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# Math Help - Logic Question

1. ## Logic Question

I'm reading a book about discrete math, and am learning about the rules of inference. Here's the problem:

Given:
i.) p--> q is false
ii.) p is true.

The book says that it cannot be concluded whether or not you can change your grade. It seems to me that it can be it can be concluded that q is false. We know that p is true and p-->q is false; it can be concluded that q is false because p-->q is false only when p is true and q is false.

The book says these p and q are propositions, NOT predicates. Am I missing something here, or is this a mistake by the author? My only guess is that the author is miswrote that p and q are propositions, and in fact is treating p and q as predicates where

q = x can change his grade

In which case "you" are in the domain of discourse, and there exists x1 in the domain of discourse such that p(x1)^not(q(x1)) is true. In this case, it would make sense to state that the conclusion is undecidable.

2. Originally Posted by mbalaban
I'm reading a book about discrete math, and am learning about the rules of inference. Here's the problem:

Given:
i.) p--> q is false
ii.) p is true.

The book says that it cannot be concluded whether or not you can change your grade. It seems to me that it can be it can be concluded that q is false. We know that p is true and p-->q is false; it can be concluded that q is false because p-->q is false only when p is true and q is false.

The book says these p and q are propositions, NOT predicates. Am I missing something here, or is this a mistake by the author? My only guess is that the author is miswrote that p and q are propositions, and in fact is treating p and q as predicates where

q = x can change his grade

In which case "you" are in the domain of discourse, and there exists x1 in the domain of discourse such that p(x1)^not(q(x1)) is true. In this case, it would make sense to state that the conclusion is undecidable.
[Edit: like this post is wrong. See below.]

My understanding is that when you take a predicate and specify a subject, it becomes a proposition. Here, I think it was intended that "you" is not the general term for "somebody" but rather the particular person, you (the reader).

I would write your (i) as

i) (p -> q) is false

just to be extra safe that nobody interprets it as

i) (p -> (q is false)) is true

I can give an illustration of why the book's conclusion is right: Say every Monday when you can access the network, you can change your grade, but only on Monday. So accessing the network is not sufficient for changing your grade, meaning (p -> q) is false. But suppose you have access to the network on all days, meaning p is true. Now given all these facts, we cannot conclude whether q is true, because we don't know what day of the week it is.

3. Perhaps another way of thinking about it is this: the fact that the implication is false doesn't imply that there is only one possible truth assignment to the constituent propositions. It only means that p true and q false is possible.

[EDIT] See posts 4 and 5 below for a correction to this post.

4. Given:
i.) p--> q is false
ii.) p is true.

The book says that it cannot be concluded whether or not you can change your grade. It seems to me that it can be it can be concluded that q is false. We know that p is true and p-->q is false; it can be concluded that q is false because p-->q is false only when p is true and q is false.
I absolutely agree with you; I don't understand why the author claims what he does.

Say every Monday when you can access the network, you can change your grade, but only on Monday. So accessing the network is not sufficient for changing your grade, meaning (p -> q) is false. But suppose you have access to the network on all days, meaning p is true. Now given all these facts, we cannot conclude whether q is true, because we don't know what day of the week it is.
I believe that whether this example applies depends on what the topic the book is teaching. If the topic is the context of reasoning, e.g., adding new, previously hidden, propositions that may change the truth value of the old ones, then yes, the example is relevant. However, if the topic teaches regular propositional logic and regular rules of inference, it is probably a stretch.

Perhaps another way of thinking about it is this: the fact that the implication is false doesn't imply that there is only one possible truth assignment to the constituent propositions. It only means that p true and q false is possible.
Again, if the topic is modal logic, possibility and necessity, then OK. It seemed to me that the OP meant regular introductory discreet math textbook.

5. On second thought, after reading emakarov's post, I'd have to say that if you look at the truth tables involved, you are forced to the conclusion that q is false. It's the only row of the table that works. p is true, q is false, and p -> q is false. So please disregard Post #3.

6. [Edit: like this post is wrong. See below.]

Hmm, here is a more "mathy" example.

Let x be a real number.

Then the statement: $x^2 = 4 \implies x=2$ is false, because we could also have $x = -2$.

Given:

(i) $x^2 = 4 \implies x=2$ is false.
(ii) $x^2 = 4$ is true.

Surely we can't conclude that $\display x \ne 2$ ? Am I missing something?

If you have a problem with "Let x be a real number" then put it this way.

a = "x is a real number"
b = " $x^2 = 4$"
p = $a \land b$
q = "x = 2"

7. I think you can conclude that $x\not=2.$ You can think of the implication $p\to q$ as $\neg p\vee q,$ right? Negate this, now, to get $\neg(\neg p\vee q)\equiv p\land\neg q.$

Therefore, technically, you don't even need the second premiss in the OP in order to conclude that p is true and q is false.

8. Originally Posted by Ackbeet
I think you can conclude that $x\not=2.$ You can think of the implication $p\to q$ as $\neg p\vee q,$ right? Negate this, now, to get $\neg(\neg p\vee q)\equiv p\land\neg q.$

Therefore, technically, you don't even need the second premiss in the OP in order to conclude that p is true and q is false.
Thanks, after looking at the truth tables, I'm given to believe that you are right, nevertheless I'm puzzled as to why my examples don't apply. Maybe something to do with an indeterminate?

Sorry for misinformation ..

Edit: I now definitely believe the discrepancy is due to indeterminates; for example there's no way to fill in values for x and still have it appear the way it did in post #6:

(i) $2^2 = 4 \implies -2=2$ is false.
(ii) $2^2 = 4$ is true.

We can therefore conclude that (-2=2) is false, and like Ackbeet said, we don't even need (ii) to arrive at this conclusion. In fact, the truth of (i) implies the truth of (ii).

So, in the example with days of the week (post #2), I got rid of one indeterminate ("To what person does 'you' refer to?") and replaced it with another indeterminate ("What day of the week is it?"). And in the example with x^2 = 4, x is an indeterminate. So I confused propositional logic with first-order logic (of predicate logic). Sorry again.


text:
## A computational approach to image retrieval

Kunt, Murat
Year:
1999
Publisher:
Lausanne, EPFL
Laboratories:

Note: The status of this file is: EPFL only

format: plain


text:
Commutative residuated lattices

Abbreviation: CRL

Definition

A commutative residuated lattice is a residuated lattice $\mathbf{L}=\langle L, \vee, \wedge, \cdot, e, \backslash, /\rangle$ such that

$\cdot$ is commutative: $xy=yx$

Remark:

Morphisms

Let $\mathbf{L}$ and $\mathbf{M}$ be commutative residuated lattices. A morphism from $\mathbf{L}$ to $\mathbf{M}$ is a function $h:L\rightarrow M$ that is a homomorphism:

$h(x\vee y)=h(x)\vee h(y)$, $h(x\wedge y)=h(x)\wedge h(y)$, $h(x\cdot y)=h(x)\cdot h(y)$, $h(x\backslash y)=h(x)\backslash h(y)$, $h(x/y)=h(x)/h(y)$, and $h(e)=e$

Example 1:

Properties

Classtype Variety Decidable Undecidable Undecidable No Unbounded Yes Yes Yes, n=2 No Yes No Yes No No

Finite members

$\begin{array}{lr} f(1)= &1\\ f(2)= &1\\ f(3)= &3\\ f(4)= &16\\ f(5)= &100\\ f(6)= &794\\ f(7)= &7493\\ f(8)= &84961\\ \end{array}$


text:
# Thread: Finding Z value.

1. ## Finding Z value.

How can i find value of z through dN/dt=0.35+0.25z , dP/dt=-13.38-0.6Z , dz/dt=0.04z? please help me.. thank you so much..

2. ## Re: Please help me finding Z value

Originally Posted by smiley188
How can i find value of z through dN/dt=0.35+0.25z , dP/dt=-13.38-0.6Z , dz/dt=0.04z? please help me.. thank you so much..
The first two equations are irrelevant. What ideas do you have for solving the third equation?


text:
If you find any mistakes, please make a comment! Thank you.

Chapter 8 Exercise A

1. Solution: Since
$$T^2(w, z) = T(z, 0) = (0, 0),$$ it follows that $G(0, T) = V$. Therefore every vector in $\mathbb{C}^2$ is a generalized eigenvector of $T$.

2. Solution: The eigenvalues of $T$ are $i$ and $-i$. Since $\mathbb{C}^2$ has dimension $2$, the generalized eigenspaces are the eigenspaces themselves.

3. Solution: We will prove $\operatorname{null} (T – \lambda I)^n = \operatorname{null} \left(T^{-1} – \dfrac{1}{\lambda} I\right)^n$ for all nonnegative integers $n$ by induction on $n$.

It is easy to check that $\operatorname{null} (T – \lambda I) = \operatorname{null} \left(T^{-1} – \dfrac{1}{\lambda} I\right)$ (see Exercise 9 in section 5C). Let $n > 1$ and assume the result holds for all nonnegative integers less than $n$.

Suppose $v \in \operatorname{null}(T – \lambda I)^n$. Then
$$(T – \lambda I)v \in \operatorname{null}\left(T – \lambda I\right)^{n-1}.$$ By the induction hypothesis $$(T – \lambda I)v \in \operatorname{null}\left(T^{-1} – \frac{1}{\lambda} I\right)^{n-1}.$$ Thus $$0 = \left(T^{-1} – \frac{1}{\lambda} I\right)^{n-1}(T – \lambda I)v = (T – \lambda I)\left(T^{-1} – \frac{1}{\lambda} I\right)^{n-1}v,$$ where the second equality follows from Theorem 1 below.

Theorem 1. Suppose $T \in \mathcal{L}(V)$ is invertible and $p, q \in \mathcal{P}(\mathbb{F})$. Then $p(T^{-1}) q(T) = q(T) p(T^{-1})$.

Proof. The key idea used here is that $T$ commutes with $T^{-1}$, even when raised to different powers.
Suppose $p(z) = \sum_{j=0}^m a_j z^j$ and $q(z) = \sum_{k=0}^n b_k z^k$ for $z \in \mathbb{F}$. Then
\begin{aligned} p\left(T\right)q\left(T^{-1}\right) &= \left(\sum_{j=0}^m a_j T^j\right)\left(\sum_{k=0}^n b_k \left(T^{-1}\right)^k\right)\\ &= \sum_{j=0}^m \sum_{k=0}^n a_j b_k T^j \left(T^{-1}\right)^k\\ &= \sum_{j=0}^m \sum_{k=0}^n b_k a_j \left(T^{-1}\right)^k T^j\\ &= \sum_{k=0}^n \sum_{j=0}^m b_k a_j \left(T^{-1}\right)^k T^j\\ &= \left(\sum_{k=0}^n b_k \left(T^{-1}\right)^k\right)\left(\sum_{j=0}^m a_j T^j\right)\\ &= q\left(T^{-1}\right)p\left(T\right) \end{aligned}.

Therefore $$\left(T^{-1} – \frac{1}{\lambda}I\right)^{n-1}v \in \operatorname{null} (T – \lambda I).$$ But
$$\operatorname{null} (T – \lambda I) = \operatorname{null} \left(T^{-1} – \frac{1}{\lambda} I\right).$$ Hence
$$\left(T^{-1} – \frac{1}{\lambda}I\right)^{n-1}v \in \operatorname{null} \left(T^{-1} – \frac{1}{\lambda} I\right)$$ and so
$$0 = \left(T^{-1} – \frac{1}{\lambda} I\right) \left(T^{-1} – \frac{1}{\lambda}I\right)^{n-1}v = \left(T^{-1} – \frac{1}{\lambda}I\right)^n v,$$ which shows that $v \in \operatorname{null} (T^{-1} – \frac{1}{\lambda}I)$. Therefore $$\operatorname{null} (T – \lambda I)^n \subset \operatorname{null} \left(T^{-1} – \frac{1}{\lambda} I\right)^n.$$ To prove the inclusion in the other direction, it suffices to repeat the same thing replacing $\left(T – \lambda I\right)$ with $\left(T^{-1} – \frac{1}{\lambda}I\right)$ and vice versa.

Now, by 8.11, we have
$$G(\lambda, T) = \operatorname{null}(T – \lambda I)^{\operatorname{dim} V} = \operatorname{null}\left(T^{-1} – \frac{1}{\lambda} I\right)^{\operatorname{dim} V} = G\left(\frac{1}{\lambda}, T^{-1}\right).$$

4. Solution: Suppose $v \in G(\alpha, T) \cap G(\beta, T)$ and suppose by contradiction that $v \neq 0$. Then $v, v$ are generalized eigenvectors corresponding to distinct generalized eigenvalues of $T$. Now 8.13 implies that $v, v$ is linearly independent, which is clearly a contradiction. Therefore $v$ must be $0$.

5. Solution: Let $a_0, a_1, \dots, a_{m-1} \in \mathbb{F}$ such that
$$0 = a_0v + a_1Tv + \dots + a_{m-1}T^{m-1}v$$ Applying $T^{m-1}$ to both sides of the equation above yields
$$0 = a_0T^{m-1}v,$$ which shows that $a_0 = 0$. Therefore
$$0 = a_1Tv + \dots + a_{m-1}T^{m-1}v.$$ Applying $T^{m-2}$ yields
$$0 = a_1T^{m-1}v,$$ which shows that $a_1 = 0$. Continuing in this fashion, we see that $a_0 = a_1 = \dots = a_m = 0$. Thus $v, Tv, T^2v, \dots, T^{m-1}v$ is linearly independent.

6. Solution: Suppose by contradiction that $S \in \mathcal{L}(\mathbb{C}^3)$ is a square root of $T$. Note that $V = \operatorname{null} T^3$. We have
\begin{aligned} V &= \operatorname{null} T^3 = \operatorname{null} R^6\\ &= \operatorname{null} R^3 = \operatorname{null} RT\\ &\subset \operatorname{null} R^2T= \operatorname{null} T^2, \end{aligned} where the third line follows by 8.4. But this a contradiction, since $T^2(z_1, z_2, z_3) = (z_3, 0, 0)$, we see that $\operatorname{null} T^2 = \{(0, 0, z): z \in \mathbb{C}\}$, then we can’t have $V \subset \operatorname{null} T^2$.

7. Solution: This follows directly from 8.19 and 5.32.

8. Solution: False. Let $V = \mathbb{C}^2$. Define $S, T \in \mathcal{L}(\mathbb{C})$ by
\begin{aligned} S(z_1, z_2) &= (0, z_1)\\ T(z_1, z_2) &= (z_2, 0). \end{aligned} Both $S$ and $T$ are nilpotent, however $S + T$ is not (its square equals the identity).

9. Solution: We have
$$\operatorname{null} (TS)^{\operatorname{dim} V} = \operatorname{null} TS(TS)^{\operatorname{dim} V} = \operatorname{null} T(ST)^{\operatorname{dim} V}S = V,$$ where the first equality follows from 8.4 and the third because $(ST)^{\operatorname{dim} V} = 0$ (by 8.18). Thus $(TS)^{\operatorname{dim} V} = 0$ and so $TS$ is nilpotent.

10. Solution: If $T$ is not nilpotent, then $\operatorname{dim} \operatorname{null} T^n < n$ and , by the same reasoning used in 8.4, it follows that $\operatorname{null} T^{n-1} = \operatorname{null} T^n$. Thus, by 8.5, we have
$$V = \operatorname{null} T^{n-1} + \operatorname{range} T^n.$$ Since $\operatorname{range} T^n \subset \operatorname{range} T^{n-1}$, we must also have
$$V = \operatorname{null} T^{n-1} + \operatorname{range} T^{n-1}.$$ Then, by the Fundamental Theorem of Linear Maps (3.22),
$$\operatorname{dim} (\operatorname{null} T^{n-1} + \operatorname{range} T^{n-1}) = \operatorname{dim} V = \operatorname{dim} \operatorname{null} T^{n-1} + \operatorname{dim} \operatorname{range} T^{n-1}.$$ 3.78 now implies that $\operatorname{null} T^{n-1} + \operatorname{range} T^{n-1}$ is a direct sum.

12. Solution: Suppose $v_1, \dots, v_n$ is such basis. Then $Nv_1 = 0$, because the the first column of the matrix has $0$ in all its entries. The definition of matrix of linear map shows that $Nv_2 \in \operatorname{span}(v_1)$. But this implies that $N^2v_2 = 0$. Similarly, $Nv_3 \in \operatorname{span}(v_1, v_2)$, so $N^3v_3 = 0$.

Continuing like this, we see that $N^j v_j = 0$, for each $j = 1, \dots, n$. Therefore $N^n = 0$ and so $N$ is nilpotent.

13. Solution: It is easy when $\mathbb{F} = \mathbb{C}$, because then $V$ has a basis consisting of eigenvectors of $N$ and for each vector $v$ in this basis we have $0 = N^{\operatorname{dim} V} v = \lambda^{\operatorname{dim} V} v$ for the corresponding eigenvalue $\lambda$, which implies that $\lambda = 0$.

More generally, without restricting $\mathbb{F}$ to $\mathbb{C}$, we will prove $N^{\operatorname{dim} V – 1} = 0$ and this fact can be used to show $N^{\operatorname{dim} V – 2} = 0$, which then can be used to show… and so on until $N^1$.

Let $\mathcal{N} = N^{\operatorname{dim} V – 1}$. Note that $\mathcal{N}$ is also normal and that $\mathcal{N}^2 = 0$. Then, for all $v \in V$,
$$||\mathcal{N}^*\mathcal{N}v||^2 = ||\mathcal{N}\mathcal{N}v|| = 0,$$ where the first equality comes from 7.20. Thus $\mathcal{N}^*\mathcal{N} = 0$. Therefore
$$||\mathcal{N}v||^2 = \langle \mathcal{N}v, \mathcal{N}v \rangle = \langle v, \mathcal{N}^*\mathcal{N}v \rangle = 0,$$ which shows that $\mathcal{N} = 0$.

14. Solution: This follows directly from 8.19 and 6.37.

15. Solution: By the same reasoning used in the proof of 8.4, it follows that $\operatorname{dim} \operatorname{null} N^{\operatorname{dim} V} \ge \operatorname{dim} V$. Thus $\operatorname{dim} \operatorname{null} N^{\operatorname{dim} V} = \operatorname{dim} V$ and so $N$ is nilpotent. We have $\operatorname{dim} V + 1$ null spaces each of different dimension. Since the sequence
$$\operatorname{dim} \operatorname{null} N^0, \operatorname{dim} \operatorname{null} N^1, \dots, \operatorname{dim} \operatorname{null} N^{\operatorname{dim} V}$$ must be sorted in strictly increasing order, the only way this can fit is if $\operatorname{dim} \operatorname{null} N^j = j$ for each $j$.

16. Solution: Obviously $V = \operatorname{range} T^0 = \operatorname{range} I$. Let $k$ be a nonnegative integer. Suppose $v \in \operatorname{range} T^{k+1}$. Then $v = T^{k+1}u$ for some $u \in V$. But then $v = T^k(Tu)$. This implies that $v \in \operatorname{range} T^k$.

17. Solution: By the Fundamental Theorem of Linear Maps (3.22), we have
$$\operatorname{dim} \operatorname{null} T^m + \operatorname{dim} \operatorname{range} T^m = \operatorname{dim} \operatorname{null} T^{m+1} + \operatorname{dim} \operatorname{range} T^{m+1},$$ which implies that $\operatorname{dim} \operatorname{null} T^m = \operatorname{dim} \operatorname{null} T^{m+1}$ (because $\operatorname{dim} \operatorname{range} T^m = \operatorname{dim} \operatorname{range} T^{m+1}$). Thus, by 8.3, for all $k > m$, we have
$$\operatorname{dim} \operatorname{null} T^m = \operatorname{dim} \operatorname{null} T^{m+k}.$$ Applying the Fundamental Theorem of Linear Maps again to $T^m$ and $T^{m+k}$ we see that
$$\operatorname{dim} \operatorname{range} T^m = \operatorname{dim} \operatorname{range} T^{m+k}.$$ Since $\operatorname{range} T^{m+k} \subset \operatorname{range} T^m$, it follows that $\operatorname{range} T^{m+k} = \operatorname{range} T^m$.

18. Solution: This follows directly from the previous exercise and 8.4.

19. Solution: This is just a matter of realizing that $\operatorname{null} T^m \subset \operatorname{null} T^{m+1}$ and $\operatorname{range} T^{m+1} \subset \operatorname{range} T^m$ and applying the Fundamental Theorem of Linear Maps.

20. Solution: By Exercise 19, $\operatorname{null} T^4 \neq \operatorname{null} T^5$. By Exercise 15, this implies that $T$ is nilpotent.

21. Solution: Let $W = \mathbb{F}^\infty \times \mathbb{F}^\infty$ and define $T \in \mathcal{L}(W)$ by
$$T\bigr((x_1, x_2, x_3, \dots), (y_1, y_2, y_3, \dots)\bigl) = \bigr((x_2, x_3, \dots), (0, y_1, y_2, y_3, \dots)\bigl),$$ that is, $T$ applies the backward shift operator (call it $B$) on the first slot and forward shift operator (call it $F$) on the second slot. Thus, for each positive integer $k$, we have $$\operatorname{null} B^k = \{(x_1, x_2, x_3, \dots) \in \mathbb{F}^\infty: x_j = 0 \text{ for all } j > k\}$$ and $$\operatorname{range} F^k = \{(x_1, x_2, x_3, \dots) \in \mathbb{F}^\infty: x_1 = x_2 = \dots = x_k = 0\}.$$ Moreover $\operatorname{range} B^k = \mathbb{F}^\infty$ and $\operatorname{null} F^k = \{0\}$. Note that $\operatorname{null} B^k \subsetneq \operatorname{null} B^{k+1}$ and $\operatorname{range} F^k \supsetneq \operatorname{range} F^{k+1}$. Thus $$\operatorname{null} T^k = \{(x, 0) \in \mathbb{F}^\infty \times \mathbb{F}^\infty: x \in \operatorname{null} B^k\}$$ and
$$\operatorname{range} T^k = \{(x, y) \in \mathbb{F}^\infty \times \mathbb{F}^\infty: y \in \operatorname{range} T^k\}.$$ Hence $\operatorname{null} T^k \subsetneq \operatorname{null} T^{k+1}$ and $\operatorname{range} T^k \supsetneq \operatorname{range} T^{k+1}$.

Linearity

1. for the ex 3, I have a convenient method , note the $\mathrm{ex} 4,5 , 3.D}$ , then how about consider
$$\mathrm{dim}\ker(\lambda_iT)^n(T^{-1}-\lambda_iI)^n$$

2. For ex $16,17,18,19$ , why not use theorem $8.2-8.4$ on $T^*$, and then consider the orthogonal complement.

1. Did you see anything inner product in this section? I have no idea what are you talking about.

Yes, you are right.

3. Any thoughts on 11?

1. Consider the counterexample, $I+N$, where $N$ is a nonzero nilpotent operator such that $N\ne 0$ and $N^2=0$. I will update details later.

1. The nilpotent operator N will have an upper triangular matrix with zeros on the diagonal. The square of I + N will be I + 2N which will be upper triangular, so all the eigenvalues are 1 by 5.32.
There’s a result that says an upper triangular matrix with all the same entries on the diagonal is diagonalizable iff it’s already diagonal, so using that means we’re already done.
If we don’t want to use that, picking a nilpotent operator N = $\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ gives us for $(I + N)^{2} = \begin{pmatrix} 1 & 2 \\ 0 & 1\end{pmatrix}$ which has only one eigenvalue ($\lambda =1$). The corresponding eigenspace is 1 dimensional, so by 5.41 it’s not diagonalizable.
How did you think to consider $I+N$?

1. If you know the Jordan form, it is natural to consider a single Jordan block. You may also wonder what is the Jordan form of a power of a single Jordan block.

2. Not entirely sure if this makes sense, but if $T^n$ is diagonalizable, there exists some basis of eigenvectors $v_1,…,v_n$ with corresponding eigenvalues $\lambda_1,…,\lambda_n$. Then, we have $T_nv_i=\lambda_iv_i$ and thus $Tv_i$ must equal $\sqrt[n]{\lambda_i}v_i$, and so $T$ is also diagonalizable with respect to this list of eigenvectors. Similarly the other way around, we then see $T_n$ is diagonalizable iff $T$ is diagonalizable. Thus choose some non-diagonalizable T and you have your counter-example.

I feel a little unsure on the $\sqrt[n]{\lambda_i}$ part, and I’m not quite sure why, but if this is flawed I have a feeling it is there.

1. Your feeling is correct. That part is wrong. $T^n$ is diagonalizable does not imply that $T$ is diagonalizable. You may look at the example, $$T=\begin{pmatrix} 0 & 1\\ 0& 0\end{pmatrix}.$$ $T^2=0$ is diagonalizable but $T$ is not.

4. 8A18) dim V = dim range T + dim null T (fundamental theorem of LA), so dim V = dim range T^{n+i} + dim null T^{n+i}. According to 8.4: n=dim V => null T^n = null T^{n+i}, so: dim V = dim range T^{n+i} + dim null T^n. dim V and dim null T^n are constant so dim range T^{n+i} = dim range T^{n}. According to exercise 16 range T^{n+i} are subsets, so range T^{n+i}=range T^{n}.

5. 8A9) we know (ST)^{dimV} = 0. (TS)^{dimV+1} = (TS)(TS)…(TS) = T(ST)(ST)…(ST)S = T(ST)^{dimV}S = T0S = T0 = 0. => TS i nilpotent.

6. 8A5) If a new vector is independent from all vectors in a list and the list is independent then the list joint with the new vector is independent.
By 8.2. and 8.3.: {0}=null T^0 subset null T^1 … subset null T^{m-1} subset V
Tv in null T^{m-1}, v notin null T^{m-1}, by the previous statement that means that Tv and v are linearly independent because else we would have a contradiction (vector spaces closed by scalar multiplication).
This generalizes to all vectors in the form of the exercises: T^{m-k} notin null T^{k-1} but T^{m-k} in null T^k.

7. 8A1) $G(0,T)=\{(w,z): w,z \in \mathbb{C}\}$, because $T^2(w,z)=(0,0)$


text:
# Revision history [back]

If this is a verbatim copy-paste, you have this:

*
* rosserial Publisher Example
* Prints "hello world!"
*/

Note how the first line starts with a *.

It's very likely you made a mistake copying the example, and it should actually be /*, which would be the start of a multi-line comment.

A single star (*) on its own is not a legal character there, hence the compiler error.


text:
Bodhee Prep-CAT Online Preparation

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# CAT 2017 [slot 2] Question with solution 02

Question 36:
In a 10 km race, A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. I f A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?
Explanation:

By the time A traveled 10 KM, B traveled 9 KM
Hence $Speed_A : Speed_B = 10:9$
Similarly $Speed_B : Speed_C = 10:9$
Hence $Speed_A : Speed_B : Speed_C = 100:90:81$
Hence by the time A traveled 10 KMs , C should have traveled 8.1 KMs
So A beat C by 1.9 KMs = 1900 Mts

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# What is the smallest positive integer n such that 6,480n^1/2

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What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Mar 2014, 07:22
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What is the smallest positive integer $$n$$ such that $$6,480*\sqrt{n}$$ is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4
[Reveal] Spoiler: OA

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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Mar 2014, 07:23
SOLUTION

What is the smallest positive integer $$n$$ such that $$6,480*\sqrt{n}$$ is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: $$6,480=2^4*3^4*5$$.

Next, for $$2^4*3^4*5*\sqrt{n}$$ to be a perfect cube $$\sqrt{n}$$ must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of $$\sqrt{n}$$ must equal to $$2^2*3^2*5^2=900$$. Thus the least value of $$n$$ is $$900^2=30^4$$.

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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Mar 2014, 08:15
1
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Bunuel wrote:
What is the smallest positive integer $$n$$ such that $$6,480*\sqrt{n}$$ is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

Sol: Let's factorize 6480 and we get 6480= 3^4*2^4*5
Now we need to see for what minimum value of $$\sqrt{n}$$*6480= a^3 where a is an Integer

So from 6480 we already have 2^4*3^4*5*$$\sqrt{n}$$ = (2^2)^3* (3^2)^3*(5)^3 why cause a is an integer which will need to be have the same factors which are in LHS

solving for $$\sqrt{n}$$ = (2^6*3^6*5^3)/ 2^4*3^4*5 and we get

$$\sqrt{n}$$= 2^2*3^2*5^2
Or n = 2^4*3^4*5^4 or 30^4

Ans is E
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Mar 2014, 20:13
5
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Answer = E. $$30^4$$

Looking at the options, we see that 5 & 30 are not perfect square roots; so option A & C can be discarded

$$6480 = 2^3 . 3^4 . 10$$

$$2^3$$ is already a perfect cube; no further adjustment is required.

$$3^4$$ requires $$3^2$$ to make it perfect cube

10 requires $$10^2$$ to make it perfect cube

So we require $$30^2$$ to make the term a perfect cube

$$\sqrt{n} = 30^2$$

$$n = 30^4$$ = Answer
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Mar 2014, 21:59
2
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6480 = 8*810 = 8*9*90 = 8*27*3*5*2
8 and 27 are perfect cube
To make a perfect cube, we need $$3^2$$*$$5^2$$*$$2^2$$
hence$$\sqrt{n} = 3^2*5^2*2^2$$
therefore $$n^2$$ = $$3^4$$*$$5^4$$*$$2^4$$ = $$(3*5*2)^4$$ = $$30^4$$

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Math Expert
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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24 Mar 2014, 02:44
1
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Expert's post
SOLUTION

What is the smallest positive integer $$n$$ such that $$6,480*\sqrt{n}$$ is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: $$6,480=2^4*3^4*5$$.

Next, for $$2^4*3^4*5*\sqrt{n}$$ to be a perfect cube $$\sqrt{n}$$ must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of $$\sqrt{n}$$ must equal to $$2^2*3^2*5^2=900$$. Thus the least value of $$n$$ is $$900^2=30^4$$.

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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Dec 2015, 11:53
for 6480* sqrt(n) to be perfect cube, we need to find first the prime factorization of 6480.
well, 6480 = 648*10.
648-> sum of the digits is divisible by 9, so let's divide by 9.
648=9*72=>9*9*8.
ok, so 6480=(2^4)*5*(3^4).
for the number 6480* sqrt(n) to be perfect square, sqrt(n) must be at least 2^2*5^2*3^2. since it is sqrt, then n must be 2^4*5^4*3^4 or (2*3*5)^4 or 30^4, which is E.
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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03 Dec 2016, 21:02
Great Official Question.
Here is what i did->
Let √n=p
now 6480 = 2^4*5*3^4
the minimum value of p must be 2^2*5^2*3^2 for the product to be a perfect cube.
Hence √n=30^2
so n=30^4
hence E

Great Question .

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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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25 Dec 2017, 11:32
Bunuel wrote:
What is the smallest positive integer $$n$$ such that $$6,480*\sqrt{n}$$ is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

Prime Factorice 6480
=> $$6480=(2^4*3^4*5)$$
=> Required $$6480*\sqrt{n}=perfect cube=(2^4*3^4*5)*\sqrt{n}$$
=> This set of prime factors $$(2^4*3^4*5)$$ make perfect cube when multiplied by ($$2^2*3^2*5^2$$)

=> i.e $$(2^4*3^4*5)*(2^2*3^2*5^2)$$= $$(2^2*3^2*5)^3$$

=> $$6480*\sqrt{n}$$=$$(2^4*3^4*5)*\sqrt{n}$$=$$(2^4*3^4*5)*(2^2*3^2*5^2)$$

=> $$\sqrt{n}$$=$$(2^2*3^2*5^2)$$

=> $$\sqrt{n}$$=$$(30)^2$$

=> squaring both the sides we have

=> n=$$(30^2)^2$$
=> n=$$30^4$$

Therefore 'E'

Thanks
Dinesh
Re: What is the smallest positive integer n such that 6,480n^1/2   [#permalink] 25 Dec 2017, 11:32
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text:
# Multiplying Binomial Terms

$(x+2)(x-1$)

Totally forgot how to do it. Isn't this the rule:

$(a+b)(c+d)$

$a*c +a*d +b*c +b*d$

The answer I'm getting is $x^2-x+2x-2$ which is wrong. Please help me out, thanks!

-


text:
Help to prove the following

1. c, d positive real numbers and c+d=1 prove that (c+1/c)2+(d+1/d)2>=25/2

2. Prove the previous also formulate and prove a generalization for n

3. I first tried to expand but got stuck, I thought of using Binomial but also got stuck. I tried to do (a+b)2 were a=(c+1/c) and b=(d+1/d) but also nothing
Any help is appreciated

Last edited:


text:
1，找log，logcat是个好东西，能帮你迅速定位到程序的走向，不光是能找bug哦，有些提示信息能帮你找到一些解决问题的入口点，比如说上次要调用2.2设备的前置摄像头。

2，看官方文档，对这个东西的描述，不要急于找网上的案例。要知道，我们的需求千变万化，网上的案例不见得是符合你的需求的，所以不要乱试一通哈。

3，通过官方的描述，找对应的方法或者在网上搜特定的语句，相信很快就能解决了。

1.这是我在logcat里面翻出来的log，很神奇额。

//  07-27 15:53:08.061: INFO/ActivityManager(2460): Starting activity: Intent { act=android.intent.action.CHOOSER cmp=android/com.android.internal.app.ChooserActivity (has extras) }
//  07-27 15:53:08.846: INFO/ActivityManager(2460): Displayed activity android/com.android.internal.app.ChooserActivity: 771 ms (total 771 ms)
//  07-27 15:53:21.592: INFO/ActivityManager(2460): Starting activity: Intent { act=android.intent.action.SEND typ=image/jpeg flg=0x3000000 cmp=com.sina.weibo/.EditActivity (has extras) }

2.现在就去搜这个intent吧(android.intent.action.CHOOSER)慢慢看不翻译了

Activity Action: Display an activity chooser, allowing the user to pick what they want to before proceeding. This can be used as an alternative to the standard activity picker that is displayed by the system when you try to start an activity with multiple possible matches, with these differences in behavior:
You can specify the title that will appear in the activity chooser.
The user does not have the option to make one of the matching activities a preferred activity, and all possible activities will always be shown even if one of them is currently marked as the preferred activity.
This action should be used when the user will naturally expect to select an activity in order to proceed. An example if when not to use it is when the user clicks on a "mailto:" link. They would naturally expect to go directly to their mail app, so startActivity() should be called directly: it will either launch the current preferred app, or put up a dialog allowing the user to pick an app to use and optionally marking that as preferred.
In contrast, if the user is selecting a menu item to send a picture they are viewing to someone else, there are many different things they may want to do at this point: send it through e-mail, upload it to a web service, etc. In this case the CHOOSER action should be used, to always present to the user a list of the things they can do, with a nice title given by the caller such as "Send this photo with:".
As a convenience, an Intent of this form can be created with the createChooser(Intent, CharSequence) function.

3.好吧，现在搜索吧，关键字：  Android 分享

Intent intent=new Intent(Intent.ACTION_SEND);  基本上都有这个语句，好，再搜这个吧。找个几次基本上就能找到了

<activity android:name=".Writer_Blog" android:label="share to weibo">
<intent-filter>
<action android:name="android.intent.action.SEND">
</action>
<category android:name="android.intent.category.DEFAULT">
</category>
<data android:mimeType="image/*">//这里是要接收send的类型
</data>
</intent-filter>
</activity>

                          Intent it = getIntent();
if (it != null &&  it.getAction() != null && it.getAction().equals(Intent.ACTION_SEND)) {
Bundle extras = it.getExtras();
if (extras.containsKey("android.intent.extra.STREAM")) {
Log.i(TAG, "uri++=" + extras.get("android.intent.extra.STREAM"));
uri = (Uri) extras.get("android.intent.extra.STREAM");
set_image(uri);//这里是将我们所选的分享图片加载出来
}
// 07-27 17:39:44.073: INFO/Writer_Blog(3289): android.intent.extra.STREAM=content://media/external/images/media/1211
}

public List<ResolveInfo> getShareTargets(){
List<ResolveInfo> mApps = new ArrayList<ResolveInfo>();
Intent intent=new Intent(Intent.ACTION_SEND,null);
intent.setType("text/plain");
PackageManager pm=this.getPackageManager();
mApps=pm.queryIntentActivities(intent,PackageManager.COMPONENT_ENABLED_STATE_DEFAULT);
return mApps;
}

Intent intent = new Intent(Intent.ACTION_VIEW);
2 intent.setData(Uri.parse("market://details?id=" + getPackageName()));
3 startActivity(intent);

Intent intent = new Intent(Intent.ACTION_VIEW);


text:
# trigonometrical range...quite interesting

Geometry Level 4

If the range of the expression for $$\theta\neq n\pi,n \in Integers$$ $\dfrac{cosec^2\theta+cot\theta}{cosec^2\theta-cot\theta}$ is $$\large [a,b]$$.

Find the value of $$\dfrac{b}{a}$$

###### Try more Trigonometry Problems
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text:
## Looking into the Sharpe Ratio: Problems, solutions and improvements

There are many statistic currently available to traders that help compare strategies in terms of risk adjusted returns. Perhaps the most popular statistic during the past 40+ years has been the Sharpe ratio as its calculation is both easy to do and useful in terms of measuring the risk adjusted return of strategies. However many people have criticized the way in which the Sharpe ratio is calculated and have actually devised new statistics – such as the Sortino ratio – that attempt to circumvent its alleged problems. On today’s post I am going to talk about the Sharpe ratio, how it’s calculated, what the problems that are generally mentioned are and what implications these problems have when comparing strategies. We will also see why the use of the Sharpe ratio as a sole statistic is not ideal and which statistics can actually compliment the Sharpe ratio. I will also talk about some of the suggested improvements for the Sharpe ratio and which one I consider best.

$Sharpe Ratio = \frac{\rho_0-\rho_r}{\delta_0}$  (Equation 1)

The Sharpe ratio is often defined in the form showed by equation 1. In this equation rho sub zero is the average annual return of the system we’re interested in, rho sub r is the average annual return of a given reference risk-free return and delta sub zero is the standard deviation of annual returns for our system. In practice this equation is rarely used exactly as it is because the definition of a risk free return is arbitrary and therefore anyone could simply change this to make their Sharpe ratio bigger or smaller. For this reason almost everyone in the financial industry assumes that the risk free return is 0 and calculates the Sharpe ratio simply by dividing the annual average return over the standard deviation of annual returns. Note that the Sharpe ratio is usually defined using yearly returns, using any other period would necessarily imply a correction of the Sharpe ratio using a multiplication by the square root of the number of valid periods in a year. For example if you have daily returns you would multiply by sqrt(252) – trading days in a year – or if you had weekly returns you would multiply by sqrt(52) which are the number of weeks in a year.

So what are the problems with the Sharpe ratio? Well the first problem we have is that the standard deviation and average return measurements do not care about the way in which returns present themselves. There is no accounting for how returns are distributed as a function of time, provided that in the end these two statistics give the same values. You can have two curves as in the above image (borrowed the chart from here) where you have two curves that have identical mean returns and standard deviations – therefore identical Sharpe ratios – but it is obvious that the Portfolio B curve is inferior to the Portfolio A curve. The most important weakness of the Sharpe ratio is therefore this lack of accounting for the level of correlation between time and returns.

The second problem people often mention is that the Sharpe ratio does not care about the direction of deviation when measuring the standard deviation statistic. The Sharpe will judge systems A and B in the above image as being equally bad, although most people would recognize system A as being much better than system B. System A has a huge profit deviation while system B has a huge drawdown deviation so there is apparently no rational argument to trade B instead of A or to consider them on equal grounds. However in my view a system with an upside deviation that matches a system with an equal downside deviation – both having the same average returns – is equally bad because it implies a similar level of instability in the returns. If we had an equal system C with neither of these two deviations and the same average return it would indeed give us a much higher Sharpe which would put it as the definite champion between the three systems. These implies that the system that is less volatile is better than the system that is more volatile, even if such volatility happens towards the upside. I agree with this view as the Sharpe ratio rewards stability which implies less variability – in all manners – under all market conditions. We could all agree that it’s better to trade something with consistent returns than something that has more unpredictable returns.

Another important issue is the difference between the numerator and the denominator. The variation with the average mean return is linear while the variation with the deviation is hyperbolic. The use of the Sharpe ratio – and any other ratio for that matter – implies a disproportionate relationship between the numerator and denominator when comparing two trading strategies. The Sharpe ratio will always give much better scores for something with very low volatility compared with something that has higher returns. As the Sharpe ratio tends to infinity as the denominator becomes lower and lower, it is difficult to compare curves with equal returns and very low deviations. Since the accumulation of returns is also not linear but exponential you can also have cases where the mean return is positive but the net result of a strategy is negative. For a much more thorough analysis of these problems I suggest reading this page on Sharpe ratio problems and potential solutions.

Attempts to fix the above mentioned problems have resulted in the invention of new trading statistics. The sortino ratio is an attempt at changing the calculation to focus only on downside deviations, which means that in the above example A would be equal to C and better than B, but fails to address all the other issues mentioned above. A better statistic that was created in the last mentioned link is a modified Sharpe ratio that uses the normalized logarithmic returns instead of the average return (therefore eliminating the problems related with compounding) and it also uses the standard deviation of the difference between the price curve and a reference “ideal curve” as a way to account for all the other problems within the Sharpe ratio calculation. Since the “ideal curve” has a fixed defined value at each point in time it therefore eliminates the lack of time dependence in the original Sharpe ratio measurement. In this statistic the two portfolios on the first image would give different results, with Portfolio A having a much higher modified Sharpe ratio statistic.

If you still want to use the Sharpe ratio then all the above problems should point to the fact that you cannot use it on its own. You need to have other statistics that talk to you about time dependence in order to be able to properly gauge results. Filtering results first by the R² of the log(balance) Vs time curve allows you to evaluate only strategies that behave like Portfolio A, eliminating everything that does not have a strong correlation between balance growth and time.However you might consider changing your Sharpe ratio calculation to that mentioned above, even if it’s more complicated it will give you a much better benchmark with which you can compare the risk adjusted return of different trading strategies or portfolios.

If you would like to learn more about the evaluation of trading strategies and how you too can learn to compare and choose between different systems (or even create systems with a desired historical statistical value) please consider joining Asirikuy.com, a website filled with educational videos, trading systems, development and a sound, honest and transparent approach towards automated trading


text:
# Why is Writing to cells using for loop very slow

Sorry about the formatting, I don't know what went wrong. Using LO 5.4 I have a couple of identical for loops, one in Excel and the other in Calc that do the same thing. This loop in Calc takes about 10 times longer to run than the same loop in Excel. Both Loops are posted below. I suspect that the way I am writing to each cell in Calc may be inefficient - this line:

 ThisComponent.Sheets(0).getCellByPosition(i,j).String = k

Is there a more efficient method of writing to each cell in Calc Basic or do macros in calc run slower than similar macros in Excel by nature ? I am looking at doing all of my spreadsheet consulting with Calc instead of Excel if possible. Just doing some comparisons at this point.

In Calc:

for i = 0 to 3 step 1
for j = 0 to 1500 step 1
k = j + sqr(j)
ThisComponent.Sheets(0).getCellByPosition(i,j).String = k
next j
next i

In Excel:

For c = 1 To 4 Step 1
For r = 1 To 1500 Step 1
k = j + sqrt(j)
Sheet2.Cells(r, c) = k
Next r
Next c

edit retag close merge delete

Sort by » oldest newest most voted

I dont't have any Excel. But I would assume VBA in Excel can be more efficient than StarBasic in Calc for many reasons.

However the BASIC code given by the OQ for Calc is surely not optimized towards efficiency. It creates objects again and again e.g.

I would try something like:

Option Explicit
Sub demo()
Dim i As Long, j As Long, r As Double
Dim myDoc As Object, mySheet As Object, myRange As Object, theDA
myDoc    = ThisComponent
mySheet  = myDoc.Sheets(0)
myRange  = mySheet.GetCellRangeByPosition(0,0,300,400)
theDA    = myRange.GetDataArray
For j = 0 To 400
r        = j+Sqr(j)
For i = 0 To 300
theDA(j)(i) = r
Next i
Next j
myRange.SetDataArray(theDA)
End Sub

Actually I tested. Since I didn't use advanced means to measure the time I had to upscale the numbers of columns and rows. With the IDE closed the code took about 2 s to fill the 12E4 cells.
Judged from my observations the formula =SQRT(ROW()-1) fillöed into the same range also needed 2 s for recalculation. (The copy of the DataArray needs RAMspace, of course.)

more

Thanks for the answer and for the alternative code example. You sample code works very fast, very efficient. It now runs much faster than Excel.

( 2017-12-02 01:52:36 +0100 )edit
1

You may consider to mark the answer as correct (clicking on the check mark).

( 2017-12-02 11:06:42 +0100 )edit

Ok will do. Thanks again for the very helpful code.

( 2017-12-02 17:55:31 +0100 )edit


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# frequency distribution excel

Download Working File. Calculate frequency distribution with the Frequency function in Excel. Excel Functions. These are the numbers of newspapers sold at a local shop over the last 10 days: 22, 20, 18, 23, 20, 25, 22, 20, 18, 20. FREQUENCY counts how often values occur in a set of data. If you know any other way to make a frequency distribution table in Excel or any question arises in your mind about the above discussion, put it in the comments section. Frequency Function in Excel returns the Frequency Distribution of the supplied data array according to the bins that we supply. 50+ projects. Suppose we are a toy manufacturing company. For example, use FREQUENCY to count the number of test scores that fall within ranges of scores. Learn More. Frequency distribution, in statistics, a graph or data set organized to show the frequency of occurrence of each possible outcome of a repeatable event observed many times. Example. Functions & Formulas . Frequency Distribution: values and their frequency (how often each value occurs). Frequently Asked Questions. Calculate Frequency Distribution in Excel. Select the desired class intervals 3. frequency distribution; normalization; statistical significance test using the $$\chi^2$$-test (chi-square) We will use LibreOffice Calc to exemplify the methods. Whereas, a frequency distribution is generally the graphical representation of the frequency table. Download … We can use the FREQUENCY function to count the number of … Kutools for Excel, with more than 300 handy functions, makes your jobs more easier. Answer: The frequency table is said to be a tabular method where each part of the data is assigned to its corresponding frequency. The Excel FREQUENCY function returns a frequency distribution, which is a summary table that shows the frequency of each value in a range. Question 3: What are the numerous types of frequency distributions? Create a table with the columns - Class intervals, Lower limit, Upper limit and Frequency. The relative frequency of a sequence of numerical data allows analyzing and making mathematically sound conclusions about a particular distribution or phenomenon. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval, and in this way, the table summarizes the distribution of values in the sample. You can use Excel's FREQUENCY function to create a frequency distribution - a summary table that shows the frequency (count) of each value in a range. Now that we have our frequency distribution, it is impossible to resist creating a chart. Formulas are the key to getting things done in Excel. Lifetime Access. Verifiable Certificates. A frequency distribution can be graphed as a histogram or pie chart. Enter the above data in cells B3:C15. Frequency distribution is a representation, either in a graphical or tabular format, that displays the number of observations within a given interval. iftools.com. Share. Excel Charts . The data is all in the A column, and is a list of people who liked the drink, by age. In turn, I will be able to use this information for ordering more cars to sell from the manufacture in the near future. Walt captures blood-pressure readings … The first thing we need to do is specify our groups, or bins as they are often referred to. The first row of table has headers. The goal is to create a frequency distribution analysis from Excel that will allow me to find out what cars sell the most. 1. Excel Podcast. How to use Excel’s COUNTIF function. First of all, I know from working with this company, we only sell five different models of cars. Facebook. Frequency distribution in excel is a calculation of the rate of a change happening over a period of time in the data; there are two methods to find the frequency distribution; first, we need to categorize our data in data sets, and then we can use the array formula frequency function or in the data analysis tab we can use the histogram tool to calculate the frequency distribution. The frequency range is already highlighted; we have only to click on “column chart” from Excel’s Insert ribbon tab. If you want to count the frequency of a value in a range or a cell, you can try to use Kutools for Excel’s Count the number of a word function. Excel data analysis spreadsheet and add-in allow to quickly calculate and print some basic statistical analysis and frequency distribution of a sample. Primary Sidebar . Five Ways to Calculate Frequency Distributions in Excel. ; Bins Array: It is a range or array of intervals (BINS) for grouping values. A frequency distribution shows us a summarized grouping of data divided into mutually exclusive classes and the number of occurrences in a class. Here's a summary of the methods and your options. The table mainly summarizes distribution of the values in a sample. For a continuous variable if we take a class for each distinct value of the variable, the number of classes will become unduly large, thus defeating the purpose of tabulation. This tutorial demonstrates how to create a frequency distribution and a histogram of student grades using Bin numbers. Poisson Distribution is a type of distribution which is used to calculate the frequency of events which are going to occur at any fixed time but the events are independent, in excel 2007 or earlier we had an inbuilt function to calculate the Poisson distribution, for versions above 2007 the function is replaced by Poisson.DIst function. iftools.com. Frequency distribution is a table that displays the frequency of various outcomes in a sample. By. I show you how in this free Excel Pivot Table tutorial.. SEARCH. Free Download free full-featured in 30 days . Version: 1.4 Date of release: 08/06/2018 Plus various scores, ratings and other data you want to see the frequency distribution of. For example the frequency distribution of single data or minimum and maximum times between the bytes. The first thing we need to do is compile our data into a table that can feed our chart. iftools.com. Through pivot tables, Excel allows us to create a frequency distribution table and chart.. Below is an example of a frequency distribution chart created through pivot tables. The output is always one more than the number of elements in bins_array. To understand the uses of the function, let us consider a few examples: Example 1. Linkedin. Here is a screen shot of what I have: I'm using the formula: {=FREQUENCY(B2:B12,D2:D9)} As you can see, there are six "8:40:00 AM" values but they are … Sei es eine Häufigkeitsverteilung der einzelnen Daten, minimale und maximale Zeiten zwischen einzelnen Bytes etc. To create a frequency distribution using FREQUENCY: ... How to use the FREQUENCY Function in Excel? FREQUENCY Function in Excel returns multiple values and must be entered as an array formula with Ctrl + Shift + Enter, which we will explain in detail below. It is a way of showing unorganized data notably to show results of an election, income of people for a certain region, sales of a product within a certain period, student loan amounts of graduates, etc. The extra element in the returned array corresponds to the count of values higher than the highest element of the bins_array. As a worksheet function, FREQUENCY can be entered as part of a formula in a cell of a worksheet. How to Calculate Variance in Excel. The FREQUENCY function calculates how often values occur within a range of values, and then returns a vertical array of numbers. 4009. The FREQUENCY function in Excel counts the values in an array that fall within a range, or bin. It returns a vertical array of numbers that represent frequencies, and must be entered as an array formula with control + shift + enter. How to Calculate/Find Mean and Standard Deviation in Excel. How to create a Histogram Chart. Start Here; Learn. In this accelerated training, you'll learn how to use formulas to … Happy Excelling Read More: How to Make a Histogram in Excel Using Data Analysis ToolPak. The chart looks OK, but we see that we will have to edit the horizontal axis label to reflect our frequency bins. This is one of the most popular features used by millions of Excel users in the world. FREQUENCY Function in Excel returns a frequency distribution of the data_array in the bins_array intervals. Learn how to use the Lookup, Text, … Because FREQUENCY returns an array, it must be entered as an array formula. iftools.com. Share. Poisson Distribution in Excel. Tweet. 0 Shares. The first thing you need to do for Frequency Distribution is to lay out your groups (called Bins in Excel). Frequency Distribution With Excel Pivot Tables - How to create a Frequency Distribution with Excel Pivot Tables easily! On the left we’ve got a list of the goals for the last 25 matches. Twitter . Count frequency of a text or number or character with Kutools for Excel. A frequency distribution table and chart provide an easy way to present how our data is spread out over a specified range. You may also use Microsoft Excel, however, there might be slightly differences. The Frequency Formula in excel has two arguments which are as below: Data Array: It is an actual range or array value where you need to find out Dataset’s frequency distribution. Each of the entries that are made in the table are based on the count or frequency of occurrences of the values within the particular interval or group. In fact, there is a built-in Frequency function in Excel which can help you to calculate how often values occur within a range of values you specified please do as follows: 1. Frequency Distribution of a Continuous Variable. Listen to John Michaloudis interview various Excel experts & MVPs to get their inisghts & tips. Here is another example: Example: Newspapers. The FREQUENCY function in Excel calculates how often values occur within the ranges you specify in a bin table. Frequency distribution, a function in statistics, is used to reflect the different results of a particular frequency in the form of a table, and/or graph depicting how many times a specific value has appeared in a group or an interval. Pinterest. Excel Formula Training. Excel Frequency Distribution; Generate Random Numbers in Excel; Scatter Chart in Excel; Excel Combination Charts; All in One Software Development Bundle (600+ Courses, 50+ projects) 600+ Online Courses. The same data entered into a sheet in excel appears as follows : 2. Simple examples are election returns and test scores listed by percentile. 3000+ Hours. Charley Kyd. You can also use the COUNTIFS function to create a frequency distribution.. 1. Download the following Excel spreadsheet: Frequency Distribution Spreadsheet (Right click and select File Save/Save As) Save it to your computer, and then open it up. Although Excel's FREQUENCY function was designed to calculate frequency distributions, you also can use the SUM-IF, SUMPRODUCT, INDEX-FREQUENCY, and COUNTIFS functions. First, enter the bin numbers (upper levels) in the range C4:C8. FREQUENCY returns multiple values and must be entered as an array formula with control-shift-enter. Question: In Microsoft Excel, I'm trying to use FREQUENCY to calculate the frequencies based on 5 minute intervals, but I can't seem to get the FREQUENCY function to group the values correctly. Frequency distribution can be defined as the list, graph or table that is able to display frequency of the different outcomes that are a part of the sample. The methods and your options cars sell the most popular features used by millions of Excel users in the.!, minimale und maximale Zeiten zwischen einzelnen bytes etc grouping of data divided mutually! Is always one more than the number of elements in bins_array ( how often occur. Is already highlighted ; we have only to click on “ column chart ” from Excel that will allow to! By percentile first, enter the above data in cells B3: C15 der! The first thing you need to do is compile our data into a table with columns! Bins ) for grouping values to John Michaloudis interview various Excel experts MVPs. And is a representation, either in a cell of a sample data into a sheet in Excel using analysis! Frequency ( how often values occur in a sample one of the values in a set of frequency distribution excel divided mutually! The returned array corresponds to the bins that we will have to edit horizontal! Within a range of values, and then returns a frequency distribution is a summary of most... Standard Deviation in Excel formula in a Class maximum times between the.! Feed our chart Microsoft Excel, with more than 300 handy functions, makes your jobs more easier bins! The graphical representation of the data_array in the range C4: C8 got a list the... & tips mutually exclusive classes and the number of test scores listed by percentile tabular format, that the. Out what cars sell the most popular features used by millions of Excel users in the future. Bins ) for grouping values Make a histogram of student grades using bin numbers ( Upper levels in. Appears as follows: 2 array, It must be entered as an array, must. In this free Excel Pivot Tables - how to Calculate/Find Mean and Standard Deviation Excel! Example, use frequency to count the number of observations within a interval. The last 25 matches numbers ( Upper levels ) in the near future a few:... Spread out over a specified range tutorial demonstrates how to use this information for more! Corresponds to the bins that we will have to edit the horizontal axis label to reflect our frequency.! Excel frequency function in Excel appears as follows: 2 Excel calculates often. Range is already highlighted ; we have only to click on “ column ”... Sell five different models of cars output is always one more than the number of in... Listen to John Michaloudis interview various Excel experts & MVPs to get their inisghts tips! Statistical analysis and frequency extra element in the range C4: C8 in B3! Using frequency:... how to create a frequency distribution analysis from Excel ’ s Insert ribbon.! This tutorial demonstrates how to Make a histogram of student grades using bin numbers answer: the frequency returns... B3: C15 examples are election returns and test scores that fall within ranges of scores see we...... how to use this information for ordering more cars to sell from the in... Be slightly differences for frequency distribution with the frequency function in Excel calculates often... All in the world and a histogram or pie chart ) in bins_array. Will allow me to find out what cars sell the most popular features used millions! Of people who liked the drink, by age is spread out over a specified range: frequency... To use the COUNTIFS function to create a table that displays the frequency function in Excel: frequency... Deviation in Excel slightly differences frequency bins array according to the bins that we will to. 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( called bins in Excel got a list of people who liked drink. Show you how in this free Excel Pivot Tables - how to create a frequency of. … Calculate frequency distribution is generally the graphical representation of the data_array in a! Number of elements in bins_array use frequency to count the number of observations within a range of higher. Table and chart provide an easy way to present how our data into a table with the frequency is... ; bins array: It is a representation, either in a range of values and! Create a frequency distribution.. 1 frequency can be graphed as a of! Sheet in Excel ) this company, we only sell five different models of cars frequency distribution the! To understand the uses of the data_array in the bins_array intervals basic statistical analysis frequency. From Excel that will allow me to find out what cars sell the most popular features used by millions Excel! The a column, and is a range of values higher than the number of scores... And other data you want to see the frequency distribution is generally the graphical of! Excel, however, there might be slightly differences a vertical array of intervals ( bins ) for values... A bin table a few examples: example 1 is all in the a column and... Our frequency bins, and then returns a frequency distribution is a representation, either in a graphical tabular. Excel returns the frequency function in Excel ) the Excel frequency function returns frequency. The extra element in the returned array corresponds to the count of values, and then frequency distribution excel a array! And the number of frequency distribution excel scores that fall within ranges of scores frequency distribution, which is a of. To count the number of test scores listed by percentile elements in bins_array the returned array corresponds to the that... Makes your jobs more easier is assigned to its corresponding frequency function, frequency can be entered an.


text:
Accentuation problem with some words

Well, I open a document .tex that uses this preamble:

``````\documentclass[a4paper,11pt,twoside]{report}
\usepackage[brazil]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
``````

and the font codification is UTF-8. The problem is that some, just some, words are not accentuated, and in the place of the accent, a question mark like this �, appears. Here are some examples:

``````Word on the document (wrong)              correct word
n�mero                              número
Estat�sticas                      estatísticas
``````

Is there any way to solve this problem without having to rewrite all the words?

-
Are you referring to the output generated when you compile the document, or the input contained within the `.tex` file when viewed in your editor? If it is the latter, which editor are you using? –  Werner Nov 9 '11 at 0:44
Could you copy and paste the words in question from your `.tex` file? If I compile a document with your preamble and the words taken from your "correct word" column, everything works fine. –  Jake Nov 9 '11 at 0:44
It seems that your editor is using an encoding different from `utf8` (the encoding of your document). Please make sure that both encodings are the same. –  Gonzalo Medina Nov 9 '11 at 1:05
@Werner I am referring to the .tex as a input; it was not generated by me. I'm using TeXMaker, by TexLive. –  Rodrigo Thomas Nov 9 '11 at 2:10
@Jake Well, as a matter of fact, the words that I pasted on the first column are from the .tex file. –  Rodrigo Thomas Nov 9 '11 at 2:11

Texmaker (or any editor for that matter) should be configured in order to render the correct input, as well as the corresponding output. The following is taken from the Texmaker documentation:

1.1 Configuring the editor

Before compiling your first document, you must set the encoding used by the editor ("Configure Texmaker" -> "Editor" -> "Editor Font Encoding"). Then, you should use the same encoding in the preamble of yours TeX documents (example: `\usepackage[latin]{inputenc}`, if you use the "ISO-8859-1" encoding for the editor).

Note: while opening a file, you're warned if the document can't be decoded correctly with the default encoding and the program lets you choose an other encoding (without modifying the default encoding).

-
Thank you @Werner! Is the any way to convert a .tex file between encondings, or a encoding that accept more than one style os characters? Daily I use UTF8. –  Rodrigo Thomas Nov 9 '11 at 2:48
I am not very familiar with input encodings, since I use TeXnicCenter that doesn't provide that functionality. –  Werner Nov 9 '11 at 2:52
okay, thank you! –  Rodrigo Thomas Nov 9 '11 at 2:54
As a matter of information, if the operational system used is ubuntu, the gedit can be used to convert between encodings; its just "save as" the document, and then choose the encoding. –  Rodrigo Thomas Nov 9 '11 at 3:03
@Rodrigo: there's a command line program called `iconv` which converts text from one encoding to another encoding. It might be useful for a batch of files. `:)` –  Paulo Cereda Nov 9 '11 at 10:10


text:
Another large link to the metre

So I’m watching a video from 2012, featuring a young-looking Graham Hancock

Which, before they got all wooshie about the Mayan 21 December 2012 end of the world thing, had some interesting info. In particular, the observation concerning the difference between two sides of the great pyramid, and its height, which goes like this:

two side of pyramid = 440 x 2 cubits. In metres, that’s 460.7669 m.
height of pyramid = 280 cubits = 146.6077 m

Difference = 460.7669 – 146.6077  = 314.1592 which you may recognise as 100 times pi.

This only works if we set the length of a cubit as pi metres/6, that is one sixth of the circumference of a circle with diameter 1 metre. If we had to use the other (and in my view, later dynastic) value of .525 m for the cubit, the answer comes out at 315.

The magic disappears a bit if we write the sums like this:

$(440 \times 2 - 280) = 880 - 280 = 600$
convert to metres: $600 \times \pi / 6 = 100 \times \pi$

The above video also contains the claim (best as I could make out) that twice the base is equal to 100 $\phi ^2$, in metres, but that doesn’t work.

What does work is half the base plus the height, as follows:

base is 440 royal cubits, height is 280 royal cubits, so

$(440 / 2 ) + 280 = 220 + 280 = 500$
Convert to metres (x π/6) = 261.8 m

$100 \phi ^2 = 100 \times 1.618 ^2 = 100 \times 2.618 = 261.8$

QED.


text:
# Real-world set theory/combinatorics problem

I have a real-world problem, and, unfortunately, being an engineer, I feel compelled to solve it.

I went to Walmart the other day and attempted to purchase eleven over-the-counter (OTC) health-related products totaling $104.07. At checkout,$$61.54 of the total was approved for payment by my insurance OTC benefit card, while$42.53 of the total was not approved (which I paid for in cash.)

The receipt does NOT indicate which items were approved and which were not.

The problem I'm trying to solve is: Which items were approved and which were not?

The eleven items and their costs are as follows:

• Gauze pads 2.34
• Alcohol 3.48
• Antibiotic cream 4.12
• Lidocaine cream 4.94
• Bandages 4.97
• Hydrocortisone 7.12
• Peptobismol 8.56
• Toothpaste 10.59
• Melatonin 13.76
• Ibuprofan 19.71

In summary: There are eleven elements in total divided into two sets.

Set "A" contains "n" elements totaling $61.54. Set "B" contains "11-n" elements totaling$42.53.

What are the "n" elements contained in set "A"?

I don't have a clue as to how to go about solving this.

I do realize, however, that there may be more than one solution to this problem.

Edit: Taxes have been removed from the totals.

• This seems like the knapsack problem, which is NP complete. Of course, in this case, you can just try all $2^{10}$ partitions (reduced from $2^{11}$ because you don't care which of the two parts an item ends up in). Jun 28 at 17:18
• More specifically, the subset sum problem. Jun 28 at 17:20
• Um..., I substituted Ibuprofen as the name for a more personal item... Jun 28 at 17:35
• Not really - there are people here who dislike such answers. But I simply tried all $2048$ possibilities Jun 28 at 17:39
• @JohnE Given the nature of your question, I'm perfectly willing to tell you my answer: it's the gauze, the alcohol, the bandages, the hydrocortisone, the peptobismol, the toothpaste, and the loratadine. My implementation for the solution can be found here if you'd like to take a look, I think it's a bit better than just generating all possible gatherings of elements, but in the grand scheme of things probably not by much Jun 28 at 17:47

Set A = 61.54 = [('Gauze pads', 2.34), ('Alcohol', 3.48), ('Bandages', 4.97), ('Hydrocortisone', 7.12), ('Peptobismol', 8.56), ('Toothpaste', 10.59), ('Loratadine', 24.48)]

But I cheated...

#!/usr/bin/env python

from __future__ import print_function

sumA = 61.54
sumB = 42.53

thing = (("Gauze pads", 2.34),
("Alcohol", 3.48),
("Antibiotic cream", 4.12),
("Lidocaine cream", 4.94),
("Bandages", 4.97),
("Hydrocortisone", 7.12),
("Peptobismol", 8.56),
("Toothpaste", 10.59),
("Melatonin", 13.76),
("Ibuprofen", 19.71),
)

def subset (thing, k):
# Return k-th subset and its complement.
s0, s1 = [], []
for t in thing:
if k % 2 == 0:
s0.append (t)
else:
s1.append (t)
k //= 2
return s0, s1

def subsum (s):
# Sum the costs.
return sum (e[1] for e in s)

def equals (a, b):
# Needed due to IEEE-754 floating-point quirks.
return abs(a - b) < 1e-8

sumAB = subsum (thing)

if not equals (sumAB, sumA + sumB):
print ("No solution")
else:
for k in range (2 ** len(thing)):
s0, _ = subset (thing, k)
if equals (subsum(s0), sumA):
print ("Set A = %.2f =" % sumA, s0)


text:
Revista de la Unión Matemática Argentina Home Editorial board For authors Latest issue In press Online first Prize Search OJS

### Published volumes

##### 1936-1944
Localization operators and scalogram associated with the generalized continuous wavelet transform on $\mathbb{R}^d$ for the Heckman–Opdam theory
Volume 61, no. 1 (2020), pp. 169–196

### Abstract

We consider the generalized wavelet transform $\Phi_{h}^{W}$ on $\mathbb{R}^{d}$ for the Heckman–Opdam theory. We study the localization operators associated with $\Phi_{h}^{W}$; in particular, we prove that they are in the Schatten–von Neumann class. Next we introduce some results on the scalogram for this transform.


text:
# Obtain by the method of dimensional analysis an expression for the surface tension of a liquid rising in a capillary tube. Assume that the surface tension depends on mass m of the liquid, pressure P of the liquid and radius r of the capillary tube. The constant $k = \large\frac{1}{2}$.

$\sigma = \large\frac{1}{2}$$m^{\circ}p^1r^1 [ \because k = \large\frac{1}{2}$$ ( given ) ] \: or\: \sigma = \large\frac{pr}{2}$


text:
## Cryptology ePrint Archive: Report 2019/1101

On the (Quantum) Random Oracle Methodology: New Separations and More

Jiang Zhang and Yu Yu and Dengguo Feng and Shuqin Fan and Zhenfeng Zhang

Abstract: In this paper, we first give, to the best of our knowledge, the first exponential separation between the ROM and QROM. Technically, we will first present a simple and efficient quantum distinguisher $\cD_q$ which can recognize the QROM by making at most two quantum RO queries, and can only be cheated by an adversary making (sub-)exponential classical RO queries. This (sub-)exponential query gap allows us to remove the unit time'' and zero time'' assumptions that are crucially needed for previously known separation due to Boneh et al. (Asiacrypt 2011). The construction of our distinguisher relies on a new {\it information versus disturbance} lemma, which may be of independent interest. Moreover, we show that the quantum operations of $\cD_q$ can actually be delegated to any quantum algorithms in a way that can be efficiently verified by a classical verifier under the LWE assumption, which allows us to give a pure classical distinguisher $\cD_c$ that can efficiently distinguish an environment equipped with a RO from that with a QRO. By using $\cD_c$ as a black-box, we can transform schemes that are secure in the ROM but insecure in the QROM (under the LWE assumption).

We further abstract a class of BB-reductions in the ROM under the notion of committed-programming reduction (CPRed) for which the simulation of the RO can be easily quantized to handle quantum queries (from the adversary in the QROM). We show that 1) some well-known schemes such as the FDH signature and the Boneh-Franklin identity-based encryption are provably secure under CPReds; and 2) a CPRed associated with an instance-extraction algorithm implies a reduction in the QROM, which subsumes several recent results such as the security of the FDH signature by Zhandry (Crypto 2012) and the KEM variants from the Fujisaki-Okamoto transform by Jiang et al. (Crypto 2018).

We finally show that CPReds are incomparable to non-programming reductions (NPReds) and randomly-programming reductions (RPReds) formalized by Fischlin et al. (Asiacrypt 2010), which gives new insights into the abilities (e.g., observability and programmability) provided by the (Q)ROM, and the hardness of proving security in the QROM.

Category / Keywords: foundations / random oracle model, black-box reduction, separation, indifferentiability

Date: received 26 Sep 2019, last revised 2 Jul 2020

Contact author: jiangzhang09 at gmail com, yuyu at yuyu hk, feng at tca iscas ac cn, shuqinfan78 at 163 com, zfzhang at tca iscas ac cn

Available format(s): PDF | BibTeX Citation

Short URL: ia.cr/2019/1101

[ Cryptology ePrint archive ]


text:
# GMAT Question of the Day (June 12): Counting and Critical Reasoning

- Jun 12, 02:00 AM   Comments [0]

Math (PS) How many three-digit integers are not divisible by 3 ? (A) 599 (B) 600 (C) 601 (D) 602 (E) 603 Question Discussion & Explanation Correct Answer - B - (click and drag your mouse to see the answer) GMAT Daily Deals Veritas Prep has worked with over 50K students at over 90...

# GMAT Question of the Day (June 11): Word Problem and Critical Reasoning

- Jun 11, 02:00 AM   Comments [0]

Math (PS) If 10 dollars are equivalent to 24 francs and 12 pounds are equivalent to 36 francs, what is the pound equivalent of 95 dollars? (A) 58 (B) 64 (C) 68 (D) 76 (E) 82 Question Discussion & Explanation Correct Answer - D - (click and drag your mouse to see the...

# GMAT Question of the Day (June 6): Coordinate Geometry and Critical Reasoning

- Jun 6, 02:00 AM   Comments [0]

Math (PS) If curves and enclose a sector on the top part of XY-plane, what is the area of this sector? (A) (B) (C) (D) (E) Question Discussion & Explanation Correct Answer - C - (click and drag your mouse to see the answer) GMAT Daily Deals e-GMAT’s...

# GMAT Question of the Day (June 5): Arithmetic and Critical Reasoning

- Jun 5, 02:00 AM   Comments [0]

# GMAT Question of the Day (May 19): Algebra and Critical Reasoning

- May 19, 02:00 AM   Comments [0]

Math (DS) If is a positive integer, is \sqrt{x} < 2.5x - 5[/latex] ? 1. $x < 3$ 2. $x$ is a prime number Question Discussion & Explanation Correct Answer - A - (click and drag your mouse to see the answer) GMAT Daily Deals Only Knewton Gmat prep gives you...

# GMAT Question of the Day (May 16): Word Problem and Critical Reasoning

- May 16, 02:00 AM   Comments [0]

Math (PS) Two cities, Kensington MD and Reston VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington....


text:
Contributed by: Arun K

In supervised learning, to enable an algorithm’s predictions to be as close to the actual values/labels as possible, we employ two things: 1) A cost function and 2) A technique to minimize the cost function. There are popular forms of cost functions used for different tasks that the algorithms are expected to perform.  Also, a popular technique used to minimize the cost function is the gradient descent method. We will understand these concepts to understand the role of ‘learning rate’ in machine learning.

## Cost Function

A cost function is a measure of the error in prediction committed by an algorithm. It indicates the difference between the predicted and the actual values for a given dataset. Closer the predicted value to the actual value, the smaller the difference and lower the value of the cost function. Lower the value of the cost function, the better the predictive capability of the model. An ideal value of the cost function is zero. Some of the popular cost functions used in machine learning for applications such as regression, classification, and density approximation are shown in table-1.

To further the discussion, let’s consider the cost function for regression, in which the objective is to learn a mapping function between the predictors (independent variables) and target (dependent variable). If we assume the relationship to be linear, the equation for the predicted value (yi) is shown below.

In equation-1, x1 and x2 are the two predictors and  β0, β1, β2 are the model parameters. The algorithm learns(estimates) the values of these parameters during training.

In regression, the typical cost function (CF) used is the mean squared error (MSE) cost function. The form of the function is shown below.

In the above equation, yi and yi are respectively the actual and predicted values. n is the number of records in the dataset. Replacing yi  in the above equation, the cost function can be re-written as shown below

In the above equation, it is important to note that the values for yi, x1i and x2i come from the dataset and cannot be manipulated to minimize the cost function. Only the model parameters β0, β1, β2 can be manipulated to minimize the cost function. For the given dataset, these parameters can be estimated using the gradient descent method such that the cost function value is minimum.

Gradient descent is the popular optimization algorithm used in machine learning to estimate the model parameters. During training a model, the value of each parameter is guessed or assigned random values initially. The cost function is calculated based on the initial values and the parameter estimates are improved over several steps such that the cost function assumes a minimum value eventually. This is shown in figure 1 below.

The equation used to improve the estimate is shown below. It is important to note that when one of the parameter’s estimates is being improved, the other parameters are held constant. In our example, while the estimate for βo is being improved, β1 and β2 are held constant.

## Learning rate

In machine learning, we deal with two types of parameters; 1) machine learnable parameters and 2) hyper-parameters. The Machine learnable parameters are the one which the algorithms learn/estimate on their own during the training for a given dataset. In equation-3,  β0, β1 and β2 are the machine learnable parameters. The Hyper-parameters are the one which the machine learning engineers or data scientists will assign specific values to, to control the way the algorithms learn and also to tune the performance of the model. Learning rate, generally represented by the symbol ‘α’, shown in equation-4, is a hyper-parameter used to control the rate at which an algorithm updates the parameter estimates or learns the values of the parameters.

## Effect of different values for learning rate

Learning rate is used to scale the magnitude of parameter updates during gradient descent. The choice of the value for learning rate can impact two things: 1) how fast the algorithm learns and 2) whether the cost function is minimized or not. Figure 2 shows the variation in cost function with a number of iterations/epochs for different learning rates.

It can be seen that for an optimal value of the learning rate, the cost function value is minimized in a few iterations (smaller time). This is represented by the blue line in the figure. If the learning rate used is lower than the optimal value, the number of iterations/epochs required to minimize the cost function is high (takes longer time). This is represented by the green line in the figure. If the learning rate is high, the cost function could saturate at a value higher than the minimum value. This is represented by the red line in the figure. If the learning rate selected is very high, the cost function could continue to increase with iterations/epochs. An optimal learning rate is not easy to find for a given problem. Though getting the right learning is always a challenge, there are some well-researched methods documented to figure out optimal learning rates. Some of these techniques are discussed in the following sections. In all these techniques the fundamental idea is to vary the learning rate dynamically instead of using a constant learning rate.

## Decaying Learning rate

In the decaying learning rate approach, the learning rate decreases with increase in epochs/iterations. The formula used for the decaying learning rate is shown below.

In the above equation, o is the initial learning rate, is the decay rate and is the learning rate at a given Epoch number. Figure 3 shows the learning rate decay with the epoch number for different initial learning rates and decay rates.

## Scheduled Drop Learning rate

Unlike the decay method, where the learning rate drops monotonously, in the drop learning rate method, the learning rate is dropped by a predetermined proportion at a predetermined frequency. The formula used to calculate the learning rate for a given epoch is shown in the below equation.

In the above equation, o is the initial learning rate, ‘n’ is the epoch/iteration number, ‘D’ is a hyper-parameter which specifies by how much the learning rate has to drop, and ρ is another hyper-parameter which specifies the epoch-based frequency of dropping the learning rate. Figure 4 shows the learning rate variation with epochs for different values of ‘D’ and ‘ρ’.

The limitation with both the decay approach and the drop approach is that they do not evaluate if decreasing the learning rate is required or not. In both the methods, the learning rate decreases irrespective of difficulty involved in minimizing the cost function.

In this approach, the learning rate increases or decreases based on the gradient value of the cost function. For higher gradient value, the learning rate will be smaller and for lower gradient value, the learning rate will be larger. Hence, the learning decelerates and accelerates respectively at steeper and shallower parts of the cost function curve. The formula used in this approach is shown in the below equation.

In the above equation, o is the initial learning rate and ‘sn is the momentum factor, which is calculated using below equation. ‘n’ is the epoch/iteration number

In the above equation, 𝛾 is a hyperparameter whose value is typically between 0.7 and 0.9. Note that in equation-7, momentum factor Sn is an exponentially weighted average of gradients. So not only the value of the current gradient is considered, but also the values of gradients from the previous epochs are considered to calculate the momentum factor. Figure 5 shows the idea behind the gradient adapted learning rate. When the cost function curve is steep, the gradient is large, and the momentum factor ‘Sn’ is larger. Hence the learning rate is smaller. When the cost function curve is shallow, the gradient is small and the momentum factor ‘Sn’ is also small. The learning rate is larger.

The gradient adapted learning rate approach eliminates the limitation in the decay and the drop approaches by considering the gradient of the cost function to increase or decrease the learning rate. This approach is widely used in training deep neural nets with stochastic gradient descent.

## Cycling Learning Rate

In this approach, the learning rate varies between a base rate and a maximum rate cyclically. Figure 6 shows the idea behind this approach. The figure shows that the learning rate varies in a triangular form between the maximum and the base rates at a fixed frequency.

It is reported [1] that other forms such as sinusoidal or parabolic too yield similar results. The frequency of variation can be adjusted by setting the value of ‘step size’.  The formula used in this approach is shown below.

In the above equation, E is the learning rates for a given epoch, E is the epoch number, max and base are respectively the maximum and the base learning rates. S is the step size. Note that the above equation valid when E>S. for E≤S below equation can be used

An important step in making this approach work is identifying the right base and the maximum learning rates. The process to identify these, referred to as ‘LR range test’ in [1] is; train the model for a few epochs allowing the learning rate to vary linearly starting from a small number. Capture the accuracy of the model for different learning rates and plot. The plot could look like the image shown in figure 7.

From the plot identify two learning rate values; 1) the value at which the accuracy starts to increase and 2) the value at which the accuracy begins to fluctuate or to decrease. The first point corresponds to the base learning rate, and the second point corresponds to the maximum learning rate. Figure 8 shows the comparison of model accuracy achieved with different learning rate approaches on CIFER-10 image dataset in a convolutional neural network (CNN). In the figure, ‘Original learning rate’ corresponds to a fixed learning rate, ‘Exponential’ corresponds to an exponential learning rate decay approach and ‘CLR’ corresponds to the cyclic learning rate approach. While the constant learning and the exponential approach seem to have taken close to 70000 iterations to achieve an accuracy of 81%, the cyclic learning rate seems to have taken close to 25000 iterations indicating close to 2.5 times quicker convergence.

In this article, the role of learning rate as a hyper-parameter is discussed highlighting the role it plays in the time taken to train a model and the prediction accuracy achieved by the model. Some of the Important aspects to remember are; using variable learning rate instead of a constant learning rate could help achieve higher accuracy in smaller training time. Some of the common techniques to vary the learning rate during the training have been elucidated. One of these or a combination of these techniques, especially the adaptive learning rate and the cyclic learning rate can be used in practical applications to train algorithms using the gradient descent method.

## References

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# Functional equation with three variables

I have a functional equation with three variables. $$f(x,y,z)$$ is a real function with three variables where y is different from z i.e., $$f(x,y,z)$$ defined only for $$y \neq z$$. This function satisfies

1. $$f(x,x,y)=0$$
2. $$f(x,y,x)=1$$
3. $$f(x,y,z)f(z,y,r)=f(x,y,r)$$

What is the general solution for $$f$$?

Or simpler:

Given $$f(x,x,y) = 0$$ and $$f(x,y,x) = 1$$, just set $$x=y$$ to see the contradiction.

There is no such function. If you put $$z=y$$ in 3) you get $$f(x,y,r)=0$$ for all $$x,y,r$$ (provided $$y \neq r$$). But this contradicts 2).

There is no solution.

Note that by 3, For all $$x,y,z$$ you have $$f(x,y,z)=f(x,y,y)f(y,y,z)=f(x,y,y) \cdot 0 =0$$

Edited If $$f(x,y,z)$$ is only defined for $$y \neq z$$ then,a solution is given by $$f(x,y,z)=\frac{x-y}{z-y}$$
• It's not 100% clear from the current phrasing of the question, but I think $f(x,y,z)$ is only defined for $(x,y,z) \in \mathbb{R}^3$ such that $y \neq z$. Hence, $f(x,y,y)$ is undefined. – JimmyK4542 May 5 '19 at 23:25
• Thank you very much. There is a more general solution for this: $f(x,y,z)=\frac{h(x)-h(y)}{h(z)-h(y)}$ or $f(x,y,z)=\frac{h(x)g(y)-g(x)h(y)}{h(z)g(y)-g(z)h(y)}$. I wonder whether there are other solutions. – user409680 May 6 '19 at 0:56


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## Nada94 2 years ago Solid Geometry, A vertical pole BP stands at one corner of a horizontal rectangular field. If AB=10m, AD=5m and the angle of elevation of P from A is 22 degrees. Calculate : (a) the height of the pole. (b) the angle of elevation of P from C. (c) the length of a diagonal of the rectangle ABCD. (d) the angle of elevation of P from D.

|dw:1357581070280:dw|

2. syhna

a) tan22 = BP / 10m tan22 (10m) =BP b) BP/5m = tanC c) $\sqrt{(10^{2}) +(5^{2})} = diagonal of the rectangle$ = sci-cal can help but i forgot mine .. not quite sure with my solutions though

indeed it's correct. I checked that with my teacher just today & what you wrote is correct. Thanks a million for your helpful comment. :)

4. UDIT123

|dw:1364296590348:dw|


text:
## Topic: Easy2Boot

Date 13/06/2020

By kike

### Subject Re: Re: iso windows 7 x64 bits pro efi

Thank you very much for your answer, I have tried and it has not worked, I have copied the file and mounted with ultraiso ..., and it keeps giving me the same errors
I have even added boot.wim and install.wim of that iso in another iso that yes is efi booteable (the one that fails me is udf booteable), to see if it succeeded but when booting it gives me error file: \ winedows \ system32 \ boot \ winload.efi
status: 0xc000000f
missing or corrupt

The problem is that I have an iso w7 that does boot with efi, but when it repairs it tells me that it is not the installed version of windows
And the iso that if you let me repair let me repair in legacy, but in efi it gives me those errors ...
Can you convert a bootable iso udf to bootable efi ??
Or some other possibility?
Where would be the tools to create an imgptn ???
Thank you so much for your help

Back


text:
Mike McCann
22-23 January 2015

Functions in R: Base

We have already seen many of the basic functions that come pre-installed with R.

sum(seq(1,100,1))
abs(-100+50)
dim(iris)
str(iris)
colnames(iris)

Functions in R: Packages

We have also seen functions that are loaded from packages.

# This will not run. tree is not loaded
tree(formula=Species ~ . -Species, data=iris)

install.packages("tree") # Install it

# Now it should work
tree(formula=Species ~ . -Species, data=iris)

Functions in R: Make your own!

It is also possible to define your own functions.

This is especially important if you are going to write the same lines of code over & over again.

• R functions are objects just like anything else.
• By default, R function arguments are lazy, i.e., they're only evaluated if they're actually used.
• Every call on a R object is almost always a function call.

Basic components of a function

• The body(), the code inside the function.
• The formals(), the “formal” argument list, which controls how you can call the function.
• The environment(), which determines how variables referred to inside the function are found.
• args() to list arguments.

Writing functions

# define a function, f
f <- function(x){
x
}

# call function f (argument=1)
f(1)

[1] 1

Functions and Environments

Variables defined inside functions exist in a different environment than the global environment.

However, if a variabe is not defined inside a function, the function will look one level above.

Example of a function

x <- 2 # variable defined outside the function

g <- function() {
y <- 1 # variable defined inside the function
c(x, y)
}

g()

[1] 2 1

Example of a "useful" function

first <- function(x, y) {
z <- x + y
return(z)
}

first(5, 7)

[1] 12

What happens if you don't write return() inside the function?

Try It!

1.) Create a function that takes in two arguments, x and y, and computes x*2 * y.

2.) Create a function that takes in three arguments, and makes a vector from the result.

3.) Create a function that counts the number of matching items. Hint: use %in% to create a logical statement.

Example of a "useful" function

add <- function(a, b){
return(a + b)
}

vector <- c(3,4,5,6)

[1] 7

What does this function return?

x <- 5
f <- function() {
y <- 10
c(x = x, y = y)
}

What does this function return?

x <- 5
f <- function() {
y <- 10
c(x = x, y = y)
}

f()

 x  y
5 10

Functions with pre-defined values

subtract <- function(a=5, b=2){
return(a-b)
}

subtract()

[1] 3

subtract(5,6)

[1] -1

Try It!

1.) Write a function that takes in a vector and multiplies the sum of the vector by 10. Return a logical statement based on whether the sum is under 1000.

2.) Write a function that calculates the mean of every column in a dataframe. Have it break gracefully if the columns are not numbers, using class(x) != “numeric”. Try your function on the iris dataset

Try It!

func3 <- function(x){
for (i in 1:ncol(x)){
if(class(x[,i]) != "numeric"){next}
if(class(x[,i]) == "numeric"){
print(mean(x[,i]))
}
}
}
func3(iris)

[1] 5.843333
[1] 3.057333
[1] 3.758
[1] 1.199333

Functions usually return the last value it computed

f <- function(x) {
if (x < 10) {
0
} else {
10
}
}
f(5)

[1] 0

f(15)

[1] 10


text:
# functional analysis – simple question about and unitary element and self-adjointness

In Murphy’s book P42, it says

If $$A$$ is a unital $$C^*$$– algebra and $$a$$ is self-adjoint in $$A$$, then $$e^{ia}$$ is unitary.

Why is the statement true?

I’ve been playing around with the involution. But I don’t see how we get $$(e^{ia})^* = e^{-ia}$$.

Thank you for your help!