CN107273336A - A kind of curve-fitting method based on least square method - Google Patents

Abstract

The present invention relates to a kind of curve-fitting method based on least square method, this method sets up linear model on the basis of nonlinear fitting curve, each term coefficient in the function of matched curve is provided using the method for iterative calculation, coefficient is substituted into the theoretical value obtained after fitting function has the characteristics of error sum of squares is minimum, this method can be by way of accurate Iterative, the accurate each term coefficient for providing fitting function, makes the error sum of squares of fitting minimum；Present invention, avoiding the minimum variation issue that least square method during Linear Transformation is present, and the curve-fitting method that provides of the present invention, fitting precision is high, and logicality is strong, it is easy to program, and method is simple, it is easy to accomplish.

Description

A kind of curve-fitting method based on least square method

Technical field

The present invention relates to a kind of curve-fitting method based on least square method, belong to numerical analysis and reliability engineering neck Domain.

Background technology

There are many functions in scientific experimentation and engineer applied, its analytical expression is ignorant.Such as, inertia device Part can be varied from the extension of period of storage, its performance and parameter, but the change procedure is only capable of the side by experimental observation Method measures a series of node x_iOn value f_i.Now the problem is that seeking to test the approximating function y (x) of point sequence, or use geometry language It is exactly to seek a curve y (x) to be fitted (smooth) this n point for speech.

Traditional curve-fitting method is to use least square method, it is desirable to

Wherein,

Then y (x) is called f (x) on weight coefficient { ω_iLeast square approximation, and deserve to be called and state criterion and seek approximating function y (x) method is least square method.In above formula, { ω_j(x) it is } collection of functions of linear independence, { a_jIt is corresponding coefficient set.

When counting the fail data of inertia type instrument, failure distribution function curve is determined according to multiple test points and is predicted Life of product.Distribution function is generally comprised：Exponential distribution, Weibull distribution, the extreme value distribution, logarithm normal distribution etc..Due to this A little functions are all nonlinear equations, when solving the design parameter of distribution function, first model are linearized, then pass through minimum Square law solves parameter.Bibliography (" accelerometer storage life evaluation " under temperature stress,《Armored force engineering institute Journal》, Vol.28No.3, P28 in 2014) in also make referrals to this method.Such as, by taking Weibull distribution as an example, its distributed model For

Wherein, t is time variable, and m and η are Stationary Parameter.

It is directly more difficult according to measured value solution parameter m and η because Weibull distribution is nonlinear equation, therefore, can be right It enters line translation.Define A=[ln (T)-I_k×1], Y=ln (- ln (1-F)), wherein, T=[t₁,t₂,…,t_n]^T, F=[f₁, f₂,…,f_n]^T, following formula can be transformed to

Ln (- ln (1-F))=mln (t)-mln (η)

It is transformed to

According to multiple test points by least square method, it can obtain

Parameter m and η can be solved according to above formula, and then provide fitting result.

With bibliography (" accelerometer storage life evaluation " under temperature stress,《Armored force engineering institute journal》, Vol.28 No.3, P28 in 2014) in table 1 exemplified by, if inertia type instrument --- accelerometer carried out under conditions of 25 DEG C from The fail data (sample size sampled every time during detection is 100) of right storage environment experiment, result of the test is

The test failure data statistics situation of table 1

Using the equation after above-mentioned change, m=1.7618596, η=49.748074 can be solved.Be illustrated in figure 2 to " * " point is the test data in table 1 in matched curve after fixed measurement sequence and its linear transformation, figure, and " o " point is fitting Value.As can be seen that the two still has larger gap.

Reason is analyzed, mainly least square method can only ensureMost It is small, and it cannot be guaranteed thatIt is minimum.

Exist for this reason, it may be necessary to study one kindCurve-fitting method under the conditions of least commitment.

The content of the invention

It is an object of the invention to the drawbacks described above for overcoming prior art, there is provided a kind of curve plan based on least square method Conjunction method, this method by way of accurate Iterative, can accurately provide each term coefficient of fitting function, make the mistake of fitting Poor quadratic sum is minimum.

What the above-mentioned purpose of the present invention was mainly achieved by following technical solution：

A kind of curve-fitting method based on least square method, comprises the following steps：

(1), preset time sequence { t_iCorresponding measurement sequence { f_i, i=1,2 ..., n；

(2) the iunction for curve F (a relevant with the time, is determined₁,a₂,…,a_j, t), wherein：a₁,a₂,…,a_jTo wait to ask The Stationary Parameter of solution, t is time variable, and j is positive integer；

(3) the Stationary Parameter a to be solved, is given₁,a₂,…,a_jInitial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)；

(4), according to the initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁,a₂,…,a_j, t) carry out linear Change is handled, and sets up linearity error model；

(5) according to measurement sequence { f_iAnd the linearity error model, Δ a is solved using least square method₁,Δa₂,…, Δa_j, and judge Δ a₁,Δa₂,…,Δa_jWhether satisfaction is required, step (8) is entered if requirement is met, otherwise into step (6)；The Δ a₁,Δa₂,…,Δa_jStationary Parameter a respectively to be solved₁,a₂,…,a_jAnd the deviation of actual value；

(6) according to the deviation delta a₁,Δa₂,…,Δa_jTo initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)It is updated；

(7) return to step (4), according to the initial value a after renewal₁₍₀₎,a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁, a₂,…,a_j, linearization process t) is carried out, linearity error model is set up；Circulate according to this, until solving obtained deviation delta a₁,Δ a₂,…,Δa_jMeet and require, into step (8)；

(8) according to the deviation delta a for meeting and requiring₁,Δa₂..., Δ a is to current initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)Enter The a that row is obtained after updating₁₍₀₎,a₂₍₀₎,…,a_j(0), Stationary Parameter a as to be solved₁,a₂,…,a_j。

In the above-mentioned curve-fitting method based on least square method, according to the initial value a in the step (4)₁₍₀₎, a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁,a₂,…,a_j, linearization process t) is carried out, the linearity error model of foundation is represented It is as follows：

In the above-mentioned curve-fitting method based on least square method, according to measurement sequence { f in the step (5)_iAnd The linearity error model, Δ a is solved using least square method₁,Δa₂,…,Δa_jSpecific method it is as follows：

According to measurement sequence { f_iAnd the linearity error model to provide n rank error formulas as follows：

Δ a is solved using least square method₁,Δa₂,…,Δa_jValue；

Wherein：

In the above-mentioned curve-fitting method based on least square method, according to the deviation delta a in the step (6)₁,Δ a₂,…,Δa_jTo initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)The method being updated is：

a_{1 (0) updates}=a₁₍₀₎+Δa₁, a_{2 (0) update}=a₂₍₀₎+Δa₂..., a_{J (0) updates}=a_j(0)+Δa_j。

Obtained deviation delta a is solved in the above-mentioned curve-fitting method based on least square method, in the step (7)₁, Δa₂,…,Δa_jRequirement is met to refer to meet simultaneously：

|Δa₁|≤1×10^-6, | Δ a₂|≤1×10^-6..., | Δ a_j|≤1×10^-6。

In the above-mentioned curve-fitting method based on least square method, meet what is required according to described in the step (8) Deviation delta a₁,Δa₂..., Δ a is to current initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)The method being updated is：

a_{1 (0) updates}=a₁₍₀₎+Δa₁, a_{2 (0) update}=a₂₍₀₎+Δa₂..., a_{J (0) updates}=a_j(0)+Δa_j。

The present invention having the beneficial effect that compared with prior art：

(1), the invention provides a kind of curve-fitting method of least square method, on the basis of nonlinear fitting curve Linear model is set up, each term coefficient in the function of matched curve is provided using the method for iterative calculation, coefficient is substituted into fitting letter The theoretical value obtained after number has the characteristics of error sum of squares is minimum, and this method can be accurate by way of accurate Iterative Each term coefficient of fitting function is really provided, makes the error sum of squares of fitting minimum.

(2), the curve-fitting method based on least square method that the present invention is provided is a kind of suitable for nonlinear function Curve matching new method, with error sum of squares it is minimum the characteristics of, it is to avoid least square method is present during Linear Transformation Minimum variation issue；

(3), the present invention solves the coefficient of fitting function using iterative calculation, and solve that nonlinear equation can not solve asks Topic；

(4), the curve-fitting method that the present invention is provided, fitting precision is high, and logicality is strong, it is easy to program, and method is simple, It is easily achieved.

Brief description of the drawings

Fig. 1 is the curve-fitting method flow chart of the invention based on least square method；

Fig. 2 is the matched curve after given measurement sequence and its linear transformation；

Fig. 3 be the embodiment of the present invention in use m of the Weibull distribution after iteration₀Change procedure figure；

Fig. 4 be the embodiment of the present invention in use η of the Weibull distribution after iteration₀Change procedure figure；

Fig. 5 be the embodiment of the present invention in use Δ m change procedure figure of the Weibull distribution after iteration；

Fig. 6 be the embodiment of the present invention in use Δ η change procedure figure of the Weibull distribution after iteration；

Fig. 7 is the matched curve in the embodiment of the present invention after iterative process.

Embodiment

The present invention is described in further detail with specific embodiment below in conjunction with the accompanying drawings：

The curve-fitting method for the least square method that the present invention is provided, sets up linear on the basis of nonlinear fitting curve Model, each term coefficient in the function of matched curve is provided using the method for iterative calculation, is obtained after coefficient is substituted into fitting function Theoretical value the characteristics of have error sum of squares minimum.

It is as shown in Figure 1 the flow chart of the curve-fitting method of the invention based on least square method, it is of the invention as seen from the figure Following steps are specifically included by the curve-fitting method based on least square method：

(1) time series { t, is obtained_iCorresponding measurement sequence { f_i, i=1,2 ..., n；

(2) the iunction for curve F (a relevant with the time, is determined₁,a₂,…,a_j, t), wherein：a₁,a₂,…,a_jTo wait to ask The Stationary Parameter of solution, t is time variable, and j is the positive integer more than or equal to 1；

(3) the Stationary Parameter a to be solved, is given₁,a₂,…,a_jInitial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)；

(4), according to the initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁,a₂,…,a_j, t) carry out linear Change is handled, and sets up linearity error model, specific as follows：

Wherein：Δa₁,Δa₂,…,Δa_jStationary Parameter a respectively to be solved₁,a₂,…,a_jWith corresponding actual value Deviation；

(5) the measurement sequence { f in step (1)_iAnd step (4) in linearity error model, using least square Method solves Δ a₁,Δa₂,…,Δa_j, the Δ a₁,Δa₂,…,Δa_jFor Stationary Parameter a to be solved₁,a₂,…,a_jWith it is true The deviation of value；Specific method is as follows：

According to measurement sequence { f_iAnd the linearity error model to provide n rank error formulas as follows：

Δ a is solved using least square method₁,Δa₂,…,Δa_jValue；

Wherein：

And judge Δ a₁,Δa₂,…,Δa_jWhether satisfaction is required, step (8) is entered if requirement is met, and is otherwise entered and is walked Suddenly (6)；

(6) according to the deviation delta a₁,Δa₂,…,Δa_jTo initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)It is updated, is specially：

a_{1 (0) updates}=a₁₍₀₎+Δa₁, a_{2 (0) update}=a₂₍₀₎+Δa₂..., a_{J (0) updates}=a_j(0)+Δa_j。

(7) repeat step (4), (5), according to the initial value a after renewal_{1 (0) updates},a_{2 (0) update},…,a_{J (0) updates}To curve matching letter Number F (a₁,a₂,…,a_j, linearization process t) is carried out, linearity error model is set up, afterwards according to measurement sequence { f_iAnd linearly miss Differential mode type, Δ a is solved using least square method₁,Δa₂,…,Δa_j, and judge Δ a₁,Δa₂,…,Δa_jWhether satisfaction is required, Require to enter step (8) if meeting, otherwise return to step (6), i.e.,：By the Δ a of solution₁,Δa₂,…,Δa_jAfter once updating Initial value a_{1 (0) updates},a_{2 (0) update},…,a_{J (0) updates}Updated again, obtain secondary renewal initial value a_{1 (0) updates 2},a_{2 (0) update 2},…, a_{J (0) updates 2}, step (4), (5) are repeated, according to the initial value a after secondary renewal_{1 (0) updates 2},a_{2 (0) update 2},…,a_{J (0) updates 2}Located Reason, is circulated according to this, until solving obtained deviation delta a₁,Δa₂,…,Δa_jMeet and require, into step (8)；

The step of this in the embodiment of the present invention solves obtained deviation delta a₁,Δa₂,…,Δa_jRequirement is met to refer to simultaneously completely Foot：

|Δa₁|≤1×10^-6, | Δ a₂|≤1×10^-6..., | Δ a_j|≤1×10^-6。

(8) the deviation delta a required according to meeting₁,Δa₂..., Δ a is to current initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)(i.e. upper one Initial value a after secondary renewal₁₍₀₎,a₂₍₀₎,…,a_j(0)) be updated after obtained a₁₍₀₎,a₂₍₀₎,…,a_j(0), it is as to be solved Stationary Parameter a₁,a₂,…,a_j。

The deviation delta a required in the step according to meeting₁,Δa₂..., Δ a is to current initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)Enter Row update method be：

a_{1 (0) updates}=a₁₍₀₎+Δa₁, a_{2 (0) update}=a₂₍₀₎+Δa₂..., a_{J (0) updates}=a_j(0)+Δa_j。

If given initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0), obtained deviation delta a is calculated for the first time₁,Δa₂..., Δ a is to meet It is required that, then by deviation delta a₁,Δa₂..., Δ a is to giving initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)It is updated, obtains to be solved determining Normal parameter a₁,a₂,…,a_j。

The data of test can be carried out curve fitting in actual applications using the inventive method, for being carried out to equipment Life appraisal, judges the term of validity and storage life with pre- measurement equipment.

Embodiment 1

With bibliography (" accelerometer storage life evaluation " under temperature stress,《Armored force engineering institute journal》, Vol.28 No.3, P28 in 2014) in table 1 exemplified by, if inertia type instrument --- accelerometer carried out under conditions of 25 DEG C from The fail data (sample size sampled every time during detection is 100) of right storage environment experiment, result of the test is

The test failure data statistics situation of table 1

Using Weibull distributionWhen, ln (- ln (1-F)) is solved by least square method Parameter m and η provide the solution of step (3) in=m ln (T)-m ln (η)：M=1.7618596, η=49.748074.Matched curve As shown in Fig. 2 " * " point is the test data in table 1 in figure, " o " point is match value.As can be seen that the two still has larger Gap.Now,

Using the curve-fitting method based on least square method of the present invention, m₀=1.7618596, η₀=49.748074 As initial value, by solving equation

Provide the variable Δ m and Δ η in step (5).

According to the inventive method, after 18 iterative process, finally there is m₀=3.3282777, η₀=32.085768, its In, m₀Iterative process as shown in figure 3, Fig. 3 be the embodiment of the present invention in use m of the Weibull distribution after iteration₀Changed Cheng Tu；η₀Iterative process as shown in figure 4, Fig. 4 be the embodiment of the present invention in use η of the Weibull distribution after iteration₀Change Procedure chart.It is illustrated in figure 5 in the embodiment of the present invention and uses Δ m change procedure figure of the Weibull distribution after iteration；Fig. 6 institutes It is shown as in the embodiment of the present invention using Δ η change procedure figure of the Weibull distribution after iteration.Δ m, Δ η pass through as seen from the figure Level off to 0 after 18 iteration, meet exhausted angle value≤1 × 10^-6Requirement.

Two parameter m₀、η₀Substitute into and be fitted in Weibull Function, as a result as shown in fig. 7, Fig. 7 is the present invention Matched curve in embodiment after iterative process, the matched curve of where the dotted line signifies that conventional method, " * " dotted line is experiment The matched curve of data, " o " dotted line is the matched curve that the inventive method is obtained, as seen from the figure the inventive method and experiment The matched curve of data coincide very much.Now,As can be seen that the value is small by one The individual order of magnitude.

It is described above, it is only an embodiment of the invention, but protection scope of the present invention is not limited thereto, and is appointed What those familiar with the art the invention discloses technical scope in, the change or replacement that can be readily occurred in, all It should be included within the scope of the present invention.

Unspecified part of the present invention belongs to general knowledge as well known to those skilled in the art.

Claims (6)

1. a kind of curve-fitting method based on least square method, it is characterised in that：Comprise the following steps：

(1), preset time sequence { t_iCorresponding measurement sequence { f_i, i=1,2 ..., n；

(2) the iunction for curve F (a relevant with the time, is determined₁,a₂,…,a_j, t), wherein：a₁,a₂,…,a_jTo be to be solved Stationary Parameter, t is time variable, and j is positive integer；

(3) the Stationary Parameter a to be solved, is given₁,a₂,…,a_jInitial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)；

(4), according to the initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁,a₂,…,a_j, t) carry out at linearisation Reason, sets up linearity error model；

(5) according to measurement sequence { f_iAnd the linearity error model, Δ a is solved using least square method₁,Δa₂,…,Δa_j, And judge Δ a₁,Δa₂,…,Δa_jWhether satisfaction is required, step (8) is entered if requirement is met, otherwise into step (6)；Institute State Δ a₁,Δa₂,…,Δa_jStationary Parameter a respectively to be solved₁,a₂,…,a_jAnd the deviation of actual value；

(6) according to the deviation delta a₁,Δa₂,…,Δa_jTo initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)It is updated；

(7) return to step (4), according to the initial value a after renewal₁₍₀₎,a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁,a₂,…, a_j, linearization process t) is carried out, linearity error model is set up；Circulate according to this, until solving obtained deviation delta a₁,Δa₂,…, Δa_jMeet and require, into step (8)；

(8) according to the deviation delta a for meeting and requiring₁,Δa₂..., Δ a is to current initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)Carry out more The a obtained after new₁₍₀₎,a₂₍₀₎,…,a_j(0), Stationary Parameter a as to be solved₁,a₂,…,a_j。

2. the curve-fitting method according to claim 1 based on least square method, it is characterised in that：The step (4) It is middle according to the initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)To iunction for curve F (a₁,a₂,…,a_j, linearization process t) is carried out, is built Vertical linearity error model is expressed as follows：

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3. the curve-fitting method according to claim 1 based on least square method, it is characterised in that：The step (5) It is middle according to measurement sequence { f_iAnd the linearity error model, Δ a is solved using least square method₁,Δa₂,…,Δa_jIt is specific Method is as follows：

According to measurement sequence { f_iAnd the linearity error model to provide n rank error formulas as follows：

<mrow> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mn>1</mn> </msub> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mn>2</mn> </msub> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mtable> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mi>n</mi> </msub> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mo>...</mo> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> </mrow> </mfrac> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mo>...</mo> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> </mrow> </mfrac> </mtd> </mtr> <mtr> <mtd> <mtable> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> </mtable> </mtd> <mtd> <mtable> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> </mtable> </mtd> <mtd> <mtable> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> </mtable> </mtd> <mtd> <mtable> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mo>...</mo> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> </mrow> </mfrac> </mtd> </mtr> </mtable> </mfenced> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;a</mi> <mn>1</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;a</mi> <mn>2</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mtable> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;a</mi> <mi>j</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>

Δ a is solved using least square method₁,Δa₂,…,Δa_jValue；

<mrow> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;a</mi> <mn>1</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;a</mi> <mn>2</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mtable> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;a</mi> <mi>j</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <msup> <mrow> <mo>(</mo> <msup> <mi>A</mi> <mi>T</mi> </msup> <mi>A</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>A</mi> <mi>T</mi> </msup> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mn>1</mn> </msub> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mo>...</mo> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mn>2</mn> </msub> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mo>...</mo> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mtable> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mi>n</mi> </msub> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mo>...</mo> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>

Wherein：

<mrow> <mi>A</mi> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mo>...</mo> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>1</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> </mrow> </mfrac> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mo>...</mo> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mn>2</mn> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> </mrow> </mfrac> </mtd> </mtr> <mtr> <mtd> <mtable> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> </mtable> </mtd> <mtd> <mtable> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> <mtr> <mtd> <mo>.</mo> </mtd> </mtr> </mtable> </mtd> <mtd> <mtable> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> </mtable> </mtd> <mtd> <mtable> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> </mtr> </mtable> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mtd> <mtd> <mo>...</mo> </mtd> <mtd> <mfrac> <mrow> <mo>&amp;part;</mo> <mi>F</mi> <mrow> <mo>(</mo> <mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>a</mi> <mrow> <mi>j</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>,</mo> <msub> <mi>t</mi> <mi>n</mi> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> </mrow> </mfrac> </mtd> </mtr> </mtable> </mfenced> <mo>.</mo> </mrow>

4. the curve-fitting method according to claim 1 based on least square method, it is characterised in that：The step (6) It is middle according to the deviation delta a₁,Δa₂,…,Δa_jTo initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)The method being updated is：

a_{1 (0) updates}=a₁₍₀₎+Δa₁, a_{2 (0) update}=a₂₍₀₎+Δa₂..., a_{J (0) updates}=a_j(0)+Δa_j。

5. the curve-fitting method according to claim 1 based on least square method, it is characterised in that：The step (7) It is middle to solve obtained deviation delta a₁,Δa₂,…,Δa_jRequirement is met to refer to meet simultaneously：

|Δa₁|≤1×10^-6, | Δ a₂|≤1×10^-6..., | Δ a_j|≤1×10^-6。

6. the curve-fitting method according to claim 1 based on least square method, it is characterised in that：The step (8) The middle deviation delta a required according to the satisfaction₁,Δa₂..., Δ a is to current initial value a₁₍₀₎,a₂₍₀₎,…,a_j(0)The side being updated Method is：

a_{1 (0) updates}=a₁₍₀₎+Δa₁, a_{2 (0) update}=a₂₍₀₎+Δa₂..., a_{J (0) updates}=a_j(0)+Δa_j。