Confusing maclaurin series problem

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SUMMARY

The discussion centers on deriving the Maclaurin series representation for arcsinh(x) based on the series expansion of sinh(x). The series for sinh(x) is given as x + x³/3! + x⁵/5! + ..., leading to the coefficients for arcsinh(x) defined as b0 = a0, b1 = 1/a1 = 1, b2 = -a2/(a1)³ = 0, and b3 = [1/(a5)⁶](2(a2)² - a1a3) = -1/6. Consequently, the series expansion for arcsinh(x) is established as arcsinh(x) = x - (1/6)x³ + ... The coefficients are derived using the nth derivative of the function at zero divided by n factorial, a fundamental concept in series expansion.

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  • Familiarity with hyperbolic functions, specifically sinh(x)
  • Knowledge of derivatives and their applications in series
  • Basic algebraic manipulation of series and factorials
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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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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:
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.
 

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