# The exponential function again

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Let us say we want to have a function such that it goes through the point (0, 1) and that it is its one derivative i.e. that y´=y.

Next, we assume it can be written as a power series.

$y = f(x) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ... = \sum\limits_{k = 0}^\infty {{a_k}{x^k}}$

The derivative of this is

$y' = f'(x) = {a_1} + 2{a_2}x + 3{a_3}{x^2} + ... = \sum\limits_{k = 1}^\infty {k{a_k}{x^{k - 1}}}$

Since y´=y for varying values of x, the two equations must be equal term by term, and thus the coefficients too. We get

$\begin{gathered} {a_0} = {a_1},\quad {a_1} = 2{a_2},\quad {a_2} = 3{a_2}, \hfill \\ ...,\quad {a_n} = 2{a_{n + 1}} \hfill \\ \end{gathered}$

Now a1 must be one since we want the function to go through the point (0, 1), and that means that a1=1 and that a1=1=2a2, and thus that a2=1/2, then we get that a2=1/2=3a3, so a3=1/6=1/3! and so on. We thus have an=1/a!. This gives us

$y = f(x) = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... = \sum\limits_{k = 0}^\infty {\frac{{{x^k}}}{{k!}}}$

I.e. the exponential function ex, as expected.

Up a level : Power Series
Previous page : Lagrange´s error term
Next page : Generalized Binomial ExpansionLast modified: Dec 28, 2023 @ 13:54