Recent content by songoku

Why can we take ##p## to be 1?

Thank you very much for the explanation JimWhoKnew, Gavran, PeroK, FactChecker

I have posted the full question so I don't know about this Sorry but why should we consider the case where the quadratic must have one solution? Using differentiation, I can get the stationary point but I still don't see what the max value of ##b## since the greatest value of ##b## for...

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there...

I understand. Thank you very much for the help and explanation DaveE, nasu, Steve4Physics

I answered A but the answer key is B. I thought since the wave is moving to the right, P started first compared to Q (the crest of P is closer to y-axis compared to Q). Why is the correct answer B? Thanks Edit: Wait, is it because the x-axis is position so the position of Q is in front of P...

I understand. Thank you very much for all the help and explanation PeroK, BvU, WWGD, DaveC426913, JimWhoKnew, Gavran

Yes. Let R = red ball and B = blue ball. Arrangement of BBRR in circle will be the same as BRRB So, the strategy is to list all the possible arrangements? I was considering another case where there are only 2 red balls and 2 blue balls so n = 4 Yes I have tried listing several possibilities...

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of...

I am very sorry for late reply Thank you very much for all the help and explanation pasmith, PeroK, FactChecker, BvU, Gavran, anuttarasammyak

Is calculating the bounds based on one - sided limit like post #24? But in post#15, the delta you take is ##|\sqrt[3]{8+\epsilon}-2|##, not the negative one? By minimum, do you mean ##\delta=\text{min}~(1, \frac{\epsilon}{4})##? Thanks

Ah my bad Oh ok, but how about something like this? That example is from textbook and there is part 2 of the working (to show that this ##\delta## works). Is this part not needed and I can just stop at part 1 after stating ##\delta=\frac{\epsilon}{4}##? I am sorry I don't get this. In post...

I am trying to but sorry I don't have idea how to continue without choosing a certain value for ##\delta##. What I practice up until now is the one which always choose the value of ##\delta##. If not choosing any value for ##\delta##, what is the continuation after ##\delta=\text{min}...

Then how to continue the proof? Let: ##\delta_1=\text{min}\{\sqrt[3] {8-\epsilon}-2, \sqrt[3] {8+\epsilon}-2\}## ##\delta_2=\text{min}\{\sqrt{4-\epsilon}-2, \sqrt {4+\epsilon}-2\}## ##\delta=\text{min} \{\delta_1, \delta_2 \}## For left limit: ##2-\delta<x<2##. How to change this into ##|x^3...

Wait, I don't need to choose which one is the minimum? But then when I want to prove it, what should I put as ##delta##? Oh I have to take one common ##\delta## and use it for left and right side limit? Thanks