Question

Let f(z) e-1/2.2 for xメ0, f(0) = 0. (a) Show that the derivative fk (0) exists for all k 21. So, f is Coo everywhere on R. b) Show that the Taylor series of f about p -0 converges everywhere on R but that it represents f only at the origin.

Answer 1

SOLUTION

Given function f(x)=e^{(-1/x^2)} for x not equal to zero and f(0)=0 that means if we substitute 0 in place of x then the power of -infinity to e will give zero.

(a). f(x)=e^{(-1/x^2)}

f⁽¹⁾(x)=e^{(-1/x^2})/2!(x³).

f⁽²⁾(x)=(1/2!)(x³(e^{(-1/x^2)}/2!(x³)-(e^{(-1/x^2})(3x²))/(x⁶)..............

Like that derivatives will go

f⁽¹⁾(0)=0,f⁽²⁾(0)=0.......

Derivative exists for all k>=1.So,f is converges to infinity that means f is C^& every where on the real axis.that means f is converges to 0 for derivative changing from 1 to infinity times from above explanation.

(b).Now,take formula for taylor series expansion

f(x)=f(p)+f^'(p)(x-p)+f^"(p)(x-p)²/2!+..........

Now,if p=0 then

f(x)=f(0)+f^'(0)(x)+(f^"(0))(x²)/2+........

Already given in the problem that f(0)=0 and we also got f^'(0)=f^"(0)=.....=0.

So,substitute these values in the above expansion

Finally we will get

f(x)=0.So,it is converges everywhere on real axis but it represents f only at origin because we got f(x)=0 and it is 0 for 0 value of x that means f(0)=0.So, x value zero means we are representing f value at origin only and that 0(x=0).