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>\label{eq:int_log}
+
</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 x^k. Calculus provides us with the formula 
    \int x^k\mathrm{d}x = \frac{x^{k+1}}{k+1}.
However, this equation is only correct for k\neq -1. If we try to take k=-1 the right hand side is meaningless because we have a zero on the denominator of the fraction, i.e. \frac{1}{0}. But in this case a separate argument gives the answer 
    \int x^{-1}\mathrm{d}x = \int \frac{1}{x}\mathrm{d}x = \ln(x).%\ln(|x|).
Our expectation is that these two formulae should be reconciled. Indeed, if we let k approach -1 in the first we should end up with the second, but that fails to happen. For each x, \lim_{k\rightarrow -1} \frac{x^{k+1}}{k+1} is undefined. You can try this on the graph below.

Main article