# Pay for best business plan online

Buy essay online cheap written in disgust of vulgar superstition In some mathematical books and articles you will see one other form of notation which saved space : What numbers stand in place of the letters? What is the general principle for computing the gcd given two numbers expressed as powers of the same primes? What is the greatest common divisor of 24 and 18 (call it G)? What is the gcd of 24, 18 and 30? How is it related to the gcd of G and 30? [This is Propositions 3 of Euclid's The ElementsBook 7.] The gcd is useful in simplifying a fraction. But when adding fractions we find equivalents with the same denominator. This time we need the smallest number into sample resume of newly graduate nurse both denominators will divide exactly, ot the lowest common multiple (lcm). Using the prime-powers decomposition of two denominators, how can we find the lcm? Use 168 and 720 as an example. What is the relationship beween gcd(a,b), lcm(a,b) and pay for best business plan online You can check your answers using the Euclid's Algorithm CF Calculator in the next section or. When we multiply gcd(a,b) by lcm(a,b) we are using each prime power in both a and b. Therefore gcd(a,b) × lcm(a,b) = a × b. This Calculator shows Eucild's Algorithm for finding the GCD of two numbers a and b ; the list of the divisors of a and b with the greatest highlighted; how Euclid's algorithm relates to the CF of a/bfor example for 45/16 : This shows that 45/16 = [2; 1, 4, 3] as a continued fraction (CF). a stylised picture of the "jigsaw of squares" for a/b where the sizes of the squares are shown followed by the number What are ways for girls to ask boys to prom? direction of the repetitions. For the example above we have: If the numerators of the fractions are all 1then the continued fraction is called a simple continued fraction. This is what we will mean when we use the term "continued fraction" - we will often abbreviate it to just CF - on this page. A more general form of continued fraction has numerators which are not one but we only touch on this topic later on this page. Usually all the numbers in a continued fraction (general or simple) will be positive although alternative forms are possible where negative whole numbers are allowed and we examine this kind later on this page. Any and every fraction, P / Q where P and Q are whole, positive numbers can be written in the form of a continued fraction as shown above using Euclid's Algortihm in any of the various forms above. We know that any proper fraction can be expressed using a finite number of terms in its continued fraction. We can write down any continued fraction such as just as a list of the numbers [a; b, c. ]. The first number, ais special as it is the whole number part of the value. The rest is written as a list with comma separators (,) like this: None of the values will be zero except possibly the first, the one before the semicolon (;).If the first is zero it means the value of the continued fraction is less than 1 . There are two forms for the fraction 7/30 namely [0; 4, 3, 2] and [0; 4, 3, 1, 1] This is true for all continued fractions. We can always write the last number, nas (n-1) + 1/1 and so change the n to n − 1 and continue the fraction by one more number, 1provided n > 1. Alternatively, if the last number is 1we can remove it by adding it to the number before in the lost; So we have the general rule: Express the following as continued fractions: 41/13 = [3; 6, 2] = [3; 6, 1, 1] 124/37 = [3; 2, 1, 5, 2] = [3; 2, 1, 5, 1, 1] 5/12 = [0; 2, 2, 2] = [0; 2, 2, 1, 1] The three rectangles in the picture are split into squares. Assuming that the smallest sized square has sides of length 1, what is the ratio of the two sides of each of the three rectangles? and form fractions from neighbouring squares. First let's look at this sample resume of newly graduate nurse of fractions formed from neighbouring square numbers in the list above. Change each of these fractions into a continued fraction. sample resume of newly graduate nurse = [1; 1, 1, 3, 2] 49/36 = [1; 2, 1, 3, 3] 81/64 = [1; 3, 1, 3, 4] 121/100 = [1; 4, 1, 3, 5] Can you spot the pattern for. So there are two "surprising results" and we can see examples of both in the tables at the start of this section. This result pay for best business plan online known to Euler (see references below). The first number can be transformed to produce the 1-complement of the fraction. This is very easily established by simple algebra since [0; a,B] = 0+1/ (a+1/B) = B/(1+aB) whereas [0;1,a-1,B] = 1/(1+1(a-1+1/B)) = 1 − B/(1+aB) The CF reversal result is proved in an amazing way by Harvey Mudd College's Professor business plan wedding planning business Mathematics: Art Benjamin in the first of these references: Counting on Continued Fractions Arthur T Benjamin, F E Su, J J Quinn Maths Mag vol 73 (2000), pages 98-104 also on Art's list of papers, preprints and books where there is a link to a PDF version to view. De Fractionibus Continuis Dissertatio written in 1737 but only published in 1744. It has been translated into English by M F Wyman and B F Wyman in An Essay on Continued FractionsMath Systems Theory vol 18 (1985, Springer-Verlag) pages 295-328 PDF A T Benjamin, J J Quinn, W Watkins, Proofs That Really Count published by The Mathematical Association of America (Aug 2003), 208 pages. This is a wonderful book on using counting methods in proofs by induction to prove many results in combinatorics and number theory which includes many Fibonacci number formulae. The authors make the proofs both easy to understand and fun too! Euler showed that the CFs [ a 1 ; a 2. a n ] and its reversal [ a n ; a n-1. a 1 ] have the same numerators but we must note that the first and last terms in both CFs are non-zero : see Davenport's book in the References section at the foot of this page. There is a pleasant surprise here since every square-root has a repeating pattern in its continued fraction . But what about continued fractions for numbers which we only have in the form of a decimal? There are two methods of converting them into continued fractions: using the decimal itself or finding a proper fraction for the decimal number. Both methods are explained here. The method is very easy on a calculator when you use the subtract button and the 1/x or x -1 button only! Input the decimal number. Write down the whole number part (before the decimal point) as the next value in the CF list. For an input value less than 1, this will be 0. Subtract it from the display to get a value less than 1. If 0 is showing, stop otherwise press the 1/x button Start again with the new number which is bigger than 1 in the display. You can try this on √2, e, π or any computed value but be careful as rounding errors will eventually build up and then the CF terms are meaningless. For example, 2·875. has a whole part of 2 so write that down to begin its continued fraction: 2·875 = [2;. Subtract the 2 to get 0.875 Press 1/x to get 1.14285714285714 Starting again: write down the whole number part: 1so the CF is now 2·875 = [2; 1. Subtract the 1 to get 0.14285714285714 Press 1/x to get 7.00000000000014 Start again: write down 7 so the CF is now 2·875 = [2; 1, 7 Subtract the 7 to get 0.00000000000014 This is practically 0 so we stop here allowing for rounding errors. 2.875 = [2; 1, 7] Checking shows that when we expand [2; 1, 7] we have 23/8 which is 2 + 7/8 = 2.875 exactly. Convert 64/29 into bullier automation nanterre university decimal fraction and approximate it by rounding it off to 3 dps. Now convert that decimal fraction to a CF using the method of the last section. Was it easy to see when to stop if the real value of our decimal was 64/29? This time, when we convert 2·875 to an equivalent fraction we get 2875/1000 and Euclid's algorithm gives: But now we turn our attention to some infinite continued fractions which, maybe surprisingly, have some very simple forms. Take, for example, √5. The nearest square below 5 is 4, so we know √5 begins with 2. Let's write: Suppose n is not a perfect square and that √n = a + x where a 2 is any square number less than n. Using the method above, what is the general formula for √n as a continued fraction without reducing the numerators to 1? There is a method which will find the sample resume of indian lawyer CF for any (whole-number) square-root and the numerators are always 1. It is not quite as simple as the method above, but it only uses the mathematics and algebra of secondary school. We will need this useful technique: If we have a fraction with (√A + B) in the denominator then the secret is to multiply top and bottom by (√A − B)that is, keep the numbers the same but just change the sign between them. If we had √A − B in the denominator then we would use √A + B instead. If you are good at algebra you will recognise that x+y and x−y are the two factors of x 2 − y 2. You can multiply out the intern record label resume to check this. For our denominator, we now have (√A + B)(√A − Analogy essay topics ideas which expands to (A − B 2 ) and, since A and B are integers, this is a whole number! Here is a worked example: Now, √n = m + 1/x where n and m are whole numbers. Step 2: Rearrange the equation of Step 1 into a form involving the square root which will appear as the denominator of a fraction: x = 1 / (√n - m) Step 3: We now have a fraction with a square-root in the denominator. Use the method above to eliminate the square-root from the denominator. In this case, multiply top and bottom by (√ n + m) and simplify. either Step 4A: stop if this expression is the square root plus the original first integer. or Step 4B: otherwise start again from Step 1 but using the expression at the end of Step 3. We will take √14 and see how we find the continued fraction [ 3; 1,2,1,6, 1,2,1,6, 1,2,1,6. ] = [ 3; 1,2,1,6 ] using the algorithm above: In order to distinguish the x's at each stage repeating the steps of the method, we will use subscripts to distinguish the different x's as x changes: x 1then x 2x 3 and so on: This method is completely general and applies to all square roots. Is there a pattern behind this table? Justin Miller (University of Arizona) has a list of several patterns within the table (see References). Can you extend his table? Can you find a single overall unifying formula? What patterns do you notice in the table of square-roots above? Four easy ones first: What is special about the first number of the continued fraction? For instance, take x 2 − 5 x − 1 = 0 Can you think of an x value for which this equation holds? We can rewrite the equation in a different way as: Suppose that x is the continued fraction [2; 2 ] = 2 + 1/(2 + 1/. ). We can write this as x = 2 + 1/x and, by multiplying both sides by x we have the quadratic equation. Repeat the above for [A; A ] by writing it as x = A + 1/x and finding a quadratic in x to solve (using the Formula). Show that one root is positive (and, if A>1 then that root is also > 1) and the other root is negative but less than 1. Find the 2 roots and a continued fraction for a root of these quadratic equations: x 2 + x = 1 x 2 − 2x = 1. x 2 + x = 1 ⇒ x 2 = −x + 1 ⇒ x = −1 + 1/x x = [-1; -1, -1, -1. ] = -[1;1, 1, 1. ] = -Phi x 2 − 2x = 1 ⇒ x 2 = 2x + 1 ⇒ x = 2 + 1/x x = [2; 2, 2, 2. ] = 1 + √2. Although the convergents of simple CFs only are the best approximations, meaning that no smaller denominator will produce a better approximation, not all the "best approximations" are in the list of convergents. The others are those with an identical initial part of the CF of the value being approximated and a final number which can be smaller or the same as the CF term. So in such a list we will find all the convergents of the CF of the value being approximated. You can experiment with the following Calculator which also allows you to find the best numerator for a given denominator in a fraction approximation of the given numerical value. As with the CF Calculator for Expressions earlier on this page you can also give an expression to be evaluated in the value box or just a decimal number. By truncating the continued fractions for Pi, we quickly find fractions that are best approximations. These include some interesting fractions as follows: 3 = 3 The nearest whole number. [ 3; 7 ] = 22/7 = 3.142857 = pi + 0.00126 This is the value everyone knows from school, 22/7. It is a good approximation for Pi, accurate to one-eighth of one percent. [ 3; 7, 15 ] = 333/106 = 3.1415094. = pi − 0.00008. [ 3; 7, 15, 1 ] = 355/113 = essay for you smart people know that time is valuable. = pi + 0.000000266. This value is easy to remember - think of the first three odd numbers written down twice: 113355, then split it in the middle to form two three-digit numbers, 113 355, and put the larger number above the smaller! [ 3; 7, 15, 1, 292 ] = 103993/33102 =3.1415926530. = pi − 0.00000000057. This is the next convergent to pi. It corresponds to a term in the CF that is a large number so it gives a particularly good approximation to pi. It is over 400 times more accurate than the previous one (355/113), but this time the numbers involved are not so easy to remember! [ 3; 7, 15, 1, 292, 1 ] = 104348/33215 = pi + 0.00000000033. The final element of this CF is 1 so the fraction is fairly close to the previous convergent. Notice that the convergents are alternately above and below the value of Pi; this is true always for all CF convergents. Continued fractions can be simplified by cutting them off after a certain number of terms. The result - a terminating continued fraction - will give a true fraction but it will only be an approximation to the full value. Here, we will use the term exact value bullier automation nanterre university the exact (irrational) value of an infinite continued fraction or the final value of a terminating continued fraction. It turns out - and we shall not prove this here - that these convergents (fractions) are "the best possible approximations". These approximations are called convergents of the continued fraction. By "best" here, we mean the convergents are proper fractions the convergents alternate between being larger and smaller than the exact value if a convergent is smaller than the exact value, no other fraction with smaller denominator is closer and smaller than the exact value if a convergent is larger than the exact value, no other fraction with smaller denominator is closer and larger than the exact value.