Question

Find the Maclaurin series of the first few terms of:

Ln(sin(x)/(x)

Answer 1

Without taking each successive derivative!

Answer 2

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

Answer 3

the first one

Answer 4

i know the series for sinx/x

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

to a certain extent yes

Answer 9

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 \(\cot x\)...

Answer 10

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

Answer 11

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

Answer 12

Im trying to figure this using the binomial series formula

Answer 13

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

Answer 14

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 \(\log(\sin(x))\)

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

hmmm ok

Answer 21

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

Answer 22

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\[\ln \left( \frac{ \sin \left( x \right) }{ x } \right)\approx-\frac{ x ^{2} }{ 6 }-\frac{ x ^{4} }{ 180 }-O \left( x ^{6} \right)\]