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MathSofiya

  • 3 years ago

\[f(x)=cosh(x)=1+0+\frac{x^2}{2!}+0+\frac{x^4}{4!}+....\] 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?

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  1. MathSofiya
    • 3 years ago
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    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

  2. TuringTest
    • 3 years ago
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    these are even values of the exponent, so we use the other formula with the 2n

  3. MathSofiya
    • 3 years ago
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    ooops

  4. MathSofiya
    • 3 years ago
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    \[\sum_{n=0}^{\infty} \frac{...x^{(2n)}}{(2n)!}\]

  5. TuringTest
    • 3 years ago
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    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

  6. TuringTest
    • 3 years ago
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    so just take out the dots lol

  7. MathSofiya
    • 3 years ago
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    oooh because their just zero's and one's ?

  8. TuringTest
    • 3 years ago
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    yup try out some values of n plug in n=0 n=1 n=2 etc. you will see you get your series

  9. MathSofiya
    • 3 years ago
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    check out my table

  10. TuringTest
    • 3 years ago
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    you do not need to represent the 0 terms

  11. TuringTest
    • 3 years ago
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    you only need to represent the resulting series

  12. MathSofiya
    • 3 years ago
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    \[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....\] ?

  13. TuringTest
    • 3 years ago
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    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

  14. TuringTest
    • 3 years ago
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    that agrees with your results doesn't it?

  15. MathSofiya
    • 3 years ago
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    makes sense

  16. TuringTest
    • 3 years ago
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    so there you have it\[\cosh x=\sum_{n=0}^\infty{x^{2n}\over (2n)!}\]a nice, simple series representation

  17. MathSofiya
    • 3 years ago
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    \[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....=\sum_{n=0}^{\infty} \frac{x^{(2n)}}{(2n)!}\] final answer. Very good, THankss!

  18. TuringTest
    • 3 years ago
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    welcome!

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