Approximation of an expression (no calculator please!)

Question

Today I had my college admission exam, It was good, but there was a question which I found a bit interesting (but unable to solve at the moment). It says,

Question: Find the positive integer which is just equal to the expression $$(1+0.0001)^{10000}$$

3

4

5

6

Well, I tried this with binomial, even with limits, but no solid deduction till yet. Any way to get out with this one?

Note: Not a homework question!

Thanks in advance.

@Paul, you might as well post that as an answer... (possibly with an estimate of the error) — A.P., Apr 21 '15 at 14:49
@Paul The point is not reputation. If you post an answer, then the OP can accept it. Otherwise it will regularly pop-up on the front page. — A.P., Apr 21 '15 at 14:54
The answer is none of the above, as the expression does not evaluate to a integer. — Ross Millikan, Apr 21 '15 at 14:56
I don't understand the question: The expression does not evaluate to an integer. — copper.hat, Apr 21 '15 at 14:57
@A.P. OK, good point. Too busy at the mo & as said, it isn't terribly close to 3! — Paul, Apr 21 '15 at 14:59
For the record, according to WolframAlpha the number in question is about $2.7181459 \dotsc$, so definitely not an integer... — A.P., Apr 21 '15 at 15:03
@RossMillikan I guess from the phrase "just equal to" they meant that we will be needing a minor approximation. Although answer approximation isn't that much minor. You can freely blame the question. — Saharsh, Apr 21 '15 at 15:31

danimal · Answer 1 · 2015-04-21T15:30:55.553

3

You can use the binomial expansion, which approximates the sum of reciprocals of the factorials: $$(1+0.0001)^{10000} = 1^{10000}+10000\times0.0001+{10000\times9999\over2}\times0.0001^2+{10000\times9999\times9998\over2\times3}\times0.0001^3+{10000\times9999\times9998\times9997\over2\times3\times4}\times0.0001^4+...\simeq1+1+1/2+1/6+1/24+1/120\simeq 3$$

Also there is the (well-known?) expression for the number $e$: $$e=\lim_{n\to\infty}(1+1/n)^n$$ and $$e\simeq 3$$

edited Apr 21 '15 at 15:30

answered Apr 21 '15 at 14:57

danimal

1,889

1

Seriously, one needs great persistence to attempt this one with the first method, I even tried it but dropped it in the mid-way thinking that this would take took long. Anyway thanks! – Saharsh Apr 21 '15 at 15:39

A.P. · Answer 2 · 2015-04-21T15:34:27.843

Observe that your number is $f(10^4)$, where $$ f(n) := \left( 1 + \frac{1}{n} \right)^n $$ and recall that $e = \lim\limits_{n \to \infty} f(n)$. Further, observe that $f(n)$ is strictly increasing for $n>1$, therefore $f(10^4) < e \approx 2.71828$.

If you don't already know that $f(n)$ is increasing, you could prove it by recalling that $\log(1 + \frac{1}{x}) = x - \frac{x^2}{2} + O(x^3)$ and noting that $$ \begin{gather} n \left( \frac{1}{n} - \frac{1}{2n^2} \right) < (n+1) \left( \frac{1}{n+1} - \frac{1}{2(n+1)^2} \right) \\ 1 - \frac{1}{2n} < 1 - \frac{1}{2(n+1)} \end{gather} $$

score 0 · Answer 3 · answered Apr 21 '15 at 15:43

0

there is no positive integer that is equal to $ (1+0.0001)^{10000}$ because $$2 < (1+0.0001)^{10000} < e < 3.$$ this follows for two reasons: (a) $(1+\frac 1 x)^x$ is an increasing function, (b) $\lim_{x \to \infty}(1+\frac 1 x)^x = e.$

answered Apr 21 '15 at 15:43

abel

29,170

Approximation of an expression (no calculator please!)

3 Answers3