Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the Maclaurin expansion $\sqrt{\cos(2x)}$ and $\tan^2x$ up to degree $4$.

I tried differentiation but it gives me something really horrible.

share|improve this question
    
Are you sure they want you to find the Maclaurin expansion of $\sqrt{\cos(2x)}$? It really is just horrible differentiation with no visible pattern, but it doesn't seem like a very useful exercise if that is the case. The one for $\tan^2(x)$ can be found more easily by simply squaring the Maclaurin expansion for $\tan(x)$. –  august Feb 5 '13 at 20:05
    
@august I have to find both up to degree 4 so squaring would be a bit of a pain, wouldn't it? –  bbr4in Feb 5 '13 at 20:14
    
@user52187: if all you need is the expression use wolframalpha.com, it saves your time –  Alex Feb 5 '13 at 21:17
    
@Alex working is needed –  bbr4in Feb 5 '13 at 22:44
    
Not really, if you only need up to degree 4 then squaring will give you the 4 terms with relative ease. –  august Feb 6 '13 at 4:57

1 Answer 1

Working up to degree $4$ is the key here; otherwise the whole affair would be hopeless. You are given the license to ignore all powers of $x$ above $4$. So, $$\cos 2x = 1-\frac{(2x)^2}{2}+\frac{(2x)^4}{24} +\dots $$ Write this as $1+u$ because $1+u$ is the kind of thing you want to have under the square root. There's no need to worry about $u^3$ and higher powers, they will contain only powers of $x$ above $4$. So, $$\sqrt{1+u}= 1+\frac12 u - \frac{1}{8}u^2 + \dots$$ And the conclusion is $$\sqrt{\cos 2x} = 1+ \frac12\left(-\frac{(2x)^2}{2}+\frac{(2x)^4}{24}\right) -\frac18 \left(-\frac{(2x)^2}{2}\right)^2 +\dots$$

With $\tan^2x$ it's easier: $$\tan^2x=(x-x^3/6+\cdots)^2 (1-x^2/2+\cdots)^{-2} = (x^2- 2x^4/6 +\cdots) (1+x^2+\cdots) $$

share|improve this answer
    
To "40 votes". Your expansion of $\cos 2x$ is clearly wrong. What you have written is expansion for $\cos x$. Similar is case with $\tan^{2}x$. Anyway the OP wants to get a maclaurin series upto $x^{4}$ and in this case it is best to do four times differentiation rather than using series manipulation techniques. –  Paramanand Singh Jul 28 '13 at 4:47
    
@ParamanandSingh Thanks, I put in the missing $2$. –  40 votes Jul 28 '13 at 4:50
    
check the expansion for $\tan^{2}x$ too. It seems this has been typed in hurry. Ideally it should be $(x - x^{3}/6 + \cdots)^{2}(1 - x^{2}/2 + \cdots)^{-2}$. –  Paramanand Singh Jul 28 '13 at 4:53
    
@ParamanandSingh Thanks again for checking! You are right about typing in a hurry. –  40 votes Jul 28 '13 at 4:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.