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Now this formula holds at t= 0, so substituting t= 0 into the above gives the rst required identity Di erentiating with respect to tas we may for t>0, gives the second required identity 2 Proceeding in a similar way as the above example, we can easily show that Fexp( 2 1 2 t)(x) = exp(1 2 x2);x2R. Now this formula holds at t= 0, so substituting t= 0 into the above gives the rst required identity Di erentiating with respect to tas we may for t>0, gives the second required identity 2 Proceeding in a similar way as the above example, we can easily show that Fexp( 2 1 2 t)(x) = exp(1 2 x2);x2R. Y *6/*7&34*5 è8 èÌ y Î Nùt !Ëb ÔùUK b{ 7 B » ² {³ Ê ² { 4³ Ê!迨 ² {³.

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