A textbook of Computer Based Numerical and Statiscal Techniques part 17 ppsx

... R.H.S.3.9 FACTORIAL NOTATIONSThe product of n consecutive factors each at a constant difference and the first factor being x iscalled a factorial function or a factorial polynomial of degree n and ... L.H.S.150 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES CorCorCorCorCorollary:ollary:ollary:ollary:ollary:To show that,()nnx∆= n! hn and 1n+∆ x(n) = 0We know that,()nx∆= ... by 1 and 2 by 1 and add them to get the sum –1 which is to bewritten below 3. The remainder –10 is the value of D.4. Add the terms of corresponding columns of (a) and get 2, –1 and 2 of (b)5....

A textbook of Computer Based Numerical and Statiscal Techniques part 2 ppsx

... problems. A majoradvantage for numerical technique is that a numerical answer can be obtained even when a problem has no analytical solution. However, result from numerical analysis is an approximation,in ... significant figures at each step of computation. At each step of computations, retain at least one more significant figurethan that given in the data, perform the last operation, and then round off.3. ... value – Approximate value A computer has a finite word length and so only a fixed number of digits are stored and used during computation. This would mean that even in storing an exact decimal...

A textbook of Computer Based Numerical and Statiscal Techniques part 3 ppsx

... 3Hence maximum percentage error (Ep)max = 11.667%.Example 26. How accurately should the length and time of vibration of a pendulum should bemeasured in order that the computed value of g is ... 4008 and 30.4647 + 0.0639 =30.5286.Example 12. Find the product of 346.1 and 865.2 and state how many figures of the result aretrustworthy, given that the numbers are correct to four significant ... ERRORS AND FLOATING POINT9Example 11. Find the relative error in calculation of 7.3420.241. Where numbers 7.342 and 0.241 arecorrect to three decimal places. Determine the smallest interval...

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A textbook of Computer Based Numerical and Statiscal Techniques part 5 ppsx

... 26 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The basic operations specified by IEEE arithmetic are first and foremost addition, subtraction,multiplication, and division. Square roots ... formulae are also available for cubic and biquadratic polynomial equations but we rarely remember them. For higher order polynomialequations and non-polynomial equations, it is difficult and in many ... a nwhere a s are constant (a 0≠ 0) and n is a positive integer, is called a polynomial in x of degreen, and the equation f (x) = 0 is called an algebraic equation of degree n. If f (x) contains...

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A textbook of Computer Based Numerical and Statiscal Techniques part 14 ppsx

... 122 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 22. Evaluate: ∆n [sin (ax + b)]Sol. We know f∆(x) = f(x + h) – f (x) therefore∆sin (ax + b) = sin [a (x + h)+b] – sin)(ax ... hD hDhD116 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 14. Evaluate the following:I. ∆2 (cos 2x)II. ∆2 (3ex)III. ∆ tan–1 xIV. ∆(x + cos x)the interval of differencing ... f(x)∴∆(e a + bx)= e a( x + 1)+b – eax + b = eax + b (e a – 1)∴∆2 (e a + bx)= ∆ (∆e a + bx) = ∆ {eax + b (e a –1)}= (e a – 1) (∆ eax + b)= (e a – 1) eax + b (e a –1)= (e a –1)2...

A textbook of Computer Based Numerical and Statiscal Techniques part 16 ppsx

... Supposefor n + 1 equidistant argument values x = a, a + h, a + 2h, , a + nh, are given.140 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. For the given data, difference table is:∆∆ ∆ ∆−−−−−−∆∆∆∆∆∆∆∆∆23432233432543542656701211 ... 0f−+×−×+=138 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES i.e., 4f(3) = 124i.e., f(3) = 31(function values are 3n type and this is not a polynomial)Example 7. Find the missing value of the data:12 ... 2212nnnan bn c n a n b n a n−−++−∆+∆+∆142 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 9. Evaluate the production of wool in the year 1935 from the given data: ( ) 1931...

A textbook of Computer Based Numerical and Statiscal Techniques part 25 ppsx

... bc x a cx ac x a bx abfa fb fcab a c b a bc c acb−+ + −+ + −+ +++−− −− −−For maximum, we havef’(x)= 0i.e., f’(x)=()()()()()()()()()()()()222xbcfa xacfb xabfcabac babc cacb−+ ... =()()()()()()()()()()()()22 22 22b c fa c a fb a b fcfa b c fb c a fc a b−+−+−−+ −+ −. Proved.Example 5. Applying Lagrange’s formula, find a cubic polynomial which approximate the followingdatax : –2 –1 2 3y(x) ... required polynomial.Example 4. Value of f(x) are given at a, b, c. Show that the maximum is obtained by x = ()()()()()()()()()()()()fa b c fb c a fc a bfa b c fb c a fc a b−+ −+...

A textbook of Computer Based Numerical and Statiscal Techniques part 26 ppsx

... differenceswhere as the difference defined earlier are called ordinary differences. Lagrange’s interpolationformula has the disadvantage that if any other interpolation point were added, the interpolationco-efficient ... (5)244 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 10. Show that the sum of Lagrangian co-efficients is unity.Sol. Let()x∏ = ()()()01 nxx xx xx−− −The reciprocal of ... 12.46xfx[Ans. 8.337]242 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 7. Find the value of x correct to one decimal place for which y = 7Givenx:1 3 4y:41219Sol. Here we use Lagrange’s...

A textbook of Computer Based Numerical and Statiscal Techniques part 29 ppsx

... co-efficients a 1, a 2 a n will produce littleerror for small x near zero. But probably substantial error near the ends of the interval268 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply ... approximation to a given function.For evaluating a function f(x) on a computer it is generally more efficient of space and timeto have an analytic approximation to f(x) rather than to store a table ... polynomial. Therefore all the interpolation formulae shouldgive the same functional value.274 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 2. Obtain the best lower degree approximation...

A textbook of Computer Based Numerical and Statiscal Techniques part 51 ppsx

... cm===GGG486 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Negative LCL being taken as zero. Also for drawing the control chart we mark the sampleNo.’s along the horizontal axis and control ... 0.13]488 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The values in above computation are expressed in units of 0.01 ounces in excess of 10ounces. The actual value of UCL = 10.2084, and LCL ... assume that the claim of the girl about her sales is justified and that the increase in salesis normally distributed with mean µ = 700 and variance σ2. This assumption is called statisticalhypothesis....

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