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Removed error in proofs.

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Luc Bijl 2024-01-03 13:46:55 +01:00
parent 4b35b10aaa
commit 201a3ed1e0
8 changed files with 64 additions and 142 deletions
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27
docs/en/mathematics/multivariable-calculus/differentation.md
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@ -22,12 +22,9 @@ $$
\partial_{12} f(P) = \partial_{21} f(P), \partial_{12} f(P) = \partial_{21} f(P),
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## Total derivatives ## Total derivatives
@ -55,12 +52,9 @@ $$
with $\nabla f(\mathbf{a})$ the gradient of $f$. with $\nabla f(\mathbf{a})$ the gradient of $f$.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## Chain rule ## Chain rule
@ -90,8 +84,7 @@ The direction of the gradient is the direction of steepest increase of $f$ at $\
*Theorem*: gradients are orthogonal to level lines and level surfaces. *Theorem*: gradients are orthogonal to level lines and level surfaces.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
let $\mathbf{r}(t) = \big(x(t),\; y(t) \big)^T$ be a parameterization of the level curve of $f$ such that $\mathbf{r}(0) = \mathbf{a}$. Then for all $t$ near $0$, $f(\mathbf{r}(t)) = f(\mathbf{a})$. Differentiating this equation with respect to $t$ using the chain rule, we obtain let $\mathbf{r}(t) = \big(x(t),\; y(t) \big)^T$ be a parameterization of the level curve of $f$ such that $\mathbf{r}(0) = \mathbf{a}$. Then for all $t$ near $0$, $f(\mathbf{r}(t)) = f(\mathbf{a})$. Differentiating this equation with respect to $t$ using the chain rule, we obtain
@ -106,8 +99,6 @@ $$
$$ $$
obtaining that $\nabla f$ is orthogonal to $\mathbf{\dot r}$. obtaining that $\nabla f$ is orthogonal to $\mathbf{\dot r}$.
</details>
<br>
## Directional derivatives ## Directional derivatives
@ -155,10 +146,6 @@ We have two interpretations:
* the composition of linear maps, * the composition of linear maps,
* the matrix multiplication of the Jacobian. * the matrix multiplication of the Jacobian.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later.
</details>
<br>
Will be added later.

35
docs/en/mathematics/multivariable-calculus/extrema.md
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@ -8,12 +8,9 @@
*Theorem*: if $f$ has local $\begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix}$ at $\mathbf{x^*} \in D$ then $\mathbf{x^*}$ is a critical point of for $f$. *Theorem*: if $f$ has local $\begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix}$ at $\mathbf{x^*} \in D$ then $\mathbf{x^*}$ is a critical point of for $f$.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## A second derivative test ## A second derivative test
@ -30,23 +27,17 @@ $$
* If $H_f(\mathbf{x^*})$ is indefinite (both positive and negative eigenvalues), then $f$ has a saddle point at $\mathbf{x^*}$. * If $H_f(\mathbf{x^*})$ is indefinite (both positive and negative eigenvalues), then $f$ has a saddle point at $\mathbf{x^*}$.
* If $H_f(\mathbf{x^*})$ is neither positive nor negative definite, nor indefinite, (eigenvalues equal to zero) this test gives no information. * If $H_f(\mathbf{x^*})$ is neither positive nor negative definite, nor indefinite, (eigenvalues equal to zero) this test gives no information.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## Extrema on restricted domains ## Extrema on restricted domains
*Theorem*: let $D \subseteq \mathbb{R}^n$ be bounded and closed ($D$ contains all boundary points). Let $f: D \to \mathbb{R}$ be continuous, then $f$ has a global maximum and minimum. *Theorem*: let $D \subseteq \mathbb{R}^n$ be bounded and closed ($D$ contains all boundary points). Let $f: D \to \mathbb{R}$ be continuous, then $f$ has a global maximum and minimum.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
**Procedure to find the global maximum and minimum**: **Procedure to find the global maximum and minimum**:
@ -68,12 +59,9 @@ $$
L(\mathbf{x}, \lambda) := f(\mathbf{x}) - \lambda g(\mathbf{x}). L(\mathbf{x}, \lambda) := f(\mathbf{x}) - \lambda g(\mathbf{x}).
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
### The general case ### The general case
@ -89,12 +77,9 @@ $$
L(\mathbf{x}, \mathbf{\lambda}) := f(\mathbf{x}) - \big\langle \mathbf{\lambda},\; \mathbf{g}(\mathbf{x}) \big\rangle. L(\mathbf{x}, \mathbf{\lambda}) := f(\mathbf{x}) - \big\langle \mathbf{\lambda},\; \mathbf{g}(\mathbf{x}) \big\rangle.
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
#### Example #### Example

14
docs/en/mathematics/multivariable-calculus/implicit-equations.md
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@ -26,12 +26,9 @@ $$
\phi' (x) = - \frac{\partial_1 f\big(x,\phi(x)\big)}{\partial_2 f\big(x,\phi(x)\big)}. \phi' (x) = - \frac{\partial_1 f\big(x,\phi(x)\big)}{\partial_2 f\big(x,\phi(x)\big)}.
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## General case ## General case
@ -56,9 +53,6 @@ $$
D \mathbf{\phi}(\mathbf{x}) = - \Big(D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) \Big)^{-1} D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big). D \mathbf{\phi}(\mathbf{x}) = - \Big(D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) \Big)^{-1} D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big).
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>

21
docs/en/mathematics/multivariable-calculus/integration.md
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@ -8,12 +8,9 @@ $$
implying that order can be interchanged, this is true for $n \in \mathbb{N}$. implying that order can be interchanged, this is true for $n \in \mathbb{N}$.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## Iteration of integrals ## Iteration of integrals
@ -23,12 +20,9 @@ $$
\iint_D f dA = \iint_R F dA, \qquad \text{where } F(\mathbf{x}) = \begin{cases} F(\mathbf{x}) \quad &\mathbf{x} \in D, \\ 0 \quad &\mathbf{x} \notin D. \end{cases} \iint_D f dA = \iint_R F dA, \qquad \text{where } F(\mathbf{x}) = \begin{cases} F(\mathbf{x}) \quad &\mathbf{x} \in D, \\ 0 \quad &\mathbf{x} \notin D. \end{cases}
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
## Coordinate transformation for integrals ## Coordinate transformation for integrals
@ -46,12 +40,9 @@ $$
with $D_\phi$ the Jacobian of $\phi$. with $D_\phi$ the Jacobian of $\phi$.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
### Example ### Example

8
docs/en/mathematics/multivariable-calculus/taylor-polynomials.md
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@ -24,18 +24,14 @@ $$
f(\mathbf{x}) = T(\mathbf{x}) + \frac{1}{(n+1)!} \partial_\mathbf{h}^{n+1} f(\mathbf{a} + \theta \mathbf{h}). f(\mathbf{x}) = T(\mathbf{x}) + \frac{1}{(n+1)!} \partial_\mathbf{h}^{n+1} f(\mathbf{a} + \theta \mathbf{h}).
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
apply Taylor’s theorem in 1D and the chain rule to the function $\phi : [0, 1] \to \mathbb{R}$ given by Apply Taylor’s theorem in 1D and the chain rule to the function $\phi : [0, 1] \to \mathbb{R}$ given by
$$ $$
\phi(\theta) := f(\mathbf{a} + \theta \mathbf{h}). \phi(\theta) := f(\mathbf{a} + \theta \mathbf{h}).
$$ $$
</details>
<br>
## Other methods ## Other methods
Creating multivariable Taylor polynomials by using 1D Taylor polynomials of the different variables and composing them. Creating multivariable Taylor polynomials by using 1D Taylor polynomials of the different variables and composing them.

36
docs/en/mathematics/ordinary-differential-equations/laplace-transform.md
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@ -18,12 +18,9 @@ $$
on the interval where both are defined. on the interval where both are defined.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
If $c \in \mathbb{R}$ then $cf$ also has a Laplace transform and, If $c \in \mathbb{R}$ then $cf$ also has a Laplace transform and,
@ -45,12 +42,9 @@ $$
on this interval on this interval
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
**More shifting**: let $a>0$, if $f$ has a Laplace transform $F$ on $s_0, \infty$ then the function $g$ given by **More shifting**: let $a>0$, if $f$ has a Laplace transform $F$ on $s_0, \infty$ then the function $g$ given by
@ -66,12 +60,9 @@ $$
on this interval. on this interval.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
**Scaling**: let $a > 0$. If $f$ has a Laplace transform $F$ on $(s_0, \infty)$ then the function $g$ given by **Scaling**: let $a > 0$. If $f$ has a Laplace transform $F$ on $(s_0, \infty)$ then the function $g$ given by
@ -87,12 +78,9 @@ $$
on this interval. on this interval.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
**Derivatives**: if $f$ has a derivative $g$ having a Laplace transform $G$ on the interval $(s_0,\infty)$ then $f$ has a Laplace transform on the same interval, and **Derivatives**: if $f$ has a derivative $g$ having a Laplace transform $G$ on the interval $(s_0,\infty)$ then $f$ has a Laplace transform on the same interval, and
@ -106,10 +94,9 @@ $$
\mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0) \mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0)
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
for large enough $s$, the case $n=1$ follows by integration by parts For large enough $s$, the case $n=1$ follows by integration by parts
$$ $$
\begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*}, \begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*},
@ -121,9 +108,6 @@ $$
\begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}. \begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}.
$$ $$
</details>
<br>
## Examples ## Examples
**Solving a second order linear ODE**: with $y: \mathbb{K} \to \mathbb{R}$ given by **Solving a second order linear ODE**: with $y: \mathbb{K} \to \mathbb{R}$ given by

19
docs/en/mathematics/ordinary-differential-equations/second-order-ode.md
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@ -24,8 +24,6 @@ Then the consequence is that the general solution is a linear space.
$(*)$ is said to have **resonance** if $f$ can be split into linearly independent terms of which at least one lies in the solution space of $(*)$. $(*)$ is said to have **resonance** if $f$ can be split into linearly independent terms of which at least one lies in the solution space of $(*)$.
<br>
### Solving homogeneous linear second-order ODEs with constant coefficients ### Solving homogeneous linear second-order ODEs with constant coefficients
Therefore solving Therefore solving
@ -52,12 +50,9 @@ $$
y(t) = (c_1 + c_2t) e^{\lambda_1 t}. y(t) = (c_1 + c_2t) e^{\lambda_1 t}.
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
will be added later. Will be added later.
</details>
<br>
#### Example #### Example
@ -73,8 +68,6 @@ $$
y(t) = e^{-2t}\big(d_1\cos 2t + d_2 \sin 2t \big), \quad d_1,d_2 \in \mathbb{R}. y(t) = e^{-2t}\big(d_1\cos 2t + d_2 \sin 2t \big), \quad d_1,d_2 \in \mathbb{R}.
$$ $$
<br>
### Solving inhomogeneous linear second-order ODEs with constant coefficients ### Solving inhomogeneous linear second-order ODEs with constant coefficients
*Theorem*: let $y_p$ be a particular solution to $(*)$. Then the general solution to $(*)$ is given by *Theorem*: let $y_p$ be a particular solution to $(*)$. Then the general solution to $(*)$ is given by
@ -85,13 +78,9 @@ $$
with $y_h$ the solution to the homegeneous case. with $y_h$ the solution to the homegeneous case.
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
let $y$ be a solution to $(*)$, then $L[y - y_p] = L[y] - L[y_p] = f - f = 0$. Therefore $y = (y - y_p) + y_p = y_h + y_p$. Let $y$ be a solution to $(*)$, then $L[y - y_p] = L[y] - L[y_p] = f - f = 0$. Therefore $y = (y - y_p) + y_p = y_h + y_p$.
</details>
<br>
#### Method of variation of parameters #### Method of variation of parameters

8
docs/en/mathematics/ordinary-differential-equations/systems-of-linear-ode.md
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@ -50,13 +50,9 @@ $$
\mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_h(t), \qquad t \in I \mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_h(t), \qquad t \in I
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
similar to 1d case, will be added later. Similar to 1d case, will be added later.
</details>
<br>
### Method of variation of parameters ### Method of variation of parameters