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 Apr 25 asked truth table of equivalence relation Feb 18 awarded Promoter Feb 18 awarded Tumbleweed Feb 11 asked transform from the replication equation to the Lotka-Volterra Feb 3 accepted explanation of slope in a graph? Feb 3 asked explanation of slope in a graph? Jan 30 accepted what does the inverse membership symbol means? Jan 29 asked what does the inverse membership symbol means? Jan 22 accepted What is the difference between a polynomial regression and a generalized linear model? Jan 22 awarded Student Jan 22 comment What is the difference between a polynomial regression and a generalized linear model? @StefanHansen, it could be something like this client_pref = c0 + c1 * year + c2 * bimonthly. For the second case it would be bodyfat=age+breast+arms. I dont know why I cannot use abline in the first case. For what some people told me is because in the first case I will have a hyperplane, but in the second one not. In the second case I got three variables so why I cannot get an hyperplane also Jan 22 comment What is the difference between a polynomial regression and a generalized linear model? @StefanHansen, I have included two examples, maybe you can give me a hint with that Jan 22 revised What is the difference between a polynomial regression and a generalized linear model? added 313 characters in body Jan 22 comment What is the difference between a polynomial regression and a generalized linear model? exactly, but I saw a case in which there were no exponents. I think that it has to be that the predictor variables have no relationship within each other, but in the second case they have. For example first case could be about prediction of consumer interest yearly and each two months; while the second one could be about weight prediction for a person taking the measures of other parts of their body Jan 22 asked What is the difference between a polynomial regression and a generalized linear model? Jan 20 awarded Scholar Jan 20 awarded Supporter Jan 20 comment Prove - In $\frac{1}{{2}^{\lambda }}\leq \frac{1}{{\lambda }^{d}}$, there exists a large value of $\lambda$, so that the equation holds for any $d$. thanks @marty cohen, I have been thinking in the first part of the solution that I mentioned, but I was missing the rest of it. It was very helpful your post Jan 20 accepted Prove - In $\frac{1}{{2}^{\lambda }}\leq \frac{1}{{\lambda }^{d}}$, there exists a large value of $\lambda$, so that the equation holds for any $d$. Jan 20 awarded Commentator