4
$\begingroup$

Let $a=3.00000000001234...$ (irrational number)

If $\overline{a}=3.00000000001$ (approximation $11$ places) then $|a-\overline{a}|<10^{-11}$

Note that the reciprocal is not satisfied:

If $\overline{a}=2.99999999998$ (approximation $0$ places) but $|a-\overline{a}|<10^{-10}$


How calculate $\pi$ to an accuracy of $10$ decimal places ?

Note that $|\pi-\overline{a}|<10^{-10}$ not guarantee the accuracy of one decimal place of $\pi$.

$\overline{a}:$ approximation

Any hints would be appreciated.

$\endgroup$
2
$\begingroup$

Hint: $\pi$ has an irrationality measure of no more than $7.6063$.

$\endgroup$
1
$\begingroup$

Your definition of number of places only has problems close to "rollovers of the odometer". I think it would be more normal to consider $11$ places satisfied when your decimal is within $\frac 5{10^{12}}$ so it rounds properly, even if the rounding propagates over many places. As $\pi \approx 3.141592653589793$ there is no rollover problem greater than one in these places. Once you find a digit other than zero or nine you are exempt from the rollover.

$\endgroup$
  • $\begingroup$ It is more normal to think about accuracy that way but it does not answer the question. $\endgroup$ – Dan Brumleve Sep 1 '13 at 6:21
0
$\begingroup$

The following idea can be generalized to get better approximations of $\pi$:

By D.P. Dalzell

$$\int_0^1\frac{x^4 (1-x)^4}{1+x^2}\,\text{d}x=\frac{22}{7}-\pi$$

$$\frac{1}{1260} = \int_0^1\frac{x^4 (1-x)^4}{2}\,\text{d}x < \int_0^1\frac{x^4 (1-x)^4}{1+x^2}\,\text{d}x < \int_0^1\frac{x^4 (1-x)^4}{1}\,\text{d}x = {1 \over 630}.$$

Thus we have $${22 \over 7} - {1 \over 630} < \pi < {22 \over 7} - {1 \over 1260}$$

Hence $3.1412... < \pi < 3.1421...$ in decimal expansion then $$\pi=3.14...$$

Accuracy of 2 decimal places.

$\endgroup$
  • $\begingroup$ In general it may take a rational approximation of accuracy $10^{-7.6063 \cdot k}$ to guarantee $k$ correct digits. See my answer. $\endgroup$ – Dan Brumleve Sep 1 '13 at 18:29
0
$\begingroup$

just an idea:

We want to aproximate $A=a_0.a_1a_2\ldots a_n \ldots$ with $B=b_0.b_1b_2\ldots b_n \ldots$.

Claim 1: If $0<A-B<10^{-n-1}$ and $b_{n+1}\neq 9$ then $a_k=b_k \; \forall k\le n$.

Claim 2: If $0<B-A<10^{-n-1}$ and $b_{n+1}\neq 0$ then $a_k=b_k \; \forall k\le n$.

Proof of claim 1: Suppose $b_k\neq a_k$ for some $k\le n$ (for the sake of simplicity I will suppose $a_n\neq b_n$ and $a_k=b_k$ if $k<n$, but the proof is similar in the general case). Then

$$ 10^{-n-1}>A-B=\sum_j (a_j -b_j)10^{-j}>10^{-n} - |a_{n+1}-b_{n+1}|10^{-n-1} - 9 \sum_{j\ge n+2} 10^{-j} \\>10^{-n} -8 * 10^{-n-1} - 10^{-n-1} =10^{-n-1} $$ A contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.