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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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45
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
@ -90,24 +84,21 @@ 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
$$ $$
\partial_1 f(\mathbf{x}) \dot x(t) + \partial_2 f(\mathbf{x}) \dot y(t) = 0, \partial_1 f(\mathbf{x}) \dot x(t) + \partial_2 f(\mathbf{x}) \dot y(t) = 0,
$$ $$
at $t=0$, we can rewrite this to at $t=0$, we can rewrite this to
$$ $$
\big\langle \nabla f(\mathbf{a}),\; \mathbf{\dot r}(0) \big\rangle = 0, \big\langle \nabla f(\mathbf{a}),\; \mathbf{\dot r}(0) \big\rangle = 0,
$$ $$
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. Will be added later.
</details>
<br>

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

14
docs/en/mathematics/multivariable-calculus/taylor-polynomials.md
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@ -24,17 +24,13 @@ $$
f(\mathbf{x}) = T(\mathbf{x}) + \frac{1}{(n+1)!} \partial_\mathbf{h}^{n+1} f(\mathbf{a} + \theta \mathbf{h}). f(\mathbf{x}) = T(\mathbf{x}) + \frac{1}{(n+1)!} \partial_\mathbf{h}^{n+1} f(\mathbf{a} + \theta \mathbf{h}).
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
apply Taylor’s theorem in 1D and the chain rule to the function $\phi : [0, 1] \to \mathbb{R}$ given by Apply Taylor’s theorem in 1D and the chain rule to the function $\phi : [0, 1] \to \mathbb{R}$ given by
$$ $$
\phi(\theta) := f(\mathbf{a} + \theta \mathbf{h}). \phi(\theta) := f(\mathbf{a} + \theta \mathbf{h}).
$$ $$
</details>
<br>
## Other methods ## Other methods

50
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
@ -106,23 +94,19 @@ $$
\mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0) \mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0)
$$ $$
<details> ??? note "*Proof*:"
<summary><em>Proof</em>:</summary>
for large enough $s$, the case $n=1$ follows by integration by parts For large enough $s$, the case $n=1$ follows by integration by parts
$$ $$
\begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*}, \begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*},
$$ $$
suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$, then by assumption suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$, then by assumption
$$ $$
\begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}. \begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}.
$$ $$
</details>
<br>
## Examples ## Examples

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