Difference between 1/x and Cosech (x) graph

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The discussion focuses on the differences between the graphs of 1/x and the hyperbolic cosecant function, csch(x). Users highlight that the two graphs are fundamentally different, with distinct shapes and behaviors. The hyperbolic cosecant function is defined as g(x) = 2/(e^x - e^{-x}), while 1/x is a simple rational function. The conversation emphasizes the importance of graphing both functions to visually compare their characteristics. Overall, the user seeks a basic explanation of these differences for someone without a strong mathematical background.
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Homework Statement


Hi
Can anyone tell me the difference between the 1/x and Cosech(x) graph?
Thanks



Homework Equations





The Attempt at a Solution

 
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You mean the hyperbolic cosecant?
g(x) = csch\;x = \frac{2}{e^x - e^{-x}}
Graph this and
f(x) = \frac{1}{x}
on your grapher and compare the table of values. Are they the same?
 
That's a very strange question! The two graphs are not at all the same! There are so many differences, what kind of answer are you looking for? HAVE you actually draw the graphs yourself?
 
HallsofIvy said:
That's a very strange question! The two graphs are not at all the same! There are so many differences, what kind of answer are you looking for? HAVE you actually draw the graphs yourself?
Yes! I have drawn them. But how would you explain the differences to someone without a mathematical understanding?
 
eumyang said:
You mean the hyperbolic cosecant?
g(x) = csch\;x = \frac{2}{e^x - e^{-x}}
Graph this and
f(x) = \frac{1}{x}
on your grapher and compare the table of values. Are they the same?
Thanks. My question is more in relation to a basic sketch and features.
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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