1

So I was thinking about this question and I came up with the question below.

Use the fact that $4\tan^{-1}(\frac{1}{5}) - \tan^{-1}(\frac{1}{239})= \frac\pi4$ and Taylor polynomials to approximate $\pi$ to $6$ decimals of accuracy. Include error estimates.

Here's my work.

The given equality is equivalent to $16\tan^{-1}(\dfrac{1}{5})-4\tan^{-1}(\dfrac{1}{239})=\pi.$ Also, we have that $|\pi -(16P_{n,0}(\dfrac{1}{5})-4P_{n,0}(\dfrac{1}{239}))|\leq |16\tan^{-1}(\dfrac{1}{5})-16P_{n,0}(\dfrac{1}{5})|+|4\tan^{-1}(\dfrac{1}{239})-4P_{n,0}(\dfrac{1}{239})|.$

We consider the errors separately. Let $f(x) = \tan^{-1}(x)$. Then by Taylor's Theorem, $16|\tan^{-1}(1/5)-P_{n,0}(1/5)|\leq \dfrac{16\cdot\displaystyle\max_{0\leq z \leq 1/5}\{|f^{(n+1)}(z)|\}}{(n+1)!}(1/5-0)^{n+1}.$ We want this error to be smaller than $5\cdot 10^{-7}.$ We can compute the Taylor series for $\tan^{-1} x$ centred at $x=0$ as follows: $\dfrac{d}{dx}\tan^{-1} (x) = \dfrac{1}{1+x^2} = \displaystyle\sum_{i=0}^\infty (-1)^i x^{2i}.$ Hence $\tan^{-1} (x) = \displaystyle\int \dfrac{d}{dx} \tan^{-1} (x)dx = \displaystyle\sum_{i=0}^\infty (-1)^i\dfrac{x^{2i+1}}{2i+1}.$

Hence the error can be simplified to $16\cdot\dfrac{(1/5)^{2n+3}}{2n+3}.$ The smallest possible value for which this is less than $5\cdot 10^{-7}$ is $n=4,$ in which case the error is smaller than $3\cdot 10^{-8}.$

We know proceed with the other value $4|\tan^{-1}(\dfrac{1}{239})-P_{n,0}(\dfrac{1}{239})| \leq 4\cdot\dfrac{(1/239)^{2n+3}}{2n+3}.$ The smallest value of $n$ for which this error is smaller than $5\cdot 10^{-7}$ is $1.$ So for this problem, the minimum value of $n$ for which the sum of both errors is less than $5\cdot 10^{-7}$ is $4.$

Hence, the desired approximation is $16P_{4,0}(\dfrac{1}{5}) - 4P_{4,0}(\dfrac{1}{239})\\ =16(0.2-\frac{0.2^3}{3}+\frac{0.2^5}{5} - \frac{0.2^7}{7}+\frac{0.2^9}{9}) - 4(\frac{1}{239}-\frac{1}{239^3}\cdot\frac{1}{3}+\frac{1}{239^5}\cdot\frac{1}{5}-\frac{1}{239^7}\cdot\frac{1}{7}+\frac{1}{239^9}\cdot\frac{1}{9})\approx 3.141593.$

Community
  • 1
  • This post and this post are strikingly similar, and posted within 6 hours of each other. – 2'5 9'2 Nov 17 '19 at 04:05
  • @alex.jordan you are indeed correct. I found that question interesting and was trying to figure out a way to reword the question so that Taylor polynomials would be useful. This question is based off of the post you linked in your comment. –  Nov 17 '19 at 04:08
  • Also, what are you trying to imply, @alex.jordan? That this is a duplicate? –  Nov 17 '19 at 04:09
  • It is good practice to link to the question that inspired your own question. As in "I was thinking about this question and it made me wonder about..." This gives people reading your question access to information that may help put your own question in context, and may even help answer your question. – 2'5 9'2 Nov 17 '19 at 04:12
  • @alex.jordan I'll edit my post. –  Nov 17 '19 at 04:12

1 Answers1

0

Do as you mention and center the Taylor series at $0$. Look at it once with $x=1/5$, and again with $x=1/239$, and combine the two results. When you look at it with $x=1/5$, you may need to look at a higher degree Taylor polynomial to get six decimal places of accuracy than you do when you look at it with $x=1/239$. In that sense, you will be looking at Taylor polyonomials plural.

2'5 9'2
  • 54,717
  • Am I supposed to find the minimum degree such that the sum of the errors is less than $5\cdot 10^{-7}$? –  Nov 17 '19 at 04:58