Integration

[Ben Tilly]

thanks yllanes

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

Um, you mean d^2y/dx^2, don't you?

FYI there is a reason for the notation, though it is mostly historical curiousity now. The idea was that "d" was an operator, which took the value an infinitesmal away and subtracted it from the original. (And at the end you'd drop any infinitesmal remainders as meaningless.) So dy/dx meant d/d(x). And the derivative of the derivative was d(d/d(x))/d(x). But since d was a linear operator, this was the same as d(d)/d(x)/d(x). Or, in shorthand, d^2y/dx^2.

As I said, this is mostly historical curiousity. But if you were trying to compute a numerical approximation to the derivative, you'd find that the formulas you'd come up with have a very close relation to what the formula says.

Cheers,
Ben

 

[andyandy]

Um, you mean d^2y/dx^2, don't you?

FYI there is a reason for the notation, though it is mostly historical curiousity now. The idea was that "d" was an operator, which took the value an infinitesmal away and subtracted it from the original. (And at the end you'd drop any infinitesmal remainders as meaningless.) So dy/dx meant d/d(x). And the derivative of the derivative was d(d/d(x))/d(x). But since d was a linear operator, this was the same as d(d)/d(x)/d(x). Or, in shorthand, d^2y/dx^2.

As I said, this is mostly historical curiousity. But if you were trying to compute a numerical approximation to the derivative, you'd find that the formulas you'd come up with have a very close relation to what the formula says.

Cheers,
Ben

yes - just another silly mistake.....i'm quite good at them

 

[andyandy]

boldly going forward to discover yet more things i don't understand....

order notation...which i thought was pretty straightforward,

until trying some questions.....

all of these are finding limits using first principles....

[latex]$ (a) lim_x_0 \frac{({1+x})^{0.5} - 1}{1- ({1-x})^{0.5}}

(b) lim_x_1 \frac{x^a - 1}{x-1}

(c) lim_x_{\pi/2-} (\pi-2x)tanx

[/latex]

(quick explanation - lim x1 means x tends to 1 i don't know the correct latex)

now, the answers.....

(a) Using the binomial expansion gives you;

[latex]$ \frac{1+x/2+O(x^2)-1}{1-(1-x/2+O(x^2))} [/latex]

and this simplifies to;

[latex]$ \frac{x/2+O(x^2)}{x/2-O(x^2)} [/latex]

but then the answer is given as

[latex]$ \frac{1+O(x)}{1+O(x)} = 1+O(x)[/latex]

so, dividing top and bottom by x/2 gives you the 1s, and reduces the order(x^2) by one....but where did the minus turn into a plus?

ok...onto (b)

the answer starts by putting x=1+h and then gives

[latex]$\frac{e^{aln(1+h)}}{h} = \frac{e^{ah+O(h^2)}-1}{h} = a+O(h) [/latex]

now, this loses me - why convert it to exponential - and how does that manipulation end up with a+O(h) ?

and lastly...

(c) the answer suggests putting x=pi/2 - h (h>0) so

[latex]$ \pi-2x(tanx) = \frac{2h}{tan(h)} [/latex]

which is fine....forget this one

 

[Yllanes]

(quick explanation - lim x1 means x tends to 1 i don't know the correct latex)

Just \to (or \rightarrow):

[latex]
\[
\lim_{x\to\infty}
\]
[/latex]

so, dividing top and bottom by x/2 gives you the 1s, and reduces the order(x^2) by one....but where did the minus turn into a plus?

Irrelevant, a O(x^b) = O(x^b). Remember the O makes sense if we are thinking about limits, no matter how big a is it will be swallowed. The sign is also irrelevant.

now, this loses me - why convert it to exponential - and how does that manipulation end up with a+O(h) ?

The expansion of the exponential and logarithms are simpler. It ends up that way because

[latex]
\[
\frac{\mathrm{e}^{ah+\mathcal O(h^2)}-1}{h} = \frac{1+ah+\mathcal O(h^2)-1}{h}
\]
[/latex]

And O(h^2)/h = O(h). In general x^a O(x^b) = O(x^a+b).

 

[andyandy]

Just \to (or \rightarrow):

[latex]
\[
\lim_{x\to\infty}
\]
[/latex]



Irrelevant, a O(x^b) = O(x^b). Remember the O makes sense if we are thinking about limits, no matter how big a is it will be swallowed. The sign is also irrelevant.


The expansion of the exponential and logarithms are simpler. It ends up that way because

[latex]
\[
\frac{\mathrm{e}^{ah+\mathcal O(h^2)}-1}{h} = \frac{1+ah+\mathcal O(h^2)-1}{h}
\]
[/latex]

And O(h^2)/h = O(h). In general x^a O(x^b) = O(x^a+b).

ok - so for (b) you're using power series expansions first for ln(1+h)= h + O(h^2)
and then the expansion of e^ah = 1+ah+0(h^2).....but what do you do with the e^O(h^2) - shouldn't you multiply this by the expansion for e^ah? Or can you just incorporate it into the O(h^2) of the expansion?

Edited to add....yes you can....because e^(very small number) will be very close to 1.....

yay! Thanks

 

[andyandy]

ok, back to the main thread title...this is an integration problem that should be easy....but doesn't seem to simplify....the question asks to first differentiate the integrand

[latex]$ I(a)=\int_0^{\infty} =dx \frac {arctan(ax)}{x(1+x^2)} [/latex]

so using the product rule for the integrand I get

[latex]$ \frac{1}{1+a^2x^2}.\frac{1}{x(1+x^2)} - \frac{arctanax(1+3x^2)}{(x+x^3)^2} [/latex]

but i fail to see how this simplifies to the given answer of

[latex]$ I'(a)= \int_0^{\infty} =dx \frac {1}{(1+a^2x^2)(1+x^2)} [/latex]

hmmmm.....what am i missing?

 

[boooeee]

ok, back to the main thread title...this is an integration problem that should be easy....but doesn't seem to simplify....the question asks to first differentiate the integrand

[latex]$ I(a)=\int_0^{\infty} =dx \frac {arctan(ax)}{x(1+x^2)} [/latex]

so using the product rule for the integrand I get

[latex]$ \frac{1}{1+a^2x^2}.\frac{1}{x(1+x^2)} - \frac{arctanax(1+3x^2)}{(x+x^3)^2} [/latex]

but i fail to see how this simplifies to the given answer of

[latex]$ I'(a)= \int_0^{\infty} =dx \frac {1}{(1+a^2x^2)(1+x^2)} [/latex]

hmmmm.....what am i missing?

Remember which variable you're differentiating by. You don't need the product rule for this one.

 

[andyandy]

Remember which variable you're differentiating by. You don't need the product rule for this one.

yes, just realised it's differentiating under an integral for I(a)....well that's the first step....now let's see if i can do it.....

ooh....

let's see.. oh you just partially differentiate with regard to a.... That appears to work....

[latex] acrtan(ax)\partial_a = \frac{x}{1+a^2x^2} [/latex]

and then that simplifies.....double yay!

 

[andyandy]

ok let's see....

[latex]$ I'(z)=\frac{2}{\sqrt{1-z^2}} arctan(\frac{1-z}{1+z})^{0.5}

where

z=cos\pi a [/latex]

now this should simplify to

[latex]$ \frac{dI}{dz} = \frac{\pi a}{sin\pi a} [/latex]

now,
[latex]$ \frac{2}{\sqrt{1-z^2}} [/latex] obviously becomes [latex]$ \frac{2}{sin\pi a} [/latex]

and the arctan bit looks like it could use an inverse simplification.....but does it really simplify to [latex]$ \frac{\pi a}{2} [/latex] ?

 

[boooeee]

You'll need a standard trig identity.

How does (sin(x/2))^2 and (cos(x/2))^2 relate to cos(x)?

 

[andyandy]

[latex]$ arctan(\frac{1-z}{1+z})^{0.5} [/latex]

right, well, this can simplify to [latex]$ arctan(\frac{sin\pi a}{1+ cos\pi a}) [/latex]



....and i assume i should be getting something of the form [latex]$ arctan(\frac{sin\pi a}{cos\pi a}) = \pi a} [/latex]


hmmmm.....what to do with that 1?

*ping*

it's the bloody half angle formulae! So that equals tan(pi a /2) and it all works....