# limit of series 1 A

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• Jan 18th 2011, 10:29 AM
HallsofIvy
Quote:

Originally Posted by transgalactic
i have dy/dx=x+y

y0=1
y1=1+x+(x^2)/2
y2=1+x+(x^2)/2+(x^3)/6
what is the limit of the series yn
?

The limit of that sequence is $\displaystyle e^x$ as you should be able to recognize the MacLaurin series for $\displaystyle e^x$
However, the derivative of $\displaystyle y= e^x$ is $\displaystyle e^x= y$, not x+ y.

If you are trying to use Picard's method to solve this equation, as Ackbeet suggested, then you are doing it wrong.

From dy/dx= x+y we get dy= (x+y)dx so that $\displaystyle y= \int (x+ y)dx$. Of course, since we don't know y but we can approximate it. If we are given y(0)= 1 (which you don't say but is implied by "y0= 1") as a first approximation, assume y(x)= 1 for all x. Then dy/dx= x+1 so that dy= (x+1)dx and [tex]y= (1/2)x^2+ x+ C[tex] and y(0)= C= 1 gives $\displaystyle y1= (1/2)x^2+ x+ 1$, as you have.

But then, $\displaystyle y2= \int x+ y1 dx= \int x+ (1/2)x^2+ x+ 1 dx= \int (1/2)x^2+ 2x+ 1 dx= (1/6)x^3+ x^2+ x+ C$ and, again, $\displaystyle y(0)= C= 1$ gives $\displaystyle y2= (1/6)x^2+ x^2+ x+ 1$ which is NOT what you have. I expect you forgot the added x.

Now, $\displaystyle y3= \int x+ y2 dx= \int x+ (1/6)x^3+ x^2+ x+ 1 dx$$\displaystyle = \int (1/6)x^3+ x^2+ 2x+ 1 dx= (1/24)x^4+ (1/3)x^3+ x^2+ x+ C and \displaystyle y(0)= C= 1 gives \displaystyle y3= (1/24)x^4+ (1/3)x^3+ x^2+ x+ 1. One more time: \displaystyle y4= \int x+ y3 dy= \int x+(1/24)x^3+ (1/3)x^3+ x^2+ x+ 1 dx$$\displaystyle = \int (1/24)x^4+ (1/3)x^3+ x^2+ 2x+ 1 dx= (1/5!)x^5+ (1/12)x^4+ (1/3)x^3+ x^2+ x+ C$ and $\displaystyle y(0)= C= 1$ gives
$\displaystyle y4= (1/5!)x^5+ (2/4!)x^4+ (2/3!)x^3+ (2/2!)x^2+ x+ 1$.

I think I would conjecture now that
$\displaystyle yn= (1/n!)x^n+ (2/(n-1)!)x^{n-1}+ \cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ x+ 1$
and I would attempt to prove that by "induction"

When n= 0, that says that y0= 1 which is true.

Now assume that, for some k, $\displaystyle yk= (1/k!)x^k+ (2/(k-1)!)x^{k-1}+\cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ x+ 1$ so that
$\displaystyle y(k+1)= \int x+ yk dx= \int (1/k!)x^k+ (2/(k-1)!)x^{k-1}+ \cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ 2x+ 1 dx$$\displaystyle = (1/(k+1)!)x^{k+1}+ (2/k!)x^k+ \cdot\cdot\cdot+ (2/4!)x^4+ (2/3!)x^3+ x^2+ x+ C$ and $\displaystyle y(k+1)(0)= C= 1$ gives
$\displaystyle y(k+1)= (1/(k+1)!)x^{k+1}+ (2/k!)x^k+ \cdot\cdot\cdot+ (2/4!)x^4+ (2/3!)x^3+ (2/2!) x^2+ x+ 1$
just as we wish.
• Jan 21st 2011, 03:25 AM
transgalactic
Quote:

Originally Posted by Ackbeet
Ok. So y(0) = 1 implies that

$\displaystyle y_{0}=1,$

$\displaystyle y_{1}=\displaystyle\int(x+1)\,dx=1+x+\frac{x^{2}}{ 2},$

$\displaystyle y_{2}=\displaystyle\int\left(x+1+x+\frac{x^{2}}{2} \right)\,dx=1+x+x^{2}+\frac{x^{3}}{6},$

$\displaystyle y_{3}=\displaystyle\int\left(x+1+x+x^{2}+\frac{x^{ 3}}{6}\right)dx=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6 }+\frac{x^{4}}{24},\dots$

That's what you got, I believe. The sequence $\displaystyle \{y_{n}\}$ converges to $\displaystyle e^{x}.$ To convince yourself of that fact, just expand $\displaystyle e^{x}$ in a Taylor series about 0, and you'll see that it's the exact same series as you're getting here.

How rigorous a proof does this require?

i dont have y2 of a exponent
y2=1+$\displaystyle \int t+1+t+\frac{t^3}{6}dt$

i get $\displaystyle t^2$
i should get $\displaystyle \frac{t^2}{2}$
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