ns36
  • ns36
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ns36
  • ns36
I'll try to make to write f(x) in the equation editor....one sec.
ns36
  • ns36
\[\sum_{0}^{\infty} ((-1)^{n} x^{2n})/((2n)!)\]
ns36
  • ns36
That's f(x), and I need to show that it's a solution to the differential equation f''(x) + f(x) = 0.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
take the second derivative of the sumation and add it to the sumation itself to see if it goes to 0
amistre64
  • amistre64
the summation is jsut a polynomial
amistre64
  • amistre64
in teh eq editor, if you type in \frac{}{} you can fill in the {}s with your top and bottom arguments
ns36
  • ns36
Oh I see, f(x) is just cos(x)! Thanks a lot!
amistre64
  • amistre64
\[\sum_{0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}\] \[D_x[\sum ]=\sum_{1}^{\infty} \frac{(-1)^{n} 2n\ x^{(2n-1)}}{(2n)!}\] \[D_{xx}[\sum ]=\sum_{2}^{\infty} \frac{(-1)^{n} 2n(2n-1)\ x^{(2n-2)}}{(2n)!}\]
amistre64
  • amistre64
lol, that too
amistre64
  • amistre64
i think i need more practice on my summation derivatives :) but your way is sufficient i believe

Looking for something else?

Not the answer you are looking for? Search for more explanations.