calculus II

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Now we can see the pattern so we can write `f^((n))(x).`

The denominator is obviously `(3-x)^(n+1)` (exponent is by 1 greater then the degree of derivation). Numerator is `n!,` hence

`f^((n))(x)=(n!)/(3-x)^(n+1)`  <-- n-th derivative` `

You can prove this formally by using mathematical induction.


Formula for Taylor series is `f(x)=sum_(n=0)^oo (f^((n))(x))/(n!)(x-a)^n` . Since in your case `a=4` you have

`f(x)=sum_(n=0)^oo ((n!)/(3-4)^(n+1))/(n!)(x-4)^n=sum_(n=0)^oo(-1)^(n+1)(x-4)^n` <-- Your Taylor series 


Again we first find several derivatives.






Here we are looking for Maclaurin series which is actually Taylor series about 0. Since `sinh0=0` and `cosh0=1` we will only be interested for odd derivatives (even derivatives contain `sinh0` which is equal to 0) and (2n+1)-th derivative is equal to `2^(2n+1).` Thus we have

`sinh2x=sum_(n=0)^oo(2^(2x+1)x^(2n+1))/((2n+1)!)` <-- Your Maclaurin series

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