nyuuzyou/wikis huggingface dataset:
Lagrange Polynomial Approximation

== Theorem ==
Let $f: D \to \R$ be $n + 1$ times differentiable in an interval $I \subseteq \R$.
Let $x_0, \dotsc, x_n \in I$ be pairwise distinct points.
Let $P$ be the Lagrange Interpolation Formula of degree at most $n$ such that:
:$\forall i \in \set {0, \dotsc, n}: \map P {x_i} = \map f {x_i}$
Let $\map R x = \map f x - \map P x$.
Then, for every $x \in I$, there exists $\xi$ in the interval spanned by $x$ and $x_i$, $i = 0, \dotsc, n$, such that:
:$\map R x = \dfrac {\paren {x - x_0} \paren {x - x_1} \dotsm \paren {x - x_n} \map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!}$
== Proof ==
This proof is similar to the proof of Taylor's theorem with the remainder in the Lagrange form, and is also based on Rolle's Theorem.
Observe that:
:$\map R {x_i} = 0$ for $i = 0, \dotsc, n$
and that:
:$R^{\paren {n + 1} } = f^{\paren {n + 1} }$
, assume that $x$ is different from all $x_i$ for $i = 0, \dotsc, n$.
Let the function $g$ be defined by:
:$\map g t = \map R t - \dfrac {\paren {t - x_0} \paren {t - x_1} \dotsm \paren {t - x_n} \map R x} {\paren {x - x_0} \paren {x - x_1} \dotsm \paren {x - x_n} }$
Then $\map g {x_i} = 0$ for $i = 0, \dotsc, n$, and $\map g x = 0$.
Denote by $J$ the interval spanned by $x$ and $x_i$, $i = 0, \dotsc, n$.
Thus $g$ has at least $n + 2$ zeros in $J$.
The Extended Rolle's Theorem is applied in $J$ successively to $g$, $g'$, $g$ and so on until $g^{\paren {n + 1} }$, which thus has at least one zero $\xi$ located between the two known zeros of $g^{\paren n}$ in $J$.
Then:
:$0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}! \map R x} {\paren {x - x_0} \paren {x - x_1} \dotsm \paren {x - x_n} }$
and the formula for $\map R x$ follows.
== Comment ==
This theorem gives an estimate for the error of the Lagrange polynomial approximation and is similar to the Mean Value Theorem and Taylor's Theorem with the remainder in Lagrange form.
Category:Real Analysis
Category:Approximation Theory
Lambert W of Non-Zero Algebraic Number is Transcendental

== Theorem ==
Let $W$ denote the (general) Lambert W function.
Let $a$ be a non-zero algebraic number.
Then, $W \left({a}\right)$ is transcendental.
== Proof ==
From the definition of Lambert W function:
:$a = W \left({a}\right) e^{W \left({a}\right)}$
 $W \left({a}\right)$ is not transcendental.
Hence, $W \left({a}\right)$ is algebraic.
From the weaker Hermite-Lindemann-Weierstrass theorem, $e^{W \left({a}\right)}$ is transcendental.
However, from the equation above, it is also equal to:
:$\dfrac a {W \left({a}\right)}$
which must be algebraic, considering the fact that $a$ and $W \left({a}\right)$ are both algebraic.
From this contradiction, it follows that $W \left({a}\right)$ must be transcendental whenever $a$ is a non-zero algebraic number.
Category:Lambert W Function
Category:Transcendental Number Theory
Lambert W of Zero is Zero

== Theorem ==
Let $W_0$ denote principal branch of the Lambert W function.
Then:
:$W_0 \left({0}\right) = 0$
== Proof ==
From the definition of the principal branch of the Lambert W function:
:$y = W_0 \left({x}\right) \iff x = y e^y$
where $x \in \left[{-\dfrac 1 e \,.\,.\, \to}\right)$ and $y \in \left[{-1 \,.\,.\, \to}\right)$.
The result follows from substituting $x = 0$ and $y = 0$.
Category:Lambert W Function
Lami's Theorem

== Theorem ==
Let $B$ be a body in static equilibrium.
Let the only forces acting on $B$ be $F_a$, $F_b$ and $F_c$.
Let $F_a$, $F_b$ and $F_c$ be represented by the vectors $V_a$, $V_b$ and $V_c$ respectively, such that the magnitudes and directions of each force corresponds to the magnitudes and directions of each vectors.
Let the directions of $V_a$, $V_b$ and $V_c$ be non-parallel.
Then $V_a$, $V_b$ and $V_c$ are coplanar and concurrent, and:
:$\dfrac {\size V_a} {\sin A} = \dfrac {\size V_b} {\sin B} = \dfrac {\size V_c} {\sin C}$
where:
:$A$, $B$ and $C$ are the angles between the directions of $V_b$ and $V_c$, $V_a$ and $V_c$, and $V_a$ and $V_b$ respectively
:$\size V_a$, $\size V_b$ and $\size V_c$ are the magnitudes of $V_a$, $V_b$ and $V_c$ respectively.
== Proof ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
Category:Lami's Theorem
Category:Statics
Lamé's Theorem

== Theorem ==
Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $c$ and $d$ be the number of digits in $a$ and $b$ respectively when expressed in decimal notation.
Let the Euclidean Algorithm be employed to find the GCD of $a$ and $b$.
Then it will take fewer than $5 \times \min \set {c, d}$ integer divisions to find $\gcd \set {a, b}$.
=== Variant: Least Absolute Remainder ===
== Proof ==
=== Lemma===
 suppose $a \ge b$. 
Then $\min \set {c, d}$ is the number of digits in $b$.
By Number of Digits in Number, we have:
:$\min \set {c, d} = \floor {\log b} + 1$
 it takes at least $5 \paren {\floor {\log b} + 1}$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.
Then we have:
For $b = 1$, both sides are equal to $1$, giving $1 > 1$, which is a contradiction.
Hence we consider $b > 1$ and take $\log$ on both sides:
However, $\dfrac 1 {\log \phi} \approx 4.785 < 5$.
This is a contradiction.
Hence the result by Proof by Contradiction.
== Examples ==
== Historical Note ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
* : $5$
* : $5$
* 
* 
Category:Euclidean Algorithm
Category:Lamé's Theorem
Lamé's Theorem/Examples/12378 and 3054

== Example of Use of Lamé's Theorem ==
The Euclidean Algorithm, when employed to find the GCD of $12378$ and $3054$, will take no more than $20$ steps.
== Proof ==
The smaller of the $2$ integers $12378$ and $3054$ has $4$ digits.
Hence by Lamé's Theorem it will take no more than $4 \times 5 = 20$ steps of the Euclidean Algorithm to terminate.
In fact, from Euclidean Algorithm: $12378$ and $3054$, it in fact takes as few as $6$ steps.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Lamé's Theorem
Lamé's Theorem/Least Absolute Remainder

== Theorem ==
Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $c$ and $d$ be the number of digits in $a$ and $b$ respectively when expressed in decimal notation.
Let the Euclidean Algorithm: Least Absolute Remainder variant be employed to find the GCD of $a$ and $b$.
Then it will in general take fewer integer divisions to find $\gcd \set {a, b}$ than it does to use the conventional form of the Euclidean Algorithm.
In fact, it will take fewer than $3 \times \min \set {c, d}$ integer divisions to find $\gcd \set {a, b}$.
== Proof ==
=== Lemma===
 suppose $a \ge b$. 
Then $\min \set {c, d}$ is the number of digits in $b$.
By Number of Digits in Number, we have:
:$\min \set {c, d} = \floor {\log b} + 1$
 it takes at least $3 \paren {\floor {\log b} + 1}$ cycles around the Euclidean Algorithm: Least Absolute Remainder variant to find $\gcd \set {a, b}$.
Then we have:
For $b = 1$, both sides are equal to $1$, giving $1 > 1$, which is a contradiction.
Hence we consider $b > 1$ and take $\log$ on both sides:
However, $\dfrac 1 {\log \paren {1 + \sqrt 2} } \approx 2.6125 < 3$.
This is a contradiction.
Hence the result by Proof by Contradiction.
== Historical Note ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Euclidean Algorithm
Category:Lamé's Theorem
Lamé's Theorem/Least Absolute Remainder/Lemma

== Theorem ==
Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let the Euclidean Algorithm: Least Absolute Remainder variant be employed to find the GCD of $a$ and $b$.
Suppose it takes $n$ cycles around the Euclidean Algorithm: Least Absolute Remainder variant to find $\gcd \set {a, b}$.
Then $\min \set {a, b} \ge P_{n + 1}$, where $P_n$ denotes the $n$-th Pell number.
== Proof ==
 suppose $a \ge b$.
Let $q_i, r_i$ be the quotients and remainders of each step of the Euclidean Algorithm: Least Absolute Remainder variant, that is:
so $r_n = \gcd \set {a, b}$.
Note that for each $3 \le k \le n$, we have:
:$r_{k - 2} = q_k r_{k - 1} + r_k$
From definition we have:
:$-\dfrac {\size {r_{k - 1}}} 2 < r_k \le \dfrac {\size {r_{k - 1}}} 2$
which implies:
Also we have:
Now we prove that $\size {r_{n - m}} \ge P_{m + 1}$ for $0 \le m < n$ by the Principle of Mathematical Induction:
=== Basis for the Induction ===
When $m = 0$, we have:
When $m = 1$, we have:
These are our base cases.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\size {r_{n - k + 2}} \ge P_{k - 1}$
:$\size {r_{n - k + 1}} \ge P_k$
for some $k$ where $2 \le k < n$.
It is to be demonstrated that:
:$\size {r_{n - k}} \ge P_{k + 1}$
=== Induction Step ===
This is our induction step:
We have:
:$r_{n - k} = q_{n - k + 2} r_{n - k + 1} + r_{n - k + 2}$
and:
:$\size {r_{n - k}} = \size {q_{n - k + 2} } \size {r_{n - k + 1} } \pm \size {r_{n - k + 2}}$
for some choice of $+$ or $-$.
By $(2)$, we have $\size {q_{n - k + 2}} \ge 2$.
For the case $\size {q_{n - k + 2}} \ge 3$:
For the case $\size {q_{n - k + 2}} = 2$, we have to show that $\size {r_{n - k} } = 2 \size {r_{n - k + 1} } + \size {r_{n - k + 2} }$.
In other words, we need to show that:
:$\size {r_{n - k} } \ne 2 \size {r_{n - k + 1} } - \size {r_{n - k + 2} }$
We have:
Therefore, in both cases, we have:
We have shown that the result for $m = k - 2$ and $m = k - 1$ implies the result for $m = k$.
Thus we have $r_{n - m} \ge P_{m + 1}$ for all $0 \le m < n$.
Since $b \ge 2 \size {r_1}$, we can apply the same argument as the induction step and obtain:
Category:Lamé's Theorem
Lamé's Theorem/Lemma

== Theorem ==
Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let the Euclidean Algorithm be employed to find the GCD of $a$ and $b$.
Suppose it takes $n$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.
Then $\min \set {a, b} \ge F_{n + 2}$, where $F_n$ denotes the $n$th Fibonacci number.
== Proof ==
, suppose $a \ge b$.
Let $q_i, r_i$ be the quotients and remainders of each step of the Euclidean Algorithm, that is:
so $r_n = \gcd \set {a, b}$.
We prove that $r_{n - m} \ge F_{m + 2}$ for $0 \le m < n$ by the Principle of Mathematical Induction:
=== Basis for the Induction ===
When $m = 0$, we have:
When $m = 1$, we have:
These are our base cases.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$r_{n - k + 2} \ge F_k$
:$r_{n - k + 1} \ge F_{k + 1}$
for some $k$ where $2 \le k < n$.
It is to be demonstrated that:
:$r_{n - k} \ge F_{k + 2}$
=== Induction Step ===
This is our induction step:
We have shown that the result for $m = k - 2$ and $m = k - 1$ implies the result for $m = k$.
Thus we have $r_{n - m} \ge F_{m + 2}$ for all $0 \le m < n$.
Hence:
Category:Lamé's Theorem
Landau's Theorem

== Theorem ==
Let $f: \C \to \C$ be a holomorphic function in the unit disk $\cmod z \le 1$.
Let $\cmod {\map {f'} 0} = 1$.
Then there exists:
:a disk $D$ of radius $B$
:an analytic function $\phi$ in $D$ such that $\map f {\map \phi z} = z$ for all $z$ in $D$
such that $L$ is an absolute constant where:
:$L > B$
where $B$ is Bloch's constant.
== Also see ==
* Definition:Landau's Constant
* Value of Landau's Constant
* Bloch's Theorem
* Definition:Bloch's Constant
* Value of Bloch's Constant
Category:Complex Analysis
Language of Predicate Logic has Unique Parsability

== Theorem ==
Each WFF of predicate logic which starts with a left bracket or a negation sign has exactly one main connective.
== Proof ==
We have the following cases:
# $\mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$.
# $\mathbf A = \paren {\mathbf B \circ \mathbf C}$ where $\circ$ is one of the binary connectives.
# $\mathbf A = \map p {t_1, t_2, \ldots, t_n}$, where $t_1, t_2, \ldots, t_n$ are terms, and $p \in \PP_n$.
# $\mathbf A = \paren {Q x: \mathbf B}$, where $\mathbf B$ is a WFF of length $k - 5$, $Q$ is a quantifier ($\forall$ or $\exists$) and $x$ is a variable.
We deal with these in turn.
Cases 1 and 2 are taken care of by Language of Propositional Logic has Unique Parsability.
Cases 3 and 4 do not start with either a left bracket or a negation sign, so do not have to be investigated. 
== Sources ==
* : $\S 2.2$: Theorem $2.2.4$
Category:Language of Predicate Logic
Category:Named Theorems
Language of Propositional Logic has Unique Parsability

== Theorem ==
The language of propositional logic $\LL_0$ has unique parsability.
== Proof ==
It is to be demonstrated that each WFF arises by a unique rule of formation from the bottom-up specification of propositional logic.
The rules $\mathbf W : TF$ and $\mathbf W : \PP_0$ need no further treatment.
From inspection of the first character it is clear that the remaining $\mathbf W : \neg$ and $\mathbf W : Op$ cannot yield the same WFF.
What remains is to establish uniqueness in applying $\mathbf W : \neg$ and $\mathbf W : Op$.
For $\mathbf W : \neg$, this means to consider:
:$\mathbf A = \neg \mathbf B = \neg \mathbf C$
from which it is immediate that $\mathbf B = \mathbf C$.
Lastly, for $\mathbf W : Op$, we have the following lemma:
=== Lemma ===
Having examined all possible combinations of rules of formation, we conclude that $\LL_0$ has unique parsability.
== Also presented as ==
Some sources only prove a subset of this result, deeming the rest of the cases a triviality not worth devoting precious space to.
Category:Language of Propositional Logic
Language of Propositional Logic has Unique Parsability/Lemma

== Lemma ==
Let $\LL_0$ be the language of propositional logic.
Let $\mathbf A$ be a WFF.
Suppose that $\mathbf A = \paren {\mathbf B \circ \mathbf C} = \paren {\mathbf D * \mathbf E}$.
Then $\mathbf B = \mathbf D$, ${\circ} = {*}$, and $\mathbf C = \mathbf E$.
== Proof ==
The WFFs $\mathbf B$ and $\mathbf D$ are strings which both start in the same place, right after the first left bracket in $\mathbf A$.
By Prefix of WFF of PropLog is not WFF, neither $\mathbf B$ nor $\mathbf D$ can be a prefix of the other.
Therefore $\mathbf B = \mathbf D$.
It follows that $* = \circ$ and $\mathbf C = \mathbf E$.
Hence the result.
== Sources ==
* : $\S 1.4$: Main Connective: Theorem $1.4.2$
Category:Language of Propositional Logic
Laplace's Equation/Examples/Electric Force in Free Space

== Example of Laplace's Equation ==
Let $R$ be a region of free space.
Let $\mathbf V$ be an electric force in a given electric field over $R$.
Then $\mathbf V$ satisfies Laplace's equation.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(i)}$
Category:Laplace's Equation
Category:Electrostatics
Laplace's Equation/Examples/Irrotational Motion of Incompressible Fluid

== Example of Laplace's Equation ==
Let $B$ be a body of incompressible fluid.
Let $\mathbf V$ be the vector field which describes the irrotational motion of $B$.
Then $\mathbf V$ satisfies Laplace's equation.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(i)}$
Category:Laplace's Equation
Category:Fluid Mechanics
Laplace's Expansion Theorem

== Theorem ==
Let $D$ be the determinant of order $n$.
Let $r_1, r_2, \ldots, r_k$ be integers such that:
:$1 \le k < n$
:$1 \le r_1 < r_2 < \cdots < r_k \le n$
Let $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be an order-$k$ minor of $D$.
Let $\map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be the cofactor of $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$.
Then:
:$\ds D = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k} \, \map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$
A similar result applies for columns.
== Proof ==
Let us define $r_{k + 1}, r_{k + 2}, \ldots, r_n$ such that:
:$1 \le r_{k + 1} < r_{k + 2} < \cdots < r_n \le n$
:$\rho = \tuple {r_1, r_2, \ldots, r_n}$ is a permutation on $\N^*_n$.
Let $\sigma = \tuple {s_1, s_2, \ldots, s_n}$ be a permutation on $\N^*_n$.
Then by Permutation of Determinant Indices we have:
We can obtain all the permutations $\sigma$ exactly once by separating the numbers $1, \ldots, n$ in all possible ways into a set of $k$ and $n - k$ numbers.
We let $\tuple {s_1, \ldots, s_k}$ vary over the first set and $\tuple {s_{k + 1}, \ldots, s_n}$ over the second set.
So the summation over all $\sigma$ can be replaced by:
:$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
:$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
:$\tuple {s_1, \ldots, s_k} = \map \sigma {u_1, \ldots, u_k}$
:$\tuple {s_{k + 1}, \ldots, s_n} = \map \sigma {u_{k + 1}, \ldots, u_n}$
Thus we get:
That last inner sum extends over all integers which satisfy:
:$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
:$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
But for each set of $u_1, \ldots, u_k$, then the integers $u_{k + 1}, \ldots, u_n$ are clearly uniquely determined.
So that last inner sum equals 1 and the theorem is proved.
The result for columns follows from Determinant of Transpose. 
== Comment ==
This gives us an expansion of the determinant $D$ in terms of $k$ specified rows.
We form all possible order-$k$ minors of $D$ which involve all of these rows, and multiply each of them by their cofactors.
The sum of these products is equal to $D$.
== Also known as ==
This theorem is also known as the Laplace cofactor expansion.
== Examples ==
== Also see ==
* Expansion Theorem for Determinants: the special case where $k = 1$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.5$: Determinants
Category:Determinants
Category:Laplace's Expansion Theorem
Laplace's Expansion Theorem/Examples/Arbitrary 3x3 Matrix

== Example of Use of Laplace's Expansion Theorem ==
Let $\mathbf A$ be the matrix defined as:
:$\mathbf A = \begin {bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end {bmatrix}$
Then $\map \det {\mathbf A}$ can be calculated using Laplace's Expansion Theorem as follows.
Expanding row $2$:
This shows that $\mathbf A$ is singular.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.5$: Determinants: Example $2.2.2$
Category:Laplace's Expansion Theorem
Laplace Transform/Examples/Example 1

== Example of Laplace Transform ==
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
:$\laptrans {4 e^{5 t} + 6 t^3 - 3 \sin 4 t + 2 \cos 2 t} = \dfrac 4 {s - 5} + \dfrac {36} {s^4} - \dfrac {12} {s^2 + 16} + \dfrac {2 s} {s^2 + 4}$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Linearity Property: $6$
Category:Examples of Laplace Transforms
Laplace Transform/Graphical Interpretation/Examples

== Examples of Graphical Interpretation of Laplace Transforms ==
The following are three examples of the contours defined by the integrand defining the Laplace transform of $\map \cos t$.
=== Laplace Transform of $\cos t$ of $3 - 9 i$ ===
=== Laplace Transform of $\cos t$ of $\dfrac 1 2 + 4 i$ ===
=== Laplace Transform of $\cos t$ of $-3 + 12 i$ ===
Category:Examples of Laplace Transforms
Laplace Transform/Graphical Interpretation/Examples/Cosine t of -3 + 12 i

== Example of Graphical Interpretation of Laplace Transforms ==
From Laplace Transform of Cosine, $\map {\laptrans {\map \cos t} } {-3 + 12 i}$ does not exist.
Notice how the contour spirals outward as $t$ increases without bound, never settling at a "stable point":
:
Category:Examples of Laplace Transforms
Laplace Transform/Graphical Interpretation/Examples/Cosine t of 3 - 9 i

== Example of Graphical Interpretation of Laplace Transforms ==
For $\map {\laptrans {\map \cos t} } {3 - 9 i}$, notice how the contour spirals towards the origin.
Intuitively, the integral converges because as $t$ increases without bound, the contour "shrinks" as it continues its path into the origin.
:
Category:Examples of Laplace Transforms
Laplace Transform/Graphical Interpretation/Examples/Cosine t of Half + 4 i

== Example of Graphical Interpretation of Laplace Transforms ==
For $\map {\laptrans {\map \cos t} } {\dfrac 1 2 + 4 i}$, though the contour is not simple, it still "shrinks" as the parameter $t$ increases without bound:
:
Category:Examples of Laplace Transforms
Laplace Transform Determination

== Solution Techniques for Laplace Transforms ==
The following techniques may be used to find the Laplace transform $\laptrans f$ of a function $f$.
=== Direct Method ===
=== Series Method ===
=== Method of Differential Equations ===
=== Differentiation with Respect to Parameter ===
=== Miscellaneous Methods ===
=== Use of Tables ===
Category:Laplace Transform Determination
Laplace Transform Determination/Differentiation with Respect to Parameter

== Solution Technique for Laplace Transform ==
To find the Laplace transform of a function $f$, one can evaluate it as follows:
Let $\map f t$ be expressible in the form:
:$\map f t = \dfrac \d {\d a} \map g {a, t}$
Then it may be possible to express:
:$\map {\dfrac \d {\d a} } {\laptrans {\map g {a, t} } } = \laptrans {\map {\dfrac \d {\d a} } {\map g {a, t} } }$
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Methods of Finding Laplace Transforms: $4$. Differentiation with respect to a parameter
Category:Laplace Transform Determination
Laplace Transform Determination/Direct Method

== Solution Technique for Laplace Transform ==
To find the Laplace transform of a function $f$, one can evaluate it directly from the definition:
:$\laptrans {\map f t} = \ds \int_0^{\to \infty} e^{-s t} \map f t \rd t$
whenever the primitive of $e^{-s t} \map f t$ can be evaluated.
== Sources ==
* : Chapter $1$: The Laplace Transform: Methods of Finding Laplace Transforms: $1$. Direct method
Category:Laplace Transform Determination
Laplace Transform Determination/Method of Differential Equations

== Solution Technique for Laplace Transform ==
To find the Laplace transform of a function $f$, one can evaluate it as follows:
:$(1): \quad$ Find a differential equation satisfied by $\map f t$
:$(2): \quad$ Evaluate the derivatives according to Higher Order Derivatives of Laplace Transform and its various instances.
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Methods of Finding Laplace Transforms: $3$. Method of differential equations
Category:Laplace Transform Determination
Laplace Transform Determination/Miscellaneous Methods

== Solution Technique for Laplace Transform ==
To find the Laplace transform of a function $f$, one can evaluate it using one of the following techniques:
==== Linear Combination of Laplace Transforms ====
==== Laplace Transform of Exponential times Function ====
==== Laplace Transform of Function of t minus a ====
==== Laplace Transform of Constant Multiple ====
==== Laplace Transform of Higher Order Derivatives ====
==== Laplace Transform of Integral ====
==== Higher Order Derivatives of Laplace Transform ====
==== Integral of Laplace Transform ====
==== Laplace Transform of Periodic Function ====
==== Initial Value Theorem of Laplace Transform ====
==== Final Value Theorem of Laplace Transform ====
== Sources ==
* : Chapter $1$: The Laplace Transform: Methods of Finding Laplace Transforms: $5$. Method of differential equations
Category:Laplace Transform Determination
Laplace Transform Determination/Series Method

== Solution Technique for Laplace Transform ==
To find the Laplace transform of a function $f$, one can evaluate it as follows:
Let $\map f t$ have a power series expansion given by:
Then the Laplace transform of $f$ can be found by taking the sum of the Laplace transforms of each term in the series:
if $\laptrans {\map f t}$ is convergent for $s > \gamma$.
== Proof ==
From Laplace Transform of Positive Integer Power:
:$\laptrans {t^n} = \dfrac {n!} {s^{n + 1} }$
Then it is seen that $\ds \laptrans {\sum_{n \mathop = 0}^\infty a_n t^n}$ is a Linear Combination of Laplace Transforms.
The result follows.
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Methods of Finding Laplace Transforms: $2$. Series method
Category:Laplace Transform Determination