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

1. JenniferSmart1

I know I'm a disappointment :(

2. TuringTest

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

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

4. hartnn

num = x^(2m)

5. JenniferSmart1

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

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

7. TuringTest

yeah, that's it @hart

8. JenniferSmart1

9. JenniferSmart1

#9

10. JenniferSmart1

how and why?

11. hartnn

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

got that ^ ?

13. JenniferSmart1

still thinking

14. JenniferSmart1

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

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

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

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

I didn't quite get that part...

19. hartnn

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

20. JenniferSmart1

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

21. JenniferSmart1

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

22. JenniferSmart1

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

23. TuringTest

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

ok

25. JenniferSmart1

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

26. TuringTest

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

*odds I mean

28. hartnn

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

29. hartnn

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

30. TuringTest

ohhhhhhhhhhhhhh

31. JenniferSmart1

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

32. JenniferSmart1

I understand the hint.... :(

33. JenniferSmart1

I don't understand I meant

34. hartnn

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

35. JenniferSmart1

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

36. hartnn

i try to bring factorial form wherever possible.

37. hartnn

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

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

39. JenniferSmart1

and yes this part makes sense

40. hartnn

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

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

42. JenniferSmart1

nevermind

43. hartnn

ohh..

44. hartnn

start from m=0

45. hartnn

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

46. JenniferSmart1

$\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ 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

*my not me

48. JenniferSmart1

soo many typos sorry

49. JenniferSmart1

2m+1

50. JenniferSmart1

LOL

51. JenniferSmart1

<---frustrated koala :S

52. hartnn

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

oh so 2m+1 would be right!

54. hartnn

yes.

55. JenniferSmart1

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

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

57. hartnn

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

$\frac{x^5}{3\cdot5}$ 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

I'm just very visual

60. hartnn

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

61. JenniferSmart1

thanks @hartnn

62. JenniferSmart1

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

63. hartnn

welcome ^_^

64. JenniferSmart1

no I think this is enough. Ok thanks

65. JenniferSmart1

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

66. JenniferSmart1

for both a_0 and a_1

67. hartnn

thats for a0 only...

68. JenniferSmart1

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

69. JenniferSmart1

70. JenniferSmart1

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

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

72. JenniferSmart1

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