anonymous
  • anonymous
Find the Maclaurin series of the first few terms of: Ln(sin(x)/(x)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Without taking each successive derivative!
anonymous
  • anonymous
\[\ln\left(\frac{\sin x}{x}\right) ~~\text{or}~~\frac{\ln\sin x}{x}~~?\]
anonymous
  • anonymous
the first one

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anonymous
  • anonymous
i know the series for sinx/x
anonymous
  • anonymous
and i know the series for Ln(1+x)...is there a way in can incorporate both?
anonymous
  • anonymous
Perhaps a slightly different route... Do you know how to integrate/differentiate a power series?
anonymous
  • anonymous
I'm not sure if what I'm thinking of will work, but it has in the past for a similar problem.
anonymous
  • anonymous
to a certain extent yes
anonymous
  • anonymous
Here's my idea. First, we differentiate the given function, the try to find the power series for it. So, if \(y=\ln\left(\dfrac{\sin x}{x}\right)\), then \(y'=\frac{\dfrac{x\cos x-\sin x}{x^2}}{\dfrac{\sin x}{x}}=\cot x-\dfrac{1}{x}\). As it turns out (with a check with Wolfram), the first term of the power series for \(\cot x\) is \(\dfrac{1}{x}\), so that would cancel out. The only obstacle here would be to then find the power series for \(\cot x\)...
anonymous
  • anonymous
well cotx is cosx/sinx...i know the series for each of those...so i would divide the series of cosx by sinx i think
anonymous
  • anonymous
If that works out, the last thing to do would be to simply integrate this series.
anonymous
  • anonymous
Im trying to figure this using the binomial series formula
anonymous
  • anonymous
i know the series for sinx.x and Ln but am trying to figure out how to put them together
anonymous
  • anonymous
at first i thought maybe you could use the series for \(\log(\frac{\sin(x)}{x})\) but now i am thinking maybe it would be easier to find the series for \(\log(\sin(x))\)
anonymous
  • anonymous
i meant "use the series for \(\frac{\sin(x)}{x}\)" sorry
anonymous
  • anonymous
know the series for sinx/x: I got 1-(x^3/3!)+(x^4+5!)-(x^6/7!)
anonymous
  • anonymous
oh the scond term is supposed to be x^2/3!
anonymous
  • anonymous
and the series of Ln(1+x)= x-(x^2/2)+(x^3/3)-(x^4/4)...+..
anonymous
  • anonymous
yeah here is an idea, maybe this will help (maybe not) take the derivative you get \[\cos(x)-\frac{1}{x}\] then use the series for \(\cos(x)\) then integrate term by term just talking off the top of my head, i am not sure that makes it easier
anonymous
  • anonymous
hmmm ok
anonymous
  • anonymous
that is not right it is \(\cot(x)-\frac{1}{2}\)
anonymous
  • anonymous
first you must find Maclaurin series for \[\frac{ \sin \left( x \right) }{ x }\approx1-\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-O \left( x ^{6} \right)\] by just dividing Maclaurin series for sin(x) by x \[\ln \left( \frac{ \sin \left( x \right) }{ x }\right)\approx \ln \left( 1-\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-O \left( x ^{6} \right) \right)\] as you know Maclaurin series for log is \[\ln \left( 1+k \right)\approx k-\frac{ k ^{2} }{ 2 }+O \left( k ^{3} \right) \] in above equation \[k =1-\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-O \left( x ^{6} \right)\] so \[\ln \left( 1+\left( -\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-O \left( x ^{6} \right) \right) \right)\approx -\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-O \left( x ^{6} \right)-\frac{ \left( -\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-O \left( x ^{6} \right) \right)^{2} }{ 2 }\] now lets write only of accuracy x^4 we will get:\[\ln \left( \frac{ \sin \left( x \right) }{ x } \right)\approx-\frac{ x ^{2} }{ 3! }+\frac{ x ^{4} }{ 5! }-\frac{ \left( -\frac{ x ^{2} }{ 3! } \right)^{2} }{ 2 }\] you get first two terms from linear approximation in k and and third term from quadratic. now if you will work out factorials formula will become\[\ln \left( \frac{ \sin \left( x \right) }{ x } \right)\approx-\frac{ x ^{2} }{ 6 }-\frac{ x ^{4} }{ 180 }-O \left( x ^{6} \right)\]

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