Find the Maclaurin series of the first few terms of:
Ln(sin(x)/(x)

- anonymous

Find the Maclaurin series of the first few terms of:
Ln(sin(x)/(x)

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- anonymous

Without taking each successive derivative!

- anonymous

\[\ln\left(\frac{\sin x}{x}\right) ~~\text{or}~~\frac{\ln\sin x}{x}~~?\]

- anonymous

the first one

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## More answers

- anonymous

i know the series for sinx/x

- anonymous

and i know the series for Ln(1+x)...is there a way in can incorporate both?

- anonymous

Perhaps a slightly different route... Do you know how to integrate/differentiate a power series?

- anonymous

I'm not sure if what I'm thinking of will work, but it has in the past for a similar problem.

- anonymous

to a certain extent yes

- 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

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

If that works out, the last thing to do would be to simply integrate this series.

- anonymous

Im trying to figure this using the binomial series formula

- anonymous

i know the series for sinx.x and Ln but am trying to figure out how to put them together

- 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

i meant "use the series for \(\frac{\sin(x)}{x}\)" sorry

- anonymous

know the series for sinx/x: I got
1-(x^3/3!)+(x^4+5!)-(x^6/7!)

- anonymous

oh the scond term is supposed to be x^2/3!

- anonymous

and the series of Ln(1+x)= x-(x^2/2)+(x^3/3)-(x^4/4)...+..

- 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

hmmm ok

- anonymous

that is not right
it is \(\cot(x)-\frac{1}{2}\)

- 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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