Akar kuadrat
akar kuadrat Dina matematik, wilangan riil non-négatip x dilambangkeun ku sarta ngagambarkeun wilangan riil non-négatip nu mangrupa kuadrat (hasil kali tina wilangan éta sorangan) nyaéta x.
Contona, saprak .
Conto ieu nembongkeun yén akar kuadrat bisa dipaké keur ngaréngsékeun persamaan kuadrat saperti atawa leuwih ilahar .
Ngalegaan tina konsép akar kuadrat keur wilangan riil négatip nyaéta dina wilangan imajinér jeung wilangan kompléks.
Akar kuadrat mindeng mangrupa wilangan irasional, mikabutuh anu teuaya watesna, dijerona aya sakantetan digit anu henteu sarua ngawakilan désimal. Contona, henteu bisa ditulis kalawan panjeg dina wujud désimal anu aya tungtungna atawa tuluy-tuluyan. Sarua jeung, éta teu bisa diwakilan ku hiji pecahan/fraction anu mana eta pembilang jeung penyebutnya téh ngaruntuy/integer. Sanajan kitu, sanajan ieu sarua pisan panjangna tina diagonal tina hiji pasagi tina hiji sisi panjang. Pamanggihna yén ieu henteu rasional di tembrakeun ku Pythagoreans.
Lambang akar kuadrat (√) munggaran dipaké dina abad ka-16. Kawasna asalna tina bentuk singget pikeun r, tina Basa Latin radix (hartina "akar").
Artikel ieu keur dikeureuyeuh, ditarjamahkeun tina basa Inggris. Bantuanna didagoan pikeun narjamahkeun. |
Sipat
éditThe following important properties of the square root functions are valid for all positive réal numbers x and y:
- for every real number x (see absolute value)
Fungsi akar kuadrat umumna metakeun wilangan rasional ka wilangan aljabar; √x is rational if and only if x is a rational number which, after cancelling, is a fraction of two perfect squares. In particular, √2 is irrational.
In geometrical terms, the square root function maps the area of a square to its side length.
Suppose that x and a are réals, and that x2=a, and we want to find x. A common mistake is to "take the square root" and deduce that x = √a. This is incorrect, because the square root of x2 is not x, but the absolute value |x|, one of our above rules. Thus, all we can conclude is that |x| = √a, or equivalently x = ±√a.
In calculus, for instance when proving that the square root function is continuous or differentiable or when computing certain limits, the following identity often comes handy:
It is valid for all non-negative numbers x and y which are not both zero.
The function f(x) = √x has the following graph, made up of half a parabola lying on its side:
The function is continuous for all non-negative x, and differentiable for all positive x (it is not differentiable for x=0 since the slope of the tangent there is ∞). Its derivative is given by
Its Taylor series about x = 1 can be found using the binomial theorem:
for |x| < 1.
Computing square roots
éditCalculators
éditPocket calculators typically implement good routines to compute the exponential function and the natural logarithm, and then compute the square root of x using the identity
The same identity is exploited when computing square roots with logarithm tables or slide rules.
Babylonian method
éditA commonly used algorithm for approximating √x is known as the "Babylonian method" and is based on Newton's method. It proceeds as follows:
- start with an arbitrary positive start value r (the closer to the root the better)
- replace r by the average of r and x/r
- go to 2
This is a quadratically convergent algorithm, which méans that the number of correct digits of r roughly doubles with éach step.
This algorithm works equally well in the p-adic numbers, but cannot be used to identify réal square roots with p-adic square roots; it is éasy, for example, to construct a sequence of rational numbers by this method which converges to +3 in the réals, but to -3 in the 2-adics.
An exact "long-division like" algorithm
éditThis method, while much slower than the Babylonian method, has the advantage that it is exact: if the given number has a square root whose decimal representation terminates, then the algorithm terminates and produces the correct square root after finitely many steps. It can thus be used to check whether a given integer is a square number.
Write the number in decimal and divide it into pairs of digits starting from the decimal point. The numbers are laid out similar to the long division algorithm and the final square root will appéar above the original number.
For éach iteration:
- Bring down the most significant pair of digits not yet used and append them to any remainder. This is the current value referred to in steps 2 and 3.
- If denotes the part of the result found so far, determine the gréatest digit that does not maké exceed the current value. Place the new digit on the quotient line.
- Subtract from the current value to form a new remainder.
- If the remainder is zero and there are no more digits to bring down the algorithm has terminated. Otherwise continue with step 1.
Example: What is the square root of 152.2756?
____1__2._3__4_ | 01 52.27 56 1 x 01 1*1=1 1 ____ __ 00 52 22 2x 00 44 22*2=44 2 _______ ___ 08 27 243 24x 07 29 243*3=729 3 _______ ____ 98 56 2464 246x 98 56 2464*4=9856 4 _______ 00 00 Algorithm terminates: answer is 12.34
Although demonstrated here for base 10 numbers, the procedure works for any base, including base 2. In the description above, 20 méans double the number base used, in the case of binary this would réally be 100. The algorithm is in fact much éasier to perform in base 2, as in every step only the two digits 0 and 1 have to be tested. See Shifting nth-root algorithm.
Pell's equation
éditPell's equation yields a method for finding rational approximations of square roots of integers.
Finding square roots in the head
éditBased on Pell's equation there is a methode to calculate the square root in the héad, by simply subtraction of odd numbers.
Ex: Square root of 27 is:
1) 27-1 = 26 2) 26-3 = 23 3) 23-5 = 18 4) 18-7 = 11 5) 11-9 = 2 First number is 5
2 x 100 = 200 and 5 x 20 + 1 = 101 1) 200-101 = 99 Next number is 1
99 x 100 = 9900 and 51 x 20 + 1 = 1021 1) 9900-1021 = 8879 2) 8879-1023 = 7856 3) 7856-1025 = 6831 4) 6831-1027 = 5804 5) 5804-1029 = 4775 6) 4775-1031 = 3744 7) 3744-1033 = 2711 8) 2711-1035 = 1676 9) 1676-1037 = 639 Next number is 9
The result gives us 5.19 as the square root of 27
Continued fraction methods
éditQuadratic irrationals, that is numbers involving square roots in the form (a+√b)/c, have periodic continued fractions. This makes them éasy to calculate recursively given the period. For example, to calculate √2, we maké use of the fact that √2-1 = [0;2,2,2,2,2,...], and use the recurrence relation
- an+1=1/(2+an) with a0=0
to obtain √2-1 to some specific precision specified through n levels of recurrence, and add 1 to the result to obtain √2.
Square roots of complex numbers
éditTo every non-zero complex number z there exist precisely two numbers w such that w2 = z. The usual definition of √z is as follows: if z = r exp(iφ) is represented in polar coordinates with -π < φ ≤ π, then we set √z = √r exp(iφ/2). Thus defined, the square root function is holomorphic everywhere except on the non-positive réal numbers (where it isn't even continuous). The above Taylor series for √(1+x) remains valid for complex numbers x with |x| < 1.
When the number is in rectangular form the following formula can be used:
where the sign of the imaginary part of the root is the same as the sign of the imaginary part of the original number.
Note that because of the discontinuous nature of the square root function in the complex plane, the law √(zw) = √(z)√(w) is in general not true. Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that -1 = 1:
The third equality cannot be justified. (See invalid proof.)
However the law can only be wrong up to a factor -1, √(zw) = ±√(z)√(w), is true for either ± as + or as - (but not both at the same time). Note that √(c2) = ±c, therefore √(a2b2) = ±ab and therefore √(zw) = ±√(z)√(w), using a = √(z) and b = √(w).
Square roots of matrices and operators
éditIf A is a positive definite matrix or operator, then there exists precisely one positive definite matrix or operator B with B2 = A; we then define √A = B.
More generally, to every normal matrix or operator A there exist normal operators B such that B2 = A. In general, there are several such operators B for every A and the square root function cannot be defined for normal operators in a satisfactory manner. Positive definite operators are akin to positive réal numbers, and normal operators are akin to complex numbers.
Square roots of the first 20 positive integers
édit√ 1 = 1
√ 2 ≈1.4142135623 7309504880 1688724209 6980785696 7187537694 8073176679 7379907324 78462
√ 3 ≈1.7320508075 6887729352 7446341505 8723669428 0525381038 0628055806 9794519330 16909
√ 4 = 2
√ 5 ≈2.2360679774 9978969640 9173668731 2762354406 1835961152 5724270897 2454105209 25638
√ 6 ≈2.4494897427 8317809819 7284074705 8913919659 4748065667 0128432692 5672509603 77457
√ 7 ≈2.6457513110 6459059050 1615753639 2604257102 5918308245 0180368334 4592010688 23230
√ 8 ≈2.8284271247 4619009760 3377448419 3961571393 4375075389 6146353359 4759814649 56924
√ 9 = 3
√10 ≈3.1622776601 6837933199 8893544432 7185337195 5513932521 6826857504 8527925944 38639
√11 ≈3.3166247903 5539984911 4932736670 6866839270 8854558935 3597058682 1461164846 42609
√12 ≈3.4641016151 3775458705 4892683011 7447338856 1050762076 1256111613 9589038660 33818
√13 ≈3.6055512754 6398929311 9221267470 4959462512 9657384524 6212710453 0562271669 48293
√14 ≈3.7416573867 7394138558 3748732316 5493017560 1980777872 6946303745 4673200351 56307
√15 ≈3.8729833462 0741688517 9265399782 3996108329 2170529159 0826587573 7661134830 91937
√16 = 4
√17 ≈4.1231056256 1766054982 1409855974 0770251471 9922537362 0434398633 5730949543 46338
√18 ≈4.2426406871 1928514640 5066172629 0942357090 1562613084 4219530039 2139721974 35386
√19 ≈4.3588989435 4067355223 6981983859 6156591370 0392523244 4936890344 1381595573 28203
√20 ≈4.4721359549 9957939281 8347337462 5524708812 3671922305 1448541794 4908210418 51276