Misplaced Pages

User talk:Hayson1991/mathpage: Difference between revisions

Article snapshot taken from Wikipedia with creative commons attribution-sharealike license. Give it a read and then ask your questions in the chat. We can research this topic together.
< User talk:Hayson1991 Browse history interactively← Previous editNext edit →Content deleted Content addedVisualWikitext
Revision as of 02:35, 11 April 2008 editHayson1991 (talk | contribs)242 editsNo edit summary← Previous edit Revision as of 02:35, 11 April 2008 edit undoHayson1991 (talk | contribs)242 editsNo edit summaryNext edit →
Line 1: Line 1:
<math>f(x) = \ln (5-x) \,</math><br /> <math>f(x) = \ln (5-x) \,</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 /> <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 /> <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 /> <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 /> <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 /> <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 /> <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 /> <math>\ln (1) = C-\sum_{n=1}^\infty \frac{(0)^n}{n} \,</math><br /><br />
<math>\ln (1) = C \,</math><br /> <math>\ln (1) = C \,</math><br /><br />
<math>C = 0 \,</math><br /> <math>C = 0 \,</math><br /><br />
<math>f(x) = -\sum_{n=1}^\infty \frac{(x-4)^n}{n} \,</math><br /> <math>f(x) = -\sum_{n=1}^\infty \frac{(x-4)^n}{n} \,</math><br /><br />

Revision as of 02:35, 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}}\,}

ln ⁡ ( 5 − x ) = C − ∑ n = 1 ∞ ( x − 4 ) n n {\displaystyle \ln(5-x)=C-\sum _{n=1}^{\infty }{\frac {(x-4)^{n}}{n}}\,}

ln ⁡ ( 5 − 4 ) = C − ∑ n = 1 ∞ ( 4 − 4 ) n n {\displaystyle \ln(5-4)=C-\sum _{n=1}^{\infty }{\frac {(4-4)^{n}}{n}}\,}

ln ⁡ ( 1 ) = C − ∑ n = 1 ∞ ( 0 ) n n {\displaystyle \ln(1)=C-\sum _{n=1}^{\infty }{\frac {(0)^{n}}{n}}\,}

ln ⁡ ( 1 ) = C {\displaystyle \ln(1)=C\,}

C = 0 {\displaystyle C=0\,}

f ( x ) = − ∑ n = 1 ∞ ( x − 4 ) n n {\displaystyle f(x)=-\sum _{n=1}^{\infty }{\frac {(x-4)^{n}}{n}}\,}