The tidiness of elementary mathematics: Unterschied zwischen den Versionen
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We all expect a certain tidiness to mathematics, and for many this provides a gratifying aesthetic pleasure. However, there are some wrinkles. One of these arises in the integral of <math>x^k</math>. Calculus provides us with the formula <math> | We all expect a certain tidiness to mathematics, and for many this provides a gratifying aesthetic pleasure. However, there are some wrinkles. One of these arises in the integral of <math>x^k</math>. Calculus provides us with the formula <math> | ||
\int x^k\mathrm{d}x = \frac{x^{k+1}}{k+1}. | \int x^k\mathrm{d}x = \frac{x^{k+1}}{k+1}. | ||
− | </math> However, this equation is only correct for <math>k\neq -1</math>. If we try to take <math>k=-1</math> the right hand side is meaningless because we have a zero on the denominator of the fraction, i.e. <math>\frac{1}{0}</math>. But in this case a separate argument gives the answer <math> | + | </math> However, this equation is only correct for <math>k\neq -1</math>. If we try to take <math>k=-1</math> the right hand side is meaningless because we have a zero on the denominator of the fraction, i.e. <math>\frac{1}{0}</math>. But in this case a separate argument gives the answer <math> |
\int x^{-1}\mathrm{d}x = \int \frac{1}{x}\mathrm{d}x = \ln(x).%\ln(|x|). | \int x^{-1}\mathrm{d}x = \int \frac{1}{x}\mathrm{d}x = \ln(x).%\ln(|x|). | ||
</math> Our expectation is that these two formulae should be reconciled. Indeed, if we let <math>k</math> approach <math>-1</math> in the first we should end up with the second, but that fails to happen. For each <math>x</math>, <math>\lim_{k\rightarrow -1} \frac{x^{k+1}}{k+1} </math> is undefined. You can try this on the graph below. | </math> Our expectation is that these two formulae should be reconciled. Indeed, if we let <math>k</math> approach <math>-1</math> in the first we should end up with the second, but that fails to happen. For each <math>x</math>, <math>\lim_{k\rightarrow -1} \frac{x^{k+1}}{k+1} </math> is undefined. You can try this on the graph below. | ||
== Main article == | == Main article == |
Version vom 8. November 2011, 23:18 Uhr
We all expect a certain tidiness to mathematics, and for many this provides a gratifying aesthetic pleasure. However, there are some wrinkles. One of these arises in the integral of . Calculus provides us with the formula However, this equation is only correct for . If we try to take the right hand side is meaningless because we have a zero on the denominator of the fraction, i.e. . But in this case a separate argument gives the answer Our expectation is that these two formulae should be reconciled. Indeed, if we let approach in the first we should end up with the second, but that fails to happen. For each , is undefined. You can try this on the graph below.