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## MathSofiya Group Title Use power series to solve the differential equation $y''+xy'+y=0$ $y=\sum_{n=0}^{\infty} a_n x^n$ $y'=\sum_{n=1}^{\infty}a_nnx^{n-1}$ $y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}$ $xy'=\sum_{n=1}^{\infty}a_nnx^n$ $\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0$ one year ago one year ago

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1. TuringTest Group Title

yep, now what's first?

2. MathSofiya Group Title

changing the sums so that they all begin with n=2 changing the last sum from n=0 to n=1 would be $a_1+\sum_{n=1}^\infty a_nx^n$

3. TuringTest Group Title

no we don't strip out the terms yet, we just do an index shift

4. TuringTest Group Title

the first order of business is to get all the x's to be to the nth power through using index shifts

5. TuringTest Group Title

the only series that has x to some power other than n is the first one how can we change the index of the first series to make the exponent on x be n?

6. MathSofiya Group Title

you mean on $y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}$

7. TuringTest Group Title

yes, we need to start from a different n to get x^n through an index shift

8. MathSofiya Group Title

so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=2-2} $-\sum_{n=0}^\infty n(n-1)a_nx^n$ negative to make up for the negative?

9. MathSofiya Group Title

nooooo

10. MathSofiya Group Title

hold on

11. MathSofiya Group Title

$\sum_{n=0}^\infty n(n+1)a_nx^n$

12. TuringTest Group Title

nice :)

13. MathSofiya Group Title

still not right

14. MathSofiya Group Title

because that would give us zero

15. TuringTest Group Title

no but closer

16. TuringTest Group Title

you didn't add 2 to every n

17. MathSofiya Group Title

$\sum_{n=0}^\infty (n+2)(n+1)a_nx^n$

18. TuringTest Group Title

and I do mean $$every$$ n (subscripts included)

19. TuringTest Group Title

again better, but...^

20. MathSofiya Group Title

$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n$

21. MathSofiya Group Title

can I add them?

22. TuringTest Group Title

now that, I do believe, is correct :)

23. MathSofiya Group Title

or multiply

24. MathSofiya Group Title

or just leave as is?

25. TuringTest Group Title

add what? oh that? I wouldn't yet

26. MathSofiya Group Title

ok

27. TuringTest Group Title

yeah leave as is asd now rewrite what you have

28. TuringTest Group Title

and*

29. TuringTest Group Title

the whole expression please I'm lazy

30. MathSofiya Group Title

$\sum_{n=0}^{\infty} (n+2)(n-1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0$

31. TuringTest Group Title

$\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0$okay next order of business is what you tried to do earlier; get all indices to start at the same n value the only problem child here is which term?

32. MathSofiya Group Title

the middle

33. TuringTest Group Title

so what do we do to fix it?

34. TuringTest Group Title

you have to be a bit clever here...

35. MathSofiya Group Title

$xa_n +\sum_{n=0}^\infty a_n(n+1)x^n$

36. TuringTest Group Title

did you try to do an index shift?

37. TuringTest Group Title

it looks like you combined the idea of the index shift with stripping out the first term

38. MathSofiya Group Title

oh my...let's try this again

39. TuringTest Group Title

it is a bit tricky so let me give you a hint...

40. TuringTest Group Title

you cannot do another index shift. If you do you will change the exponent on x, which means we are screwed again, so we need to do something else to get this to start from n=0

41. TuringTest Group Title

we can't really strip out a term because we don't have this series defined for n<1, but ask yourself, if it was defined at n=0, what would the first term be?

42. MathSofiya Group Title

just have the addition of an a_n up front like in the last example?

43. TuringTest Group Title

no, answer my question above and you will see why...

44. MathSofiya Group Title

a 1

45. MathSofiya Group Title

times a_n

46. TuringTest Group Title

no, be more careful

47. TuringTest Group Title

what would be the first term of$\sum_{n=0}^\infty a_n nx^n$

48. MathSofiya Group Title

zero

49. TuringTest Group Title

right, so...

50. TuringTest Group Title

$0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n$so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.

51. TuringTest Group Title

so we can go ahead and just change the starting point of that series without further manipulation

52. TuringTest Group Title

always watch for that, it comes in handy

53. MathSofiya Group Title

oh ok uhm.....I'll revisit the last example later, because the I'm not convinced that we needed to do a index shift (or was it stripping of a term)

54. TuringTest Group Title

first to get the x^n on all terms we did an index shift (where we add or subtract from each n in the summand) next to get all the starting points for each series the same, we strip out terms/or do the trick we just did they are different tricks, and the order in which you do them is imortant

55. MathSofiya Group Title

ok

56. MathSofiya Group Title

$\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0$

57. TuringTest Group Title

it is important for you to try to understand why the index shift requires you to add/subtract each n in the summand while stripping out a term does not

58. TuringTest Group Title

good, so now make it all into one summation...

59. MathSofiya Group Title

$\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0$

60. TuringTest Group Title

nice, so what will you set to zero to find the recurrence relation?

61. TuringTest Group Title

(I am going to type minimally now as I am eating dinner at the same time :)

62. MathSofiya Group Title

food comes first in my book... :P Ok let's see here...

63. TuringTest Group Title

I can eat and respond with "yes" or "no" pretty well, no worries :)

64. MathSofiya Group Title

$(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}$

65. MathSofiya Group Title

si?

66. TuringTest Group Title

si, pero si quieras podrias factor la $$-a_n$$

67. MathSofiya Group Title

ok $a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}$

68. TuringTest Group Title

and now comes the tedious part that you can do while I eat plug in n=0,1,2,3 and check what results from that recurrence relation until you notice a pattern like last time so just cranks it out; plug in n=0,n=1,n=2 and if a_2 can be written in terms of a_0 or something like last time, do it. That is how we find the pattern.

69. TuringTest Group Title

go ahead and list out the first 5 or so and let's see if we see a relation

70. MathSofiya Group Title

$n=1:a_3=0$ $n=2:a_4=-a_2\frac{1}{12}$ $n=3:a_5=-a_3\frac{2}{20}$ $n=4:a_6=-a_4\frac{3}{30}$ $n=5:a_7=-a_5\frac{4}{42}$

71. TuringTest Group Title

what about n=0 ? we need to start from n=0 since our series starts from n=0 this time (last time it started from n=1)

72. MathSofiya Group Title

$n=0:a_2=-a_0\frac{-1}{2}$

73. MathSofiya Group Title

$n=0:a_2=a_0\frac{1}{2}$

74. MathSofiya Group Title

whatcha eating?

75. TuringTest Group Title

torta de pollo=chicken sandwich on a french sort of bread

76. TuringTest Group Title

con chipotle

77. MathSofiya Group Title

nice

78. MathSofiya Group Title

oh yeah...the pattern

79. TuringTest Group Title

so you did well in your subs I think, but now you must try to always sub all the way back as far as you can..

80. TuringTest Group Title

yeah, the pattern write each on as some finite number of unknowns, in this case everything in terms of $$a_0$$ and $$a_3$$

81. TuringTest Group Title

each one*

82. MathSofiya Group Title

in the previous example you seemed to have simply switched the $a_{odd}$ to zero and the $a_{even}$ to $a_0$

83. MathSofiya Group Title

take you're time...i'm gonna watch a few more mins of my episode

84. TuringTest Group Title

it's not different, it's the same as last time odd subscript gives zero even has a pattern reducible to a_0

85. TuringTest Group Title

okay I got a fairly simple answer, let me see if I can confirm it...

86. TuringTest Group Title

stupid wolfram is of no help :/

87. MathSofiya Group Title

$n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}$ $n=1:a_3=0$ $n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}$ $n=3:a_5=0$ $n=4:a_6=-a_4\frac{3}{30}=\frac{a_0}{5(3!)}$ $n=5:a_7=0$

88. MathSofiya Group Title

how did I do?

89. TuringTest Group Title

almost, I actually realized that I made a mistake because I think I see a mistake from you so you did as well as me I'd say :D

90. TuringTest Group Title

let me try on paper again...

91. TuringTest Group Title

here I think I can demonstrate the benefit of not multiplying out terms and writing things over-explicitly to find patterns...

92. MathSofiya Group Title

ok

93. TuringTest Group Title

$n=0:a_2=\frac{a_0}{2\cdot 1}$$n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}$now reevaluate the pattern

94. MathSofiya Group Title

$n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}$ $n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}$

95. TuringTest Group Title

so try to do the same thing as last time name all even numbers and make a new series with index k

96. TuringTest Group Title

then represent the coefficients and plug that in for $$a_n$$

97. MathSofiya Group Title

n=2k $a_n=a_{2k}=\frac{a_0}{k!}$

98. TuringTest Group Title

is it over k! ?

99. MathSofiya Group Title

hhm...you know I thought about 2k!, but wouldn't that me too much? $a_n=a_{2k}=\frac{a_0}{2k!}$

100. MathSofiya Group Title

*be

101. TuringTest Group Title

well that is the answer I get... so unless you have a way of confirming it apart from this I say that is correct btw you should put the (2k)! with parentheses to indicate that it is not 2(k!)

102. MathSofiya Group Title

$a_n=a_{2k}=\frac{a_0}{(2k)!}$

103. TuringTest Group Title

yeah, the sub it into the original guess for y

104. TuringTest Group Title

then*

105. MathSofiya Group Title

$y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}$

106. TuringTest Group Title

...and that's what I got :D I sure think it's right as it comes out quite nicely, as opposed to wolf's answer http://www.wolframalpha.com/input/?i=y''%2Bxy'%2By%3D0&t=crmtb01

107. MathSofiya Group Title

I forgot, what did you intend to study? Electrical Engineering or Mathematics?

108. TuringTest Group Title

it definitely has the e^(-x^2) think in it though just by glancing, so I am even more confident now

109. MathSofiya Group Title

why the e^?

110. TuringTest Group Title

Electrical engineering, though I admit I vacillate at times We'll seee how I feel after a year in uni

111. TuringTest Group Title

oh because the tayor expansion of e^(ax^2) always has x^(2k)/(2k)! in it (try the expansion yourself and see why, it's not that hard if you remember Taylor series)

112. MathSofiya Group Title

don't have the brain cells left to do so, but I trust your reasoning

113. TuringTest Group Title

haha, well thanks. I hope I'm right. Nos vemos!

114. MathSofiya Group Title

see ya!

115. MathSofiya Group Title

my spanish sucks as you might be able to tell

116. TuringTest Group Title

Nos vemos ~ we'll see each other

117. MathSofiya Group Title

I was right!

118. TuringTest Group Title

sweet :)

119. MathSofiya Group Title

we've made a mistake

120. MathSofiya Group Title

$n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}$ is 24 not 12...so the pattern is probably incorrect too

121. TuringTest Group Title

let's see...

122. TuringTest Group Title

and why should 24 be 12 exactly?

123. MathSofiya Group Title

because we need 12 4*3 is 12

124. MathSofiya Group Title

I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one

125. TuringTest Group Title

you dropped a 2...

126. MathSofiya Group Title

where?

127. MathSofiya Group Title

$a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}$

128. MathSofiya Group Title

oh I see

129. TuringTest Group Title

you wrote$n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}$$n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}$but$a_2=\frac{a_0}2$and$\frac1{12}=\frac1{2(3!)}$so 1)$a_4=a_2\frac1{2(3!)}=\frac{a_0}2\cdot\frac1{2(3!)}=\frac{a_0}{4!}$and that minus sign you put that minus sign does not belong

130. MathSofiya Group Title

I did not pick up on the$a_2=\frac{a_0}{2}$ idea...i just randomly subbed a_0 for a_2 :/

131. MathSofiya Group Title

I'll finish y''=y tomorrow...I got it except for the parts stated above

132. TuringTest Group Title

once you find a base formula in terms of $$a_0$$ or whatever the lowest you can express is, milk it to find the pattern

133. MathSofiya Group Title

ok

134. TuringTest Group Title

and like I say, sometime the writing out each term, even if it's just multiplying by one, helps make the pattern obvious g'night then!

135. MathSofiya Group Title

g'night

136. mahmit2012 Group Title

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137. mahmit2012 Group Title

|dw:1348252918564:dw|

138. mahmit2012 Group Title

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139. mahmit2012 Group Title

|dw:1348253097421:dw|

140. mahmit2012 Group Title

|dw:1348253165705:dw|

141. MathSofiya Group Title

@TuringTest would this statement about striping a term be true (generally speaking) $0+a_1+\sum_{n=2}^\infty a_n nx^n$

142. TuringTest Group Title

yep, exactly

143. TuringTest Group Title

oh no, you dropped an x

144. TuringTest Group Title

at n=1 the exponent on x is 1

145. TuringTest Group Title

$\sum_{n=0}^\infty a_nnx^n=a_0(0)x^0+\sum_{n=1}^\infty a_nnx^n=0+a_1(1)x^1+\sum_{n=2}^\infty a_nx^n$$=a_1x+\sum_{n=2}^\infty a_nx^n$

146. TuringTest Group Title

easier example, stripping out terms for the sum of the first n integers$\sum_{i=1}^n i=1+\sum_{i=2}^n=1+2+\sum_{i=3}^n i=1+2+3+\sum_{i=4}^n i$

147. MathSofiya Group Title

much better

148. TuringTest Group Title

thought so, probably should have done that a while ago, hehe :P

149. MathSofiya Group Title

Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though

150. MathSofiya Group Title

cool thanks !

151. TuringTest Group Title

Awesome, always a pleasure :D

152. MathSofiya Group Title

@TuringTest shouldn't $a_4$ have a minus sign? I know you said it shouldn't but $a_2=\frac{a_0}{2}$ and now we have $a_4=-a_2\frac{1}{12}=-\frac{a_0}{24}$

153. TuringTest Group Title

hm... I guess you may be right

154. MathSofiya Group Title

yuuss!!!!

155. MathSofiya Group Title

^^^^my way of saying: yes!

156. MathSofiya Group Title

so it's an alternating series? for 2k+1?

157. MathSofiya Group Title

I mean 2k

158. TuringTest Group Title

sorry I was double-checking this whole time yeah it alternates, so you know the only change we're going to make?

159. MathSofiya Group Title

let's see if I remember, can I cheat?

160. TuringTest Group Title

cheating would be having me tell you, any way you wanna figure out how to make a series change sign every other term is up to you :)

161. MathSofiya Group Title

cheating would be using a the google search engine, but let's see if I can figure it out on my own without google.com

162. TuringTest Group Title

good idea brb

163. MathSofiya Group Title

k

164. MathSofiya Group Title

argh! $\sum_{n=0}^{\infty}x^n=0,x,x^2,etc.$ $\sum_{n=0}^{\infty}(-1)x^n=0,-x,-x^2$

165. TuringTest Group Title

what can we do to -1 to make it even?

166. TuringTest Group Title

sorry, positive*

167. MathSofiya Group Title

$(-1)^n$

168. TuringTest Group Title

yes

169. MathSofiya Group Title

AWESOME!

170. TuringTest Group Title

very awesome!!!!!!!!!!!!

171. MathSofiya Group Title

hey I got 240 for a_6

172. TuringTest Group Title

you love multiplying the numbers out, I just call it 6!

173. MathSofiya Group Title

no that's 720

174. TuringTest Group Title

haha whatever I'm not even checking if you are having a problem with the constants again give me a min and I'll brb again to help

175. MathSofiya Group Title

sounds good... sorry don't mean to be nitpicking :S

176. TuringTest Group Title

oh no, I;m glad you pointed it out, but.... must... eat

177. MathSofiya Group Title

dude go for it LOL I'll see if I can figure this pattern out

178. MathSofiya Group Title

I think this is incorrect $a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}$ because $a_{n+2}=\frac{-a_nn-1}{(n+2)(n+1)}\neq-a_n\frac{n-1}{(n+2)(n+1)}$ It should be $-a_n\frac{(n+1)}{(n+2)(n+1)}$

179. MathSofiya Group Title

$-\frac{a_n}{n+2}$

180. MathSofiya Group Title

$a_{n+2}=-\frac{a_n}{n+2}$

181. TuringTest Group Title

$(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}$is what you wrote, let's see if that is correct...

182. TuringTest Group Title

$(n+2)(n+1)a_{n+2}+a_n(n+1)=0\implies a_{n+2}=-a_n\frac{n+1}{(n+2)(n+1)}$aw crap you're right again

183. MathSofiya Group Title

yuuus!!!

184. TuringTest Group Title

well, being right that we were wrong is always a little bittersweet, haha gotta do the pattern over, but at least it's only$a_{n+2}=-\frac{a_n}{n+2}$which should be easier I would think

185. MathSofiya Group Title

Yeah, let's see if I can figure the pattern out... (btw I did yoga today...feeling more alert, or it could be plain coincidence =)

186. TuringTest Group Title

well you're catching all my mistakes, so it seems to be working :)

187. TuringTest Group Title

the pattern seems harder to write concisely for me now... though it seems obvious

188. MathSofiya Group Title

$a_2=-\frac{a_0}{2}=-\frac{a_0}{2!}$ $a_3=-\frac{a_1}{3}$ $a_4=-\frac{a_2}{4}=\frac{a_0}{3!}$ $a_5=-\frac{a_3}{5}$ $a_6=-\frac{a_4}{6}=-\frac{a_0}{36}$ what's withe the 36?

189. TuringTest Group Title

it's like the odds get cut out of the factorials and I'm trying to figure out how to write that same for the evens

190. MathSofiya Group Title

yeah it's quite strange, but the formula is correct, right? could we have still made an error there?

191. TuringTest Group Title

a mistake in the recursion relation? we could have... to be safe we ought to take it from the top, but I can't do that, I have to leave...

192. MathSofiya Group Title

sure I'll start from the beginning and post it in a new post...this one is getting quite long

193. TuringTest Group Title

good idea I am thinking$a_{2k}=(-1)^k\frac{a_0}{2^kk!}$and do not yet have an expression for the odds

194. MathSofiya Group Title

oh ok, I'll see ya trmw then

195. TuringTest Group Title

hasta luego

196. MathSofiya Group Title

hasta luego mi amigo!