Finding the nth term of the maclaurin's expansion of sqrt(1+x)

In summary, the nth term of the Maclaurin's expansion of sqrt(1+x) can be represented as (-1)^n*(2n)!/((1-2n)*(n!)^2*4^n), where n is the term number. Alternatively, the nth term can also be expressed as 1+x/2-x/8+... with a formula for the nth term provided after the first few terms. The pattern for the numerators can be described as a double factorial, written as n!. The (1-2n) in the denominator changes signs between n=0 and n=1, cancelling the sign change from the (-1)^n term.
  • #1
rock.freak667
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[SOLVED] Finding the nth term of the maclaurin's expansion of sqrt(1+x)

Homework Statement



Find the nth term of the Maclaurin's expansion of [itex]\sqrt{1+x}[/itex]

Homework Equations


The Attempt at a Solution



so far I've expanded and gotten

[tex]\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{2!4}+\frac{3x^3}{3!8}-\frac{15x^4}{4!16}+\frac{105x^5}{5!32}[/tex]

so far I've gotten that part of the nth term should be

[tex]\frac{x^n}{n!2^n}[/tex]

but what I do not know is how to deal with the + and - signs as well as the numbers in th numerators...please help me
 
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  • #2
(-1)^n should cover the + and -
as for the numbers, I can't help you. I'm tired. I'll be back after I work on it.
 
  • #3
I think you know the pattern for the numerators. You just don't know the name for it. Look up double factorials written n!.
 
  • #4
But the (-1)^n doesn't work for the 2nd term or after
 
  • #5
You know what, I'll give you an answer. (-1)^n*(2n)!/((1-2n)*(n!)^2*4^n). I didn't work that out, I looked it up because I got annoyed with the problem. Notice the (1-2n) in the denominator changes signs between n=0 and n=1 cancelling the sign change from the (-1)^n. Getting one formula that works for all n can be more of an art than a science. I would consider it perfectly legit to write 1+x/2-x/8+... and then give a formula for the nth term after the first few. It could probably be written more simply than the one fits all version.
 

FAQ: Finding the nth term of the maclaurin's expansion of sqrt(1+x)

How do I find the nth term of the Maclaurin's expansion of √(1+x)?

The nth term of the Maclaurin's expansion of √(1+x) can be found by using the general formula for the nth term of a Maclaurin's series, which is given by:
an = (f(n)(0) / n!) * xn
In this case, f(x) = √(1+x) and f(n)(x) represents the nth derivative of f(x).

What is the Maclaurin's expansion of √(1+x)?

The Maclaurin's expansion of √(1+x) is given by:
√(1+x) = 1 + (1/2)x - (1/8)x2 + (1/16)x3 - (5/128)x4 + ...

What is the significance of the Maclaurin's expansion of √(1+x)?

The Maclaurin's expansion of √(1+x) is significant because it allows us to approximate the value of √(1+x) for any value of x, without having to use a calculator. It is also useful in solving various mathematical problems and in understanding the behavior of functions near the point of expansion.

How many terms should I use to get an accurate approximation of √(1+x)?

The number of terms needed to get an accurate approximation of √(1+x) depends on the value of x. As a general rule, the more terms you use, the more accurate the approximation will be. However, for values of x close to 0, using 2-3 terms is usually sufficient to get a good approximation.

Can the Maclaurin's expansion of √(1+x) be used for values of x other than 0?

Yes, the Maclaurin's expansion of √(1+x) can be used for any value of x within the interval of convergence, which in this case is -1 < x < 1. For values of x outside this interval, the series will either diverge or give inaccurate results.

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