The binomial theorem

[Rothschild]

You are trying to sound smart. You asks questions you don't understand so you naturally cannot understand the answers. Then instead of cutting your losses you continue to ask questions in an attempt to seem even smarter, not realising that you looked like a fool before you pressed "New Topic".

But is it really about looking like a fool or isn't it about learning? I did not pretend to know everything which is why I ask questions. Its not like you are actually helping me to understand you only try to patronise me.

[Skola]

I don't understand the motivation behind this whole maths topic businessYou're just trying to irritate us, or are you genuinely asking for advice?If it's the former, then you've succeeded, the latter, as is mentioned above, there are far more reliable sources of info that VIPf*cking2

[Bacardi_Up]

Ashman is on some hype.OK yes we get it. You are educating yourself good for you BUT.... many man on here are in Uni or have graduated so impressed the crowd is not.BTW I hope you aren't picking out random chapters from a maths book to carry on a pretense that you in education.

How is it pretense? I am in education and have had university offers there is no pretense my son.

we know u r on an access course...we also know this maths is beyond that level of learning...so cut it out.Peace

It's like dude opened up C3 book and waved his finger blindly over the index. Then hit new topic.

[Captain Planet]

1 + nx + [n(n-1)/2!](x^2) + [n(n-1)(n-2)/3!](x^3) etc.Now buy a f*ck*ng maths book you poor bastard.

That is for (1+x)^n and if n<0 then |x|<1. So its not just one formula you have to remember. But can you explain why its true?

If -1<n<1 then you can use your head to work it out.It merely diverges instead of converge.But on a real why are you doing this do you not have professionals that you can ask this stuff?Join a maths forum there are millions.Just stop trying to sound smart because we have already established our opinions on you, due to the actions you demonstrate.

so if n=1/3 how would you use your head to work it out? To be honest I ask a question and you give a confusing answer then tell me that I'm trying to sound smart?

Gosh, stick it in the formula you twat.But I'm finished with you, you obviously are just seeking attention from a forum.

[Rothschild]

1 + nx + [n(n-1)/2!](x^2) + [n(n-1)(n-2)/3!](x^3) etc.Now buy a f*ck*ng maths book you poor bastard.

That is for (1+x)^n and if n<0 then |x|<1. So its not just one formula you have to remember. But can you explain why its true?

If -1<n<1 then you can use your head to work it out.It merely diverges instead of converge.But on a real why are you doing this do you not have professionals that you can ask this stuff?Join a maths forum there are millions.Just stop trying to sound smart because we have already established our opinions on you, due to the actions you demonstrate.

so if n=1/3 how would you use your head to work it out? To be honest I ask a question and you give a confusing answer then tell me that I'm trying to sound smart?

Gosh, stick it in the formula you twat.But I'm finished with you, you obviously are just seeking attention from a forum.

You obviously dont want the discussion to go any further in case you can no longer answer. Why do you keep up the false pretense that you know the answers?

[A Bit Better]

Lol at Ashman asking exhausting questions and then saying people can't answer his questions.You are a douche Ashman. A douche.

[Rothschild]

Lol at Ashman asking exhausting questions and then saying people can't answer his questions.You are a douche Ashman. A douche.

You are a coward mate a coward. Someone that reads my topics then only answers when it suits him but evades further questions put to you.If you want to battle, we will battle mate!

[DazLeR]

someone probs sed it already but just find out on ur calculater man like if its to the power of 10 on ur calc press 0Ncr10 for first term 1Ncr10 second etc havent done it since last yr but i think thats right

[Lord Flashheart]

Remove his new topic priviliges already

[Brandnew]

c how he selectively ignores posts...

[Rothschild]

someone probs sed it already but just find out on ur calculater man like if its to the power of 10 on ur calc press 0Ncr10 for first term 1Ncr10 second etc havent done it since last yr but i think thats right

But if its to the power of 1/3 then you will get error? You dont know what you're on about mate.

[VENOM]

Q

[VENOM]

actually ur chatting sh*t bcos when you equate the theorem the conundrum rule states that This equation gives the rest mass of an object which has an arbitrary amount of momentum and energy. The interpretation of this equation is that the rest mass is the relativistic length of the energy-momentum four-vector.If the equation E = mc2 is used with the rest mass or invariant mass of the object, the E given by the equation will be the rest energy of the object, and will change according to the object's internal energy, heat and sound and chemical binding energies (all of which must be added or subtracted from the object), but will not change with the object's overall motion (in the case of systems, the motion of its center of mass). However, if a system is closed, its invariant mass does not vary between observers, and is also constant and conserved.If the equation E = mc2 is used with the relativistic mass of the object, the energy will be the total energy of the object, which is conserved so long as no energy is added to or subtracted from the object, but which, like the kinetic energy, depends on the velocity of the object, and is different in different inertial frames.Mass-Velocity RelationshipIn developing special relativity, Einstein found that the kinetic energy of a moving body is K.E. = \frac{m_0 c^2}\sqrt{1-\frac{v^2}{c^2}} - m_0 c^2,with v the velocity, and m0 the rest mass.He included the second term on the right to make sure that for small velocities, the energy would be the same as in classical mechanics: K.E. = \frac{1}{2}m_0 v^2 + ... Without this second term, there would be an additional contribution in the energy when the particle is not moving.i found that the total momentum of a moving particle is: P = \frac{m_0 v}\sqrt{1-\frac{v^2}{c^2}}. and it is this quantity which is conserved in collisions. The ratio of the momentum to the velocity is the relativistic mass, m. m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}And the relativistic mass and the relativistic kinetic energy are related by the formula: K.E. = m c^2 - m_0 c^2. \,i wanted to omit the unnatural second term on the right-hand side, whose only purpose is to make the energy at rest zero, and to declare that the particle has a total energy which obeys: E = m c^2 \,which is a sum of the rest energy m0c2 and the kinetic energy. This total energy is mathematically more elegant, and fits better with the momentum in relativity. But to come to this conclusion, Einstein needed to think carefully about collisions. This expression for the energy implied that matter at rest has a huge amount of energy, and it is not clear whether this energy is physically real, or just a mathematical artifact with no physical meaning.In a collision process where all the rest-masses are the same at the beginning as at the end, either expression for the energy is conserved. The two expressions only differ by a constant which is the same at the beginning and at the end of the collision. Still, by analyzing the situation where particles are thrown off a heavy central particle, it is easy to see that the inertia of the central particle is reduced by the total energy emitted. This allowed Einstein to conclude that the inertia of a heavy particle is increased or diminished according to the energy it absorbs or emits.[edit] Relativistic mass Main article: mass in special relativityAfter i first made his proposal, it became clear that the word mass can have two different meanings. The rest mass is what Einstein called m, but others defined the relativistic mass with an explicit index: m_{\mathrm{rel}} = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\,\, . This mass is the ratio of momentum to velocity, and it is also the relativistic energy divided by c2 (it is not Lorentz-invariant, in contrast to m0). The equation E = mrelc2 holds for moving objects. When the velocity is small, the relativistic mass and the rest mass are almost exactly the same.E = mc2 either means E = m0c2 for an object at rest, or E = mrelc2 when the object is moving.Also i used velocity- and direction-dependent mass concepts (longitudinal and transverse mass) in my 2005 electrodynamics paper and in another paper in2006 However, in my first paper on E = ^HG i treated m as what would now be called the rest mass. Some eedyats claim that i did not like the idea of "relativistic mass." so other physicists say "mass", they are usually talking about rest mass, since if they meant "relativistic mass", they would just say "energy". Low-speed expansionWe can rewrite the expression for the energy as a Taylor series: E = m_0 c^2 \left[1 + \frac{1}{2} \left(\frac{v}{c}\right)^2 + \frac{3}{8} \left(\frac{v}{c}\right)^4 + \frac{5}{16} \left(\frac{v}{c}\right)^6 + \ldots \right]. For speeds much smaller than the speed of light, higher-order terms in this expression get smaller and smaller because v / c is small. For low speeds we can ignore all but the first two terms: E \approx m_0 c^2 + \frac{1}{2} m_0 v^2 . The total energy is a sum of the rest energy and the Newtonian kinetic energy.The classical energy equation ignores both the m0c2 part, and the high-speed corrections. This is appropriate, because all the high-order corrections are small. Since only changes in energy affect the behavior of objects, whether we include the m0c2 part makes no difference, since it is constant. For the same reason, it is possible to subtract the rest energy from the total energy in relativity. By considering the emission of energy in different frames, Einstein could show that the rest energy has a real physical meaning.The higher-order terms are extra correction to Newtonian mechanics which become important at higher speeds. The Newtonian equation is only a low-speed approximation, but an extraordinarily good one. All of the calculations used in putting astronauts on the moon, for example, could have been done using Newton's equations without any of the higher-order corrections.Q.E.D.

[Guest Esquilax]

einsteinz a prik an so iz ur mum fam

[Roy Batty]

What the f*ck is this complete and utter nonsense you are posting? "what if it is 1/3?"Faahckin 'ell mate.

[Rothschild]

What the f*ck is this complete and utter nonsense you are posting? "what if it is 1/3?"Faahckin 'ell mate.

Yes to the power of 1/3.{a+B}^1/3 = X such that x^3=a+b where X is a power series involving terms a and b.

[Snoop]

Remove his new topic priviliges already

[Rothschild]

Remove his new topic priviliges already

Then I will simply switch accounts.

[Guest Esquilax]

a=(b+4) x b^(a3+B)tbh