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I came across a question in a workbook that asked me to find the general expression of the function $f(x)=\frac{1}{\sqrt{x}}$ centred at $x=4$.

I approached this problem by first finding its fourth-degree taylor series (the problem did not specify the degree of the taylor series, just the expression of the general term).

$$f(x)=\frac{1}{\sqrt{x}}\Longrightarrow f(4)=\frac{1}{\sqrt{4}}=\frac{1}{2}$$ $$f'(x)=\frac{-1}{2x^\frac{3}{2}}\Longrightarrow f'(4)=\frac{-1}{2{\sqrt{4}}^3}=\frac{-1}{16}$$ $$f''(x)=\frac{3}{4x^{\frac{5}{2}}}\Longrightarrow f''(4)=\frac{3}{4{\sqrt{4}}^5}=\frac{3}{128}$$ $$f'''(x)=\frac{-15}{8x^{\frac{7}{2}}}\Longrightarrow f'''(4)=\frac{-15}{8{\sqrt{4}}^7}=\frac{-15}{1024}$$ $$f^{(4)}(x)=\frac{105}{16x^{\frac{9}{2}}}\Longrightarrow f^{(4)}(4)=\frac{105}{16{\sqrt{4}}^9}=\frac{105}{8192}$$

As such, the fourth-degree taylor series may be written as:

$$P_4(x)=\frac{1}{2}-\frac{1}{16}(x-4)+\frac{3}{256}(x-4)^{2}-\frac{15}{6144}(x-4)^{3}+\frac{105}{196608}(x-4)^4$$

However, this is where the problem appears. I am struggling to find a representation for the numerator part of the coefficients. The pattern that I found seems to be $1\times3\times5...\times(2n-1)$. However, I have no idea how I may represent this algebraically.

As such, the best general term expression that I can do is $\sum_{n=0}^{\infty} {\frac{(-1)^{n}}{n!}}⋅{\frac{1⋅3⋅5⋅...⋅(2n-1)}{2^{n}⋅2^{2n+1}}}⋅(x-4)^{n}$, which I am not quite satisfied with. Is there a better way to show this?

Golden_Ratio
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Cold_Spaghetti
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1 Answers1

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Your expression looks essentially correct. You can use the double factorial for conciseness: $$(2n-1)!!=1\times 3\times ...\times(2n-1),$$

with the convention $(-1)!!=1.$

Note you can also express the series in terms of combinatorial coefficients:

$$\sum_{n=0}^\infty {-1/2 \choose n}\frac{1}{2^{2n+1}}(x-4)^n.$$

Golden_Ratio
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