Confusing maclaurin series problem

In summary, the series representation for arcsinh(x) can be found by using the Maclaurin series expansion of sinh(x) and finding the coefficients using the formula of the nth derivative of f(0) divided by n factorial. This process involves taking the derivative multiple times and simplifying the terms to find the coefficients. Another way to derive it is by substituting (-x) for every x in the series for e^x and e^(-x), then simplifying and dividing by two. This should result in the same series expansion for arcsinh(x).
  • #1
demonelite123
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Given that sinh(x) = (e^x - e^-x) / 2, find a series representation for arcsinh(x).

So my book did a Maclaurin series expansion of sinh(x) = x + x^3 / 3! + x^5 / 5! + ...

Then it said: the inverse will have some series expansion which we will write as arcsinh(x) = b0 + b1 x + b2 x^2 + b3 x^3 + ... where the "b"s are coefficients.

we label the coeficients in the series expansion of sinh(x) as a_j. we find that,
b0 = a0, b1 = 1 / a1 = 1, b2 = -a2/(a1)^3 = 0, b3 = [1 / (a5)^6] (2(a2)^2 - a1a3) = -1/6

therefore it follows that arcsinh(x) = x - (1/6)x^3 + ...

how did they get the coefficients? i have no idea where they came from.
 
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  • #2
They way they did it is using the maclaurin series expansion. basically, this coefficient is defined as

[the nth derivative of f(0)] divided by [n factorial]

your book should explain how this is derived, but basically it deals with taking the derivative many times so terms disappearHere is a non-taylor series way to derive it. This is what i thought you originally asked since i didnt read your post.
You know that the series for e^x
e^x = (sigma sign) x^n/n!

Therefore, if you substitute (-x) for every x

e^(-x) = (sigma sign) (-1)^n * x^n/n!looking at the original problem:sinh(x) = (e^x - e^-x) / 2

e^x - e^(-x) = [(sigma sign) x^n/n!] - [(sigma sign) (-1)^n * x^n/n!]

Write out these terms. Every other term cancels, and the others are multiplied by two
i don't feel like writing it out, so you can do so yourself

you should get
e^x - e^(-x) = (a bunch of terms all starting with a 2)

then, you divide both sides by two

(e^x - e^-x) / 2 = (those same terms without that 2)

This should be the answer the book has.
 
Last edited:
  • #3
when i did it your way, all i got was sinh(x) back again which is x + x^3 / 3! + x^5 / 5! + ...

it seems like all you did was convert e^x and e^(-x) into series form and using sinh(x) = (e^x - e^-x) / 2 you got back the exact definition of the series form of sinh(x). the question asks for the series expansion of arcsinh(x) given the series form of sinh(x). I'm curious on how my book did this problem with the coefficients thing. could you explain a little more? i still have no idea what my book was doing.
 

1. What is a Maclaurin series?

A Maclaurin series is a special type of power series that represents a function as a sum of infinite terms. It is centered at zero and is used to approximate functions in calculus and other mathematical fields.

2. Why are Maclaurin series sometimes confusing?

Maclaurin series can be confusing because they involve infinite terms and their accuracy may depend on the function being approximated. Additionally, determining the convergence of a Maclaurin series can be challenging.

3. How do I find the Maclaurin series of a function?

To find the Maclaurin series of a function, you can use the Taylor series formula and plug in x=0. This will give you the Maclaurin series. Alternatively, you can use known Maclaurin series and manipulate them to fit the function you want to approximate.

4. Can Maclaurin series be used for any function?

No, Maclaurin series can only be used for functions that are infinitely differentiable at x=0. This means that the function must have derivatives of all orders at that point.

5. How do I know if a Maclaurin series is accurate?

The accuracy of a Maclaurin series is dependent on the function it is approximating. It is important to check the remainder term of the series to determine how close the approximation is to the actual value of the function. Additionally, using more terms in the series can increase the accuracy.

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