Method Error
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Method Error Definition Physics
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What Is Method Error In Physics
Answers Community Guidelines Leaderboard Knowledge Partners Points & Levels Blog Safety Tips Science & Mathematics Physics Next What is the definition of method error in physics? Follow 1 answer 1 Report Abuse Are you sure you want to delete this answer? Yes No Sorry, something has gone wrong. Trending Now Tim McGraw Israel Houghton Aly Raisman Sarah Palin method error examples Conor McGregor iPhone 7 Kim Kardashian Psoriatic Arthritis Symptoms Meg White Credit Cards Answers Best Answer: The term 'method error' is not a standard physics term. It may be used by examiners though when marking examination papers and indicates the candidate has used an incorrect method in a calculation even though, by accident, the answer is correct. Source(s): Steve4Physics · 4 years ago 0 Thumbs up 1 Thumbs down Comment Add a comment Submit · just now Asker's rating Report Abuse Add your answer What is the definition of method error in physics? Add your answer Source Submit Cancel Report Abuse I think this question violates the Community Guidelines Chat or rant, adult content, spam, insulting other members,show more I think this question violates the Terms of Service Harm to minors, violence or threats, harassment or privacy invasion, impersonation or misrepresentation, fraud or phishing, show more Additional Details If you believe your intellectual property has been infringed and would like to file a complaint, please see our Copyright/IP Policy Repo
rounding and truncation are introduced. It is important to have a notion of their nature and their order. A newly developed method is worthless without an error analysis. Neither does it make sense to
Instrument Error
use methods which introduce errors with magnitudes larger than the effects to be personal error measured or simulated. On the other hand, using a method with very high accuracy might be computationally too expensive to absolute error justify the gain in accuracy. Contents 1 Accuracy and Precision 2 Absolute Error 3 Relative Error 4 Sources of Error 4.1 Truncation Error 4.2 Roundoff Error Accuracy and Precision[edit] Measurements and calculations https://answers.yahoo.com/question/index?qid=20121216122155AAZsNe7 can be characterized with regard to their accuracy and precision. Accuracy refers to how closely a value agrees with the true value. Precision refers to how closely values agree with each other. The following figures illustrate the difference between accuracy and precision. In the first figure, the given values (black dots) are more accurate; whereas in the second figure, the given values are more precise. The https://en.wikibooks.org/wiki/Numerical_Methods/Errors_Introduction term error represents the imprecision and inaccuracy of a numerical computation. Accuracy Precision Absolute Error[edit] Absolute Error is the magnitude of the difference between the true value x and the approximate value xa, Therefore absolute error=[x-xa] The error between two values is defined as ϵ a b s = ∥ x ~ − x ∥ , {\displaystyle \epsilon _{abs}=\left\|{\tilde {x}}-x\right\|\quad ,} where x {\displaystyle x} denotes the exact value and x ~ {\displaystyle {\tilde {x}}} its approximation. Relative Error[edit] The relative error of x ~ {\displaystyle {\tilde {x}}} is the absolute error relative to the exact value. Look at it this way: if your measurement has an error of ± 1 inch, this seems to be a huge error when you try to measure something which is 3 in. long. However, when measuring distances on the order of miles, this error is mostly negligible. The definition of the relative error is ϵ r e l = ∥ x ~ − x ∥ ∥ x ∥ . {\displaystyle \epsilon _{rel}={\frac {\left\|{\tilde {x}}-x\right\|}{\left\|x\right\|}}\quad .} Sources of Error[edit] In a numerical computation, error may arise because of the following reasons: Truncation error Roundoff error Truncation Error[edit] Truncation error r
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Suppose we have an approximation of the root xn which has an error of (r - xn). What is the error of the next approximation xn + 1 found after one iteration of Newton's method? Suppose r is the actual root of f(x). Then from the Taylor series, we have that: where ξ ∈ [r, xn]. Note, however, that f(r) = 0, so if we set the left-hand side to zero and divide both sides by f(1)(xn), we get: We can bring the first two terms to the left-hand side and multiple each side by -1. For the next step, I will group two of the terms on the terms on the left-hand side: Note that the object in the parentheses on the left-hand side is, by definition, xn + 1 (after all, xn + 1 = xn - f(xn)/f(1)(xn) ), and thus we have: But the left hand side is the error of xn + 1, and therefore we see that error is reduced by a scalar multiple of the square of the previous error. To demonstrate this, let us find the root of f(x) = ex - 2 starting with x0 = 1. We note that the 1st and 2nd derivatives of f(x) are equal, so we will approximate ½f(2)(ξ)/f(1)(xn) by ½. Table 1 shows the Newton iterates, their absolute errors, and the approximation of the error based on the square previous error. Table 1. Newton iterates in finding a root of f(x) = ex - 2. n xn errn = ln(2) - xn ½ errn - 12 01.0-3.069 ⋅ 10-1N/A 10.735758882342885-4.261 ⋅ 10-2 -4.708 ⋅ 10-2 20.694042299918915-8.951 ⋅ 10-4 -9.079 ⋅ 10-4 30.693147581059771-4.005 ⋅ 10-7 -4.006 ⋅ 10-7 40.693147180560025-8.016 ⋅ 10-14-8.020 ⋅ 10-14 Note that the error at the nth step is very closely approximated by the error of the (n - 1)th step. Now, in reality, we do not know what the actual error is (otherwise, we wouldn't be using Newton's method, would we?