anonymous
  • anonymous
Find the first few terms of the Maclaurin series of: sin(sqrtx)/(sqrtx)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
im not sure which method to use to solve this....Taking each derivative is way too long
anonymous
  • anonymous
By any chance, are you given the series for \(\dfrac{\sin x}{x}\)? I'm not sure if there is one, but if there is, there's a quick way to an answer.
anonymous
  • anonymous
Actually, you might not even need that. Do you have the series for sine?

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anonymous
  • anonymous
|dw:1377545838357:dw|
anonymous
  • anonymous
I think it'd be easier to show what to do given the sigma notation: \[\sin x=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\] So, you have \[\begin{align*}\frac{\sin x}{x}&=\frac{1}{x}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n} \end{align*}\] Now replace \(x\) with \(\sqrt x\).
anonymous
  • anonymous
i.e. \(\sqrt x=x^{1/2}\), for simplification purposes
anonymous
  • anonymous
@SithsAndGiggles Very elegant way to solve it!
anonymous
  • anonymous
ok i see what you did...i orginally took the series of sin(x) and square rooted every x in the series, and then divided by square root of x to complete it..
anonymous
  • anonymous
@CarlosGP, thanks! I'd hate to have to find multiple derivatives...
anonymous
  • anonymous
@requiem, that sounds like the same process. I personally prefer the (imo) more concise sigma notation.
anonymous
  • anonymous
so what i got as the series is:
anonymous
  • anonymous
|dw:1377546611383:dw|
anonymous
  • anonymous
Yes, that's correct: http://www.wolframalpha.com/input/?i=Sum%5B%28-1%29%5Ek%2F%282k%2B1%29%21*x%5Ek%2C%7Bk%2C0%2CInfinity%7D%5D
anonymous
  • anonymous
Here are the first 4 terms: http://www.wolframalpha.com/input/?i=Sum%5B%28-1%29%5Ek%2F%282k%2B1%29%21*x%5Ek%2C%7Bk%2C0%2C3%7D%5D
anonymous
  • anonymous
ok thanks siths!
anonymous
  • anonymous
You're welcome!

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