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# Higher Mathematics Teaching Curriculum Model Based on Lagrangian Mathematical Model

###### Akzeptiert: 30 Mar 2022
Zeitschriftendaten
Format
Zeitschrift
eISSN
2444-8656
Erstveröffentlichung
01 Jan 2016
Erscheinungsweise
2 Hefte pro Jahr
Sprachen
Englisch
Introduction

Reliability theory is developing and improving rapidly. It has been paid more attention to geotechnical engineering and has been widely used [1]. Scholars go through the linear transformation method and the Rosenblatt transformation. Scholars have done a lot of research to consider the correlation between random variables. Some scholars have introduced the correlated variable independence method based on Cholesky decomposition into the stochastic response surface method. But the technique only works in the case of correlated standard variables. The most widely used is the Rackwitz-Fiessler transform. The transformation takes the marginal cumulative distribution function and covariance matrix of random variables as basic information. This paper adopts the Rackwitz-Fiessler probability transformation and merges this transformation with the Lagrange multiplier method. At the same time, this paper proposes a primary reliability method based on Rackwitz-Fiessler probability transformation and the Lagrange multiplier method [2]. In this paper, the rational polynomial method is introduced into the derivation of the inverse function expression of the distribution function. Finally, we demonstrate the effectiveness of the method proposed in this paper through a numerical example analysis.

Rackwitz-Fiessler transform

We use the principle of equal probability transformation to introduce the standard normal random variable Y = (y1, y2, ⋯, yn), as follows [3]: ${Φ(yi)=FI(xi)yi=Φ−1[Fi(xi)](i=1,⋯,n)$ \left\{ {\matrix{ {\Phi \left( {{y_i}} \right) = {F_I}\left( {{x_i}} \right)} \hfill \cr {{y_i} = {\Phi ^{ - 1}}\left[ {{F_i}\left( {{x_i}} \right)} \right]} \hfill \cr } } \right.\left( {i = 1, \cdots ,\,n} \right)

Φ( ) and Φ−1 ( ) are the standard normal cumulative distribution function and the inverse of the cumulative distribution function. We find the following rules based on the Rackwitz-Fiessler distribution theory [4]. When Y is a joint average random vector, we can derive the joint probability density function of X by using the implicit function derivation rule: $fx(x)=fx1(x1)fx2(x2)⋯fxn(xn)×φn(y,ρ0)φ(y1)φ(y1)⋯φ(yn)$ {f_x}\left( x \right) = {f_{x1}}\left( {{x_1}} \right){f_{x2}}\left( {{x_2}} \right) \cdots {f_{xn}}\left( {{x_n}} \right) \times {{{\varphi _n}\left( {y,{\rho _0}} \right)} \over {\varphi \left( {{y_1}} \right)\varphi \left( {{y_1}} \right) \cdots \varphi \left( {{y_n}} \right)}}

φ ( ) is the probability density function of the standard normal distribution. φn (y, ρ0) is a vector with zero equivalents, a matrix of unit standard deviations, and a correlation coefficient [5]. The matrix is the n dimensional standard normal distribution of ρ0 · ρ0 = [ρ0,ij] is the correlation coefficient matrix of the joint average random vector Y. The correlation coefficient ρij of the element ρ0,ij of ρ0 and X has the following relationship: $ρij=∫−∞∞∫−∞∞(xi−μiσi)(xj−μjσj)φ2(yi,yj,ρ0,ij)dyidyj$ {\rho _{ij}} = \int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\left( {{{{x_i} - {\mu _i}} \over {{\sigma _i}}}} \right)\left( {{{{x_j} - {\mu _j}} \over {{\sigma _j}}}} \right){\varphi _2}\left( {{y_i},{y_j},{\rho _{0,ij}}} \right)d{y_i}d{y_j}} }

φ2 (yi, yj, ρ0,ij is a two-dimensional standard normal distribution with zero equivalence vector, unit standard deviation matrix, and correlation coefficient matrix ρ0,ij. Analytic relations for ρij and ρ0,ij are difficult to give [6]. Through a large number of numerical studies, we have provided an empirical formula for calculating the equivalent correlation coefficient between two types of probability distributions with a total of 10 probability distributions: $ρ0,ij=Fρij$ {\rho _{0,ij}} = F{\rho _{ij}}

The coefficient F ≥ 1 in the formula. It is the correlation coefficient ρij and the marginal distribution FXi (xi). We can calculate it by the two-dimensional Gaussian-Hermite integral [7]. Then we use the nonlinear root-finding method to find the correlation coefficient ρ0,ij. The correlation coefficient matrix ρ0 = [ρ0,ij] of which we obtain the standard average random vector Y is symmetric. It can perform Cholesky decomposition. $ρ0=L0L0T$ {\rho _0} = {L_0}L_0^T

In the formula L0 is the lower triangular decomposition matrix obtained by the Cholesky decomposition of the correlation coefficient matrix ρ0. So we can get the positive transformation form of Rackwitz-Fiessler as follows: $u=L0−1Y=L0−1{y1y2⋮yn}=L0−1{Φ−1[F1(x1)]Φ−1[F2(x2)]⋮Φ−1[Fn(xn)]}$ u = L_0^{ - 1}Y = L_0^{ - 1}\left\{ {\matrix{ {{y_1}} \cr {{y_2}} \cr \vdots \cr {{y_n}} \cr } } \right\} = L_0^{ - 1}\left\{ {\matrix{ {{\Phi ^{ - 1}}\left[ {{F_1}\left( {{x_1}} \right)} \right]} \cr {{\Phi ^{ - 1}}\left[ {{F_2}\left( {{x_2}} \right)} \right]} \cr \vdots \cr {{\Phi ^{ - 1}}\left[ {{F_n}\left( {{x_n}} \right)} \right]} \cr } } \right\}

Lagrange Multiplier Method for Reliability Analysis
The geometric meaning of reliability indicators

We can obtain the following theorem according to the first-order second-moment theory of reliability analysis [89]. We can prove that this distance is an essential indicator in reliability analysis: reliability indicator β. We therefore assume that the limit state equation with n standard variables is expressed as follows: $z=g(x1,x2,⋯,xn)$ z = g\left( {{x_1},{x_2}, \cdots ,{x_n}} \right)

After normalizing the normal variables, we have the following relationship: $yi=(xi−μxi)/σxi$ {y_i} = \left( {{x_i} - {\mu _{xi}}} \right)/{\sigma _{xi}}

μxi, σxi are the mean and standard deviation of the variable, respectively [10]. Then the mathematical model of the expected average space calculation of the reliability index can be expressed as follows: ${β=min(∑i=1nyi2)1/2z=g(x1,x2,⋯,xn)=0$ \left\{ {\matrix{ {\beta = \min {{\left( {\sum\limits_{i = 1}^n {y_i^2} } \right)}^{1/2}}} \hfill \cr {z = g\left( {{x_1},{x_2}, \cdots ,{x_n}} \right) = 0} \hfill \cr } } \right.

If each random variable follows a general distribution, we can perform Gaussian transformation.

Lagrange Multiplier Method for Reliability Analysis

Suppose the inverse matrix of the lower triangular decomposition matrix L0 obtained by the Cholesky decomposition of the formula is expressed as follows: $L0−1={a11a12a22⋮a1na2n⋯ann}$ L_0^{ - 1} = \left\{ {\matrix{ {{a_{11}}} & {} & {} & {} \cr {{a_{12}}} & {{a_{22}}} & {} & {} \cr {} & \vdots & {} & {} \cr {{a_{1n}}} & {{a_{2n}}} & \cdots & {{a_{nn}}} \cr } } \right\}

The formula (6) can be rewritten as: ${u1u2⋮un}=L0−1{y1y2⋮yn}={a11y1∑i=12ai2yi⋮∑i=1nainyi}$ \left\{ {\matrix{ {{u_1}} \cr {{u_2}} \cr \vdots \cr {{u_n}} \cr } } \right\} = L_0^{ - 1}\left\{ {\matrix{ {{y_1}} \cr {{y_2}} \cr \vdots \cr {{y_n}} \cr } } \right\} = \left\{ {\matrix{ {{a_{11}}{y_1}} \cr {\sum\limits_{i = 1}^2 {{a_{i2}}{y_i}} } \cr \vdots \cr {\sum\limits_{i = 1}^n {{a_{in}}{y_i}} } \cr } } \right\}

Therefore, the standard regular space mathematical model of structural reliability index with correlation variables can be obtained from equation (9) as follows: ${β=min(∑i=1nui2)1/2=min[(a11y1)2+(∑i=1nainyi)2+⋯+(∑i=1nainyi)2]1/2z=g(x1,x2,⋯,xn)=0$ \left\{ {\matrix{ {\beta = \min {{\left( {\sum\limits_{i = 1}^n {u_i^2} } \right)}^{1/2}} = \min {{\left[ {{{\left( {{a_{11}}{y_1}} \right)}^2} + {{\left( {\sum\limits_{i = 1}^n {{a_{in}}{y_i}} } \right)}^2} + \cdots + {{\left( {\sum\limits_{i = 1}^n {{a_{in}}{y_i}} } \right)}^2}} \right]}^{1/2}}} \hfill \cr {z = g\left( {{x_1},{x_2}, \cdots ,{x_n}} \right) = 0} \hfill \cr } } \right.

$Ψ=(x1,x2,⋯,xn)=β2=∑i=1nui2$ \Psi = \left( {{x_1},{x_2}, \cdots ,{x_n}} \right) = {\beta ^2} = \sum\limits_{i = 1}^n {u_i^2} , then the formula (12) essentially becomes the problem of solving the extremum value of the function Ψ = (x1, x2,⋯, xn) under the condition z = g(x1, x2,⋯, xn) = 0. Therefore, according to the Lagrange multiplier method, the operation can be constructed as follows: $G(x1,x2,⋯,xn)=∑i=1nui2+λg(x1,x2,⋯,xn)$ G\left( {{x_1},{x_2},\cdots,{x_n}} \right) = \sum\limits_{i = 1}^n {u_i^2} + \lambda g\left( {{x_1},{x_2},\cdots,{x_n}} \right)

In the formula, ui is calculated according to procedure (11). λ is a constant. We take the first-order partial derivative of equation (13) with respect to x1, x2,⋯, xn. We make it zero [11]. Then we combine it with the equation g(x1, x2,⋯, xn) = 0 to get the following formula: ${Ψxn(xn)+λgxn(x1,x2,⋯,xn)=0g(x1,x2,⋯,xn)=0$ \left\{ {\matrix{ {{\Psi _{xn}}\left( {{x_n}} \right) + \lambda {g_{xn}}\left( {{x_1},\,{x_2},\,\cdots,\,{x_n}} \right) = 0} \hfill \cr {g\left( {{x_1},\,{x_2},\,\cdots,\,{x_n}} \right) = 0} \hfill \cr } } \right.

x1, x2,⋯, xn and λ can be solved by equation system (14), then x1, x2,⋯, xn is the coordinate (checkpoint coordinate) when the structural reliability index is the minimum value [12]. We solve the reliability index β of the structure from $(∑i=1nui2)1/2$ {\left( {\sum\limits_{i = 1}^n {u_i^2} } \right)^{1/2}} .

Rational Polynomial Method for Derivative Solving

Ψ / ∂xi in normal and lognormal distributions can be found directly. Since expressions for other distribution factor functions Φ−1 [F(x)] are difficult to give, it is difficult to solve the derivatives instantly. Because it is difficult for us to solve the derivative of the implicit or complex explicit function directly, we can use the rational polynomial method. The reasonable polynomial method is divided into the 3-point formulated, and the 5-point prepared [1314]. Since the 5-point formula has the best accuracy, the calculation example in this paper is solved by the 5-point formula method.

The value of the derivative of the function z = g(x1, x2,⋯, xn) at x* needs to be solved further. First, assume that the 5 x(i) (i = 1, 2, ⋯, 5) are known values. Where x(i) ∈ [x* − 3σx, x* + 3σx]. At this point we get 5 function values: $z11=g(x1(1),x2,⋯,xn),z12=g(x1(2),x2,⋯,xn),⋯,z15=g(x 1(5),x2,⋯,xn)$ {z_{11}} = g\left( {x_1^{\left( 1 \right)},{x_2}, \cdots ,{x_n}} \right),\,{z_{12}} = g\left( {x_1^{(2)},{x_2}, \cdots ,{x_n}} \right), \cdots ,{z_{15}} = g\left( {x{\kern 1pt} _1^{\left( 5 \right)},{x_2}, \cdots ,{x_n}} \right)

The function of the function concerning the variable x can be approximated by the rational polynomial method as a function continued fraction form: $z=g(x1(1),x2,⋯,xn)≈a1+x−x(1)a2+x−x(2)a3+⋯+x−x(4)a5$ z = g\left( {x_1^{\left( 1 \right)},{x_2}, \cdots ,{x_n}} \right) \approx {a_1} + {{x - {x^{\left( 1 \right)}}} \over {{a_2} + {{x - {x^{\left( 2 \right)}}} \over {{a_3} + \cdots + {{x - {x^{\left( 4 \right)}}} \over {{a_5}}}}}}}

(16) can also be written as: $z=g(x1,x2,⋯,xn)≈ϕ1(x)$ z = g\left( {{x_1},{x_2}, \cdots ,{x_n}} \right) \approx {\phi _1}\left( x \right)

In: $ϕi(x)=ai+(x−x(i))/ϕi+1(x)(i=1,2,⋯,4)$ \matrix{ {{\phi _i}\left( x \right) = {a_i} + \left( {x - {x^{\left( i \right)}}} \right)/{\phi _{i + 1}}\left( x \right)} \hfill \cr {\left( {i = 1,2, \cdots ,4} \right)} \hfill \cr } $ϕ5(x)=a5$ {\phi _5}\left( x \right) = {a_5}

The coefficient a1, a2,⋯, a5 in the above formula can be obtained from Table 1.

Rational polynomial coefficients for variable x

a1 a2 a3 a4 a5
a1 = z11 / / / /
a21 = z12 $a2=x(2)−x(1)a21−a1$ {a_2} = {{{x^{\left( 2 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{21}} - {a_1}}} / / /
a31 = z13 $a32=x(3)−x(1)a31−a1$ {a_{32}} = {{{x^{\left( 3 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{31}} - {a_1}}} $a3=x(3)−x(2)a31−a2$ {a_3} = {{{x^{\left( 3 \right)}} - {x^{\left( 2 \right)}}} \over {{a_{31}} - {a_2}}} / /
a41 = z14 $a42=x(4)−x(1)a41−a1$ {a_{42}} = {{{x^{\left( 4 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{41}} - {a_1}}} $a43=x(4)−x(2)a42−a2$ {a_{43}} = {{{x^{\left( 4 \right)}} - {x^{\left( 2 \right)}}} \over {{a_{42}} - {a_2}}} $a4=x(4)−x(3)a43−a3$ {a_4} = {{{x^{\left( 4 \right)}} - {x^{\left( 3 \right)}}} \over {{a_{43}} - {a_3}}} /
a51 = z15 $a52=x(5)−x(1)a51−a1$ {a_{52}} = {{{x^{\left( 5 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{51}} - {a_1}}} $a53=x(5)−x(2)a52−a2$ {a_{53}} = {{{x^{\left( 5 \right)}} - {x^{\left( 2 \right)}}} \over {{a_{52}} - {a_2}}} $a54=x(5)−x(3)a53−a3$ {a_{54}} = {{{x^{\left( 5 \right)}} - {x^{\left( 3 \right)}}} \over {{a_{53}} - {a_3}}} $a5=x(5)−x(4)a54−a4$ {a_5} = {{{x^{\left( 5 \right)}} - {x^{\left( 4 \right)}}} \over {{a_{54}} - {a_4}}}

Then from equations (16) to (19), we can get: $∂z∂x1=ϕ1(x1)=ϕ2(x1)−(x1−x1(1))ϕ2(x1)[ϕ2(x1)]2$ {{\partial z} \over {\partial {x_1}}} = {\phi _1}\left( {{x_1}} \right) = {{{\phi _2}\left( {{x_1}} \right) - \left( {{x_1} - x_1^{\left( 1 \right)}} \right){\phi _2}\left( {{x_1}} \right)} \over {{{\left[ {{\phi _2}\left( {{x_1}} \right)} \right]}^2}}} $ϕi′(x1)=ϕi+1(x1)−(x−x1(i))ϕi+1′(x1)[ϕi+1(x1)]2(i=1,2,⋯,4)$ \matrix{ {\phi _i^\prime\left( {{x_1}} \right) = {{{\phi _{i + 1}}\left( {{x_1}} \right) - \left( {x - x_1^{\left( i \right)}} \right)\phi _{i + 1}^\prime\left( {{x_1}} \right)} \over {{{\left[ {{\phi _{i + 1}}\left( {{x_1}} \right)} \right]}^2}}}} \hfill \cr {\left( {i = 1,2, \cdots ,4} \right)} \hfill \cr } Where $ϕi′(x1)$ \phi _i^\prime\left( {{x_1}} \right) is the first derivative of the function ϕi(x1) with respect to the independent variable x1. $ϕ5′(x1)=0$ \phi _5^\prime\left( {{x_1}} \right) = 0 can be obtained from equation (19). Then $ϕ4′(x1)=1/a5$ \phi _4^\prime\left( {{x_1}} \right) = 1/{a_5} can be obtained from $ϕ5′(x1)=0$ \phi _5^\prime\left( {{x_1}} \right) = 0 and equation (21). And so on to get the value of $ϕ3′(x1)$ \phi _3^\prime\left( {{x_1}} \right) , $ϕ2′(x1)$ \phi _2^\prime\left( {{x_1}} \right) .

Example analysis

The limit state equation $g(x)=x12−2x2,x1$ g\left( x \right) = x_1^2 - 2{x_2},{x_1} follows a lognormal distribution. Its mean and coefficient of variation are μx1 = 10, δx1 = 0.2, x2 respectively. It follows an extreme value type I distribution [1516]. Its mean and coefficient of variation are μx2 = 20, δx2 = 0.25, x1 respectively. The correlation coefficient ρ12 = 0.5 with x12. The empirical coefficient between the correlation coefficient ρ0,12 and ρ12 after the normalization of the quantity is expressed as follows $F=1.029+0.001ρ12+0.014δx2+0.004ρ122+0.233δx22−0.197ρ12δx2$ F = 1.029 + 0.001{\rho _{12}} + 0.014{\delta _{x2}} + 0.004\rho _{12}^2 + 0.233\delta _{x2}^2 - 0.197{\rho _{12}}{\delta _{x2}}

We can get F = 1.0239 by substituting ρ12 and 2 δx2 into equation (22). So we can get ρ0,12 = F ρ12 = 0.512. The lower triangular decomposition matrix L0 and its inverse matrix L-10 obtained by the Cholesky decomposition of the correlation coefficient matrix ρ0 are respectively expressed as follows

$L0=[1.0000.5120.859]$ {L_0} = \left[ {\matrix{ {1.00} & 0 \cr {0.512} & {0.859} \cr } } \right]

$L0−1=[1.000−0.59601.1641]$ L_0^{ - 1} = \left[ {\matrix{ {1.00} & 0 \cr { - 0.5960} & {1.1641} \cr } } \right]

We transform x1 and x2 into standard standardized variables with equal probability: $y1=[ln x1−ln(μx1/1+δx12)]/ln(1+δx12)$ {y_1} = \left[ {\ln \,{x_1} - \ln \left( {{\mu _{x1}}/\sqrt {1 + \delta _{x1}^2} } \right)} \right]/\sqrt {\ln \left( {1 + \delta _{x1}^2} \right)} $y2=Φ−1{exp[−exp(−α(x2−v))]}$ {y_2} = {\Phi ^{ - 1}}\left\{ {\exp \left[ { - \exp \left( { - \alpha \left( {{x_2} - v} \right)} \right)} \right]} \right\}

In the formula, the distribution parameter $α=π/(6σx2)=0.2565$ \alpha = \pi /\left( {\sqrt 6 {\sigma _{x2}}} \right) = 0.2565 , v = μx2 − 0.5770 / a = 17.7498 can be obtained from a procedure (11): ${u1u2}=L0−1{y1y2}={y1−0.5960y1+1.1641y2}$ \left\{ {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right\} = L_0^{ - 1}\left\{ {\matrix{ {{y_1}} \cr {{y_2}} \cr } } \right\} = \left\{ {\matrix{ {{y_1}} \hfill \cr { - 0.5960{y_1} + 1.1641{y_2}} \hfill \cr } } \right\}

Ψx1 (x1, x2) and Ψx2 (x1, x2) are solved using the rational polynomial 5-point method. The general method solves the structure-function function partial derivatives [17]. The final calculation results are shown in Table 2.

Comparison of calculation results between this method and other methods

Calculation method Checkpoint Function function Reliable indicator
x1 x2 g(x) β
Nataf transform 6.4348 20.7033 −3.6335E-05 2.6822
Linear Rackwitz-Fiessler Transform 6.4366 20.7148 −2.0729E-06 2.6822
Extended Algorithm
Normal distribution transformation 6.4622 20.8801 −1.3240E-06 2.6604
The method of this paper 6.4314 20.6815 1.2120E-07 2.6865

It can be seen from Table 2 that:

The verification points and reliability indexes obtained by the method in this paper are consistent with the calculation results of other methods.

The absolute value of the function obtained by the method in this paper and the absolute value of the function calculated by other methods are relatively small. This shows that this paper's calculation method has higher accuracy than other methods.

Conclusion

The Rackwitz-Fiessler transform and the Lagrange multiplier method are straightforward to calculate the structural reliability. Its calculation result is reasonable. Engineers may not program after obtaining the correlation coefficient ρ0,ij. In this paper, the use of spreadsheets can be achieved using Solver. Rackwitz-Fiessler transforms, and the Lagrange multiplier method does not need to calculate the equivalent average mean and standard deviation of variables when calculating structural reliability. There is no need for trial calculation and iterative calculation in the computation of this model. Engineering examples show the effectiveness of the method and the accuracy of the results. Therefore, the calculation method proposed in this paper has specific engineering practical significance. This model is a more straightforward calculation method worth promoting.

#### Comparison of calculation results between this method and other methods

Calculation method Checkpoint Function function Reliable indicator
x1 x2 g(x) β
Nataf transform 6.4348 20.7033 −3.6335E-05 2.6822
Linear Rackwitz-Fiessler Transform 6.4366 20.7148 −2.0729E-06 2.6822
Extended Algorithm
Normal distribution transformation 6.4622 20.8801 −1.3240E-06 2.6604
The method of this paper 6.4314 20.6815 1.2120E-07 2.6865

#### Rational polynomial coefficients for variable x

a1 a2 a3 a4 a5
a1 = z11 / / / /
a21 = z12 a2=x(2)x(1)a21a1 {a_2} = {{{x^{\left( 2 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{21}} - {a_1}}} / / /
a31 = z13 a32=x(3)x(1)a31a1 {a_{32}} = {{{x^{\left( 3 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{31}} - {a_1}}} a3=x(3)x(2)a31a2 {a_3} = {{{x^{\left( 3 \right)}} - {x^{\left( 2 \right)}}} \over {{a_{31}} - {a_2}}} / /
a41 = z14 a42=x(4)x(1)a41a1 {a_{42}} = {{{x^{\left( 4 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{41}} - {a_1}}} a43=x(4)x(2)a42a2 {a_{43}} = {{{x^{\left( 4 \right)}} - {x^{\left( 2 \right)}}} \over {{a_{42}} - {a_2}}} a4=x(4)x(3)a43a3 {a_4} = {{{x^{\left( 4 \right)}} - {x^{\left( 3 \right)}}} \over {{a_{43}} - {a_3}}} /
a51 = z15 a52=x(5)x(1)a51a1 {a_{52}} = {{{x^{\left( 5 \right)}} - {x^{\left( 1 \right)}}} \over {{a_{51}} - {a_1}}} a53=x(5)x(2)a52a2 {a_{53}} = {{{x^{\left( 5 \right)}} - {x^{\left( 2 \right)}}} \over {{a_{52}} - {a_2}}} a54=x(5)x(3)a53a3 {a_{54}} = {{{x^{\left( 5 \right)}} - {x^{\left( 3 \right)}}} \over {{a_{53}} - {a_3}}} a5=x(5)x(4)a54a4 {a_5} = {{{x^{\left( 5 \right)}} - {x^{\left( 4 \right)}}} \over {{a_{54}} - {a_4}}}

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