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March 12th, 2017, 08:33 PM   #11
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How do you know the Taylor series for $\cosh x$?

$\dfrac{e^x + e^{-x} -2}{x^2} = \left(\dfrac{\sinh u}{u}\right)^2$, where $u = \dfrac{x}{2}$.
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March 12th, 2017, 08:50 PM   #12
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How do you know the power series for $e^x$?
It's a standard result. It might even be your definition of the function.
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March 13th, 2017, 01:24 AM   #13
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Its usual proof assumes (amongst other things) that its derivative is $e^x$.

By definition, that derivative is $\displaystyle \lim_{h\to0}\frac{e^{x + h} - e^x}{h} = e^x \lim_{h\to0}\frac{e^h - 1}{h}$.

Thus, its proof rests on knowing a limit that is very similar to (in fact, equivalent to) the one you're trying to evaluate.
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March 13th, 2017, 03:03 AM   #14
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Originally Posted by skipjack View Post
Its usual proof assumes (amongst other things) that its derivative is $e^x$.

By definition, that derivative is $\displaystyle \lim_{h\to0}\frac{e^{x + h} - e^x}{h} = e^x \lim_{h\to0}\frac{e^h - 1}{h}$.

Thus, its proof rests on knowing a limit that is very similar to (in fact, equivalent to) the one you're trying to evaluate.
The same criticism can be made for L'hospital's rule which is why 90% of the time it is used it doesn't actually apply since. Take for example
$$\lim_{x \rightarrow 0} \frac{\sin x}{x}$$

which many would claim can be computed using L'Hospital. However, using L'Hospital requires knowing the derivative of $\sin x$ exists and you may recognize that the limit we are computing is precisely this derivative at $x=0$.

In fact, if you apply L'Hospital to the given example, this will be exactly the logical flaw you run into in the 2nd application. The limit you are computing is actually the definition of the derivative of a function you must assume the derivative of in order to apply it. It is completely circular.
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March 13th, 2017, 04:18 AM   #15
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The difference being that it isn't necessary to know the limit to compute the derivative. For example, if your definition is that $f(x)=e^x$ is the inverse of the function $\ln{(x)} = \int_{1}^x \frac1t \,\mathrm dt$, we get the derivative via $$\begin{align*}
y &= f(x) &\implies g(y) &= x \\
\implies g'(y)y' &= 1 &\implies y' = f'(x) &= \frac1{g'(y)} = \frac1{g'\big(f(x)\big)}
\end{align*}$$
Or, if one defines $f(x)=e^x$ as the function that is it's own derivative,we get the limit from the derivative and not vice versa.
Even if your definition is that $$f(x)=e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n$$
The derivative comes without knowledge of the limit (although in this case we don't need either the derivative or the limit for this problem).

Last edited by v8archie; March 13th, 2017 at 04:51 AM.
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March 15th, 2017, 06:38 AM   #16
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How do you know the Taylor series for $\cosh x$?
I was responding to the OP, not the matter of the digression.
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