Bruker:Karl Erik/Problem1 - Sideversjonshistorikk

Karl Erik: Tømmer siden

2011-02-04T16:33:02Z

Tømmer siden

← Eldre sideversjon		Sideversjonen fra 4. feb. 2011 kl. 16:33
Linje 1:		Linje 1:
	~~1a)~~
	~~----~~
	For every x, there exists a d(x) such that <tex>y^{(n)}(x)-y^{(n-1)}(x)=d(x)</tex> by the assumptions of the problem. As y is smooth, so are its derivatives, so d(x) is a difference between two smooth functions and hence smooth. We will now write d=d(x) and y=y(x) for simplicity. Note that d'=<tex>y'-y)'=y''-y'=d</tex>, so d'=d, which can be easily solved for d to obtain <tex>d=Ce^x</tex>. Hence <tex>y'-y=Ce^x</tex>, which is a simple differential equation which yields <tex>y=(Cx+D)e^x</tex>. By insertion we see that these are solutions for all choices of C, D, so these are all the solutions.


	~~1b)~~
	~~----~~
	As in the previous part, we see that there exists a function r(x)=r (Again we are not saying that r(x) is constant, but simplifying notation.) such that <tex>y^{(n+1)}=ry^{(n)}</tex>. We see then that r is smooth except possibly where y=0. Note also that in this case <tex>y^{(m)}=0</tex> for all m.

	We work now in <tex>Y=\mathbb {R}</tex> <tex>-y^{-1}(\{0\})</tex>. As <tex>\{0\}</tex> is closed, this is an open set, so we have that r is smooth on this open set, and we can solve the differential equation here, as <tex>y\not = 0</tex>. Then <tex>r'=\left ( \frac {y'} {y} \right ) ' = \frac{y''y-y' \cdot y'} {y^2} = \frac {r^2y^2- (ry)^2} {y^2} = 0</tex>, so r must be constant, and hence <tex>y'=Ry</tex> for some constant R, which is easily solved to yield <tex>y=Ce^{Rx}</tex> for all <tex>x \in Y</tex>.

	Now, assuming y is not constantly equal to 0, Y is nonempty, so select some p in Y. Let <tex>A=(-\infty, p] \cap y^{-1} (\{0 \})</tex> and <tex>B=[p, \infty) \cap y^{-1} (\{0 \})</tex>. Either y is never equal to zero, or one of these are nonempty, so either sup A or inf B exists. Without loss of generality, we suppose sup A exists. Then this is a limit point of Y, so we can find a sequence <tex>c_n \in Y</tex> converging to <tex>\sup A</tex>. But it is also a limit point of A, so we can find a sequence <tex>a_n \in A</tex> converging to <tex>\sup A</tex>. However the former implies that <tex>y(\sup A)=\lim y(y_n) \not = 0</tex> (as <tex>y(y_n)=Ce^{Ry_n}</tex>, which (as C is assumed to be nonzero) only tends to zero if <tex>Ry_n</tex> tends to negative infinity, which is not the case as <tex>y_n</tex> has a real (finite) limit), and the latter implies that <tex>y(\sup A)=\lim y(a_n)=0</tex>, so we have a contradiction. Hence y is either always equal to zero, or never equal to zero, which implies that it is one of the solutions we have found.

	~~Hence <tex>y=Ce^{Rx}</tex> are all the solutions. By insertion all choices of C, R yield valid solutions.~~

	~~Huskepå~~
	~~----~~

	Nå kom det en revisjon nettopp der de sa at den geometriske rekka måtte være ikke-triviell, det vil si med alle ledd forskjellige fra null. I så fall blir løsningen mye enklere i og med at vi kan se bortifra divisjon på null og alt det der, så vi kan egentlig kutte ned på et avsnitt eller to. Vi må også huske på å nevne at C, R skal være forskjellige fra null.

Karl Erik på 4. feb. 2011 kl. 11:40

2011-02-04T11:40:25Z

← Eldre sideversjon		Sideversjonen fra 4. feb. 2011 kl. 11:40
Linje 13:		Linje 13:

	Hence <tex>y=Ce^{Rx}</tex> are all the solutions. By insertion all choices of C, R yield valid solutions.		Hence <tex>y=Ce^{Rx}</tex> are all the solutions. By insertion all choices of C, R yield valid solutions.

			Huskepå
			----

			Nå kom det en revisjon nettopp der de sa at den geometriske rekka måtte være ikke-triviell, det vil si med alle ledd forskjellige fra null. I så fall blir løsningen mye enklere i og med at vi kan se bortifra divisjon på null og alt det der, så vi kan egentlig kutte ned på et avsnitt eller to. Vi må også huske på å nevne at C, R skal være forskjellige fra null.

Karl Erik på 3. feb. 2011 kl. 23:52

2011-02-03T23:52:26Z

← Eldre sideversjon		Sideversjonen fra 3. feb. 2011 kl. 23:52
Linje 8:		Linje 8:
	As in the previous part, we see that there exists a function r(x)=r (Again we are not saying that r(x) is constant, but simplifying notation.) such that <tex>y^{(n+1)}=ry^{(n)}</tex>. We see then that r is smooth except possibly where y=0. Note also that in this case <tex>y^{(m)}=0</tex> for all m.		As in the previous part, we see that there exists a function r(x)=r (Again we are not saying that r(x) is constant, but simplifying notation.) such that <tex>y^{(n+1)}=ry^{(n)}</tex>. We see then that r is smooth except possibly where y=0. Note also that in this case <tex>y^{(m)}=0</tex> for all m.

	We work now in <tex>Y=\mathbb {R}</tex> <tex>-y^{-1}(\{0\})</tex>. As <tex>\{0\}</tex> is closed, this is an open set, so we have that r is smooth on this open set, and we can solve the differential equation here, as <tex>y\not = 0</tex>. Then <tex>r'=\left ( \frac {y'} {y} \right ) ' = \frac{y''y-y' \cdot y'} {y^2} = \frac {r^2y^2- (ry)^2} {y^2} = 0</tex>, so r must be constant, and hence <tex>y'=Ry</tex> for some constant R, which is easily solved to yield <tex>y=Ce^{Rx}</tex> for all <tex>x \in Y</tex>. ~~But~~ y is ~~continuous~~, and ~~hence there~~ is ~~no place where~~ it can ~~equal~~ 0 (as <tex>Ce^{Rx} ~~\not = 0~~</tex> ~~for any x~~, ~~and~~ only tends to zero ~~as Rx~~ tends to negative infinity), so <tex>y=Ce^{Rx}</tex> ~~for~~ all x. ~~It is easily verified that these are~~ all ~~solutions~~, ~~so they are all the~~ solutions.		We work now in <tex>Y=\mathbb {R}</tex> <tex>-y^{-1}(\{0\})</tex>. As <tex>\{0\}</tex> is closed, this is an open set, so we have that r is smooth on this open set, and we can solve the differential equation here, as <tex>y\not = 0</tex>. Then <tex>r'=\left ( \frac {y'} {y} \right ) ' = \frac{y''y-y' \cdot y'} {y^2} = \frac {r^2y^2- (ry)^2} {y^2} = 0</tex>, so r must be constant, and hence <tex>y'=Ry</tex> for some constant R, which is easily solved to yield <tex>y=Ce^{Rx}</tex> for all <tex>x \in Y</tex>.

			Now, assuming y is not constantly equal to 0, Y is nonempty, so select some p in Y. Let <tex>A=(-\infty, p] \cap y^{-1} (\{0 \})</tex> and <tex>B=[p, \infty) \cap y^{-1} (\{0 \})</tex>. Either y is never equal to zero, or one of these are nonempty, so either sup A or inf B exists. Without loss of generality, we suppose sup A exists. Then this is a limit point of Y, so we can find a sequence <tex>c_n \in Y</tex> converging to <tex>\sup A</tex>. But it is also a limit point of A, so we can find a sequence <tex>a_n \in A</tex> converging to <tex>\sup A</tex>. However the former implies that <tex>y(\sup A)=\lim y(y_n) \not = 0</tex> (as <tex>y(y_n)=Ce^{Ry_n}</tex>, which (as C is assumed to be nonzero) only tends to zero if <tex>Ry_n</tex> tends to negative infinity, which is not the case as <tex>y_n</tex> has a real (finite) limit), and the latter implies that <tex>y(\sup A)=\lim y(a_n)=0</tex>, so we have a contradiction. Hence y is either always equal to zero, or never equal to zero, which implies that it is one of the solutions we have found.

			Hence <tex>y=Ce^{Rx}</tex> are all the solutions. By insertion all choices of C, R yield valid solutions.

Karl Erik på 3. feb. 2011 kl. 22:52

2011-02-03T22:52:01Z

← Eldre sideversjon		Sideversjonen fra 3. feb. 2011 kl. 22:52
Linje 1:		Linje 1:
	~~Problem~~ 1		1a)
			----
			For every x, there exists a d(x) such that <tex>y^{(n)}(x)-y^{(n-1)}(x)=d(x)</tex> by the assumptions of the problem. As y is smooth, so are its derivatives, so d(x) is a difference between two smooth functions and hence smooth. We will now write d=d(x) and y=y(x) for simplicity. Note that d'=<tex>y'-y)'=y''-y'=d</tex>, so d'=d, which can be easily solved for d to obtain <tex>d=Ce^x</tex>. Hence <tex>y'-y=Ce^x</tex>, which is a simple differential equation which yields <tex>y=(Cx+D)e^x</tex>. By insertion we see that these are solutions for all choices of C, D, so these are all the solutions.


			1b)
	----		----
			As in the previous part, we see that there exists a function r(x)=r (Again we are not saying that r(x) is constant, but simplifying notation.) such that <tex>y^{(n+1)}=ry^{(n)}</tex>. We see then that r is smooth except possibly where y=0. Note also that in this case <tex>y^{(m)}=0</tex> for all m.

			We work now in <tex>Y=\mathbb {R}</tex> <tex>-y^{-1}(\{0\})</tex>. As <tex>\{0\}</tex> is closed, this is an open set, so we have that r is smooth on this open set, and we can solve the differential equation here, as <tex>y\not = 0</tex>. Then <tex>r'=\left ( \frac {y'} {y} \right ) ' = \frac{y''y-y' \cdot y'} {y^2} = \frac {r^2y^2- (ry)^2} {y^2} = 0</tex>, so r must be constant, and hence <tex>y'=Ry</tex> for some constant R, which is easily solved to yield <tex>y=Ce^{Rx}</tex> for all <tex>x \in Y</tex>. But y is continuous, and hence there is no place where it can equal 0 (as <tex>Ce^{Rx} \not = 0</tex> for any x, and only tends to zero as Rx tends to negative infinity), so <tex>y=Ce^{Rx}</tex> for all x. It is easily verified that these are all solutions, so they are all the solutions.

Karl Erik på 3. feb. 2011 kl. 22:15

2011-02-03T22:15:21Z

← Eldre sideversjon		Sideversjonen fra 3. feb. 2011 kl. 22:15
Linje 1:		Linje 1:
	Problem 1		Problem 1
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Karl Erik: Ny side: Problem 1 ---

2011-02-03T22:15:11Z

Ny side: Problem 1 ---

Ny side

Problem 1
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