The first four nonzero terms of the series for f(x) from the taylor series is F(x)=5/3-5/9(x-2)+5/27(x-2)²-5/81(x-2)³+....
Given that,
We have to get the first four nonzero terms of the series for f(x), centering at the specified value of a, use the definition of a Taylor series. (Separate your responses into a list using commas.)
f(x)=5/1+x and a=2
We know that,
Take
f(x)=5/1+x and a=2
f(2)=5/1+2=5/3
f'(x)=-5/(1+x)² f'(2)=-5/(1+2)²=-5/9 (differentiating)
f''(x)=10/(1+x)³ f''(2)=10/(1+2)³=10/27
f'''(x)=-30/(1+x)⁴ f'''(2)=-30/(1+2)⁴=-30/81=-10/27
Using the taylor series
F(x)=f(a)-(x-a)f'(a)+(x-a)²/2!f''(a)-(x-a)³/3!f'''(a)+....
F(x)=f(2)-(x-2)f'(2)+(x-2)²/2f''(2)-(x-2)³/3!f'''(2)+....
F(x)=5/3-(x-2)(5/9)+(x-2)²/2(10/27)-(x-2)³/6(-10/27)+....
F(x)=5/3-5/9(x-2)+5/27(x-2)²-5/81(x-2)³+....
Therefore, The first four nonzero terms of the series for f(x) from the taylor series is F(x)=5/3-5/9(x-2)+5/27(x-2)²-5/81(x-2)³+....
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