## myininaya 5 years ago calculus dream problem? (fg)'=f'g' and (fg)''=f''g'' I don't think this what you guys are looking for but here what i did

1. watchmath

again? :D

2. anonymous

watchmath: ur problem has become famous..:D

3. myininaya

4. myininaya

5. myininaya

have you done it?

6. watchmath

Check it here. It is solution for f. The g will be of the same form, but then the coefficients need to be adjusted.

7. myininaya

8. watchmath

:D :D

9. myininaya

i feel like i really dont sleep when i dream about math problems

10. myininaya

i have to look at this later but im very interested thanks for the link

11. myininaya

pdf*

12. watchmath

you are welcome :)

13. anonymous

i got as far as g = C e^ int ( f' / ( f' - f)

14. anonymous

i think, why did someone say , let F = e ^f, and G = e^g

15. anonymous

then ln F = f , and ln G = f ,

16. watchmath

Yes, that's good laplace. Then basically the idea is that from that g we can express, g' and g'' in terms of g. By doing that, then we have an equation in terms of f,f' and f'' only.

17. watchmath

I mean after you find $$g=Ce^{\int f'/(f'-f)}$$

18. watchmath

Laplace, are you cantorset?

19. anonymous

yes, but dont tell anyone

20. watchmath

:D :D

21. anonymous

my account has been suspended,

22. anonymous

im here to do math, not to debate with admin :)

23. anonymous

so from that g , whats g ' , its ...

24. watchmath

do you see the pdf file that I attached above?

25. anonymous

g ' = f' ( f' - f) C e ^ ( int f' / f' - f ) ?

26. watchmath

correct, which is f'g/(f'-f)

27. anonymous

oh im going backwards,

28. anonymous

this problem has many twists and turns

29. anonymous

one sec , im going to read

30. watchmath

there is some little mistake whenn I said that if f'=f then f'g=0 it should be then fg'=0. But similar to what I wrote we will arrive a t contradiction.

31. anonymous

you can fix it and send me new pdf?

32. anonymous

or you already updated? i dont know how to write pdf's, thats pretty cool

33. watchmath

hold on

34. anonymous

i also want to solve that torus problem,

35. watchmath

Here is the updated version:

36. watchmath

Are you teaching as a TA laplace?

37. anonymous

no, i tutor at the moment online

38. anonymous

a TA is what? a teacher assistant?

39. watchmath

Yes, usually if you take a graduate program in math you are a TA as well.

40. anonymous

i should,

41. anonymous

sorry was away, i enjoy math. but sometimes i cant remember the last problem i was working on. some people say its ok to forget all the loose ends problems, and start fresh. dunno. i have various interests in math, linear algebra, diff eq. they always end up in an interesting problem

42. watchmath

43. anonymous

thanks

44. anonymous

whats an easy way to draw a graph and upload it? microsoft paint?

45. watchmath

is it a graph of a well defined function? if it is you can plot it at http://fooplot.com and save it as an image

46. anonymous

and how would you upload it, whats a good site

47. anonymous

i use sendspace.com, not sure how long they keep it up

48. watchmath

where do you want to upload the picture?

49. anonymous

well i can draw problems and solve them, and you can look over

50. anonymous

www.sendspace.com , is a website that you can upload anything

51. anonymous

ok heres a question i was working on, i got no solution 2/ln 3 * ln ( x - 5) = log base 3 ( x-2) + log base 3 (5x) , i made the right side a product and then changed base

52. watchmath

so the left hand side is $$\frac{2}{\ln 3\ln(x-5)}$$

53. anonymous

nope, ln (x-5) in numerator

54. anonymous

i know, its ambiguous when you have a bad student :)

55. anonymous

im careful, i use CAS (computer algebra software)

56. anonymous

maple, did you know that maple cant find tan (105 degrees) ,

57. anonymous

tan ( 105* Pi/180) , its an exact angle, i want to complain to the makers

58. watchmath

Oh ok, then by the change of base formula we have $$2\log_3(x-5)=\log_3((x-2)(5x))$$ So $$(x-5)^2=(x-2)(5x)$$ $$x^2-10x+255=5x^2-10x$$ $$4x^2=225$$ $$x=\pm\frac{15}{2}$$ only the postive one works

59. watchmath

sorry ..... It should be 4x^2=25 So x=5/2 then no solution as well

60. anonymous

hah, yeah

61. anonymous

weird to give a student

62. anonymous

you changed the left side to log base 3

63. anonymous

2/ln 3 * ln ( x - 5) = log base 3 ( x-2) + log base 3 (5x) , what happened to your left side?

64. anonymous

your left side is log _3 (x-5) / log _3 ln 3 * 2/ln 3

65. watchmath

remember that $\log_ba=\frac{\log_c a}{\log_cb}$ Where you can choose your c as you like. Choosince c to be the natural number e, then the left side is exactly $$\log_3((x-5))$$

66. anonymous

right, i followed the rule

67. anonymous

but you have 2/ln 3 *log _3 (x-5) / log _3 (ln 3)

68. anonymous

and whats log _3 of ln 3 ? log base 3 of ln 3

69. watchmath

In that case I thin we can't simplify it nicely.

70. anonymous

its funny you did that and still got right answer. my instinct would be to change the right side to ln instead of log base 3

71. anonymous

so it must be the case that ln 3 * log_3 ln 3 = 1 ?

72. watchmath

no

73. watchmath

it is equal to ln(ln 3)

74. anonymous

oh i think i see it now

75. anonymous

for the left side we have 2/( ln 3) * ln (x-5) = 2/ln 3 * log_3 (x-5) / log _3 (e) = 2 *log_3 (x-5) * 1/ ( ln 3 * log _3 (e) = log _3 (x-5)^2 * 1/ ( log _3 e^(ln 3) ) = log _3 (x-5)^2* 1 / ( log _3 3 ) = log_3 (x-5)^2 * 1