Maclaurin Series, expressing 2^x as a M series.

[Disowned]

Okay I was given this problem as a challenge question. It simply says expressing 2^x power as a Maclaurin Series. At first, following an example given by my instructor, I thought that by examining the function as I took multiple derivatives I could find a pattern.

By as you can imagine taking multiple derivatives of an exponential function is anything but pretty. So instead I thought to express the function as a logarithm and set 2^x = n, from there I could rewrite it as Log₂ n = x. But I'm having trouble visualizing that as a sum at 0 to infinity. Anyone got any hints?

 

[dynamicsolo]

Well, taking higher derivatives of a^x isn't all that awful... but there is a short cut.

Write 2 as e^{ln 2}. Then our function becomes

y = 2^x = (e^{ln 2})^x = e^{(ln 2) \cdot x}

using properties of exponents. (This is a fairly standard "trick".)

Now call u = (ln 2) · x and use it in the Maclaurin series for e^u.

 

[Disowned]

THAT. IS. BRILLIANT! Thank you, I can use this info.

 

[dynamicsolo]

Yes... yes, it is rather... not that I invented the idea...

The same trick is used to obtain the rule for differentiating exponential functions:

\frac{d}{dx} a^x = \frac{d}{dx} (e^{ln a})^x = \frac{d}{dx} e^{(ln a) \cdot x}<br /> = (ln a) \cdot e^{(ln a) \cdot x} [by the Chain Rule]

= (ln a) \cdot a^x

(a useful differentiation rule to remember; you can see immediately that it gives \frac{d}{dx}e^x = e^x as expected)

So, by extension, we have for f(x) = a^x ,

\frac{d^n}{dx^n} a^x = (ln a)^n \cdot a^x

which would also give you the Maclaurin series you were after.