daviddoji/CalculusWithJuliaNotes.jl
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Update 40 & 41

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some typos.
This commit is contained in:
Fang Liu
2023-05-21 11:43:35 +08:00
parent 6c8b7dda94
commit eaac42f6dd
2 changed files with 11 additions and 11 deletions
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8
quarto/integrals/partial_fractions.qmd
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@@ -86,7 +86,7 @@ $$
a(x) Q(x) + b(x) q(x) = 1.
$$
Then dividing by $q^k(x)Q(x)$ gives the decomposition
Then dividing by $q(x)^kQ(x)$ gives the decomposition
$$
@@ -175,7 +175,7 @@ $$
\frac{Ax + B}{((ax)^2 \pm 1)^j}
$$
This can be done by finding a $d$ so that $a(x-d)^2 + b(x-d) + c = dx^2 + e = e((\sqrt{d/e}x^2 \pm 1)$.
This can be done by finding a $d$ so that $a(x-d)^2 + b(x-d) + c = dx^2 + e = e((\sqrt{d/e}x)^2 \pm 1)$.
The integrals of the type $Ax/((ax)^2 \pm 1)$ can completed by $u$-substitution, with $u=(ax)^2 \pm 1$.
@@ -210,7 +210,7 @@ In [Bronstein](http://www-sop.inria.fr/cafe/Manuel.Bronstein/publications/issac9
\begin{align*}
\int \frac{A}{(x-a)^j} &= \frac{A}{1-j}\frac{1}{(x-a)^{1-j}}\\
\int \frac{A}{(x-a)^j} &= \frac{A}{1-j}\frac{1}{(x-a)^{j-1}}\\
\int \frac{A}{x-a} &= A\log(x-a)\\
\int \frac{Bx+C}{x^2 + bx + c} &= \frac{B}{2} \log(x^2 + bx + c) + \frac{2C-bB}{\sqrt{4c-b^2}}\cdot \arctan\left(\frac{2x+b}{\sqrt{4c-b^2}}\right)\\
\int \frac{Bx+C}{(x^2 + bx + c)^j} &= \frac{B' x + C'}{(x^2 + bx + c)^{j-1}} + \int \frac{C''}{(x^2 + bx + c)^{j-1}}
@@ -466,7 +466,7 @@ answ = 3
radioq(choices, answ, keep_order=true)
```
##### Question
###### Question
The partial fraction decomposition, as presented, factors the denominator polynomial into linear and quadratic factors over the real numbers. Alternatively, factoring over the complex numbers is possible, resulting in terms like: