find a power series representation for the following function

d/dx ln(5-x) = -1/5(1-(x/5))

so,
$$-\int\limits_{}^{}\frac{dx}{5(1-\frac{x}{5})} = \sum_{n=0}^{\infty} \int\limits_{}^{}(\frac{x}{5})^{n}dx$$

=

$$\ln(5-x) + c = \sum_{n=0}^{\infty} \frac{5^{n}x^{-n+1}}{n(1-n)}$$
by setting x = 0,
c = -ln(5)

so,

$$\ln(5-x) = \ln(5) + \sum_{n=0}^{\infty} \frac{5^{n}x^{-n+1}}{n(1-n)}$$

what did I do wrong it says this is incorrect.

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Question

find a power series representation for the following function

d/dx ln(5-x) = -1/5(1-(x/5))

so,
$$-\int\limits_{}^{}\frac{dx}{5(1-\frac{x}{5})}  = \sum_{n=0}^{\infty} \int\limits_{}^{}(\frac{x}{5})^{n}dx$$

=

$$\ln(5-x) + c = \sum_{n=0}^{\infty} \frac{5^{n}x^{-n+1}}{n(1-n)}$$
by setting x = 0,
c = -ln(5)

so,

$$\ln(5-x) = \ln(5) + \sum_{n=0}^{\infty} \frac{5^{n}x^{-n+1}}{n(1-n)}$$

what did I do wrong it says this is incorrect.




=

australopithecus · Answer

sorry it should be

$$\ln(5-x)  = \ln(5)- \sum_{n=0}^{\infty} \frac{5^{n}x^{-n+1}}{n(1-n)}$$

I missed the negative

australopithecus · Answer

ugh I took the integral wrong its way too hot out to be doing math ugh

australopithecus · Answer

$$\ln(5-x) =\ln(5) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{5^{n+1}(n+1)}$$

australopithecus · Answer

there it should be fixed now but it is still probably wrong

anonymous · Answer

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