anonymous
  • anonymous
\[f(x)=cosh(x)=1+0+\frac{x^2}{2!}+0+\frac{x^4}{4!}+....\] Maclaurin's I'm having some trouble writing this in summation form \[\sum_{n=0}^{\infty} \frac{f^(0)}{n!}x^(n)\] I know that: All even #'s can be written as 2n All odd #'s can be written as 2n+1 \[\sum_{n=0}^{\infty} \frac{...x^{(2n+1)}}{(2n+1)!}\] I think? so how do I write the \[f^n (0)\] correctly?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
here is my table n f^n x f^n 0 0 coshx 1 1 sinhx 0 2 coshx 1 3 sinhx 0 4 coshx 1
TuringTest
  • TuringTest
these are even values of the exponent, so we use the other formula with the 2n
anonymous
  • anonymous
ooops

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anonymous
  • anonymous
\[\sum_{n=0}^{\infty} \frac{...x^{(2n)}}{(2n)!}\]
TuringTest
  • TuringTest
besides that you have written all there is to the series; there is nothing in front of each term that you are missing in your sigma notation
TuringTest
  • TuringTest
so just take out the dots lol
anonymous
  • anonymous
oooh because their just zero's and one's ?
TuringTest
  • TuringTest
yup try out some values of n plug in n=0 n=1 n=2 etc. you will see you get your series
anonymous
  • anonymous
check out my table
TuringTest
  • TuringTest
you do not need to represent the 0 terms
TuringTest
  • TuringTest
you only need to represent the resulting series
anonymous
  • anonymous
\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....\] ?
TuringTest
  • TuringTest
your table is for the coefficients, and it should tell you that you are keeping only even exponents, hence you use 2n for the exponent and factorial
TuringTest
  • TuringTest
that agrees with your results doesn't it?
anonymous
  • anonymous
makes sense
TuringTest
  • TuringTest
so there you have it\[\cosh x=\sum_{n=0}^\infty{x^{2n}\over (2n)!}\]a nice, simple series representation
anonymous
  • anonymous
\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....=\sum_{n=0}^{\infty} \frac{x^{(2n)}}{(2n)!}\] final answer. Very good, THankss!
TuringTest
  • TuringTest
welcome!

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