Find n given that $\lim\limits_{x\rightarrow0} \frac{1-\sqrt{\cos2x}.\sqrt[3]{\cos3x}.\sqrt[4]{\cos4x}...\sqrt[n]{\cos{nx}}}{x^2} = 10$

Question

I'm trying to solve this rather interesting problem. We have been given that $\lim\limits_{x\rightarrow0} \frac{1-\sqrt{\cos2x}.\sqrt[3]{\cos3x}.\sqrt[4]{\cos4x}...\sqrt[n]{\cos{nx}}}{x^2}$ = 10 and we are required to find n.

This is the $\frac{0}{0}$ form, so we can use L'Hôpital's rule. After taking the derivatives of the numerator and the denominator separately, the problem becomes-

$\lim\limits_{x\rightarrow0} \frac{-\frac{d}{dx}\sqrt{\cos2x}.\sqrt[3]{\cos3x}.\sqrt[4]{\cos4x}...\sqrt[n]{\cos{nx}}}{2x}$ = 10

$\lim\limits_{x\rightarrow0} \frac{-\frac{d}{dx}\prod_{i = 2}^{n} (\cos ix)^\frac{1}{i}}{2x}$ = 10

Now, the numerator looks like a pretty difficult expression to differentiate. Here, I decided to simplify the expression in the numerator first.

Let y= $\prod_{i = 2}^{n} (\cos ix)^\frac{1}{i}$

Now, we can simplify the expression by taking the natural logarithm of both sides.

$\log(y)$= $\log(\prod_{i = 2}^{n} (\cos ix)^\frac{1}{i})$

Now we can use the property of logarithms to simplify the expression, $\log(xy)=\log(x)+\log(y)$

$\log(y)$= $\sum_{i = 2}^{n} \frac{1}{i}.\log(\cos ix)$

This is where I ran out of ideas to simplify this expression any further. Any ideas would be appreciated.

Answer 1

You proceed further by implicitly differentiating: $\dfrac{dy}{dx}(\log y)=\dfrac1y \dfrac{dy}{dx}$.

$$\dfrac1y \dfrac{dy}{dx}=\sum_{i=2}^n\frac1i\frac{1}{\cos ix}(-i\sin ix)=-\sum_{i=2}^n\tan ix.$$ Therefore, $\dfrac{dy}{dx}=y'=-y\sum_{i=2}^n\tan ix.$ Now, as $x\to 0$, $\sum_{i=2}^n\tan ix\to 0$ and $y\to 1$, so $y'\to 0$.

Thus, we have again got a $\dfrac00$ form, $\displaystyle \frac{y\sum_{i=2}^n\tan ix}{2x}$, on our hands. Differentiating again, we get $$\lim_{x\to 0}\frac{y'\sum_{i=2}^n\tan ix+y\tfrac{d}{dx}\left(\sum_{i=2}^n\tan ix\right)}{2}=\lim_{x\to 0}\frac{-y\left(\sum_{i=2}^n\tan ix\right)^2+y\left(\sum_{i=2}^ni\sec^2 ix\right)}{2}$$$$=\dfrac{\sum_{i=2}^ni}{2}=\dfrac{n^2+n-2}{4}=10 \ (given).$$ Therefore, we get $n=6, -7.$ But since $n$ is the index of a sum, it must be a natural number. Thus, $n=6$.

EDIT: As per user @Vishu's comment, $\displaystyle \lim_{x\to 0}\frac{y\sum_{i=2}^n\tan ix}{2x}$ can be simplified more easily as $\displaystyle \lim_{x\to 0}\dfrac y2\sum_{i=2}^n\frac{\tan ix}{x}=\displaystyle \lim_{x\to 0}\dfrac y2\sum_{i=2}^ni\frac{\tan ix}{ix}=\displaystyle \lim_{x\to 0}\dfrac y2\sum_{i=2}^n i$ by using the fact that $\lim_{u\to 0}\dfrac{\tan u}{u}=1.$

Answer 2

Using the fact that: $$\cos(nx)\sim 1-\frac{n^2}{2}x^2$$ we have: $$\frac{1-\sqrt{\cos2x}.\sqrt[3]{\cos3x}.\sqrt[4]{\cos4x}...\sqrt[n]{\cos{nx}}}{x^2}\sim\frac{1-\left((1-2x^2)^{\frac12}\dots\left(1-\frac{n^2}{2}x^2\right)^{\frac{1}{n}}\right)}{x^2}$$ Now, remember that: $$(1-f(x))^{\alpha}\sim1-\alpha f(x)$$ when $f(x) \to 0$, so: $$\frac{1-\left((1-2x^2)^{\frac12}\dots\left(1-\frac{n^2}{2}x^2\right)^{\frac{1}{n}}\right)}{x^2}\sim\frac{\left(\sum_{i=2}^{n}\frac{1}{i}\cdot\frac{i^2}{2}\right)x^2}{x^2}=\sum_{i=2}^{n}\frac{i}{2}=\frac12\cdot\sum_{i=2}^{n}i=$$ $$=\frac{n\cdot(n+1)}4-\frac12$$ Thus, we have: $$\frac{n\cdot(n+1)}{4}-\frac12=10 \leftrightarrow n=6$$

Answer 3

Note : I realized that this answer is probably not recommended in an exam, but in any case it's useful to know.

One can use the "multiple product" rule for differentiation. Let's write $f'$ for $\frac{df}{dx}$ for this answer. We know that $(fg)' = f'g+fg'$ is the product rule, but as an extension (by induction), one can show that $$(f_1f_2\ldots f_m)' = \sum_{j=1}^m f_1f_2\ldots f_{j-1}f_j'f_{j+1}\ldots f_m$$ for functions $f_1,\ldots,f_m$. This rule is usable in your situation, instead of taking the logarithm. (Roughly, you sum over all possible products of $m$ functions where exactly one of the functions is differentiated and the others are not).

However, you need to look ahead. When you get to $$ \lim_{x \to 0} \frac{-\frac {d}{dx} \prod_{j=2}^{n} (\cos jx)^{\frac 1j}}{2x} $$ You know that, in hindsight, this needs to equal $10$. However, the denominator evaluates to $0$ at $x=0$, so you actually need to differentiate the numerator once more as well, because you need another round of L'Hospital's rule to make the $2x$ into $2$ and make the denominator well defined and get a limit which is not equal to $0$ or $+\infty$, but some finite number.

It turns out that, similar to the multinomial theorem, one has a rule for differentiating a product of functions as many times as necessary. It's the "ultimate" product rule.

"Product rule" : If $f_1,\ldots,f_m$ are functions and $L$ is the number of times you need to differentiate $F = f_1\ldots f_m$ then the $L$th derivative of $f$ equals $$ F^{(L)} = \sum_{k_1+k_2+\ldots+k_m = L, k_i \geq 0} \binom{L}{k_1,k_2,\ldots,k_m}f_1^{(k_1)}f_2^{(k_2)}\ldots f_m^{(k_m)} $$ where $\binom{L}{k_1,k_2,\ldots,k_m}$ is the multinomial coefficient.

See here for details on this rule.

What you need is to differentiate the numerator $\prod_{j=2}^{n} (\cos jx)^{\frac 1j}$ two times. According to this rule, you will obtain $$ \sum_{k_1+k_2+\ldots+k_n = 2, k_i \geq 0} \binom{2}{k_1,k_2,\ldots,k_n}\left((\cos 2x)^{\frac 12}\right)^{(k_1)}\ldots \left((\cos nx)^{\frac 1n}\right)^{(k_n)} $$

Thankfully, this is not too difficult when you realize that only one of two things can happen.

• Either there are two different $i \neq j$ such that $k_i=k_j=1$ and the others are $0$.

• There is a single $i$ for which $k_i=2$ and the others are $0$.

In the first case, the first derivative of $(\cos ix)^{\frac{1}{i}}$ is $$ \left((\cos ix)^{1/i}\right)' = (\sin ix)(-\cos^{1/(i-1)}(ix)) $$ which equals $0$ when evaluated at $0$. So this case gives $0$.

In the second case, we differentiate again using the product rule to get $$ \left((\cos ix)^{1/i}\right)'' = (\cos^{1/(i-2)}(ix))(-(i-1)\sin^2(ix)-i \cos^2(ix)) $$

at $0$, this equals $-i$. The multinomial coefficients are all equal to $1$ in this case, so we land up with $$ \left(\prod_{j=2}^{n} (\cos jx)^{\frac 1j}\right)'' (0) = -2-3-\ldots-n = 1-\frac{n(n+1)}{2} $$ Plugging this back into the original L'Hospital rule, the limit of the quantity you want as a function of $n$, is then equal to $$ \frac{\frac{n(n+1)}{2}-1}{2} $$

(the $2$ denominator comes from differentiating the $2x$ at the bottom during the $2$nd L'Hospital round).

If this is equated to $10$, then you get that $$ \frac{n(n+1)}{2} = 21 \implies n=6 $$ as desired.

This method is the "brute force" way of extending the original post's approach. It requires one heavy tool, but that tool finds a lot of usage in analysis of products in the future (and can be made slightly easier if one uses the cosine Taylor series) and I think it's a nice thing to know at the high school level.

Answer 4

Using $y=e^{\ln(y)}$, you can differentiate

$$y'=(e^{\ln(y)})'=e^{\ln(y)}\cdot (\ln(y))'=y\cdot(\sum_{i=2}^n \frac{1}{i}\ln(\cos(ix)))'=(\sum_{i=2}^n \frac{1}{i}\frac{-i\sin(ix)}{\cos(ix)})\cdot y$$ $$=-y\cdot \sum_{i=2}^n \tan(ix)$$

So now your limit becomes

$$\lim_{x\to 0} \frac{y\cdot \sum_{i=2}^n \tan(ix)}{2x}$$

Applying L'Hospital's rule once more (as this limit is still of type $\frac{0}{0}$) gives a derivative of $2$ in the denominator. For the numerator the drivative is

$$y'\cdot \sum_{i=2}^n \tan(ix)+y\cdot (\sum_{i=2}^n \tan(ix))'=y'\cdot \sum_{i=2}^n \tan(ix)+y\cdot (\sum_{i=2}^n \frac{i}{\cos^2(ix)})$$

Plugging in our expression for $y'$, we get

$$-y\cdot \sum_{i=2}^n \tan(ix)\cdot \sum_{i=2}^n \tan(ix)+y\cdot (\sum_{i=2}^n \frac{i}{\cos^2(ix)})$$

As $x\to 0$, it is easy to see that $y\to 1$, and that the entire first term here goes to $0$. $\cos(ix)\to 1$ for $x\to 0$, and thus we are overall left with:

$$\frac{\sum_{i=2}^n i}{2}=10 \implies \frac{\frac{1}{2}\cdot(n^2+n-2)}{2}=10\implies n^2+n-42=0 \implies n=6 \space\text{or}\space n=-7$$

$n\in \mathbb{N}$, so we have $n=6$.