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Find $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$.
I wrote this proof:

Let $f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$|\sqrt{e}-P_N(\frac{1}{2})|=|R_N(\frac{1}{2})| < \frac{1}{100}$

From the Taylor theorem we get that there exists $0<c<\frac{1}{2}$ such that:
$|R_N(\frac{1}{2})| = |\frac{f^{(N+1)}(c)}{(N+1)!2^N}| = \frac{e^c}{(N+1)!2^N} \leq \frac{e}{(N+1)!2^N}\leq \frac{3}{(N+1)!2^N}\leq \frac{3}{(N+1)!}<10^{-2}$

hence for $N=5$ we get $720>300$ and the inequality holds.
Therefore:

$P_5(\frac{1}{2})=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{6\cdot8}+\frac{1}{16\cdot 24}+\frac{1}{32\cdot 120} = \frac{32\cdot120+16\cdot 120+4\cdot 120+4\cdot 20+2\cdot 5}{32\cdot 120} = \frac{n}{m}$

Did I get it right? Was there a simpler method?

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Yes, there is a simpler method using CF (Continued Fractions). Do you happen to know them ? Here is the result (explanations just after). $$\underbrace{\frac{28}{17}}_{\approx \ 1.6471} < \underbrace{\sqrt{e}}_{\approx \ 1.6487} < \underbrace{\frac{33}{20}}_{= \ 1.65}$$

Explanations : A continued fraction is an expression of the form:

$$a=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{a_4+\cfrac{1}{a_5+\cfrac{1}{a_6+\cdots}}}}}}$$

Here, we rely on this formula:

$$e^{1/2}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{5+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{13+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}}}}}}}}}$$

(recalled for example in this answer to a question where you will find other things about $\sqrt{e}$) with (excepted for the first two $a_k$) a periodicity with $a_k=1$, but for terms $a_{3k+4}=4k+5$...

It is not the place here to recall the theory of continued fractions. It suffices to say that we work with convergents, the $n$th convergent being the rational number obtained by stopping at the $n$th level, erasing everything situated to the right and the bottom of $a_n$ ; here are the first convergents in our case :

$$c_0=1,\ c_1=1+\cfrac{1}{1}=2, \ c_2=1+\cfrac{1}{1+\cfrac{1}{1}}=\frac32,$$

$$c_3=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1}}}=\frac53,\ c_4=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{5}}}}=\frac{28}{17},... $$

It is known that it is among these convergents that one can find the closest rational approximations of the given real number. Moreover, you may have observed that the sequence $c_n$ is alternated around its limit here $\sqrt{e}$ (at one time below, at next time above, etc.). Therefore, we can achieve closer and closer "bracketings" of $\sqrt{e}$ by rational approximations.

Remark: all $e^{\frac{1}{n}}, \ n>1$ have the same kind of continued fractions: see here.

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  • $\begingroup$ I have attempted to be more precise in my answer. Any comment ? $\endgroup$
    – Jean Marie
    Jun 13 at 15:00
  • $\begingroup$ I haven't studied about this, but your answer was very interesting and educating! Thank you $\endgroup$
    – Lilo
    Jun 13 at 23:12
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Your solution is correct, but unnecessarily complicated. Also, there seems to be an error in your expression for $\left\lvert{R_N\left(\tfrac12\right)}\right\rvert,$ which doesn't invalidate your numerical answer, but adds to the complication of finding it.

You're using Taylor's theorem with the Lagrange form of the remainder. Surely the expression should be: $$ \left\lvert{R_N\left(\frac12\right)}\right\rvert = \left\lvert{\frac{f^{(N+1)}(c)}{(N+1)!\cdot2^{N+1}}}\right\rvert $$ Having $2^{N+1}$ rather than $2^N$ in the denominator makes things slightly easier. Also, you have gone rather too far in your simplifying approximations; the approach is valid, but the calculation is harder than it needs to be.

Continuing in a simpler way, using the fact that $e < 4,$ so $\sqrt{e} < 2$: $$ \frac{e^c}{(N+1)!\cdot2^{N+1}} < \frac{\sqrt{e}}{(N+1)!\cdot2^{N+1}} < \frac{1}{(N+1)!\cdot2^N}. $$

Therefore, all we need is to find $N$ large enough that: $$ (N+1)!\cdot2^N > 100. $$ So we don't need $N = 5$; it is good enough to take $N = 3.$ Then we find: $$ \sqrt{e} - \frac{79}{48} < \frac{1}{192}. $$

We could get the same value for $\tfrac{n}{m}$ even more simply, without using the remainder expression for Taylor's theorem - it's quite easy to get it wrong, and I'm not sure I haven't! - instead just using the infinite series for $\sqrt{e}$: \begin{align*} \sqrt{e} & = 1 + \frac12 + \frac18 + \frac1{48} + \frac1{384}\left(1 + \frac1{5\cdot2} + \frac1{5\cdot6\cdot2^2} + \cdots\right) \\ & < \frac{79}{48} + \frac1{384}\left(1 + \frac1{2} + \frac1{2^2} + \cdots\right) \\ & = \frac{79}{48} + \frac1{192}, \end{align*} an approximation which is still more than good enough $\ldots$ indeed, the accuracy is just the same! (I miscalculated at first.) I presume this is just a coincidence.

(Continued fraction methods, as used in the two previous answers, give more accurate approximations than these - in fact, the most accurate approximations possible - but we don't need that level of precision for this question.)

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  • $\begingroup$ Thank you very much for your descriptive answer! $\endgroup$
    – Lilo
    Jun 13 at 23:11
  • $\begingroup$ Would the downvoter care to explain? $\endgroup$ Jun 14 at 16:51
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You can just compute the continued fraction getting $[1;1,1,1,5,1,1,9,1,1,13,\ldots]$. If you stop after the $5$ this evaluates to $\frac {28}{17}$, which is off by less than $0.0017$

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  • $\begingroup$ $23/14$ also works and is simpler. $\endgroup$
    – lhf
    Jun 13 at 14:18
  • $\begingroup$ @lhf: I tried stopping before the $5$ and got $/frac 53$, which isn't good enough. Usually stopping just before a big number means you are very close, but $5$ isn't so big. You are right that is good enough. $\endgroup$ Jun 13 at 14:21

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