Yeo Doglas (Singapore, Independent Researcher)
Published in matho.philica.com Abstract I have derived a theorem similar to that of Maclaurin’s expansion in usage, but very much different from it in other ways. Its fundamental principle is entirely different from that of Maclaurin’s expansion. This is a more mathematical paper, than scientific. Article body
Theorem: f(x+nh) = f(x) + h f'(x+h) + h f'(x+2h) + h f' (x+3h) + … + h f' [x+(n-1)h] Where h is small, and n is a large integer. (n, h, and thus nh are constants) The derivation of this theorem is based on calculus, and is not very difficult, but it is too tedious to be listed down here. This theorem looks similar to Maclaurin's expansion, but is different in the following ways: 1) Maclaurin's expansion is an infinite series, while this is finite. 2) Maclaurin's expansion requires the second derivative, and 3 ^{rd} derivative, and so on. This theorem only requires the first derivative. 3) The Maclaurin's series is only accurate for small values of x, unless the full series is to be expanded. This theorem is accurate for large values of x and nh, the only requirement is for h to be small. 4) In the Maclaurin's series, the coefficient is calculated using the binomial theorem, with factorials that can turn out to be quite large. In this theorem, the coefficient is always h, which can be chosen arbitrarily. It is similar to Maclaurin's expansion in the following ways: 1) Both can be used to find f(x) of any function. 2) Both are only approximations, unless the full series is calculated. Applications of this theorem: 1) Find the increase in the value of the function f(x), when x is increased by a certain constant. The constant need not be small. In fact, it can be very large. For example, Let f(x)=x_{}^{2} To find f(x+100), Firstly, let x=10. Thus, nh=10 Let h=10^-5 And n=10^7 LHS=f(10+100)= (10+100)^2=12100
RHS= f(x) +h f'(x+h) + h f'(x+2h) + … + h f'[x+(n-1)h] = 10^2 + 2h[x+h+x+2h+x+3h+…+x+(n-1)h] = 100 + 2h [ (n-1)/2 * (20+nh) ] ——————————- (arithmetic series) = 100+ 12000 =12100 This shows that the above expansion is very accurate. (Thus Increase in f(x) = 12100 - 1000) Hence, we can find the increase in f(x), when x is increased by a certain amount nh, which can be larger than x. If h is chosen to be very small, the approximation will be very accurate. This is more convenient than the Maclaurin's expansion, as only the first derivative need to be found. 2) The second application is to find f(x) itself, like in the Maclaurin's expansion. Using the previous example of f(x)=x^2, To find f(25), Simply let n=1000, and h=0.025 Hence nh = 25 f(x+25) = f(x) + h f' (x+25) + h f' (x+2h) + … + h f' [x+(n-1)h] = f(x) + 2h [ (n-1)/2 * (2x+25) ] ——————- using formula for arithmetic series Letting x=0, f(25) = 2(0.025)(12487.5) = 624.375 This is pretty close to the value of 625. Greater accuracy could be obtained by using smaller values of h, but making sure that nh=25. Using this theorem, it is possible to obtain accurate values of both f(x) and f(x+nh), and only the first derivative is required. Furthermore, in simple functions such as x^2, the sum of the series is an algebraic progression, which can be greatly simplified. Using the above method, similar expansions for sin x, cos x, e^x, ln (1+x), and (1+x)^n can be derived.
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**Peer-review ratings** (from 4 reviews, where a score of 100 represents the ‘average’ level)**:** Originality = **92.19**, importance = **10.94**, overall quality = **15.62** This Article was published on 24th July, 2006 at 11:56:44 and has been viewed 6125 times.
1 Peer review [reviewer #8814] added 25th July, 2006 at 18:03:12 A math article that says the derivation “is not very difficult, but it is too tedious to be listed down here.” is neither math nor an article. I would guess (Why do I have to guess?) that the result follows from using the trapezoid rule to integrate f’. The derivation could easily be written down, but would show the triviality of the result. The expansion is virtually useless, because it would generally be easier to calculate f(x+nh) than n first derivatives. In contrast to the Maclaurin expansion, all n derivatives would have to be calculated to find f(x+nh).
Originality: **5**, Importance: **1**, Overall quality: **1**
2 added 20th August, 2006 at 15:33:53 The theorem was actually derived using the fundamental definition of the derivative:
f’(x)=lim [f(x+h)-f(x)]/h
as you can see, the idea of ‘h’ in ‘nh’ comes from here.
This theorem is no doubt less useful than Maclaurin’s theorem, but it could be useful in cases whereby only the first derivative is known. Maclaurin’s theorem requires the use of all derivatives up to the nth derivative,
3 Peer review [reviewer #82274] added 30th August, 2006 at 23:31:32 It is really too bad that the presentation is so awful. Was this submitted before the LaTeX system started? Also, perhaps English is not the first language of the author, but it needs drastic improvement.
Originality: **4**, Importance: **1**, Overall quality: **1**
4 Peer review [reviewer #10906] added 3rd October, 2006 at 03:56:18 O.K., I’ll believe this is original. Despite the author’s claim that it comes from the definition of the derivative, I agree with the first reviewer that it would be easiest to derive it using the Fundamental Theorem of Calculus and the Trapezoidal Rule. (So why not use Simpson’s Rule to make the obvious improvement?)
It’s not that useful a result computationally, and the exposition makes it hard to follow by anyone who does not already know Calculus.
Originality: **3**, Importance: **1**, Overall quality: **1**
5 Peer review [reviewer #31786] added 4th October, 2006 at 17:58:12 I agree with the comments of the previous reviewers.
Additionally, I also question the claim of the author that his expression has a larger domain of validity than a McLaurin expansion. That is comparing apples to oranges, as the author’s expression is fundamentally a quadrature formula as opposed to a real series expansion.
Originality: **3**, Importance: **1**, Overall quality: **2** |