How do you find a power series converging to f(x)=int arcsint/t dt from [0,x] and determine the radius of convergence?
1 Answer
f(x) = x+x^3/18+(3x^5)/200+(5x^7)/784+(35x^9)/10458 + ...
Explanation:
Let us start by using a well known series that is easily shown using a Binomial Series expansion
1/sqrt(1-x^2) = 1+x^2/2+(3x^4)/8+(5x^6)/16+(35x^8)/128 + ...
Then we can form the Maclaurin Series for arcsinx# via integration, as:
arcsinx = int_0^x 1/sqrt(1-t^2) \ dt
" " = int_0^x {1+t^2/2+(3t^4)/8+(5t^6)/16+(35t^8)/128 + ... } \ dt
" " = [t+t^3/6+(3t^5)/40+(5t^7)/112+(35t^9)/1162 + ... ]_0^x
" " = x+x^3/6+(3x^5)/40+(5x^7)/112+(35x^9)/1162 + ...
So now we can re-write
f(x) = int_0^x \ arcsint/t \ dt
" " = int_0^x \ {t+t^3/6+(3t^5)/40+(5t^7)/112+(35t^9)/1162 + ...}/t \ dt
" " = int_0^x \ 1/t \ {t+t^3/6+(3t^5)/40+(5t^7)/112+(35t^9)/1162 + ...} \ dt
" " = int_0^x \ {1+t^2/6+(3t^4)/40+(5t^6)/112+(35t^8)/1162 + ...} \ dt
" " = [t+t^3/18+(3t^5)/200+(5t^7)/784+(35t^9)/10458 + ...]_0^x
" " = x+x^3/18+(3x^5)/200+(5x^7)/784+(35x^9)/10458 + ...
