Urea, also called carbamide, the diamide of carbonic acid. Its formula is H2NCONH2. Urea has important uses as a fertilizer and feed supplement, as well as a starting material for the manufacture of plastics and drugs. It is a colourless, crystalline substance that melts at 132.7° C (271° F) and decomposes before boiling.

Urea is the chief nitrogenous end product of the metabolic breakdown of proteins in all mammals and some fishes. The material occurs not only in the urine of all mammals but also in their blood, bile, milk, and perspiration. In the course of the breakdown of proteins, amino groups (NH2) are removed from the amino acids that partly comprise proteins. These amino groups are converted to ammonia (NH3), which is toxic to the body and thus must be converted to urea by the liver. The urea then passes to the kidneys and is eventually excreted in the urine.

Urea was first isolated from urine in 1773 by the French chemist Hilaire-Marin Rouelle. Its preparation by the German chemist Friedrich Wöhler from ammonium cyanate in 1828 was the first generally accepted laboratory synthesis of a naturally occurring organic compound from inorganic materials. Urea is now prepared commercially in vast amounts from liquid ammonia and liquid carbon dioxide. These two materials are combined under high pressures and elevated temperatures to form ammonium carbamate, which then decomposes at much lower pressures to yield urea and water.

Because its nitrogen content is high and is readily converted to ammonia in the soil, urea is one of the most concentrated nitrogenous fertilizers. An inexpensive compound, it is incorporated in mixed fertilizers as well as being applied alone to the soil or sprayed on foliage. With formaldehyde it gives methylene–urea fertilizers, which release nitrogen slowly, continuously, and uniformly, a full year’s supply being applied at one time. Although urea nitrogen is in nonprotein form, it can be utilized by ruminant animals (cattle, sheep), and a significant part of these animals’ protein requirements can be met in this way. The use of urea to make urea–formaldehyde resin is second in importance only to its use as a fertilizer. Large amounts of urea are also used for the synthesis of barbiturates.

Urea reacts with alcohols to form urethanes and with malonic esters to give barbituric acids. With certain straight-chain aliphatic hydrocarbons and their derivatives, urea forms crystalline inclusion compounds, which are useful for purifying the included substances.

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Βεν Γ. Κυθισ wrote:Do you mean rather than ?it should be (a^m)n^a.

DANG IT! I forgot to move the variables around! And I forgot to confirm it with a calculator or *my miiiiind*.

Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to thefixed pointof the cosine function,

That makes so much sense! I also just tried

and it converged to its fixed point, 0. Tan has a fixed point but it is not attractive; sinh becomes a vertical line on 0 which means there is no attractive point, 0; cosh has no fixed point but it converges to ∞; 0 is tanh's attractive fixed point; cot has no attractive point due to it being tan with the x axis shifted; sec is related to tan because it shares half of its discontinuous points with tan suggesting it has no attractive point, and it does have no attractive point; csc is sec with the x axis shifted so it has no fixed point.]]>Welcome to the forum.

Yes, you should be doing your chem engg project. How long did this take?

forum rules wrote:

No Swearing or Offensive Topics. Young people use these forums, and should not be exposed to crudeness.

My first reaction was to delete it completely but then we lose all your posts. I think you should do the right thing and edit it yourself to something more wholesome.

Thanks,

Bob

]]>Mersenne Number= 2^11 -1

Factors = 23 and 89

2^11 -1 takes eleven goes to get to zero if I continually minus one and divide by 2. Watch what happens to 23 and 89 acting as remainders as I do the same to them.

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

8(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

They both take eleven goes to get to zero too.

Working backwards it can be seen that this must be the case. Starting with zero I multiply by 2 and add one continually until I get to 2^11 -1. The same must be done to the** remainders **which will then become 23 and 89 or zero, the factors of 2^11 -1.

Let the number of goes an odd number, y, takes to get to zero by continually minusing one and dividing by 2, (adding y when even) =z.

The z for 2^11 -1, 23 and 89 =11.

**Whatever z equals for y, 2^z -1 must be factorable by y. **When we use the method for 2^z -1, we never add 2^z -1 as it never equals an even number. This could potentially alter remainders for potential factors, but for Mersenne Numbers, this is not the case.

Does anyone know what “z” a composite can’t be?

]]>You can record such a parameterization of solutions.

]]>Did you want me to provide a solution or what? No square brackets (to hide) on my Kindle so you just have to close your eyes.

Also no diagram making, sorry.

This is 3D so I'll describe the diagram.

Let TB be the tower in the 'z' direction and BE be an easterly direction with the first observation at E (x direction).

Draw an oblique line NES to indicate the North/South direction (y) with ES= 42.4 ft.

As TEB = 45, BE = TB = h, the height of the tower.

And as TSB= 30, BS = root 3 h.

So in triangle BES we have BE = h, BS = root 3 h and ES = 42.4 with a right angle at E.

So it would be easy to use Pythag to get h (approx) 30 ft I think. But here's a construction for it. I have no paper nor instruments handy so the scale might not work.

Draw a verticle line BE' in the top right corner of your page. I'll make it 5cm to represent 10 ft. Construct a right angle at E' and extend the line right across the paper.

On a fresh sheet, construct an equilateral triangle sides 5 cm. Construct a perpendicular bisector so the height = 5 root 3 and set this as a radius. On the first diagram, centre B, make an arc to cut the horizontal line at S'

The triangle BE'S' is the right shape for BES but needs to be scaled up.

On E'S' produced Mark X so that E'X = 21.2 cm.

Draw XS parallel to BE' with BS' extended crossing it at S. SE parallel to S'E' will fix E and hence the correct size for triangle BES. h = BE.

Bob

]]>If I minus 1 and divide by 2 it will take p goes to get to zero.

Example:

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

Similarly if 2^p -1 has a factor of the form, n, it should take p goes to get to zero. Only, if minus 1 divided by 2 results in an even number, I should be allowed to add n to the number to continue. As this is what happens to n, as a remainder for 2^p -1, as each go takes place.

Examples:

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

Both 23 and 89 take eleven goes to get to zero, as they must do, as they are factors of 2^11 -1, and 2^11 -1 takes eleven goes to get to zero also.

]]>Yes that is it. Or to make it easier it is the sum of the last two numbers in the sequence.

Hi Mcbattle, I've just finished an inductive proof of your claim: Let

, and if . Then, and

for all integers .

(the 2nd statement is needed to complete the induction step.)

Furthermore, you need

I guess the easiest way is to use the Euler-Binnet formula; with a little more work, it can be proved without it.

Regards,

zahlenspieler

These coefficients are part of a more general phenomenon called the Savitzky-Golay filter in numerical analysis. In fact the traditional five-point 'stencil' reads:where . You can derive this result simply by playing around with the Taylor series for with . It's a little tedious, but you can see the derivation here: https://en.wikipedia.org/wiki/Five-poin … he_formulaThere are analogous formulae for higher order derivatives too, and several published papers about the error term.

That is very informative, zeta. Thanks for the reply.

I wonder if there is a way to calculate the midpoint differentials by 4 points around it:

dy/dx ]x=0

ddy/dx/dx ]x=0

given y[-0.5], y[0.5], y[-1.5] & y[1.5]

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