is an odd function. $\endgroup$ Jun 11, 2013 at 7:03
  • $\begingroup$ @Amzoti: Sorry, I misunderstood. As to the name ome uses, the answer is easy, use the notation of your book/course. What I was commenting about was that in estimate, we should take advantage of the fact the fourth derivative is $0$. $\endgroup$ Jun 11, 2013 at 16:23
  • $\begingroup$ @AndréNicolas: Yes, I actually decided to clean up both and took advantage of the fourth derivative being $0$. Thanks for clarifying! Regards $\endgroup$
    – Amzoti
    Jun 11, 2013 at 16:38

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