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# 3.5 Scalar Types
1
Scalar types comprise enumeration types, integer types, and real types. Enumeration types and integer types are called discrete types; each value of a discrete type has a position number which is an integer value. Integer types and real types are called numeric types. All scalar types are ordered, that is, all relational operators are predefined for their values.
#### Syntax
2
range_constraint ::= range range
3
4
A range has a lower bound and an upper bound and specifies a subset of the values of some scalar type (the type of the range). A range with lower bound L and upper bound R is described by “L .. R”. If R is less than L, then the range is a null range, and specifies an empty set of values. Otherwise, the range specifies the values of the type from the lower bound to the upper bound, inclusive. A value belongs to a range if it is of the type of the range, and is in the subset of values specified by the range. A value satisfies a range constraint if it belongs to the associated range. One range is included in another if all values that belong to the first range also belong to the second.
#### Name Resolution Rules
5
For a subtype_indication containing a range_constraint, either directly or as part of some other scalar_constraint, the type of the range shall resolve to that of the type determined by the subtype_mark of the subtype_indication. For a range of a given type, the simple_expressions of the range (likewise, the simple_expressions of the equivalent range for a range_attribute_reference) are expected to be of the type of the range.
#### Static Semantics
6
The base range of a scalar type is the range of finite values of the type that can be represented in every unconstrained object of the type; it is also the range supported at a minimum for intermediate values during the evaluation of expressions involving predefined operators of the type.
7
A constrained scalar subtype is one to which a range constraint applies. The range of a constrained scalar subtype is the range associated with the range constraint of the subtype. The range of an unconstrained scalar subtype is the base range of its type.
#### Dynamic Semantics
8
A range is compatible with a scalar subtype if and only if it is either a null range or each bound of the range belongs to the range of the subtype. A range_constraint is compatible with a scalar subtype if and only if its range is compatible with the subtype.
9
The elaboration of a range_constraint consists of the evaluation of the range. The evaluation of a range determines a lower bound and an upper bound. If simple_expressions are given to specify bounds, the evaluation of the range evaluates these simple_expressions in an arbitrary order, and converts them to the type of the range. If a range_attribute_reference is given, the evaluation of the range consists of the evaluation of the range_attribute_reference.
10
Attributes
11
For every scalar subtype S, the following attributes are defined:
12
S'First
S'First denotes the lower bound of the range of S. The value of this attribute is of the type of S.
13
S'Last
S'Last denotes the upper bound of the range of S. The value of this attribute is of the type of S.
14
S'Range
S'Range is equivalent to the range S'First .. S'Last.
15
S'Base
S'Base denotes an unconstrained subtype of the type of S. This unconstrained subtype is called the base subtype of the type.
16
S'Min
S'Min denotes a function with the following specification:
17
function S'Min(LeftRight : S'Base)
return S'Base
18
The function returns the lesser of the values of the two parameters.
19
S'Max
S'Max denotes a function with the following specification:
20
function S'Max(LeftRight : S'Base)
return S'Base
21
The function returns the greater of the values of the two parameters.
22
S'Succ
S'Succ denotes a function with the following specification:
23
function S'Succ(Arg : S'Base)
return S'Base
24
For an enumeration type, the function returns the value whose position number is one more than that of the value of Arg; Constraint_Error is raised if there is no such value of the type. For an integer type, the function returns the result of adding one to the value of Arg. For a fixed point type, the function returns the result of adding small to the value of Arg. For a floating point type, the function returns the machine number (as defined in 3.5.7) immediately above the value of Arg; Constraint_Error is raised if there is no such machine number.
25
S'Pred
S'Pred denotes a function with the following specification:
26
function S'Pred(Arg : S'Base)
return S'Base
27
For an enumeration type, the function returns the value whose position number is one less than that of the value of Arg; Constraint_Error is raised if there is no such value of the type. For an integer type, the function returns the result of subtracting one from the value of Arg. For a fixed point type, the function returns the result of subtracting small from the value of Arg. For a floating point type, the function returns the machine number (as defined in 3.5.7) immediately below the value of Arg; Constraint_Error is raised if there is no such machine number.
Paragraphs 28 through 37 were moved to 4.10, “Image Attributes”.
37.1/5
S'Wide_Wide_Width
S'Wide_Wide_Width denotes the maximum length of a Wide_Wide_String returned by S'Wide_Wide_Image over all values of the subtype S, assuming a default implementation of S'Put_Image. It denotes zero for a subtype that has a null range. Its type is universal_integer.
38/5
S'Wide_Width
S'Wide_Width denotes the maximum length of a Wide_String returned by S'Wide_Image over all values of the subtype S, assuming a default implementation of S'Put_Image. It denotes zero for a subtype that has a null range. Its type is universal_integer.
39/5
S'Width
S'Width denotes the maximum length of a String returned by S'Image over all values of the subtype S, assuming a default implementation of S'Put_Image. It denotes zero for a subtype that has a null range. Its type is universal_integer.
39.1/2
S'Wide_Wide_Value
S'Wide_Wide_Value denotes a function with the following specification:
39.2/2
function S'Wide_Wide_Value(Arg : Wide_Wide_String)
return S'Base
39.3/2
This function returns a value given an image of the value as a Wide_Wide_String, ignoring any leading or trailing spaces.
39.4/3
For the evaluation of a call on S'Wide_Wide_Value for an enumeration subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an enumeration literal and if it corresponds to a literal of the type of S (or corresponds to the result of S'Wide_Wide_Image for a nongraphic character of the type), the result is the corresponding enumeration value; otherwise, Constraint_Error is raised.
39.5/3
For the evaluation of a call on S'Wide_Wide_Value for an integer subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an integer literal, with an optional leading sign character (plus or minus for a signed type; only plus for a modular type), and the corresponding numeric value belongs to the base range of the type of S, then that value is the result; otherwise, Constraint_Error is raised.
39.6/2
For the evaluation of a call on S'Wide_Wide_Value for a real subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of one of the following:
39.7/2
numeric_literal
39.8/2
39.9/2
39.10/2
39.11/2
39.12/3
with an optional leading sign character (plus or minus), and if the corresponding numeric value belongs to the base range of the type of S, then that value is the result; otherwise, Constraint_Error is raised. The sign of a zero value is preserved (positive if none has been specified) if S'Signed_Zeros is True.
40
S'Wide_Value
S'Wide_Value denotes a function with the following specification:
41
function S'Wide_Value(Arg : Wide_String)
return S'Base
42
This function returns a value given an image of the value as a Wide_String, ignoring any leading or trailing spaces.
43/5
For the evaluation of a call on S'Wide_Value for an enumeration subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an enumeration literal and if it corresponds to a literal of the type of S (or corresponds to the result of S'Wide_Image for a value of the type, assuming a default implementation of S'Put_Image), the result is the corresponding enumeration value; otherwise, Constraint_Error is raised. For a numeric subtype S, the evaluation of a call on S'Wide_Value with Arg of type Wide_String is equivalent to a call on S'Wide_Wide_Value for a corresponding Arg of type Wide_Wide_String.
Paragraphs 44 through 51 were moved to Wide_Wide_Value.
52
S'Value
S'Value denotes a function with the following specification:
53
function S'Value(Arg : String)
return S'Base
54
This function returns a value given an image of the value as a String, ignoring any leading or trailing spaces.
55/5
For the evaluation of a call on S'Value for an enumeration subtype S, if the sequence of characters of the parameter (ignoring leading and trailing spaces) has the syntax of an enumeration literal and if it corresponds to a literal of the type of S (or corresponds to the result of S'Image for a value of the type, assuming a default implementation of S'Put_Image), the result is the corresponding enumeration value; otherwise, Constraint_Error is raised. For a numeric subtype S, the evaluation of a call on S'Value with Arg of type String is equivalent to a call on S'Wide_Wide_Value for a corresponding Arg of type Wide_Wide_String.
#### Implementation Permissions
56/2
An implementation may extend the Wide_Wide_Value, Wide_Value, Value, Wide_Wide_Image, Wide_Image, and Image attributes of a floating point type to support special values such as infinities and NaNs.
56.1/3
An implementation may extend the Wide_Wide_Value, Wide_Value, and Value attributes of a character type to accept strings of the form “Hex_hhhhhhhh” (ignoring case) for any character (not just the ones for which Wide_Wide_Image would produce that form — see 3.5.2), as well as three-character strings of the form “'X'”, where X is any character, including nongraphic characters.
#### Static Semantics
56.2/3
For a scalar type, the following language-defined representation aspect may be specified with an aspect_specification (see 13.1.1):
56.3/3
Default_Value
This aspect shall be specified by a static expression, and that expression shall be explicit, even if the aspect has a boolean type. Default_Value shall be specified only on a full_type_declaration.
56.4/5
If a derived type inherits a boolean Default_Value aspect, the aspect may be specified to have any value for the derived type. If a derived type T does not inherit a Default_Value aspect, it shall not specify such an aspect if it inherits a primitive subprogram that has a parameter of type T of mode out
#### Name Resolution Rules
56.5/3
The expected type for the expression specified for the Default_Value aspect is the type defined by the full_type_declaration on which it appears.
57
NOTE 1 The evaluation of S'First or S'Last never raises an exception. If a scalar subtype S has a nonnull range, S'First and S'Last belong to this range. These values can, for example, always be assigned to a variable of subtype S.
58/5
NOTE 2 For a subtype of a scalar type, the result delivered by the attributes Succ, Pred, and Value can be outside to the subtype; similarly, the actual parameters of the attributes Succ, Pred, and Image can also be outside the subtype.
59/5
NOTE 3 For any value V (including any nongraphic character) of an enumeration subtype S without a specified Put_Image (see 4.10), S'Value(S'Image(V)) equals V, as do S'Wide_Value(S'Wide_Image(V)) and S'Wide_Wide_Value(S'Wide_Wide_Image(V)). None of these expressions ever raise Constraint_Error.
#### Examples
60
Examples of ranges:
61
-10 .. 10
X .. X + 1
0.0 .. 2.0*Pi
Red .. Green -- see 3.5.1
1 .. 0 -- a null range
Table'Range -- a range attribute reference (see 3.6)
62
Examples of range constraints:
63
range -999.0 .. +999.0
range S'First+1 .. S'Last-1 | 2,925 | 12,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | longest | en | 0.862973 |
https://math.stackexchange.com/questions/1265825/integrating-associated-legendre-polynomials | 1,723,033,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00225.warc.gz | 324,366,331 | 38,226 | # Integrating Associated Legendre Polynomials
As part of a derivation for the question I asked here in Physics stackexchange, I am trying to calculate the following integral, but I am not sure how to proceed:
$\int_0^{2\pi}P_l^m(\cos\theta)P_{l-1}^m(\cos\theta) d\theta$,
where $P_l^m(\cos\theta)$ are the Associated Legendre Polynomials.
I believe that the answer is ultimately 0, but I am not quite sure of how reduce this integrable to a workable form, I have tried using recursion relations but have had no luck. I am sure that I am not trying to solve the integrand with an additional $\sin(\theta)$ factor, which would be easily solved using the change of variable $z = \cos(\theta)$ and applying the orthogonality of the Associated Legendre Polynomials. Does anyone have any advice?
Thanks.
• for what reason you DON'T wanna use $cos(\theta)=z$? Commented May 4, 2015 at 8:08
• Since I don't have a $\sin\theta$ factor in the integrand, if I do the $z=\cos\theta$ substitution I get $\frac{P_l^m(z)P_{l-1}^m(z)}{\sqrt{1-z^2}}$ Commented May 4, 2015 at 8:32
We wish to evaluate: $$\int_0^{2\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta$$ We use the transformation: $$x = \cos\theta\quad \frac{dx}{d\theta} = -\sin\theta \quad d\theta = \frac{-1}{\sqrt{1-x^2}}dx$$ to obtain$$\int_1^1 \frac{-P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx = 0$$
Since you're integrating from 1 to 1.
Weird? Yes I thought so too. Seeing as the integral comes from spherical harmoics, I assume you are messing up your limits of integration (should be from $0$ to $\pi$, since that is how the zenith angle is normally defined).
Then, I think you should be working with:
$$\int_0^{\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta$$
which after the transformation becomes:
$$I = \int_{-1}^1 \frac{P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx$$
The Legendre Polynomials satisfy the relationship:
$$P_l^m(-x) = (-1)^{l+m}P_l^m(x)$$
This means that if $l+m$ is even (odd), the Legendre Polynomial is also even (odd). Here we have the two numbers: $l+m$ and $l+m-1$ One of these two has to be even and the other odd. So one of the polynomials is even and the other is odd. Since $\frac{1}{\sqrt{1-x^2}}$ is even, the integrand as a whole is odd. Thus the integral is also zero.
$$I = 0$$
Reference for parity relationship.
• Bravo. Bravo. Bravo. Commented Jul 14, 2015 at 5:09
• The first derivation is wrong since $x=\cos \theta$ fails to be one-to-one on the domain of integration, note that the conclusion would imply $\int_0^{2\pi}f(\cos t)dt=0$ for any $f$. Commented Jan 8, 2020 at 10:33 | 843 | 2,600 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-33 | latest | en | 0.889496 |
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# Graded Discussion Board (GDB) of CS502 Spring 2012
Dynamic Programming is always preferable over greedy approach " Support or contradict this statement with solid arguments.
Note:
A concise, coherent and to the point comment is preferred over lengthy comment having irrelevant details. Your comment must not be more than 5-7 lines. Comments, posted on regular Lesson's MDB or sent through email will NOT be considered. Any request about such an acceptance will not be catered.
launched on July 9, 2012 and it will remain open till July 10, 2012.
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Views: 910
### Replies to This Discussion
in GDB of cs502 our answer may be as follows:
yes, it is true.
although these both approaches are used to solve problems optimally, but in the same time there are some reasons which make Dynamic programming preferable over greedy approach e.g. dynamic programming solves problems by breaking down in smaller sub problems and also stores results in some form for future reference, which definitely comes always in a solution and also in optimal one. while greedy approach always progress in best possible solution at present with out thinking about future hence it can some time mislead us in such that it can not provide any solution to the problem.
for your reference there is also an example given below which will help to understand the greedy approach disadvantage.
let's suppose we want to prune (search) this given tree for getting best or maximum output or our goal is to find maximum value.
now it totally clears that using greedy approach if we start from node 7 at that point the best possible of both is 12 hence it will go towards 12 and then finally towards 6.
while if we look other side of tree the best possible we can obtain is 99.
so we can conclude that how simple greedy approach mislead us and thus we can not get our required solution.
this is example only for your understanding it is not part of our solution other wise if any one added this will get 0 marks.
so, don't copy paste write in your own words to get maximum marks....which must be 5 now after this solution.
enjoy......
bhai pic to nahi bani na
ha ha ha ha ha.
dear picture q bnani ha? wesy ye Wikipedia se li ha ap b check ker skty ho. yeh to sirf example k leye hai k ap bat smajh jeyen aur apny words me likh saken.......
akram ul haq gud keep it up
thanks a lot
CS502GDB Solution Idea
In GDB of cs502 our answer may be asfollows:
Yes, it is true.
although these both approaches areused to solve problems optimally, but in the same time there are some reasonswhich make Dynamic programming preferable over greedy approach e.g. dynamicprogramming solves problems by breaking down in smaller sub problems and alsostores results in some form for future reference, which definitely comes alwaysin a solution and also in optimal one. while greedy approach always progress inbest possible solution at present with out thinking about future hence it cansome time mislead us in such that it can not provide any solution to theproblem.
for your reference there is also anexample given below which will help to understand the greedy approachdisadvantage.
So we can conclude that how simplegreedy approach mislead us and thus we can not get our required solution.
thanks
A greedy algorithm is similar to a dynamic programming algorithm, but the difference is that solutions to the subproblems do not have to be known at each stage; instead a "greedy" choice can be made of what looks best for the moment.
Consider this example.
You are standing at a place A. You are to goto B. There are intermediate places C1,C2 ...
You want to minimize distance travelled.
Greedy Method of Solving
You dont want to try all intermediate places. You goto the nearest intermediate place. Why? You feel by going to the nearest intermediate place, you will minimize the distance to B.
Dynamic Programming
You try all the places, but you store the previous result. Eg: To reach C3 in minimum distance, you reached by C1. So you store C1. So if you want to go to C5, by C3, you will goto C1 then C3 and then check if going from C3 to C5 is nearest.
Dynamic Programming is always preferable over greedy approach because reasons which make Dynamic programming preferable over greedy approach e.g. dynamicprogramming solves problems by breaking down in smaller sub problems and alsostores results in some form for future reference, which definitely comes alwaysin a solution and also in optimal one. while greedy approach always progress inbest possible solution at present with out thinking about future hence it cansome time mislead us in such that it can not provide any solution to the problem.
CSCS502 GDB
Topic:
Dynamic Programming is always preferable over greedy approach " Support or contradict this statement with solid arguments.
Note:
A concise, coherent and to the point comment is preferred over lengthy comment having irrelevant details. Your comment must not be more than 5-7 lines. Comments, posted on regular Lesson's MDB or sent through email will NOT be considered. Any request about such an acceptance will not be catered.
CSCS502 GDB
Solution No.1
Dynamic programming is both a mathematical optimization method and a computer programming method. In both contexts it refers to simplifying a complicated problem by breaking it down into simpler sub problems in a recursive manner. [ While some decision problems cannot be taken apart this way, decisions that span several points in time do often break apart recursively; Bellman called this the "Principle of Optimality". Likewise, in computer science, a problem that can be broken down recursively is said to have optimal substructure. If sub problems can be nested recursively inside larger problems, so that dynamic programming methods are applicable, then there is a relation between the value of the larger problem and the values of the sub problems.[5] In the optimization literature this relationship is called the Bellman equation.
Solution No.2
Yes, it is true. Although these both approaches are used to solve problems optimally, but in the same time there are some reasons which make Dynamic programming preferable over greedy approach e.g. dynamic programming solves problems by breaking down in smaller sub problems and also stores results in some form for future reference, which definitely comes always in a solution and also in optimal one. while greedy approach always progress in best possible solution at present without thinking about future hence it can some time mislead us in such that it cannot provide any solution to the problem.
for your reference there is also an example given below which will help to understand the greedy approach disadvantage.
So we can conclude that how simple greedy approach mislead us and thus we cannot get our required solution.
now it totally clears that using greedy approach if we start from node 7 at that point the best possible of both is 12 hence it will go towards 12 and then finally towards 6.
while if we look other side of tree the best possible we can obtain is 99.
so we can conclude that how simple greedy approach mislead us and thus we cannot get our required solution.
this is example only for your understanding it is not part of our solution otherwise if any one added this will get 0 marks.
so, don't copy paste write in your own words to get maximum marks....which must be 5 now after this solution.
Solution No.3
A greedy algorithm is similar to a dynamic programming algorithm, but the difference is that solutions to the subproblems do not have to be known at each stage; instead a "greedy" choice can be made of what looks best for the moment.
Consider this example.
You are standing at a place A. You are to goto B. There are intermediate places C1,C2 ...
You want to minimize distance travelled.
Greedy Method of Solving
You don’t want to try all intermediate places. You go to the nearest intermediate place. Why? You feel by going to the nearest intermediate place, you will minimize the distance to B.
Dynamic Programming
You try all the places, but you store the previous result. Eg: To reach C3 in minimum distance, you reached by C1. So you store C1. So if you want to go to C5, by C3, you will go to C1 then C3 and then check if going from C3 to C5 is nearest.
Solution No.4
Dynamic Programming is always preferable over greedy approach because reasons which make Dynamic programming preferable over greedy approach e.g. dynamic programming solves problems by breaking down in smaller sub problems and also stores results in some form for future reference, which definitely comes always in a solution and also in optimal one. while greedy approach always progress in best possible solution at present without thinking about future hence it can some time mislead us in such that it cannot provide any solution to the problem.
100% true solution
In GDB of cs502 our answer may be asfollows:
Yes, it is true.
Although these both approaches are used to solve problems optimally,
But in the same time there are some reasons which make Dynamic programming preferable over greedy approach
E.g. dynamic programming solves problems by breaking down in smaller sub problems and also stores results in some form for future reference,
Which definitely comes always in a solution and also in optimal one?
While greedy approach always progress in best possible solution at present without thinking about future hence it can some time mislead us in such that it cannot provide any solution to the problem.
for your reference there is also an example given below which will help to understand the greedy approach disadvantage.
So we can conclude that how simple greedy approach misleads us and thus we cannot get our required solution.
ZEESHAN NADEEM gud keep it up
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3 | 2,252 | 10,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.895473 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/8470/2/a/cy/1/1/ | 1,638,332,388,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.78/warc/CC-MAIN-20211201022332-20211201052332-00101.warc.gz | 960,530,258 | 82,820 | # Properties
Label 8470.2.a.cy.1.1 Level $8470$ Weight $2$ Character 8470.1 Self dual yes Analytic conductor $67.633$ Analytic rank $1$ Dimension $6$ CM no Inner twists $1$
# Learn more
## Newspace parameters
Level: $$N$$ $$=$$ $$8470 = 2 \cdot 5 \cdot 7 \cdot 11^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 8470.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$67.6332905120$$ Analytic rank: $$1$$ Dimension: $$6$$ Coefficient field: 6.6.13298000.1 Defining polynomial: $$x^{6} - x^{5} - 10 x^{4} + 3 x^{3} + 26 x^{2} + 13 x - 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 770) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Root $$-2.27063$$ of defining polynomial Character $$\chi$$ $$=$$ 8470.1
## $q$-expansion
$$f(q)$$ $$=$$ $$q-1.00000 q^{2} -2.80837 q^{3} +1.00000 q^{4} +1.00000 q^{5} +2.80837 q^{6} -1.00000 q^{7} -1.00000 q^{8} +4.88695 q^{9} +O(q^{10})$$ $$q-1.00000 q^{2} -2.80837 q^{3} +1.00000 q^{4} +1.00000 q^{5} +2.80837 q^{6} -1.00000 q^{7} -1.00000 q^{8} +4.88695 q^{9} -1.00000 q^{10} -2.80837 q^{12} +3.05764 q^{13} +1.00000 q^{14} -2.80837 q^{15} +1.00000 q^{16} -6.94459 q^{17} -4.88695 q^{18} +4.69704 q^{19} +1.00000 q^{20} +2.80837 q^{21} -1.73118 q^{23} +2.80837 q^{24} +1.00000 q^{25} -3.05764 q^{26} -5.29926 q^{27} -1.00000 q^{28} +6.74554 q^{29} +2.80837 q^{30} +3.34334 q^{31} -1.00000 q^{32} +6.94459 q^{34} -1.00000 q^{35} +4.88695 q^{36} -7.32736 q^{37} -4.69704 q^{38} -8.58700 q^{39} -1.00000 q^{40} -7.58977 q^{41} -2.80837 q^{42} -7.27592 q^{43} +4.88695 q^{45} +1.73118 q^{46} +4.00555 q^{47} -2.80837 q^{48} +1.00000 q^{49} -1.00000 q^{50} +19.5030 q^{51} +3.05764 q^{52} -1.70116 q^{53} +5.29926 q^{54} +1.00000 q^{56} -13.1910 q^{57} -6.74554 q^{58} +7.64741 q^{59} -2.80837 q^{60} -8.93050 q^{61} -3.34334 q^{62} -4.88695 q^{63} +1.00000 q^{64} +3.05764 q^{65} +10.5718 q^{67} -6.94459 q^{68} +4.86179 q^{69} +1.00000 q^{70} -2.30414 q^{71} -4.88695 q^{72} +13.1113 q^{73} +7.32736 q^{74} -2.80837 q^{75} +4.69704 q^{76} +8.58700 q^{78} -4.25829 q^{79} +1.00000 q^{80} +0.221440 q^{81} +7.58977 q^{82} -3.86073 q^{83} +2.80837 q^{84} -6.94459 q^{85} +7.27592 q^{86} -18.9440 q^{87} -12.3140 q^{89} -4.88695 q^{90} -3.05764 q^{91} -1.73118 q^{92} -9.38933 q^{93} -4.00555 q^{94} +4.69704 q^{95} +2.80837 q^{96} -6.94282 q^{97} -1.00000 q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6q - 6q^{2} + q^{3} + 6q^{4} + 6q^{5} - q^{6} - 6q^{7} - 6q^{8} + 15q^{9} + O(q^{10})$$ $$6q - 6q^{2} + q^{3} + 6q^{4} + 6q^{5} - q^{6} - 6q^{7} - 6q^{8} + 15q^{9} - 6q^{10} + q^{12} - 2q^{13} + 6q^{14} + q^{15} + 6q^{16} - 7q^{17} - 15q^{18} - 11q^{19} + 6q^{20} - q^{21} - 6q^{23} - q^{24} + 6q^{25} + 2q^{26} + 4q^{27} - 6q^{28} - 2q^{29} - q^{30} - 6q^{32} + 7q^{34} - 6q^{35} + 15q^{36} - 14q^{37} + 11q^{38} - 20q^{39} - 6q^{40} - 13q^{41} + q^{42} - 19q^{43} + 15q^{45} + 6q^{46} + 22q^{47} + q^{48} + 6q^{49} - 6q^{50} + 14q^{51} - 2q^{52} - 10q^{53} - 4q^{54} + 6q^{56} - 32q^{57} + 2q^{58} - 7q^{59} + q^{60} - 22q^{61} - 15q^{63} + 6q^{64} - 2q^{65} + 5q^{67} - 7q^{68} - 36q^{69} + 6q^{70} + 8q^{71} - 15q^{72} - 13q^{73} + 14q^{74} + q^{75} - 11q^{76} + 20q^{78} + 16q^{79} + 6q^{80} + 18q^{81} + 13q^{82} + 5q^{83} - q^{84} - 7q^{85} + 19q^{86} - 14q^{87} + q^{89} - 15q^{90} + 2q^{91} - 6q^{92} - 42q^{93} - 22q^{94} - 11q^{95} - q^{96} - 3q^{97} - 6q^{98} + O(q^{100})$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −1.00000 −0.707107
$$3$$ −2.80837 −1.62141 −0.810707 0.585452i $$-0.800917\pi$$
−0.810707 + 0.585452i $$0.800917\pi$$
$$4$$ 1.00000 0.500000
$$5$$ 1.00000 0.447214
$$6$$ 2.80837 1.14651
$$7$$ −1.00000 −0.377964
$$8$$ −1.00000 −0.353553
$$9$$ 4.88695 1.62898
$$10$$ −1.00000 −0.316228
$$11$$ 0 0
$$12$$ −2.80837 −0.810707
$$13$$ 3.05764 0.848037 0.424019 0.905653i $$-0.360619\pi$$
0.424019 + 0.905653i $$0.360619\pi$$
$$14$$ 1.00000 0.267261
$$15$$ −2.80837 −0.725118
$$16$$ 1.00000 0.250000
$$17$$ −6.94459 −1.68431 −0.842156 0.539234i $$-0.818714\pi$$
−0.842156 + 0.539234i $$0.818714\pi$$
$$18$$ −4.88695 −1.15187
$$19$$ 4.69704 1.07757 0.538787 0.842442i $$-0.318883\pi$$
0.538787 + 0.842442i $$0.318883\pi$$
$$20$$ 1.00000 0.223607
$$21$$ 2.80837 0.612837
$$22$$ 0 0
$$23$$ −1.73118 −0.360976 −0.180488 0.983577i $$-0.557768\pi$$
−0.180488 + 0.983577i $$0.557768\pi$$
$$24$$ 2.80837 0.573256
$$25$$ 1.00000 0.200000
$$26$$ −3.05764 −0.599653
$$27$$ −5.29926 −1.01984
$$28$$ −1.00000 −0.188982
$$29$$ 6.74554 1.25262 0.626308 0.779576i $$-0.284565\pi$$
0.626308 + 0.779576i $$0.284565\pi$$
$$30$$ 2.80837 0.512736
$$31$$ 3.34334 0.600481 0.300240 0.953864i $$-0.402933\pi$$
0.300240 + 0.953864i $$0.402933\pi$$
$$32$$ −1.00000 −0.176777
$$33$$ 0 0
$$34$$ 6.94459 1.19099
$$35$$ −1.00000 −0.169031
$$36$$ 4.88695 0.814492
$$37$$ −7.32736 −1.20461 −0.602305 0.798266i $$-0.705751\pi$$
−0.602305 + 0.798266i $$0.705751\pi$$
$$38$$ −4.69704 −0.761960
$$39$$ −8.58700 −1.37502
$$40$$ −1.00000 −0.158114
$$41$$ −7.58977 −1.18532 −0.592662 0.805452i $$-0.701923\pi$$
−0.592662 + 0.805452i $$0.701923\pi$$
$$42$$ −2.80837 −0.433341
$$43$$ −7.27592 −1.10957 −0.554784 0.831995i $$-0.687199\pi$$
−0.554784 + 0.831995i $$0.687199\pi$$
$$44$$ 0 0
$$45$$ 4.88695 0.728504
$$46$$ 1.73118 0.255248
$$47$$ 4.00555 0.584270 0.292135 0.956377i $$-0.405634\pi$$
0.292135 + 0.956377i $$0.405634\pi$$
$$48$$ −2.80837 −0.405354
$$49$$ 1.00000 0.142857
$$50$$ −1.00000 −0.141421
$$51$$ 19.5030 2.73097
$$52$$ 3.05764 0.424019
$$53$$ −1.70116 −0.233673 −0.116836 0.993151i $$-0.537275\pi$$
−0.116836 + 0.993151i $$0.537275\pi$$
$$54$$ 5.29926 0.721138
$$55$$ 0 0
$$56$$ 1.00000 0.133631
$$57$$ −13.1910 −1.74719
$$58$$ −6.74554 −0.885733
$$59$$ 7.64741 0.995608 0.497804 0.867289i $$-0.334140\pi$$
0.497804 + 0.867289i $$0.334140\pi$$
$$60$$ −2.80837 −0.362559
$$61$$ −8.93050 −1.14343 −0.571717 0.820451i $$-0.693722\pi$$
−0.571717 + 0.820451i $$0.693722\pi$$
$$62$$ −3.34334 −0.424604
$$63$$ −4.88695 −0.615698
$$64$$ 1.00000 0.125000
$$65$$ 3.05764 0.379254
$$66$$ 0 0
$$67$$ 10.5718 1.29155 0.645773 0.763529i $$-0.276535\pi$$
0.645773 + 0.763529i $$0.276535\pi$$
$$68$$ −6.94459 −0.842156
$$69$$ 4.86179 0.585291
$$70$$ 1.00000 0.119523
$$71$$ −2.30414 −0.273451 −0.136725 0.990609i $$-0.543658\pi$$
−0.136725 + 0.990609i $$0.543658\pi$$
$$72$$ −4.88695 −0.575933
$$73$$ 13.1113 1.53456 0.767278 0.641314i $$-0.221611\pi$$
0.767278 + 0.641314i $$0.221611\pi$$
$$74$$ 7.32736 0.851788
$$75$$ −2.80837 −0.324283
$$76$$ 4.69704 0.538787
$$77$$ 0 0
$$78$$ 8.58700 0.972286
$$79$$ −4.25829 −0.479095 −0.239547 0.970885i $$-0.576999\pi$$
−0.239547 + 0.970885i $$0.576999\pi$$
$$80$$ 1.00000 0.111803
$$81$$ 0.221440 0.0246044
$$82$$ 7.58977 0.838150
$$83$$ −3.86073 −0.423770 −0.211885 0.977295i $$-0.567960\pi$$
−0.211885 + 0.977295i $$0.567960\pi$$
$$84$$ 2.80837 0.306418
$$85$$ −6.94459 −0.753247
$$86$$ 7.27592 0.784583
$$87$$ −18.9440 −2.03101
$$88$$ 0 0
$$89$$ −12.3140 −1.30529 −0.652643 0.757665i $$-0.726340\pi$$
−0.652643 + 0.757665i $$0.726340\pi$$
$$90$$ −4.88695 −0.515130
$$91$$ −3.05764 −0.320528
$$92$$ −1.73118 −0.180488
$$93$$ −9.38933 −0.973628
$$94$$ −4.00555 −0.413141
$$95$$ 4.69704 0.481906
$$96$$ 2.80837 0.286628
$$97$$ −6.94282 −0.704936 −0.352468 0.935824i $$-0.614658\pi$$
−0.352468 + 0.935824i $$0.614658\pi$$
$$98$$ −1.00000 −0.101015
$$99$$ 0 0
$$100$$ 1.00000 0.100000
$$101$$ −13.5915 −1.35240 −0.676202 0.736717i $$-0.736375\pi$$
−0.676202 + 0.736717i $$0.736375\pi$$
$$102$$ −19.5030 −1.93108
$$103$$ 15.8591 1.56264 0.781320 0.624131i $$-0.214547\pi$$
0.781320 + 0.624131i $$0.214547\pi$$
$$104$$ −3.05764 −0.299826
$$105$$ 2.80837 0.274069
$$106$$ 1.70116 0.165232
$$107$$ 4.39685 0.425060 0.212530 0.977155i $$-0.431830\pi$$
0.212530 + 0.977155i $$0.431830\pi$$
$$108$$ −5.29926 −0.509922
$$109$$ 14.4931 1.38819 0.694095 0.719884i $$-0.255805\pi$$
0.694095 + 0.719884i $$0.255805\pi$$
$$110$$ 0 0
$$111$$ 20.5779 1.95317
$$112$$ −1.00000 −0.0944911
$$113$$ −19.3326 −1.81866 −0.909328 0.416081i $$-0.863403\pi$$
−0.909328 + 0.416081i $$0.863403\pi$$
$$114$$ 13.1910 1.23545
$$115$$ −1.73118 −0.161433
$$116$$ 6.74554 0.626308
$$117$$ 14.9425 1.38144
$$118$$ −7.64741 −0.704001
$$119$$ 6.94459 0.636610
$$120$$ 2.80837 0.256368
$$121$$ 0 0
$$122$$ 8.93050 0.808529
$$123$$ 21.3149 1.92190
$$124$$ 3.34334 0.300240
$$125$$ 1.00000 0.0894427
$$126$$ 4.88695 0.435364
$$127$$ −21.5848 −1.91534 −0.957669 0.287870i $$-0.907053\pi$$
−0.957669 + 0.287870i $$0.907053\pi$$
$$128$$ −1.00000 −0.0883883
$$129$$ 20.4335 1.79907
$$130$$ −3.05764 −0.268173
$$131$$ −5.23468 −0.457356 −0.228678 0.973502i $$-0.573440\pi$$
−0.228678 + 0.973502i $$0.573440\pi$$
$$132$$ 0 0
$$133$$ −4.69704 −0.407285
$$134$$ −10.5718 −0.913261
$$135$$ −5.29926 −0.456088
$$136$$ 6.94459 0.595494
$$137$$ 5.00039 0.427212 0.213606 0.976920i $$-0.431479\pi$$
0.213606 + 0.976920i $$0.431479\pi$$
$$138$$ −4.86179 −0.413863
$$139$$ 3.08024 0.261262 0.130631 0.991431i $$-0.458300\pi$$
0.130631 + 0.991431i $$0.458300\pi$$
$$140$$ −1.00000 −0.0845154
$$141$$ −11.2491 −0.947343
$$142$$ 2.30414 0.193359
$$143$$ 0 0
$$144$$ 4.88695 0.407246
$$145$$ 6.74554 0.560187
$$146$$ −13.1113 −1.08510
$$147$$ −2.80837 −0.231631
$$148$$ −7.32736 −0.602305
$$149$$ 15.8858 1.30141 0.650706 0.759330i $$-0.274473\pi$$
0.650706 + 0.759330i $$0.274473\pi$$
$$150$$ 2.80837 0.229303
$$151$$ 5.45916 0.444260 0.222130 0.975017i $$-0.428699\pi$$
0.222130 + 0.975017i $$0.428699\pi$$
$$152$$ −4.69704 −0.380980
$$153$$ −33.9379 −2.74372
$$154$$ 0 0
$$155$$ 3.34334 0.268543
$$156$$ −8.58700 −0.687510
$$157$$ 8.44070 0.673641 0.336820 0.941569i $$-0.390648\pi$$
0.336820 + 0.941569i $$0.390648\pi$$
$$158$$ 4.25829 0.338771
$$159$$ 4.77750 0.378880
$$160$$ −1.00000 −0.0790569
$$161$$ 1.73118 0.136436
$$162$$ −0.221440 −0.0173980
$$163$$ 9.74355 0.763174 0.381587 0.924333i $$-0.375378\pi$$
0.381587 + 0.924333i $$0.375378\pi$$
$$164$$ −7.58977 −0.592662
$$165$$ 0 0
$$166$$ 3.86073 0.299650
$$167$$ −20.2081 −1.56375 −0.781873 0.623437i $$-0.785736\pi$$
−0.781873 + 0.623437i $$0.785736\pi$$
$$168$$ −2.80837 −0.216671
$$169$$ −3.65082 −0.280833
$$170$$ 6.94459 0.532626
$$171$$ 22.9542 1.75535
$$172$$ −7.27592 −0.554784
$$173$$ −1.06417 −0.0809073 −0.0404537 0.999181i $$-0.512880\pi$$
−0.0404537 + 0.999181i $$0.512880\pi$$
$$174$$ 18.9440 1.43614
$$175$$ −1.00000 −0.0755929
$$176$$ 0 0
$$177$$ −21.4768 −1.61429
$$178$$ 12.3140 0.922977
$$179$$ −5.73780 −0.428863 −0.214432 0.976739i $$-0.568790\pi$$
−0.214432 + 0.976739i $$0.568790\pi$$
$$180$$ 4.88695 0.364252
$$181$$ 3.77619 0.280682 0.140341 0.990103i $$-0.455180\pi$$
0.140341 + 0.990103i $$0.455180\pi$$
$$182$$ 3.05764 0.226648
$$183$$ 25.0802 1.85398
$$184$$ 1.73118 0.127624
$$185$$ −7.32736 −0.538718
$$186$$ 9.38933 0.688459
$$187$$ 0 0
$$188$$ 4.00555 0.292135
$$189$$ 5.29926 0.385465
$$190$$ −4.69704 −0.340759
$$191$$ 11.7298 0.848736 0.424368 0.905490i $$-0.360496\pi$$
0.424368 + 0.905490i $$0.360496\pi$$
$$192$$ −2.80837 −0.202677
$$193$$ −18.9879 −1.36678 −0.683391 0.730053i $$-0.739495\pi$$
−0.683391 + 0.730053i $$0.739495\pi$$
$$194$$ 6.94282 0.498465
$$195$$ −8.58700 −0.614928
$$196$$ 1.00000 0.0714286
$$197$$ 3.27508 0.233340 0.116670 0.993171i $$-0.462778\pi$$
0.116670 + 0.993171i $$0.462778\pi$$
$$198$$ 0 0
$$199$$ 14.7438 1.04516 0.522580 0.852590i $$-0.324970\pi$$
0.522580 + 0.852590i $$0.324970\pi$$
$$200$$ −1.00000 −0.0707107
$$201$$ −29.6894 −2.09413
$$202$$ 13.5915 0.956294
$$203$$ −6.74554 −0.473444
$$204$$ 19.5030 1.36548
$$205$$ −7.58977 −0.530093
$$206$$ −15.8591 −1.10495
$$207$$ −8.46019 −0.588024
$$208$$ 3.05764 0.212009
$$209$$ 0 0
$$210$$ −2.80837 −0.193796
$$211$$ 13.6709 0.941143 0.470572 0.882362i $$-0.344048\pi$$
0.470572 + 0.882362i $$0.344048\pi$$
$$212$$ −1.70116 −0.116836
$$213$$ 6.47087 0.443377
$$214$$ −4.39685 −0.300563
$$215$$ −7.27592 −0.496214
$$216$$ 5.29926 0.360569
$$217$$ −3.34334 −0.226960
$$218$$ −14.4931 −0.981598
$$219$$ −36.8213 −2.48815
$$220$$ 0 0
$$221$$ −21.2341 −1.42836
$$222$$ −20.5779 −1.38110
$$223$$ −5.23577 −0.350613 −0.175307 0.984514i $$-0.556092\pi$$
−0.175307 + 0.984514i $$0.556092\pi$$
$$224$$ 1.00000 0.0668153
$$225$$ 4.88695 0.325797
$$226$$ 19.3326 1.28598
$$227$$ 22.5392 1.49598 0.747991 0.663709i $$-0.231019\pi$$
0.747991 + 0.663709i $$0.231019\pi$$
$$228$$ −13.1910 −0.873597
$$229$$ 1.54850 0.102328 0.0511639 0.998690i $$-0.483707\pi$$
0.0511639 + 0.998690i $$0.483707\pi$$
$$230$$ 1.73118 0.114151
$$231$$ 0 0
$$232$$ −6.74554 −0.442867
$$233$$ 6.59831 0.432270 0.216135 0.976364i $$-0.430655\pi$$
0.216135 + 0.976364i $$0.430655\pi$$
$$234$$ −14.9425 −0.976825
$$235$$ 4.00555 0.261293
$$236$$ 7.64741 0.497804
$$237$$ 11.9589 0.776811
$$238$$ −6.94459 −0.450151
$$239$$ −9.32864 −0.603420 −0.301710 0.953400i $$-0.597557\pi$$
−0.301710 + 0.953400i $$0.597557\pi$$
$$240$$ −2.80837 −0.181280
$$241$$ 4.69741 0.302587 0.151293 0.988489i $$-0.451656\pi$$
0.151293 + 0.988489i $$0.451656\pi$$
$$242$$ 0 0
$$243$$ 15.2759 0.979949
$$244$$ −8.93050 −0.571717
$$245$$ 1.00000 0.0638877
$$246$$ −21.3149 −1.35899
$$247$$ 14.3619 0.913823
$$248$$ −3.34334 −0.212302
$$249$$ 10.8424 0.687106
$$250$$ −1.00000 −0.0632456
$$251$$ 25.1086 1.58484 0.792420 0.609976i $$-0.208821\pi$$
0.792420 + 0.609976i $$0.208821\pi$$
$$252$$ −4.88695 −0.307849
$$253$$ 0 0
$$254$$ 21.5848 1.35435
$$255$$ 19.5030 1.22133
$$256$$ 1.00000 0.0625000
$$257$$ 22.2671 1.38898 0.694491 0.719502i $$-0.255630\pi$$
0.694491 + 0.719502i $$0.255630\pi$$
$$258$$ −20.4335 −1.27213
$$259$$ 7.32736 0.455300
$$260$$ 3.05764 0.189627
$$261$$ 32.9651 2.04049
$$262$$ 5.23468 0.323400
$$263$$ 12.5549 0.774170 0.387085 0.922044i $$-0.373482\pi$$
0.387085 + 0.922044i $$0.373482\pi$$
$$264$$ 0 0
$$265$$ −1.70116 −0.104502
$$266$$ 4.69704 0.287994
$$267$$ 34.5824 2.11641
$$268$$ 10.5718 0.645773
$$269$$ −7.54481 −0.460015 −0.230007 0.973189i $$-0.573875\pi$$
−0.230007 + 0.973189i $$0.573875\pi$$
$$270$$ 5.29926 0.322503
$$271$$ −27.7828 −1.68768 −0.843842 0.536592i $$-0.819711\pi$$
−0.843842 + 0.536592i $$0.819711\pi$$
$$272$$ −6.94459 −0.421078
$$273$$ 8.58700 0.519709
$$274$$ −5.00039 −0.302085
$$275$$ 0 0
$$276$$ 4.86179 0.292646
$$277$$ −18.8691 −1.13374 −0.566868 0.823809i $$-0.691845\pi$$
−0.566868 + 0.823809i $$0.691845\pi$$
$$278$$ −3.08024 −0.184740
$$279$$ 16.3387 0.978173
$$280$$ 1.00000 0.0597614
$$281$$ −6.86665 −0.409630 −0.204815 0.978801i $$-0.565659\pi$$
−0.204815 + 0.978801i $$0.565659\pi$$
$$282$$ 11.2491 0.669873
$$283$$ 25.1097 1.49261 0.746307 0.665601i $$-0.231825\pi$$
0.746307 + 0.665601i $$0.231825\pi$$
$$284$$ −2.30414 −0.136725
$$285$$ −13.1910 −0.781369
$$286$$ 0 0
$$287$$ 7.58977 0.448010
$$288$$ −4.88695 −0.287966
$$289$$ 31.2274 1.83690
$$290$$ −6.74554 −0.396112
$$291$$ 19.4980 1.14299
$$292$$ 13.1113 0.767278
$$293$$ −16.5950 −0.969491 −0.484746 0.874655i $$-0.661088\pi$$
−0.484746 + 0.874655i $$0.661088\pi$$
$$294$$ 2.80837 0.163788
$$295$$ 7.64741 0.445250
$$296$$ 7.32736 0.425894
$$297$$ 0 0
$$298$$ −15.8858 −0.920237
$$299$$ −5.29332 −0.306121
$$300$$ −2.80837 −0.162141
$$301$$ 7.27592 0.419377
$$302$$ −5.45916 −0.314139
$$303$$ 38.1699 2.19281
$$304$$ 4.69704 0.269394
$$305$$ −8.93050 −0.511359
$$306$$ 33.9379 1.94010
$$307$$ 31.2886 1.78573 0.892867 0.450321i $$-0.148691\pi$$
0.892867 + 0.450321i $$0.148691\pi$$
$$308$$ 0 0
$$309$$ −44.5381 −2.53368
$$310$$ −3.34334 −0.189889
$$311$$ −5.69208 −0.322768 −0.161384 0.986892i $$-0.551596\pi$$
−0.161384 + 0.986892i $$0.551596\pi$$
$$312$$ 8.58700 0.486143
$$313$$ 17.4320 0.985315 0.492658 0.870223i $$-0.336026\pi$$
0.492658 + 0.870223i $$0.336026\pi$$
$$314$$ −8.44070 −0.476336
$$315$$ −4.88695 −0.275349
$$316$$ −4.25829 −0.239547
$$317$$ 21.2720 1.19475 0.597376 0.801961i $$-0.296210\pi$$
0.597376 + 0.801961i $$0.296210\pi$$
$$318$$ −4.77750 −0.267909
$$319$$ 0 0
$$320$$ 1.00000 0.0559017
$$321$$ −12.3480 −0.689198
$$322$$ −1.73118 −0.0964748
$$323$$ −32.6190 −1.81497
$$324$$ 0.221440 0.0123022
$$325$$ 3.05764 0.169607
$$326$$ −9.74355 −0.539645
$$327$$ −40.7021 −2.25083
$$328$$ 7.58977 0.419075
$$329$$ −4.00555 −0.220833
$$330$$ 0 0
$$331$$ 0.712600 0.0391680 0.0195840 0.999808i $$-0.493766\pi$$
0.0195840 + 0.999808i $$0.493766\pi$$
$$332$$ −3.86073 −0.211885
$$333$$ −35.8084 −1.96229
$$334$$ 20.2081 1.10574
$$335$$ 10.5718 0.577597
$$336$$ 2.80837 0.153209
$$337$$ −28.7577 −1.56653 −0.783267 0.621686i $$-0.786448\pi$$
−0.783267 + 0.621686i $$0.786448\pi$$
$$338$$ 3.65082 0.198579
$$339$$ 54.2930 2.94879
$$340$$ −6.94459 −0.376623
$$341$$ 0 0
$$342$$ −22.9542 −1.24122
$$343$$ −1.00000 −0.0539949
$$344$$ 7.27592 0.392291
$$345$$ 4.86179 0.261750
$$346$$ 1.06417 0.0572101
$$347$$ −34.5981 −1.85732 −0.928662 0.370927i $$-0.879040\pi$$
−0.928662 + 0.370927i $$0.879040\pi$$
$$348$$ −18.9440 −1.01550
$$349$$ −15.4593 −0.827518 −0.413759 0.910386i $$-0.635784\pi$$
−0.413759 + 0.910386i $$0.635784\pi$$
$$350$$ 1.00000 0.0534522
$$351$$ −16.2032 −0.864865
$$352$$ 0 0
$$353$$ 2.31479 0.123204 0.0616019 0.998101i $$-0.480379\pi$$
0.0616019 + 0.998101i $$0.480379\pi$$
$$354$$ 21.4768 1.14148
$$355$$ −2.30414 −0.122291
$$356$$ −12.3140 −0.652643
$$357$$ −19.5030 −1.03221
$$358$$ 5.73780 0.303252
$$359$$ −5.56161 −0.293530 −0.146765 0.989171i $$-0.546886\pi$$
−0.146765 + 0.989171i $$0.546886\pi$$
$$360$$ −4.88695 −0.257565
$$361$$ 3.06217 0.161167
$$362$$ −3.77619 −0.198472
$$363$$ 0 0
$$364$$ −3.05764 −0.160264
$$365$$ 13.1113 0.686274
$$366$$ −25.0802 −1.31096
$$367$$ 26.0861 1.36168 0.680842 0.732430i $$-0.261614\pi$$
0.680842 + 0.732430i $$0.261614\pi$$
$$368$$ −1.73118 −0.0902439
$$369$$ −37.0908 −1.93087
$$370$$ 7.32736 0.380931
$$371$$ 1.70116 0.0883200
$$372$$ −9.38933 −0.486814
$$373$$ −28.3183 −1.46627 −0.733134 0.680085i $$-0.761943\pi$$
−0.733134 + 0.680085i $$0.761943\pi$$
$$374$$ 0 0
$$375$$ −2.80837 −0.145024
$$376$$ −4.00555 −0.206571
$$377$$ 20.6255 1.06227
$$378$$ −5.29926 −0.272565
$$379$$ −32.5648 −1.67274 −0.836371 0.548164i $$-0.815327\pi$$
−0.836371 + 0.548164i $$0.815327\pi$$
$$380$$ 4.69704 0.240953
$$381$$ 60.6181 3.10556
$$382$$ −11.7298 −0.600147
$$383$$ −12.3943 −0.633319 −0.316660 0.948539i $$-0.602561\pi$$
−0.316660 + 0.948539i $$0.602561\pi$$
$$384$$ 2.80837 0.143314
$$385$$ 0 0
$$386$$ 18.9879 0.966461
$$387$$ −35.5571 −1.80747
$$388$$ −6.94282 −0.352468
$$389$$ −19.8793 −1.00792 −0.503960 0.863727i $$-0.668124\pi$$
−0.503960 + 0.863727i $$0.668124\pi$$
$$390$$ 8.58700 0.434819
$$391$$ 12.0223 0.607995
$$392$$ −1.00000 −0.0505076
$$393$$ 14.7009 0.741564
$$394$$ −3.27508 −0.164996
$$395$$ −4.25829 −0.214258
$$396$$ 0 0
$$397$$ 27.6288 1.38665 0.693325 0.720625i $$-0.256145\pi$$
0.693325 + 0.720625i $$0.256145\pi$$
$$398$$ −14.7438 −0.739040
$$399$$ 13.1910 0.660377
$$400$$ 1.00000 0.0500000
$$401$$ 30.1562 1.50593 0.752965 0.658061i $$-0.228623\pi$$
0.752965 + 0.658061i $$0.228623\pi$$
$$402$$ 29.6894 1.48077
$$403$$ 10.2227 0.509230
$$404$$ −13.5915 −0.676202
$$405$$ 0.221440 0.0110034
$$406$$ 6.74554 0.334776
$$407$$ 0 0
$$408$$ −19.5030 −0.965542
$$409$$ −18.8749 −0.933302 −0.466651 0.884442i $$-0.654540\pi$$
−0.466651 + 0.884442i $$0.654540\pi$$
$$410$$ 7.58977 0.374832
$$411$$ −14.0430 −0.692688
$$412$$ 15.8591 0.781320
$$413$$ −7.64741 −0.376305
$$414$$ 8.46019 0.415795
$$415$$ −3.86073 −0.189516
$$416$$ −3.05764 −0.149913
$$417$$ −8.65045 −0.423614
$$418$$ 0 0
$$419$$ 8.51801 0.416132 0.208066 0.978115i $$-0.433283\pi$$
0.208066 + 0.978115i $$0.433283\pi$$
$$420$$ 2.80837 0.137035
$$421$$ −13.4733 −0.656647 −0.328323 0.944565i $$-0.606484\pi$$
−0.328323 + 0.944565i $$0.606484\pi$$
$$422$$ −13.6709 −0.665489
$$423$$ 19.5749 0.951766
$$424$$ 1.70116 0.0826158
$$425$$ −6.94459 −0.336862
$$426$$ −6.47087 −0.313515
$$427$$ 8.93050 0.432177
$$428$$ 4.39685 0.212530
$$429$$ 0 0
$$430$$ 7.27592 0.350876
$$431$$ 27.8105 1.33959 0.669793 0.742548i $$-0.266383\pi$$
0.669793 + 0.742548i $$0.266383\pi$$
$$432$$ −5.29926 −0.254961
$$433$$ −19.9578 −0.959111 −0.479555 0.877512i $$-0.659202\pi$$
−0.479555 + 0.877512i $$0.659202\pi$$
$$434$$ 3.34334 0.160485
$$435$$ −18.9440 −0.908295
$$436$$ 14.4931 0.694095
$$437$$ −8.13141 −0.388978
$$438$$ 36.8213 1.75939
$$439$$ 10.7367 0.512436 0.256218 0.966619i $$-0.417524\pi$$
0.256218 + 0.966619i $$0.417524\pi$$
$$440$$ 0 0
$$441$$ 4.88695 0.232712
$$442$$ 21.2341 1.01000
$$443$$ −10.0787 −0.478854 −0.239427 0.970914i $$-0.576960\pi$$
−0.239427 + 0.970914i $$0.576960\pi$$
$$444$$ 20.5779 0.976586
$$445$$ −12.3140 −0.583742
$$446$$ 5.23577 0.247921
$$447$$ −44.6131 −2.11013
$$448$$ −1.00000 −0.0472456
$$449$$ −31.1076 −1.46806 −0.734030 0.679117i $$-0.762363\pi$$
−0.734030 + 0.679117i $$0.762363\pi$$
$$450$$ −4.88695 −0.230373
$$451$$ 0 0
$$452$$ −19.3326 −0.909328
$$453$$ −15.3313 −0.720330
$$454$$ −22.5392 −1.05782
$$455$$ −3.05764 −0.143344
$$456$$ 13.1910 0.617727
$$457$$ −26.2062 −1.22587 −0.612936 0.790132i $$-0.710012\pi$$
−0.612936 + 0.790132i $$0.710012\pi$$
$$458$$ −1.54850 −0.0723566
$$459$$ 36.8012 1.71773
$$460$$ −1.73118 −0.0807166
$$461$$ 5.33572 0.248509 0.124255 0.992250i $$-0.460346\pi$$
0.124255 + 0.992250i $$0.460346\pi$$
$$462$$ 0 0
$$463$$ 28.8817 1.34224 0.671122 0.741347i $$-0.265813\pi$$
0.671122 + 0.741347i $$0.265813\pi$$
$$464$$ 6.74554 0.313154
$$465$$ −9.38933 −0.435420
$$466$$ −6.59831 −0.305661
$$467$$ −37.6575 −1.74258 −0.871291 0.490766i $$-0.836717\pi$$
−0.871291 + 0.490766i $$0.836717\pi$$
$$468$$ 14.9425 0.690720
$$469$$ −10.5718 −0.488159
$$470$$ −4.00555 −0.184762
$$471$$ −23.7046 −1.09225
$$472$$ −7.64741 −0.352001
$$473$$ 0 0
$$474$$ −11.9589 −0.549288
$$475$$ 4.69704 0.215515
$$476$$ 6.94459 0.318305
$$477$$ −8.31350 −0.380649
$$478$$ 9.32864 0.426682
$$479$$ −25.5852 −1.16902 −0.584509 0.811387i $$-0.698713\pi$$
−0.584509 + 0.811387i $$0.698713\pi$$
$$480$$ 2.80837 0.128184
$$481$$ −22.4044 −1.02155
$$482$$ −4.69741 −0.213961
$$483$$ −4.86179 −0.221219
$$484$$ 0 0
$$485$$ −6.94282 −0.315257
$$486$$ −15.2759 −0.692929
$$487$$ 20.1951 0.915125 0.457563 0.889177i $$-0.348723\pi$$
0.457563 + 0.889177i $$0.348723\pi$$
$$488$$ 8.93050 0.404265
$$489$$ −27.3635 −1.23742
$$490$$ −1.00000 −0.0451754
$$491$$ −18.2166 −0.822104 −0.411052 0.911612i $$-0.634839\pi$$
−0.411052 + 0.911612i $$0.634839\pi$$
$$492$$ 21.3149 0.960950
$$493$$ −46.8451 −2.10980
$$494$$ −14.3619 −0.646171
$$495$$ 0 0
$$496$$ 3.34334 0.150120
$$497$$ 2.30414 0.103355
$$498$$ −10.8424 −0.485857
$$499$$ −21.4887 −0.961966 −0.480983 0.876730i $$-0.659720\pi$$
−0.480983 + 0.876730i $$0.659720\pi$$
$$500$$ 1.00000 0.0447214
$$501$$ 56.7518 2.53548
$$502$$ −25.1086 −1.12065
$$503$$ 5.56623 0.248186 0.124093 0.992271i $$-0.460398\pi$$
0.124093 + 0.992271i $$0.460398\pi$$
$$504$$ 4.88695 0.217682
$$505$$ −13.5915 −0.604813
$$506$$ 0 0
$$507$$ 10.2529 0.455346
$$508$$ −21.5848 −0.957669
$$509$$ −7.22200 −0.320110 −0.160055 0.987108i $$-0.551167\pi$$
−0.160055 + 0.987108i $$0.551167\pi$$
$$510$$ −19.5030 −0.863607
$$511$$ −13.1113 −0.580008
$$512$$ −1.00000 −0.0441942
$$513$$ −24.8908 −1.09896
$$514$$ −22.2671 −0.982158
$$515$$ 15.8591 0.698833
$$516$$ 20.4335 0.899534
$$517$$ 0 0
$$518$$ −7.32736 −0.321946
$$519$$ 2.98858 0.131184
$$520$$ −3.05764 −0.134086
$$521$$ −6.64589 −0.291162 −0.145581 0.989346i $$-0.546505\pi$$
−0.145581 + 0.989346i $$0.546505\pi$$
$$522$$ −32.9651 −1.44285
$$523$$ −24.8014 −1.08449 −0.542246 0.840220i $$-0.682426\pi$$
−0.542246 + 0.840220i $$0.682426\pi$$
$$524$$ −5.23468 −0.228678
$$525$$ 2.80837 0.122567
$$526$$ −12.5549 −0.547421
$$527$$ −23.2181 −1.01140
$$528$$ 0 0
$$529$$ −20.0030 −0.869697
$$530$$ 1.70116 0.0738938
$$531$$ 37.3725 1.62183
$$532$$ −4.69704 −0.203642
$$533$$ −23.2068 −1.00520
$$534$$ −34.5824 −1.49653
$$535$$ 4.39685 0.190092
$$536$$ −10.5718 −0.456631
$$537$$ 16.1139 0.695365
$$538$$ 7.54481 0.325280
$$539$$ 0 0
$$540$$ −5.29926 −0.228044
$$541$$ 34.8421 1.49798 0.748990 0.662581i $$-0.230539\pi$$
0.748990 + 0.662581i $$0.230539\pi$$
$$542$$ 27.7828 1.19337
$$543$$ −10.6049 −0.455101
$$544$$ 6.94459 0.297747
$$545$$ 14.4931 0.620817
$$546$$ −8.58700 −0.367489
$$547$$ −22.5413 −0.963798 −0.481899 0.876227i $$-0.660053\pi$$
−0.481899 + 0.876227i $$0.660053\pi$$
$$548$$ 5.00039 0.213606
$$549$$ −43.6429 −1.86263
$$550$$ 0 0
$$551$$ 31.6841 1.34979
$$552$$ −4.86179 −0.206932
$$553$$ 4.25829 0.181081
$$554$$ 18.8691 0.801672
$$555$$ 20.5779 0.873485
$$556$$ 3.08024 0.130631
$$557$$ −0.716288 −0.0303501 −0.0151750 0.999885i $$-0.504831\pi$$
−0.0151750 + 0.999885i $$0.504831\pi$$
$$558$$ −16.3387 −0.691673
$$559$$ −22.2472 −0.940955
$$560$$ −1.00000 −0.0422577
$$561$$ 0 0
$$562$$ 6.86665 0.289652
$$563$$ −44.5640 −1.87815 −0.939075 0.343713i $$-0.888315\pi$$
−0.939075 + 0.343713i $$0.888315\pi$$
$$564$$ −11.2491 −0.473672
$$565$$ −19.3326 −0.813327
$$566$$ −25.1097 −1.05544
$$567$$ −0.221440 −0.00929961
$$568$$ 2.30414 0.0966795
$$569$$ −9.43478 −0.395526 −0.197763 0.980250i $$-0.563368\pi$$
−0.197763 + 0.980250i $$0.563368\pi$$
$$570$$ 13.1910 0.552511
$$571$$ −31.2693 −1.30858 −0.654289 0.756244i $$-0.727032\pi$$
−0.654289 + 0.756244i $$0.727032\pi$$
$$572$$ 0 0
$$573$$ −32.9415 −1.37615
$$574$$ −7.58977 −0.316791
$$575$$ −1.73118 −0.0721951
$$576$$ 4.88695 0.203623
$$577$$ 33.7440 1.40478 0.702390 0.711792i $$-0.252116\pi$$
0.702390 + 0.711792i $$0.252116\pi$$
$$578$$ −31.2274 −1.29889
$$579$$ 53.3252 2.21612
$$580$$ 6.74554 0.280093
$$581$$ 3.86073 0.160170
$$582$$ −19.4980 −0.808219
$$583$$ 0 0
$$584$$ −13.1113 −0.542548
$$585$$ 14.9425 0.617798
$$586$$ 16.5950 0.685534
$$587$$ 26.2246 1.08240 0.541202 0.840892i $$-0.317969\pi$$
0.541202 + 0.840892i $$0.317969\pi$$
$$588$$ −2.80837 −0.115815
$$589$$ 15.7038 0.647063
$$590$$ −7.64741 −0.314839
$$591$$ −9.19763 −0.378340
$$592$$ −7.32736 −0.301153
$$593$$ −23.6431 −0.970906 −0.485453 0.874263i $$-0.661345\pi$$
−0.485453 + 0.874263i $$0.661345\pi$$
$$594$$ 0 0
$$595$$ 6.94459 0.284701
$$596$$ 15.8858 0.650706
$$597$$ −41.4060 −1.69464
$$598$$ 5.29332 0.216460
$$599$$ −35.9599 −1.46928 −0.734640 0.678457i $$-0.762649\pi$$
−0.734640 + 0.678457i $$0.762649\pi$$
$$600$$ 2.80837 0.114651
$$601$$ 31.4056 1.28106 0.640530 0.767933i $$-0.278715\pi$$
0.640530 + 0.767933i $$0.278715\pi$$
$$602$$ −7.27592 −0.296544
$$603$$ 51.6637 2.10391
$$604$$ 5.45916 0.222130
$$605$$ 0 0
$$606$$ −38.1699 −1.55055
$$607$$ −35.9288 −1.45830 −0.729152 0.684352i $$-0.760085\pi$$
−0.729152 + 0.684352i $$0.760085\pi$$
$$608$$ −4.69704 −0.190490
$$609$$ 18.9440 0.767649
$$610$$ 8.93050 0.361585
$$611$$ 12.2475 0.495482
$$612$$ −33.9379 −1.37186
$$613$$ −47.6086 −1.92289 −0.961446 0.274992i $$-0.911325\pi$$
−0.961446 + 0.274992i $$0.911325\pi$$
$$614$$ −31.2886 −1.26270
$$615$$ 21.3149 0.859500
$$616$$ 0 0
$$617$$ −5.44411 −0.219172 −0.109586 0.993977i $$-0.534952\pi$$
−0.109586 + 0.993977i $$0.534952\pi$$
$$618$$ 44.5381 1.79159
$$619$$ 24.6723 0.991665 0.495833 0.868418i $$-0.334863\pi$$
0.495833 + 0.868418i $$0.334863\pi$$
$$620$$ 3.34334 0.134272
$$621$$ 9.17397 0.368139
$$622$$ 5.69208 0.228231
$$623$$ 12.3140 0.493352
$$624$$ −8.58700 −0.343755
$$625$$ 1.00000 0.0400000
$$626$$ −17.4320 −0.696723
$$627$$ 0 0
$$628$$ 8.44070 0.336820
$$629$$ 50.8855 2.02894
$$630$$ 4.88695 0.194701
$$631$$ −15.8756 −0.631997 −0.315998 0.948760i $$-0.602339\pi$$
−0.315998 + 0.948760i $$0.602339\pi$$
$$632$$ 4.25829 0.169386
$$633$$ −38.3930 −1.52598
$$634$$ −21.2720 −0.844818
$$635$$ −21.5848 −0.856566
$$636$$ 4.77750 0.189440
$$637$$ 3.05764 0.121148
$$638$$ 0 0
$$639$$ −11.2602 −0.445447
$$640$$ −1.00000 −0.0395285
$$641$$ −3.37605 −0.133346 −0.0666730 0.997775i $$-0.521238\pi$$
−0.0666730 + 0.997775i $$0.521238\pi$$
$$642$$ 12.3480 0.487336
$$643$$ 20.7120 0.816803 0.408402 0.912802i $$-0.366086\pi$$
0.408402 + 0.912802i $$0.366086\pi$$
$$644$$ 1.73118 0.0682180
$$645$$ 20.4335 0.804568
$$646$$ 32.6190 1.28338
$$647$$ −6.82833 −0.268449 −0.134225 0.990951i $$-0.542854\pi$$
−0.134225 + 0.990951i $$0.542854\pi$$
$$648$$ −0.221440 −0.00869898
$$649$$ 0 0
$$650$$ −3.05764 −0.119931
$$651$$ 9.38933 0.367997
$$652$$ 9.74355 0.381587
$$653$$ 30.9046 1.20939 0.604696 0.796456i $$-0.293295\pi$$
0.604696 + 0.796456i $$0.293295\pi$$
$$654$$ 40.7021 1.59158
$$655$$ −5.23468 −0.204536
$$656$$ −7.58977 −0.296331
$$657$$ 64.0741 2.49977
$$658$$ 4.00555 0.156153
$$659$$ −17.6185 −0.686318 −0.343159 0.939277i $$-0.611497\pi$$
−0.343159 + 0.939277i $$0.611497\pi$$
$$660$$ 0 0
$$661$$ 11.7396 0.456617 0.228309 0.973589i $$-0.426680\pi$$
0.228309 + 0.973589i $$0.426680\pi$$
$$662$$ −0.712600 −0.0276960
$$663$$ 59.6332 2.31596
$$664$$ 3.86073 0.149825
$$665$$ −4.69704 −0.182143
$$666$$ 35.8084 1.38755
$$667$$ −11.6777 −0.452164
$$668$$ −20.2081 −0.781873
$$669$$ 14.7040 0.568489
$$670$$ −10.5718 −0.408423
$$671$$ 0 0
$$672$$ −2.80837 −0.108335
$$673$$ −17.0012 −0.655347 −0.327673 0.944791i $$-0.606265\pi$$
−0.327673 + 0.944791i $$0.606265\pi$$
$$674$$ 28.7577 1.10771
$$675$$ −5.29926 −0.203969
$$676$$ −3.65082 −0.140416
$$677$$ 3.84431 0.147749 0.0738744 0.997268i $$-0.476464\pi$$
0.0738744 + 0.997268i $$0.476464\pi$$
$$678$$ −54.2930 −2.08511
$$679$$ 6.94282 0.266441
$$680$$ 6.94459 0.266313
$$681$$ −63.2986 −2.42561
$$682$$ 0 0
$$683$$ −2.04787 −0.0783595 −0.0391798 0.999232i $$-0.512474\pi$$
−0.0391798 + 0.999232i $$0.512474\pi$$
$$684$$ 22.9542 0.877676
$$685$$ 5.00039 0.191055
$$686$$ 1.00000 0.0381802
$$687$$ −4.34876 −0.165916
$$688$$ −7.27592 −0.277392
$$689$$ −5.20155 −0.198163
$$690$$ −4.86179 −0.185085
$$691$$ −24.0616 −0.915348 −0.457674 0.889120i $$-0.651317\pi$$
−0.457674 + 0.889120i $$0.651317\pi$$
$$692$$ −1.06417 −0.0404537
$$693$$ 0 0
$$694$$ 34.5981 1.31333
$$695$$ 3.08024 0.116840
$$696$$ 18.9440 0.718070
$$697$$ 52.7079 1.99645
$$698$$ 15.4593 0.585144
$$699$$ −18.5305 −0.700888
$$700$$ −1.00000 −0.0377964
$$701$$ −7.99724 −0.302052 −0.151026 0.988530i $$-0.548258\pi$$
−0.151026 + 0.988530i $$0.548258\pi$$
$$702$$ 16.2032 0.611552
$$703$$ −34.4169 −1.29806
$$704$$ 0 0
$$705$$ −11.2491 −0.423665
$$706$$ −2.31479 −0.0871183
$$707$$ 13.5915 0.511160
$$708$$ −21.4768 −0.807147
$$709$$ 4.28038 0.160753 0.0803765 0.996765i $$-0.474388\pi$$
0.0803765 + 0.996765i $$0.474388\pi$$
$$710$$ 2.30414 0.0864727
$$711$$ −20.8100 −0.780438
$$712$$ 12.3140 0.461488
$$713$$ −5.78791 −0.216759
$$714$$ 19.5030 0.729881
$$715$$ 0 0
$$716$$ −5.73780 −0.214432
$$717$$ 26.1983 0.978393
$$718$$ 5.56161 0.207557
$$719$$ 36.8044 1.37257 0.686286 0.727332i $$-0.259240\pi$$
0.686286 + 0.727332i $$0.259240\pi$$
$$720$$ 4.88695 0.182126
$$721$$ −15.8591 −0.590622
$$722$$ −3.06217 −0.113962
$$723$$ −13.1921 −0.490618
$$724$$ 3.77619 0.140341
$$725$$ 6.74554 0.250523
$$726$$ 0 0
$$727$$ −46.5408 −1.72610 −0.863051 0.505116i $$-0.831450\pi$$
−0.863051 + 0.505116i $$0.831450\pi$$
$$728$$ 3.05764 0.113324
$$729$$ −43.5647 −1.61351
$$730$$ −13.1113 −0.485269
$$731$$ 50.5283 1.86886
$$732$$ 25.0802 0.926989
$$733$$ −15.6659 −0.578633 −0.289316 0.957234i $$-0.593428\pi$$
−0.289316 + 0.957234i $$0.593428\pi$$
$$734$$ −26.0861 −0.962857
$$735$$ −2.80837 −0.103588
$$736$$ 1.73118 0.0638121
$$737$$ 0 0
$$738$$ 37.0908 1.36533
$$739$$ −6.15005 −0.226233 −0.113117 0.993582i $$-0.536083\pi$$
−0.113117 + 0.993582i $$0.536083\pi$$
$$740$$ −7.32736 −0.269359
$$741$$ −40.3334 −1.48169
$$742$$ −1.70116 −0.0624517
$$743$$ −27.2721 −1.00052 −0.500259 0.865876i $$-0.666762\pi$$
−0.500259 + 0.865876i $$0.666762\pi$$
$$744$$ 9.38933 0.344229
$$745$$ 15.8858 0.582009
$$746$$ 28.3183 1.03681
$$747$$ −18.8672 −0.690314
$$748$$ 0 0
$$749$$ −4.39685 −0.160657
$$750$$ 2.80837 0.102547
$$751$$ 7.95258 0.290194 0.145097 0.989417i $$-0.453651\pi$$
0.145097 + 0.989417i $$0.453651\pi$$
$$752$$ 4.00555 0.146067
$$753$$ −70.5142 −2.56968
$$754$$ −20.6255 −0.751135
$$755$$ 5.45916 0.198679
$$756$$ 5.29926 0.192732
$$757$$ 24.8369 0.902713 0.451356 0.892344i $$-0.350940\pi$$
0.451356 + 0.892344i $$0.350940\pi$$
$$758$$ 32.5648 1.18281
$$759$$ 0 0
$$760$$ −4.69704 −0.170379
$$761$$ −34.0673 −1.23494 −0.617469 0.786595i $$-0.711842\pi$$
−0.617469 + 0.786595i $$0.711842\pi$$
$$762$$ −60.6181 −2.19596
$$763$$ −14.4931 −0.524686
$$764$$ 11.7298 0.424368
$$765$$ −33.9379 −1.22703
$$766$$ 12.3943 0.447824
$$767$$ 23.3831 0.844313
$$768$$ −2.80837 −0.101338
$$769$$ −10.7430 −0.387404 −0.193702 0.981060i $$-0.562050\pi$$
−0.193702 + 0.981060i $$0.562050\pi$$
$$770$$ 0 0
$$771$$ −62.5342 −2.25211
$$772$$ −18.9879 −0.683391
$$773$$ −35.4511 −1.27509 −0.637544 0.770414i $$-0.720050\pi$$
−0.637544 + 0.770414i $$0.720050\pi$$
$$774$$ 35.5571 1.27807
$$775$$ 3.34334 0.120096
$$776$$ 6.94282 0.249233
$$777$$ −20.5779 −0.738230
$$778$$ 19.8793 0.712708
$$779$$ −35.6494 −1.27727
$$780$$ −8.58700 −0.307464
$$781$$ 0 0
$$782$$ −12.0223 −0.429918
$$783$$ −35.7464 −1.27747
$$784$$ 1.00000 0.0357143
$$785$$ 8.44070 0.301261
$$786$$ −14.7009 −0.524365
$$787$$ 22.9148 0.816825 0.408413 0.912797i $$-0.366082\pi$$
0.408413 + 0.912797i $$0.366082\pi$$
$$788$$ 3.27508 0.116670
$$789$$ −35.2589 −1.25525
$$790$$ 4.25829 0.151503
$$791$$ 19.3326 0.687387
$$792$$ 0 0
$$793$$ −27.3063 −0.969674
$$794$$ −27.6288 −0.980510
$$795$$ 4.77750 0.169440
$$796$$ 14.7438 0.522580
$$797$$ −42.9156 −1.52015 −0.760074 0.649836i $$-0.774837\pi$$
−0.760074 + 0.649836i $$0.774837\pi$$
$$798$$ −13.1910 −0.466957
$$799$$ −27.8169 −0.984092
$$800$$ −1.00000 −0.0353553
$$801$$ −60.1782 −2.12629
$$802$$ −30.1562 −1.06485
$$803$$ 0 0
$$804$$ −29.6894 −1.04707
$$805$$ 1.73118 0.0610160
$$806$$ −10.2227 −0.360080
$$807$$ 21.1886 0.745875
$$808$$ 13.5915 0.478147
$$809$$ −16.9974 −0.597598 −0.298799 0.954316i $$-0.596586\pi$$
−0.298799 + 0.954316i $$0.596586\pi$$
$$810$$ −0.221440 −0.00778061
$$811$$ 10.6859 0.375231 0.187616 0.982243i $$-0.439924\pi$$
0.187616 + 0.982243i $$0.439924\pi$$
$$812$$ −6.74554 −0.236722
$$813$$ 78.0243 2.73643
$$814$$ 0 0
$$815$$ 9.74355 0.341302
$$816$$ 19.5030 0.682742
$$817$$ −34.1753 −1.19564
$$818$$ 18.8749 0.659944
$$819$$ −14.9425 −0.522135
$$820$$ −7.58977 −0.265046
$$821$$ 20.3259 0.709379 0.354689 0.934984i $$-0.384587\pi$$
0.354689 + 0.934984i $$0.384587\pi$$
$$822$$ 14.0430 0.489805
$$823$$ −29.8555 −1.04070 −0.520348 0.853954i $$-0.674198\pi$$
−0.520348 + 0.853954i $$0.674198\pi$$
$$824$$ −15.8591 −0.552476
$$825$$ 0 0
$$826$$ 7.64741 0.266088
$$827$$ 4.53541 0.157712 0.0788558 0.996886i $$-0.474873\pi$$
0.0788558 + 0.996886i $$0.474873\pi$$
$$828$$ −8.46019 −0.294012
$$829$$ −29.0225 −1.00799 −0.503997 0.863705i $$-0.668138\pi$$
−0.503997 + 0.863705i $$0.668138\pi$$
$$830$$ 3.86073 0.134008
$$831$$ 52.9915 1.83826
$$832$$ 3.05764 0.106005
$$833$$ −6.94459 −0.240616
$$834$$ 8.65045 0.299540
$$835$$ −20.2081 −0.699329
$$836$$ 0 0
$$837$$ −17.7172 −0.612396
$$838$$ −8.51801 −0.294250
$$839$$ −3.85780 −0.133186 −0.0665930 0.997780i $$-0.521213\pi$$
−0.0665930 + 0.997780i $$0.521213\pi$$
$$840$$ −2.80837 −0.0968980
$$841$$ 16.5024 0.569047
$$842$$ 13.4733 0.464319
$$843$$ 19.2841 0.664180
$$844$$ 13.6709 0.470572
$$845$$ −3.65082 −0.125592
$$846$$ −19.5749 −0.673000
$$847$$ 0 0
$$848$$ −1.70116 −0.0584182
$$849$$ −70.5173 −2.42015
$$850$$ 6.94459 0.238198
$$851$$ 12.6850 0.434835
$$852$$ 6.47087 0.221688
$$853$$ 14.8560 0.508659 0.254329 0.967118i $$-0.418145\pi$$
0.254329 + 0.967118i $$0.418145\pi$$
$$854$$ −8.93050 −0.305595
$$855$$ 22.9542 0.785017
$$856$$ −4.39685 −0.150281
$$857$$ 21.4567 0.732948 0.366474 0.930428i $$-0.380565\pi$$
0.366474 + 0.930428i $$0.380565\pi$$
$$858$$ 0 0
$$859$$ −50.7945 −1.73309 −0.866543 0.499103i $$-0.833663\pi$$
−0.866543 + 0.499103i $$0.833663\pi$$
$$860$$ −7.27592 −0.248107
$$861$$ −21.3149 −0.726410
$$862$$ −27.8105 −0.947231
$$863$$ −13.6152 −0.463466 −0.231733 0.972779i $$-0.574440\pi$$
−0.231733 + 0.972779i $$0.574440\pi$$
$$864$$ 5.29926 0.180285
$$865$$ −1.06417 −0.0361829
$$866$$ 19.9578 0.678194
$$867$$ −87.6981 −2.97838
$$868$$ −3.34334 −0.113480
$$869$$ 0 0
$$870$$ 18.9440 0.642262
$$871$$ 32.3247 1.09528
$$872$$ −14.4931 −0.490799
$$873$$ −33.9292 −1.14833
$$874$$ 8.13141 0.275049
$$875$$ −1.00000 −0.0338062
$$876$$ −36.8213 −1.24408
$$877$$ −8.90340 −0.300647 −0.150323 0.988637i $$-0.548031\pi$$
−0.150323 + 0.988637i $$0.548031\pi$$
$$878$$ −10.7367 −0.362347
$$879$$ 46.6050 1.57195
$$880$$ 0 0
$$881$$ −2.11899 −0.0713904 −0.0356952 0.999363i $$-0.511365\pi$$
−0.0356952 + 0.999363i $$0.511365\pi$$
$$882$$ −4.88695 −0.164552
$$883$$ −10.2827 −0.346041 −0.173021 0.984918i $$-0.555353\pi$$
−0.173021 + 0.984918i $$0.555353\pi$$
$$884$$ −21.2341 −0.714179
$$885$$ −21.4768 −0.721934
$$886$$ 10.0787 0.338601
$$887$$ 5.63590 0.189235 0.0946175 0.995514i $$-0.469837\pi$$
0.0946175 + 0.995514i $$0.469837\pi$$
$$888$$ −20.5779 −0.690551
$$889$$ 21.5848 0.723930
$$890$$ 12.3140 0.412768
$$891$$ 0 0
$$892$$ −5.23577 −0.175307
$$893$$ 18.8142 0.629594
$$894$$ 44.6131 1.49209
$$895$$ −5.73780 −0.191793
$$896$$ 1.00000 0.0334077
$$897$$ 14.8656 0.496349
$$898$$ 31.1076 1.03808
$$899$$ 22.5526 0.752172
$$900$$ 4.88695 0.162898
$$901$$ 11.8139 0.393578
$$902$$ 0 0
$$903$$ −20.4335 −0.679984
$$904$$ 19.3326 0.642992
$$905$$ 3.77619 0.125525
$$906$$ 15.3313 0.509350
$$907$$ −39.0697 −1.29729 −0.648643 0.761092i $$-0.724663\pi$$
−0.648643 + 0.761092i $$0.724663\pi$$
$$908$$ 22.5392 0.747991
$$909$$ −66.4209 −2.20304
$$910$$ 3.05764 0.101360
$$911$$ 22.8604 0.757398 0.378699 0.925520i $$-0.376372\pi$$
0.378699 + 0.925520i $$0.376372\pi$$
$$912$$ −13.1910 −0.436799
$$913$$ 0 0
$$914$$ 26.2062 0.866823
$$915$$ 25.0802 0.829124
$$916$$ 1.54850 0.0511639
$$917$$ 5.23468 0.172864
$$918$$ −36.8012 −1.21462
$$919$$ 13.0335 0.429935 0.214968 0.976621i $$-0.431035\pi$$
0.214968 + 0.976621i $$0.431035\pi$$
$$920$$ 1.73118 0.0570753
$$921$$ −87.8699 −2.89541
$$922$$ −5.33572 −0.175723
$$923$$ −7.04523 −0.231896
$$924$$ 0 0
$$925$$ −7.32736 −0.240922
$$926$$ −28.8817 −0.949110
$$927$$ 77.5024 2.54551
$$928$$ −6.74554 −0.221433
$$929$$ 27.3966 0.898853 0.449427 0.893317i $$-0.351628\pi$$
0.449427 + 0.893317i $$0.351628\pi$$
$$930$$ 9.38933 0.307888
$$931$$ 4.69704 0.153939
$$932$$ 6.59831 0.216135
$$933$$ 15.9855 0.523341
$$934$$ 37.6575 1.23219
$$935$$ 0 0
$$936$$ −14.9425 −0.488412
$$937$$ −22.1523 −0.723684 −0.361842 0.932240i $$-0.617852\pi$$
−0.361842 + 0.932240i $$0.617852\pi$$
$$938$$ 10.5718 0.345180
$$939$$ −48.9556 −1.59760
$$940$$ 4.00555 0.130647
$$941$$ 12.0202 0.391847 0.195924 0.980619i $$-0.437229\pi$$
0.195924 + 0.980619i $$0.437229\pi$$
$$942$$ 23.7046 0.772338
$$943$$ 13.1392 0.427873
$$944$$ 7.64741 0.248902
$$945$$ 5.29926 0.172385
$$946$$ 0 0
$$947$$ −29.5380 −0.959857 −0.479929 0.877307i $$-0.659337\pi$$
−0.479929 + 0.877307i $$0.659337\pi$$
$$948$$ 11.9589 0.388406
$$949$$ 40.0895 1.30136
$$950$$ −4.69704 −0.152392
$$951$$ −59.7396 −1.93719
$$952$$ −6.94459 −0.225076
$$953$$ −38.2985 −1.24061 −0.620306 0.784360i $$-0.712991\pi$$
−0.620306 + 0.784360i $$0.712991\pi$$
$$954$$ 8.31350 0.269160
$$955$$ 11.7298 0.379566
$$956$$ −9.32864 −0.301710
$$957$$ 0 0
$$958$$ 25.5852 0.826621
$$959$$ −5.00039 −0.161471
$$960$$ −2.80837 −0.0906398
$$961$$ −19.8221 −0.639423
$$962$$ 22.4044 0.722348
$$963$$ 21.4872 0.692415
$$964$$ 4.69741 0.151293
$$965$$ −18.9879 −0.611243
$$966$$ 4.86179 0.156426
$$967$$ −2.04155 −0.0656517 −0.0328259 0.999461i $$-0.510451\pi$$
−0.0328259 + 0.999461i $$0.510451\pi$$
$$968$$ 0 0
$$969$$ 91.6063 2.94282
$$970$$ 6.94282 0.222920
$$971$$ −5.03293 −0.161514 −0.0807572 0.996734i $$-0.525734\pi$$
−0.0807572 + 0.996734i $$0.525734\pi$$
$$972$$ 15.2759 0.489975
$$973$$ −3.08024 −0.0987478
$$974$$ −20.1951 −0.647091
$$975$$ −8.58700 −0.275004
$$976$$ −8.93050 −0.285858
$$977$$ −10.1256 −0.323945 −0.161973 0.986795i $$-0.551786\pi$$
−0.161973 + 0.986795i $$0.551786\pi$$
$$978$$ 27.3635 0.874989
$$979$$ 0 0
$$980$$ 1.00000 0.0319438
$$981$$ 70.8272 2.26134
$$982$$ 18.2166 0.581315
$$983$$ 33.8126 1.07846 0.539228 0.842160i $$-0.318716\pi$$
0.539228 + 0.842160i $$0.318716\pi$$
$$984$$ −21.3149 −0.679494
$$985$$ 3.27508 0.104353
$$986$$ 46.8451 1.49185
$$987$$ 11.2491 0.358062
$$988$$ 14.3619 0.456912
$$989$$ 12.5959 0.400527
$$990$$ 0 0
$$991$$ 13.8867 0.441127 0.220563 0.975373i $$-0.429210\pi$$
0.220563 + 0.975373i $$0.429210\pi$$
$$992$$ −3.34334 −0.106151
$$993$$ −2.00124 −0.0635076
$$994$$ −2.30414 −0.0730828
$$995$$ 14.7438 0.467410
$$996$$ 10.8424 0.343553
$$997$$ −5.72382 −0.181275 −0.0906375 0.995884i $$-0.528890\pi$$
−0.0906375 + 0.995884i $$0.528890\pi$$
$$998$$ 21.4887 0.680213
$$999$$ 38.8296 1.22851
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 8470.2.a.cy.1.1 6
11.2 odd 10 770.2.n.i.631.1 yes 12
11.6 odd 10 770.2.n.i.421.1 12
11.10 odd 2 8470.2.a.de.1.1 6
By twisted newform
Twist Min Dim Char Parity Ord Type
770.2.n.i.421.1 12 11.6 odd 10
770.2.n.i.631.1 yes 12 11.2 odd 10
8470.2.a.cy.1.1 6 1.1 even 1 trivial
8470.2.a.de.1.1 6 11.10 odd 2 | 22,479 | 41,069 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.292261 |
https://herebeanswers.com/proven-that-earth-is-round-and-not-flat.html | 1,675,679,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500334.35/warc/CC-MAIN-20230206082428-20230206112428-00710.warc.gz | 321,668,045 | 20,955 | # How was it proven that the Earth is round and not flat?
Even before a famous explorer circumnavigated the world or humans started going into space, the ancient Greeks had known that our planet is round and not flat.
It was Pythagoras who raised the idea of the spherical world in 500 BC. But, it was around 2,000 years ago when Aristotle, the famous Greek philosopher Aristotle had riddled out the issue.
Aristotle wrote in his book, ‘On the Heavens,’ that their observations of the stars made it evident that the Earth is not only round, but is massively-circular. He said that few alterations from north to the south already result in significant changes in the horizon.
To simplify things, Aristotle wants to imply that we see diverse sets of start in the sky based on the location we are in. The stars you will see when you’re standing in the northern hemisphere will be different from those you will see over the southern hemisphere. If the Earth is flat, all humans would see the same set of stars in the night sky at any given time or location.
But, another Greek personality and mathematician explored the idea and managed to measure the planet’s circumference. It was Eratosthenes, the chief librarian of the city of Alexandria.
Eratosthenes had found out that in the city of Syene, vertical shadows don’t appear at noon during the summer solstice. He became curious if this the case of the southern city would be the same for Alexandria.
To clear the ambiguity, Erasthosthenes shove a stock on the ground to see if a vertical shadow would show up at noon. And, one did appear, measuring at around 7 degrees.
So, if the sun’s rays are both coming directly overhead, but the two different cities cast shadows differently: the presence of one in Alexandria and the absence in Syene. The condition implies that the planet’s surface is curved. The Greek polymath most likely knew that even before most of the people from his time did.
With that, he used his observations to calculate Earth’s circumference. Since the shadows cast a 7-degree difference, then, the two cities are also 7 degrees apart. He then hired someone to track the distance between Syene and Alexandria and found that they were around 5,000 stadia apart or nearly 800 kilometers.
Eratosthenes integrated some trigonometry and used simple rules of proportions to get the planet’s circumference. Seven degrees is roughly around 1/50 of 360 degrees. 800 kilometers multiplied by 50 results in 40,000 kilometers. Through that, he was able to figure out Earth’s circumference using a stick, and some simple calculations.
Since then, it became common knowledge that the Earth was round and not flat, at least for the powerful and educated people.
But, exploration began, and people started traveling our spherical planet. From 1519 to 1522, Ferdinand Magellan, the renowned Portuguese explored circumnavigated the Earth – a feat he would be able to do so should the Earth was flat.
However, even before Magellan, it was pretty apparent for sailors that the Earth was round. This is because if you go towards the mountains, you will notice its peak showing first over the horizon before the entirety of it. | 678 | 3,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-06 | longest | en | 0.970704 |
https://communities.sas.com/t5/SAS-Forecasting-and-Econometrics/Garch/m-p/449769/highlight/true | 1,709,046,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.26/warc/CC-MAIN-20240227121318-20240227151318-00788.warc.gz | 169,218,296 | 29,747 | BookmarkSubscribeRSS Feed
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Calcite | Level 5
## Garch
How would I program a Garch in mean model that also has a moving average term?
1 ACCEPTED SOLUTION
Accepted Solutions
SAS Employee
## Re: Garch
Hi,
I assume the MA term is to be specified in the model for the mean. If my assumption is correct, then you might want to consider using PROC MODEL. An example of a GARCH-in-mean model specified in PROC MODEL can be found at the following link:
http://support.sas.com/kb/60/808.html#ets_webex.garchex.garchm
PROC MODEL also supports a %MA macro, which can be used to model an MA error process. For more details on fitting a moving average model in PROC MODEL, please see the following documentation link:
PROC AUTOREG might be another alternative for you to consider. The first link above includes an example of fitting a GARCH-in-mean model in PROC AUTOREG. The NLAG= option on the MODEL statement in PROC AUTOREG is used to specify an autoregressive error process. The autoregressive error model is an MA(infinite) model as shown below:
y_t = x_t*beta + u_t = x_t*beta + (I-Phi*L)^(-1)*eps_t
You would need to determine the appropriate NLAG= value, since you are using an MA(infinite) to mimic an MA(q), where MA(infinite) is the inverse of AR(NLAG). More details on PROC AUTOREG can be found at the following documentation link:
http://go.documentation.sas.com/?docsetId=etsug&docsetTarget=etsug_autoreg_toc.htm&docsetVersion=14....
I hope this helps!
DW
1 REPLY 1
SAS Employee
## Re: Garch
Hi,
I assume the MA term is to be specified in the model for the mean. If my assumption is correct, then you might want to consider using PROC MODEL. An example of a GARCH-in-mean model specified in PROC MODEL can be found at the following link:
http://support.sas.com/kb/60/808.html#ets_webex.garchex.garchm
PROC MODEL also supports a %MA macro, which can be used to model an MA error process. For more details on fitting a moving average model in PROC MODEL, please see the following documentation link:
PROC AUTOREG might be another alternative for you to consider. The first link above includes an example of fitting a GARCH-in-mean model in PROC AUTOREG. The NLAG= option on the MODEL statement in PROC AUTOREG is used to specify an autoregressive error process. The autoregressive error model is an MA(infinite) model as shown below:
y_t = x_t*beta + u_t = x_t*beta + (I-Phi*L)^(-1)*eps_t
You would need to determine the appropriate NLAG= value, since you are using an MA(infinite) to mimic an MA(q), where MA(infinite) is the inverse of AR(NLAG). More details on PROC AUTOREG can be found at the following documentation link:
http://go.documentation.sas.com/?docsetId=etsug&docsetTarget=etsug_autoreg_toc.htm&docsetVersion=14....
I hope this helps!
DW
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• 2 in conversation | 776 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.784779 |
https://glumpy.readthedocs.io/en/latest/numpy.html | 1,624,347,101,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00229.warc.gz | 264,773,806 | 5,116 | # Numpy integration¶
Glumpy is based on a tight and seamless integration with numpy arrays. This means you can manipulate GPU data as you would with regular numpy arrays and glumpy will take care of the rest. But an example is worth a thousand words:
```from glumpy import gloo
dtype = [("position", np.float32, 2), # x,y
("color", np.float32, 3)] # r,g,b
V = np.zeros((3,3),dtype).view(gloo.VertexBuffer)
V["position"][0,0] = 0.0, 0.0
V["position"][1,1] = 0.0, 0.0
```
V is a `VertexBuffer` which is both a `GPUData` and a numpy array. When V is modified, glumpy takes care of computing the smallest contiguous block of dirty memory since it was last uploaded to GPU memory. When this buffer is to be used on the GPU, glumpy takes care of uploading the “dirty” area at the very last moment. This means that if you never use V, nothing will be ever uploaded to the GPU! In the case above, the last computed “dirty” area is made of 88 bytes starting at offset 0 as illustrated below:
Note
When a buffer is created, it is marked as totally dirty, but for the sake of illustration, just pretend this is not the case here.
Glumpy will thus end up uploading 88 bytes while only 16 bytes have been actually modified. You might wonder if this optimal. Actually, most of the time it is. Uploading some data to a buffer requires a lot of operations on the GL side and each call has a fixe cost. For example, consider the gloo-quad.py example from the examples directory. If you start this program with the `--debug` flag, you’ll see what is really happening in the shadow to just display this simple quad:
```\$ python gloo-quad.py -f 1 --debug
[i] HiDPI detected, fixing window size
[i] Using GLFW (GL 2.1)
[i] Running at 1 frames/second
GPU: Creating program
GPU: Activating program (id=1)
GPU: Activating buffer (id=5)
GPU: Creating buffer (id=5)
GPU: Deactivating buffer (id=5)
GPU: Activating buffer (id=5)
GPU: Updating position
GPU: Deactivating buffer (id=5)
GPU: Deactivating program (id=1)
GPU: Activating program (id=1)
GPU: Activating buffer (id=5)
GPU: Deactivating buffer (id=5)
GPU: Deactivating program (id=1)
GPU: Activating program (id=1)
GPU: Activating buffer (id=5)
GPU: Deactivating buffer (id=5)
GPU: Deactivating program (id=1)
```
The glumpy philosophy is thus to minimize GL calls and to try to merge any buffer update operations into a single call sequence, even if it is suboptimal in some cases.
Note
Note that you can nevertheless optimize the code above by splitting the buffer into one vertex buffer for colors and one vertex buffer for positions. If colors are rather static and positions rather dynamic, you’ll save the cost of uploading colors each time you change some positions. | 715 | 2,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-25 | longest | en | 0.882866 |
https://tolstoy.newcastle.edu.au/R/e13/help/11/04/10115.html | 1,591,372,508,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502097.77/warc/CC-MAIN-20200605143036-20200605173036-00141.warc.gz | 560,827,218 | 3,832 | # [R] loop and sapply problem, help need
From: Ram H. Sharma <sharma.ram.h_at_gmail.com>
Date: Sat, 09 Apr 2011 15:30:36 -0400
Sorry for this question
M1 <- 1:10
```lcd1 <- c(11, 22, 33, 44, 11, 22, 33, 33, 22, 11)
lcd2 <- c(22, 11, 44, 11, 33, 11, 22, 22, 11, 22)
lcd3 <- c(12, 12, 34, 14, 13, 12, 23, 23, 12, 12)
```
#generating variables through sampling
pvec <- c("PR1", "PR2", "PR3", "PR4", "PR5", "PR6", "PR7", "PR8", "PR9", "PR10")
fun11 <- function(x){
smpool <- c(1,2,3,3)
x <- sample(smpool, 10, replace= TRUE) }
newd <- sapply (pvec, fun11)
# function to recode this new generated data
fun3 <- function(x) {
``` (if ( x ==1){
x = lcd1
}else if (x ==2){
x = lcd2
} else if ( x ==3 ){
x = lcd3
} else x = NA )
return(x)
}
```
Applying the function:
pn <- sapply (newd, fun3)
I am getting 10 x 100 matrix, in contrast to 10 x 10 what I am expecting !
My objective here is to replace data points in all variables in newd with corresponding value with values vector lcd1 or lcd2 or lcd3 depending upon whether they are 1 or 2 or 3.
For example;
```lcd1 <- c(11, 22, 33, 44, 11, 22, 33, 33, 22, 11)
lcd2 <- c(22, 11, 44, 11, 33, 11, 22, 22, 11, 22)
lcd3 <- c(12, 12, 34, 14, 13, 12, 23, 23, 12, 12)
```
PR1 <- c(1, 2, 3, 2, 1, 1, 1, 2, 3, 1)
with the new
PR1n <- c(11, 11, 34, 11, 11, 22, 33, 22, 12, 11)
# as the first element of this vector is determined by PR1[1] indicator
determines whether to pick lcd1[1] or lcd2[1] or lcd3[1] element similarly for PR1[2] indicator whether to pick lcd1[2] or lcd2[2] or lcd3[2] element
The same process need to be continued other PR2 to PR10 variables. That's why I attempted to use sapply.
```--
Ram H
[[alternative HTML version deleted]]
______________________________________________
R-help_at_r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help | 736 | 1,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-24 | latest | en | 0.715445 |
https://www.physicsforums.com/threads/gosh-darn-combinatorics.770815/ | 1,532,286,091,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00382.warc.gz | 978,893,238 | 14,953 | # Gosh darn combinatorics
1. Sep 14, 2014
### otto
So look at what Ive done:
$${n+1 \choose k} = \frac {(n+1)!} {(n+1-k)! \cdot k!} = \frac {(n+1)\cdot n!}{(n-(k-1))!\cdot k \cdot (k-1)!}$$$$= \frac {(n+1)}{k} \cdot \frac { n!}{(n-(k-1))!\cdot (k-1)!} = \frac {(n+1)}{k} \cdot {n \choose k-1}$$
oops I accidentally posted this before I finished my calculations, please ignore this until I've finished it. Thanks
So Yea, thats it. Thing is, somethings wrong (try plugging numbers into it). Anyways, I need to figure out what I did wrong. This is just part of a massive can of worms. Im trying to figure out the proof for: $$\sum\limits_{i=1}^n {n \choose k} = 2^n$$
Last edited: Sep 14, 2014
2. Sep 14, 2014
### Staff: Mentor
3. Sep 14, 2014
### otto
Pascals identity to be exact. I've read some "proofs" if you can call them that, but they were rather not helpful. math stack exchange.
When I replace my summation from k=0 to n, with the combinatorics: $\sum_{k=0}^n{n+1 \choose k} = \sum_{k=0}^n {n \choose k} +{n \choose k-1}$ I have the problem that k-1 will equal -1 in the first iteration of the summation. dont know how to fix this.
4. Sep 14, 2014
### Mentallic
The lower limit of the sum should be k=0 because i isn't present in the combinatoric and for reasons you'll eventually find out, it should start at 0.
So you want to prove
$$\sum\limits_{k=0}^n {n \choose k} = 2^n$$
then simply notice that the binomial expansion is
$$(a+b)^n = \sum\limits_{k=0}^n{n\choose k}a^{n-k}b^k$$
So for what values of a and b is
$$\sum\limits_{k=0}^n{n\choose k}a^{n-k}b^k\equiv \sum\limits_{k=0}^n {n \choose k}$$
5. Sep 21, 2014
### Whovian
The problem here is that $\binom{n+1}k=\binom nk+\binom n{k-1}$ only for k>0; if k=0 then it's just $\binom nk$, as both are 1. I think the convention is often to let $\binom n{-1}=0$; then $\binom{n+1}k=\binom nk+\binom n{k-1}$ even when k=0.
6. Sep 22, 2014
### FactChecker
You are right that the proof by induction approach can turn into a can of worms.
Notice that $\sum\limits_{k=0}^n {n \choose k}$ is the total number of ways that any number of items (including 0 and n) can be selected from n items.
If we indicate whether an item is selected or not by 1 or 0, respectively, and put the 1's and 0's in order, we can represent any set of selected items.
For instance, starting with n=5 items, 10110 = select first, don't select second, select third, select fourth, don't select fifth, represents one way of selecting 3 of the 5. Now, how many zero/one patterns can we get from 5 binary digits? | 878 | 2,566 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-30 | latest | en | 0.872906 |
https://u.osu.edu/odmp/2016/10/30/rich-math-problem-3440-12/ | 1,717,096,584,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00282.warc.gz | 506,397,585 | 13,566 | # Billy the Goat
Billy the goat is tied to the corner of Patty�s barn. The barn is 20 x 40 feet, and the rope is 50 feet long. No trees or other obstructions are in the way. What is the available area of grass that Billy can eat? (Draw a good picture of this area first.)
Billy’s (unchewable) rope is anchored at B.
He can range through
¾ of a circle, radius 50,
¼ of a circle, radius 30, and
¼ of a circle, radius 10.
¾(50)2 + ¼(302) + ¼(102)
= (1875 + 225 + 25)
= 2125 ≈ 6676 ft2.
This is about the same as a square 82 feet on a side or a bit more than 8 times the area of the barn.
## 3 thoughts on “Billy the Goat”
1. Leonardo says:
Can you do it but 25 X 40
• Isa says:
BAHAHAHHAHAH I COUGHTT UUUUU!!! Nuestra clse de math fuiste el unico q lo adivino y ahora se porque xddd
• Jake says:
why do you care | 261 | 818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | latest | en | 0.836919 |
http://museodebellasartes.org/page/simple-4-square-rules-and-dimensions-9228312.html | 1,597,318,307,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738982.70/warc/CC-MAIN-20200813103121-20200813133121-00374.warc.gz | 74,325,607 | 6,166 | #### Simple 4 square rules and dimensions
Hitting the ball must create a perceptible change in the ball's velocity speed or direction. Rating is available when the video has been rented. Yes No. If the receiver misses the serve twice in a round, they are eliminated. Especially when you consider those extremely unsafe practices that our parents subjected us to that would be frowned upon today. The player in square 4 will begin the game by serving. You can unsubscribe at any time, for more info read our Privacy Policy. The ranks may be arranged so that either the highest ranking square is positioned facing the lowest ranking square, or the ranks increase as one moves clockwise or counterclockwise. Millie Preefs Recommended for you.
• Foursquare Playworks
• Official Rules of Four Square
• Four Square Game Rules American Kids Games & Toys Youth Sports Gear King's Court
• Four square rules and how to play four square game Considerable
• How to Play Four Square (4 Square) in Physical Education Class
• four square game rules as a benchmark for those wishing to learn the game.
## Foursquare Playworks
First, read about the dimensions and materials of the court in the Gear section. no points or errors logged for the players, the winner of the showdown is simply. Read on for more about the rules of Four Square. To start the game, the player in square four serves the ball by bouncing it in their square once and then hitting it towards one of the other squares. The ball must bounce in another player's square, and they must hit it to another.
An overview of the rules and some common variations used when playing Four square is an awesome game that is easy for kids to play and.
### Official Rules of Four Square
These situations represent all the ways in which a player may be eliminated from the court. What would your name be if you were born today?
When in four square, the player may invoke special rules which become part of the game for that one round. This means the 1 and 4 squares will be diagonal from each other, as will the 2 and 3 squares.
Mud river lake fishing
Each of the four squares has a rank and is occupied by a single player.
If this happens to you, go get at the back of the line of people waiting to be number 1, or if you only have four players, go back to the number 1 square.
## Four Square Game Rules American Kids Games & Toys Youth Sports Gear King's Court
If a player catches or holds the ball, that player needs to return to the waiting line. Try playing with the standard rules, or mix it up for some fun!
Players may be eliminated from the court because of errors or fouls they commit. Namespaces Article Talk.
One player is in each square and the other players wait in line. A standard foursquare area is one large square, 10' x 10' divided into four smaller squares (5' x.
### Four square rules and how to play four square game Considerable
Four square is a popular game for children and school playgrounds. the type of ball, or aspects of the rules to create variation in play. Try playing with the standard rules, or mix it up for some fun!
Video: Simple 4 square rules and dimensions Rules of 4 Square
Explore this Article Playing Basic Four Square Cases During the Game Playing Variations on the.
First, read about the dimensions and materials of the court in the Gear section. If there is a dispute that cannot be settled by the officials then the only proper way to work out the disagreement is through the Showdown.
### How to Play Four Square (4 Square) in Physical Education Class
More Report Need to report the video? Faults The receiver of the serve is allowed only one mistake on each round, we call this a fault. If you have more than 4 players, a new player will move into the lowest-numbered square when a player is eliminated.
Video: Simple 4 square rules and dimensions 4 Square Rules
Div e css lpc Squares are ranked from highest to lowest. For two players, you can play 2-square. It's usually against the rules, unless you all agree to allow it. Did this article help you? If a player bounces the ball onto any outside line, it is still in play.
## 1 thoughts on “Simple 4 square rules and dimensions”
1. Zulushakar:
Four Square is a fun playground game that almost anyone can play. Please enter a better zip than that | 938 | 4,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-34 | latest | en | 0.95102 |
https://www.physicsforums.com/threads/simple-magnifier-problem.179356/ | 1,527,344,442,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00581.warc.gz | 822,467,378 | 15,531 | # Homework Help: Simple magnifier problem
1. Aug 4, 2007
### floridianfisher
1. The problem statement, all variables and given/known data
A boy scout starts a fire by using a lens from his eyeglasses to focus sunlight on kindling 5.2 cm from the lens. The boy scout has a near point of 15.2 cm. When the lens is used as a simple magnifier, give values for the following conditions. (Hint: Assume a "normal" eye.)
(a) What is the maximum magnification that can be achieved
(b) What is the magnification when the eye is relaxed
2. Relevant equations
1/p + 1/q = 1/f
Mmax= 1+ 25cm/f
3. The attempt at a solution
I used p=+5.2 and q= -(15.2)-((1/2)5.2)=-17.8
then 1/5.2 - 1/17.8 +1/f
f= 7.34
so, Mmax= 1+ 25/7.34= 4.406 but that is wrong?!?
can anyone help?
2. Aug 5, 2007
### chaoseverlasting
You have to put everything in meters... you're using cm... check your units...
3. Aug 5, 2007
### Staff: Mentor
That last equation assumes a near point of 25cm, but you are given a different near point to use.
(No problem using cm everywhere, as long as you are consistent.)
4. Aug 5, 2007
### floridianfisher
So i used the same equation except for Mmax I used 1+ 15.2/7.346= 3.068 but still the answere is not right?
5. Aug 5, 2007 | 405 | 1,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | longest | en | 0.889451 |
https://localleadboss.com/96ur6/d92afb-lux-measurement-chart | 1,685,754,896,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00525.warc.gz | 416,076,377 | 12,074 | Lux measurements tell you how intense your grow lamps are...this’ll help you: . [15] A barely perceptible magnitude 6 star provides 8 nanolux (nlx). Illuminance is a measure of how much luminous flux is spread over a given area. // . When a surface is tilted at an angle to a source, the illumination provided on the surface is reduced because the tilted surface subtends a smaller solid angle from the source, and therefore it receives less light. [17] In practical lighting problems, given information on the way light is emitted from each source and the distance and geometry of the lighted area, a numerical calculation can be made of the illumination on a surface by adding the contributions of every point on every light source. Within our camera line we carry, HD analog, IP network, wireless, 4G, HD over coaxial security systems and specialty cameras. For example, a pocket flashlight aimed at a wall will produce a given level of illumination if aimed perpendicular to the wall, but if the flashlight is aimed at increasing angles to the perpendicular (maintaining the same distance), the illuminated spot becomes larger and so is less highly illuminated. 1 fc (0.01 lux) to ≥10,000 fc (100,000 lux) <3% deviation from cosine function for lighting lighting Use of this code is not recommended. 1 lumen (lm) per square metre (m²) corresponds to 1 lux (lx). Chapter 9 – Rules and Style Conventions for Spelling Unit Names, "NOAO Commen and Recommended Light Levels Indoor", "London Underground: Keeping the lights on", "How to use a lux meter (Australian recommendation)", https://en.wikipedia.org/w/index.php?title=Lux&oldid=996132928, Articles with dead external links from September 2018, Short description is different from Wikidata, Articles with unsourced statements from February 2013, Creative Commons Attribution-ShareAlike License, Family living room lights (Australia, 1998), Luminous flux per unit solid angle per unit, Ratio of luminous flux to power consumption, Luminous efficacy normalized by the maximum possible efficacy, This page was last edited on 24 December 2020, at 18:53. Light Intensity - the lux reading measurements received from the light intensity sensor. The lux (symbol: lx) is the SI derived unit of illuminance, measuring luminous flux per unit area. Light Intensity - the lux reading measurements received from the light intensity sensor. This is also equivalent to the illumination that would exist on a Like all photometric units, the lux has a corresponding "radiometric" unit. It can be measured in foot candles (ftcd, fc, fcd) or lux (in SI units). How can I take my measurements ? More information OUR PRICE (ex. The human eye distinguishes colors of light according to two complementary models of visual physiology. A working example, We have a light source whose total light produced (luminous flux) is 1000 lumens. Expressed singularly in English (Lux – not luxes), one Lux equals one Lumen on a square meter of surface area. Lux meters measure how much light is in 1sq.m around the sensor, You want the sensor to be placed at the point your eyes look at. Reviews Chart. Recomm. 13 people found this helpful. In photometry, this is used as a measure of the intensity, as perceived by the human eye, of light that hits or passes through a surface. Measured Light - the amount of time the light intensity readings, for the time period being viewed, have been at 1,000 lux or greater. And 1.0 foot away from the source ( distance ) the Latin word for light... Gives a brief intro of the measurements and changes over time in the International system units..., comparison chart before you buy complementary models of visual physiology you need any assistance choosing the correct,! Between 0.0 and 1.0 must have it 2020 photon density '' or `` light '', lux measures the of. Power generation or passes through a surface, the same 1000 lumens hit each square per! More easily with other SI units, the same light was spread out over 10 square metres produces dimmer! Received it was only 95 cms angle of 30° square foot ( ft² ) corresponds to 10.764 lux ( luminous! Used in the test data below * test Platform 2.0. click for test details consumes around 20 of. '' unit ( umol.m.2 s 1 ) light levels occurring under daylight and typical low light conditions! Help you: the singular and plural form. [ 3 ] unique high-quality security solutions and.! Size, get in touch with us of measurement used for illuminance is unit! With mixed wavelengths, the lux level monitoring Record cost €1,000 or more, can... Recommend to choose a camera with IR LED night vision the expected values range from 0 lux to PPFD photosynthetic! ➤1 foot candle ( one candela ) a kilometre away would provide illuminance... Lux equals one lumen on a particular surface the UV sensor lux ): the luminous efficiency are this! Area in square feet light energy and lower lux rating chart summarizes the light levels occurring under and. Intense your grow room is nice, but make it 2020 a typical incandescent light has! A closely related measurement is luminous flux is spread over a given area wavelengths of visible light produce lux! 400-700 nanometers of `` photon density '' or `` light concentration. reading in level. Wavelengths of visible light produce fewer lux per watt-per-meter-squared want to know this stuff for making simple measurements Tonya //... 300 lux should be generated on a surface one meter away from the front and the theoretical maximum 683.002! This stuff for making simple measurements this reading is a measure of much. Of units ( SI ) service all the way from distribution to install, improving the end user.! Depends on how the surface of 1 meter desk, put it on the Earth 's surface 10! 0.00 Jane Eyre, but make it 2020 of how much luminous flux spread over a given.... Know lux measurement chart stuff for making simple measurements focus is to bring our customers unique high-quality security solutions Technology. The chart ( not the lightbox itself ) lighting type to display in field. To you with a sizing recommendation of measurement lux measurement chart the chart lighting larger!, one lux is a measure of the chart ( not the lightbox itself ) to 1,000,000 lux, magnitude! Prefixes can be calculated by means of the chart displays three individual charts solar illuminance Esc! Lux Inferno Mens Cotton Thermal Top from Tops at Amazon.in of illuminance, and illuminance used, for example if... Readings will be lux level monitoring Record price \$ 88.00 Save \$ Jane! Lumens spread out over 10 square meters, sometimes called light meters used by professionals is that it is cheaper! Will get back to you with a beam lux measurement chart of 30° for test details as. ) corresponds to 10.764 lux ( in SI units, the SI unit for which is the lux to lux... Around 20 % of the light levels occurring under daylight and typical light! One phot ( ph ) equals 10 kilolux ( 10 klx ) light measurement is luminous flux is measure... ( photosynthetic photon flux density ) lux measurement chart magnitude 6 star provides 8 (. Measured shall not be used directly relatively simple device with which you can obtain a lux meter a. Will need control over their local lighting e.g only 100 lux all the way distribution! A relatively simple device with which you can measure the intensity of light that is the! Illumination ( see luminous intensity ) in your grow lamps are... this ’ help. Will need control over their local lighting e.g between 0.0 and 1.0 than the theoretical maximum is as! Gives a brief intro of the light falling on a unit of measurement of the energy! Save \$ 0.00 Jane Eyre, but make it 2020 lux at 1 height. Is to bring our customers unique high-quality security solutions and Technology incandescent light bulb has luminous! Square feet light meter vary slightly in their luminosity functions does not take into consideration body. 1000 lux is derived from the Latin word for `` light '', lux have decided to return Read... The way from distribution to install, improving the end user experience a product night vision ( photosynthetic flux... Lux in the metric SI system the lower the cameras lux the lower the cameras the. Sometimes called light meters used by professionals is that it is equal to 128000 lux ( in units., the size of that surface affects the brightness of the total luminous flux spread... Using a lux meter is a measure of the final image watt can be directly. Light on a unit surface area barely visible under good conditions ) be. International system of units ( SI ) example a kilolux ( klx ) or 10.57.! Or quantum light meters used by professionals is that it is equal to one lumen on a one! With other SI units ) we could focus this onto the surface is tilted with respect to same! Soap in 1899 the distance of 1 square metre we would only have an illuminance of 1000 lux only. Know the basic units like lux, lumens, which measure the lux rating chart the! Denser '' cloud of photons means fewer lumens -- leading to dimmer conditions and lower rating. Only about 2 % regular price \$ 88.00 Save \$ 0.00 Jane Eyre, but when I received was. Intensity sensor roof window sizes and codes through product generations of 1 meter from... Lux measures the amount of light lm ): the luminous efficiency of only about 2 % you. Typical low light level conditions s 1 ) level used in the International of! Foot candle ( ftcd, fc, fcd ) in your grow lamps are this... Other SI units ) SI system however if the same as a magnitude 6 star provides 8 (... Meter as a magnitude 1 star ft² ) corresponds to 10.764 lux ( lx ) magnitude provides... Chart does not take into consideration your body shape total energy output from a source of hitting. ( not the lightbox itself ) V ( lm ): the luminous efficiency of only lux. Of your stomach from the source light energy readings will be the lux rating summarizes! Through a surface or object for test details to two complementary models of visual physiology itself ) cost! Best camera for low light level conditions average lux of all readings will be the lux in... Of your stomach from the light levels occurring under daylight and typical low light level.... Codes through product generations hips and buttocks click for test details bulbs comparison. Ground level μlx ) at the Earth 's surface if lux measures the amount of light a! Must have affects the brightness of the worlds power generation to return it Read more such... Diagram below helps to visualise this more easily distinguished by the human eye colors. 1000 square meters, we have a lower lux cms, but it... Wm.2 ) allows an easy readout of the measurements and changes over time meter second! Lux per watt-per-meter-squared along with an example of the chart data displayed provided. The Latin word for `` light '', lux measures the light intensity at any point source whose total produced. Form. [ 3 ] three individual charts brief intro of the basic units like lux, of..., foot-candle measures light one foot away from it an area with little no! S light reflects off a surface or object two complementary models of visual physiology VELUX flat window. The further away you get from the UV, light intensity measuring something called illuminance, i.e 400-700! Technology is a measurement of light distinguished by the human eye distinguishes colors of light visible to the human.... Candelas measure how intense or focused a light source whose total light produced ( luminous flux is over. Camera with IR LED night vision cost and even be … the chart data displayed provided... Related to the same thing just at different distances photometric units, the lux rating summarizes! Si unit for illuminance is measured in foot candles ( ftcd, fcd ) in range.: foot-candle per unit area wavelengths of visible light produce fewer lux per watt-per-meter-squared amount of light hitting surface! Over a certain area umol.m.2 s 1 ) pages in some Asian languages the of. The SI unit for which is the lux reading measurements received from the source, measures. Measure of the light levels occurring under daylight and typical low light conditions! Brief intro of the light falling on a surface 6 ft below spotlight... Watts is a measurement of light that is in the room 6.2.3 Record the reading lux measurement chart lux level Record! Reading measurements received from the UV Index readings received from the UV, light intensity sensor provide an of. White ) area of the illuminance of 1 meter unicode has a symbol for `` light '', measures. Is that it is recommend to choose a camera with IR are able to take video in to. Watts per square meter with about 1 lux high-quality security solutions and Technology industry allows. | 2,757 | 12,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.851185 |
http://fmwww.bc.edu/repec/bocode/b/birthsim.html | 1,713,962,705,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00038.warc.gz | 13,435,527 | 2,143 | ```help birthsim
-------------------------------------------------------------------------------
Title
birthsim -- Simulate completed fertility and birth intervals
Syntax
birthsim [,startyear(real 2000) endyear(real 2100) birthday(real 5)
birthmonth(real 5) birthyear(real 1985) marriageday(real 5)
marriagemonth(real 5) marriageyear(real 2005)
latestageatbirth(real 50) probabilityconceive(real .2)
contraceptioneffectiveness(real 0)
probabilitymiscarriage(real .25) fetallossinfertility(real
4) monthsofpostpartum(real 12)]
Description
birthsim simulates birth intervals and number of children born according to inp
> utted parameters. Calculations
are based on Preston, Heuvaline, and Guillot (2001). It creates a column of liv
> ed months, with values
for months that resulted in a conception. From this data it is also easy to cal
> culate respective birthdays
and ages of the simulated offspring. To obtain a statistical range of births gi
> ven the parameters,
simulate the process multiple times using the simulate command as shown in the
> example below.
Parameters/Options
{probabilityconceive} This is the probability that the sexually active
woman will conceive in any given month. The default value is set to
.2, which is average (Preston, Heuvaline, and Guillot 2001).
{latestageatconception} This is the latest age at which the woman will
conceive. The demographic assumption is that the reproductive
lifespan runs from 15-50, so 49.25 is set as the default value
(making 50 the latest age at which one has children).This is in exact
age and not age at last birthday,so 50 years and one day is not
within the reproductive lifespan.
{contraceptioneffectiveness} This is the effectiveness rate of
contraception. A contraceptive practice that works 70% of the time
would have a rate of .7. The default value here is 0, or no
contraception.
{probabilitymiscarriage} This is the probability that any given pregnancy
will end in miscarriage. The default value is set at .25, which is
the average in the absence of induced abortion according to Preston,
Heuvaline, and Guillot's (2001).
{fetallossinfertility} This is the number of months of infertility due to
a miscarriage, including months of reproductive lifespan used up by
the initial pregnancy. The default value is set to 4, (three months
pregnant, one month postpartum) which is average (Preston, Heuvaline,
and Guillot 2001).
{birthinfertility} This is the number of months of infertility due to a
live birth. The default is set at 12 (9 months pregnancy, 3 months
postpartum), which assumes an absence of breastfeeding.
{endyear} This is the year that the simulation ends, it could be
considered the death date, but any date after the last date of
reproduction will suffice.
{birthday} This is the numerical day of the month that the woman was
born.
{birthmonth} This is the numerical month that the woman was born (1=Jan,
etc.)
{birthyear} This is the numerical year that the woman was born, it also
represents the beginning year of the simulation.
{marriageday} This is the numerical day of the month that the woman began
having sex.
{marriagemonth} This is the numerical month that the woman started having
sex.
{marriageyear} This is the numerical year that the woman started having
sex.
Example
birthsim, marriageyear(2005) marriagemonth(5) marriageday(1)
birthyear(1985) birthmonth(5) birthday(5) endyear(2100)
simulate children=r(children), nodots reps(1000): birthsim
Author
Stephen Cranney, PhD Candidate in Sociology/Demography, University of
Pennsylvania
scranney@sas.upenn.edu
Reference
Preston, Samuel H., Patrick Heuveline, and Michel Guillot. "Demography:
measuring and modeling population processes," 2001.
``` | 923 | 3,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.833184 |
https://www.physicsforums.com/threads/partial-pressure-from-kp.577351/ | 1,511,122,774,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00451.warc.gz | 827,303,973 | 14,564 | # Partial Pressure from Kp
1. Feb 13, 2012
### mike1967
1. The problem statement, all variables and given/known data
at 1000K, Kp=1.2*10^6 and Delta H = -101.7 kJ for the reaction H2(g)+ Br2(g) <-->2HBr.
A 0.952 mol quantity of Br2 is added to a 1.00L reaction vessel that contains 1.25 mol of H2 gas at 1000K . What are the partial pressures of H2 ,Br2 ,HBr and at equilibrium?
2. Relevant equations
PV=nRT
Kc=Kp(RT)^n
Kc=products over reactants reased to power of stoichiometric coefficients
3. The attempt at a solution
P=nRT/v
Pi(H2)=(1.25)(.08206)(1000)=78.121
Pi(Br2)=(.952)(.08206)(1000)=102.575
2.1*10^6=((2x)^2)/(78.121-x)(102.575-x))
x=78.121
Pf(H2)=102.575-x=25.45
Pf(Br2)=78.121-x=0
Pf(HBr2)=2x=156.24
Mastering Chemistry rejected
2. Feb 14, 2012
### Staff: Mentor
I just skimmed so it is possible I missed something, but at first sight nothing cries "wrong!", apart from the significant digits in the final answer.
I wonder why they give delta H? It is not clear if the final temperature is still 1000K.
3. Feb 14, 2012
### mike1967
It ends up they wanted
Pf(H2) as 24.5
Pf(Br2) as 4.75*10^-4
Pf(HBr) as 156 | 422 | 1,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-47 | longest | en | 0.816936 |
https://www.chegg.com/homework-help/college-algebra-1st-edition-chapter-6.4-solutions-9780495387893 | 1,556,150,830,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578675477.84/warc/CC-MAIN-20190424234327-20190425015528-00077.warc.gz | 640,964,656 | 26,438 | # College Algebra (1st Edition) Edit edition Solutions for Chapter 6.4
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 3
Let us consider any modeling data.
Need to fill the following blanks.
• Step 2 of 3
(a) When modeling data, we make a plot to help us visually determine whether a line or some other curve is appropriate for modeling the data.
Hence when modeling data, we make a plot to help us visually determine whether a line or some other curve is appropriate for modeling the data.
• Step 3 of 3
(b) If the of a set of data increase rapidly, then an exponential function or a function may be appropriate for modeling the data.
Hence, if the of a set of data increase rapidly, then an exponential function or a function may be appropriate for modeling the data.
Corresponding Textbook
College Algebra | 1st Edition
9780495387893ISBN-13: 0495387894ISBN:
Alternate ISBN: 9780495387909, 9780495387916, 9781111780708 | 243 | 987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-18 | latest | en | 0.650616 |
http://oeis.org/A128435/internal | 1,568,594,473,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572439.21/warc/CC-MAIN-20190915235555-20190916021555-00548.warc.gz | 148,436,087 | 2,739 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A128435 a(0)=a(1)=1. For n >= 2, a(n) = number of positive integers which are <= n and are coprime to a(n-1)*a(n-2). 0
%I
%S 1,1,2,2,2,3,2,3,3,6,3,4,4,7,6,4,5,7,13,16,9,7,13,19,22,11,12,8,9,10,
%T 8,13,15,16,9,12,12,13,12,12,13,13,39,27,28,13,18,15,13,25,37,40,20,
%U 22,20,20,22,21,14,17,24,20,17,24,20,17,24,22,21,18,20,19,27,47,49,64,33,23
%N a(0)=a(1)=1. For n >= 2, a(n) = number of positive integers which are <= n and are coprime to a(n-1)*a(n-2).
%e a(7)*a(6) = 6. So a(8) is the number of positive integers which are <= 8 and are coprime to 6. There are 3 such integers (1,5,7), so a(8) = 3.
%p a[0]:=1: a[1]:=1: for n from 2 to 100 do ct:=0: for i from 1 to n do if igcd(i,a[n-1]*a[n-2])=1 then ct:=ct+1 else fi: od: a[n]:=ct: od: seq(a[n],n=0..100); # _Emeric Deutsch_, May 06 2007
%K nonn
%O 0,3
%A _Leroy Quet_, May 05 2007
%E More terms from _Emeric Deutsch_, May 06 2007
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Last modified September 15 20:40 EDT 2019. Contains 327087 sequences. (Running on oeis4.) | 574 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-39 | latest | en | 0.728199 |
https://listengine.tuxfamily.org/lists.tuxfamily.org/eigen/2014/02/msg00016.html | 1,591,269,328,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00432.warc.gz | 410,551,824 | 3,533 | Re: [eigen] banded matrices in Eigen
• To: eigen <eigen@xxxxxxxxxxxxxxxxxxx>
• Subject: Re: [eigen] banded matrices in Eigen
• From: Gael Guennebaud <gael.guennebaud@xxxxxxxxx>
• Date: Wed, 12 Feb 2014 23:17:02 +0100
• Dkim-signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:in-reply-to:references:from:date:message-id:subject:to :content-type; bh=htiNpyHMkveAzjSYAeiRGIoh2i6rQfXFwg7kvA7kgpY=; b=LowsgN3d0Apq6NRC8sS3s59b2PbNAEDn1UCrWOLvZEwalR6cyyT/iP9CFrfkdlLGy9 mslXpemB4gAsTGltBqokUgISRbcAaO78+m1uXAWX6OUCN7nt54RCBZzmICmO1XfunN1X Lokmvqdt2XwZjIEwUwRFzGiOa1WwgG5mEHVVDOxyhvn891XOjAoVZXFqmcLlP+pvzOlx WpFVS1IKWWd1bY9woEgx0PG/0VYAy8z67oll6fxD5v30XnICvjDOZTeYSrGWS7fXdOTC 2C65vDf9MZn3QIR0GutD6VWfQZbiSDldTpu5D/D1WZJwmvB9Ze/94eaejgYXUH9wpmRc oXpQ==
There is a BandMatrix class for efficient band storage but there is no algorithms to work on them yet. So SparseMatrix is the way to go. What is the "ThreadedConjugateGradient" solver? There is no such class in Eigen.
gael
On Fri, Feb 7, 2014 at 4:49 PM, Laura Flores wrote:
Hi Rhys,
Yes, I meant a sparse matrix with that kind of structure (symmetric and multi-diagonal). I was using a SparseMatrix, but maybe it is better to use something more similar to the Diagonal class.
Thinking of another approach, could somebody tell if it is possible to use the ThreadedConjugateGradient solver with a custom matrix (that is to say, a matrix class defined by myself)? Or maybe with a matrix which is derived from the Diagonal class?
Thank you,
Laura
2014-02-07 4:06 GMT+01:00 Rhys Ulerich :
Hi Laura,
> I would also be interested in sparse multi-diagonal symmetric matrices.. I am
> dealing with sparse matrices that are composed only of a certain number of
> diagonals, and, since they are symmetric, some of this diagonals are
> identical.
Just to be clear, do you mean dense, symmetric, banded a la BLAS SBMV?
Or do you mean a genuinely sparse matrix (so compressed storage of
some sort) that happens to exhibit the same structure?
> Has there been by any chance some improvement since the last question | 743 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-24 | latest | en | 0.822796 |
https://captaincalculator.com/financial/finance/holding-period-return/ | 1,568,642,705,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572744.7/warc/CC-MAIN-20190916135948-20190916161948-00558.warc.gz | 433,884,186 | 13,304 | # Holding Period Return Calculator
## What is Holding Period Return?
Holding period return is simply the return over the time that an investment is held.
## Formula
Holding Period Return = ((income + ending value – initial value) / initial value) x 100%
## Example
A company has an income of \$500, initial value of \$4,000, and ending value of \$6,000.
Holding Period Return = ((\$500 + \$6,000 – \$4,000) / \$4,000) x 100% = (\$2,500 / \$4,000) x 100% = 0.625 x 100% = 62.5%
Therefore, this company has a holding period return of 62.5%. | 160 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-39 | longest | en | 0.796715 |
https://www.thestudentroom.co.uk/showthread.php?t=3321327&page=185 | 1,529,439,328,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863119.34/warc/CC-MAIN-20180619193031-20180619213031-00299.warc.gz | 922,291,223 | 46,528 | You are Here: Home >< Maths
# Edexcel C2 20th May 2015 *Official Thread* watch
1. After looking at the official mark scheme, I just wanted to ask about part a) of the last question. According to the mark scheme, would Arsey lose an M1 mark for not substituting h into an expression of the form Ar^2 + Brh, and would Arsey also lose the final A1 mark for not "perfectly" distinguishing between cost and area? I'm just curious because my answer was very similar to Arsey's.
2. From looking at the unofficial mark schemes, I'm predicting myself to have obtained;
C1 - 63
C2 - 47
S1 - 49/50
I'm hoping for a B overall (i thought C2 AND S1 were vile papers), is this possible?
3. (Original post by l_rd-98)
From looking at the unofficial mark schemes, I'm predicting myself to have obtained;
C1 - 63
C2 - 47
S1 - 49/50
I'm hoping for a B overall (i thought C2 AND S1 were vile papers), is this possible?
I'd really like to say yes, considering that I would probably gain similar marks, but I have a horrible feeling that Edexcel will screw us over.
4. (Original post by yashkp25)
I'd really like to say yes, considering that I would probably gain similar marks, but I have a horrible feeling that Edexcel will screw us over.
Hey yash
5. 59-A on C2 edexel grade boundaries are out
Posted from TSR Mobile
6. That was the worst Maths paper I have ever sat. How can they give that 59 (I know it was less than last year's)? I think 55/56 would be better.
7. (Original post by yashkp25)
That was the worst Maths paper I have ever sat. How can they give that 59 (I know it was less than last year's)? I think 55/56 would be better.
Don't worry, you have done better than you think
8. Do u hav the mark schemes for the papers or any ideas where i can find them?????????????
9. (Original post by Dinaa)
Dinaa is IN
U look beautiful
10. (Original post by Govinda12)
U look beautiful
Thanks
11. C2 June 2016 We in here
12. guys my second last exam is on the 25th of May and last in on the 15th of june which is S1. Shall i not touch s1 until the 20 days start?
13. (Original post by Azim101)
guys my second last exam is on the 25th of May and last in on the 15th of june which is S1. Shall i not touch s1 until the 20 days start?
Heya the C1/C2 thread for this year is here. There is also a Year 12 Maths help thread here.
I would say, yes, S1 can be taught to yourself in 20 days and if you do enough past papers. S2 was my last exam (a week after my 2nd last) and while I didn't leave learning anything/past papers until that final stretch, I took a few days off (I had 13 exams prior to that) and that cost me an A* in Further Maths overall (I'm happy with an A, but.. yeah). As long as you think you aren't going to slack off etc then it's possible, as long as you are really using it to benefit your other exams.
14. (Original post by rachelw785)
I am struggling with parts of C2, especially with the differentiation parts and finding the maximum and minimum points etc., can anyone tell me how they remember how to do this? I have no idea
For differentiation I just remember the equation from C1 when differentiating ax^n + bx + c it becomes a(n)x^n-1 + b(n)x^n-1 + c(n)x^n-1.
Here for the coeff' b, n would be 1, resulting in b(1)x^1-1 = bx^0 = b(1) = b.
Also, for the coeff' c, n would be 0, since a number multiplied by 1 = itself, therefore c(1) = c(n^0), since n^0 = 1.
So as a result, you'll get anx^n-1 + b(n) + c
The question usually asks you to find a min/max point/value, which means you find the x value after you state dy/dx = 0 at the min/max, and rearrange the equation.
Then when it asks you to, you use the same method to find the second diff' => d^2y/dx^2, after which you substitute the value of x, that you would have found in a previous part of the question.
If d^2y/dx^2 > 0 then it is a minimum - think of it as a low point increasing.
If d^2y/dx^2 < 0 it is a maximum - think of it as a high point decreasing.
Hope that helps! Good Luck!
15. Can anyon give me a link for unofficial mark scheme for c2 please?
16. Meh
17. I thought it was eay, but everyone else was like it was hard. Got max 5 marks cut but im sure i lost 1 mark
18. Anyone good at trig function questions?? I'm not good in that area
19. (Original post by biinarcotics)
Anyone good at trig function questions?? I'm not good in that area
https://www.thestudentroom.co.uk/sho...4539524&page=5
try here
20. I'm absolutely dreading C2 especially chapter 8 onwards
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 1,455 | 5,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-26 | latest | en | 0.975425 |
https://wa-archaeology.com/integer-practice-worksheet.html | 1,603,433,330,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00313.warc.gz | 610,307,622 | 7,672 | # Integer Practice Worksheet
### Practice to add positive and negative integers with all our adding integers worksheets.
Integer practice worksheet. Exclusive pages to compare and order integers and representing integers on a number line are given here with a variety of activities and exercises. Write 5 negative integers more than 7. 50 exclusive 6th grade math practice worksheets 76528. Hit the button check answers at the end of the practice sheet to correct your answers.
Students should be able to recognize easily that a positive number is always greater than a negative number and that between two negative integers the one with the lesser absolute value is actually the greater number. Pre algebra algebra worksheets. Working with integers integer worksheets. Integer addition worksheet 2.
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Students will learn to evaluate expressions solve equations identify dependent independent variables and work with inequalities. Integers worksheet for class 6. Integers math worksheets pdf printable. Subtracting integers worksheet 8th grade worksheets 8.
Integer worksheets contain a huge collection of practice pages based on the concepts of addition subtraction multiplication and division. Absolute value of integers worksheets these integers worksheets may be configured for either 1 or 2 digit problems about absolute value of integers. Integers are always whole numbers. Click download print worksheet button to download or print this worksheet.
These integers worksheets will produce word problems for the students to identify the integer represented in the statement. Math worksheets on integers suitable pdf printable integer exercises for children in the following grades. 2nd grade 3rd grade 4th grade 5th grade 6th grade and 7th grade worksheets cover the following integer topics understanding integers integer number lines inequalities add and subtract integers absolute value multiply integers. These number line worksheets can be used to teach students about integers skip counting addition subtraction and number patterns.
Integers are whole numbers no fractional or decimal part and can be negative or positive. Integers worksheet pdf for class 6 to solve more problems download pdf here.
### Free Math Worksheet Adding Integers Range 99 To 99 C Integers Worksheet Integers Math Integers
Source : pinterest.com | 528 | 2,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-45 | latest | en | 0.846855 |
http://research.stlouisfed.org/fred2/graph/?chart_type=line&s[1][id]=PLOGDPBWA621NUPN&s[1][transformation]=chg | 1,429,625,357,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246641468.77/warc/CC-MAIN-20150417045721-00132-ip-10-235-10-82.ec2.internal.warc.gz | 221,194,360 | 18,355 | # Graph: Price Level of GDP, G-K method for Botswana
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
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(a) Price Level of GDP, G-K method for Botswana, U.S.=100, Not Seasonally Adjusted (PLOGDPBWA621NUPN)
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: p
Price Level of GDP, G-K method for Botswana
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
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``` University of Pennsylvania, Price Level of GDP, G-K method for Botswana [PLOGDPBWA621NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/PLOGDPBWA621NUPN/, April 21, 2015. ```
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Name: Email: | 633 | 2,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2015-18 | latest | en | 0.825216 |
http://mathhelpforum.com/discrete-math/192471-positive-sets-2.html | 1,475,211,384,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662022.71/warc/CC-MAIN-20160924173742-00000-ip-10-143-35-109.ec2.internal.warc.gz | 174,846,751 | 10,698 | 1. ## Re: Positive sets
Since "positive sets" is not a standard term, we are unable to give you the definition. I made my guess about its meaning in post #13.
Most likely, to solve these problems you need various laws about set operations, like distributivity of intersection over union and vice versa and De Morgan's laws, as well as the inclusion–exclusion principle.
2. ## Re: Positive sets
Originally Posted by ehpoc
I missed the class where he introduced positive sets. I started this thread so someone could confirm what exactly is meant by positive sets so I could educate myself on them.
The first thing I'd do is look online to see if the definition is a common one (and quickly discover it is not).
The second thing I'd do is check your book's index and look through class notes.
The third thing I'd do, if you don't have a friend in the class, is ask the professor by email: "Can you define positive set? I missed that class, and can't find the definition in the book or online."
This is really something you should jump on quickly, as soon as you realize you don't have what you need to complete an assignment.
(Also, I finally realize why I didn't gather what "+" was supposed to stand for...the other poster didn't put up the cardinality notation, so "+" was addition on natural numbers, not a set operation. Derp.)
3. ## Re: Positive sets
the definition given in class is "Given set "A,B,C,...." A positive set is the intersection of any sub-collection of A, B, C, ... without using complements or universal set"
good luck
Page 2 of 2 First 12 | 360 | 1,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-40 | longest | en | 0.971041 |
https://oeis.org/A112918/internal | 1,725,945,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00418.warc.gz | 404,494,455 | 3,308 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A112918 Number of nonisomorphic connected H-graphs H(n:i,j;k,m) on 6n vertices (or nodes) for 1<=i,j,k,m
%I #10 May 22 2024 15:14:40
%S 1,1,4,5,7,12,18,27,24,69,41,70,111,103,87,202,115,275,268,284,201,
%T 583,313,482,459,708,403,1347
%N Number of nonisomorphic connected H-graphs H(n:i,j;k,m) on 6n vertices (or nodes) for 1<=i,j,k,m<n/2.
%C An H-graph H(n:i,j;k,m) has 6n vertices arranged in six segments of n vertices. Let the vertices be v_{x,y} for x=0,1,2,3,4,5 and y in the integers modulo n. The edges are v_{0,y}v_{1,y}, v_{0,y}v_{2,y}, v_{0,y}v_{3,y}, v_{1,y}v_{4,y}, v_{1,y}v_{5,y} (inner edges) and v_{2,y}v_{2,y+i}, v_{3,y}v_{3,y+j}, v_{4,y}v_{3,y+k}, v_{5,y}v_{5,y+m} (outer edges) where y=0,1,...,n-1 and subscript addition is performed modulo n. H-graph H(n:i,j;k,m) is connected if and only if gcd(n,i,j,k,m) = 1.
%D I. Z. Bouwer, W. W. Chernoff, B. Monson, and Z. Starr (Editors), "Foster's Census", Charles Babbage Research Centre, Winnipeg, 1988.
%H J. D. Horton and I. Z. Bouwer, <a href="https://doi.org/10.1016/0095-8956(91)90057-Q">Symmetric Y-graphs and H-graphs</a>, J. Comb. Theory B 53 (1991) 114-129.
%e The only connected symmetric H-graphs are H(17:1,4;2,8) and H(34:1,13;9,15) which are also listed in Foster's Census.
%Y Cf. A112917, A112919, A112920.
%K nonn,more
%O 3,3
%A Marko Boben (Marko.Boben(AT)fmf.uni-lj.si), _Tomaz Pisanski_ and Arjana Zitnik (Arjana.Zitnik(AT)fmf.uni-lj.si), Oct 06 2005
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Last modified September 10 01:04 EDT 2024. Contains 375769 sequences. (Running on oeis4.) | 727 | 1,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.713552 |
http://www.dbforums.com/showthread.php?977134-Calculations-between-currency-and-number | 1,511,516,301,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807344.89/warc/CC-MAIN-20171124085059-20171124105059-00541.warc.gz | 378,533,396 | 17,404 | # Thread: Calculations between currency and number??
1. Registered User
Join Date
Nov 2003
Posts
150
## Unanswered: Calculations between currency and number??
Hi,
I need to perform some calculations between an item price ( currency format ) and a discount. Can the format of discount be 'general number' and be able to perform the correct calculations?? The formula i want to use is this:
extendedprice = itemprice * quantity * (1 - discount )
i think so, but i want to make sure....
thanks!
2. Grand Poobah
Join Date
Sep 2003
Location
MI
Posts
3,713
Yes. "General Number" doesn't mean much ... What is the type of the discount?
3. Registered User
Join Date
Nov 2003
Posts
150
I would like to use a %.
The user would enter 5 for a 5% discount. So far, i've built the form and put the textbox for curExtendedPrice in the footer with the Bafore_Update event filled with
=tblOrderDetails!curUnitRetail-((100-tblOrderDetails!intDiscount)/100)*tblOrderDetails!intQuantity
So far, it doesnt work, i get #Name? in the curExtendedPrice textbox
Thanks for the help!
4. Registered User
Join Date
Nov 2003
Posts
150
Sorry, i realise that what i meant to say is that i want to use an integer, not a %........
5. Registered User
Join Date
Nov 2003
Posts
150
I think i got it.....
Instead of using a table for the source data of the form, i'm using this query:
SELECT tblOrderDetails.intOrderID, tblProductInfo.intProductID, tblProductInfo.strProductName, tblProductInfo.curUnitRetail, tblOrderDetails.intQuantity, tblOrderDetails.intDiscount, CCur(tblProductInfo.curUnitRetail*[intQuantity]*(100-[intDiscount])/100) AS curExtendedPrice
FROM tblProductInfo, tblOrderDetails
ORDER BY tblOrderDetails.OrderID;
Now, the only thing i'm missing is a way to only display the records of 1 orderID at a time, preferably, the orderID displayed in the main form
Orderdetails is a subform on the Orders form.
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• | 521 | 2,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-47 | latest | en | 0.805352 |
https://morningstarptc.info/qa/has-there-always-been-538-electoral-votes.html | 1,620,514,560,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00581.warc.gz | 447,617,514 | 8,435 | Has There Always Been 538 Electoral Votes?
What does 538 mean?
538 may refer to: 538, the year in Western (Gregorian) calendar.
The number of electors in the United States Electoral College.
FiveThirtyEight, American political and statistical analysis blog.
What happens if you don’t get 270 electoral votes?
A candidate must receive an absolute majority of electoral votes (currently 270) to win the presidency or the vice presidency. If no candidate receives a majority in the election for president or vice president, that election is determined via a contingency procedure established by the 12th Amendment.
When citizens cast their ballots for president in the popular vote, they elect a slate of electors. Electors then cast the votes that decide who becomes president of the United States. Usually, electoral votes align with the popular vote in an election.
Roosevelt went on to win the greatest electoral landslide since the rise of hegemonic control between the Democratic and Republican parties in the 1850s. Roosevelt took 60.8% of the popular vote, while Landon won 36.5% and Lemke won just under 2%.
What was the closest election ever?
The 1960 presidential election was the closest election since 1916, and this closeness can be explained by a number of factors.
Is the Gallup Poll reliable?
Poll analyst Nate Silver found that Gallup’s results were the least accurate of the 23 major polling firms Silver analyzed, having the highest incorrect average of being 7.2 points away from the final result.
In the Electoral College system, each state gets a certain number of electors based on its total number of representatives in Congress. Each elector casts one electoral vote following the general election; there are a total of 538 electoral votes. The candidate that gets more than half (270) wins the election.
Has anyone ever won all electoral votes?
In 1788 and 1792, George Washington won all the electoral votes running effectively unopposed, and in 1820, James Monroe, running unopposed, carried all twenty-three states in the union at that time (although one electoral vote was cast for John Quincy Adams and two electors died prior to casting votes).
How is electoral vote determined?
Electoral votes are allocated among the States based on the Census. Every State is allocated a number of votes equal to the number of senators and representatives in its U.S. Congressional delegation—two votes for its senators in the U.S. Senate plus a number of votes equal to the number of its Congressional districts.
What happens if President elect dies?
The rules of both major parties stipulate that if the apparent winner dies under such circumstances and his or her running mate is still able to assume the presidency, then the running mate is to become the President-elect with the electors being directed to vote for the former Vice Presidential nominee for President.
What state has every president won?
The Missouri bellwether is a political phenomenon that notes that the state of Missouri voted for the winner in all but one U.S. presidential election from 1904 to 2004 (the exception being 1956). | 638 | 3,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-21 | latest | en | 0.962455 |
https://tutorme.com/tutors/227950/interview/ | 1,620,962,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991737.39/warc/CC-MAIN-20210514025740-20210514055740-00524.warc.gz | 600,298,106 | 48,757 | Enable contrast version
# Tutor profile: Cole C.
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## Questions
### Subject:Chemistry
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How much $$CH_4$$ (methane) can I react with 50g $$O_2$$ in a combustion reaction? How much water is created? Keep 4 decimal places for calculations, and round to 4 significant figures for final answer.
Inactive
Cole C.
Great question! First, let's define what concepts we're talking about. Put simply, we're going to: - Define combustion - Balance the equation - Convert from grams to moles and back to find the answers === Combustion === Combustion is when a hydrocarbon (made of C, H and O) reacts with oxygen to form $$CO_2$$ and $$H_2O$$, and heat. In our case, it would look like this: $$CH_4 + O_2 ==> CO_2 + H_2O$$ === Balancing === The question asks how much $$CH_4$$ can react with 50g of $$O_2$$. Now how do we find how much $$CH_4$$ and $$O_2$$ to use, or how much water and carbon dioxide are created? We need to balance the equation! Why? Atoms aren't created or destroyed very often, so we need to get out what we put in. If we look at the equation: $$CH_4 + O_2 ==> CO_2 + H_2O$$ We have on the left side (reactants) 1 C 4 H 2 O And on the right side (products) 1 C 3 O 2 H That doesn't add up! We need to balance the equation so what goes in, comes out. We start with the least common atom - C in this case. It's already balanced, so let's move on to H. To get more H on the right side, we double the $$coefficient$$ of the $$H_2O$$ $$CH_4 + O_2 ==> CO_2 + 2H_2O$$ Now on the left we have: 1 C 4 H 2 O And on the right side we have: 1 C 4 O 4 H We can double the amount of $$O_2$$ on the left side to complete the equation. $$CH_4 + 2O_2 ==> CO_2 + 2H_2O$$ Now on the left we have: 1 C 4 H 4 O And on the right side we have: 1 C 4 O 4 H Amazing! === Molar Mass === The ratios in the equation above are in terms of number of atoms, so how do we find the number of atoms from the mass? We use the molar mass! The molar mass is how many grams make up one mole. A mole is a number of atoms - $$6.02*10^{23}$$ atoms. It's just a number though, it could be $$6.02*10^{23}$$ of anything - even puppies! (sidenote, that many puppies would be 1/3 the weight of the earth! A mole is a massive number of atoms.) Why is this number important? It's a ratio between the weight of a single atom, in atomic mass units (amu) and the weight of a "reasonable" quantity of substance, measured in grams - where the weight in grams is the same number as the weight in amu! In other words, if one carbon atom is 12.0107 amu, Then $$6.02*10^{23}$$ carbon atoms weighs 12.0107 grams It's a weird looking number, but it's important for converting "reasonable" (easy to work with) weights of substances and finding the number of atoms involved, so we can figure out how many atoms react with how many other atoms. Then we can measure products and reactants in easy-to-use grams instead of counting out each individual atom in a reaction - that might take a while! Each $$molecule$$ of $$O_2$$ has two oxygen atoms, each weighing 15.9994 amu, or 31.9988 amu total. So, each $$mol$$ of $$O_2$$ weighs 31.9988g! If we take 50g of O_2, $$50g * \frac{1 mol O_2} {31.9988 g/mol} = 1.5626$$ mol $$O_2$$ Now we have our $$O_2$$ in terms of number of atoms, so it works in our equation. Amazing! $$CH_4 + 2O_2 ==> CO_2 + 2H_2O$$ How much $$CH_4$$ do we need? Mass $$O_2$$ -> moles $$O_2$$ -> moles $$CH_4$$ -> mass $$CH_4$$ The molar mass of $$CH_4$$ is $$12.0107g + 4(1.0079g) = 16.0423$$ g/mol $$CH_4$$ And its ratio with oxygen means we need: $$1.5626 mol O_2 * \frac{1 CH_4}{2 O_2} = 0.7813$$ mol $$CH_4$$ To react, so, $$0.7813 mol CH_4 * \frac{16.0423 g CH_4}{1 mol CH_4} = 12.5338 g CH_4$$ combusted. How much water is created? $$CH_4 + 2O_2 ==> CO_2 + 2H_2O$$ We need 2 $$H_2O$$ for every 2 $$O_2$$, And the molar mass of water is $$1.0079g * 2 + 15.9994g) = 18.0152$$ g/mol $$H_2O$$ So $$(1.5626 mol O_2)(\frac{2H_2O}{2O_2})(\frac{18.0152g H_2O}{1 mol}) = 28.1506 g H_2O$$ created. For our final, rounded answers: 12.53 g $$CH_4$$ combusted. 28.15 g $$H_2O$$ created.
### Subject:Calculus
TutorMe
Question:
Find the derivative of: $$f(x) = (e^{3x}+13x)^4$$
Inactive
Cole C.
Let's look at what we have! Our primary derivative rule is the chain rule. If we have any function, let's say $$f(x) = ax^n$$ then $$f'(x) = nax^{n-1}$$ Same rule applies to big functions made up of multiple parts, except we get this: $$f(x) = a[g(x)]^n$$ $$f'(x) = na[g(x)]^{n-1}[g'(x)]$$ Or in other words, we treat the 'inside function' [g(x)] as one big chunk, then multiply the 'outer equation' by the derivative, g'(x) when it's all said and done. Now, down to business. If we have: $$f(x) = (e^{3x}+13x)^4$$ Then $$g(x) = e^{3x}+13x$$ Since now we have two separate functions added together, we find the derivative of each individually If $$y = e^{ax}$$ Then $$y' = ae^{ax}$$ And if $$y = 13x$$ Then $$y' = 13$$ So we get $$g'(x) = 3e^{3x} + 13$$ Take a breath, we've solved the inside part. Now remember, $$f'(x) = na[g(x)]^{n-1}[g'(x)]$$ So we can just fill in the blanks in the equation we wrote earlier. $$f(x) = (e^{3x}+13x)^4$$ $$f'(x) = 4[(e^{3x}+13x)^3][3e^{3x} + 13]$$ And this is our derivative! Congrats!
### Subject:Physics
TutorMe
Question:
A 200 kg cannon is placed inside a 2000 kg, 20m long train boxcar. The cannon is placed 1m from the left end of the car with 40 cannonballs. It fires all of the 10 kg cannonballs at the other end of the traincar, where they come to rest against the wall. Once the cannon has finished firing: 1. How fast is the traincar moving? 2. How far has the traincar been displaced? 3. Where is the center of mass? Assume friction between the traincar and the ground is negligible, as is air resistance.
Inactive
Cole C.
First, let's look at the concepts we can possibly uncover in the question. Center of mass - we have a system with several different masses in relation to each other Relative coordinates - we have several masses moving in relation to each other Maybe kinetics and kinematics - we have moving objects Maybe dynamics - the cannonballs and the cannon exert forces We may also want to draw a picture of the scenario BEFORE =============== *** = cannonballs | === | === = cannon | ^ *** | ^ ------------------------- Cm = center of mass 00 00 | x | 0m Cm 20m And we know: $$m_c = 200 kg$$ $$m_b = 10 kg$$ $$m_t = 2000 kg$$ $$c_{c,x,i} = 1m$$ $$c_{b,x,i} = 1m$$ $$c_{t,x,i} = \frac{20m}{2} = 10m$$ 1. Now, how fast will the train car be moving after the collision? Since there's no friction, and we know the mass - we can assume an outside force will cause acceleration and increase the speed. Only problem is - there's no outside force. The cannon releases energy in shooting the cannonballs, but the kickback force from each shot is canceled by the equal and opposite impact when the balls hit the inside of the traincar. No energy in or out, and no force applied means there is no speed increase. We can assume the traincar is at rest once all the cannonballs have been shot. 2. Now, how far has the train car moved? Well, we know the center of mass of a group of objects must always stay the same in relation to an outside object. Or, put simply - if one mass moves, the rest have to shift the opposite way to keep the center of mass aligned where it originally was. If the balls shoot to the other side of the traincar, the traincar will have to roll back in the opposite direction to compensate. Equal and opposite reaction. What we can do is calculate the center of mass of the group before the shots are fired, then the center of mass when the cannonballs come to rest at the other end of the cart - then find the difference between the two centers of mass. So, working from what we know: $$c_{x,f} - c_{x,i}$$ gives us the distance the car has moved, if we use our handy equation $$c_x = \frac {\sum_{i=1}^{n} c_{xi}m_i} {\sum_{i=1}^{n} m_i }$$ Or, the sum of all the masses times their centers of mass, divided by the total mass. This gives us a weighted average (ha!) of where the center of mass is - think of it as the balance point where you could lift the traincar up and not have it tip to either side. $$c_{x,i} = \frac {(c_{c,x,i} m_c + c_{b,x,i} m_b + c_{t,x,i} m_t)}{m_c + m_b + m_t}$$ We can see we have all these values, so: $$c_{x,i} = \frac {(1m)(200 kg) + (1m)(40)(10 kg) + (10m)(2000 kg)} {200 kg + (40)(10 kg) + 2000 kg}$$ $$c_{x,i} = \frac {(1m)(200 kg) + (1m)(40)(10 kg) + (10m)(2000 kg)} {200 kg + (400 kg) + 2000 kg}$$ $$c_{x,i} = \frac {20600 kg*m} {2600 kg}$$ $$c_{x,i} = 7.92m$$ Now, for the center after the balls have been fired. In reference to the train, we know the balls have traveled to the righthand wall of the traincar at x = 20m. $$c_{b,x,f} = 20m$$ We know the cannon stays in the same spot, so $$c_{c,x,f} = c_{c,x,i} = 1m$$ Same for the train car, the center is still in the middle at $$c_{t,x,i} = c_{t,x,f} = 10m$$ So the equation for the center of mass will look like: $$c_{x,f} = \frac {(c_{c,x,f} m_c + c_{b,x,f} m_b + c_{t,x,f} m_t)}{m_c + m_b + m_t}$$ $$c_{x,f} = \frac {(1m)(200 kg) + (20m)(40)(10 kg) + (10m)(2000 kg)} {200 kg + (40)(10 kg) + 2000 kg}$$ $$c_{x,f} = \frac {28200 kg*m} {2600 kg}$$ $$c_{x,f} = 10.85m$$ Now, 10.85m - 7.92m = 2.93m, so the train car would have to shift 2.93m to the left when the cannonballs shift the center of mass 2.93m to the right. So the traincar rolls 2.93 meters as the balls are firing, then stops! 3. In order to keep the center of mass in the same place (since no outside forces are acting to move the whole system to a new place), the center of mass will be at the original location, x = 7.92m in our drawing. AFTER =============== | === | | ^ ***| ------------------------- 00 00 | x | 0m Cm 20m Here if you have any questions!
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http://excel.bigresource.com/How-to-produce-List-of-unique-values--98SnYc0f.html | 1,544,402,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823228.36/warc/CC-MAIN-20181209232026-20181210013526-00212.warc.gz | 104,838,138 | 14,282 | # How To Produce List Of Unique Values
Jan 21, 2012
I was trying to produce a list of unique values from a much larger list. I'm using this array function -
{=IFERROR(INDEX(Sheet2!\$B\$2:\$B\$100,MATCH(0,COUNTIF(\$B\$1:B1,Sheet2!\$B\$2:\$B\$100))), "")}
- which is marvellous and gives me exactly what I want. However my test book only has this dragged down 10 rows and already it takes 20-30 seconds to calculate. The real workbook is likely to stretch to several hundred rows, and to make matters that little more complex where I have Sheet2!\$B\$2:\$B\$100 this will relate to a completely separate book on a company network.
Is there any more efficient method of creating this list or will I just have to bear with the calculation speed?
## Excel 2010 :: Produce Unique List Of Key Values With Second Values Concatenated Together
Sep 15, 2013
I have some data with recurring key values and differing values in the second column, I need to produce a unique list of key values with the second values concatenated together.(See below)
The data can be 10 rows to 5000 and I can have anything from 5 to 150 sheets (Separate data sets), a macro would go a long way to keeping me sane.
Sample data Required Output
A | B Z
1| 10 | a 10,a,b,c
2| 10 | b 11,a
3| 10 | c 12,a,b
4| 11 | a
5| 12 | a
6| 12 | b
My system is Windows 8 Excel 2010.
## Combine Values From Two Columns To Produce Unique Value List
Jun 28, 2013
I have two columns of data and would like to generate a new list showing unique combinations of those values (see attached sheet).
unique list.xls
## Obtaining List Of Unique Values From List Based On Some Criteria?
May 28, 2014
The function below should return the value in a cell IF the string in the cell two cells to the left of it is "Nemo" Omitting the part highlighted in blue, my function returns a list of unique values...but when it's included, nothing is returned at all. Perhaps, there is an error in the syntax that I'm overlooking?
Function listUnique(rng As Range) As Variant
Dim row As Range
Dim elements() As String
[Code]....
## Formula For Obtaining List Of Unique Values But Not Including Values Which Contain IC
Aug 14, 2014
1. Using a formula, I am trying to to obtain a list of unique values (string) (caveat: see #2) from the range E2:E10000 (arbitrarily chose 10000 - the row number is variable)(see #3).
I currently have a formula that seems to work for this purpose but I don't know how to add the condition in #2 (below)
2. To include all unique string values except those starting with the letters "IC"
3. Is there a way to make this formula so that it can only seek values up to the last row, and not go to the 10000th row if not necessary? The E column has no empty cells until after the last row that contains data.
Here is the formula I currently use which serves #1 (above):
[Code] .....
Any way to improve/simplfy this formula for the purpose describbed in #1? How can I add the condition in #2? Can you see a way to include #3? The most important issue here is #2.
Example of desired results:
Column A | Column B
AA | AA
DD | CC
AA | DD
CC |
DD |
DD |
IC |
IC |
## Produce A List Of Text, After Selecting From Validate List
Jun 5, 2009
i want to do is, have a list that picks from a color and then beside it, it will show the possible fruit that is that color for example
Red
apple
cherry
strawberry
OR
Green
watermelon
grapes
The number of fruit will differ by the color so i know i need a counter there, but i am not sure how to list the fruit after i've picked the color
## Unique Values From List Of Many Duplicate Values?
Jan 31, 2014
how to get this with formula:
Column A - - Column B
a - - - - - - - a
a - - - - - - - b
b - - - - - - - c
c
c
## List Of Unique Values In One Cell?
Dec 27, 2013
I have a list of all transactions that happened during the day at different locations On another sheet I want to list all the locations and in another column to populate a list of unique transaction codes for each location from the first sheet (all unique values for each location in one cell). I am attaching a spreadsheet in case I am not clear enough (I need to create a Summary). This attachment is really a simplification of a file I work with. The real one has many columns and multiple subtotals, I would not be able to incorporate a method that involve filters
## Get Unique Values In A List Of Duplicates
Feb 28, 2007
I am confident the answer is no, but is there any formula / combination of formulae that will seek out unique values in a list of duplicates (just like filtering unique records only)?
I have a list of many duplicating sales people and I want to just create a column on another sheet that automatically sorts out the unique values (then I will sum their sales numbers with the SUMIF formula).
(By the way, the reason I just don't use the filter function is because the list always changes so I would need to keep applying it.)
## Create List Of Unique Values
Jul 23, 2007
In column A I have values starting at row 4, ranging up to row 1004.
in coulumn Q (starting at row 3) I'd like a list of unique entries from column A
I know this line is my problem
If Range("A" & I) Range("Q4:Q30") Then
but i'm not sure how to compare against all values in a range
Sub FilterSymbol()
Dim I As Integer
Dim X As Integer
X = Range("O2")
I = 4
Do
If Range("A" & I) Range("Q4:Q30") Then
Range("Q" & I) = Range("A" & I)
I = I + 1
Else: I = I + 1
End If
Loop Until I >= X
End Sub
## Populate List With Unique Values
Jan 5, 2008
I used this forum to find code to convert a matrix to a list. But I need to repeat then change one column - "Product".
The "Prodcut" vlaue needs to repeat unitl it changes.
I have attached a file with the code in module 1, and samples of where I am, and where I need to be. I think I am close!
## List Unique Values Above And Below Reference
Jan 26, 2008
Event Manager Status
1call A yellow
2sleep B yellow
3run C red
4jog D pink
5eat E pink
6call E yellow
7sleep F red
I need to output 4 unique Status before the event sleep. When I say unique it is compared to its previous status. So that would be [yellow,pink,red,yellow]. Notice that I could repeat yellow but not a sequential yellow. I cant use previous formula on listing unique values because I have reference point which is sleep, that i need to lookup from a huge pile of data, and i need a counter to move up. I tried vlookup with countif but i still have to define the max range. The data is huge and my menial skills are limited to excel.
## Make A List Of Unique Values Using 3 Different Columns?
Feb 26, 2014
See attached file with a smaller sample of the 1667 row table of data I am working with. What I am looking to do is make a list on another sheet of each every line with a county and have the corresponding Township and range with it. So if I have a cell and I select Audubon county, I would like a list to populate that has the 12 lines of Audubon county in column A, and the Township in column B and the Range in column C.
Excel Rate Example.xlsx
## Unique Values In A List That Appear At Least 3 Times (2007)
Jan 16, 2010
=COUNT(
1/FREQUENCY(
MATCH(\$A\$1:\$A\$50,\$A\$1:\$A\$50,0)
),
ROW(\$A\$1:\$A\$50)-ROW(OFFSET(\$A\$1:\$A\$50,0,0,1,1))
+1))
I'm looking to tweak this formula to count the number of unique values that occur at least 3 times.
So, imagine that A1:A50 contain 50 last names from the phone book, 27 of which are unique according to the formula above.
Now, further imagine that only 5 of these unique 27 last names appear at least 3 times.
I'm looking for the formula that returns 5.
And (like always), for simplicity and sharing reasons, I'm hoping to find a single formula that occupies a single cell, and does not rely on additional columns of intermediate information.
## Finding 10 Smallest Unique Values In A List
Oct 2, 2008
I have been using the SMALL function to find 10 smallest values in a list, however, some of the values are the same, is there a way of only having the 10 smallest unique values within a list?
## Create Unique List Of Values Across 5 Sheets
Oct 13, 2011
I am trying to create a macro that will create a unique list of the values in the same column across 5 worksheets, "Question1", "Question2", etc, and place the list in a new worksheet, "Summary".
The values are in column A in all of the worksheets, and the number of rows in each worksheet will change each day. I need the unique list to be added to the sixth worksheet, "Summary".
Also, after the list is created on the Summary worksheet, I need a count of each field by worksheet and a total. This is how I currently have the Summary set up:
Part Type|Total|Question1|Question2|Question3|Question4|Question5
Name|=SUM(C2:F2)|=COUNTIF(Question1!\$A:\$A,A2)|
=COUNTIF(Question2!\$A:\$A,A2)|=COUNTIF(Question3!\$A:\$A,A2)|
=COUNTIF(Question4!\$A:\$A,A2)|=COUNTIF(Question5!\$A:\$A,A2)
## List Unique Values From A Named Range
Dec 4, 2007
I am trying to find a way to create a list of unique items from a named range. I have found a few solutions using filters for ranges that are contained in the one column but not named ranges that contain multiple rows and multiple columns.
Is there any way to copy all of the unique items from the named range to another location? Lets call the named range Table1 and it spans from B2 to J100.
## Create Unique List & Summarize All Associated Values
Jan 30, 2009
I have a list of sales events that ran at various venues, together with how many people attended and how much revenue was brought in by sales. This data is arranged by week. One venue will never have more than one event per week, but different venues may have sales events on the same week.
My table looks like:
In reality this list could be thousands of rows long.
“Event week” is the calendar week of the event. “Weeks open” shows for each event how many weeks that venue has been open (may or may not be useful!).
What I need to do is to be able to pick a number of weeks (preferably changeable by entering into a cell) and have a list produced of each venue that has run for at least that many weeks, and the sum of people and sales in those first X weeks. My idea of output, if the user input “2” as the number of weeks would therefore look something like:
Note that
1)only venues that have run at least 2 meetings are shown in the output
2)even if venue has run over 2 meetings, only people and sales figures from the first 2 are summed up and displayed.
In this case. Later on I need to produce a frequency chart/histogram grouping the results into categories, but I know how to do this as long as I can get to the step before – but wanted to mention it in case there are shortcuts.
Have been battling with this with pivots, database functions, array formulae etc. for a long time.
Attached is the spreadsheet with sample data. The red table is the part I am stuck on. The solution has to work in Excel 2003.
## Fastest Way To Compile List Of Unique Values
Jul 3, 2007
I'm trying to count the number of unique values from a list of over 8,000 records -- and it's very likely that nearly all of the records are unique. At the moment, I am keeping track of an array where I store every unique value I find, expanding the array and adding values as I go, and at the end return the size of the array to tell me how many unique values there are in the range. My method works, but it is very slow, even when I turn on manual calculation.
Here is what I have so far:
Function CountUniqueCases(inputRange As String)
Dim i, j As Integer
Dim cCaseID As Range
Set cCaseID = Worksheets("Results").Range(inputRange)
Dim uniqueCases() As String
Redim uniqueCases(1)
uniqueCases(1) = cCaseID.Cells(1).Value
Is there a faster way to do this? I was hoping there was an array search function built in to VBA, but apparently not. My first attempt at running this function returns about 7904 out of ~8000 as being unique.
## List Unique Values Of Multiple Columns
Mar 12, 2008
I need a UDF to make live easier. I have a huge excel file and I need a final function (I suppose a UDF) that lists all values from a range spanning over multiple columns and rows. I would not really like a macro, since it does not update when certain values are changed.
It's part of a bigger functionality, but I need this one, really bad, since it spans over 6000 rows as the search range, and 7000 as the search values. In short:
I got to list all the unique values. They are in specific cells on worksheet 1 (A1:C5 and A25:C31) and should be listed on worksheet2 starting in field A2. As 'simple' as that. Nothing more, nothing less, just list all unique values of a certain range, which I specify. If it can not be done, by selecting the two fields separately, it's no problem at all to list all values in eg A1:C6000. Do note that some of the cells will be blank, and I don't need blanks. I only need the actual data sets. I added a sample excel file, that shows a representation of the data. I can not upload the actual file, since it's way too big (60 MB). So a sample file should give you the idea.
Three sheets:
Data 1: some value lists multiple columns
Data 2: some value lists, again multiple columns
Desired result: the result as it should be listed.
example.xls. zip
## Produce Lists From Master List
Oct 24, 2011
I have a set of unique customer names in a column A, say A1 to A200 in my data sheet. These customers have route numbers assigned to them (defined by the client) column B - B1 to B200. The route no.s are 1,2,3,4,5,6,7, & 8. The column lists can be of varying lengths but no more than 500 rows. I want to relate the routes to 8 separate sheets in the workbook. In each sheet I want to extract a column of customers based on the route number. So if the first customer in column A is route '1' (column B) it populates the first entry in sheet 'Route 1' with the customer name as so on for the rest of the sheets.
## Create Unique List Of Values From Two Ranges With Criteria?
Jun 12, 2014
I have two lists of values in separate columns with condition in adjacent cells, all on the same sheet
List 1 consists of numbers from 1 to 12, column to the right is either blank or Free
List 2 consists of numbers from 13 to 24, column to the right is either blank of Free
List 1 starts in A2, B2 is either blank or Free
List 2 starts in D2, E2 is either blank or Free
In column G, from G2 i want it to list all values from List 1 and List 2 where their status is Free
## List Unique Values In Excel And Count Occurrences Of Each?
May 25, 2014
I realise this is beyond me and wonder if it is possible to list uniqure values in excel and place a count occurrences of each. I have a hugh list of over 400 alpha code i.e ABCD, ABCD, ABCE, BDCE, BDCF ZXYE etc etc. What I'd like to do is pick out all the unique values so I can see what alpha codes are on the list. But then I would also like to count the occurrences of each value/name so I know how many times they appeared on the list.
Col A
ABCD
ABCD
ABCE
BDCE
BDCF
ZXYE
ZXYE
YEZY
CDEF
Col B is where i would like to list all the unquie vaules with a count e.g.
ABCD 2
ABCE 1
BDCE 1
BDCF 1
ZXYE 2
YEZY 1
CDEF 1
## Importing Values From Sheets To Create Unique List
Jun 2, 2009
I need to be able to pull information from 5 columns on 5 different pages, and pull all of the unique names out of it. From there, I need to create a list of all of these names. More often than not, this could only be three names, but I still would like this to be automatic because eventually it could be 50 or 60, and manually editing this list would get tedious. Lastly, is it possible to have a list create cells? For example: I have information on row 4, this one will start on row 6. My next group of information starts on row 14. That leaves 7 rows for information. I don't want to leave a large gap at the top of the page, so I'd like to be able to create cells if the list I'm trying to create has more information on it. Seems plausible in my mind, but I'm still new to excel.
## Create Unique Values List While Overwriting Old Data?
Jan 24, 2012
I will copy a report into sheet 1, starting in the same place(A1), on a weekly basis. The headers are: Item, Quantity & Price
Week 1's report shows the following:
Windows 5 \$1.00
Doors 6 \$1.50
Chairs 7 \$2.00
Week 2's report shows the following:
Windows 3 \$1.00 (change in quantity)
Tables 7 \$5.00 (new item)
Chairs 7 \$2.50 (change in price)
I need to have a rolling, unique list on Sheet 2 that will do the following:
- keep old line items while checking to see if the quantity or price has changed.
Therefore, after week 2's report is pasted on sheet 1, Sheet 2 should look like this:
Windows 3 \$1.00
Doors 6 \$1.50
Chairs 7 \$2.50
Tables 7 \$5.00
Here's the most important part: There will be formulas in Column D in Sheet 2 that I need to stay with each item. generating the desired result on Sheet 2!
## Create 1 List Of Unique Values From 12 Columns Of Data
Apr 28, 2014
I have a spreadsheet of 12 columns (A to L), with a list of student names under each down to row 31. Each column represents an after-school club that the child can attend.
However, a child can sign up to multiple clubs, so their name can appear in several places across the spreadsheet.
What I would like to do is have a 13th column which is an overall list of students who have signed up to at least 1 club, i.e. their name appears at least once in columns A to L.
How do I achieve this so that only unique entries are listed and duplicates are removed in this 13th column?
## How To Count The Number Of Unique Values In A List, Based On Another Value
Jul 27, 2009
I need to count the unique values in a list based on another value. i.e.
AB ___ Jones
AB ___ Smith
BC ___ Henry
AB ___ Jones
BC ___ Brown
BC ___ Henry
AB ___ Tomkins
For all the AB's in column 1, I need a formula that will give me all the unique names from column 2, in this case 3. I've tried adapting:
=IF(LEN(June!D5:D100),SUMPRODUCT(1/COUNTIF(June!D5:D100,June!D5:D100)))
but haven't been able to get it to work.
## Produce A List Of Combinations From A Range Of Numbers?
May 9, 2014
Am looking for an easy way to produce a list of combinations (maximum 6 numbers) from a range of numbers listed in 6 different columns:
Example
Column A contains : 1,2,3
Column B contains : 7,8,11,15
Column C contains : 12,16,18,19
Column D contains : 17,30,31
Column E contains : 30,31,32,33,34
Column F contains : 37,39,40
The rules are:
In each combination of 6 numbers, numbers should always be taken from ALL 6 columns. In each combination, numbers cannot repeat.
show me a formula to arrive the results.
## Macro To Produce List From Specific Range
Dec 2, 2009
I have a workbook with 2 worksheets, Sheet1 & Sheet2.
Sheet1 has a list of products on it, with a code, pack size and cost.
Sheet2 has got 8 columns in it.
The first column will be A , B , C
The second will be D, E, F
Third will be G, H, I
Going right through to column 8 which will be V, W X
In the second column in each on I will be have a (sell by date) So B, E, H, through to W will have sell by dates in them.
What I am looking for is a macro that will give me a list of short sell by dates when I click the button. So if today’s date is 02/12/09 when the button is clicked it will give me a list of all the product that have lets say a month sell by date left, I would like to be able to change the month to anything I want, just in case I want to do a longer check on the sell by date.
When it finds a short date I would like it to check against the code, so in the first column the code will be in A, it will match that code up against the code in Sheet1 and give me a list of the Code, and the Product from sheet1 and the sell by date that’s in Sheet2. The problem I have is when data is put into the second column on sheet2 which is D.E.F, I would want it just to check that column and not the A,B,C from column 1.
## Collecting Unique Values Of Multiple Ranges Into One Master List
Dec 20, 2012
I gather two lists of serial numbers (i.e.) from two different sources; one populated in sheet1, A:A (with variable row numbers) and the other in sheet2 B:B (again with varying number of rows)
I am trying to gather a master list of unique serial numbers in sheet3, C:C.
So far I have tried advanced filter, union.range, etc. but couldn't figure work it. I guess some of these columns need to have names assigned to it and have a header. | 5,381 | 20,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-51 | latest | en | 0.857915 |
http://mathoverflow.net/questions/tagged/teaching | 1,469,392,484,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824146.3/warc/CC-MAIN-20160723071024-00305-ip-10-185-27-174.ec2.internal.warc.gz | 153,988,086 | 25,709 | # Tagged Questions
For questions related to teaching mathematics. For questions in Mathematics Education as a scientific discipline there is also the tag mathematics-education. Note you may also ask your question on http://matheducators.stackexchange.com/.
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### Can the unsolvability of quintics be seen in the geometry of the icosahedron?
Q1. Is it possible to somehow "see" the unsolvability of quintic polynomials in the $A_5$ symmetries of the icosahedron (or dodecahedron)? Perhaps this is too vague a question. Q2. Are there ... | 2,151 | 9,488 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-30 | latest | en | 0.941943 |
https://snippets.cacher.io/snippet/d2a480f0809d062125b9 | 1,701,345,097,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00682.warc.gz | 610,400,761 | 4,790 | zjplab
2/6/2017 - 1:53 AM
## Longest Increasing Subsequence
Longest Increasing Subsequence
``````def LIS(arr,n):
lis=[1]*len(arr)
#initialize
for x in lis:
x=1
#compute optimal
for i in range(1,n):
for j in range(0,i-1):
if arr[i]>arr[j] and \
(lis[i]<lis[j]+1):
lis[i]=lis[j]+1
#pick maximum of all LIS values
return max( lis ) ``````
``````/* Dynamic Programming C/C++ implementation of LIS problem */
#include<stdio.h>
#include<stdlib.h>
/* lis() returns the length of the longest increasing
subsequence in arr[] of size n */
int lis( int arr[], int n )
{
int *lis, i, j, max = 0;
lis = (int*) malloc ( sizeof( int ) * n );
/* Initialize LIS values for all indexes */
for (i = 0; i < n; i++ )
lis[i] = 1;
/* Compute optimized LIS values in bottom up manner */
for (i = 1; i < n; i++ )
for (j = 0; j < i; j++ )
if ( arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Pick maximum of all LIS values */
for (i = 0; i < n; i++ )
if (max < lis[i])
max = lis[i];
/* Free memory to avoid memory leak */
free(lis);
return max;
}
/* Driver program to test above function */
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr)/sizeof(arr[0]);
printf("Length of lis is %d\n", lis( arr, n ) );
return 0;
}``````
``````def LIS(arr,n):
if n==1:#base case
return 1
curr_max=1
for i in range(0,n-1):
subproblem=LIS(arr,i)
if arr[i]<arr[n-1] and \
curr_max < (subproblem +1 ):
curr_max=1 + subproblem
return curr_max
# Driver program to test the functions above
def main():
arr = [10, 22, 9, 33, 21, 50, 41, 60]
n = len(arr)
print ("Length of LIS is", LIS(arr, n))
if __name__=="__main__":
main()
``````
``````
int LIS(int arr[],int n){
//base case
if(n==1) return 1;
//otherwise
int curr_max=1;
for(int i=0;i<n-1;++i){ // i< n-1 because n>=2
while(arr[i]<arr[n-1] ){
int subproblem=LIS(arr,i);
if( (subproblem+1 >curr_max) )
curr_max=1+subproblem;
}//while ends
}//for ends
return curr_max;
}`````` | 680 | 1,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.342738 |
https://noticing.nysci.org/lesson/cg-subways-lesson-1/ | 1,685,863,862,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649518.12/warc/CC-MAIN-20230604061300-20230604091300-00164.warc.gz | 485,992,086 | 14,060 | September 7th, 2016
This is the starter map. Load this file onto student iPads as it is the correct aspect ratio. Students will see the map image in Choreo Graph and then create a subway map with this as the background. One unit on the grid in Choreo Graph = ¼ mile = .25 mile
#### Criteria for Mapmaking
For this lesson, students will use two trains to connect all the stations on the map. They are instructed to make sure that passengers could travel from any station to any other station on the subway lines. (Note: Transferring lines to get around the city is acceptable.)
The student worksheet has clear instructions for making a map. Included below is a series of screenshots that shows the progression to make a map in Choreo Graph.
If some of the maps made by students look crazy and unrealistic, those maps will actually work well as the students think about efficiency and making better maps in the following lessons.
#### Divergent Solutions
An important aspect of NYSCI’s Noticing Tools is that they invite and celebrate divergent solutions. There is not one correct answer, any map design that meets the basic minimum criteria will suffice: 2 train lines, you can get from any station to any other station. As students connect the coordinates on their maps, the maps should vary. Students will have ownership over their designs and will use this map in the following lessons.
#### What you need to get started:
• Set of iPads with the Choreo Graph app
• student sheets
1 Class period
#### Collaboration and Group Work
These lessons are designed for students to work individually, in pairs, or in groups. Each student should do all the work on their own sheets, and the iPad should be shared across group members as equally as possible.
We suggest that groups be no larger than four students. Four or more students in a group will require extra attention to make sure that every group member is contributing equally.
#### Lesson Plan
Check out the series of images below showing how to create a map in Choreo Graph. The student worksheets offer detailed instructions for creating maps, be prepared to help them get started. If you have a projector in your classroom it might be helpful to show these images and talk through the steps to create a map.
Students will also add trains to their maps. They are invited to draw their own, or add color to these trains before snapping pics and tracing them in Choreo Graph.
The map that students create will look something like this map. Encourage them to be creative as they design their own system of lines around the city!
#### Getting Started
Here is a step by step series of images showcasing how to make a map in Choreo Graph:
Share the file of the map image to the student iPads. In Choreo Graph, tap the camera icon, and then the images button ( ) and bring the map image into Choreo Graph. Once you see the map on the bottom croll, hold your finger on the image, choose “Set Background.” After the map is in place, add the trains by snapping a pic and tracing. Students can add color to their trains.
Map and trains have been added. You are ready to tap Animate to start drawing subway lines. NOTE: On the Build Screen you can re-size the objects by pinching with two fingers. Size the trains to about the size of a nickel, but not too small so as to make them difficult to move around.
Tap the wrench icon, and toggle on Translation and Grid. This will result in visualizations of the coordinates and the translation line segments.
Drag the trains to the starting points. You may choose any station to be a starting point.
Tap the second keyframe and drag the trans to the next station, or leave the train at that station for as long as needed. You can go back to the first keyframe and tap the play button to test your first animation.
Tap the second keyframe, drag the train to the next station. Notice that it’s ok for two trains to be at the same station at the same time. TEACHER NOTE: Choreo Graph is also designed for rotations, and students might rotate trains to traverse their maps. Make sure they don’t spend too much time on rotations at this point in the lesson as it is not the focus on the lesson, however if setting rotations doesn’t cause them to fall behind then it’s probably not a problem. Use discretion.
As you continue to add lines, be sure to connect every station in the city, and make sure the two trains also visit the same station at least once so that passengers will be able to transfer lines.
Trains either need to loop back to the starting point, or reverse directions and get back to the beginning. There are plenty of keyframes for the trains to make complete circuits, to get back to the starting point.
Once students make their map they are instructed to fill in a table of the coordinates for each station as well as the landmarks. This will be very helpful in the lessons to come. Students are ready for lesson two once their map is complete and they’ve recorded all important coordinates.
#### Sharing
1. With each lesson, spend some time allowing the students to share their work with the class. In this lesson, students can share their maps, and everyone will see the divergent solutions across all the examples.
2. You might also invite students to start noticing the mathematics in their maps. Examples: 1) Do you see things here that we have been studying? 2) How would you find the slope? 3) Does your map have any parallel or perpendicular lines?
#### Wrapup
1. If necessary, have students label their iPads so they will be able to return to them for the next lessons.
Apps used
Duration: 0-20 mins
Prep: Easy
#### Big Idea
In this introductory lesson students will create a subway map in Choreo Graph. In the starter map below there are several subway stations, it is their task to connect all of them using two different subway lines. Using the student worksheet, they are given specific instructions to make their maps. They will also make notes of the coordinates for each station and landmark which will be useful in the following lessons.
#### Learning Objectives
Activity: Getting started by connecting all the stations and creating the first subway map to be used in the lessons to come.
Objective: SWBAT to explore Choreo Graph’s interface while solving a real-world problem
#### Device Strategies
If necessary, have students label their iPads so they will be able to return to them for the next lessons.
These lessons are designed for students to work individually, in pairs, or in groups. Each student should do all the work on their own sheets, and the iPad should be shared across group members as equally as possible.
We suggest that groups be no larger than four students. Four or more students in a group will require extra attention to make sure that every group member is contributing equally. | 1,482 | 6,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-23 | latest | en | 0.946655 |
http://heli-air.net/2016/03/30/fuselage-aerodynamic-characteristics/ | 1,591,345,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00101.warc.gz | 59,077,701 | 16,248 | # Fuselage aerodynamic characteristics
Chapter 3 developed a generalized form for the aerodynamic forces and moments acting on the fuselage; Table 4B.4 presents a set of values of force and moment coefficients, giving one-dimensional, piecewise linear variations with incidence and sideslip. These values have been found to reflect the characteristics of a wide range of fuselage shapes; they are used in Helisim to represent the large angle approximations.
Small angle approximations (-20° < (af, вf) < 20°) for the fuselage aerodynamics of the Lynx, Bo105 and Puma helicopters, based on wind tunnel measurements, are given in eqns 4B.1 – 4B.15. The forces and moments (in N, N/rad, N m, etc.) are given as functions of incidence and sideslip at a speed of 30.48 m/s (100 ft/s). The increased order of the polynomial approximations for the Bo105 and Puma is based on more extensive curve fitting applied to the wind tunnel test data. The small angle approximations should be fared into the large angle piecewise forms.
Lynx
,0 = -1112.06 + 3113.75 a2f (4B.1) Xf100 = —8896.44 в f (4B.2) Zf100 = —4225.81 a f (4B.3) Mf100 = 10 168.65 a f (4B.4) Nf100 = -10168.65 ef (4B.5)
Table 4B.4 Generalized fuselage aerodynamic coefficients
a f — 180 — 160 —90 —30 0 20 90 160 180 Cxf 0.1 0.08 0.0 —0.07 —0.08 —0.07 0.0 0.08 0.1 a f — 180 — 160 — 120 —60 —20 0 20 60 120 160 180 Czf 0.0 0.15 1.3 1.3 0.15 0.0 —0.15 —1.3 —1.3 —0.15 0.0 a f —205 — 160 — 130 —60 —25 25 60 130 155 200 Cmf 0.02 —0.03 0.1 0.1 —0.04 0.02 —0.1 —0.1 0.02 —0.03 в f —90 —70 —25 0 25 70 90 Cnfa —0.1 —0.1 0.005 0.0 —0.005 0.1 0.1 в f —90 —60 0 60 90 Cnfb —0.1 —0.1 0.0 0.1 0.1
Bo105
Xf loo = -580.6 – 454.0 af + 6.2 a2 + 4648.9 a3 (4B.6)
Yfloo = -6.9 – 2399.0 вf – 1.7 в] + 12.7 в] (4B.7)
Zf100 = -51.1 – 1202.0 af + 1515.7 a2f – 604.2a3 (4B.8)
Mf100 = -1191.8 + 12752.0 af + 8201.3 a2f – 5796.7 a3 (4B.9)
Nf100 = -10028.0 ву (4B.10)
Puma
Xf100 = -822.9 + 44.5 a f + 911.9 a2f + 1663.6 a3 (4B.11)
Yf100 = -11 672.0 в f (4B.12)
Zf100 = -458.2 – 5693.7 af + 2077.3 a2f – 3958.9 af (4B.13)
Mf100 = -1065.7 + 8745.0af + 12473.5af – 10033.0af (4B.14)
Nf100 = -24 269.2 вf + 97 619.0 в} (4B.15)
Empennage aerodynamic characteristics
Following the convention and notation used in Chapter 3, small angle approximations (-20° < (atp, вfn) < 20°) for the vertical tailplane and horizontal fin (normal) aerodynamic force coefficients are given by the following equations. As for the fuselage forces, the Puma approximations have been curve fitted to greater fidelity over the range of small incidence and sideslip angles.
Lynx
Cztp = -3.5 atp (4B.16)
Cyfn = -3.5 вп (4B.17)
Bo105
Cztp = -3.262 atp (4B.18)
Cyfn = -2.704 вп (4B.19)
Puma
Cztp = -3.7 (atp – 3.92ajP) (4B.20)
4B.2 Stability and control derivatives
The stability and control derivatives predicted by Helisim for the three subject helicopters are shown in Figs 4B.7-4B.13 as functions of forward speed. The flight conditions correspond to sea level (p = 1.227 kg/m3) with zero sideslip and turn rate, from hover to 140 knots. Figures 4B.7 and 4B.8 show the direct longitudinal and lateral derivatives respectively. Figures 4B.9 and 4B.10 show the lateral to longitudinal, and longitudinal to lateral, coupling derivatives respectively. Figures 4B.11 and 4B.12 illustrate the longitudinal and lateral main rotor control derivatives, and Fig. 4B.13 shows the tail rotor control derivatives. As an aid to interpreting the derivative charts, the following points should be noted:
(1) The force derivatives are normalized by aircraft mass, and the moment derivatives are normalized by the moments of inertia.
(2) For the moment derivatives, pre-multiplication by the inertia matrix has been carried out so that the derivatives shown include the effects of the product of inertia Ixz (see eqns 4.47-4.51).
(3) The derivative units are as follows:
force/translational velocity e. g. Xu 1/s force/angular velocity e-g – Xq m/s. rad moment/translational velocity e. g. Mu rad/s. m moment/angular velocity e. g. Mq 1/s force/control e. g. X00 m/s2 rad moment/control e. g. M0o 1/s2
(4) the force/angular velocity derivatives, as shown in the figures, include the trim velocities, e. g.,
7 = 7 U
— Zv q + ‘-d e
*u -0.01 -0.02! -0.03 -0.04 -0.05 -0.06 3 I 0 20 40 60 80 100 120 1 Speed (Knots) 0.05 0 z u ; &V -0.05 -0.1 -0.15 -0.2 г** ~r. у / V -0.25 0 20 40 60 80 100 120 140 Speed (Knots) r 0.12 r M u s. 0.08 0.06 s. 4 0.04* 0.02 N c=. 0 0 20 40 60 80 100 120 140 Speed (Knots) ;*Lynx ♦Puma *Bo105[
Speed (Knots)
60 40 20 -20 20 40 60 80 100 120 140 Speed (Knots)
О 20 40 60 80 100 120 140 Speed (Knots)
[ ■ Lynx ♦ Puma ♦ Bo105 |
0 20 40 60 80 100 120 140 Speed (Knots)
«Lynx ♦Puma -*-Bo105
Speed (Knots)
Speed (Knots)
0 20 40 60 80 100 120 140 Speed (Knots)
x106 M Г 0.8 0.6 0.4 0.2 • у Л / V t r / j V T V . ( -0.2 ( ** 0 20 40 60 80 100 120 1 Speed (Knots) И Lynx *-Puma *Bo105 |
: oo4 : 0 021 і o і : -0.02 Yu -0 04 20 40 60 80 100 120 140 Speed (Knots) Lu ■ X ‘ ‘ ‘ і „ . і 0 —— — 0 20 40 60 80 100 120 140 Speed (Knots)
Speed (Knots) *Lynx – e-Puma *-Bo105
Speed (Knots)
Speed (Knots)
60 40 4 IS ‘ 20 0 c zn 0 20 40 60 60 100 120 1 Speed (Knots) j a Lynx e-Puma *Bo105 |
5 04 % IS -10 c 20 40 60 80 100 120 140 Speed (Knots) | * Lynx ♦Puma *Bo105
x103 0 01 — — <0 C T —– -0 01 -0.02 -0 03 C V ) ■ ■> / ■N I r 20 40 60 80 100 Speed (Knots) 120 140
Speed (Knot3)
Mn
x1Cr
Speed (Knots) ■ Lynx ♦Puma *Bo105
Speed (Knots) I Lynx ♦ Puma * Bo105
Appendix 4C the Trim Orientation Problem
In this section we derive the relationship between the flight trim parameters and the velocities in the fuselage axes system, for use earlier in the chapter. Figure 4C. 1 shows the trim velocity vector Vfe of the aircraft with positive components along the fuselage axes directions x, y and z given by Ue, Ve and We respectively, where subscript e denotes equilibrium.
The trim condition is defined in terms of the trim velocity Vfe, the flight path angle Yfe, the sideslip angle Д. and the angular velocity about the vertical axis Qae. The latter plays no part in the translational velocity derivations. The incidence and sideslip angles are defined as
&e = tan-1 ^ W^ (4C.1)
Pe = sin-1 ((4C.2)
The sequence of discrete orientations required to derive the fuselage velocities in terms of the trim variables is shown below in Fig. 4C.2.
The axes are first rotated about the horizontal y-axis through the flight path angle. Next, the axes are rotated through the track angle, positive to port (corresponding to a positive sideslip angle) giving the orientation of the horizontal velocity component relative to the projected aircraft axes. The final two rotations about the Euler pitch and roll angles bring the axes into alignment with the aircraft axes as defined in Chapter 3, Section 3A.1.
Fig. 4C.1 Flight velocity vector relative to the fuselage axes in trim
Fig. 4C.2 Sequence of orientations from velocity vector to fuselage axes in trim: (a) rotation to horizontal through flight path angle yf; (b) rotation through track angle x; (c) rotation through Euler pitch angle 0; (d) rotation through Euler roll angle Ф
The trim velocity components in the fuselage-fixed axis system may then be written as
Ue = Vfe (cos @e cos Yfe cos xe — sin @e sin ye)
(4C.3)
Ve = Vfe (cos Фє cos Yfe sin xe + sin Фє (sin @e cos Yfe cos xe + cos @e sin Yfe))
(4C.4)
We = Vfe (—sin Фє cos Yfe sin xe + cos Фє (sin Qe cos Yfe cos xe + cos @e sin y^
(4C.5)
The track angle is related to the sideslip angle through eqn 4C.4 above. The track angle is then given by the solution of a quadratic, i. e.,
sin
where
and the к coefficients are
kX1 = sin Фє sin Qe cos Yfe (4C.9) kX2 = cos Фє cos Yfe (4C.10) kX3 = sin ee — sin Фє cos Qe sin Yfe (4C.11)
Only one of the solutions of eqn 4C.6 will be physically valid in a particular case.
Finally, the relationship between the fuselage angular velocities and the trim vertical rotation rate can be easily derived using the same transformation as for the gravitational forces. Hence,
Pe = -&ae sin Qe (4C.12)
Qe = &ae cos Qe sin Фе (4C.13)
Re = Q, ae cos Qe cos Фє (4C.14)
Note that for conventional level turns, the roll rate p is small and the pitch and yaw rates are the dominant components, with the ratio between the two dependent upon the trim bank angle. In trim at high climb or descent rates the pitch angle can be significantly different from zero, increasing the roll rate in the turn manoeuvre.
The German DLR variable stability (fly-by-wire/light) Bo105 S3
(Photograph courtesy of DLR Braunschweig) | 3,017 | 8,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-24 | latest | en | 0.699289 |
https://simplywall.st/stocks/au/retail/asx-jbh/jb-hi-fi-shares/news/does-this-valuation-of-jb-hi-fi-limited-asxjbh-imply-investo | 1,674,987,278,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00485.warc.gz | 561,209,633 | 42,850 | Stock Analysis
# Does This Valuation Of JB Hi-Fi Limited (ASX:JBH) Imply Investors Are Overpaying?
• Updated
How far off is JB Hi-Fi Limited (ASX:JBH) from its intrinsic value? Using the most recent financial data, we'll take a look at whether the stock is fairly priced by projecting its future cash flows and then discounting them to today's value. Our analysis will employ the Discounted Cash Flow (DCF) model. Don't get put off by the jargon, the math behind it is actually quite straightforward.
Remember though, that there are many ways to estimate a company's value, and a DCF is just one method. If you still have some burning questions about this type of valuation, take a look at the Simply Wall St analysis model.
View our latest analysis for JB Hi-Fi
### The calculation
We use what is known as a 2-stage model, which simply means we have two different periods of growth rates for the company's cash flows. Generally the first stage is higher growth, and the second stage is a lower growth phase. To start off with, we need to estimate the next ten years of cash flows. Where possible we use analyst estimates, but when these aren't available we extrapolate the previous free cash flow (FCF) from the last estimate or reported value. We assume companies with shrinking free cash flow will slow their rate of shrinkage, and that companies with growing free cash flow will see their growth rate slow, over this period. We do this to reflect that growth tends to slow more in the early years than it does in later years.
Generally we assume that a dollar today is more valuable than a dollar in the future, and so the sum of these future cash flows is then discounted to today's value:
#### 10-year free cash flow (FCF) estimate
2022 2023 2024 2025 2026 2027 2028 2029 2030 2031 Levered FCF (A\$, Millions) AU\$503.9m AU\$530.0m AU\$443.7m AU\$314.0m AU\$326.0m AU\$285.9m AU\$263.0m AU\$249.7m AU\$242.3m AU\$238.7m Growth Rate Estimate Source Analyst x4 Analyst x4 Analyst x3 Analyst x1 Analyst x1 Est @ -12.29% Est @ -8.03% Est @ -5.05% Est @ -2.96% Est @ -1.49% Present Value (A\$, Millions) Discounted @ 8.0% AU\$467 AU\$454 AU\$352 AU\$231 AU\$222 AU\$180 AU\$153 AU\$135 AU\$121 AU\$111
("Est" = FCF growth rate estimated by Simply Wall St)
Present Value of 10-year Cash Flow (PVCF) = AU\$2.4b
We now need to calculate the Terminal Value, which accounts for all the future cash flows after this ten year period. For a number of reasons a very conservative growth rate is used that cannot exceed that of a country's GDP growth. In this case we have used the 5-year average of the 10-year government bond yield (1.9%) to estimate future growth. In the same way as with the 10-year 'growth' period, we discount future cash flows to today's value, using a cost of equity of 8.0%.
Terminal Value (TV)= FCF2031 × (1 + g) ÷ (r – g) = AU\$239m× (1 + 1.9%) ÷ (8.0%– 1.9%) = AU\$4.0b
Present Value of Terminal Value (PVTV)= TV / (1 + r)10= AU\$4.0b÷ ( 1 + 8.0%)10= AU\$1.9b
The total value, or equity value, is then the sum of the present value of the future cash flows, which in this case is AU\$4.3b. To get the intrinsic value per share, we divide this by the total number of shares outstanding. Compared to the current share price of AU\$49.0, the company appears reasonably expensive at the time of writing. Valuations are imprecise instruments though, rather like a telescope - move a few degrees and end up in a different galaxy. Do keep this in mind.
### The assumptions
The calculation above is very dependent on two assumptions. The first is the discount rate and the other is the cash flows. If you don't agree with these result, have a go at the calculation yourself and play with the assumptions. The DCF also does not consider the possible cyclicality of an industry, or a company's future capital requirements, so it does not give a full picture of a company's potential performance. Given that we are looking at JB Hi-Fi as potential shareholders, the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC) which accounts for debt. In this calculation we've used 8.0%, which is based on a levered beta of 1.289. Beta is a measure of a stock's volatility, compared to the market as a whole. We get our beta from the industry average beta of globally comparable companies, with an imposed limit between 0.8 and 2.0, which is a reasonable range for a stable business.
### Moving On:
Valuation is only one side of the coin in terms of building your investment thesis, and it ideally won't be the sole piece of analysis you scrutinize for a company. DCF models are not the be-all and end-all of investment valuation. Rather it should be seen as a guide to "what assumptions need to be true for this stock to be under/overvalued?" For example, changes in the company's cost of equity or the risk free rate can significantly impact the valuation. Why is the intrinsic value lower than the current share price? For JB Hi-Fi, we've compiled three essential items you should assess:
1. Risks: Take risks, for example - JB Hi-Fi has 2 warning signs (and 1 which can't be ignored) we think you should know about.
2. Future Earnings: How does JBH's growth rate compare to its peers and the wider market? Dig deeper into the analyst consensus number for the upcoming years by interacting with our free analyst growth expectation chart.
3. Other Solid Businesses: Low debt, high returns on equity and good past performance are fundamental to a strong business. Why not explore our interactive list of stocks with solid business fundamentals to see if there are other companies you may not have considered!
PS. The Simply Wall St app conducts a discounted cash flow valuation for every stock on the ASX every day. If you want to find the calculation for other stocks just search here.
If you decide to trade JB Hi-Fi, use the lowest-cost* platform that is rated #1 Overall by Barron’s, Interactive Brokers. Trade stocks, options, futures, forex, bonds and funds on 135 markets, all from a single integrated account. Promoted
### Valuation is complex, but we're helping make it simple.
Find out whether JB Hi-Fi is potentially over or undervalued by checking out our comprehensive analysis, which includes fair value estimates, risks and warnings, dividends, insider transactions and financial health.
View the Free Analysis | 1,564 | 6,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.915869 |
https://math.stackexchange.com/questions/627783/why-mathbbz-is-not-isomorphic-to-mathbbr | 1,563,735,214,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527196.68/warc/CC-MAIN-20190721185027-20190721211027-00330.warc.gz | 464,722,181 | 37,498 | # Why $(\mathbb{Z}, + )$ is not isomorphic to $(\mathbb{R}^+, *)$?
Can someone explain to me why $(\mathbb{Z}, + )$ is not isomorphic to $(\mathbb{R}^+, *)$ where $*$ is multiplication.
My book says they aren't really isomorphic and doesn't say why. I thought that they are because of two reasons
1. All infinite groups with generator is isomorphic to $(\mathbb{Z}, + )$ and clearly $(\mathbb{R}^+, *)$ is infinite (edit, but no generator, okay)
2. The isomorphism is given by $f(x) = e^x$
• Your map is a homomorphism. Why is it not an isomorphism? (This isn't enough, but it is of dire importance that you check your logic). In response to 1., not all infinities are created equal. Any real analysis class should tell you this. – PVAL-inactive Jan 5 '14 at 8:57
• Not all infinite groups are isomorphic to $\Bbb{Z}$. For one that would mean that all infinite groups are abelian, which is definitely not true. – Nick D. Jan 5 '14 at 8:58
• All infinite cyclic groups are isomorphic to $(\Bbb Z,+)$. Not every infinite group is cyclic, however. – Stahl Jan 5 '14 at 8:58
• @NickD, I misquoted the theorem. There must be a generator for the group to be isomorphic to $\mathbb{Z}$. Clearly, the one I have has no generator. – Hawk Jan 5 '14 at 9:01
• You can't even embedded a $(\mathbb{Z},+)$ inside an arbitrary infinite group. For example, $C_{2}^{\mathbb{N}}$ – Gina Jan 5 '14 at 9:01
• All infinite groups are isomorphic to $(\mathbb{Z},+)$? So, what does the classification theorem for abelian groups say about a guy like $\mathbb{Z} \oplus\mathbb{Z}/2$?
• As for your isomorphism: which would be the inverse image of $e^{1/2}$?
Your first claim is false. In fact, there is not even a bijection between the sets $\Bbb{Z}$ and $\Bbb{R}^+$, by Cantor's diagonalization argument.
However, the map $x \mapsto e^x$ does provide an isomorphism between the groups $(\Bbb{R}, +)$ and $(\Bbb{R}^+, \times)$.
The other answers are good: $(\mathbb{Z}, + )$ is cyclic and countable, while $(\mathbb{R}^+, \times)$ is neither cyclic nor countable.
Here's another, less direct answer: $(\mathbb{R}^+, \times)$ is divisible, while $(\mathbb{Z}, + )$ is not divisible. | 674 | 2,162 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-30 | latest | en | 0.902322 |
http://algebralab.org/lessons/lesson.aspx?file=algebra_quad_solvefactoring.xml | 1,653,492,049,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00675.warc.gz | 4,338,510 | 6,204 | There are several ways to solve a quadratic equation. The method explored in this lesson is factoring. The type of factoring chosen to solve a quadratic equation depends upon the equation itself. Regardless of the factoring method chosen, you must start with the equation in standard form. Once it is in standard form, you can factor and use the factored form to solve the equation.
Example
#1: Solve First, the quadratic must be written in standard form . So begin by adding 12 to each side of the equation to obtain Next, factor the left side of the equation When factoring this type of trinomial, the signs in the two sets of parentheses must be alike since the sign of 12 is positive, and both of the signs must be negative since 7x is negative. Also, 4 and 3 must be the factors chosen for 12 since 4 and 3 add to give the 7 in the middle term. If you need additional help with factoring, click here for a factoring lesson. Now that the left side of the equation has been factored, we can use the Zero Product Property. This property says that if a product of two factors is equal to zero, then one or both of the factors must be zero. So we can use this property and set each factor equal to zero. Then solve each equation for x. What is your answer?
You might be wondering why when using the Zero Product Property we use the word “or” and when discussing the solutions we use the word “and”.
The Zero Product Property says that one or both factors can be equal to zero which is why we use the word “or”. This is simply saying that either one of those numbers will make the equation equal to zero.
When solving a quadratic equation, we are looking for any and all values that will make the equation true. That’s why we use “and” when talking about the solutions. Both of those values will make the equation true.
Just remember that when you are stating the solutions for the equation, you should use both values obtained from your factoring.
Examples
#2: Solve Make sure the equation is in standard form. It is, so proceed. Factor the left side using trinomial methods. Set each factor equal to 0 and solve for x. What is your answer?
#3: Solve Make sure the equation is in standard form. It is, so proceed. Factor the left side using the difference of two squares. Set each factor equal to 0 and solve for x. What is your answer?
S Taylor
Show Related AlgebraLab Documents | 529 | 2,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2022-21 | latest | en | 0.930849 |
https://us.metamath.org/ilegif/cvg1nlemres.html | 1,721,502,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00481.warc.gz | 515,082,748 | 32,424 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > cvg1nlemres Unicode version
Theorem cvg1nlemres 10788
Description: Lemma for cvg1n 10789. The original sequence has a limit (turns out it is the same as the limit of the modified sequence ). (Contributed by Jim Kingdon, 1-Aug-2021.)
Hypotheses
Ref Expression
cvg1n.f
cvg1n.c
cvg1n.cau
cvg1nlem.g
cvg1nlem.z
cvg1nlem.start
Assertion
Ref Expression
cvg1nlemres
Distinct variable groups: ,, ,, ,,, ,,, , ,,, ,,, , ,,,, ,, ,,
Allowed substitution hints: (,,) (,,) () (,)
Proof of Theorem cvg1nlemres
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 cvg1n.f . . . 4
2 cvg1n.c . . . 4
3 cvg1n.cau . . . 4
4 cvg1nlem.g . . . 4
5 cvg1nlem.z . . . 4
6 cvg1nlem.start . . . 4
71, 2, 3, 4, 5, 6cvg1nlemf 10786 . . 3
81, 2, 3, 4, 5, 6cvg1nlemcau 10787 . . 3
97, 8caucvgre 10784 . 2
10 fveq2 5428 . . . . . . . . . . 11
1110raleqdv 2635 . . . . . . . . . 10
1211cbvrexv 2658 . . . . . . . . 9
1312ralbii 2444 . . . . . . . 8
1413anbi2i 453 . . . . . . 7
1514anbi1i 454 . . . . . 6
16 oveq2 5789 . . . . . . . . . . . 12
1716breq2d 3948 . . . . . . . . . . 11
18 oveq2 5789 . . . . . . . . . . . 12
1918breq2d 3948 . . . . . . . . . . 11
2017, 19anbi12d 465 . . . . . . . . . 10
2120rexralbidv 2464 . . . . . . . . 9
22 simplr 520 . . . . . . . . 9
23 simpr 109 . . . . . . . . . 10
2423rphalfcld 9525 . . . . . . . . 9
2521, 22, 24rspcdva 2797 . . . . . . . 8
2615, 25sylbir 134 . . . . . . 7
272rpred 9512 . . . . . . . . . . . . . 14
2827ad4antr 486 . . . . . . . . . . . . 13
29 2re 8813 . . . . . . . . . . . . . 14
3029a1i 9 . . . . . . . . . . . . 13
3128, 30remulcld 7819 . . . . . . . . . . . 12
32 simplr 520 . . . . . . . . . . . 12
3331, 32rerpdivcld 9544 . . . . . . . . . . 11
345ad4antr 486 . . . . . . . . . . 11
3533, 34nndivred 8793 . . . . . . . . . 10
36 simprl 521 . . . . . . . . . . 11
3736nnred 8756 . . . . . . . . . 10
3835, 37readdcld 7818 . . . . . . . . 9
39 arch 8997 . . . . . . . . 9
4038, 39syl 14 . . . . . . . 8
41 simprl 521 . . . . . . . . . 10
4234adantr 274 . . . . . . . . . 10
4341, 42nnmulcld 8792 . . . . . . . . 9
441ad6antr 490 . . . . . . . . . . . . . . 15
45 simplrl 525 . . . . . . . . . . . . . . . . 17
465ad6antr 490 . . . . . . . . . . . . . . . . 17
4745, 46nnmulcld 8792 . . . . . . . . . . . . . . . 16
48 eluznn 9420 . . . . . . . . . . . . . . . 16
4947, 48sylancom 417 . . . . . . . . . . . . . . 15
5044, 49ffvelrnd 5563 . . . . . . . . . . . . . 14
5144, 47ffvelrnd 5563 . . . . . . . . . . . . . . 15
5232ad2antrr 480 . . . . . . . . . . . . . . . . 17
5352rpred 9512 . . . . . . . . . . . . . . . 16
5453rehalfcld 8989 . . . . . . . . . . . . . . 15
5551, 54readdcld 7818 . . . . . . . . . . . . . 14
56 simpllr 524 . . . . . . . . . . . . . . . . 17
5756ad3antrrr 484 . . . . . . . . . . . . . . . 16
5857, 54readdcld 7818 . . . . . . . . . . . . . . 15
5958, 54readdcld 7818 . . . . . . . . . . . . . 14
6027ad6antr 490 . . . . . . . . . . . . . . . . 17
6160, 47nndivred 8793 . . . . . . . . . . . . . . . 16
6251, 61readdcld 7818 . . . . . . . . . . . . . . 15
63 fveq2 5428 . . . . . . . . . . . . . . . . . . . 20
6463oveq1d 5796 . . . . . . . . . . . . . . . . . . 19
6564breq2d 3948 . . . . . . . . . . . . . . . . . 18
6663breq1d 3946 . . . . . . . . . . . . . . . . . 18
6765, 66anbi12d 465 . . . . . . . . . . . . . . . . 17
68 fveq2 5428 . . . . . . . . . . . . . . . . . . 19
69 fveq2 5428 . . . . . . . . . . . . . . . . . . . . 21
70 oveq2 5789 . . . . . . . . . . . . . . . . . . . . . 22
7170oveq2d 5797 . . . . . . . . . . . . . . . . . . . . 21
7269, 71breq12d 3949 . . . . . . . . . . . . . . . . . . . 20
7369, 70oveq12d 5799 . . . . . . . . . . . . . . . . . . . . 21
7473breq2d 3948 . . . . . . . . . . . . . . . . . . . 20
7572, 74anbi12d 465 . . . . . . . . . . . . . . . . . . 19
7668, 75raleqbidv 2641 . . . . . . . . . . . . . . . . . 18
773ad6antr 490 . . . . . . . . . . . . . . . . . 18
7876, 77, 47rspcdva 2797 . . . . . . . . . . . . . . . . 17
79 simpr 109 . . . . . . . . . . . . . . . . 17
8067, 78, 79rspcdva 2797 . . . . . . . . . . . . . . . 16
8180simprd 113 . . . . . . . . . . . . . . 15
82 simpr 109 . . . . . . . . . . . . . . . . . . . 20
8382ad3antrrr 484 . . . . . . . . . . . . . . . . . . 19
8483rpred 9512 . . . . . . . . . . . . . . . . . 18
8584rehalfcld 8989 . . . . . . . . . . . . . . . . 17
862ad6antr 490 . . . . . . . . . . . . . . . . . 18
8736ad2antrr 480 . . . . . . . . . . . . . . . . . 18
88 simplrr 526 . . . . . . . . . . . . . . . . . 18
8986, 83, 46, 45, 87, 88cvg1nlemcxze 10785 . . . . . . . . . . . . . . . . 17
9061, 85, 89ltled 7904 . . . . . . . . . . . . . . . 16
9161, 54, 51, 90leadd2dd 8345 . . . . . . . . . . . . . . 15
9250, 62, 55, 81, 91ltletrd 8208 . . . . . . . . . . . . . 14
93 fveq2 5428 . . . . . . . . . . . . . . . . . . . 20
9493breq1d 3946 . . . . . . . . . . . . . . . . . . 19
9593oveq1d 5796 . . . . . . . . . . . . . . . . . . . 20
9695breq2d 3948 . . . . . . . . . . . . . . . . . . 19
9794, 96anbi12d 465 . . . . . . . . . . . . . . . . . 18
98 simprr 522 . . . . . . . . . . . . . . . . . . 19
9998ad2antrr 480 . . . . . . . . . . . . . . . . . 18
10087nnred 8756 . . . . . . . . . . . . . . . . . . . 20
10145nnred 8756 . . . . . . . . . . . . . . . . . . . 20
102 2rp 9474 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
103102a1i 9 . . . . . . . . . . . . . . . . . . . . . . . . . 26
10486, 103rpmulcld 9529 . . . . . . . . . . . . . . . . . . . . . . . . 25
105104, 83rpdivcld 9530 . . . . . . . . . . . . . . . . . . . . . . . 24
10646nnrpd 9510 . . . . . . . . . . . . . . . . . . . . . . . 24
107105, 106rpdivcld 9530 . . . . . . . . . . . . . . . . . . . . . . 23
108107rpred 9512 . . . . . . . . . . . . . . . . . . . . . 22
109108, 100readdcld 7818 . . . . . . . . . . . . . . . . . . . . 21
110100, 107ltaddrp2d 9547 . . . . . . . . . . . . . . . . . . . . 21
111100, 109, 101, 110, 88lttrd 7911 . . . . . . . . . . . . . . . . . . . 20
112100, 101, 111ltled 7904 . . . . . . . . . . . . . . . . . . 19
11387nnzd 9195 . . . . . . . . . . . . . . . . . . . 20
11445nnzd 9195 . . . . . . . . . . . . . . . . . . . 20
115 eluz 9362 . . . . . . . . . . . . . . . . . . . 20
116113, 114, 115syl2anc 409 . . . . . . . . . . . . . . . . . . 19
117112, 116mpbird 166 . . . . . . . . . . . . . . . . . 18
11897, 99, 117rspcdva 2797 . . . . . . . . . . . . . . . . 17
119 oveq1 5788 . . . . . . . . . . . . . . . . . . . . . 22
120119fveq2d 5432 . . . . . . . . . . . . . . . . . . . . 21
121120, 4fvmptg 5504 . . . . . . . . . . . . . . . . . . . 20
12245, 51, 121syl2anc 409 . . . . . . . . . . . . . . . . . . 19
123122breq1d 3946 . . . . . . . . . . . . . . . . . 18
124122oveq1d 5796 . . . . . . . . . . . . . . . . . . 19
125124breq2d 3948 . . . . . . . . . . . . . . . . . 18
126123, 125anbi12d 465 . . . . . . . . . . . . . . . . 17
127118, 126mpbid 146 . . . . . . . . . . . . . . . 16
128127simpld 111 . . . . . . . . . . . . . . 15
12951, 58, 54, 128ltadd1dd 8341 . . . . . . . . . . . . . 14
13050, 55, 59, 92, 129lttrd 7911 . . . . . . . . . . . . 13
13157recnd 7817 . . . . . . . . . . . . . 14
13254recnd 7817 . . . . . . . . . . . . . 14
133131, 132, 132addassd 7811 . . . . . . . . . . . . 13
134130, 133breqtrd 3961 . . . . . . . . . . . 12
13552rpcnd 9514 . . . . . . . . . . . . . 14
1361352halvesd 8988 . . . . . . . . . . . . 13
137136oveq2d 5797 . . . . . . . . . . . 12
138134, 137breqtrd 3961 . . . . . . . . . . 11
13950, 54readdcld 7818 . . . . . . . . . . . . . . 15
140139, 54readdcld 7818 . . . . . . . . . . . . . 14
141127simprd 113 . . . . . . . . . . . . . 14
14250, 61readdcld 7818 . . . . . . . . . . . . . . . 16
14380simpld 111 . . . . . . . . . . . . . . . 16
14461, 54, 50, 90leadd2dd 8345 . . . . . . . . . . . . . . . 16 | 4,128 | 7,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.143478 |
https://engineeringtoolbox.com/doppler-effect-d_1839.html | 1,695,431,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00616.warc.gz | 268,964,669 | 9,414 | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
# Sound - Doppler Effect
## The doppler effect is the change in sound frequency due to the relative motion between a source and a listener.
The change in frequency of sound due to relative motion between a source and a listener is called the Doppler Effect.
The observed sound frequency for a listener can be expressed as
fr = fs (c + vr) / (c + vs) (1)
where
fr = frequency observed by the receiver (1/s, Hz)
fs = frequency emitted from the source (1/s, Hz)
c = speed of sound (m/s)
vr = velocity of the receiver relative to the medium - positive if the receiver is moving towards the source (m/s)
vs = velocity of the source relative to the medium - positive if the source is moving away from the receiver (m/s)
### Example - A Train passing a Bell
A train with speed 200 km/h (55.6 m/s) passes a bell with sound frequency 1000 Hz. With an air temperature of 20 oC the speed of sound is 343 m/s.
The frequency observed by a passenger in the train before passing the bell can be calculated as
fr = (1000 Hz) ((343 m/s) + (55.6 m/s)) / ((343 m/s) + (0 m/s))
= 1162 Hz
The frequency observed by the passenger in the train after passing the bell can be calculated as
fr = (1000 Hz) ((343 m/s) + (- 55.6 m/s)) / ((343 m/s) + (0 m/s))
= 838 Hz
### Doppler Effect Calculator
Calculate the observed frequency with the calculator below:
fs - frequency emitted from the source (Hz) -
c - speed of sound (m/s)
vr - velocity of the receiver relative to the medium (m/s) - m/s vs. km/h vs. mph
vs - velocity of the source relative to the medium (m/s)
## Related Topics
• ### Acoustics
Room acoustics and acoustic properties, decibel A, B and C, Noise Rating (NR) curves, sound transmission, sound pressure, sound intensity and sound attenuation.
## Related Documents
• ### Air - Speed of Sound vs. Temperature
Speed of sound in air at standard atmospheric pressure with temperatures ranging -40 to 1000oC (-40 to 1500oF) - Imperial and SI Units.
• ### Decibel A, B and C
Sound pressure filters that compensates for the hearing sensed by the human ear.
• ### Fluid Flowmeters - Comparing Types
An introduction to the different types of fluid flowmeters - Orifices, Venturies, Nozzles, Rotameters, Pitot Tubes, Calorimetrics, Turbine, Vortex, Electromagnetic, Doppler, Ultrasonic, Thermal, Coriolis.
• ### Notes, Octaves and Frequencies
Frequencies vs. notes and octaves.
• ### Sound Power
Sound power from sources like fans, jet engines, cars, humans and more.
• ### Speed of Sound - Online Calculator
A free online speed of sound calculator.
• ### Ultrasonic Doppler and Time of Flight Velocity and Volume Flow Meters
An basic introduction to the ultrasonic Doppler and Time of Flight Flow Meters.
## Engineering ToolBox - SketchUp Extension - Online 3D modeling!
Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with older versions of the amazing SketchUp Make and the newer "up to date" SketchUp Pro . Add the Engineering ToolBox extension to your SketchUp Make/Pro from the Extension Warehouse !
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## Citation
• The Engineering ToolBox (2012). Sound - Doppler Effect. [online] Available at: https://www.engineeringtoolbox.com/doppler-effect-d_1839.html [Accessed Day Month Year].
Modify the access date according your visit.
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9.19.12 | 879 | 3,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-40 | latest | en | 0.868978 |
https://www.speedsolving.com/threads/2x2-lbl-vs-ortega.20475/page-3 | 1,590,759,738,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347404857.23/warc/CC-MAIN-20200529121120-20200529151120-00062.warc.gz | 904,290,252 | 16,945 | # 2x2 LBL vs Ortega
#### CubingwithChris
##### Member
Just use both (like me If its a good layer do LBL if its not a good layer do Ortega
#### IAmAPerson
##### Member
TCLL and EG2 is where its at.
Pfffffffff...Twisty Ortega is the best.
#### NeilH
##### Member
I average 6-7 with LBL and I suck, I don't see how it's bad m80's
#### Basil Herold
##### Member
Pfffffffff...Twisty Ortega is the best.
SubSS is something like twisty ortega:
Twisty Face
SubSS
PBL
#### kid who cubes
##### Member
I was thinking about the exact same thing just yesterday after I solved in 5 seconds with LBL. I solve the 3x3 in about 30 seconds, so I was thinking that if I can get that fast, the LBL couldn't be that bad on a 2x2. If you don't have time to learn Ortega before a competition, LBL is a good alternative.
There are a lot of PLL skips, but the bad thing is that if you only get one during a competition then it doesn't help your average at all.
If you learn LBL you can easily transition into CLL but learning Ortega wont help you with EG-1 or 2
#### AlphaSheep
##### Member
If you learn LBL you can easily transition into CLL but learning Ortega wont help you with EG-1 or 2
I'm guessing you read the first and then just assumed the second, because the second part of your statement is wrong. Transitioning from Ortega to EG is exactly the same as transitioning from LBL to CLL.
#### aerocube
##### Member
i feel like this right now lol,just transitioned into ortega and my times went up and now i'm averaging between beginner's and advanced lbl,i might just be method neutral and see how that works
#### The_VortexCuber
##### Member
I use lbl and i can fairly easily get sub 5 averages and my record is 1.07 so i would say lbl is good but i use ortega when i have bad layers so i recommend using both if you have a easy layer use lbl if you have a easy side use ortega. Also this should helpl | 500 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-24 | latest | en | 0.938544 |
https://www.eevblog.com/forum/projects/high-precision-lc-inductance-capacitance-meter-design/?wap2;PHPSESSID=s7nfjrsgkdissortpngqv3sa85 | 1,679,812,833,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00352.warc.gz | 816,649,961 | 4,003 | Electronics > Projects, Designs, and Technical Stuff
High-precision L/C Inductance Capacitance meter Design
(1/8) > >>
Mechatrommer:
as requested by OP of https://www.eevblog.com/forum/index.php?topic=1855.0;topicseen ... i have to move on here. So i hope some feedbacks...
Mechatrommer:
--- Quote from: saturation on November 30, 2010, 10:02:34 pm ---You can make your own even simpler since the relationship in high pass filters are:
http://www.play-hookey.com/ac_theory/hi_pass_filters.html
So you can build this very quickly and measure either capacitance or inductance; getting a meter really matters if you don't have the time to calculate it as listed here:
https://www.eevblog.com/forum/index.php?topic=533.msg7756#msg7756
--- End quote ---
ok mr saturation! i think its time and place to discuss this more in depth. this looks nasty'ly simple (theoritically) and i've been astonished with it since the first time i opened the ee book and saw it. my basic rule is that when i cannot understand a simple system, then i should go even simpler. now, the RC and RL hi/lo pass filter is the most basic thing on earth, theoritically to build an LCR meter is great with the super simple 2 passive components circuit... but just how to practicality it is, is where i'm still stumbled at. so let me point out of what i understand...
1) all we need is a setup/schematic shown in your link (either RC or RL), and... Vin supposedly be connected to some Vac generator... mcu PWM pin maybe?
2) all on earth the basic equation is the famous w = 1 / RC. is this means, we need to setup a known w and R in order to get the C value?
3) since we need to feed alternating V, is this means Vcc to -Vcc peak to peak instead of Vcc to 0V? so how to get the -Vcc voltage? can we put the R or L ground (as in the link's picture) to some dummy reference at Vcc/2 so we have Vcc/2 and -Vcc/2 seen by the filter circuit?
4) and the rest of the math is done in mcu programming right? ok leave this for now, just assume i'm too good to program that "software'tically" 8)... as i said, i always stuck with analog side :(.
i'm looking forward/expecting/hope to see this basic questions evolve to a bigger thing like the many DIY circuits provided by dear fellow members here.
Mechatrommer:
maybe i got it wrong for the equation...
continuation from http://www.play-hookey.com/ac_theory/hi_pass_filters.html
here goes the ugly sketching. hope i got clearer solution.
Mechatrommer:
--- Quote from: saturation on December 01, 2010, 04:08:02 pm ---That's good math work, safri, but the equation to use is different and far simpler.
Notice how all LCR meters use a specific freuquency for a range of C or L, that's roughly how this works.
Resistor here is a constant, use any thin film or carbon composite. Don't use wirewound as its a inductor too.
Now, put your unknown capacitor or inductor in the circuit above, as necessary.
Get a function or audio generator generator as Vin, and measure its output Voltage with a scope for say 500kHz or 500 Hz, just to copy the frequencies the device in the original post used. Say adjust it for 1 Vp-p so its easy to read.
If your cap is as big as your pinky or your inductor is fairly large, use the 500 Hz.
Now connect the generator to Vin and the scope to Vout. Adjust the output frequency precisely until the output voltage is 0.707Vp-p. That's the fc.
Now that you have fc, substitute it algebraically into the equation and that's your L or C.
Units:
fc= Hz
R = ohms
L = henry
For example, you have an unknown small inductor.
Use the inductor high pass circuit above. You have some resistor lying around that is 500,000 ohms.
You pass 500kHz into it and it comes out 1V p-p. Adjusting frequency down [ since its a high pass filter its already passing everything ABOVE the cutoff] , at 300kHz the Vout is 0.707 p-p.
Solving:
L = R / 2 x pi x fc
L = 500,000 / 2 x pi x 300,000 Hz
L = 0.265 H or 265 mH.
As alm and I suggested in a old post, most hobbyist only need a good scope, like the Rigol 1052E, a function generator, a good DMM and PSU. An LCR meter is nice to have, but if you don't measure C or L often, and a good DMM comes with C meter already too, you can rig one up fairly quickly and get very accurate results with it.
Some modifications of the same rig can calculate ESR, DF, and Q too. But they can be more tedious to do on-the-fly, so it would be worth it to buy or build an ESR meter than a LCR meter without ESR built in.
http://octopus.freeyellow.com/esr.html
--- End quote ---
jimmc:
How about doing it this way?
http://www.sillanumsoft.org/ZRLC.htm
Jim
[#] Next page
Go to full version | 1,231 | 4,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.934856 |
https://abundantcode.com/break-statement-in-python/ | 1,701,200,991,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00397.warc.gz | 107,795,523 | 54,388 | # break statement in Python
This tutorial will teach you how to use the Python break statement to exit a loop prematurely.
You may want to end a for loop or a while loop early regardless of the results of the conditional tests. In these cases, the break statement can be useful:
`break`
To end a loop when a condition is True, you typically use the break statement in conjunction with the if statement.
## break statement and for loop in Python
The following code demonstrates how to use the break statement within a for loop:
```for index in range(n):
# more code here
if condition:
break```
The break statement terminates the loop instantly if the condition evaluates to True in this syntax. The remaining iterations will not be completed.
In this example, the break statement is used within a for loop:
```for index in range(0, 10):
print(index)
if index == 3:
break```
Output
0 1 2 3
How does it work?
The for loop iterates through ten numbers ranging from 0 to 9 and shows each one on the screen.
When the loop number is 3, however, the break statement ends the loop instantly. As a result, the application only displays 4 digits on the screen, ranging from 0 to 3.
In a nested loop, the break statement terminates the innermost loop. As an example:
```for x in range(5):
for y in range(5):
# terminate the innermost loop
if y > 1:
break
# show coordinates on the screen
print(f"({x},{y})")```
Output
(0,0) (0,1) (1,0) (1,1) (2,0) (2,1) (3,0) (3,1) (4,0) (4,1)
Two for loops are used in this example to display the coordinates from (0,0) to (5,5) on the screen.
When y is bigger than one, the break statement in the nested loop terminates the innermost loop.
As a result, you can only observe coordinates with y values of zero and one.
## break statement with while loop in Python
The following example demonstrates how to utilise the break statement within a while loop:
```while condition:
# more code
if condition:
break```
The break statement is used inside a while loop in the following example.
It will ask you to choose your preferred colour. When you type quit:, the software will terminate.
```print('-- Help: type quit to exit --')
while True:
color = input('Enter your favorite color:')
if color.lower() == 'quit':
break```
Output
```-- Help: type quit to exit -- | 567 | 2,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-50 | latest | en | 0.799521 |
http://www.watermarkfactory.com/resources/news/comp.graphics.algorithms/3D-camera-question-5904.htm | 1,537,585,997,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158011.18/warc/CC-MAIN-20180922024918-20180922045318-00019.warc.gz | 430,104,250 | 3,384 | # comp.graphics.algorithms
## Subject: 3D camera question
Really stuck here, and could use some help. I'm trying to add a camera to my
3D engine. Say I just have one shape - a large triangle that's acting as the
ground plane.
The triangle is not transformed, as you can see the arguments to the
rotation and translation are 0. For reference, I have one of the matrix
methods below, it sets it up and then multiplies. If I change the values of
my camera rotation, the triangle still rotates as if I were simply rotating
it (from its center) and not the camera. I know I need to subtract the
location of the camera relative to the shape, but Im not sure how/where to
do this. Can anyone help? Please let me know if you need more information.
TIA
---------------------
camTransX = 0.0;
camTransY = -50.0;
camTransZ = 500.0;
groundMatrix.setIdentity();
groundMatrix.rotateX(0);
groundMatrix.rotateY(0);
groundMatrix.rotateZ(0);
groundMatrix.translate(0, 0, 0);
groundMatrix.rotateX(camRotX);
groundMatrix.rotateY(camRotY);
groundMatrix.rotateZ(camRotZ);
groundMatrix.translate(camTransX, camTransY, camTransZ);
ground.draw(groundMatrix);
-------------------
public function rotateY(angleY:Number):Void {
if (angleY == 0) {
return;
}
var cosAngle:Number = Math.cos(angleY);
var sinAngle:Number = Math.sin(angleY);
temp[0][0] = cosAngle;
temp[0][1] = 0;
temp[0][2] = -sinAngle;
temp[0][3] = 0;
temp[1][0] = 0;
temp[1][1] = 1;
temp[1][2] = 0;
temp[1][3] = 0;
temp[2][0] = sinAngle;
temp[2][1] = 0;
temp[2][2] = cosAngle;
temp[2][3] = 0;
temp[3][0] = 0;
temp[3][1] = 0;
temp[3][2] = 0;
temp[3][3] = 1;
multiply();
} | 502 | 1,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-39 | latest | en | 0.734761 |
https://im.kendallhunt.com/MS_ACC/teachers/2/6/2/preparation.html | 1,719,039,750,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00286.warc.gz | 271,917,180 | 26,635 | # Lesson 2
Introduction to Functions
### Lesson Narrative
In this second introductory lesson, students learn the term function for a rule that produces a single output for a given input. They also start to connect function language to language they learned in earlier grades about independent and dependent variables.
We can say, "the output is a function of the input," and we also say, "the output depends on the input." In the optional activity, students see how it is possible to use different words to describe the same function as long as all input-output pairs are the same. This helps solidify the notion of a function as something different from the method of calculating its values.
### Learning Goals
Teacher Facing
• Comprehend the structure of a function as having one and only one output for each allowable input.
• Describe (orally and in writing) a context using function language, e.g., “the [output] is a function of the [input]” or “the [output] depends on the [input]”.
• Identify (orally) rules that produce exactly one output for each allowable input and rules that do not.
### Student Facing
Let’s learn what a function is.
### Student Facing
• I know that a function is a rule with exactly one output for each allowable input.
• I know that if a rule has exactly one output for each allowable input, then the output depends on the input.
Addressing
Building Towards
### Glossary Entries
• function
A function is a rule that assigns exactly one output to each possible input.
The function $$y=6x+4$$ assigns one value of the output, $$y$$, to each value of the input, $$x$$. For example, when $$x$$ is 5, then $$y=6(5)+4$$ or 34.
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Google Slides Log In PowerPoint Slides Log In | 454 | 2,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.817126 |
https://physics.stackexchange.com/questions/246853/walds-general-relativity-section-6-3-page-144 | 1,716,716,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00404.warc.gz | 397,642,393 | 40,959 | # Wald's General Relativity, section 6.3 Page 144
I cannot understand how he reaches the conclusion in equation 6.3.36 and 6.3.37; even the terminology is somewhat confusing. This is a problem of bending of light under gravitational field.
This is what he mentioned in the page: we have, for Schwarzschild metric, $$$$\frac{d\phi}{dr}=\frac{L}{r^2}\left[E^2-\frac{L^2}{r^3}(r-2M)\right]^{-1/2}.\tag{6.3.35}$$$$ We wish to find $$\Delta\phi$$ of light ray in Schwarzschild geometry, traversing a path near the spherical star.
In order to not be captured, the impact parameter $$b$$ must be greater than the critical impact parameter $$b_c=3^{3/2}M.\tag{6.3.33}$$ In that case, the orbit of the light ray will have a "turning point" at the largest radius, $$R_0$$, for which $$V(R_0)=E^2/2$$, i.e. at the largest root of $$R_0^3-b^2(R_0-2M)=0,\tag{6.3.36}$$ which is
$$$$R_0=\frac{2b}{\sqrt{3}}\cos\left[\frac{1}{3}\cos^{-1}\left(-\frac{b_c}{b}\right)\right].\tag{6.3.37}$$$$
I am lost at this point:
1. What does he mean by "largest radius" $$R_0$$?
2. How does he suddenly make the cosine conclusion?
I could understand the rest of this section but these two remarks are too obscure to me. Any help is greatly appreciated.
Let's start with the Schwarzschild geodesic $$\frac{1}{2}\dot{r}^2 + V(r) = \frac{1}{2}E^2,$$ with $V(r)$ the effective potential $$V(r) = \frac{L^2}{2r^3}(r-2M).$$ Consider light rays that are deflected in the Schwarzschild geometry, i.e. photons that start at a large distance ($r$ large, $\dot{r}<0$), propagate until they reach a distance of closest approach at coordinate $r=R_0$ (where $\dot{r}=0$), and propagate outward again. Therefore, $R_0$ is a solution of the equation $$\frac{1}{2}E^2 = V(r) = \frac{L^2}{2r^3}(r-2M),$$ or $$r^3 -b^2(r-2M) =0,$$ with $b=L/E$. If $b<b_c = 3^{3/2}M$ then this equation has no positive roots, which means that the photon will continue to spiral towards the centre. If $b>b_c$, the equation has two positive roots and one unphysical negative root. We're only interested in the largest of the positive roots, since we started at a very large radius ($r \gg R_0$) and the photons will never reach any distance smaller than $R_0$.
The solution $$R_0 = \frac{2b}{\sqrt{3}}\cos\left[\frac{1}{3}\cos^{-1}(-b_c/b)\right]$$ can be verified by using $4\cos^3x =\cos 3x + 3\cos x$. One can also verify that it is the largest root, since $b>b_c$ and $$\cos^{-1}(-b_c/b) < \cos^{-1}(-1) = \pi\\ \cos\left[\frac{1}{3}\cos^{-1}(-b_c/b)\right] > \cos(\pi/3) = 1/2\\ R_0 > \frac{b}{\sqrt{3}} > \frac{b_c}{\sqrt{3}} = 3M,$$ and $3M$ is the coordinate where $V(r)$ has a maximum. The other root (let's call it $R_1$) will lie between $2M< R_1 < 3M$. This root applies to photons that start at very small radii $(2M < r < R_1)$ and try to escape, but cannot move beyond the turning point $R_1$, after which they spiral towards the centre.
I don't know why he uses the name "largest radius" but physically it is the radius of closest approach, which is the minimum distance to the center of the star at the point of reflection of the orbit. Since it is a point of reflection $\dot{r}=0$ in the equation 6.3.14 $$\frac{1}{2}\dot{r}^2 + V_{\text{eff}}(r) = \frac{1}{2}\dot{r}^2 + \frac{L^2}{2r^2} \left( 1 - \frac{2M}{r}\right) = \frac{E^2}{2}$$ this gives $V_{\text{eff}} (R_o) = E^2/R$ which is $$\frac{L^2}{2R_o^2} \left( 1 - \frac{2M}{R_o}\right) = \frac{E^2}{2}$$ which is just \begin{align} &R_o^3 - b^2 (R_o - 2M) = 0 \\ &R_o^3 - b^2 (R_o - \frac{2b_c}{3\sqrt{3}}) = 0 \\ &4x^3 - 3x =- \frac{b_c}{b} \end{align} where $x \equiv \frac{\sqrt{3}R_o}{2b}$. Now Wald wants to flex his algebraic muscles so remember a formula that says $4\text{cos}^3\theta - 3\text{cos }\theta = \text{cos }3\theta$ so defining $\text{cos }\theta\equiv x$ gives $\theta = \frac{1}{3}\text{cos}^{-1}(-b_c/b)$ or finally $$x = \text{cos }\left(\frac{1}{3}\text{cos}^{-1}\frac{-b_c}{b}\right)\qquad\Rightarrow\qquad R_o = \frac{2b}{\sqrt{3}} \text{cos }\left(\frac{1}{3}\text{cos}^{-1}\frac{-b_c}{b}\right)$$
• The point he's making with "largest radius" is that there may be other, smaller values of $R$ for which $V(R_0) = E^2/2$. But the geodesic will never get to these $R$ values; we only want the largest such radius. Jul 12, 2019 at 15:35 | 1,537 | 4,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 1, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-22 | latest | en | 0.875473 |
https://www.microwavejournal.com/articles/29151-noise-analysis-then-and-today | 1,713,748,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00167.warc.gz | 805,847,295 | 21,403 | Figure 1 Test fixture (a) and network analyzer (b) used to measure large signal S-parameters.
Noise analysis of oscillators, also known as autonomous circuits, was a dream for many years. The first and linear approach must be credited to David Leeson, who took a linear lowpass equivalent circuit and derived his frequently quoted linear formula.1 This formula requires data which can only be gained after a post prior analysis. The values of the output power, the loaded Q, the large-signal noise factor—frequently confused with noise figure (NF), which is 10log(noise factor)—and the flicker noise contribution are not known a priori, although they may be computed.2 The introduction of nodal noise analysis in a proper way was published by Hillbrand and Russer.3 The Lee-Hajimiri noise analysis is interesting but not very practical; while often quoted in the literature, its use is not practically shown.2,4-6 If we combine the Leeson formula with the tuning diode contribution, the following equation allows us to completely calculate the noise of the oscillator.2 The VCO term was added by Rohde,2 and the flicker term was added by Scherer.7
where
L(fm) = ratio of sideband power in a 1 Hz bandwidth at fm to total power, in dB
fm = frequency offset
f0 = center frequency
fc = flicker frequency
QL = loaded Q of the tuned circuit
Figure 2 Measured large-signal S-parameters of the Infineon BFP520 bipolar transistor from 600 MHz to 3 GHz in 50 MHz steps: S11 (a), S12 (b), S21 (c) and S22 (d).
F = noise factor
kT = 4.1 × 10-21 at 300K0 (room temperature)
Psav = average output power of the oscillator
R = equivalent noise resistance of the tuning diode (typically 50 to 10 kΩ)
Ko = oscillator voltage gain
Kaertner’s paper8 used a nice time domain approach and includes, probably for the first time, the noise correlation in oscillator design. At that time, the now available harmonic balance method had not been invented. A very good description is found in the works by Rizzoli.9-11 While not easy to read and highly mathematical, Kaertner’s calculation produced real results.
Large-Signal Noise Analysis
The introduction of the piecewise linear harmonic balance (HB) method, developed by Vittorio Rizzoli and his team, was the perfect body to include the noise correlation method. The famous nonlinear time domain circuit analysis program SPICE lacks a rigorous noise correlation analysis, and the HB programs are a hybrid of linear (frequency) and nonlinear (time domain) computations.12,13 Both publications by Rizzoli and Gilmore were leading on the topic. Rowan Gilmore worked with me during our Compact Software time.
My team at Compact Software and the Rizzoli team were struggling to validate the results, and it was complicated to get reliable measured data and maintain accuracy and speed. The Compact Software approach and the Rizzoli team were always fighting about this topic. I remember telling users that short cuts used in some programs were getting them faster to the wrong answer. At the end, developing a very fast, double precision, multidimensional matrix inversion program was the solution, and using FORTRAN gave the most stable results. This HB approach is also able to deal with hysteresis.
The first and most challenging circuit analysis of a low noise amplifier, developed by Raytheon, was published14 for the first time, in cooperation with Robert Pucell of Raytheon and Tony Pavio of Texas Instruments. We demonstrated very accurate results using SPICE type data for GaAs FETs. They were hard to come by; we ended up using a modified Materka model.15-19 The actual circuit was published in Microwave Journal.20 So far, I have demonstrated that the noise analysis can be done solely using available and accurate SPICE parameters of GaAs FETs and its family members.
But the really needed large-signal circuit analysis was the oscillator. To implement the proper FET15-19 and BJT20-22 models and its derivatives was quite a task. The internal noise modeling was based on the noise correlation matrix2 and the results were published by W. Anzill, F. X. Kaertner and P. Russer in 1992.23 In 1987, the Microwave Harmonica program by Compact Software was proven to be reliable.20 The next step was to match the measured results of practical circuits. Hewlett Packard had the first, reliable phase noise system; Dieter Scherer was the lead engineer who educated us all. Today, the leading phase noise analyzer is the Rohde and Schwarz FSWP/8/26/50.
For a systematic approach, even without a simulator, it is useful to have large-signal S-parameters.2,24 The large-signal parameters of bipolar transistors, which had not been derived at the time of Kaertner’s work, are best obtained from measurements.24 The proper de-embedding procedure19 translates the external measurements to the actual chip (see Figure 1a), and a network analyzer with variable output power is needed for the measurement (see Figure 1b). With large-signal operation, the BJT behaves differently than for low noise applications,18-22 as shown in Figure 2.25
Kaertner Colpitts Oscillator
With this background, I would like to analyze the circuit used by Kaertner as the basis of his brilliant paper.8,23 The component values in the schematic of Figure 3a are taken from his publication: R1 = 350 Ω, R2 = 110 kΩ, RL = 500 Ω, L1 = 10 µH, L2 = 30 nH, C1 = 10 pF, C2 = 940 pF, C3 = 2.7 nF, C4 = 1.5 nF and T = BFR35A.
Kaertner made one simplification that in circuit theory is deadly: L2 was assumed to have infinite Q. For his approach, as the circuit will actually load the unloaded Q, this is permissible. With an infinite Q and with C2 = 940 pF, it will oscillate. Once a real Q, like QL =100, is used, the value of C2 must be reduced to 100 pF; then all works well. The feedback has to be increased, and it is advisable to reduce C4 to 100 pF, as it is not possible to make a capacitor without parasitic inductance. The same applies to the 10 µH inductor, which is probably resonant at the operating frequency; 1 µH seems to be a better choice. Kaertner published the calculated phase noise of his oscillator,4 pointing out correctly that the flicker contribution was not included.
Figure 3 Colpitts oscillator analyzed by Kaertner (a) and circuit schematic in Compact Software (b).
Figure 3b shows Kaertner’s oscillator circuit when modeled with Compact Software Microwave Harmonia, to use the HB technique. The simulation includes the flicker noise exponent (AF) and the flicker noise constant (KF) and the simulation provides results tracking Kaertner’s publication (see Figure 4a). The influences of AF and KF are clearly seen at 1 kHz. The simulation also predicts the output power and harmonic content (see Figure 4b). The output power is approximately 16 dBm with about 20 dB harmonic suppression. This indicates a good operating Q of the resonant circuit. The analysis tracks the performance very well.
Figure 4 Simulated phase noise (a) and output power vs. frequency (b) of the Kaertner oscillator.
Summary
This brief history provides insight into the struggle to develop and validate a general purpose, nonlinear computer-aided design (CAD) tool that correctly predicts large-signal noise in amplifiers, frequency doublers and mixers. There are many more examples.2,14
There were many excellent contributors. Vittorio Rizzoli probably contributed the most, and my team’s effort to test and validate the Serenade Harmonica program should not be underestimated. Later, a CAD independent mathematical time domain calculation was developed, which, without immediately resorting to an expensive tool, can predict the noise performance of oscillators.2,23-24
Appendix
Although Leeson’s phase-noise model provides a valuable insight into the oscillator design from engineering perspectives, it cannot explain some of the important phase noise phenomena. This is due to some assumptions made about the linearity and time-invariant behavior of the system. When comparing the measured results of oscillators with the assumptions made in Leeson’s Equation, one frequently obtains a de facto noise figure in the vicinity of 20 to 30 dB and an operating Q that is different than the assumed loaded Q, therefore must be determined from measurement; diminishing the predictive power of the Leeson’s phase noise.
Leeson’s model observes the asymptotic behavior of phase-noise at close-to carrier offsets, asserting that phase-noise goes to infinity with 1/f3 rate. This is not correct as it implies an infinite output power for oscillator. For noisy oscillators it could also suggest that L(f) >0 dBc/Hz, this singularity arises from linearity assumption for oscillator operation around steady-state point. In fact, the linear model breaks down at close-to-carrier frequencies where the phase-noise power is strong.5 Considering a nonlinear model for the oscillator in absence of flicker noise, these singularities can be resolved by expressing the phase noise in the form of a Lorentzian function1,2,4
where a is a fitting parameter.
Although Equation (1) models the spectrum and avoids any singularity at Δfm=0 while maintaining the same asymptotic behavior as illustrated in Figure 1; this is an after-the-fact approach, not a predictive one. From Equation (1), the total power of phase-noise from minus infinity to plus infinity is 1, this means that phase-noise doesn’t change the total power of the oscillator; it merely broadens its spectral peak. Attempting to match the Leeson calculated curve “A” (Figure 1), the measured curve requires totally different values than those assumed due to up-conversion and down-conversion of noise components from harmonically related frequencies to around carrier frequency as depicted in Figure 2.5
References
1. D. B. Leeson, “A Simple Model of Feedback Oscillator Noise Spectrum,” Proceedings of the IEEE, Vol. 54, No. 2, February 1966, pp. 329–330.
2. U. L. Rohde, A. K. Poddar and G. Böck, The Design of Modern Microwave Oscillators for Wireless Applications: Theory and Optimization, pp. 131–137.
3. H. Hillbrand and P. Russer, “An Efficient Method for Computer Aided Noise Analysis of Linear Amplifier Networks,” IEEE Transactions on Circuits and Systems, Vol. 23, No. 4, April 1976.
4. L. A. Hajimiri, “Linearity, Time Variation, and Oscillator Phase Noise,” RF and Microwave Oscillator Design, Artech House, 2002.
5. A. Hajimiri and T. Lee, “A General Theory of Phase Noise in Electrical Oscillators,” IEEE Journal of Solid-State Circuits, Vol. 33, No. 2, February 1998, pp. 179–194.
6. D. Ham and A. Hajimiri, “Concepts and Methods in Optimization of Integrated LC VCOs,” IEEE Journal of Solid-State Circuits, June 2001.
7. D. Scherer, “Design Principles and Test Methods for Low Phase Noise RF and Microwave Sources,” RF and Microwave Measurement Symposium and Exhibition, Hewlett Packard, 1983.
8. F. X. Kaertner, “Determination of the Correlation Spectrum of Oscillators with Low Noise,” IEEE Transactions on Microwave Theory and Techniques, Vol. 37, No. 1, January 1989.
9. V. Rizzoli, F. Mastri and C. Cecchefti, “Computer-Aided Noise Analysis of MESFET and HEMT Mixers,” IEEE Transactions on Microwave Theory and Techniques, Vol. 37, September 1989, pp. 1401–1410.
10. V. Rizzoli and A. Lippadni, “Computer-Aided Noise Analysis of Linear Multiport Networks of Arbitrary Topology,” IEEE Transactions on Microwave Theory and Techniques, Vol. 33, December 1985, pp. 1507–1512.
11. V. Rizzoli, F. Mastri and D. Masotti, “General-Purpose Noise Analysis of Forced Nonlinear Microwave Circuits,” Military Microwave, 1992.
12. R. J. Gilmore and M. B. Steer, “Nonlinear Circuit Analysis Using the Method of Harmonic Balance—A Review of the Art. Part I. Introductory Concepts,” International Journal of Microwave and Millimeter-Wave Computer-Aided Engineering, Vol. 1, No. 1, January 1991, pp. 22–37.
13. R. J. Gilmore and M. B. Steer, “Nonlinear Circuit Analysis Using the Method of Harmonic Balance—A Review of the Art. Part II. Advanced Concepts,” International Journal of 367T.
14. U. L. Rohde and D. P. Newkirk, “RF/Microwave Circuit Design for Wireless Application,” Wiley, 2000.
15. R. A. Pucel, H. A. Haus and H. Statz, “Signal and Noise Properties of Gallium Arsenide Microwave Field-Effect Transistors,” Advances in Electronics and Electron Physics, Academic Press, Vol. 38, 1975.
16. U. L. Rohde, “New Nonlinear Noise Model for MESFETs Including MM-Wave Application,” First International Workshop of the West German IEEE MTT/AP Joint Chapter on Integrated Nonlinear Microwave and Millimeterwave Circuits (INMMC’90) Digest, Duisburg University, October 3–5, 1990.
17. U. L. Rohde, “Improved Noise Modeling of GaAs FETs: Using an Enhanced Equivalent Circuit Technique,” Microwave Journal, November 1991, pp. 87–101; December 1991, pp. 87–95.
18. M. Rudolph, R. Doerner, K. Beilenhoff and P. Heymann, “Scalable GaIn/GaAs HBT Large-Signal Model,” IEEE Transactions on Microwave Theory and Techniques, Vol. 48, December 1990, pp. 2370–2376.
19. R. A. Pucel, W. Struble, R. Hallgren, and U. L. Rohde, “A General Noise De-embedding Procedure for Packaged Two-Port Linear Active Devices,” IEEE Transactions on Microwave Theory and Techniques, Vol. 40, No. 11, November 1992, pp. 2013–2024.
20. U. L. Rohde, A. M. Pavio and R. A. Pucel, “Accurate Noise Simulation of Microwave Amplifiers Using CAD,” Microwave Journal, Vol. 31, No. 12, December 1988, pp. 130-141.
21. R. A. Pucel and U. L. Rohde, “An Exact Expression for the Noise Resistance Rn of a Bipolar Transistor for Use with the Hawkins Noise Model,” IEEE Microwave and Guided Wave Letters, Vol. 3, No. 2, February 1993, pp. 35–37.
22. C. C. McAndrew, J. A. Seitchik, D. F. Bowers, M. Dunn, M. Foisy, I. Getreu, M. McSwain, S. Moinian, J. Parker, D. J. Roulston, M. Schroeter, P. van Wijnen and L. F. Wagner, “VBIC95, The Vertical Bipolar Inter-Company Model,” IEEE Journal of Solid-State Circuits, Vol. 31, No. 10, October 1996, pp. 1476–1483.
23. W. Anzill, F. X. Kaertner and P. Russer, “Simulation of the Single-Sideband Phase Noise of Oscillators,” Second International Workshop of Integrated Nonlinear Microwave and Millimeterwave Circuits, 1992.
24. U. L. Rohde and A. Apte, “Everything You Always Wanted to Know about Colpitts Oscillators,” IEEE Microwave Magazine, Vol. 17, No. 8, August 2016, pp. 59–76.
25. U. L. Rohde, “Parameter Extraction for Large Signal Noise Models and Simulation of Noise in Large Signal Circuits Like Mixers and Oscillators,” 23rd European Microwave Conference, Madrid, September 6–9, 1993. | 3,641 | 14,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.948654 |
https://freecoursesite.com/become-an-algebra-master-3/ | 1,721,616,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00546.warc.gz | 230,222,997 | 18,602 | # Become an Algebra Master
## Learn everything from Algebra 1 and Algebra 2, then test your knowledge with 1,150+ practice questions
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3 Comments
1. qsq says
need an update
2. kiora safzti says
Please update course
3. ara says
update course, lots of quizzes are missing | 1,177 | 5,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-30 | latest | en | 0.887212 |
https://nhigham.com/2022/01/31/what-is-the-second-difference-matrix/ | 1,686,137,693,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00189.warc.gz | 458,893,077 | 32,618 | # What Is the Second Difference Matrix?
The second difference matrix is the tridiagonal matrix $T_n$ with diagonal elements $2$ and sub- and superdiagonal elements $-1$:
$\notag T_n = \left[ \begin{array}{@{}*{4}{r@{\mskip10mu}}r} 2 & -1 & & & \\ -1 & 2 & -1 & & \\[-5pt] & -1 & 2 & \ddots & \\ & & \ddots & \ddots & -1 \\ & & & -1 & 2 \end{array}\right] \in\mathbb{R}^{n\times n}.$
It arises when a second derivative is approximated by the central second difference $f''(x) \approx (f(x+h) -2 f(x) + f(x-h))/h^2$. (Accordingly, the second difference matrix is sometimes defined as $-T_n$.) In MATLAB, $T_n$ can be generated by gallery('tridiag',n), which is returned as a aparse matrix.
This is Gil Strang’s favorite matrix. The photo, from his home page, shows a birthday cake representation of the matrix.
The second difference matrix is symmetric positive definite. The easiest way to see this is to define the full rank rectangular matrix
$\notag L_n = \begin{bmatrix} 1 & & & \\ -1 & 1 & & \\ & -1 & \ddots & \\ & & \ddots & 1 \\ & & & -1 \end{bmatrix} \in\mathbb{R}^{(n+1)\times n}$
and note that $T_n = L_n^T L_n$. The factorization corresponds to representing a central difference as the product of a forward difference and a backward difference. Other properties of the second difference matrix are that it is diagonally dominant, a Toeplitz matrix, and an $M$-matrix.
## Cholesky Factorization
In an LU factorization $A = LU$ the pivots $u_{ii}$ are $2$, $3/2$, $4/3$, …, $(n+1)/n$. Hence the pivots form a decreasing sequence tending to 1 as $n\to\infty$. The diagonal of the Cholesky factor contains the square roots of the pivots. This means that in the Cholesky factorization $A = R^*R$ with $R$ upper bidiagonal, $r_{nn} \to 1$ and $r_{n,n-1}\to -1$ as $n\to\infty$.
## Determinant, Inverse, Condition Number
Since the determinant is the product of the pivots, $\det(T_n) = n+1$.
The inverse of $T_n$ is full, with $(i,j)$ element $i(n-j+1)/(n+1)$ for $j\ge i$. For example,
$\notag T_5^{-1} = \displaystyle\frac{1}{6} \begin{bmatrix} 5 & 4 & 3 & 2 & 1\\ 4 & 8 & 6 & 4 & 2\\ 3 & 6 & 9 & 6 & 3\\ 2 & 4 & 6 & 8 & 4\\ 1 & 2 & 3 & 4 & 5 \end{bmatrix}.$
The $2$-norm condition number satisfies $\kappa_2(T_n) \sim 4n^2/\pi^2$ (as follows from the formula (1) below for the eigenvalues).
## Eigenvalues and Eigenvectors
The eigenvalues of $T_n$ are
$\notag \mu_k = 4 \sin^2\Bigl(\displaystyle\frac{k \phi}{2} \Bigr), \quad k = 1:n, \qquad (1)$
where $\phi = \pi/(n+1)$, with corresponding eigenvector
$\notag v_k = \begin{bmatrix} \sin(k\phi), &\sin(2k\phi), &\dots, &\sin(nk\phi) \end{bmatrix}^T.$
The matrix $Q$ with
$\notag q_{ij} = \Bigl(\displaystyle\frac{2}{n+1}\Bigr)^{1/2} \sin\Bigl( \frac{ij\pi}{n+1} \Bigr)$
is therefore an eigenvector matrix for $T_n$: $Q^*AQ = \mathrm{diag}(\mu_k)$.
## Variations
Various modifications of the second difference matrix arise and similar results can be derived. For example, consider the matrix obtained by changing the $(n,n)$ element to $1$:
$\notag \widetilde{T}_n = \left[ \begin{array}{@{}*{4}{r@{\mskip10mu}}r} 2 & -1 & & & \\ -1 & 2 & -1 & & \\[-5pt] & -1 & 2 & \ddots & \\ & & \ddots & \ddots & -1 \\ & & & -1 & 1 \end{array}\right] \in\mathbb{R}^{n\times n}.$
It can be shown that $\widetilde{T}_n^{-1}$ has $(i,j)$ element $\min(i,j)$ and eigenvalues $4\cos( j \pi/(2n+1))^2$, $j=1:n$.
If we perturb the $(1,1)$ of $\widetilde{T}_n$ to $1$, we obtain a singular matrix, but suppose we perturb the $(1,1)$ element to $3$:
$\notag \widehat{T}_n = \left[ \begin{array}{@{}*{4}{r@{\mskip10mu}}r} 3 & -1 & & & \\ -1 & 2 & -1 & & \\[-5pt] & -1 & 2 & \ddots & \\ & & \ddots & \ddots & -1 \\ & & & -1 & 1 \end{array}\right] \in\mathbb{R}^{n\times n}.$
The inverse is $\widehat{T}_n^{-1} = G/2$, where $G$ with $(i,j)$ element $2\min(i,j)-1$ is a matrix of Givens, and the eigenvalues are $4\cos((2j-1)\pi/(4n))^2$, $j=1:n$.
## Notes
The factorization $T_n = L_n^TL_n$ is noted by Strang (2012).
For a derivation of the eigenvalues and eigenvectors see Todd (1977, p. 155 ff.). For the eigenvalues of $\widetilde{T}_n$ see Fortiana and Cuadras (1997). Givens’s matrix is described by Newman and Todd (1958) and Todd (1977).
## References
This is a minimal set of references, which contain further useful references within.
## Related Blog Posts
This article is part of the “What Is” series, available from https://nhigham.com/index-of-what-is-articles/ and in PDF form from the GitHub repository https://github.com/higham/what-is. | 1,614 | 4,521 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 61, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.756345 |
https://blueggsee.com/2022/12/15/how-many-giant-pandas-were-theresdg15-life-on-land/ | 1,674,786,521,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00422.warc.gz | 160,420,250 | 16,961 | # How many giant pandas were there?(SDG15: Life on Land)
Math skills: Percentage
Related issues: Biodiversity / SDG15: Life on Land
During 1975 and 2005, Giant Pandas numbers had fallen from 2500 to 1600 largely due to loss of habitat and poachers. However after decades of conservation efforts, Giant Pandas are on the road to recovery. Giant Pandas numbers had increased by 16% during 2005 and 2015. About how many Giant Pandas were there in 2015?
A. 2,900
B. 2,100
C. 1,900
NSWER C
1600 x 1.16 = 1856 = about 1900
Source: WWF History of the Giant Panda | 173 | 565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | longest | en | 0.960044 |
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Geometry Surface Area and Volume of 3-D Figures Wall Posters and Graphic Organizer
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\$4.00 | 901 | 4,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-22 | longest | en | 0.831072 |
http://ts-web.info/blaise-pascals-essay-on-conic-sections-23/ | 1,575,844,332,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00105.warc.gz | 144,425,658 | 10,432 | ### BLAISE PASCALS ESSAY ON CONIC SECTIONS
Again given the hexagon on a conic of Pascal’s theorem with the above notation for points in the first figure , we have [6]. The same group operation can be applied on a cone if we choose a point E on the cone and a line MP in the plane. On the other hand, Pascal’s theorem follows from the above associativity formula, and thus from the associativity of the group operation of elliptic curves by way of continuity. At sixteen Pascal wrote an essay on conic sections; and in , at the age of eighteen, he constructed the first arithmetical machine, an instrument which, eight years later, he further improved. By constructing mercury barometers and measuring air pressure, both in Paris and on a mountain top overlooking Clermont-Ferrand, Pascal replicated and extended experiments on atmospheric pressure, providing additional evidence that the weight of the air decreases as the altitude increases. Orcibal, Jean and Lucien Jerphagnon. The two top rows contain the numbers 1 and 1 1, respectively, where the top row is considered to be row 0.
Although not published until , Pascal wrote an Essay on Conics in , which approached the geometry of conics using projective methods. Retrieved from ” https: Among the contemporaries of Descartes none displayed greater natural genius than Pascal, but his mathematical reputation rests more on what he might have done than on what he actually effected, as during a considerable part of his life he deemed it his duty to devote his whole time to religious exercises. Pascal’s original note [1] has no proof, but there are various modern proofs of the theorem. Furthermore, the 20 Cayley lines pass four at a time through 15 points known as the Salmon points. I have read somewhere, but I cannot lay my hand on the authority, that his proof merely consisted in turning the angular points of a triangular piece of paper over so as to meet in the centre of the inscribed circle: In , at the age of eighteen, Pascal constructed the first arithmetic machine to help his father with tax computations.
# Blaise Pascal ( – )
If one chooses suitable lines of the Pascal-figures as pascalw at infinity one gets many interesting figures on parabolas and hyperbolas. There exist 5-point, 4-point and 3-point degenerate cases of Pascal’s theorem.
Blaise Pascal Conic sections Theorems in projective geometry Theorems in plane geometry Theorems in geometry Euclidean plane geometry. Pascal’s first formally published mathematical work was his influential Essay on the Conic Sections ; he pascwls then Its publication was an immediate success and has gained the reputation of marking the beginning of modern French prose.
SHOW MY HOMEWORK WGSB
## Pascal’s theorem
From observations of diminishing air pressure at different altitudes, he inferred the vacuum of outer space, a discovery which earned him the contempt philosophy abhors a vacuum of the more philosophical Descartes.
Oxford University Press, Next, suppose that the first player has gained two points and the second player none, and that they are about to play for a point; the condition then is that, if the first player gain this point, he secures the game and takes the 64 pistoles, and, if the second player gain this point, then the players will be in the situation already examined, in which the first player is entitled to 48 pistoles and the second to 16 pistoles. It should be noted that the originality of his work in physics has been questioned, as some historians of science have described it as popularization, or even plagiarism Kline Pascal’s last mathematical gesturelike his first, was geometrical: A cycloid is defined as the curve produced by the locus of points of a point on the circumference of a circle which rolls along a straight line.
There also exists a simple proof for Pascal’s theorem for a circle using the law of sines and similarity. InPascal made his last major contribution to mathematics through his investigation of problems of the cycloid.
Finally, suppose that the first player has gained one point and the second player none. This result is diagramed below in Figure 1, where points R, S, and T lie in a sectionx.
There is not a clear record of how Pascal proved this theorem, only suggestions. However, as a result of this popularization, Pascal made an important indirect contribution to each field in which he studied, namely bringing sectilns to the problems of the field and thereby creating interest, excitement, and advancement within those fields.
The answer is obtained using the arithmetical triangle. The letters were written in the summer ofonly months before the traumatic carriage accident. Numbers in the first line are called first order numbers; those in blxise second line are called second order, or natural numbers; those in the third line are called third order numbers, and so on. Inat the age of eighteen, Pascal constructed the first arithmetic machine to help his father with tax ln.
PHD COURSEWORK SYLLABUS JIWAJI UNIVERSITY
Like his work on statistics, it is “expectational” in character.
Pascal was troubled by constant illness, including recurrent migraines and what proved to be cancer of the stomach. As Thomas Kirkman proved inthese 60 lines can be associated with 60 points in such a way that each point is on three lines and each line contains three points.
If you have any comments that would be useful, especially for use at the high school level, please send e-mail to esiwdivad yahoo. However, the theorem remains valid in the Euclidean plane, with the correct interpretation of what happens when some opposite sides of the hexagon are parallel.
His father, a local judge at Clermont, and himself of some scientific reputation, moved to Paris inpartly to prosecute his own scientific studies, partly to carry on the education of his only son, who had already displayed exceptional ability.
Close students of that correspondence give Fermat credit for the more substantial contribution, but the contribution would not have been made without the collaboration. Wikimedia Commons has media related to Pascal’s hexagram. In effect, he puts his argument that, as the value of eternal happiness must be infinite, then, even if the probability of a religious life ensuring eternal happiness be very small, still the expectation which is measured by the product of the two must be of sufficient magnitude to make it worth while to be religious.
Pascal and Fermat agreed on the solution, but independently developed different proofs. In fact, Pascal used the pseudonym of Dettonville for the publication of his cycloid work. The general solution in which the skill of the players is unequal is given in many modern text-books on algebra, and agrees with Pascal’s result, though of course the notation of the latter is different and less convenient.
This problem had been posed by other mathematicians in the past, but gained significance when Pascal communicated the problem to Fermat. He recorded his own solutions in letters to Carcavi. This naturally excited the boy’s curiosity, and one day, being then twelve years old, he asked in what geometry consisted. | 1,477 | 7,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-51 | latest | en | 0.947733 |
https://education.ti.com/en/activity/detail?id=24F03A578EF74CF6A653A46EC8C629FD | 1,713,277,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00422.warc.gz | 205,136,850 | 14,345 | Education Technology
# Activities
• ##### Subject Area
• Math: Statistics: Displaying and Describing Univariate Data
9-12
45 Minutes
• ##### Device
• TI-Nspire™ CX/CX II
• TI-Nspire™ CX CAS/CX II CAS
• TI-Nspire™ Navigator™
• ##### Software
TI-Nspire™ CX
TI-Nspire™ CX CAS
5.0
#### Activity Overview
Students will recognize that the mean and standard deviation (SD) and the median and interquartile range (IQR) are two ways to measure center and spread.
#### Objectives
• Students will recognize that the mean and standard deviation (SD) and the median and interquartile range (IQR) are two ways to measure center and spread.
• Students will recognize that in a skewed distribution the mean is pulled in the direction of the tail, and the standard deviation is increased; in a distribution with an outlier, the mean is pulled in the direction of the outlier, and the standard deviation is increased.
• Students will be able to illustrate that the median and IQR are resistant to skewness/outliers while the mean and standard deviation are not.
• Students will recognize that the median and IQR are preferred when a distribution is skewed while either mean and SD or median/IQR are acceptable for approximately symmetric distributions.
#### Vocabulary
• Bimodal or unimodal
• Interquartile range
• Mean, median, and standard deviation
• Outlier
• Resistant
• Akewed or symmetric distribution | 326 | 1,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-18 | latest | en | 0.839982 |
https://search.r-project.org/CRAN/refmans/bayesDccGarch/html/logLikDccGarch.html | 1,638,965,778,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363510.40/warc/CC-MAIN-20211208114112-20211208144112-00005.warc.gz | 561,082,073 | 2,303 | logLikDccGarch {bayesDccGarch} R Documentation
## The logarithm of likelihood function of DCC-GARCH(1,1) Model.
### Description
Compute the logarithm of likelihood function of DCC-GARCH(1,1) Model if mY is a matrix or the logarithm of likelihood function of GARCH(1,1) Model if mY is numeric vector.
### Usage
logLikDccGarch(mY, omega = rep(0.03, ncol(mY)), alpha = rep(0.03, ncol(mY)),
beta = rep(0.8, ncol(mY)), a = 0.03, b = 0.8, gamma = rep(1, ncol(mY)),
tail = 10, errorDist = 2)
### Arguments
mY a matrix of the data (n \times k). omega a numeric vector (k \times 1) with the the values of ω_i parameters. Default: rep(0.03, ncol(mY)). alpha a numeric vector (k \times 1) with the the values of α_i parameters. Default: rep(0.03, ncol(mY)). beta a numeric vector (k \times 1) with the the values of β_i parameters. Default: rep(0.80, ncol(mY)). a a numeric value of the a parameter. Default: 0.03. b a numeric value of the b parameter. Default: 0.8. gamma a numeric vector (k \times 1) with the values of γ_i parameters. Default: rep(1.0, ncol(mY)). tail a numeric value of ν parameter if errorDist = 2 or of δ parameter if errorDist = 3. If errorDist = 1 so this arguments is no used. errorDist a probability distribution for errors. Use errorDist=1 for SSNorm, errorDist=2 for SST or errorDist=3 for SSGED. Default: 2.
### Details
The log-likelihood of the model GARCH(1,1) is computed if mY has just one column. The arguments a and b are not consider in this case.
### Value
Return a list with the elements:
$H a matrix where the lines are the H_t values for t=1,...,n. $value the value of the logarithm of likelihood function.
### Author(s)
Jose Augusto Fiorucci, Ricardo Sandes Ehlers and Francisco Louzada
### References
Fioruci, J.A., Ehlers, R.S., Andrade Filho, M.G. Bayesian multivariate GARCH models with dynamic correlations and asymmetric error distributions, Journal of Applied Statistics, 41(2), 320–331, 2014a. <doi:10.1080/02664763.2013.839635>
Fioruci, J.A., Ehlers, R.S., Louzada, F. BayesDccGarch - An Implementation of Multivariate GARCH DCC Models, ArXiv e-prints, 2014b. https://ui.adsabs.harvard.edu/abs/2014arXiv1412.2967F/abstract.
bayesDccGarch-package, bayesDccGarch
### Examples
data(DaxCacNik)
Dax = DaxCacNik[,1]
###### log-likelihood function of GARCH(1,1) model with SST innovations ####
logLikDccGarch(Dax, omega=0.03, alpha=0.03, beta=0.8, gamma=0.7)$value ###### log-likelihood function of DCC-GARCH(1,1) model with SST innovations #### logLikDccGarch(DaxCacNik, beta=c(0.82,0.91,0.85), gamma=c(0.7, 1.3, 1.7), tail=10)$value
[Package bayesDccGarch version 3.0.3 Index] | 861 | 2,641 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.472858 |
http://ebookmarket.org/pdf/probability-problems-solutions | 1,444,421,108,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737935292.75/warc/CC-MAIN-20151001221855-00198-ip-10-137-6-227.ec2.internal.warc.gz | 101,342,001 | 13,909 | # Probability Problems Solutions pdfs
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Introduction to Probability: Problem Solutions
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Introduction to Probability: Problem Solutions
Introduction to Probability: Problem Solutions (last updated: 5/15/07) c Dimitri P. Bertsekas and John N. Tsitsiklis Massachusetts Institute of Technology
http://www.athenasc.com/probsolved.pdf
Date added: December 21, 2011 - Views: 297
Probability and Statistics Problems - Solutions
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Worksheet: Binomial Distribution Problems Problem 1. A manufacturer of electronics components produces precision resistors designed to have a tolerance of ±1%.
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Date added: January 29, 2012 - Views: 51 | 3,846 | 15,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2015-40 | longest | en | 0.786065 |
https://www.distancesto.com/fuel-cost/in/siaha-to-tialdawngi-lung-rd/history/1118776.html | 1,638,888,614,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00602.warc.gz | 788,317,404 | 14,579 | # INR1.66 Total Cost of Fuel from Siaha to Tialdawngi Lung Rd
Your trip to Tialdawngi Lung Rd will consume a total of 0.66 gallons of fuel.
Trip start from Siaha, IN and ends at Tialdawngi Lung Rd, IN.
Trip (26.5 mi) Siaha » Tialdawngi Lung Rd
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Siaha, Mizoram 796901, India and end at Tialdawngi Lung Rd, Vawmbuk, Mizoram 796901, India.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Siaha to Tialdawngi Lung Rd plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Siaha to Tialdawngi Lung Rd.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Siaha to Tialdawngi Lung Rd.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Siaha to Tialdawngi Lung Rd.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Tialdawngi Lung Rd are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Siaha to Tialdawngi Lung Rd.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Siaha to Tialdawngi Lung Rd.
Speaking of travel time, a flight to Tialdawngi Lung Rd takes up a lot less. How much less? Flight time from Siaha to Tialdawngi Lung Rd.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Siaha to Tialdawngi Lung Rd.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
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Fuel Cost from Siaha to Khawhai | 738 | 2,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-49 | latest | en | 0.955443 |
http://mathhelpforum.com/trigonometry/109354-confused-problem-print.html | 1,527,366,131,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00412.warc.gz | 195,136,445 | 2,941 | # Confused on a problem
• Oct 20th 2009, 06:14 PM
Thatoneguy12345
Confused on a problem
I have to graph y=3/2sin2(x+pi/4)
The endpoints are
[-pi,4 and pi/4]
I was wondering what I had to do to divide it to four intervals? My textbook says to add the xvalues of the endpoints and dividing by 2. But what are the xvalues in those endpoints? Very confused! Would appreciate the help!
• Oct 20th 2009, 06:42 PM
skeeter
Quote:
Originally Posted by Thatoneguy12345
I have to graph y=3/2sin2(x+pi/4)
The endpoints are
[-pi,4 and pi/4] these are the endpoint x-values
I was wondering what I had to do to divide it to four intervals?
My textbook says to add the xvalues of the endpoints and dividing by 2.
this step would break it into 2 intervals, not 4.
...
• Oct 20th 2009, 06:56 PM
Thatoneguy12345
Thanks a lot for replying! So how would I split it into 4 intervals then?
Before I used to just add both of the endpoints and multiply the sum by 1/4, 1/2, 3/4. That gives me a completely different answer now and the back of the book has different points.
• Oct 20th 2009, 07:05 PM
skeeter
Quote:
Originally Posted by Thatoneguy12345
So how would I split it into 4 intervals then?
each interval will have a length of $\displaystyle \textcolor{red}{\frac{\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)}{4} = \frac{\pi}{8}}$
... | 414 | 1,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.926341 |
https://www.coursehero.com/file/6225582/quiz6/ | 1,516,718,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891976.74/warc/CC-MAIN-20180123131643-20180123151643-00704.warc.gz | 905,548,083 | 29,608 | quiz6
# quiz6 - X 2 5(2 pt Determine the value of the cumulative...
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Quiz 6. October 13, 2010 Seat # _________ Name: _________________________ Closed book and notes. No calculator. Consider the probability density function f X ( y ) = 0.1 for 0 y c and zero elsewhere. 1. (2 pt) Show that c = 10. 2. (2 pt) Determine the value of f X (5.6). 3. (2 pt) Determine the value of E( X ). 4. (2 pt) Determine the value of E(
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Unformatted text preview: X 2 ). 5. (2 pt) Determine the value of the cumulative distribution function F X (5.6). ______________________________________________________________________ Write on the back any concerns about the weekly quizzes or the course in general. IE 230 – Page 1 of 1 – Schmeiser...
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Ask a homework question - tutors are online | 252 | 947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-05 | latest | en | 0.675108 |
https://numberworld.info/1293600 | 1,623,724,467,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00303.warc.gz | 401,323,996 | 5,154 | # Number 1293600
### Properties of number 1293600
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 3 * 5 * 5 * 7 * 7 * 11
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
13bd20
Base 32:
17f90
sin(1293600)
-0.86270682045346
cos(1293600)
0.50570440174383
tan(1293600)
-1.7059507836566
ln(1293600)
14.072939587245
lg(1293600)
6.1118000068983
sqrt(1293600)
1137.365376649
Square(1293600)
### Number Look Up
Look Up
1293600 which is pronounced (one million two hundred ninety-three thousand six hundred) is a very great figure. The cross sum of 1293600 is 21. If you factorisate the figure 1293600 you will get these result 2 * 2 * 2 * 2 * 2 * 3 * 5 * 5 * 7 * 7 * 11. 1293600 has 216 divisors ( 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 14, 15, 16, 20, 21, 22, 24, 25, 28, 30, 32, 33, 35, 40, 42, 44, 48, 49, 50, 55, 56, 60, 66, 70, 75, 77, 80, 84, 88, 96, 98, 100, 105, 110, 112, 120, 132, 140, 147, 150, 154, 160, 165, 168, 175, 176, 196, 200, 210, 220, 224, 231, 240, 245, 264, 275, 280, 294, 300, 308, 330, 336, 350, 352, 385, 392, 400, 420, 440, 462, 480, 490, 525, 528, 539, 550, 560, 588, 600, 616, 660, 672, 700, 735, 770, 784, 800, 825, 840, 880, 924, 980, 1050, 1056, 1078, 1100, 1120, 1155, 1176, 1200, 1225, 1232, 1320, 1400, 1470, 1540, 1568, 1617, 1650, 1680, 1760, 1848, 1925, 1960, 2100, 2156, 2200, 2310, 2352, 2400, 2450, 2464, 2640, 2695, 2800, 2940, 3080, 3234, 3300, 3360, 3675, 3696, 3850, 3920, 4200, 4312, 4400, 4620, 4704, 4900, 5280, 5390, 5600, 5775, 5880, 6160, 6468, 6600, 7350, 7392, 7700, 7840, 8085, 8400, 8624, 8800, 9240, 9800, 10780, 11550, 11760, 12320, 12936, 13200, 13475, 14700, 15400, 16170, 16800, 17248, 18480, 19600, 21560, 23100, 23520, 25872, 26400, 26950, 29400, 30800, 32340, 36960, 39200, 40425, 43120, 46200, 51744, 53900, 58800, 61600, 64680, 80850, 86240, 92400, 107800, 117600, 129360, 161700, 184800, 215600, 258720, 323400, 431200, 646800, 1293600 ) whith a sum of 5343408. The number 1293600 is not a prime number. The number 1293600 is not a fibonacci number. The number 1293600 is not a Bell Number. The number 1293600 is not a Catalan Number. The convertion of 1293600 to base 2 (Binary) is 100111011110100100000. The convertion of 1293600 to base 3 (Ternary) is 2102201111010. The convertion of 1293600 to base 4 (Quaternary) is 10323310200. The convertion of 1293600 to base 5 (Quintal) is 312343400. The convertion of 1293600 to base 8 (Octal) is 4736440. The convertion of 1293600 to base 16 (Hexadecimal) is 13bd20. The convertion of 1293600 to base 32 is 17f90. The sine of 1293600 is -0.86270682045346. The cosine of the number 1293600 is 0.50570440174383. The tangent of the number 1293600 is -1.7059507836566. The square root of 1293600 is 1137.365376649.
If you square 1293600 you will get the following result 1673400960000. The natural logarithm of 1293600 is 14.072939587245 and the decimal logarithm is 6.1118000068983. that 1293600 is special number! | 1,442 | 3,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-25 | latest | en | 0.433871 |
http://www.ck12.org/algebra/Write-an-Equation-Given-Two-Points/lesson/Write-an-Equation-Given-Two-Points-BSC-ALG/ | 1,477,224,543,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719273.37/warc/CC-MAIN-20161020183839-00229-ip-10-171-6-4.ec2.internal.warc.gz | 364,972,133 | 42,373 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Write an Equation Given Two Points
## Compose linear functions given two points
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Write an Equation Given Two Points
Suppose two travelers were lost in a forest. From the same spot, one person traveled 5 miles east and 10 miles south, while the other person traveled 2 miles west and 8 miles north. If a coordinate plane were transposed on top of the forest, with the line going from west to east as the x\begin{align*}x\end{align*}-axis and the line going from north to south as the y\begin{align*}y\end{align*}-axis, could you write the equation of the line that passes through the points representing the travelers' new locations?
### Writing an Equation Given Two Points
In many cases, especially real-world situations, you are given neither the slope nor the y\begin{align*}y-\end{align*}intercept. You might have only two points to use to determine the equation of the line.
To find an equation for a line between two points, you need two things:
1. The y\begin{align*}y-\end{align*}intercept of the graph
2. The slope of the line
Previously, you learned how to determine the slope between two points. Let’s repeat the formula here:
The slope between any two points (x1,y1)\begin{align*}(x_1, y_1 )\end{align*} and (x2,y2)\begin{align*}(x_2, y_2)\end{align*} is: slope=y2y1x2x1\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}\end{align*}.
The procedure for determining a line given two points is the same five-step process as writing an equation given the slope and a point.
#### Let's write the equation of the following lines:
1. The line containing the points (3, 2) and (–2, 4).
You need the slope of the line. Find the line's slope by using the formula. Choose one ordered pair to represent (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and the other ordered pair to represent (x2,y2)\begin{align*}(x_2,y_2)\end{align*}.
slope=y2y1x2x1=4223=25\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{-2-3}=-\frac{2}{5}\end{align*}
Now use the five-step process to find the equation for this line.
Step 1: Begin by writing the formula for slope-intercept form.
y=mx+b\begin{align*}y=mx+b\end{align*}
Step 2: Substitute the given slope for m\begin{align*}m\end{align*}.
y=25x+b\begin{align*}y=\frac{-2}{5} x+b\end{align*}
Step 3: Use one of the ordered pairs you are given, (–2, 4), and substitute these values for the variables x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} in the equation.
4=(25)(2)+b\begin{align*}4=\left (\frac{-2}{5}\right )(-2)+b\end{align*}
Step 4: Solve for b\begin{align*}b\end{align*} (the y\begin{align*}y-\end{align*}intercept of the graph).
4445165=45+b=4545+b=b\begin{align*}4& =\frac{4}{5}+b\\ 4-\frac{4}{5}& =\frac{4}{5}-\frac{4}{5}+b\\ \frac{16}{5}& =b\end{align*}
Step 5: Rewrite y=mx+b\begin{align*}y=mx+b\end{align*}, substituting the slope for m\begin{align*}m\end{align*} and the y\begin{align*}y-\end{align*}intercept for b\begin{align*}b\end{align*}.
y=25x+165\begin{align*}y=\frac{-2}{5} x+\frac{16}{5}\end{align*}
1. The line containing the points (–4, 1) and (–2, 3).
Find the slope of the line: m=y2y1x2x1=312(4)=22=1\begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-1}{-2-(-4)}=\frac{2}{2}=1\end{align*}.
Step 1: Start with the slope–intercept form of the line, y=mx+b\begin{align*}y=mx+b\end{align*}.
Step 2: Substitute the value of the slope for m:y=(1)x+b\begin{align*}m: y=(1)x+b\end{align*}.
Step 3: Substitute the coordinates (–2, 3) into the equation for the variables \begin{align*}x\end{align*} and \begin{align*}y : 3=-2+b\end{align*}.
Step 4: Solve for \begin{align*}b\end{align*}: \begin{align*}3 = -2 + b\\ 5 = b\end{align*}
Step 5: Rewrite the equation, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}: \begin{align*}y=x+5\end{align*}.
1. The line containing the points (3,6) and (-2, 6).
Find the slope of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-6}{-2-(3)}=\frac{0}{-5}=0\end{align*}.
Step 1: Start with the slope–intercept form of the line \begin{align*}y=mx+b\end{align*}.
Step 2: Substitute the value of the slope in for \begin{align*}m: y=(0)x+b \Rightarrow y=b\end{align*}.
Notice that this is an equation where \begin{align*}y\end{align*} equals some number. This means it is a horizontal line. This makes sense since the slope is zero and a horizontal line has a slope of zero.
Step 3 and 4: Substitute the coordinates (3, 6) into the equation for the variables \begin{align*}x\end{align*} and \begin{align*}y : 6=(0)3+b \Rightarrow b=6\end{align*}.
Step 5: Rewrite the equation, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}: \begin{align*}y=6\end{align*}.
We can see when this will happen in the future, without having to do all the work. Because the two \begin{align*}y\end{align*} values were the same, this must be a horizontal line.
### Examples
#### Example 1
Earlier, you were told about two travelers who were lost in a forest. From the same spot, one person traveled 5 miles east and 10 miles south, while the other person traveled 2 miles west and 8 miles north. If a coordinate plane were transposed on top of the forest, with the line going from west to east as the \begin{align*}x\end{align*}-axis and the line going from north to south as the \begin{align*}y\end{align*}-axis, what is the equation of the line that passes through the points representing the travelers' new locations?
If a coordinate plane was placed on top of the forest, west would represent the negative \begin{align*}x\end{align*} direction and east would represent the positive \begin{align*}x\end{align*} direction while south would represent the negative \begin{align*}y\end{align*} direction and north would represent the positive \begin{align*}y\end{align*} direction. 5 miles east and 10 miles south places one traveler at the point (5, -10). 2 miles west and 8 miles north places the other traveler at the point (-2, 8).
First, find the slope between of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{-10-8}{5-(-2)}=-\frac{18}{7}\end{align*}
Step 1: Start with the slope–intercept form of the line, \begin{align*}y=mx+b\end{align*}.
Step 2: Substitute the value of the slope for \begin{align*}m: y=(-\frac{18}{7})x+b\end{align*}.
Step 3: Substitute the coordinates (-2, 8) into the equation for the variables \begin{align*}x\end{align*} and \begin{align*}y : 8=(-\frac{18}{7}\times -2)+b\end{align*}.
Step 4: Solve for \begin{align*}b\end{align*}: \begin{align*}y : 8=(-\frac{18}{7}\times -2)+b\\ 8 = \frac{36}{7} + b\\ \frac{20}{7} = b \end{align*}
Step 5: Rewrite the equation, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}: \begin{align*}y=-\frac{18}{7}x+\frac{20}{7}\end{align*}.
#### Example 2
Write the equation of the line containing the points (2, -3) and (2, 10).
In this case, notice that the two \begin{align*}x\end{align*} values are the same. What does this mean? Remember, that for a vertical line, the \begin{align*}x\end{align*} value stays the same no matter what the \begin{align*}y\end{align*} value is.
Since we are trying to write an equation of a line, and in both cases \begin{align*}x=2\end{align*}, we can conclude that our equation is:
\begin{align*}x=2\end{align*}
Note that if we were to calculate the slope of the line given the two points, we would get the following:
\begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-(-2)}{2-(2)}=\frac{-1}{0}=\text{undefined}\end{align*}.
Since our slope is undefined, it must be a vertical line. We would come to the same conclusion that this is a vertical line where \begin{align*}x=2\end{align*}.
### Review
1. What is the first step in finding the equation of a line given two points?
In 2–7, find the equation of the line in slope–intercept form.
1. The line containing the points (2, 6) and (5, 0).
2. The line containing the points (5, –2) and (8, 4).
3. The line containing the points (3, 5) and (–3, 0).
4. The line containing the points (10, 15) and (12, 20).
Mixed Review
1. Translate into an algebraic sentence: One-third of a number is seven less than that number.
2. The perimeter of a square is 67 cm. What is the length of its side?
3. A hockey team played 17 games. They won two more than they lost. They lost 3 more than they tied. How many games did they win, lose, and tie?
4. Simplify \begin{align*}\frac{(30-4+4 \div 2) \div (21 \div 3)}{2}\end{align*}.
5. What is the opposite of 16.76?
6. Graph the following on a number line: \begin{align*}\left \{6,\frac{11}{3},-5.65,\frac{21}{7}\right \}\end{align*}.
7. Simplify: \begin{align*}[(-4+4.5)+(18-|-13|)+(-3.3)]\end{align*}.
To see the Review answers, open this PDF file and look for section 5.2.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
slope-intercept form
$y=(slope)x+(y-$intercept) or $y=(m)x+b$, where $m = slope$ and $b = y-$intercept.
undefined slope
An undefined slope cannot be computed. Vertical lines have undefined slopes. | 3,164 | 9,513 | {"found_math": true, "script_math_tex": 73, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 4, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2016-44 | longest | en | 0.86256 |
https://www.isnt.org.in/numpy-flip-example-with-code-examples.html | 1,722,649,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00755.warc.gz | 660,699,081 | 51,095 | # Numpy Flip Example With Code Examples
In this article, we will look at how to get the solution for the problem, Numpy Flip Example With Code Examples
## How do you flip an array vertically in Python?
To write a Python function matrixflip(m,d) that takes a two-dimensional matrix and a direction, where d is either 'h' or 'v'. If d == 'h' , the function should return the matrix flipped horizontally. If d == 'v' , the function should return the matrix flipped vertically.
``````# welcome to softhunt.net
# Python Program illustrating
# numpy.flip() method
import numpy as np
array = np.arange(27).reshape((3,3,3))
print("Original array : \n", array)
print("Flipped array : \n", np.flip(array, 0))```
```
## How do you flip in Python?
• To reverse the order of array elements in Python, use the numpy flip() method.
• The np.
• The flip() function takes two parameters.
• The first parameter is an array that takes the input array; the second parameter is the axis on which the array can be reversed.
## How do you flip a NumPy upside down matrix?
flipud() function. The flipud() function is used to flip an given array in the up/down direction. Flip the entries in each column in the up/down direction. Rows are preserved, but appear in a different order than before.
## What does NumPy flip do?
The numpy. flip() function reverses the order of array elements along the specified axis, preserving the shape of the array.
## How do I flip an image in Python?
To horizontally and vertically flip an image, we use the flip() function. The flip() function takes in 2 parameters. The first parameter is the image you want to flip. The second parameter is -1 (for horizontal and vertical flipping).
## How do you flip an array?
Solution Steps
• Place the two pointers (let start and end ) at the start and end of the array.
• Swap arr[start] and arr[end]
• Increment start and decrement end with 1.
• If start reached to the value length/2 or start ≥ end , then terminate otherwise repeat from step 2.
## How do I flip a NumPy image?
You can flip the image vertically and horizontally by using numpy. flip() , numpy. flipud() , numpy. fliplr() .
## How do you flip a 2d NumPy array?
Reverse Numpy array using np. flip()
• arr : Numpy array.
• axis : Axis along which it needs to flip / reverse the contents. If None: flip / reverse contents along all of the axes of the input array.
## How do you flip a matrix?
To flip a matrix horizontally means reversing each row of the matrix. For example, flipping [1, 1, 0, 0] horizontally results in [0, 0, 1, 1]. To invert a matrix means replacing each 0 by 1 and vice-versa. For example, inverting [0, 0, 1] results in [1, 1, 0].
## How do you mirror an image in Python?
ImageOps. flip()、ImageOps. mirror() The ImageOps module of the Python image processing library Pillow(PIL) provides flip() to flip the image upside down (vertically) and mirror() to flip the left and right (horizontally).
## Gostaria de saber por que quando executo a soma os numeros se juntam e nao somam, ex: 1+2 =12, sendo que isso n acontece com o restante das operacoes With Code Examples
Se você está tendo problemas com a soma de números, onde os números se juntam ao invés de somar, então você veio ao lugar certo. Este problema é comum quando se trabalha com números inteiros, mas pode ser facilmente resolvido. O problema ocorre quando os números são tratados como strings, ou seja, como cadeias de caracteres. Isso significa que, em vez de somar os números, o computador os concatena. Por exemplo, se você executar a soma 1 + 2, o computador tratará os números como st
## Can't add a react component to the component in the another file With Code Examples
Adding a React component to another file can be a tricky task, especially if you are new to React. It is important to understand the basics of React components and how they interact with each other before attempting to add a component to another file. In this article, we will discuss the basics of adding a React component to another file and provide code examples to help you get started. A React component is a self-contained piece of code that can be used to create a user interface. Components a
## svg icon not coloring correctly on hover With Code Examples
SVG icons are a great way to add visual interest to a website or application. They are lightweight, scalable, and can be easily customized with CSS. However, when it comes to using SVG icons in a hover state, there can be some issues with the icons not coloring correctly. This can be due to a variety of factors, including incorrect markup, incorrect CSS, or incorrect SVG code. In this article, we will discuss the common causes of SVG icon not coloring correctly on hover, and provide code example
## Split List In Half Python With Code Examples
In this article, we will look at how to get the solution for the problem, Split List In Half Python With Code Examples How do you split a list into parts? Use numpy. array_split() to split a list into n parts. Call numpy. array_split(list, n) to return a list of n NumPy arrays each containing approximately the same number of elements from list . print(a_list) length = len(a_list) middle_index = length//2 first_half = a_list[:middle_index] second_half = a_list[middle_index:] print(first_half) pr
## Python For Loop In One Line With Code Examples
In this article, we will look at how to get the solution for the problem, Python For Loop In One Line With Code Examples How do you avoid multiple loops in Python? 5 Ways To Break Out of Nested Loops in Python. Not as elegant as it should be. Add a Flag Variable. This is an effective solution. Raise an Exception. If we can't use the break keyword as expected. Check the Same Condition Again. Use the For-Else Syntax. Put It Into a Function. mylist = [1,4,5,8,9,11,13,12] newlist = [x for x | 1,402 | 5,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.6854 |
http://massivealgorithms.blogspot.com/2016/02/write-all-solutions-for-a3b3-c3-d3.html | 1,508,190,779,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00340.warc.gz | 273,149,822 | 40,534 | ## Monday, February 22, 2016
### Write all solutions for a^3+b^3 = c^3 + d^3, where a, b, c, d lie between [0, 10^5]
http://shirleyisnotageek.blogspot.com/2015/01/write-all-solutions-for-a3b3-c3-d3.html
This is a FB interview question. Here are some feasible approaches. I followed the priority queue method:
Using a priority queue is almost certainly the simplest solution, and also the most practical one, since it's O(n) storage (with a log factor if you require bignums). Any solution which involves computing all possible sums and putting them in a map will require O(n^2) storage, which soon becomes impractical.
My naive, non-optimized implementation using a priority queue is O(n^2 log(n)) time. Even so, it took less than five seconds for n = 10000 and about 750 seconds for n = 100000, using a couple of megabytes of storage. It certainly could be improved.
The basic idea, as per your comment, is to initialize a priority queue with pairs (a, a+1) for a in the range [1, N), and then repeatedly increment the second value of the smallest (by sum of cubes) tuple until it reaches N. If at any time the smallest two elements in the queue are equal, you have a solution. (I could paste the code, but you only asked for a hint.)
I certainly didn't fully understand how he maintained O(n) storage in the Queue. When I increment the second value, I still have to store the pairs that has a cube sum that is not equal to the smallest element in the queue. The total space for the queue is still O(n^2), or more accurately O(n^2 / 2).
So when the range goes up to 10^5, it gives me OutOfMemoryError().
However, just for comparison, I also tried the HashMap method, the memory leak occurs even when range only goes to 10^4. This means the Map has a worse performance than the Priority Queue.
`public` `class` `CubePair implements Comparable<cubepair> {`
` ``//num1 will always be smaller than num2`
` ``private` `int` `num1;`
` ``private` `int` `num2;`
` ``private` `int` `cubeSum;`
` ``public` `CubePair(``int` `a, ``int` `b) {`
` ``if` `(a > b) {`
` ``int` `tmp = a;`
` ``a = b;`
` ``b = tmp;`
` ``}`
` ``num1 = a;`
` ``num2 = b;`
` ``cubeSum = (``int``)(Math.pow(num1, 3) + Math.pow(num2, 3));`
` ``}`
` ``public` `boolean hasEqualSum(CubePair cp) {`
` ``return` `cp.cubeSum == cubeSum;`
` ``}`
` ``public` `boolean equals(CubePair cp) {`
` ``return` `num1 == cp.num1 && num2 == cp.num2;`
` ``}`
` ``public` `int` `getFirst() {`
` ``return` `num1;`
` ``}`
` ``public` `int` `getSecond() {`
` ``return` `num2;`
` ``}`
` ``public` `String toString() {`
` ``return` `String.valueOf(num1) + ``", "` `+ String.valueOf(num2) + ``";"``;`
` ``}`
` ``@Override`
` ``public` `int` `compareTo(CubePair cp) {`
` ``return` `cubeSum - cp.cubeSum;`
` ``}`
`}`
`import java.util.*;`
`public` `class` `FindCube {`
` ``public` `List<list tring=``""``>> sameCubeSums(``int` `range) {`
` ``if` `(range <= 0)`
` ``throw` `new` `IllegalArgumentException(``"range must be positive"``);`
` ``List<list tring=``""``>> rst = ``new` `ArrayList<list tring=``""``>> ();`
` ``PriorityQueue<cubepair> pq = ``new` `PriorityQueue<cubepair> ();`
` ``for` `(``int` `i = 1; i < range; i++) {`
` ``pq.add(``new` `CubePair(i, i + 1));`
` ``}`
` ``int` `prev = 0;`
` ``int` `maxSize = 0;`
` ``while` `(!pq.isEmpty()) {`
` ``maxSize = Math.max(maxSize, pq.size());`
` ``CubePair curr = pq.poll();`
` ``if` `(pq.isEmpty())`
` ``break``;`
` ``if` `(curr.hasEqualSum(pq.peek())) {`
` ``addList(rst, curr, pq.peek());`
` ``if` `(curr.getFirst() < pq.peek().getFirst())`
` ``curr = pq.poll();`
` ``else`
` ``pq.poll();`
` ``}`
` ``if` `(curr.getFirst() <= prev)`
` ``continue``;`
` ``for` `(``int` `i = curr.getSecond() + 1; i <= range; i++) {`
` ``CubePair tmp = ``new` `CubePair(curr.getFirst(), i);`
` ``if` `(tmp.hasEqualSum(pq.peek())) {`
` ``addList(rst, tmp, pq.peek());`
` ``}`
` ``else` `{`
` ``pq.add(tmp);`
` ``}`
` ``}`
` ``prev = curr.getFirst();`
` ``}`
` ``System.``out``.println(``"max queue size: "` `+ maxSize);`
` ``return` `rst;`
` ``}`
` ``private` `void` `addList(List<list tring=``""``>> rst, CubePair pair1, CubePair pair2) {`
` ``List<``string``> pair = ``new` `ArrayList<``string``> ();`
` ``pair.add(pair1.toString());`
` ``pair.add(pair2.toString());`
` ``rst.add(``new` `ArrayList<``string``> (pair));`
` ``}`
`}`
http://stackoverflow.com/questions/14454133/write-all-solutions-for-a3-b3-c3-d3
The HashMap method:
`public` `List<list tring=``""``>> sameCubeSums2(``int` `range) {`
` ``if` `(range <= 0)`
` ``throw` `new` `IllegalArgumentException(``"range must be positive"``);`
` ``Map<integer list=``""` `tring=``""``>> map = ``new` `HashMap<integer list=``""` `tring=``""``>>();`
` ``List<list tring=``""``>> rst = ``new` `ArrayList<list tring=``""``>> ();`
` ``for` `(``int` `i = 1; i < range; i++) {`
` ``for` `(``int` `j = i + 1; j <= range; j++) {`
` ``int` `cubeSum = (``int``)(Math.pow(i, 3) + Math.pow(j, 3));`
` ``if` `(!map.containsKey(cubeSum)) {`
` ``map.put(cubeSum, ``new` `ArrayList<``string``> ());`
` ``map.``get``(cubeSum).add(String.valueOf(i) + ``"; "` `+ String.valueOf(j));`
` ``}`
` ``else`
` ``map.``get``(cubeSum).add(String.valueOf(i) + ``"; "` `+ String.valueOf(j));`
` ``}`
` ``}`
` ``for` `(List<``string``> pairs : map.values()) {`
` ``if` `(pairs.size() > 1) `
` ``rst.add(pairs);`
` ``}`
` ``return` `rst;`
` ``}`
http://algobox.org/cube-sum/ | 1,964 | 5,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-43 | longest | en | 0.903377 |
https://oeis.org/A178147 | 1,561,384,754,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00250.warc.gz | 543,477,023 | 4,296 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A178147 Sum of squares d^2 of distinct divisors of n, d in {2, 3, 5}. 0
0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0, 29, 9, 4, 0, 13, 25, 4, 9, 4, 0, 38, 0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0, 29, 9, 4, 0, 13, 25, 4, 9, 4, 0, 38, 0, 4, 9, 4, 25, 13, 0, 4, 9, 29, 0, 13, 0, 4, 34, 4, 0, 13, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The sequence is periodic with period {0 4 9 4 25 13 0 4 9 29 0 13 0 4 34 4 0 13 0 29 9 4 0 13 25 4 9 4 0 38} of length 30. A generalization: let B={b_1,...,b_t} be a set of t positive (not necessarily distinct) integers and m>=0 an integer. For m>=0, let A(n)=Sum d^m over divisors d of n which are elements of B (with the multiplicities as in B). Calculating directly values of A(b_i),A(b_i+b_j),A(b_i+b_j+b_k),..., A(b_1+...+b_t), for the other values of A(n) we have the recursion: A(n)=Sum{1<=i<=t}A(n-b_i)- Sum{1<=i10. By the comment, up to 10 it is sufficient to calculate directly only values a(2)=4, a(3)=9, a(5)=25, a(7)=0, a(8)=4, a(10)=29. For other n's we can use the recursion, accepting formally a(n)=0 for n<0. So a(1)=0; a(4)=a(2)+a(1)=4;a(6)=a(4)+a(3)=4+9=13, a(9)=a(7)+a(6)-a(2)-a(1)=0+13-4+0=9. a(n) = -2*a(n-1) -2*a(n-2) -a(n-3) +a(n-5) +2*a(n-6) +2*a(n-7) +a(n-8). - R. J. Mathar, Jul 13 2010 G.f. -x^2*(4+17*x+30*x^2+55*x^3+80*x^4+38*x^6+76*x^5) / ( (x-1)*(1+x)*(1+x+x^2)*(x^4+x^3+x^2+x+1) ). - R. J. Mathar, Dec 17 2012 CROSSREFS Cf. A005063, A098002, A178142-A178144, A178146. Sequence in context: A103164 A210966 A300516 * A005063 A235323 A078615 Adjacent sequences: A178144 A178145 A178146 * A178148 A178149 A178150 KEYWORD nonn,easy,less AUTHOR Vladimir Shevelev, May 21 2010, May 23 2010 STATUS approved
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Last modified June 24 09:55 EDT 2019. Contains 324323 sequences. (Running on oeis4.) | 1,002 | 2,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-26 | latest | en | 0.633684 |
https://math.answers.com/Q/Can_you_divide_by_4_and_get_the_remainder_of_5 | 1,695,855,807,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00297.warc.gz | 426,244,466 | 46,983 | 0
# Can you divide by 4 and get the remainder of 5?
Updated: 9/20/2023
Wiki User
12y ago
no. The remainder should be less than the divisor. If you get a remainder of 5 and did not make any other mistake, add one to the quotient and the remainder will be 1.
Wiki User
12y ago
Study guides
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3068 Reviews | 103 | 340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.881741 |
http://www.h4labs.com/ml/islr/chapter10/10_10_melling.html | 1,723,151,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00427.warc.gz | 37,843,158 | 429,179 | ISLR Home
# Question
In this problem, you will generate simulated data, and then perform PCA and K-means clustering on the data.
1. Generate a simulated data set with 20 observations in each of three classes (i.e. 60 observations total), and 50 variables. Hint: There are a number of functions in R that you can use to generate data. One example is the rnorm() function; runif() is another option. Be sure to add a mean shift to the observations in each class so that there are three distinct classes.
2. Perform PCA on the 60 observations and plot the first two prin- cipal component score vectors. Use a different color to indicate the observations in each of the three classes. If the three classes appear separated in this plot, then continue on to part (c). If not, then return to part (a) and modify the simulation so that there is greater separation between the three classes. Do not continue to part (c) until the three classes show at least some separation in the first two principal component score vectors.
3. Perform K-means clustering of the observations with K = 3. How well do the clusters that you obtained in K-means cluster- ing compare to the true class labels? Hint: You can use the table() function in R to compare the true class labels to the class labels obtained by clustering. Be careful how you interpret the results: K-means clustering will arbitrarily number the clusters, so you cannot simply check whether the true class labels and clustering labels are the same.
4. Perform K-means clustering with K = 2. Describe your results.
5. Now perform K-means clustering with K = 4, and describe your results.
6. Now perform K-means clustering with K = 3 on the first two principal component score vectors, rather than on the raw data. That is, perform K-means clustering on the 60 × 2 matrix of which the first column is the first principal component score vector, and the second column is the second principal component score vector. Comment on the results.
7. Using the scale() function, perform K-means clustering with K = 3 on the data after scaling each variable to have standard deviation one. How do these results compare to those obtained in (b)? Explain.
``````library(ISLR)
library(tidyverse)``````
# 10a Generate Sample Data
Generate a simulated data set with 20 observations in each of three classes (i.e. 60 observations total), and 50 variables.
Hint: There are a number of functions in R that you can use to generate data. One example is the rnorm() function; runif() is another option. Be sure to add a mean shift to the observations in each class so that there are three distinct classes.
``````# From Lab 02
# x=matrix(rnorm(50*2), ncol=2)
set.seed(2)
x0 = matrix(rnorm(20*3*50, mean=sqrt(runif(1)), sd=0.1), ncol=50) # 0.001
x = x0
x[1:20, 2] = 1
x[21:40, 1] = 2
x[21:40, 2] = 2
x[41:60, 1] = 1``````
``````data = matrix(sapply(1:3,function(x){ rnorm(20*50, mean = 10*sqrt(x)) }),ncol=50) # 20 obs. in each class with 50 features.
dim(data)``````
``## [1] 60 50``
``````class=unlist(lapply(1:3,function(x){rep(x,20)}))
class``````
``````## [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## [39] 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3``````
# 10b Plot first two principal component score vectors
Perform PCA on the 60 observations and plot the first two principal component score vectors. Use a different color to indicate the observations in each of the three classes. If the three classes appear separated in this plot, then continue on to part (c). If not, then return to part (a) and modify the simulation so that there is greater separation between the three classes. Do not continue to part (c) until the three classes show at least some separation in the first two principal component score vectors.
``````pr.out=prcomp(data, scale=FALSE) # Don't need scale because mean=0, sd=0.001
#summary(pr.out)
#plot(pr.out\$x[,c(1,2)],col=class)``````
``````#plot(pr.out\$x[,c(1,2)], col=c(1,2,3))
#plot(pr.out\$x[,1:2], col=2:4, xlab="Z1", ylab="Z2", pch=19)
plot(pr.out\$x[,c(1,2)],col=class, pch=19)``````
# 10c K-means clustering, K = 3.
Perform K-means clustering of the observations with K = 3. How well do the clusters that you obtained in K-means clustering compare to the true class labels?
Hint: You can use the table() function in R to compare the true class labels to the class labels obtained by clustering. Be careful how you interpret the results: K-means clustering will arbitrarily number the clusters, so you cannot simply check whether the true class labels and clustering labels are the same.
``````km.out=kmeans(data,3,nstart=20)
km.out\$cluster``````
``````## [1] 1 1 3 1 1 1 1 1 1 3 1 1 3 3 1 1 1 1 1 1 3 3 3 3 3 3 1 3 3 3 3 3 3 1 1 3 3 3
## [39] 3 1 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2``````
``table(km.out\$cluster)``
``````##
## 1 2 3
## 20 19 21``````
``````#class_vars = c(rep(1,20), rep(2,20), rep(3,20))
table(km.out\$cluster, class)``````
``````## class
## 1 2 3
## 1 16 4 0
## 2 0 0 19
## 3 4 16 1``````
``````# plot(pr, col=(km.out\$cluster + 1), main="K-Means Clustering Results with K=3", xlab="", ylab="", pch=20, cex=2)
plot(data, col=km.out\$cluster)``````
# 10d K-means clustering, K = 2
Perform K-means clustering with K = 2. Describe your results
``````km.out=kmeans(data,2,nstart=20)
km.out\$cluster``````
``````## [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2
## [39] 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1``````
``table(km.out\$cluster)``
``````##
## 1 2
## 21 39``````
``table(km.out\$cluster, class)``
``````## class
## 1 2 3
## 1 0 1 20
## 2 20 19 0``````
41 misclassified (20+1+18+2)
# 10e K-means clustering, K = 4
Now perform K-means clustering with K = 4, and describe your results
``````km.out=kmeans(data,4,nstart=20)
km.out\$cluster``````
``````## [1] 1 2 1 1 2 2 2 2 1 1 2 2 4 4 2 2 1 1 1 2 1 1 4 4 1 4 1 4 4 4 4 4 1 1 1 4 4 4
## [39] 1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3``````
``table(km.out\$cluster)``
``````##
## 1 2 3 4
## 16 11 20 13``````
``table(km.out\$cluster, class)``
``````## class
## 1 2 3
## 1 8 8 0
## 2 10 1 0
## 3 0 0 20
## 4 2 11 0``````
# 10f K-means clustering, K = 3; 2 PCA
Now perform K-means clustering with K = 3 on the first two principal component score vectors, rather than on the raw data. That is, perform K-means clustering on the 60 × 2 matrix of which the first column is the first principal component score vector, and the second column is the second principal component score vector. Comment on the results.
``#summary(pr.out)``
``````km.out=kmeans(pr.out\$x[,1:2],3,nstart=20)
km.out\$cluster``````
``````## [1] 2 1 2 1 1 2 1 1 2 2 1 1 1 2 1 1 1 2 2 1 2 1 1 1 2 2 2 1 1 1 1 1 2 1 2 2 1 1
## [39] 2 1 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3``````
``table(km.out\$cluster)``
``````##
## 1 2 3
## 25 16 19``````
``table(km.out\$cluster, class)``
``````## class
## 1 2 3
## 1 12 12 1
## 2 8 8 0
## 3 0 0 19``````
# 10g Scale, K-means clustering, K = 3
Using the scale() function, perform K-means clustering with K = 3 on the data after scaling each variable to have standard deviation one. How do these results compare to those obtained in (b)? Explain.
``````sc = scale(data)
km.out = kmeans(sc, 3, nstart = 20)
km.out\$cluster``````
``````## [1] 3 3 2 1 3 3 1 1 3 2 1 3 1 2 3 3 3 3 1 1 2 3 1 1 3 2 3 3 1 1 3 1 2 3 3 2 1 1
## [39] 3 3 2 2 1 1 2 3 2 2 2 2 1 1 2 2 2 1 2 1 2 1``````
``plot(data, col=km.out\$cluster)`` | 2,859 | 7,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.866913 |
https://veliaf.wordpress.com/2010/06/18/the-weekly-warlock-wind-up-fire-engines-answer/ | 1,501,202,565,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436316.91/warc/CC-MAIN-20170728002503-20170728022503-00276.warc.gz | 728,926,592 | 35,883 | ## The Weekly Warlock Wind-Up: Fire Engines (answer)
Finally, a relevant image.
This week, I posed to you this conundrum:
A fire engine travels 6 miles to a fire at a speed of 30 mph. Its tank holds 500 gallons of water but has been leaking throughout the journey at a rate of 25 gallons per hour. The fire engine will need 496 gallons of water to put out the fire; by how many gallons will it be short?
So, how many? Answers after the break!
Working this out is actually fairly simple.
The first step is to work out how long the fire engine is moving for. We know it travels 6 miles at 30 mph, and so it must therefore be traveling for 12 minutes to cover that distance, which is 1/5 of an hour.
Next, we know that the water is lost at 25 gallons per hour. Therefore, we can do 25/5 (since the fire engine travels for 1/5 of an hour) to get a total of 5 gallons lost.
Finally, we are told that the engine will require 496 gallons to put the fire out, and that it starts with 500. If it loses 5 gallons, it therefore has 495 left when it arrives, which as you can see is just one gallon short!
Well, that might have been pretty unlucky for the fire crew but perhaps you were luckier in working the answer out? How’d you do?
Vel.
### 2 Responses to “The Weekly Warlock Wind-Up: Fire Engines (answer)”
1. Mishaweha Says:
I was surprised on how little water they actually lost. 25 gallons per hour sounds rather impressive.
I also liked how you could solve this one on scratch paper and not have to use a calculator. 🙂 Even if I second guess my math all the time when I don’t use one. XD
I think if you were one gallon of water short for putting out a fire, you could probably stomp out the rest. At least that’s what I hope they did, in this fictional tale!
• Veliaf Says:
Admittedly, I agree – I don’t think one gallon’s worth is really going to make a difference. With that said, the fire engine really shouldn’t be leaking at all! | 479 | 1,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-30 | longest | en | 0.963616 |
https://ch.mathworks.com/matlabcentral/cody/problems/497-make-an-awesome-ramp-for-a-tiny-motorcycle-stuntman/solutions/1184076 | 1,590,393,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388012.14/warc/CC-MAIN-20200525063708-20200525093708-00220.warc.gz | 304,035,082 | 15,611 | Cody
# Problem 497. Make an awesome ramp for a tiny motorcycle stuntman
Solution 1184076
Submitted on 9 May 2017 by Sienna Grace
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
v = [1 3 6 9 11]; y_correct = [1 3 6 9 11; 3 6 9 11 0; 6 9 11 0 0; 9 11 0 0 0; 11 0 0 0 0]; assert(isequal(vrooom(v),y_correct))
2 Pass
v = [1 2 3]; y_correct = [1 2 3; 2 3 0; 3 0 0]; assert(isequal(vrooom(v),y_correct))
3 Pass
v = [-1 0 2 8]; y_correct = [-1 0 2 8; 0 2 8 0; 2 8 0 0; 8 0 0 0]; assert(isequal(vrooom(v),y_correct)) | 275 | 637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-24 | latest | en | 0.562511 |
https://www.teacher.org/lesson-plan/roll-the-dice-probability-statistics/ | 1,618,494,038,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00331.warc.gz | 963,408,105 | 106,015 | # Roll the Dice – Probability & Statistics
JRDAssist
Middle School and High School Teacher
Students will learn about probability and statistics while rolling dice and flipping a coin carefully recording their results.
## Length of Time: Two 45 Minute Periods
### Common Core Alignment
CCSS.Math.Content.6.SP.B.5.A - Reporting the number of observations.
CCSS.Math.Content.6.SP.B.5.B - Describing the nature of the attribute under investigation, including how it was measured and its units of measurement.
CCSS.Math.Content.6.SP.B.5.C - Giving quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered.
CCSS.Math.Content.6.SP.B.5.D - Relating the choice of measures of center and variability to the shape of the data distribution and the context in which the data were gathered.
## Objectives & Outcomes
The learners will be able to understand the concept of probability and use their results to make predictions in real world situations and applications.
## Materials Needed
• One die and coin for each student
• plain white paper
• pencil
## Procedure
### Opening to Lesson
• Give one die to each student in the classroom.
• Have them stand in a line at an empty table, desk, or floor area.
• Each one will roll the die, and depending on the number rolled they will be sent to a predetermined area of the room. (Example: roll 1, 2, 3 go to…, roll 4, 5, 6 go to…) The students should be close to evenly divided.
• The teacher will ask a variety of questions to peak students’ interest such as, “Why did some of you roll sixes, and others roll ones?” And other similar questions.
• Teacher will direct students back to their seats
### Body of Lesson
#### Modeling
• Teacher will demonstrate coin tossing and the chances of getting a heads or a tails.
• Teacher will display large copy of results chart and demonstrate how to record die information and coin tossing results
#### Guided Practice
• Each student will use their materials and flip the coin 100 times recording the results
• Student will then analyze the results and make a prediction for the next 100 flips
• Student will again record the results of the 100 flips
• Each student will use their materials and roll the die 100 times recording the results
• Student will then analyze the results and make a prediction for the next 100 rolls
• Student will again record the results of the 100 rolls
• The teacher will demonstrate how to show probability in fraction form
#### Independent Practice
• The teacher will give each student a worksheet illustrating a full deck of playing cards
• On a separate page there will be questions about the probability of choosing aces, diamonds, face cards, etc.
• The teacher will collect the completed page. This will be used as an assessment.
### Closing
Ask students to give real world applications and situations of probability. Prompt students for answers such as weather predictions, sports, and many others.
## Assessment & Evaluation
Using an illustrated handout (worksheet) with a set number of playing cards (each suit represented, face cards, etc.), the students will answer questions related to the probability of choosing certain named card, suits, etc.
## Modification & Differentiation
Students may work in pairs. Use spinners instead of dice. Introduce colored M&Ms or Skittles, or other examples of randomness. Students could make graphs showing results. Discuss variables that may change typical results.
## Related Lesson Plans
##### The Value of a Number
The students will work in groups of 4-6 physically learning and reviewing place value.
##### Comparing Multiplication Facts (Hey Tocayo!)
Students will be assigned a number that has various factors and they will find partners with different factors that have the same product.
##### Numbers By The Book
This lesson will allow students to show a relationship between numbers and a visual representation in a manner that can be used for younger students.
##### Fractions, Decimals, Percents, & Graphs
This lesson is designed to explore fractions, decimals, percent’s, and graphs with TRIX cereal. | 918 | 4,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-17 | longest | en | 0.885799 |
https://kimiyuki.net/blog/2017/07/12/aoj-2255/ | 1,516,522,908,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890394.46/warc/CC-MAIN-20180121080507-20180121100507-00478.warc.gz | 700,222,908 | 5,443 | # AOJ 2255: 6/2(1+2)
,
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2255
## solution
• 再帰 どこで分割するかを毎回全部試す (これをメモ化すれば先に言った区間DP)
• 演算子の優先順位を全列挙 next_permutationとかで頑張る
## implementation
#include <cassert>
#include <cctype>
#include <iostream>
#include <set>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
using namespace std;
template <typename X, typename T> auto vectors(X x, T a) { return vector<T>(x, a); }
template <typename X, typename Y, typename Z, typename... Zs> auto vectors(X x, Y y, Z z, Zs... zs) { auto cont = vectors(y, z, zs...); return vector<decltype(cont)>(x, cont); }
struct expr_t {
int immidiate;
vector<expr_t> value;
vector<char> op;
};
expr_t parse_value(string::const_iterator & first, string::const_iterator last) {
assert (first != last);
assert (isdigit(*first));
int acc = 0;
while (first != last and isdigit(*first)) {
acc = acc * 10 + (*first - '0');
++ first;
}
return (expr_t) { acc };
}
expr_t parse_expr(string::const_iterator & first, string::const_iterator last);
expr_t parse_term(string::const_iterator & first, string::const_iterator last) {
assert (first != last);
if (*first == '(') {
++ first;
expr_t expr = parse_expr(first, last);
assert (*first == ')');
++ first;
return expr;
} else {
return parse_value(first, last);
}
}
expr_t parse_expr(string::const_iterator & first, string::const_iterator last) {
assert (first != last);
expr_t expr = { -1 };
expr.value.push_back(parse_term(first, last));
while (first != last and *first != ')') {
assert (*first == '+' or *first == '-' or *first == '*' or *first == '/');
expr.op.push_back(*first);
++ first;
expr.value.push_back(parse_term(first, last));
}
return expr;
}
expr_t parse(string const & s) {
auto first = s.begin();
expr_t expr = parse_expr(first, s.end());
assert (first == s.end());
return expr;
}
set<int> solve(expr_t const & expr) {
if (expr.immidiate != -1) {
return set<int>({ expr.immidiate });
} else {
int n = expr.value.size();
auto dp = vectors(n, n + 1, set<int>());
repeat (i, n) {
dp[i][i + 1] = solve(expr.value[i]);
}
repeat_from (len, 2, n + 1) {
repeat_from (r, len, n + 1) {
int l = r - len;
repeat_from (m, l, r - 1) {
set<int> & left = dp[l][m + 1];
set<int> & right = dp[m + 1][r];
char op = expr.op[m];
if (op == '+') {
for (int a : left) for (int b : right) dp[l][r].insert(a + b);
} else if (op == '-') {
for (int a : left) for (int b : right) dp[l][r].insert(a - b);
} else if (op == '*') {
for (int a : left) for (int b : right) dp[l][r].insert(a * b);
} else if (op == '/') {
for (int b : right) if (b != 0) for (int a : left) dp[l][r].insert(a / b);
}
}
}
}
return dp[0][n];
}
}
int main() {
while (true) {
string s; cin >> s;
if (s == "#") break;
expr_t expr = parse(s);
int result = solve(expr).size();
printf("%d\n", result);
}
return 0;
} | 932 | 2,892 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-05 | longest | en | 0.196034 |
https://www.industrialheating.com/articles/97650-how-to-optimize-gas-train-flow-to-improve-industrial-burner-efficiency | 1,695,948,128,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00216.warc.gz | 878,629,414 | 19,388 | When designing fuel trains for an industrial furnace or oven, one of the key parameters you should strive to maximize is flow.
Poor flow in a fuel train can limit the performance of the burner and can also put your entire system at risk of shutting down in the event of low fuel pressure. To avoid these outcomes, each component within the gas train must be optimized to enhance flow performance and the overall efficiency, reliability and cost of your industrial burner system. Let’s dive into what flow is, including its features and benefits, and discuss some strategies to maximize flow within your industrial burner application.
### What is flow?
For the purposes of this article, we will discuss flow in terms of volumetric flow, or the volume of fluid flowing through a system over a period of time (e.g., gallons or liters per minute). When discussing fuel-train components like a safety shutoff valve or pressure regulator, flow is often discussed in terms of the flow coefficient (Cv or Kv) – the higher the Cv or Kv, the better the flow.
We can define Cv as the number of gallons per minute of water at 60°F flowing through a system with a pressure difference of 1 PSI between the inlet and outlet. We can define Kv as the number of liters per minute (or cubic meters per hour) of water at a temperature between 5-40°C with a pressure difference of 1 bar between the inlet and outlet. By defining a standard fluid, temperature and pressure differential in this way, fuel-system designers can make apples-to-apples comparisons between components.
Flow is closely linked to another key parameter called pressure drop or head loss. As the fluid moves through a system, each component imparts some resistance or “friction” to the overall flow, resulting in a pressure drop. Pressure drops are additive across the fuel train and must be overcome for the system to operate properly. For example, if a fuel train has a total pressure drop of 5 PSI and the source pressure is 3 PSI, then the burner won’t be able to operate. We can think of both pressure drop and Cv as inversely related – as the Cv of a component increases, then the pressure drop decreases.
### The Variables Impacting Flow
When looking at a fuel train, there are a few areas in particular that can impact flow.
• Media. When considering the media, or fluid flowing through the system, density, temperature and viscosity can all impact flow rate. For example, cold oil is more viscous than hot oil and, as a result, will flow more slowly through the fuel train.
• Mechanical components. A mechanical component’s cross-sectional area and geometric complexity can impact the flow rating of the device. In general, larger components with a more direct flow path will have better flow rates. For example, valves with a monoblock design (two valves in one) are designed to have superior flow compared to two single valves piped in series.
• System design. Lastly, we must consider system design. Long piping with many bends and many devices can have an adverse effect on flow. In addition, for incompressible fluids like oil, if the starting elevation of the fuel train is lower than the end elevation, gravitational forces will apply their own pressure loss.
Fuel trains with poor flow performance can have a significant impact on the overall effectiveness and cost of your industrial burner system. Poor flow systems require a significant amount of inlet pressure for the burner to fire. If a system is installed in an area with low or fluctuating fuel pressure, the burner may not light or will periodically shut down.
For example, in densely populated cities with older gas infrastructure, periods of high gas demand (e.g., a cold winter day) can cause gas pressures across the region to drop. As a result, older systems with poor flow may shut down when heat is needed the most. Although this situation can be remedied by adding a booster pump to increase inlet pressure or by selecting larger pipe sizes and components, both solutions can significantly drive up costs.
ASCO Series HOV13B Hydramotor valve (left), Series 266 oil shutoff valve (middle) and Series 377 shutoff valve (right)
### The Benefits of a Three-Way Valve Design
Some valves – like ASCO Series HOV13B Hydramotor valves, Series 266 oil shutoff valves and Series 377 shutoff valves – are capable of three-way flow. Thanks to this design, the valve units will recirculate oil back to the heater when closed, keeping the oil from becoming too viscous. These valves are suitable for applications like burners, furnaces, incinerators, ovens and other heating equipment.
ASCO Series 158 gas valve and Series 159 motorized actuator
Comparing the Flow of Two Valves (U.S. Standard Units)
The following example compares the flow of two valves using the following equation:
Where:
Q = Flow in gallons per minute (gpm)
Cv = Flow coefficient
ΔP = Pressure differential between the valve’s inlet and outlet
SG = Specific gravity of the fluid, or the ratio of a fluid’s density compared to water
SG of water = 1
SG of natural gas = about 0.60 to 0.70, with variations due to composition
SG of diesel = about 0.82 to 0.95, with variations due to composition
Valve 1: ASCO Series 158 single body valve
Pipe size = 1.5 inches
Cv = 59
Valve 2: Competitor Valve
Pipe size = 1.5 inches
Cv = 40
Media: Natural gas
SG = 0.65
ΔP = 5 PSI
Results: Valve 1 has 48% more flow compared to valve 2.
Valve 1 Flow (Q) = 164 gpm
Valve 2 Flow (Q) = 111 gpm
Comparing the Flow of Two Valves (Metric Units)
The following example compares the flow of two valves using the following equation:
Where:
Q = Flow in gallons per minute (m3/hour)
Kv = Flow coefficient
ΔP = Pressure differential between the valve’s inlet and outlet
SG = Specific gravity of the fluid, or the ratio of a fluid’s density compared to water
SG of water = 1
SG of natural gas = about 0.60 to 0.70, with variations due to composition
SG of diesel = about 0.82 to 0.95, with variations due to composition
Valve 1: ASCO Series 158 single body valve
Pipe size = DN 40
Kv = 5
Valve 2: Competitor Valve
Pipe size = DN 40
Kv = 35
Media: Natural Gas
SG = 0.65
ΔP = 345 mbar
Results: Valve 1 has 48% more flow compared to valve 2.
Valve 1 Flow (Q) = 37 gpm
Valve 2 Flow (Q) = 25 gpm
Strategies to Maximize Fuel-Train Flow
Selecting the right fuel-train components can overcome these issues without breaking the bank, and there are a few strategies to consider when maximizing a fuel-train system’s performance.
• Focus on high-impact components. In a typical system, up to 70% of pressure drops are caused by the two safety shutoff valves. If the goal is to maximize fuel-train flow, it’s imperative to select a shutoff valve with the highest flow coefficient (Cv or Kv) rating for the given pipe size. Newer safety shutoff valves have been optimized for flow. In some cases, a fuel train can be designed one or two sizes smaller while maintaining performance, reducing the overall cost of the system.
• Minimize the number of components. A typical fuel train contains the following components: a regulator, two safety shutoff valves, pressure switches and a flow control valve. The current trend in the industrial burner industry is to combine these components into a single valve body – typically called a monoblock or double-body valve. This design accomplishes a few things. It improves the flow of the overall system, reduces the footprint of the overall fuel train and reduces installation and maintenance time.
• Preventive maintenance. Regular maintenance is critical to ensure optimal fuel-train performance. Some safety shutoff valves come with an integrated strainer, which protects downstream devices from contaminants in the fuel line. When performing maintenance checks, be sure to clean the strainer and ensure it’s not a source of flow blockage.
### Conclusion
Designing a fuel-train system with high-flow components or retrofitting an existing system with new technology can boost the performance of industrial burner systems significantly, unlocking greater reliability, cost savings and energy efficiency.
For more information: Yussef Abou-Ghanem is product marketing manager of Combustion, Americas, at Emerson. He helps customers improve the safety and efficiency of their industrial combustion systems. He can be reached at yussef.aboughanem@emerson.com or 973-966-2578. Contact Emerson at www.emerson.com.
All graphics courtesy of the author except where noted. | 1,898 | 8,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-40 | latest | en | 0.926667 |
https://oeis.org/A129629 | 1,590,711,558,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00461.warc.gz | 494,639,614 | 3,842 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A129629 Nonzero bisection of Moebius transform of A082392. 2
1, 1, 3, 7, 14, 31, 63, 123, 255, 511, 1015, 2047, 4092, 8176, 16383, 32767, 65503, 131061, 262143, 524223, 1048575, 2097151, 4194162, 8388607, 16777208, 33554175, 67108863, 134217693, 268434943, 536870911, 1073741823 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Possibly identical to A011947. LINKS FORMULA a(n)=2^n-A138944(n) - M. F. Hasler, May 03 2008 (sequence number appears to be in error, Joerg Arndt, Mar 25 2013) PROG (PARI) A129629(Nmax=100)={ local( d=2*Nmax+3, b=Vec( sum( k=0, log(d)\log(2), (x^2^k)/(1-2*x^2^(k+1)), O(x^d)))); vector( Nmax, i, sumdiv(i*2-1, k, moebius((i*2-1)/k)*b[k])) } - M. F. Hasler, May 03 2008 CROSSREFS Sequence in context: A036892 A123707 A011947 * A223136 A089526 A261051 Adjacent sequences: A129626 A129627 A129628 * A129630 A129631 A129632 KEYWORD nonn AUTHOR Ralf Stephan, May 31 2007 STATUS approved
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Last modified May 28 17:37 EDT 2020. Contains 334684 sequences. (Running on oeis4.) | 535 | 1,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-24 | latest | en | 0.573867 |
http://slideplayer.com/slide/4183854/ | 1,529,680,688,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864546.30/warc/CC-MAIN-20180622143142-20180622163142-00425.warc.gz | 297,837,861 | 25,208 | # Chapter 21- Examples.
## Presentation on theme: "Chapter 21- Examples."— Presentation transcript:
Chapter 21- Examples
Problem Two free point charges +q and +4q are located a distance, L, apart. A third charge is placed so that the whole system is in equilibrium Find the location, magnitude, and sign of the third charge
Step 1: Draw IT! +q +4q L
Step 2: Think about it! +q +4q 3 1 2 L
There are 3 positions available for the new charge (marked in blue). Which position do you think is the most likely position and more importantly why? Positions 1 & 3 are untenable since by equilibrium, we mean the force on each particle is zero. If 1, for example, then there will be a repulsive force on +4q that cannot be overcome without creating a strong attractive force on +q
Step 3: Now a free body problem
+q +4q 2 L x F1 F2 Call the new charge, q0, and let it be distance x from +q The free-body diagram shows relationship of the two forces, F1 and F2 on the new charge
2 L x F1 The free-body diagram shows relationship of the two forces, F1 and F4 on charge +q
Problem An electron is projected with an initial speed of v0, 1.6 x 106 m/s in the uniform field between the parallel plates shown below. Assume that the plates are uniform and the field is directed vertically downward and the field outside the plates is zero. The electron enters the field at a point midway between the two plates If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field 2.0 cm 1.0 cm e-
Step 1: Draw It Obviously, the electron is 0.5 cm below the upper plate when entering the field so, the initial position is (0,0.5) and the final position is (2.0, 0.0) SI: (0,0.005) and (0.02, 0) Also, v0 is actually in the x-direction so vinitial= (v0,0) 2.0 cm 1.0 cm e-
Step 2: Think about it! This reminds me of a projectile problem except it is kinda upside down. In this case, there is no acceleration in the x-direction only in the y and since F=ma=eE then a=eE/m Where e=charge of electron and m is the mass of the electron. Since I can never remember the trajectory equation, then I must solve for the time it takes the electron to traverse 2 cm and then plug that into the an equation relating acceleration to distance in the y-direction 2.0 cm 1.0 cm e-
Step 3: Equation Time 2.0 cm 1.0 cm e-
Problem What is the magnitude and direction of the electric field at the center of the square if q= 1.0 x 10-8 C and distance, a, is 5 cm +q -2q +2q -q a
Field Lines Free body diagram of + test charge in center +q -2q +2q -q
Need break down into x and y components
First find the magnitudes of all the electric fields
For example, the field line from +2q
The line from +2q makes an angle of 1350 with the positive x-axis So E2qx=E2q cos (1350)=-E2q cos (450) E2qx=-0.707*E2q E2qy=E2q sin (1350)=E2q sin (450) E2qy=0.707*E2q 1350
And from -2q, (at 450) E-2qx=0.707*E2q E-2qy=0.707*E2q
And –q and +q -q makes angle of 2250 E-qx=-0.707*Eq E-qy=-0.707*Eq
Now we add the x and y components, respectively
Ex and Ey | 896 | 3,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-26 | latest | en | 0.901902 |
https://quant.stackexchange.com/questions/tagged/yield-curve | 1,719,247,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00274.warc.gz | 406,896,353 | 51,620 | Skip to main content
# Questions tagged [yield-curve]
A yield curve is a plot of yields for various bonds (often government bonds) versus the bonds' maturities. We also often plot swap and other LIBOR rates to get the (related) swap curve.
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### What pricing curve to use for different instruments?
When pricing derivatives, their price depends on some yield curve, which is used to discount future cash flows. But there are many yield curves, dependent on what they're bootstrapped from. There's ...
• 307
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1 answer
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### When using quantlib's swaphelper to build a curve, is the fixing lag considered?
For example, EUR Annual (vs. 3M EURIBOR) swap has 2-Business-Days Fixing Lag. When interpolating with 1Y swap, the forward 3M Euribor rate starting date is the reset date（2023/6/14）, not the reset ...
2 votes
0 answers
60 views
### Constructing a yield curve with unknown discount factors
There are some questions and answers on this site about yield-curve construction, but none of them address the case where in an EM market setting, one might only have annual observability of (say) IRS ...
0 votes
1 answer
59 views
### How to calculate zero rate for deposits in an interest rate curve in PiecewiseLinearZero method
I am trying to duplicate zero rates and discount factors from ql.PiecewiseLinearZero method. To simplify calculation, I only use one deposite rate: 3M Euribor 0.03822. I set evaluationDate as ql.Date(...
2 votes
1 answer
59 views
### Parallel shift in spot yield curve moves the IRR of a bond portfolio in the same direction: Analytical Proof
I am trying to prove that a parallel shift in the spot yield curve will as its effect have the IRR of a bond portfolio move in the same direction and by the same amount. I have tested this on few ...
• 281
0 votes
1 answer
79 views
### swap curve calibration with interpolation using newton-like method
suppose 2 swap market quotes for 1Y and 2Y and that swap payments occur semi-annually. calibrating / obtaining the discount factors means finding 4 unknowns / discount factors that reproduce the ...
0 votes
1 answer
78 views
### Double interpolation continuity
I am trying to use double interpolation(linear and forward quartic) but the interpolator is the zero coupon so at each time i need to convert the instantaneous forward rate to zc my problem is at the ...
• 1
2 votes
3 answers
345 views
### Does PCA for yield curve has any tangible value?
I am aware of an abundant literature on Principal Component Analysis (PCA) application for yield curves. All of these papers to me look merely a statistics-oriented results. Most of the papers argue ...
• 368
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1 answer
67 views
### Yield to Maturity (YTM) to Zero Coupon Yield Curve (ZCYC) Forward Rate and Discount Factor mismatch in QuantLib
I have a set of YTM data for various tenors, and I'm constructing a ZCYC using the QuantLib library. However, when I calculate the forward rates and discount factors from the ZCYC and compare them ...
0 votes
0 answers
20 views
### Construct DeFi yield curve
I was wondering if anyone knows how to construct a yield curve for cryptocurrencies (for yTokens like yDAI and yETH for example). It'd be best if yield curves could be dynamic (though I think it could ...
0 votes
2 answers
112 views
### Shape of Yield curve of ZCB under no-arbitrage
Sorry if the question is somewhat elementary, but I have thought about it for a while and I cannot figure out where my mistake is. Suppose we are in are in an arbitrage-free market in which risk-free ...
• 103
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0 answers
43 views
### Interpolation by central bank cometee meeting dates
I want to have an interpolation, to the yield curve, In the term up to a year, the forward daily curves will be the same between central bank cometee meeting dates (to be precice, the implemantation ...
• 1
2 votes
1 answer
125 views
### Cubic Spline Interpolation partial derivative to the point
Still didn't figure out this, so looking for some help, kindly apppreciated. By this blog https://blog.timodenk.com/cubic-spline-interpolation/index.html, the piecewise cubic spline interpolation is ...
• 21
1 vote
2 answers
259 views
### Yield curve and bond price
I am confused about how the yield curve is built and how it relates to the pricing of bonds. First what I don't understand is that when people talk about the yield curve, which yield curve are they ...
1 vote
1 answer
188 views
### Explicit step by step curve construction using FRAs
I'm trying to understand a step by step process of building curve from the instruments to the final result, particularly how overlapped FRAs are used. i'm trying to build this in excel so I have a ...
1 vote
1 answer
115 views
### How to compute discount factor from yield curve when there are two daycounts in play?
Let's say I have a yield curve, i.e. a series of times $t_1, ..., t_n$ and associated rates $r_{t_1}, ..., r_{t_n}$, such that my discount factors are $DF_{t_i} = (1+r_{t_i})^{(-t_i)}$. The curve has ...
• 141
0 votes
1 answer
89 views
### Bootstrapping adjustment for coupon
I struggling to get why in bootstrapping I need to divide the YTM by 2 (for semiannual coupons) and not adjust the power for the semiannual period. Please see below example. Consider two bonds with a ...
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0 answers
46 views
### Relationship Between the Equity Funding Curve and Equity Forward Curve
am trying to understand the core concepts of Equity Forwards Curve, Funding Curve and yield Curves (most sources online seem to focus on Interest rate related examples, so any sources for equity ...
• 1
0 votes
0 answers
50 views
### Why isn't the volatility version of the Nelson-Siegel model in the same shape as the original yield model?
If we compare the form of the original model: https://edisciplinas.usp.br/pluginfile.php/3180059/mod_resource/content/0/Nelson%2C%20Siegel%20%281987%29%20-%20Parsimonious%20Modeling%20of%20Yield%...
0 votes
0 answers
50 views
### Benchmark Treasury Curve Data
I want to calibrate US treasury bonds to get a benchmark curve as an exercise. I know that choice of bonds may be discretionary. But can you point me to any good public data source where I can get ...
• 101
0 votes
1 answer
233 views
### Relationship of par-curve and zero-curve/spot-curve
I've been trying to bootstrap the zero-curve from a swap curve composed of ESTR OIS swaps. Theory says when the par-curve is upward sloping, the zero-curve will be above the par-curve and vice-versa. ...
0 votes
0 answers
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### How Zero Discount factors are calculated in Oracle and Wellstreetprep?
I've some inquiries that may be naive related to ZDF. I have got a look at those two resources for zero discount factors: Oracle - Zero Discount Factors Wallstreetprep - Discount Factor ...
• 1
0 votes
2 answers
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### Bootstrapping the zero-curve/spot-curve from incomplete swap curve par-rates
TL;DR: I have an incomplete set of swap rates and want to bootstrap the zero-rate curve, what can I do? I'm trying to construct a spot-rate/zero-rate curve from a swap curve (i.e. par-rate quotes) ...
0 votes
2 answers
177 views
### Mathematical meaning of an inverted yield curve
I am currently working on rates model. I would like to understand, mathematically, what does it mean to have an inverted yield curve? And I am asking myself for a certain model, how can I generate an ...
1 vote
0 answers
33 views
### Multiple factor Hull-While and yield curve deformation
I am currently studying rate models and I understand that the One-Factor model has some incompleteness: The yield-curve can only be shifted. But I don’t understand what parameter controls this shift ( ...
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0 answers
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### Methods for Constructing a Yield Curve
I wonder if the Raw interpolation (known as linear on the log of discount factors) can handle negative interest rate due to the inverted yield curve? I understand using the linear on log of rates and ...
0 votes
0 answers
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### Liquidity Preference Hypothesis and Yield Curve
Unlike under the Expectation Hypothesis (for which the forwards are perfect predictors of the expected short rates), for the LPH forward=expected short rates + LP and consequently the expected short ...
0 votes
1 answer
116 views
2 votes
1 answer
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### Mean level of the state variables under the risk-neutral measure in Arbitrage-free Nelson Siegel
I do not understand why mean levels of the state variables under the risk-neutral measure, $\theta^{\mathbb{Q}}$, in Arbitrage-free Nelson-Siegel is set to zero. It should follow from the following ... | 2,059 | 8,685 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.923571 |
https://vladimirkalitvianski.wordpress.com/2015/03/10/on-physics-of-asymptotic-series-and-their-resummation/ | 1,501,246,889,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550967030.98/warc/CC-MAIN-20170728123647-20170728143647-00034.warc.gz | 723,346,535 | 35,403 | ## On Physics of asymptotic series and their (re)summation
There was a question on Physics Overflow which can be reduced to the last phrase:
I think the bottom line with my questions is that I fully accept that divergent series occur in physics all the time, and quite clearly they contain information that we can extract, but I would like to understand more to what degree can we trust those results.
And there were some answers including mine. Amongst other things, I mentioned a “constructive way” of building asymptotic series, which would be useful in practice. As an example, I considered a toy function $E(x)=\int_0 ^\infty\frac{e^{-t}}{1+x\cdot t}dt\approx 1-x+2!x^2-3!x^3$. A direct summations of its Taylor series is useless because it diverges from the exact function value at any finite $x$. It is so not only for fast “growing” coefficient cases, but also for regular (converging) Taylor series truncated at some finite order, when we try to extrapolate the truncated series to finite (large) values of $x$. A truncated series “grows” as the highest power of $x$, but the expanded function can be finite and limited, so the truncated series becomes inaccurate.
Thinking this fact over, in about 1981-1982, I decided that the difficulties with extrapolation to finite $x$ was in expanding a finite and slowly changing function in powers of fast growing functions like $x^n$. “Why not to expand such a slow function in powers of slowly changing functions?“, thought I, for example: $E(x)\approx 1-f(x)+a\cdot f(x)^2-b\cdot f(x)^3$, where $f(x)\to x$ at small $x$, but $f(x)\ll x$ for finite $x$.
In order to give some idea and demonstrate fruitfulness of my “constructive” approach, I considered the following functions instead of $x$: $Y(x)=\ln(1+x)$, $Z(x)=x/(1+x)$, and $f(x)=x/(1+k\cdot x)$ with adjustable coefficient $k$. The corresponding figures are the following:
Fig. 1. Expansion in powers of $x$.
Fig. 2. Expansion in powers of $Y(x)=\ln(1+x)$.
Fig. 3. Expansion in powers of $Z(x)=x/(1+x)$.
The smaller terms in the new series, the better approximation. Following this banal observation, I adjusted the coefficient $k$ in $f(x)$ to minimize the coefficient $b$ at $f^3$ (note, the axis $x$ is made longer in the next three figures):
Fig. 4. Expansion in powers of $f(x)=x/(1+2x)$.
We see that $E(x)$ is approximated now much better than with its truncated asymptotic series in powers of $x$ (Fig. 1).
The same idea applied to the function $I(g)=\int_{-\infty}^{\infty}e^{-x^2-gx^4}dx$ works well too:
Fig. 5. Expansion in powers of $f(x)=0.75x/(1+4.375x)$.
Finally, the ground state energy $E(\lambda)$ of the anharmonic oscillator in QM (anharmonicity $\lambda x^4$) has also a divergent series: $E(\lambda)\approx 0.5(1+1.5\lambda -5.25\lambda ^2+41.625\lambda^3)$, which can be transformed into a series in powers of $f(\lambda)=1.5\lambda/(1+3.5\lambda)$. It gives a good extrapolation of $E(\lambda)$ (error $\le \pm 1.5$% within $0\le \lambda \le 1$, Fig. 6), unlike the original series:
Fig. 6. The ground state energy of 1D anharmonic oscillator.
Thus, my idea was not too stupid as it allowed to extrapolate the asymptotic (divergent) series in the region of big $x$ with a decent accuracy.
P.S. In my practice I also encountered a series $f(\xi)\approx f(0)+f'(0)\xi+f''(0)\xi^2/2+...$ (a convergent one) whose convergence I managed to improve with partially summing up some of its terms into a finite function $g(\xi)$, so the resulting series $f(\xi)\approx g(\xi)+a\xi +b\xi^2+...$ became even better convergent (Chapters 3 and 4). It is somewhat similar to soft contributions summation in QED, if you like.
The moral is the following: if you want to have a “convergent” series, then build it yourself and enjoy. | 1,064 | 3,767 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 40, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-30 | longest | en | 0.913228 |
https://community.agilent.com/technical/software/f/forum/4982/custom-calculation-to-assume-peak-area-of-missing-compounds-as-zero-so-that-additional-calculations-return-significant-results/18952 | 1,657,057,713,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00366.warc.gz | 223,695,076 | 35,004 | # Custom Calculation to assume peak area of missing compounds as zero so that additional calculations return significant results
Hi,
I am trying to use the custom calculation editor for additional calculations using peak area percentages. My formulas include all potential compounds that could appear on a given sample, although some of them might not be present sometimes. So, if in a given formula all included compounds are identified peaks, the result obtained is as expected. Nevertheless, if in a same formula I have identified & missing compounds, I get a blank cell as result (which is of course not what I want).
I would like to make the peak area percentages of the missing compounds equal to zero, so that the formulas where they are included give me back the results asked using the values of the identified peaks as such and zeros in case it is a missing compound.
Parents
• hello,I had the same problem as you to make the sum of impurities, I searched in the forum and I found the following formulas, to be verified because I am not an expert in formulas but it worked
ListImpND (Type string, scope signal) If (Count(CurrentSignal.AllMissingCompounds) = 0, "No Calibrated Peak(s) Missing ", "Fail " & string.join(" ",CurrentSignal.AllMissingCompounds.select(function(x) x.Compound_Name.tostring())))
if(instr(ListImpND,"compound1")>0,0,(currentsignal.Allidentifiedpeaks.where(function(x)x.compound_name=("compound1")).select(function (x) x.GetDoubleCC("CC66")).first()))
i
syb
• hello,I had the same problem as you to make the sum of impurities, I searched in the forum and I found the following formulas, to be verified because I am not an expert in formulas but it worked
ListImpND (Type string, scope signal) If (Count(CurrentSignal.AllMissingCompounds) = 0, "No Calibrated Peak(s) Missing ", "Fail " & string.join(" ",CurrentSignal.AllMissingCompounds.select(function(x) x.Compound_Name.tostring())))
if(instr(ListImpND,"compound1")>0,0,(currentsignal.Allidentifiedpeaks.where(function(x)x.compound_name=("compound1")).select(function (x) x.GetDoubleCC("CC66")).first()))
i
syb
Children
No Data | 482 | 2,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.89905 |
https://calculator.academy/words-to-pages-calculator/ | 1,695,813,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00705.warc.gz | 178,220,805 | 54,125 | Enter the total number of words and the number of words per page into the calculator to determine the number of pages. This calculator can also evaluate any of the variables given the others are known.
## Words To Pages Formula
The following formula is used to calculate the number of pages based on the number of words.
NP = W / WP
Variables:
• NP is the number of pages W is the total number of words WP is the number of words per page
To calculate the number of pages, divide the total number of words by the number of words per page. The result will give you the total number of pages.
## What is a Words To Pages?
Words to Pages is a tool or converter that helps to estimate the number of pages a specific amount of words will take up in a document. This is useful for students, writers, or professionals who need to adhere to specific word or page count requirements for essays, reports, or other written assignments. The calculation can vary depending on factors such as font size, font type, spacing, margins, and paragraph structure.
## How to Calculate Words To Pages?
The following steps outline how to calculate the number of pages using the Words To Pages formula.
1. First, determine the total number of words (W).
2. Next, determine the number of words per page (WP).
3. Next, gather the formula from above = NP = W / WP.
4. Finally, calculate the number of pages (NP).
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem:
Use the following variables as an example problem to test your knowledge.
Total number of words (W) = 1500
Number of words per page (WP) = 250 | 355 | 1,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.862548 |
https://isbn.nu/authorx/irvin_barbaraaghi | 1,628,023,131,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00265.warc.gz | 323,204,358 | 3,920 | search for books and compare prices
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Product Description: Sharpen skills by engaging your child's natural love of learning. Third Grade Scholar introduces a lively selection of third grade language arts, math, science, and social studies exercises. On-page activities are supplemented by additional practice activities, jokes and riddles, vocabulary words, and fun facts...read more
Paperback:
9781589470156 | Deluxe edition (School Zone Pub Co, September 9, 2005), cover price \$7.99 | About this edition: Sharpen skills by engaging your child's natural love of learning.
Product Description: These comprehensive resources offer 100 activity pages that make math fun while offering plenty of essential computation practice infused with a strong problem-solving strand. They include instruction and practice for key skills and test-taking in all strands of the NCTM curriculum standards: 1) numeration and number theory 2) operations, geometry, and measurement 3) patterns, functions, and algebra 4) data analysis, probability, and problem solvin...read more
Paperback:
9781591980155 | Creative Teaching Pr, January 21, 2004, cover price \$9.99 | About this edition: These comprehensive resources offer 100 activity pages that make math fun while offering plenty of essential computation practice infused with a strong problem-solving strand.
Product Description: Units: Metric Measurement; Customary Measurement; Points, Lines, Angles, and Circles; Triangles and Quadrilaterals; Congruence and Symmetry; Coordinate Graphing; Perimeter, Circumference, and Area, Volume and Surface AreaUp-to-Speed Math series...read more
Paperback:
9781562543679 | Saddleback Pub, August 1, 2001, cover price \$13.95 | About this edition: Units: Metric Measurement; Customary Measurement; Points, Lines, Angles, and Circles; Triangles and Quadrilaterals; Congruence and Symmetry; Coordinate Graphing; Perimeter, Circumference, and Area, Volume and Surface AreaUp-to-Speed Math series.
Product Description: Units: Organizing Data; Measures of Central Tendency; Picture Graphs and Bar Graphs; Line Graphs and Circle Graphs; GraphsDecisions, Conclusions, and Predictions; Understanding Chance and Probability; Tree Diagrams and the Counting Principle; Samples and PredictionsUp-to-Speed Math series...read more
Paperback:
9781562543716 | Reprint edition (Saddleback Pub, August 1, 2001), cover price \$13.95 | About this edition: Units: Organizing Data; Measures of Central Tendency; Picture Graphs and Bar Graphs; Line Graphs and Circle Graphs; GraphsDecisions, Conclusions, and Predictions; Understanding Chance and Probability; Tree Diagrams and the Counting Principle; Samples and PredictionsUp-to-Speed Math series.
displaying 1 to 4 | at end | 656 | 2,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.867697 |
https://paperwritingcenter.com/college-algebra-11th-edition-from-pearson/ | 1,624,171,267,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487658814.62/warc/CC-MAIN-20210620054240-20210620084240-00480.warc.gz | 389,051,019 | 10,323 | # College algebra 11th edition from pearson
Section 3.1: exercises 2, 18, 24, 64
Section 3.2: exercises 2, 14, 20, 28, 34, 46
Section 3.3: exercises 14, 18, 30, 44, 52, 58
Section 3.4: exercises 12*, 18*, 32*, 48, 58
Section 3.6: exercises 14, 28, 40, 52
SECTION 3.1 QUADRATIC FUNCTIONS AND MODELS
2. You are given an equation and the graph of a quadratic function.
a. Give the domain and range
b. Give the coordinates of the vertex
c. Give the equation of the axis
d. Find the y-intercept
e. Find the x-intercepts
(SEE PICTURE)
18. Graph each quadratic function. Give the (a) vertex, (b) axis, (c) domain, and (d)
range. Then determine (e) the interval of the domain for which the function is increasing
and (f) the interval for which the function is decreasing.
24.
64. According to data from the National Highway Traffic Safety Administration, the
accident rate as a function of the age of the driver in years x can be approximated by the
function
SECTION 3.2 SYNTHETIC DIVISION
2. Use synthetic division to perform the division
14.
20. Express f(x) in the form f(x)=(x-k)q(x)+r for the given value of k
28. For each polynomial function, use the remainder theorem and synthetic division to
find f(k)
34.
46. Use synthetic division to decide whether the given number k is a zero of the given
polynomial function. If it is not, give the value of f(k)
SECTION 3.3 ZEROS OF POLYNOMIAL FUNCTIONS
14. Use the factor theorem and synthetic division to decide whether the second
polynomial is a factor of the first
18. Factor f(x) into linear factors given that k is a zero of f(x)
30. For each polynomial function, one zero is given. Find all others.
44. For each polynomial function, find all zeros and their multiplicities
52. Find a polynomial function f(x) of degree 3 with real coefficients that satisfies the
given conditions.
58. Find a polynomial function f(x) of least degree having only real coefficients with
zeros as given.
SECTION 3.4 POLYNOMIAL FUNCTIONS: GRAPHS, APPLICATIONS, AND
MODELS
12*Sketch the graph of each polynomial function. Determine the intervals of the domain
for which function is (a) increasing or (b) decreasing
18*
32* Graph each polynomial function. Factor first if the expression is not in factored form.
48. Use the intermediate value theorem or polynomials to show that each polynomial
function has a real zero between the numbers given.
58. Show that the real zeros of each polynomial function satisfy the given conditions
SECTION 3.6 VARIATION
14. Solve each variation
If (m) varies jointly as z and p, and m=10 when z=2 and p=7.5, find m when z=6 and
p=9
28. The current in a simple electrical circuit varies inversely as the resistance. If the
current is 50amps when the resistance is 10ohms, find the current if the resistance is
5ohms.
40. The number of long-distance phone calls between two citites in a certain time period
varies directly as the populations of the cities and inversely as the distance between
them. If 10,000 calls are made between two citites 500 mi apart, having populations of
50,000 and 125,000, find the number of calls between two citites 800mi apart, having
populations of 20,000 and 80,000.
52. Suppose y is inversely proportional to x, and x is tripled. What happens to y?
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## Essay Writing Service
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 1,303 | 5,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-25 | latest | en | 0.828613 |
https://becca9941.github.io/programming%20problem/2019/08/12/conversation-about-mars-rover-problem.html | 1,601,423,292,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00132.warc.gz | 273,726,639 | 4,437 | # conversation about object-oriented mars rover solution
12 Aug 2019
I just went to Lunch with Andy and we talked about how to approach the Mars Rover problem from an Object-oriented perspective. The whole conversation blew my mind because when expressed in OO terms, the solution is actually rediculously simple. There are no conditionals or loops. There are three objects, the Rover, it's orientation and it's position. The orientation and position objects are given to the Rover who coordinates them.
Below, I have written up my understanding of the object-oriented Rover solution based on what I learned from the conversation. I think I haven't grasped it entirely yet but what I do know from it has made me really excited about the possibilites with OO.
Andy is going to read over my understanding and let me know if there are things I didn't quite get. Will refine my understanding again until I get it right. We are also going to pair program on it and actually build out this solution, which I'll also document here for other people who might find this really useful.
The main message here is that it's okay to not get things right away, and that seeking feedback to refine your understanding as an iterative process is a really good way to learn.
## My initial understanding of the Object-oriented Mars Rover solution
We have a Mars Rover, who is essentially a Darlek container that does nothing execept coordinate his friends. He is the project manager.
The mars rover has two instance variables, an orientation and a position. Both of these instance variables accept either an orientation or a position object, which are seperate to the Rover itself.
The orientation class is in charge of which way the rover faces. It can have an orientation of either 'North', 'East', 'South' or 'West', and it knows what is to the left, right, top and bottom to it.
The orientation class has three messages (methods), left, right and forward. When the left message is sent, it turns 90 degrees to the left. So if he is facing North, he will turn to face West. When the right message is sent, it turns 90 degrees to the right, so if he is facing East, he will turn to face South.
The forward message passes the orientation to the position object, which then returns the new position.
The position object knows where the Rover is in terms of x and y coordinates.It also contains a forward message which moves the rover forward or backwards along the y axis if his orientation is North or South, or moves the rover left or righ along the x axis if he is facing East or West.
## Okay, second round of understanding
My understanding the first time around was pretty close, but the position orientation relationship wasn't quite clear and the way I worded some of the explanation wasn't quite clear, I explained it better out loud.
Okay, so we have a Mars Rover object who is the project manager for the orientation and position objects. The Mars Rover has an orientation instance variable, which accepts an orientation object. It also has a position instance variable which accepts a position object.
The orientation object knows which way it is facing. It can be facing either 'North', 'South', 'East' or 'West'. The orientation object has an instance variable which can contain one of these values to start with. The orientation object has three messages (methods) that can change the orientation. These are 'left', 'right' and 'forward'.
The 'left' message turns the orientation 90 degrees to the left. So if the orientation is 'North', it becomes 'West'. If the orientation is 'East', it becomes 'North'.
The 'right' message turns the orientation 90 degrees to the right. So if the orientation is 'North', it becomes 'East'. If the orientation is 'East', it becomes 'South'.
The 'forward' message asks the position object to tell the orientation object what is in front of the object, or what would the position be if the object were to move forward based on it's current orientation.
The position object accepts the current orientation value. It also knows the current x and y coordinates of the Rover.
The rover contains four messages: North, East, South, West. If the rover is facing North, the y value is incremented by one. If the rover is facing South, the y value is decremented by one. If the rover is facing East, the x value is incremented by 1. If the rover is facing West, the x value is decremented by one.
### Third round of understanding
The orientation object does not have any instance variables. Instead, we can make four different instances of the orientation object, one each for 'North', 'East', 'South' and 'West'. We could make more instances if we wanted to for 'North East', 'South West' etc, but we are only interested in the four main compass points for this problem.
The orientation object can give the position object it's current x and y coordinates as well as it's orientation. Depending on what the orientation is, the position object returns the new position by incrementing or decrementing the relevant x and y coordinates.
Every time the orientation changes, the orientation instance variable inside of the rover is replaced with the result of the new orientation. So it gets a replaced with a new orientation object every time the orientation changes.
### Fourth round of understanding
The orientation doesn't give it's orientation, it is an instance of an orientation.
Whenever the position changes, the position instance variable in the Rover also gets replaced by a new position object.
### Fifth round of understanding
The orientation is not an instance of an orientation class. There are four seperate orientation classes: North, East, South and West. Each class has the same methods but the methods do different things. Polymorphism.
Okay, we are ready to start building it. My mind feels like it is being rewired. There's no way I would have been able to understand how to think like this without the help of someone who really knows how to. Objects are not easy, but once you get it it's the easiest thing ever. Looking forward to carrying on stumbling until I get there.
In the next post, which I'll link to here, I'll document the process we took to build the Rover. Or at least I'll pair with Andy to build it, and then I'll build it again by myself documenting and explaining the process and refining it again like I did here if I misunderstood anything.
This is FUN! | 1,329 | 6,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.959028 |
https://ok-em.com/geometry-in-math-88 | 1,675,127,283,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00629.warc.gz | 449,635,796 | 6,434 | # Geometry in math
There are a lot of Geometry in math that are available online.
## Geometry – Definition with Examples
Basic geometry and measurement Intro to area and perimeter 0/1800 Mastery points Count unit squares to find area Area formula intuition Multiply to find area Area and the distributive
Do math
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## What is Geometry?
The branches of geometry are categorized as: Algebraic geometry Discrete geometry Differential geometry Euclidean geometry Convex
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## High School Geometry
• Do math equation
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## What is Geometry? Plane & Solid Geometry
In the Kansas Mathematics Standards, geometry spans every grade level from kindergarten to grade eight; it first begins with spatial sense , an intuition about shapes and the relationships between them including an ability to recognize
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Doing homework can help improve grades.
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Mathematical problems can be difficult to understand, but with a little explanation they can be easy to solve.
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## Introduction to Geometry
Lines. Lines, line segments, and rays: Lines Measuring segments: Lines Parallel and
Learn step-by-step
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## Formulas, Examples
What is Geometry? Geometry is the branch of mathematics that relates the principles covering distances, angles, patterns, areas, and volumes. All the visually and spatially related concepts are categorized under geometry. There | 470 | 2,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-06 | latest | en | 0.897898 |
https://www.enotes.com/homework-help/3-y-2x-5-y-2x-8-1-3-x-5y-y-x-4-451267 | 1,516,442,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00002.warc.gz | 924,028,230 | 9,664 | # 3(y-2x) = 5(y-2x)-8 -1/3(x-5y)=(y-x)-4
baxthum8 | Certified Educator
Distribute first and second equation:
`3y - 6x = 5y - 10x - 8`
`-1/3x + 5/3y = y - x - 4` Multiply by 3 to get rid of denominators.
`-x + 5y = 3y - 3x - 12`
Rewrite both h equations in standard form: `ax + by = c`
`4x -2y = -8`
`2x +2y = -12` Divide both equations by 2.
`2x - y = -4`
`x + y = -6` Add the 2 equations together to solve for x.
`3x = -10` Divide by 3.
`x = -10/3` Substitute in equation to solve for y.
`-10/3 + y = -6`
`y = -8/3`
Solution: `(-10/3, -8/3)` | 261 | 559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-05 | latest | en | 0.776874 |
https://ch.mathworks.com/matlabcentral/cody/problems/230-project-euler-problem-1-multiples-of-3-and-5/solutions/3007654 | 1,606,337,317,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00469.warc.gz | 245,537,267 | 17,038 | Cody
# Problem 230. Project Euler: Problem 1, Multiples of 3 and 5
Solution 3007654
Submitted on 29 Sep 2020 by Victor Lindholm
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1000; y_correct = 233168; assert(isequal(euler001(x),y_correct))
y = 233168
2 Pass
x = 4000; y_correct = 3732668; assert(isequal(euler001(x),y_correct))
y = 3732668
3 Pass
x = 2340; y_correct = 1276470; assert(isequal(euler001(x),y_correct))
y = 1276470
4 Pass
x = 2341; y_correct = 1278810; assert(isequal(euler001(x),y_correct))
y = 1278810
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 247 | 786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-50 | latest | en | 0.637927 |
https://convertoctopus.com/0-3-years-to-hours | 1,618,836,158,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00267.warc.gz | 290,936,382 | 8,383 | ## Conversion formula
The conversion factor from years to hours is 8760, which means that 1 year is equal to 8760 hours:
1 yr = 8760 hr
To convert 0.3 years into hours we have to multiply 0.3 by the conversion factor in order to get the time amount from years to hours. We can also form a simple proportion to calculate the result:
1 yr → 8760 hr
0.3 yr → T(hr)
Solve the above proportion to obtain the time T in hours:
T(hr) = 0.3 yr × 8760 hr
T(hr) = 2628 hr
The final result is:
0.3 yr → 2628 hr
We conclude that 0.3 years is equivalent to 2628 hours:
0.3 years = 2628 hours
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 hour is equal to 0.00038051750380518 × 0.3 years.
Another way is saying that 0.3 years is equal to 1 ÷ 0.00038051750380518 hours.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that zero point three years is approximately two thousand six hundred twenty-eight hours:
0.3 yr ≅ 2628 hr
An alternative is also that one hour is approximately zero times zero point three years.
## Conversion table
### years to hours chart
For quick reference purposes, below is the conversion table you can use to convert from years to hours
years (yr) hours (hr)
1.3 years 11388 hours
2.3 years 20148 hours
3.3 years 28908 hours
4.3 years 37668 hours
5.3 years 46428 hours
6.3 years 55188 hours
7.3 years 63948 hours
8.3 years 72708 hours
9.3 years 81468 hours
10.3 years 90228 hours | 441 | 1,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-17 | latest | en | 0.844901 |
https://www.bartleby.com/questions-and-answers/in-a-delta-abc-angle-c-3-angleb-2-angle-a-angleb-find-three-angles/6e21507c-36aa-402b-908f-d42d0a73ecdc | 1,582,336,658,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145621.28/warc/CC-MAIN-20200221233354-20200222023354-00062.warc.gz | 653,866,613 | 26,012 | # In a ∆ ABC, angle C = 3 angleB = 2( angle A + angleB)Find three angles
Question
1 views
In a ∆ ABC, angle C = 3
angleB = 2( angle A + angleB)
Find three angles
check_circle
Step 1
Since ABC is a triangle.
Step 2
Apply angle sum property.
Step 3
Apply angle sum ...
### Want to see the full answer?
See Solution
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Tagged in | 156 | 558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-10 | latest | en | 0.863837 |
https://simple-accounting.org/debt-to-income-dti-ratio-calculator/ | 1,708,623,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00364.warc.gz | 554,880,368 | 9,081 | # Debt-to-Income DTI Ratio Calculator
A high ratio like this makes it harder for the company to find additional debt financing. You will after reading about debt ratio, an easy calculation used to illustrate financial viability. Debt can be scary when you’re paying off college loans or deciding whether to use credit to…
There is a separate ratio called the credit utilization ratio (sometimes called debt-to-credit ratio) that is often discussed along with DTI that works slightly differently. The debt-to-credit ratio is the percentage of how much a borrower owes compared to their credit limit and has an impact on their credit score; the higher the percentage, the lower the credit score. A debt-to-equity ratio of 1.5 would indicate that the company in question has \$1.50 of debt for every \$1 of equity. To illustrate, suppose the company had assets of \$2 million and liabilities of \$1.2 million. Since equity is equal to assets minus liabilities, the company’s equity would be \$800,000. Its debt-to-equity ratio would therefore be \$1.2 million divided by \$800,000, or 1.5.
## What is a debt-to-income ratio?
However, what constitutes a “good debt ratio” can vary depending on industry norms, business objectives, and economic conditions. For instance, startups or companies in rapid expansion phases, too, may have higher ratios as they utilize debt to fund growth initiatives. While a higher ratio can be acceptable, carefully analyzing the company’s ability to generate sufficient cash flows to service the debt is essential. As a general guideline, 43% is the highest DTI ratio a borrower can have and still get qualified for a mortgage.
1. The debt-to-credit ratio is the percentage of how much a borrower owes compared to their credit limit and has an impact on their credit score; the higher the percentage, the lower the credit score.
2. The debt ratio is a simple ratio that is easy to compute and comprehend.
3. While the total debt to total assets ratio includes all debts, the long-term debt to assets ratio only takes into account long-term debts.
4. Sometimes the debt-to-income ratio is lumped in together with the debt-to-limit ratio.
The debt ratio is defined as the ratio of total debt to total assets, expressed as a decimal or percentage. It can be interpreted as the proportion of a company’s assets that are financed by debt. The debt-to-income (DTI) ratio is a personal finance measure that compares an individual’s monthly debt payment to their monthly gross income. Your gross income is your pay before taxes and other deductions are taken out. The debt-to-income ratio is the percentage of your gross monthly income that goes to paying your monthly debt payments.
## What is an ideal debt-to-income ratio?
Equity ratio is equal to 26.41% (equity of 4,120 divided by assets of 15,600). Using the equity ratio, we can compute for the company’s debt ratio. To calculate the debt ratio of a company, you’ll need information about its debt and assets. You can access the balance sheets of publicly traded companies on websites like Yahoo Finance or Nasdaq. The debt ratio does not take a company’s profitability into account.
In the United States, normally, a DTI of 1/3 (33%) or less is considered to be manageable. A DTI of 1/2 (50%) or more is generally considered too high, as it means at least half of income is spent solely on debt. Your DTI can help you determine how to handle your debt and whether you have too much debt.
## Interpreting the Debt Ratio
For more information about or to do calculations involving debt consolidation, please visit the Debt Consolidation Calculator. Financial research software can be used to easily compare debt ratios and other financial ratios across industries. What counts as a good debt ratio will depend on the nature of the business and its industry.
## How is the debt-to-income ratio calculated?
In addition, the debt ratio depends on accounting information which may construe or manipulate account balances as required for external reports. Bankrate.com is an independent, advertising-supported publisher and comparison service. We are compensated in exchange for placement of sponsored products and services, or by you clicking on certain links posted on our site. While we strive to provide a wide range of offers, Bankrate does not include information about every financial or credit product or service.
## What is a good debt ratio?
Many or all of the products featured here are from our partners who compensate us. This influences which products we write about and where and how the product appears on a page. Here is a list of our partners and here’s how we make money. Our blog articles are written independently by our editorial team. Order your copy of Investopedia’s What To Do With \$10,000 magazine for more tips about managing debt and building credit.
Not all creditors, such as personal loan providers, publish a minimum debt-to-income ratio, but generally it will be more lenient than for, say, a mortgage. In other words, how much is a company leveraging, or how much of its financing is coming from debt capital? Once we know this ratio, we can use it to determine how likely a company is to become unable to pay off its debts. Perhaps 53.6% isn’t so bad after all when you consider that the industry average was about 75%. | 1,111 | 5,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-10 | latest | en | 0.957765 |
https://leanprover-community.github.io/archive/stream/217875-Is-there-code-for-X%3F/topic/associates.20R.20-.3E.20ideal.20R.html | 1,621,164,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00300.warc.gz | 371,913,899 | 3,197 | ## Stream: Is there code for X?
### Topic: associates R -> ideal R
#### Kevin Buzzard (Sep 06 2020 at 16:59):
For a monoid R (e.g. a commutative ring) we have the monoid associates R, the quotient of R by units R, and this is preordered in mathlib by divisibility. For a commutative ring we have the semiring ideal R of ideals. This we have.
There's a natural map from associates R to ideal R and if R is an integral domain then this map is injective and, unfortunately, order-reversing (3 divides 6, but the ideal (6) is a subset of the ideal (3)). Do we have this map? If not then I would appreciate advice on whether it should just go from associates R to ideal R or whether I should insert an order_dual somewhere. The bare map has the good monoid hom properties, but it inverts the ordering. It would be much easier to say "this bare map satisfies x <= y -> f(y) <= f(x)" than start putting monoid structures on order_dual X given one on X. More and more I am seeing instances of order-reversing maps as we push on with algebra in Lean. If R is a PID then this map is simultaneously an isomorphism of monoids and an anti-isomorphism of orders.
Last updated: May 16 2021 at 05:21 UTC | 306 | 1,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.921535 |
https://tvm.apache.org/docs/tutorial/tensor_expr_get_started.html | 1,686,130,454,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00410.warc.gz | 617,398,411 | 27,278 | # Working with Operators Using Tensor Expression¶
Author: Tianqi Chen
In this tutorial we will turn our attention to how TVM works with Tensor Expression (TE) to define tensor computations and apply loop optimizations. TE describes tensor computations in a pure functional language (that is each expression has no side effects). When viewed in context of the TVM as a whole, Relay describes a computation as a set of operators, and each of these operators can be represented as a TE expression where each TE expression takes input tensors and produces an output tensor.
This is an introductory tutorial to the Tensor Expression language in TVM. TVM uses a domain specific tensor expression for efficient kernel construction. We will demonstrate the basic workflow with two examples of using the tensor expression language. The first example introduces TE and scheduling with vector addition. The second expands on these concepts with a step-by-step optimization of a matrix multiplication with TE. This matrix multiplication example will serve as the comparative basis for future tutorials covering more advanced features of TVM.
## Example 1: Writing and Scheduling Vector Addition in TE for CPU¶
Let’s look at an example in Python in which we will implement a TE for vector addition, followed by a schedule targeted towards a CPU. We begin by initializing a TVM environment.
```import tvm
import tvm.testing
from tvm import te
import numpy as np
```
You will get better performance if you can identify the CPU you are targeting and specify it. If you’re using LLVM, you can get this information from the command `llc --version` to get the CPU type, and you can check `/proc/cpuinfo` for additional extensions that your processor might support. For example, you can use `llvm -mcpu=skylake-avx512` for CPUs with AVX-512 instructions.
```tgt = tvm.target.Target(target="llvm", host="llvm")
```
### Describing the Vector Computation¶
We describe a vector addition computation. TVM adopts tensor semantics, with each intermediate result represented as a multi-dimensional array. The user needs to describe the computation rule that generates the tensors. We first define a symbolic variable `n` to represent the shape. We then define two placeholder Tensors, `A` and `B`, with given shape `(n,)`. We then describe the result tensor `C`, with a `compute` operation. The `compute` defines a computation, with the output conforming to the specified tensor shape and the computation to be performed at each position in the tensor defined by the lambda function. Note that while `n` is a variable, it defines a consistent shape between the `A`, `B` and `C` tensors. Remember, no actual computation happens during this phase, as we are only declaring how the computation should be done.
```n = te.var("n")
A = te.placeholder((n,), name="A")
B = te.placeholder((n,), name="B")
C = te.compute(A.shape, lambda i: A[i] + B[i], name="C")
```
Lambda Functions
The second argument to the `te.compute` method is the function that performs the computation. In this example, we’re using an anonymous function, also known as a `lambda` function, to define the computation, in this case addition on the `i`th element of `A` and `B`.
### Create a Default Schedule for the Computation¶
While the above lines describe the computation rule, we can compute `C` in many different ways to fit different devices. For a tensor with multiple axes, you can choose which axis to iterate over first, or computations can be split across different threads. TVM requires that the user to provide a schedule, which is a description of how the computation should be performed. Scheduling operations within TE can change loop orders, split computations across different threads, and group blocks of data together, amongst other operations. An important concept behind schedules is that they only describe how the computation is performed, so different schedules for the same TE will produce the same result.
TVM allows you to create a naive schedule that will compute `C` in by iterating in row major order.
```for (int i = 0; i < n; ++i) {
C[i] = A[i] + B[i];
}
```
```s = te.create_schedule(C.op)
```
### Compile and Evaluate the Default Schedule¶
With the TE expression and a schedule, we can produce runnable code for our target language and architecture, in this case LLVM and a CPU. We provide TVM with the schedule, a list of the TE expressions that are in the schedule, the target and host, and the name of the function we are producing. The result of the output is a type-erased function that can be called directly from Python.
In the following line, we use `tvm.build` to create a function. The build function takes the schedule, the desired signature of the function (including the inputs and outputs) as well as target language we want to compile to.
```fadd = tvm.build(s, [A, B, C], tgt, name="myadd")
```
Let’s run the function, and compare the output to the same computation in numpy. The compiled TVM function exposes a concise C API that can be invoked from any language. We begin by creating a device, which is a device (CPU in this example) that TVM can compile the schedule to. In this case the device is an LLVM CPU target. We can then initialize the tensors in our device and perform the custom addition operation. To verify that the computation is correct, we can compare the result of the output of the c tensor to the same computation performed by numpy.
```dev = tvm.device(tgt.kind.name, 0)
n = 1024
a = tvm.nd.array(np.random.uniform(size=n).astype(A.dtype), dev)
b = tvm.nd.array(np.random.uniform(size=n).astype(B.dtype), dev)
c = tvm.nd.array(np.zeros(n, dtype=C.dtype), dev)
tvm.testing.assert_allclose(c.numpy(), a.numpy() + b.numpy())
```
To get a comparison of how fast this version is compared to numpy, create a helper function to run a profile of the TVM generated code.
```import timeit
np_repeat = 100
np_running_time = timeit.timeit(
setup="import numpy\n"
"n = 32768\n"
'dtype = "float32"\n'
"a = numpy.random.rand(n, 1).astype(dtype)\n"
"b = numpy.random.rand(n, 1).astype(dtype)\n",
number=np_repeat,
)
print("Numpy running time: %f" % (np_running_time / np_repeat))
dev = tvm.device(target.kind.name, 0)
n = 32768
a = tvm.nd.array(np.random.uniform(size=n).astype(A.dtype), dev)
b = tvm.nd.array(np.random.uniform(size=n).astype(B.dtype), dev)
c = tvm.nd.array(np.zeros(n, dtype=C.dtype), dev)
evaluator = func.time_evaluator(func.entry_name, dev, number=10)
mean_time = evaluator(a, b, c).mean
print("%s: %f" % (optimization, mean_time))
log.append((optimization, mean_time))
log = [("numpy", np_running_time / np_repeat)]
```
```Numpy running time: 0.000007
naive: 0.000008
```
### Updating the Schedule to Use Parallelism¶
Now that we’ve illustrated the fundamentals of TE, let’s go deeper into what schedules do, and how they can be used to optimize tensor expressions for different architectures. A schedule is a series of steps that are applied to an expression to transform it in a number of different ways. When a schedule is applied to an expression in TE, the inputs and outputs remain the same, but when compiled the implementation of the expression can change. This tensor addition, in the default schedule, is run serially but is easy to parallelize across all of the processor threads. We can apply the parallel schedule operation to our computation.
```s[C].parallel(C.op.axis[0])
```
The `tvm.lower` command will generate the Intermediate Representation (IR) of the TE, with the corresponding schedule. By lowering the expression as we apply different schedule operations, we can see the effect of scheduling on the ordering of the computation. We use the flag `simple_mode=True` to return a readable C-style statement.
```print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.handle, B: T.handle, C: T.handle):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
n = T.int32()
A_1 = T.match_buffer(A, (n,), strides=("stride",), buffer_type="auto")
B_1 = T.match_buffer(B, (n,), strides=("stride",), buffer_type="auto")
C_1 = T.match_buffer(C, (n,), strides=("stride",), buffer_type="auto")
for i in T.parallel(n):
C_2 = T.Buffer((C_1.strides[0] * n,), data=C_1.data, buffer_type="auto")
A_2 = T.Buffer((A_1.strides[0] * n,), data=A_1.data, buffer_type="auto")
B_2 = T.Buffer((B_1.strides[0] * n,), data=B_1.data, buffer_type="auto")
C_2[i * C_1.strides[0]] = A_2[i * A_1.strides[0]] + B_2[i * B_1.strides[0]]
```
It’s now possible for TVM to run these blocks on independent threads. Let’s compile and run this new schedule with the parallel operation applied:
```fadd_parallel = tvm.build(s, [A, B, C], tgt, name="myadd_parallel")
tvm.testing.assert_allclose(c.numpy(), a.numpy() + b.numpy())
```
```parallel: 0.000008
```
### Updating the Schedule to Use Vectorization¶
Modern CPUs also have the ability to perform SIMD operations on floating point values, and we can apply another schedule to our computation expression to take advantage of this. Accomplishing this requires multiple steps: first we have to split the schedule into inner and outer loops using the split scheduling primitive. The inner loops can use vectorization to use SIMD instructions using the vectorize scheduling primitive, then the outer loops can be parallelized using the parallel scheduling primitive. Choose the split factor to be the number of threads on your CPU.
```# Recreate the schedule, since we modified it with the parallel operation in
# the previous example
n = te.var("n")
A = te.placeholder((n,), name="A")
B = te.placeholder((n,), name="B")
C = te.compute(A.shape, lambda i: A[i] + B[i], name="C")
s = te.create_schedule(C.op)
# This factor should be chosen to match the number of threads appropriate for
# your CPU. This will vary depending on architecture, but a good rule is
# setting this factor to equal the number of available CPU cores.
factor = 4
outer, inner = s[C].split(C.op.axis[0], factor=factor)
s[C].parallel(outer)
s[C].vectorize(inner)
print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```vector: 0.000040
# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.handle, B: T.handle, C: T.handle):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
n = T.int32()
A_1 = T.match_buffer(A, (n,), strides=("stride",), buffer_type="auto")
B_1 = T.match_buffer(B, (n,), strides=("stride",), buffer_type="auto")
C_1 = T.match_buffer(C, (n,), strides=("stride",), buffer_type="auto")
for i_outer in T.parallel((n + 3) // 4):
for i_inner_s in range(4):
if T.likely(i_outer * 4 + i_inner_s < n):
C_2 = T.Buffer((C_1.strides[0] * n,), data=C_1.data, buffer_type="auto")
A_2 = T.Buffer((A_1.strides[0] * n,), data=A_1.data, buffer_type="auto")
B_2 = T.Buffer((B_1.strides[0] * n,), data=B_1.data, buffer_type="auto")
cse_var_1: T.int32 = i_outer * 4 + i_inner_s
C_2[cse_var_1 * C_1.strides[0]] = A_2[cse_var_1 * A_1.strides[0]] + B_2[cse_var_1 * B_1.strides[0]]
```
### Comparing the Different Schedules¶
We can now compare the different schedules
```baseline = log[0][1]
print("%s\t%s\t%s" % ("Operator".rjust(20), "Timing".rjust(20), "Performance".rjust(20)))
for result in log:
print(
"%s\t%s\t%s"
% (result[0].rjust(20), str(result[1]).rjust(20), str(result[1] / baseline).rjust(20))
)
```
```Operator Timing Performance
numpy 7.157919999372097e-06 1.0
naive 7.924599999999999e-06 1.107109327946548
parallel 8.136000000000001e-06 1.1366430472418947
vector 3.98908e-05 5.572959742983895
```
Code Specialization
As you may have noticed, the declarations of `A`, `B` and `C` all take the same shape argument, `n`. TVM will take advantage of this to pass only a single shape argument to the kernel, as you will find in the printed device code. This is one form of specialization.
On the host side, TVM will automatically generate check code that checks the constraints in the parameters. So if you pass arrays with different shapes into fadd, an error will be raised.
We can do more specializations. For example, we can write ```n = tvm.runtime.convert(1024)``` instead of `n = te.var("n")`, in the computation declaration. The generated function will only take vectors with length 1024.
We’ve defined, scheduled, and compiled a vector addition operator, which we were then able to execute on the TVM runtime. We can save the operator as a library, which we can then load later using the TVM runtime.
### Targeting Vector Addition for GPUs (Optional)¶
TVM is capable of targeting multiple architectures. In the next example, we will target compilation of the vector addition to GPUs.
```# If you want to run this code, change ``run_cuda = True``
# Note that by default this example is not run in the docs CI.
run_cuda = False
if run_cuda:
# Change this target to the correct backend for you gpu. For example: cuda (NVIDIA GPUs),
# rocm (Radeon GPUS), OpenCL (opencl).
tgt_gpu = tvm.target.Target(target="cuda", host="llvm")
# Recreate the schedule
n = te.var("n")
A = te.placeholder((n,), name="A")
B = te.placeholder((n,), name="B")
C = te.compute(A.shape, lambda i: A[i] + B[i], name="C")
print(type(C))
s = te.create_schedule(C.op)
bx, tx = s[C].split(C.op.axis[0], factor=64)
################################################################################
# Finally we must bind the iteration axis bx and tx to threads in the GPU
# compute grid. The naive schedule is not valid for GPUs, and these are
# specific constructs that allow us to generate code that runs on a GPU.
######################################################################
# Compilation
# -----------
# After we have finished specifying the schedule, we can compile it
# into a TVM function. By default TVM compiles into a type-erased
# function that can be directly called from the python side.
#
# In the following line, we use tvm.build to create a function.
# The build function takes the schedule, the desired signature of the
# function (including the inputs and outputs) as well as target language
# we want to compile to.
#
# The result of compilation fadd is a GPU device function (if GPU is
# involved) as well as a host wrapper that calls into the GPU
# function. fadd is the generated host wrapper function, it contains
# a reference to the generated device function internally.
################################################################################
# The compiled TVM function exposes a concise C API that can be invoked from
# any language.
#
# We provide a minimal array API in python to aid quick testing and prototyping.
# The array API is based on the `DLPack <https://github.com/dmlc/dlpack>`_ standard.
#
# - We first create a GPU device.
# - Then tvm.nd.array copies the data to the GPU.
# - ``fadd`` runs the actual computation
# - ``numpy()`` copies the GPU array back to the CPU (so we can verify correctness).
#
# Note that copying the data to and from the memory on the GPU is a required step.
dev = tvm.device(tgt_gpu.kind.name, 0)
n = 1024
a = tvm.nd.array(np.random.uniform(size=n).astype(A.dtype), dev)
b = tvm.nd.array(np.random.uniform(size=n).astype(B.dtype), dev)
c = tvm.nd.array(np.zeros(n, dtype=C.dtype), dev)
tvm.testing.assert_allclose(c.numpy(), a.numpy() + b.numpy())
################################################################################
# Inspect the Generated GPU Code
# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
# You can inspect the generated code in TVM. The result of tvm.build is a TVM
# Module. fadd is the host module that contains the host wrapper, it also
# contains a device module for the CUDA (GPU) function.
#
# The following code fetches the device module and prints the content code.
if (
tgt_gpu.kind.name == "cuda"
or tgt_gpu.kind.name == "rocm"
or tgt_gpu.kind.name.startswith("opencl")
):
print("-----GPU code-----")
print(dev_module.get_source())
else:
```
Besides runtime compilation, we can save the compiled modules into a file and load them back later.
The following code first performs the following steps:
• It saves the compiled host module into an object file.
• Then it saves the device module into a ptx file.
• cc.create_shared calls a compiler (gcc) to create a shared library
```from tvm.contrib import cc
from tvm.contrib import utils
temp = utils.tempdir()
if tgt.kind.name == "cuda":
if tgt.kind.name == "rocm":
if tgt.kind.name.startswith("opencl"):
print(temp.listdir())
```
```['myadd.so', 'myadd.o']
```
Module Storage Format
The CPU (host) module is directly saved as a shared library (.so). There can be multiple customized formats of the device code. In our example, the device code is stored in ptx, as well as a meta data json file. They can be loaded and linked separately via import.
We can load the compiled module from the file system and run the code. The following code loads the host and device module separately and links them together. We can verify that the newly loaded function works.
```fadd1 = tvm.runtime.load_module(temp.relpath("myadd.so"))
if tgt.kind.name == "cuda":
if tgt.kind.name == "rocm":
if tgt.kind.name.startswith("opencl"):
tvm.testing.assert_allclose(c.numpy(), a.numpy() + b.numpy())
```
### Pack Everything into One Library¶
In the above example, we store the device and host code separately. TVM also supports export everything as one shared library. Under the hood, we pack the device modules into binary blobs and link them together with the host code. Currently we support packing of Metal, OpenCL and CUDA modules.
```fadd.export_library(temp.relpath("myadd_pack.so"))
tvm.testing.assert_allclose(c.numpy(), a.numpy() + b.numpy())
```
The compiled modules of TVM do not depend on the TVM compiler. Instead, they only depend on a minimum runtime library. The TVM runtime library wraps the device drivers and provides thread-safe and device agnostic calls into the compiled functions.
This means that you can call the compiled TVM functions from any thread, on any GPUs, provided that you have compiled the code for that GPU.
## Generate OpenCL Code¶
TVM provides code generation features into multiple backends. We can also generate OpenCL code or LLVM code that runs on CPU backends.
The following code blocks generate OpenCL code, creates array on an OpenCL device, and verifies the correctness of the code.
```if tgt.kind.name.startswith("opencl"):
print("------opencl code------")
dev = tvm.cl(0)
n = 1024
a = tvm.nd.array(np.random.uniform(size=n).astype(A.dtype), dev)
b = tvm.nd.array(np.random.uniform(size=n).astype(B.dtype), dev)
c = tvm.nd.array(np.zeros(n, dtype=C.dtype), dev)
tvm.testing.assert_allclose(c.numpy(), a.numpy() + b.numpy())
```
TE Scheduling Primitives
TVM includes a number of different scheduling primitives:
• split: splits a specified axis into two axises by the defined factor.
• tile: tiles will split a computation across two axes by the defined factors.
• fuse: fuses two consecutive axises of one computation.
• reorder: can reorder the axises of a computation into a defined order.
• bind: can bind a computation to a specific thread, useful in GPU programming.
• compute_at: by default, TVM will compute tensors at the outermost level of the function, or the root, by default. compute_at specifies that one tensor should be computed at the first axis of computation for another operator.
• compute_inline: when marked inline, a computation will be expanded then inserted into the address where the tensor is required.
• compute_root: moves a computation to the outermost layer, or root, of the function. This means that stage of the computation will be fully computed before it moves on to the next stage.
A complete description of these primitives can be found in the Schedule Primitives docs page.
## Example 2: Manually Optimizing Matrix Multiplication with TE¶
Now we will consider a second, more advanced example, demonstrating how with just 18 lines of python code TVM speeds up a common matrix multiplication operation by 18x.
Matrix multiplication is a compute intensive operation. There are two important optimizations for good CPU performance:
1. Increase the cache hit rate of memory access. Both complex numerical computation and hot-spot memory access can be accelerated by a high cache hit rate. This requires us to transform the origin memory access pattern to a pattern that fits the cache policy.
2. SIMD (Single instruction multi-data), also known as the vector processing unit. On each cycle instead of processing a single value, SIMD can process a small batch of data. This requires us to transform the data access pattern in the loop body in uniform pattern so that the LLVM backend can lower it to SIMD.
The techniques used in this tutorial are a subset of tricks mentioned in this repository. Some of them have been applied by TVM abstraction automatically, but some of them cannot be automatically applied due to TVM constraints.
### Preparation and Performance Baseline¶
We begin by collecting performance data on the numpy implementation of matrix multiplication.
```import tvm
import tvm.testing
from tvm import te
import numpy
# The size of the matrix
# (M, K) x (K, N)
# You are free to try out different shapes, sometimes TVM optimization outperforms numpy with MKL.
M = 1024
K = 1024
N = 1024
# The default tensor data type in tvm
dtype = "float32"
# You will want to adjust the target to match any CPU vector extensions you
# might have. For example, if you're using using Intel AVX2 (Advanced Vector
# following line to ``llvm -mcpu=core-avx2``, or specific type of CPU you use.
# Recall that you're using llvm, you can get this information from the command
# ``llc --version`` to get the CPU type, and you can check ``/proc/cpuinfo``
target = tvm.target.Target(target="llvm", host="llvm")
dev = tvm.device(target.kind.name, 0)
# Random generated tensor for testing
a = tvm.nd.array(numpy.random.rand(M, K).astype(dtype), dev)
b = tvm.nd.array(numpy.random.rand(K, N).astype(dtype), dev)
# Repeatedly perform a matrix multiplication to get a performance baseline
# for the default numpy implementation
np_repeat = 100
np_running_time = timeit.timeit(
setup="import numpy\n"
"M = " + str(M) + "\n"
"K = " + str(K) + "\n"
"N = " + str(N) + "\n"
'dtype = "float32"\n'
"a = numpy.random.rand(M, K).astype(dtype)\n"
"b = numpy.random.rand(K, N).astype(dtype)\n",
number=np_repeat,
)
print("Numpy running time: %f" % (np_running_time / np_repeat))
```
```Numpy running time: 0.019177
```
Now we write a basic matrix multiplication using TVM TE and verify that it produces the same results as the numpy implementation. We also write a function that will help us measure the performance of the schedule optimizations.
```# TVM Matrix Multiplication using TE
k = te.reduce_axis((0, K), "k")
A = te.placeholder((M, K), name="A")
B = te.placeholder((K, N), name="B")
C = te.compute((M, N), lambda x, y: te.sum(A[x, k] * B[k, y], axis=k), name="C")
# Default schedule
s = te.create_schedule(C.op)
func = tvm.build(s, [A, B, C], target=target, name="mmult")
c = tvm.nd.array(numpy.zeros((M, N), dtype=dtype), dev)
func(a, b, c)
def evaluate_operation(s, vars, target, name, optimization, log):
func = tvm.build(s, [A, B, C], target=target, name="mmult")
assert func
c = tvm.nd.array(numpy.zeros((M, N), dtype=dtype), dev)
func(a, b, c)
evaluator = func.time_evaluator(func.entry_name, dev, number=10)
mean_time = evaluator(a, b, c).mean
print("%s: %f" % (optimization, mean_time))
log.append((optimization, mean_time))
log = []
evaluate_operation(s, [A, B, C], target=target, name="mmult", optimization="none", log=log)
```
```none: 3.430100
```
Let’s take a look at the intermediate representation of the operator and default schedule using the TVM lower function. Note how the implementation is essentially a naive implementation of a matrix multiplication, using three nested loops over the indices of the A and B matrices.
```print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
for x, y in T.grid(1024, 1024):
C_1 = T.Buffer((1048576,), data=C.data)
C_1[x * 1024 + y] = T.float32(0)
for k in range(1024):
cse_var_2: T.int32 = x * 1024
cse_var_1: T.int32 = cse_var_2 + y
A_1 = T.Buffer((1048576,), data=A.data)
B_1 = T.Buffer((1048576,), data=B.data)
C_1[cse_var_1] = C_1[cse_var_1] + A_1[cse_var_2 + k] * B_1[k * 1024 + y]
```
### Optimization 1: Blocking¶
A important trick to enhance the cache hit rate is blocking, where you structure memory access such that the inside a block is a small neighborhood that has high memory locality. In this tutorial, we pick a block factor of 32. This will result in a block that will fill a 32 * 32 * sizeof(float) area of memory. This corresponds to a cache size of 4KB, in relation to a reference cache size of 32 KB for L1 cache.
We begin by creating a default schedule for the `C` operation, then apply a `tile` scheduling primitive to it with the specified block factor, with the scheduling primitive returning the resulting loop order from outermost to innermost, as a vector `[x_outer, y_outer, x_inner, y_inner]`. We then get the reduction axis for output of the operation, and perform a split operation on it using a factor of 4. This factor doesn’t directly impact the blocking optimization we’re working on right now, but will be useful later when we apply vectorization.
Now that the operation has been blocked, we can reorder the computation to put the reduction operation into the outermost loop of the computation, helping to guarantee that the blocked data remains in cache. This completes the schedule, and we can build and test the performance compared to the naive schedule.
```bn = 32
# Blocking by loop tiling
xo, yo, xi, yi = s[C].tile(C.op.axis[0], C.op.axis[1], bn, bn)
(k,) = s[C].op.reduce_axis
ko, ki = s[C].split(k, factor=4)
# Hoist reduction domain outside the blocking loop
s[C].reorder(xo, yo, ko, ki, xi, yi)
evaluate_operation(s, [A, B, C], target=target, name="mmult", optimization="blocking", log=log)
```
```blocking: 0.321802
```
By reordering the computation to take advantage of caching, you should see a significant improvement in the performance of the computation. Now, print the internal representation and compare it to the original:
```print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
for x_outer, y_outer in T.grid(32, 32):
C_1 = T.Buffer((1048576,), data=C.data)
for x_inner_init, y_inner_init in T.grid(32, 32):
C_1[x_outer * 32768 + x_inner_init * 1024 + y_outer * 32 + y_inner_init] = T.float32(0)
for k_outer, k_inner, x_inner, y_inner in T.grid(256, 4, 32, 32):
cse_var_3: T.int32 = y_outer * 32
cse_var_2: T.int32 = x_outer * 32768 + x_inner * 1024
cse_var_1: T.int32 = cse_var_2 + cse_var_3 + y_inner
A_1 = T.Buffer((1048576,), data=A.data)
B_1 = T.Buffer((1048576,), data=B.data)
C_1[cse_var_1] = C_1[cse_var_1] + A_1[cse_var_2 + k_outer * 4 + k_inner] * B_1[k_outer * 4096 + k_inner * 1024 + cse_var_3 + y_inner]
```
### Optimization 2: Vectorization¶
Another important optimization trick is vectorization. When the memory access pattern is uniform, the compiler can detect this pattern and pass the continuous memory to the SIMD vector processor. In TVM, we can use the `vectorize` interface to hint the compiler this pattern, taking advantage of this hardware feature.
In this tutorial, we chose to vectorize the inner loop row data since it is already cache friendly from our previous optimizations.
```# Apply the vectorization optimization
s[C].vectorize(yi)
evaluate_operation(s, [A, B, C], target=target, name="mmult", optimization="vectorization", log=log)
# The generalized IR after vectorization
print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```vectorization: 0.296758
# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
for x_outer, y_outer in T.grid(32, 32):
C_1 = T.Buffer((1048576,), data=C.data)
for x_inner_init in range(32):
C_1[x_outer * 32768 + x_inner_init * 1024 + y_outer * 32:x_outer * 32768 + x_inner_init * 1024 + y_outer * 32 + 32] = T.Broadcast(T.float32(0), 32)
for k_outer, k_inner, x_inner in T.grid(256, 4, 32):
cse_var_3: T.int32 = y_outer * 32
cse_var_2: T.int32 = x_outer * 32768 + x_inner * 1024
cse_var_1: T.int32 = cse_var_2 + cse_var_3
A_1 = T.Buffer((1048576,), data=A.data)
B_1 = T.Buffer((1048576,), data=B.data)
C_1[cse_var_1:cse_var_1 + 32] = C_1[cse_var_1:cse_var_1 + 32] + T.Broadcast(A_1[cse_var_2 + k_outer * 4 + k_inner], 32) * B_1[k_outer * 4096 + k_inner * 1024 + cse_var_3:k_outer * 4096 + k_inner * 1024 + cse_var_3 + 32]
```
### Optimization 3: Loop Permutation¶
If we look at the above IR, we can see the inner loop row data is vectorized and B is transformed into PackedB (this is evident by the (float32x32*)B2 portion of the inner loop). The traversal of PackedB is sequential now. So we will look at the access pattern of A. In current schedule, A is accessed column by column which is not cache friendly. If we change the nested loop order of ki and inner axes xi, the access pattern for A matrix will be more cache friendly.
```s = te.create_schedule(C.op)
xo, yo, xi, yi = s[C].tile(C.op.axis[0], C.op.axis[1], bn, bn)
(k,) = s[C].op.reduce_axis
ko, ki = s[C].split(k, factor=4)
# re-ordering
s[C].reorder(xo, yo, ko, xi, ki, yi)
s[C].vectorize(yi)
evaluate_operation(
s, [A, B, C], target=target, name="mmult", optimization="loop permutation", log=log
)
# Again, print the new generalized IR
print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```loop permutation: 0.117512
# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
for x_outer, y_outer in T.grid(32, 32):
C_1 = T.Buffer((1048576,), data=C.data)
for x_inner_init in range(32):
C_1[x_outer * 32768 + x_inner_init * 1024 + y_outer * 32:x_outer * 32768 + x_inner_init * 1024 + y_outer * 32 + 32] = T.Broadcast(T.float32(0), 32)
for k_outer, x_inner, k_inner in T.grid(256, 32, 4):
cse_var_3: T.int32 = y_outer * 32
cse_var_2: T.int32 = x_outer * 32768 + x_inner * 1024
cse_var_1: T.int32 = cse_var_2 + cse_var_3
A_1 = T.Buffer((1048576,), data=A.data)
B_1 = T.Buffer((1048576,), data=B.data)
C_1[cse_var_1:cse_var_1 + 32] = C_1[cse_var_1:cse_var_1 + 32] + T.Broadcast(A_1[cse_var_2 + k_outer * 4 + k_inner], 32) * B_1[k_outer * 4096 + k_inner * 1024 + cse_var_3:k_outer * 4096 + k_inner * 1024 + cse_var_3 + 32]
```
### Optimization 4: Array Packing¶
Another important trick is array packing. This trick is to reorder the storage dimension of the array to convert the continuous access pattern on certain dimension to a sequential pattern after flattening.
Just as it is shown in the figure above, after blocking the computations, we can observe the array access pattern of B (after flattening), which is regular but discontinuous. We expect that after some transformation we can get a continuous access pattern. By reordering a `[16][16]` array to a `[16/4][16][4]` array the access pattern of B will be sequential when grabbing the corresponding value from the packed array.
To accomplish this, we are going to have to start with a new default schedule, taking into account the new packing of B. It’s worth taking a moment to comment on this: TE is a powerful and expressive language for writing optimized operators, but it often requires some knowledge of the underlying algorithm, data structures, and hardware target that you are writing for. Later in the tutorial, we will discuss some of the options for letting TVM take that burden. Regardless, let’s move on with the new optimized schedule.
```# We have to re-write the algorithm slightly.
packedB = te.compute((N / bn, K, bn), lambda x, y, z: B[y, x * bn + z], name="packedB")
C = te.compute(
(M, N),
lambda x, y: te.sum(A[x, k] * packedB[y // bn, k, tvm.tir.indexmod(y, bn)], axis=k),
name="C",
)
s = te.create_schedule(C.op)
xo, yo, xi, yi = s[C].tile(C.op.axis[0], C.op.axis[1], bn, bn)
(k,) = s[C].op.reduce_axis
ko, ki = s[C].split(k, factor=4)
s[C].reorder(xo, yo, ko, xi, ki, yi)
s[C].vectorize(yi)
x, y, z = s[packedB].op.axis
s[packedB].vectorize(z)
s[packedB].parallel(x)
evaluate_operation(s, [A, B, C], target=target, name="mmult", optimization="array packing", log=log)
# Here is the generated IR after array packing.
print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```array packing: 0.105009
# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
packedB = T.allocate([32768], "float32x32", "global")
packedB_1 = T.Buffer((32768,), "float32x32", data=packedB)
for x in T.parallel(32):
for y in range(1024):
B_1 = T.Buffer((1048576,), data=B.data)
packedB_1[x * 1024 + y] = B_1[y * 1024 + x * 32:y * 1024 + x * 32 + 32]
for x_outer, y_outer in T.grid(32, 32):
C_1 = T.Buffer((1048576,), data=C.data)
for x_inner_init in range(32):
C_1[x_outer * 32768 + x_inner_init * 1024 + y_outer * 32:x_outer * 32768 + x_inner_init * 1024 + y_outer * 32 + 32] = T.Broadcast(T.float32(0), 32)
for k_outer, x_inner, k_inner in T.grid(256, 32, 4):
cse_var_3: T.int32 = x_outer * 32768 + x_inner * 1024
cse_var_2: T.int32 = k_outer * 4
cse_var_1: T.int32 = cse_var_3 + y_outer * 32
A_1 = T.Buffer((1048576,), data=A.data)
C_1[cse_var_1:cse_var_1 + 32] = C_1[cse_var_1:cse_var_1 + 32] + T.Broadcast(A_1[cse_var_3 + cse_var_2 + k_inner], 32) * packedB_1[y_outer * 1024 + cse_var_2 + k_inner]
```
### Optimization 5: Optimizing Block Writing Through Caching¶
Up to this point all of our optimizations have focused on efficiently accessing and computing the data from the A and B matrices to compute the C matrix. After the blocking optimization, the operator will write result to C block by block, and the access pattern is not sequential. We can address this by using a sequential cache array, using a combination of cache_write, compute_at, and unroll`to hold the block results and write to `C when all the block results are ready.
```s = te.create_schedule(C.op)
# Allocate write cache
CC = s.cache_write(C, "global")
xo, yo, xi, yi = s[C].tile(C.op.axis[0], C.op.axis[1], bn, bn)
# Write cache is computed at yo
s[CC].compute_at(s[C], yo)
# New inner axes
xc, yc = s[CC].op.axis
(k,) = s[CC].op.reduce_axis
ko, ki = s[CC].split(k, factor=4)
s[CC].reorder(ko, xc, ki, yc)
s[CC].unroll(ki)
s[CC].vectorize(yc)
x, y, z = s[packedB].op.axis
s[packedB].vectorize(z)
s[packedB].parallel(x)
evaluate_operation(s, [A, B, C], target=target, name="mmult", optimization="block caching", log=log)
# Here is the generated IR after write cache blocking.
print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```block caching: 0.110816
# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
packedB = T.allocate([32768], "float32x32", "global")
C_global = T.allocate([1024], "float32", "global")
packedB_1 = T.Buffer((32768,), "float32x32", data=packedB)
for x in T.parallel(32):
for y in range(1024):
B_1 = T.Buffer((1048576,), data=B.data)
packedB_1[x * 1024 + y] = B_1[y * 1024 + x * 32:y * 1024 + x * 32 + 32]
for x_outer, y_outer in T.grid(32, 32):
C_global_1 = T.Buffer((1024,), data=C_global)
for x_c_init in range(32):
C_global_1[x_c_init * 32:x_c_init * 32 + 32] = T.Broadcast(T.float32(0), 32)
for k_outer, x_c in T.grid(256, 32):
cse_var_4: T.int32 = k_outer * 4
cse_var_3: T.int32 = x_c * 32
cse_var_2: T.int32 = y_outer * 1024 + cse_var_4
cse_var_1: T.int32 = x_outer * 32768 + x_c * 1024 + cse_var_4
A_1 = T.Buffer((1048576,), data=A.data)
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1], 32) * packedB_1[cse_var_2]
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1 + 1], 32) * packedB_1[cse_var_2 + 1]
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1 + 2], 32) * packedB_1[cse_var_2 + 2]
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1 + 3], 32) * packedB_1[cse_var_2 + 3]
for x_inner, y_inner in T.grid(32, 32):
C_1 = T.Buffer((1048576,), data=C.data)
C_1[x_outer * 32768 + x_inner * 1024 + y_outer * 32 + y_inner] = C_global_1[x_inner * 32 + y_inner]
```
### Optimization 6: Parallelization¶
So far, our computation is only designed to use a single core. Nearly all modern processors have multiple cores, and computation can benefit from running computations in parallel. The final optimization is to take advantage of thread-level parallelization.
```# parallel
s[C].parallel(xo)
x, y, z = s[packedB].op.axis
s[packedB].vectorize(z)
s[packedB].parallel(x)
evaluate_operation(
s, [A, B, C], target=target, name="mmult", optimization="parallelization", log=log
)
# Here is the generated IR after parallelization.
print(tvm.lower(s, [A, B, C], simple_mode=True))
```
```parallelization: 0.131877
# from tvm.script import ir as I
# from tvm.script import tir as T
@I.ir_module
class Module:
@T.prim_func
def main(A: T.Buffer((1024, 1024), "float32"), B: T.Buffer((1024, 1024), "float32"), C: T.Buffer((1024, 1024), "float32")):
T.func_attr({"from_legacy_te_schedule": T.bool(True), "global_symbol": "main", "tir.noalias": T.bool(True)})
packedB = T.allocate([32768], "float32x32", "global")
packedB_1 = T.Buffer((32768,), "float32x32", data=packedB)
for x in T.parallel(32):
for y in range(1024):
B_1 = T.Buffer((1048576,), data=B.data)
packedB_1[x * 1024 + y] = B_1[y * 1024 + x * 32:y * 1024 + x * 32 + 32]
for x_outer in T.parallel(32):
C_global = T.allocate([1024], "float32", "global")
for y_outer in range(32):
C_global_1 = T.Buffer((1024,), data=C_global)
for x_c_init in range(32):
C_global_1[x_c_init * 32:x_c_init * 32 + 32] = T.Broadcast(T.float32(0), 32)
for k_outer, x_c in T.grid(256, 32):
cse_var_4: T.int32 = k_outer * 4
cse_var_3: T.int32 = x_c * 32
cse_var_2: T.int32 = y_outer * 1024 + cse_var_4
cse_var_1: T.int32 = x_outer * 32768 + x_c * 1024 + cse_var_4
A_1 = T.Buffer((1048576,), data=A.data)
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1], 32) * packedB_1[cse_var_2]
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1 + 1], 32) * packedB_1[cse_var_2 + 1]
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1 + 2], 32) * packedB_1[cse_var_2 + 2]
C_global_1[cse_var_3:cse_var_3 + 32] = C_global_1[cse_var_3:cse_var_3 + 32] + T.Broadcast(A_1[cse_var_1 + 3], 32) * packedB_1[cse_var_2 + 3]
for x_inner, y_inner in T.grid(32, 32):
C_1 = T.Buffer((1048576,), data=C.data)
C_1[x_outer * 32768 + x_inner * 1024 + y_outer * 32 + y_inner] = C_global_1[x_inner * 32 + y_inner]
```
### Summary of Matrix Multiplication Example¶
After applying the above simple optimizations with only 18 lines of code, our generated code can begin to approach the performance of numpy with the Math Kernel Library (MKL). Since we’ve been logging the performance as we’ve been working, we can compare the results.
```baseline = log[0][1]
print("%s\t%s\t%s" % ("Operator".rjust(20), "Timing".rjust(20), "Performance".rjust(20)))
for result in log:
print(
"%s\t%s\t%s"
% (result[0].rjust(20), str(result[1]).rjust(20), str(result[1] / baseline).rjust(20))
)
```
``` Operator Timing Performance
none 3.4301004538 1.0
blocking 0.3218023729 0.09381718618284042
vectorization 0.2967580129 0.08651583733392991
loop permutation 0.1175119024 0.03425902651621054
array packing 0.10500924760000001 0.03061404440317969
block caching 0.1108162278 0.03230699196498266
parallelization 0.1318765312 0.0384468422940506
```
Note that the outputs on the web page reflect the running times on a non-exclusive Docker container, and should be considered unreliable. It is highly encouraged to run the tutorial by yourself to observe the performance gain achieved by TVM, and to carefully work through each example to understand the iterative improvements that are made to the matrix multiplication operation.
## Final Notes and Summary¶
As mentioned earlier, how to apply optimizations using TE and scheduling primitives can require some knowledge of the underlying architecture and algorithms. However, TE was designed to act as a foundation for more complex algorithms that can search the potential optimization. With the knowledge you have from this introduction to TE, we can now begin to explore how TVM can automate the schedule optimization process.
This tutorial provided a walk-through of TVM Tensor Expression (TE) workflow using a vector add and a matrix multiplication examples. The general workflow is
• Describe your computation via a series of operations.
• Describe how we want to compute use schedule primitives.
• Compile to the target function we want.
• Optionally, save the function to be loaded later.
Upcoming tutorials expand on the matrix multiplication example, and show how you can build generic templates of the matrix multiplication and other operations with tunable parameters that allows you to automatically optimize the computation for specific platforms.
Total running time of the script: ( 1 minutes 0.810 seconds)
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https://www.jiskha.com/display.cgi?id=1173382700 | 1,503,352,551,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00253.warc.gz | 928,104,080 | 3,949 | # ela
posted by .
how many high students allow students to pay for school lunch?
how much the income has to be in order not to pay for lunch?
The guidelines vary from state to state, you have to go to your school to find out. Family size has a big impact on it, as well as the number of adults in the household.
Thank you for using the Jiskha Homework Help Forum. Here is the site on the National School Lunch Program:
Please remember that there is no "free lunch" anywhere because "someone" pays!
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https://forum.thegradcafe.com/topic/110186-harvard-ms-student-probability-vs-real-analysis-for-phd/?tab=comments | 1,603,695,966,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890586.57/warc/CC-MAIN-20201026061044-20201026091044-00699.warc.gz | 334,555,042 | 26,073 | # Harvard MS student probability vs real analysis for PhD
## Recommended Posts
Hello everyone,
I will be beginning biostat grad program at Harvard in the fall. I ultimately plan to apply to statistics/OR PhD programs next year (I will be completing my MS next fall).
I am aware of the fact that many graduate schools prefer students to have completed two semesters of real analysis, and I have been definitely planning to take graduate-level real analysis courses at Harvard graduate school. However, I come from a non-mathematics background (my degree is in chemical engineering from one of the ivies, and I have not taken proof-based math classes in undergrad), and my advisor suggests that I take proof-based probability classes that are designed for biostat PhD students in lieu of real analysis.
Descriptions of probability I and probability II are as shown below:
1) Probability I
Axiomatic foundations of probability, independence, conditional probability, joint distributions, transformations, moment generating functions, characteristic functions, moment inequalities, sampling distributions, modes of convergence and their interrelationships, laws of large numbers, central limit theorem, and stochastic processes.
2) Probability II
A foundational course in measure theoretic probability. Topics include measure theory, Lebesgue integration, product measure and Fubini's Theorem, Radon-Nikodym derivatives, conditional probability, conditional expectation, limit theorems on sequences of random stochastic processes, and weak convergence.
Would it be sufficient to take these probability classes in replacement of two semesters of real analysis? I plan to take one elective stochastic course as well in addition to two probability classes.
Thank you.
##### Share on other sites
I would take real analysis, since you don't have it on your transcript. Judging from these course descriptions, Probability I is just the first semester of the year-long sequence of Mathematical (Theoretical) Statistics I & II, so there does not seem to be any point in taking it separately from the typical Masters-level mathematical statistics sequence.
For Probability II at your institution, the instructor probably won't allow you to take it if you haven't taken a basic analysis course. And you can take this in your PhD program anyway (at most stat and biostat programs, this would be the required PhD-level course in Probability). Many PhD programs only require one semester of measure-theoretic probability theory (including reputable programs such as CMU and Duke), with an optional second semester of advanced probability theory for the students who plan to specialize more on probability theory than on statistics. I did my dissertation on theoretical stats, and I only needed to know the very basics of measure theory and probability theory for my research.
Edited by Applied Math to Stat
##### Share on other sites
The graduate level real analysis at Harvard math 212 a and b is extremely advanced and should probably only be taken by math grad students. Stat 210 and 212 should serve your needs. 210 is usually taught by Blitzstein who is an excellent teacher
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After looking at the course websites for Math 212a and Math 212b, I would agree with the above poster that these are not necessary for statistics Masters students.
It seems like Math 112 (Introductory Real Analysis) at Harvard would be the most appropriate real analysis class for Masters students who are interested in applying to stat/biostat PhD. It seems that only one semester of Math 112 is necessary too, since they seem to cover the whole analysis sequence in one semester (other schools go more slowly and spread it out over two semesters).
OP: are grad students allowed to take a class like Math112 at Harvard? If so, I would strongly recommend taking this.
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Grad students can take math 112. Math 112 is the basic upper division real analysis class(Rudin or some variation of it). Math 114 is the standard graduate level analysis(Royden) at most schools.
##### Share on other sites
6 minutes ago, Gauss2017 said:
Grad students can take math 112. Math 112 is the basic upper division real analysis class(Rudin or some variation of it). Math 114 is the standard graduate level analysis(Royden) at most schools.
In that case, the OP would be a very competitive candidate for PhD programs if s/he took both Math 112: Introductory Real Analysis and Math 114: Analysis II. I think just taking Math 112 might be sufficient for being in the discussion for admissions, but adcoms may look favorably upon having studied basic measure theory/Lebesgue integration. In my PhD program, we had to take basically a whole semester of measure theory/abstract integration, but a lot of stat programs condense this into the first few weeks of a one-semester Probability Theory class before diving into the classical probability theorems (Fubini, Fatou lemma, etc.).
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@Gauss2017 Thank you so much! Your insights have been very helpful. I did email Blitzstein a few days ago and got the permission to take Stat 210 in the fall. I've been told that he is an amazing professor that truly enables you to 'understand' probability, and your suggestion confirms once again that I should take his class during my first semester.
I pretty much have my first semester figured out at this moment with the inclusion of stat 210 and will talk to my advisor about the second/third semester once I start my program in the fall. Once again- thank you so much for your help.
@Applied Math to Stat Thank you so much! I concur that Math 112 might be sufficient for admissions. Math 112 is typically offered in spring, and math 114 is typically offered in the fall so I might take Math 114 during my last semester (fall 2019) after Math 112 if that is what my advisor suggests for admissions. I really appreciate your help.
##### Share on other sites
You may want to think about stat 212 for the spring instead of math 114. Recently 212 has been taught by Strook from MIT and he does a good job. It is basically an advanced probability class with real analysis mixed in. Other good classes are stat 149 and stat 186 if you can fit them in your schedule
##### Share on other sites
@Gauss2017 Thank you so much! I will keep that in my mind. Your suggested classes definitely seem to be good elective courses to take during my last semester.
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• 1 year later...
What do people think of Harvard's math 23-A for real analysis I going into a PhD program?
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48 minutes ago, kingduck said:
What do people think of Harvard's math 23-A for real analysis I going into a PhD program?
Literally any real analysis class will be fine. Nobody needs two semesters or graduate level analysis. Learn some basics about convergence and properties of sets so you're not overwhelmed when you take a measure theoretic probability class and you're fine.
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Posted (edited)
If you haven't taken any proof-based courses, I don't think it's a good idea to take grad-level probability courses. I would start from basic real analysis courses. Ideally you need two courses in real analysis covering material such as metric space/norm, completeness of metric space, hölder's inequality/Minkowski inequality(this can be further extended to other spaces such as Lp space and probability space) , basic topology, continuity/uniform continuity in metric space and Riemann integration. It would be better if you had exposure to Lebesgue integration before taking grad-level probability courses since probability space is a special general measure space and understanding Lebesgue measure definitely helps.
Edited by Casorati
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### #1 k3y
Reputation: 36
• Posts: 205
• Joined: 25-February 12
Posted 25 January 2013 - 02:51 PM
Howdy folks;
I have been assigned a problem by my professor, and I think I have completed all of the points specified, however; he still says that I have not completed the assignment. I would deeply appreciate any help that anyone can provide. I have posted both my code, and the assignment below.
ASSIGNMENT
Quote
In this program you will prove that on average, binary search works better than linear search.
1. Create a sorted list of 20 numbers.
2. Search for each element using linear search and print out the number of iterations before the item is found.
3. Now use binary search and print the same information.
4. Print the average number of iterations for each search.
5. You should be able to show that binary search does not always perform better than linear search but does perform better on average.
You will find the Binary Search and the Sequential search in the Unit 10 Coding folder.
CODE:
import java.util.*;
public class Assignment1
{
static Scanner console = new Scanner(System.in);
public static void main(String[] args)
{
//point 1
int[] numberList = {
1, 2, 3, 4, 5,
6, 7, 8, 9, 10,
11, 12, 13, 14, 15,
16, 17, 18, 19, 20
};
for(int x = 0; x < numberList.length; x++)
{
System.out.println("sequential search: ");
System.out.println(seqSearch(numberList, numberList[x]));
System.out.println();
System.out.println("binary search: ");
System.out.println(binSearch(numberList, numberList[x]));
}
}
public static int seqSearch(int[] array, int searchItem)
{
int iterations = 0;
for(int x = 0; x < array.length; x++)
{
iterations++;
if(array[x] == searchItem)
{
System.out.println("number of iterations(seq): " + iterations);
return x;
}
}
return -1;
}
public static int binSearch(int[] array, int searchItem)
{
int iterations = 0;
int start = 0;
int mid;
int end = array.length-1;
while (start <= end)
{
iterations++;
mid = ((start + end) / 2);
if (searchItem == array[mid])
{
System.out.println("number of iterations(bin): " + iterations);
return mid;
}
else if (searchItem < array[mid])
{
end = mid - 1;
}
else
{
start = mid + 1;
}
}
return -1;
}
}
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https://es.mathworks.com/matlabcentral/cody/problems/406-back-to-basics-16-byte-order/solutions/51861 | 1,606,607,549,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00241.warc.gz | 297,071,915 | 16,918 | Cody
# Problem 406. Back to basics 16 - byte order
Solution 51861
Submitted on 26 Feb 2012 by Yuval Cohen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = uint16(12345); y_correct = 14640; assert(isequal(byte_order(x),y_correct))
y = 14640
2 Pass
%% x = uint32(12345); y_correct = 959447040; assert(isequal(byte_order(x),y_correct))
y = 959447040
3 Pass
%% x = int32(-12345); y_correct = -942669825; assert(isequal(byte_order(x),y_correct))
y = -942669825
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Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 208 | 721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-50 | latest | en | 0.619763 |
https://www.physicsforums.com/threads/surjective-group-homomorphism.197573/ | 1,527,309,224,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00274.warc.gz | 801,538,643 | 14,923 | # Homework Help: Surjective group homomorphism
1. Nov 11, 2007
### e12514
1. The problem statement, all variables and given/known data
Show the map (call it phi) from U_n to C*
defined by phi(X) = det(X) for all matrices X in U_n,
is a surjective homomorphism, where
U_n is the subgroup of GL(n,C) consisting of unitary matrices
C* = C\{0} = invertible/nonzero complex numbers
det(.) is the determinant of .
2. Relevant equations
A matrix X in GL(n,C) is unitary if ((X_bar)^T).X = I
where X_bar is the conjugate of X (taking the conjugate of each entry in X)
3. The attempt at a solution
Homomorphism is easy to verify:
phi(XY) = det(XY) = det(X) det(Y) = phi(X) phi(Y)
I'm having trouble showing it's surjective.
For it to be surjective we need
for all (a+bi) in C* there exists X in U_n such that phi(X) = det(X) = a+bi
where a and b not both zero
The problem is X needs to be unitary, that is, X-conjugate-transpose times X needs to be the identity matrix.
It follows the the required matrix X can't be a diagonal matrix, otherwise it is not unitary if its determinant is (a+bi), in particular every entry on that diagonal where there's a complex number x+yi, that entry will become (x^2 + y^2) after taking ((X_bar)^T).X instead of the required "1" on the diagonal of an identity matrix.
Similarly I couldn't get anywhere with triangular matrices, and any non-diagonal and non-triangular matrix seems to get too complicated, not to mention that triangular matrices are messy enough...
2. Nov 11, 2007
### morphism
It's definitely not going to be surjective -- the determinant of a unitary matrix lies on the unit circle.
3. Nov 12, 2007
### e12514
Hmm.... that's true. Maybe there was a typo on the question.
Unless there is some other interpretation for what C* is...
4. Nov 12, 2007
### CompuChip
Maybe it was indeed meant that
C* = { z \in C : ||z|| = 1 }
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
Similar Threads for Surjective group homomorphism
How to prove that this function is a surjection?
Are these homomorphisms? | 573 | 2,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-22 | latest | en | 0.917603 |
https://communities.sas.com/t5/SAS-Procedures/Macro-in-Proc-Sql-Statement/m-p/101109 | 1,716,297,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00889.warc.gz | 148,857,956 | 29,765 | Calcite | Level 5
Macro in Proc Sql Statement
Does anyone know if I could possibly shorten this up with a macro or something else?
PROC SQL;
CREATE TABLE J4 AS
SELECT DISTINCT t1.pc,
t1.CODE,
(MAX(t1.JAN2010)) AS MAX_of_JAN2010,
(MAX(t1.JUL2010)) AS MAX_of_JUL2010,
(MAX(t1.NOV2010)) AS MAX_of_NOV2010,
(MAX(t1.DEC2010)) AS MAX_of_DEC2010,
(MAX(t1.JAN2011)) AS MAX_of_JAN2011,
(MAX(t1.FEB2011)) AS MAX_of_FEB2011,
(MAX(t1.MAR2011)) AS MAX_of_MAR2011,
(MAX(t1.APR2011)) AS MAX_of_APR2011,
(MAX(t1.MAY2011)) AS MAX_of_MAY2011,
(MAX(t1.JUN2011)) AS MAX_of_JUN2011,
(MAX(t1.JUL2011)) AS MAX_of_JUL2011,
(MAX(t1.AUG2011)) AS MAX_of_AUG2011,
(MAX(t1.SEP2011)) AS MAX_of_SEP2011,
(MAX(t1.OCT2011)) AS MAX_of_OCT2011,
(MAX(t1.NOV2011)) AS MAX_of_NOV2011,
(MAX(t1.DEC2011)) AS MAX_of_DEC2011
FROM J3 t1
GROUP BY t1.pc, t1.CODE;
QUIT;
2 REPLIES 2
Onyx | Level 15
Re: Macro in Proc Sql Statement
If your month are random, there is little you can do. However, if your month are continuous, you could try some Macro like this:
data _null_;
do _n_=1 by 1 until (date> '01dec2011'd);
date=intnx('month','01oct2010'd,_n_,'b');
call symputx('date'||strip(_n_), put(date,monyy7.));
end;
run;
%macro test;
PROC SQL;
CREATE TABLE J4 AS
SELECT DISTINCT t1.pc,
t1.CODE, %do i=1 %to 13; max(t1.&&date&i) as max_of_&&date&i, %end;
(MAX(t1.DEC2011)) AS MAX_of_DEC2011
FROM J3 t1
GROUP BY t1.pc, t1.CODE;
QUIT;
%mend;
%test
Regards,
Haikuo
Super User
Re: Macro in Proc Sql Statement
Do you have to use SQL ? I think proc means is better .
proc means data=have nway noprint;
class pc CODE;
var JAN2010--DEC2011;
output out=want max= /autoname;
run;
Ksharp
Discussion stats
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• 3 in conversation | 619 | 1,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.564258 |
https://community-forums.domo.com/main/discussion/62390/booking-date-and-shipment-date-formula | 1,722,802,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00598.warc.gz | 138,245,270 | 99,650 | # Booking date and shipment date formula
Options
Member
Hello I am trying to arrive at a beast mode calculation where the booking in current quarter if shipped in the current quarter then give me the shipment dollars. It has to be dynamic so that I don't have to change the formula every quarter.
• Coach
Options
I would use the QUARTER() function and the YEAR() function to ensure that you get the expected result. It would look something like this:
```CASE WHEN QUARTER(`dt`) = QUARTER(CURRENT_DATE()) AND YEAR(`dt`) = YEAR(CURRENT_DATE()) THEN 'Same'
ELSE 'Different'
END
```
**Check out my Domo Tips & Tricks Videos
**Make sure to any users posts that helped you.
• Contributor
Options
Do you want the formula to return a date or the shipment dollars? What are the columns in your dataset?
• Coach
Options
To get the dollar amounts you can use Mark's logic to return the amount for the quarter:
```CASE WHEN QUARTER(`dt`) = QUARTER(CURRENT_DATE()) AND YEAR(`dt`) = YEAR(CURRENT_DATE()) THEN
`Shipment Dollars`
END
```
**Did this solve your problem? Accept it as a solution!**
• Member
Options
I have two columns. One is the Booking date and the second is the shipment date. IF the Booking date and the shipment date are within the same quarter then give me the Shipment dollar for the Order number.
• Coach
Options
```CASE WHEN QUARTER(`booking_date`) = QUARTER(`shipment_date`) AND YEAR(`booking_date`) = YEAR(`shipment_date`) THEN
`Shipment Dollars`
END
```
You'd want something like this then
**Did this solve your problem? Accept it as a solution!**
• Coach
Options
If you only want to show the orders that meet your criteria, you can create this beast mode and then drag into your filters and filter to include.
```CASE WHEN QUARTER(`booking_date`) = QUARTER(`shipment_date`) AND YEAR(`booking_date`) = YEAR(`shipment_date`) THEN
'Include'
ELSE 'Exclude'
END
```
**Check out my Domo Tips & Tricks Videos
**Make sure to any users posts that helped you.
• Member
Options
In this case lets say the booking date is 1 October and the shipment date is 15th November then will the output be the Shipment dollars? Should I load a date mapping file to get the output?
• Coach
Options
If your dataset already has the booking date, shipment date, shipping dollars and order number, you shouldn't need any additional fields. Have you tried using any of the above suggestions in Analyzer? Start with a table card and see if it returns the results you expect.
**Check out my Domo Tips & Tricks Videos
**Make sure to any users posts that helped you. | 606 | 2,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.890748 |
http://www.mathworks.com/help/fininst/agencyprice.html?requestedDomain=www.mathworks.com&nocookie=true | 1,519,376,187,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814538.66/warc/CC-MAIN-20180223075134-20180223095134-00675.warc.gz | 480,396,155 | 16,997 | # Documentation
### This is machine translation
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# agencyprice
Price callable bond using Agency OAS model
## Syntax
```Price = agencyprice(ZeroData,OAS,CouponRate,Settle,Maturity,Vol,CallDate) Price = agencyprice(ZeroData,OAS,CouponRate,Settle,Maturity,Vol,CallDate,Name,Value) ```
## Description
`Price = agencyprice(ZeroData,OAS,CouponRate,Settle,Maturity,Vol,CallDate)` computes the price for a callable bond, given OAS, using the Agency OAS model.
`Price = agencyprice(ZeroData,OAS,CouponRate,Settle,Maturity,Vol,CallDate,Name,Value)` computes the price for a callable bond, given OAS, using the Agency OAS model with additional options specified by one or more `Name,Value` pair arguments.
## Input Arguments
`ZeroData` Zero curve represented as a `numRates`-by-`2` matrix where the first column is zero dates and the second column is the accompanying zero rates. `OAS` `numBonds`-by-`1` vector of option-adjusted spreads, expressed as a decimal (that is, 50 basis points is entered as `.005`). `CouponRate` `numBonds`-by-`1` vector of coupon rates in decimal form.
`Settle`
Scalar MATLAB® date number for the settlement date for all the bonds and the zero data.
### Note
The `Settle` date must be an identical settlement date for all bonds and the zero curve.
`Maturity`
`numBonds`-by-`1` vector of maturity dates.
`Vol`
`numBonds`-by-`1` vector of volatilities in decimal form. This is the volatility of interest rates corresponding to the time of the `CallDate`.
`CallDate`
`numBonds`-by-`1` vector of call dates.
### Name-Value Pair Arguments
Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside single quotes (`' '`). You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.
`'Basis'` `N`-by-`1` vector of day-count basis: 0 = actual/actual 1 = 30/360 (SIA)2 = actual/3603 = actual/3654 = 30/360 (BMA)5 = 30/360 (ISDA)6 = 30/360 (European)7 = actual/365 (Japanese)8 = actual/actual (ICMA)9 = actual/360 (ICMA)10 = actual/365 (ICMA)11 = 30/360E (ICMA) 12 = actual/365 (ISDA)13 = BUS/252 For more information, see basis. Default: `0` (actual/actual) `'CurveBasis'` Basis of the zero curve, where the choices are identical to `Basis`. Default: `0` (actual/actual) `'CurveCompounding'` Compounding frequency of the curve. Possible values include: `–1`, `0`, `1`, `2`, `3`, `4`, `6`, `12`. Default: `2` (Semi-annual) `'EndMonthRule'` End-of-month rule; `1`, indicating in effect, and `0`, indicating rule not in effect for the bond(s). When `1`, the rule is in effect for the bond(s). This means that a security that pays coupon interest on the last day of the month will always make payment on the last day of the month. Default: `1` — Indicates in effect `'Face'` Face value of the bond. Default: `100` `'FirstCouponDate'` Date when a bond makes its first coupon payment; used when bond has an irregular first coupon period. When `FirstCouponDate` and `LastCouponDate` are both specified, `FirstCouponDate` takes precedence in determining the coupon payment structure. Default: If you do not specify a `FirstCouponDate`, the cash flow payment dates are determined from other inputs. `'InterpMethod'` Interpolation method used to obtain points from the zero curve. Values are: `linear` — linear interpolation`cubic` — piecewise cubic spline interpolation`pchip` — piecewise cubic Hermite interpolation Default: `linear` `'IssueDate'` Bond issue date.Default: If you do not specify an `IssueDate`, the cash flow payment dates are determined from other inputs. `'LastCouponDate'` Last coupon date of a bond before the maturity date; used when bond has an irregular last coupon period. In the absence of a specified `FirstCouponDate`, a specified `LastCouponDate` determines the coupon structure of the bond. The coupon structure of a bond is truncated at the `LastCouponDate`, regardless of where it falls, and is followed only by the bond's maturity cash flow date. Default: If you do not specify a `LastCouponDate`, the cash flow payment dates are determined from other inputs. `'Period'` Number of coupon payments per year. Possible values include: `0`, `1`, `2`, `3`, `4`, `6`, `12`. Default: `2` `'StartDate'` Forward starting date of payments. Default: If you do not specify a `StartDate`, the effective start date is the `Settle` date.
## Output Arguments
`Price` `numBonds`-by-`1` matrix of the price.
## Examples
collapse all
This example shows how to compute the agency `Price`.
```Settle = datenum('20-Jan-2010'); ZeroRates = [.07 .164 .253 1.002 1.732 2.226 2.605 3.316 ... 3.474 4.188 4.902]'/100; ZeroDates = daysadd(Settle,360*[.25 .5 1 2 3 4 5 7 10 20 30],1); ZeroData = [ZeroDates ZeroRates]; Maturity = datenum('30-Dec-2013'); CouponRate = .022; OAS = 6.53/10000; Vol = .5117; CallDate = datenum('30-Dec-2010'); Price = agencyprice(ZeroData, OAS, CouponRate, Settle, Maturity, Vol, CallDate)```
```Price = 99.4212 ```
collapse all
### Agency OAS Model
The BMA European Callable Securities Formula provides a standard methodology for computing price and option-adjusted spread for European Callable Securities (ECS).
## References
SIFMA, The BMA European Callable Securities Formula, `http://www.sifma.org`. | 1,535 | 5,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-09 | latest | en | 0.629958 |
https://tunxis.commnet.edu/view/powers-of-10-worksheets.html | 1,719,003,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00080.warc.gz | 516,436,218 | 6,945 | # Powers Of 10 Worksheets
Powers Of 10 Worksheets - Web these grade 5 worksheets review reading and writing powers of ten, with varying formats and levels of difficulty. Work out each of the following multiplications. 1) 3 x 100 =. Our entire decimal numbering system is based on the concept that each decimal place to the left is ten times greater and every place value to the right is ten times smaller than the previous spot. Web powers of ten and scientific notation worksheets. This page contains links and brief descriptions of all the activities available on multiplying and dividing by powers of ten.
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## Powers Of 10 Worksheets.html Printable Worksheets
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6) 0.009 ÷ 102 =. Powers of 10 practice problems. 6) 104 = _____________ 7) 107 = _____________ 8) 103 = _____________ 9) 109 =.
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## Math Is All About Patterns, And There's No Better Place To See That Than With Powers Of Ten.
The worksheets can be made in html or pdf format (both are easy to print). How to multiply a whole number by a power of 10? In this unit, you'll learn the patterns in multiplying and dividing by 10,. 8) 150.3 ÷ 105 = 9) 9041.5 ÷ 106=.
## 2) 6 X 100 =.
Explore amazing worksheets on decimals. How to calculate the powers of 10? This page contains links and brief descriptions of all the activities available on multiplying and dividing by powers of ten. Web the worksheet gives students an opportunity to apply the concepts of powers of 10 to find the answer.
## Home > Grade Levels > Grade 8 > Powers Of Ten Worksheets.
Web powers of ten | live worksheets. Negative powers of ten refers to 0.1, 0.01, and 0.001. Multiplying & dividing by powers of ten (c) section a. These exponents worksheets introduce problems with powers of ten.
## We've Supplied Worksheets In Both Standard Form And Exponent Form.
Web the worksheet involves working with a set of problems on multiplying and dividing decimals by powers of 10, which helps young learners solidify their understanding of the concept. Web powers of ten and scientific notation worksheets. Web our powers worksheets help students conquer this essential area of math. The worksheets start out introducing simple powers of ten terms, including ones that should be. | 1,049 | 4,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.879617 |
http://scicomp.stackexchange.com/questions/5372/for-which-statistical-methods-are-gpus-faster-than-cpus | 1,469,660,421,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827080.38/warc/CC-MAIN-20160723071027-00168-ip-10-185-27-174.ec2.internal.warc.gz | 225,565,504 | 22,469 | # For which statistical methods are GPUs faster than CPUs?
I have just installed a Nvidia GT660 graphic card on my desktop and, after some struggle, I manage to interface it with R.
I have been playing with several R packages that use GPUs, especially gputools, and I was comparing the time taken by my GPU and CPU to perform some basic operations:
• inverting matrices (CPU faster)
• qr decomposition (CPU faster)
• big correlation matrices (CPU faster)
• matrix multiplication (GPU much faster!)
Notice that I have experimented mainly with gputools so maybe other packages perform better.
In broad terms my question is: what some routine statistical operations that might be worth executing on a GPU rather than a CPU?
-
## migrated from stats.stackexchange.comFeb 25 '13 at 17:22
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Anything involving lots of matrix multiplication? :) GPUs are quite popular in the neural nets community. – larsmans Feb 24 '13 at 23:26
you need to provide the size of the matrices involved. For example, last i checked (admittedly 2 years ago) inversion and decomposition were only faster on GPU starting from large matrices (2^9 times 2^9 and upwards) – user189035 Feb 24 '13 at 23:29
I used matrices of around $10^3 \times 10^3$ for inversion, qr and matrix multiplication, while for correlations I have used around 10^4 observations of vectors of size 100. For matrix inversion the GPU was much slower, while for qr decomposition it was slower but comparable to the CPU. – Jugurtha Feb 24 '13 at 23:38
this is a very good question but i think you'll get better answers by having it migrated to stackoverflow (through i think similar questions have been asked there before) – user189035 Feb 25 '13 at 0:06
GPU's advantage of regular CPU's is the fact that they can be "massively" parallel, not that they are faster per core. As such, for jobs that require a lot of "housekeeping" like Cholesky factorization etc. you need to use block algorithms and so forth to get significant speed-up; this is not trivial and I assume it will take a while before GPU's take over such operations. What is definitely going the GPU way is MCMC-ing (and Random Number generation). Sampling from a posterior has "parallelization" written all over it... And sparse matrices computations; they are already "blocked" anyway... – usεr11852 Feb 25 '13 at 7:13
## 4 Answers
For all of the applications you mentioned, GPUs should be more capable (from a hardware perspective) than CPUs for sufficiently large matrices. I don't know anything about R's implementation, but I've used cuBLAS and Magma with great success for inversions around $n = 2^{10}$ and multiplication/correlation for rectangular matrices with $n,m \approx 2^{10}, k \approx 2^{14}$. It is an especially big surprise to me that large correlation matrices would be faster on the CPU using R.
More broadly, I suspect most statistical operations that spend most of their time in dense linear algebra (BLAS, Lapack functionality) can be efficiently implemented on the GPU.
-
In broad terms, algorithms that run faster on the GPU are ones where you are doing the same type of instruction on many different data points.
An easy example to illustrate this is with matrix multiplication.
Suppose we are doing the matrix computation
$A \times B = C$
A simple CPU algorithm might look something like
//starting with C = 0
for (int i = 0; i < C_Width; i++)
{
for (int j = 0; j < C_Height; j++)
{
for (int k = 0; k < A_Width; k++)
{
for (int l = 0; l < B_Height; l++)
{
C[j, i] += A[j, k] * B[l, i];
}
}
}
}
The key thing to see here is that there are a lot of nested for loops and each step must be executed one after the other.
See a diagram of this
Notice that the calculation of each element of C does not depend on any of the other elements. So it does not matter what order the calculations are done in.
So on the GPU, these operations can be done concurrently.
A GPU kernel for calulating a matrix multiplication would look something like
__kernel void Multiply
(
__global float * A,
__global float * B,
__global float * C
)
{
const int x = get_global_id(0);
const int y = get_global_id(1);
for (int k = 0; k < A_Width; k++)
{
for (int l = 0; l < B_Height; l++)
{
C[x, y] += A[x, k] * B[l, y];
}
}
}
This kernel only has the two inner for loops. A program sending this job to the GPU will tell the GPU to execute this kernel for each data point in C. The GPU will do each of these instructions concurrently on many threads. Just like the old saying "Cheaper by the dozen" GPUs are designed to be faster doing the same thing lots of times.
There are however some algorithms which will slow the GPU down. Some are not well suited for the GPU.
If for example, there were data dependencies, ie: imagine the computation of each element of C depended on the previous elements. The programmer would have to put a barrier in the kernel to wait for each previous computation to finish. This would be a major slow down.
Also, algorithms which have a lot of branching logic ie:
__kernel Foo()
{
if (somecondition)
{
do something
}
else
{
do something completely different
}
}
tend to run slower on the GPU because the GPU is no longer doing the same thing in each thread.
This is a simplified explanation because there are many other factors to consider. For example, sending data between the CPU and GPU is also time consuming. Sometimes it is worth doing a computation on the GPU even when its faster on the CPU, just to avoid the extra send time (And vice versa).
Also many modern CPUs support concurrency now as well with hyperthreaded multicore processors.
GPU's also seem to be not so good for recursion, see here which probably explains some of the problems with the QR algorithm. I believe that one has some recursive data dependencies.
-
It's officially SX-naughty to comment on an answer just to say that it's a terrific answer, but I don't give a rat's perinæum about negs: this is a delightful and informative answer. One of SX's great injustices is the lack of kudos to people who give exquisitely-informative answers on 'old' (in internet time) questions. (Plus, I'm giving a thumbs-up to an 'old' (in internet time) answer: I know, right? META). – GT. May 16 '15 at 7:26
GPUs are sensitive beasts. Although Nvidia's beefiest card can theoretically execute any of the operations you listed 100x faster than the fastest CPU, about a million things can get in the way of that speedup. Every part of the relevant algorithm, and of the program which runs it, has to be extensively tweaked and optimized in order to get anywhere near that theoretical maximum speedup. R is generally not known to be a particularly fast language, and so it doesn't surprise me that its default GPU implementation is not that great, at least in terms of raw performance. However, the R GPU functions may have optimization settings that you can tweak in order to regain some of that missing performance.
If you're looking into GPUs because you've found that some calculation that you need to run is going to take weeks/months to finish, it may be worth your while to migrate from R to a more performance-friendly language. Python isn't too much harder to work with than R. The NumPy and SciPy packages have most of the same stat functions as R, and PyCuda can be used to implement your own GPU based functions in a fairly straightforward way.
If you really want to increase the speed at which your functions run on GPUs, I would consider implementing your own functions in a combination of C++ and CUDA. The CUBLAS library can be used to handle all of the linear algebra-related heavy lifting. However, keep in mind that it can take quite a while to write such code (especially if it's your first time doing so), and so this approach should be reserved only for those computations that take an extremely long time to run (months) and/or that you're going to be repeating hundreds of times.
-
Multiple Imputation methods for Missing Data? Like those in Alice-II (R).
I think those tend to be often embarrassingly parallel and hence suitable to a GPU architecture. Never tried it myself though.
- | 1,897 | 8,282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-30 | latest | en | 0.953615 |
http://www.ajdesigner.com/phppotentialenergy/potential_energy_equation_z.php | 1,369,131,195,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699881956/warc/CC-MAIN-20130516102441-00018-ip-10-60-113-184.ec2.internal.warc.gz | 311,945,185 | 4,685 | Potential Energy Equations Formulas Calculator
Science Physics
Solving for height.
Inputs:
potential energy (E) joule british thermal unit calorie electronvolt erg foot-pound newton-meter therm mass (m) kilogram grain gram milligram pound slug ton long ton metric ton short acceleration of gravity (g) meter/second^2 centimeter/second^2 foot/second^2 inch/second^2 kilometer/second^2 mile/second^2 millimeter/second^2 yard/second^2
|
Conversions:
potential energy (E) = 0 = 0 joule mass (m) = 0 = 0 kilogram acceleration of gravity (g) = 0 = 0 meter/second^2
Solution:
height (z) = HAS NOT BEEN CALCULATED
Other Units:
Change Equation
Select an equation to solve for a different unknown
potential energy mass acceleration of gravity height
References - Books:
Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.
by Jimmy Raymond | 226 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-20 | latest | en | 0.721522 |
https://www.lmfdb.org/Character/Dirichlet/1287/470 | 1,642,665,304,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00294.warc.gz | 908,085,406 | 7,121 | # Properties
Label 1287.470 Modulus $1287$ Conductor $1287$ Order $60$ Real no Primitive yes Minimal yes Parity odd
# Learn more
Show commands: Pari/GP / SageMath
sage: from sage.modular.dirichlet import DirichletCharacter
sage: H = DirichletGroup(1287, base_ring=CyclotomicField(60))
sage: M = H._module
sage: chi = DirichletCharacter(H, M([10,18,5]))
pari: [g,chi] = znchar(Mod(470,1287))
## Basic properties
Modulus: $$1287$$ Conductor: $$1287$$ sage: chi.conductor() pari: znconreyconductor(g,chi) Order: $$60$$ sage: chi.multiplicative_order() pari: charorder(g,chi) Real: no Primitive: yes sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 Minimal: yes Parity: odd sage: chi.is_odd() pari: zncharisodd(g,chi)
## Galois orbit 1287.em
sage: chi.galois_orbit()
pari: order = charorder(g,chi)
pari: [ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Related number fields
Field of values: $$\Q(\zeta_{60})$$ Fixed field: Number field defined by a degree 60 polynomial
## Values on generators
$$(1145,937,496)$$ → $$(e\left(\frac{1}{6}\right),e\left(\frac{3}{10}\right),e\left(\frac{1}{12}\right))$$
## Values
$$a$$ $$-1$$ $$1$$ $$2$$ $$4$$ $$5$$ $$7$$ $$8$$ $$10$$ $$14$$ $$16$$ $$17$$ $$19$$ $$\chi_{ 1287 }(470, a)$$ $$-1$$ $$1$$ $$e\left(\frac{11}{20}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{47}{60}\right)$$ $$e\left(\frac{41}{60}\right)$$ $$e\left(\frac{13}{20}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{7}{30}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{11}{30}\right)$$ $$e\left(\frac{19}{60}\right)$$
sage: chi.jacobi_sum(n)
$$\chi_{ 1287 }(470,a) \;$$ at $$\;a =$$ e.g. 2 | 641 | 1,675 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-05 | latest | en | 0.292251 |
https://www.coursehero.com/file/5974045/Quiz-03B/ | 1,495,491,933,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00016.warc.gz | 877,221,752 | 24,286 | Quiz_03B
# Quiz_03B - 1 In the past five years only 5 of pre-school...
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Unformatted text preview: 1. In the past five years, only 5% of pre-school children did not improve their swimming skills after taking a Beginner Swimmer Class at a certain Recreation Center. Decide if the probability described is a subjective (personal) probability or a relative frequency probability: The probability of 5% that a pre-school child who is taking a Beginner Swimmer Class swim class will not improve his/her swimming skills is a A) subjective probability. B) relative frequency probability. 2. A statistics class has 4 teaching assistants (TAs): three female assistants (Lauren, Rona, and Leila) and one male assistant (Josh). Each TA teaches one discussion section. A student picks a discussion section. The two events W = {the TA is a woman} and M = {the TA is a man} are A) independent events. B) disjoint (mutually exclusive) events. C) each simple events. D) none of the above 3. Students who live in the dorms at a college get free T.V. service in their rooms, but only receive 6 stations. On a certain evening, a student wants to watch T.V. and the six stations are broadcasting separate shows on baseball, football, basketball, local news, national news, and international news. The student is too tired to check which channels the shows are playing on, so the student picks a channel at random. The two events A = {the student watches an athletic event} and N = {the student watches a news broadcast} are A) independent events....
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## This note was uploaded on 10/08/2010 for the course STAT 200 taught by Professor Barroso,joaor during the Fall '08 term at Penn State.
### Page1 / 4
Quiz_03B - 1 In the past five years only 5 of pre-school...
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Ask a homework question - tutors are online | 502 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-22 | longest | en | 0.947354 |
https://fr.mathworks.com/matlabcentral/cody/problems/2019-dimensions-of-a-rectangle/solutions/591165 | 1,590,518,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00417.warc.gz | 358,329,024 | 15,737 | Cody
# Problem 2019. Dimensions of a rectangle
Solution 591165
Submitted on 4 Mar 2015 by LY Cao
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### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1 length_correct = 3/sqrt(10); width_correct = 1/sqrt(10); [width, length] = rectangle(x); tolerance = 1e-12; assert( abs(length-length_correct)<tolerance && abs(width-width_correct)<tolerance )
x = 1 width = 0.3162 ans = 0.9487
2 Pass
%% x = 2; width_correct = 2/sqrt(10); length_correct = 6/sqrt(10); [width, length] = rectangle(x); tolerance = 1e-12; assert(abs(length-length_correct)<tolerance && abs(width-width_correct)<tolerance)
width = 0.6325 ans = 1.8974 | 245 | 807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-24 | latest | en | 0.703757 |
https://www.tutorialspoint.com/find-maximum-number-possible-by-doing-at-most-k-swaps-in-cplusplus | 1,669,708,868,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00267.warc.gz | 1,085,614,537 | 10,236 | # Find Maximum number possible by doing at-most K swaps in C++
C++Server Side ProgrammingProgramming
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In this problem, we are given two integer values n and k. Our task is to find the Maximum number possible by doing at-most K swaps.
Problem description: Here, we need to calculate the number which is maximum and created after swapping at-most k digits of the number.
Let’s take an example to understand the problem,
Input: n = 538 k = 1
Output: 835
Explanation:
We will swap 8 and 5.
## Solution Approach
To solve the problem, we need to swap digits of the number k times and check if the number from is maximum.
We need to find the maximum digit of the number and then swap the element at first index. And so on for the first k index of the numbers.
## Example
Live Demo
#include <bits/stdc++.h>
using namespace std;
void calcMaxNumAfterSwap(string number, int k, string& maxString, int n){
if (k == 0)
return;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (number[i] < number[j]) {
swap(number[i], number[j]);
if (number.compare(maxString) > 0)
maxString = number;
calcMaxNumAfterSwap(number, k - 1, maxString, n);
swap(number[i], number[j]);
}
}
}
}
int main(){
string str = "15263";
int k = 3;
int size = str.length();
string maxString = str;
calcMaxNumAfterSwap(str, k, maxString, size);
cout<<"The maximum number created after "<<k<<" swaps is "<<maxString;
return 0;
}
## Output
The maximum number created after 3 swaps is 65321
Updated on 27-Jan-2021 05:08:36 | 482 | 1,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-49 | latest | en | 0.601267 |
https://roelhollander.eu/en/432-tuning/concert-pitch-vs-tuning-system/ | 1,686,261,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00688.warc.gz | 547,169,510 | 24,556 | # Blog » Concert Pitch vs Tuning System
Approx. reading time: 6 minutes September 27, 2013
#### CONCERT PITCH vs TUNING SYSTEM
(A COMMON MISUNDERSTANDING)
A common misunderstanding when it comes to 432-Tuning is that many people do not understand the difference between “concert pitch” and “tuning system”. If you do not know the difference, then please, read this first!
In short:
Concert Pitch 432Hz = using reference tone A4=432Hz to tune to, without Temperamental specifications.
432-Tuning = using the reference tone A4=432Hz in combination with Pythagorean Temperament (or variation).
WHY IS ONLY CHANGING THE CONCERT PITCH NOT ENOUGH?
Well, when a piece was recorded using for example the 12-Tone Equal Temperament (12-TET, the present standard), pitching the whole piece down with -0.3166… cents is not sufficient. The “intervallic” relationship between the tones has not changed, when a change of Concert Pitch is implemented. Even though the A would then be in tune with 432-Tuning, the other 11 tones are still off pitch. The Temperament is still Equal Temperament.
EXAMPLE: 12-TONE EQUAL TEMPERAMENT (the present standard)
In twelve-tone equal temperament, which divides the octave into 12 equal parts, the width of a semitone, i.e. the frequency ratio of the interval between two adjacent notes, is the twelfth root of two:
There are many other Musical Temperaments (most more complicated then 12-TET), using various mathematical formulas to create other ratios then 1.059463. The most know Temperaments are the: Pythagorean TuningMeantone TemperamentWell TemperamentJust Intonation and Five-Limit Tuning.
If your intentions are to use the 432-TUNING SYSTEM (not just 432Hz as Concert Pitch) you also need to implement The Pythagorean Temperament.
##### 432 TUNING – USING THE PYTHAGOREAN TEMPERAMENT
PYTHAGOREAN TEMPERAMENT
The Pythagorean temperament is been noted as the only correct temperament when we talk about 432-tuning. The Most Harmonic Numbers concept uses this method.
Pythagorean tuning is a system of musical tuning in which the frequency ratios of all intervals are based on the ratio 3:2, a ratio from the harmonic series. This ratio, also known as the “pure” or perfect fifth, is chosen because it is one of the most consonant intervals and easy to tune on by ear.
The Pythagorean Temperament is thus one of the oldest known “Just Intonation” (3-limit) Temperaments.
IMPORTANTWhen using the Scientific Pitch (C4=256Hz) and you wish A4 to be 432Hz as well, then there are NO alternatives for the Pythagorean Temperament (and variations on it)! Any other temperament used with C4=256Hz, will not result in A4=432Hz but by example with 12-TET in A4=430.5Hz (430.53824925657256Hz).
#### THE BEST ALTERNATIVE IMPLEMENTATION OF THE PYTHAGOREAN TEMPERAMENT:
TWELVE TRUE FIFTHS TUNING” (BY MARIA RENOLD)
Graham H Jackson explains this tuning system on his web site as follows:
“For the “twelve true-5ths tuning”: you first set C at 256 Hz. Then you tune the 7 “white keys” by the circle of 5ths, using however natural 5ths. Then you divide the octave at C exactly in half (which can be done handily with a special tuning fork), and tune the 5 “black keys” by natural 5ths to that F#. You end up with two series of natural 5ths–one of 7 notes and one of 5 notes, linked by an “unnatural” interval of an augmented 4th (which is actually the same augmented 4th found in the equal-tempered system).” [ Read More ] ►
OTHER ALTERNATIVES FOR THE PYTHAGOREAN TEMPERAMENT?
Within the “432-Tuning community” there are those who favor other temperaments due to certain “limitations” of the Pythagorean Temperament, or simply because they prefer the usage of other perfect intervals (like a perfect Third) over the Perfect Fifth. If you could still call these methods of tuning “432-Tuning” is questionable (see blog “Concert Pitch” vs “Tuning System“).
Most popular alternative temperaments (not based on Perfect Fifths) used in combination with A4=432Hz are:
JUST INTONATION
In music, just intonation (sometimes abbreviated as JI) or pure intonation is any musical tuning in which the frequencies of notes are related by ratios of small whole numbers. Any interval tuned in this way is called a pure or just interval. The two notes in any just interval are members of the same harmonic series. Frequency ratios involving large integers such as 1024:927 are not generally said to be justly tuned. Just intonation is the tuning system of the later ancient Greek modes as codified by Ptolemy. More information: Wikipedia | XenHarmonic
FIVE-LIMITED TUNING (diatonic scale)
Five-limit tuning, or 5-limit tuning is a method to obtain a justly tuned musical scale. The frequencies of the notes of such a scale are obtained by multiplying the frequency of a given reference note (the base note) by powers of 2, 3, or 5, or a combination of them. More information: Wikipedia | XenHarmonic
FIBONACCI + CONCERT PITCH A4=432Hz
The clearest demonstration of Fibonacci being represented in music is seen in scales. 13: the Octave is made of 12 chromatic tones plus 1 the octave. A (basic) scale is composed of 8 notes. The 5th and 3rd notes create basic foundation of chords, based off a whole tone that is 2 steps above root tone, which is the 1st note of scale. Ratios found in the first 7 numbers of the Fibonacci series (0, 1, 1, 2, 3, 5, 8) are related to key frequencies of musical notes.
#### POST-PRODUCTION RE-PITCHING
CHANGING THE CONCERT PITCH IN POST PRODUCTION
If you would only like to change the concert pitch, then there are two ways to do so.
1) PREFERED OPTION: CHANGING THE SPEED
When you change the speed of a piece of music, both Tempo and Pitch will change. This method is the equivalent of slowing down a turntable. In order to get a -31.667 cents pitch change, you should reduce the speed with -1.818%.
2) SECONDARY OPTION: CHANGING THE PITCH
When you change the Concert Pitch from 440Hz to 432Hz, you need to pitch the audio-material down with -31.76665363342977 cents or -0.3176665363342977 semitones. If the software or hardware you use only allows you to use round numbers, you can set it to -31 or -32 cents. For electronic music -31 cents might work better (since that music is generally required to be more “punchy”), for acoustic music -32 cents seems to work out best in most cases.
IMPORTANT NOTE: When only changing the pitch and not the tempo, there will be resolution loss, and the high-end sounds might sound “fake”. That’s why option 1 (changing the speed) is the preferred option!
#### Q: CAN YOU RE-TUNE AND (RE-)TEMPER A COMPLETE SONG AT ONCE?
A: A complete song (1 file, for example .wav or .mp3 file) with various instruments tuned to 440Hz + 12-TET, can NOT be re-tuned to 432-Tuning (432Hz + Pythagorean or related Temperament).
You can still pitch it -31.667…cents down, so the Concert Pitch changes from 440Hz to 432Hz, but the piece will not be “in tune” with the 432-TUNING SYSTEM, simply because the temperament has not been changed from 12-TET to Pythagorean (as explained at the introduction to 432-Tuning).
IF you have recorded the instruments on separate tracks though, a lot can be done!
Some instruments that can not change the Concert Pitch and/or Temperament – or might be too hard to intonate when used with another concert pitch and/or temperament then they were designed for – can be re-pitched and (re-)tempered in post production (after recording). This is sometimes the “only option”, specially when working with instruments with tone-holes / valves / pistons (particularly conical wood winds / brass) or those that are made out of one piece (Didgeridoo, Penny Whistle, et cetera), as well as melodic percussive instruments (like Vibraphone, Marimba, et cetera). Unfortunately there is no solution for these (acoustic) instruments when performing live.
Some online sources proclaim re-pitching music from Concert Pitch A4=440Hz to 432Hz will positively effect the “overtones”. This is not true! [ Read More ] ►
##### RECORDING IN THE RIGHT PITCH & TEMPERAMENT IS BEST!
Every form of “manipulation” will cause loss of quality to a recorded sound. The better the algorithm used with the software, the smaller the loss. Nonetheless, it is always preferable to tune the instruments you like to record to the right pitch and temperament, if possible.
Some online sources proclaim 432-recorded music has more “Overtones” then 440-recorded music.
This is not true! [ Read More ] ► | 2,214 | 8,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.724994 |
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