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Revision as of 02:35, 11 April 2008 editHayson1991 (talk | contribs)242 editsNo edit summary← Previous edit Revision as of 02:41, 15 April 2008 edit undoHayson1991 (talk | contribs)242 editsNo edit summaryNext edit →
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<math>f(x) = \ln (5-x) \,</math><br /><br /> <math>f(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \,</math><br /><br />
<math>\frac{d}{dx} f(x) = -\frac{1}{5-x} = -\frac{1}{1-(x-4)} = -\sum_{n=0}^\infty (x-4)^n \,</math><br /><br />
<math>\int df(x) = \int\left(-\sum_{n=0}^\infty (x-4)^n\right)dx \,</math><br /><br />
<math>f(x) = C-\sum_{n=0}^\infty \frac{(x-4)^{n+1}}{n+1} \,</math><br /><br />
<math>f(x) = C-\sum_{n=1}^\infty \frac{(x-4)^n}{n} \,</math><br /><br />
<math>\ln (5-x) = C-\sum_{n=1}^\infty \frac{(x-4)^n}{n} \,</math><br /><br />
<math>\ln (5-4) = C-\sum_{n=1}^\infty \frac{(4-4)^n}{n} \,</math><br /><br />
<math>\ln (1) = C-\sum_{n=1}^\infty \frac{(0)^n}{n} \,</math><br /><br />
<math>\ln (1) = C \,</math><br /><br />
<math>C = 0 \,</math><br /><br />
<math>f(x) = -\sum_{n=1}^\infty \frac{(x-4)^n}{n} \,</math><br /><br />

Revision as of 02:41, 15 April 2008

f ( x ) = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! {\displaystyle f(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n}}{(2n)!}}\,}