mm-2609 === Subject: Re: On Math > Mathematics is made up of a bunch of garbage > ideas in a world made up of pure reality. Make up your mind, please. Is the world made up or is it reality? David Ames === Subject: RE: Cartesian Equation Question I'm assuming a cartesian equation is, as posted in the other response, one in terms of y and x only, with no parameters. To this effect, you've gotten one, albeit a function of y as opposed to x, but a cartesian equation nonetheless. If you want a gemoetric interpretation of the formula you have, then it is the parabola opening to the right, intersecting at y=3 and y=1. So, if you want to try to break it into 2 square root functions, go ahead, but this should suffice for whatever application you've got. ------= Binary Usenet downloading made easy =--------- -= Get GrabIt for free from http://www.shemes.com/ =- === Subject: Re: Cartesian Equation Question >I'm assuming a cartesian equation is, as posted in the other response, one in terms of y and x only, with no parameters. To this effect, you've gotten one, albeit a function of y as opposed to x, but a cartesian equation nonetheless. If you want a gemoetric interpretation of the formula you have, then it is the parabola opening to the right, intersecting at y=3 and y=1. So, if you want to try to break it into 2 square root functions, go ahead, but this should suffice for whatever application you've got. >------= Binary Usenet downloading made easy =--------- >-= Get GrabIt for free from http://www.shemes.com/ =- Mike, it would be helpful to others if you would post context along with your replies. I see several of your posts this morning where I have no idea what you are replying to. Not everyone keeps complete threads for days on end. --Lynn === Subject: Re: Cartesian Equation Question I just realized that now. Oops. I was using GrabIt at the time, which is Sorry about that. I also notice now that it adds something to the end of my post, of which I do not approve. SHould be better now. Mike >>I'm assuming a cartesian equation is, as posted in the other response, one >>in terms of y and x only, with no parameters. To this effect, you've >>gotten one, albeit a function of y as opposed to x, but a cartesian >>equation nonetheless. If you want a gemoetric interpretation of the >>formula you have, then it is the parabola opening to the right, >>intersecting at y=3 and y=1. So, if you want to try to break it into 2 >>square root functions, go ahead, but this should suffice for whatever >>application you've got. >>------= Binary Usenet downloading made easy =--------- >>-= Get GrabIt for free from http://www.shemes.com/ =- > Mike, it would be helpful to others if you would post context along > with your replies. I see several of your posts this morning where I > have no idea what you are replying to. Not everyone keeps complete > threads for days on end. > --Lynn === Subject: Re: Cartesian Equation Question <0Jlmf.76492$ki.5758@pd7tw2no> You shouldn't be top-posting, either. (Your response should be after the relevant part of the post you're quoting.) I'm top-posting here. > I just realized that now. Oops. I was using GrabIt at the time, which is > Sorry about that. I also notice now that it adds something to the end of my > post, of which I do not approve. SHould be better now. I hope so. (I'm not top-posting here.) --- Christopher Heckman >>I'm assuming a cartesian equation is, as posted in the other response, one >>in terms of y and x only, with no parameters. To this effect, you've >>gotten one, albeit a function of y as opposed to x, but a cartesian >>equation nonetheless. If you want a gemoetric interpretation of the >>formula you have, then it is the parabola opening to the right, >>intersecting at y=3 and y=1. So, if you want to try to break it into 2 >>square root functions, go ahead, but this should suffice for whatever >>application you've got. >> >> >>------= Binary Usenet downloading made easy =--------- >>-= Get GrabIt for free from http://www.shemes.com/ =- > Mike, it would be helpful to others if you would post context along > with your replies. I see several of your posts this morning where I > have no idea what you are replying to. Not everyone keeps complete > threads for days on end. > --Lynn === Subject: Re: On Math Hmm, answer me something: Where ISN'T Mathematics in the real world? You do realize that 15 years ago, 2 day forecasts weren't nearly as predictable as the 5-day forecasts are now, right? That is only partially due to the fact that computers have evolved, but more so to do with the fact that the mathematics behind it has evolved so much more. So, your reasons for thinking that mathematicians constantly spend their time working on irrelevant or trivial problems is entirely unfounded and without proof. Or, perhaps, you DO have proof? If so, where do you get this proof? Are you yourself a mathematician? If so, then you would be guilty of the sins that you claim that all mathematicians are; or you are not guilty of them, or you aren't a mathematicion. Therefore, you are either not a mathematician, a hypocrite, or one guilty of making gross generalizations; neither of the latter 2 having any place in this group. While some problems may not be solved as quickly as you like, they are being solved. You want them done faster? Do them yourself and stop criticising the people who got technology and science where it is today. ------= Binary Usenet downloading made easy =--------- -= Get GrabIt for free from http://www.shemes.com/ =- === Subject: Re: Please help me find solution to this calculus problem Ok, your answer is right, but your second term ends up going to 1, so the straight area is just 1/e. As for plotting the rectangle in Mathematica, you could try plotting -Ln[x] and (x)*(-Ln[x]) on the same graph using the Plot function, then comparing where the second function peaks to the -Ln[x] graph. But, as for the area, it is 1/e. Hope it helps ------= Binary Usenet downloading made easy =--------- -= Get GrabIt for free from http://www.shemes.com/ =- === Subject: Re: Small Question About Complex Numbers Technically, there is no flaw in your logic, however, there is an omission. In your very last step, where you state that sqrt(1)=1, you have omitted the fact that (-1)^2=1, and therefore sqrt(1)=+/-1, and as we know that i^2 is not equal to 1 (By definition), then it is equal to -1. So, not so much a flaw, as an omission, but one that is entirely understandable, as we are so used to only worrying about the positive square roots of something. Hope this helps ------= Binary Usenet downloading made easy =--------- -= Get GrabIt for free from http://www.shemes.com/ =- === Subject: Re: dy/dy In Liebniz notation (dy/dx), dy/dx can literally be translated as The rate of change of y with respect to the rate of change of x, and if x is an independant variable, then dx/dx is 1 (You can also treat dx/dx as a fraction, and reduce it to lowest terms by dividing the top and bottom by dx, thus getting 1). As for gradient, that essentially only applies to a 3-dimensional surface, plotted as z=f(x,y). Gradient is actually a vector in 3-space that points in the direction of steepest increase (In the positive z direction) of the curve f(x,y) at a point (a,b). The gradient is given by: Gradient(f)=(Df/Dx,Df/Dy) where Df/Dx and Df/Dy represent the partial derivatives of f with respect to x and y respectively. It should also be noted that the gradient is a vector, pointing in the direction stated above. That is probably more information that you were after, but hope some of it helps. ------= Binary Usenet downloading made easy =--------- -= Get GrabIt for free from http://www.shemes.com/ =- === Subject: Re: The prize 10 000 new conjecture Aside from the obvious fact that it'll never have a solution (The top is always even, the bottom is always odd at some point), here is a semi-informal/intuitive proof: 2^n/9x=y, for n,x,y in N, cannot have solutions for odd x (Obviously). Therefore, we are restricted to letting x be even. THerefore, x=2m, for some m in N. We see that 2^n/9x=2^n/(9*2m)=2^(n-1)/9*m which again, has no solutions for m odd, therefore m is even again therefore let m = 2t, and repeat as above to get that 2^(n-1)/9*m=2^(n-2)/9*l which once agian.....And so on. Therefore, our factor of x must be a power of 2 for even the possiblity of solution to arise. So, x cannot be larger than 2^(n-4) for us to have a solution in the integers (Positive or negative), this is because if we let x=2^(n-3) then 2^n=x*2^3=x*8 Aside from the obvious fact that it'll never have a solution (The top is always even, the bottom is always odd at some point), here is a semi-informal/intuitive proof: > 2^n/9x=y, for n,x,y in N, cannot have solutions for odd x (Obviously). Therefore, we are restricted to letting x be even. THerefore, x=2m, for some m in N. We see that 2^n/9x=2^n/(9*2m)=2^(n-1)/9*m which again, has no solutions for m odd, therefore m is even again therefore let m = 2t, and repeat as above to get that 2^(n-1)/9*m=2^(n-2)/9*l which once agian.....And so on. Therefore, our factor of x must be a power of 2 for even the possiblity of solution to arise. So, x cannot be larger than 2^(n-4) for us to have a solution in the integers (Positive or negative), this is because if we let x=2^(n-3) then 2^n=x*2^3=x*8 This could be extended to a rigorous proof through induction and some thought, thought I don't dignify this problem with. Whoever wants a counter example to this is just wasting people's time. > ------= Binary Usenet downloading made easy =--------- > -= Get GrabIt for free from http://www.shemes.com/ =- === Subject: 1 + 1/x = x How would one find the solution to: 1 + 1/x = x === Subject: Re: 1 + 1/x = x Mr. Patel: The equation you give may be considered to be the definition of the famous Golden Number or Golden Mean. It is solved by multiplying through by x to get the quadratic x^2 - x - 1 = 0 which leads to the solutions x = (1 + sqrt(5))/2 = 1.61803398... and x = (1 - sqrt(5))/2 = 0.61803398... in which the letter x is usually replaced by the Greek letter gamma (which my keyboard can't print). You can see by the defining equation that gamma = 1/gamma +- 1. (I have to write +- when I mean plus or minus; again, my keyboards limits me.) Grover Hughes > How would one find the solution to: > 1 + 1/x = x === Subject: Re: 1 + 1/x = x > Mr. Patel: > The equation you give may be considered to be the definition of the famous > Golden Number or Golden Mean. It is solved by multiplying through by x > to get the quadratic > x^2 - x - 1 = 0 > which leads to the solutions > x = (1 + sqrt(5))/2 = 1.61803398... > and > x = (1 - sqrt(5))/2 = 0.61803398... > in which the letter x is usually replaced by the Greek letter gamma No; usually, it's the letter phi (which looks like a circle with a slash through it). Gamma is usually reserved for the Euler-Mascheroni constant lim (1 + 1/2 + ... + 1/n - ln n) as n goes to infinity, or approximately 0.577... --- Christopher Heckman > How would one find the solution to: > 1 + 1/x = x === Subject: Re: 1 + 1/x = x > How would one find the solution to: > 1 + 1/x = x Multiply both sides by x: x + 1 = x^2 x^2 - x - 1 = 0 Solve with the quadratic formula. x=1.618 or x=-0.618 Nicholas Sherlock === Subject: Re: 1 + 1/x = x > How would one find the solution to: > 1 + 1/x = x > Multiply both sides by x: > x + 1 = x^2 > x^2 - x - 1 = 0 > Solve with the quadratic formula. > x=1.618 or x=-0.618 > Nicholas Sherlock sides by the variable itself (not reversible)? Doing this creates extra solutions right? === Subject: Re: 1 + 1/x = x > How would one find the solution to: > 1 + 1/x = x >> Multiply both sides by x: >> x + 1 = x^2 >> x^2 - x - 1 = 0 >> Solve with the quadratic formula. >> x=1.618 or x=-0.618 > sides by the variable itself (not reversible)? Doing this creates extra > solutions right? No, put the negative and positive answers back into the original equation to try them out. They are both valid solutions. Nicholas Sherlock === Subject: Re: basic statistics query (and sanity check!) I'd sure like to meet your wife's teacher and do a little betting on coin flips er sumpin...... hang in there with your (and her) 1/6, or else all I've ever learned about probabilities has been time wasted... Grover Hughes > My wife is doing a college course part of which is a GCSE in maths, > whiuch historically has been a sticky subkect for her. Me, on the > other hand, always found maths my easiest subject, and I ended up with > s staticstics degree! > That degree however WAS 21 years ago now however and I accept that I > may now be totally and utterly confused today... so I turn to your > good selves for some feedback. > So... here is the question verbatim from her assigment sheet: > a) 1250 tickets are sold in a raffle. > They are individually placed in a big hat. If you only draw one ticket > what is the probability that your number will be drawn first? > [My wife answered 1/1250, and was marked correct on this. I agree!] > b)A fair die is rolled 2 times and each time it shows a 4. What is the > probability that the die will show a 4 again on the third roll? > well??? > (scroll down) > My wife answered 1/6. That is also my answer on the basis that is it a > FAIR die (as stated in the question) and that each roll is therefore a > random event, unaffected by anything that has happened previously. > However... her LECTURER told her it was NOT 1/6 but..... ONE! ie > 1/1! > Because , he says, HISTORICAL evidence shows that it MUST be a 4! > Surely that cannot be right? Surely it IS 1/6? > HEELLLPPPP! My sanity is at stake! > didds2 === Subject: Re: basic statistics query (and sanity check!) > My wife is doing a college course part of which is a GCSE in maths, > whiuch historically has been a sticky subkect for her. Me, on the > other hand, always found maths my easiest subject, and I ended up with > s staticstics degree! > That degree however WAS 21 years ago now however and I accept that I > may now be totally and utterly confused today... so I turn to your > good selves for some feedback. > So... here is the question verbatim from her assigment sheet: > a) 1250 tickets are sold in a raffle. > They are individually placed in a big hat. If you only draw one ticket > what is the probability that your number will be drawn first? > [My wife answered 1/1250, and was marked correct on this. I agree!] > b)A fair die is rolled 2 times and each time it shows a 4. What is the > probability that the die will show a 4 again on the third roll? > well??? > (scroll down) > My wife answered 1/6. That is also my answer on the basis that is it a > FAIR die (as stated in the question) and that each roll is therefore a > random event, unaffected by anything that has happened previously. > However... her LECTURER told her it was NOT 1/6 but..... ONE! ie > 1/1! > Because , he says, HISTORICAL evidence shows that it MUST be a 4! > Surely that cannot be right? Surely it IS 1/6? > HEELLLPPPP! My sanity is at stake! > didds2 Of course you are right, if it was stated as a fair die as opposed to a die - then you might have some debate. In fact, how could it be a fair die if it always comes up 4! Bill === Subject: Re: f(x) = 6x^4 ln4x^3 > f(x) = 6x^4 ln4x^3 > find f'(x) is the ansewr 24x^3(ln 4x^3) + 6x^4/4x63 * 12x^2 ? Yes, except that you accidentally mis-typed in a couple of places-- you spelled answer as ansewr, and you failed to press the shift key when you meant to print a caret after the fourth x to show exponentiation. Glad to see I'm not the only one who can't always type correctly! Good luck in the future! Grover Hughes