Question

write the formula for expansion of tan detta in a series of ascending power of detta ?​

Answer 1

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-……….