Question

Calculus,

Can someone explain to me with a simple example: what do we use mac lauren series FOR ?

Answer 1

maclauren series is a very special case of taylor series. we use taylor series to find the expansion of a function.

Answer 2

https://www.khanacademy.org/math/integral-calculus/sequences-series-approx-calc/maclaurin-taylor/v/maclauren-and-taylor-series-intuition

Answer 3

What is "mac lauren series"?

Maclaurin Series are Taylor series expansions of a function about 0,
\[f(x)=f(0) + f \prime(0)x + (f \prime \prime(0)/2!) x ^{2} + ... + (f ^{n}(0)/n!)x^{n} + ...\]
Well that isn't very helpful I guess. What does your book say? Does it have example problems? This must be one of those questions that try to get you to focus your prior learning and what you just read so it is more important for you to think about it than to just find the correct answer.

My book focuses on cosine and sine functions and then shows how they are related to the Maclaurin Series for
\[e ^{ix}\]
function. I am guessing you may be just starting out with differential equations because using the Maclauren series expansion of \[e ^{ix}\]leads to Euler's Method of solving Ordinary Differential equations.

I hope this helps you focus your attention toward what the book or your instructor wanted you to notice.

Answer 4

MacLaurin series is a math technique to express a function in terms of an infinite series.
For example,
\(f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\frac{x^8}{8!}+...\)
By itself, it may not look very useful.
However, we often encounter integrals of simple functions which do not have an explicit solution. This is one case where MacLaurin series comes to our help.
Take the example of the well-known integral related to the normal distribution,
\(\int_0^x e^{-t^2}dt\)
which has no explicit solution.
However, if we expand \(e^{-x^2}\) and integrate term by term, we get a new series that represents the integral, for any given value of x.
Example:
\(f(x)=e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}....\)
Integrating term by term, we get
\(F(x)=\int_0^x e^{-t^2}dt=x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\frac{x^9}{216}-...\)
which represents the integral which has no explicit solution!

Answer 5

@mathmate what is the difference in taylor and maclauren series ?

Answer 6

I know that we use the one to evaluate at "0" and the other one to evaluate at other points ... I mean what is the difference in terms of practical usage like the one you just demonstrated above ?

Answer 7

It is useful for:
1. You can use them to prove trig identities, for example, to prove that:
\[\cos(x)+i\sin(x)=e^{ix}\]
2. it is used to approximate functions with polynomials, and so then we can use this idea to find local minimas and local maximas on a surface/function etc.
That's all I can think of currently

Answer 8

MacLaurin Series is a special case of the Taylor series.
While Taylor series allows us to expand the function about any point, MacLaurin's series expands only about x=0. This suits most purposes, except when x=0 is a vertical asymptote which makes it inconvenient for expansion.
An example is to expand log(x), which is not possible with MacLaurin series, but possible with Taylor series about x=1.
With the above-mentioned understanding, we usually use the two terms almost interchangeably.
See below for expansion of log(x).
http://www.math.com/tables/expansion/log.htm