# Question about deriving Maclaurin Series

• Potatochip911
In summary, the series for ##\sin{(\ln{(1+x)})}## can be found using the Maclaurin series expansion for sin x and the Maclaurin series expansion for ln(1+x), as was done in the problem. No software is necessary for this. You should make a point of memorising the power series for sin x and ln(1+x), and for other elementary functions. These basic series will allow you to calculate the series for more complicated examples.
Potatochip911

## Homework Statement

As I've been going through examples in my textbook they are becoming increasingly lengthy to compute and thus I have resorted to using software to complete the task. For example when computing the series for ##\sin{(\ln{(1+x)})}## $$\ln{(1+x)}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \\ \sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}+... \\ \sin{(\ln{(1+x)})}=\sin{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})}\\ =(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})-\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^3}{3!}+\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^5}{5!}$$

## Homework Equations

3. The Attempt at a Solution [/B]
This would take a ridiculously long time to calculate by hand (for me at least) and I was just curious as to whether or not it's normal to encounter difficult expressions to evaluate by hand when finding Maclaurin series.

Potatochip911 said:
This would take a ridiculously long time to calculate by hand (for me at least) and I was just curious as to whether or not it's normal to encounter difficult expressions to evaluate by hand when finding Maclaurin series.

The reason for setting problems like this is to familiarise you with the math and to help you to develop skill in computation. To use software at this stage of the game is counterproductive.

In your problem, you should have recognised that you need to find the series for sin x and the series for ln(1+x), and that you can get the final series by substituting one into the other, as was done above. No software is necessary for this. You have however missed a step: the final series should have the form $a_0 + a_1 x + a_2 x^2 + \cdots$.

In a test or exam, you would be required (unless stated otherwise) to derive the series for sin x and for ln(1+x).

The object of these exercises is 1. to develop skills, thinking and calculational; 2. to impress the theory into your mind by means of concrete examples; 3. to add to your repertoire of basic information that any professional must have if he is to practice his profession. In this case, you should make a point of memorising the power series for sin x and ln(1+x), and for other elementary functions. These basic series will allow you to calculate the series for more complicated examples.

Do not use software for solving your homework problems. It defeats the object of the exercise and is counter productive.

MarcusAgrippa said:
The reason for setting problems like this is to familiarise you with the math and to help you to develop skill in computation. To use software at this stage of the game is counterproductive.

In your problem, you should have recognised that you need to find the series for sin x and the series for ln(1+x), and that you can get the final series by substituting one into the other, as was done above. No software is necessary for this. You have however missed a step: the final series should have the form $a_0 + a_1 x + a_2 x^2 + \cdots$.

In a test or exam, you would be required (unless stated otherwise) to derive the series for sin x and for ln(1+x).

The object of these exercises is 1. to develop skills, thinking and calculational; 2. to impress the theory into your mind by means of concrete examples; 3. to add to your repertoire of basic information that any professional must have if he is to practice his profession. In this case, you should make a point of memorising the power series for sin x and ln(1+x), and for other elementary functions. These basic series will allow you to calculate the series for more complicated examples.

Do not use software for solving your homework problems. It defeats the object of the exercise and is counter productive.
I probably should have worded my question differently, I am doing everything in the problem up until the last part I left up there, at that stage I am then plugging it into the software to expand the function.

Potatochip911 said:
I probably should have worded my question differently, I am doing everything in the problem up until the last part I left up there, at that stage I am then plugging it into the software to expand the function.

It shouldn't take very long to expand the series by hand up to terms in ##x^4## or ##x^5##.

Potatochip911 said:

## Homework Statement

As I've been going through examples in my textbook they are becoming increasingly lengthy to compute and thus I have resorted to using software to complete the task. For example when computing the series for ##\sin{(\ln{(1+x)})}## $$\ln{(1+x)}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \\ \sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}+... \\ \sin{(\ln{(1+x)})}=\sin{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})}\\ =(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})-\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^3}{3!}+\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^5}{5!}$$

## Homework Equations

3. The Attempt at a Solution [/B]
This would take a ridiculously long time to calculate by hand (for me at least) and I was just curious as to whether or not it's normal to encounter difficult expressions to evaluate by hand when finding Maclaurin series.

In this case, the direct approach is easier than you may imagine. Let ##f(x) = \sin(\ln(x+1))## and ##g(x) = \cos(\ln(x+1))##. Then we have
$$f'(x) = \frac{g(x)}{x+1}\\ f''(x) = \frac{-f(x)-g(x)}{(x+1)^2} \\ f'''(x) = \frac{g(x) + 3 f(x)}{(x+1)^3}\\ f^{(4)}(x) = \frac{-10 f(x)}{(x+1)^4} \\ \vdots$$
and so forth. So, getting a general formula for the ##n##th derivative ##f^{(n)}(x)## should not be too hard. From that you can quite easily get the Maclaurin series
$$f(0) + f'(0) x + \frac{1}{2!} f''(0) x^2 + \frac{1}{3!} f'''(0) x^3 + \cdots \\$$

Potatochip911, PeroK and MarcusAgrippa
Potatochip911 said:

## Homework Statement

As I've been going through examples in my textbook they are becoming increasingly lengthy to compute and thus I have resorted to using software to complete the task. For example when computing the series for ##\sin{(\ln{(1+x)})}## $$\ln{(1+x)}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \\ \sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}+... \\ \sin{(\ln{(1+x)})}=\sin{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})}\\ =(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})-\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^3}{3!}+\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^5}{5!}$$

## Homework Equations

3. The Attempt at a Solution [/B]
This would take a ridiculously long time to calculate by hand (for me at least) and I was just curious as to whether or not it's normal to encounter difficult expressions to evaluate by hand when finding Maclaurin series.
You're probably thinking of truncating the series for log after a few terms, multiplying everything out, and then adding them together, right? If you did that, it would be quite tedious and long. Instead, calculate each term one by one, identifying where all the different contributions come from. For example, consider the ##x^4## term. The third term in the expansion of sin (log series) is
$$\frac{1}{5!} \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \right)^5 = \frac{x^5}{5!} \left(1 - \frac{x}{2} + \frac{x^2}{3} -\frac{x^3}{4} + \cdots\right)^5.$$ From this, you can see that this term will contribute only terms of ##x^5## and higher, so for the ##x^4## term, we can ignore it.

The second term is
$$-\frac{x^3}{3!} \left(1 - \frac{x}{2} + \frac{x^2}{3} -\frac{x^3}{4} + \cdots\right)^3,$$ so it will contribute terms of order ##x^3## and higher. For the ##x^4## term specifically, it comes from the product of ##x^3## and the linear term in the expansion of ##(1-x/2+x^2/3-\cdots)^3##, which you can determine using the binomial expansion. The contribution to the ##x^4## term is therefore ##-\frac{1}{3!} x^3 \times (-\frac 32 x) = \frac 14 x^4##. When you combine this result with the ##x^4## term from the first term in the sin expansion, you see the ##x^4## term vanishes.

It gets more complicated as you go to higher-order terms, but to figure out the first few terms of the expansion, it's not too bad.

Potatochip911 and SammyS
vela said:
You're probably thinking of truncating the series for log after a few terms, multiplying everything out, and then adding them together, right? If you did that, it would be quite tedious and long. Instead, calculate each term one by one, identifying where all the different contributions come from. For example, consider the ##x^4## term. The third term in the expansion of sin (log series) is
$$\frac{1}{5!} \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \right)^5 = \frac{x^5}{5!} \left(1 - \frac{x}{2} + \frac{x^2}{3} -\frac{x^3}{4} + \cdots\right)^5.$$ From this, you can see that this term will contribute only terms of ##x^5## and higher, so for the ##x^4## term, we can ignore it.

The second term is
$$-\frac{x^3}{3!} \left(1 - \frac{x}{2} + \frac{x^2}{3} -\frac{x^3}{4} + \cdots\right)^3,$$ so it will contribute terms of order ##x^3## and higher. For the ##x^4## term specifically, it comes from the product of ##x^3## and the linear term in the expansion of ##(1-x/2+x^2/3-\cdots)^3##, which you can determine using the binomial expansion. The contribution to the ##x^4## term is therefore ##-\frac{1}{3!} x^3 \times (-\frac 32 x) = \frac 14 x^4##. When you combine this result with the ##x^4## term from the first term in the sin expansion, you see the ##x^4## term vanishes.

It gets more complicated as you go to higher-order terms, but to figure out the first few terms of the expansion, it's not too bad.
If we are only getting terms with a degree less than or equal to 5 isn't it easier to quickly derive $$(ax+bx^2+cx^3+dx^4)^3=(3(a^2c+ab^2))x^5+(3a^2b)x^4+a^3x^3$$ which gives all the terms with a degree of 5 or less.

Try it and find out. As Marcus says above, one of the reasons for this type of problem is to develop your skills in doing these types of calculations.

Potatochip911 said:
If we are only getting terms with a degree less than or equal to 5 isn't it easier to quickly derive $$(ax+bx^2+cx^3+dx^4)^3=(3(a^2c+ab^2))x^5+(3a^2b)x^4+a^3x^3$$ which gives all the terms with a degree of 5 or less.

It would be quicker to use the method in Post #5. With ##f(x) = \sin(\ln(x+1))## and ##g(x) = \cos(\ln(x+1))## we have the ##n##th derivative of ##f## as
$$f^{(n)}(x) = \frac{A_n f(x) + B_n g(x)}{(x+1)^n},$$
where ##A_n, B_n## are constants and ##A_1 = 0, B_1 = 1##. It follows easily that the ##A, B## obey a simple recursion scheme:
$$A_{n+1} = -n A_n - B_n, \; \; B_{n+1} = A_n - n B_n$$
Using this, we can easily and quickly get ##f^{(n)}(0) = B_n## for ##n## up to 10 or more with little effort (although the magnitudes of ##A,B## grow large quickly as ##n## increases, so a calculator becomes a necessity). You can keep the numerical growth under control by looking instead at ##\alpha_n = A_n / n!## and ##\beta_n = B_n / n!##. These satisfy
$$\alpha_{n+1} = -\frac{n}{n+1} \alpha_n - \frac{1}{n+1} \beta_n, \;\; \beta_{n+1} = \frac{1}{n+1} \alpha_n - \frac{n}{n+1} \beta_n,$$
with ##\alpha_1 = 0, \beta_1 = 1##. The quantity ##\beta_n## is the coefficient of ##x^n## in the Maclaurin expansion of ##f##. Thus, getting directly the first 10 or so terms of the Maclaurin expansion is relatively straightforward.

Potatochip911

## 1. What is a Maclaurin Series and why is it important in mathematics?

A Maclaurin Series is a type of power series expansion that is used to approximate a function as a polynomial. It is important in mathematics because it allows us to express complicated functions in a simpler form, making them easier to manipulate and analyze.

## 2. How do you derive a Maclaurin Series?

To derive a Maclaurin Series, we use the Taylor Series expansion formula, which involves taking derivatives of the function at a specific point (usually 0) and plugging them into the formula. This results in a polynomial expression that approximates the original function near that point.

## 3. What is the difference between a Taylor Series and a Maclaurin Series?

A Taylor Series is a more general form of a power series expansion that can be centered at any point on the function's domain, while a Maclaurin Series is specifically centered at 0. This means that a Maclaurin Series is a special case of a Taylor Series.

## 4. What are some common functions that can be expressed as Maclaurin Series?

Some common functions that can be expressed as Maclaurin Series include trigonometric functions, exponential functions, and logarithmic functions. These series can be used to approximate these functions to a certain degree of accuracy.

## 5. How can Maclaurin Series be used in real-world applications?

Maclaurin Series can be used in a variety of real-world applications, such as in physics, engineering, and finance, to approximate and analyze functions. They are also used in numerical methods for solving differential equations and other mathematical problems.

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