# Macluarin series of 2^x

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1. Mar 18, 2016

### nfcfox

1. The problem statement, all variables and given/known data
Find the Maclaurin series for f(x) by any method.
f(x)=2^x

2. Relevant equations
d/dx(b^x)= ln(b)b^x

3. The attempt at a solution
Ok so I basically took the derivative about 3 or so times and came out with ∑ n=0 to ∞ of ((ln(2))^n(something has to go here))/n!

This much I have right and all of my answer choices have that, I just don't know what to put in terms of x to satisfy the series. Please help!

2. Mar 18, 2016

### Guneykan Ozgul

You need to put x^n for n'th term in the series.

3. Mar 18, 2016

### nfcfox

Right but doesn't that not satisfy the derivatives? Like the first derivative or n=1 is ln(2)2^x but plugging in n=1 to the series with that would give ln(2)x which isn't the same thing...

4. Mar 18, 2016

### nfcfox

Never mind man I'm thinking about this in a way of just plugging numbers in. I just realized the Maclaurin series states it will be (x-a)^n or just x^n. Thanks for you help!

5. Mar 18, 2016

### Guneykan Ozgul

Ooh, the first derivative should be equal to sum of the derivatives of the each term in the series. So the sum should satisfy the series.

6. Mar 18, 2016

### nfcfox

Ok thanks man. Series has overall been very shaky for me. I really don't understand what any of this means.

7. Mar 18, 2016

### Guneykan Ozgul

Maybe this is useful.

8. Mar 18, 2016

### Ray Vickson

$2^x = e^{\ln(2) x} = e^{cx}$, where $c = \ln(2)$.

9. Mar 18, 2016

### nfcfox

What's the point of doing that?

10. Mar 18, 2016

### Ray Vickson

I am not allowed to tell you, since I would be doing the problem for you. However, be assured, it makes everything really, really simple (depending, of course, on what you know already).

11. Mar 18, 2016

### nfcfox

How does it make it easier? We already solved the problem; before, I didn't understand the actual equation.

12. Mar 18, 2016

### Staff: Mentor

In many of these kinds of problems of finding the Maclaurin series for some function, there are two ways to go. One way is using the definition of the Maclaurin series with many derivatives of the given function. The other way uses shortcuts based on the previously found results. If it's at all feasible, the second method is to be preferred, as it's much less work.

For example, given $f(x) = \frac 1 {1 - x}$, you can find the coefficients of the Maclaurin series from f(0), f'(0), f''(0), etc, or you can simply use polynomial long division to arrive at $\frac 1 {1 - x} = 1 + x + x^2 + x^3 + \dots + x^n + \dots$.

13. Mar 18, 2016

### Isaac0427

Maclaurin series for $e^x$:
$x^0 / 0! + x^1 / 1! + x^2 / 2! + x^3 / 3! + x^4 / 4! ...$
Can you rewrite $2^n$ (I use n as x is already used) as some form of $e^x$? What would x be in terms of constants and n?

14. Mar 18, 2016

### nfcfox

Wait so 1/(1-x) gets you that sequence? I'm so confused right now.

15. Mar 18, 2016

### HallsofIvy

Staff Emeritus
Do you know what a "geometric series" is?

16. Mar 18, 2016

### Staff: Mentor

1/(1 - x) gets you that series, yes. The sequence is {1, x, x2, x3, ..., xn, ...}. The series is the sum of the terms in the sequence. It's important to make sure you understand the difference between these two words: sequence and series.

17. Mar 18, 2016

### Ray Vickson

IF you know the Maclaurin series ford the exponential (which I hoped/assumed you do) then it is a snap: $e^y = 1 + y + y^2/2! + y^3/3! + \cdots$ $=\sum_{n=0}^{\infty} y^n/n!$. Put $y = \ln(2) x$ and you are done, with next to zero work.

However, I see that my suggestion in #10 appeared after posts from you that indicated you had already solved the problem; that type of thing happens to me quite often on this Forum, where several posts in a thread appear only after I have composed a message and pressed the "enter" key.