With the help of Maclaurin serial expansion lncos(9x) , find the first four terms that are different from zero ?

Is there anyone who can solve this question?

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With the help of Maclaurin serial expansion lncos(9x) , find the first four terms that are different from zero ?

Is there anyone who can solve this question?

Is there anyone who can solve this question?

I am sure there are many people who can answer this question!

First, it is a "McLaurin series", not "serial". And to continue the English lesson, one "answers" questions and "solves" problems!

Now the the McLaurin series of a function, f(x), is the power series where the coefficient of $x^n$ is $\frac{\frac{d^n f(0)}{dx^n}}{n!}$.

Here, $f(x)= ln(cos(9x))$. $f(0)= ln(cos(0))= ln(1)= 0$. $\frac{df}{dx}= \frac{1}{cos(9x)}\left(-9 sin(9x)\right)$ so that $\frac{df}{dx}(0)= \frac{1}{1}(0)= 0$.

To find the third coefficient, take the second derivative and evaluate it at x= 0.

First, it is a "McLaurin series", not "serial". And to continue the English lesson, one "answers" questions and "solves" problems!

Now the the McLaurin series of a function, f(x), is the power series where the coefficient of $x^n$ is $\frac{\frac{d^n f(0)}{dx^n}}{n!}$.

Here, $f(x)= ln(cos(9x))$. $f(0)= ln(cos(0))= ln(1)= 0$. $\frac{df}{dx}= \frac{1}{cos(9x)}\left(-9 sin(9x)\right)$ so that $\frac{df}{dx}(0)= \frac{1}{1}(0)= 0$.

To find the third coefficient, take the second derivative and evaluate it at x= 0.

- HallsofIvy
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