.Expand the following in powers of x cos invers 1x^2-1/x^2+1
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Answer:
f(x)=tan-¹(x)
Here (x)' indicates derivative of x with respect to x.
The maclaurin series of f(x) is defined by
f(x)=f(0)+[xf'(0)]/1! + [x²f’’(0)]/2! +[x³f’’’(0)]/3!+……
Here f'(x)=1/(1+x²)
f''(x)=-2x/(1+x²)²
f’’'(x)=-2[ {(1+x²)²(x)'-x[(1+x²)²]'}/[(1+x²)²]²]
f'''(x)=-2[{(1+x²)²(1)-x[2(1+x²)(2x)]}/[(1+x²)⁴]
f'''(x)=-2[{(1+x²)²-4x²(1+x²)}]/(1+x²)⁴
f'''(0)=-2[{(1+0)²-4(0²)(1+0)²}]/(1+0²)⁴
f'''(0)=-2[1–0]/1=-2
f(0)=tan-¹(0)=0
f'(0)=1/(1+0²)=1
f''(0)=-2(0)/(1+0²)²=0
Substituting the obtained values in the above Maclaurin series expansion we get
Tan-¹(x)=0+x(1)/1!+0*(x²/2!)+(-2)(x³/3!)+…….
tan-¹(x)=x+0-(2/6)x³+……
tan-¹(x)=x-x³/3 + x^5/5-……….
Step-by-step explanation:
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