rm WeaveSupport
This commit is contained in:
@@ -14,19 +14,6 @@ using Roots
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using Polynomials # some name clash with SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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fig_size=(800, 600)
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const frontmatter = (
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title = "Curve Sketching",
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description = "Calculus with Julia: Curve Sketching",
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tags = ["CalculusWithJulia", "derivatives", "curve sketching"],
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);
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nothing
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```
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---
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@@ -611,4 +598,3 @@ choices = [
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answ = 1
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radioq(choices, answ)
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```
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@@ -15,17 +15,7 @@ using SymPy
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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using DataFrames
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const frontmatter = (
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title = "Derivatives",
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description = "Calculus with Julia: Derivatives",
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tags = ["CalculusWithJulia", "derivatives", "derivatives"],
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);
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fig_size=(800, 600)
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nothing
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```
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@@ -1645,4 +1635,3 @@ The limit of a composition (under assumptions on ``v``):
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]
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radioq(choices, 3, keep_order=true)
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```
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@@ -13,19 +13,6 @@ using SymPy
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using Roots
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "The first and second derivatives",
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description = "Calculus with Julia: The first and second derivatives",
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tags = ["CalculusWithJulia", "derivatives", "the first and second derivatives"],
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);
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nothing
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```
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---
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@@ -1048,4 +1035,3 @@ choices = ["As ``x^3`` has no extrema at ``x=0``, neither will ``f``",
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"As ``x^4`` is of higher degree than ``x^3``, ``f`` will be ``U``-shaped, as ``x^4`` is."]
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radioq(choices, 1)
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```
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@@ -13,18 +13,6 @@ using Roots
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using SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "Implicit Differentiation",
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description = "Calculus with Julia: Implicit Differentiation",
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tags = ["CalculusWithJulia", "derivatives", "implicit differentiation"],
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);
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nothing
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```
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---
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@@ -13,21 +13,6 @@ using SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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using Roots
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fig_size=(800, 600)
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const frontmatter = (
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title = "L'Hospital's Rule",
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description = "Calculus with Julia: L'Hospital's Rule",
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tags = ["CalculusWithJulia", "derivatives", "l'hospital's rule"],
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);
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nothing
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```
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---
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@@ -259,6 +244,7 @@ A first proof of L'Hospital's rule takes advantage of Cauchy's [generalization](
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```{julia}
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#| echo: false
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#| cache: true
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using Roots
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let
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## {{{lhopitals_picture}}}
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@@ -836,4 +822,3 @@ choices = [
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answ = 1
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radioq(choices, answ)
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```
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@@ -14,19 +14,6 @@ using TaylorSeries
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using DualNumbers
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "Linearization",
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description = "Calculus with Julia: Linearization",
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tags = ["CalculusWithJulia", "derivatives", "linearization"],
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);
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nothing
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```
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---
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@@ -861,4 +848,3 @@ n=100
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val = exp(-n*(n-1)/2/365)
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numericq(val, 1e-2)
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```
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@@ -16,18 +16,8 @@ using Roots
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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using Printf
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using SymPy
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fig_size = (800, 600)
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const frontmatter = (
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title = "The mean value theorem for differentiable functions.",
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description = "Calculus with Julia: The mean value theorem for differentiable functions.",
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tags = ["CalculusWithJulia", "derivatives", "the mean value theorem for differentiable functions."],
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);
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nothing
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```
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@@ -707,4 +697,3 @@ L"The squeeze theorem applies, as $0 < g(x) < x$",
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answ = 3
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radioq(choices, answ)
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```
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@@ -14,19 +14,6 @@ using Roots
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using SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "Derivative-free alternatives to Newton's method",
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description = "Calculus with Julia: Derivative-free alternatives to Newton's method",
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tags = ["CalculusWithJulia", "derivatives", "derivative-free alternatives to newton's method"],
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);
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nothing
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```
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---
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@@ -614,4 +601,3 @@ choices = [
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answ = 3
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radioq(choices, answ, keep_order=true)
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```
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@@ -13,20 +13,6 @@ using SymPy
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using Roots
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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fig_size = (800, 600)
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const frontmatter = (
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title = "Newton's method",
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description = "Calculus with Julia: Newton's method",
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tags = ["CalculusWithJulia", "derivatives", "newton's method"],
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);
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nothing
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```
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---
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@@ -14,18 +14,6 @@ using SymPy
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using Roots
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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const frontmatter = (
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title = "Numeric derivatives",
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description = "Calculus with Julia: Numeric derivatives",
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tags = ["CalculusWithJulia", "derivatives", "numeric derivatives"],
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);
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nothing
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```
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---
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@@ -383,4 +371,3 @@ fp_(h) = 3*32h^2 - 62
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c = 2
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numericq(fp_(2))
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```
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@@ -13,20 +13,6 @@ using Roots
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using SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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fig_size = (800, 600)
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frontmatter = (
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title = "Optimization",
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description = "Calculus with Julia: Optimization",
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tags = ["CalculusWithJulia", "derivatives", "optimization"],
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);
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nothing
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```
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---
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@@ -1478,4 +1464,3 @@ d(a) = (a-x(a))^2 + (a^2 - x(a)^2)^2
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a = find_zero(d', 1)
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numericq(a)
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```
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@@ -13,19 +13,6 @@ using Roots
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using SymPy
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```
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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fig_size=(800, 600)
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const frontmatter = (
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title = "Related rates",
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description = "Calculus with Julia: Related rates",
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tags = ["CalculusWithJulia", "derivatives", "related rates"],
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);
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nothing
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```
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---
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@@ -814,4 +801,3 @@ choices = [
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answ=1
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radioq(choices, answ)
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```
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@@ -16,15 +16,7 @@ using Unitful
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```{julia}
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#| echo: false
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#| results: "hidden"
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using CalculusWithJulia.WeaveSupport
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using Roots
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fig_size = (800, 600)
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const frontmatter = (
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title = "Taylor Polynomials and other Approximating Polynomials",
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description = "Calculus with Julia: Taylor Polynomials and other Approximating Polynomials",
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tags = ["CalculusWithJulia", "derivatives", "taylor polynomials and other approximating polynomials"],
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);
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nothing
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```
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@@ -1259,5 +1251,3 @@ scatter!(cps, h1.(cps), markersize=5, marker=:square)
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```
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Again by Rolle's theorem, between any pair of adjacent zeros $\xi^1_i, \xi^1_{i+1}$ there must be a zero $\xi^2_i$ of $h''(x)$. So there are $n-1$ zeros of $h''(x)$. Continuing, we see that there will be $n+1-3$ zeros of $h^{(3)}(x)$, $n+1-4$ zeros of $h^{4}(x)$, $\dots$, $n+1-(n-1)$ zeros of $h^{n-1}(x)$, and finally $n+1-n$ ($1$) zeros of $h^{(n)}(x)$. Call this last zero $\xi$. It satisfies $x_0 \leq \xi \leq x_n$. Further, $0 = h^{(n)}(\xi) = f^{(n)}(\xi) - g^{(n)}(\xi)$. But $g$ is a degree $n$ polynomial, so the $n$th derivative is the coefficient of $x^n$ times $n!$. In this case we have $0 = f^{(n)}(\xi) - f[x_0, \dots, x_n] n!$. Rearranging yields the result.
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