Mathematics, 26.07.2019 04:30 anaiyamills

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). what is the radius of convergence? (note: you don’t need to use the ratio test here because we know the radius of convergence of the series p∞ n=0 x n .) (b) use part (a) to find a power series for f(x) = x ln(1 − x). (c) by putting x = 1 2 in your result from part (a), express ln 2 as the sum of an infinite series

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Answer from: liana8

The power series of f(x) = ln(1 - x) is $\rm -\sum^{\infty}_{n=1}\dfrac{x^{n} }{n}$ , the power series of xln(1 - x) is $\rm -\sum^{\infty}_{n=1}\dfrac{x^{n+1} }{n}$ and the value of ln(2) is $\rm \sum^{\infty}_{n=0}\dfrac{1}{n2^n}$ .

Given :

f(x) = ln (1−x)

a) The integration of 1/(1 - x) is given by:

$\rm \int \dfrac{1}{1-x}dx=-ln|1-x| + C$

When |x| >1 :

$\dfrac{1}{1-x} = \sum^{\infty}_{n=0} x^n$

Now, integrate on both sides in the above equation.

$\rm -ln(1-x) = C+\sum^{\infty}_{n=0}\dfrac{x^{n+1} }{n+1}$ --- (1)

Now, at (x = 0) the above expression becomes:

$\rm -ln(1-0) = C+\sum^{\infty}_{n=0}\dfrac{0^{n+1} }{n+1}$

By simplifying the above expression in order to get the value of C.

C = 0

Now, substitute the value of C in expression (1).

$\rm ln(1-x) =-\sum^{\infty}_{n=0}\dfrac{x^{n+1} }{n+1}$

Now, by shifting the index the above expression becomes:

$\rm ln(1-x) =-\sum^{\infty}_{n=1}\dfrac{x^{n} }{n}$

b) Now, multiply by 'x' in the above expression in order to get the power series of (x ln(1 - x)).

$\rm xln(1-x) =-\sum^{\infty}_{n=1}\dfrac{x^{n+1} }{n}$

c) Now, substitute the value x = 1/2 in the expression (1).

$\rm ln2 = \sum^{\infty}_{n=0}\dfrac{1}{n2^n}$

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Answer from: lizzie3545

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For |x| , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

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

If x=0 , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

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

We can shift the index to simplify the sum slightly.

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

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by .

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

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

Answer from: Quest

answer

a the answer hopes this

step-by-step explanation:

Which table shows a proportional relationship?

Answer from: Quest

"it's not the heat, it's the humidity". that's a partly valid phrase you may have heard in the summer, but it's actually both. the heat index, also known as the apparent temperature, is what the temperature feels like to the human body when relative humidity is combined with the air temperature. this has important considerations for the human body's comfort. when the body gets too hot, it begins to perspire or sweat to cool itself off. if the perspiration is not able to evaporate, the body cannot regulate its temperature. evaporation is a cooling process. when perspiration is evaporated off the body, it effectively reduces the body's temperature. when the atmospheric moisture content (i.e. relative humidity) is high, the rate of perspiration from the body decreases. in other words, the human body feels warmer in humid conditions. the opposite is true when the relative humidity decreases because the rate of perspiration increases. the body actually feels cooler in arid conditions. there is direct relationship between the air temperature and relative humidity and the heat index, meaning as the air temperature and relative humidity increase (decrease), the heat index increases (decreases).

step-by-step explanation: