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We know for $$f(x)=\dfrac14(\sin^4x+\cos^4x)~~~;~~~g(x)=\dfrac16(\sin^6x+\cos^6x)$$ have $f(x)-g(x)=\dfrac{1}{12}$. My question is

Are there other real $p$ and $q$ such that $$f(x)=\dfrac1p(\sin^px+\cos^px)~~~;~~~g(x)=\dfrac1q(\sin^qx+\cos^qx)$$ give us $f(x)-g(x)=C$ for a real constant $C$?

I had some idea but they were not useful. Thanks.

Nosrati
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  • Well, obviously... $p=4$ and $q=6$. And $p=q=2$. –  Mar 09 '17 at 19:46
  • Do you mean to find all such $p$ and $q$? –  Mar 09 '17 at 19:47
  • Im not sure about $p=-q$! – Nosrati Mar 09 '17 at 19:49
  • Nope, my mistake. I edited my comment. –  Mar 09 '17 at 19:50
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    I can't speak to real $p$ and $q$, but for other integers: Consider that e.g. $\cos(x) = \frac12(e^{ix}+e^{-ix})$, so $\cos^n(x)=\frac1{2^n}\sum_{k=0}^n{n\choose k}e^{(n-2k)ix}$. You can find a similar formula for $\sin^n(x)$ and sum these; what you're looking for are then identities between different $n$. There is some cancellation of coefficients for all $n$, but enough don't cancel that you should be able to show inequality for most if not all other integers. – Steven Stadnicki Mar 09 '17 at 20:05
  • @MyGlasses which question is correct? The one in the title, or the one in the question (if either). Not that if it's the one in the title, the answer is no: consider p = 2, q = 1. – Turambar Mar 09 '17 at 20:07
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    (And by basic theorems for Fourier series, if the coefficients aren't individually equal for these Laurent polynomials in e^{ix}, then the functions can't be equal.) – Steven Stadnicki Mar 09 '17 at 20:07
  • If the expression is constant, then $(-1)^p=(-1)^q$. –  Mar 09 '17 at 20:09
  • Maybe differentiate w.r.t. $x$? – Yuriy S Mar 09 '17 at 20:16
  • After some quick computing: apparently, $(4,6)$ is the only pair of positive integers satisfying $p < q \le 200$ with the required property. – Olod Mar 09 '17 at 20:16
  • @Olod I'm thinking about $200$. Sorry, Is your result obtained by a software. – Nosrati Mar 09 '17 at 20:57
  • Yes, the value of a particular function, say $h_{p,q}$ has been tried at 20 random points. Only the pair $(4,6)$ worked. – Olod Mar 09 '17 at 21:04

4 Answers4

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Let $k \geqslant 1$ be a natural number and let $$ f_k(x)=\frac 1k \left( \cos^k x+\sin^k x\right). $$ It is easy to see that if $k \geqslant 5,$ then $$ f_k^{(4)}(0)=3k-2 $$ (a math software can help a little here). Further, $$ f_1^{(4)}(0)=1, f_2^{(4)}(0)=0, f_3^{(4)}(0)=7, f_4^{(4)}(0)=16. $$ Thus we see that whenever $k < m$ are natural numbers, then $$ f_k^{(4)}(0)=f_m^{(4)}(0) \iff (k,m)=(4,6). $$

Olod
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Yes, $p=q=2$, since then: $$f(x)=\tfrac12(\sin^2x+\cos^2x)=\tfrac12=g(x)$$ and so $f(x)-g(x)=0$.

  • While true, I don't consider this answer useful or insightful. I feel like this was a case that was merely not mentioned in the question due to its triviality. – Turambar Mar 09 '17 at 19:52
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    If however not a desired answer, it should've been mentioned in the question. –  Mar 09 '17 at 19:53
  • I don't disagree with that, but it seems that the question isn't complete yet given the discrepancies between the title and the body. If we're talking trivial solutions, you could have at least generalized to p = q. – Turambar Mar 09 '17 at 20:08
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If we consider the derivative of $f(x)-g(x)$ then in order to be constant the result must be equal to 0. So after doing some operations $$sin^{q-2}(x)[1-sin^{p-q}(x)]=cos^{q-2}(x)[1-cos^{p-q}(x)]$$ Therefore if it is equal to 0 within any value, taking the sine and cosine of $30$ degrees and calling $q-2=a$ and $p-q=b$ after some operations: $$2^b=\frac{\sqrt{3}^{a+b}-1}{\sqrt{3}^{a}-1}$$ Since both sides must be integers for positive integer values of a and b then $a=2n$ and $b=2m$ studying the result the only possible integer solution of that is $n=1$ and $m=1$ therefore $q=4$ and $p=6$ which we know it works.

J. P. C.
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Let $p,q$ be integers. If $p=q$, one has nothing to do. Suppose $p<q$. Let $$F(x)=\dfrac1p(\sin^px+\cos^px)-\dfrac1q(\sin^qx+\cos^qx)$$ and then \begin{eqnarray} F'(x)&=&\sin^{p-1}x\cos x-\cos^{p-1}x\sin x-\sin^{q-1}x\cos x+\cos^{q-1}x\sin x\\ &=&\sin x\cos x(\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x). \end{eqnarray} If $f(x)-g(x)$ is constant, then $F'(x)\equiv0$ and hence $$ \sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x\equiv0.$$ Let $$ h(x)=\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x $$ and then \begin{eqnarray} h'(x)&=&(p-2)\sin^{p-3}x\cos x+(p-2)\cos^{p-3}x\sin x-(q-2)\sin^{q-3}x\cos x-(q-2)\cos^{q-3}x\sin x\\ &=&\sin x\cos x\big\{(p-2)\big[\sin^{p-4}x+\cos^{p-4}x\big]-(q-2)\big[\sin^{q-4}x+\cos^{q-4}x\big]\big]. \end{eqnarray} Since $h(x)\equiv0$, $h'(x)\equiv0$. Clearly if $p=2$, then $q=2$. Suppose $p>2$. Let $x=\frac{\pi}{4}$ and then one has $$ (\frac{\sqrt{2}}{2})^{p}=\frac{q-2}{p-2}(\frac{\sqrt{2}}{2})^{q}$$ or $$ 2^{\frac12(q-p)}=\frac{q-2}{p-2}. $$ Let $p-2=2m,q-2=2n$ ($m<n$) and then one has $$ 2^{n-m}=\frac{n}{m} $$ from which one has $n=m2^r$ ($r>0$). So $$ 2^{m(2^r-1)}=2^r$$ or $$ m(2^r-1)=r. $$ Noting if $r>1$, $2^r-1>r$ and hence $m(2^r-1)>r$, one must have $r=1$ and hence $m=1,n=2$. Thus $p=4,q=6$.

xpaul
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