frac{x^2 - y^2}{1!} + frac{x^4 - y^4}{2!} + f | vedclass

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Question

(B) We know that the exponential series is given by $e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$. Substituting $z = x^2$,

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$e^{x^2} - e^{y^2}$

Vedclass · Answer

(B) We know that the exponential series is given by $e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$. Substituting $z = x^2$,we get $e^{x^2} = 1 + \frac{x^2}{1!} + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \infty$. Similarly,substituting $z = y^2$,we get $e^{y^2} = 1 + \frac{y^2}{1!} + \frac{y^4}{2!} + \frac{y^6}{3!} + \dots \infty$. Subtracting the two series: $e^{x^2} - e^{y^2} = (1 - 1) + \frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty$. Thus,the given expression is equal to $e^{x^2} - e^{y^2}$.

$\frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty = $

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