View topic - e<2.8 • Art of Problem Solving

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Author	Message
nayel Posts: 1388 Location: Dhaka Blog: View Blog	Posted: May 05, 2007, 2:07 am • # 1 Prove that $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots <2.8$ _________________ Samin Riasat Ever experienced zero gravity?

Kouichi Nakagawa

Posts: 1340
Location: Japan (日本)
Japan

Posted: May 05, 2007, 3:37 am • # 2

all solutions that arent hidden will be rated spam by me :wink:

$6!=720>625=5^{4}, \ k!>5^{k-2}\ (k\ge6)$
$\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}=\frac{13}{60}=\frac{325}{1500}<\frac{372}{1500}=\frac{31}{125}=\frac{1}{5}+\frac{1}...$
So,
$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\cdots$
$< 1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}+\frac{1}{5^{4}}+\cdots$
$=1+\frac{1}{2}+\sum_{n=0}^\infty\frac{1}{5^{n}}$
$=1+\frac{1}{2}+\frac{1}{1-\frac{1}{5}}$
$=\frac{3}{2}+\frac{5}{4}=\frac{11}{4}=2.75<2.8$

Last edited by Kouichi Nakagawa on May 05, 2007, 7:07 am, edited 1 time in total.

mathwhiz23 Posts: 37 Location: Troy, Michigan	Posted: May 05, 2007, 6:57 am • # 3 nice solution.[/youtube] _________________ I have been sucking at math all year....!

nayel Posts: 1388 Location: Dhaka Blog: View Blog	Posted: May 05, 2007, 7:19 am • # 4 nice! my solution is almost the same. i use the fact that $n!>2^{n}$ , for $n\ge 4$ . _________________ Samin Riasat Ever experienced zero gravity?

mets501

Posts: 71
Location: Queens, NY

Posted: May 05, 2007, 6:24 pm • # 5

Better solution :-)

The well-known Maclaurin series for $e^{x}$ is
$e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots$
Substituting 1 for $x$ , you get
$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots$

It follows (as $e\approx 2.718 < 2.8$ ) that
$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots <2.8$

Q.E.D.

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kohjhsd Posts: 1063 Location: MA	Posted: May 05, 2007, 6:55 pm • # 6 mets501 wrote: ... (as $e\approx 2.718 < 2.8$ ) ...... But we are proving $e<2.8$ . _________________ N

13375P34K43V312

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Location: Lexington, MA
United States

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Posted: May 05, 2007, 7:10 pm • # 7

mets501 wrote:

Better solution :-)

um that's like saying well i cheated and knew the exact value for e and it's less than 2.8 so i win

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mets501

Posts: 71
Location: Queens, NY

Posted: May 05, 2007, 9:19 pm • # 8

13375P34K43V312 wrote:

um that's like saying well i cheated and knew the exact value for e and it's less than 2.8 so i win

Yes, but that is all the problem is asking. If the problem was to prove that $\pi$ is greater than $3$ , is it not significantly rigorous to write $\pi \approx 3.1415926$ , so therefore $\pi > 3$ ?

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kohjhsd

Posts: 1063
Location: MA
Korea, Republic of

Posted: May 05, 2007, 9:32 pm • # 9

mets501 wrote:

13375P34K43V312 wrote:

um that's like saying well i cheated and knew the exact value for e and it's less than 2.8 so i win

Yes, but that is all the problem is asking. If the problem was to prove that $\pi$ is greater than $3$ , is it not significantly rigorous to write $\pi \approx 3.1415926$ , so therefore $\pi > 3$ ?

I think you thought it was showing that the expression on the LHS equals $e$ , but the title of this thread is "e<2.8" so it assumes you know that, I guess. But I am understanding your logic as well. What if we change the problem then, to:

Quote:

Prove that
$e < 2.8.$

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me@home

Posts: 2362
Location: Portland, OR - Check out the Oregon Forum!!!
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Posted: May 05, 2007, 9:51 pm • # 10

In conclusion, this problem is too hard for HSB and the intended solution is of the style mentioned by Nakagawa and (of course) nayel, which are pretty nice. In fact calculating $e$ works on this identity (usually) so it is pointless to replace the lhs with $e$ .

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Hamster1800

Posts: 1736
United States

Posted: May 06, 2007, 3:27 pm • # 11

me@home wrote:

I disagree
Hidden Text

Suppose we already know that $e<3$
Let's recall from our calculus that $|R_{n-1}(x)| \leq |\frac{f^{(n)}(x)(x-c)^{n}}{n!}|$ (This is called Lagrange Error if you haven't had calculus yet).
Let's consider the 5th degree polynomial
$R_{5}(1) \leq \frac{e}{5!}< \frac{3}{120}$

Thus, $e < 1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{3}{120}< 2.8$

So in other words, we see that if we know a bound on $e$ , we can use it to prove a sharper bound on $e$ .

But obviously the other solutions are better.

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Kouichi Nakagawa Posts: 1340 Location: Japan (日本)	Posted: May 06, 2007, 4:34 pm • # 12 I was separated from a problem about the Napier Number, but remembered that there were the following problems by a problem about circle ratio. Prove that pi is greater than 3.05. (Tokyo University entrance examination/science 2003, Problem 6) That is to say Prove that $\pi>3.05$ Maybe, This problem may be introduced somewhere.

mathwhiz23 Posts: 37 Location: Troy, Michigan	Posted: May 06, 2007, 7:24 pm • # 13 i luv the solutions....beautiful. but dont u think that nayel's is not for hsb. idk. _________________ I have been sucking at math all year....!

mathwhiz23 Posts: 37 Location: Troy, Michigan	Posted: May 06, 2007, 7:35 pm • # 14 sry. i meant mets501's solution. great idea, though. _________________ I have been sucking at math all year....!

Ravi B Posts: 2520 Location: New York City	Posted: May 06, 2007, 7:37 pm • # 15 Another method is to use to use the inequality $e < \left( 1+\frac{1}{n}\right)^{n+1},$ where $n$ is a positive integer. Then plug in say $n = 20$ .

nayel

Posts: 1388
Location: Dhaka
Bangladesh

Blog: View Blog

Posted: May 07, 2007, 12:58 am • # 16

okey the solutions are nice. but i just wanted a simple and elementary proof of the inequality $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots <2.8$ without assuming that the $LHS=e$ , assuming even that you don't know what $e$ is, because i think everyone knows that. and if you know it, the problem is obvious.

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kunny

Posts: 10841
Location: Japan
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Posted: May 07, 2007, 1:52 am • # 17

Kouichi Nakagawa wrote:

I was separated from a problem about the Napier Number, but remembered that there were the following problems by a problem about circle ratio.

Prove that pi is greater than 3.05.
(Tokyo University entrance examination/science 2003, Problem 6)

That is to say Prove that $\pi>3.05$

Maybe, This problem may be introduced somewhere.

You can see here.

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1081288167&t=17871

Kouichi Nakagawa

Posts: 1340
Location: Japan (日本)
Japan