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The Taylor Series of f(x) = exp(-x^2) at x = 0 is 1-x^2...
Why is this?
The formula for Taylor Series is f(x) = f(0) + (x/1!)(f'(0)) + (x^2/2!)(f''(0)) + ...
and f'(x) = -2x(exp(-x^2)) therefore f'(0) = 0???
Can someone please explain why it is 1-x^2???
Why is this?
The formula for Taylor Series is f(x) = f(0) + (x/1!)(f'(0)) + (x^2/2!)(f''(0)) + ...
and f'(x) = -2x(exp(-x^2)) therefore f'(0) = 0???
Can someone please explain why it is 1-x^2???