Aktuelle Version vom 11. November 2011, 00:04 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 $\int x^{-1}\mathrm{d}x = \int \frac{1}{x}\mathrm{d}x = \ln(x).$

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.

This vingette explores this issue.

Main article

Exceptional cases such as this are particularly problematic when we attempt to automate knowledge, that is when we seek to devise a computer algebra system which will symbolically manipulate expressions. If we want to integrate $x^{k}$ with respect to $x$ we certainly have two choices

Ignore the special case $k=-1$ .
Ask the user if $k=-1$ .

Both of these approaches are problematic. $k=-1$ is only one point out of an infinity of other values for $k$ . However, in mathematics such exceptional cases occur rather more often than you might expect if you just stuck a pin in the number line, or even if you chose an integer $k$ at random. Asking the user can also be problematic as they many not yet know the value of $k$ , or the integral might occur deep within some other function and the user may have no idea what the question is asking! Automatically using the formula might result in an error creeping into a much longer calculation. Mistakes like this are very hard to track down and can be serious. These issues are addressed more fully by [2].

What we would really like is a formula that works for all $k$ . Here we are going to consider the comment in [2] that really, at a symbolic level, the answer to $\int x^k\mathrm{d}x$ might better be written as $\int x^k\mathrm{d}x = \frac{x^{k+1}-1}{k+1}+c.$

First consider the (easy) case $k\neq -1$ . If we fix $k$ and differentiate then we obtain $\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^{k+1}-1}{k+1} = x^k.$ So $\frac{x^{k+1}-1}{k+1}=\frac{x^{k+1}}{k+1}-\frac{1}{k+1}$ is an anti-derivative of $x^k$ . Really we have chosen a slightly different constant of integration.

Whereas, for $k=-1$ we originally had the form $\frac{x^{0}}{0}=\frac{1}{0}$ , the new formula gives $\frac{x^{0}-1}{0}=\frac{0}{0}$ . To understand what is going on we need to look at this expression for $k$ close to $-1$ and take a limit, i.e. $\lim_{k\rightarrow -1} \frac{x^{k+1}-1}{k+1}.$ If we can make sense of this limit, then we will define the value of the expression at $k=-1$ by the value of the limit, thus removing the exceptional case in a meaningful way. This sort of thing is a standard mathematician’s move.

Euler’s formula for $e^x = \lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n$

To consider the special case $k=-1$ we shall need to make use of a special formula discovered by Euler.

$e^x = \lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n.$

If we consider the Taylor series for $e^x$ , on the left hand side we have

$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$

On the right, we can expand out the product using the Binomial Theorem

$\left( 1+\frac{x}{n}\right)^n = \sum_{i=0}^n \frac{n(n-1)\cdots(n-i+1)}{i!}\left(\frac{x}{n}\right)^i = \sum_{i=0}^n \frac{x^i}{i!} \frac{n(n-1)\cdots(n-i+1)}{n^i}.$

Consider the coefficients of $x^i$ . This equals $\frac{1}{i!} \frac{n(n-1)\cdots(n-i+1)}{n^i} = \frac{1}{i!} \frac{n}{n}\frac{n-1}{n} \cdots \frac{n-i+1}{n}.$ As $n\rightarrow \infty$ each term here containing $n$ converges to $1$ giving $\lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n = \sum_{i=0}^\infty \frac{x^i}{i!},$ or exactly the Taylor series for $e^x$ .

You might think this argument has made somewhat cavalier use of limits. The following argument, adapted from [1], is close to Euler’s original. He was even more bold in his use of limits!

Euler unhesitatingly accepts the existence of both infinitely small and infinitely large numbers, and uses them to such effect that the modern reader’s own hesitation must be tinged with envy. [1]

Thinking about $e^x$ near $x=0$ , and its gradient, we see that, for infinitely small $\epsilon$ $e^\epsilon = 1 + \epsilon$ Let $x$ be any given number then $N=x/\epsilon$ is infinitely large. So $e^x = e^{N\epsilon} = (e^\epsilon)^N = (1+\epsilon)^N = \left( 1 +\frac{x}{N} \right)^N.$ Written in terms of limits, the infinitely large $N$ would give us our formula.

In this applet move the slider so that $n \rightarrow \infty$ .

The awkward integral when $k=-1$

We shall make use of Euler’s limit to evaluate the awkward integral when $k=-1$ .

Let $x = \left( 1+\frac{y}{n}\right)^n.$

Rearranging this we have $x^{\frac{1}{n}} = 1 + \frac{y}{n}$ and then $y = n \left(x^{\frac{1}{n}} -1\right).$

Let $m=\frac{1}{n}$ so $y = \frac{1}{m} \left(x^{m}-1\right)=\frac{x^m-1}{m}.$

Now, set $m=k+1$ , to give $y =\frac{x^{k+1}-1}{k+1}$ and we are coming somewhat closer to the right hand side of the formula for our integral.

If $k\rightarrow -1$ , then $m\rightarrow 0$ and $n=\frac{1}{m}\rightarrow \infty$ . But if $n\rightarrow \infty$ then we may apply to and conclude that $x=e^y$ . Taking the inverse gives $y=\ln(x)$ . Hence $\lim_{k\rightarrow -1} \frac{x^{k+1}-1}{k+1} = \ln(x)$ and so is indeed valid for all $k$ , including the special case $k=-1$ , when we define the value of in this case by the limit $\lim_{k\rightarrow -1}$ . This limit happens to give the right answer to the integral for $k=-1$ .

We can see this limit in the applet below. Mve the slider so that $k \rightarrow -1$ .

The mathematical moral

They key to all of this is to choose the constant of integration in so that the resulting formula also holds in the awkward limiting case, $k=-1$ . Like many tricks in mathematics it is a technique which occasionally has useful applications elsewhere.

There is a deep consistency to elementary mathematics, sometimes we just have to search to find it.

References

[1] C. H. Edwards. The Historical Development of the Calculus. Springer-Verlag, 1979.

[2] D. R. Stoutemyer. Crimes and misdemeanors in the computer algebra trade. Notices of the American Mathematical Society, 38(7):778-785, September 1991.

@@ Zeile 10: / Zeile 10: @@
 == Main article ==
-== Applets ==
+Exceptional cases such as this are particularly problematic when we attempt to automate knowledge, that is when we seek to devise a ''computer algebra system'' which will symbolically manipulate expressions. If we want to integrate <math>x^{k}</math> with respect to <math>x</math> we certainly have two choices
-In this applet move the slider so that <math>k \rightarrow -1</math>.
+# Ignore the special case <math>k=-1</math>.
+# Ask the user if <math>k=-1</math>.
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+Both of these approaches are problematic. <math>k=-1</math> is only one point out of an infinity of other values for <math>k</math>. However, in mathematics such exceptional cases occur rather more often than you might expect if you just stuck a pin in the number line, or even if you chose an integer <math>k</math> at random. Asking the user can also be problematic as they many not yet know the value of <math>k</math>, or the integral might occur deep within some other function and the user may have no idea what the question is asking! Automatically using the formula  might result in an error creeping into a much longer calculation. Mistakes like this are very hard to track down and can be serious.  These issues are addressed more fully by [2].
-In this applet move the slider so that <math>n \rightarrow \infty</math>.
+What we would really like is a formula that works for all <math>k</math>. Here we are going to consider the comment in [2] that really, at a symbolic level, the answer to <math>\int x^k\mathrm{d}x </math> might better be written as <math>
+    \int x^k\mathrm{d}x = \frac{x^{k+1}-1}{k+1}+c.
+</math>
+First consider the (easy) case <math>k\neq -1</math>. If we fix <math>k</math> and differentiate then we obtain <math>\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^{k+1}-1}{k+1} = x^k.</math> So <math>\frac{x^{k+1}-1}{k+1}=\frac{x^{k+1}}{k+1}-\frac{1}{k+1}</math> is an anti-derivative of <math>x^k</math>. Really we have chosen a slightly different constant of integration.
+Whereas, for <math>k=-1</math> we originally had the form <math>\frac{x^{0}}{0}=\frac{1}{0}</math>, the new formula gives <math>\frac{x^{0}-1}{0}=\frac{0}{0}</math>. To understand what is going on we need to look at this expression for <math>k</math> close to <math>-1</math> and take a limit, i.e.  <math>\lim_{k\rightarrow -1} \frac{x^{k+1}-1}{k+1}.</math> If we can make sense of this limit, then we will define the value of the expression at <math>k=-1</math> by the value of the limit, thus removing the exceptional case in a meaningful way. This sort of thing is a standard mathematician&rsquo;s move.
+= Euler&rsquo;s formula for <math>e^x = \lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n</math> =
+To consider the special case <math>k=-1</math> we shall need to make use of a special formula discovered by Euler.
+<math> e^x = \lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n.</math>
+If we consider the Taylor series for <math>e^x</math>, on the left hand side we have
+<math>e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots </math>
+On the right, we can expand out the product using the Binomial Theorem
+<math>\left( 1+\frac{x}{n}\right)^n
+= \sum_{i=0}^n \frac{n(n-1)\cdots(n-i+1)}{i!}\left(\frac{x}{n}\right)^i
+= \sum_{i=0}^n \frac{x^i}{i!} \frac{n(n-1)\cdots(n-i+1)}{n^i}.</math>
+Consider the coefficients of <math>x^i</math>. This equals <math>\frac{1}{i!} \frac{n(n-1)\cdots(n-i+1)}{n^i} = \frac{1}{i!} \frac{n}{n}\frac{n-1}{n} \cdots \frac{n-i+1}{n}.</math> As <math>n\rightarrow \infty</math> each term here containing <math>n</math> converges to <math>1</math> giving <math>\lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n = \sum_{i=0}^\infty \frac{x^i}{i!},</math> or exactly the Taylor series for <math>e^x</math>.
+You might think this argument has made somewhat cavalier use of limits. The following argument, adapted from [1], is close to Euler&rsquo;s original. He was even more bold in his use of limits!
+<blockquote>Euler unhesitatingly accepts the existence of both infinitely small and infinitely large numbers, and uses them to such effect that the modern reader&rsquo;s own hesitation must be tinged with envy. [1]
+</blockquote>
+Thinking about <math>e^x</math> near <math>x=0</math>, and its gradient, we see that, for infinitely small <math>\epsilon</math> <math>e^\epsilon = 1 + \epsilon
+</math> Let <math>x</math> be any given number then <math>N=x/\epsilon</math> is infinitely large. So <math>e^x = e^{N\epsilon}  =  (e^\epsilon)^N
+ =  (1+\epsilon)^N
+ =  \left( 1 +\frac{x}{N} \right)^N.
+</math> Written in terms of limits, the infinitely large <math>N</math> would give us our formula.
+<p>In this applet move the slider so that <math>n \rightarrow \infty</math>.
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+= The awkward integral when <math>k=-1</math> =
+We shall make use of Euler&rsquo;s limit to evaluate the awkward integral when <math>k=-1</math>.
+Let <math> x = \left( 1+\frac{y}{n}\right)^n.</math>
+Rearranging this we have <math>x^{\frac{1}{n}} = 1 + \frac{y}{n}</math> and then <math>y = n \left(x^{\frac{1}{n}} -1\right).</math>
+Let <math>m=\frac{1}{n}</math> so <math>y = \frac{1}{m} \left(x^{m}-1\right)=\frac{x^m-1}{m}.</math>
+Now, set <math>m=k+1</math>, to give <math>y =\frac{x^{k+1}-1}{k+1}</math> and we are coming somewhat closer to the right hand side of the formula for our integral.
+If <math>k\rightarrow -1</math>, then <math>m\rightarrow 0</math> and <math>n=\frac{1}{m}\rightarrow \infty</math>. But if <math>n\rightarrow \infty</math> then we may apply  to  and conclude that <math>x=e^y</math>. Taking the inverse gives <math>y=\ln(x)</math>. Hence <math>\lim_{k\rightarrow -1} \frac{x^{k+1}-1}{k+1} = \ln(x)</math> and so  is indeed valid for all <math>k</math>, including the special case <math>k=-1</math>, when we define the value of  in this case by the limit <math>\lim_{k\rightarrow -1}</math>. This limit happens to give the right answer to the integral for <math>k=-1</math>.
+We can see this limit in the applet below.  Mve the slider so that <math>k \rightarrow -1</math>.
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+= The mathematical moral =
+They key to all of this is to choose the constant of integration in  so that the resulting formula also holds in the awkward limiting case, <math>k=-1</math>. Like many tricks in mathematics it is a technique which occasionally has useful applications elsewhere.
+There is a deep consistency to elementary mathematics, sometimes we just have to search to find it.
+= References =
+[1] C. H. Edwards. The Historical Development of the Calculus. Springer-Verlag, 1979.
+[2] D. R. Stoutemyer. Crimes and misdemeanors in the computer algebra trade. Notices of the American Mathematical Society, 38(7):778-785, September 1991.

The tidiness of elementary mathematics: Unterschied zwischen den Versionen

Aktuelle Version vom 11. November 2011, 00:04 Uhr

Inhaltsverzeichnis

Main article

Euler’s formula for $e^x = \lim_{n\rightarrow\infty} \left( 1+\frac{x}{n}\right)^n$

The awkward integral when $k=-1$

The mathematical moral

References

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