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gAr wrote:

Hi,

Next integral:

hi gar

hi ganesh

Perhaps this might be helpful:

Zhylliolom wrote:

Hi,

I wasn't able to solve this using substitution, but here's a solution I came up with:

Ahhh, I see it now! Thanks, Bob!

zetafunc wrote:

#4

Hi,

gAr wrote:

Hi gAr,

ganesh wrote:

Hi,

M # 444. Two right circular cylinders of equal volume have their heights in the ratio 1:2. Find the ratio of their radii.

Hi ganesh,

Hi,

.

#4242. If the numerator of a fraction is increased by 300% and the denominator is increased by 500%, the resultant fraction is

. What was the original fraction?

zetafunc wrote:

ganesh wrote:Whats so special about the TN

1.444667861..................??????Is it known if this number (i.e. e^(1/e)) is transcendental?

It has not been proven to be transcendental, I don't think.

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Good to know! x)

Yeah, I've been a been lacking in the uni department lately, because I got a temporary job, so I had less time to study, but it's pretty good otherwise!

Hi Bob!

How've you been?

Except not really.

Here's a link relevant to the original post: Intersection of polar curves

So, another equation that would give you a solution would be

bobbym wrote:

The double integral you posted in post #9,

that one does not converge.

There shouldn't minuses in those exponents.

Cześć!

Well, one sure way to get a definitive result is to go through all the possibilities, which are not many, since you just have to choose which operation gets performed before which. 4 operations means there are 4!=24 different orders to do the operations in. So, for example, you could label each operation with a number 1 through 4, then list out all the permutations of the sequence 1,2,3,4 and perform the operations in the order signified by each permutation. E.g., 3241 would mean you'd group them like 5-((2*(6-4))+2).

Hi Bob,

You can superscript the "\prime" to have it look like normal, like so:

Hmm, out Intro to NA professor mentioned that trick with the example of the integral

It's quite brilliant.

Also, yes, if you ran it to a[0], you would get an approximation of ln(9/8). This is also, I think, the only value of a[0] for which the sequence a[n] converges.

bobbym wrote:

Here I suggest you use your own head.

f[x_] := x^6 + x^5 - 5 x^4 - 4 x^3 + 6 x^2 + 3 x - 1

x =0.

x = x - f[x]/f'[x]

Well, of course, Newton's method has a number of conditions to be met so that it would converge.

Also, isolating the zeroes might not be a bad idea.

Agnishom wrote:

Something like this?

I do not understand. The NewtonRaphson function does not take f as an argument in his code.

It uses f as a global variable, assigned a value outside of the subroutine, so, to call it for a function, you'd have to set f to that function before calling NewtonRaphson. I assume it works without much problem, but that is a silly thing to do...