Pi (mathematical constant)/Proofs/Student level proof that 22 over 7 exceeds π: Difference between revisions
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We work out the following integral: | We work out the following integral: | ||
:<math> | :<math> | ||
I \equiv \int_0^1 \frac{t^4(1-t)^4}{1+t^2} \, \mathrm{d}t | I \equiv \int_0^1 \frac{t^4(1-t)^4}{1+t^2} \, \mathrm{d}t | ||
</math> | </math> | ||
It is possible to divide polynomials in a manner that is analogous to long division of decimal numbers. By doing this it, can be shown that | |||
:<math> | :<math> | ||
\frac{t^4(1-t)^4}{1+t^2} = \frac{t^8-4t^7+6t^6-4t^5+t^4}{1+t^2}= t^6 -4t^5 +5t^4 -4t^2 +4 - \frac{4}{1+t^2} | \frac{t^4(1-t)^4}{1+t^2} = \frac{t^8-4t^7+6t^6-4t^5+t^4}{1+t^2}= t^6 -4t^5 +5t^4 -4t^2 +4 - \frac{4}{1+t^2} | ||
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where −4 is the remainder of the polynomial division. | where −4 is the remainder of the polynomial division. | ||
Using: | |||
:<math> | :<math> | ||
\int_0^1 t^n \, \mathrm{d}t = \frac{1}{n+1} | \int_0^1 t^n \, \mathrm{d}t = \frac{1}{n+1} | ||
</math> | </math> | ||
for n=6, 5, 4, 2, and 0 | for n=6, 5, 4, 2, and 0, respectively, one obtains | ||
:<math> | :<math> | ||
\int_0^1 (t^6 -4t^5 +5t^4 -4t^2 +4) \, \mathrm{d}t = \frac{1}{7} - \frac{4}{6} + \frac{5}{5} - \frac{4}{3} +\frac{4}{1} = \frac{22}{7} | \int_0^1 (t^6 -4t^5 +5t^4 -4t^2 +4) \, \mathrm{d}t = \frac{1}{7} - \frac{4}{6} + \frac{5}{5} - \frac{4}{3} +\frac{4}{1} = \frac{22}{7} | ||
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t = \tan(x) \, | t = \tan(x) \, | ||
</math> | </math> | ||
The integrand (expression under the integral) of the integral ''I'' is everywhere positive on the integration interval [0, 1] and, remembering that an integral can be defined as a sum of integrand values, it follows that the integral | The integrand (expression under the integral) of the integral ''I'' is everywhere positive on the integration interval [0, 1] and, remembering that an integral can be defined as a sum of integrand values, it follows that the integral <math>I\,</math> is positive. Finally, | ||
:<math> | :<math> | ||
0 < I = \frac{22}{7} - \pi \quad \Longrightarrow \quad \frac{22}{7} > \pi | 0 < I = \frac{22}{7} - \pi \quad \Longrightarrow \quad \frac{22}{7} > \pi | ||
</math> | </math> | ||
which was to be proved. | which was to be proved. |
Revision as of 05:55, 16 September 2009
We work out the following integral:
It is possible to divide polynomials in a manner that is analogous to long division of decimal numbers. By doing this it, can be shown that
where −4 is the remainder of the polynomial division.
Using:
for n=6, 5, 4, 2, and 0, respectively, one obtains
The following holds
The latter integral is easily evaluated by making the substitution
The integrand (expression under the integral) of the integral I is everywhere positive on the integration interval [0, 1] and, remembering that an integral can be defined as a sum of integrand values, it follows that the integral is positive. Finally,
which was to be proved.