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...Guneykan Ozgul said:You need to put x^n for n'th term in the series.
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!Guneykan Ozgul said:You need to put x^n for n'th term in the series.
Ok thanks man. Series has overall been very shaky for me. I really don't understand what any of this means.Guneykan Ozgul said: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.
nfcfox said:Homework Statement
Find the Maclaurin series for f(x) by any method.
f(x)=2^x
Homework Equations
d/dx(b^x)= ln(b)b^x
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!
What's the point of doing that?Ray Vickson said:##2^x = e^{\ln(2) x} = e^{cx}##, where ##c = \ln(2)##.
nfcfox said:What's the point of doing that?
How does it make it easier? We already solved the problem; before, I didn't understand the actual equation.Ray Vickson said: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).
Ray Vickson said:##2^x = e^{\ln(2) x} = e^{cx}##, where ##c = \ln(2)##.
nfcfox said:What's the point of doing that?
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.nfcfox said:How does it make it easier? We already solved the problem; before, I didn't understand the actual equation.
Wait so 1/(1-x) gets you that sequence? I'm so confused right now.Mark44 said: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##.
Do you know what a "geometric series" is?nfcfox said:Wait so 1/(1-x) gets you that sequence? I'm so confused right now.
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.nfcfox said:Wait so 1/(1-x) gets you that sequence? I'm so confused right now.
nfcfox said:How does it make it easier? We already solved the problem; before, I didn't understand the actual equation.
The Maclaurin series of 2^x is a mathematical representation of the function 2^x using a series of terms that involve powers of x. It is named after Scottish mathematician Colin Maclaurin who first developed it.
The Maclaurin series of 2^x is derived by using the Taylor series expansion for the function 2^x, centered at x=0. This involves finding the derivatives of the function at x=0 and plugging them into the general form of the Taylor series.
The general form of the Maclaurin series of 2^x is given by: 2^x = 1 + x*ln(2) + (x^2/2!)*(ln(2))^2 + (x^3/3!)*(ln(2))^3 + ... + (x^n/n!)*(ln(2))^n + ...
The Maclaurin series of 2^x is significant because it allows us to approximate the value of 2^x for any value of x, by using a finite number of terms in the series. This can be useful in various mathematical and scientific calculations.
The accuracy of the Maclaurin series of 2^x depends on the number of terms used in the series. The more terms included, the closer the approximation will be to the actual value of 2^x. However, as x gets larger, more terms are needed for an accurate approximation.