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185 votes

Why not include as a requirement that all functions must be continuous to be differentiable?

Because that suggests that there might be functions which are discontinuous at $a$ for which it is still true that the limit$$\lim_{t\to0}\frac{f(a+t)-f(a)}t$$exists. Besides, why add a condition ...
José Carlos Santos's user avatar
167 votes
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Why does L'Hopital's rule fail in calculating $\lim_{x \to \infty} \frac{x}{x+\sin(x)}$?

Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rule says is that IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists ...
Barry Cipra's user avatar
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161 votes
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Why not include as a requirement that all functions must be continuous to be differentiable?

Definitions tend to be minimalistic, in the sense that they don't include unnecessary/redundant information that can be derived as a consequence. Same reason why, for example, an equilateral triangle ...
dxiv's user avatar
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154 votes
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What is the probability that a point chosen randomly from inside an equilateral triangle is closer to the center than to any of the edges?

You are right to think of the probabilities as areas, but the set of points closer to the center is not a triangle. It's actually a weird shape with three curved edges, and the curves are parabolas. ...
Zubin Mukerjee's user avatar
133 votes

Is there any integral for the Golden Ratio?

Potentially interesting: $$\log\varphi=\int_0^{1/2}\frac{dx}{\sqrt{x^2+1}}$$ Perhaps also worthy of consideration: $$\arctan \frac{1}{\varphi}=\frac{\int_0^2\frac{1}{1+x^2}\, dx}{\int_0^2 dx}=\frac{...
πr8's user avatar
  • 10.8k
127 votes
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Is there a symbol for "taking a derivative of something"?

You would denote the derivative of $5x^3+7x^2+4x+9$ as $$\frac{d}{dx}(5x^3+7x^2+4x+9)$$ That is the only notation I've ever seen unless the expression is expressed as a function.
Franklin Pezzuti Dyer's user avatar
124 votes

$100$-th derivative of the function $f(x)=e^{x}\cos(x)$

HINT: $e^x\cos x$ is the real part of $y=e^{(1+i)x}$ As $1+i=\sqrt2e^{i\pi/4}$ $y_n=(1+i)^ne^{(1+i)x}=2^{n/2}e^x\cdot e^{i(n\pi/4+x)}$ Can you take it from here?
lab bhattacharjee's user avatar
122 votes
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Why it is important to write a function as sum of even and odd functions?

When I was a high school student I thought that the even/odd decomposition you write about seemed kind of peculiar and not so fundamental. After learning more mathematics I realized the method behind ...
KCd's user avatar
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119 votes
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Why can't calculus be done on the rational numbers?

$\newcommand{\QQ}{\mathbb{Q}}$ Derivatives don't really go wrong, it's antiderivatives. (EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on ...
Henry Swanson's user avatar
119 votes
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Why do engineers use derivatives in discontinuous functions? Is it correct?

In general, computing the extrema of a continuous function and rounding them to integers does not yield the extrema of the restriction of that function to the integers. It is not hard to construct ...
Servaes's user avatar
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112 votes
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If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?

We can write: $$1+ x + x^2 = \frac{1-x^3}{1-x}$$ Therefore: $$f(x) = \frac{1-x}{1-x^3} $$ We can then expand this in powers of $x$: $$f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$$ which is valid for $...
Count Iblis's user avatar
  • 10.4k
111 votes

How can a "proper" function have a vertical slope?

The tangent line is simply an ideal picture of what you would expect to see if you zoom in around the point. $\hspace{8em}$ Hence, the vertical tangent line to the graph $y = \sqrt[3]{x}$ at $(0,0)$ ...
Sangchul Lee's user avatar
109 votes
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Is it possible for the derivative of a function to grow arbitrarily faster than the function itself?

Consider the differential equation $$ \frac{f'}{f} = g $$ where $g$ is the fast-growing function you want. For instance, for $g(x) = e^x$ (and say the initial condition $f(0) =1$) you get $$f(x) = e^{...
Clement C.'s user avatar
  • 67.4k
108 votes
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The deep reason why $\int \frac{1}{x}\operatorname{d}x$ is a transcendental function ($\log$)

I'll try to give a soft answer to what I see as the spirit of the question, which is not why you get exactly log, but why the behaviour is different when integrating $x^{k}$ for $k=-1$. The way I see ...
Anonymous's user avatar
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107 votes
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$100$-th derivative of the function $f(x)=e^{x}\cos(x)$

Find fewer order derivatives: \begin{align} f'(x)&=&e^x (\cos x -\sin x)&\longleftarrow&\\ f''(x)&=&e^x(\cos x -\sin x -\sin x -\cos x) \\ &=& -2e^x\sin x&\...
choco_addicted's user avatar
106 votes
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Why is Euler's number $2.71828$ and not anything else?

$\sum\frac1{n!}$ is not that special. $\lim_{n\to\infty}\left(1+\frac1n\right)^n$ is not really special. $f'(x)=f(x)$ is a very simple differential equation, but unremarkable, really. $\ln (x)$ is ...
Arthur's user avatar
  • 200k
100 votes
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Why can't the second fundamental theorem of calculus be proved in just two lines?

The problem with your proof is the assertion Now $dF$ is just the small change in $F$ and $f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis. That is indeed intuitively ...
Ethan Bolker's user avatar
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98 votes
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Motivation for the rigour of real analysis

In general, the push for rigor is usually in response to a failure to be able to demonstrate the kinds of results one wishes to. It's usually relatively easy to demonstrate that there exist objects ...
Stella Biderman's user avatar
97 votes

Why and How do certain manipulations in indefinite integrals "just work"?

Many otherwise-mysterious tricks in integrals involving trigonometric functions can be explained by expressing the trig functions in terms of exponentials, as in $\cos(x)=(e^{ix}+e^{-ix})/2$. The ...
paul garrett's user avatar
  • 52.6k
94 votes

Why can't calculus be done on the rational numbers?

This is a slightly softer answer. You can 'do calculus' in-so-far as you can define the derivative and perhaps compute some things. But you'll get no theorems out: the main interval theorems (the ...
Will R's user avatar
  • 8,996
94 votes

Intuition behind this interesting calculus result?

Yes, this was considered a pretty strange phenomenon when Torricelli first constructed such an example in 1643. (Torricelli himself found it so incredible that he offered two different proofs that the ...
hmakholm left over Monica's user avatar
93 votes
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How can a "proper" function have a vertical slope?

No, we don't need two vertical points. By the same idea, if the graph of a function $f$ has an horizontal tangent line somewhere, then there has to be two points of the graph of $f$ with the same $y$ ...
José Carlos Santos's user avatar
93 votes
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Why does a polynomial with real, simple roots change its sign between its roots?

Firstly, to simplify the problem, start by re-numbering all the $a_{i}$’s from least to greatest. Think of the behavior at $x=a_n$ notice how the polynomial will look like $$(\text{pos numb})(\text{...
Snacc's user avatar
  • 2,085
89 votes

Is there any integral for the Golden Ratio?

In this answer, it is shown that $$ \int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi} $$
robjohn's user avatar
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89 votes
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What is $\, _4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)$?

A complete answer now. If we exploit the identities $$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{1}$$ $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{2}\sum_{n\geq 1}\frac{4^n ...
Jack D'Aurizio's user avatar
88 votes

What does Big O actually tell you?

Reading between the lines, I think you may be misunderstanding Big O analysis as being a replacement for benchmarking. It's not. An engineer still needs to benchmark their code if they want to know ...
Aurast's user avatar
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85 votes
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Mathematical symbol for 'slightly greater than'?

More often it is used as $b=a+\epsilon$ where $\epsilon$ normally stands for a small positive quantity. That provides b slightly greater than a. Similarly $-\epsilon$ for slightly below.
AHusain's user avatar
  • 5,183
83 votes
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Can a function have two derivatives?

The $(\cdot)^{\frac{1}{4}}$ operation has to be understood as a function. A function can only have one image for any argument. Depending upon how you interpret the fourth root, the image could be ...
Shubham Johri's user avatar

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