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In mathematics, the method of equating the coefficients is a way of solving a functional equation of two expressions such as polynomials for a number of unknown parameters. It relies on the fact that two expressions are identical precisely when corresponding coefficients are equal for each different type of term.
unknown ∞. infinity sign 1655 ... 1684 (deriving from use of colon to denote fractions, dating back to 1633) ...
A continued fraction is a mathematical expression that can be written as a fraction with a denominator that is a sum that contains another simple or continued fraction. Depending on whether this iteration terminates with a simple fraction or not, the continued fraction is finite or infinite .
The rate of convergence depends on the absolute value of the ratio between the two roots: the farther that ratio is from unity, the more quickly the continued fraction converges. When the monic quadratic equation with real coefficients is of the form x 2 = c, the general solution described above is useless because division by zero is not well ...
By considering the complete quotients of periodic continued fractions, Euler was able to prove that if x is a regular periodic continued fraction, then x is a quadratic irrational number. The proof is straightforward. From the fraction itself, one can construct the quadratic equation with integral coefficients that x must satisfy.
Suppose Q(x) = (x − α) r S(x) and S(α) ≠ 0, that is α is a root of Q(x) of multiplicity r. In the partial fraction decomposition, the r first powers of (x − α) will occur as denominators of the partial fractions (possibly with a zero numerator).
A simple fraction (also known as a common fraction or vulgar fraction) [n 1] is a rational number written as a/b or , where a and b are both integers. [9] As with other fractions, the denominator (b) cannot be zero. Examples include 1 / 2 , − 8 / 5 , −8 / 5 , and 8 / −5 .
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
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