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What is the newton's generalised binomial theorem?
It's better to think about the ordinary binomial theorem first. Consider a binomial (x + y), and raising it to a power, say squaring it. (x + y)^2 = (x + y)(x + y) = x^2 + 2xy + y^2 Now try cubing it. (x + y)^3 = (x + y)(x + y)(x + y) = x^3 + 3x^2 y + 3xy^2 + y^3 It becomes very tedious to do this. The binomial theorem allows us to expand binomial expressions to a power very quickly. The generalised binomial theorem is, as it says, 'generalised' - the 'original' binomial theorem only allows us to expand binomial expressions to a power which is a whole number (0, 1, 2, 3 ... etc) but not numbers such as 1/2, 1/3 or -1. Newton's generalised binomial theorem allows us to expand binomial expressions for any _rational_ power. (that is any number which can be expressed as a ratio of two integers - not something horrible like the cube root of three) So now we can expand things like (x + y)^0.5, (1 - x)^-1 and all that malarky - this has some fairly deep significances, such as allowing numerical approximations of surds and bears relevance to some power series. For example, take (1 - x)^-4, using Newton's generalised binomial theorem it can be seen that (1 - x)^-4 = 1 + 4x + 10x^2 + 20x^3 ... Each expansion for a rational exponent of the binomial expressions creates an infinite series. The actual calculations are best left to a site which can show you the mathematical notation, but if you can do the normal binomial theorem - the nuances of this one will be easy to grap.
What is the unit for the Universal Law of Gravity?
The Universal Law of Gravitation is a force equation, therefore it should have units of Newtons.
How is poisson distribution related to binomial distribution?
The Poisson distribution is a limiting case of the binomial distribution when the number of trials is very large and the probability of success is very small. The Poisson distribution is used to model the number of occurrences of rare events in a fixed interval of time or space, while the binomial distribution is used to model the number of successful outcomes in a fixed number of trials.
What is the power formula for radioactivity?
The power formula for radioactivity is given by P = λ*N, where P is the power, λ is the decay constant, and N is the number of radioactive atoms. This formula represents the rate at which energy is released by radioactive decay.
What does a gradient of 0.5 show on a log log graph?
If y=xn, then log y =nlogx and n indicates the power in the power function. If one has a set of data [x,y] and if a plot of logy vs logx yields a straight line or one reasonably so, then the slope (gradient) of the line reveals the power relation between x and y
Define binomial theorem?
The binomial theorem describes the algebraic expansion of powers of a binomial: that is, the expansion of an expression of the form (x + y)^n where x and y are variables and n is the power to which the binomial is raised. When first encountered, n is a positive integer, but the binomial theorem can be extended to cover values of n which are fractional or negative (or both).
How does the equation 2 raised to the power of log n equal n?
When the equation 2 raised to the power of log n is simplified, it equals n.
What is n raised to the power n-1?
n-1 = 1/n
What is Cardinal number raised to power 30?
It is n^30 where n is the cardinal number.
What is a binomial coefficient?
A binomial coefficient is a coefficient of any of the terms in the expansion of the binomial (x+y)^n.
Validity of binomial expansion for n less than 0?
The binomial expansion is valid for n less than 1.
Why is 7 to the power of 0 1?
Why is 7^0 = 1 Algebraic proof. Let 'n' be any value Let 'n be raised to the power of 'a' Hence n^a Now if we divide n^a by n^a we have n^a/n^a and this cancels down to '1' Or we can write n^(a)/n^(a) = n^(a-a) = n^(0) , hence it equals '1' Remember when the lower /denominating index is a negative power ,when raised above the division line.
The power expended with a barbell is raised 2.0m in 2s is?
n nk
What is binomial expansion theorem?
We often come across the algebraic identity (a + b)2 = a2 + 2ab + b2. In expansions of smaller powers of a binomial expressions, it may be easy to actually calculate by working out the actual product. But with higher powers the work becomes very cumbersome.The binomial expansion theorem is a ready made formula to find the expansion of higher powers of a binomial expression.Let ( a + b) be a general binomial expression. The binomial expansion theorem states that if the expression is raised to the power of a positive integer n, then,(a + b)n = nC0an + nC1an-1 b+ nC2an-2 b2+ + nC3an-3 b3+ ………+ nCn-1abn-1+ + nCnbnThe coefficients in each term are called as binomial coefficients and are represented in combination formula. In general the value of the coefficientnCr = n!r!(n-r)!It may be interesting to note that there is a pattern in the binomial expansion, related to the binomial coefficients. The binomial coefficients at the same position from either end are equal. That is,nC0 = nCn nC1 = nCn-1 nC2 = nCn-2 and so on.The advantage of the binomial expansion theorem is any term in between can be figured out without even actually expanding.Since in the binomial expansion the exponent of b is 0 in the first term, the general term, term is defined as the (r+1)th b term and is given by Tr+1 = nCran-rbrThe middle term of a binomial expansion is [(n/2) + 1]th term if n is even. If n is odd, then terewill be two middle terms which are [(n+1)/2]th and [(n+3)/2]th terms.
What is the commutator of the operator x with the momentum operator p raised to the power of n?
The commutator of the operator x with the momentum operator p raised to the power of n is ih-bar times n times p(n-1), where h-bar is the reduced Planck constant.
How do you find the coefficients of the terms in the binomial expansion?
The coefficient of x^r in the binomial expansion of (ax + b)^n isnCr * a^r * b^(n-r)where nCr = n!/[r!*(n-r)!]
What is n3 to the 43 power?
To calculate ( n^3 ) raised to the 43rd power, you can use the exponentiation rule which states that ( (a^m)^n = a^{m \cdot n} ). Therefore, ( (n^3)^{43} = n^{3 \cdot 43} = n^{129} ). Thus, ( n^3 ) to the 43rd power is ( n^{129} ).
