text:
# Can one tumble past a reach weapon

Consider the following situation:

target is using a reach weapon (a glaive). Is it possible for character to move to C1 and attack target avoiding an attack of opportunity from the reach weapon by tumbling past the reach weapon?

Character is a 10th level rogue with a +15 in Tumble and Skill Mastery in Tumble, so she should be able to do the tumble without having to roll for it.

Second question: character has the Spring Attack feat. Is she able to attack target (using Tumble) and move back to her starting position without provoking an attack of opportunity? She has a movement of 30.

• Since the two questions are so related I thought it was better to ask them in one question. However, if necessary I can split the question in two separate question. – Pantalaimon Aug 20 '15 at 14:03

# Getting past reach? This is what the skill Tumble is for...

At least, it's a really important part of what the skill Tumble is for.

Although Ernir's answer accurately quotes the skill Tumble from the Player's Handbook, the Rules Compendium makes the tumbling past part of this use of the skill Tumble a bit clearer:

If you succeed on a DC 15 Tumble check, you can move past opponents without provoking attacks of opportunity. You must make a Tumble check to avoid provoking attacks of opportunity from an opponent as you move by that foe, adding 2 to the DC for each opponent after the first. Failure means you provoke attacks of opportunity normally. If two opponents must be passed at the same time, the tumbler chooses which one to check against first.

Bear in mind that movement is not plotted then taken, heedless of consequences or changing battlefield conditions. With a deeply literal definition of move past, according to the diagram, the character would move from A1 to B1 to C1 to D2 to E1 to F2, provoking an attack of opportunity from the target during B1 to C1 but being unable to make a Tumble check to avoid the attack of opportunity until he moved from E1 to F2. This would make for a lot of dead rogues as it would render this use of the Tumble skill pointless, the character being unable to make Tumble skill checks until he was, in some DM-defined way, past the creature.

Instead, movement is completed square by square. When the character decides to move, he moves from A1 to B1, entering a threatened area and not provoking an attack of opportunity from the target. The the character can decide to continue moving from B1 to C1 and provoke an attack of opportunity from the target or use the skill Tumble to get past that target, avoiding that and future movement-generated attacks of opportunity in the same vein from that target this round.1

So when the text of either description says move past or move by, that's shorthand for move within or exit an opponent's threatened area. Since entering a threatened area doesn't provoke attacks of opportunity, after entering and upon moving within or exiting the threatened area, the character is moving past the target, i.e. past the area the target threatens with its great big pole arm.2

That use of the skill Tumble really is to give the character a chance to avoid attacks of opportunity from the target for the character's movement.

## ...But the skill Tumble is unnecessary if the character has the feat Spring Attack and wants to attack the target

The premium edition of the Player's Handbook (2012) updates the benefit of the feat Spring Attack:

When using the attack action with a melee weapon, you can split your move action in that round in order to move both before and after the attack, provided that the total distance moved is not greater than your speed. Moving this way does not provoke an attack of opportunity from the defender you attack, though it might provoke attacks of opportunity from other creatures, if appropriate. You can't use this feat if you are wearing heavy armor.

You must move at least 5 feet both before and after you make your attack in order to utilize the benefits of Spring Attack. (100)

While this update could've used another editorial pass,3 this makes it clear that, as per the diagram, the character can use part of his speed to move into a position to attack the target, make a standard attack against the target, then move back to his starting position—or another position—using the remainder of his speed, all without provoking attacks of opportunity from the target for the character's movement and without using the skill Tumble.

1 Were the character then to try to move through the target's square, that's a whole 'nother thing.
2 Of course, its possible he's also moving past the target's threatened area into another area the target threatens. Luckily for the character, the character's movement this way can only provoke an attack of opportunity from the target once per round.
3 Two uses of in order and a utilize? Ew.

Yes, you can Tumble past a reach weapon.

Let's take a look at what the Tumble skill says:

[DC 15] Tumble at one-half speed as part of normal movement, provoking no attacks of opportunity while doing so. Failure means you provoke attacks of opportunity normally. Check separately for each opponent you move past, in the order in which you pass them (player’s choice of order in case of a tie). Each additional enemy after the first adds +2 to the Tumble DC.

So if you can make a DC 15 Tumble skill check, your normal (half-speed) movement does not provoke attacks of opportunity. Particularly, note that this is per opponent, making no distinctions about their reach or threatened area.

It is also possible to Spring Attack around a reach weapon.

Let's also take a look at Spring Attack:

When using the attack action with a melee weapon, you can move both before and after the attack, provided that your total distance moved is not greater than your speed. Moving in this way does not provoke an attack of opportunity from the defender you attack, though it might provoke attacks of opportunity from other creatures, if appropriate.

Again, this works on a per-creature basis, in this case that of the defender. When Spring Attacking, the defender doesn't get an AoO, period. There is not even a need to Tumble, which is one difference between Spring Attack and the somewhat similar Flyby Attack.

• But you cannot combine the two with a speed of 30, because moving at half speed means you can only move 3 squares and you won't end back up at your starting position. – Erik Aug 20 '15 at 14:28
• Thanks. As for the Tumble, I probably agree with you, although technically you are not 'passing' the opponent. – Pantalaimon Aug 20 '15 at 14:29
• As for the Spring Attack, I thought the AoO bit concerned the moving back part of the movement which would normally provoke an AoO. This would mean that you could also move to e.g. D0 without provoking an AoO as long as you do a Spring Attack? – Pantalaimon Aug 20 '15 at 14:31
• @Erik 'as part of normal movement' suggests that you can combine normal movement with Tumble; in that case you can move 5 ft normally; tumble 5 ft and then do the same in reverse. Right? – Pantalaimon Aug 20 '15 at 14:33
• @Erik You can use accelerated tumble to move more than three squares. You also can tumble, then swim, then hustle (normal combat movement), and then jump all as parts of one move action provided you have enough speed. – annoying imp Aug 20 '15 at 15:09

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text:
## Statement of Shannon’s coding theorem:

Let $$\Sigma_1, \Sigma_2$$ denote two finite alphabets and let $$\Sigma_1^*$$ and $$\Sigma_2^*$$ denote the set of all finite words from these alphabets. Now, let’s suppose that $$X$$ is a random variable taking values in $$\Sigma_1$$ and let $$f$$ be a uniquely-decodable code from $$\Sigma_1^*$$ to $$\Sigma_2^*$$:

$$f: \Sigma_1^* \rightarrow \Sigma_2^*$$

where $$\Sigma_1 \subset \Sigma_1^*$$ and $$\lvert \Sigma_2 \rvert = a$$.

If we define $$S$$ as an auxiliary random variable which measures the length of the codeword $$f(x)$$:

$$\forall X \sim \Sigma_1, S = l(f(X))$$

then $$f$$ has the minimal expected word length for $$X$$ if it satisfies:

$$\frac{H(X)}{\log_2 a} \leq \mathbb{E}[S] < \frac{H(X)}{\log_2 a} + 1$$

## Shannon’s coding theorem:

Assuming that $$|\Sigma_1| = n$$, let $$s_i$$ denote the word length of each $$x_i \in \Sigma_1$$. In particular, let’s choose $$s_i$$ so that:

$$s_i := \lceil -\log_a p_i \rceil$$

which implies that:

$$-\log_a p_i \leq s_i < -\log_a p_i + 1$$

and so we have:

$$p_i \approx a^{-s_i}$$

$$a^{-s_i} \leq p_i$$

which implies that:

$$\sum_{i=1}^n a^{-s_i} \leq \sum_{i=1}^n p_i = 1$$

and by Kraft’s inequality there exists a prefix-free code with these code lengths.

It follows that the minimal $$S$$ satisfies:

$$\mathbb{E}[S] = \sum_{i} p_i \cdot s_i < \sum_i p_i (-\log_a p_i + 1) = - \sum_i p_i \frac{\log_2 p_i}{\log_2 a} + 1$$

which gives us the right hand side of the inequality:

$$\mathbb{E}[S] < \frac{H(X)}{\log_2 a} + 1$$

and we also have:

$$\mathbb{E}[S] \geq = - \sum_i p_i \frac{\log_2 p_i}{\log_2 a}$$

from which we may deduce (3):

$$\frac{H(X)}{\log_2 a} \leq \mathbb{E}[S] < \frac{H(X)}{\log_2 a} + 1$$

which simplifies to:

$$H(X) \leq \mathbb{E}[S] < H(X) + 1$$

when $$a=2$$.

## References:

1. C.E. Shannon, “A Mathematical Theory of Communication”, Bell System Technical Journal, vol. 27, pp. 379–423, 623-656, July, October, 1948.

2. David J. C. MacKay. Information Theory, Inference, and Learning Algorithms Cambridge: Cambridge University Press, 2003.


text:
# Tutorial: Tree-LSTM in DGL¶

Author: Zihao Ye, Qipeng Guo, Minjie Wang, Jake Zhao, Zheng Zhang

In this tutorial, you learn to use Tree-LSTM networks for sentiment analysis. The Tree-LSTM is a generalization of long short-term memory (LSTM) networks to tree-structured network topologies.

The Tree-LSTM structure was first introduced by Kai et. al in an ACL 2015 paper: Improved Semantic Representations From Tree-Structured Long Short-Term Memory Networks. The core idea is to introduce syntactic information for language tasks by extending the chain-structured LSTM to a tree-structured LSTM. The dependency tree and constituency tree techniques are leveraged to obtain a ‘’latent tree’‘.

The challenge in training Tree-LSTMs is batching — a standard technique in machine learning to accelerate optimization. However, since trees generally have different shapes by nature, parallization is non-trivial. DGL offers an alternative. Pool all the trees into one single graph then induce the message passing over them, guided by the structure of each tree.

## The task and the dataset¶

The steps here use the Stanford Sentiment Treebank in dgl.data. The dataset provides a fine-grained, tree-level sentiment annotation. There are five classes: Very negative, negative, neutral, positive, and very positive, which indicate the sentiment in the current subtree. Non-leaf nodes in a constituency tree do not contain words, so use a special PAD_WORD token to denote them. During training and inference their embeddings would be masked to all-zero.

The figure displays one sample of the SST dataset, which is a constituency parse tree with their nodes labeled with sentiment. To speed up things, build a tiny set with five sentences and take a look at the first one.

import dgl
from dgl.data.tree import SST
from dgl.data import SSTBatch

# Each sample in the dataset is a constituency tree. The leaf nodes
# represent words. The word is an int value stored in the "x" field.
# The non-leaf nodes have a special word PAD_WORD. The sentiment
# label is stored in the "y" feature field.
trainset = SST(mode='tiny')  # the "tiny" set has only five trees
tiny_sst = trainset.trees
num_vocabs = trainset.num_vocabs
num_classes = trainset.num_classes

vocab = trainset.vocab # vocabulary dict: key -> id
inv_vocab = {v: k for k, v in vocab.items()} # inverted vocabulary dict: id -> word

a_tree = tiny_sst[0]
for token in a_tree.ndata['x'].tolist():
print(inv_vocab[token], end=" ")

Out:

Preprocessing...
Dataset creation finished. #Trees: 5
the rock is destined to be the 21st century 's new  conan '' and that he 's going to make a splash even greater than arnold schwarzenegger , jean-claud van damme or steven segal .

## Step 1: Batching¶

Add all the trees to one graph, using the batch() API.

import networkx as nx
import matplotlib.pyplot as plt

graph = dgl.batch(tiny_sst)
def plot_tree(g):
# this plot requires pygraphviz package
pos = nx.nx_agraph.graphviz_layout(g, prog='dot')
nx.draw(g, pos, with_labels=False, node_size=10,
node_color=[[.5, .5, .5]], arrowsize=4)
plt.show()

plot_tree(graph.to_networkx())

You can read more about the definition of batch(), or skip ahead to the next step: .. note:

**Definition**: :func:~dgl.batch unions a list of :math:B
:class:~dgl.DGLGraph\ s and returns a :class:~dgl.DGLGraph of batch
size :math:B.

- The union includes all the nodes,
edges, and their features. The order of nodes, edges, and features are
preserved.

- Given that you have :math:V_i nodes for graph
:math:\mathcal{G}_i, the node ID :math:j in graph
:math:\mathcal{G}_i correspond to node ID
:math:j + \sum_{k=1}^{i-1} V_k in the batched graph.

- Therefore, performing feature transformation and message passing on
the batched graph is equivalent to doing those
on all DGLGraph constituents in parallel.

- Duplicate references to the same graph are
treated as deep copies; the nodes, edges, and features are duplicated,
and mutation on one reference does not affect the other.
- The batched graph keeps track of the meta
information of the constituents so it can be
:func:~dgl.batched_graph.unbatch\ ed to list of DGLGraph\ s.

## Step 2: Tree-LSTM cell with message-passing APIs¶

Researchers have proposed two types of Tree-LSTMs: Child-Sum Tree-LSTMs, and $$N$$-ary Tree-LSTMs. In this tutorial you focus on applying Binary Tree-LSTM to binarized constituency trees. This application is also known as Constituency Tree-LSTM. Use PyTorch as a backend framework to set up the network.

In N-ary Tree-LSTM, each unit at node $$j$$ maintains a hidden representation $$h_j$$ and a memory cell $$c_j$$. The unit $$j$$ takes the input vector $$x_j$$ and the hidden representations of the child units: $$h_{jl}, 1\leq l\leq N$$ as input, then update its new hidden representation $$h_j$$ and memory cell $$c_j$$ by:

$\begin{split}i_j & = & \sigma\left(W^{(i)}x_j + \sum_{l=1}^{N}U^{(i)}_l h_{jl} + b^{(i)}\right), & (1)\\ f_{jk} & = & \sigma\left(W^{(f)}x_j + \sum_{l=1}^{N}U_{kl}^{(f)} h_{jl} + b^{(f)} \right), & (2)\\ o_j & = & \sigma\left(W^{(o)}x_j + \sum_{l=1}^{N}U_{l}^{(o)} h_{jl} + b^{(o)} \right), & (3) \\ u_j & = & \textrm{tanh}\left(W^{(u)}x_j + \sum_{l=1}^{N} U_l^{(u)}h_{jl} + b^{(u)} \right), & (4)\\ c_j & = & i_j \odot u_j + \sum_{l=1}^{N} f_{jl} \odot c_{jl}, &(5) \\ h_j & = & o_j \cdot \textrm{tanh}(c_j), &(6) \\\end{split}$

It can be decomposed into three phases: message_func, reduce_func and apply_node_func.

Note

apply_node_func is a new node UDF that has not been introduced before. In apply_node_func, a user specifies what to do with node features, without considering edge features and messages. In a Tree-LSTM case, apply_node_func is a must, since there exists (leaf) nodes with $$0$$ incoming edges, which would not be updated with reduce_func.

import torch as th
import torch.nn as nn

class TreeLSTMCell(nn.Module):
def __init__(self, x_size, h_size):
super(TreeLSTMCell, self).__init__()
self.W_iou = nn.Linear(x_size, 3 * h_size, bias=False)
self.U_iou = nn.Linear(2 * h_size, 3 * h_size, bias=False)
self.b_iou = nn.Parameter(th.zeros(1, 3 * h_size))
self.U_f = nn.Linear(2 * h_size, 2 * h_size)

def message_func(self, edges):
return {'h': edges.src['h'], 'c': edges.src['c']}

def reduce_func(self, nodes):
# concatenate h_jl for equation (1), (2), (3), (4)
h_cat = nodes.mailbox['h'].view(nodes.mailbox['h'].size(0), -1)
# equation (2)
f = th.sigmoid(self.U_f(h_cat)).view(*nodes.mailbox['h'].size())
# second term of equation (5)
c = th.sum(f * nodes.mailbox['c'], 1)
return {'iou': self.U_iou(h_cat), 'c': c}

def apply_node_func(self, nodes):
# equation (1), (3), (4)
iou = nodes.data['iou'] + self.b_iou
i, o, u = th.chunk(iou, 3, 1)
i, o, u = th.sigmoid(i), th.sigmoid(o), th.tanh(u)
# equation (5)
c = i * u + nodes.data['c']
# equation (6)
h = o * th.tanh(c)
return {'h' : h, 'c' : c}

## Step 3: Define traversal¶

After you define the message-passing functions, induce the right order to trigger them. This is a significant departure from models such as GCN, where all nodes are pulling messages from upstream ones simultaneously.

In the case of Tree-LSTM, messages start from leaves of the tree, and propagate/processed upwards until they reach the roots. A visualization is as follows:

DGL defines a generator to perform the topological sort, each item is a tensor recording the nodes from bottom level to the roots. One can appreciate the degree of parallelism by inspecting the difference of the followings:

print('Traversing one tree:')
print(dgl.topological_nodes_generator(a_tree))

print('Traversing many trees at the same time:')
print(dgl.topological_nodes_generator(graph))

Out:

Traversing one tree:
(tensor([ 2,  3,  6,  8, 13, 15, 17, 19, 22, 23, 25, 27, 28, 29, 30, 32, 34, 36,
38, 40, 43, 46, 47, 49, 50, 52, 58, 59, 60, 62, 64, 65, 66, 68, 69, 70]), tensor([ 1, 21, 26, 45, 48, 57, 63, 67]), tensor([24, 44, 56, 61]), tensor([20, 42, 55]), tensor([18, 54]), tensor([16, 53]), tensor([14, 51]), tensor([12, 41]), tensor([11, 39]), tensor([10, 37]), tensor([35]), tensor([33]), tensor([31]), tensor([9]), tensor([7]), tensor([5]), tensor([4]), tensor([0]))
Traversing many trees at the same time:
(tensor([  2,   3,   6,   8,  13,  15,  17,  19,  22,  23,  25,  27,  28,  29,
30,  32,  34,  36,  38,  40,  43,  46,  47,  49,  50,  52,  58,  59,
60,  62,  64,  65,  66,  68,  69,  70,  74,  76,  78,  79,  82,  83,
85,  88,  90,  92,  93,  95,  96, 100, 102, 103, 105, 109, 110, 112,
113, 117, 118, 119, 121, 125, 127, 129, 130, 132, 133, 135, 138, 140,
141, 142, 143, 150, 152, 153, 155, 158, 159, 161, 162, 164, 168, 170,
171, 174, 175, 178, 179, 182, 184, 185, 187, 189, 190, 191, 192, 195,
197, 198, 200, 202, 205, 208, 210, 212, 213, 214, 216, 218, 219, 220,
223, 225, 227, 229, 230, 232, 235, 237, 240, 242, 244, 246, 248, 249,
251, 253, 255, 256, 257, 259, 262, 263, 267, 269, 270, 271, 272]), tensor([  1,  21,  26,  45,  48,  57,  63,  67,  77,  81,  91,  94, 101, 108,
111, 116, 128, 131, 139, 151, 157, 160, 169, 173, 177, 183, 188, 196,
211, 217, 228, 247, 254, 261, 268]), tensor([ 24,  44,  56,  61,  75,  89,  99, 107, 115, 126, 137, 149, 156, 167,
181, 186, 194, 209, 215, 226, 245, 252, 266]), tensor([ 20,  42,  55,  73,  87, 124, 136, 154, 180, 207, 224, 243, 250, 265]), tensor([ 18,  54,  86, 123, 134, 148, 176, 206, 222, 241, 264]), tensor([ 16,  53,  84, 122, 172, 204, 239, 260]), tensor([ 14,  51,  80, 120, 166, 203, 238, 258]), tensor([ 12,  41,  72, 114, 165, 201, 236]), tensor([ 11,  39, 106, 163, 199, 234]), tensor([ 10,  37, 104, 147, 193, 233]), tensor([ 35,  98, 146, 231]), tensor([ 33,  97, 145, 221]), tensor([ 31,  71, 144]), tensor([9]), tensor([7]), tensor([5]), tensor([4]), tensor([0]))

Call prop_nodes() to trigger the message passing:

import dgl.function as fn
import torch as th

graph.ndata['a'] = th.ones(graph.number_of_nodes(), 1)
graph.register_message_func(fn.copy_src('a', 'a'))
graph.register_reduce_func(fn.sum('a', 'a'))

traversal_order = dgl.topological_nodes_generator(graph)
graph.prop_nodes(traversal_order)

# the following is a syntax sugar that does the same
# dgl.prop_nodes_topo(graph)

Note

Before you call prop_nodes(), specify a message_func and reduce_func in advance. In the example, you can see built-in copy-from-source and sum functions as message functions, and a reduce function for demonstration.

## Putting it together¶

Here is the complete code that specifies the Tree-LSTM class.

class TreeLSTM(nn.Module):
def __init__(self,
num_vocabs,
x_size,
h_size,
num_classes,
dropout,
pretrained_emb=None):
super(TreeLSTM, self).__init__()
self.x_size = x_size
self.embedding = nn.Embedding(num_vocabs, x_size)
if pretrained_emb is not None:
print('Using glove')
self.embedding.weight.data.copy_(pretrained_emb)
self.dropout = nn.Dropout(dropout)
self.linear = nn.Linear(h_size, num_classes)
self.cell = TreeLSTMCell(x_size, h_size)

def forward(self, batch, h, c):
"""Compute tree-lstm prediction given a batch.

Parameters
----------
batch : dgl.data.SSTBatch
The data batch.
h : Tensor
Initial hidden state.
c : Tensor
Initial cell state.

Returns
-------
logits : Tensor
The prediction of each node.
"""
g = batch.graph
g.register_message_func(self.cell.message_func)
g.register_reduce_func(self.cell.reduce_func)
g.register_apply_node_func(self.cell.apply_node_func)
# feed embedding
g.ndata['h'] = h
g.ndata['c'] = c
# propagate
dgl.prop_nodes_topo(g)
# compute logits
h = self.dropout(g.ndata.pop('h'))
logits = self.linear(h)
return logits

## Main Loop¶

Finally, you could write a training paradigm in PyTorch.

from torch.utils.data import DataLoader
import torch.nn.functional as F

device = th.device('cpu')
# hyper parameters
x_size = 256
h_size = 256
dropout = 0.5
lr = 0.05
weight_decay = 1e-4
epochs = 10

# create the model
model = TreeLSTM(trainset.num_vocabs,
x_size,
h_size,
trainset.num_classes,
dropout)
print(model)

# create the optimizer
lr=lr,
weight_decay=weight_decay)

def batcher(dev):
def batcher_dev(batch):
batch_trees = dgl.batch(batch)
return SSTBatch(graph=batch_trees,
wordid=batch_trees.ndata['x'].to(device),
label=batch_trees.ndata['y'].to(device))
return batcher_dev

batch_size=5,
collate_fn=batcher(device),
shuffle=False,
num_workers=0)

# training loop
for epoch in range(epochs):
g = batch.graph
n = g.number_of_nodes()
h = th.zeros((n, h_size))
c = th.zeros((n, h_size))
logits = model(batch, h, c)
logp = F.log_softmax(logits, 1)
loss = F.nll_loss(logp, batch.label, reduction='sum')
loss.backward()
optimizer.step()
pred = th.argmax(logits, 1)
acc = float(th.sum(th.eq(batch.label, pred))) / len(batch.label)
print("Epoch {:05d} | Step {:05d} | Loss {:.4f} | Acc {:.4f} |".format(
epoch, step, loss.item(), acc))

Out:

TreeLSTM(
(embedding): Embedding(19536, 256)
(dropout): Dropout(p=0.5, inplace=False)
(linear): Linear(in_features=256, out_features=5, bias=True)
(cell): TreeLSTMCell(
(W_iou): Linear(in_features=256, out_features=768, bias=False)
(U_iou): Linear(in_features=512, out_features=768, bias=False)
(U_f): Linear(in_features=512, out_features=512, bias=True)
)
)
Epoch 00000 | Step 00000 | Loss 433.6387 | Acc 0.3077 |
Epoch 00001 | Step 00000 | Loss 247.8803 | Acc 0.7326 |
Epoch 00002 | Step 00000 | Loss 737.2761 | Acc 0.4249 |
Epoch 00003 | Step 00000 | Loss 315.7477 | Acc 0.7509 |
Epoch 00004 | Step 00000 | Loss 154.1384 | Acc 0.8352 |
Epoch 00005 | Step 00000 | Loss 289.3680 | Acc 0.7766 |
Epoch 00006 | Step 00000 | Loss 193.6976 | Acc 0.8425 |
Epoch 00007 | Step 00000 | Loss 95.8642 | Acc 0.8938 |
Epoch 00008 | Step 00000 | Loss 72.3045 | Acc 0.9231 |
Epoch 00009 | Step 00000 | Loss 60.1748 | Acc 0.9341 |

To train the model on a full dataset with different settings (such as CPU or GPU), refer to the PyTorch example. There is also an implementation of the Child-Sum Tree-LSTM.

Total running time of the script: ( 0 minutes 2.204 seconds)

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text:
The Mathematical Description of a Wave Video Lessons

Concept

# Problem: A transverse harmonic wave travels on a rope according to the following expression:y(x,t) = 0.12sin(2.1x + 17.6t)9) In what direction is the wave traveling?10) On the same rope, how would increasing the wavelength of the wave change the period of oscillation?

###### FREE Expert Solution

9)

For a wave moving in the positive x-direction:

${\mathbf{y}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{,}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{Asin}}{\mathbf{\left(}}{\mathbf{kx}}{\mathbf{-}}{\mathbf{\omega t}}{\mathbf{\right)}}$

99% (360 ratings)
###### Problem Details

A transverse harmonic wave travels on a rope according to the following expression:

y(x,t) = 0.12sin(2.1x + 17.6t)

9) In what direction is the wave traveling?

10) On the same rope, how would increasing the wavelength of the wave change the period of oscillation?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the The Mathematical Description of a Wave concept. You can view video lessons to learn The Mathematical Description of a Wave. Or if you need more The Mathematical Description of a Wave practice, you can also practice The Mathematical Description of a Wave practice problems.


text:
# Fight Finance

#### CoursesTagsRandomAllRecentScores

The below three graphs show probability density functions (PDF) of three different random variables Red, Green and Blue. Let $P_1$ be the unknown price of a stock in one year. $P_1$ is a random variable. Let $P_0 = 1$, so the share price now is \$1. This one dollar is a constant, it is not a variable.

Which of the below statements is NOT correct? Financial practitioners commonly assume that the shape of the PDF represented in the colour:


text:
## Little Elephant and Lemonade : CodeChef Problem LELEMON

Problem Statement:
Little Elephant likes lemonade.

When Little Elephant visits any room, he finds the bottle of the lemonade in that room that contains the greatest number of litres of lemonade and drinks it all.

There are n rooms (numbered from 0 to n-1), each contains $C_i$ bottles. Each bottle has a volume (in litres). The first room visited by Little Elephant was $P_0$th, the second $P_1$th, …, the m-th $P_{m-1}$th room. Note that Little Elephant may visit a room more than once.

Find for Little Elephant the total volume of lemonade he has drunk.

## Hello Hello : CodeChef Problem HELLO

Problem Statement:
Chef talks a lot on his mobile phone. As a result he exhausts his talk-value (in Rokdas) very quickly. One day at a mobile recharge shop, he noticed that his service provider gives add-on plans which can lower his calling rates (Rokdas/minute). One of the plans said “Recharge for 28 Rokdas and enjoy call rates of 0.50 Rokdas/min for one month”. Chef was very pleased. His normal calling rate is 1 Rokda/min. And he talked for 200 minutes in last month, which costed him 200 Rokdas. If he had this plan activated, it would have costed him: 28 + 0.5*200 = 128 Rokdas only! Chef could have saved 72 Rokdas. But if he pays for this add-on and talks for very little in the coming month, he may end up saving nothing or even wasting money. Now, Chef is a simple guy and he doesn’t worry much about future. He decides to choose the plan based upon his last month’s usage.

There are numerous plans. Some for 1 month, some for 15 months. Some reduce call rate to 0.75 Rokdas/min, some reduce it to 0.60 Rokdas/min. And of course each of them differ in their activation costs. Note – If a plan is valid for M months, then we must pay for (re)activation after every M months (once in M months). Naturally, Chef is confused, and you (as always) are given the task to help him choose the best plan.


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# On Geometric Brownian motion and Itô's formula

Let $$S_t$$ be a geometric brownian motion such as $$d S(t) = rS(t)dt +\sigma S(t)dW(t),$$ where $$W$$ is a standard Brownian motion.

With Itô's lemma and formulas $$(dt)^2=dtdW_t=dW_tdt=0$$ and $$(dW_t)^2=dt$$, we can show that

$$(dS_t)^2=\sigma^2 S_t^2 dt$$ Problem:

At the beginning of a demonstration, the author of an article uses the following equality: $$(\int^T_0 dS_t)^2=\int^T_0 (dS_t)^2$$ I can't see how he got such a result knowing that I find with Itô's lemma that : $$(\int^T_0 dS_t)^2=\int^T_0 (dS_t)^2+2(S_0^2-S_0S_T+\int^T_0 S_tdS_t)$$ Help me see a little more clearly. Thank you in advance.

• This is Ito isometry. Take a look at en.wikipedia.org/wiki/Itô_isometry. I think you are missing the expectation operations. – ilovevolatility Mar 25 at 14:57
• Exactly, I think the author forgot to put the average operator. – M. A. Kacef Mar 25 at 15:04


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# American Institute of Mathematical Sciences

January  2011, 15(1): 293-308. doi: 10.3934/dcdsb.2011.15.293

## Analysis of a delayed free boundary problem for tumor growth

 1 Department of Mathematics, Zhaoqing University, Zhaoqing, 526061, China

Received  December 2009 Revised  February 2010 Published  October 2010

In this paper we study a delayed free boundary problem for the growth of tumors. The establishment of the model is based on the diffusion of nutrient and mass conservation for the two process proliferation and apoptosis(cell death due to aging). It is assumed the process of proliferation is delayed compared to apoptosis. By $L^p$ theory of parabolic equations and the Banach fixed point theorem, we prove the existence and uniqueness of a local solutions and apply the continuation method to get the existence and uniqueness of a global solution. We also study the asymptotic behavior of the solution, and prove that in the case $c$ is sufficiently small, the volume of the tumor cannot expand unlimitedly. It will either disappear or evolve to a dormant state as $t\rightarrow\infty.$
Citation: Shihe Xu. Analysis of a delayed free boundary problem for tumor growth. Discrete & Continuous Dynamical Systems - B, 2011, 15 (1) : 293-308. doi: 10.3934/dcdsb.2011.15.293
##### References:

show all references

##### References:
 [1] Hoang The Tuan. On the asymptotic behavior of solutions to time-fractional elliptic equations driven by a multiplicative white noise. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020318 [2] Touria Karite, Ali Boutoulout. Global and regional constrained controllability for distributed parabolic linear systems: RHUM approach. Numerical Algebra, Control & Optimization, 2020  doi: 10.3934/naco.2020055 [3] Wei Feng, Michael Freeze, Xin Lu. On competition models under allee effect: Asymptotic behavior and traveling waves. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5609-5626. doi: 10.3934/cpaa.2020256 [4] Huiying Fan, Tao Ma. Parabolic equations involving Laguerre operators and weighted mixed-norm estimates. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5487-5508. doi: 10.3934/cpaa.2020249 [5] Yichen Zhang, Meiqiang Feng. A coupled $p$-Laplacian elliptic system: Existence, uniqueness and asymptotic behavior. Electronic Research Archive, 2020, 28 (4) : 1419-1438. doi: 10.3934/era.2020075 [6] Yongxiu Shi, Haitao Wan. Refined asymptotic behavior and uniqueness of large solutions to a quasilinear elliptic equation in a borderline case. Electronic Research Archive, , () : -. doi: 10.3934/era.2020119 [7] Zhenzhen Wang, Tianshou Zhou. Asymptotic behaviors and stochastic traveling waves in stochastic Fisher-KPP equations. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020323 [8] Scipio Cuccagna, Masaya Maeda. A survey on asymptotic stability of ground states of nonlinear Schrödinger equations II. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020450 [9] Luca Battaglia, Francesca Gladiali, Massimo Grossi. Asymptotic behavior of minimal solutions of $-\Delta u = \lambda f(u)$ as $\lambda\to-\infty$. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 681-700. doi: 10.3934/dcds.2020293 [10] Zhilei Liang, Jiangyu Shuai. Existence of strong solution for the Cauchy problem of fully compressible Navier-Stokes equations in two dimensions. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020348 [11] Thabet Abdeljawad, Mohammad Esmael Samei. Applying quantum calculus for the existence of solution of $q$-integro-differential equations with three criteria. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020440 [12] Lin Shi, Xuemin Wang, Dingshi Li. Limiting behavior of non-autonomous stochastic reaction-diffusion equations with colored noise on unbounded thin domains. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5367-5386. doi: 10.3934/cpaa.2020242 [13] Mengni Li. Global regularity for a class of Monge-Ampère type equations with nonzero boundary conditions. Communications on Pure & Applied Analysis, 2021, 20 (1) : 301-317. doi: 10.3934/cpaa.2020267 [14] Adam Glick, Antonio Mastroberardino. Combined therapy for treating solid tumors with chemotherapy and angiogenic inhibitors. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020343 [15] Emre Esentürk, Juan Velazquez. Large time behavior of exchange-driven growth. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 747-775. doi: 10.3934/dcds.2020299 [16] Vivina Barutello, Gian Marco Canneori, Susanna Terracini. Minimal collision arcs asymptotic to central configurations. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 61-86. doi: 10.3934/dcds.2020218 [17] Neng Zhu, Zhengrong Liu, Fang Wang, Kun Zhao. Asymptotic dynamics of a system of conservation laws from chemotaxis. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 813-847. doi: 10.3934/dcds.2020301 [18] Yi An, Bo Li, Lei Wang, Chao Zhang, Xiaoli Zhou. Calibration of a 3D laser rangefinder and a camera based on optimization solution. Journal of Industrial & Management Optimization, 2021, 17 (1) : 427-445. doi: 10.3934/jimo.2019119 [19] Jun Zhou. Lifespan of solutions to a fourth order parabolic PDE involving the Hessian modeling epitaxial growth. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5581-5590. doi: 10.3934/cpaa.2020252 [20] Haiyu Liu, Rongmin Zhu, Yuxian Geng. Gorenstein global dimensions relative to balanced pairs. Electronic Research Archive, 2020, 28 (4) : 1563-1571. doi: 10.3934/era.2020082

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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 8.1: Exponential Properties Involving Products

Difficulty Level: Basic Created by: CK-12
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Practice Exponential Properties Involving Products
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What if you rolled a die and got an integer between 1 to 6? Try multiplying the square of the integer by the cube of the integer. Now take the square of the result. If the original integer were represented by the variable x\begin{align*}x\end{align*}, what would you have after you performed these operations? How would you find the exponent in your answer? In this Concept, you'll learn how to simplify exponential expressions by using the exponential properties involving products so that you can correctly solve this type of problem.

### Exponential Properties Involving Products

In this Concept, you will be learning what an exponent is and about the properties and rules of exponents. You will also learn how to use exponents in problem solving.

Definition: An exponent is a power of a number that shows how many times that number is multiplied by itself.

An example would be 23\begin{align*}2^3\end{align*}. You would multiply 2 by itself 3 times: 2×2×2\begin{align*}2 \times 2 \times 2\end{align*}. The number 2 is the base and the number 3 is the exponent. The value 23\begin{align*}2^3\end{align*} is called the power.

#### Example A

Write in exponential form: α×α×α×α\begin{align*}\alpha \times \alpha \times \alpha \times \alpha\end{align*}.

Solution: You must count the number of times the base, α\begin{align*}\alpha\end{align*}, is being multiplied by itself. It’s being multiplied four times so the solution is α4\begin{align*}\alpha^4\end{align*}.

Note: There are specific rules you must remember when taking powers of negative numbers.

(negative number)×(positive number)(negative number)×(negative number)=negative number=positive number\begin{align*}(negative \ number) \times (positive \ number) &= negative \ number\\ (negative \ number) \times (negative \ number) &= positive \ number\end{align*}

For even powers of negative numbers, the answer will always be positive. Pairs can be made with each number and the negatives will be cancelled out.

(2)4=(2)(2)(2)(2)=(2)(2)(2)(2)=+16\begin{align*}(-2)^4 = (-2)(-2)(-2)(-2) = (-2)(-2) \cdot (-2)(-2) = +16\end{align*}

For odd powers of negative numbers, the answer is always negative. Pairs can be made but there will still be one negative number unpaired, making the answer negative.

(2)5=(2)(2)(2)(2)(2)=(2)(2)(2)(2)(2)=32\begin{align*}(-2)^5 = (-2)(-2)(-2)(-2)(-2) = (-2)(-2) \cdot (-2)(-2) \cdot (-2) = -32\end{align*}

When we multiply the same numbers, each with different powers, it is easier to combine them before solving. This is why we use the Product of Powers Property.

Product of Powers Property: For all real numbers χ,χnχm=χn+m\begin{align*}\chi, \chi^n \cdot \chi^m = \chi^{n+m}\end{align*}.

#### Example B

Multiply χ4χ5\begin{align*}\chi^4 \cdot \chi^5\end{align*}.

Solution: χ4χ5=χ4+5=χ9\begin{align*}\chi^4 \cdot \chi^5 = \chi^{4+5} = \chi^9\end{align*}

Note that when you use the product rule you DO NOT MULTIPLY BASES.

#### Example C

222345\begin{align*}2^2 \cdot 2^3 \neq 4^5\end{align*}

Another note is that this rule APPLIES ONLY TO TERMS THAT HAVE THE SAME BASE.

The Power of a Product

\begin{align*}& (x^4)^3 = x^4 \cdot x^4 \cdot x^4 \qquad \qquad 3 \ \text{factors of} \ x \ \text{to the power} \ 4.\\ & \underbrace{(x \cdot x \cdot x \cdot x}_{x^4}) \cdot \underbrace{(x \cdot x \cdot x \cdot x}_{x^4}) \cdot \underbrace{(x \cdot x \cdot x \cdot x}_{x^4})=\underbrace{(x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x)}_{x^{12}}\end{align*}

This situation is summarized below.

Power of a Product Property: For all real numbers \begin{align*}\chi\end{align*}:

\begin{align*}(\chi^n)^m = \chi^{n \cdot m}\end{align*}

The Power of a Product Property is similar to the Distributive Property. Everything inside the parentheses must be taken to the power outside. For example, \begin{align*}(x^2y)^4=(x^2)^4 \cdot (y)^4=x^8y^4\end{align*}. Watch how it works the long way.

\begin{align*}\underbrace{(x \cdot x \cdot y)}_{x^2y} \cdot \underbrace{(x \cdot x \cdot y)}_{x^2y} \cdot \underbrace{(x \cdot x \cdot y)}_{x^2y} \cdot \underbrace{(x \cdot x \cdot y)}_{x^2y}=\underbrace{(x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y)}_{x^8y4}\end{align*}

The Power of a Product Property does not work if you have a sum or difference inside the parenthesis. For example, \begin{align*}(\chi+\gamma)^2 \neq \chi^2 + \gamma^2\end{align*}. Because it is an addition equation, it should look like \begin{align*}(\chi+\gamma)(\chi+\gamma)\end{align*}.

#### Example D

Simplify \begin{align*}(\chi^2)^3\end{align*}.

Solution: \begin{align*}(\chi^2)^3= \chi^{2\cdot 3} = \chi^6\end{align*}

### Guided Practice

1. Show that \begin{align*}2^2 \cdot 3^3 \neq 6^5\end{align*}.

2. Simplify \begin{align*}(\chi^3\cdot \chi^4)^2\end{align*}.

Solutions:

1. Evaluate each side separately, to show that they are not equal:

\begin{align*}2^2 \cdot 3^3 &= (2\cdot 2)\cdot (3\cdot 3 \cdot 3)=4\cdot 27=108 \\ 6^5 &= 6\cdot 6\cdot 6\cdot 6\cdot 6=7776 \end{align*}

Since \begin{align*}108 \neq 7776\end{align*}, this means that \begin{align*}2^2 \cdot 3^3 \neq 6^5\end{align*}.

2. \begin{align*}\left(\chi^3\cdot \chi^4\right)^2=\left(\chi^{3+4}\right)^2=\left(\chi^7\right)^2=\chi^{7\cdot 2}=\chi^{14} \end{align*}

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Exponent Properties Involving Products (14:00)

1. Consider \begin{align*}a^5\end{align*}. a. What is the base? b. What is the exponent? c. What is the power? d. How can this power be written using repeated multiplication?

Determine whether the answer will be positive or negative. You do not have to provide the answer.

1. \begin{align*}-(3^4)\end{align*}
2. \begin{align*}-8^2\end{align*}
3. \begin{align*}10 \times (-4)^3\end{align*}
4. What is the difference between \begin{align*}-5^2\end{align*} and \begin{align*}(-5)^2\end{align*}?

Write in exponential notation.

1. \begin{align*}2 \cdot 2\end{align*}
2. \begin{align*}(-3)(-3)(-3)\end{align*}
3. \begin{align*}y \cdot y \cdot y \cdot y \cdot y\end{align*}
4. \begin{align*}(3a)(3a)(3a)(3a)\end{align*}
5. \begin{align*}4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\end{align*}
6. \begin{align*}3x \cdot 3x \cdot 3x\end{align*}
7. \begin{align*}(-2a)(-2a)(-2a)(-2a)\end{align*}
8. \begin{align*}6 \cdot 6 \cdot 6 \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y\end{align*}

Find each number.

1. \begin{align*}1^{10}\end{align*}
2. \begin{align*}0^3\end{align*}
3. \begin{align*}7^3\end{align*}
4. \begin{align*}-6^2\end{align*}
5. \begin{align*}5^4\end{align*}
6. \begin{align*}3^4 \cdot 3^7\end{align*}
7. \begin{align*}2^6 \cdot 2\end{align*}
8. \begin{align*}(4^2)^3\end{align*}
9. \begin{align*}(-2)^6\end{align*}
10. \begin{align*}(0.1)^5\end{align*}
11. \begin{align*}(-0.6)^3\end{align*}

Multiply and simplify.

1. \begin{align*}6^3 \cdot 6^6\end{align*}
2. \begin{align*}2^2 \cdot 2^4 \cdot 2^6\end{align*}
3. \begin{align*}3^2 \cdot 4^3\end{align*}
4. \begin{align*}x^2 \cdot x^4\end{align*}
5. \begin{align*}x^2 \cdot x^7\end{align*}
6. \begin{align*}(y^3)^5\end{align*}
7. \begin{align*}(-2 y^4)(-3y)\end{align*}
8. \begin{align*}(4a^2)(-3a)(-5a^4)\end{align*}

Simplify.

1. \begin{align*}(a^3)^4\end{align*}
2. \begin{align*}(xy)^2\end{align*}
3. \begin{align*}(3a^2b^3)^4\end{align*}
4. \begin{align*}(-2xy^4z^2)^5\end{align*}
5. \begin{align*}(3x^2 y^3) \cdot (4xy^2)\end{align*}
6. \begin{align*}(4xyz) \cdot (x^2y^3) \cdot (2yz^4)\end{align*}
7. \begin{align*}(2a^3b^3)^2\end{align*}
8. \begin{align*}(-8x)^3(5x)^2\end{align*}
9. \begin{align*}(4a^2)(-2a^3)^4\end{align*}
10. \begin{align*}(12xy)(12xy)^2\end{align*}
11. \begin{align*}(2xy^2)(-x^2y)^2(3x^2y^2)\end{align*}

Mixed Review

1. How many ways can you choose a 4-person committee from seven people?
2. Three canoes cross a finish line to earn medals. Is this an example of a permutation or a combination? How many ways are possible?
3. Find the slope between (–9, 11) and (18, 6).
4. Name the number set(s) to which \begin{align*}\sqrt{36}\end{align*} belongs.
5. Simplify \begin{align*}\sqrt{74x^2}\end{align*}.
6. 78 is 10% of what number?
7. Write the equation for the line containing (5, 3) and (3, 1).

### Vocabulary Language: English Spanish

Exponent

Exponent

Exponents are used to describe the number of times that a term is multiplied by itself.
Power of a Product Property

Power of a Product Property

For all real numbers $\chi$, $(\chi^n)^m = \chi^{n \cdot m}$.
Product of Powers Property

Product of Powers Property

For all real numbers $\chi$, $\chi^n \cdot \chi^m = \chi^{n+m}$.
Base

Base

When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Power

Power

The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power".

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## On homogeneous totally disconnected 1-dimensional spaces.(English)Zbl 0861.54028

A separable metrizable space $$X$$ is almost 0-dimensional if $$X$$ has a basis $${\mathcal B}$$ such that $$X \smallsetminus \overline B$$ is a union of clopen sets for each $$B\in {\mathcal B}$$. It is known that a topologically complete almost 0-dimensional space has dimension at most 1. One-dimensional examples include “complete Erdös space” (the set of irrational points in Hilbert space), the set of endpoints $$E(L)$$ of the Lelek fan, and the set of endpoints of the Julia set of the exponential map. In this paper, the authors provide a characterization of $$E(L)$$ in terms of a metric and a sequence of disjoint clopen coverings of $$X$$. The result is applied to show (among other things) that the three examples mentioned above are all homeomorphic.

### MSC:

 54F65 Topological characterizations of particular spaces 54H20 Topological dynamics (MSC2010) 54E35 Metric spaces, metrizability

### Keywords:

totally disconnected; homogeneous; complete
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# An algebra problem by Phudish Prateepamornkul

Algebra Level 3

$\begin{eqnarray} ax+by &=& 40 \\ ax^2+by^2 &=& 110 \\ ax^3+by^3 &=& 310 \\ ax^4+by^4 &=& 890 \end{eqnarray}$

Let $$a,b,x,$$ and $$y$$ be real variables satisfying the system of equations above.

Find the value of $$ax^5+by^5$$.

×


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# How to show that the spherical metric satisfies the triangle inequality?

For $x,y\in \mathbb R^n$ define $$d(x,y)={\|x-y\| \over \sqrt{1+\|x\|^2} \sqrt{1+\|y\|^2}}$$ Here $\|x\|$ is the euclidean norm of a vector. How to prove that $d$ (the spherical metric) is indeed a metric?

Progress so far:

1. $d(x,y)\ge 0$ is obvious.
2. $d(x,y) =0 \iff \|x-y\|=0 \iff x=y$, so the positivity holds.
3. $d(x,y) = d(y,x)$ is clear from the formula, so symmetry holds.

But I am having difficulties with the triangle inequality. Writing it out in coordinates leads to a complicated inequality with square roots all over the place in denominators. Is there a better way?

-
Cauchy Schwartz inequality ? –  Maman Jan 3 at 23:35
Haven't tried that, how would it look?does it work? –  Bozo Vulicevic Jan 3 at 23:42
@BozoVulicevic Why don't you look it up before asking such a question. Don't say "does it work?" because it sounds like you are expecting everyone to answer your question as a response to nagging, without doing work yourself. Try it and see. –  Suzu Hirose Jan 4 at 0:08
@Suzu Hirose I apologize for not being honest, i have tried that to no avail, like many other techniques as well. None have worked and i apologize again for not displaying my efforts for vividly for the rest to see. What does not work is not worth mentioning i believe. A lack of effort on my part is not present. –  Bozo Vulicevic Jan 4 at 0:44
I would use stereographic projection: introduce 3D coordinates $u,v,w$ related to $x,y$ by the projection formulas; check that the distance function you have corresponds to the usual Euclidean distance in 3D space; conclude. –  Normal Human Jan 4 at 1:01

## 1 Answer

The quickest way I know is to "cheat"${}^*$ with Stereographic projection. Introduce the map $F:\mathbb R^n\to\mathbb R^{n+1}$ defined by $F( x)=( z,t)\in\mathbb R^n\times \mathbb R$ with $$z=\frac{x}{1+\|x\|^2},\quad t= \frac{\|x\|^2}{1+\|x\|^2}$$ (This is a projection onto the sphere $\|z\|^2+(t-1/2)^2=1/4$, but this fact isn't needed.) Direct computation shows that $$\|F(x)-F(y)\|^2 = \frac{\|x\|^2}{(1+\|x\|^2)^2}+\frac{\|y\|^2}{(1+\|y\|^2)^2} - \frac{2x\cdot y}{(1+\|x\|^2)(1+\|y\|^2)} + \frac{1}{(1+\|x\|^2)^2}+\frac{1}{(1+\|y\|^2)^2} - \frac{2}{(1+\|x\|^2)(1+\|y\|^2)}$$ which simplifies to $$\frac{1}{1+\|x\|^2}+\frac{1}{1+\|y\|^2} - \frac{2x\cdot y}{(1+\|x\|^2)(1+\|y\|^2)} - \frac{2}{(1+\|x\|^2)(1+\|y\|^2)}$$ and subsequently to $$\frac{2+\|x\|^2+\|y\|^2- 2x\cdot y -2}{(1+\|x\|^2)(1+\|y\|^2)} = {\|x-y\|^2 \over ( 1+\|x\|^2)\,(1+\|y\|^2)}$$ Thus, $\|F(x)-F(y)\|=d(x,y)$, and the triangle inequality for $d$ follows from the triangle inequality for the Euclidean distance in $\mathbb R^{n+1}$.

$(*)$ I think this is not much cheating, because what use is this metric to us without knowing its relation to the sphere?

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[–] 6 points7 points  (0 children)

I've used BlueGround to find a (furnished) rental in Paris and I've been super happy with them. They are setup for medium to long term rentals and work well with expats. The rent also include the utilities (which is a big deal from a paperwork point of view).

Going through an agency will require a lot of documentation which may be difficult to procure depending on your situation (several years of french income tax receipts, long term employment contract, etc...).

Also, you should know that you will need (a) a french telephone number, (b) a french bank account and (c) a street address in France for many everyday tasks. Might as well get started on that early. And note that (a) requires (b) and (c), and (b) requires (a) and (c).

Good luck!

[–] 1 point2 points  (0 children)

Thanks for sharing this. Great work.

True, I'm not a fan of phonetic modes, but as far as phonetic modes are concerned, this is a good one :)

I really like your treatment of vowels. They make sense to me.

You introduced some interesting innovations that I think work really well: treatment of "liaisons" and punctuation for foreign words.

Not so sure about the numeric mode. Doesn't seem to be related to the language, and there are other numeric modes existing. Not clear why this one is needed.

It seems to me you have too many glyphs for /s/ and /z/. I understand the need of the nuquerna versions to facilitate placement of tehta, but _thule_ for /s/ and _anto_ for /z/ just seem weird to me, and you don't really provide guidance on when one would use one or the other. I would consider either removing _thule_ and _anto_ altogether, or restricting their usage (perhaps _thule_ for soft-c or c-cedilla?).

Why not _ungwe_ rather than _anga_ for hard /g/?

You could consider using _sa-rince_ for final-s or compounds like "x".

Just to make sure I interpreted things correctly:

"xavier": anga, esse, a, ampa, i, short-carrier, e

"axe": short-carrier, a, calma, silme

"porte": parma, o, romen, tinco

"ail": short-carrier, a, yanta

[–] 19 points20 points  (0 children)

It's a synonym for const customError = response.error as any as Error;

You sometimes need to convert to any as an intermediate step when the types do not overlap sufficiently.

Sometimes this rule can be too conservative and will disallow more complex coercions that might be valid. If this happens, you can use two assertions, first to any (or unknown, which we’ll introduce later), then to the desired type: const a = (expr as any) as T;

[–] 0 points1 point  (0 children)

Be my guest! Let me know if you have any questions or need any help.

BTW, I should have mentioned that the the code display/editing is done with CodeMirror (optionally: if CodeMirror not available, it falls back to using a boring text area).

Also, your project is making me think that perhaps I should have an autoRun option so that changes in the code playground are reflected immediately without having to press the Run button... 🤔

[–] 8 points9 points  (0 children)

Super interesting idea!

Completely agree that interactive documentation would be much beneficial.

Your solution is interesting, and it might even be overpowered for what you're trying to do. The references to 'spreadsheet' confused me at first, and I would have just called it reactive programming. But I get that this is part of a bigger project. If I did understand right, you also have your own language (.ellx?) that you're using to write those reactive examples. The use of single curly brackets to denote reactive expressions is a little weird and it would probably need to be worked around for more complex Javascript examples, but I suppose that can work for simple examples.

Very cool project! Thanks for sharing.

For my own needs, I've been working on a web component along the same lines: specify some HTML, CSS and Javascript, wrap it with this <code-playground> component and it becomes an interactive (and editable) code section in your web page.

javascript <code-playground> <div slot="javascript"> console.log("Hello world") </div> </code-playground>

See here for an example: each of the code sections (on a dark background) are editable: modify them, then press the Run button to see the updated example.

I also have a markdown pre-processor that turns triple tick fence sections into playground automatically. For example, the source code for the example above is context-menu.md and it also displays in GitHub directly, but as a plain, non-interactive, readme file.

[–] 1 point2 points  (0 children)

"external" meaning intended to be used as a public API, as opposed to something that is only used for implementation.

"reachable" meaning following the exported symbols for that module, and the symbols they reference. That means also that if you have a directory with a bunch of files in it, only the files relevant to the specified module are documented.

Basically, it reads all the files in a given directory and builds a gigantic graph of all the symbols and their relationships (actually, Typedoc does that), then it finds the module (or modules) you specify and document the public symbols following that graph.

If you import something and not export it, it will only get documented if something else that is exported references it. I currently don't have an option to skip documenting things that are referenced but not explicitly exported.

[–] 0 points1 point  (0 children)

I've built a tool to generate external documentation. You can see an example of the documentation it generates here.

The tool is called grok and is available on GitHub. It is built on the low level data generated by TypeDoc. You can specify which "module" to have documented. Only symbols reachable by the module are documented.

Maybe you'll find grok suitable for what you're looking for, or it will help you understand how to use the JSON generated by Typedoc.

[–] 1 point2 points  (0 children)

Use “thi’s” (apostrophe between _I_ and _S_) instead of “this”: this will separate the sa-rince.

[–] 1 point2 points  (0 children)

Something like this:

\left.
\begin{matrix}
(E_F-E_i)_{n-side}=kT\ln\left(\frac{n_p}{n_i} \right) \\
(E_F-E_i)_{n-side}=kT\ln\left(\frac{p_p}{n_i} \right)
\end{matrix}
\right\rbrace
\rightarrow
V_{bi} = \frac{kT}{q}\ln\left(\frac{n_np_p}{n_i^2} \right)

The trick is to use a \left.\right\rbrace around a matrix environment.

[–] 2 points3 points  (0 children)

“Writing in the Elvish language” can be interpreted in different ways.

First, there are multiple “elvish languages”, including Quenya and Sindarin. You may want to translate an English phrase into one of these languages and the r/quenya or r/sindarin sub may help with that, but that really depends on the phrase, as our knowledge of these languages is still incomplete.

Second, Tengwar is an ”elvish” writing system. It can be used to write multiple languages, including Quenya, Sindarin, English and others. Different languages use different “conventions” (called “modes”), and once you have a translation you can use Tecendil to do a transcription in Tengwar and the r/tengwar subreddit will be happy to double-check it for you. Transcriptions by Tecendil in Quenya and Sindarin are very reliable, but transcriptions in English are sometimes more a matter of opinion/preferences.

[–] 0 points1 point  (0 children)

It's 128 higher in decimal. 0x80 in hexadecimal is 128 in decimal.

[–] 0 points1 point  (0 children)

I'm not sure exactly what you're trying to do, but I should perhaps point out that in Tecendil you can also select the tengwar just like you would select text and paste them in a text editor. If you apply the font, they will display as expected.

[–] 0 points1 point  (0 children)

Tengwar Annatar does not support uppercase tengwar.

[–] 0 points1 point  (0 children)

Are you on macOS or Windows?

- on Windows: you can type the Unicode characters by using alt+the unicode code, so "alt+e080" for uppercase tinco

- on macOS: System Preferences > Keyboard > Show Input menu in menu bar, then Input Menu > Show Emoji & Symbols, then Gear menu > Customize List > Code Tables > Unicode

[–] 3 points4 points  (0 children)

Telcontar does have the "uppercase" tengwar. That is indeed what I use on Tecendil.

The uppercase tengwar are at unicode code points 128 higher than their lowercase counterpart, so uppercase tinco is at U+E080.

[–] 0 points1 point  (0 children)

Sorry for the delay, this had dropped off my todo list. Tecendil has now been updated with that change. Note that it was already handling the grace accent as a long-carrier vowel, but now it will also do so for the acute accent.

[–] 0 points1 point  (0 children)

Tecendil has been updated to fix the transcription for Joseph.

And also to use plain _silme_ for "s" and _extended formen_ for "ph".

[–] 0 points1 point  (0 children)

Sure. I will get to it next week.

[–] 0 points1 point  (0 children)

I'm not sure what should happen in this case. Just ignore the accent and handle it as "Luthien"?

[–] 0 points1 point  (0 children)

This is awesome. I'm going to spend more time looking at it and thinking it over but I think this looks great.

The liaison feel weird (a word preceded by a liaison will "look" different than the same word without), but I don't know how else to handle it in a phonemic mode. Maybe with a "ring below" mark on the liaison tengwa?

One small nit, I think you have a typo in "Esse: /s/". Shouldn't that be "Esse: /z/"?

[–] 0 points1 point  (0 children)

😀 yes, it's a deliberate choice. It just thought it looked better. And apparently there is some evidence of Tolkien writing Quenya using the classical mode, so if it was good enough for Tolkien...

[–] 0 points1 point  (0 children)

[–] 2 points3 points  (0 children)

To translate into Quenya you need to learn the Quenya language, just like you would need to learn Italian to translate something in English into Italian.

There are some resources on the sidebar of places that can help you learn Quenya.

Once you have something translated into Quenya, you can learn to write it into the Tengwar writing system, or use Tecendil to do the transcription for you.

Note that Tecendil does not translate, it simply transcribes what you type into the Tengwar writing system (Tengwar is not a language, it's a way of writing *any* language).

I believe that you were given a translation of “ Umin ruce i lemiello q'uina. Umin ruce i vadhiello erezea abareze” but that you're not sure of the validity of the translation. Someone who is more an expert on Quenya than me might be able to confirm or correct this translation.

Good luck.


text:
Discrete Gaussian is a cornerstone of many lattice-based cryptographic constructions. Aiming at the orthogonal lattice of a vector, we propose a discrete Gaussian rejection sampling algorithm, by modifying the dynamic programming process for subset sum problems. Within O(nq2) time, our algorithm generates a distribution statistically indistinguishable from discrete Gaussian at width s>ω(log n). Moreover, we apply our sampling algorithm to general high-dimensional dense lattices, and orthogonal lattices of matrices $\matA\in\Z_q^{O(1)\times n}$. Compared with previous polynomial-time discrete Gaussian samplers, our algorithm does not rely on the short basis.

IEICE TRANSACTIONS on Fundamentals of Electronics, Communications and Computer Sciences
Cryptanalysis of Widely-used Hash Function Standards and Beyond
Cryptology

Xiao, D, Yu, Y, & Bi, J. (2018). A new discrete gaussian sampler over orthogonal lattices. IEICE TRANSACTIONS on Fundamentals of Electronics, Communications and Computer Sciences, (11), 1880–1887.


text:
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# Inequalities trick

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Updated on: 26 Apr 2018, 03:34
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I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

$$f(x) = (x-a)(x-b)(x-c)(x-d) < 0$$

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. $$a<b<c<d$$

So for f(x) < 0 consider "-" curves and the ans is: $$(a < x < b)$$, $$(c < x < d)$$
and for f(x) > 0 consider "+" curves and the ans is: $$(x < a)$$, $$(b < x < c)$$, $$(d < x)$$

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Attachment:

1.jpg [ 6.73 KiB | Viewed 72353 times ]

Attachment:

Untitled.png [ 12.08 KiB | Viewed 12750 times ]

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Originally posted by gurpreetsingh on 16 Mar 2010, 10:11.
Last edited by Bunuel on 26 Apr 2018, 03:34, edited 1 time in total.
Edited.
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17 Oct 2010, 16:14
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

I don't understand this part alone. Can you please explain?
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17 Oct 2010, 17:19
Dreamy wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

I don't understand this part alone. Can you please explain?

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.

f(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve
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22 Oct 2010, 06:33
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Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:

doc.jpg [ 7.9 KiB | Viewed 71405 times ]

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
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22 Oct 2010, 11:45
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if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

in case when factors are divided then the numerator will contain = sign

like for (x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

correct me if i am wrong
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11 Mar 2011, 06:29
VeritasPrepKarishma wrote:
vjsharma25 wrote:

How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors?

Check out my post above for explanation.

I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values.
What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change?
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11 Mar 2011, 06:49
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vjsharma25 wrote:
VeritasPrepKarishma wrote:
vjsharma25 wrote:

How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors?

Check out my post above for explanation.

I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values.
What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change?

I always struggle with this as well!!!

There is a trick Bunuel suggested;

(x+2)(x-1)(x-7) < 0

Here the roots are; -2,1,7
Arrange them in ascending order;

-2,1,7; These are three points where the wave will alternate.

The ranges are;
x<-2
-2<x<1
1<x<7
x>7

Take a big value of x; say 1000; you see the inequality will be positive for that.
(1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side.

Graph is +ve after 7.
Between 1 and 7-> -ve
between -2 and 1-> +ve
Before -2 -> -ve

Since the inequality has the less than sign; consider only the -ve side of the graph;

1<x<7 or x<-2 is the complete range of x that satisfies the inequality.
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11 Mar 2011, 19:57
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vjsharma25 wrote:

I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values.
What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change?

Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0
Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2.

Now let me add another factor: (x+8)(x+2)(x-1)(x-7)
Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes.

So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

e.g. (x+2)(x-1)(7 - x)<0

Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
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07 Aug 2011, 07:24
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".

yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a

This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma

However, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.

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08 Aug 2011, 11:59
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Asher wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".

yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a

This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma

However, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.

When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.

Similarly for (x-a)^2(x-b) > 0, x > b

As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.

$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
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08 Aug 2011, 22:22
Quote:
When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.

Similarly for (x-a)^2(x-b) > 0, x > b

Thanks Karishma for the explanation. Hope you wouldn't mind clarifying a few more doubts.

Firstly, in the above case, since x>b could we say that everything with be positive.

would the graph look something like this: positive.. b..postive.. a.. positive

On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative.

would the graph look something like this: negative.. b..postive.. a.. postive

Am i right?

If it is not too much of a trouble, could you please show the graphical representation.

problems with \sqrt{x}.. this is all i could find (googled actually ):

1. √(-x+4) ≤ √(x)
2. x^\sqrt{x} =< (\sqrt{x})^x

{P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems}
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09 Aug 2011, 03:28
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Asher wrote:

Firstly, in the above case, since x>b could we say that everything with be positive.

would the graph look something like this: positive.. b..postive.. a.. positive

On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative.

would the graph look something like this: negative.. b..postive.. a.. postive

So when you have $$(x - a)^2(x - b) < 0$$, you ignore x = a and just plot x = b.
It is positive in the rightmost region and negative on the left. So the graph looks like this:

negative ... b ... positive

Am i right?

If it is not too much of a trouble, could you please show the graphical representation.

problems with \sqrt{x}.. this is all i could find (googled actually ):

1. √(-x+4) ≤ √(x)
2. x^\sqrt{x} =< (\sqrt{x})^x

{P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems}

Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression.
e.g. $$(x-4)^2(x - 9)(x+11) < 0$$
We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this:

positive... -11 ... negative ... 9 ... positive

Since we need the region where x is negative, we get -11 < x < 9.
Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression.

I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them.
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10 Aug 2011, 07:00
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hey ,
can u please tel me the solution for this ques

a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?

ans -11
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10 Aug 2011, 17:01
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WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.

1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

example -
(x+2)(x-1)(7 - x)<0

Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.

2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)

We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0

3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

BUT if it is a fraction the denominator in the solution will not have = SIGN

example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.

$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.

Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate

<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.

I will save this future references....

Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post

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11 Aug 2011, 22:57
1
3
sushantarora wrote:
hey ,
can u please tel me the solution for this ques

a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?

ans -11

Please put questions in new posts. Put it in the same post only if it is totally related or a variation of the question that we are discussing.

Now for the solution:

There are 42 cars on the lot. 1/7 of sports cars and 1/2 of luxury cars have sunroofs. This means that 1/7 of number of sports cars and 1/2 of number of luxury cars should be integers (You cannot have 1.5 cars with sunroofs, right?)
We want to minimize the sunroofs. Since 1/2 of luxury cars have sunroofs and only 1/7 of sports cars have them, it will be good to have fewer luxury cars and more sports cars. Best would be to have all sports cars. But, the question says there are some of each kind at any time. So let's say there are 2 luxury cars (since 1/2 of them should be an integer value). But 1/7 of 40 (the rest of the cars are sports cars) is not an integer number. Let's instead look for the multiple of 7 that is less than 42. The multiple of 7 that is less than 42 is 35. So we could have 35 sports cars. But then, 1/2 of 7 (since 42 - 35 = 7 are luxury cars) is not an integer. The next smaller multiple of 7 is 28. This works. 1/2 of 14 (since 42 - 28 = 14 are luxury cars) is 7. So we can have 14 luxury cars and 28 sports cars. That is the maximum number of sports cars that we can have.
1/7 of 28 sports cars = 4 cars have sunroofs
1/2 of 14 luxury cars = 7 cars have sunroofs

So at least 11 cars will have sunroofs.
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22 Jul 2012, 03:03
VeritasPrepKarishma wrote:
mrinal2100: Kudos to you for excellent thinking!

Correct me if I'm wrong.
If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.
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23 Jul 2012, 03:13
2
Stiv wrote:
VeritasPrepKarishma wrote:
mrinal2100: Kudos to you for excellent thinking!

Correct me if I'm wrong.
If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.

x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined.
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25 Jul 2012, 22:21
pavanpuneet wrote:
Hi Karishma,

Just for my reference, say if the equation was (x+2)(x-1)/(x-4)(x-7) and the question was for what values of x is this expression >0, then the roots will be -2,1,4,7 and by placing on the number line and making the extreme right as positive...

----(-2)----(1)----(4)---(7)----then x>7, 1<x<4 and x<-2...Please confirm..
However, is say it was >=0 then x>7, 1<=x<4 and x<=-2; given that the denominator cannot be zero. Please confirm

Yes, you are right in both the cases.

Also, if you want to verify that the range you have got is correct, just plug in some values to see. Put x = 0, the expression is -ve. Put x = 2, the expression is positive.
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26 Jul 2012, 07:59
Thankyou Karishma.

Further, say if the same expression was (x+2)(1-x)/(x-4)(x-7) and still the question was for what values of x is the expression positive, then ... make it x-1 and with the same roots, have the rightmost as -ve. Then we look for the +ve intervals and check for those intervals if the expression is positive. for examples, in this case, -2<x<1 and 4<x<7 both depict positive interval but only first range satisfies the condition. Please confirm

However, if for the same equation as mentioned, say the expression was (x+2)(x-1)/(x-4)(x-7) >0 and then we were asked to give the range where this is valid, then we would also multiply the -ve sign and make is <0 and then make the range after extreme right root -ve and provide all the intervals where it is negative. Please confirm
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26 Jul 2012, 09:47
VeritasPrepKarishma wrote:

When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.
.

IMHO, it should be x<b and also x is not equal to a .
so, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0
correct me, if I am wrong

my question is -
do we always have a sequence of + and - from rightmost to the left side. I mean is it possible to have + and then + again ?
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Re: Inequalities trick   [#permalink] 26 Jul 2012, 09:47

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text:
## What is the dot product of these two vectors?

What is the dot product of the magnitude A = (2.0x^2)i - (4x)j + (5)k and the magnitude B = (11.0)i + (2.5x)j?

So basically A dot B


text:
# What is the solubility (in M) of PbCl2 in a 0.15 M solution of HCl? The Ksp of PbCl2 is 1.6 × 10-5?

Nov 9, 2015

Reaction:

$P b C {l}_{2} r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 C {l}^{-}$; ${K}_{s p} = 1.6 \times {10}^{- 5}$

Solubility of $P b C {l}_{2}$ $=$ $7.11 \times {10}^{- 4}$ $m o l \cdot {L}^{- 1}$.

#### Explanation:

Now here ${K}_{s p} = \left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2}$. In pure water, we would write ${K}_{s p} = \left(S\right) {\left(2 S\right)}^{2} = 4 {S}^{3}$, where $S$ is the solubility of lead chloride.

However, here the concentration of chloride anion has been (artificially) increased by the presence of hydrochloric acid, which gives stoichiometric quantities of $C {l}^{-}$. So $\left[C {l}^{-}\right]$ $=$ (0.15*mol*L^(-1) + the twice the solubility of lead chloride).

Our revised ${K}_{s p}$ expression is ${K}_{s p} = \left\{S\right\} {\left\{0.15 + 2 S\right\}}^{2}$. Now we can (reasonably) make the approximation, $\left\{0.15 + 2 S\right\} \cong 0.15$; we have to justify this approximation later.

So now, ${K}_{s p} \cong \left[S\right] {\left[0.15\right]}^{2}$, and $S = {K}_{s p} / {\left(0.15\right)}^{2}$ $=$ $\frac{1.6 \times {10}^{- 5}}{0.15} ^ 2$ $=$ $7.11 \times {10}^{- 4}$ $m o l \cdot {L}^{- 1}$.

This value is indeed small compared to 0.15. We could make a 2nd approximation and plug this $S$ value back in the first equation but I am not going to bother. Would the solubility of lead chloride be greater in pure water? Why or why not?


text:
## Distributions with Hierarchical models

This problem concerns three variables, X, Y, and W. The pair of random variables (X,Y) is
described by a hierarchical model; X~$N(\mu, \tau^2)$ and Y|X ~ $N(x, \sigma^2)$. The third variable is defined in terms of the first two that is $W = Y - X$.

1) Find the joint distribution of X and Y.

2) Find the joint distribution of X and W.

3) Are X and W independent?

4) What is the distribution of W?


text:
# Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Miscellaneous Exercise 1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(I) Choose the correct option from the given alternatives:

Question 1.
Let f(1) = 3, f'(1) = $$-\frac{1}{3}$$, g(1) = -4 and g'(1) = $$-\frac{8}{3}$$. The derivative of $$\sqrt{[f(x)]^{2}+[g(x)]^{2}}$$ w.r.t. x at x = 1 is
(a) $$-\frac{29}{15}$$
(b) $$\frac{7}{3}$$
(c) $$\frac{31}{15}$$
(d) $$\frac{29}{15}$$
(d) $$\frac{29}{15}$$

Question 2.
If y = sec(tan-1 x), then $$\frac{d y}{d x}$$ at x = 1, is equal to
(a) $$\frac{1}{2}$$
(b) 1
(c) $$\frac{1}{\sqrt{2}}$$
(d) 2
(c) $$\frac{1}{\sqrt{2}}$$

Question 3.
If f(x) = $$\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)$$, which of the following is not the derivative of f(x)?
(a) $$\frac{2 \cdot 4^{x} \log 4}{1+4^{2 x}}$$
(b) $$\frac{4^{x+1} \log 2}{1+4^{2 x}}$$
(c) $$\frac{4^{x+1} \log 4}{1+4^{4 x}}$$
(d) $$\frac{2^{2(x+1)} \log 2}{1+2^{4 x}}$$
(c) $$\frac{4^{x+1} \log 4}{1+4^{4 x}}$$

Question 4.
If xy = yx, then$$\frac{d y}{d x}$$ = _______
(a) $$\frac{x(x \log y-y)}{y(y \log x-x)}$$
(b) $$\frac{y(y \log x-x)}{x(x \log y-y)}$$
(c) $$\frac{y^{2}(1-\log x)}{x^{2}(1-\log y)}$$
(d) $$\frac{y(1-\log x)}{x(1-\log y)}$$
(b) $$\frac{y(y \log x-x)}{x(x \log y-y)}$$

Question 5.
If y = sin (2 sin-1 x), then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}$$
(b) $$\frac{2+4 x^{2}}{\sqrt{1-x^{2}}}$$
(c) $$\frac{4 x^{2}-1}{\sqrt{1-x^{2}}}$$
(d) $$\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$$
(a) $$\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}$$

Question 6.
If y = $$\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)+\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{x}{\sqrt{1-x^{2}}}$$
(b) $$\frac{1-2 x}{\sqrt{1-x^{2}}}$$
(c) $$\frac{1-2 x}{2 \sqrt{1-x^{2}}}$$
(d) $$\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$$
(c) $$\frac{1-2 x}{2 \sqrt{1-x^{2}}}$$

Question 7.
If y is a function of x and log(x + y) = 2xy, then the value of y'(0) = _______
(a) 2
(b) 0
(c) -1
(d) 1
(d) 1

Question 8.
If g is the inverse of function f and f'(x) = $$\frac{1}{1+x^{7}}$$, then the value of g'(x) is equal to:
(a) 1 + x7
(b) $$\frac{1}{1+[g(x)]^{7}}$$
(c) 1 + [g(x)]7
(d) 7x6
(c) 1 + [g(x)]7

Question 9.
If $$x \sqrt{y+1}+y \sqrt{x+1}=0$$ and x ≠ y, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{1}{(1+x)^{2}}$$
(b) $$-\frac{1}{(1+x)^{2}}$$
(c) (1 + x)2
(d) $$-\frac{x}{x+1}$$
(b) $$-\frac{1}{(1+x)^{2}}$$

Question 10.
If y = $$\tan ^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)$$, where -a < x < a, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{x}{\sqrt{a^{2}-x^{2}}}$$
(b) $$\frac{a}{\sqrt{a^{2}-x^{2}}}$$
(c) $$-\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
(d) $$\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
(c) $$-\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
[Hint: Put x = a cos θ]

Question 11.
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), then $$\left[\frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}$$ = _______
(a) $$\frac{8 \sqrt{2}}{a \pi}$$
(b) $$-\frac{8 \sqrt{2}}{a \pi}$$
(c) $$\frac{a \pi}{8 \sqrt{2}}$$
(d) $$\frac{4 \sqrt{2}}{a \pi}$$
(a) $$\frac{8 \sqrt{2}}{a \pi}$$

Question 12.
If y = a cos (log x) and $$A \frac{d^{2} y}{d x^{2}}+B \frac{d y}{d x}+C=0$$, then the values of A, B, C are _______
(a) x2, -x, -y
(b) x2, x, y
(c) x2, x, -y
(d) x2, -x, y
(b) x2, x, y

(II) Solve the following:

Question 1.

Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
Solution:

Question 2.
The values of f(x), g(x), f'(x) and g'(x) are given in the following table:

Match the following:

Solution:

Question 3.
Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.

(i) The derivative of f[g(x)] w.r.t. x at x = 0 is _______
(ii) The derivative of g[f(x)] w.r.t. x at x = 0 is _______
(iii) The value of $$\left[\frac{d}{d x}\left[x^{10}+f(x)\right]^{-2}\right]_{x=1}$$ is _______
(iv) The derivative of f[(x+g(x))] w.r.t. x at x = 0 is _______
Solution:

Question 4.
Differentiate the following w.r.t. x:
(i) $$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$
Solution:
Let y = $$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$
Put x = cos θ, Then θ = cos-1 x and

(ii) $$\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]$$
Solution:
Let y = $$\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]$$
Put x = cos θ. Then θ = cos-1 x and

(iii) $$\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]$$
Solution:

(iv) $$\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$$
Solution:
Let y = $$\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$$
Put x = cos θ. Then θ = cos-1 x and

(v) $$\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)+\cot ^{-1}\left(\frac{1-10 x^{2}}{7 x}\right)$$
Solution:

(vi) $$\tan ^{-1}\left[\sqrt{\frac{\sqrt{1+x^{2}+x}}{\sqrt{1+x^{2}}-x}}\right]$$
Solution:

Question 5.
(i) If $$\sqrt{y+x}+\sqrt{y-x}=c$$, show that $$\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$$
Solution:
$$\sqrt{y+x}+\sqrt{y-x}=c$$
Differentiating both sides w.r.t. x, we get

(ii) If $$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$, then show that $$\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$$
Solution:
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$
$$y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$$
Differentiating both sides w.r.t. x, we get

(iii) If x sin(a + y) + sin a cos(a + y) = 0, then show $$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$$
Solution:
x sin(a + y) + sin a . cos (a + y) = 0 ….. (1)
Differentiating w.r.t. x, we get

(iv) If sin y = x sin(a + y), then show that $$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$$
Solution:

(v) If x = $$e^{\frac{x}{y}}$$, then show that $$\frac{d y}{d x}=\frac{x-y}{x \log x}$$
Solution:
x = $$e^{\frac{x}{y}}$$
$$\frac{x}{y}$$ = log x …..(1)
y = $$\frac{x}{\log x}$$

(vi) If y = f(x) is a differentiable function of x, then show that $$\frac{d^{2} x}{d y^{2}}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^{2} y}{d x^{2}}$$
Solution:
If y = f(x) is a differentiable function of x such that inverse function x = f-1(y) exists,

Question 6.
(i) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$$
Solution:

(ii) Differentiate $$\log \left[\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}-x}\right]$$ w.r.t. cos(log x)
Solution:

(iii) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^{2}}}{2 \sqrt{1+x^{2}}}}\right)$$
Solution:

Question 7.
(i) If y2 = a2 cos2x + b2 sin2x, show that $$y+\frac{d^{2} y}{d x^{2}}=\frac{a^{2} b^{2}}{y^{3}}$$
Solution:
y2 = a2 cos2x + b2 sin2x …… (1)
Differentiating both sides w.r.t. x, we get

(ii) If log y = log(sin x) – x2, show that $$\frac{d^{2} y}{d x^{2}}+4 x \frac{d y}{d x}+\left(4 x^{2}+3\right) y=0$$
Solution:

(iii) If x = a cos θ, y = b sin θ, show that $$a^{2}\left[y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]+b^{2}=0$$
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

(iv) If y = A cos(log x) + B sin(log x), show that x2y2 + xy1 + y = o.
Solution:
y = A cos (log x) + B sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(v) If y = A emx + B enx, show that y2 – (m + n) y1 + (mn) y = 0.
Solution:
y = A emx + B enx
Differentiating w.r.t. x, we get