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

Question

I'm trying to find the second Taylor polynomial ( T_2(x) ) for the function ( f(x) = e^{x^2} ) centered at ( a = 0 ). However, I'm stuck because I calculated the first derivative, ( f'(x) = 2x e^{x^2} ), which equals zero at ( a = 0 ). How can I proceed to find the second Taylor polynomial in this case?

Anonymous · Accepted Answer

by definition, the correct mclaurin series expansion for the function f(x) = e^(x) is 
∞
∑ [ x^(n) / (n!) ] 
n = 0
we have e^( x^2 ) ......call t = x^2 
e^(t)
a)
n = the number of derivatives & number of factorials
f^(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)-----an
0------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 
=======
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Anonymous · Answer

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

Anonymous

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

2 Answers

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