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Is there a common function with the power series

$$ \dfrac{1}{2}x^2 + \dfrac{3}{8}x^4 + \dfrac{2}{7}x^6 + c_8x^8 + \ldots$$

where $c_8$ is in the 0.2 - 0.3 range?

I'm getting a match for some numerical data and it is likely from a very common function but I can't figure out which.

graphs: below g and g' are two functions that I am trying to match (g is the one with the power series above, $g(x)' =\bigg( 2-\dfrac{4}{x+2}\bigg)$, $f_1 = \sqrt(x^2 + 1) - 1$ and $f_2 = \dfrac{1}{\sqrt{1-x^2}} - 1$ as suggested by some of the answerers.

enter image description here

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    What about this one?$$\frac1{\sqrt{1-x^2}}-1=\frac{x^2}{2}+\frac{3 x^4}{8}+\frac{5 x^6}{16}+\frac{35 x^8}{128}+O(x^{10})$$ – 2017-01-04
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    ooooohhh. that looks promising – 2017-01-04
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    hmmm not quite. – 2017-01-04
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    @JasonS I do believe it is equivalent to what I meant to write. – 2017-01-04
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    Haha, all very close functions. – 2017-01-05
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    One way to test the above suggestion is to note that if $y=(1-x^2)^{-1/2}-1$, then $(y+1)^{-2}=1-x^2$. So if you plot $(y+1)^{-2}$ versus $x^2$, you should get a linear relation. – 2017-01-05

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A quick binomial:

$$-1+(1-x^2)^{-1/2}=\frac12x^2+\frac38x^4+\frac{5}{16}x^6+\frac{35}{128}x^8+\dots$$

Close, but not quite right? Well, we can fix this:

$$-1+(1-x^2)^{-1/2}-\frac3{112}x^6-a_8x^8=\frac12x^2+\frac38x^4+\frac27x^6+c_8x^8+\dots$$

where $0.2<\frac{35}{128}-a_8<0.3$

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    I'm 100% sure about the first two coefficients, and the third is between 0.28 and 0.29, most likely 2/7 but not 100% sure. – 2017-01-04
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    @JasonS I used a calculator to get that one, since my scientific calculator returns fractions in simplest form if the numbers are reasonable. – 2017-01-04
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    $$\sqrt{1 + x^2} = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \frac{1}{16}x^6 + O(x^8),$$ you may have thought of $$(1 - x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \frac{5}{16}x^6 + O(x^8).$$ – 2017-01-04
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    @DanielFischer Likely, my typos. – 2017-01-04
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    @OlivierOloa You mean the $x^8$ coefficient? I am pretty sure it is right, but I will double check since you mentioned it – 2017-01-04
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    @OlivierOloa Ah, you were right, there does happen to be typo. Thank you for bringing it to my attention!. – 2017-01-05
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    @OlivierOloa I need to go back and relearn my binomial expansion! Goodness grief...what am I doing. – 2017-01-05
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    i figured out the answer with a lot of grungy algebra (and the benefit of a bunch of other ugly modeling equations it was too difficult to post in my original question, sorry) – 2017-01-13
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I sharpened my pencil and went through a bunch of grungy algebra in some modeling equations for the data in question, to show that my function is actually

$$\frac{2x^2}{4-3x^2} = \frac{1}{2}x^2 + \frac{3}{8}x^4 + \frac{9}{32}x^6 + \frac{27}{128}x^8 + \dots + \frac{1}{2}\left(\frac{3}{4}\right)^{k-1}x^{2k} + \dots$$

I goofed on the $x^6$ term; $2/7 \approx 0.285714$ whereas $9/32 = 0.28125$.

(as I said in another comment, I was sure it was between 0.28 and 0.29)

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    Haha, but how'd you come up with it? – 2017-01-13
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    it's from some matrix math, part of it is proprietary for some engineering work I'm doing, and it's about 3 pages of derivation that took me several hours to work through; even the problem statement would probably take me an hour to typeset in MathJax. :( so I just posted the part I needed. – 2017-01-14
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    Haha, ok, sounds interesting. :-) – 2017-01-14