Understanding the Maclaurin Series of cos(x) for Use in Functions

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The discussion focuses on understanding the Maclaurin series for cos(x) and its application in deriving the series for the function h(x) = (cos(3x) - 1)/x^2. The Maclaurin series for cos(x) is expressed as a power series involving even powers of x and alternating signs. To find the Maclaurin series for h(x), the known series for cos(x) is utilized, leading to a simplification that excludes the constant term. The final goal is to express h(x) in standard form, highlighting the relationship between the two series. This approach emphasizes the utility of known series in deriving related functions.
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I've looked at a number of explanations for the Maclaurin series of cos(x) yet none have given an easily understood answer, i was wondering if anyone has a way of explaining it when it is used as only a part of a function

eg. use that Maclaurin series of cos(x) to obtain the Macalurin series for

h(x)=(cos(3x)-1)/x^2
 
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The formula for Maclaurin series is
f(x)=\sum_{k=0}^\infty \frac{x^k}{k!}f^{(k)}(x)
as can be easily derived several ways
-integration by parts
-formal operator manipulation
-matching derivatives at zero
-complex contour integration
This can be applied to cos(x) as
\cos(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos^{(k)}(0)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos(k\pi/2)=\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}(-1)^k
where the index is changed in the final step to indicate odd terms are 0
Often a known Maclaurin series can be used to find a closely related one.
Just apply the known series
h(x)=(cos(3x)-1)/x^2
h(x)=\frac{1}{x^2}\left(\sum_{k=0}^\infty \frac{(3x)^{2k}}{(2k)!}(-1)^k-1\right)
simplify that into standard form.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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