hcju/proofwikips huggingface dataset:
Complex Sequence is Cauchy iff Convergent/Lemma 1
{{refactor|Extract this into a second definition of Definition:Complex Cauchy Sequence.}} == Lemma for Complex Sequence is Cauchy iff Convergent ==  Let $\sequence {z_n}$ be a complex sequence. Let $\mathcal N$ be the domain of $\sequence {z_n}$. Let $x_n = \Re \paren {z_n}$ for every $n \in \mathcal N$. Let $y_n = \Im \paren {z_n}$ for every $n \in \mathcal N$. Then $\sequence {z_n}$ is a (complex) Cauchy sequence  $\sequence {x_n}$ and $\sequence {y_n}$ are (real) Cauchy sequences.  == Proof == === Necessary Condition === Let $\sequence {z_n}$ be a Cauchy sequence. This means that, for a given $\epsilon > 0$: :$\exists N: \forall m, n \in \mathcal N: m, n \ge N: \cmod {z_n - z_m} < \epsilon$ We have, for every $m, n \ge N$: {{begin-eqn}} {{eqn | l = \cmod {x_n − x_m}       | r = \cmod {\Re \paren {z_n − z_m} }       | c =  }} {{eqn | o = \le       | r = \cmod {z_n − z_m}       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {x_n}$ is a Cauchy sequence by definition. A similar argument shows that $\sequence {y_n}$ is a Cauchy sequence. {{qed|lemma}} === Sufficient Condition === Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences. This means for $\sequence {x_n}$ that, for a given $\epsilon > 0$: : $\exists N_1: \forall m, n \in \mathcal N: m, n \ge N_1: \cmod {x_n - x_m} < \dfrac \epsilon 2$ Also, for $\sequence {y_n}$: : $\exists N_2: \forall m, n \in \mathcal N: m, n \ge N_2: \cmod {y_n - y_m} < \dfrac \epsilon 2$ Let $N = \max \paren {N_1, N_2}$. Let $i = \sqrt {-1}$ denote the imaginary unit. We have, for every $m, n \ge N$: {{begin-eqn}} {{eqn | l = \cmod {z_n − z_m}       | r = \cmod {x_n + i y_n − \paren {x_m + i y_m} }       | c = as $z_n = \Re \paren {z_n} + i \Im \paren {z_n}$ and $z_m = \Re \paren {z_m} + i \Im \paren {z_m}$ }} {{eqn | r = \cmod {x_n − x_m + i \paren {y_n − y_m} }       | c =  }} {{eqn | o = \le       | r = \cmod {x_n − x_m} + \cmod {i\paren {y_n − y_m} }       | c = Triangle Inequality for Complex Numbers }} {{eqn | r = \cmod {x_n − x_m} + \cmod {y_n − y_m}       | c = {{Defof|Complex Modulus}} }} {{eqn | o = <       | r = \frac \epsilon 2 + \cmod {y_n − y_m}       | c = since $\cmod {x_n - x_m} < \dfrac \epsilon 2$ }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = since $\cmod {y_n - y_m} < \dfrac \epsilon 2$ }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {z_n}$ is a Cauchy sequence by definition.
Field Adjoined Set/Corollary
== Corollary to Field Adjoined Set == Let $F$ be a field. Let $S \subseteq F$ be a subset of $F$. Ket $K \le F$ be a subfield of $F$.  Let $A = K \left[{X_1, \ldots, X_n}\right]$ be the ring of polynomial functions in $n$ indeterminates over $K$.  Let $B = K \left({X_1, \ldots, X_n}\right)$ be the field of rational functions in $n$ indeterminates over $K$.  Let $\alpha_1, \ldots, \alpha_n \in F$. Then: :$(1): \quad x \in K \left[{\alpha_1, \ldots, \alpha_n}\right] \iff x = f \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $f \in A$ :$(2): \quad x \in K \left({\alpha_1, \ldots, \alpha_n}\right) \iff x = f \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $f \in B$ :$(3): \quad x \in K \left[{S}\right] \iff x \in K \left[{\alpha_1, \ldots, \alpha_n}\right]$ for some $\alpha_1, \ldots, \alpha_n \in S$ :$(4): \quad x \in K \left({S}\right) \iff x \in K \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $\alpha_1, \ldots, \alpha_n \in S$  == Proof ==
Rule of Conjunction/Proof Rule
== Proof Rule == The Rule of Conjunction is a valid deduction sequent in propositional logic. As a proof rule it is expressed in the form:  :If we can conclude both $\phi$ and $\psi$, we may infer the compound statement $\phi \land \psi$.  It can be written: :$\ds {\phi \qquad \psi \over \phi \land \psi} \land_i$ === Tableau Form === {{:Rule of Conjunction/Proof Rule/Tableau Form}} == Explanation == {{:Rule of Conjunction/Explanation}} == Also known as == {{:Rule of Conjunction/Also known as}} == Also see == * This is a rule of inference of the following proof system: ** Definition:Hilbert Proof System/Instance 2 * Rule of Simplification == Technical Note == {{:Rule of Conjunction/Proof Rule/Technical Note}} == Sources == * {{BookReference|Introduction to Symbolic Logic|1959|A.H. Basson|author2 = D.J. O'Connor|ed = 3rd|edpage = Third Edition|prev = Modus Ponendo Ponens/Proof Rule|next = Definition:Axiom (Formal Systems)}}: $\S 4.2$: The Construction of an Axiom System: $RST \, 4$ * {{BookReference|Logic: Techniques of Formal Reasoning|1964|Donald Kalish|author2 = Richard Montague|prev = Rule of Simplification/Proof Rule|next = Rule of Addition/Proof Rule}}: $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$ * {{BookReference|Beginning Logic|1965|E.J. Lemmon|prev = Definition:Disjunct|next = Rule of Conjunction/Sequent Form/Formulation 1/Proof 1}}: $\S 1.3$: Conjunction and Disjunction * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Rule of Simplification|next = Rule of Addition/Proof Rule}}: $3.1$: Formal Proof of Validity * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Rule of Addition/Proof Rule|next = Definition:Logical Equivalence}}: $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $9$ * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Definition:Valid Argument|next = Rule of Simplification/Proof Rule}}: $\S 1.2.1$: Rules for natural deduction
Rule of Simplification/Proof Rule
== Proof Rule == The Rule of Simplification is a valid deduction sequent in propositional logic. As a proof rule it is expressed in either of the two forms:  :$(1): \quad$ If we can conclude $\phi \land \psi$, then we may infer $\phi$. :$(2): \quad$ If we can conclude $\phi \land \psi$, then we may infer $\psi$.  It can be written: :$\ds {\phi \land \psi \over \phi} \land_{e_1} \qquad \qquad {\phi \land \psi \over \psi} \land_{e_2}$ === Tableau Form === {{:Rule of Simplification/Proof Rule/Tableau Form}} == Explanation == {{:Rule of Simplification/Explanation}} == Also known as == {{:Rule of Simplification/Also known as}} == Also see == * Rule of Conjunction == Technical Note == {{:Rule of Simplification/Proof Rule/Technical Note}} == Sources == * {{BookReference|Introduction to Logic and to the Methodology of Deductive Sciences|1946|Alfred Tarski|ed = 2nd|edpage = Second Edition|prev = Definition:Statement Variable|next = Law of Identity}}: $\S \text{II}.12$: Laws of sentential calculus * {{BookReference|Logic: Techniques of Formal Reasoning|1964|Donald Kalish|author2 = Richard Montague|prev = Definition:Language of Propositional Logic|next = Rule of Conjunction/Proof Rule}}: $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$ * {{BookReference|Beginning Logic|1965|E.J. Lemmon|prev = Rule of Exportation/Forward Implication/Formulation 1/Proof|next = Rule of Simplification/Sequent Form/Formulation 1/Form 1/Proof 1}}: $\S 1.3$: Conjunction and Disjunction * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Destructive Dilemma/Formulation 2|next = Rule of Conjunction/Proof Rule}}: $3.1$: Formal Proof of Validity * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Hypothetical Syllogism/Formulation 1|next = Rule of Addition}}: $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $7$ * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Rule of Conjunction/Proof Rule|next = Definition:Tableau Proof (Formal Systems)}}: $\S 1.2.1$: Rules for natural deduction
Quadratic Residue/Examples/11
== Example of Quadratic Residues ==  The set of quadratic residues modulo $11$ is: :$\set {1, 3, 4, 5, 9}$  {{OEIS|A010375}} == Proof == To list the quadratic residues of $11$ it is enough to work out the squares $1^2, 2^2, \ldots, 10^2$ modulo $11$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod {11} }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {11} }} {{eqn | l = 3^2       | o = \equiv       | r = 9       | rr= \pmod {11} }} {{eqn | l = 4^2       | o = \equiv       | r = 5       | rr= \pmod {11} }} {{eqn | l = 5^2       | o = \equiv       | r = 3       | rr= \pmod {11} }} {{eqn | l = 6^2       | o = \equiv       | r = 3       | rr= \pmod {11} }} {{eqn | l = 7^2       | o = \equiv       | r = 5       | rr= \pmod {11} }} {{eqn | l = 8^2       | o = \equiv       | r = 9       | rr= \pmod {11} }} {{eqn | l = 9^2       | o = \equiv       | r = 4       | rr= \pmod {11} }} {{eqn | l = 10^2       | o = \equiv       | r = 1       | rr= \pmod {11} }} {{end-eqn}} So the set of quadratic residues modulo $11$ is: :$\set {1, 3, 4, 5, 9}$ The set of quadratic non-residues of $11$ therefore consists of all the other non-zero least positive residues: :$\set {2, 6, 7, 8, 10}$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Definition:Quadratic Residue|next = Quadratic Residue/Examples/3}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Law of Quadratic Reciprocity|next = Definition:Quadratic Residue}}: Chapter $7$: Patterns in Numbers: Gauss
Equation of Catenary/Formulation 1
== Curve == Consider a '''catenary'''. Let a cartesian plane be arranged so that the y-axis passes through the lowest point of the catenary.  The '''catenary''' is described by the equation: :$y = \dfrac {e^{ax} + e^{-ax}} {2 a} = \dfrac {\cosh a x} a$ where $a$ is a constant. The lowest point of the catenary is at $\left({0, \dfrac 1 a}\right)$.  == Proof == {{:Equation of Catenary/Formulation 1/Proof}} == Historical Note == {{:Definition:Catenary/Historical Note}} == Lingustic Note == {{:Definition:Catenary/Linguistic Note}}
Principle of Conservation of Energy
== Physical Law == Let $B$ be a body in a force field. Let $B$ be subject only to forces within that force field. Let $P$ be the potential energy of $B$. Let $K$ be the kinetic energy of $B$. Then: :$P + K = \text{constant}$ == Historical Note == {{:Principle of Conservation of Energy/Historical Note}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Speed of Body under Free Fall from Height/Proof 2|next = Kinetic Energy of Motion}}: $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Kinetic Energy of Motion
== Physical Law == Let $B$ be a body of mass $m$ moving with speed $v$ such that $v$ is very much less than the speed of light. Then its kinetic energy $K$ is given by: :$K = \dfrac {m v^2} 2$ == Proof ==  == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Principle of Conservation of Energy|next = Potential Energy of Position}}: $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Potential Energy of Position
== Physical Law == Let $B$ be a body of mass $m$ in a gravitational field. Let the Acceleration Due to Gravity be of constant magnitude $g$. Let $B$ be at a height $h$ above a reference level $h_0$. Then the potential energy $P$ of $B$ due to the position of $B$ relative to $h_0$ is given by: :$P = m g h$ == Proof ==  == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Kinetic Energy of Motion|next = Speed of Body under Free Fall from Height/Proof 3}}: $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Equation of Confocal Ellipses/Formulation 2
== Definition ==  The equation: :$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$c$ is a (strictly) positive constant :$a$ is a (strictly) positive parameter such that $a > c$ defines the set of all confocal ellipses whose foci are at $\tuple {\pm c, 0}$.  == Proof == Let $a$ and $c$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $E$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ As $a > c$ it follows that: :$a^2 > c^2$ and so: :$a^2 - c^2 > 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ From Equation of Ellipse in Reduced Form, this is the equation of an ellipse in reduced form. Thus: :$\tuple {\pm a, 0}$ are the positions of the vertices of $E$ :$\tuple {0, \pm b}$ are the positions of the covertices of $E$ From Focus of Ellipse from Major and Minor Axis: :$\tuple {\pm c, 0}$ are the positions of the foci of $E$. Hence the result.  == Also see == * Equation of Confocal Conics * Equation of Confocal Hyperbolas == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Equation of Confocal Conics/Formulation 2|next = Equation of Confocal Hyperbolas/Formulation 2}}: $1$: The Nature of Differential Equations: Miscellaneous Problems for Chapter $1$: $6$
Equation of Confocal Hyperbolas/Formulation 2
== Definition ==  The equation: :$\dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$c$ is a (strictly) positive constant :$a$ is a (strictly) positive parameter such that $a < c$ defines the set of all confocal hyperbolas whose foci are at $\tuple {\pm c, 0}$.  == Proof == Let $a$ and $c$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $H$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ As $a < c$ it follows that: :$a^2 < c^2$ and so: :$a^2 - c^2 < 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$ Thus from Equation of Hyperbola in Reduced Form, $H$ defines an hyperbola where: :$\tuple {\pm a, 0}$ are the positions of the vertices of $H$ :the transverse axis of $H$ has length $2 a$ :the conjugate axis of $H$ has length $2 b$ From Focus of Hyperbola from Transverse and Conjugate Axis :$\tuple {\pm c, 0}$ are the positions of the foci of $H$. Hence the result.  == Also see == * Equation of Confocal Conics * Equation of Confocal Ellipses == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Equation of Confocal Ellipses/Formulation 2|next = Confocal Conics are Self-Orthogonal}}: $1$: The Nature of Differential Equations: : Miscellaneous Problems for Chapter $1$: $6$
Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor
== Proof Technique ==  The technique to solve a linear first order ordinary differential equation in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ It immediately follows from Integrating Factor for First Order ODE that: :$e^{\int \map P x \rd x}$ is an integrating factor for $(1)$. Multiplying it by: :$e^{\int \map P x \rd x}$ to reduce it to a form: :$\dfrac {\d y} {\d x} e^{\int \map P x \rd x} y = e^{\int \map P x \rd x} \map Q x$ is known as '''Solution by Integrating Factor'''. It is remembered by the procedure: :''Multiply by $e^{\int \map P x \rd x}$ and integrate.''  == Examples == {{:Solution by Integrating Factor/Examples}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Linear First Order ODE/y' + (y over x) = 3 x|next = Solution to Linear First Order Ordinary Differential Equation/Proof 1}}: $\S 2.10$
Ohm's Law
== Physical Law == Let a resistor $r$ oppose a current $I$ with resistance $R$. Then the drop in electromotive force produced by $r$ is given by: :$E_R = R I$ {{Namedfor|Georg Simon Ohm|cat = Ohm, Georg}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Pursuit Curve of Boat in River|next = Mathematician:Georg Simon Ohm}}: $\S 2.13$: Simple Electric Circuits
Drop in EMF caused by Inductance is proportional to Rate of Change of Current
== Physical Law == The drop in EMF $E_L$ caused by an inductor is proportional to the rate of change of current $I$ through the inductor: :$E_L = L \dfrac {\d I} {\d t}$ The constant of proportion $L$ is known as the inductance of the inductor. == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Mathematician:Georg Simon Ohm|next = Drop in EMF caused by Capacitance is proportional to Accumulated Charge}}: $\S 2.13$: Simple Electric Circuits
Drop in EMF caused by Capacitance is proportional to Accumulated Charge
== Physical Law == The drop in EMF $E_C$ caused by a capacitor is proportional to the electric charge $Q$ that has accumulated on the capacitor: :$E_C = \dfrac 1 C Q$ The reciprocal of the constant of proportion $\dfrac 1 C$ is known as the capacitance of the capacitor. == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Drop in EMF caused by Inductance is proportional to Rate of Change of Current|next = Definition:Electric Current}}: $\S 2.13$: Simple Electric Circuits
Kirchhoff's Voltage Law
== Physical Law == The algebraic sum of the electromotive forces around a closed electrical circuit is equal to zero. {{Namedfor|Gustav Robert Kirchhoff|cat = Kirchhoff}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Definition:Electric Current|next = Mathematician:Gustav Robert Kirchhoff}}: $\S 2.13$: Simple Electric Circuits
Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1
== Lemma ==  :$\set {\pm 1}$ is a normal subgroup of $G$.  == Proof == {{begin-eqn}} {{eqn| ll= \forall g \in G:      | l = g \circ 1 \circ g^{-1}      | r = g \circ g^{-1}      | c = {{Defof|Identity Element}} }} {{eqn| r = 1      | c = {{Defof|Inverse Element}} }} {{end-eqn}} {{begin-eqn}} {{eqn| ll= \forall g \in G:      | l = \paren {g \circ \paren {-1} \circ g^{-1} }^2      | r = g \circ \paren {-1} \circ g^{-1} \circ g \circ \paren {-1} \circ g^{-1}      | c =  }} {{eqn| r = g \circ \paren {-1} \circ \paren {-1} \circ g^{-1}      | c = {{Defof|Inverse Element}} }} {{eqn| r = g \circ g^{-1}      | c = Order of $-1$ is $2$ }} {{eqn| r = 1      | c = {{Defof|Inverse Element}} }} {{end-eqn}} So: : $g \circ \paren {-1} \circ g^{-1} = 1$ or: : $g \circ \paren {-1} \circ g^{-1} = -1$ Suppose $g \circ \paren {-1} \circ g^{-1} = 1$. Then: {{begin-eqn}} {{eqn| l = g \circ \paren {-1} \circ g^{-1}      | r = 1      | c =  }} {{eqn| l = g^{-1} \circ g \circ \paren {-1} \circ g^{-1} \circ g      | r = g^{-1} \circ 1 \circ g      | c = applying same action on both sides }} {{eqn| l = -1      | r = 1      | c = {{Defof|Inverse Element}}, {{Defof|Identity Element}} }} {{end-eqn}} which is a contradiction. So it is the other case: : $g \circ \paren {-1} \circ g^{-1} = -1$ Therefore: : $\forall g \in G: g \circ \set {\pm 1} \circ g^{-1} = \set {\pm 1}$ By definition, $\set {\pm 1}$ is a normal subgroup.
Method of Undetermined Coefficients/Sine and Cosine
== Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be a linear combination of sine and cosine: :$\map R x = \alpha \sin b x + \beta \cos b x$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$.  Let $\map R x = \alpha \sin b x + \beta \cos b x$. Consider the auxiliary equation to $(1)$: :$(2): \quad m^2 + p m + q = 0$ There are two cases which may apply. === $i b$ is not Root of Auxiliary Equation === First we investigate the case where $i b$ is not a root of the auxiliary equation to $(1)$. {{:Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation}} === $i b$ is Root of Auxiliary Equation === Now suppose that $(1)$ is of the form $y'' + b^2 y = A \sin b x + B \cos b x$. Thus one of the $i b$ is one of the roots of the auxiliary equation to $(1)$. From Linear Second Order ODE: $y'' + k^2 y = 0$ the general solution to $(2)$ is: :$y = C_1 \sin b x + C_2 \cos b x$ and it can be seen that an expression of the form $A \sin b x + B \cos b x$ is already a particular solution of $(2)$. Thus we have: :$\paren {q - b^2}^2 + b^2 p^2 = 0$. But using the Method of Undetermined Coefficients in the above manner, this would result in an attempt to calculate: :$\dfrac {\alpha \paren {q - b^2} + \beta b p} {\paren {q - b^2}^2 + b^2 p^2}$ and: :$\dfrac {\beta \paren {q - b^2} - \alpha b p} {\paren {q - b^2}^2 + b^2 p^2}$ both of which are are undefined. {{:Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Method of Undetermined Coefficients/Exponential|next = Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation/Trigonometric Form}}: $\S 3.18$: The Method of Undetermined Coefficients
Method of Undetermined Coefficients/Polynomial
== Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be a polynomial in $x$: :$\displaystyle \map R x = \sum_{j \mathop = 0}^n a_j x^j$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$.  Let $\displaystyle \map R x = \sum_{j \mathop = 0}^n a_j x^j$. Assume that there is a particular solution to $(1)$ of the form: :$\displaystyle y_p = \sum_{j \mathop = 0}^n A_j x^j$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = \sum_{j \mathop = 1}^n j A_j x^{j - 1}       | c = Power Rule for Derivatives }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = \sum_{j \mathop = 2}^n j \paren {j - 1} A_j x^{j - 2}       | c = Power Rule for Derivatives }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 2}^n j \paren {j - 1} A_j x^{j - 2} + p \sum_{j \mathop = 1}^n j A_j x^{j - 1} + q \sum_{j \mathop = 0}^n A_j x^j       | r = \sum_{j \mathop = 0}^n a_j x^j       | c =  }} {{end-eqn}} The coefficients $A_0$ to $A_n$ can hence be solved in terms of $a_0$ to $a_n$ using the techniques of simultaneous equations. The general form is tedious and unenlightening to evaluate.  == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Linear Second Order ODE/y'' - 7 y' - 5 y = x^3 - 1|next = Method of Undetermined Coefficients/Exponential}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.3$ Particular solution: polynomial $\map f x$ * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation/Trigonometric Form|next = Linear Second Order ODE/y'' + 3 y' - 10 y = 6 exp 4 x}}: $\S 3.18$: The Method of Undetermined Coefficients