# Tag Info

14

After having given him the square root function definition, it may be helpful to show the following: $$\sqrt{(-3)^2}=\sqrt{9}=\sqrt{3^2}=3\neq -3.$$ Although well-definedness may be a bit of a heavy topic for a 14-year old, it should be very easy to explain why you cannot have ambiguous definitions in mathematics.

7

This is about the difference between a function on one variable and an equation with one variable. A function should only have one result for each argument, whereas an equation can have more than one solution, or no solutions at all. Let's consider a very simple function, $f(x) = x + 1$. For example, $f(3) = 4$, $f(4) = 5$. If $x$ doesn't change, neither ...

7

I think the kid has misunderstood something his teacher said about cancelling. A math student of any age is bound to misunderstand his teacher at some point or other. When I was in high school, long, long ago, Mr. Jones was fond of saying that such and such equation has no real solutions, equations like, say, $x^2 + 9 = 0$. I thought he meant such an ...

6

Give him the definition of the square root function : for $x\in\mathbf{R}_{+}$, the number $\sqrt{x}$ is by definition the unique number $y\in\mathbf{R}_{+}$ such that $y^2 = x$.

5

Explain how the squaring erases the information about the sign of $-3$. The square root operation only sees a 9 coming in, it can't know any more that it was made by squaring a $-3$ or a $+3$. So by convention it is defined to always give the positive answer. We could also define it to always give the negative one (although it would be clumsy), but what we ...

5

He (and his teacher, if the latter did indeed say that) may be right. After all what does the expression "the negative square root" mean. Many people use it, and it makes sense, and the negative square root of $(-3)^2$ is indeed $-3$. So I believe a proper explanation should take into account that the terms "negative square root" and "square root" do ...

5

There have been some comments about this requiring memorization or it being different from the way the word is used in normal conversation, but this simply isn't true. It's English. You can walk the dog only if you have a leash. Therefore, if you can walk the dog, it follows logically that you must have a leash. Can walk the dog implies have a leash. You ...

5

Actually your question is the combination of two questions. About your first question: You certainly know the Euclidean norm, $$\left\|v\right\|_2 = \sqrt{\sum_{k=1}^n \left|v_k\right|^2} = \left(\sum_{k=1}^n \left|v_k\right|^2\right)^{\frac{1}{2}}$$ derived from the Euclidean scalar product. Note that I used absolute value bars here, which are superfluous ...

4

$v\otimes w$ is either of the following: The rank 1 linear mapping $V^*\to V$ given by $\alpha\mapsto v.\langle \alpha,w\rangle$ where $\langle\;,\;\rangle:V^*\times V\to \mathbb F$ is the duality. This is the easiest to visualize. The decomposable bilinear form $V^*\times V^*\to \mathbb F$ given by $(\alpha,\beta)\mapsto \langle\alpha,v\rangle ... 4 If he is struggling with math, he may find a graphic argument more helpful. Draw a graph of$f(x) = x^2$and show him that every positive element on the$y$axis has two (symmetric) preimages. Which one we choose to invert the function is indeed (mostly) irrelevant so it is a matter of convention: we like continuous functions, so we want to pick one side of ... 4 There is an unspoken agreement that$\sqrt{x}$means the principal square root of$x$. The principal square root of a positive real number is a positive real number, and the principal square root of a negative real number is a positive imaginary number, though of course a negative imaginary number is just as valid a square root of a negative real number. To ... 4 For a student struggling with math, first I would get the student comfortable with problems like$\sqrt{25}=5$(square root of a specific positive number is positive). Also for these initial problems the student should understand that we mean the positive square root (i.e. we would not say that$\sqrt{25}$is$\pm 5$; nor that it can be$-5$if we want). ... 4 Here is an argument: Technical modification. Fix any$s_1 > s_0$, and let $$H_0 (x) = \int_{0}^{x} e^{-s_1 t}h(t) \, dt \quad \text{and} \quad H(x) = e^{s_1 x}H_0 (x).$$ Then$H_0(x)$is differentiable and$H_0(x) = o(1)$as$x \to \infty$. Then for any$s > s_1, we have $$\int_{0}^{R} e^{-st} h(t) \, dt = \left[ e^{-(s-s_1)t} H_0(t) ... 3 The assumption is equivalent to$$ \int_0^\infty e^{-sx}\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{1} $$for all s\gt0 Since$$ \begin{align} \int_0^\infty\left(e^{-x}-e^{-2x}\right)^n\,\mathrm{d}x &=\int_0^\infty e^{-(n-1)x}\left(1-e^{-x}\right)^n\,\mathrm{d}e^{-x}\\ &=\int_0^1t^{n-1}(1-t)^n\,\mathrm{d}t\\ &=\frac{n!(n-1)!}{(2n)!}\tag{2} ... 3 Many issues I had at a similar age were rectified after being taught by a teacher about order of operations. First Parentheses, Division, Multiplication, Addition and finally subtraction. It might be useful to introduce the kid to a certain "order" to which you must comply if you want to simplify a formula which might include exponentiation and taking ... 3 By definition, the square root ofx$(if it exists) is the (unique) non-negative number$y \geq 0$such that$x = y^2$. The square root only exists if$x \geq 0$, and in this case it always exists and is unique. If$y \geq 0$then it is true that$\sqrt{y^2} = y$. More generally, $$\sqrt{y^2} = |y|.$$ For a more complete understanding, your ... 3 It's not quite right to say that the square and the square root cancel each other (in fact, this example demonstrates that nicely!) For this problem, we have$\sqrt{(-3)^2} = \sqrt{9} = 3$. But why are these the right two steps? Because we evaluate functions from the inside out. Thus, the first step is to square the$-3$. The second step is a matter of ... 3 I understand "A only if B" as "A can be true only in those possible worlds where B is true also". 3 It might help if you have an example. http://calculus.subwiki.org/wiki/Derivative_of_differentiable_function_need_not_be_continuous for instance. What happens here is that the function oscillates faster and faster, so the derivative jumps up and down more and more and never settles on a value as$x\rightarrow 0$. 3 Let's just think about norms in the plane. The ordinary Euclidean norm is$(x^2 + y^2)^{1/2}$. The unit circle for that norm is the ordinary unit circle. There are other useful norms. One is$|x| + |y|$. Its "unit circle" is the square with vertices$(\pm 1, 0)$and$(0, \pm 1)$. These norms are part of a family of norms defined by$\|(x,y)\|_p = (|x|^p + ...

2

I’ve not dealt much with the concept, but an example that I find helpful is $\Bbb R^2$ with the product partial order, $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. Then it’s not hard to check that $\langle x_0,y_0\rangle\prec\langle x_1,y_1\rangle$ iff $x_0<x_1$ and $y_0<y_1$. Consider, for instance, the set ...

2

To get a broader context for the well above relation you can consult any introductory text on domain theory. However, the context of continuity spaces is a bit different, and I prefer to have the intuition for Flagg's value quantales come directly from their intended role. So, the way I think about the well above relation is that it solves some nasty ...

2

Think Lagrange interpolation and Taylor's formula. In fact, the best way is to look at linear interpolation as the most basic Hermite (and second most basic Lagrange) and make the connection. Suppose you want to approximate a differentiable function $f:\mathbb R\to\mathbb R$ and you give yourself two nodes: $x_0$ and $x_1$, with $x_1\neq x_0$. The main ...

2

Both kinds of interpolation formulas rely on the superposition principle (the sum of the effect of individuals causes is the effect of the sum of the causes), and achieve a decomposition such that every point brings its own contribution. Actually, you form a basis of polynomials and linear combinations thereof. In the case of Lagrange, consider the special ...

2

Recall that at every point $p \in P$ the vertical tangent space $V_p P$ can be identified with the Lie algebra of the structure group via $(\Psi_p)': \mathfrak{g} \to T_pP$ where the group action in $P$ is denoted by $\Psi: P \times G \to P$. If $\omega$ is your connection form, then the evaluation $\omega(X) \in \mathfrak{g}$ is nothing else as the vertical ...

2

Formally, Birkhoff's ergodic theorem states that for an integrable function $f$, the time average equals the space average: $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1} f(T^i(x)) = \int_S f(x) d\mu(x).$$ Here, $(S,\mathcal{S})$ is a measurable space, $T : S \rightarrow S$ is an ergodic map for the measure $\mu$, and $f \in L^1$. Example 1 Suppose ...

2

Your thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.

2

Remember we are working with principal bundles. Thus, contrary to what other people seem to think, a curvature $2$-form is not the same thing as a Riemann curvature tensor. However, we can get a Riemann curvature tensor from it, in specific cases. We shall discuss that construction momentarily. Let $P$ be a principal $G$-bundle with projection $\pi:P\to B$. ...

1

Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas. Proposition P could stand for "Someone who was born in Dallas" Proposition Q could stand for "Someone who is texan" Truth table for implication is: So, in this example is easy to check line by line ...

1

This question has been asked and answered on MathOverflow. I have replicated David Speyer's accepted answer below. Here is some nonsense that I find useful: On a complex manifold, \frac{\mbox{locally constant functions}}{\mbox{smooth functions}} \approx \frac{\mbox{locally constant functions}}{\mbox{holomorphic functions}} \cdot ...

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