What is the Solution to Part B of the Charge Distribution Integral Homework?

[binbagsss]

Homework Statement

part b of below
[/B]

Homework Equations

$(1+x)^{1/2}=1+\frac{1}{2}x-\frac{x^{2}}{8}+...$

The Attempt at a Solution

[/B]
$\int\limits^{\Lambda}_{-\Lambda} \frac{dy}{\sqrt{r^2+y^2}}=log(\lambda+\sqrt{\lambda^2+r^2}) - log(-\lambda+\sqrt{\lambda^2+r^2})$
$= log(\lambda(1+\sqrt{1+\frac{r^2}{\lambda^2}}))-log(\lambda(-1+\sqrt{1+\frac{r^2}{\lambda^2}}) )$
$\approx log(\lambda(1+1+\frac{r^2}{2\lambda^2}-\frac{r^4}{8\lambda^4}) )- log(\lambda(\frac{r^2}{2\lambda^2}-\frac{r^4}{8\lambda^4}))$

which is off track... thanks

 

[Orodruin]

Hint: $\log(x) + \log(y) = \log(xy)$

 

[binbagsss]

Hint: $\log(x) + \log(y) = \log(xy)$

huh? i used that already. oh my expansion need to use a taylor to get the right order

Hint: $\log(x) + \log(y) = \log(xy)$

ok ,
so do i need to consider the bit I am applying the logarithm to, expand this out and then apply the logarithm.
i.e. so i need to consider $\frac{\Lambda(1+\Lambda(1+\frac{r^2}{\Lambda^2})}{\Lambda(-1+\Lambda(1+\frac{r^2}{\Lambda^2})}=(1+\Lambda(1+\frac{r^2}{\Lambda^2})(-1+\Lambda(1+\frac{r^2}{\Lambda^2}))^{-1}$, and get this to first order in $x$but now do I expand out the square roots first in both terms, using $(1+x^2)^{1/2} \approx 1 + x^2/2 +...$ and then use $(1+x)^{-1} \approx...$ or do I do it the other way around?
how do I know which is correct?

 

[Orodruin]

huh? i used that already

Not shown in the OP, so how can I know if you used it or not?

I suggest that you use it in the other direction. Expanding the square roots is fine.

 

[binbagsss]

Not shown in the OP, so how can I know if you used it or not?

I suggest that you use it in the other direction. Expanding the square roots is fine.

apologies yes you did- how do you know however, that you are doing something wrong if you use the way I intially tried? since it's an identity I don't understand why you would have to select and how the other could be wrong? thanks

 

[Orodruin]

I don't understand, what do you think is wrong?

 

[binbagsss]

I don't understand, what do you think is wrong?

you said hint
as in me trying to use $log(x)+log(y)$ and expand the expressions $x$ and $y$ is not correct but rather I need to consider expanding $x/y$ , how can that be when it is just an identity?
was this not the idea behind your post 2

 

[Orodruin]

I did not say it was not correct. I said there is an easy way to do it and a hard way to do it. The easy way to do it is to expand everything on the form log(xy) to log(x)+log(y).

 

[NFuller]

I would suggest that before using the expansion you should use the log identity mentioned by Orodruin. This will cancel the first $\lambda$ terms, although I would find it more straight foreword to use the quotient rule to do this. Once you have everything as a single fraction, expand the numerator and simplify the first term. What does that give you?

 

[Orodruin]

I would suggest that before using the expansion you should use the log identity mentioned by Orodruin. This will cancel the first $\lambda$ terms, although I would find it more straight foreword to use the quotient rule to do this. Once you have everything as a single fraction, expand the numerator and simplify the first term. What does that give you?

Again, as stated in #8, this is the long way around. I strongly advise against going to a single log expression.

 

[binbagsss]

Again, as stated in #8, this is the long way around. I strongly advise against going to a single log expression.

wait I'm confused again, i did the start the expansion with two log terms not a single one?

 

[Orodruin]

wait I'm confused again, i did the start the expansion with two log terms not a single one?

Yes, and going to a single one is going to make it more complicated. The more straightforward way to do it is to further expand the logs that you have.

 

[binbagsss]

Yes, and going to a single one is going to make it more complicated. The more straightforward way to do it is to further expand the logs that you have.

is your post 2/4 not telling me to do the opposite of what I started?

So you want me to include higher order terms to what I did in my OP:

$\approx log(\lambda(1+1+\frac{r^2}{2\lambda^2}-\frac{r^4}{8\lambda^4}) )- log(\lambda(\frac{r^2}{2\lambda^2}-\frac{r^4}{8\lambda^4}))$?

 

[Orodruin]

No, I want you to expand those logs.

 

[binbagsss]

No, I want you to expand those logs.

oh right, using (1+x) \approx ...

but the final expression has log? this would get rid of them?

 

[vela]

He's saying use: log(ab)-log(ac) = log(a)+log(b)-log(a)-log(c)

 

[binbagsss]

No, I want you to expand those logs.

He's saying use: log(ab)-log(ac) = log(a)+log(b)-log(a)-log(c)

ok so I now have
$log \Lambda + log (2+ \frac{r^2}{2\Lambda^2}) - log \Lambda - log(\frac{r^2}{2\Lambda^2})$
so the \lambda term cancels, as you can see obviously by considering the log as a single expression initially, but I am stil far from the answer- of the wrong order in $x=\frac{r}{\Lambda}$

 

[Orodruin]

He's saying use: log(ab)-log(ac) = log(a)+log(b)-log(a)-log(c)

Not only that. There are also the manipulations that will leave only logs on the sought form, logs on the form log(1+x) that can be series expanded, and constant logs.

 

[Orodruin]

ok so I now have
$log \Lambda + log (2+ \frac{r^2}{2\Lambda^2}) - log \Lambda - log(\frac{r^2}{2\Lambda^2})$
so the \lambda term cancels, as you can see obviously by considering the log as a single expression initially, but I am stil far from the answer- of the wrong order in $x=\frac{r}{\Lambda}$

Use log(xy) = log(x)+log(y) more times!

 

[binbagsss]

Use log(xy) = log(x)+log(y) more times!

hmmm something like $log (\frac{r}{\sqrt{2}\Lambda}(\frac{2\sqrt{2}r}{\Lambda}+\frac{r}{\sqrt{2}\Lambda}))-log(\frac{r}{\sqrt{2}\Lambda}\frac{r}{\sqrt{2}\Lambda})$

$log(\frac{r}{\sqrt{2}\Lambda})+log(\frac{2\sqrt{2}r}{\Lambda}+\frac{r}{\sqrt{2}\Lambda}) -log(\frac{r}{\sqrt{2}\Lambda})-log(\frac{r}{\sqrt{2}\Lambda}) = log(\frac{2\sqrt{2}r}{\Lambda}+\frac{r}{\sqrt{2}\Lambda})-log(\frac{r}{\sqrt{2}\Lambda})$ ?

what should I have done here instead?

 

[Orodruin]

No, you are just overcomplicating things now. Just keep expanding from what you had.

 

[Eclair_de_XII]

Forgive me for being off-topic, but is it not easier to just use trigonometric substitution: $y=r⋅tan(\theta)$ and prove that the integral diverges on the interval $(-\frac{\pi}{2},\frac{\pi}{2})$?

 

[vela]

The OP is working on part (b).

 

[binbagsss]

No, you are just overcomplicating things now. Just keep expanding from what you had.

there's nothing to talk to unless you factorise something?

 

[Orodruin]

And you already had perfectly fine factors. There was no need to overcomplicate it. Why you would start introducing factors of $\sqrt 2$ I really cannot understand. Just use the rule and in the logs where you have two terms factor out the leading one (as it is, not with any weird factors of $\sqrt 2$ - how else are you going to get a factor 1+x?).

 

[binbagsss]

And you already had perfectly fine factors. There was no need to overcomplicate it. Why you would start introducing factors of $\sqrt 2$ I really cannot understand. Just use the rule and in the logs where you have two terms factor out the leading one (as it is, not with any weird factors of $\sqrt 2$ - how else are you going to get a factor 1+x?).

ok
, but why would you want to seek the form log (1+x) when we don't want to expand things since we want log (x) ... or that's ok for the O(x^2) terms I guesss, where we have $x = \frac{r}{\lambda}$?

So how about:$log 2 + log (1+x^2) - log ( \frac{x^2}{2}) = 2 log 2 + log(1+x') - log(x')$

but I'm still stuck with $log(x')$ , where $x'=\frac{r^2}{\lambda^2}$, the log 2 is a constant as required, the $log(1+x')$, once expanded, will take the form of constant terms plus $O(\frac{r^2}{\lambda^2})$so that's fine, its just the log(x') term..?

 

[Orodruin]

but I'm still stuck with $log(x')$ , where $x'=\frac{r^2}{\lambda^2}$

Again, use $\log(ab) = \log(a) + \log(b)$.