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December 27, 2009

Find the Maclaurin Series for sqrt(x+1)

Filed under: Uncategorized — Tags: — aspensmonster @ 7:41 am

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This problem was given on a Calculus II review a few semesters back. Our class spent all 50 minutes trying to develop an expression for the given Maclaurin series, but could not find one. The best we could do was tack on an additional infinite series to the expressions we did develop. One all-nighter and a bunch of caffeine later, I found an expression.

Problem statement:

Find the Maclaurin series for f(x) = \sqrt{x+1}


First, we will make a table of the values for f^{n}(x) and f^{n}(0):

(Noting that LaTeX tables are not supported with standard WP)

n f^{n}(x) f^{n}(0)
0 \left(x+1\right)^{1/2} \frac{1}{1}
1 \frac{1}{2}\left(x+1\right)^{-1/2} \frac{1}{2}
2 \frac{-1}{4}\left(x+1\right)^{-3/2} \frac{-1}{4}
3 \frac{3}{8}\left(x+1\right)^{-5/2} \frac{3}{8}
4 \frac{-15}{16}\left(x+1\right)^{-7/2} \frac{-15}{16}
5 \frac{105}{32}\left(x+1\right)^{-9/2} \frac{105}{32}
6 \frac{-945}{64}\left(x+1\right)^{-11/2} \frac{-945}{64}
7 \frac{10395}{128}\left(x+1\right)^{-13/2} \frac{10395}{128}
8 \frac{-135135}{256}\left(x+1\right)^{-15/2} \frac{-135135}{256}

The denominator of this expression is clear: 2^{n}. However, the numerator is being a pain in the ass right off the bat:

  • The first two values (of the fraction as a whole, not just the numerator) are positive before alternating sign
  • The first three values (of the numerator) have magnitude 1
  • The remaining values (of the numerator) follow a seemingly benign sequence: 3, 15, 105, 945, 10395, 135135

To make this easier to analyze, ignore signs for now and simply focus on magnitudes. In doing this, notice the following pattern:

n numerator magnitude factored
0 1 1
1 1 1
2 1 1
3 3 1*3
4 15 1*3*5
5 105 1*3*5*7
6 945 1*3*5*7*9
7 10395 1*3*5*7*9*11
8 135135 1*3*5*7*9*11*13

Isn’t that sexy? It’s behaving in a factorial manner once we get past the first two values, but only with odd numbers! This right here is what made finding the solution an all-nighter. I tried for several hours to develop an expression that yielded such behavior. I failed. Instead I went to Google and found, via Wolfram Mathworld, that this sort of thing is called a double factorial:

  \left(n\right)!! \equiv \begin{cases} n\left(n-2\right)...5\cdot 3\cdot 1 & n > 0, odd \\ n\left(n-2\right)...6\cdot 4\cdot 2 & n > 0, even \\ 1 & \text{n=-1,0} \end{cases}  

So… we could try to define our numerator as \left(2n-3\right)!!. We do this because we want the first three values to equate to one. When n is one or two, we get the expected “ones” at the beginning of the sequence. When n is three or greater, the pattern of 3, 15, 105 … holds. However, there is a problem if we try to take care of the zero case: n!! is undefined for -3. So the question becomes: How can we take care of the zero case while also ensuring that the pattern is taken care of? Another question becomes: Why the fuck is this problem so difficult?

Well, we could do this:


Which takes care of that pesky zero case while also maintaining the pattern. As it turns out, there is another expression that can help us simplify this expression even further:

  =[\left(2n-1\right)\left(2n-3\right)\cdots 1][2n][2\left(n-1\right)][2\left(n-2\right)]\cdots 2\left(1\right) \\  
  =[\left(2n-1\right)\left(2n-3\right)\cdots 1][2n\left(2n-2\right)\left(2n-4\right)\cdots 2] \\   
  =2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)\left(2n-4\right)\cdots 2\left(1\right) \\   

Which gives: \left(2n-1\right)!! = \frac{\left(2n\right)!}{2^{n}n!}

So… we now have the numerator figured out; the “ones” all work and the benign sequence still follows. Which means we have the entire f^{n}\left(0\right) fraction figured out as well:

\frac{\left(2n-3\right)!!}{2^{n}} \equiv \frac{\frac{\left(2n-1\right)!!}{2n-1}}{2^{n}} \equiv \frac{\frac{\left(2n\right)!}{2^{n}n!\left(2n-1\right)}}{2^{n}} \equiv \frac{\left(2n\right)!}{4^{n}n!\left(2n-1\right)}

So… we have an expression that solves two of our three gripes with this problem:

  • The first three values of the numerator have magnitude 1
  • The remaining values of the numerator follow a seemingly benign sequence: 3, 15, 105, 945, 10395, 135135

Like I said, lets continue to ignore signs for now. Plugging them in at this point would still only give you the first term being negative with the rest being positive anyway. Now that we have the fraction taken care of, there is nothing left to do but add the x^{n} and \frac{1}{n!} and \left(-1\right)^{n} into the mix to complete the sequence that this infinite series expands to:

a_{n} = \frac{\left(-1\right)^{n}\left(2n\right)!}{4^{n}n!^{2}\left(2n-1\right)}x^{n}

\left(x+1\right)^{1/2} = \displaystyle\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}\left(2n\right)!}{4^{n}n!^{2}\left(2n-1\right)}x^{n} = \displaystyle\sum_{n=0}^{\infty} a_{n}

But alas, the signs are wrong. From n=0 to infinity, we have negative, negative, positive, negative, postive, negative, etc. What we need to do is swap these. The key? Change \left(2n-1\right) in the denominator to \left(1-2n\right) . This maintains the magnitudes and provides the needed swap of signs. It also gives the final answer:

\left(x+1\right)^{1/2} = \displaystyle\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}\left(2n\right)!}{4^{n}n!^{2}\left(1-2n\right)}x^{n}

Isn’t math awesome?


  1. great!!

    Comment by moonhee — December 5, 2010 @ 12:40 pm

  2. Extremely good!

    Comment by Sam — September 5, 2011 @ 3:27 pm

  3. I don’t fully understand how you’d educed your numerator to be: (2n – 3)!!

    Comment by David — February 2, 2014 @ 3:24 am

  4. Well, when I was first tackling the problem… five years ago… man does time fly! The internet continues to amaze me and I wouldn’t be surprised if this page gets views perpetually. Anyway…

    I wanted to get a handle on the values for the numerator and denominator of f^(n)(0) and work my way from there to a solution. Looking at the numerator values, I factored them and found the pattern in the table. Since the numerator consisted of double factorials of odd integers, I wanted an expression that would generate odd integers, such as:


    However, if I just used this, I’d get something like this:

    n | (2n-1) | (2n-1)!!
    0 -1 1
    1 1 1
    2 3 3
    3 5 15

    The problem of course is that the actual sequence for sqrt(x+1) had THREE “one’s,” not just two. For example, for n = 3, the value in our original table for the numerator of f^(3)(0) was 3, NOT 15. If I left it as (2n-1)!! the sequence would be off. I knew I wanted to manipulate the (2n-1)!! expression to give me THREE “one’s” rather than two. So I decided to shift the (2n-1)!! pattern down one index by using (2n-3)!!, like so:

    n | (2n-3) | (2n-3)!!
    0 -3 ???? [(-3)!! is not defined]
    1 -1 1
    2 1 1
    3 3 3

    Notice how now, n=3 aligns with the correct numerator value for f^(3)(0) of 3, rather than 15. The problem of course is that (-2)!! is not defined. So, is there a way to make sure that the numerator is defined for all n >= 0 while also maintaining the pattern? Yep. Just do this:

    (2n-1)!! / (2n-1)

    This gives the following:

    n | (2n-1) | (2n-1)!! | ((2n-1)!! / (2n-1))
    0 -1 1 -1 [Don’t forget about that negative sign :D]
    1 1 1 1
    2 3 3 1
    3 5 15 3
    4 7 105 15

    Another way of “proving” this visually is realizing what the expression expands to:

    (2n-1)!! = (2n-1) * (2n-3) * (2n-5) * (2n – 7) * …

    And so, ((2n-1)!!/(2n-1)) is another way of saying

    (2n-3)!! = (2n-3) * (2n-5) * (2n-7) * …

    in a way that our expression is defined for all n in our series (n >= 0). The whole point is really just that (2n-3)!! is not defined if you start your index at n=0. You could try to simply state that (-3)!! is also 1 by definition, but then you’d have to change your definitions for everything else too (for regular factorials, and so on). Down that path be dragons. The definitions are the way they are for a reason 😀

    Comment by aspensmonster — February 2, 2014 @ 4:11 am

  5. You’re a genius man. I had the same problem in my class. After 20 minutes or so, my professor frantically wrote up an answer, but I knew it was wrong and didn’t produce the correct result. So, after trying for hours to figure it out on my own, I also ended up turning to the internet. Your result came up in the top.

    Your explanation is fantastic. I don’t know how you ended up figuring this all out. I thought I loved math… but I guess I am wrong.

    Thanks for posting this.

    Comment by Mark — March 20, 2014 @ 9:45 pm

  6. Of course you still love math! Math is awesome, even when we don’t get it. I wracked my brain over this for quite a while, and still find myself doubting the answer occasionally. I know I followed the process, and found some helpful relations to reduce the double factorials, but it certainly seems like this particular mathematical construct touches on some relatively high-level branches of mathematics. Trying to confirm my answer with wolfram alpha leads me down a rabbit hole, from “binomial coefficient” to “Sterling Numbers” to the “Pochhammer Symbol,” whatever the hell that is. The initial numerical expansion provided by wolfram alpha matches my results — in the “Series expansion at x=0” section– but I would only really trust the expertise of a professional mathematician for confirming all the fancy formalisms of pure mathematics.

    Comment by aspensmonster — March 21, 2014 @ 10:03 pm

  7. Just wondering what would happen if you switched the signs… sqrt(1-x). Would the signs then flip flop?

    Comment by BigLouDotCom — March 21, 2014 @ 8:45 pm

  8. It’s been forever since I first looked at this (and every few months I get to revisit it :D) but… I’d suspect that, if the signs were swapped, it wouldn’t be possible. Not strictly in the Real domain, anyway. When it came time to evaluate f^(n)(x) at x=0, you’d get a negative number under the roots, which isn’t defined in the Real domain. My intuition would tell me just to tack on an “i” beside the “x(n)” in the final expression, but seeing as I was an engineering major and not a math major, I’m sure there’s something tricky about complex arguments in Taylor/Maclaurin series that I’ll miss. On the other hand… wolframalpha does seem to be suggesting that simply tacking an “i” next to the x^n is a sufficient start, though it throws additional sign issues into the mix. See: . From the looks of things –and some quick back of the envelope mental math– it looks like every coefficient from n=1 onward is negative. Don’t quote me on it, but I suspect that if you simply removed the (-1)^n and changed x^n to i(x)^n in the final answer, it would be correct.

    Comment by aspensmonster — March 21, 2014 @ 9:50 pm

  9. (2n-3)!! is actually defined to be equal to -1, realizing this allows you to solve this slightly faster. Personally I would recommend just bringing all factors of 1/2 to the denominator and then factoring (-1) out of the first few expressions with negative signs in them, it is quite easy to see then that the expression is [(-1)^{n-1}(2n-3)!!]/[2^{n}*n!]*x^n when starting at n=0 of course.

    Comment by Mathstudent — May 26, 2015 @ 1:35 am

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