Express the function e^z as an infinite series. Use Euler's equation (2) and his Section 116 infinitesimal methods to find an infinite series expression for e^z.

This is equation 2

a? = (1+ kz/j)' = 1+ =kz+ 1 (j -1) 2,2+ 1 (j - 1) (j - 2) k323 1 1 . 2 . j 1 . 2j . 3j 1(j - 1) (j -2) (j -3) + K + 24 + .... (2) 1 . 2j . 3j . 4jSection 116. Since j is infinitely large, J - 1 j - 2 = 1, . . = 1, J - 3 = 1, and so forth. J It follows that J - 1 j - 2 J - 3 = = = 2j and so forth. When we substitute 2 3j 4j 4 kz k2 22 k323 K424 these values [into equation (2)], we obtain 1+ 1 + + 1 . 2 1 . 2 . 3 + 1 . 2 . 3 . 4 + .... This equation expresses a relationship between the numbers a and k, since when we let z = 1, we have k a = 1+ + + ... 1 . 2 1 . 2 .3 (3)