Find the second Taylor polynomial T2(x) for f(x)=e^x^2 based at a=0?

Question

I'm stuck because f'(x)=2x(e^x^2), which is zero at a=0. How do I get around that?

Dr. Juanita Wolff · Accepted Answer

by definition, the correct mclaurin series expansion for the function f(x) = e^(x) is ∞∑ [ x^(n) / (n!) ] n = 0we have e^( x^2 ) ......call t = x^2 e^(t)a)n = the number of derivatives & number of factorialsf^(n)(x) = is first original and the derived equations. f^(n)(0) = plugging 0 in the original equation and derived equations.an = is coefficient numbers of the variables. AS TABLE:n ---- f^(n)() ----f^n(0)-----an0------e^(t)-----1-------(1)/0! ( 0! is equal to one )1------e^(t)-----1--------(1/1! 2------e^(t)-----1--------1/2! 3------e^(t)-----1--------1/3!) e^(t) = 1 t^0 / 0! + [ t^1 / 1! ] + [ 1 t^2 / 2! ] + ...e^(t) = 1 * 1/ 1+ [ t / 1 ] + [ t^2 / 2 ] + ...e^(t) = 1 + t + (1/2) t^2 + ...when t = x^2 e^(x^2) = 1 + x^2 + (1/2) ( x^2 )^2 + ...e^(x^2) = 1 + x^2 + (1/2) x^4 + ...we have 2x * e^( x^2 ) 2x * e^(x^2) = 2x [ 1 + x^2 + (1/2) ( x^2 )^2 + ... ]2x * e^(x^2) = 2x + 2x^3 + x^5 + ...T_2(x) = 2x^3 =======freee to e-mail if have a question

Dr. Juanita Wolff · Answer

e^z =1+z+z^2/2+z^3/6+z^4/24+.ñput z=x^2 and you get e^x^2 = 1+x^2+x^4/2+x^6/6+x^8/24In an other wayf(x) = f(0) +xf´(0)+x^2f´´(0)/2f(0)=1 ,f´(0)=0 f´´(x) = 2[e^x^2+2x^^2 e^x^2) =2 at x=0sof(x) = 1 +x^2

Dr. Katlynn Swaniawski II

Find the second Taylor polynomial T2(x) for f(x)=e^x^2 based at a=0?

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