## JenniferSmart1 Group Title how do I write this in sum form? $y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)$ one year ago one year ago

1. JenniferSmart1 Group Title

I know I'm a disappointment :(

2. TuringTest Group Title

no you're not, I don't have it either$y=a_0\sum_{n=0}^\infty\frac{x^{2n}}{}$darn I know there is a way to do the denom, but I can't think of it... I have seen final answers for power series DE's written as you have them though, so the summation notation is not necesarry

3. hartnn Group Title

i would have written denominator of the form 2^m m!

4. hartnn Group Title

num = x^(2m)

5. JenniferSmart1 Group Title

this is the whole (series?) $a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7$

6. hartnn Group Title

$$\large y=a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}$$ doesn't this work ??

7. TuringTest Group Title

yeah, that's it @hart

8. JenniferSmart1 Group Title

9. JenniferSmart1 Group Title

#9

10. JenniferSmart1 Group Title

how and why?

11. hartnn Group Title

2 = 2 (1) = 2^1 (1!) 2.4 = 2.2(1.2) = 2^2 (2!) 2.4.6 = 2.2.2 (1.2.3) = 2^3 (3!) in general denominator = (2^n n!)

12. hartnn Group Title

got that ^ ?

13. JenniferSmart1 Group Title

still thinking

14. JenniferSmart1 Group Title

ok from the top. so you look at this and you think..........? What is your thought process? $a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7$

15. hartnn Group Title

my 1st thought : did u change the Q ? where does these a1's jump in..... ? anyways, i would first factor out a0 and a1....

16. JenniferSmart1 Group Title

that's the whole sum (series?) what I wrote is just the a_1, then I added the a_0 It's from a previous problem http://openstudy.com/users/jennifersmart1#/updates/50d63d0fe4b0d6c1d541ef3a

17. hartnn Group Title

ok, so did u get how , after factoring a0, u get $$\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}$$ now your next step is to find sum for a1, right ?

18. JenniferSmart1 Group Title

I didn't quite get that part...

19. hartnn Group Title

which part ? x^(2m) or 2^m or m! ??

20. JenniferSmart1 Group Title

ok I'm back. I can make sense of the answer but how do I come up with this on my own.... =(

21. JenniferSmart1 Group Title

$y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)$ $\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}$ Let's look at this

22. JenniferSmart1 Group Title

so as m increases ... that makes sense Problem is will I be able to do this on my own?

23. TuringTest Group Title

Well as you saw, I had forgotten about the 2^m for the product of only the even numbers in the factorial, but this is kind of a tricky pattern to write if you ask me. They are not all this bad. You can just reason to yourself about needing a 2 for each of the m elements in the factorial to make each factor double, and therefor even.

24. JenniferSmart1 Group Title

ok

25. JenniferSmart1 Group Title

ok I'll try writing the sum a_1 let me know how I did

26. TuringTest Group Title

I'm not sure I know that one... I actually remember trying to come up with this before, a summation representing the product of evens and could not. Maybe I'll be more lucky this time.

27. TuringTest Group Title

*odds I mean

28. hartnn Group Title

product of odds is not difficult...u'll get it turing....

29. hartnn Group Title

hint : 1/(1.3.5) = 2.4 /(5!)

30. TuringTest Group Title

ohhhhhhhhhhhhhh

31. JenniferSmart1 Group Title

=$a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9$ =$a_1 \left(x^1+\frac{x^3}{3}+\frac{x^5}{3\cdot5}+\frac{x^7}{3\cdot5\cdot7}+\frac{x^9}{3\cdot5\cdot7\cdot9}+\frac{x^{10}}{3\cdot5\cdot7\cdot9\cdot10}\right)$ = ok let's start with the exponents....or where would I look first?

32. JenniferSmart1 Group Title

I understand the hint.... :(

33. JenniferSmart1 Group Title

I don't understand I meant

34. hartnn Group Title

1/(1.3.5) = 2.4 /(5!) = 2^2 . (1.2) / (5!) = 2^2 (2!)/(5!) got that ?

35. JenniferSmart1 Group Title

so how ....Ohhh by having 2x4 in the numerator your eliminating the evens in the denominator!!!!

36. hartnn Group Title

i try to bring factorial form wherever possible.

37. hartnn Group Title

for 1.3.5 i needed 2.4 to get 5! so i multiplied and divided by 2.4 now from 2.4, i get 1.2=2! by factoring out 2, twice.

38. JenniferSmart1 Group Title

$\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ doesn't m=3 for this?

39. JenniferSmart1 Group Title

and yes this part makes sense

40. hartnn Group Title

why m=3 ? i would say m=2 5 = 2m+1 so, the term will be x^(2m+1) 2^m m!/ (2m+1)! got that ?

41. JenniferSmart1 Group Title

no I mean this is the 3rd ....whatchamacallit? "m" is from 0 to infinity ...and this would be the third term?

42. JenniferSmart1 Group Title

nevermind

43. hartnn Group Title

ohh..

44. hartnn Group Title

start from m=0

45. hartnn Group Title

m=0 <-1st term m=1 <-2nd term m=2 <-3rd term

46. JenniferSmart1 Group Title

$\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ my point is....where does me thought process continue from here...how do I change these numbers to variables. ok so were working on the third term now. where m=2

47. JenniferSmart1 Group Title

*my not me

48. JenniferSmart1 Group Title

soo many typos sorry

49. JenniferSmart1 Group Title

2m+1

50. JenniferSmart1 Group Title

LOL

51. JenniferSmart1 Group Title

<---frustrated koala :S

52. hartnn Group Title

what i do is always take 3rd or 4th term., bring it in factorial form, take lowest number =m all other numbers in terms of m....

53. JenniferSmart1 Group Title

oh so 2m+1 would be right!

54. hartnn Group Title

yes.

55. JenniferSmart1 Group Title

where would it go? $\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ shall we put it somewhere in the numerator?

56. JenniferSmart1 Group Title

sorry koalas are fairly slow...I'm still thinking

57. hartnn Group Title

all this exercise was to find the general term , which we found as x^(2m+1) 2^m m!/ (2m+1)! right ? now just put a summation sign b4 this, with limits from 0 to infinity.

58. JenniferSmart1 Group Title

$\frac{x^5}{3\cdot5}$ m=2 $\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ $\frac{x^{(2m+1)}2^mm!}{(2m+1)!}$ ok I can see it now

59. JenniferSmart1 Group Title

I'm just very visual

60. hartnn Group Title

finally it will be $$\large a_1[\sum\limits _0^\infty\frac{x^{(2m+1)}2^mm!}{(2m+1)!}]$$

61. JenniferSmart1 Group Title

thanks @hartnn

62. JenniferSmart1 Group Title

do either of you have a similar problem somewhere? LOL I wanna try this again

63. hartnn Group Title

welcome ^_^

64. JenniferSmart1 Group Title

no I think this is enough. Ok thanks

65. JenniferSmart1 Group Title

does this make sense to anyone? $\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}$

66. JenniferSmart1 Group Title

for both a_0 and a_1

67. hartnn Group Title

thats for a0 only...

68. JenniferSmart1 Group Title

It's the solution in the back of my book for y''-xy'-y=0

69. JenniferSmart1 Group Title

70. JenniferSmart1 Group Title

I think I wrote it out correctly, and the the sums we wrote seem right too...but somehow he took those two and made one compact solution

71. hartnn Group Title

u mean in the last image ? he just wrote solution of initial value problem.... with a1=0

72. JenniferSmart1 Group Title

ohhhhhhh....durrrr sorry. Ok I got now haha