If [tex]f(x)=\ln(5-x)[/tex], then
[tex]f(0)=\ln5[/tex]
[tex]f'(x)=-\dfrac1{5-x}=\dfrac1{x-5}\implies f'(0)=-\dfrac15[/tex]
[tex]f''(x)=-\dfrac1{(x-5)^2}\implies f''(0)=-\dfrac1{25}[/tex]
[tex]f'''(x)=\dfrac2{(x-5)^3}\implies f'''(0)=-\dfrac2{125}[/tex]
[tex]f^{(4)}(x)=-\dfrac6{(x-5)^4}\implies f^{(4)}(0)=-\dfrac6{625}[/tex]
So the first five terms of the series expansion around [tex]x=0[/tex] are
[tex]\ln5-\dfrac x5-\dfrac{x^2}{25}-\dfrac{2x^3}{125}-\dfrac{6x^4}{625}[/tex]
The first five non-zero terms of power series representation centered at [tex]x = 0[/tex] are contained in [tex]p_{5}(x) = 1.609 + 0.2\cdot x -0.02\cdot x^{2}+(2.667\times 10^{-3})\cdot x^{3}-(4\times 10^{-4})\cdot x^{4}+(6.4\times 10^{-5})\cdot x^{5}[/tex].
Natural logarithms are trascendent functions and transcendent functions can be approximated by Taylor polynomials, a fifth-order Taylor polynomial is represented by the following expression:
[tex]p_{5}(x) = f(a) +\frac{f'(a)}{1!}\cdot (x-a)+\frac{f''(a)}{2!} \cdot (x-a)^{2} +\frac{f'''(a)}{3!}\cdot (x-a)^{3} + \frac{f^{(4)}(a)}{4!}\cdot (x-a)^{4}+\frac{f^{(5)}(a)}{5!}\cdot (x-a)^{5}[/tex] (1)
Where [tex]a[/tex] is the initial point. If we know that [tex]a = 0[/tex] and [tex]f(x) = \ln (5-x)[/tex], then the Taylor polynomial is:
[tex]f'(x) = (5-x)^{-1}[/tex], [tex]f''(x) = -(5-x)^{-2}[/tex], [tex]f'''(x) = 2\cdot (5-x)^{-3}[/tex], [tex]f^{(4)} = -6\cdot (5-x)^{-4}[/tex], [tex]f^{(5)}(x) = 24\cdot (5-x)^{-5}[/tex][tex]p_{5}(x) = \ln 5 +\frac{x}{5}+\left(\frac{1}{2} \right)\cdot \left(-\frac{1}{25} \right) \cdot x^{2}+\left(\frac{1}{6} \right)\cdot \left(\frac{2}{125} \right)\cdot x^{3} +\left(\frac{1}{24} \right)\cdot \left(-\frac{6}{625} \right)\cdot x^{4} + \left(\frac{1}{120} \right)\cdot \left(\frac{24}{3125} \right)\cdot x^{5}[/tex]
[tex]p_{5}(x) = 1.609 + 0.2\cdot x -0.02\cdot x^{2}+(2.667\times 10^{-3})\cdot x^{3}-(4\times 10^{-4})\cdot x^{4}+(6.4\times 10^{-5})\cdot x^{5}[/tex]
The first five non-zero terms of power series representation centered at [tex]x = 0[/tex] are contained in [tex]p_{5}(x) = 1.609 + 0.2\cdot x -0.02\cdot x^{2}+(2.667\times 10^{-3})\cdot x^{3}-(4\times 10^{-4})\cdot x^{4}+(6.4\times 10^{-5})\cdot x^{5}[/tex].
We kindly invite to check this question on power series: https://brainly.com/question/16407904
