I came across this very complex equation (calculating the Gaussian curvature of a surface): \begin{align*} 1 \not\equiv &-\frac{m}{2}\Bigl(\frac{3}{2}C+Su^{-1}-Tu^{-1}+2+Qu^{-1}\Bigr)\\ &\qquad\times(u^3C+Su^2-Tu^2)^{(-2u-2Q-5S+5T)/(6uC+4S-4T)} \\ &+\frac{m}{2}\Bigl[u^3(4C^2-3C)+u^2\Bigl(\frac{1}{3}TC-\frac{1}{3}SC-2S+2T-3QC\Bigr)\\ &\qquad\qquad+u(4TS-2S^2-2T^2-2QS+2QT)\Bigr]\\ &\qquad\times(u^3C+Su^2-Tu^2)^{(-2u-2Q-9S+9T-6uC)/(6uC+4S-4T)} \end{align*} I should somehow be able to prove that the right side of the equation is not identically equal to $1$. I would need some rule or something that "by eye" makes it obvious that it can not be $1$ (i.e. is not a constant function equal to $1$). $T$, $S$, $Q$, $C$ and $m$ are all constants (all the constants can not be zero, $m>2$ and must be integer, $T$ can not be equal to $S$ ); the only variable is $u$.

I thought that to be worth 1 certainly the functions that are "exponents" must vanish, so they should become zero, but the numerators of the two "exponents" being different, if one is zero (for some value of the constants) the other can not zero and therefore the formula will never be constant equal to 1.

Tips?

  • By "cannot be 1," do you mean that it is never equal to 1 for any value of u, or that it is not identically 1 for any choice of constants? – Gabe K Nov 8 at 15:37
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    I made an edit to try to clarify, changing the statement to $1 \not\equiv \cdots$ and writing "the right side is not identically equal to 1". Feel free to roll back or edit further if this is not satisfactory. – Nate Eldredge Nov 8 at 17:18
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    Well, for starters it can't be identically equal to any constant unless its derivative with respect to u is identically equal to zero. What restrictions on the constants do you find when you impose that condition? – Jeanne Clelland Nov 8 at 18:53
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    It seems to me that no attempt has been made by the OP to simplify the expression. There are a number of obvious things: like cancelling the common $u$ factor in the "exponent",... I am coming around to the view that this question is not appropriate to MO. – José Figueroa-O'Farrill Nov 8 at 19:23
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    @exxxit8 - Maybe my comment wasn't clear enough. If you differentiate your expression with respect to u, you'll obtain an expression involving polynomial and exponential terms in u. In order for this function to be identically zero, all the coefficients of the functionally distinct terms in this expression must vanish. This should give you several expressions involving your constants that must all vanish, and that will put restrictions on the values of your constants. – Jeanne Clelland Nov 8 at 23:16
up vote 5 down vote accepted

First of all, we introduce $R = T-S$, since $T$ and $S$ only appear in that combination. Secondly, we introduce the notation $$ \alpha = \frac{-2 u - 2Q + 5 R}{6 C u - 4 R} $$ which is the first exponent: the second exponent is $\alpha -1$.

If I have not made an error, I find that the RHS to the expression in the post simplifies to the following: $$ \frac{m}{2} (C u - R)^{\alpha-1} u^{2\alpha -1} \left( \tfrac52 C(C-2) u^2 + (\tfrac{17}{6} CR + 4 R - 4 CQ) u + 3 R(Q-R)\right), $$ which we can rewrite in a more suggestive form as follows: $$ \frac{m}{2} (C u^3 - Ru^2)^{\alpha-1} \left( \tfrac52 C(C-2) u^3 + (\tfrac{17}{6} CR + 4 R - 4 CQ) u^2 + 3 R(Q-R)u\right), $$ which has the form $$ \frac{m}{2} P_1(u)^{\alpha -1} P_2(u) $$ where $P_1$ and $P_2$ are two cubic polynomials in $u$. Your conditions say that neither $C$ nor $R$ can be zero, so that $P_1(u)$ is not zero for all $u$. Hence for this expression to be equal to $1$ for all $u$, it must be the case that $$ P_1(u)^{1-\alpha} = \frac{m}{2} P_2(u) $$ Since $C \neq 0$, there are two possibilities: either $\alpha = 0$ and then $P_1/P_2 = m/2$ for all $u$, or else $\alpha = 1$ and then we must ensure that $m/2 P_2(u)=1$ for all $u$. It follows from the expression for $\alpha$ above that it cannot be equal to zero (unless perhaps you are allowed to take the limit $C \to \infty$), so the only possibility is $\alpha = 1$.

This condition fixes $C = -1/3$ and $Q = 9R/2$, and plugging these back into the expression for $P_2$ we see that it is not constant.

I don't discard having made calculational errors, but I think the idea is sound.

  • Now that you’ve answered the question, have you changed your mind from mathoverflow.net/questions/314847/… that the question is not appropriate for MO? Your answer is very nice, but certainly makes the question seem not research level. – LSpice Nov 8 at 22:23
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    @LSpice I have not changed my opinion, but I could not resist the temptation to bring order into chaos :) I'm a little out of touch with MO as of late, but back in the day I would have expected this question to have been closed as "too localized" in a very short time. – José Figueroa-O'Farrill Nov 8 at 22:29
  • @José Figueroa-O'Farrill - Thank you very much – exxxit8 Nov 9 at 7:18
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    @exxxit8 - for the exponent, if $C=0$ then $\alpha$ is a constant only for some fixed value of $u$ – Alexander Pigazzini Nov 9 at 17:25
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    @exxxit8 Since $C\neq 0$ the polynomial $P_1$ has nonzero top coefficient so it is truly a cubic. Now you have a cubic polynomial to some power has to be equal to a polynomial $P_2$ which has degree at most $3$. Comparing the highest order term, the power has to be $0$ (in which case $P_2$ is actually a constant polynomial) or $1$. – José Figueroa-O'Farrill Nov 9 at 17:28

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