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<math>\frac{x^2}{a^3-x^3} = \frac{\frac{x^2}{a^3}}{1-\frac{x^3}{a^3}}</math> <math>f(x) = \ln (5-x) \,</math><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 />
<math>\int df(x) = \int\left(-\sum_{n=0}^\infty (x-4)^n\right)dx \,</math><br />
<math>f(x) = C-\sum_{n=0}^\infty \frac{(x-4)^{n+1}}{n+1} \,</math><br />
<math>f(x) = C-\sum_{n=1}^\infty \frac{(x-4)^n}{n} \,</math><br />

Revision as of 02:31, 11 April 2008

f ( x ) = ln ⁡ ( 5 − x ) {\displaystyle f(x)=\ln(5-x)\,}
d d x f ( x ) = − 1 5 − x = − 1 1 − ( x − 4 ) = − ∑ n = 0 ∞ ( x − 4 ) n {\displaystyle {\frac {d}{dx}}f(x)=-{\frac {1}{5-x}}=-{\frac {1}{1-(x-4)}}=-\sum _{n=0}^{\infty }(x-4)^{n}\,}
∫ d f ( x ) = ∫ ( − ∑ n = 0 ∞ ( x − 4 ) n ) d x {\displaystyle \int df(x)=\int \left(-\sum _{n=0}^{\infty }(x-4)^{n}\right)dx\,}
f ( x ) = C − ∑ n = 0 ∞ ( x − 4 ) n + 1 n + 1 {\displaystyle f(x)=C-\sum _{n=0}^{\infty }{\frac {(x-4)^{n+1}}{n+1}}\,}
f ( x ) = C − ∑ n = 1 ∞ ( x − 4 ) n n {\displaystyle f(x)=C-\sum _{n=1}^{\infty }{\frac {(x-4)^{n}}{n}}\,}