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How To Find The Difference Quotient Of A Function

Expression in calculus

In single-variable calculus, the difference quotient is ordinarily the name for the expression

f ( ten + h ) f ( x ) h {\displaystyle {\frac {f(x+h)-f(x)}{h}}}

which when taken to the limit as h approaches 0 gives the derivative of the function f.[1] [2] [3] [iv] The name of the expression stems from the fact that it is the quotient of the difference of values of the role by the difference of the corresponding values of its argument (the latter is (x + h) - x = h in this case).[5] [half-dozen] The divergence caliber is a measure of the average rate of change of the function over an interval (in this case, an interval of length h).[vii] [viii] : 237 [nine] The limit of the divergence quotient (i.e., the derivative) is thus the instantaneous charge per unit of modify.[9]

By a slight alter in notation (and viewpoint), for an interval [a, b], the difference quotient

f ( b ) f ( a ) b a {\displaystyle {\frac {f(b)-f(a)}{b-a}}}

is called[5] the mean (or average) value of the derivative of f over the interval [a, b]. This name is justified by the mean value theorem, which states that for a differentiable function f, its derivative f′ reaches its mean value at some point in the interval.[5] Geometrically, this difference quotient measures the slope of the secant line passing through the points with coordinates (a, f(a)) and (b, f(b)).[10]

Difference quotients are used as approximations in numerical differentiation,[8] but they have besides been subject of criticism in this application.[11]

Departure quotients may too notice relevance in applications involving Time discretization, where the width of the fourth dimension pace is used for the value of h.

The difference quotient is sometimes also chosen the Newton quotient [ten] [12] [xiii] [14] (after Isaac Newton) or Fermat's departure quotient (afterwards Pierre de Fermat).[15]

Overview [edit]

The typical notion of the difference caliber discussed above is a particular case of a more full general concept. The main vehicle of calculus and other higher mathematics is the part. Its "input value" is its argument, usually a point ("P") expressible on a graph. The difference between two points, themselves, is known as their Delta (ΔP), as is the difference in their office result, the detail note being adamant past the direction of formation:

  • Forward deviation:  ΔF(P) = F(P + ΔP) − F(P);
  • Primal difference:  δF(P) = F(P + ½ΔP) − F(P − ½ΔP);
  • Backward difference: ∇F(P) = F(P) − F(P − ΔP).

The general preference is the forward orientation, as F(P) is the base, to which differences (i.east., "ΔP"due south) are added to it. Furthermore,

The role difference divided by the point difference is known as "divergence quotient":

Δ F ( P ) Δ P = F ( P + Δ P ) F ( P ) Δ P = F ( P + Δ P ) Δ P . {\displaystyle {\frac {\Delta F(P)}{\Delta P}}={\frac {F(P+\Delta P)-F(P)}{\Delta P}}={\frac {\nabla F(P+\Delta P)}{\Delta P}}.\,\!}

If ΔP is infinitesimal, then the difference quotient is a derivative, otherwise it is a divided difference:

If | Δ P | = ι : Δ F ( P ) Δ P = d F ( P ) d P = F ( P ) = G ( P ) ; {\displaystyle {\text{If }}|\Delta P|={\mathit {\iota }}:\quad {\frac {\Delta F(P)}{\Delta P}}={\frac {dF(P)}{dP}}=F'(P)=M(P);\,\!}
If | Δ P | > ι : Δ F ( P ) Δ P = D F ( P ) D P = F [ P , P + Δ P ] . {\displaystyle {\text{If }}|\Delta P|>{\mathit {\iota }}:\quad {\frac {\Delta F(P)}{\Delta P}}={\frac {DF(P)}{DP}}=F[P,P+\Delta P].\,\!}

Defining the point range [edit]

Regardless if ΔP is infinitesimal or finite, at that place is (at least—in the case of the derivative—theoretically) a point range, where the boundaries are P ± (0.5) ΔP (depending on the orientation—ΔF(P), δF(P) or ∇F(P)):

LB = Lower Boundary;   UB = Upper Boundary;

Derivatives tin exist regarded as functions themselves, harboring their ain derivatives. Thus each function is home to sequential degrees ("higher orders") of derivation, or differentiation. This property can be generalized to all divergence quotients.
As this sequencing requires a corresponding boundary splintering, it is practical to break up the signal range into smaller, equi-sized sections, with each department being marked past an intermediary point (P i ), where LB = P 0 and UB = P ń , the nth point, equaling the caste/order:

          LB =  P0          = P0          + 0ΔoneP     = Pń          − (Ń-0)Δ1P;         P1          = P0          + 1ΔoneP     = Pń          − (Ń-1)Δ1P;         P2          = P0          + 2ΔiP     = Pń          − (Ń-2)ΔoneP;         P3          = P0          + 3ΔaneP     = Pń          − (Ń-three)ΔoneP;             ↓      ↓        ↓       ↓        Pń-3          = P0          + (Ń-three)Δ1P = Pń          − 3Δ1P;        Pń-2          = P0          + (Ń-2)ΔaneP = Pń          − 2ΔaneP;        Pń-1          = P0          + (Ń-1)Δ1P = Pń          − 1ΔaneP;   UB = Pń-0          = P0          + (Ń-0)Δ1P = Pń          − 0ΔaneP = Pń;        
          ΔP = ΔaneP = P1          − P0          = Pii          − Pane          = P3          − P2          = ... = Pń          − Pń-1;        
          ΔB = UB − LB = Pń          − P0          = ΔńP = ŃΔoneP.        

The primary difference quotient (Ń = i) [edit]

Δ F ( P 0 ) Δ P = F ( P n ´ ) F ( P 0 ) Δ north ´ P = F ( P 1 ) F ( P 0 ) Δ ane P = F ( P one ) F ( P 0 ) P 1 P 0 . {\displaystyle {\frac {\Delta F(P_{0})}{\Delta P}}={\frac {F(P_{\astute {n}})-F(P_{0})}{\Delta _{\acute {due north}}P}}={\frac {F(P_{one})-F(P_{0})}{\Delta _{1}P}}={\frac {F(P_{1})-F(P_{0})}{P_{1}-P_{0}}}.\,\!}

As a derivative [edit]

The departure quotient as a derivative needs no explanation, other than to point out that, since P0 essentially equals P1 = P2 = ... = Pń (as the differences are infinitesimal), the Leibniz annotation and derivative expressions do not distinguish P to P0 or Pń:
d F ( P ) d P = F ( P 1 ) F ( P 0 ) d P = F ( P ) = G ( P ) . {\displaystyle {\frac {dF(P)}{dP}}={\frac {F(P_{ane})-F(P_{0})}{dP}}=F'(P)=Yard(P).\,\!}

There are other derivative notations, but these are the well-nigh recognized, standard designations.

As a divided deviation [edit]

A divided difference, however, does require further elucidation, as it equals the average derivative between and including LB and UB:
P ( t n ) = L B + T Due north 1 U T 1 Δ B = U B U T T N U T i Δ B ; . ( P ( 1 ) = L B , P ( u t ) = U B ) . F ( P a ~ ) = F ( 50 B < P < U B ) = T North = one U T = F ( P ( t n ) ) U T . {\displaystyle {\begin{aligned}P_{(tn)}&=LB+{\frac {TN-ane}{UT-1}}\Delta B\ =UB-{\frac {UT-TN}{UT-1}}\Delta B;\\[10pt]&{}\qquad {\color {white}.}(P_{(1)}=LB,\ P_{(ut)}=UB){\color {white}.}\\[10pt]F'(P_{\tilde {a}})&=F'(LB<P<UB)=\sum _{TN=1}^{UT=\infty }{\frac {F'(P_{(tn)})}{UT}}.\end{aligned}}}
In this interpretation, Pã represents a office extracted, average value of P (midrange, but usually not exactly midpoint), the detail valuation depending on the part averaging it is extracted from. More formally, Pã is constitute in the mean value theorem of calculus, which says:
For any function that is continuous on [LB,UB] and differentiable on (LB,UB) in that location exists some Pã in the interval (LB,UB) such that the secant joining the endpoints of the interval [LB,UB] is parallel to the tangent at Pã.
Essentially, Pã denotes some value of P between LB and UB—hence,
P a ~ := Fifty B < P < U B = P 0 < P < P north ´ {\displaystyle P_{\tilde {a}}:=LB<P<UB=P_{0}<P<P_{\acute {n}}\,\!} P_{{\tilde  {a}}}:=LB<P<UB=P_{0}<P<P_{{\acute  {n}}}\,\!
which links the mean value event with the divided difference:
D F ( P 0 ) D P = F [ P 0 , P 1 ] = F ( P 1 ) F ( P 0 ) P ane P 0 = F ( P 0 < P < P 1 ) = T N = 1 U T = F ( P ( t due north ) ) U T , = D F ( L B ) D B = Δ F ( L B ) Δ B = F ( U B ) Δ B , = F [ L B , U B ] = F ( U B ) F ( Fifty B ) U B Fifty B , = F ( L B < P < U B ) = G ( Fifty B < P < U B ) . {\displaystyle {\begin{aligned}{\frac {DF(P_{0})}{DP}}&=F[P_{0},P_{1}]={\frac {F(P_{1})-F(P_{0})}{P_{i}-P_{0}}}=F'(P_{0}<P<P_{ane})=\sum _{TN=1}^{UT=\infty }{\frac {F'(P_{(tn)})}{UT}},\\[8pt]&={\frac {DF(LB)}{DB}}={\frac {\Delta F(LB)}{\Delta B}}={\frac {\nabla F(UB)}{\Delta B}},\\[8pt]&=F[LB,UB]={\frac {F(UB)-F(LB)}{UB-LB}},\\[8pt]&=F'(LB<P<UB)=G(LB<P<UB).\end{aligned}}} {\begin{aligned}{\frac  {DF(P_{0})}{DP}}&=F[P_{0},P_{1}]={\frac  {F(P_{1})-F(P_{0})}{P_{1}-P_{0}}}=F'(P_{0}<P<P_{i})=\sum _{{TN=1}}^{{UT=\infty }}{\frac  {F'(P_{{(tn)}})}{UT}},\\[8pt]&={\frac  {DF(LB)}{DB}}={\frac  {\Delta F(LB)}{\Delta B}}={\frac  {\nabla F(UB)}{\Delta B}},\\[8pt]&=F[LB,UB]={\frac  {F(UB)-F(LB)}{UB-LB}},\\[8pt]&=F'(LB<P<UB)=G(LB<P<UB).\end{aligned}}
As there is, by its very definition, a tangible deviation betwixt LB/P0 and UB/Pń, the Leibniz and derivative expressions do require divarication of the part argument.

Higher-order difference quotients [edit]

2nd society [edit]

Δ 2 F ( P 0 ) Δ 1 P 2 = Δ F ( P 0 ) Δ ane P = Δ F ( P 1 ) Δ 1 P Δ F ( P 0 ) Δ 1 P Δ 1 P , = F ( P 2 ) F ( P 1 ) Δ one P F ( P 1 ) F ( P 0 ) Δ 1 P Δ one P , = F ( P 2 ) 2 F ( P 1 ) + F ( P 0 ) Δ one P two ; {\displaystyle {\begin{aligned}{\frac {\Delta ^{ii}F(P_{0})}{\Delta _{1}P^{2}}}&={\frac {\Delta F'(P_{0})}{\Delta _{ane}P}}={\frac {{\frac {\Delta F(P_{1})}{\Delta _{one}P}}-{\frac {\Delta F(P_{0})}{\Delta _{i}P}}}{\Delta _{ane}P}},\\[10pt]&={\frac {{\frac {F(P_{two})-F(P_{ane})}{\Delta _{1}P}}-{\frac {F(P_{1})-F(P_{0})}{\Delta _{1}P}}}{\Delta _{1}P}},\\[10pt]&={\frac {F(P_{2})-2F(P_{1})+F(P_{0})}{\Delta _{1}P^{2}}};\end{aligned}}}
d 2 F ( P ) d P ii = d F ( P ) d P = F ( P one ) F ( P 0 ) d P , = d G ( P ) d P = Thousand ( P i ) Chiliad ( P 0 ) d P , = F ( P 2 ) 2 F ( P i ) + F ( P 0 ) d P two , = F ( P ) = Grand ( P ) = H ( P ) {\displaystyle {\brainstorm{aligned}{\frac {d^{2}F(P)}{dP^{2}}}&={\frac {dF'(P)}{dP}}={\frac {F'(P_{1})-F'(P_{0})}{dP}},\\[10pt]&=\ {\frac {dG(P)}{dP}}={\frac {G(P_{1})-G(P_{0})}{dP}},\\[10pt]&={\frac {F(P_{2})-2F(P_{ane})+F(P_{0})}{dP^{two}}},\\[10pt]&=F''(P)=Yard'(P)=H(P)\terminate{aligned}}}
D 2 F ( P 0 ) D P 2 = D F ( P 0 ) D P = F ( P 1 < P < P 2 ) F ( P 0 < P < P 1 ) P 1 P 0 , . F ( P 1 ) F ( P 0 ) P 1 P 0 , = F [ P 0 , P 1 , P 2 ] = F ( P 2 ) 2 F ( P 1 ) + F ( P 0 ) ( P one P 0 ) two , = F ( P 0 < P < P 2 ) = T Northward = i F ( P ( t n ) ) U T , = G ( P 0 < P < P 2 ) = H ( P 0 < P < P 2 ) . {\displaystyle {\begin{aligned}{\frac {D^{2}F(P_{0})}{DP^{2}}}&={\frac {DF'(P_{0})}{DP}}={\frac {F'(P_{1}<P<P_{ii})-F'(P_{0}<P<P_{1})}{P_{1}-P_{0}}},\\[10pt]&{\color {white}.}\qquad \neq {\frac {F'(P_{1})-F'(P_{0})}{P_{1}-P_{0}}},\\[10pt]&=F[P_{0},P_{1},P_{2}]={\frac {F(P_{ii})-2F(P_{1})+F(P_{0})}{(P_{1}-P_{0})^{2}}},\\[10pt]&=F''(P_{0}<P<P_{2})=\sum _{TN=1}^{\infty }{\frac {F''(P_{(tn)})}{UT}},\\[10pt]&=G'(P_{0}<P<P_{2})=H(P_{0}<P<P_{2}).\end{aligned}}}

Tertiary order [edit]

Δ iii F ( P 0 ) Δ ane P 3 = Δ 2 F ( P 0 ) Δ 1 P two = Δ F ( P 0 ) Δ one P = Δ F ( P 1 ) Δ 1 P Δ F ( P 0 ) Δ 1 P Δ 1 P , = Δ F ( P 2 ) Δ 1 P Δ F ( P ane ) Δ 1 P Δ 1 P Δ F ( P 1 ) Δ 1 P Δ F ( P 0 ) Δ one P Δ i P Δ 1 P , = F ( P 3 ) 2 F ( P 2 ) + F ( P one ) Δ ane P 2 F ( P 2 ) 2 F ( P 1 ) + F ( P 0 ) Δ 1 P 2 Δ 1 P , = F ( P iii ) 3 F ( P 2 ) + 3 F ( P 1 ) F ( P 0 ) Δ 1 P 3 ; {\displaystyle {\begin{aligned}{\frac {\Delta ^{3}F(P_{0})}{\Delta _{ane}P^{3}}}&={\frac {\Delta ^{two}F'(P_{0})}{\Delta _{1}P^{2}}}={\frac {\Delta F''(P_{0})}{\Delta _{1}P}}={\frac {{\frac {\Delta F'(P_{1})}{\Delta _{1}P}}-{\frac {\Delta F'(P_{0})}{\Delta _{1}P}}}{\Delta _{1}P}},\\[10pt]&={\frac {{\frac {{\frac {\Delta F(P_{2})}{\Delta _{1}P}}-{\frac {\Delta F'(P_{1})}{\Delta _{1}P}}}{\Delta _{i}P}}-{\frac {{\frac {\Delta F'(P_{one})}{\Delta _{1}P}}-{\frac {\Delta F'(P_{0})}{\Delta _{1}P}}}{\Delta _{1}P}}}{\Delta _{1}P}},\\[10pt]&={\frac {{\frac {F(P_{iii})-2F(P_{ii})+F(P_{1})}{\Delta _{1}P^{2}}}-{\frac {F(P_{2})-2F(P_{1})+F(P_{0})}{\Delta _{1}P^{2}}}}{\Delta _{1}P}},\\[10pt]&={\frac {F(P_{iii})-3F(P_{2})+3F(P_{ane})-F(P_{0})}{\Delta _{ane}P^{3}}};\end{aligned}}}
d iii F ( P ) d P 3 = d ii F ( P ) d P 2 = d F ( P ) d P = F ( P i ) F ( P 0 ) d P , = d 2 M ( P ) d P ii = d G ( P ) d P = G ( P 1 ) G ( P 0 ) d P , . = d H ( P ) d P = H ( P ane ) H ( P 0 ) d P , = G ( P 2 ) 2 1000 ( P 1 ) + G ( P 0 ) d P 2 , = F ( P three ) 3 F ( P 2 ) + 3 F ( P ane ) F ( P 0 ) d P 3 , = F ( P ) = One thousand ( P ) = H ( P ) = I ( P ) ; {\displaystyle {\brainstorm{aligned}{\frac {d^{3}F(P)}{dP^{iii}}}&={\frac {d^{2}F'(P)}{dP^{2}}}={\frac {dF''(P)}{dP}}={\frac {F''(P_{1})-F''(P_{0})}{dP}},\\[10pt]&={\frac {d^{two}G(P)}{dP^{2}}}\ ={\frac {dG'(P)}{dP}}\ ={\frac {Grand'(P_{ane})-K'(P_{0})}{dP}},\\[10pt]&{\color {white}.}\qquad \qquad \ \ ={\frac {dH(P)}{dP}}\ ={\frac {H(P_{1})-H(P_{0})}{dP}},\\[10pt]&={\frac {G(P_{two})-2G(P_{1})+M(P_{0})}{dP^{two}}},\\[10pt]&={\frac {F(P_{three})-3F(P_{two})+3F(P_{1})-F(P_{0})}{dP^{three}}},\\[10pt]&=F'''(P)=G''(P)=H'(P)=I(P);\end{aligned}}}
D three F ( P 0 ) D P three = D 2 F ( P 0 ) D P 2 = D F ( P 0 ) D P = F ( P 1 < P < P 3 ) F ( P 0 < P < P 2 ) P i P 0 , . F ( P 1 ) F ( P 0 ) P 1 P 0 , = F ( P 2 < P < P three ) F ( P 1 < P < P 2 ) P 1 P 0 F ( P i < P < P ii ) F ( P 0 < P < P 1 ) P 1 P 0 P one P 0 , = F ( P two < P < P 3 ) 2 F ( P i < P < P two ) + F ( P 0 < P < P 1 ) ( P 1 P 0 ) 2 , = F [ P 0 , P one , P 2 , P 3 ] = F ( P 3 ) 3 F ( P 2 ) + 3 F ( P 1 ) F ( P 0 ) ( P 1 P 0 ) 3 , = F ( P 0 < P < P 3 ) = T Due north = ane U T = F ( P ( t n ) ) U T , = M ( P 0 < P < P 3 ) = H ( P 0 < P < P 3 ) = I ( P 0 < P < P 3 ) . {\displaystyle {\begin{aligned}{\frac {D^{three}F(P_{0})}{DP^{iii}}}&={\frac {D^{2}F'(P_{0})}{DP^{two}}}={\frac {DF''(P_{0})}{DP}}={\frac {F''(P_{1}<P<P_{three})-F''(P_{0}<P<P_{2})}{P_{1}-P_{0}}},\\[10pt]&{\colour {white}.}\qquad \qquad \qquad \qquad \qquad \ \ \neq {\frac {F''(P_{1})-F''(P_{0})}{P_{1}-P_{0}}},\\[10pt]&={\frac {{\frac {F'(P_{2}<P<P_{3})-F'(P_{1}<P<P_{2})}{P_{ane}-P_{0}}}-{\frac {F'(P_{ane}<P<P_{2})-F'(P_{0}<P<P_{ane})}{P_{1}-P_{0}}}}{P_{1}-P_{0}}},\\[10pt]&={\frac {F'(P_{ii}<P<P_{3})-2F'(P_{1}<P<P_{2})+F'(P_{0}<P<P_{1})}{(P_{1}-P_{0})^{2}}},\\[10pt]&=F[P_{0},P_{1},P_{ii},P_{three}]={\frac {F(P_{3})-3F(P_{2})+3F(P_{1})-F(P_{0})}{(P_{1}-P_{0})^{3}}},\\[10pt]&=F'''(P_{0}<P<P_{3})=\sum _{TN=ane}^{UT=\infty }{\frac {F'''(P_{(tn)})}{UT}},\\[10pt]&=K''(P_{0}<P<P_{3})\ =H'(P_{0}<P<P_{3})=I(P_{0}<P<P_{3}).\terminate{aligned}}}

Nthursday order [edit]

Δ n ´ F ( P 0 ) = F ( n ´ 1 ) ( P 1 ) F ( north ´ 1 ) ( P 0 ) , = F ( n ´ 2 ) ( P 2 ) F ( n ´ 2 ) ( P 1 ) Δ 1 P F ( due north ´ 2 ) ( P 1 ) F ( n ´ 2 ) ( P 0 ) Δ 1 P , = F ( n ´ 3 ) ( P 3 ) F ( n ´ 3 ) ( P 2 ) Δ i P F ( n ´ 3 ) ( P 2 ) F ( n ´ iii ) ( P 1 ) Δ 1 P Δ one P . F ( n ´ three ) ( P ii ) F ( n ´ 3 ) ( P 1 ) Δ one P F ( n ´ iii ) ( P ane ) F ( n ´ 3 ) ( P 0 ) Δ one P Δ one P , = {\displaystyle {\begin{aligned}\Delta ^{\acute {n}}F(P_{0})&=F^{({\astute {north}}-1)}(P_{1})-F^{({\acute {n}}-1)}(P_{0}),\\[10pt]&={\frac {F^{({\acute {northward}}-2)}(P_{ii})-F^{({\acute {due north}}-2)}(P_{1})}{\Delta _{1}P}}-{\frac {F^{({\acute {n}}-2)}(P_{i})-F^{({\acute {due north}}-2)}(P_{0})}{\Delta _{1}P}},\\[10pt]&={\frac {{\frac {F^{({\astute {due north}}-3)}(P_{three})-F^{({\acute {n}}-3)}(P_{ii})}{\Delta _{i}P}}-{\frac {F^{({\astute {n}}-3)}(P_{2})-F^{({\astute {northward}}-3)}(P_{1})}{\Delta _{ane}P}}}{\Delta _{1}P}}\\[10pt]&{\color {white}.}\qquad -{\frac {{\frac {F^{({\acute {n}}-iii)}(P_{ii})-F^{({\astute {due north}}-3)}(P_{ane})}{\Delta _{i}P}}-{\frac {F^{({\acute {n}}-3)}(P_{ane})-F^{({\acute {north}}-3)}(P_{0})}{\Delta _{1}P}}}{\Delta _{1}P}},\\[10pt]&=\cdots \end{aligned}}}
Δ n ´ F ( P 0 ) Δ i P north ´ = I = 0 N ´ ( 1 Northward ´ I ) ( N ´ I ) F ( P 0 + I Δ 1 P ) Δ 1 P due north ´ ; n ´ F ( P n ´ ) Δ i P n ´ = I = 0 Due north ´ ( one I ) ( N ´ I ) F ( P northward ´ I Δ i P ) Δ 1 P northward ´ ; {\displaystyle {\begin{aligned}{\frac {\Delta ^{\acute {due north}}F(P_{0})}{\Delta _{1}P^{\acute {northward}}}}&={\frac {\sum _{I=0}^{\acute {N}}{-1 \choose {\astute {N}}-I}{{\acute {N}} \cull I}F(P_{0}+I\Delta _{1}P)}{\Delta _{1}P^{\acute {northward}}}};\\[10pt]&{\frac {\nabla ^{\acute {n}}F(P_{\astute {n}})}{\Delta _{ane}P^{\acute {n}}}}\\[10pt]&={\frac {\sum _{I=0}^{\acute {N}}{-1 \cull I}{{\acute {Due north}} \cull I}F(P_{\astute {n}}-I\Delta _{1}P)}{\Delta _{ane}P^{\acute {northward}}}};\end{aligned}}}
d n ´ F ( P 0 ) d P n ´ = d northward ´ 1 F ( P 0 ) d P n ´ one = d northward ´ ii F ( P 0 ) d P northward ´ 2 = d northward ´ three F ( P 0 ) d P due north ´ three = = d due north ´ r F ( r ) ( P 0 ) d P n ´ r , = d n ´ 1 Yard ( P 0 ) d P northward ´ 1 = d n ´ 2 Grand ( P 0 ) d P n ´ ii = d due north ´ 3 G ( P 0 ) d P n ´ iii = = d due north ´ r G ( r 1 ) ( P 0 ) d P due north ´ r , . = d n ´ 2 H ( P 0 ) d P n ´ two = d northward ´ 3 H ( P 0 ) d P n ´ 3 = = d north ´ r H ( r 2 ) ( P 0 ) d P north ´ r , . = d n ´ 3 I ( P 0 ) d P n ´ 3 = = d due north ´ r I ( r 3 ) ( P 0 ) d P n ´ r , = F ( n ´ ) ( P ) = G ( n ´ i ) ( P ) = H ( northward ´ 2 ) ( P ) = I ( n ´ 3 ) ( P ) = {\displaystyle {\begin{aligned}{\frac {d^{\acute {n}}F(P_{0})}{dP^{\astute {n}}}}&={\frac {d^{{\astute {n}}-1}F'(P_{0})}{dP^{{\acute {n}}-i}}}={\frac {d^{{\astute {north}}-2}F''(P_{0})}{dP^{{\astute {n}}-2}}}={\frac {d^{{\acute {n}}-iii}F'''(P_{0})}{dP^{{\astute {n}}-three}}}=\cdots ={\frac {d^{{\astute {n}}-r}F^{(r)}(P_{0})}{dP^{{\acute {n}}-r}}},\\[10pt]&={\frac {d^{{\astute {north}}-1}K(P_{0})}{dP^{{\acute {n}}-1}}}\\[10pt]&={\frac {d^{{\astute {due north}}-2}Yard'(P_{0})}{dP^{{\acute {northward}}-two}}}=\ {\frac {d^{{\acute {northward}}-3}G''(P_{0})}{dP^{{\acute {n}}-3}}}=\cdots ={\frac {d^{{\acute {n}}-r}One thousand^{(r-one)}(P_{0})}{dP^{{\astute {due north}}-r}}},\\[10pt]&{\color {white}.}\qquad \qquad \qquad ={\frac {d^{{\acute {n}}-2}H(P_{0})}{dP^{{\acute {n}}-ii}}}=\ {\frac {d^{{\acute {n}}-3}H'(P_{0})}{dP^{{\acute {n}}-3}}}=\cdots ={\frac {d^{{\acute {n}}-r}H^{(r-2)}(P_{0})}{dP^{{\acute {n}}-r}}},\\&{\colour {white}.}\qquad \qquad \qquad \qquad \qquad \qquad \ =\ {\frac {d^{{\acute {n}}-3}I(P_{0})}{dP^{{\acute {northward}}-3}}}=\cdots ={\frac {d^{{\acute {n}}-r}I^{(r-three)}(P_{0})}{dP^{{\acute {north}}-r}}},\\[10pt]&=F^{({\acute {n}})}(P)=G^{({\astute {n}}-1)}(P)=H^{({\acute {n}}-2)}(P)=I^{({\acute {north}}-three)}(P)=\cdots \end{aligned}}}
D northward ´ F ( P 0 ) D P n ´ = F [ P 0 , P 1 , P 2 , P iii , , P north ´ 3 , P n ´ two , P n ´ i , P n ´ ] , = F ( n ´ ) ( P 0 < P < P n ´ ) = T Due north = 1 U T = F ( n ´ ) ( P ( t n ) ) U T = F ( n ´ ) ( 50 B < P < U B ) = Grand ( due north ´ 1 ) ( L B < P < U B ) = {\displaystyle {\brainstorm{aligned}{\frac {D^{\acute {n}}F(P_{0})}{DP^{\acute {n}}}}&=F[P_{0},P_{ane},P_{ii},P_{3},\ldots ,P_{{\acute {north}}-3},P_{{\acute {n}}-2},P_{{\acute {northward}}-1},P_{\acute {n}}],\\[10pt]&=F^{({\acute {n}})}(P_{0}<P<P_{\acute {northward}})=\sum _{TN=i}^{UT=\infty }{\frac {F^{({\acute {n}})}(P_{(tn)})}{UT}}\\[10pt]&=F^{({\acute {n}})}(LB<P<UB)=G^{({\acute {n}}-1)}(LB<P<UB)=\cdots \end{aligned}}} {\brainstorm{aligned}{\frac  {D^{{\acute  {due north}}}F(P_{0})}{DP^{{\astute  {n}}}}}&=F[P_{0},P_{1},P_{2},P_{3},\ldots ,P_{{{\acute  {n}}-3}},P_{{{\astute  {n}}-2}},P_{{{\astute  {north}}-ane}},P_{{\acute  {n}}}],\\[10pt]&=F^{{({\astute  {due north}})}}(P_{0}<P<P_{{\astute  {north}}})=\sum _{{TN=1}}^{{UT=\infty }}{\frac  {F^{{({\astute  {north}})}}(P_{{(tn)}})}{UT}}\\[10pt]&=F^{{({\astute  {n}})}}(LB<P<UB)=G^{{({\astute  {northward}}-1)}}(LB<P<UB)=\cdots \end{aligned}}

Applying the divided difference [edit]

The quintessential application of the divided difference is in the presentation of the definite integral, which is nothing more a finite difference:

L B U B G ( p ) d p = L B U B F ( p ) d p = F ( U B ) F ( L B ) , = F [ 50 B , U B ] Δ B , = F ( L B < P < U B ) Δ B , = G ( L B < P < U B ) Δ B . {\displaystyle {\brainstorm{aligned}\int _{LB}^{UB}G(p)\,dp&=\int _{LB}^{UB}F'(p)\,dp=F(UB)-F(LB),\\[10pt]&=F[LB,UB]\Delta B,\\[10pt]&=F'(LB<P<UB)\Delta B,\\[10pt]&=\ Yard(LB<P<UB)\Delta B.\end{aligned}}} {\brainstorm{aligned}\int _{{LB}}^{{UB}}G(p)\,dp&=\int _{{LB}}^{{UB}}F'(p)\,dp=F(UB)-F(LB),\\[10pt]&=F[LB,UB]\Delta B,\\[10pt]&=F'(LB<P<UB)\Delta B,\\[10pt]&=\ Grand(LB<P<UB)\Delta B.\end{aligned}}

Given that the mean value, derivative expression form provides all of the same information as the classical integral note, the mean value form may be the preferable expression, such as in writing venues that only support/have standard ASCII text, or in cases that just require the average derivative (such as when finding the average radius in an elliptic integral). This is peculiarly true for definite integrals that technically have (eastward.m.) 0 and either π {\displaystyle \pi \,\!} or 2 π {\displaystyle ii\pi \,\!} as boundaries, with the same divided difference establish equally that with boundaries of 0 and π ii {\displaystyle {\begin{matrix}{\frac {\pi }{2}}\finish{matrix}}} (thus requiring less averaging attempt):

0 ii π F ( p ) d p = 4 0 π 2 F ( p ) d p = F ( 2 π ) F ( 0 ) = 4 ( F ( π 2 ) F ( 0 ) ) , = 2 π F [ 0 , 2 π ] = 2 π F ( 0 < P < ii π ) , = 2 π F [ 0 , π 2 ] = 2 π F ( 0 < P < π 2 ) . {\displaystyle {\begin{aligned}\int _{0}^{2\pi }F'(p)\,dp&=four\int _{0}^{\frac {\pi }{2}}F'(p)\,dp=F(two\pi )-F(0)=four(F({\begin{matrix}{\frac {\pi }{2}}\terminate{matrix}})-F(0)),\\[10pt]&=2\pi F[0,2\pi ]=2\pi F'(0<P<2\pi ),\\[10pt]&=two\pi F[0,{\begin{matrix}{\frac {\pi }{2}}\terminate{matrix}}]=2\pi F'(0<P<{\begin{matrix}{\frac {\pi }{2}}\terminate{matrix}}).\end{aligned}}}

This also becomes particularly useful when dealing with iterated and multiple integrals (ΔA = AU − AL, ΔB = BU − BL, ΔC = CU − CL):

C L C U B L B U A L A U F ( r , q , p ) d p d q d r = T C = i U C = ( T B = 1 U B = ( T A = one U A = F ( R ( t c ) : Q ( t b ) : P ( t a ) ) Δ A U A ) Δ B U B ) Δ C U C , = F ( C L < R < C U : B L < Q < B U : A L < P < A U ) Δ A Δ B Δ C . {\displaystyle {\begin{aligned}&{}\qquad \int _{CL}^{CU}\int _{BL}^{BU}\int _{AL}^{AU}F'(r,q,p)\,dp\,dq\,dr\\[10pt]&=\sum _{T\!C=ane}^{U\!C=\infty }\left(\sum _{T\!B=1}^{U\!B=\infty }\left(\sum _{T\!A=i}^{U\!A=\infty }F^{'}(R_{(tc)}:Q_{(tb)}:P_{(ta)}){\frac {\Delta A}{U\!A}}\correct){\frac {\Delta B}{U\!B}}\correct){\frac {\Delta C}{U\!C}},\\[10pt]&=F'(C\!L<R<CU:BL<Q<BU:AL<P<\!AU)\Delta A\,\Delta B\,\Delta C.\end{aligned}}} {\begin{aligned}&{}\qquad \int _{{CL}}^{{CU}}\int _{{BL}}^{{BU}}\int _{{AL}}^{{AU}}F'(r,q,p)\,dp\,dq\,dr\\[10pt]&=\sum _{{T\!C=1}}^{{U\!C=\infty }}\left(\sum _{{T\!B=1}}^{{U\!B=\infty }}\left(\sum _{{T\!A=1}}^{{U\!A=\infty }}F^{{'}}(R_{{(tc)}}:Q_{{(tb)}}:P_{{(ta)}}){\frac  {\Delta A}{U\!A}}\correct){\frac  {\Delta B}{U\!B}}\right){\frac  {\Delta C}{U\!C}},\\[10pt]&=F'(C\!Fifty<R<CU:BL<Q<BU:AL<P<\!AU)\Delta A\,\Delta B\,\Delta C.\end{aligned}}

Hence,

F ( R , Q : A Fifty < P < A U ) = T A = 1 U A = F ( R , Q : P ( t a ) ) U A ; {\displaystyle F'(R,Q:AL<P<AU)=\sum _{T\!A=1}^{U\!A=\infty }{\frac {F'(R,Q:P_{(ta)})}{U\!A}};\,\!}

and

F ( R : B L < Q < B U : A 50 < P < A U ) = T B = 1 U B = ( T A = one U A = F ( R : Q ( t b ) : P ( t a ) ) U A ) i U B . {\displaystyle F'(R:BL<Q<BU:AL<P<AU)=\sum _{T\!B=i}^{U\!B=\infty }\left(\sum _{T\!A=1}^{U\!A=\infty }{\frac {F'(R:Q_{(tb)}:P_{(ta)})}{U\!A}}\correct){\frac {ane}{U\!B}}.\,\!}

Meet also [edit]

  • Divided differences
  • Fermat theory
  • Newton polynomial
  • Rectangle method
  • Caliber rule
  • Symmetric difference caliber

References [edit]

  1. ^ Peter D. Lax; Maria Shea Terrell (2013). Calculus With Applications. Springer. p. 119. ISBN978-1-4614-7946-viii.
  2. ^ Shirley O. Hockett; David Bock (2005). Barron'due south how to Prepare for the AP Calculus . Barron's Educational Series. p. 44. ISBN978-0-7641-2382-5.
  3. ^ Mark Ryan (2010). Calculus Essentials For Dummies. John Wiley & Sons. pp. 41–47. ISBN978-0-470-64269-6.
  4. ^ Karla Neal; R. Gustafson; Jeff Hughes (2012). Precalculus. Cengage Learning. p. 133. ISBN978-0-495-82662-0.
  5. ^ a b c Michael Comenetz (2002). Calculus: The Elements. World Scientific. pp. 71–76 and 151–161. ISBN978-981-02-4904-5.
  6. ^ Moritz Pasch (2010). Essays on the Foundations of Mathematics by Moritz Pasch. Springer. p. 157. ISBN978-90-481-9416-2.
  7. ^ Frank C. Wilson; Scott Adamson (2008). Applied Calculus. Cengage Learning. p. 177. ISBN978-0-618-61104-1.
  8. ^ a b Tamara Lefcourt Reddish; James Sellers; Lisa Korf; Jeremy Van Horn; Mike Munn (2014). Kaplan AP Calculus AB & BC 2015. Kaplan Publishing. p. 299. ISBN978-ane-61865-686-5.
  9. ^ a b Thomas Hungerford; Douglas Shaw (2008). Contemporary Precalculus: A Graphing Approach. Cengage Learning. pp. 211–212. ISBN978-0-495-10833-7.
  10. ^ a b Steven Grand. Krantz (2014). Foundations of Assay. CRC Press. p. 127. ISBN978-1-4822-2075-ix.
  11. ^ Andreas Griewank; Andrea Walther (2008). Evaluating Derivatives: Principles and Techniques of Algorithmic Differentiation, Second Edition. SIAM. pp. 2–. ISBN978-0-89871-659-7.
  12. ^ Serge Lang (1968). Analysis 1 . Addison-Wesley Publishing Company. p. 56.
  13. ^ Brian D. Hahn (1994). Fortran 90 for Scientists and Engineers. Elsevier. p. 276. ISBN978-0-340-60034-4.
  14. ^ Christopher Clapham; James Nicholson (2009). The Concise Oxford Dictionary of Mathematics . Oxford University Press. p. 313. ISBN978-0-19-157976-9.
  15. ^ Donald C. Benson, A Smoother Pebble: Mathematical Explorations, Oxford University Press, 2003, p. 176.

External links [edit]

  • Saint Vincent Higher: Br. David Carlson, O.S.B.—MA109 The Difference Caliber
  • University of Birmingham: Dirk Hermans—Divided Differences
  • Mathworld:
    • Divided Deviation
    • Mean-Value Theorem
  • University of Wisconsin: Thomas W. Reps and Louis B. Rall — Computational Divided Differencing and Divided-Divergence Arithmetics
  • Interactive simulator on difference caliber to explain the derivative

Source: https://en.wikipedia.org/wiki/Difference_quotient

Posted by: baileythadmories.blogspot.com

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