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Let $m,n\in \mathbb{N}$ such that $m<n$. Define the function $\phi$ on $(0,\pi/2)$ as follow:

$$\phi(x)=\frac{\int_{0}^{x} \sin^{m}{t}\,dt}{\int_{0}^{x} \sin^{n}{t}\,dt}$$

Is $\phi$ increasing?

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1  
What have you tried? Where are you stuck? –  Zev Chonoles Feb 25 '13 at 17:53
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You can differentiate this with the fundamental theorem of calculus. Then it is not too difficult to study the sign. You'll have to factor out $\sin^mx\sin^nx$. –  1015 Feb 25 '13 at 17:55
    
@julien I think the last sentence of your hint is misleading. I'm trying not to give too big a hint, but maybe "You can differentiate this using the fundamental theorem of calculus. Combine the resulting numerator into one big integral and it's not too difficult to study its sign." –  David Speyer Feb 25 '13 at 18:29
    
@DavidSpeyer Yes, with your comment added to mine, I think the OP should be able to do it. Once the numerator is written as a single big integral, the sign of the integrand can be determined, for instance, by factoring out $\sin^nx\sin^mx$. –  1015 Feb 25 '13 at 18:35
    
@julien Fair enough. After walking away from the keyboard, I figured out why you phrased it that way. I'll leave it alone now. –  David Speyer Feb 25 '13 at 18:41

1 Answer 1

This problem is related to the following discrete property:

Let $a_i,b_i$ be positive real numbers for $i=1,2\dots$, such that the sequence $\dfrac{a_i}{b_i}$ is increasing. Then the sequence $\dfrac{\sum_{i=1}^k a_i}{\sum_{i=1}^k b_i}$ is also increasing. Furthermore, for any $r<k$ we have that: $$\frac{\sum_{i=1}^ra_i}{\sum_{i=1}^rb_i} < \frac{\sum_{i=1}^ka_i}{\sum_{i=1}^kb_i}$$

Now let $0<x<y<\dfrac{\pi}{2}$. For each $k$ we consider a partition of $[0,y]$ consisting of $k$ subintervals, such that $x$ is an endpoint of one of these subintervals, and in each subinterval we choose some $t_i$. As usual $\Delta t_i$ denotes the length of the subinterval in which $t_i$ is. Then we have Riemann sums for this partition of $[0,y]$: $$R_1(k;y)=\sum_{i=1}^k(\sin^m{t_i})\Delta t_i$$ $$R_2(k;y)=\sum_{i=1}^k(\sin^n{t_i})\Delta t_i$$

Assume that $r$ of this same subintervals form a partition of the interval $[0,x]$. Then we have Riemann sums for this partition of $[0,x]$: $$R_1(r;x)=\sum_{i=1}^r(\sin^m{t_i})\Delta t_i$$ $$R_2(r;x)=\sum_{i=1}^r(\sin^n{t_i})\Delta t_i$$

In the interval $[0,\frac{\pi}{2}]$ we have that $\dfrac{\sin^mt}{\sin^nt}=\sin^{m-n}t$ is an increasing function if $m>n$ (so this assumption must be the other way round). By the property mentioned above we have that for any $k$ and $r$ defined as above:

$$\frac{R_1(r;x)}{R_2(r;x)}<\frac{R_1(k;y)}{R_2(k;y)}$$ Taking limit for the Riemann sums as the $\Delta t_i\to 0$ on both sides of the inequality we conclude that $\phi(x)<\phi(y)$.

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