# Tag Info

114

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 examples. However, your particular function is convex on the domain $k>0$. In this case the extremum is at one or both of the two integers nearest to the ...

66

Yes. There is a geometric explanation. For simplicity, let me take $x=0$ and $h=1$. By the Fundamental Theorem of Calculus (FTC), $$f(1)=f(0)+\int_{0}^{1}dt_1\ f'(t_1)\ .$$ Now use the FTC for the $f'(t_1)$ inside the integral, which gives $$f'(t_1)=f'(0)+\int_{0}^{t_1}dt_2\ f''(t_2)\ ,$$ and insert this in the previous equation. We then get $$f(1)=f(0)+... 51 Let's consider the following, very simple, differential equation: f'(x) = g(x), where g(x) is some given function. The solution is, of course, f(x) = \int g(x) dx, so for this specific equation the question you're asking reduces to the question of "which simple functions have simple antiderivatives". Some famous examples (such as g(x) = e^{... 39 The main question here seems to be "why can we differentiate a function only defined on integers?". The proper answer, as divined by the OP, is that we can't--there is no unique way to define such a derivative, because we can interpolate the function in many different ways. However, in the cases that you are seeing, what we are really interested ... 35 Here is a heuristic argument which I believe naturally explains why we expect the factor \frac{1}{k!}. Assume that f is a "nice" function. Then by linear approximation,$$ f(x+h) \approx f(x) + f'(x)h. \tag{1} $$Formally, if we write D = \frac{\mathrm{d}}{\mathrm{d}x}, then the above may be recast as f(x+h) \approx (1 + hD)f(x). Now ... 35 Compare Differential Equations to Polynomial Equations. Polynomial Equations are, arguably, much, much more simple. The solution space is smaller, and the fundamental operations that build the equations (multiplication, addition and subtraction) are extremely simple and well understood. Yet (and we can even prove this!) there are Polynomial Equations for ... 28 The polynomials$$p_k(h):=\frac{h^k}{k!}$$have two remarkable properties: they are derivatives of each other, p_{k+1}'(h)=p_k(h), their n^{th} derivative at h=0 is \delta_{kn} (i.e. 1 iff n=k, 0 otherwise). For this reason, they form a natural basis to express a function in terms of the derivatives at a given point: if you form a linear ... 26 You're correct that it doesn't really make sense to write \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} unless we already know the limit exists, but it's really just a grammar issue. To be precise, you could first say that the difference quotient can be re-written \frac{f(x+h)-f(x)}{h}=2x+h, and then use the fact that \lim\limits_{h\to 0}x=x and \lim\... 20 No. The derivative is defined as$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$This is a limit of real numbers, hence if it exists it is real. 18 Computable Functions are Rare When stating mathematical problems, we usually state them in terms of elementary functions, but most certainly computable functions, because those are the only ones we know how to write down in finite space! Because our brains can only explicitly conceptualize the computable functions, we have an innate bias towards thinking ... 18 Set a = f(0) = g(0) and b = f(1) = g(1). f and g are strictly increasing from [0, 1] to [a, b] and therefore invertible. If we define the function$$ h: [a, b]\to \Bbb R, \, h(t) = f^{-1}(t) - g^{-1}(t) $$then h(a) = h(b) = 0 and Rolle's theorem shows that for some c \in (a, b)$$ 0 = h'(c) = \frac{1}{f'(f^{-1}(c))} - \frac{1}{g'(g^{-1}(c))...

15

I think you have kind of hit it on the head when you say that every time we can solve a differential equation, it is an algebraic coincidence. There is simply no good reason why any random equation should have a solution, let alone a nice or basic one. The thought might come about as a result of having been taught, at school or early undergraduate level, ...

14

Consider the function $$f_p(x)=\frac{x^p-1}p.$$ It has the derivative $$f'_p(x)=x^{p-1}$$ and is such that $f_p(1)=0$ and $f_p(0)=-\dfrac1p.$ Now if you let $p$ tend to $0$, you have that $$\lim_{p\to0}f_p(x)=\ln(x)$$ and $$\lim_{p\to0}f'_p(x)=\frac1x.$$ Below, a pencil of curves for various positive and negative $p$. Also consider the inverse of this ...

14

The better way, for me, is as follows: $$f(x)=|\sin(x)|=\sqrt{\sin^2(x)}$$ Now, differentiate both sides to get $$f'(x)=\frac{1}{2\sqrt{\sin^2(x)}}\cdot2\sin(x)\cos(x)=\frac{\sin(2x)}{2|\sin(x)|}$$ Therefore, $$\big(|\sin(x)|\big)'=\frac{\sin(2x)}{2|\sin(x)|}, \ \ \ \ x \neq k\pi, k\in \mathbb{Z}$$ Appendum: This approach can easily be extended to a ...

14

I've thought about this for a few days now, I didn't originally intend to answer my own question but it seems best to write this as an answer rather than add to the question. I think there is nice interpretation in the following: $$f(x) = \lim_{h \to 0} \frac{e^{h f(x)}-1}{h}$$ also consider the Abel shift operator $$e^{h D_x}f(x) = f(x+h)$$ from the ...

13

You are confusing a mathematical model of the system with the system itself. The map is not the territory. Obviously in the real system both $n$ and $k$ must be integers. On the other hand, the math formula for the execution time is a perfectly good function for any real (or even complex!) values of $n$ and $k$ except when $k = 0$. So you can certainly find ...

12

Write $$f(x) = \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} = \left( \int_{0}^{1} e^{xs} \, \mathrm{d}s \right)^{\frac{1}{x}}.$$ Now let $0 < x < y$ be arbitrary and write $p = \frac{y}{x} > 1$. Then by the Jensen's inequality applied to the strictly convex function $\varphi(t) = t^p$ over $[0, \infty)$, we get f(x)^{y} = \varphi\left( \int_{... 12 There is a nice trick using multivariable calculus that somehow is more natural: if you write f(y, z) = y^z and g(x) = (x, x) for the diagonal map, then x^x = f(g(x)). Now the differential of f at a point (y, z) is (z y^{z-1}, y^z \log y)^T and the differential of g is just (1,1), so by the chain rule the derivative of x^x is x x^{x-1} \... 11 If f is periodic with period T, then f' is also periodic with period T, because, if f is differentiable at x,\begin{align}f'(x+T)&=\lim_{h\to0}\frac{f(x+T+h)-f(x+T)}h\\&=\lim_{h\to0}\frac{f(x+h)-f(x)}h\\&=f'(x).\end{align}And \{x\}' is periodic with period 1 (although its domain is not \Bbb R). 11 One can solve it recursively. For example, let f_n(x) be such that\left(x-c_1\frac{d}{dx}\right)^nf_n(x)=0.$$Then one needs to find f_{n+1}(x) such that$$\left(x-c_1\frac{d}{dx}\right)^{n+1}f_{n+1}(x)=0\Leftrightarrow \left(x-c_1\frac{d}{dx}\right)^n\left[\left(x-c_1\frac{d}{dx}\right)f_{n+1}(x)\right]=0\Leftrightarrow \left(x-c_1\frac{d}{...

11

Consider the function: $$f(\lambda,x) = \exp(\lambda x)$$ Then the function you want to differentiate n times w.r.t. x is $\frac{\partial^2 f}{\partial\lambda^2}$ at $\lambda = 1$. So, we want to evaluate: $$\frac{\partial^n}{\partial x^n}\frac{\partial^2 f}{\partial\lambda^2}$$ We can then interchange the order of differentiation to write this as: $$\... 11 I think an analogy with computer science may provide some insight. There are extremely simple programs that produce solutions of extraordinary complexity. The famous Rule 30 in cellular automata is the prime example: With a handful of bytes, one can write a deterministic program whose output is "as complex as possible," that is, it passes all ... 11 No, it is not true. Take f_n(x)=\sqrt{\frac1n+\left(x-\frac12\right)^2}. Then each f_n is a C^\infty function. But (f_n)_{n\in\Bbb N} converges uniformly to f\colon[0,1]\longrightarrow\Bbb R with f(x)=\left|x-\frac12\right|, which is not differentiable. 10 f(x) = -ce^x, c > 0 This isn't a particularly exciting answer, but it is the correct one. All functions that are their own derivatives are of the form f(x) = ce^x, c \in \mathbb{R}, as explained in this question: Prove that C\exp(x) is the only set of functions for which f(x) = f'(x) 10 So there are two inequalities to be proved. You can use that \sqrt{1+x^6} \leq \sqrt{2} for all x \in [-1,1] for the upper bound, as it follows \int_{[-1,1]} \sqrt{1+x^6} dx\leq \int_{[-1,1]} \sqrt{2} dx\leq 2 \sqrt{2}. The lower bound follows very similarly. 10 Just take any continuous function with its value changed at a single point. For example let f:[0, 1] \to \mathbb R be any continuous function and define \tilde{f}: [0, 1] \to \mathbb R by$$ \tilde{f}(x) = \begin{cases} f(x) + 1 & \text{if $x = 1/2$}, \\ f(x) & \text{if $x \neq 1/2$}. \end{cases}$$Then since f and \tilde{f} differ by only a ... 10 We don't do it by choice. We observe nature and notice that things are governed by differential equations. Suppose you let the water out of your bath tub. Initially the water leaves very quickly as the pressure is high. But as the water level drops, the pressure also drops and the water leaves slower. The rate of water leaving is related to the state of how ... 10 Hint : Show inductively that$$f'(x)=\left[\cos(x)\right]\times\left[\cos(\sin(x))\right]\times\left[\cos(\sin(\sin(x))\right]\times\left[...\right]\times\left[\cos(\sin(\sin(\sin(...(x)))))\right]

9

The absolute value function is continuous so has an antiderivative. The antiderivative is differentiable at 0, but its derivative (the absolute value function) is not.

Only top voted, non community-wiki answers of a minimum length are eligible