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Find the first five non-zero terms of the power series representation centered at 0 for the function?

f(x)=ln(6-x)



f(x)= ______+_______+________+________+_______…


what is the radius of convergence?


R=?

Verified Answer :

f(x) = ln(6−x) --------> f(0) = ln(6) 
f'(x) = −1/(6−x) -----> f'(0) = −1/6 
f''(x) = −1/(6−x)² -----> f''(0) = −1/6² 
f'''(x) = −2/(6−x)³ -----> f'''(0) = −2/6³ 
f⁽⁴⁾(x) = −6/(6−x)⁴ -----> f⁽⁴⁾(0) = −6/6⁴ 

f(x) = ln(6) − (1/6) x − (1/6²)/2! x² − (2/6³)/3! x³ − (6/6⁴)/4! x⁴ − ... − ((n−1)!/6^n)/n! x^n − ... 
f(x) = ln(6) − x/6 − x²/(26²) − x³/(36³) − x⁴/(46⁴) − . . . − x^n/(n6^n) − . . . 
f(x) = ln(6) − x/6 − x²/72 − x³/648 − x⁴/5184 − . . . − x^n/(n6^n) − . . . 

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Alternate method: 

Use series for ln(1−x) centred at x = 0 
ln(1−x) = −x − x²/2 − x³/3 − x⁴/4 − . . . − x^n/n − . . . 

f(x) = ln(6−x) 
f(x) = ln(6(1−x/6)) 
f(x) = ln(6) + ln(1−x/6) 
f(x) = ln(6) − x/6 − (x/6)²/2 − (x/6)³/3 − (x/6)⁴/4 − . . . − (x/6)^n/n − . . . 
f(x) = ln(6) − x/6 − x²/(26²) − x³/(36³) − x⁴/(46⁴) − . . . − x^n/(n6^n) − . . . 
f(x) = ln(6) − x/6 − x²/72 − x³/648 − x⁴/5184 − . . . − x^n/(n6^n) − . . .

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