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ns36

  • 2 years ago

Help with sequences and series: Show that the function f(x) = sum from 0 to infinity ((-1)^n * x^(2n))/((2n)!) is a solution of the differential equation f''(x) + f(x) = 0.

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  1. ns36
    • 2 years ago
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    I'll try to make to write f(x) in the equation editor....one sec.

  2. ns36
    • 2 years ago
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    \[\sum_{0}^{\infty} ((-1)^{n} x^{2n})/((2n)!)\]

  3. ns36
    • 2 years ago
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    That's f(x), and I need to show that it's a solution to the differential equation f''(x) + f(x) = 0.

  4. amistre64
    • 2 years ago
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    take the second derivative of the sumation and add it to the sumation itself to see if it goes to 0

  5. amistre64
    • 2 years ago
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    the summation is jsut a polynomial

  6. amistre64
    • 2 years ago
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    in teh eq editor, if you type in \frac{}{} you can fill in the {}s with your top and bottom arguments

  7. ns36
    • 2 years ago
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    Oh I see, f(x) is just cos(x)! Thanks a lot!

  8. amistre64
    • 2 years ago
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    \[\sum_{0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}\] \[D_x[\sum ]=\sum_{1}^{\infty} \frac{(-1)^{n} 2n\ x^{(2n-1)}}{(2n)!}\] \[D_{xx}[\sum ]=\sum_{2}^{\infty} \frac{(-1)^{n} 2n(2n-1)\ x^{(2n-2)}}{(2n)!}\]

  9. amistre64
    • 2 years ago
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    lol, that too

  10. amistre64
    • 2 years ago
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    i think i need more practice on my summation derivatives :) but your way is sufficient i believe

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