Let f(t) = _ k=1 ¹⁰ ( e^ -4^k t^2 / 2 (2^k t) ) . The value of _ x 0 ₀^x (f(t)) dt x^4 x is equal to

Step-by-step solution

First, simplify the expression for (f(t)) : (f(t)) = _ k=1 ¹⁰ ( - 4^k t^2 2 + ( (2^k t)) ) Using the Maclaurin series expansion for ( u) up to u^4 : ( u) = - ( u) = - (1 - u^2 2 + u^4 24 - ) Using (1+v) v - v^2 2 , where v = - u^2 2 + u^4 24 : ( u) - (- u^2 2 + u^4 24 ) + 1 2 (- u^2 2 )^2 = u^2 2 - u^4 24 + u^4 8 = u^2 2 + u^4 12 Substitute u = 2^k t : ( (2^k t)) 4^k t^2 2 + 16^k t^4 12 The t^2 terms cancel out perfectly: - 4^k t^2 2 + ( (2^k t)) 16^k t^4 12 Summing this geometric progression from k=1 to 10 : (f(t)