A calculus problem by Akshat Sharda

Calculus Level 3

$\large I_{n}=\int^{\pi}_{0}\dfrac{\sin \left\{ \left(n+\frac{1}{2}\right)x\right\} }{\sin \frac{x}{2} } dx, \quad n\in \mathbb{W}$

Find the value of $\displaystyle \frac{1}{\pi}\sum^{100}_{n=1}I_{n}$ .

The answer is 100.

2 solutions

Mark Hennings
Aug 24, 2017

Since $\sin(n+\tfrac12)x - \sin(n-\tfrac12)x = 2\cos nx \sin\tfrac12x$ , we see that $I_n - I_{n-1} \; = \; 2\int_0^\pi \cos nx\,dx \; = \; \tfrac{2}{n}\Big[\sin nx \Big]_0^\pi \; = \; 0$ so that $I_n = I_0 = \pi$ for all integers $n$ . Thus the answer is $\boxed{100}$ .

D G
Aug 24, 2017

You should be careful with your parentheses- it wasn't clear if the $x$ was inside the $sin$ .

Well, convention allows for a certain degree of flexibility here. We generally write $\sin 2x$ without using brackets, for example. If a sine is to be multiplied by a function, convention puts the multiplication term at the front. We would write $x\sin 2$ instead of the much more clumsy $(\sin2)x$ .

The trigonometric functions are so commonly used that they are given a good deal of slack. For example, the fact that $\sin^2x$ and $\sin^{-1}x$ mean very different things represent that we agree to work with these functions in context.

I would prefer to write $\sin(n+1)x$ instead of the bracket-intensive $\sin((n+1)x)$ , and my mathematical colleagues over many years have been of the same mind. Akshat has now edited the problem, using curly braces $\{\,\}$ , so a whole new cause for confusion has been raised. Does he now mean the sine of the fractional part of $(n+\tfrac12)x$ , or just the sine of $(n+\tfrac12)x$ ?

Mark Hennings - 3 years, 9 months ago

A calculus problem by Akshat Sharda

The answer is 100.

2 solutions

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