Integration

[Yllanes]

with my sub - i integrated the erf(x) with limits 0 and (x/(2sqrt(kt)))
to get u
can't you do this?

Of course you can, but you is a greater probability of error . The actual operations are just the same, but you are writing the formula in all its gory detail from the beginning, which is error prone. Notice that you have to apply the fundamental theorem of calculus, which tells you that if [latex]$F(x)=\int_0^x f(t) \ \mathrm{d}t$[/latex], then F'(c)=f(c) (provided certain smoothness conditions are satisfied). Doing it your way you do not have [latex]$\int_0^x$[/latex], but [latex]$\int_0^{g(x)}$[/latex]. Having a function of x in the limits of integration is not a problem (you just apply the chain rule), but can be confusing if you are not very confident. The best way is to use the chain rule from the beginning and keep the expressions you have to differentiate as simple as possible.

 

[andyandy]

Of course you can, but you is a greater probability of error . The actual operations are just the same, but you are writing the formula in all its gory detail from the beginning, which is error prone. Notice that you have to apply the fundamental theorem of calculus, which tells you that if [latex]$F(x)=\int_0^x f(t) \ \mathrm{d}t$[/latex], then F'(c)=f(c) (provided certain smoothness conditions are satisfied). Doing it your way you do not have [latex]$\int_0^x$[/latex], but [latex]$\int_0^{g(x)}$[/latex]. Having a function of x in the limits of integration is not a problem (you just apply the chain rule), but can be confusing if you are not very confident. The best way is to use the chain rule from the beginning and keep the expressions you have to differentiate as simple as possible.

gotcha. Simple is definitly best

 

[Unnamed]

with my sub - i integrated the erf(x) with limits 0 and (x/(2sqrt(kt)))
to get u
can't you do this?

You can't really integrate [latex]\int e^{-u^2}\,du[/latex] (you were probably thinking of e^u). That's why the error function was created, to give a name to that integral. If you try the substitution v = -u², then you don't have the required dv = -2u du anywhere.

 

[Yllanes]

with my sub - i integrated the erf(x) with limits 0 and (x/(2sqrt(kt)))
to get u
can't you do this?

You can't really integrate [latex]\int e^{-u^2}\,du[/latex] (you were probably thinking of e^u). That's why the error function was created, to give a name to that integral. If you try the substitution v = -u², then you don't have the required dv = -2u du anywhere.

Hmm... I misread andyandy's post... I thought he meant he had substituted x/2sqrt(kt) in the limits of integration and then differentiated, applying the FTC. That's what I said was possible. Obtaining a closed form for the integral of exp(-x^2) isn't. The integral [latex]$\int_0^\infty \mathrm{e}^{-x^2}\ \mathrm{d}x$[/latex] can be solved and its value is sqrt(pi)/2, but with an arbitrary upper limit you can't.

 

[andyandy]

You can't really integrate [latex]\int e^{-u^2}\,du[/latex] (you were probably thinking of e^u). That's why the error function was created, to give a name to that integral. If you try the substitution v = -u², then you don't have the required dv = -2u du anywhere.

oh dear....no you can't can you?
*sigh*

must try harder

 

[andyandy]

hmmmmm

limits.....

here is the question and the answer (taken from wikibooks http://en.wikibooks.org/wiki/Calculus/Limits)

4) Show that the limit of sin(1/x) as x approaches 0 does not exist.

Suppose the limit exists and is L. We will proceed by contradiction. Assume that L =! 1 , the case for L = 1 is similar. Choose ε = L - 1, then for every δ > 0, there exists a large enough n such that

[latex] $
0<x_0= \frac{1}{\pi/2-2\pi n} < \delta $[/latex]
, but | sin(1 / x0) − l | = | 1 − l | = ε a contradiction.

it's specifically this step
[latex] $
0<x_0= \frac{1}{\pi/2-2\pi n} < \delta $[/latex]
that i don't understand....why do you choose that particular fraction?

and why do you choose ε = L - 1?

thanks

 

[Yllanes]

[latex] $
0<x_0= \frac{1}{\pi/2-2\pi n} < \delta $[/latex]
that i don't understand....why do you choose that particular fraction?

and why do you choose ε = L - 1?

thanks

This proof shows that, for each delta, there are infinite x0 < delta such that sin(1/x0)=1. If the limit were, say, 0.75, then there are points arbitrarily small such that |f(x)-L| = 0.25. So you cannot guarantee that you will get closer than 0.25 from the limit, no matter how close to zero you go.

If L=0.75 is the limit and you give me an epsilon, say, 0.1, then I should be able to get a delta such that all x smaller than delta give |f(x)-0.75| < 0.1. But I can't, because there always are x0 with |f(x0)-0.75|=0.25>1. Same for any value of L =/= 1.

 

[andyandy]

[latex] $
0<x_0= \frac{1}{\pi/2-2\pi n} < \delta $[/latex]


so by setting Xo as the above, you make sin(1/x0) as 1 (for n=integer) and however small Xo gets, n=integer will always = 1

ok...that helps..

 

[andyandy]

a couple more....

specifically Riccati and homogenous equations....

for both of them i'm working through how the formulas are derrived....

for the riccati equation

when

[latex]

$ \frac {dy}{dx} = f(x,y) [/latex]


you can approximate f(x,y) while keeping x constant to get,
[latex]
f(x,y) = P(x) + Q(x)y + R(x)y^2 +....

[/latex]

so if we consider a second degree approximation, with y* a particular solution of

[latex]$ \frac{dy}{dx} = P(x) + Q(x)y + R(x)y^2

$[/latex]

with a new function defined by

[latex] $ z= \frac{1}{y-y*}

$ [/latex]

then we should arrive at

[latex] $ \frac{dz}{dx} = -(Q(x) + 2y*R(x))z - R(x)

$ [/latex]

which you can then use to solve equations of that type. I can't work out how to arrive at the end equation for dz/dx though....

i differentiated z to get dz/dy which using the chain rule (dz/dy dy/dx = dz/dx) I thought should have got me the given equation...but it doesn't!


The second question i have concerns homogenous equations

where if f(x,y) is homogenous and we use the sub z=y/x
then the explanation given is that......
[latex]$

f(x,y) = f(1,z) = F(z) $

and since

$ \frac {dy}{dx} = x\frac{dz}{dx} + z $

then

$ {x}\frac{dz}{dx} + z = f(1,z) $

[/latex]

but why does [latex]$ \frac {dy}{dx} = {x}\frac{dz}{dx} + z $ [/latex] ?

thanks

 

[andyandy]

ooh! I think i got the second one....

if the sub is y/x = z

then y = xz

so

xz - y = 0

and then dz/dx using implicit differentiation gives;

x dz/dx + 1 z - 0 = 0

which is the required form.....

is that right?

 

[Yllanes]

xz - y = 0

and then dz/dx using implicit differentiation gives;

x dz/dx + 1 z - 0 = 0

which is the required form.....

is that right?

Almost.

xz-y=0

So

[latex]
$$
0=\frac{\mathrm{d}}{\mathrm{d}x}(xz-y)=x\frac{\mathrm{d}z}{\mathrm{d}x}+1\cdot z - \frac{\mathrm{d}y}{\mathrm{d}x} = 0
$$
[/latex]

You can't say dy/dx=0! This gives the relation you are looking for.


As for the Riccati equation, best to do the deduction in two steps. We have an equation of the form [latex]\footnotesize$y' = p(x) + q(x) y + r(r) y^2$[/latex]. Now this is not solvable in general, but we can do it if we already know a particular solution. I will call this particular solution y^*. We define the new variable u = y - y*. If we do this we will lose the term p(x) and get a Bernoulli equation, which is solvable. You can prove that u' = (q(x) + 2r(x) y^*(x))u + r(x) u². In an intermediary step you will recover the original equation for y* and you must introduce the condition that y* is a solution to get rid of that term. Now, the equation we get for u is a Bernoulli equation. A Bernoulli equation is of the form

y' = a(x) y + b(x) y^p (p real >1)

You can prove as an additional (easy) excercise that this equation is converted into a linear one with the change z = y^1-p (hence z = u^-1 = 1/(y-y*) in our case).

As an example, try to solve

(1+x³) y ' + 2xy²+ x²y + 1 = 0

To do this, search for a particular solution of the form y* = Ax. Once found, follow the method. The general solution should be y = (1-C x) / (C +x^2).

 

[andyandy]

....ok,
by the substitution

z=1/(y-y*)

we get

y = y* + 1/z

and

y' = -z'/z^2

which we can sub into the original equation to give;

[latex]

$\frac{-z'}{z^2}= p(x) + q(x)({y* + \frac {1}{z}}) + r(x)({y* + \frac {1}{z}})^2 $

[/latex]

now, i can get the required result by multiplying out the brackets, then equating the numerical terms p(x) + q(x)y* + r(x)y*^2 to = 0 thus leaving the equation as multipliable by z^2 to get the required final form.....

but, can i equate the numerical terms to zero and eliminate them like that? I didn't think i should be able to, but it seems to work.....


....wow.....i need to get to sleep! Just realised it's 02.15am.....i get kinda obsessed with maths sometimes

 

[Yllanes]

now, i can get the required result by multiplying out the brackets, then equating the numerical terms p(x) + q(x)y* + r(x)y*^2 to = 0 thus leaving the equation as multipliable by z^2 to get the required final form.....

but, can i equate the numerical terms to zero and eliminate them like that? I didn't think i should be able to, but it seems to work.....

Not quite, it's the same mistake as the one you had earlier with the homogenous equation (I guess due to the late hour of your post...)

we get

y = y* + 1/z

and

y' = -z'/z^2

y*' =/= 0. You are almost there, however, you just have to do y' = y*' - z'/z^2. You must then eliminate

y*'(x) - (p(x) + q(x)y*(x) + r(x)y*(x)^2)

which is zero for all x (just as cos^2 x + sin^2 x - 1 would be zero for all x).

 

[andyandy]

Not quite, it's the same mistake as the one you had earlier with the homogenous equation (I guess due to the late hour of your post...)



y*' =/= 0. You are almost there, however, you just have to do y' = y*' - z'/z^2. You must then eliminate

y*'(x) - (p(x) + q(x)y*(x) + r(x)y*(x)^2)

which is zero for all x (just as cos^2 x + sin^2 x - 1 would be zero for all x).

ok gotcha - cheers

 

[andyandy]

so where
{[latex]x_n[/latex]} is a sequence of positive numbers, and assuming
[latex]$\lim_{n\to \infty}\frac{x_{n+1}}{x_{n}}=L [/latex]

we are asked to show that

[latex]$\lim_{n\to \infty}(x_n)^{1/n} =L [/latex]

This is a question from the online maths resource, SOS maths http://www.sosmath.com/calculus/sequence/hardlim/hardlim.html


I've played around but can't figure out the best approach - whether you can use the geometric mean to help, if you need to use error definition, if you need to use the sandwich theorum or if you can just manipulate it....any prods in the right direction?

 

[andyandy]

okaykokey.....

integration by differentiation of a parameter.....

[latex]$ \frac{d}{du}\int_{a(u)}^{b(u)}dx f(x,u) = f(b,u)\frac{db}{du} - f(a,u)\frac{da}{du}+\int_{a(u)}^{b(u)}dx\frac {\partial f}{\partial u} [/latex]

now given

[latex]$ y(t)=\frac{1}{w}\int_{a}^{t}dx f(x)sinw(t-x)

y(0) = 0
y'(0)=0

[/latex]


the question asks for
[latex]$ \frac {dy}{dt}

$\frac{d^2 y}{d y^2} [/latex]

now......i'm a bit stuck....

help appreciated - i'll have a think and post what i've got so far......

 

[Yllanes]

[latex]\[y(t)=\frac{1}{w}\int_{a}^{t}\mathrm{d}x\ f(x)\sin(t-w)\]
[/latex]

sin(t-w) does not depend on the dummy integration variable x, is this correct or a typo (w instead of x)?

Anyway, the formula that started your post is just a direct generalisation of the Fundamental Theorem of Calculus, do they give it any special name?

 

[andyandy]

sin(t-w) does not depend on the dummy integration variable x, is this correct or a typo (w instead of x)?

Anyway, the formula that started your post is just a direct generalisation of the Fundamental Theorem of Calculus, do they give it any special name?

sorry typo - it should be sinw(t-x)

i'll ammend it in the original post.

in the text, which gives an overview of integration methods, it first briefly outlines integration by parts, then change of variables, then differentiation of a parameter.....
and for the parameter method it just gives the equation i posted with which to work with.

the answer for dy/dt is given as

[latex]$ \frac{d}{dy}=\int_{a}^{t}dx f(x)cosw(t-x)

[/latex]

which is clearly just the last part of the equation - so the first two terms on the right must equal zero....but i'm a bit confused by how the parameters are set - ie what a and b are....

 

[Yllanes]

sorry typo - it should be sinw(t-x)

Ah, now it makes more sense.

in the text, which gives an overview of integration methods, it first briefly outlines integration by parts, then change of variables, then differentiation of a parameter.....
and for the parameter method it just gives the equation i posted with which to work with.

OK. As an exercise, it may be a good idea to derive that formula from the FThC, it's just a straightforward application of the chain rule.

which is clearly just the last part of the equation - so the first two terms on the right must equal zero....but i'm a bit confused by how the parameters are set - ie what a and b are....

In your example, b(t) = t. And a(t) = a = constant, so its derivative is identically zero and there goes one of the terms. Also, the f(x,u) from the formula is your f(x) sin w(t-x). The second term tells you to evaluate it at x = t, so you get f(t) sin w(t-t) = 0.

 

[andyandy]

Ah, now it makes more sense.

OK. As an exercise, it may be a good idea to derive that formula from the FThC, it's just a straightforward application of the chain rule.

In your example, b(t) = t. And a(t) = a = constant, so its derivative is identically zero and there goes one of the terms. Also, the f(x,u) from the formula is your f(x) sin w(t-x). The second term tells you to evaluate it at x = t, so you get f(t) sin w(t-t) = 0.

thanks yllanes

yes that makes sense....i'll see if i can get the right answer for dy^2/dx^2