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https://www.tradersinsight.news/ibkr-quant-news/neural-network-in-python-part-viii/
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# Neural Network In Python – Part VIII
Contributor:
QuantInsti
Visit: QuantInsti
In the previous installment, the author discussed Importing the dataset.
Preparing the dataset
dataset[‘H-L’] = dataset[‘High’] – dataset[‘Low’]
dataset[‘O-C’] = dataset[‘Close’] – dataset[‘Open’]
dataset[‘3day MA’] = dataset[‘Close’].shift(1).rolling(window = 3).mean()
dataset[’10day MA’] = dataset[‘Close’].shift(1).rolling(window = 10).mean()
dataset[’30day MA’] = dataset[‘Close’].shift(1).rolling(window = 30).mean()
dataset[‘Std_dev’]= dataset[‘Close’].rolling(5).std()
dataset[‘RSI’] = talib.RSI(dataset[‘Close’].values, timeperiod = 9)
dataset[‘Williams %R’] = talib.WILLR(dataset[‘High’].values, dataset[‘Low’].values, dataset[‘Close’].values, 7)
We then prepare the various input features which will be used by the artificial neural network learning for making the predictions. We define the following input features:
• High minus Low price
• Close minus Open price
• Three day moving average
• Ten day moving average
• 30 day moving average
• Standard deviation for a period of 5 days
• Relative Strength Index
• Williams %R
dataset[‘Price_Rise’] = np.where(dataset[‘Close’].shift(-1) > dataset[‘Close’], 1, 0)
We then define the output value as price rise, which is a binary variable storing 1 when the closing price of tomorrow is greater than the closing price of today.
dataset = dataset.dropna()
Next, we drop all the rows storing NaN values by using the dropna() function.
X = dataset.iloc[:, 4:-1]
y = dataset.iloc[:, -1]
We then create two data frames storing the input and the output variables. The dataframe ‘X’ stores the input features, the columns starting from the fifth column (or index 4) of the dataset till the second last column. The last column will be stored in the dataframe y, which is the value we want to predict, i.e. the price rise.
Splitting the dataset
split = int(len(dataset)*0.8)
X_train, X_test, y_train, y_test = X[:split], X[split:], y[:split], y[split:]
In this part of the code, we will split our input and output variables to create the test and train datasets. This is done by creating a variable called split, which is defined to be the integer value of 0.8 times the length of the dataset.
We then slice the X and y variables into four separate data frames: Xtrain, Xtest, ytrain and ytest. This is an essential part of any machine learning algorithm, the training data is used by the model to arrive at the weights of the model. The test dataset is used to see how the model will perform on new data which would be fed into the model. The test dataset also has the actual value for the output, which helps us in understanding how efficient the model is. We will look at the confusion matrix later in the code, which essentially is a measure of how accurate the predictions made by the model are.
In the next installment, the author will demonstrate how Feature Scaling
Visit https://www.quantinsti.com/ for ready-to-use functions as applied in trading and data analysis.
##### Disclosure: Interactive Brokers
Information posted on IBKR Traders’ Insight that is provided by third-parties and not by Interactive Brokers does NOT constitute a recommendation by Interactive Brokers that you should contract for the services of that third party. Third-party participants who contribute to IBKR Traders’ Insight are independent of Interactive Brokers and Interactive Brokers does not make any representations or warranties concerning the services offered, their past or future performance, or the accuracy of the information provided by the third party. Past performance is no guarantee of future results.
This material is from QuantInsti and is being posted with permission from QuantInsti. The views expressed in this material are solely those of the author and/or QuantInsti and IBKR is not endorsing or recommending any investment or trading discussed in the material. This material is not and should not be construed as an offer to sell or the solicitation of an offer to buy any security. To the extent that this material discusses general market activity, industry or sector trends or other broad based economic or political conditions, it should not be construed as research or investment advice. To the extent that it includes references to specific securities, commodities, currencies, or other instruments, those references do not constitute a recommendation to buy, sell or hold such security. This material does not and is not intended to take into account the particular financial conditions, investment objectives or requirements of individual customers. Before acting on this material, you should consider whether it is suitable for your particular circumstances and, as necessary, seek professional advice.
In accordance with EU regulation: The statements in this document shall not be considered as an objective or independent explanation of the matters. Please note that this document (a) has not been prepared in accordance with legal requirements designed to promote the independence of investment research, and (b) is not subject to any prohibition on dealing ahead of the dissemination or publication of investment research.
Any trading symbols displayed are for illustrative purposes only and are not intended to portray recommendations.
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# square root of 036
## Square Root of 0.036 (√.036)
The reason the square root of 0.036 is greater than 0.036 is because when you take the square root of the dividend (√9), the decrease of the dividend is smaller than the decrease of the divisor when you take the square root of the divisor (√250). This would not be the case if the whole number in front of the decimal point wasn''t 0.
## What is the square root of 0.36
In an A.P, if the common difference is 4 and sum up to n term is 88 and 8th term is 16, then the value of ''n'' will be a square root of which number Medium. View solution > The area (in sq. units) of the part of the circle x 2 + y 2 = 3 6, which is outside the parabola y 2 = 9 x is: Medium. View solution >
## Square Root of 36 (√36)
Square root of 36 definition The square root of 36 in mathematical form is written with the radical sign like this √36. We call this the square root of 36 in radical form. The square root of 36 is a quantity (q) that when multiplied by itself will equal 36. √ 36 = q × q = q 2
## Square Root of 36 How to Find the Square Root of 36
The square root of 36 is 6. It is the positive solution of the equation x 2 = 36. The number 36 is a perfect square. Square Root of 36: 6
## What is the square root of 0.36, Socratic
Since √36 = ± 6, we know that 6 is the square factor of 36, and we know √36 will be greater in magnitude than √0.36. Therefore, we can try 0.6: (0.6) (0.6) = 0.36. This means √0.36 = ± 0.6
## Square Root of 36 EasyCalculation
Perfect √49. The square root of 36 is an integer 6, which is hence termed as a perfect square. The nearest previous perfect square is 25 and the nearest next perfect square is 49 .
## Find the square root of .036 upto threeplace of decimal
Find the square root of .036 upto threeplace of decimal Get the answers you need, now!
## Calculate the Square Root square root of 0.0036, Mathway
Calculate the Square Root square root of 0.0036. √0.0036 0.0036. The result can be shown in multiple forms. Exact Form: √0.0036 0.0036. Decimal Form: 0.06 0.06.
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https://www.reddit.com/r/haskell/comments/1cyajx/the_mother_of_all_monads_school_of_haskell/
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Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
38
Posted by5 years ago
Archived
If you aren't that familiar with continuations, prepare to have your mind blown by this tutorial.
edit: sorry .. forgot the link, thanks DR6
91% Upvoted
Sort by
level 1
6 points · 5 years ago
level 1
6 points · 5 years ago
I believe `Cont` is actually overkill for this, you can get polymorphic `do` notation with just the reader monad.
Basically, you replace `f a` with `Monad f -> f a`, where `Monad f` is the dictionary.
level 2
9 points · 5 years ago · edited 5 years ago
Cont uses the existing monad definition and do notation. What you propose is
`````` newtype AMonad m a = AMonad {runAMonad :: Monad m => m a}
return x = AMonad (return x)
``````
which although it does not require `m` to be a monad, does require that to run it. On the other hand, `ContT` only requires `m` be pointed (have a `return` function) and only uses bind for lifting.
EDIT: perhaps another way to say this is "what do you mean by just the reader monad". Fully expand this type and you get:
`````` newtype AMonad m a = AMonad ((forall b. b -> m b) -> (forall b c. m b -> (b -> m c) -> m c) -> m a)
``````
which is a bunch uglier than
`````` newtype Codensity m a = Codensity (forall r. (a -> m r) -> m r)
``````
but, more than that, you might wonder which is bigger. Well, one can go from `Codensity m a` to `AMonad m a`
`````` codensityToAMonad (Codensity f) = AMonad \$ \r b -> (f r)
``````
but, you can't go the other way without a monad constraint. If `m` is a monad, one can define
`````` aMonadToCodensity (AMonad f) = Codensity \$ (>>=) (f return (>>=))
``````
note that
`````` aMonadToCodensity . codensityToAMonad \$ (Codensity f)
= Codensity \$ (>>=) (f return)
``````
which is not necessarily the same thing as `Codensity f`
`````` codensityToAMonad . aMonadToCodensity \$ AMonad f
= codensityToAMonad \$ Codensity \$ (>>=) (f return (>>=))
= AMonad \$ \r b -> (f return (>>=)) >>= r
``````
which if `r` is the same thing iff `AMonad` can only be called with the actual monad instance. Going from `AMonad` to codensity and back gets you back to something that when giving (the one singled out unique) monad instances is what you started with. Take from that what you will.
There is a ton of confusion over the nature of the "mother of all monad" argument. I strongly sugest people read the original work by Filinski. However, the intuition that
`````` x <=> (x >>=)
``````
is I think really easy to get a grasp on. A monad is something that can be bound. `Codensity` is the type of everything after the first argument in bind. Binding with return is identity (monad law). Bind associativity does not matter (monad law).
level 3
3 points · 5 years ago
I'm not really following... Dan's post deals with the question - if we couldn't overload `do` notation and had to just pick a `Monad` to use in `do`, which one would it be? `AMonad` seems smaller than `ContT` in the sense that it is literally just dictionary passing--it is 'just syntax'. `ContT` seems too big, since you can write `callCC` and probably do other weird stuff. Do I have the wrong idea here? And where does `Codensity` come in? That isn't even mentioned in Dan's post. I know `Codensity` right associates binds, and it's a `Monad` regardless of `f`, but what does it have to do with overloading `do` notation?
Btw, that is a good suggestion to read the original Filinski.
level 4
3 points · 5 years ago · edited 5 years ago
Codensity is just `Cont` where the return type is universal.
`````` Condensity m a = forall r. ContT r m a = forall r. Cont (m r) a
``````
Is `AMonad` really just syntax? What happens if you pass it something else? Like, what happens if you pass it in `const Nothing` as the first argument?
`````` isJustWeird = AMonad (\r b -> case r undefined of
Nothing -> Just True
Just _ -> Just False)
setBit boll (AMonad m) = AMonad (\r b -> if bool then (m r b) else (m (const Nothing) b))
``````
I assert that these two functions let you use `AMonad` as if it were `ReaderT Bool` so long as the underlying monad is the maybe monad (you can generalize to `MaybeT`). Now, doesn't that make you suspicious that just maybe (pun not intended) `AMonad` gives you "extra stuff". (Clearly this is avoided if we define `AMonad` with typeclasses, but that defeats the point).
Now, you might be right that `AMonad` is still smaller. Maybe my observations about converting between them support that view. But I don't think it is obvious by any means. And, here is the crucial thing--they have different requirements
• To get "into" Codensity/Cont requires `>>=` (and nothing else)
• To get "out of" Codensity/Cont requires `return` (and nothing else)
• To get into AMonad requires nothing
• To get out of AMonad requires both `return` and `>>=`.
To me, that `lift :: m a -> ContT r m a` is just `>>=` (wrapped in newtypes) and that `runCont :: ContT r m r -> m r` is just `return` seems signifigant. With `Cont` these are the only times you use bind and return.
`AMonad` might be a reader monad, but it is a very special reader monad (one with a very different shaped monad instance). It doesn't get you out of using the monad you started with. `Cont` does. That is why languages with continuations let you "represent" all the monads in direct style, but languages with just configuration parameters don't.
level 4
3 points · 5 years ago
`Codensity` is just a more principled alternative to `ContT` in this setting.
With `ContT` you can ignore the current continuation. With `Codensity` it is the only way to get the final `r` as you know nothing else about it.
level 1
3 points · 5 years ago
I don't like this whole discussion. `Cont` isn't the "mother of all monads" in any interesting sense. At best it's the mother of all do-notations, but that's about it. To do this at all, you still need monad instances for everything involved, you just don't need support for do notation for all monads. But you still have to define return and bind and that takes all the oomph out of this "mother of all monads" thing.
level 2
4 points · 5 years ago
Cont is the mother of all monad, in that any monad that supports delimited continuations allows you to use any monad in direct style (hence adding continuations to a CBV language lets you use monads). In Haskell this is do notation, but in other languages it is the built in semi-colon.
level 3
2 points · 5 years ago
Like GP, I detect a whiff of selling Cont as a universal monad the way a limit is an ultracone. That Cont is just ultrado leaves me overwhelmingly underwhelmed.
level 4
1 point · 5 years ago
I don't think it is just about `do`--the statement is a more something like
"Proposition: for any monad `m` and any value `x` of type `m a` is uniquely charachterized by what happens when you bind a continuation to it. `forall r. Cont (m r) a` is the type of `(x >>=)`. Further, `Cont j` is a monad for all `j`.
Corollary 1: Any monadic language providing operations equivalent in power to those of the `Cont` monad is able to embed any monad (encodable in that language) in direct style.
Corallary 2: Do notation over the `Cont` monad is able to embed any monad (encodable in Haskell) in direct style. "
Does Cayley's theorem leaved you underwhelmed also?
level 5
1 point · 5 years ago · edited 5 years ago
"Proposition: for any monad m and any value x of type m a is uniquely charachterized by what happens when you bind a continuation to it.
A value of any type is "uniquely characterized" by what happens when you bind a continuation to it, if you really want to push this CT-ish idea.
forall r. Cont (m r) a is the type of (x >>=).
Overlooking the fact that you've switched the type of `x` from `m a` to `a`, I observe, again, that there's nothing inherently monadic about this typing judgment. It's an undistinguished instance of the type swizzling embodied in `let f = flip id in (x `f`) :: forall r. Cont r a`.
Further, Cont j is a monad for all j.
So is `State s` for all types `s`. Your point being?
level 6
1 point · 5 years ago · edited 5 years ago
A value of any type is "uniquely characterized" by what happens when you bind a continuation to it, if you really want to push this CT-ish idea.
Yes. Except that we are talking about a different bind. You have just repeated my claim but in the specific case of the identity monad. Perhaps if I had written "is uniquely charachterized by what happens when you `>>=` a continuation to it" this would have been clearer?
Overlooking the fact that you've switched the type of x from m a to a, I observe, again, that there's nothing inherently monadic about this typing judgment. It's an undistinguished instance of the type swizzling embodied in let f = flip id in (x `f`) :: forall r. Cont r a.
I don't think I switched the types. What are you talking about?
So is State s for all types s. Your point being?
In general I know of no monad morphism `m -> State (m r)`.
level 3
4 points · 5 years ago
Thank you for repeating what I said?
level 4
[deleted]
-3 points · 5 years ago
I guess some people don’t really know what a monad is, yet think they do. (Dunning-Kruger effect being at it again.)
It’s more than do notation, guys.
level 5
3 points · 5 years ago
sigfpe (the person who originally commented on `Cont` being the mother of all monads) is well aware of what a monad is. I just think he was being sensationalist.
level 3
3 points · 5 years ago
I don't know how good the examples in the article are.
The types are left off everything, but when we look at the type of `i`:
``````i :: m a -> Cont (m r) a
``````
So when I go to embed multiple different monads at the same time in this 'mother of all monads', it isn't going to work out. Something like:
``````do x <- i [1, 2] ; y <- i Nothing ; z <- i ask ; return (x + y + z)
``````
is a type error. So this isn't making use of a single `Cont R` that many monads can be embedded into, it's picking a different `R` for each monad.
It's kind of like saying that `Codensity` is the mother of all monads, because you can embed `m` into `Codensity m`. But `Codensity` isn't a monad; `Codensity m` is a monad for each choice of `m`, and it's potentially a different monad for each choice of `m`.
I don't recall the Filinski paper having this flaw, but it's been a long time since I read it.
level 4
1 point · 5 years ago · edited 5 years ago
Embedding multiple monads does not work out for the same reason you can't (easily) use multiple monads at the same time in Haskell.
Filinski's later paper on representing layered monad showed how to do just that.
level 5
2 points · 5 years ago
Embedding multiple monads does not work out for the same reason you can't (easily) use multiple monads at the same time in Haskell.
This isn't a very satisfying answer. I skimmed the Filinski paper, and it also looks like he kind of papers over the problem by using some `Any` type that you can manipulate in not-very-principled ways, like coercing an `m a` into, and then getting it back out to do useful work on the values you've hidden in it. And I would assume that any native continuations are expected to behave the same way.
But this is not what I'd expect out of a principled, 'this one monad subsumes all other monads.'
Filinski's later paper on representing layered monad showed how to do just that.
And that uses a language with an effect system, with an effect per embedded monad. That's cool, but it also doesn't sound like one single monad subsuming all others, which is how this is usually sold.
level 6
1 point · 5 years ago · edited 5 years ago
Filinski does sketchy things to get delimited control (the `Cont` monad) to type check. This does not have to be the case in languages with proper (substructural polymorphic) type systems for continuations (the indexed continuation monad in Haskell is a good start).
The underlying lambda terms should be fine (actually I think some bugs exist in Filinski's encoding, but one can fix them).
Perhaps "the single monad subsuming all others" is the wrong way to sell it (although, it is probably the case for "the mother of all indexed monads"). Instead, why don't you think of it with this analogy:
Cayley: Every Group is a subgroup of a canonical group of permutations.
Yoneda: Every Category is "a subcategory" of a canonical functor category.
level 7
2 points · 5 years ago
I dare you to make this precise category-theoretically. No hand-waving, squinting, or same-diff "what ever that means".
It's easy to invoke Cayley and Yoneda, much harder to get the analogies right. See for instance [1]: "Indeed, the obvious generalisation of Cayley’s theorem to monoids ... is false."
Because of this and other reasons I give below, I don't think you can do it.
level 8
1 point · 5 years ago · edited 5 years ago
I agree. It is hard to get this exactly right. So hard that it has not been published, and were I to figure it out reddit would not be the place I would post it.
On the other hand, the not "entirely complete" but the claim that `>>=` and `(\$ return)` are maps to and from the `Codensity` monad and that `>>=` is a section of `return` is trivial to prove. Also, `>>=` is a monoid morphism. We don't have the way in which this construction is universal worked, but it seems obvious to me that there is some way in which it is.
By the way, that blog post generalizes Cayley in a way that is obviously wrong. Instead the correct claim is that "Every monoid is a sub monoid of the set of functions (with composition) on that monoid"--not the "Symmetric monoid" which has far too much structure! Generalizing Cayley to embed any monad into a group can't be right. The correct generalization is easy to prove and uses exactly the same argument as one used to prove Cayley.
Not that the last bit matters. My claim is not that they are the same thing. It is that they are analogously mathematically interesting. This is a new subject. We don't have the benefit of a century of work on making it sound nice and working out all the kinks.
This is not an argument about category theory. It is not a fully categorical result. It is a concrete result in the subject of the algebra of functional programs. As a researcher I would like to fully categorify it but that has not yet happened. Yet, despite claims in this thread, I think it is both a robust and profound result, one who's implications are not fully known, but which has already lead to asymptotic improvements in real programs.
level 1
1 point · 5 years ago
Hooray, sigfpe is on SoH! I think this is based on an old article on his blog, but hopefully he'll make some more cool stuff!
level 2
Original Poster1 point · 5 years ago
Yes it's based on a 2008 article. It's great to be able to play with the live examples.
level 1
1 point · 5 years ago · edited 5 years ago
Can someone explain why
``````ex2 = do
a <- return 1
b <- cont (\fred -> 10)
return \$ a+b
test2 = runCont ex2 show
main = print test2
``````
Doesn't compile? I don't think I'm really getting how these continuations work. I realize I don't understand why the "escape" text doesn't include anything about the "1" from `a <- return 1`. I guess I thought continuations were basically lambdas, but this clearly isn't the case.
level 2
4 points · 5 years ago
This is because `Cont` is
`````` newtype Cont r a = Cont ((a -> r) -> r)
``````
so the return type of the continuation must match the type returned
But, `Cont` could be defined
`````` newtype ContI r s a = ContI ((a -> s) -> r)
``````
`````` class IMonad m where
return :: a -> m r r a
(>>=) :: m r s a -> (a -> m s t b) -> m r t b
``````
and then your code would typecheck just fine. Actually, almost all monads have more general indexed versions. For example, state should have the form
`````` newtype StateI r s a = StateI (r -> (a,s))
``````
which allows you to vary the type of the state (just as `ContI` allows for varying the type of the continuation).
level 2
2 points · 5 years ago
Maybe because `cont` parameter must have type
``````(Int -> String) -> String
``````
? `fred` is a continuation with composed `show`, that is whole monad experession is equivalent to
``````continuation = \b -> 1 + b
test2 = (\fred -> fred 10) \$ (show . continuation)
``````
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Good morning!
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# Good morning! - PowerPoint PPT Presentation
Good morning!. Recursive Algorithms. Dr. Jeyakesavan Veerasamy jeyv@utdallas.edu. Example: Gift box!. Example: Gift box!. Example: Gift box!. Value of gift box : Equation?. Example: Treasure hunt!. What is recursion?. Popular in math definitions Inductive proof. Example: factorial(n).
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### Recursive Algorithms
Dr. Jeyakesavan Veerasamy
jeyv@utdallas.edu
What is recursion?
• Popular in math definitions
• Inductive proof
Example: factorial(n)
• Non-recursive solution
How does recursion work?
• Stack memory
• How much stack memory is needed?
How to understand recursion?
• Method invocations & returns diagram
• Example: factorial(5)
How to understand recursion?
• Method invocations & returns diagram
• Example: fibonacci(n)
How to understand recursion?
• Tree diagram
• Example: fibonacci(n)
How to analyze recursion?
• Recurrence relation & Time complexity
Example: Knapsack problem
• Item weights: 40, 10, 46, 23, 22, 16, 27, 6
• Instance #1: Target : 50
• Instance #2: Target: 60
• Instance #3: Target: 70
How to make recursion efficient?
• Parameters
• Tail recursion
Conclusions
• Recursion is one of the difficult concepts to understand, perhaps it is not that intuitive.
• As per a few mathematicians & CS folks, it is one of the most beautiful concepts!
• While it is not used much in commercial applications, it certainly puts your logical thinking skills to work!
• It is easy to remove tail recursion, but all others are lot harder to remove.
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# The length of rectangular garden is 5 meters more than its width.If the perimeter is 130 meters,find its dimensions
2
by bej
## Answers
2014-08-02T22:37:36+08:00
width is 30 m and the length is 35 meters
2014-08-02T22:38:12+08:00
Let x be the width of the garden
x+5 - length
P=130m
Solve for x:
2x+2(x+5)=130
2x+2x+10=130
4x+10-10=130-10
4x/4=120/4
x=30 m
Substituting the value of x, Then,
Width - 30 m
Length - 35 m
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for Learning Math and Coding
Piano
Grid: Tics Lines: Width px Hash Lines: Width px Labels: Font px Trace Lines: Robot 1: Width px Robot 2: Width px Robot 3: Width px Robot 4: Width px
Labels: x-axis y-axis Show Grid Grid: 12x12 inches 24x24 inches 36x36 inches 72x72 inches 96x96 inches 192x192 inches Quad: 4 Quadrants 1 Quadrant 1&4 Quadrants Units: US Customary Metric
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Graphing Linear Equations with Robots 2: Slope & y-intercept
Graphing Linear Equations with Robots 2: Slope & y-intercept
Given y = -2/3x + 6. Have the tickbot graph the line described by this equation by plotting a point and using the slope to find additional points. Your graph must go through the y-intercept and contain three additional points not including the y-intercept. The points should lie on both sides of the y-intercept (in Quadrant II and I). Have the last point of the line be (6,2).
Load Blocks Symbol Symbol+Word Word Hardware LArduino Workspace Show Ch Console Save File
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# Bayesian information criterion (BIC) on KDE?
by davipatti Last Updated October 29, 2018 16:19 PM
Consider two datasets, $$X$$ and $$Y$$. Both have 2 dimensions with $$a$$ and $$b$$ samples respectively.
I would like to test whether one kernel density estimate (KDE) on the concatenated data ($$XY$$, shape $$(n+m, 2)$$) is a better model of the data than two separate KDEs on each individual dataset. (I'll denote a KDE trained on dataset $$X$$ as $$kde_X$$).
Using something like the bayesian information criterion (BIC) to compare the one-KDE vs. the two-KDE setup seems attractive. With $$n$$ samples, $$k$$ parameters and likelihood $$\mathcal{L}$$:
$$BIC = ln(n)k - 2ln(\mathcal{L})$$
Then, for the one-KDE setup I would compute BIC with:
• $$n$$ = $$a + b$$
• $$k$$ = 1
• $$\mathcal{L}$$ = $$P(XY | kde_{XY})$$
And for the two-KDE setup I would compute BIC with:
• $$n = a + b$$
• $$k = 2$$
• $$\mathcal{L} = P(X | kde_{X}) \times P(Y | kde_{Y})$$
My question is: Is this a fair way to compare these two model setups? I am more used to seeing BIC being used to compare parametric models where $$k$$ really is a number of parameters, as opposed to here, where I'm using $$k$$ as the number of KDEs.
Furthermore: Is there a better way to test whether to model the two datasets jointly or separately?
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### Create an Account
Home / Questions / An investment offers \$9900 per year for 13 years with the first payment occurring one year...
# An investment offers \$9900 per year for 13 years with the first payment occurring one year from now Assume the required return is 11 percent What is the value of the investment today
An investment offers \$9,900 per year for 13 years, with the first payment occurring one year from now. Assume the required return is 11 percent.
What is the value of the investment today? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
Present value \$
What would the value be if the payments occurred for 38 years? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
Present value \$
What would the value be if the payments occurred for 73 years? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
Present value \$
What would the value be if the payments occurred forever? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
Present value
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Hailey has a summer job at the water slide park. She earns \$9.50 an hour as a lifeguard, but never works more than 25 hours in a week. She determines that her salary is modeled by the function s = 9.5h. What is the domain of this function in this situation? A) s ≤ 237.50 B) all real numbers C) {0 ≤ h ≤ 25} D) {0 ≥ h ≥ 25}
Answer: C) {0 ≤ h ≤ 25}
Step-by-step explanation:
• The domain of a function is the set of all values for independent variable for which the function must be defined.
Given: The amount earned by Hailey for one hour work =\$9.50
She determines that her salary is modeled by the function .
Here h is the independent variable such that 's' is dependent on 'h'.
Since she never works more than 25 hours in a week.
Therefore, the maximum hours she will work = 25
Hence, the domain of this function in this situation is {0 ≤ h ≤ 25}, where h is number of hours she works.
Related Questions
Posted in Mathematics
If a polynomial is divided by (x-a) and the remainder equals zero then (x -a) is a factor of the polynomial. true or false
The Factor theorem states that for any polynomial f(x) if f(c)=0 then x-c is a factor of the polynomial f(x).
If any polynomial f(x) is divided by x-a and remainder is 0 that means f(a)= 0 .In other words we can say x-a is a factor of the polynomial f(x).So the statement :If a polynomial is divided by (x-a) and the remainder equals zero then (x -a) is a factor of the polynomial is True.
Posted in Mathematics
4. Which of the following sets of numbers is a Pythagorean triple? A. 4, 5, 6 B. 8, 15, 17 C. 5, 12, 14 D. 8, 24, 25 Suppose a triangle has side lengths 6, 7, and 10. Which of the following best describes this triangle? A. obtuse B. acute C. right D. Cannot be determined.
Here are the answers to the given questions above:
4. When we say pythagorean triple, this would consists of three positive integers a, b, and c, such that a2 + b2 = c2. So for the given numbers above, the set that is considered a pythagorean triple would be B. 8, 15, 17.
5. The correct answer for this question would be option D. It cannot be determined whether the triangle is obtuse, acute or right if the given values are the side lengths. It should be angles. Hope this answer helps.
Posted in Mathematics
For f(x)=2x+1 and g(x)=x^2-7, find (f-g)(x)
I hope this helps you
(f-g)(x)=2x+1-(x^2-7)
(f-g)(x)=2x+1-x^2+7
(f-g)(x)= -x^2+2x+8
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Posted in Mathematics
A quadratic equation is shown below: 9x2 - 36x + 36 = 0 Part A: Describe the solution(s) to the equation by just determining the discriminant. Show your work. (3 points) Part B: Solve 2x2 - 9x + 7 = 0 using an appropriate method. Show the steps of your work and explain why you chose the method used.(4 points)
For part A:
9x² - 36x + 36 = 0
9(x² - 4x + 4) = 0
9(x-2)(x-2) = 0
Radicand = Discriminant = 36² - 4(9)(36) = 0
there is 1 real zero that repeats itself)
zero is x=2.
Use factoring. Factors of 7 are 1 and 7
2x²-9x+7 = 0
(2x-7)(x-1) = 0
2x-7 = 0 (x-1)=0
2x = 7x=1
x = 7/2
Your solutions are x = 1 and x = 7/2
I hope my answers helped you
Posted in Mathematics
Anastasia uses the equation p = 0.7(rh + b) to estimate the amount of take-home pay, p, for h hours worked at a rate of r dollars per hour and any bonus received, b. What is an equivalent equation solved for h? h = (– b)÷ r h = – b ÷ r h = ÷ r – b h = ÷ r
For the answer to the question above, first we must enlist the given
p = take home pay
r = rate of dollars per hour
h = number of hours worked
b = bonus
So now we can find h and let us start with the formula.
p = 0.7(rh + b)
p = 0.7*r*h + 0.7b
p - 0.7b = 0.7rh
So the answer for this problem is
h = (p-0.7b)/0.7r
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Posted in Mathematics
The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is the greater integer, which system of inequalities could represent the values of a and b? A. a + b ≥ 30 b ≥ a + 10 B.a + b ≥ 30 b ≤ a – 10 c. a + b ≤ 30 b ≥ a + 10 D. a + b ≤ 30 b ≤ a – 10
I believe the correct answer from the choices listed above is option A. If b is the greater integer, then the system of inequalities that could represent the values of a and b would be a + b ≥ 30 and b ≥ a + 10. Hope this answers the question. Have a nice day.
Posted in Mathematics
Jasmine wants to use her savings of \$1,128 to buy video games and music CDs. The total price of the music CDs she bought was \$72. The video games cost \$43 each. What is the maximum number of video games that Jasmine can buy with her savings? 24 25 26 27
For the answer to the question above, she can buy 24 video games
Because she bought a CD on her savings, we must deduce the cost of the CD from her 1,128 savings
1128 - 72=1056
Then just divide the difference to how much each videogame cost
1056 / 43 = 24
Posted in Mathematics
1. Which set of ordered pairs in the form of (x,y) does not represent a function of x? (1 Point) A. (1, 1.5), (2, -1.5), (3, 1.5), (4, 1.5) B. (0, 1.5), (3, 2.5), (1, 3.3), (1, 4.5) C. (1, 1.5), (-1, 1.5), (2, 2.5), (-2, 2.5) D. (1, 1.5), (-1, -1.5), (2, 2.5), (-2, 2.5)
Based on the given set of ordered pairs above, the set of ordered pairs in the form of (x,y) that does not represent a function of x would be option B. Why is this so? if you notice, the number 1 in the x is being repeated so this means this is not a function. Hope this answer helps.
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Posted in Mathematics
Which test point holds true for y − 2x ≤ 1? (0, 2) (-2, 4) (1, 4) (5, 0)
So in order to find the correct answer, we can just easily plug in the values to check which ordered pair matches the given inequality above. So based on my solutions, the correct answer would be the last pair.
y − 2x ≤ 1
0 - 2(5)
≤ 1
0-10 ≤ 1
-10 ≤ 1
Posted in Mathematics
A triangle has side lengths measuring 3x cm, 7x cm, and h cm. Which expression describes the possible values of h, in cm? 4x < h < 10x 10x < h < 4x h = 4x h = 10x
In a triangle sum of two sides , should always be greater than the third side.
Sides of the given triangle are, 3 x cm, 7 x cm and h cm.
1.
3 x + 7 x > h
→ 10 x > h
2.
3 x+ h > 7 x
→h > 7 x -3 x
→h> 4 x
3.
7 x + h > 3 x, which will be always true.
Combining 1, 2 and 3 we get
→4 x < h < 10 x
Option A
Posted in Mathematics
Drey makes \$10 an hour plus \$15 an hour for every hour of overtime. Overtime hours are any hours more than 40 hours for the week. Part A: Create an equation that shows the amount of wages earned, W, for working x hours in a week when there is no overtime. Part B: Create an equation that shows the amount of wages earned, S, for working y hours of overtime. Hint: Remember to include in the equation the amount earned from working 40 hours. Part C: Drey earned \$490 in 1 week. How many hours (regular plus overtime) did he work?
For the answer to the question above, let us solve the first for
Part A. \$10 x 40 hours= \$400 (\$10x = W) X= 40 hours a week
Part b. \$10 x 40 hours = \$400 + \$15 x Y=S (\$10x=\$400 +\$15y=S) Y= number of hours overtime. Since he gets \$15 for working more than 40 hours you take the number of hours over time and multiply it by 15.
Part c. Now that he made \$490 in 1 week you know that he worked for 40 hours at a rate of \$10 an hour. But that's only \$400, so we need to figure out 15 times what is equal to 90. 15 times six is equal to 90 which means that Drey worked for 46 hours
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Posted in Mathematics
What is f(x) = 2x2 + 28x – 5 written in vertex form?
To convert a quadratic from y = ax2 + bx + c form to vertex form, y = a(x - h)2+ k, you use the process of completing the square. We do as follows:
f(x) = 2x^2 + 28x – 5
f(x) = 2(x^2 + 14x) – 5
f(x) = 2(x^2 + 14x + 49) – 5 - 98
f(x) = 2(x + 7)^2 – 103
Hope this answers the question. Have a nice day.
Posted in Mathematics
Given f(x)=17-x^2, what is the average rate of change in f(x) over the interval [1, 5]?
For the answer to the question above, I'll show my to solution to the answer below.
average rate of change is
(f(5) - f(1))/(5 - 1) =
(17 - 5^2 - (17 - 1^2))/4 =
(17 - 25 - 17 + 1)/4 =
So the answer for this problem is
-6
I hope my answer helped you. Feel free to ask for more questions.
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Match Term Definition Line Segment A) a series of points that extend in two directions without end Plane B) two lines that intersect at 90° angles Perpendicular Lines C) lines that lie in the same plane and do not intersect Line D) a flat surface that extends infinitely and has no thickness Parallel Lines E) part of a line that has two endpoints
The correct matches are as follows:
Line Segment
E) part of a line that has two endpoints
Plane
D) a flat surface that extends infinitely and has no thickness
Perpendicular Lines
B) two lines that intersect at 90° angles
Line
A) a series of points that extend in two directions without end
Parallel Lines
C) lines that lie in the same plane and do not intersect
Hope this answers the question. Have a nice day.
Posted in Mathematics
What is the solution to the system of equations? y = 1.5x – 4 y = –x a. (–1.6, 1.6) b. (–1.5, 1.5) c. (1.5, –1.5) d. (1.6, –1.6)
I believe the correct answer from the choices listed above is option D. The solution to the system of equations given above would be (1.6, –1.6). We can use the substitution method to solve this. We do as follows:
-x = 1.5x – 4
4 = 2.5x
x = 1.6
y = -x = -1.6
Posted in Mathematics
A certain forest covers an area of 2,000 square kilometers. Suppose that each year this area decreases by 6%. What is the function that best represents the area of the forest each year and how much area remains after 12 years? Round your answer to the nearest square kilometer. Hint: Use the formula, f(x) = P(1 + r)x
For the answer to the question above, I think the formula you is wrong. The x must be in exponential form. It must look like this
f(x) = P(1 + r)^x
Now we can solve.
Given:
P = 2000
r = -0.06
f(x) = 2000 (1 - 0.06)^x
= 2000 (0.94)^x
2000 (0.94)¹² =
2000 (0.476) =
951.8 square kilometers.
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Posted in Mathematics
Which of the following statements is true of a reflection? -It is an isometry. -It slides a plane. -The transformation is done over a line of reflection. -It flips a plane about a fixed line. I-t changes the size of the figure being reflected.
I think the correct answers from the choices listed above are the first and the fourth option. The statements that are true about a reflection would be that it is an isometry and the it flips a plane about a fixed line. reflection over a line k (notation rk) is a transformation in which each point of the original figure (pre-image) has an image that is the same distance from line k.
Posted in Mathematics
Which ordered pair is a solution of the equation 2x − y = 9 (-4,1) (-2,5) (5,1) (6,-3)
In order to confirm if which ordered pair is a solution of the given equation above, we can just simply plug in the values.
So based on my solution, the correct answer would be the third option: (5,1)
So let us try to plug in the values.
2(5) - 1 = 9
10 -1 = 9
9 = 9
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What is the range of the function f(x) = 3x2 + 6x – 9
I have a solution here however with a slight change in the equation:
f(x)=3x^2-6x+1
My solution is:
The domain is all real numbers--there are no restrictions like a square root or variable in the denominator
this is a U shape parabola and you want to find the bottom point, it is at the vertex
x=-b/2a =- (-6) /2 (3) =1
f(1) =3(1)^2 -6(1) +1 =3-6+1 =-2
the minimum is (1, -2)
so the range is all real numbers >= -2
By examining my solution, you could just answer the problem on your own! I hope it helps!
Posted in Mathematics
Solve the equation for y 2x+6y=13
So here is how we are going to solve for y for the given equation above:
2x+6y=13
Next, transfer 2x to the right side and it will look like this:
6y=13-2x
Now, divide both sides by 6, and it will look like this:
y=13-2x
--------
6
So, this is the final answer for y.
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Posted in Mathematics
What is another way to write 250,000 ml?
Another way to write it would be 250 liters.
hope this helps you
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Posted in Mathematics
What interval includes all possible values of x, where –3(6 – 2x) ≥ 4x + 12? (–∞, –3] [–3, ∞) (–∞, 15] [15, ∞)
The solution to the problem is as follows:
solve for x :
−3(6−2x)≥4x+12
−18+6x≥4x+12
2x≥30
x≥15
So the answer is D ) 15 to +infinity
I hope my answer has come to your help. God bless and have a nice day ahead!
Posted in Mathematics
How to solve this problem 14-6= 10-6=?
I really don't think you wrote the question properly. Or maybe you did, but this question doesn't make any sense! :/
Posted in Mathematics
Algebraic terms are separated by ______. Choose all that apply. 1)= 2)+ 3)x 4)÷ 5)-
1)=
2)+
5)-
Explanation
When letter or numbers are separated by multiplication or division can be combined to make one term.
In an algebra equation, terms are separated by an addition sign, subtraction sign or an equal sign. For example,
3a + 7b = 21.
This equation have 3 terms.
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# love.math.colorToBytes
Available since LÖVE 11.3 This function is not supported in earlier versions.
Converts a color from 0..1 to 0..255 range.
## Function
### Synopsis
```rb, gb, bb, ab = love.math.colorToBytes( r, g, b, a )
```
### Arguments
`number r`
Red color component.
`number g`
Green color component.
`number b`
Blue color component.
`number a (nil)`
Alpha color component.
### Returns
`number rb`
Red color component in 0..255 range.
`number gb`
Green color component in 0..255 range.
`number bb`
Blue color component in 0..255 range.
`number ab (nil)`
Alpha color component in 0..255 range or nil if alpha is not specified.
## Notes
Here's implementation for 11.2 and earlier.
```function love.math.colorToBytes(r, g, b, a)
if type(r) == "table" then
r, g, b, a = r[1], r[2], r[3], r[4]
end
r = floor(clamp01(r) * 255 + 0.5)
g = floor(clamp01(g) * 255 + 0.5)
b = floor(clamp01(b) * 255 + 0.5)
a = a ~= nil and floor(clamp01(a) * 255 + 0.5) or nil
return r, g, b, a
end
```
Where `clamp01` is defined as follows
```local function clamp01(x)
return math.min(math.max(x, 0), 1)
end
```
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Search
## Search Loci: Convergence:
Keyword
Random Quotation
The modern, and to my mind true, theory is that mathematics is the abstract form of the natural sciences; and that it is valuable as a training of the reasoning powers not because it is abstract, but because it is a representation of actual things.
In N. Rose, Mathematical Maxims and Minims, Raleigh NC: Rome Press Inc., 1988.
See more quotations
# Euler Squares
## Sudoku Skills
Skills learned in solving Sudoku puzzles can be applied to solving the Euler square. However, this is not an easy method, and is probably best applied after beginning with the two methods above. Randomly place pegs on diagonals or rows, and then continue by choosing a spot on the array and eliminating already used colors and shapes for that row and column. Now chose wisely from the remaining pegs, adjusting when necessary. At times the last color or last shape can be placed by checking which row or column does not yet have that attribute. This solution method was more random and did not evince diagonalization or symmetry (see Figure 7).
Figure 7: Using Sudoku skills to place the last red peg
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## Learning Objectives
• Describe the categories of electromagnetic energy, and be able to place these categories relative to each other in wavelength and energy.
• Describe the wavelengths of visible light and name a color for a given wavelength range.
• Convert wavelengths of electromagnetic energy between commonly used units (meters, micrometers, nanometers, Angstroms).
• None
## Key Terms
• Wavelength
• Frequency
• Visible, Infrared, Ultraviolet light
# Guided Inquiry
## Watch:
A tour of the Electromagnetic spectrum:
And a crash course on light (relevant portion is 0-4 minutes):
After reviewing the videos, look closely at the following diagrams of the electromagnetic spectrum:
Figure 3.2.1 The Electromagnetic Spectrum. National Aeronautics and Space Administration, Science Mission Directorate. (2010). Introduction to the Electromagnetic Spectrum. Retrieved March 9, 2019, from NASA Science website: http://science.nasa.gov/ems/01_intro and https://smd-prod.s3.amazonaws.com/science-pink/s3fs-public/thumbnails/image/EMS-Introduction_0.jpeg
Figure 3.2.2. The Electromagnetic Spectrum. OpenStax, College Physics. OpenStax CNX. Mar 4, 2019 http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@14.48 Figure 24.9. Also at: https://courses.lumenlearning.com/physics/chapter/24-3-the-electromagnetic-spectrum/
3.2.1. Does ultraviolet light have longer or shorter wavelengths than infrared light?
3.2.2. Do X-rays have larger or smaller energy than visible light?
3.3.3. A Raman spectrum includes a peak at 10 micrometers wavelength. In which region is this peak located?
Figure 3.2.3. The visible light spectrum, showing wavelength ranges for colors in nanometers. From: OpenStax, College Physics. OpenStax CNX. Mar 4, 2019 http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@14.48, Figure 24.16.
By SpiggetOwn work, CC BY-SA 3.0, Link Figure 3.2.4 The visible light spectrum.
3.2.4. What is the range of wavelengths (in nanometers) for the color green, according to Figure 3.2.3?
3.2.5. What color is a typical laser pointer with a wavelength of 650 nm? Hint: Figure 3.2.4 might be easier to use for this question.
## Frequency and Wavelength
The number of crests that pass a given point within one second is described as the frequency of the wave. One wave—or cycle—per second is called a Hertz (Hz), after Heinrich Hertz who established the existence of radio waves. A wave with two cycles that pass a point in one second (as shown for the top wave in Figure 3.2.5) has a frequency of 2 Hz.
Electromagnetic waves have crests and troughs similar to (but not identical in behavior to) those of ocean waves. The distance between crests is the wavelength.
Figure 3.2.5. Frequency and wavelength of a wave. From National Aeronautics and Space Administration, Science Mission Directorate. (2010). Anatomy of an Electromagnetic Wave. Retrieved March 9, 2019, from NASA Science website: http://science.nasa.gov/ems/02_anatomy
## Energy
An electromagnetic wave can also be described in terms of its energy—in units of measure called electron volts (eV). An electron volt is the amount of kinetic energy needed to move an electron through one volt potential. Moving along the spectrum from long to short wavelengths, energy increases as the wavelength shortens. Consider a jump rope with its ends being pulled up and down. More energy is needed to make the rope have more waves.
Figure 3.2.6. Higher and lower energy waves. From National Aeronautics and Space Administration, Science Mission Directorate. (2010). Anatomy of an Electromagnetic Wave. Retrieved March 9, 2019, from NASA Science website: http://science.nasa.gov/ems/02_anatomy
3.2.6. Does a wave with a high frequency have a high energy? Explain your answer.
Following from the answer above, we can write an equation to describe the relationship between wavelength and frequency as:
Equation 3.2.1 c = νλ
The speed of light (electromagnetic energy) is a constant c = 3.00 × 108 m/s. Wavelength is represented by lambda (λ), and frequency is represented by the Greek letter “nu” or ν. Sometimes “f” is used to replace nu.
3.2.7. Calculate the wavelength of a 1530-kHz AM radio signal.
Solution
Rearranging gives:
For the ν = 1530 kHz AM radio signal:
λ=3.00×108 m/s / 1530×103 cycles/s =196 m
## Summary
The electromagnetic spectrum is divided into different categories of light energy: gamma rays, X-rays, ultraviolet, visible, and infrared light, microwaves, and radio waves. The energy and wavelength of electromagnetic energy changes across the spectrum, with gamma rays having short wavelength and highest energies, and radio waves having longest wavelengths and lowest energies. Humans can detect only a narrow range of electromagnetic energy with our eyes, called visible light. Energy and frequency of light are positively correlated with each other, but are negatively correlated with wavelength.
## References
Lumen Learning Physics Textbook: https://courses.lumenlearning.com/physics/
National Aeronautics and Space Administration, Science Mission Directorate. (2010). Introduction to the Electromagnetic Spectrum. Retrieved March 9, 2019, from NASA Science website: http://science.nasa.gov/ems/01_intro
National Aeronautics and Space Administration, Science Mission Directorate. (2010). Anatomy of an Electromagnetic Wave. Retrieved March 9, 2019, from NASA Science website: http://science.nasa.gov/ems/02_anatomy
OpenStax, College Physics. OpenStax CNX. Mar 4, 2019 http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@14.48
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Welcome to Scribd. Sign in or start your free trial to enjoy unlimited e-books, audiobooks & documents.Find out more
Standard view
Full view
of .
Contoh 7.2:
7.5 FUNGSI KEANGGOTAAN
a. Representasi Linear
Contoh 7.3:
Contoh 7.4:
Contoh 7.5:
Contoh 7.6:
NORMAL
Bahu
Contoh 7.7:
Contoh 7.8:
(i) Kurva PI
Contoh 7.9:
(ii) Kurva BETA
PAROBAYA
Contoh 7.10:
(iii) Kurva GAUSS
PENGENDARA BERESIKO TINGGI (dalam umur)
7.6 OPERATOR DASAR ZADEH UNTUK OPERASI HIMPUNAN FUZZY
7.6.1 Operator AND
Contoh 7.11:
7.6.2 Operator OR
Contoh 7.12:
7.6.3 Operator NOT
Contoh 7.13:
7.7 PENALARAN MONOTON
Contoh 7.14:
TINGGI
7.8 FUNGSI IMPLIKASI
7.8 SISTEM INFERENSI FUZZY 7.8.1 Metode Tsukamoto
Contoh 7.15:
7.8.2 Metode Mamdani
4. Penegasan (defuzzy)
Contoh 7.16:
7.8.3 Metode Sugeno
Contoh 7.17
7.9 BASISDATA FUZZY
NIP Nama Tgl Lahir Th. Masuk Gaji/bl (Rp)
NIP Nama Umur (th) Masa Kerja (th)* Gaji/bl
MUDA PAROBAYA TUA
BARU LAMA
RENDAH SEDANG TINGGI
0 of .
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# Difference between revisions of "Math 543: Advanced Probability 1"
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## Catalog Information
### Title
Advanced Probability 1.
3
### Prerequisite
Math 314 or equivalent.
### Recommended
Math 341, Stat 441(?); or equivalents.
### Description
Foundations of the modern theory of probability with applications. Probability spaces, random variables, independence, conditioning, expectation, generating functions, and Markov chains.
## Desired Learning Outcomes
This should be an advanced course in probability and,therefore, clearly distinguishable from an introductory course like Math 431. Furthermore, it is supposed to be a course in the modern theory of probability, which suggests that it should be based on Kolmogorov's measure-theoretic approach, or something equivalent.
### Prerequisites
The official prerequisite is multivariable calculus. Other prior courses that will contribute to student success include:
• an introductory course in probability;
• a course in rigorous mathematical reasoning;
• an introductory course in analysis.
### Minimal learning outcomes
Outlined below are topics that all successful Math 543 students should understand well. As evidence of that understanding, students should be able to demonstrate mastery of all relevant vocabulary, familiarity with common examples and counterexamples, knowledge of the content of the major theorems, understanding of the ideas in their proofs, and ability to make direct application of those results to related problems.
1. Probability spaces
• Sigma-algebras and Borel sets
• Kolmogorov axioms
• Carathéodory's measure extension theorem
• Lebesgue-Stieltjes measure
2. Random variables
• Measurable maps
• Distributions and distribution functions
3. Independence
• Of events and classes of events
• Of random variables
• The Borel-Cantelli lemmas
4. Expectation
• Of arbitrary nonnegative random variables
• Of integrable real-valued random variables
• Of compositions
• Monotone convergence theorem
• Uniform integrability and dominated convergence
5. Conditioning
• Probability conditioned on a non-null set
• Expectation conditioned on a sigma-algebra
• Expectation conditioned on a random variable
• Bayes' formula
• Regular conditional distributions
6. Generating functions
7. Discrete Markov chains
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Questions
# Remember concepts with our Masterclasses.
80k Users
60 mins Expert Faculty Ask Questions
a
The ray gets totally internally reflected at face CD.
b
The ray comes out through face AD
c
The angle between the incident ray and the emergent ray is 90º
d
The angle between the incident ray and the emergent ray is 120º
Check Your Performance Today with our Free Mock Tests used by Toppers!
detailed solution
Correct option is A
Applying Snell’s law at face AB, we get
(where, ${\mathrm{\theta }}_{\mathrm{c}}$ is the critical angle)
From geometry, angle of incidence at Q is 45º. At face CD angle of incidence is greater than ${\mathrm{\theta }}_{\mathrm{c}}$. Therefore, the rays gets totally internally reflected at face CD. Hence, option (a) is correct. From geometry, angle of incidence at R is 30º. At face AD angle of incidence is less than ${\mathrm{\theta }}_{\mathrm{c}}$So, the ray comes through face AD. Hence option (b) is correct. Applying Snell’s law at face AD, we get$\mathrm{e}={\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)={60}^{\circ }$
The angle between the incident ray and the emergent ray is 90º.
Hence, option (c) is correct and option (d) is incorrect
+91
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https://en.khanacademy.org/math/cc-fourth-grade-math/imp-fractions-2/imp-fractions-with-denominators-of-10-and-100/v/decompose-hundredths-on-number-line
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Main content
# Decomposing hundredths on number line
Sal decomposes 26/100 into tenths and hundredths using a number line.
## Want to join the conversation?
• 0.1 is in tenths and 0.01 is in hundrethes
(10 votes)
• why is the universe?
(7 votes)
• I don't understand!
(3 votes)
• How do you decompose from hundredths to tenths?
(2 votes)
• upvote this
(2 votes)
• A few hours ago I was typing an essay on a Khan academy answer box. I usually write it in a google text box but I decided otherwise for no reason at all. I also type these texts at a google search after I search for something else. I am choosing to stop here because I'm getting bored of typing. (I actually think it's fun but I stopped because I wanted to. )
(0 votes)
• Seems hard when looking at it.
(0 votes)
• this look easy but it mabye id not
(0 votes)
• This from twenty years ago
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• 3 months ago now.
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• i DO NOT get this
he talks to fast lol
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• I had to speed it up because he was slow
(1 vote)
## Video transcript
- [Voiceover] So we have an example exercise from our Khan Academy exercises. This is the decomposed fractions. with denominators of 100. Use the following number line to complete the equation. The equation says 26-hundredths is equal to star-tenths plus six-hundredths. And they give us this number line and on this number line-- Let's see, this is zero right over here, zero-tenths, which is the same thing as zero, and it goes all the way to 26-hundredths. Let's see if that makes sense. Well, you see between zero-tenth and one-tenth, they split into 10 equal sections. One, two, three, four, five, six, seven, eight, nine, 10. So if you divide a tenth into tenths, each of these is going to be 100. A hundredth is one-tenth of a tenth. So that's 100, 200, 300, 400, five, six, seven, eight, nine, 10-hundredths, 11, 12, 13, 14, 15, 16, 17, 18, 19 20-hundredths. 21, 22, 23, 24, 25, 26-hundredths. So that's exactly what they wrote there. So they say 26-hundredths, which is this thing right over here, is equal to star-tenths, which is really what's being depicted in magenta, plus six-hundredths. The six-hundredths is this green right over here. Adding the six-hundredths. Adding one, two, three, four, five, six-hundredths. So what is this right over here? This magenta or this pink arrow? Well we could say that's 20-hundredths. If you count it that's 10, 20-hundredths. Or, you could see, well that's just two-tenths. So 26-hundredths is the same thing as two-tenths plus six-hundredths. And it gets you right over there. So this right over here is going to be two-tenths.
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# compact inclusion of domains of unbounded operators
Let $L$ be a positive self-adjoint operator defined densely on $L^2(M)$ where $M$ is a compact manifold. Also, let $\mathcal{D}(L) \subset H^1(M)$. It is known that $\mathcal{D}(L) \subset \mathcal{D}((L)^{1/2})$ as a continuous inclusion. I am trying to see whether this inclusion is also compact. I suspect that this has something to do with Rellich's theorem and interpolation spaces, but cannot work out a rigorous proof. Any help would be appreciated.
Edit: I can also work out that $L$ has compact resolvent.
• Are $\mathcal{D}(L)$, $\mathcal{D}(L^{1/2})$ equipped with their graph norms, or something else? – Nate Eldredge Apr 23 '15 at 23:44
• @NateEldredge I think $\mathcal{D}((L)^{s})$ is getting the norm from $H^s$. – anonymous Apr 23 '15 at 23:50
• How do we know that $\mathcal{D}(L^{1/2}) \subset H^{1/2}$? It isn't obvious to me. – Nate Eldredge Apr 23 '15 at 23:51
• @NateEldredge By interpolation, $\mathcal{D}(L^s) = [L^2, \mathcal{D}(L)]_s$? – anonymous Apr 24 '15 at 0:46
The mention of $M$ and $H^1(M)$ is irrelevant. The information on the resolvent implies that the spectrum of $L$ is discrete with a sequence of eigenvalues which increases to infinity. By the spectral theorem, the underlying Hilbert space can then be identified with $\ell^2$ and the other two spaces with weighted versions thereof. It is then transparent that the inclusion is compact as required.
I think I have the idea for an answer, but I would really appreciate people's opinion on this. Here is what I think: let $R(\lambda)$ denote the resolvent $(\lambda + L)^{-1}$. Then, $R(\lambda)^{1/2} : \mathcal{D}(L^{1/2}) \to \mathcal{D}(L)$ is continuous, hence maps bounded sets in $\mathcal{D}(L^{1/2})$ to bounded sets in $\mathcal{D}(L)$. Also, $R(\lambda)^{1/2} : \mathcal{D}(L^{1/2}) \to \mathcal{D}(L^{1/2})$ is compact, as $R(\lambda) : \mathcal{D}(L^{1/2}) \to \mathcal{D}(L^{1/2})$ is compact, and fractional powers of compact operators are compact. So bounded sets in $\mathcal{D}(L)$ are precompact in $\mathcal{D}(L^{1/2})$.
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# Strength and direction of an electric field at a particular point problem
Hello all - this is my first time giving this forum a try. I've been using another site for a while but it's just too spotty and not enough folks there care to give decent responses, so here I am :)
## Homework Statement
http://img684.imageshack.us/i/27p28.jpg/
What are the strength and direction of the electric field at the position indicated by the dot in the figure?
a) Give your answer in component form. (Assume that x-axis is horizontal and points to the right, and y-axis points upward.) Express your answer in terms of the unit vectors i and j.
b) Give your answer as a magnitude and angle measured cw [Edit: ClockWise?] from the positive x-axis.
c) Give theta, the degrees clockwise from the positive x axis.
## Homework Equations
E_1x = E_1*cos(theta) = k*(q1 / r1^2) * cos(theta)
At least, I believe this is the main equation you need...?
## The Attempt at a Solution
The book I'm using is confusing about this. I've tried the idea of calculating the fields for each of the points (the two positive charges and the electric charge), but as far as the angles go and how the negative charge will pull away from the rest of those charges...I start to get a headache. How does that factor in with the rest of the angles? Does it, perhaps, pull away straight up and leave the x components of the other one to push it out? Why am I not getting the right answer for that then when I try it, in such a case... Does anyone have an idea?
ideasrule
Homework Helper
The book I'm using is confusing about this. I've tried the idea of calculating the fields for each of the points (the two positive charges and the electric charge), but as far as the angles go and how the negative charge will pull away from the rest of those charges...I start to get a headache. How does that factor in with the rest of the angles? Does it, perhaps, pull away straight up and leave the x components of the other one to push it out?
Yes, the electric field of the negative charge points up at P. However, you don't need to worry about this. Since Ex=kq/r^2 * cos(theta), you can simply insert the negative charge into q and get the right answer. There's nothing special in the equation that makes it only work for positive charges.
Why am I not getting the right answer for that then when I try it, in such a case... Does anyone have an idea?
Can you show all your work? It might just be an algebra mistake.
Well I actually just think I might be on the right track for these parts at least...though honestly I'm really quite confused still, having not done anything like this in years (you'll see that from my work, I'm sure...). My work:
<Note: I've labeled the charges as such: E1 is for the field from the lower left charge, E2 from the upper left, and E3 from the upper right>
E_netx = E_1x + E_2x
E_1x = 9*10^9 * 5*10^-9 / 0.02^2 = 112500 N/C
E_2x = 9*10^9 * 10*10^-9 * cos(63.4degrees) / 0.045^2 = 1990 N/C
Therefore E_netx = 114490 N/C
E_nety = E_2y + E_3y
E_2y = 9*10^9 * 10*10^-9 * cos(26.6) / 0.045^2 = 3974 N/C
E_3y = 9*10^9 * (-5*10^-9) * cos(0) / 0.04^2 = -28125
Therefore E_nety = 3974 - 28125 = -24151 N/C
....and honestly I have no idea what to do with this information o.o I'd need to find the overall angle as to where the direction of that field would go, and for some reason I'm just not getting that. How horribly wrong am I, and what can I do to correct myself and push forth?
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# Lattice Mult. V2
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### Lattice Mult. V2
1. 1. How to do: Lattice Multiplication Created by Lilian Patrick 2009
2. 2. What is lattice multiplication and when would we use it? Lattice multiplication is a way to multiply larger numbers (bigger than a 1-digit number times another 1-digit number) You can use lattice multiplication with any big multiplication problem as long as you can draw a big enough grid!
3. 3. Sounds great! How do we get started?
4. 4. Let’s start with a 2- digit by 1-digit example… If you want to multiply: 42 x 9 First, you have to make a 2 by 1 grid (because it’s a 2-digit by 1-digit problem)
5. 5. Next, in each box, you need to draw a diagonal line from the upper right hand corner through the lower left hand corner. The line should go past the lower left hand corner to the outside of your grid.
6. 6. Put the 2-digit number across the top of your grid and the 1-digit number on the right side of your grid. 4 2 9
7. 7. Then, multiply the number on the top of each column with the number to the right of that row. 4 2 4x9 2x9 9
8. 8. The digit in the tens place of the product goes to the left of the diagonal. The digit in the ones place of the product goes to the right of the diagonal. *If there are no tens or no ones in the product, don’t forget to put a 0! 4 2 3 1 6 8 9
9. 9. Now, you’re finished with the numbers 42 and 9. Be careful not to use these numbers in the next step! 4 2 3 1 6 8 9
10. 10. Finally, you need to ADD the numbers inside the grid along the diagonals. Start with the diagonal on the right. Write the answer in the space outside the grid. 4 2 3 1 3 + nothing = 6 8 9 3 6+1= 8 + nothing = 7 8
11. 11. Your answer is the number outside of the grid. 4 2 3 1 6 8 9 3 7 8 42 x 9 = 378
12. 12. Now let’s try a 2-digit by 2- digit multiplication problem… Let’s multiply: 54 x 97 Start by making your grid (a 2 by 2 grid this time):
13. 13. Add the diagonals: 5 4 And the numbers: 9 Since they’re both 2-digits, it doesn’t matter which one goes on the top and 7 which one goes to the right of the grid.
14. 14. Multiply the number at the top of the column with the side of the row. 5 4 5 4 4x9 5x9 9 4 3 5 6 9 5x7 4x7 7 3 2 8 7 5
15. 15. Finally, add the numbers along the diagonals. If the sum is greater than 10, carry the 1 into the next diagonal. The 1 is carried from the previous diagonal: 5 4 5 + 2 + 6 = 13 The 3 goes at the bottom of that diagonal. The 1 gets added to the next 1 1 4 3 diagonal. 5 5 6 9 3 2 8 7 5 2 3 8 54 x 97 = 5,238
16. 16. Now, try these two examples on your own. The grids are drawn for you. Don’t go on to the next slide until you’ve given these problems a try. Look back to previous slides if you need help. 1.) 26 x 4 = 2.) 31 x 62 = 2 6 3 1 4 6 2
17. 17. See if you got the answers! 1.) 2.) 3 1 2 6 1 0 1 0 1 2 1 6 6 8 1 8 4 4 0 0 6 2 2 9 0 4 2 2 26 x 4 = 104 31 x 62 = 1,922
18. 18. Now that you’ve learned lattice multiplication, you can practice with any numbers of your choice! Good luck!
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Home MathLab Write a Function That Find the Volume of a 3D Rectangle in Matlab
# Write a Function That Find the Volume of a 3D Rectangle in Matlab
There is a function behind every Matlab code that performs a computational task, and here is how you can create your own function that will help you go faster while working with Matlab.
We will be creating a simple one-line code function in this post, just to help you have a glimpse into how creating functions in Matlab
## Write a function in Matlab
Here is what you do when writing your own function in Matlab.
Open a new function file
Here is what appear on the next window
In Matlab, the section after the % is considered as a comment (the section that appears in green in the window above)
Here is how we need to structure our Matlab function
[php]function [v] = vol(L,W,H)
% L is the length of the 3D rectangle
% W is the width of the 3D rectangle
% H is the height of the 3D rectangle
v= L*W*H;
end[/php]
You need to copy the whole code above and paste it in the file, Save the file with the name vol.m and make sure not to change the directory Matlab will guide you to.
Here is how it will look
## How to use the Matlab function we have just created
Every time you will need to compute the such a volume, simply type the following code
`vol(L,W,H)`
where L, W, And H will be real numbers.
#### Example
Let’s say, we need the volume of a 3D rectangle with L, W and H being respectively 10, 3, 15, we can simply type:
`vol(10,3,15)`
And Matlab will call the function vol, compute the volume of the 3D rectangle and display the following.
If you are working in an exercise where you will often need this volume, you can just use this function instead of actually computing manually the volume each time you need it. It might not look very important in this exercise, but when you have a complex operation you need to perform often in Matlab, creating a function doing it automatically will be time-efficient and necessary.
Using this technique, you can create more complex functions in Matlab and have an easier life while using them. Learn more about functions in Matlab.
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# Formula:DLMF:25.13:E2
$\displaystyle {\displaystyle \PeriodicZeta@{x}{s} = \frac{\EulerGamma@{1-s}}{(2 \cpi)^{1-s}} \* \left( \expe^{\cpi \iunit (1-s)/2} \HurwitzZeta@{1-s}{x} + \expe^{\cpi \iunit (s-1)/2} \HurwitzZeta@{1-s}{1-x} \right) }$
## Constraint(s)
$\displaystyle {\displaystyle 0 < x < 1}$ &
$\displaystyle {\displaystyle \realpart{s} > 1}$
## Proof
We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.
## Symbols List
& : logical and
$\displaystyle {\displaystyle F}$ : periodic zeta function : http://dlmf.nist.gov/25.13#E1
$\displaystyle {\displaystyle \Gamma}$ : Euler's gamma function : http://dlmf.nist.gov/5.2#E1
$\displaystyle {\displaystyle \pi}$ : ratio of a circle's circumference to its diameter : http://dlmf.nist.gov/5.19.E4
$\displaystyle {\displaystyle \mathrm{e}}$ : the base of the natural logarithm : http://dlmf.nist.gov/4.2.E11
$\displaystyle {\displaystyle \mathrm{i}}$ : imaginary unit : http://dlmf.nist.gov/1.9.i
$\displaystyle {\displaystyle \zeta}$ : Hurwitz zeta function : http://dlmf.nist.gov/25.11#E1
$\displaystyle {\displaystyle \Re {z}}$ : real part : http://dlmf.nist.gov/1.9#E2
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# EPSG:1051
## Lambert Conic Conformal (2SP Michigan)
### Attributes
Data source: EPSG
Information source: EPSG guidance note #7-2, http://www.epsg.org
Revision date: 2018-08-29
### Formula
```Note: These formulas have been transcribed from EPSG Guidance Note #7-2. Users are encouraged to use that document rather than the text which follows as reference because limitations in the transcription will be avoided.
To derive the projected Easting and Northing coordinates of a point with geographical coordinates (lat,lon) the formulas for the one standard parallel case are:
E = EF + r sin(theta)
N = NF + rF - r cos(theta)
where
m = cos(lat)/(1 - e^2 sin^2(lat))^0.5 for m1, lat1, and m2, lat2 where lat1 and lat2 are the latitudes of the two standard parallels.
t = tan(pi/4 - lat/2)/[(1 - e sin(lat))/(1 + e sin(lat))]^(e/2) for t1, t2, tF and t using lat1, lat2, latF and lat respectively.
n = (loge(m1) - loge(m2))/(loge(t1) - loge(t2))
F = m1/(n t1^n)
r = a K F t^n for rF and r, where rF is the radius of the parallel of latitude of the false origin and K is the ellipsoid scaling factor.
theta = n(lon - lon0)
The reverse formulas to derive the latitude and longitude of a point from its Easting and Northing values are:
lat = pi/2 - 2atan{t'[(1 - esin(lat))/(1 + esin(lat))]^(e/2)}
lon = theta'/n +lon0
where
r' = +/-[(E - EF)^2 + {rF - (N - NF)}^2]^0.5 , taking the sign of n
t' = (r'/(aKF))^(1/n)
theta' = atan2 [(E- EF),(rF - (N- NF))] (see GN7-2 implementation notes in preface for atan2 convention)
and n, F, and rF are derived as for the forward calculation.
Note that the formula for lat requires iteration. First calculate t' and then a trial value for lat* using
lat = π/2-2atan(t'). Then use the full equation for lat substituting the trial value into the right hand side of the equation. Thus derive a new value for lat. Iterate the process until lat does not change significantly. The solution should quickly converge, in 3 or 4 iterations.```
### Example
```For Projected Coordinate System NAD27 / Michigan Central
Parameters:
Ellipsoid Clarke 1866, a = 6378206.400 metres = 20925832.16 US survey feet
1/f = 294.97870
then e = 0.08227185 and e^2 = 0.00676866
First Standard Parallel = 44°11'00"N = 0.771144641 rad
Second Standard Parallel = 45°42'00"N = 0.797615468 rad
Latitude False Origin = 43°19'00"N = 0.756018454 rad
Longitude False Origin = 84°20'00"W = -1.471894336 rad
Easting at false origin = 2000000.00 US survey feet
Northing at false origin = 0.00 US survey feet
Ellipsoid scaling factor = 1.0000382
Forward calculation for:
Latitude = 43°45'00.00"N = 0.763581548 rad
Longitude = 83°10'00.00"W = -1.451532161 rad
first gives :
m1 = 0.718295175 m2 = 0.699629151
t = 0.429057680 tF = 0.433541026
t1 = 0.424588396 t2 = 0.409053868
n = 0.706407410 F = 1.862317735
r = 21436775.51 rF = 21594768.40
theta = 0.014383991
Then Easting X = 2308335.75 US survey feet
Northing Y = 160210.48 US survey feet
Reverse calculation for same easting and northing first gives:
theta' = 0.014383991 r' = 21436775.51
t' = 0.429057680
Then Latitude = 43°45'00.000"N
Longitude = 83°10'00.000"W```
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# Generating Random Numbers Between 1 And 25: Methods And Applications
//
Thomas
Explore various methods for generating random numbers between 1 and 25, from using generators to mathematical formulas, and discover their wide range of applications in simulating events and improving security measures.
## Methods for Generating Random Numbers
Random numbers play a crucial role in various fields such as cryptography, statistics, and gaming. There are several methods for generating random numbers, each with its own advantages and limitations. Two common methods include using a Random Number Generator (RNG) and employing mathematical formulas.
### Using a Random Number Generator
A Random Number Generator is a software algorithm or hardware device that generates a sequence of numbers that lack any pattern or predictability. RNGs are commonly used in computer programming, simulations, and gaming applications. They come in two main types: true random number generators and pseudo-random number generators. True RNGs generate numbers based on physical processes such as atmospheric noise or radioactive decay, ensuring true randomness. On the other hand, pseudo-RNGs use mathematical algorithms to produce sequences of numbers that appear random but are deterministic in nature.
### Using Mathematical Formulas
Another method for generating random numbers is through mathematical formulas. These formulas take a seed value as input and produce a sequence of numbers based on complex mathematical calculations. While mathematical formulas can provide a deterministic way to generate random numbers, they may not always be truly random and can be prone to biases or patterns. However, they are often used in statistical simulations, modeling, and cryptographic applications.
• When using a Random Number Generator, it is important to consider factors such as the quality of randomness, speed of generation, and seed selection.
• Mathematical formulas offer a deterministic approach to generating random numbers but may lack true randomness.
• Both methods have their own strengths and weaknesses, making it essential to choose the right approach based on the specific requirements of the application.
## Applications of Random Numbers
### Simulating Random Events
In the realm of computer simulations and modeling, random numbers play a crucial role in simulating random events. These events can range from weather patterns and traffic flow to the outcomes of complex systems like financial markets or biological processes. By incorporating random numbers into these simulations, researchers and analysts can generate realistic scenarios that closely mimic real-world conditions.
One common application of simulating random events is in the field of Monte Carlo simulation. This technique involves using random numbers to model the behavior of complex systems and calculate the likelihood of different outcomes. For example, in finance, Monte Carlo simulations are used to estimate the risk and return of investment portfolios by generating thousands of random scenarios and analyzing the results.
Another important application of random numbers in simulating random events is in the field of gaming and entertainment. Video game developers use random numbers to create unpredictable gameplay experiences, ensuring that each playthrough is unique and engaging. Similarly, online casinos rely on random number generators to ensure fair and unbiased outcomes for their games.
Overall, simulating random events using random numbers allows researchers, analysts, and developers to explore complex systems, make informed decisions, and create engaging experiences for users.
### Encryption and Security
Random numbers also play a critical role in encryption and security systems, where unpredictability is key to protecting sensitive information from unauthorized access. By incorporating random numbers into cryptographic algorithms, security experts can enhance the strength of encryption and prevent attackers from deciphering encrypted data.
One common of random numbers in encryption is in the generation of cryptographic keys. These keys are used to encrypt and decrypt data, and their security relies on the randomness of the numbers used to create them. By using truly random numbers, rather than predictable sequences, encryption keys become virtually impossible to crack through brute force attacks.
Random numbers are also used in security protocols like SSL/TLS to establish secure connections between clients and servers. By generating random numbers during the handshake process, these protocols ensure that each connection is unique and protected from eavesdropping or data tampering.
In summary, the applications of random numbers in encryption and security are essential for safeguarding sensitive information, securing online transactions, and protecting digital communications from malicious actors. By leveraging the power of randomness, encryption algorithms can create robust defenses against cyber threats and ensure the confidentiality and integrity of data.
## Challenges of Generating Random Numbers
### Bias in Random Number Generation
When it comes to generating random numbers, one of the biggest challenges is dealing with bias. Bias in random number generation can lead to skewed results and inaccurate simulations. This can be particularly problematic in applications where true randomness is essential, such as cryptography or statistical analysis.
One common source of bias is the method used to generate random numbers. Some algorithms may inadvertently introduce patterns or correlations that result in non-random outcomes. For example, a poorly designed random number generator may produce numbers that are more likely to fall within a certain range, leading to biased results.
To address bias in random number generation, it is crucial to carefully evaluate the algorithms and methods used. By using well-tested and validated random number generators, developers can minimize the risk of bias and ensure that the generated numbers are truly random.
Another important factor to consider is the seed value used to initialize the random number . The seed is the starting point for generating random numbers, and different seed values can lead to different sequences of numbers. If the seed is not chosen carefully, it can introduce bias into the generated numbers.
### Seed Selection Issues
Choosing a suitable seed value is critical to ensuring the randomness of the generated numbers. A common mistake is using a static or predictable seed, such as the current time or a fixed value. This can result in predictable sequences of numbers that are not truly random.
To address seed selection issues, developers should use a secure and unpredictable source of entropy to initialize the random number generator. This could include using system-level sources of randomness, such as hardware random number generators or operating system-provided entropy sources.
Additionally, it is important to periodically reseed the random number generator with fresh entropy to prevent predictability and ensure randomness. By carefully managing the seed value and entropy sources, developers can mitigate seed selection issues and improve the quality of generation.
In conclusion, bias in random number generation and seed selection issues are significant challenges that must be addressed to ensure the reliability and accuracy of random numbers. By understanding these challenges and implementing best practices for generating random numbers, developers can enhance the security and effectiveness of their applications.
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CC-MAIN-2024-26
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https://blog.csdn.net/PoPoQQQ/article/details/47980565
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BZOJ 2654 tree 二分答案+Kruskal
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 100100
using namespace std;
struct edge{
int x,y,z,col;
friend istream& operator >> (istream &_,edge &e)
{
return scanf("%d%d%d%d",&e.x,&e.y,&e.z,&e.col),++e.x,++e.y,_;
}
}edges[M];
bool Compare1 (const edge &e1,const edge &e2)//白边优先
{
if( e1.z!=e2.z )
return e1.z<e2.z;
return e1.col<e2.col;
}
bool Compare2 (const edge &e1,const edge &e2)//黑边优先
{
if( e1.z!=e2.z )
return e1.z<e2.z;
return e1.col>e2.col;
}
int n,m,need;
namespace Union_Find_Set{
int fa[M],rank[M];
void Initialize()
{
memset(fa,0,sizeof fa);
memset(rank,0,sizeof rank);
}
int Find(int x)
{
if(!fa[x]||fa[x]==x)
return fa[x]=x;
return fa[x]=Find(fa[x]);
}
void Union(int x,int y)
{
x=Find(x);y=Find(y);
if(x==y) return ;
if(rank[x]>rank[y])
swap(x,y);
if(rank[x]==rank[y])
++rank[y];
fa[x]=y;
}
}
bool Check(int x)//need小于最少边数返回1 大于最大反回0
{
using namespace Union_Find_Set;
int i,min_cnt=0,max_cnt=0,re=0;
for(i=1;i<=m;i++)
if(edges[i].col==0)
edges[i].z+=x;
Initialize();
sort(edges+1,edges+m+1,Compare1);
for(i=1;i<=m;i++)
{
int x=Find(edges[i].x);
int y=Find(edges[i].y);
if(x==y) continue;
Union(x,y);
if(edges[i].col==0)
++max_cnt;
}
Initialize();
sort(edges+1,edges+m+1,Compare2);
for(i=1;i<=m;i++)
{
int x=Find(edges[i].x);
int y=Find(edges[i].y);
if(x==y) continue;
Union(x,y);
re+=edges[i].z;
if(edges[i].col==0)
++min_cnt;
}
for(i=1;i<=m;i++)
if(edges[i].col==0)
edges[i].z-=x;
if(need<min_cnt)
return 1;
if(need>max_cnt)
return 0;
throw re-need*x;
}
void Bisection()
{
int l=-101,r=101;
while(r-l>1)
{
int mid=l+r>>1;
if( Check(mid) )
l=mid;
else
r=mid;
}
Check(l);
Check(r);
}
int main()
{
int i;
cin>>n>>m>>need;
for(i=1;i<=m;i++)
cin>>edges[i];
try
{
Bisection();
}
catch(int ans)
{
cout<<ans<<endl;
return 0;
}
printf("%d\n",1/0);
return 0;
}
©️2019 CSDN 皮肤主题: 大白 设计师: CSDN官方博客
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| 3.15625
| 3
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CC-MAIN-2020-05
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latest
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en
| 0.117032
|
http://www.meta-numerics.net/Documentation/html/e929f8d3-cd60-eae3-7d8e-79037a1309f2.htm
| 1,524,611,048,000,000,000
|
text/html
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crawl-data/CC-MAIN-2018-17/segments/1524125947421.74/warc/CC-MAIN-20180424221730-20180425001730-00127.warc.gz
| 447,874,893
| 4,240
|
SquareMatrix.Eigensystem Method
Computes the eigenvalues and eigenvectors of the matrix.
Namespace: Meta.Numerics.Matrices
Assembly: Meta.Numerics (in Meta.Numerics.dll) Version: 4.0.5.0 (4.0.5.0)
Syntax
public ComplexEigensystem Eigensystem()
#### Return Value
Type: ComplexEigensystem
A representation of the eigenvalues and eigenvectors of the matrix.
Remarks
For a generic vector v and matrix M, Mv = u will point in some direction with no particular relationship to v. The eigenvectors of a matrix M are vectors z that satisfy Mz = λz, i.e. multiplying an eigenvector by a matrix reproduces the same vector, up to a prortionality constant λ called the eigenvalue.
For v to be an eigenvector of M with eigenvalue λ, (M - λI)z = 0. But for a matrix to anihilate any non-zero vector, that matrix must have determinant, so det(M - λI)=0. For a matrix of order N, this is an equation for the roots of a polynomial of order N. Since an order-N polynomial always has exactly N roots, an order-N matrix always has exactly N eigenvalues.
Since a polynomial with real coefficients can still have complex roots, a real square matrix can nonetheless have complex eigenvalues (and correspondly complex eigenvectors). However, again like the complex roots of a real polynomial, such eigenvalues will always occurs in complex-conjugate pairs.
Although the eigenvalue polynomial ensures that an order-N matrix has N eigenvalues, it can occur that there are not N corresponding independent eigenvectors. A matrix with fewer eigenvectors than eigenvalues is called defective. Like singularity, defectiveness represents a delecate balance between the elements of a matrix that can typically be disturbed by just an infinitesimal perturbation of elements. Because of round-off-error, then, floating-point algorithms cannot reliably identify defective matrices. Instead, this method will return a full set of eigenvectors, but some eigenvectors, corresponding to very nearly equal eigenvalues, will be very nearly parallel.
While a generic square matrix can be defective, many subspecies of square matrices are guaranteed not to be. This includes Markov matrices, orthogonal matrices, and symmetric matrices.
Determining the eigenvalues and eigenvectors of a matrix is an O(N3) operation. If you need only the eigenvalues of a matrix, the Eigenvalues method is more efficient.
| 539
| 2,373
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| 3.484375
| 3
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CC-MAIN-2018-17
|
latest
|
en
| 0.837314
|
https://g.vovososo.com/search?q=RMS%20calculator
| 1,627,132,905,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00053.warc.gz
| 293,865,661
| 54,924
|
搜索结果
RMS calculator
This calculator calculates the RMS Voltage value from either the peak voltage, the peak-to-peak voltage, or the average voltage.
其他用户还问了以下问题
Root Mean Square Calculator is an online statistics tool for data analysis programmed to calculate the RMS or Root Mean Square or Quadratic Mean for set of ...
This RMS Voltage calculator helps to find the RMS voltage value from the known values of either peak voltage, peak-to-peak voltage or average voltage.
This RMS voltage calculator can be used to determine the root mean square (RMS) voltage values of the most frequently employed periodic waveforms; ...
RMS stands for Root Mean Square. RMS is a tool which allows us to use the DC power equations, namely: P=IV=I*I/R, with AC waveforms, ...
Sine wave with DC offset： Vrms= *sqrt(Vdc2+Vpk2/2)
Half rectified wave： Vrms= Vpk/2
Full rectified wave： Vrms= Vpk/sqrt(2)
Half sine with duration T and frequency f： Vrms= Vpk*sqrt(f*T/2)
Compute random acceleration, velocity, and displacement values from a breakpoint table with Vibration Research's Random RMS Calculator.
RMS voltage calculator. Examples; Random. Have a question about using Wolfram|Alpha?Contact Pro Premium Expert Support » · Give us your feedback ».
In mathematics, the root mean square (abbreviated RMS or rms) is a statistical measure of the magnitude of a varying quantity. It is also known as the quadratic ...
2020年8月11日 — How to calculate the root mean square by hand? · The first step is to find the square or each number. · Sum those squared numbers: · Divide 100 by ...
Online calculator and formulas for calculating the rms value of a sinusoidal ... The rms value is defined as a DC voltage value with the same thermal effect ...
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| 3.015625
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.717385
|
https://ncertlibrary.com/coordinate-geometry/
| 1,716,873,762,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00895.warc.gz
| 360,312,210
| 8,560
|
# Fundamentals of Coordinate Geometry | Concepts, Coordinate Graph, Quadrants
A plane is any flat surface that can go on infinitely in both directions. Coordinate geometry or analytical geometry is the link between algebra and geometry through graphs having curves and lines. It provides geometric aspects in algebra and enables them to solve geometric problems. Get the detailed information about the coordinate graph, all four quadrants, coordinates of points, others in the following sections.
## Coordinate Geometry Definition
Coordinate geometry is one of the branches of geometry where the position of a point is defined using coordinates. Using the coordinate geometry, you can calculate the distance between two points, find coordinates of a point, plot ordered pairs, and others. The basic terms of coordinate geometry for class 8 students are listed below.
• Coordinate Geometry Definition
• Coordinates of a Coordinate Geometry
• Coordinate Plane
### What is meant by Coordinate and Coordinate Plane?
A coordinate plane is a two-dimensional plane created by the intersection of two axes names horizontal axis (x-axis) and the vertical axis (y-axis). These lines are perpendicular to each other and meet at the point called origin or zero. the axes divide the coordinate plane into four equal sections, and each section is known as the quadrant. The number line which is having quadrants is also known as the cartesian plane.
A set of values that represents the exact position on the coordinate plane is called coordinates. Usually, it is a pair of numbers on the graph denoted as (x, y). Here, x is called the x coordinate, y is called the y coordinate.
• Quadrant 1: In this quadrant both x and y are positive. The point is represented as (+x, +y).
• Quadrant 2: X-axis is negative and the y-axis is positive. So, the point is shown as (-x, +y).
• Quadrant 3: Here, both x and y axes are negative. The point in Q3 is represented as (-x, -y).
• Quadrant 4: In this quadrant, x is positive, and y is negative. Coordinates are (+x, -y).
### How to Plot Coordinates of a Point on Graph?
Following are the simple steps to plot coordinates of a point on a graph. Have a look at them and check out how to plot a graph, represent ordered points, and identify signs of axes of a point.
• First of all, take a point which is having both x coordinate ad y coordinate.
• And know the signs of each value in the given point.
• Using those signs, identify under which quadrant the point falls.
• From the respective axes, take those numbers and put a dot on the graph.
### Linear Equation
The general form of a line in the coordinate geometry is Ax + By + C = 0
Where A is the coefficient of x
B is the coefficient of y
And C is the constant value.
Intercept form of a line is y = mx + c. Where (x, y) is a point on the line and m is the slope.
### Graph of Area vs. Side of a Square
To plot a graph of the square area and square side, you need to have the square area for each side length. Measure the square side length each time, find the area by performing the square of side length. Note down those values and take side length on the x-axis, area on the y-axis. As the square side is always positive, the points will automatically come in the first quadrant. Mark points such as side length as x-coordinate and area as y-coordinate for each point in the graph. Join those points to make a graph of area vs square side length.
### Graph of Distance vs. Time
Drawing the graph for area vs side of a square and distance vs time is the same. Here, we are checking the distance traveled by an object in a certain amount of time, what happens when a change happens in either time or distance. Make sure that, the unit of both time and distance must be the same, if not convert them into the same unit. Take distance on the y-axis and time on the x-axis, so the x coordinate will the time, and the y coordinate will be the distance. Mark those points in the first quadrant and join them to draw a graph of distance vs time.
### Example Questions
Example 1.
Plot that the points A (0, 0), B (1, 1), C (2, 2), D (3, 3) and show that these points form a line?
Solution:
Given Points are A (0, 0), B (1, 1), C (2, 2), D (3, 3)
Point A (0, 0) is the origin.
From the graph, we can say that the points form a straight line and that line passes through the origin.
Example 2.
Plot each of the following points on a graph?
a. (5, 2) b. (8, 0) c. (-5, -2) d. (9, -1)
Solution:
Given points are (5, 2), (8, 0), (-5, -2), (9, -1)
Coordinates of the point (5, 2) both abscissa and ordinate are positive so the point lies in the first quadrant. On the x-axis, take 5 units to the right of the y-axis and then on the y-axis, take 2 units above the x-axis. Therefore, we get the point (5, 2).
Coordinates of the point (8, 0) both abscissa and ordinate are positive so the point lies in the first quadrant. On the x-axis take 8 units and take 0 units on the y-axis to get the point (8, 0).
For the point (-5, -2), both abscissa and ordinate are negative so the point lies in the third quadrant. Take -5 units on the x-axis and take -2 units on the y-axis to obtain the point (-5, -2).
The point (9, -1) lies in the fourth quadrant because x-coordinate is positive and y-coordinate is negative. To plot this point, take 9 units on the x-axis and take -1 units on the y-axis.
Example 3.
Draw the graph of the linear equation y = x + 1?
Solution:
Given linear equation is y = x + 1
The given equation is in the form of y = mx + c
slope m = 1, and constant c = 1
By using the trial and error method, find the value of y for each value of x.
If x = 0, y = 0 + 1, then y = 1
If x = 1, then y = 1 + 1 = 2
If x = 2, then y = 2 + 1 = 3
x 0 1 2
y 1 2 3
Plot the graph for the points mentioned in the above table.
Mark the points (0, 1), (1, 2), (2, 3) on the graph.
Join those points to get a line equation.
### FAQs on Coordinate Geometry
1. Why do we need coordinate geometry?
Coordinate geometry has various applications in real life. Some places where we use coordinate geometry is in integration, in digital devices such as mobiles, computes, in aviation to determine the position and location of airplane accurately in GPS, and to map the geographical locations using longitudes and latitudes.
2. Who is the father of Coordinate Geometry?
The father of coordinate geometry is Rene Descartes.
3. What is the name of horizontal and vertical lines that are drawn to find out the position of any point in the Cartesian plane?
The name of horizontal and vertical lines that are drawn to find out the position of any point in the Cartesian plane is determined by the x-axis and y-axis respectively.
4. What is Abscissa and Ordinates in Coordinate Geometry?
Abscissa and Ordinates are used to identify the position of a point on the graph. The horizontal value or x-axis is called the abscissa and the vertical line or the y-axis is called the ordinate. For example, in an ordered pair (1, 8), 2 is abscissa and 8 is ordinate.
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|
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| 4.71875
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https://www.physicsforums.com/threads/elementary-problems-that-still-baffle-me.169364/
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# Elementary problems that still baffle me
1. May 8, 2007
### Gib Z
Well There are numerous problems, I'll post up a few at a time. This is part of a larger assignment and I am so ashamed to say that I don't have any idea on some of them.
1) Factor $x^8+2x^4y^4+9y^8$
and $a^4+b^4+c^2 - 2(a^2b^2 +a^2c+b^2c)$
I thought maybe it was some perfect square, the first one. I noticed the powers were quite coincidental so i let u=x^4 and v=y^4, but it seems there are no numbers that multiply to 9 and add to 2...I think thats irreducible. The second one I hoped I would notice its some expansion of something familiar, but I didnt get it..
Edit: I just tried Mathematicia and get $(x^4-2x^2y^2+3y^4)(x^4+2x^2y^2+3y^4) \mbox{and} (a^2-2ab+b^2-c)(a^2+2ab+b^2-c)$ But I want to know how to get them manually!
2) Prove $20^{22}-17^{22} +4^{33} -1$ is divisible is 174.
I think i might need some mod arithmetic or something here, but I haven't done that before. Also, I noticed the differences, 20-17, 4-1, are the same, 3, which 174 is divisible by. And perhaps even more useless, that 174 is a 17 and 4 put together, which are in there...I tried expanding 20^22-17^22 and 4^33 -1 by $x^n-1=(x-1)(1+x+x^2+x^3\cdots +x^{n-2}+x^{n-1})$ and the similar one for x^n-y^n, but no go.
Edit: Once again mathematicia says the actualy number is 40768477197265827835864143774, and when divided by 174 gives 234301593087734642734851401, but I want to know how to do it.
3) Prove that if p,q,r and s are odd integers, then $x^{10}+px^9-qx^7+rx^4-s=0$ has no integer roots.
Just no idea on that one, none at all.
4) For which real values of b do the equations $x^3+bx^2+2bx-1=0 \mbox{and} x^2+(b-1)x+b=0$ have a common root?
I tried using the quadratic formula on the quadratic, and the discriminant was b^2-6b+1, which doesn't seem to help me in any way.
5) Prove $a^4+b^4+c^4 >= a^2bc+b^2ac+c^2ab$ for all real a,b,c.
Now usually when I encounter these type of questions It always happens to work that I minus the RHS from both sides, and the LHS can be factorized into some perfect square hence more or equal to zero. But I can't factorize it this time!
I know that if I get desperate I would use some multivariable calculus and find the global extrema of the function, But I would like to know how to do this with precalc methods thanks..
6) Prove that if real numbers a, b and c satisfy $a+b+c > 0, ab+ac+bc>0, abc > 0$ then each a, b and c are positive.
I can't believe I can't do these...Sorry for the hassle guys, thanks!
Last edited: May 8, 2007
2. May 8, 2007
OK, so I don't know much about rigorous mathematical proofs but I can kinda intuitively prove 6).
$$abc > 0$$ says that
-all 3 terms are positive
or
-one is positive and two are negative.
Let's suppose 2 of them are negative.
$$a + b + c > 0$$ says that the positive one is greater than the |sum of the negative ones|.
If we now look at $$ab + ac + bc > 0$$, only one of the terms is positive. Furthermore it's the product of the two smaller numbers. So it clearly can't be true for our initial assumption of 1 positive, 2 negative.
So they must be all positive.
3. May 8, 2007
### Gib Z
I don't get how you finished it... Lets just say we choose b and c to be the negative ones. then a(b+c) + bc > 0 ----> -a|b+c| > -bc ----> a|b+c| > bc
But that doesn't get me the answer! the negative ones arent "smaller" is magnitude, just negative. Thanks for the help so far.
4. May 9, 2007
When you've got $$-a|b+c| > -bc$$, if you want to remove the minus signs you have to flip the inequality sign. So you end up with $$a|b+c| < bc$$.
OK, now consider $$a + b + c > 0 \Rightarrow a > |b + c|$$.
We had $$a|b+c| < bc$$, so now that we know that $$|b + c| < a$$, we can replace $$a$$ with $$|b + c|$$ and our inequality will still have to be true.
This says $$|b + c|^2 < bc$$. Expanding out and taking all the terms to one side we get the following contradiction: $$b^2 + bc + c^2 < 0$$
Last edited: May 9, 2007
5. May 9, 2007
### Gib Z
Ok I get most of it, but the "Expanding out" bit confused me. Did you use the triangle inequality or something? If you did, i dont think it could work because |b+c| </= |b|+|c|, so that so the directions of the arrows dont match up if you get what i mean.
6. May 9, 2007
### Hurkyl
Staff Emeritus
Well, the right hand side is even.
7. May 9, 2007
### nrqed
If we focus on x as the unknown, notice that the polynomial contains x^8, x^4 and an x independent piece. This shows that you may use the quadratic formula for the variable x^4. This trick woul dwork also if you had an x independent term, a x^3 and a x^6, etc. Notice that it works also for the second polynomial since it contains a^4, a^2 and a independent terms.
8. May 9, 2007
### VietDao29
For the first one, you can try to complete the square, like this:
x8 + 2x4y4 + 9y8
= (x4 + 3y4)2 - 4x4y4.
Hopefully, I think you can go from here, right? :)
The second one can be done in the same manner, notice that:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc.
When you see something to the power of an even number (i.e 2n), and something else to the power of half of that number (i.e, n), and some have the factor 2 in front of it. You should think about completing the square.
9. May 9, 2007
### NateTG
Hmm... 174=2*3*29 so if you show that
$$20^{22}-17^{22}+4^{33}-1$$
is divisible by 2,3, and 29, then you're done.
Each of those is individually easy using modular arithmetic.
10. May 9, 2007
### Gib Z
Thanks so much for the help so far guys.
Hurkyl -
If even, the 4 first terms of the polynomial must evalute to an even number as well. this implies s is even, but s is odd. Contradiction.
If odd, the the 4 first terms also evaluate to an even number, same contradiction. Thanks for that hint :D
nrged- I tried the quadratic formula but I also get a discriminant term that is not a perfect square and I can't simplify it, so it doesn't help me.
VietDao29- Completing the square is what I should have thought of first :) Thanks for that, Differences of 2 squares after that :) As for the expansion of the (a+b+c)^2, would it be better to just know that or should I have worked it out this time?
NateTG- I haven't done modular arithmetic before, But ill go on wikipedia and try it now. Thanks
11. May 9, 2007
### Hurkyl
Staff Emeritus
Actually, you have. Even/odd is the simplest nontrivial example of modular arithmetic.
12. May 10, 2007
### Gib Z
I know what it is now, But I can't apply it :'( So its like the remainder, but the powers on the numbers confuse me...
13. May 10, 2007
### NateTG
Let's try it this way:
Let's say we want to know what the remainder is when we divide
$17^{22}$ by $29$
We know that
$$17^{22}=17^{16}*17^4*17^2$$
(The exponents on the RHS are powers of 2 which will be convenient in a moment.)
Now, it's easy to work out that
$$17*17=289=29*9+28$$
and then
$$17^4=\left(17^2\right)^2=(29*9+28)^2=(29*9^2)*29+(2*9*28)*29+(28*28)=(n)*29+(28*28)$$
We really don't care about $n$ since we're only looking for a remainder when dividing by 29, so that leaves
$$28*28=784=(27)29+1$$
So
$$17^4=(m)29+1$$
now,
$$17^8=\left(17^4\right)^2=\left((m)29+1\right)^2=(29m^2+2m)29+1=(o)29+1$$
and similarly
$$17^16=\left(17^8\right)^2=(p)29+1$$
substitute those in in the equation:
$$17^{22}=17^{16}*17^4*17^2$$
$$17^{22}=\left(29(p)+1\right)*\left(29(n)+1\right)*\left(29(9)+28\right)=\left(29(q)+28\right)$$
So, clearly the remainder of $17^{22}$ divided by $$29$$ will be $28$.
This may have looked like a lot of work, but really, it was more effort to write stuff out than it was to do the multiplications ($17*17,28*28,1*1 \rm{and} 1*1*28$), or find the remainders.
A similar process will get you the remainders for the other terms, which can then be added together.
14. May 14, 2007
### Gib Z
Thanks tonnes for that guys, especially NateTG :)
I used that method to prove the remainders for the division by 29. For 20^22 I got 20 if i remember, and for 4^33 i got 9 i think.
For divisibly by 2, that can use simpler arguments as you know.
For divisibility by 3, I sort of noticed that the bases in question, 20, 17 and 4, and only 1 off multiples of 3. So basically I used the method NateTG posted, but it was somewhat quicker.
eg $$(3+1)^{33} = ...\mbox{some terms with a factor of 3} ... + 1^{33}$$
When divided by 3, it obviously gives a remainder of 1. :D Similarly for the others. Thanks for the help on that one!
So now its just problems 4 and 5 left.
15. May 16, 2007
### Gib Z
I just worked out another even easier way to see the divisibility by 3. It by expansion of x^n - y^n = (x-y)(blah blah...) - because in both cases 20-17 and 4-1, its 3. :) Since I haven't learned mod arithmetic yet, maybe there was some other way I could have seen divisibility by 29? Just might be easier thats all.
For number 4, I thought maybe solve the cubic with Cardano's method, but Im not meant to know that in class either...
16. Jun 21, 2007
### Gib Z
Ok guys I just found the easiest and expected way I was meant to work out number 2 (the divisibility by 174 one). Basically we show its divisible by 3 and 58. We do 3 via the expansion in the post just before this one, and we do 58 like this:
$$20^{22}-17^{22} + 4^{33} - 1 = 20^{22} + 4^{33} - (17^{22}+1) = 400^{11} + 64^{11} - (289^{11} + 1^{11})$$
Then use the expansion (valid for odd n):
$$x^n+y^n=(x+y)(x^{n-1} - yx^{n-2} + y^2x^{n-3}....+y^{n-1})$$
The first factor, the (x+y) becomes 464 for the first pair, and thats divisible by 58. The 2nd factor, (x+y) becomes 290, gives is also divisble :)
But now I have only 3 days to solve number 4, any ideas because I am fully stumped..
17. Aug 12, 2007
### povatix
These are great problems, but what about the others(3-6)?,it would be nice to know how to solve them :tongue:
18. Aug 12, 2007
### learningphysics
For part 4, assume both equations to be true. try to eliminate terms to get simpler equations (eliminate high order terms)... sort of like the way you'd solve a system of linear equations...
19. Aug 12, 2007
### Kummer
$$a^4+b^4+c^4 \geq a^2b^2+b^2c^2+a^2c^2$$
But,
$$a^2b^2+b^2c^2+a^2c^2\geq a^2bc+ab^2c+abc^2$$
Q.E.D.
20. Aug 13, 2007
### povatix
thanks for the help guys
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Discussion in 'General Electronics Chat' started by Nathan Hale, Feb 21, 2015.
1. ### Nathan Hale Thread Starter Active Member
Oct 28, 2011
125
2
Hi folks! hope all is well.
I opened up my wife's broken CD player / radio a few days ago.
It is the exact one as shown in this link below.
http://www.amazon.com/GPX-BC232R-CD-PLAYER-BOOMBOX/dp/B00T0GFTN8/ref=sr_1_3?ie=UTF8&qid=1424578507&sr=8-3&keywords=GPX BC232R
The sticker on the back of the device says that it needs 6 "C" batteries of 1.5 volts each. well that would mean this thing takes 9 volts to run. BUT...
When i opened up the radio and connected my Oscilloscope to the wires that are on the other end of the transformer it is showing a peak voltage of about 17.8 volts!
When i took out the O scope probes and stuck a multimeter that measures AC voltage, it said that there were 12.45 volts!!!Here is the exact kind of multimeter i used.
http://www.amazon.com/gp/product/B000EVYGZA/ref=oh_aui_detailpage_o05_s02?ie=UTF8&psc=1
a) Why was my o scope showing a peak voltage of 17.8 volts? shouldn't it have said 9 volts?
b) why was the hand held meter showing 12.45 volts? shouldn't it have said 9 volts also?
c) is the voltage needed to run the device different if it is running on DC or AC?
Thank you.
Jan 15, 2015
963
232
a) Why was my o scope showing a peak voltage of 17.8 volts?
b) why was the hand held meter showing 12.45 volts?
c) is the voltage different if it is running on DC or AC?
With the scope you are seeing the AC peak which is 1.414 times the RMS value which is the DC value. 12.45 * 1.414 = 17.6. Note in your image the Vrms value is 12.8 volts.
Possibly it has a regulator in there or just runs fine on 12 VDC or the 9 VDC the batteries provide. Either way, I do not see a problem at this point. Makes sense to me anyway.
Ron
3. ### WBahn Moderator
Mar 31, 2012
17,757
4,800
The 17.8V is the peak amplitude of a sinewave.
The 12.45V is the RMS voltage of that sinewave. The RMS (root-mean-square) is the "effective" voltage in terms of the ability of the voltage waveform to deliver power to a resistive load. For a sinewave, the relationship is Vpeak = sqrt(2)·Vrms.
The point that you are measuring at is the input point for AC power. That almost certain then gets rectified and regulated down to something below 9V. The 9V from the battery probably also gets regulated down to that same lower voltage.
4. ### Nathan Hale Thread Starter Active Member
Oct 28, 2011
125
2
Thank you! Learned something new today. But...Why would the guy who designed the radio put a transformer that is spitting out 12. 5 V instead of the required 9V? Don't you think he would have worried about the extra voltage busting something?
5. ### Nathan Hale Thread Starter Active Member
Oct 28, 2011
125
2
Aha! That answers my question. thank you.
6. ### WBahn Moderator
Mar 31, 2012
17,757
4,800
That input voltage is unregulated and the voltage output of walwart-type supplies tends to be quite a bit higher when unloaded than when loaded. The circuit needs a stable voltage and the regulating circuitry needs some "overhead" voltage in order to accomplish that. The regulator is designed to tolerate a pretty significant variation in input voltage and still work properly. That's not to say that you WANT the input voltage to be too much higher than the minimum that the regulator circuit needs because, if it is a linear regulator, that just means more power that has to be dissipated. In today's world of SMPS (switched-mode power supplies) this is less of an issue.
Jan 15, 2015
963
232
Also, just as a side note, the DMM you are using is a good general purpose meter. I doubt it is a true RMS responding meter but rather an average responding RMS indicating meter. So while the meter does not respond to the RMS value it should do OK with a good sine wave as far as what it indicates.
Yeah, I figure the voltage from the transformer is rectified and there will be a small drop and later maybe regulated.
What exactly is the problem with the baby boom box. I actually have a similar looking one.
Ron
Jan 15, 2015
963
232
The similar looking unit I have laying here is an old Sony. The one I have does not use a wall wart but has a standard 120 VAC power cord. I assumed the original poster's box also ran on mains power sans a wall wart? Wall wart in the box.
Ron
9. ### Nathan Hale Thread Starter Active Member
Oct 28, 2011
125
2
mine doesnt have a wal wart. it has a standard 120 vac power cord.the problem was the CD player.it wouldn't work when u stuck in a cd and hit play.
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# 992 Christmas Factor Tree
Artificial Christmas trees have to be assembled. Sometimes the assembly is easy, and sometimes it is frustrating.
This Christmas tree puzzle can be solved using LOGIC and an ordinary multiplication table, but there’s a good chance it will frustrate you. Go ahead and try to solve it!
Print the puzzles or type the solution in this excel file: 10-factors-986-992
The number 992 also can make a nice looking, well-balanced factor tree:
992 is the product of two consecutive numbers: 31 × 32 = 992.
Because of that fact, 992 is the sum of the first 31 EVEN numbers:
2 + 4 + 6 + 8 + 10 + . . . + 54 + 56 + 58 + 60 + 62 = 992
992 is palindrome 212 in BASE 22 because 2(22²) + 1(22) + 2(1) = 922. That was a lot of 2’s and 1’s in that fun fact!
• 992 is a composite number.
• Prime factorization: 992 = 2 × 2 × 2 × 2 × 2 × 31, which can be written 992 = 2⁵ × 31
• The exponents in the prime factorization are 5 and 1. Adding one to each and multiplying we get (5 + 1)(1 + 1) = 6 × 2 = 12. Therefore 992 has exactly 12 factors.
• Factors of 992: 1, 2, 4, 8, 16, 31, 32, 62, 124, 248, 496, 992
• Factor pairs: 992 = 1 × 992, 2 × 496, 4 × 248, 8 × 124, 16 × 62, or 31 × 32
• Taking the factor pair with the largest square number factor, we get √992 = (√16)(√62) = 4√62 ≈ 31.49603
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# Re: [HACKERS] Multi column range partition table
```On Wed, Jul 12, 2017 at 12:54 AM, Dean Rasheed <dean.a.rash...@gmail.com> wrote:
> On 11 July 2017 at 13:29, Ashutosh Bapat
> <ashutosh.ba...@enterprisedb.com> wrote:
>> + <para>
>> + Also note that some element types, such as <literal>timestamp</>,
>> + have a notion of "infinity", which is just another value that can
>> + be stored. This is different from <literal>MINVALUE</> and
>> + <literal>MAXVALUE</>, which are not real values that can be stored,
>> + but rather they are ways of saying the value is unbounded.
>> + <literal>MAXVALUE</> can be thought of as being greater than any
>> + other value, including "infinity" and <literal>MINVALUE</> as being
>> + less than any other value, including "minus infinity". Thus the range
>> + <literal>FROM ('infinity') TO (MAXVALUE)</> is not an empty range; it
>> + allows precisely one value to be stored — the timestamp
>> + "infinity".
>> </para>
>>
>> The description in this paragraph seems to be attaching intuitive
>> meaning of word "unbounded" to MAXVALUE and MINVALUE, which have
>> different intuitive meanings of themselves. Not sure if that's how we
>> should describe MAXVALUE/MINVALUE.
>>
>
> I'm not sure I understand your point. MINVALUE and MAXVALUE do mean
> unbounded below and above respectively. This paragraph is just making
> the point that that isn't the same as -/+ infinity.
>```
```
What confuses me and probably users is something named min/max"value"
is not a value but something lesser or greater than any other "value".
The paragraph above explains that <literal>FROM ('infinity') TO
(MAXVALUE)</> implies a partition with only infinity value in there.
What would be the meaning of <literal>FROM (MINVALUE) TO ('minus
infinity')</>, would that be allowed? What would it contain esp. when
the upper bounds are always exclusive?
>
>> Most of the patch seems to be replacing "content" with "kind",
>> RangeDatumContent with PartitionRangeDatumKind and RANGE_DATUM_FINITE
>> with PARTITION_RANGE_DATUM_VALUE. But those changes in name don't seem
>> to be adding much value to the patch. Replacing RANGE_DATUM_NEG_INF
>> and RANGE_DATUM_POS_INF with PARTITION_RANGE_DATUM_MINVALUE and
>> PARTITION_RANGE_DATUM_MAXVALUE looks like a good change in line with
>> MINVALUE/MAXVALUE change. May be we should reuse the previous
>> variables, enum type name and except those two, so that the total
>> change introduced by the patch is minimal.
>>
>
> No, this isn't just renaming that other enum. It's about replacing the
> boolean "infinite" flag on PartitionRangeDatum with something that can
> properly enumerate the 3 kinds of PartitionRangeDatum that are allowed
> (and, as noted above "finite"/"infinite isn't the right terminology
> either).
Right. I think we need that change.
> Putting that new enum in parsenodes.h makes it globally
> available, wherever the PartitionRangeDatum structure is used. A
> side-effect of that change is that the old RangeDatumContent enum that
> was local to partition.c is no longer needed.
Hmm, I failed to notice the changes in _out, _equal, _read functions.
The downside is that enum can not be used for anything other than
partitioning. But I can not imagine where will we use it though.
>
> RangeDatumContent wouldn't be a good name for a globally visible enum
> of this kind because that name fails to link it to partitioning in any
> way, and could easily be confused as having something to do with RTEs
> or range types. Also, the term "content" is more traditionally used in
> the Postgres sources for a field *holding* content, rather than a
> field specifying the *kind* of content. On the other hand, you'll note
> that the term "kind" is by far the most commonly used term for naming
> this kind of enum, and any matching fields.
Ok.
>
> IMO, code consistency and readability takes precedence over keeping
> patch sizes down.
--
Best Wishes,
Ashutosh Bapat
EnterpriseDB Corporation
The Postgres Database Company
--
Sent via pgsql-hackers mailing list (pgsql-hackers@postgresql.org)
To make changes to your subscription:
http://www.postgresql.org/mailpref/pgsql-hackers
```
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# DC Permanent Magnet Motors
© Brooke Clarke 2013 - 2016
Background
Testing DC PM Motors (separate Web page)
Basic Theory
Mabuchi Model Numbers
Motors
1.5 - 3 Volt Toy Motor
Silly Happy Crazy Birthday Card Motor
Floppy
Wireless Power Transfer
Pulleys & Gears
Beams
MicroRAX
MakerBeam
OpenBeam
Phidgets
Electronic Speed Controls (ESC) & Related Controls
Chassis
Lego Wheels, Tires, Couplers & Motors
Tamiya RC
Grasshopper
Heli-Max RC Quad Helicopter w/TV Camera
FPV Cameras
Lighthouse
Solar Showcase 360° Turntable
Related
## Background
On my web pages there are a number of antique DC and AC motors. I sometimes get technical questions about them. I'm also interested in very efficient electrical circuits, like the Joule Thief that can light a LED with a dead battery or Miller Solar Engine (MSE) that can be used to power a small DC motor.
DC PM motors make use of magnets and testing them requires special flywheels.
## Basic Theory
June 2015 - I'm interested in the DC PM motors with a 3.17mm (1/8") shaft because they are reasonably priced (\$3 to \$11) and are commonly used in RC vehicles that are about 1 foot in size. The shaft diameter of a motor is related to it's power output, for 1/8" shafts that amounts to about 50 to 200 Watts (746 Watts is 1 horsepower).
For a given shaft size the armature and frame magnetics will be about the same. The main variation is in what size wire is used to wind the armature and that relates to how many turns will fit.
The best short paper on permanent magnet DC motors is "Characteristics of DC Motors.pdf" by Rutgers Prof. Dunbar P. Birnie.
### Mabuchi Model Numbers
Mabuchi Motors - Technical Guide - Part Number List
Decoding Mabuchi part numbers: L1L2-N1N2N3L3L4-N4N5TTTT
L1 Shape: Flat (F), Round (R), Square (S)
L2 Brush: Metal: (A, E), Precious Metal (F), Carbon (C, H, K, S, T, Z)
N1 Armature Diameter: small lto large (J, K, M, N, 1, 2, 3, 4, 5, 6, 7, 8)
N2 Housing Length code:
N3 Number of Armature Poles: 3 (0), 5 (5), 6 (6), 7 (7), 8 (8), 10 (A), 12 (B)
L3 Magnet: C Isotropic (R), Ring Isotropic (T), Anisotropic or Rare-Earth (P, S, V, W), Rubber (C, E, F)
L4 Custom Design: Encoder (W)
N4 wire diameter tenth of mm: If number tenths of mm, if letter A = 1mm
N5 wire diameter hundredths of mm: 0.01mm (1), 0.05mm (D), 0.055mm (Z)
TTT Turns per slot: might be 2, 3 or 4 digits.
Notes:
Stall Torque and Stall Current are proportional to 1/(wire_diameter^2) - I tried this on some data and at first glance I don't see the correlation.
Pwr out (Watts) = [N (rpm) * T (mN-m)]/9550
The 4 cardinal performance points are: No load RPM and Current, Stall Torque and Current.
### Units
K is the relationship between applied voltage (Volts) and the speed in radians (Wiki) per second (symbol w). Since speed is commonly given as Revolutions per Minute that needs to get converted.
RPM / 60 = Rev Per Second (RPS).
RPS * 2 * PI = Radians per second
For example a For the Mabuchi RS-545SH: motor that turns 6800 RPM at 12 Volts has w = 2* PI * 6800 / 60 = 712 radians per second.
The electrical circuit looks like a constant voltage drive, routed through the internal resistance of the armature to a virtual battery whose back EMF (Vbemf) is determined by the motor RPM.
Vbemf = K * 2* PI* Rev/Sec
or
K = Voltage / (radians per sec) , = 12 / 712 = 0.01685
The electrical equation is:
Vdrive = I * Rinternal +Ebemf
Then:
Tstall = K * V / R
"Turns" refers to the number of times a wire is wound around one of the poles of the armature.
For example the Gilbert DC Motor (used in Erector Sets) has a 3-pole armature and each pole maybe a couple dozen turns.
The force on a wire in a magnetic filed depends on the magnetic field strength and the current in the wire. Since the magnetic field is fixed for a given motor the torque depends on the number of turns and the current.
### For example
The Mabuchi RS-550PC (see below) we know at 7.2 V drive the no load speed is supposed to be 13500 RPM.
So K = 2 * PI * 13500 / 60 = 1414
The stall torque is given as 3700 g-cm with a current of 83 Amps. The back EMF at stall is zero, so
R = V / I = 7.2 V / 83A = 0.0867 or 86.7 milli Ohms.
### Measuring R Directly
Note: A direct measurement of R is very difficult. If a DC measurement is made there will be errors because any dissimilar metals will generate DC voltages that are unknown, see my web page on Kelvin (4-wire) resistance measurements. The 50 milli Ohm limit of the HP 34401 DMM is very close to the resistance of the RS-550PC motor.
A direct measurement using an AC test method will have an error because of the inductance of the windings. See the HP 4328 Milliohmmetere web page for an example.
It may be possible to make a swept frequency impedance measurement using the HP 4395A and by fitting the data to an equivalent circuit separate the real and imaginary parts. This may also be possible using an LCR meter that has a series equivalent circuit representation, like the HP 4274 & HP 4275 LCR meters.
### Back EMF (Wiki)
When the armature is stalled there is no back EMF, but as soon as the armature is turning it's generating a back EMF that subtracts from the applied voltage thus reducing the armature current. Vbatt -Vemf = Iarm * Rarm.
Note if the motor has it's shaft turned by an external torque then it will act as a generator and the voltage is exactly the same as the back EMF (assuming the load resistance is very high, i.e. not using the generator to power something, just using a high impedance meter to read the voltage).
Back EMF [Volts] is proportional to (magnetic field strength[Webers]) * (total number of wires[no units]) * (rotational speed[RPM]) / 60 [convert RPM to RevPerSec]
Note: If the power circuit to the motor is opened and a high input impedance voltmeter is across the motor terminals it will read the back EMF.
Tried this on an RS-555 motor where the supply was set to 18.0 Volts. With the power applied the DMM reads 18.0 volts and when the power is disconnected the voltmeter reads about 17.9 Volts and as the motor spins down the voltage drops.
### Stall Current
For the Mabuchi RS-545SH:
Nominal spec: 6800 RPM @ 12 Volts, K= 0.01685
The no load current in in the range 0.25 to 0.3 Amp, Torque (no load) = 0.25A * 0.001765 Volts/RPM =
Voltage Stall Amps No Load Amps Rstall Ohms 3 1.18 0.25 2.54 4 1.60 0.26 2.50 5 2.02 0.28 2.47 6 2.32 0.28 2.58 7 2.78 0.29 2.51 8 3.25 0.29 2.46 9 3.6 0.28 2.50 10 3.7 0.29 2.70
The motor resistance is about 2.5 Ohms including the wires from the power supply.
### Starting Current
In June 2015 I got some RS-775 type motors that have a 5.0mm shaft diameter to work with a family of flywheel/pulleys. When powered by an HP/Agilent/Keysight E3617A power supply (0 to 60 Volts at up to 1 Amp) the motors will not start spinning. But on power supply, like the HP 6038A that can supply higher currents the motor spins up.
### Self Starting
A 2-pole motor may not start when power is applied and will turn in either direction depending on the direction of the initial manual spin. Motors with 3 or more poles will self start and run in the same direction for the same input polarity.
### Efficiency
Efficiency is a measure of the power output compared to the power input.
When a motor is running with no load it has zero efficiency because there's no useful power out. So the lower the power consumed when the motor is running at no load the better.
Note: At "no load" there really is a small load because of bearing friction and wind loss. There are some Youtube videos where the presenter was proud of the high power consumed by his motor when running at "no load", that's a big mistake.
May of the YouTube home built DC motors that claim "over unity" or "zero point energy" are being evaluated on their electrical input power and electrical output power and the mechanical output power is never evaluated. One of the presenters went to some effort to describe how to make a small dynamometer but never showed it being used on any of his "motors". So, the so called "motors" are not really motors but rather DC to DC converters and so are misnamed "motors". Note that measuring electrical power when the current is either DC or a sine wave is straight forward, but when it's a more complex waveform it's difficult to make power measurements. Many of these "motors" can be seen as works of dynamic art and are fun to look at when they are running.
## Motors
### Pager Motor
This Miller Solar Engine (MSE) circuit runs the motor for about 10 seconds. The times vary depending on how bright the light is.
On a cloudy day it might be six minutes between runs of 6 seconds, and in direct noon sun it might be 11 second runs every 10 seconds.
The circuit works by the solar panel (the back of the panel has the MSE ciruict shown in the photo) charges up a 0.33F super cap and when the 1371 gets to it's trip voltage a pin is grounded turning on a transistor that connects the super cap to the motor. The 1371 has it's own capacitor with isolation diode so that it will remain on even after the super cap voltage goes below the trip point.
I expect with some glue circuitry a number of these can be connected in series to provide higher voltages and larger value super caps can be used.
### Maxon A-max
26mm dix x 57mm overall
Motor body length: 44.65mm
2.97mm dia shaft with flat x 10.73mm long
Six threaded holes on shaft end, size & location TBD.
Motor starts to turn at 0.07V and draws 12ma.
V ma RPM 0.07 12 1.0 19 1.5 22 3.0 32
6x
Maxon makes some very high end motors with much better specifications than you get from a pager motor. Instead of the under a dollar that a new pager motor costs, these go for about \$100 each new and are not often on the surplus market. Supposed to have very low no load power and be very efficient.
### Mabuchi RS-550PC
Mabuchi RS-550PC permenant magnet DC carbon brush motor & DT2234A+ digital Optical tachometer Aluminum foil and painters blue tape wrapped around gear for tachometer. Two RS-550PC Motors coupled for measurements Blue tape on coupling to stop reflections, triangle is reflective strip. Dual motor test done using: HP 6038A 0 - 60 V 0 - 10 A Power Supply Two RS-550PC Motors coupled so No. 1 is motor and No. 2 is generator High Power Load from Tamaya Grasshopper RC car Mechanical Speed Control Optical Digital Tachometer DT2234A+
Volts In Amps In Watts In RPM Volts Out Amps Out Watts Out R In R Load 3 0.64 1.9 869 1.96 0 4.7 1.25 3.8 476 0.22 0.037 0.008 2.4 5.8 4 0.74 3.0 1295 2.96 0 5.4 1.54 6.2 721 0.32 0.048 0.015 2.6 6.6 5 0.73 3.7 1661 3.75 0 6.8 1.80 9.0 895 0.42 0.043 0.018 2.8 9.7 6 0.77 4.6 2075 4.70 0 7.8 2.09 12.5 1112 0.61 0.042 0.025 2.9 14.4 7 0.61 4.3 2589 5.86 0 11.5 2.33 16.3 1354 0.64 0.042 0.027 3.0 15.2 8 0.43 3.4 3014 6.80 0 18.6 2.38 19.0 1609 0.71 0.055 0.039 3.4 12.9 9 0.44 4.0 3516 7.92 0 20.5 2.64 23.8 1946 0.85 0.062 0.053 3.4 13.7 10 0.44 4.4 4086 9.06 0 22.7 2.85 28.5 2202 0.90 0.074 0.067 3.5 12.2 11 0.47 5.2 4468 10.06 0 23.4 3.15 34.7 2432 1.01 0.06 0.061 3.5 16.8 12 0.56 6.8 4954 11.05 0 21.3 3.42 41.0 2652 1.08 0.084 0.091 3.5 12.9
RPM = 390.51 * Vin with motor #2 as open circuit load.
This is a carbon brush permanent magnet DC motor (Data sheet).
Some specs:
Operating Voltage 6.0 - 9.6, Nominal 7.2 VDC
No Load Speed 15,300 RPM @ 1.4 Amps
At Max Efficiency: 13, 540 RPM, 10.8 Amp 41.7 mN*m, 425 g*cm, 59.1 Watts
Stall Torque 363 mN*m, 3700 g*cm @ 83 Amp
25mm diameter x 75mm overall length
3.17mm OD x 8.7mm shaft.
2x M3 mth holes 24.3mm apart?
### Mabuchi RS-555PH-3255
Rated voltage - 12 volts DC
No load current - 360 ma.
Rated load torque - 380.3 gr/cm
Stall torque - 1360.3 gr/cm
Stall current - 11.8 amps
Total length - 3 1/4" with shaft and contacts
Motor body - 2 1/4"
Diameter - 1 3/8"
Shaft length - 1/2" long, flat one side
Shaft diameter - 1/8"
### Electronic Gold Mine Motors
p/n
Description
\$
Photo
Body
Shaft
Holes
dia
mm
Len
mm
dia
mm
Len
mm
Spacing
G16126
Speed Regulated Prec DC Motor
1.49
29.78
24.34
1.96
5.25
6x<<M2
cass thrd
20.73?
G16260
Bertsch Micro Motor
3.49
G16280
RF-520C-17410 Efficient DC Motor
Label:
RF-520C-17410
04-06-06
12V 5500rpm
1.99
32.88
27.92
1.45
8.31
6x<M3
26.48
G16580
Enclosed 12VDC Motor
1.49
28.55
41.58
2.04
15.63
2x<M3
36.2?
G17687 2 Motor Drive CD Head Assy
Two motors:
Spindle: BCD3B, 05UD8X31A
24.0 DIA X 11.0
SHFT: 1.96 DIA x 7.7
6xcass thrd
25.84 DIA X 9.91
gear
x 2.95
Optical Assembly
The DC motor does the coarse track positioning
and the Electromagnetic motor does the fine
positioning (movement parallel to rod).
It's also possible to move the lens
up and down for focus.
Mirrors in Optical Assembly
1. Laser Dide
2. half silvered mirror
3. 100% silvered mirror
4. Lens on black plastic frame
5. photo diode
G18050
Super Efficient Solar Motor
Label:
6V. M9I60U22-1 CW
81212-
Sanko Electric Co. Ltd.
1.49
31.53
26.00
1.92
8.80
6x<M3
26.385dia?
G18186
Our Best DC Motor Assortment
1 ea: "RF-520C-17410" G16280
1 ea: "CDX9" G?
1ea: "MH415A"
1ea: "SDC2L"
1ea: "30011 Johnson 371421"
4.00
G18524
Very Efficient Pancake Motor w/Pulley
Cassette motor?
1.49
24.93
10.54
1.16
2.45
6x<<M2
cass thrd
18.64?
G18945
Massive 12VDC Motor
1.49
44.64
77.88
4.00
4.00
9.06
20.6
M4?
30.33?
G18984
Heavy Duty DC Motor
2.49
37.41
58.81
3.13
3.21
13.75
8.48
M3
25.67?
G19249
Micor Pancake Vibrator Motor
Might be brushelss, Allegro A1442 - AN205049
Microchip TD6505 is a 3-phase fan controller with PWM input.
2.49
G19306
Wild Rat/Mouse
3196580 Toy vehicle having resilient supports and self-contained drive means, Robert G Rakestraw, July 26, 1965, 446/484 74/84R 180/7.1 446/3 -
3530617 Vibration driven vehicle, Earle M Halvorson, Sep 29, 1970,
446/484 74/84S 180/7.1 446/3 446/230 0
5.95
G19426
Pager Motor Assortment (3)
4.99
G19468
Swiss Moxon Mini DC Gearhead Motor
41.012.012.00.00.16
17.95
G4586
Mini Cassette Assy
Single motor very much like G18524
2.95
G6622
Small DC Motor Assortment (3)
1ea: "MH415A"
1ea: "CDX9"
1ea: ?
2.00
G9322
Type 3 DC Robot Motor
1.49
27.61
43.67
2.23
13.28
<M3
16.5?
GP21
Robot Motors (3)
1ea: "28002 374941"
1ea: "MH415A"
2.49
Below here not Electronic Goldmine
### 1.5 - 3 Volt Toy Motor
\$12
10 each
2 Permanent magnets & 3-pole armature
11.92
(9.92 a. flats)
16.81
0.95
3.47
2x<cass scrw
8.71?
### Silly Happy Crazy Birthday Card Motor
5139454 Greeting card with movable parts, Michael L. Earnest, Aug 18, 1992,
p/n: 110677
V 08
http://www.ebay.com/usr/dorfmunger
Maxon Motor A-max 12mm (104092A)
• -.75w
• -3 V
• -diameter: 12 mm
• -length: max 20.6 mm
• -no load speed: 13900 rpm
• -stall torque: 1.58 mNm
• -no load current: 21 mA
• -starting current: 789mA
• terminal resistance: 3.8 ohm
• max permissible speed: 12700 rpm
• max continuous current: 464 mA
• max continuous torque: .931 mNm
• max power output at nominal voltage: 565 mW
• max efficiency: 70.6%
• -diameter: 13 mm
• -length: 16 mm
• -motor shaft length: 11 mm
• -max motor shaft diameter: 1.5 mm
• -reduction: 4.1:1
• -reduction absolute: 57/14
• -number of stages: 1
• max continuous torque: .2 Nm
• intermittently permissible torque at gear output: .3 Nm
• max efficiency: 91%
• weight: 11 g
• avg backlash no load: 1.0
• mass inertia: .025 gcm2
\$20/ea
12.94
38.92
2.74
8.31
2x<M3
9.35?
Fasttech 12.780.02
1.5V DC Micro Motor
7.97
(6.05 across flat)
16.07
0.97
4.53
2x<M2
5.72?
Fasttech 12.780.00
6V DC Gear Motor
Label:
Type: 25GA310-100
DC: 6 V
Speed: 50 RPM/Min
Date: 2012.12.15
24.81
46.57
3.95
8.00
2xM3
16.77?
Floppy Disk Drive
(free defective drive)
Top
Bottom spindle motor is center plate
Separated major components
Spindle Brushless DC Motor (Wiki) assy top
Ribbon cable pin #1 at top, Vcc (#6) at bottom
Bottom
Parts salvaged from mother board:
Stepper motor
LED
Dual push button switch
Single push button switch
4.0 MHz resonator for microcontroller
47uF 16V cap(meas: 56uF 1.7 Ohm)
200uH inductor
Opto interrupter
Inside Spindle Motor 12 Poles (coils)
13 poles on PM in lid?
80mm x 80mm Computer Fan
Brushless DC Motor (BLDC)
The fan blade does not want to just
come off like the floppy disk magnet
above.
There are 4 coils and a 4 pin IC marked: ES211 114009
http://ardupilot.com/
1/16 Brushless 4WD Monster Beatle w/ 25Amp System (USA Warehouse)
Hobby King
2040 Brushless Inrunner Motor 4800kv
25A ESC out of stock
1/10 Quanum Skull Crusher 2WD Brushless Monster Truck (RTR)
Other DC Motor web pages:
A.C. Gilbert Erector Set Motor
Wonder Electric Model Boat Motor
Z Electric Co. model motor
Knapp Small DC Motor
Electromagnetic Toy Engine
MESCO 1011 Electric engine
Weeden DC 2-pole Electro-magnetic Machine
Outrunner Brushless DC motor (airplane)
The back plate with 4 tapped holes is stationary,
the silver body and shaft (far end) turn with the propeller.
Note maybe 14 poles?
eBay: A2212 1000Kv Brushless Outrunner Motor For Airplane
Wireless Power Transfer
5 - 12 VDC Input
5 VDC Output up to 0.5 Amps
eBay: Wireless Charging Charger Module Power Supply Coil For Cell Phone DIY TX and RX
6V GA25Y3/0-46.8 156RPM
25
39
3.95
9.78
x2
6V 25GA370-1.30 30 RPM
2012.05.08
25
56
3.96
7.78
x2
6V GB37Y3530-43.8 114 RPM
130815
37
55.24
5.91
15.34
x6
Mabuchi 12V Motor - RS-545SH - Vibrator / Massager / Vacuum Motor - 7130 RPM
this into gyroscope.
• Brand new motor produced by Mabuchi
• Model: RS-545SH-3045
• Operates on 5 - 15 VDC (12V Nominal)
• No load speed: 8,698 RPM
• No load current: 340 mA
• Full load (maximum efficiency @ 12VDC) performance characteristics:
• Speed: 7,130 RPM
• Current: 1533 mA
• Torque: 175.3 g-cm (17.18 mN-m)
• Power output: 11.2 Watts
• 7.0 A stall current @ 12 V
• Stall torque 972.9 g-cm (95.35 mN-m)
• 5 pole armature
• Can be operated in either direction, simply reverse power supply polarity
• Dimensions (not including shaft): 50 mm long x 36 mm diameter
• D-Type shaft dimensions: 16 mm long x 3.2 mm diameter
• Weight: 5.2 oz. (147 g)
• Comes with solder eyelets for power connection
• Two threaded mounting holes on 25 mm centers on drive end
• These motors are used in hair dryers, vibrators, vacuum cleaners, massagers, etc.
35.65
56.53
3.15
15.7
x2
25mm c-c
Mabuchi RS-553SA motor w/ 3.17mm shaft eBay seller: szbaby11
Voltage: 120 -220 VDC
speed: 10,000 - 20,000 RPM
Shaft 3.17mm dia x 8mm long
M3 mtn screws on 25mm bolt circle
Weight: 177g
shown with one of my flywheels
draws low current at 65 VDC (max voltage of my E3617A bench supply)
## Pulleys & Gears
If you know of a source for gears that fit these motor shafts please let me know.
All the suppliers web pages I've seen have a photo but no dimensions for the shaft hole. They do tell you the make or model of the vehicle they fit but NO engineering information.
## Beams & Construction Supplies
T-Slot beams are handy for making stuff.
Comparison of the starter Kits.
Brand Weight Size Longest Beam End Section MicroRAX 2# (includes extras) 2" dia x 3' tube 35-3/8" (898.5mm) MakerBeam 3 # 4 oz 11.5" x 9" x 1" box 11.8" (300mm) OpenBeam 2# 9.5" x 7" x 2" 6" (152mm)
MicorRAX is the lightest weight (strength), OpenBeam is stronger and MakerBeam is strongest of these three (intended for making 3D printers?)
### MicroRAX
MicroRAX - 10x10mm beam, uses M3 fittings but with special narrow nuts (Starter Kit received 11 July 2013)
The 3 beams that come as part of the starter kit are 35-3/8" (898.5mm) long.
I think the Harbor Freight Mini Cut Off Saw will make cutting the beam easy.
2" dia x 3' tube 2 pounds Inside tube Starter Kit + some extras Right Angle bracket Uses standard M3 screws & lock washers. Need 2mm hex bit (not the 1.5mm bit that was also shipped) Starter Kit (3 beams not shown) \$80 Nut plates instead of standard nuts to fit into the small T-slot. Extra Parts in addition to the starter kit \$40 Tri-Corner PCB Standoff Kit Mounting Bracket for 1/4" Joining Plate Truss Style Joining Plate Corner Style
### MakerBeam
MakerBeam.EU - is a 10 x 10mm beam, uses M3 fittings - from Europe (Starter Kit on order)
Like:
When joining two beams to make a right angle corner they fit well. Haven't checked the angle.
Metal brackets.
You can attach nuts to screws and brackets then slide into beam.
Good variety of brackets.
Dislike:
11.5" x 9" x 1" box 3 lbs 4 oz Contents of Starter Kit Corner Joint showing special square headed M3 screw End of Beam with screw
### OpenBeam
OpenBeamUSA - is a 15x15mm beam, uses M3 fittings - hard to get (Received 11 July 2013) \$80
comes with:
2 packages of 8 plastic L Brackets
1 packages of 8 plastic T Brackets
1 bundle of 12 each 6" Beams
A bag of screws
A bag of nuts
A 1mm ball end hex wrench
Like:
Uses what appear to be standard M3 hex head screws and nuts.
The corners of the beam are designed to be the rails for a linear motion system.
You can attach nuts to screws and brackets then slide into beam.
Dislike:
Plastic brackets with sloppy holes.
Burs where beam has been cut.
9.5" x 7" x 2" 2 pound box
80 20 Inc - offers 20x20mm T-slot extrusions and larger sizes, also inch sizes
T-Slots - offers 2x2" or 50x50mm and larger extrusions
Parco Inc - 10x10mm - Joining Plates - fractional fasteners -
MiniTec - they have an 11x19mm (1 slot) & 19x19 (2 slots), not the slot on each face like the above products.
MakerSlide - for linear motion (related to Vslot and/or V rail?)
Faztek - useless web pages
Tamiya Pulley Unit Set TAM70121 - (Kit on order)
Tamiya Pulley (L) Set TAM70141 - (Kit on order)
Tamiya Pulley (S) Set TAM70140 - (Kit on order)
### Phidgets
T-Slot Primer - very innovative ways of making connections, much more so than the above.
20, 30 & 40mm beams, but expensive
## Electronic Speed Controls (ESC) & Related Controls
Very small DC motors can be controlled using transistors but larger motors run with much higher currents and need high current H bridges.
The input to the ESC is typically the pulse width modulated output from an Remote Control (RC) receiver, but some allow for other input formats like serial bytes or packets.
Dimension Engineering - Sabertooth dual 25A motor driver - Kangaroo X2 Motion Controller -
Pro-Bot -BeeBotLogo language bots
Duratrax - DTXM1200 Electronic Speed Controller (on order 11 July 2013)
G.T.Power RC Servo and ESC Tester (AsiaTees Retail store selling G.T. Power equipment) (on order 11 July 2013)
Duratrax Sprint 1 ESC w/Reverse opened Top Blue 4 MHz resonator at right for PIC, rectangular hole pattern for switch? at lower right IRFZ34N HEXFET: 40 milliOhms, 29 Amps (4ea at top, left 2 in parallel, right 2 in parallel) IRL3103 HSXFET: 12 milliOhms, 64Amps (4ea at bottom, left 2 in parallel, right 2 in parallel) Duratrax Sprint 1 ESC w/Reverse opened Bottom Wire connections at bottom left to right: Red Batt-, Blue Motor, White Motor (center big trace), Black Batt-, red/black On/Off sw (BA05 regulator), blk/red/wht servo input 18-pin PIC micro controller Video of the above setup The tester has three modes: 1.Manual: turning the knob moves the servo and changes the motor speed from full reverse - stop - full forward. 2.Neutral: the servo is in the center position and the motor is stopped. 3.Sweep: the servo moves back and forth and the motor cycles between full reverse, stop, full forward, stop, etc. The Electronic Speed Control (ESC) has a 5 VDC output on it's servo cable that powers the tester, so no additional power supply is needed. That same method probably will power a RC receiver with 5 VDC derived from the 7.2 VDC main battery. There is also an On/Off switch attached to the ESC which would also power down the receiver. In the early days of RC instead of the ESC they used a high power wire wound potent driven by a servo. 540-13.5T 3150 Kv Brushless DC Motor & HW 120A Sensored ESC + misc supplies The black stick with red & black wires is a capacitor too big to fit into the ESC. Dual Motor ESC for Tamiya Bullhead Truck
## Lego
Lego has provisions for adding DC PM Motors and this includes shafts, gears, etc.
This was later upgraded to include controllers and sensors.
### Before Controllers there were Motors
Long before controllers there were many electric motors.
First Generation 3-C Cell Battery Box & Motors (connection cables missing) Battery box accepts 2 or 3 prong electrical plug. Battery Box may be x468 2-prong electrical plug connects to battery box and this motor. 8055 motor, different battery box (the 8055 was a long 3 AA box). Maybe 6216m motor These motors have metal strips on bottom and do NOT accept either 2 or 3 prong plugs. Note 3-hole plug and socket. 4.5 Volt system. Note: 2-prong plug and socket, will also fit 3-hole socket.
### RCX
This was the first controller and used an 8 bit microcontroller.
First LEGO Brick controller MINDSTORMS® SDK2.5 Official Lego Mindstorms site no longer has RCX stuff 6 AA cells: 9 Volts IR Tower used as link between computer COM port and the brick Sensor This might be either 71427 or 43362? Lego brick with electrical connections, uses bricks as electrical connection, not prongs. The black LEGO electrical 2x2 blocks are for RCX motors. (Wiki) may be the 2838c01 9 Volt system
### NXT
This is the next upgrade from the RCX controllers. A more powerful uC and improved motors, sensors, etc.
The LEGO NXT series uses a (non standard) 6P6C modular connector. (Wiki)
The NXT mechanical system is different than the classic Lego bumps.
The NXT motors are called "servo motors" because they contain a sensor.
This is really a "sensored" motor in the parlance of the RC world, whether it be a burshed or brushless motor.
NXT schematics
Deriving power from NXT® motor port A by Philo
NXT "servo Motor" 53787 This is what the RC community would call a brushed sensored gear head motor. It contains two stages of planetary gears. The top and bottom three hole groups are on the standard 8mm Lego bump pitch. The "X" hole in the red motor output is standard Lego. The four holes on the red motor output are NXT. Offset notch 6P6C connector
### Power Functions
This is different from NXT. Power Functions motors are not part of the NXT system.
The GlideWheel PF Motor controller by MindSensors.com allows a NXT controller to operate Power Functions motors.
The Glide Wheel has the mechanical interface so it can be connected to the drive face of a PF motor and it contains the in phase and quadrature rotation sensor needed to make what Lego calls a "servo motor". That's to say the Glide Wheel when attached to a PF motor makes the motor appear to the NXT system as a "servo motor". Note: you can install the Glide Wheel on a motor or after some gearing (but be sure that the direction of rotation is the same).
Power Functions™ presentation by Philo & Didier Enjary
6-AA Battery box, Motors w/attached cables Gears, couplings, shafts, etc. 9 Volt system The motor may be the 58120cx1 Lego 8882 motor Note: 4 electrical connections (why?) not aligned with brick bumps. 39mm dia x 48mm long May have 2 stages of planetary gear reduction. Note: No classical Lego bumps.
### EV3
This is the next generation due to be released Fall 2013
Mindstorms EV3 (Wiki)
### Wheels & Tires
Lego has a standardized axle system that goes all the way back to the first generation motors. I started looking at Lego because they offer a lot of choices in wheels and tires that can
be used for non Lego vehicles.
LEGO Bulk Lot of Wheels Axles Tires Rims 100 Pieces (on order 11 July 2013) 14mm dia x 9mm wide NOTE: These wheel DO NOT accept standard LEGO axles! Lego Wheels and Tires Large 56 x 28 ZR * NEW Dark Red (on order 11 July 2013) Lego Gold Wheels & Balloon Tires 43.2 mm 9441 9444 70502 Ninjago Chima (on order 11 July 2013) 4 Lego NXT Technic Wheels TIRES RIMS 37 x 22 (on order 11 July 2013) LEGO 24x43 wheel & tire (Lego size width x radius?) 24mm wide x 82mm dia Wheel: 24mm w x 43mm OD Lego Wheel & Tire (Lego size? 11x21?) 11mm wide x 42mm dia. Wheel: 7mm w x 17mm OD
## Tamiya RC
These are two very old RC vehicles. They both use a number of common parts. I think they were sold without the RC transmitter, receiver and servos since they have different brands for these items. Electronic Speed Controls (ESC) were not available then so a Mechanical Speed Control (MSC) based on a servo was used to move a switch along with a power resistor. There are three speeds: stop, medium and full.
A problem with the MSC is that the contacts do not always make a good electrical connection and at some throttle positions nothing happens.
Maybe disassembling and cleaning will fix this.
### Bullhead Truck 4X4X4 (4 wheel drive - 2 motors & 4 wheel steering)
Tamiya still offers this 1/10 4x4x4 Monster RC Bullhead 58535 & a kit RB889
Two DC motors each 35.5mm dia. Airtronics Tx: AV2R, Rx: 92123 & Servos:.
75.830 MHz.
Airtronics Avenger AV2R 2-Channel Remote Control Transmitter Spectrum Plot from HP 4395A Black platform removed holds Mechanical Speed Control servo and switch deck. The power resistor is to the right. Dark blue item is battery. Servo just to right of battery is for front and rear steering. The Tamiya TEU-106BK ESC is used in the newer versions of this truck. It has separate outputs for two motors. There are other ESCs that also have dual outputs. The RC channel 82 receiver (hanging over side above) does not accept today's 3-wire servo connectors that have the tab. There are no tabs on these servo plugs and the wire colors are different. So the new Dual Motor ESC (below) does not work. Rx is loose (rubber bands disentigrated) sitting on top. Each motor is mounted to a gearbox/differential assembly. Dual Motor ESC for Tamiya Bullhead Truck
### "limited release" 49501 Grasshopper II Super G (93002?)
Same vintage (1992?) as Bullhead Truck above and uses a similar Mechanical Speed Control (MSC) setup with a power resistor.
Why didn't they put a drivers head on top of the MSC?
Date on back of RC transmitter 1992. Maybe that's the vintage of these vehicles? Mechanical Speed Control (MSC) parts removed for ESC The MSC has been replaced with a Duratrx Sprint ESC w/reverse There's a problem with the ESC not going into set-up mode. 15 Aug 2013.
I received the replacement Sprint II ESC and although it does the correct start up calibration did not run well.
See: Duratrax Sprint 2 ESC Problem when Starting
The problem was with the new 3500 mAh Ni-MH batteries not strong enough to supply the current needed by the motor.
The original batteries were Ni-Cad and in the 1200 mAh range.
What's needed is a way to test the battery to see how many flash amps it can supply.
See my comparison test of a bunch of 7.2/8.4 Volt batteries for Maximum Available Current
## Heli-Max 1SQ Vcam Quad Helicopter
Tower Hobbies carries these.
This is a complete Ready To Fly (RTF) quad copter. It's very small and light and about \$130 for the set.
When flying the Heli-Max it's best to apply full throttle to get the chopper up in the air quickly, rather than trying to slowly raise it up.
If the assent is too slow the chopper tends to move horizontally and soon encounters an obstacle.
A similar product is the Crazyflie Nano Quadcopter except that this kit does not have a regular remote, but instead needs a computer and game controller. It also does not have an on board camera and uses a lower capacity battery.
The motors and props may be the same? no.
The motors are 7mm dia x 20 mm long and have a shaft dia of 0.93mm.
There appear to be 4 different props, White CW, White CCW, Black CW and Black CCW.
The white props are on the front and the black on the back. But does it matter where the CW and CCW props go?
The camera (right center) can be manually tilted up or down. Li-Po battery and USB charger Note: left stick down, Mode 2 (half circle at bottom center (press down on right stick to toggle), all four trims are centered, power on prior to connecting battery to Heli-Max. Pressing Video button controls video camera. When red LED is flashing it's recording. Since there's no calendar/clock all the videos will have a bogus time stamp, but they do have unique file names.
The Whirlwind Quad Copter is about \$70 retail, but does not have a TV camera.
## FPV Cameras
There are two types of camera systems. Recording and First Person Viewpoint (FPV). Recording systems might use an SD memory card and FPV systems some type of transmitter that can handle video bandwidths.
uThere - Ruby Flight Control & On Screen Display (OSD) - Ready To Fly (RTF) planes w/Ruby -
DIY Drones - APM:Rover Overview
## Lighthouse
This is from eBay July 2019 with title: "LED Solar Powered Lighthouse Statue Rotating Garden Yard Outdoor Lighting Decor" under \$15 delivered. About 10-1/2" tall.
Solar Path light plus rotating reflector. The motor that drives the rotating reflector seems to be the motor that's used in electric clocks.
First saw this on YouTube BigClive: Really interesting motor inside a solar ornament, 2017-
But this appears to be an improved version" The roof comes off with a slight CCW wist and has the On-Off switch and battery compartment. No screwdriver needed to access battery.
Fig 1 taken after putting the AAA battery on charger.
The motor does not start when the sun goes down. The light turns on but the motor does not have enough torque to get the reflector turning. The LED and wires are not touching the reflector so the problem is with the motor. See Fig 2 the armature was assembled with only 2 magnets when there are spaces for 4 magnets. I think that's the problem. Note I have removed the armature from its normal place and turned it upside down so you can see the magnets.
Fig 1 Fig 2 Armature with only 2 of the 4 magnets that are needed.
## Solar Showcase 360° Turntable
Got this because it uses a clock type DC motor similar to the lighthouse (above). I'm hoping that I can use this motor in the lighthouse, i.e. that it has more power (magnets).
It does have all 4 magnets and an extra gear so might work in the light house. The motor body for the turntable is much larger than the lighthouse motor. Maybe the extra gear and armature can be swapped?
No, when the 4 magnet armature is installed in the lighthouse motor it does not run at all. . . . bummer.
Fig 1 Clear plastic turntable. Fig 2 Four solar panels wired in parallel. Fig 3 Note 3 gears for turntable, but only 2 for lighthouse. Fig 4 Individual components instead (?) of an IC. Fig 5 Lighthouse 2-magnet armature on left Turntable 4-magnet armature on right.
## Related
Flywheel Pulleys for small DC PM Motors with shaft diameters of: 1.0, 2.0, 3.17, 4.0 & 5.0mm.
Gilbert (Erector Set) D.C. Motor -
Wonder Electric Motor 10:1 Gearing
Z Electric Co. 3-pole DC motor
Knapp Small DC Motor
Erector Set - including a very early (unknown) A.C. electric motor
Electro-Magnetic Toy Engine - sounds like a gas or steam engine
Build it YOURSELF!, a REAL ELECTRIC MOTOR - Kit to make a 2 pole D.C. motor & modern version made today & Science First demonstration motor
MESCO 1011 Toy Engine - works like a steam engine
Weeden D.C. Generator -a 2 pole D.C. generator (bought with title DC Motor)
No. 6 Dry Cell - was commonly used to power small permanent magnet DC motors
Self Winding Clock Co. "Western Union" clocks - uses electromagnetic motor to wind spring page 1, page 2 - use 3VDC solenoid and ratchet gear to wind spring
FastTech - low cost DC motors, some geared
The common Drill Press uses T-slots for attaching accessories to the table.
Some 19" Equipment Racks use T-slot technology.
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https://www.gizmodo.com.au/2014/05/heres-the-best-explanation-of-the-monty-hall-problem-yet/
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# Here’s The Best Explanation Of The Monty Hall Problem Yet
To sign up for our daily newsletter covering the latest news, features and reviews, head HERE. For a running feed of all our stories, follow us on Twitter HERE. Or you can bookmark the Gizmodo Australia homepage to visit whenever you need a news fix.
The Monty Hall Problem is a fantastic probability brain teaser based on the television game show Let’s Make a Deal and this video is the best explanation of it you’re likely to find. The problem is simple. You’re on a game show, and you’re given the choice of three doors: behind one door is a car; behind the others, goats.
You choose one door — say, number 1 — and the host (who knows what’s behind the doors) opens another door — say number 3 — which has a goat behind it. You’re then allowed to stick with the door you picked, or choose the other one — number 2 in the example.
The question is: should you switch your choice? The answer is simultaneously blindingly simple and fiendishly counterintuitive. But the above video might be your best chance yet at wrapping your head around it. [YouTube]
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https://www.allaboutcircuits.com/technical-articles/linear-amplitude-modulation-applications/
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Technical Article
# Linear Amplitude Modulation Applications
April 04, 2016 by Donald Krambeck
## In this article, we'll delve into amplitude modulation and what it's used for. Also, we will differentiate between communication methods that do and do not use modulation.
We delved into amplitude modulation and what it's used for. Also, we'll differentiate between communication methods that do and do not use modulation.
A communication that does not use amplitude modulation is baseband communication. Baseband is a signal that has a narrow frequency range and requires no modulation, i.e. without any shift in the range of frequencies of the signal. On the other hand, a communication that does use modulation is a carrier communication. Carrier communications, or carrier, for short, uses modulation to shift the waveform signal (usually sinusoidal) through its amplitude or frequency. Generally, the carrier wave has a higher frequency than the input signal.
### Baseband and Carrier Communication
When talking about amplitude modulation, the term baseband is used to define a band of frequencies of a signal delivered by the source or transducer. In telephone communications, this same baseband is known as the audio band, or band of voice signals. Moving to televisions, the baseband is a video band that occupies space from the 0 to 4.3 MHz range.
As mentioned earlier, baseband signals can be transmitted without being modulated, that is, without a shift if the range of frequencies of the signal of any sort. In baseband communications, the signals have sizeable power at low frequencies, which means that they cannot be transmitted over a radio but are better suited to be transmitted over a pair of wires, coaxial or optical fibers. Two examples are telephone communications that use short-haul pulse-code-modulation (PCM) between two exchanges, and optical fibers that use long-distance PCM. Using modulation is very beneficial in utilizing the broad spectrum of frequencies available. By modulating numerous baseband signals and shifting their spectra to separate non-overlapping bands, one can use all of the available bandwidth through frequency division multiplexing (FDM). Sending signals over a long distance also requires modulation in order to shift the signal spectrum to higher frequencies in order to ensure an efficient power radiation using antennas of a rather large size.
Switching to communications that do use modulation to shift a signal’s frequency spectrum is known as carrier communication. In carrier communications, parameters (amplitude, frequency, or phase) of a certain sinusoidal carrier signal, of high frequency, is altered proportionately to the baseband signal, m(t). From modulation the signal carrier we are provided with amplitude modulation (AM), frequency modulation (FM), or phase modulation (PM). Frequency and phase modulation are actually quite similar and are classified under the category of angle modulation. This type of modulation is used to transmit analog as well as digital baseband signals. Below is an illustration of the three different types of signals after being modulated.
### Analog Modulation Techniques: Double Sideband
Amplitude modulation can be distinguished by knowing that the amplitude, A, of the carier $$A\cos(\omega_{c}t + \theta_{c})$$ varies in proportion to the message signal (baseband) m(t), or the modulating signal. Both the frequency and the phase have constant value, and we can assume that $$\theta_{c} = 0$$. By making the carrier amplitude, A, directly proportional to the modulating signal m(t), the modulated signal is represented now by$$m(t)\cos(\omega_{c}t)$$ as illustrated in Figure 1.2. This specific type of modulation will only shift the spectrum of m(t) to the carrier frequency (Fig 1.2 a).
Hence, if
$$m(t)\Leftrightarrow M(\omega)$$
then
$$m(t=\cos(\omega_{t}t)) \Leftrightarrow \frac{1}{2}[M(\omega + \omega_{c})+ M(\omega - \omega_{c})]$$ (1.2)
Knowing that $$M(\omega - \omega_{c})$$ is $$M(\omega)$$ shifted to the right by a phase of $$\omega_{c}$$ and that $$M(\omega + \omega_{c})$$ is $$M(\omega)$$ shifted to the left by a phase of $$\omega_{c}$$ will provide for a better understanding of this modulation. Having stated that, the procedure of modulation will shift the signal to the left and right by $$\omega_{c}$$. If we denote the bandwidth of the m(t) to be B Hz, then, from Fig. 1.2 c, the bandwidth of the modulated signal must be 2B Hz. Another observation is that if the modulated signal's spectrum is centered at $$\omega_{c}$$, it will have two parts: one that lies about the frequency, $$\omega_{c}$$, also known as the upper sideband (USB), and a portion that lies below the frequency $$\omega_{c}$$, also known as the lower sideband (LSB). Likewise, the spectrum centered at $$-\omega_{c}$$ has an USB and LSB too. This modulated signal does not contain a distinct component of the carrier frequency $$\omega_{c}$$. Due to this, this type of analog modulation technique is denoted as double-sideband suppressed carrier (DSB-SC) modulation.
In Fig. 1.2 c, $$\omega_{c}$$ \geq 2piB if it wants to avoid an overlap of the spectra that is centered at $$\omega_{c} and$$-\omega_{c}$$. However, if$$\omega_{c} < 2piB$$, then the spectra overlap and the information held in m(t) is lost in the modulation process, which in turn makes it impossible to reacquire m(t) from the modulated signal$$m(t)\cos(\omega_{c}t)$$. ### DSB-SC Demodulation If the original signal m(t) wants to be recovered from the modulated signal, it is absolutely necessary to retranslate the spectrum over its original position. This process of recovering the signal from the original modulated signal is known as demodulation, or detection. Looking at Fig. 1.2 c, one can see that the modulated signal spectrum is shifted to the left and right by$$\omega_{c}$$(also multiplied by one-half), the spectrum in Fig 1.2 d is obtained. This spectrum contains the desired baseband spectrum plus an additional spectrum at$$\pm 2\omega_{c}$$, however this is an unwanted spectrum. This spectrum can easily be suppressed by a low pass filter. This process is most similar and almost identical to modulation, consisting of multiplication of the incoming modulated signal$$m(t)\cos(\omega_{c}t)$$by a carrier of$$\cos(\omega_{c}t)$$that is followed by a low pass filter, as illustrated in Fig 1.2 e. This conclusion can be verified in the time domain by observing that the signal e(t) in Fig 1.2 e is:$$e(t)= m(t)\cos^2(\omega_{c}t)=\frac{1}{2}[m(t)+m(t)\cos(2\omega_{c}t)]$$(1.2 a) Thus, the Fourier transform of the signal e(t) can be denoted as$$E(\omega)=\frac{1}{2}M(\omega)+\frac{1}{4}[M(\omega + 2\omega_{c})+ M(\omega - 2\omega_{c})]$$(1.2 b) The above equations show that the signal, e(t), consists of two parts:$$\frac{1}{2}m(t)$$and$$\frac{1}{2}m(t)\cos(2\omega_{c}t)$$with their spectra shown in Fig 1.2 d. This second component's spectrum, being that it is a modulated signal with frequency$$2\omega_{c}$$, is centered at$$\pm 2\omega_{c}$$. This is why this component is suppressed by the low pass filter. The final desired component$$\frac{1}{2}M(\omega)$$, is a low pass spectrum that is centered at$$\omega = 0$$, which means that it will pass through the filter unharmed. This results in the output$$\frac{1}{2}m(t)$$. The inconvenient fraction of 1/2 can be removed in the output by using a carrier of$$2\cos(\omega_{c}t).
There is a possible form of low pass filter characteristics that are denoted as dotted lines in Fig 1.2 d. This method described of recovering the baseband signal is called synchronous detection, or coherent detection, where the same frequency and phase as the carrier is used for modulating the signals. Hence, for demodulation, a local carrier at the receiver in frequency and phase coherence with the carrier used at the modulator needs to be generated first.
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# Most efficient worker time [max/min problem]
A student wrote us with the following (lightly edited):
Hello Matheno!
It’s me again. I am so grateful find you to help me about my question. I have another 2 questions that i want to ask you. I am a little bit curious and i hope that you can teach me with details on this question. Thanks Matheno!
We’ll post her other question as a separate topic.
Hi! Welcome back and thanks for the new questions.
First, let’s translate the question “At what time during the morning is the worker performing most efficiently?” into what I like to call Calculus language.
• How are we measuring efficiency?
There are many ways to measure efficiency, but based on the function you are given, an efficient worker will likely be one who produces more units per hour, and therefore you need to find the time for which the rate of production r(t) of the worker is at its maximum.
The rate of production r(t) is just the derivative of Q(t)
r(t) = Q'(t)
and is a measure of the speed of an average worker in \, \tfrac{\text{units}}{\text{hour }} . So the question you’re trying to answer in more Calculus type language is:
**
What is the maximum of r(t) = Q'(t) for t\gt 0?
**
I’m assuming that since this is a question about finding a maximum point, you have covered maxima and minima in your course. I won’t give you the solution outright, but I will outline the steps you need to take, which you can find here on Matheno: Maxima & Minima - Matheno.com | Matheno.com
1. Find the derivative of the function you are trying to maximize. You are trying to maximize r(t), so in this case you want to calculate the derivative r'(t).
2. Then find all the times t>0 for which r'(t) = 0. You may know this as finding the critical points. Normally you need to look for all the places where r'(t) is undefined as well, but since Q(t) is a polynomial, all of its derivatives are well-defined everywhere.
3. Finally, you need to show that the time(s) you found in part 2 result in either a maximum or a minimum. One way to do this is by checking the sign of r'(t) for times less than the critical point and times greater than the critical point.
Since you’re being asked for a supporting graph or diagram to help explain your solution, it might not be a bad idea to head over to Desmos.com, and reason about this question using a graph. Here’s my graph to get you started: Worker efficiency. We’re interested in finding the point on Q(t) where the slope of the tangent line is at its steepest (positive) slope.
As always, let me know if you have further questions about this problem, or if you want to check your solution with us!
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2014-02-13T03:09:06-05:00
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
For example, exc. 12;
6/9 = (x + 1)/12 => 2/3 = (x + 1)/12 => 24 = 3(x+1) => x +1 = 8 => x =7.
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Q: What is hypotheses?
A: A hypothesis is a tentative statement about the relationship between two or more variables.
Original conversation
User: What is hypotheses?
Question
Updated 7/31/2014 11:44:21 AM
Flagged by Wallet.ro [7/31/2014 11:44:21 AM], Edited by sujaysen [7/31/2014 2:05:18 PM]
Rating
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A hypothesis is a tentative statement about the relationship between two or more variables.
Confirmed by sujaysen [7/31/2014 2:05:19 PM], Rated good by sujaysen
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1
## Chapter 8: Dynamics of Open Chains (Part 1 of 2)
6 个视频 （总计 34 分钟）, 3 个阅读材料, 7 个测验
6 个视频
Lagrangian Formulation of Dynamics (Chapter 8 through 8.1.2, Part 2 of 2)5分钟
Understanding the Mass Matrix (Chapter 8.1.3)5分钟
Dynamics of a Single Rigid Body (Chapter 8.2, Part 1 of 2)6分钟
Dynamics of a Single Rigid Body (Chapter 8.2, Part 2 of 2)4分钟
Newton-Euler Inverse Dynamics (Chapter 8.3)5分钟
3 个阅读材料
Welcome to Course 3, Robot Dynamics10分钟
How to Make This Course Successful5分钟
Chapter 8 through 8.31 小时 30 分
7 个练习
Lecture Comprehension, Lagrangian Formulation of Dynamics (Chapter 8 through 8.1.2, Part 1 of 2)6分钟
Lecture Comprehension, Lagrangian Formulation of Dynamics (Chapter 8 through 8.1.2, Part 2 of 2)6分钟
Lecture Comprehension, Understanding the Mass Matrix (Chapter 8.1.3)6分钟
Lecture Comprehension, Dynamics of a Single Rigid Body (Chapter 8.2, Part 1 of 2)8分钟
Lecture Comprehension, Dynamics of a Single Rigid Body (Chapter 8.2, Part 2 of 2)8分钟
Lecture Comprehension, Newton-Euler Inverse Dynamics (Chapter 8.3)10分钟
Chapter 8 through 8.3, Dynamics of Open Chains1 小时 30 分
2
## Chapter 8: Dynamics of Open Chains (Part 2 of 2)
4 个视频 （总计 16 分钟）, 1 个阅读材料, 6 个测验
4 个视频
Dynamics in the Task Space (Chapter 8.6)1分钟
Constrained Dynamics (Chapter 8.7)4分钟
Actuation, Gearing, and Friction (Chapter 8.9)6分钟
1 个阅读材料
Chapters 8.5-8.7 and 8.91小时
5 个练习
Lecture Comprehension, Forward Dynamics of Open Chains (Chapter 8.5)4分钟
Lecture Comprehension, Dynamics in the Task Space (Chapter 8.6)2分钟
Lecture Comprehension, Constrained Dynamics (Chapter 8.7)4分钟
Lecture Comprehension, Actuation, Gearing, and Friction (Chapter 8.9)4分钟
Chapter 8.5-8.7 and 8.9, Dynamics of Open Chains40分钟
3
## Chapter 9: Trajectory Generation (Part 1 of 2)
3 个视频 （总计 12 分钟）, 1 个阅读材料, 4 个测验
3 个视频
Point-to-Point Trajectories (Chapter 9 through 9.2, Part 2 of 2)3分钟
Polynomial Via Point Trajectories (Chapter 9.3)2分钟
1 个阅读材料
Chapter 9 through 9.31小时
4 个练习
Lecture Comprehension, Point-to-Point Trajectories (Chapter 9 through 9.2, Part 1 of 2)8分钟
Lecture Comprehension, Point-to-Point Trajectories (Chapter 9 through 9.2, Part 2 of 2)2分钟
Lecture Comprehension, Polynomial Via Point Trajectories (Chapter 9.3)6分钟
Chapter 9 through 9.3, Trajectory Generation1 小时 45 分
4
## Chapter 9: Trajectory Generation (Part 2 of 2)
3 个视频 （总计 17 分钟）, 1 个阅读材料, 4 个测验
3 个视频
Time-Optimal Time Scaling (Chapter 9.4, Part 2 of 3)6分钟
Time-Optimal Time Scaling (Chapter 9.4, Part 3 of 3)4分钟
1 个阅读材料
Chapter 9.41小时
4 个练习
Lecture Comprehension, Time-Optimal Time Scaling (Chapter 9.4, Part 1 of 3)6分钟
Lecture Comprehension, Time-Optimal Time Scaling (Chapter 9.4, Part 2 of 3)10分钟
Lecture Comprehension, Time-Optimal Time Scaling (Chapter 9.4, Part 3 of 3)6分钟
Chapter 9.4, Trajectory Generation1小时
## 关于 Modern Robotics: Mechanics, Planning, and Control 专项课程
This Specialization provides a rigorous treatment of spatial motion and the dynamics of rigid bodies, employing representations from modern screw theory and the product of exponentials formula. Students with a freshman-level engineering background will quickly learn to apply these tools to analysis, planning, and control of robot motion. Students' understanding of the mathematics of robotics will be solidified by writing robotics software. Students will test their software on a free state-of-the-art cross-platform robot simulator, allowing each student to have an authentic robot programming experience with industrial robot manipulators and mobile robots without purchasing expensive robot hardware. It is highly recommended that Courses 1-6 of the Specialization are taken in order, since the material builds on itself....
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# Department of Information and Computing Sciences
## Onderwijs Informatica en Informatiekunde
Vak-informatie Informatica en Informatiekunde
# Seminar Pattern recognition
Te lang geleden voor docent- en roosterinformatie
Contents: The course Pattern Recognition is about the classification and analysis of patterns. As an important example we will look at patterns in images and music notation, but the methods are generally applicable. There are numerous applications of pattern recognition techniques, such as industrial inspection (e.g. quality control of materials), biomedical inspection (e.g. chromosome analysis), remote sensing (earth observation), astronomy (galaxy research), and security (e.g. fingerprint and handwriting analysis). Because more and more measured data is generated, the need for automatic analysis also increases. We will look at patterns in two ways: as a collection of features (such as color and image gradient direction) that occur with a certain probability, and as a configuration of geometric primitives (such as points, lines, regions). The two corresponding ways of pattern recognition are statistical and geometrical pattern recognition. Literature: For statistical pattern recognition: Duda, Hart, Stork: Pattern Classification (2nd ed), John Wiley 2001. For geometrical pattern recognition: articles will be made available. Course form: Seminar. After a few introductory lectures by the lecturer, about statistical pattern recognition, and geometrical shape recognition, students present specific articles and book chapters. Exam form: The grade depends on the given presentations (40%) summaries (40%) and a small exam (20%). Minimum effort to qualify for 2nd chance exam: Om aan de aanvullende toets te mogen meedoen is ontbreken van ten hoogte 1 toetsactiviteit toegestaan. Description: Topics that are treated are: probability density functions, Bayesian decision theory, feature space, supervised and unsupervised classification, parametric and non-parametric decision models, geometric patterns, shape similarity measures. Neural networks are not treated to avoid overlap with the course Neural Networks, logic and reasoning aspects are not treated to avoid overlap with the course Probabilistic Reasoning.
wijzigen?
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Search a number
143465466 = 23191258469
BaseRepresentation
bin10001000110100…
…01101111111010
3100222221210121210
420203101233322
5243211343331
622122543550
73361302444
oct1043215772
9328853553
10143465466
1173a89801
1240067bb6
1323951811
14150a7494
15c8dd446
hex88d1bfa
143465466 has 16 divisors (see below), whose sum is σ = 302032800. Its totient is φ = 45304848.
The previous prime is 143465461. The next prime is 143465471. The reversal of 143465466 is 664564341.
It is an interprime number because it is at equal distance from previous prime (143465461) and next prime (143465471).
It is not an unprimeable number, because it can be changed into a prime (143465461) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 629121 + ... + 629348.
It is an arithmetic number, because the mean of its divisors is an integer number (18877050).
Almost surely, 2143465466 is an apocalyptic number.
143465466 is an abundant number, since it is smaller than the sum of its proper divisors (158567334).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
143465466 is a wasteful number, since it uses less digits than its factorization.
143465466 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1258493.
The product of its digits is 207360, while the sum is 39.
The square root of 143465466 is about 11977.7070426689. The cubic root of 143465466 is about 523.4989212801.
The spelling of 143465466 in words is "one hundred forty-three million, four hundred sixty-five thousand, four hundred sixty-six".
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https://jetcareers.com/forums/threads/la-special-flight-rules-area.80494/
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# LA Special Flight Rules Area
#### Zero1Niner
##### Well-Known Member
OK. Got a debate going on this subject. What is the designation of the airspace corridor for the Los Angeles Special Flight Rules Area to transit the LA class Bravo from the south to Santa Monica? Is it E or B, and what cloud clearances are required?
Call Pat Carey and ask him. He designed it.
It is class E because you do not need a clearance to enter. The VFR mini route is class B. The VFR cloud minimums would fall under the Day/Night VFR minimums depending on the condition of the flight.
The VFR cloud minimums would fall under the Day/Night VFR minimums depending on the condition of the flight.
Are you suggesting that VFR weather minimums in class B or E airspace depend on whether it is day or night?
"I'm pretty sure"(without looking it up because I simply can't get up from my comfortable seat) that the class E vfr weather minimums change between the day/night. Or maybe that's class G. I'm hopeless without the Good Book, the FAR/AIM.
OOOO Seanie Boy! Class E only goes to 5,1,1,1 after 10,000...no night/day changes...That my friend is G.
See you soon man!
Seems nuts to me that it would be 1 mile and clear of clouds within that tight a\$\$ corridor. Its GOT to be E cloud clearance requirements. Plus, your not talking to anyone(although you are self anouncing)...seems less and less like B to me.
Seems nuts to me that it would be 1 mile and clear of clouds within that tight a\$\$ corridor...seems less and less like B to me.
Class B airspace requires 3 sm visibility for VFR ops. You guys have read the FAR, right? :bandit:
Read what it says about this on the TAC. It is B. It says you must be VFR to use it. You can not have 2 airspaces in the same spot. And this area is B from SFC/100. As a refrence look over to KVNY. Notice that VNY is [-30] and BUR is [30/48]. So VNY is below 3000 (2999) and BUR is 3000/4800. No need to ask Pat, just look for yourself.
As for the 3SM and COC, if you can't maintain the 132 radial off of SMO while remaining COC, the you can't do it in VFR. BTW, I have sent CFI canidates to Pat for checkrides, and he will tell you this almost exactly verbatum. The reason this corridor was created was to free up radio traffic. It gets realy busy out here in LA, as many of you well know. Somedays when the marine layer is thick, you can't get a word in with SoCal. Everyone wants to go and fly, so they get an IFR climb to VFR on top and cancel. I used to get the "SADDE" climb out of SMO all the time to go and teach over Simi. I think I could probably still tell the guys in the tower to just give me the squak code and read back the clearance w/out hearing it, and get it right. In June, you can almost get all of the "ref to insts." done with a student if they fly twice a week in a SADDE climb.
Although his analogy to the class C overlying KVNY is irrelevant, mshunter is correct; LASFR is class B airspace. (Check out 14 CFR 93.93)
Thank you also, mshunter, for your interesting story about instructing in SoCal.
Although his analogy to the class C overlying KVNY is irrelevant, mshunter is correct; LASFR is class B airspace. (Check out 14 CFR 93.93)
Thank you also, mshunter, for your interesting story about instructing in SoCal.
Not sure of the irrevelance? It was to point out that you can not have two airspaces in the same place, and was to prove that point. hence, class "E" or "G" in the middle of "B." Or "D" of VNY @ 3000 and "C" of BUR at also @ 3000. Bothe airspaces would be at 3000, causing conflict. If you couuld shed some more light on the irrevelance? I'm not trying to be rude, I'm just trying to learn.
What class airspace would you say you are in just north of KASH at 2600 MSL?
Second, what about federal airways that penetrate class B or C airspace? Victor airways are class E by definition, so what type of airspace are you in in that case?
...I don't know the answer either, but I suspect in the first case Nashua has an LOA with Boston approach.
Looking at that chart... this may be the stupidest thing that I have ever said but is it possible that there was a mis-print? Reallly, I see no lines of seperation that would allow the class D to go up to 2700 w/o interfering with the class C that starts at 2500. I'm confused man, being a pilot is ridiculous :banghead:
Looking at that chart... this may be the stupidest thing that I have ever said but is it possible that there was a mis-print? Reallly, I see no lines of seperation that would allow the class D to go up to 2700 w/o interfering with the class C that starts at 2500. I'm confused man, being a pilot is ridiculous :banghead:
I would have to agree. I would definatly like to see the FAA's explaination of this. I can't see how a mistake as big as this would be out there. Everything I was tought throughout my training, and what I now teach, says that there is no way two airspaces can exist in two places at the same time. But as Bernoulli Fan has clearly shown us, ASH and MHT both control the same airspace. Imagine how hard this must bo to coridinate between the two towers.
As far as the Victor airways penetrating B, C or D, I'd have to see one with an MEA of lower than the top of someones airspace. Yes it's possible to be VFR, so no MEA applies, but you would still need a transition through "XXX" airspace when controled. I don't think LAX would allow you to transit through their airspace very eaisly even if on a victor airway. And I don't remember reading anywhere where it says that an airway is is always, or has to be E. As far as the PHAK ch. 14 goes (8083-25A), if it's not already something else, THEN, it's E "Class E airspace extends upward from either the surface or a designated altitude to the [FONT=Helvetica,Helvetica][FONT=Helvetica,Helvetica]14-3 [/FONT][/FONT]overlying or adjacent controlled airspace." So if it's designated B,C,or D, then and only then, is it E. This may sound like gibberish, it's late, and I just finished a night flight, so I am tired. I'll look at again tomorow and see if it makes any sense:crazy:.
ASH and MHT both control the same airspace. Imagine how hard this must bo to coridinate between the two towers.
In practice, Nashua tower instructs pilots to remain at or below 2500, and Boston approach instructs them to remain at or above 2500. Manchester tower wisely stays out of it (outer ring...). Again, I believe they have an LOA.
And I don't remember reading anywhere where it says that an airway is is always, or has to be E.
AIM 3-2-6 (e) (5): The Federal airways are Class E airspace areas and, unless otherwise specified, extend upward from 1,200 feet to, but not including, 18,000 feet MSL.
I guess the depiction of class B, C, or D airspace could be construed to be "otherwise depicted" (the airway would then begin above the other airspace), but I have not been able to find anything specifying that.
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Strange results
Strange results
(OP)
Hi everyone,
I've a problem on my model.
I installed a new ansys version (2019 R1) in a new PC.
I insert a plastic material data and start to do a calculation:
Young Modulus: 17000 MPa
Poisson ratio: 0.35
mm/mm MPa
0 11
0,00129 60
0,00340 95
0,00888 140,5
0,01399 158,3
0,01798 169,1
0,0240 171,5
0,02883 171,9
0,02954 172
This is what i'm doing and i have a metal pin inserted in two metal bushings comolded with structure in plastic material. (With metal i mean "Structural steel" of Ansys)
Contacts are:
- pin and bushes FRICTIONAL
- bushes and hole BONDED
I applied a positive Load of 2kN on Y on the pin and i see a strange behaviour:
According to you what's wrong in the model? According to me i need to have a compression tension below the bushes, and lower tension on bushes.
Thank you!
RE: Strange results
In my opinion the contact definitions are correct. What does the stress-strain curve look like?
There is nothing preventing the pin sliding in x direction. You should apply the load slowly
so the friction should keep it in place.
RE: Strange results
I recreated your model by scaling dimensions by eye. It seems to run fine and the stresses look reasonable. One thing, you say you need compression below the bushes, but show a plot of von Mises stress. You do realize that von Mises stress is always positive? Check the definition and you'll see it does a square root sum of squares thing, so negatives vanish. Take at look at sigma y and see if that makes sense. Also, your peak stress seems high in comparison to the rest of your stresses. take a look at your mesh. If its too course, you'll get faceting effects between the pin and the bush that will give high stress.
Rick Fischer
Principal Engineer
Argonne National Laboratory
RE: Strange results
(OP)
Hi,
thanks L_K and rickfischer51 to reply.
I think that there's a problem on material data input. I inserted data (Stress strain) on "Multilinear isotropic hardening" (Plasticity), is it correct?
The problem is that in steel bushes there's too high stress in comparison with palastic material in contact with bushes.
I tried to make analysis on this model with every component in structural steel and i found something more "Clear". Picture below:
Here bushes are less compressed and it's correct according to me.
I have a dubt to section where i inserted material data.
Thank you
RE: Strange results
Because of the plasticity, it is important to ramp up the load in small increments, first substep should be in the elastic region.
RE: Strange results
Quote:
Young Modulus: 17000 MPa
should it not be 170.000 MPa ???
RE: Strange results
(OP)
Hi L_K,
i inserted some steps:
step X Y Z
0, = 0, 0, = 0,
1, 0, 125, 0,
2, = 0, 250, = 0,
3, = 0, 500, = 0,
4, = 0, 1000, = 0,
5, = 0, 2000, = 0,
But solution is the same. Bushes with a high stress. :(
RE: Strange results
(OP)
Hi klaus.,
no it's correct because it's a plastic structure.
RE: Strange results
WB would step the load automatically, it think, unless it was told not to, at which point it would throw a warning.
I think the data is input correctly, but be aware for MISO you're inputting plastic strain.
What kind of plastic? 17000 Mpa is pretty stiff for a thermoplastic unless it is filled with a lot of fiber. An unfilled thermoplastic should be in the range 2200 to 3200 Mpa, unless its an olefin, in which case its even lower. Long glass fiber reinforcement would typically be 10000 to 12000 Mpa. Also, first line of your material data is 0,11. If I'm interpreting this correctly, at zero plastic strain, the stress is 11 Mpa, or about 1600 psi. That is your yield strength which seems awfully low, especially if its filled with the glass fiber necessary to give you that high elastic modulus.
Rick Fischer
Principal Engineer
Argonne National Laboratory
RE: Strange results
(OP)
Yes it's glass fiber.
But according to you why bushes in structural steel has higher stress in comparison with plastic surface?
Thank you
RE: Strange results
What material model for the pin and bushings? Linear elastic? If so, the steel can have a very high stress because in your simulation, it never yields, but the plastic yields and that limits the stress in the plastic components. Also, our stress levels are somewhat similar, except your max stress in the bushing is much higher than mine. Again, take a look at your mesh. if it is coarse, you may be getting contact at just a few nodes on the contact surfaces, and that will drive your stress up.
I think you have problem with your material data in that it doesnt represent your material very well. As an example, I found a 50% GF nylon 6 with E = 16 GPa an Sy = 225 Mpa (this is dry as molded). Your yield strength is 11 MPa. Bilinear and multilinear material curves in FEA are an idealization, and care is needed when taking data values from test data and assigning them to a finite element material model.
Rick Fischer
Principal Engineer
Argonne National Laboratory
RE: Strange results
(OP)
Hi rickfischer51,
I used for pin and bushings just "Structural Steel NL" got in "Engineering Data" of Mechanical.
I also tried to set for 50& GF "Bilinear Isotropic Hardening", but i didn't have different behaviour.
Tangent modulus is given by the slope from first point (Strain 0, Stress 11MPa) and last one (Strain 0.02954, Stress 172 MPa).
According to you is it correct?
Thank you
RE: Strange results
Miso is in terms of plastic strain, but biso is not. The first point in your biso model is (0,0), the second is (11/17000,11). But I think you are missing the point. In your miso curve, the first point is (0,11). The next point is (.00129,60). That means there is plastic strain for any value greater than zero, so 11 Mpa is the yield point in your model. That is way to low for a glass filled polymer. You have inadvertently made your material way too soft.
Rick Fischer
Principal Engineer
Argonne National Laboratory
RE: Strange results
Rick Fischer
Principal Engineer
Argonne National Laboratory
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# Euclidean subspace: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
Three one-dimensional subspaces (red, green and blue lines) of R2.
In linear algebra, a Euclidean subspace (or subspace of Rn) is a set of vectors that is closed under addition and scalar multiplication. Geometrically, a subspace is a flat in n-dimensional Euclidean space that passes through the origin. Examples of subspaces include the solution set to a homogeneous system of linear equations, the subset of Euclidean space described by a system of homogeneous linear parametric equations, the span of a collection of vectors, and the null space, column space, and row space of a matrix.[1]
In abstract linear algebra, Euclidean subspaces are important examples of vector spaces. In this context, a Euclidean subspace is simply a linear subspace of a Euclidean space.
## Note on vectors and Rn
In mathematics, Rn denotes the set of all vectors with n real components:
$\textbf{R}^n = \left\{(x_1, x_2, \ldots, x_n) : x_1,x_2,\ldots,x_n \in \textbf{R} \right\}$[2]
Here the word vector refers to any ordered list of numbers. Vectors can be written as either ordered tuples or as columns of numbers:
$(x_1, x_2, \ldots, x_n) = \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right]$[3]
Geometrically, we regard vectors with n components as points in an n-dimensional space. That is, we identify the set Rn with n-dimensional Euclidean space. Any subset of Rn can be thought of as a geometric object (namely the object consisting of all the points in the subset). Using this mode of thought, a line in three-dimensional space is the same as the set of points on the line, and is therefore just a subset of R3.
## Definition
A Euclidean subspace is a subset S of Rn with the following properties:
1. The zero vector 0 is an element of S.
2. If u and v are elements of S, then u + v is an element of S.
3. If v is an element of S and c is a scalar, then cv is an element of S.
There are several common variations on these requirements, all of which are logically equivalent to the list above.[4] [5]
Because subspaces are closed under both addition and scalar multiplication, any linear combination of vectors from a subspace is again in the subspace. That is, if v1, v2, ..., vk are elements of a subspace S, and c1, c2, ..., ck are scalars, then
c1 v1 + c2 v2 + · · · + ck vk
is again an element of S.
## Geometric description
Three two-dimensional subspaces of R3. The center point is the zero vector.
Geometrically, a subspace of Rn is simply a flat through the origin, i.e. a copy of a lower dimensional (or equi-dimensional) Euclidean space sitting in n dimensions. For example, there are four different types of subspaces in R3:
1. The singleton set { (0, 0, 0) } is a zero-dimensional subspace of R3.
2. Any line through the origin is a one-dimensional subspace of R3.
3. Any plane through the origin is a two-dimensional subspace of R3.
4. The entire set R3 is a three-dimensional subspace of itself.
In n-dimensional space, there are subspaces of every dimension from 0 to n.
The geometric dimension of a subspace is the same as the number of vectors required for a basis (see below).
## Systems of linear equations
The solution set to any homogeneous system of linear equations with n variables is a subspace of Rn:
\left\{ \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right] \in \textbf{R}^n : \begin{alignat}{6} a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = 0& \ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = 0& \ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && \vdots\,& \ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = 0& \end{alignat} \right\}
For example, the set of all vectors (x, y, z) satisfying the equations
$x + 3y + 2z = 0 \;\;\;\;\text{and}\;\;\;\; 2x - 4y + 5z = 0$
is a one-dimensional subspace of R3. More generally, that is to say that given a set of n, independent functions, the dimension of the subspace in Rk will be the dimension of the null set of A, the composite matrix of the n functions.
### Null space of a matrix
In linear algebra, a homogeneous system of linear equations can be written as a single matrix equation:
$A\textbf{x} = \textbf{0}$
The set of solutions to this equation is known as the null space of the matrix. For example, the subspace of R3 described above is the null space of the matrix
A = \left[ \begin{alignat}{3} 1 && 3 && 2 &\\ 2 && \;\;-4 && \;\;\;\;5 &\end{alignat} \,\right]\text{.}
Every subspace of Rn can be described as the null space of some matrix (see algorithms, below).
## Linear parametric equations
The subset of Rn described by a system of homogeneous linear parametric equations is a subspace:
\left\{ \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right] \in \textbf{R}^n : \begin{alignat}{7} x_1 &&\; = \;&& a_{11} t_1 &&\; + \;&& a_{12} t_2 &&\; + \cdots + \;&& a_{1m} t_m & \ x_2 &&\; = \;&& a_{21} t_1 &&\; + \;&& a_{22} t_2 &&\; + \cdots + \;&& a_{2m} t_m & \ \vdots \,&& && \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; & \ x_n &&\; = \;&& a_{n1} t_1 &&\; + \;&& a_{n2} t_2 &&\; + \cdots + \;&& a_{nm} t_m & \ \end{alignat} \text{ for some } t_1,\ldots,t_m\in\textbf{R} \right\}
For example, the set of all vectors (x, y, z) parameterized by the equations
$x = 2t_1 + 3t_2,\;\;\;\;y = 5t_1 - 4t_2,\;\;\;\;\text{and}\;\;\;\;z = -t_1 + 2t_2$
is a two-dimensional subspace of R3.
### Span of vectors
In linear algebra, the system of parametric equations can be written as a single vector equation:
\left[ \begin{alignat}{1} x& \\ y& \\ z& \end{alignat}\,\right] \;=\; t_1 \!\left[ \begin{alignat}{1} 2& \\ 5& \\ -1& \end{alignat}\,\right] + t_2 \!\left[ \begin{alignat}{1} 3& \\ -4& \\ 2& \end{alignat}\,\right]
The expression on the right is called a linear combination of the vectors (2, 5, -1) and (3, −4, 2). These two vectors are said to span the resulting subspace.
In general, a linear combination of vectors v1, v2, . . . , vk is any vector of the form
$t_1 \textbf{v}_1 + \cdots + t_k \textbf{v}_k\text{.}$
The set of all possible linear combinations is called the span:
$\text{Span} \{ \textbf{v}_1, \ldots, \textbf{v}_k \} = \left\{ t_1 \textbf{v}_1 + \cdots + t_k \textbf{v}_k : t_1,\ldots,t_k\in\mathbf{R} \right\}$
If the vectors v1,...,vk have n components, then their span is a subspace of Rn. Geometrically, the span is the flat through the origin in n-dimensional space determined by the points v1,...,vk.
Example
The xz-plane in R3 can be parameterized by the equations
$x = t_1, \;\;\; y = 0, \;\;\; z = t_2$
As a subspace, the xz-plane is spanned by the vectors (1, 0, 0) and (0, 0, 1). Every vector in the xz-plane can be written as a linear combination of these two:
$(t_1, 0, t_2) = t_1(1,0,0) + t_2(0,0,1)\text{.}\,$
Geometrically, this corresponds to the fact that every point on the xz-plane can be reached from the origin by first moving some distance in the direction of (1, 0, 0) and then moving some distance in the direction of (0, 0, 1).
### Column space and row space
A system of linear parametric equations can also be written as a single matrix equation:
\textbf{x} = A\textbf{t}\;\;\;\;\text{where}\;\;\;\;A = \left[ \begin{alignat}{2} 2 && 3 & \\ 5 && \;\;-4 & \\ -1 && 2 & \end{alignat} \,\right]\text{.}
In this case, the subspace consists of all possible values of the vector x. In linear algebra, this subspace is known as the column space (or image) of the matrix A. It is precisely the subspace of Rn spanned by the column vectors of A.
The row space of a matrix is the subspace spanned by its row vectors. The row space is interesting because it is the orthogonal complement of the null space (see below).
### Independence, basis, and dimension
The vectors u and v are a basis for this two-dimensional subspace of R3.
In general, a subspace of Rn determined by k parameters (or spanned by k vectors) has dimension k. However, there are exceptions to this rule. For example, the subspace of R3 spanned by the three vectors (1, 0, 0), (0, 0, 1), and (2, 0, 3) is just the xz-plane, with each point on the plane described by infinitely many different values of t1, t2, t3.
In general, vectors v1,...,vk are called linearly independent if
$t_1 \textbf{v}_1 + \cdots + t_k \textbf{v}_k \;\ne\; u_1 \textbf{v}_1 + \cdots + u_k \textbf{v}_k$
for (t1, t2, ..., tk) ≠ (u1, u2, ..., uk).[6] If v1, ..., vk are linearly independent, then the coordinates t1, ..., tk for a vector in the span are uniquely determined.
A basis for a subspace S is a set of linearly independent vectors whose span is S. The number of elements in a basis is always equal to the geometric dimension of the subspace. Any spanning set for a subspace can be changed into a basis by removing redundant vectors (see algorithms, below).
Example
Let S be the subspace of R4 defined by the equations
$x_1 = 2 x_2\;\;\;\;\text{and}\;\;\;\;x_3 = 5x_4$
Then the vectors (2, 1, 0, 0) and (0, 0, 5, 1) are a basis for S. In particular, every vector that satisfies the above equations can be written uniquely as a linear combination of the two basis vectors:
$(2t_1, t_1, 5t_2, t_2) = t_1(2, 1, 0, 0) + t_2(0, 0, 5, 1)\,$
The subspace S is two-dimensional. Geometrically, it is the plane in R4 passing through the points (0, 0, 0, 0), (2, 1, 0, 0), and (0, 0, 5, 1).
## Algorithms
Most algorithms for dealing with subspaces involve row reduction. This is the process of applying elementary row operations to a matrix until it reaches either row echelon form or reduced row echelon form. Row reduction has the following important properties:
1. The reduced matrix has the same null space as the original.
2. Row reduction does not change the span of the row vectors, i.e. the reduced matrix has the same row space as the original.
3. Row reduction does not affect the linear dependence of the column vectors.
### Basis for a row space
Input An m × n matrix A.
Output A basis for the row space of A.
1. Use elementary row operations to put A into row echelon form.
2. The nonzero rows of the echelon form are a basis for the row space of A.
See the article on row space for an example.
If we instead put the matrix A into reduced row echelon form, then the resulting basis for the row space is uniquely determined. This provides an algorithm for checking whether two row spaces are equal and, by extension, whether two subspaces of Rn are equal.
### Subspace membership
Input A basis {b1, b2, ..., bk} for a subspace S of Rn, and a vector v with n components.
Output Determines whether v is an element of S
1. Create a (k + 1) × n matrix A whose rows are the vectors b1,...,bk and v.
2. Use elementary row operations to put A into row echelon form.
3. If the echelon form has a row of zeroes, then the vectors {b1, ..., bk, v} are linearly dependent, and therefore vS .
### Basis for a column space
Input An m × n matrix A
Output A basis for the column space of A
1. Use elementary row operations to put A into row echelon form.
2. Determine which columns of the echelon form have pivots. The corresponding columns of the original matrix are a basis for the column space.
See the article on column space for an example.
This produces a basis for the column space that is a subset of the original column vectors. It works because the columns with pivots are a basis for the column space of the echelon form, and row reduction does not change the linear dependence relationships between the columns.
### Coordinates for a vector
Input A basis {b1, b2, ..., bk} for a subspace S of Rn, and a vector vS
Output Numbers t1, t2, ..., tk such that v = t1b1 + ··· + tkbk
1. Create an augmented matrix A whose columns are b1,...,bk , with the last column being v.
2. Use elementary row operations to put A into reduced row echelon form.
3. Express the final column of the reduced echelon form as a linear combination of the first k columns. The coefficients used are the desired numbers t1, t2, ..., tk. (These should be precisely the first k entries in the final column of the reduced echelon form.)
If the final column of the reduced row echelon form contains a pivot, then the input vector v does not lie in S.
### Basis for a null space
Input An m × n matrix A.
Output A basis for the null space of A
1. Use elementary row operations to put A in reduced row echelon form.
2. Using the reduced row echelon form, determine which of the variables x1, x2, ..., xn are free. Write equations for the dependent variables in terms of the free variables.
3. For each free variable xi, choose a vector in the null space for which xi = 1 and the remaining free variables are zero. The resulting collection of vectors is a basis for the null space of A.
See the article on null space for an example.
### Equations for a subspace
Input A basis {b1, b2, ..., bk} for a subspace S of Rn
Output An (nk) × n matrix whose null space is S.
1. Create a matrix A whose rows are b1, b2, ..., bk.
2. Use elementary row operations to put A into reduced row echelon form.
3. Let c1, c2, ..., cn be the columns of the reduced row echelon form. For each column without a pivot, write an equation expressing the column as a linear combination of the columns with pivots.
4. This results in a homogeneous system of nk linear equations involving the variables c1,...,cn. The (nk) × n matrix corresponding to this system is the desired matrix with nullspace S.
Example
If the reduced row echelon form of A is
\left[ \begin{alignat}{6} 1 && 0 && -3 && 0 && 2 && 0 \ 0 && 1 && 5 && 0 && -1 && 4 \ 0 && 0 && 0 && 1 && 7 && -9 \ 0 && \;\;\;\;\;0 && \;\;\;\;\;0 && \;\;\;\;\;0 && \;\;\;\;\;0 && \;\;\;\;\;0 \end{alignat} \,\right]
then the column vectors c1, ..., c6 satisfy the equations
\begin{alignat}{1} \textbf{c}_3 &= -3\textbf{c}_1 + 5\textbf{c}_2 \ \textbf{c}_5 &= 2\textbf{c}_1 - \textbf{c}_2 + 7\textbf{c}_3 \ \textbf{c}_6 &= 4\textbf{c}_2 - 9\textbf{c}_3 \end{alignat}\text{.}
It follows that the row vectors of A satisfy the equations
\begin{alignat}{1} x_3 &= -3x_1 + 5x_2 \ x_5 &= 2x_1 - x_2 + 7x_3 \ x_6 &= 4x_2 - 9x_3 \end{alignat}\text{.}
In particular, the row vectors of A are a basis for the null space of the corresponding matrix.
## Operations on subspaces
In R3, the intersection of two-dimensional subspaces is one-dimensional.
### Intersection
If U and V are subspaces of Rn, their intersection is also a subspace:
$U \cap V = \left\{ \textbf{x}\in\textbf{R}^n : \textbf{x}\in U\text{ and }\textbf{x}\in V \right\}$
The dimension of the intersection satisfies the inequality
$\dim(U) + \dim(V) - n \leq \dim(U \cap V) \leq \min(\dim U,\,\dim V)\text{.}$
The minimum is the most general case[7], and the maximum only occurs when one subspace is contained in the other. For example, the intersection of two-dimensional subspaces in R3 has dimension one or two (with two only possible if they are the same plane). The intersection of three-dimensional subspaces in R5 has dimension one, two, or three, with most pairs intersecting along a line.
The codimension of a subspace U in Rn is the difference n − dim(U). Using codimension, the inequality above can be written
$\max(\text{codim } U,\,\text{codim } V) \leq \text{codim}(U \cap V) \leq \text{codim}(U) + \text{codim}(V) \text{.}$
### Sum
If U and V are subspaces of Rn, their sum is the subspace
$U + V = \left\{ \textbf{u} + \textbf{v} : \textbf{u}\in U\text{ and }\textbf{v}\in V \right\}\text{.}$
For example, the sum of two lines is the plane that contains them both. The dimension of the sum satisfies the inequality
$\max(\dim U,\dim V) \leq \dim(U + V) \leq \dim(U) + \dim(V)\text{.}$
Here the minimum only occurs if one subspace is contained in the other, while the maximum is the most general case.[8] The dimension of the intersection and the sum are related:
$\dim(U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$
### Orthogonal complement
The orthogonal complement of a subspace U is the subspace
$U^\bot = \left\{\textbf{x}\in\textbf{R}^n : \textbf{x} \cdot \textbf{u}=0\text{ for every }\textbf{u}\in U \right\}$
Here x · u denotes the dot product of x and u. For example, if U is a plane through the origin in R3, then U is the line perpendicular to this plane at the origin.
If b1, b2, ..., bk is a basis for U, then a vector x is in the orthogonal complement of U if and only if it is orthogonal to each bi. It follows that the null space of a matrix is the orthogonal complement of the row space.
The dimension of a subspace and its orthogonal complement are related by the equation
$\dim(U) + \dim(U^\bot) = n$
That is, the dimension of U is equal to the codimension of U. The intersection of U and U is the origin, and the sum of U and U is all of Rn
Orthogonal complements satisfy a version of De Morgan's laws:
$(U + V)^\bot = U^\bot \cap V^\bot\;\;\;\;\text{and}\;\;\;\;(U \cap V)^\bot = U^\bot + V^\bot\text{.}$
In fact, the collection of subspaces of Rn satisfy all of the axioms for a Boolean algebra, with intersection as AND, sum as OR, and orthogonal complement as NOT.
## Notes
1. ^ Linear algebra, as discussed in this article, is a very well-established mathematical discipline for which there are many sources. Almost all of the material in this article can be found in Lay 2005, Meyer 2001, and Strang 2005.
2. ^ This equation uses set-builder notation. The same notation will be used throughout this article.
3. ^ To add to the confusion, there is also an object called a row vector, usually written [x1 x2 ··· xn]. Some books identify ordered tuples with row vectors instead of column vectors.
4. ^ The requirement that S contains the zero vector is equivalent to requiring that S is nonempty. (Once S contains any single vector v it must contain 0v by property 3, and therefore must contain the zero vector.)
5. ^ The second and third requirements can be combined into the following statement: If u and v are elements of S and b and c are scalars, then bu + cv is an element of S.
6. ^ This definition is often stated differently: vectors v1,...,vk are linearly independent if t1v 1 + ··· + tkvk0 for (t1, t2, ..., tk) ≠ (0, 0, ..., 0). The two definitions are equivalent.
7. ^ That is, the intersection of generic subspaces U, VRn has dimension dim(U) + dim(V) − n, or dimension zero if this number is negative.
8. ^ That is, the sum of two generic subspaces U, VRn has dimension dim(U) + dim(V), or dimension n if this number exceeds n.
## References
### Textbooks
• Axler, Sheldon Jay (1997), Linear Algebra Done Right (2nd ed.), Springer-Verlag, ISBN 0387982590
• Lay, David C. (August 22, 2005), Linear Algebra and Its Applications (3rd ed.), Addison Wesley, ISBN 978-0321287137
• Meyer, Carl D. (February 15, 2001), Matrix Analysis and Applied Linear Algebra, Society for Industrial and Applied Mathematics (SIAM), ISBN 978-0898714548
• Poole, David (2006), Linear Algebra: A Modern Introduction (2nd ed.), Brooks/Cole, ISBN 0-534-99845-3
• Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
• Leon, Steven J. (2006), Linear Algebra With Applications (7th ed.), Pearson Prentice Hall
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```from copy import deepcopy
from positive_definite_system import positive_definite_system
from utilities import mat_mult, mat_minus
def additional_algorithm(verbose, a, rhs, scaling_limit, lam_iters, mat_iters):
a_copy, rhs_copy = deepcopy(a), deepcopy(rhs)
sol1 = positive_definite_system(verbose, a, rhs, scaling_limit, lam_iters, mat_iters)
rhs_arr = [[el] for el in rhs_copy]
sol1_arr = [[el] for el in sol1]
res_arr = mat_mult(a_copy, sol1_arr)
res_arr = mat_minus(rhs_arr, res_arr)
res = [el[0] for el in res_arr]
a_copy2, res_copy = deepcopy(a_copy), deepcopy(res)
sol2 = positive_definite_system(verbose, a_copy, res, scaling_limit, lam_iters, mat_iters)
rhs_arr = [[el] for el in res_copy]
sol2_arr = [[el] for el in sol2]
res_arr = mat_mult(a_copy2, sol2_arr)
res_arr = mat_minus(rhs_arr, res_arr)
res = [el[0] for el in res_arr]
sol3 = positive_definite_system(verbose, a_copy2, res, scaling_limit, lam_iters, mat_iters)
for i in range(len(a)):
sol3[i] += sol1[i] + sol2[i]
return sol3
if __name__ == "__main__":
import sys
a, rhs = [[3.0, 3.0, 5.0], [3.0, 5.0, 9.0], [5.0, 9.0, 17.0]], [24.0, 40.0, 74.0]
x = [1.0, 2.0, 3.0]
lam_iters, mat_iters, scaling_limit = 30, 25, 10
verbose = False
sol = additional_algorithm(verbose, a, rhs, scaling_limit, lam_iters, mat_iters)
print('\n')
for i in range(len(a)):
print('x{0:d} = {1:7.4f} (exact {2:7.4f})'.format((i + 1), sol[i], x[i]))
# Typical results:
#
# x1 = 1.0009 (exact 1.0000)
# x2 = 1.9964 (exact 2.0000)
# x3 = 3.0017 (exact 3.0000)
```
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Lecture15_Part6
# Lecture15_Part6 - Discussion Faithful to our principal If...
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How can we use our decider for ? The answer here is more difficult. The new decider should first modify the input TM , M , Discussion TM A so the modified TM, , accepts, whenever TM M halts . Since M is a part of the input, the modification must be apart of the computation . 21 1 M Faithful to our principal “ If it ain’t broken don’t fix it” , the modified TM keeps M as a subroutine, and the idea is quite simple: Discussion Let and be the accepting and rejecting states of TM M , respectively. In the modified TM,
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# Content Area: Math
Topic: Multiplication/Division Fact Families
Standards:
CCSS.MATH.CONTENT.3.OA.A.4
Determine the unknown whole number in a multiplication or division equation relating three
whole numbers. For example, determine the unknown number that makes the equation true in
each of the equations 8 × ? = 48, 5 = _ ÷ 3, 6 × 6 = ?
CCSS.MATH.CONTENT.3.OA.C.7
Fluently multiply and divide within 100, using strategies such as the relationship between
multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of
Objective
This lesson focuses on helping students understand the connection between multiplication and
division by introducing fact families. At the end of this lesson, the students will be able to
complete a fact family for a given number
Rationale
This lesson is important because it is the transition for students which uses their knowledge of
multiplication and connects it to the new concept of division. Division is used in everyday life
quite often and is a crucial component of understanding for higher level math.
Pre Assessment
Since I am in the class quite often, I feel pretty confident that the students understand
multiplication at a conceptual level, even if they do not have all of the facts memorized. Going
into the lesson, this understanding is all that I need in order to make the jump to show the
connection between multiplication and division with a manipulative. During my first example, if
there are students who are struggling with multiplication concepts, I will make sure to clear up
the misunderstandings.
Summative Assessment
At the end of the lesson, I will collect record sheet that I have provided for every student to view
their work.
Connections
In my introduction, I will begin by reviewing the concept of equal groups multiplication (ex:4x3 is
4 groups of 3). I will use a manipulative and tell students that we can use what we already know
about multiplication to help us to understand division.
Introduction
To begin, I will start by introducing a problem I have. Tonight, I am going to make pizza for 4
people. There are 12 pieces of pizza total. I know that 3x4 makes 12 and I need to know how I
can use this information to help me divide the pizza
To begin, I will model how to find the first fact family for students using the unifix cubes. I will
then give 12 cubes to each student and ask them to think of a different fact family that could be
made with those 12. They will record this on their sheet and then we will share out as they
finish. I will then have them go with another student and combine their unifix cubes. This time
with 324 they will list all the possible combinations they can find. If time permits or some groups
finish faster than others, I will allow students to form a group of 4 with 48 cubes. This allows the
students to differentiate if it is easier for some than others. During this time, I will be walking
around and asking questions to check for understanding and stretch the thinking of the
students.
Closure
For the last minute or so I will bring students in for a quick discussion on their learning. Since
this is small group work and there is not a lot of time we will not discuss their answers but we
will discuss what we have learned and make sure that students understand the connection
Student Thinking
Students are required to think about multiplication and division conceptually in this lesson with
the use of manipulatives. As students group up with each other, more option become available
and the task increases in complexity. They are able to display this thinking on their worksheet
and through discussion.
Formative Assessment
As the students work in groups I will be walking around and listening in while also checking their
answers. If I hear or see anything that might be a misunderstanding or needs clarification I will
be able to jump and and help at that point.
Accommodations
This lesson is designed so that students are able to get the same content regardless of their
pace. If students are fast, their content will increase in complexity, but the basic ideas will
remain the same. The lesson is also designed so that the students will be in groups with people
close to their level since the groups will be made based on when they finish the first problem.
Due to this I will be able to help students in the slower groups more if they need it. Also, there
are a couple of students that I know may struggle to start. From my experience, If I am able to
give these students personalized attention to help them understand the concept they are able to
work on their own to finish the rest.
Technology
Technology is not really used to enhance this lesson since I am teaching from the corner of the
classroom and there are not really any options connected to what I am going to teach.
Logistics
Directions will be given before I begin to teach in order to help students understand my
expectations. I will also give directions about volume and how to work well in a group with
students before we move to group work. Finally, I will remind students to turn in their work with
their names on it at the end.
For this lesson all that I will need is the worksheets and unifix cubes. Students will need a
pencil.
## The lesson will take about 15 minutes
5 minutes: Introductions
2 minutes: work on own (12)
7 minutes: Group work (24/48)
1 minutes: Final Discussion
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 Help Me ! » Logistic Growth Functions » 2012-12-20 09:54:11
Bit646
Replies: 1
Hello!
I am struggling to understand logistic growth functions. It seems that I am given very few values to derive a solution, there are 5-10 different formulas to derive the solution out there and none of them are really connecting for me. Let me explain:
Lets take the formula: ln(y)-ln(1-y)=kt+C
I don't understand how I can solve anything when my problem only provides the following information:
Initial Population of: 250,000,000
Infected: 10,000
Rate of infection (k): .0075
Find the time that the infection reaches 50%, when will it reach 80%?
Ok, so......
Where do I input 250,000,000? What about 10,000? What is (y)? What is C?
When I tried to isolate (t) to determine the amount of time I got -.00000000000047
Another example is using the same variables, but having the equation in exponential form of:
y(t)=e^kt+C/(1+e^kt+C)
Shuffle my population values and rate of infection if you like. If I could see one example of this problem done out completely I would be set for life. That is how I learn, but I can't find an example anywhere.
I would appreciate any help you can provide, but I am very visual. I have 7 pages of someone talking about the function and it is like reading Greek.
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You are on page 1of 2
# TUGAS PERPINDAHAN PANAS
## EVALUASI DOUBLE PIPE HE
Ketentuan pengerjaan:
Tugas ini digunakan sebagai pengganti presensi minggu ke-13.
Tugas ini adalah tugas kelompok. Setiap kelompok mengerjakan 1 soal sesuai nomor kelompok.
Jawaban di-upload via klasiber dengan format NoKelompok_Tugas1, jangka waktu
pengumpulan adalah mulai hari Jumat, 18 Desember 2015 sampai dengan hari Minggu, 27
Desember 2015 jam 12 siang WIB.
Pengumpulan jawaban adalah per-kelompok.
Sebagai bukti keikutsertaan dalam pengerjaan, jawaban yang dikumpulkan harus disertai
dengan scan tanda tangan masing-masing anggota kelompok.
Tidak ada toleransi untuk pengumpulan di luar batas waktu upload.
Daftar kelompok dapat dilihat di lampiran
1. O-xylene coming from storage at 100 oF is to be heated to 150 oF by cooling 18000 lb/hr of n-
butyl alcohol from 170 to 140 oF. Available for the purpose are five 20-ft hairpin double pipe
exchangers with annuli and pipes each connected in series. The exchangers are 3-by-2-in.IPS.
What is (a) the dirt factor, (b) the pressure drops?
2. 7000 lb/hr of aniline is to be heated from 100 to 150 oF by cooling 10000 lb/hr of toluene with
an initial temperature of 185 oF in 2-by-1-in IPS double pipe hairpin exchangers 15 ft long.
Pressure drops of 10 psi are allowable, and a dirt factor of 0.005 is required. (a) How many
hairpin sections are required? (b) How shall they be arranged? (c) What is the final dirt factor?
3. 6330 lb/hr of toluene is cooled from 160 to 100 oF by heating amyl acetate from 90 to 100 oF
using 15 ft hairpins. The exchangers are 2-by-1½-in. IPS. Allowing 10 psi pressure drops and
providing a minimum dirt factor of 0.004 (a) how many hairpins are required, (b) how shall
they be arranged, and (c) what is the final dirt factor?
4. 100000 lb/hr of nitrobenzene is to be cooled from 325 to 275 oF by benzene heated from 100
to 300 oF. Twenty foot hairpins of 4- by-3-in.IPS double pipe will be employed, and pressure
drops of 10 psi are permissible. A minimum dirt factor of 0.004 is required. (a) How many
hairpins are required? (b) How shall they be arranged? (c) What is the final dirt factor?
5. Ethylene glycol coming from storage at 95 oF is to be heated to 175 oF by cooling 15000 lb/hr
of glycerol 100% from 300 to 150 oF. Available for the purpose are four 20-ft hairpin double
pipe exchangers with annuli and pipes each connected in series. The exchangers are 2½ -by-1
¼ -in.IPS. What is (a) the dirt factor, (b) the pressure drops?
6. 15000 lb/hr of acetic acid is to be heated from 98 to 125 oF by cooling 20000 lb/hr of ethyl
acetate with an initial temperature of 150 oF in 2-by-1 ¼ -in IPS double pipe hairpin exchangers
15 ft long. Pressure drops of 10 psi are allowable, and a dirt factor of 0.005 is required. (a)
How many hairpin sections are required? (b) How shall they be arranged? (c) What is the final
dirt factor?
7. 7600 lb/hr of acetic acid is cooled from 167 to 105 oF by heating ethyl alcohol 100% from 95
to 105 oF using 15 ft hairpins. The exchangers are 2-by-1½-in. IPS. Allowing 10 psi pressure
drops and providing a minimum dirt factor of 0.004 (a) how many hairpins are required, (b)
how shall they be arranged, and (c) what is the final dirt factor?
8. 100000 lb/hr of glycerol 100% is to be cooled from 430 to 225 oF by ethylene glycol heated
from 102 to 290 oF. Twenty foot hairpins of 4- by-3-in.IPS double pipe will be employed, and
pressure drops of 10 psi are permissible. A minimum dirt factor of 0.004 is required. (a) How
many hairpins are required? (b) How shall they be arranged? (c) what is the final dirt factor?
9. N-Butyl acetate coming from storage at 93 oF is to be heated to 180 oF by cooling 50000 lb/hr
of n-butyl alcohol from 200 to 135 oF. Available for the purpose are four 20-ft hairpin double
pipe exchangers with annuli and pipes each connected in series. The exchangers are 3 -by-2 -
in.IPS. What is (a) the dirt factor, (b) the pressure drops?
10. 80000 lb/hr of acetone is to be cooled from 130 to 100 oF by water heated from 95 to 125 oF.
Twenty foot hairpins of 3- by-2-in.IPS double pipe will be employed, and pressure drops of 10
psi are permissible. A minimum dirt factor of 0.004 is required. (a) How many hairpins are
required? (b) How shall they be arranged? (c) what is the final dirt factor?
## DAFTAR KELOMPOK PERPINDAHAN PANAS KELAS D
1 2 3
AHMAD PRASETIO USMAN DARMANTO MUKTAFA AKMAL
KHOIRUZZAKI AL HUSSEIN P AGENG PAMBUDI ANDRI SETIYOKO
WAHYU HIDAYAT ARIO KUSUMO NUGROHO TRIWIBOWO ALAN K
IBNU RASYID RAFSAN YANI SAWALUDIN
DWIKY MUHAMMAD IRSYAD BAYU DWI ATMAJA SUNARTO HARIO SATMOKO
DIAN MAULANA FITRIANTO
4 5 6
WILLY AGUSTIAN ZANO ZAKI YAKOB S ANNISA SHOLICHA HIDAYAT
MUHAMMAD IRFAN ILMI MUH IRWIN SARPIN MILLA NADIA
RIF'ANUL ILMI N RIDHO HALIM NOVA ARINA SILFIYAH
VIKI ROLA S. WAHYU SULISTYAWAN PUTRI MUTIA HERMAN
WAHYU HAMAM HAUFI MUHAMAD RIVAN FAUZI RIKA DESY ARTI
KARUNIA RAHMAWATI
7 8 9
SHINTA WIDYA WIDARYANTO ARI SISWARI TANTRI YUNITA
ANISA MAULIDIA NADHYA CHAIRIZA FITRI ISMA WULANSARI
WAHYU DWI UNTARI PRIMADONA CIKO KARISA ATIK TUTUR KASIH
LENY ISLAMI SARI TRI ASMAWATI NUR LAILY FARIKHA
DEASY RAHMAWAHIDA A MAULIDA HASANAH LU'LU'IL MAKNUN
10
GITA YUNITA SRI PAMBAYUN
DEA TRI ANGGRAENI
NURHAYATI
SITI KHODIJAH
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# What is a dispersion relation?
Well, that's a bit of a tricky question from a standing start. Essentially, it's something that tells you about how a specific type of wave (like a sound waves, or an electromagnetic wave, etc.) will act in a certain medium. So, before going any further, what does one mean by a 'wave'? Here's some maths... (don't be alarmed!) A 'wave' is something which looks like this:
exp(i(k x - ω t))
What does that mean? Well, exp is an exponential. k is one over the wavelength, and ω is the frequency (or the frequency times a constant, strictly). If this doesn't make sense to you, keep reading; I'll try to provide some intuition through examples. A dispersion relation is an equation that will relate ω and k. i, here, is the square root of minus one, and its presence indicates we're looking for waves - things which go up and down.
So, why care about a dispersion relation? Well, in a lot of physical cases, you can't just bung in any values of ω and k - the actual stuff your wave is travelling through has something to say about that. Let's look at a concrete example.
# A concrete example
Consider speaking down a long, rectangular air duct. Air ducts don't like to transmit every sound (making your voice sound 'flatter' and more boring, unless your voice makes the metal vibrate and add tinny sounds of its own). If we do some maths, we get the dispersion relation for a rectangular waveguide:
ω2 - cs2k2 = cs2 a-2(n2 + m2)
Here, the geometry of the problem has taken the usual dispersion relationship for speaking in air (the left hand side) and added in a constraint of these numbers n and m on the right - these numbers are integers, at least one.
That wasn't very interesting.
# An interesting example - the Plateau-Rayleigh Instability
This is going to be more involved. See, earlier, when I said the presence if 'i' indicated we were looking for waves? We aren't necessarily going to find waves - at least not ones that travel. Some waves (called instabilities) don't travel at all, but just grow. Finding instabilities and understanding them is a fundamental part of applied mathematics.
Consider a cylinder of fluid, surrounded by nothing (air is close to being nothing). Imagine, for example, you've got some golden syrup, and you're pouring it nice and slow over some delicious waffles. Well, the vibrations from your hand (and the air currents in the room, and even thermal fluctuations) are going to make ti-iny waves on the surface of that cylinder. Let's see what happens.
For simplicity's sake, say that the long cylinder (which, without the waves, would be radius a) isn't moving, and initially has a tiny wave on its surface - to remind us that it's tiny, let's say ε is a really tiny thing, and so we write:
radius = a( 1 + ε exp(i(kx - ω t)))
All we've done is say, we have a cylinder; it has a little wave on its surface; for simplicity we're going to assume it's not initially moving (although it doesn't matter, actually). In principle, if we felt like doing some work, then we could apply our knowledge of fluid dynamics to remove everything from the problem except k and ω. Well, (skipping ahead) these waves aren't going to grow. We find that for real of k (here, you can pleasingly read 'real' to mean 'allowed'), ω is imaginary. For completeness, I include Lord Rayleigh's original result here (although it's nasty):
i ω = actually, this is too nasty to bother. Something 'real'
So what? You say. Well, this is a basic example of a stability argument. If ω is imaginary, i ω is real, and that means the size of our disturbance can grow. If our waves don't propagate, but stay where they are and grow exponentially with time, we know that a system is unstable (and won't occur in nature - or, not for long).
# But my syrup pouring showed no signs of being unstable!
These instabilities need time to grow. Try pouring it from a metre onto a plate on the floor, and watch it separate into droplets before hitting the plate.
# Summary
A dispersion relationship is something that tells us how waves behave in a medium or situation; it also tells us if a system is unstable.
At least, unless we have a continuous spectrum. But I don't understand them yet.
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A jet airplain travelling at the sp...
Question
# A jet airplain travelling at the speed of ( 500 mathrm{km} / mathrm{hr} ) ejects its products of combination at the speed of ( 1500 mathrm{km} / mathrm{hr} ) relative to jet plane. What is the speed of later wrt observer on the ground?
JEE/Engineering Exams
Physics
Solution
26
4.0 (1 ratings)
Let speed = 500km/hr
Work speed ( =x mathrm{km} / mathrm{hr} )
Relative speed ( =500+x ) [both in opposite direction]
So (,500+x=1500 )
[
x=1000 mathrm{km} / mathrm{hr}
]
Deposit direction SO, reject = - 1000km/hr
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[Edu-sig] Re: Teaching Middle-School Math with Python
Kirby Urner pdx4d@teleport.com
Tue, 10 Oct 2000 15:50:01 -0700
```> Also, I'd think you'd _want_ to cover car and cdr when approaching
> programming in Scheme
>
>Why? That would be silly. Teaching in middle school and even high school
>our objectives are to produce thinking kids who can use some programming
>tool not finished Python/Scheme programmers.
I'll have to look at those docs again. Seems to me that using recursion
as your primary means of looping means you're very often writing stuff
like:
(define (Reduce procedure mylist)
(cond
[(= (length mylist) 1)(car mylist)]
[(procedure (car mylist)(Reduce procedure (cdr mylist)))]
))
Usage:
> (Reduce + '(1 2 3 4))
10
> (Reduce * '(1 2 3 4))
24
In Python:
def Reduce(proc, mylist):
if len(mylist)==1: return mylist[0]
else: return apply(proc, (mylist[0], Reduce(proc, mylist[1:])))
Usage:
>>> import operator
10
>>> Reduce(operator.mul,(1,2,3,4))
24
(Note: for illustration purposes only -- Python has a built-in
reduce function that already does the above).
... just back from the docs. Looks like you're using 'first' and 'rest'
in place of 'car' and 'cdr'. That's good. 'car' and 'crd' are silly
names. But I'm not finding where 'first' and 'rest' get defined.
They're not MzScheme primitives i.e. I get an error message when I
try using either one.
Anyway, not to worry, I'll find this missing puzzle piece eventually.
>Absolutely. Model and view is for us who know that Model is about
>functions, relations, compositions, domain, range, sets, ... (see TLL).
>View is about the stupid i/o crap that most languages impose on kids.
>
>-- Matthias
Yeah, we don't want to start off on the wrong foot with a lot of
stupid i/o crap, I agree.
Thanks for the dialog. Sounds like we're not in any serious disagreement
at this point.
Kirby
```
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6 August, 09:26
# Gale's Shipping Company charges \$3.00 plus an additional \$5.00 per pound to ship a package. Peeta's Shipping Company charges \$12.00 plus an additional \$2.00 per pound to ship a package. Write a system of linear equations that shows the cost in dollars, y, to ship a package of x pounds for each company. I) Write Gale's equation in the answer space below.
+3
1. 6 August, 09:43
0
Gale's Shipping Company Cost, y=\$ (3+5x)
Peeta's Shipping Cost, y=\$ (12+2x)
Step-by-step explanation:
Let the number of pounds being shipped=x
Gale's Shipping Company charges \$3.00 plus an additional \$5.00 per pound to ship a package.
Gale's Shipping Company Cost = \$3.00 + (\$5.00 X Number of Pounds)
Gale's Shipping Company Cost, y=\$ (3+5x)
Peeta's Shipping Company charges \$12.00 plus an additional \$2.00 per pound to ship a package.
Peeta's Shipping Cost = \$12.00 + (\$2.00 X Number of Pounds)
Peeta's Shipping Cost, y=\$ (12+2x)
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Question and answer
A point has no physical characteristics. True or False
TRUE. A point has no physical characteristics.
Expert answered|Wallet.ro|Points 616|
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Asked 6/3/2014 7:39:06 AM
Updated 6/3/2014 7:43:35 AM
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Edited by jeifunk [6/3/2014 7:43:34 AM], Confirmed by jeifunk [6/3/2014 7:43:35 AM]
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# How to check shape of NumPy array
You can check the shape of the NumPy array with the following code. If you want to learn Python, I highly recommend reading This Book.
## Example 1
```import numpy as np
a = np.array([[5,10,15],[20,25,30]])
print(a.shape)```
Output:
`(2, 3)`
## Example 2
```import numpy as np
a = np.array([[5,1]])
print(a.shape)```
Output:
`(1, 2)`
## Example 3
```import numpy as np
a = np.array([5,10,15])
print(a.shape)```
Output:
`(3,)`
## Example 4
```import numpy as np
a = np.array([[5,10,15,20],[20,25,30,50],[35,40,45,60]])
print(a.shape)```
Output:
`(3, 4)`
## Example 5
```import numpy as np
a = np.array([[[5,10,15,20],[20,25,30,50],[35,40,45,60]]])
print(a.shape)```
Output:
`(1, 3, 4)`
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Number 130706
Properties of number 130706
Cross Sum:
Factorization:
Divisors:
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):
1fe92
Base 32:
3vki
sin(130706)
-0.037638504173851
cos(130706)
-0.99929142045929
tan(130706)
0.037665192958979
ln(130706)
11.780705805215
lg(130706)
5.1162955241305
sqrt(130706)
361.53284774692
Square(130706)
Number Look Up
Look Up
130706 (one hundred thirty thousand seven hundred six) is a impressive figure. The cross sum of 130706 is 17. If you factorisate the figure 130706 you will get these result 2 * 65353. 130706 has 4 divisors ( 1, 2, 65353, 130706 ) whith a sum of 196062. The number 130706 is not a prime number. 130706 is not a fibonacci number. 130706 is not a Bell Number. The figure 130706 is not a Catalan Number. The convertion of 130706 to base 2 (Binary) is 11111111010010010. The convertion of 130706 to base 3 (Ternary) is 20122021222. The convertion of 130706 to base 4 (Quaternary) is 133322102. The convertion of 130706 to base 5 (Quintal) is 13140311. The convertion of 130706 to base 8 (Octal) is 377222. The convertion of 130706 to base 16 (Hexadecimal) is 1fe92. The convertion of 130706 to base 32 is 3vki. The sine of the figure 130706 is -0.037638504173851. The cosine of the number 130706 is -0.99929142045929. The tangent of the figure 130706 is 0.037665192958979. The root of 130706 is 361.53284774692.
If you square 130706 you will get the following result 17084058436. The natural logarithm of 130706 is 11.780705805215 and the decimal logarithm is 5.1162955241305. I hope that you now know that 130706 is special figure!
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Regular version of the site
Bachelor 2019/2020
## Algorithms and Data Structures 2
Type:
Area of studies: Applied Mathematics and Information Science
When: 2 year, 1 module
Mode of studies: distance learning
Instructors: Boris I. Goldengorin, Мельников Евгений Юрьевич
Language: English
ECTS credits: 4
### Course Syllabus
#### Abstract
The objective of this course is to make students familiar with polynomially solvable (tractable) and intractable (NP-complete and NP-hard) problems including well known problems in combinatorial optimization and computer science (e.g. the traveling salesman problem and its polynomially solvable special cases; generalized p-median problem and its applications in data analysis; max-clique, basic models of integer linear programming models (discrete optimization models) and enumeration (branch-and-bound type) algorithms for solving general mixed-integer linear programming problems as well as its special cases (minimum spanning tree and its variations, linear assignment, and traveling salesman problem). The algorithmic line of this course will be represented by an introduction to the maximization of submodular set functions (discrete analog of convex functions), and their applications to data analysis problems. We conclude this course with an overview of time and space complexities for polynomially solvable and NP-hard problems. Pre-requisites: Discrete Mathematics 1, Algorithms and Data Structures 1.
#### Learning Objectives
• Students will learn basic concepts of computational complexities for problems and algorithms (both exact and heuristic), which are necessary for further learning computational methods, algorithms for big data, algorithms and computational complexity, machine learning, operations research, game theory, and combinatorial optimization.
• Students will acquire skills in using algorithms and data structures 2 models and methods to formalize and solve applied problems.
#### Expected Learning Outcomes
• Students will learn basic notions of algorithms and data structures to solve intractable problems in computer science.
• Students will be able to find mathematical structures defined on graphs and Booleans (Hasse diagrams).
• Students will acquire basic skills to recognize and prove the time and space complexities for exact and heuristic algorithms.
#### Course Contents
• The Minimum Spanning Tree (Kruskal and Prim’s Algorithms) 1-Tree, and Linear Assignment Problems (Hungarian Algorithm).
• The Traveling Salesman Problem (TSP): Symmetric and Asymmetric Cases of TSP.
• The Branch and Bound Algorithms for the Symmetric TSP (STSP).
• The Branch and Bound Algorithms for the Asymmetric TSP (ATSP).
• Upper, Lower, and Bottleneck Tolerances in Combinatorial Optimization.
• Upper Tolerances Based Algorithm for the ATSP.
• Lower Tolerances Based Algorithm for the ATSP.
• Data Correcting Approach for Real-Valued Functions.
• Data Correcting Approach for ATSP.
• Heuristics for solving the TSP: greedy (nearest neighbor), tolerance based greedy, nearest insertion, 2, 3-opt.
• Introduction to Submodular Functions: Local, Global Maxima on Hasse Diagram and Their Components.
• Supermodular Functions Applied to the Simple Plant (Uncapacitated Facility) Location Problem (SPLP).
• Cherenin-Khachaturov’s Theorem. Excluding vs Preservation Rules. Dichotomy (Preliminary Preservation) Algorithm.
• Branch-and-Bound (BnB) Algorihm for Supermodular Function Minimization Problem and Its Applications to the SPLP.
• Non-binary branching rules applied to the pseudo-Boolean formulation of the SPLP. The branching rule: Make Quadratic Terms Linear.
• The Quadratic Cost Partition Problem (QCP) - an example for maximization of a submodular set function.
• Six equivalent definitions of submodular set functions.
• Cherenin’s Theorem (quasi-concavity of submodular functions).
• The greedy algorithm for submodular set functions.
• Boolean and pseudo-Boolean polynomials.
• Truncation Theorem for the p-Median problem in pseudo-Boolean formulation.
• Overview of BnB type Algorithms: Branch and Cut, Cutting Planes, Branch and Cut and Price, Data Correcting Algorithms, Tolerance Based Branch and Bound, Climer and Zhang's cut-and-solve algorithms.
• The theory of NP-Completeness; Relationship between P and NP classes; Polynomial transformations from a language (problem) to another language (problem).
• Cook, Karp, and Levin’s Theorem (sketch of the proof). Algorithm for proving NP-Completeness.
• Six basic NP-Complete problems: 3-Satisfiability, 3 Dimensional Matching, Vertex Cover, Clique, Hamiltonian cycle (circuit).
• Overview of the Algorithms and Data Structures 2 course.
#### Assessment Elements
• Homework assignments
• Midterm Test
Midterm tеst will be announced at least one week in advance.
• Exam
• Student’s Presentation
at most 10 min = 8min + 2 min for questions. The presentations will be based on team (individual) reports.
#### Interim Assessment
• Interim assessment (1 module)
0.4 * Exam + 0.2 * Homework assignments + 0.3 * Midterm Test + 0.1 * Student’s Presentation
#### Recommended Core Bibliography
• Cormen, T. H., Leiserson, C. E., Rivest, R. L., Stein, C. Introduction to Algorithms (3rd edition). – MIT Press, 2009. – 1292 pp.
• Fisher, M. L., Nemhauser, Â. G. L., & Wolsey, L. A. (1978). An analysis of approximations for maximizing submodular set functions - 1. CORE Discussion Papers RP. https://doi.org/10.1007/BF01588971
• Goldengorin, B., & Pardalos, P. M. (2012). Data Correcting Approaches in Combinatorial Optimization. New York: Springer. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=538347
• N. Chiba, & T. Yamanouchi. (n.d.). [7] M.R. Garey and D.S. Johnson. Computers and Intractability: A Guide to the Theory of. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsbas&AN=edsbas.C8155A74
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Solve. 1) z^4 - 8z^2 + 12 = 0
Not my Question
Flag Content
# Question : Solve. 1) z^4 - 8z^2 + 12 = 0 : 2157367
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve.
1) z4 - 8z2 + 12 = 0
A) ±√(6), ±√(2)
B) √(6), √(2)
C) √(-6), √(-2)
D) √(-6), √(2)
2) (2m - 3)2 - 10(2m - 3) + 24 = 0
A) - (1/2), - (3/2)
B) (7/2), (9/2)
C) - (7/2), - (9/2)
D) (1/3), (3/2)
3) (2p + 6)2 = 2(2p + 6) + 35
A) - (1/2), (11/2)
B) (1/2), - (11/2)
C) - (13/2), - (1/2)
D) - (13/6), (1/2)
4) √(x) = 20 - x
A) ±√(20)
B) ±9
C) 16
D) -25, -16
5) m2/5 + 2m1/5 - 8 = 0
A) 32
B) -32, 1024
C) 1024, 1,048,576
D) 32, -1024
6) 12 = (m2 + 5m)2 + (m2 + 5m)
A) 4, -1, (-5 - √(37) /2) , (-5 + √(37) /2)
B) -2, -1, (-5 - √(37) i/2) , (-5 + √(37) i/2)
C) -4, -1
D) -4, -1, (-5 - √(37)/2) , (-5 + √(37)/2)
7) x - 5√(x) - 24 = 0
A) -64, 64, -3i, 3i
B) -64, -9, 9, 64
C) 64
D) -8i, 8i
8) x2/5 - 7x1/5 + 10 = 0
A) ±1, 3125
B) 32, 3125
C) ±32, ±3125
D) -32, 625
Solve the problem.
9) A stuntman jumps from a rooftop 360 ft off the ground. How long will it take him, falling freely, to reach the ground? (Use the formula s = 16t2)
10) Your company uses the quadratic model y = -11x2 + 350x to represent how many units (y) of a new product will be sold (x) weeks after its release. How many units can you expect to sell in week 9?
A) 3249 units
B) 4041 units
C) 3051 units
D) 2259 units
11) A grasshopper is perched on a reed 5 inches above the ground. It hops off the reed and lands on the ground about 7.9 inches away. During its hop, its height is given by the equation h = -0.3x2 + 1.75x + 5, where x is the distance in inches from the base of the reed, and h is in inches. How far was the grasshopper from the base of the reed when it was 3.75 inches above the ground? Round to the nearest tenth.
A) 0.8 in.
B) 7.9 in.
C) 6.5 in.
D) 0.6 in.
12) The length of a rectangle is three inches more than the width. The area of the rectangle is 418 inches. Find the width of the rectangle.
A) 10 in.
B) 11 in.
C) 22 in.
D) 19 in.
13) A rectangular garden has dimensions of 16 feet by 13 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 492 square feet?
A) 8.5 ft
B) 6 ft
C) 7 ft
D) 8 ft
14) The area of a square is 81 square centimeters. If the same amount is added to one dimension and removed from the other, the resulting rectangle has an area 9 square centimeters less than the area of the square. How much is added and subtracted?
A) 9 cm
B) 3 cm
C) 12 cm
D) 4 cm
15) A ladder is resting against a wall. The top of the ladder touches the wall at a height of 6 ft. Find the length of the ladder if the length is 2 ft more than its distance from the wall.
A) 8 ft
B) 12 ft
C) 6 ft
D) 10 ft
16) A lot is in the shape of a right triangle. The shorter leg measures 150 m. The hypotenuse is 50 m longer than the length of the longer leg. How long is the longer leg?
A) 150 m
B) 200 m
C) 250 m
D) 300 m
17) A picture 10 inches by 14 inches is to be mounted on a piece of matboard so that there is an even amount of mat all around the picture. How wide will the border be if the area of the mounted picture is 252 square inches?
A) 4 in.
B) 2 in.
C) 4.5 in.
D) 3 in.
18) A rug is to fit in a room so that a border of even width is left on all four sides. If the room is 23 feet by 25 feet and the area of the rug is 288 square feet, how wide will the border be?
A) 4.5 ft
B) 6 ft
C) 3.5 ft
D) 5.5 ft
State whether the function is linear or quadratic.
19) f(x) = 4x - 9
A) Linear
20) f(x) = 4x + (1.9)2
A) Linear
21) f(x) = 48x + 6x2
A) Linear
22) f(x) = 4x
A) Linear
23) f(x) = 7 + 8x2
A) Linear
24) f(x) = 4
A) Linear
25) f(x) = 0.6x2 - 10x + 9
A) Linear
26) f(x) = 9x2 - 18x
A) Linear
27) f(x) = -4x - 5
A) Linear
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https://au.mathworks.com/matlabcentral/cody/problems/1430-create-an-n-by-n-null-matrix-and-fill-with-ones-certain-positions/solutions/2651861
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Cody
# Problem 1430. Create an n-by-n null matrix and fill with ones certain positions
Solution 2651861
Submitted on 4 Jul 2020 by Paul Derwin
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
n = 1; mat=[1 1]; ok = 1; assert(isequal(FillWithOnes(n,mat),ok))
2 Pass
n = 2; mat=[1 2; 2 1]; ok = [0 1; 1 0]; assert(isequal(FillWithOnes(n,mat),ok))
3 Pass
n=3; mat=[1 3; 3 2; 3 3]; ok = [0 0 1; 0 0 0; 0 1 1]; assert(isequal(FillWithOnes(n,mat),ok))
4 Pass
n=4; mat=[1 3; 3 2; 3 3]; ok = [0 0 1 0; 0 0 0 0; 0 1 1 0; 0 0 0 0]; assert(isequal(FillWithOnes(n,mat),ok))
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http://mathhelpforum.com/advanced-algebra/189867-ring-integers.html
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# Thread: Ring of Integers
1. ## Ring of Integers
There's a question that says, "Prove that the ring of integers O in the quadratic integer ring Q$\displaystyle \sqrt{2}$ is a Euclidean Domain.
What do they mean by ring of integers O?
I know Q$\displaystyle \sqrt{2}$={q + r$\displaystyle \sqrt{2}$;q,r in Q}
... so would the ring of integers be O = {q + 0$\displaystyle \sqrt{2}$;q in Z} or do they mean algebraic integers or something?
2. ## Re: Ring of Integers
Originally Posted by gummy_ratz
There's a question that says, "Prove that the ring of integers O in the quadratic integer ring Q$\displaystyle \sqrt{2}$ is a Euclidean Domain.
What do they mean by ring of integers O?
I know Q$\displaystyle \sqrt{2}$={q + r$\displaystyle \sqrt{2}$;q,r in Q}
... so would the ring of integers be O = {q + 0$\displaystyle \sqrt{2}$;q in Z} or do they mean algebraic integers or something?
Since $\displaystyle 2\equiv 2\text{ mod }4$ what they mean by $\displaystyle \mathcal{O}$ is the set $\displaystyle \mathbb{Z}[\sqrt{2}]$. This is a Euclidean domain with the usual norm $\displaystyle N(a+b\sqrt{2})=a^2+2b^2$. Can you prove it?
3. ## Re: Ring of Integers
Ohh okay, I think I can prove that. In another class I proved that Z[$\displaystyle \sqrt{-2}$] was a Euclidean Domain using the same norm, so I'm guessing it'll be pretty similar. Does the fact that they mentioned Q[$\displaystyle \sqrt{2}$] change anything, or do I just procede the same as if they said Z[$\displaystyle \sqrt{2}$] right off the bat?
4. ## Re: Ring of Integers
Originally Posted by gummy_ratz
Ohh okay, I think I can prove that. In another class I proved that Z[$\displaystyle \sqrt{-2}$] was a Euclidean Domain using the same norm, so I'm guessing it'll be pretty similar. Does the fact that they mentioned Q[$\displaystyle \sqrt{2}$] change anything, or do I just procede the same as if they said Z[$\displaystyle \sqrt{2}$] right off the bat?
Right off the bat!
5. ## Re: Ring of Integers
hmmm, wouldn't the standard norm for Z$\displaystyle \sqrt{2}$ be $\displaystyle a^2-2b^2$? Because if an element in Z$\displaystyle \sqrt{2}$ is $\displaystyle x=a+b\sqrt{2}$, then x multiplied by the complement of x would be $\displaystyle a^2-2b^2$.
6. ## Re: Ring of Integers
the co-domain of a euclidean function has to be a subset of the positive integers, so that we can use it to find a finite decreasing chain of "remainders".
7. ## Re: Ring of Integers
Right, I forgot the absolute value! So it's |a^2 - 2b^2|.
8. ## Re: Ring of Integers
well, that is positive, but your "norm" doesn't correspond very well to the notion of "distance from 0". for example, your norm makes 1+√2 the same "distance" away from 0 as 1.
use drexel28's norm, i assure you, he is experienced in such matters.
9. ## Re: Ring of Integers
Hmmm, okay, I'm just trying to understand where it came from. Because in all of the examples in class, we found the norm by multiplying x by its complement. And in my book they just say the norm in general is a^2 - Db^2. Here they seem to use the absolute value: The integers with the square root of 2 adjoined is a Euclidean domain « Project Crazy Project .
10. ## Re: Ring of Integers
The OP is actually correct, I misread $\displaystyle \sqrt{2}$ as $\displaystyle \sqrt{-2}$ even though I put $\displaystyle \sqrt{2}$ twice! It is a ED with norm $\displaystyle N(a+b\sqrt{2})=|a^2-2b^2|$.
Deveno, thanks for the vote of confidence though--it just seems this time I was a little off!
11. ## Re: Ring of Integers
Okay, cool. Thanks! And just one more thing, is it necessary to show N(a) < N(ab), or is it sufficient to just show N(r) < N(a)? Because on that website they don't mention N(a) < N(ab), but I see it in some definitions.
12. ## Re: Ring of Integers
Originally Posted by gummy_ratz
Okay, cool. Thanks! And just one more thing, is it necessary to show N(a) < N(ab), or is it sufficient to just show N(r) < N(a)? Because on that website they don't mention N(a) < N(ab), but I see it in some definitions.
What you are referring to is sometimes called the '$\displaystyle d$-inequality' There is no need to show it. It's a cool fact though that any Euclidean domain $\displaystyle (E,d)$ admits a Euclidean function (function which allows the Division Algorithm to work) also admits a Euclidean function $\displaystyle \widetilde{d}$ which satisfies the $\displaystyle d$-inequality. It is usually given by $\displaystyle \displaystyle \widetilde{d}(a)=\min_{b\ne 0}d(ab)$.
13. ## Re: Ring of Integers
Originally Posted by Drexel28
The OP is actually correct, I misread $\displaystyle \sqrt{2}$ as $\displaystyle \sqrt{-2}$ even though I put $\displaystyle \sqrt{2}$ twice! It is a ED with norm $\displaystyle N(a+b\sqrt{2})=|a^2-2b^2|$.
Deveno, thanks for the vote of confidence though--it just seems this time I was a little off!
you and me both! perhaps we should start a club: the "Little Bit Off" club. yearly fees could be \$9.985
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http://docs.spatial.ca/v1.0/UserGuide/function_lookuprange.html
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Name
lookuprange — Extract a value from a list using a range lookup.
Syntax
```lookuprange(`value`, `rangelist`, `valuelist`)
```
The LOOKUPRANGE function has the following argument:
`value`
The value to use for the range lookup.
`rangelist`
The list of numeric values that are the range boundaries.
`valuelist`
The list of values, one matching each range, that are options that can be returned.
Description
This function extracts a value by comparing the numeric value to a list of range breaks, and returning a value from a corresponding position in the value list.
The value argument must be a numeric value. This value will be compared against the numbers in the range list.
The rangelist argument must be a list of numeric values sorted in to ascending order.
The value argument will be compared to each range list value in turn. The search will stop at the last entry where the key value is less than or equal to the list value, or at the end of the list. If the key is greater than the last entry in the list, then the last entry will be used. The value to returned is from the matching position in the value list.
The number of values in the result list must be the same as the number of values in the valuelist.
Also see the tablerange function which behave similarly but with data tables.
Examples
Formula
Description
Result
lookuprange(0,{1,2,3,4},{'a','b','c','d'})
Search the value list for the last entry that is less than or equal to 2, and return the corresponding value from the result list. 0 is less than the first value list entry, so the first result entry is returned.
'a'
lookuprange(2,{1,2,3,4},{'a','b','c','d'})
Search the value list for the last entry that is less than or equal to 2, and return the corresponding value from the result list. 2 is equal to the second value list entry, so the second result entry is returned.
'b'
lookuprange(99,{1,2,3,4},{'a','b','c','d'})
Search the value list for the last entry that is less than or equal to 99, and return the corresponding value from the result list. 99 is greater than the last entry in the value list, so the last result entry is returned.
'd'
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http://cn.metamath.org/mpeuni/axpre-sup.html
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > axpre-sup Structured version Visualization version GIF version
Theorem axpre-sup 10028
Description: A nonempty, bounded-above set of reals has a supremum. Axiom 22 of 22 for real and complex numbers, derived from ZF set theory. Note: The more general version with ordering on extended reals is axsup 10151. This construction-dependent theorem should not be referenced directly; instead, use ax-pre-sup 10052. (Contributed by NM, 19-May-1996.) (Revised by Mario Carneiro, 16-Jun-2013.) (New usage is discouraged.)
Assertion
Ref Expression
axpre-sup ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦 < 𝑥) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧)))
Distinct variable group: 𝑥,𝑦,𝑧,𝐴
Proof of Theorem axpre-sup
Dummy variables 𝑤 𝑣 𝑢 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 elreal2 9991 . . . . . . 7 (𝑥 ∈ ℝ ↔ ((1st𝑥) ∈ R𝑥 = ⟨(1st𝑥), 0R⟩))
21simplbi 475 . . . . . 6 (𝑥 ∈ ℝ → (1st𝑥) ∈ R)
32adantl 481 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑥 ∈ ℝ) → (1st𝑥) ∈ R)
4 fo1st 7230 . . . . . . . . . . . 12 1st :V–onto→V
5 fof 6153 . . . . . . . . . . . 12 (1st :V–onto→V → 1st :V⟶V)
6 ffn 6083 . . . . . . . . . . . 12 (1st :V⟶V → 1st Fn V)
74, 5, 6mp2b 10 . . . . . . . . . . 11 1st Fn V
8 ssv 3658 . . . . . . . . . . 11 𝐴 ⊆ V
9 fvelimab 6292 . . . . . . . . . . 11 ((1st Fn V ∧ 𝐴 ⊆ V) → (𝑤 ∈ (1st𝐴) ↔ ∃𝑦𝐴 (1st𝑦) = 𝑤))
107, 8, 9mp2an 708 . . . . . . . . . 10 (𝑤 ∈ (1st𝐴) ↔ ∃𝑦𝐴 (1st𝑦) = 𝑤)
11 r19.29 3101 . . . . . . . . . . . 12 ((∀𝑦𝐴 𝑦 < 𝑥 ∧ ∃𝑦𝐴 (1st𝑦) = 𝑤) → ∃𝑦𝐴 (𝑦 < 𝑥 ∧ (1st𝑦) = 𝑤))
12 ssel2 3631 . . . . . . . . . . . . . . . . 17 ((𝐴 ⊆ ℝ ∧ 𝑦𝐴) → 𝑦 ∈ ℝ)
13 ltresr2 10000 . . . . . . . . . . . . . . . . . . . 20 ((𝑦 ∈ ℝ ∧ 𝑥 ∈ ℝ) → (𝑦 < 𝑥 ↔ (1st𝑦) <R (1st𝑥)))
14 breq1 4688 . . . . . . . . . . . . . . . . . . . 20 ((1st𝑦) = 𝑤 → ((1st𝑦) <R (1st𝑥) ↔ 𝑤 <R (1st𝑥)))
1513, 14sylan9bb 736 . . . . . . . . . . . . . . . . . . 19 (((𝑦 ∈ ℝ ∧ 𝑥 ∈ ℝ) ∧ (1st𝑦) = 𝑤) → (𝑦 < 𝑥𝑤 <R (1st𝑥)))
1615biimpd 219 . . . . . . . . . . . . . . . . . 18 (((𝑦 ∈ ℝ ∧ 𝑥 ∈ ℝ) ∧ (1st𝑦) = 𝑤) → (𝑦 < 𝑥𝑤 <R (1st𝑥)))
1716exp31 629 . . . . . . . . . . . . . . . . 17 (𝑦 ∈ ℝ → (𝑥 ∈ ℝ → ((1st𝑦) = 𝑤 → (𝑦 < 𝑥𝑤 <R (1st𝑥)))))
1812, 17syl 17 . . . . . . . . . . . . . . . 16 ((𝐴 ⊆ ℝ ∧ 𝑦𝐴) → (𝑥 ∈ ℝ → ((1st𝑦) = 𝑤 → (𝑦 < 𝑥𝑤 <R (1st𝑥)))))
1918imp4b 612 . . . . . . . . . . . . . . 15 (((𝐴 ⊆ ℝ ∧ 𝑦𝐴) ∧ 𝑥 ∈ ℝ) → (((1st𝑦) = 𝑤𝑦 < 𝑥) → 𝑤 <R (1st𝑥)))
2019ancomsd 469 . . . . . . . . . . . . . 14 (((𝐴 ⊆ ℝ ∧ 𝑦𝐴) ∧ 𝑥 ∈ ℝ) → ((𝑦 < 𝑥 ∧ (1st𝑦) = 𝑤) → 𝑤 <R (1st𝑥)))
2120an32s 863 . . . . . . . . . . . . 13 (((𝐴 ⊆ ℝ ∧ 𝑥 ∈ ℝ) ∧ 𝑦𝐴) → ((𝑦 < 𝑥 ∧ (1st𝑦) = 𝑤) → 𝑤 <R (1st𝑥)))
2221rexlimdva 3060 . . . . . . . . . . . 12 ((𝐴 ⊆ ℝ ∧ 𝑥 ∈ ℝ) → (∃𝑦𝐴 (𝑦 < 𝑥 ∧ (1st𝑦) = 𝑤) → 𝑤 <R (1st𝑥)))
2311, 22syl5 34 . . . . . . . . . . 11 ((𝐴 ⊆ ℝ ∧ 𝑥 ∈ ℝ) → ((∀𝑦𝐴 𝑦 < 𝑥 ∧ ∃𝑦𝐴 (1st𝑦) = 𝑤) → 𝑤 <R (1st𝑥)))
2423expd 451 . . . . . . . . . 10 ((𝐴 ⊆ ℝ ∧ 𝑥 ∈ ℝ) → (∀𝑦𝐴 𝑦 < 𝑥 → (∃𝑦𝐴 (1st𝑦) = 𝑤𝑤 <R (1st𝑥))))
2510, 24syl7bi 245 . . . . . . . . 9 ((𝐴 ⊆ ℝ ∧ 𝑥 ∈ ℝ) → (∀𝑦𝐴 𝑦 < 𝑥 → (𝑤 ∈ (1st𝐴) → 𝑤 <R (1st𝑥))))
2625impr 648 . . . . . . . 8 ((𝐴 ⊆ ℝ ∧ (𝑥 ∈ ℝ ∧ ∀𝑦𝐴 𝑦 < 𝑥)) → (𝑤 ∈ (1st𝐴) → 𝑤 <R (1st𝑥)))
2726adantlr 751 . . . . . . 7 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ (𝑥 ∈ ℝ ∧ ∀𝑦𝐴 𝑦 < 𝑥)) → (𝑤 ∈ (1st𝐴) → 𝑤 <R (1st𝑥)))
2827ralrimiv 2994 . . . . . 6 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ (𝑥 ∈ ℝ ∧ ∀𝑦𝐴 𝑦 < 𝑥)) → ∀𝑤 ∈ (1st𝐴)𝑤 <R (1st𝑥))
2928expr 642 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑥 ∈ ℝ) → (∀𝑦𝐴 𝑦 < 𝑥 → ∀𝑤 ∈ (1st𝐴)𝑤 <R (1st𝑥)))
30 breq2 4689 . . . . . . 7 (𝑣 = (1st𝑥) → (𝑤 <R 𝑣𝑤 <R (1st𝑥)))
3130ralbidv 3015 . . . . . 6 (𝑣 = (1st𝑥) → (∀𝑤 ∈ (1st𝐴)𝑤 <R 𝑣 ↔ ∀𝑤 ∈ (1st𝐴)𝑤 <R (1st𝑥)))
3231rspcev 3340 . . . . 5 (((1st𝑥) ∈ R ∧ ∀𝑤 ∈ (1st𝐴)𝑤 <R (1st𝑥)) → ∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣)
333, 29, 32syl6an 567 . . . 4 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑥 ∈ ℝ) → (∀𝑦𝐴 𝑦 < 𝑥 → ∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣))
3433rexlimdva 3060 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) → (∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦 < 𝑥 → ∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣))
35 n0 3964 . . . . . 6 (𝐴 ≠ ∅ ↔ ∃𝑦 𝑦𝐴)
36 fnfvima 6536 . . . . . . . . 9 ((1st Fn V ∧ 𝐴 ⊆ V ∧ 𝑦𝐴) → (1st𝑦) ∈ (1st𝐴))
377, 8, 36mp3an12 1454 . . . . . . . 8 (𝑦𝐴 → (1st𝑦) ∈ (1st𝐴))
38 ne0i 3954 . . . . . . . 8 ((1st𝑦) ∈ (1st𝐴) → (1st𝐴) ≠ ∅)
3937, 38syl 17 . . . . . . 7 (𝑦𝐴 → (1st𝐴) ≠ ∅)
4039exlimiv 1898 . . . . . 6 (∃𝑦 𝑦𝐴 → (1st𝐴) ≠ ∅)
4135, 40sylbi 207 . . . . 5 (𝐴 ≠ ∅ → (1st𝐴) ≠ ∅)
42 supsr 9971 . . . . . 6 (((1st𝐴) ≠ ∅ ∧ ∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣) → ∃𝑣R (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 ∧ ∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢)))
4342ex 449 . . . . 5 ((1st𝐴) ≠ ∅ → (∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣 → ∃𝑣R (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 ∧ ∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢))))
4441, 43syl 17 . . . 4 (𝐴 ≠ ∅ → (∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣 → ∃𝑣R (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 ∧ ∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢))))
4544adantl 481 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) → (∃𝑣R𝑤 ∈ (1st𝐴)𝑤 <R 𝑣 → ∃𝑣R (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 ∧ ∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢))))
46 breq2 4689 . . . . . . . . . . . 12 (𝑤 = (1st𝑦) → (𝑣 <R 𝑤𝑣 <R (1st𝑦)))
4746notbid 307 . . . . . . . . . . 11 (𝑤 = (1st𝑦) → (¬ 𝑣 <R 𝑤 ↔ ¬ 𝑣 <R (1st𝑦)))
4847rspccv 3337 . . . . . . . . . 10 (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 → ((1st𝑦) ∈ (1st𝐴) → ¬ 𝑣 <R (1st𝑦)))
4937, 48syl5com 31 . . . . . . . . 9 (𝑦𝐴 → (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 → ¬ 𝑣 <R (1st𝑦)))
5049adantl 481 . . . . . . . 8 ((𝐴 ⊆ ℝ ∧ 𝑦𝐴) → (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 → ¬ 𝑣 <R (1st𝑦)))
51 elreal2 9991 . . . . . . . . . . . . 13 (𝑦 ∈ ℝ ↔ ((1st𝑦) ∈ R𝑦 = ⟨(1st𝑦), 0R⟩))
5251simprbi 479 . . . . . . . . . . . 12 (𝑦 ∈ ℝ → 𝑦 = ⟨(1st𝑦), 0R⟩)
5352breq2d 4697 . . . . . . . . . . 11 (𝑦 ∈ ℝ → (⟨𝑣, 0R⟩ < 𝑦 ↔ ⟨𝑣, 0R⟩ < ⟨(1st𝑦), 0R⟩))
54 ltresr 9999 . . . . . . . . . . 11 (⟨𝑣, 0R⟩ < ⟨(1st𝑦), 0R⟩ ↔ 𝑣 <R (1st𝑦))
5553, 54syl6bb 276 . . . . . . . . . 10 (𝑦 ∈ ℝ → (⟨𝑣, 0R⟩ < 𝑦𝑣 <R (1st𝑦)))
5612, 55syl 17 . . . . . . . . 9 ((𝐴 ⊆ ℝ ∧ 𝑦𝐴) → (⟨𝑣, 0R⟩ < 𝑦𝑣 <R (1st𝑦)))
5756notbid 307 . . . . . . . 8 ((𝐴 ⊆ ℝ ∧ 𝑦𝐴) → (¬ ⟨𝑣, 0R⟩ < 𝑦 ↔ ¬ 𝑣 <R (1st𝑦)))
5850, 57sylibrd 249 . . . . . . 7 ((𝐴 ⊆ ℝ ∧ 𝑦𝐴) → (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 → ¬ ⟨𝑣, 0R⟩ < 𝑦))
5958ralrimdva 2998 . . . . . 6 (𝐴 ⊆ ℝ → (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 → ∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦))
6059ad2antrr 762 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑣R) → (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 → ∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦))
6152breq1d 4695 . . . . . . . . . . . . . 14 (𝑦 ∈ ℝ → (𝑦 <𝑣, 0R⟩ ↔ ⟨(1st𝑦), 0R⟩ <𝑣, 0R⟩))
62 ltresr 9999 . . . . . . . . . . . . . 14 (⟨(1st𝑦), 0R⟩ <𝑣, 0R⟩ ↔ (1st𝑦) <R 𝑣)
6361, 62syl6bb 276 . . . . . . . . . . . . 13 (𝑦 ∈ ℝ → (𝑦 <𝑣, 0R⟩ ↔ (1st𝑦) <R 𝑣))
6451simplbi 475 . . . . . . . . . . . . . . 15 (𝑦 ∈ ℝ → (1st𝑦) ∈ R)
65 breq1 4688 . . . . . . . . . . . . . . . . 17 (𝑤 = (1st𝑦) → (𝑤 <R 𝑣 ↔ (1st𝑦) <R 𝑣))
66 breq1 4688 . . . . . . . . . . . . . . . . . 18 (𝑤 = (1st𝑦) → (𝑤 <R 𝑢 ↔ (1st𝑦) <R 𝑢))
6766rexbidv 3081 . . . . . . . . . . . . . . . . 17 (𝑤 = (1st𝑦) → (∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢 ↔ ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢))
6865, 67imbi12d 333 . . . . . . . . . . . . . . . 16 (𝑤 = (1st𝑦) → ((𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) ↔ ((1st𝑦) <R 𝑣 → ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢)))
6968rspccv 3337 . . . . . . . . . . . . . . 15 (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → ((1st𝑦) ∈ R → ((1st𝑦) <R 𝑣 → ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢)))
7064, 69syl5 34 . . . . . . . . . . . . . 14 (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → (𝑦 ∈ ℝ → ((1st𝑦) <R 𝑣 → ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢)))
7170com3l 89 . . . . . . . . . . . . 13 (𝑦 ∈ ℝ → ((1st𝑦) <R 𝑣 → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢)))
7263, 71sylbid 230 . . . . . . . . . . . 12 (𝑦 ∈ ℝ → (𝑦 <𝑣, 0R⟩ → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢)))
7372adantr 480 . . . . . . . . . . 11 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (𝑦 <𝑣, 0R⟩ → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → ∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢)))
74 fvelimab 6292 . . . . . . . . . . . . . . . 16 ((1st Fn V ∧ 𝐴 ⊆ V) → (𝑢 ∈ (1st𝐴) ↔ ∃𝑧𝐴 (1st𝑧) = 𝑢))
757, 8, 74mp2an 708 . . . . . . . . . . . . . . 15 (𝑢 ∈ (1st𝐴) ↔ ∃𝑧𝐴 (1st𝑧) = 𝑢)
76 ssel2 3631 . . . . . . . . . . . . . . . . . . . . . 22 ((𝐴 ⊆ ℝ ∧ 𝑧𝐴) → 𝑧 ∈ ℝ)
77 ltresr2 10000 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑦 ∈ ℝ ∧ 𝑧 ∈ ℝ) → (𝑦 < 𝑧 ↔ (1st𝑦) <R (1st𝑧)))
7876, 77sylan2 490 . . . . . . . . . . . . . . . . . . . . 21 ((𝑦 ∈ ℝ ∧ (𝐴 ⊆ ℝ ∧ 𝑧𝐴)) → (𝑦 < 𝑧 ↔ (1st𝑦) <R (1st𝑧)))
79 breq2 4689 . . . . . . . . . . . . . . . . . . . . 21 ((1st𝑧) = 𝑢 → ((1st𝑦) <R (1st𝑧) ↔ (1st𝑦) <R 𝑢))
8078, 79sylan9bb 736 . . . . . . . . . . . . . . . . . . . 20 (((𝑦 ∈ ℝ ∧ (𝐴 ⊆ ℝ ∧ 𝑧𝐴)) ∧ (1st𝑧) = 𝑢) → (𝑦 < 𝑧 ↔ (1st𝑦) <R 𝑢))
8180exbiri 651 . . . . . . . . . . . . . . . . . . 19 ((𝑦 ∈ ℝ ∧ (𝐴 ⊆ ℝ ∧ 𝑧𝐴)) → ((1st𝑧) = 𝑢 → ((1st𝑦) <R 𝑢𝑦 < 𝑧)))
8281expr 642 . . . . . . . . . . . . . . . . . 18 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (𝑧𝐴 → ((1st𝑧) = 𝑢 → ((1st𝑦) <R 𝑢𝑦 < 𝑧))))
8382com4r 94 . . . . . . . . . . . . . . . . 17 ((1st𝑦) <R 𝑢 → ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (𝑧𝐴 → ((1st𝑧) = 𝑢𝑦 < 𝑧))))
8483imp 444 . . . . . . . . . . . . . . . 16 (((1st𝑦) <R 𝑢 ∧ (𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ)) → (𝑧𝐴 → ((1st𝑧) = 𝑢𝑦 < 𝑧)))
8584reximdvai 3044 . . . . . . . . . . . . . . 15 (((1st𝑦) <R 𝑢 ∧ (𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ)) → (∃𝑧𝐴 (1st𝑧) = 𝑢 → ∃𝑧𝐴 𝑦 < 𝑧))
8675, 85syl5bi 232 . . . . . . . . . . . . . 14 (((1st𝑦) <R 𝑢 ∧ (𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ)) → (𝑢 ∈ (1st𝐴) → ∃𝑧𝐴 𝑦 < 𝑧))
8786expcom 450 . . . . . . . . . . . . 13 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → ((1st𝑦) <R 𝑢 → (𝑢 ∈ (1st𝐴) → ∃𝑧𝐴 𝑦 < 𝑧)))
8887com23 86 . . . . . . . . . . . 12 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (𝑢 ∈ (1st𝐴) → ((1st𝑦) <R 𝑢 → ∃𝑧𝐴 𝑦 < 𝑧)))
8988rexlimdv 3059 . . . . . . . . . . 11 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (∃𝑢 ∈ (1st𝐴)(1st𝑦) <R 𝑢 → ∃𝑧𝐴 𝑦 < 𝑧))
9073, 89syl6d 75 . . . . . . . . . 10 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (𝑦 <𝑣, 0R⟩ → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → ∃𝑧𝐴 𝑦 < 𝑧)))
9190com23 86 . . . . . . . . 9 ((𝑦 ∈ ℝ ∧ 𝐴 ⊆ ℝ) → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧)))
9291ex 449 . . . . . . . 8 (𝑦 ∈ ℝ → (𝐴 ⊆ ℝ → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧))))
9392com3l 89 . . . . . . 7 (𝐴 ⊆ ℝ → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → (𝑦 ∈ ℝ → (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧))))
9493ad2antrr 762 . . . . . 6 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑣R) → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → (𝑦 ∈ ℝ → (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧))))
9594ralrimdv 2997 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑣R) → (∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢) → ∀𝑦 ∈ ℝ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧)))
96 opelreal 9989 . . . . . . . 8 (⟨𝑣, 0R⟩ ∈ ℝ ↔ 𝑣R)
9796biimpri 218 . . . . . . 7 (𝑣R → ⟨𝑣, 0R⟩ ∈ ℝ)
9897adantl 481 . . . . . 6 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑣R) → ⟨𝑣, 0R⟩ ∈ ℝ)
99 breq1 4688 . . . . . . . . . . 11 (𝑥 = ⟨𝑣, 0R⟩ → (𝑥 < 𝑦 ↔ ⟨𝑣, 0R⟩ < 𝑦))
10099notbid 307 . . . . . . . . . 10 (𝑥 = ⟨𝑣, 0R⟩ → (¬ 𝑥 < 𝑦 ↔ ¬ ⟨𝑣, 0R⟩ < 𝑦))
101100ralbidv 3015 . . . . . . . . 9 (𝑥 = ⟨𝑣, 0R⟩ → (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ↔ ∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦))
102 breq2 4689 . . . . . . . . . . 11 (𝑥 = ⟨𝑣, 0R⟩ → (𝑦 < 𝑥𝑦 <𝑣, 0R⟩))
103102imbi1d 330 . . . . . . . . . 10 (𝑥 = ⟨𝑣, 0R⟩ → ((𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧) ↔ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧)))
104103ralbidv 3015 . . . . . . . . 9 (𝑥 = ⟨𝑣, 0R⟩ → (∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧) ↔ ∀𝑦 ∈ ℝ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧)))
105101, 104anbi12d 747 . . . . . . . 8 (𝑥 = ⟨𝑣, 0R⟩ → ((∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧)) ↔ (∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧))))
106105rspcev 3340 . . . . . . 7 ((⟨𝑣, 0R⟩ ∈ ℝ ∧ (∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧))) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧)))
107106ex 449 . . . . . 6 (⟨𝑣, 0R⟩ ∈ ℝ → ((∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧)) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧))))
10898, 107syl 17 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑣R) → ((∀𝑦𝐴 ¬ ⟨𝑣, 0R⟩ < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 <𝑣, 0R⟩ → ∃𝑧𝐴 𝑦 < 𝑧)) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧))))
10960, 95, 108syl2and 499 . . . 4 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) ∧ 𝑣R) → ((∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 ∧ ∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢)) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧))))
110109rexlimdva 3060 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) → (∃𝑣R (∀𝑤 ∈ (1st𝐴) ¬ 𝑣 <R 𝑤 ∧ ∀𝑤R (𝑤 <R 𝑣 → ∃𝑢 ∈ (1st𝐴)𝑤 <R 𝑢)) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧))))
11134, 45, 1103syld 60 . 2 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅) → (∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦 < 𝑥 → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧))))
1121113impia 1280 1 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦 < 𝑥) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧)))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 383 ∧ w3a 1054 = wceq 1523 ∃wex 1744 ∈ wcel 2030 ≠ wne 2823 ∀wral 2941 ∃wrex 2942 Vcvv 3231 ⊆ wss 3607 ∅c0 3948 ⟨cop 4216 class class class wbr 4685 “ cima 5146 Fn wfn 5921 ⟶wf 5922 –onto→wfo 5924 ‘cfv 5926 1st c1st 7208 Rcnr 9725 0Rc0r 9726
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# 6.11. Line Chart¶
## 6.11.1. Rationale¶
• Show linear relation of two variables
## 6.11.2. Syntax¶
import matplotlib.pyplot as plt
x = [1, 2, 3, 4]
y = [1, 2, 3, 4]
plt.plot(x, y)
plt.show()
## 6.11.4. Single Plot¶
Listing 6.77. Vectorized Operations
import matplotlib.pyplot as plt
import numpy as np
np.random.seed(0)
x = np.arange(0, 10)
y = np.random.randint(0, 10, size=10)
plt.plot(x, y)
plt.show()
Listing 6.78. Universal Function
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 10, 1000)
y = np.sin(x)
plt.plot(x, y)
plt.show()
## 6.11.5. Multiple Plots¶
import matplotlib.pyplot as plt
x1 = [1, 2, 3, 4]
y1 = [1, 2, 3, 4]
x2 = [1, 2, 3, 4]
y2 = [4, 3, 3, 2]
plt.plot(x1, y1)
plt.plot(x2, y2)
plt.show()
Listing 6.79. Universal Function
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 10, 1000)
y1 = np.sin(x)
y2 = np.cos(x)
plt.plot(x, y1)
plt.plot(x, y2)
plt.show()
Listing 6.80. Inlined Universal Function
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 10, 1000)
plt.plot(x, np.sin(x))
plt.plot(x, np.cos(x))
plt.show()
Listing 6.81. Vectorized Operation
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 2, 100)
plt.plot(x, x)
plt.plot(x, x**2)
plt.plot(x, x**3)
plt.show()
Listing 6.82. Universal Function and Vectorized Operation
import matplotlib.pyplot as plt
import numpy as np
np.random.seed(0)
noise = np.random.normal(0.0, 0.1, size=1000)
x1 = np.linspace(0, 2*np.pi, 1000)
y1 = np.sin(x1) + noise
x2 = np.linspace(2*np.pi, 3*np.pi, 20)
y2 = np.sin(x2)
plt.plot(x1, y1)
plt.plot(x2, y2, linestyle='--')
plt.show()
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: Krishna
## how we calculate of Sand, cement and aggregate of M20 .
M20 (1 cement :1.5 sand :3 stone/brick aggregate). To determine the proportions you have to perform mix design, for this you have to find out the sp.gr. of cement, CA, FA, and water cement ratio, type of exposure, Maximum size of aggregate etc.
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## How To Calculate Number Of Bricks, Cement And Sand For .
Now we Calculate Mortar quantity of cement and sand Ratio is 1:6 Mortar quantity is 0.2305 (Wet Condition) we calculate Dry Value Dry Mortar Value is = 0.2305 x 1.33 = 0.306565 m3 Cement Calculation Volume = Volume of Mortar x (Ratio of Cement/ Sum of ...
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## How To Calculate Cement, Sand, amp Aggregate Quantity In 1 .
How to calculate quantity of cement, sand and aggregate in 1 cubic metre. Tips for basic estimation and calculation of building materials quantity.. ... Point Should Be Know Before Estimating Density of Cement = 1440 kg/m3 Sand Density = 1450 ...
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: Civil Engineers
## How to Calculate Quantity for Cement, Sand amp Aggregate .
1/2/2017018332How to Calculate Quantity for Cement, Sand amp Aggregate. In This channel You can Learn about Civil Engineering Update Videos which are using generally in civil Engineering. So please subscribe our channel for daily Update .
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## Quantity of Cement and Sand Calculation in Mortar
3/11/2014018332Calculation of quantity of cement mortar in brickwork and plaster: For the calculation of cement mortar, let us assume that we use 1m 3 of cement mortar. Procedure for calculation is: 1. Calculate the dry volume of materials required for 1m 3 cement mortar.
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## How to Calculate CEMENT, SAND Quantity for .
Cement is considered as good binding material with low cost and easy availability. Points should be remembered while calculating the quantity of cement, sand for plastering work For wall plastering, the ratio of Cement : Sand = 1 : 6 For ceiling plastering, the ratio ...
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## Quantity of sand and cement, How to calculate
Ensure you are choosing a good quality of cement and pure sand, it is essential to achieve the desired result We hope that you learn how to Quantity of sand and cement for a job Now you are able to start, have a great and successful rendering job
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## How to calculate the cement, sand, and stone in stone .
Suppose you want to know about the quantities of cement, sand and stone in the case of RRMasonry in cement mortar 1:6, we can calculate like this for one unit of 10 cu.m: RANDUM RUBBLE STONE 10 cum Bond stone 1 cu. cement mortar 1:6: 3.4cum which...
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## How to Calculate Cement, Sand and Coarse Aggregate .
Concrete Ingredients Calculation For Cement, Sand and Coarse Aggregate. This is a Volumetric Calculation. Assuming we need 2 m 3 of concrete for M20 Concrete Mix, (Mix Ratio, M20 = 1 : 1.5 : 3) Total Part of the Concrete = 1+1.5+3 = 5.5 Parts Therefore,
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## How to calculate Cement, Sand, and Aggregate for M20 .
10/2/2018018332in this Video Lecture today I will teach you How to calculate Cement, Sand, and Aggregate for M20 concrete For Reading Article click on given link: https://c...
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## How to calculate cement, sand and coarse aggregate for .
How to calculate cement, sand and coarse aggregate for concrete M15 Mix Ratio – 1 : 2 : 4 The mix ratio denotes the following 1 – Cement 2 – Sand (Fine aggregate – 2 Times of Cement Quantity) 4 – Blue metal (Coarse aggregate – 4 Times of Cement ...
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## Cement and Sand ratio for brickwork. How to Calculate .
We can divide the calculation into three parts to find out the amount of cement, sand, and bricks required. Given that the thickness is 230 mm for 1 cum brickwork and the ratio for cement mortar for brickwork is .
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## Calculation of Cement and Sand Quantity for Plastering
14/10/2019018332Calculation of Cement, Sand Quantity for Plastering How Much Cement, Sand And Water Is Required For 12mm Thick Plastering Plastering is done on the walls to remove surface imperfections, maintain line level and alignment. In addition to that it also acts as a ...
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## Calculate Quantities of Materials for Concrete Cement, .
1/4/2015018332Quantities of materials for concrete such as cement, sand and aggregates for production of required quantity of concrete of given mix proportions such as 1:2:4 (M15 ... This method is based on the principle that the volume of fully compacted concrete is equal to the ...
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## How to calculate cement, sand amp aggregate in CFT .
If you want to know the quantity of cement, sand, aggregate in CFT(Cubic feet) then first find out the quantity in Cubic metre(m179) then convert in CFT. 1 Cubic metre(m179) = 35.31 Cubic feet(CFT) Ratio 1: 1.5 : 3 is known as M20 grade concrete. Rati...
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## How To Calculate Cement, Sand Quantity For Plastering
Quantity Of Cement amp Sand Required For 100 Sq.ft Plastering Work Quantity Of Cement, Sand, amp Aggregates For 1000 sq ft Slab How To Calculate Quantities Of Materials For Different Mix Ratio Of Concrete Calculation Of Bricks In A Wall How To Calculate
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## HOW TO CALCULATE QUANTITY FOR CEMENT, SAND .
In this Article today I will teach you that how to calculate quantity of cement, sand and aggregate in concrete column. after reading this article you will be able to find out the materials in slab m beam and also for Concrete road. We used here the same formula for ...
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## How to calculate Quantity of Cement Sand and Aggregate .
31/5/2019018332Please LIKE and share the video to support ... Why Beam has more Depth than Width Why depth of beam has more than its width Depth of Beam Duration: 5:35. Civil Engineers 4,742 views
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# Mcgraw Calculus And Vectors Pdf
File Name: mcgraw calculus and vectors .zip
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## Calculus and Vectors 12
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Chapter 5. Mark the point s of intersection. Equations of Lines and Planes. Obviously, simply copying them will get you nowhere. Chapter 1 Function Transformations 0 of Buy.
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Complete market without risk-neutral measure
Let $$\mathcal{M}$$ be a one-period model with $$\Omega=\{\omega_1,\omega_2\}$$ and $$S_t^0=1$$ for $$t=0,1$$.
Find a $$D$$ such that $$S^d$$, $$d=1,...,D$$ yields a complete market without a risk-neutral measure. Why does this not contradict the second fundamental theorem of asset pricing?
So I guess this is due to the fact that it second fundamental theorem of asset pricing needs the market to be arbitrage-free. However, I was not able to construct such a model. Can someone help?
Consider a one period model with one risk-free asset that yields r (in your case, r=0 since $$S_t^{0}=1$$ for t=0,1), and one risky asset, so D=1. Now, this market is complete if and only if every contingent claim C is attainable (i.e, hedgeable). Let $$C(w_1)$$ and $$C(w_2)$$ the payoff of the Contingent claim in states $$w_1$$ and $$w_2$$, respectively. Let's call $$a:=S_1^1(w_1)$$ and $$b:=S_1^1(w_2)$$ To build a portfolio $$\xi_0$$, $$\xi_1$$ that replicates this contingent claim, we have to solve the system of equations:
$$(1+r)\xi_0 + \xi_1a = C(w_1)$$ $$(1+r)\xi_0 + \xi_1b = C(w_2)$$
This system has a unique solution when the determinant is not 0, i.e when $$(1+r)b - (1+r)a \ne 0$$, that is, when $$(1+r) \ne 0$$ and $$b \ne a$$.
Then, if $$b>a$$ and $$(1+r) \ne 0$$, the market is complete. However, if $$S_0^1 < a$$ (or $$S_0^1 > b$$), the market has an arbitrage opportunity, by buying $$S^1$$ (resp, selling $$S^1$$), borrowing from the risk-free asset and selling at t=1.
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Question
# The cost of a notebook is Rs 5 more than that of a pen and the total cost of 4 pens and 5 notebooks is Rs 88. Find the cost of one pen and one notebook.
Solution
## Let the cost of one notebook be x. So, the cost of one pen is x − 5. So, cost of 5 notebooks = 5x Cost of 4 pens = 4(x − 5) = 4x − 20 Total cost = Rs 88 ⇒ 4x − 20 + 5x = 88 ⇒ 9x = 108 ⇒ x = 12 ∴ Cost of one notebook = x = Rs 12 and the cost of one pen = x − 5 = 12 − 5 = Rs 7MathematicsMathematics(2013)Standard VIII
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# How to calculate break even point per unit
Learn how to calculate the break-even point per unit with this step-by-step guide. Understand the importance of break-even analysis and how it can help your business make informed decisions.
break-even point, per unit, calculation, break-even analysis, fixed costs, variable costs, contribution margin, profit margin
## Introduction
Understanding the break-even point per unit is essential for any business owner. It helps you to determine the minimum number of units you need to sell to cover your costs and make a profit. In this article, we will show you how to calculate the break-even point per unit with a step-by-step guide.
## Why is Break-Even Analysis Important?
Break-even analysis is a financial tool that helps businesses to make informed decisions. It allows you to determine the point at which your total revenue equals your total costs. This information is crucial for setting prices, determining production levels, and evaluating the profitability of your business.
## Step 1: Define your Fixed Costs
Fixed costs are expenses that do not vary with the level of production. Examples of fixed costs include rent, salaries, and insurance. To calculate your break-even point per unit, you need to know your total fixed costs.
## Step 2: Identify your Variable Costs
Variable costs are expenses that vary with the level of production. Examples of variable costs include raw materials, labor, and shipping. To calculate your break-even point per unit, you need to know your variable costs per unit.
## Step 3: Determine your Contribution Margin per Unit
Your contribution margin per unit is the amount of revenue that is left over after variable costs have been subtracted. To calculate your contribution margin per unit, subtract your variable costs per unit from your selling price per unit.
## Step 4: Calculate your Break-Even Point per Unit
To calculate your break-even point per unit, divide your total fixed costs by your contribution margin per unit. This will give you the number of units you need to sell to cover your fixed costs.
## Step 5: Interpret your Break-Even Point
Your break-even point per unit represents the minimum number of units you need to sell to cover your costs. Any units sold beyond this point will contribute to your profit margin. Use your break-even point to make informed decisions about pricing, production levels, and sales strategies.
## Example Calculation
Let’s consider an example. A business has fixed costs of \$10,000 and variable costs of \$5 per unit. The selling price per unit is \$10. To calculate the break-even point per unit, we need to determine the contribution margin per unit.
Contribution Margin per Unit = Selling Price per Unit - Variable Costs per Unit Contribution Margin per Unit = \$10 - \$5 Contribution Margin per Unit = \$5
Using this information, we can calculate the break-even point per unit.
Break-Even Point per Unit = Fixed Costs / Contribution Margin per Unit Break-Even Point per Unit = \$10,000 / \$5 Break-Even Point per Unit = 2,000 units
This means that the business needs to sell 2,000 units to cover its fixed costs. Any units sold beyond this point will contribute to the profit margin.
## Conclusion
Calculating the break-even point per unit is a crucial step for any business owner. It helps you to make informed decisions about pricing, production levels, and sales strategies. By following the steps outlined in this guide, you can easily calculate your break-even point per unit and use this information to grow your business.
Learn how to calculate the break-even point per unit with this step-by-step guide. Understand the importance of break-even analysis and how it can help your business make informed decisions.
break-even point, per unit, calculation, break-even analysis, fixed costs, variable costs, contribution margin, profit margin
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# How many pounds do an eagle weigh?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
#### exists and is an alternate of .
15
1 person found this useful
# How many pounds does a gallon weigh?
That depends on the type material you are referring to. Also, theseare two different types of measurements. Gallons is a measure ofvolume but the other is weight or mass. Plea
# How many pounds does a gallbladder weigh?
ABOUT 12 OUNCES NOT MORE THAN 14
# How many pounds does Jupiter weigh?
9,0000000 billion lbs
# How many pounds does Saturn weigh?
\n. \n 44 pounds exact
# How many pounds does a horse weigh?
There are many different sizes of horses and ponies.. The general weight for a standard sized, mature horse is normally around 1000 to 1100 pounds.
# If you weigh 50kgs how many pounds do you weigh?
110 pounds, just multiply the weight of kgs by 0.45 and you will get it in pounds
# How much does a bald eagle weigh in pounds?
it weighs up to 20 pounds
164.02 pounds.
# How many pounds does a puma weigh?
A puma weighs 227 pounds.
12 Pounds
# How many pounds does paper weigh?
wat its labeled as when you by it
# How many pounds does a snake weigh?
Answer: 60 POUNDS Answer: That depends entirely upon the snake. Some snakes are about 6 inches long and not much wider than a worm! Others reach nigh on 30 feet i
# How many pounds do you have to weigh to be fat?
it depends on your muscle mass and height bodybuilders might be 5'8 and 260 lbs and not have an ounce of fat on there bodies yeah, that's right. it all depends on your body
# How many pounds does a notebook weigh?
It weighs ~1 or 2 pounds jeez people and their logic these days
# How many pounds is a bald eagle?
how many pound is a bald eagles
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# In triangle ABC, angle C is 90 degrees. CH – height, AB = 49 cos A 6/7 Find AH
By the definition of cosine, we have:
AC = AB * cosA
AC = 479 * 6/7 = 42
According to metric ratios in a right-angled triangle (the square of the leg is equal to the product of the hypotenuse and the projection of this leg to the hypotenuse) AH is the projection of the leg AC:
AC ^ 2 = AH * AB
AH = AC ^ 2 / AB
AH = 42 ^ 2/49 = 36
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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# Solution - A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC - CBSE Class 10 - Mathematics
ConceptCircles Examples and Solutions
#### Question
A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC
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Solution for question: A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC concept: Circles Examples and Solutions. For the course CBSE
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## Archive for October 18th, 2009
### Here we go again: We now get a trio of PDVSA bonds
October 18, 2009
In Spanish here
No sooner had we settled the Venezuela 2019 and 2024 bonds, which investors received last Tuesday, that PDVSA announced last Friday that it would issue three bonds in a combo maturing in 2014, 2015 and 2016 with coupons of 4.9%, 5% and 5.125% respectively. Each buyer will buy a combo composed of \$1,300 of the 2014, \$1,300 of the 2015 and \$40o of the 2016. They will be sold to investors at a price of 138% in Bolivars but at Bs. 2.15.
Thus, while the whole of emerging markets countries issued around US\$ 20 billion in September and not one issue exceeded US\$ 2 billion, Venezuela began October by issuing US\$ 5 billion and now is ready to issue another \$3 billion. Of course, PDVSA could decide to issue more if it wants. Except…
that it would have to change the terms if it wants the issue to be succesful, because as it stands, buying dollars via the PDVSA bonds is more expensive than going to the parallel swap market.
Let me remind you how this works (Version for Dummies here) :
1) You buy \$3,000 in bonds denominated in \$ for Bs. In this case you pay \$3000 x Bs. 2.15 x 1.38= Bs. 8,901 for the combo
2) The bonds are not worth \$3,000 in the international markets, they sell at a discount, i.e. less than 100%. Let’s assume the combo (more on this later) sells at an average of 55% of its face value, then you get \$1,650 for your bonds if you sell them right away.
3) Since you paid Bs. 8.901, then each \$ cost you Bs. 5.39 (8,901/1,650) which is according to the swap market value, which I am not allowed to print, more expensive than doing a permuta swap according to this page.
So, whether it is a good deal or not boils down to what is the value of the bonds in the international markets. Since the bonds don’t exist yet, we have to use other existing bonds as a guide or a guesstimate or what their price should be.
PDVSA has four bonds out there: the 2011 zero coupon Petrobono (zero coupon means no interest, you get 100% of face value on maturity) a 5.25% coupon bond maturing in 2017, a 5.375% coupon bond maturing in 2027 and a 5.5% coupon bond maturing in 2037.
The new PDVSA bonds are closer in nature to the 2011 Petrobono because in contrast to the 2017, 2027 and 2037 issues, which were issued under New York law, these bonds are all issued by PDVSA under Venezuelan law. What this means is that foreign investors don’t like them as much. For some reason, they don’t trust our Courts as much as those in New York. Wonder why?
Because the new bonds mature after 2011, the first thing we can say is that they should all yield at least the same as the 2011 bond, if not more.
Why?
Well, usually (there are exceptions that I will not go into) the longer maturity of the bond the more investors want to be paid for it. This can be understood simply as saying the longer I have to look into the future, the fuzzier it is, the harder it gets to predict.
In PDVSA’s case, if you are uncomfortable that it will pay you in 2011, imagine if you had to wait three more years! Or five!
Thus, a first approximation would be to say they all yield the same as the Petrobono 2011, which on Friday was yielding 16.8%. This is the most optimistic case. If this were the case, using Excel you can calculate that the 2014 bond would be worth 60.8%, the 2015 56.4% and the 2016 53%. Which means that you would get for the combo \$1,735.8 or you would be buying each dollar at Bs. 5.14. Does not sound that attractive, right? It gets worse…
Because this is the best case scenario. In reality these bonds should not only yield more, but when they begin trading they will yield even more as the market absorbs them. As an example, the 2019 and 20124 bonds issued two weeks ago, have always yielded at least 1% above similar Venezuela bonds and are still above them.
Additionally, the longer bonds should yield more, even if the difference is small. Using as a guide Venezuela’s bonds, the 2016 yields 0.7% more than the 2014.
Thus, I think you can assume safely:
1) The 2014 will yield at least 1% more than the 2011 because of markets and you have to add 0.35% per year. Thus the 2014 should be at 16.8%+1% (markets)+1.05% (longer maturity 0.35% per year) for a total yield of 18.85%
2) The 2015 should be 0.35% above that or at 19.2%
3) The 2016 should be 0.35% above that or at 19.55%
This gives a value for each dollar of Bs. 5.69 per US\$, 13.8% above the close on Friday.
Definitely not worth it.
Why did they do this? Because PDVSA thinks the bonds should trade higher. They are probably assuming they should trade with Venezuela’s bonds or its own New York issued bonds. But markets say this is not the case.
Thus, unless PDVSA changes the 138% factor to make it more attractive, the issue should fail. People are not dumb.
Unless there is a hidden agenda somewhere and there is always an edge with the robolution when it comes to these bond issues. I haven’t figured out where this one could be. Maybe PDVSA will start buying the Petrobonos like crazy tomorrow and drive down the yield sharply to make the comparison more attractive (Remember I am assuming Friday’s prices in all this, if the Petrobono goes up in price the numbers would improve)
We shall see, I will keep you posted. For now, stay away from this.
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# Dividing Whole Numbers Word Problems Worksheets
A Reasonable Phone numbers Worksheet may help your youngster become more knowledgeable about the concepts behind this proportion of integers. Within this worksheet, students will be able to resolve 12 distinct issues associated with rational expression. They may learn to multiply several phone numbers, class them in sets, and find out their goods. They may also process simplifying logical expressions. When they have learned these concepts, this worksheet is a beneficial device for advancing their studies. Dividing Whole Numbers Word Problems Worksheets.
## Reasonable Figures really are a ratio of integers
There are 2 types of phone numbers: rational and irrational. Rational numbers are described as entire phone numbers, while irrational amounts tend not to repeat, and have an endless amount of digits. Irrational amounts are no-absolutely no, low-terminating decimals, and rectangular roots which are not excellent squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To define a logical amount, you need to realize such a rational number is. An integer is a complete quantity, and a logical number is actually a proportion of two integers. The rate of two integers will be the variety on top split from the variety at the base. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They can be manufactured in a small fraction
A reasonable quantity features a denominator and numerator that are not no. This means that they are often indicated as a fraction. Together with their integer numerators and denominators, rational numbers can also have a adverse importance. The unfavorable value ought to be positioned on the left of along with its definite worth is its extended distance from zero. To simplify this case in point, we will claim that .0333333 is really a small fraction that could be created as a 1/3.
As well as negative integers, a reasonable variety can also be produced into a small fraction. For example, /18,572 can be a reasonable quantity, although -1/ will not be. Any portion consisting of integers is reasonable, so long as the denominator will not include a and will be created being an integer. Also, a decimal that ends in a point is yet another logical variety.
## They make perception
Even with their label, realistic figures don’t make significantly feeling. In math, they are solitary organizations having a special length about the number line. Which means that whenever we add up one thing, we could order the size by its proportion to the original number. This holds correct even when there are unlimited logical numbers between two certain numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the time period of a pearl, for instance, we might add up its width. An individual pearl weighs twenty kilograms, which is actually a logical number. Additionally, a pound’s weight means ten kilograms. Therefore, we should certainly divide a lb by 15, with out be worried about the length of an individual pearl.
## They are often indicated as a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal number can be created being a several of two integers, so 4 times 5 is equal to eight. A similar dilemma involves the repeated fraction 2/1, and each side needs to be separated by 99 to obtain the proper solution. But how do you make the conversion? Here are some good examples.
A logical variety can also be developed in various forms, which includes fractions along with a decimal. A good way to symbolize a logical quantity in the decimal is usually to separate it into its fractional equal. There are three ways to break down a logical quantity, and each of these ways brings its decimal equivalent. One of these techniques would be to split it into its fractional counterpart, and that’s what’s called a terminating decimal.
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# Talk:Kernel (statistics)
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## The figure with all the kernels in a common coordinate frame seems to have a bug
There's a curve on this plot that seems to be labeled "exponential". This curve does not look exponential to me. And nor does an exponential kernel appear to be defined in the body of the article. —Preceding unsigned comment added by 151.203.248.95 (talk) 21:49, 16 April 2009 (UTC)
## Epanechnikov Kernel
The Epanechnikov Kernel shown on the page is only correct for 1 dimensional problems.
There is a generalization for n dimensions, but the prefactor has to change cf. Simon, Optimal State Estimation, p.472
There should be explicit warning about the dimensional restrictions of the form of the kernel to avoid misuse.
Thanks, Frank —Preceding unsigned comment added by 76.124.111.215 (talk) 04:41, 9 November 2008 (UTC)
## A question from 167.191.250.81
triweight--is that the same as tricube? Which I understood to be:
$K(u)=\left(1-\left|u^3\right|\right)^3$
## Why is this approach "non-parametric"
I have to admit I'm not as familiar with this subject as I ought to be. The stuff I remember about non-parametric statistics includes things like the rank-sum test, which really does not make any assumptions about the distribution of the two samples beyond the null hypothesis (both samples drawn from identical populations?).
After reading this article and a couple of related articles, it appears that the "kernel" is a sort of prior distribution. So why are these methods called non-parametric? And why isn't that question addressed somewhere in Wikipedia? Or have I just been reading the wrong articles? DavidCBryant 16:43, 11 August 2007 (UTC)
Actually, this question is addressed in Wikipedia. See Non-parametric statistics. Solarapex 23:14, 10 October 2007 (UTC)
## Munaf Kernel is incorrect?
Something is wrong with this kernel. I could not find any references to this kernel. The current formula states:
$K(u)=\frac{45}{64}(1-u^2)^4\ 1_{(|u|\leq1)}$
But it does not satisfy the requirement:
$\int K(u) du = 1$
The one that satisfies this requirement is:
$K(u)=\frac{315}{256}(1-u^2)^4\ 1_{(|u|\leq1)}$
Otherwise, the Munaf kernel should be (to follow the ratio):
$K(u)=\frac{45}{64}(1-u^4)^2\ 1_{(|u|\leq1)}$
### Generalization
In general, Epanechnikov, Quartic, Triweight, and Munaf can be generalized as:
$K(u)=B (1-u^{2m})^n\ 1_{(|u|\leq1)}$,
where
$B = \frac{1}{2 \times \sum_{j=0}^{n} C_j^n \frac{(-1)^j}{2mj + 1} }$,
$C_j^n$ - the number of combinations.
I haven't seen this in books. So I have to warn you, this may be original research. Solarapex 23:14, 10 October 2007 (UTC)
Assuming this was incorrectly copied from a correct source, my money is on the formula (45/65)(1-u^4)^2. In light of the lack of sources and importance ("Munaf kernel" only gets a Google hit on this page), I've removed it from the article. --Lambiam 11:07, 29 October 2007 (UTC)
I cannot find the above generalization, but I have found two generalizations in (http://www.ssc.wisc.edu/~bhansen/718/NonParametrics1.pdf) and (http://www2.math.cycu.edu.tw/TEACHER/MSYANG/yang-pdf/yang-n-56-mean-shift.pdf). From one source: ks(u) = (2s + 1)!! / (2^(s+1) * s!) * (1 - u^2)^s * indicator_func(|u| <= 1). The other source is somewhat similar.
## Proposed merge of Kernel (statistics) and Kernel smoother
Isn't a merge with Kernel density estimation more appropriate? --Lambiam 20:11, 16 March 2008 (UTC)
I think they are different things, although perhaps the redundancy between them could be reduced somehow. --Zvika (talk) 05:32, 17 March 2008 (UTC)
The procedure of kernel density estimation clearly is a form of kernel smoothing. Could you give a hint what you see as an essential difference? --Lambiam 19:30, 17 March 2008 (UTC)
It seems to me that both the setting and the applications are different. In one case we have want to estimate a function from noisy measurements, and in the other we want to estimate a pdf from iid realizations of its random variable. The technique itself is similar. However, I am not an expert on this, so barring any further resistance, go ahead and do what you think is right. --Zvika (talk) 07:03, 18 March 2008 (UTC)
In kernel density estimation you are actually trying to estimate the number of measurements inside each window, when with kernel smoother you are estimating the values of data points (and not the number of data points). In some sense, both techniques are similar (density estimation is a particular case of kernel smoother, when all the measurement points has the value 1). Anyway, I think that kernel density estimation is an important topic, and it should have its own article --Anry11 (talk) 17:13, 18 March 2008 (UTC)
There is also the page density estimation to consider for overlap. Melcombe (talk) 11:56, 9 April 2008 (UTC)
## Missing inverse in the definition
The last line of the definition states that "If K is a kernel, then so is the function K* defined by K*(u) = λ−1K(λu), where λ > 0." Isn't the λ "inside" the function supposed to be inverted as well? I'm pretty sure that this is the case but since I haven't written here before I don't dare to change it myself. —Preceding unsigned comment added by 217.10.116.172 (talk) 21:57, 15 April 2008 (UTC)
No, I think the definition is correct as it is: this is the way a scaling factor is used. However, please don't hesitate in the future to make changes yourself (even if you are not sure of them); in the worst case, someone will come along and correct you. --Zvika (talk) 04:05, 16 April 2008 (UTC)
This was indeed an error. I've fixed it. --Lambiam 05:06, 24 April 2008 (UTC)
You're right. I must have been confused that day :) --Zvika (talk) 10:46, 24 April 2008 (UTC)
## Exponential Kernel
$K(u) = e^{-u} 1_{u > 0}$. It's referred in the article, but there's no any information on it. Needs a graph. Solar Apex (talk) 07:14, 24 October 2010 (UTC)
## Optimal Smoothing/Bandwidth for Kernels?
I have read (see pages 31-32 or so of http://books.google.com/books?id=7WBMrZ9umRYC&printsec=frontcover#v=snippet&q=Optimal%20Smoothing&f=false) that the optimal smoothing parameter can be verified if the distribution is known. But there are a couple commonly used widths for normal kernels:
$(\frac{4}{3n})^(1/5) * \sigma$
where sigma is either the standard deviation or for a more robust estimation it is the median of absolute differences divided by 0.6745.
Can someone please create a section for optimal kernel widths assuming different kernels (as the above only applies to normal distributions) — Preceding unsigned comment added by 150.135.222.234 (talk) 20:02, 2 April 2013 (UTC)
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## Jordan Curve Theorem Statement
The Jordan curve theorem asserts that every Jordan curve divides the plane into an interior region bounded by the curve and an exterior region containing all of the nearby and far away exterior points so that any continuous path connecting a point of one region to a point of the other intersects with that loop somewhere. Lemma 41 i Bd roC r for all a.
Pdf A Nonstandard Proof Of The Jordan Curve Theorem
### E Aii exactly one of r as has bounded complement.
Jordan curve theorem statement. A so-called Jordan curve it holds that the complement has exactly two. GENERAL I ARTICLE Proof of Jordan Curve Theorem Let f be a simple closed curve in E2 and r OOEA be the components of E2 – r. S 1 R 2.
I If E I-. Cal theorems of mathematics the Jordan curve theorem. Finally a simple path or closed curve is polygonal if it is the union of a finite number of line segments called edges.
The Jordan curve theorem states that every simple closed curve has a well-defined inside and outside. The other component which we. S 1 R 2.
The celebrated theorem of Jordan states that every simple closed curve in the plane separates the complement into two connected nonempty sets. In topology the Jordan curve theorem asserts that every Jordan curve a plane simple closed curve divides the plane into an interior region bounded by the curve and an exterior region containing all of the nearby and far away exterior points. Not sure whether youd consider it.
Definitions and the statement of the Jordan theorem. It is a plane curve that is not necessarily smooth nor algebraic. The Jordan Curve theorem is actually pretty easy to prove if you assume the curve is smooth or piecewise linear.
The statement of the Jordan curve theorem may seem obvious at first but it is a rather difficult theorem to prove. It was easy to establish the result for simple curves such as polygonal lines but the problem came in generalising it for all kind of curves which included nowhere differentiable curves such as the Koch snowflake. The statement of the Jordan curve theorem seems obvious but it was a very difficult theorem to prove.
Now as r is topologically closed each r 0. It states that a simple closed curve ie a closed curve which does not cross itself always separates the plane E2 into two pieces. Openness of r 0.
The Jordan curve theorem asserts that every Jordan curve divides the plane into an interior region bounded by the curve and an exterior region containing all of the nearby and far away exterior points so that any continuous path connecting a point of one region to a point of the other intersects with that loop somewhere. Any Jordan curve divides its complement in the plane into two connected components. The Jordan curve theorem states the following.
The Jordan Curve Theorem via the Brouwer Fixed Point Theorem The goal of the proof is to take Moises intuitive proof and make it simplershorter. In topology the Jordan curve theorem asserts that every Jordan curve a plane simple closed curve divides the plane into an interior region bounded by the curve and an exterior region containing all of the nearby and far away exterior points. A Jordan curve or a simple closed curve in the plane R 2 is the image C of an injective continuous map of a circle into the plane φ.
One of these connected components which we call the interior component is bounded. This includes nowhere-differentiable curves like the boundary of the Koch snowflake and even wilder curves which cant even be drawn by hand like Mariano says. Assures us that A is a countable set.
An interior region and an exterior. A Jordan arc in the plane is the image of an injective continuous map of a closed and bounded interval a b into the plane. Jordan Curve Theorem Pro only the first part of the full theorem but this time in a fairly precise version For each subset of the plane that is homeomorphic to the circle which comes to the same thing as being the image of the circle under a continuous injection.
Bernard Bolzano was the first to formulate a precise conjecture observing that it was not a self-evident statement but that it required a proof. The theorem states that every continuous loop where a loop is a closed curve in the Euclidean plane which does not intersect itself a Jordan curve divides the plane into two disjoint subsets the connected components of the curves complement a bounded region inside the curve and an unbounded region outside of it each of which has the original curve as its boundary. Jordan Curve Theorem Any continuous simple closed curve in the plane separates the plane into two disjoint regions the inside and the outside.
Definitions and the statement of the Jordan theorem. For example it is easy to see that the unit cir cle 8 1 xiy E C. The difficulty arises when you try to handle the general case.
The Jordan Curve Theorem It is established then that every continuous closed curve divides the plane into two regions one exterior one interior. Jordans theorem on group actions characterizes primitive groups containing a large p-cycle. A Jordan arc in the plane is the image of an injective continuous map of a closed and bounded interval a b into the plane.
X2y2 1 separates the plane into. A Jordan curve or a simple closed curve in the plane R 2 is the image C of an injective continuous map of a circle into the plane φ. The result was first stated as a theorem in Camille Jordans famous textbook Cours dAnalyze de lÉcole Polytechnique.
An endpoint of an edge is called a vertex. The full-fledged Jordan curve theorem states that for any simple closed curve C in the plane the complement R2 nC has exactly two connected components. Jordans lemma is a bound for the error term in applications of the residue theorem.
Complex Analysis Jordan Curve Theorem Professor Tao S Proof Mathematics Stack Exchange
2 Examples Of Jordan Curves The Jordan Curve Tessellates The Space Download Scientific Diagram
Pdf A Proof Of The Jordan Curve Theorem
Jordan Curve Theorem A Simple Closed Curve Cuts Its Interior From Its Exterior Ppt Download
Algebraic Topology Why Is The Jordan Curve Theorem Not Obvious Mathematics Stack Exchange
Jordan Curve Theorem 3d Warehouse
Graph Theory Definition Of A Curve Jordan Curve Theorem Mathematics Stack Exchange
Gt Geometric Topology Nice Proof Of The Jordan Curve Theorem Mathoverflow
Pdf A Proof Of The Jordan Curve Theorem
Complex Analysis Jordan Curve Theorem Professor Tao S Proof Mathematics Stack Exchange
Proof Of The Jordan Curve Theorem The Jordan Curve Theorem States That A Non Self Intersecting Continuous Loop In The Plane I C A Jordan Curve Divides The Plane Into An Interior Region And An
Gn General Topology Quantitative Winding Number Mathoverflow
Jordan Curve An Overview Sciencedirect Topics
A Proof Of The Jordan Curve Theorem Via The Brouwer Fixed Point Theorem Semantic Scholar
File Jordan Curve Theorem Svg Wikipedia
General Topology Proving Interiorality And Exteriorality Of Curves Using The Jordan Curve Lemma Mathematics Stack Exchange
File Jordan Curve Theorem Svg Wikipedia
Why Did The Jordan Curve Theorem Need Proof Is It Not Obvious Enough Quora
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# Infix to Postfix Conversion
• 01-02-2013, 09:32 PM
David M.
Infix to Postfix Conversion
I'm trying to convert a given infix expression into postfix. Right now I'm trying to get a simple test case working. So far it just accepts integer inputs, then converts the expression into postfix notation and stores it in an ArrayList for simplifying later. I have the simplification working but my conversion does something weird. If I take the expression 4*(2+3), it should come out "4 2 3 + *" but my converter is throwing the left (opening) parenthesis into the equation so it looks like "4 2 3 + ( *" and I can't figure out why. Here's the code I'm using:
Code:
```import java.util.ArrayList; import java.util.Scanner; import java.util.Stack; public class Converter { public static void main(String[ ] args) { Scanner in = new Scanner(System.in); System.out.println("Enter an infix expression"); String input = in.nextLine(); ArrayList<String> postfixEquation = new ArrayList<String>(); Stack<String> stack = new Stack<String>(); String curNum = ""; String curToken = ""; int curTokenPrecedence = 0; for(int i = 0; i < input.length(); i++) { curToken = input.substring(i, i+1); if(isInteger(curToken)) //If the current token is a number curNum += curToken; //Add it to the curNumber string else //Not a number { if(curNum != "") //If there is a number in curNum { postfixEquation.add(curNum); //Add the current number to the postfix string curNum = ""; //Reset curNum } curTokenPrecedence = getPrecedence(curToken); //Get the current token's precedence //If there is no number on the stack or the token's operator is greater than the first operator on the stack if(stack.empty() || curTokenPrecedence > getPrecedence(stack.peek()) || curToken.equals("(")) stack.push(curToken); else //Current token's precedence less than token on the stack { //Until there is no operator on the stack or an operator with lower precedence while(!stack.empty() && curTokenPrecedence <= getPrecedence(stack.peek())) postfixEquation.add(stack.pop()); //Pop operators off the stack stack.push(curToken); } } if(curToken.equals(")")) { stack.pop(); //Discard the right parenthesis while(!stack.empty()) { if(stack.peek().equals("(")) { stack.pop(); break; } postfixEquation.add(stack.pop()); } } } if(curNum != "") postfixEquation.add(curNum); //Pop remaining operators off the stack while(!stack.empty()) postfixEquation.add(stack.pop()); for(int i = 0; i < postfixEquation.size(); i++) System.out.println(postfixEquation.get(i)); in.close(); } //Returns whether the given string is an integer or an operator public static boolean isInteger(String num) { try { Integer.parseInt(num); return true; //Number } catch(NumberFormatException e) { return false; //Operator } } //Returns precedence for the given operator public static int getPrecedence(String operator) { int precedence = 0; switch(operator) { case "!": precedence = 4; case "^": precedence = 3; break; case "*": case "/": precedence = 2; break; case "+": case "-": precedence = 1; break; case "(": precedence = 0; break; } return precedence; } }```
• 01-02-2013, 09:39 PM
KevinWorkman
Re: Infix to Postfix Conversion
Have you stepped through this with a debugger, or at least added some print statements, to figure out what's going on? Debugging your code is an important lesson that every programmer has to learn.
Hint: Don't use == or != with Strings. Do a search on this forum or on google for an explanation of what to use instead, and why.
• 01-03-2013, 12:09 AM
David M.
Re: Infix to Postfix Conversion
I had been running through this section of code:
Code:
```if(curToken.equals(")")) { stack.pop(); //Discard the right parenthesis while(!stack.empty()) { if(stack.peek().equals("(")) { stack.pop(); break; } postfixEquation.add(stack.pop()); } }```
I just ran through everything and found the problem was in this comparison:
Code:
```while(!stack.empty() && curTokenPrecedence <= getPrecedence(stack.peek())) postfixEquation.add(stack.pop()); //Pop operators off the stack```
Since I hadn't assigned a precedence to ')' it got the default 0. This was <= everything on the stack so the stack was emptied. I fixed it by assigning ')' to a precedence higher than anything else.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > gsumpt Structured version Visualization version Unicode version
Theorem gsumpt 17672
Description: Sum of a family that is nonzero at at most one point. (Contributed by Stefan O'Rear, 7-Feb-2015.) (Revised by Mario Carneiro, 25-Apr-2016.) (Revised by AV, 6-Jun-2019.)
Hypotheses
Ref Expression
gsumpt.b
gsumpt.z
gsumpt.g
gsumpt.a
gsumpt.x
gsumpt.f
gsumpt.s supp
Assertion
Ref Expression
gsumpt g
Proof of Theorem gsumpt
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 gsumpt.f . . . 4
2 gsumpt.x . . . . 5
32snssd 4108 . . . 4
41, 3feqresmpt 5933 . . 3
54oveq2d 6324 . 2 g g
6 gsumpt.b . . 3
7 gsumpt.z . . 3
8 eqid 2471 . . 3 Cntz Cntz
9 gsumpt.g . . 3
10 gsumpt.a . . 3
111, 2ffvelrnd 6038 . . . . . . . 8
12 eqidd 2472 . . . . . . . 8
13 eqid 2471 . . . . . . . . . 10
146, 13, 8elcntzsn 17057 . . . . . . . . 9 Cntz
1511, 14syl 17 . . . . . . . 8 Cntz
1611, 12, 15mpbir2and 936 . . . . . . 7 Cntz
1716snssd 4108 . . . . . 6 Cntz
18 eqid 2471 . . . . . . 7 mrClsSubMnd mrClsSubMnd
19 eqid 2471 . . . . . . 7 s mrClsSubMnd s mrClsSubMnd
208, 18, 19cntzspan 17560 . . . . . 6 Cntz s mrClsSubMnd CMnd
219, 17, 20syl2anc 673 . . . . 5 s mrClsSubMnd CMnd
226submacs 16690 . . . . . . . 8 SubMnd ACS
23 acsmre 15636 . . . . . . . 8 SubMnd ACS SubMnd Moore
249, 22, 233syl 18 . . . . . . 7 SubMnd Moore
2511snssd 4108 . . . . . . 7
2618mrccl 15595 . . . . . . 7 SubMnd Moore mrClsSubMnd SubMnd
2724, 25, 26syl2anc 673 . . . . . 6 mrClsSubMnd SubMnd
2819, 8submcmn2 17557 . . . . . 6 mrClsSubMnd SubMnd s mrClsSubMnd CMnd mrClsSubMnd CntzmrClsSubMnd
2927, 28syl 17 . . . . 5 s mrClsSubMnd CMnd mrClsSubMnd CntzmrClsSubMnd
3021, 29mpbid 215 . . . 4 mrClsSubMnd CntzmrClsSubMnd
31 ffn 5739 . . . . . . 7
321, 31syl 17 . . . . . 6
33 simpr 468 . . . . . . . . . 10
3433fveq2d 5883 . . . . . . . . 9
3524, 18, 25mrcssidd 15609 . . . . . . . . . . 11 mrClsSubMnd
36 fvex 5889 . . . . . . . . . . . 12
3736snss 4087 . . . . . . . . . . 11 mrClsSubMnd mrClsSubMnd
3835, 37sylibr 217 . . . . . . . . . 10 mrClsSubMnd
3938ad2antrr 740 . . . . . . . . 9 mrClsSubMnd
4034, 39eqeltrd 2549 . . . . . . . 8 mrClsSubMnd
41 eldifsn 4088 . . . . . . . . . . 11
42 gsumpt.s . . . . . . . . . . . 12 supp
43 fvex 5889 . . . . . . . . . . . . . 14
447, 43eqeltri 2545 . . . . . . . . . . . . 13
4544a1i 11 . . . . . . . . . . . 12
461, 42, 10, 45suppssr 6965 . . . . . . . . . . 11
4741, 46sylan2br 484 . . . . . . . . . 10
487subm0cl 16677 . . . . . . . . . . . 12 mrClsSubMnd SubMnd mrClsSubMnd
4927, 48syl 17 . . . . . . . . . . 11 mrClsSubMnd
5049adantr 472 . . . . . . . . . 10 mrClsSubMnd
5147, 50eqeltrd 2549 . . . . . . . . 9 mrClsSubMnd
5251anassrs 660 . . . . . . . 8 mrClsSubMnd
5340, 52pm2.61dane 2730 . . . . . . 7 mrClsSubMnd
5453ralrimiva 2809 . . . . . 6 mrClsSubMnd
55 ffnfv 6064 . . . . . 6 mrClsSubMnd mrClsSubMnd
5632, 54, 55sylanbrc 677 . . . . 5 mrClsSubMnd
57 frn 5747 . . . . 5 mrClsSubMnd mrClsSubMnd
5856, 57syl 17 . . . 4 mrClsSubMnd
598cntzidss 17069 . . . 4 mrClsSubMnd CntzmrClsSubMnd mrClsSubMnd Cntz
6030, 58, 59syl2anc 673 . . 3 Cntz
61 ffun 5742 . . . . 5
621, 61syl 17 . . . 4
63 snfi 7668 . . . . 5
64 ssfi 7810 . . . . 5 supp supp
6563, 42, 64sylancr 676 . . . 4 supp
66 fex 6155 . . . . . 6
671, 10, 66syl2anc 673 . . . . 5
68 isfsupp 7905 . . . . 5 finSupp supp
6967, 45, 68syl2anc 673 . . . 4 finSupp supp
7062, 65, 69mpbir2and 936 . . 3 finSupp
716, 7, 8, 9, 10, 1, 60, 42, 70gsumzres 17621 . 2 g g
72 fveq2 5879 . . . 4
736, 72gsumsn 17665 . . 3 g
749, 2, 11, 73syl3anc 1292 . 2 g
755, 71, 743eqtr3d 2513 1 g
Colors of variables: wff setvar class Syntax hints: wi 4 wb 189 wa 376 wceq 1452 wcel 1904 wne 2641 wral 2756 cvv 3031 cdif 3387 wss 3390 csn 3959 class class class wbr 4395 cmpt 4454 crn 4840 cres 4841 wfun 5583 wfn 5584 wf 5585 cfv 5589 (class class class)co 6308 supp csupp 6933 cfn 7587 finSupp cfsupp 7901 cbs 15199 ↾s cress 15200 cplusg 15268 c0g 15416 g cgsu 15417 Moorecmre 15566 mrClscmrc 15567 ACScacs 15569 cmnd 16613 SubMndcsubmnd 16659 Cntzccntz 17047 CMndccmn 17508 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1677 ax-4 1690 ax-5 1766 ax-6 1813 ax-7 1859 ax-8 1906 ax-9 1913 ax-10 1932 ax-11 1937 ax-12 1950 ax-13 2104 ax-ext 2451 ax-rep 4508 ax-sep 4518 ax-nul 4527 ax-pow 4579 ax-pr 4639 ax-un 6602 ax-inf2 8164 ax-cnex 9613 ax-resscn 9614 ax-1cn 9615 ax-icn 9616 ax-addcl 9617 ax-addrcl 9618 ax-mulcl 9619 ax-mulrcl 9620 ax-mulcom 9621 ax-addass 9622 ax-mulass 9623 ax-distr 9624 ax-i2m1 9625 ax-1ne0 9626 ax-1rid 9627 ax-rnegex 9628 ax-rrecex 9629 ax-cnre 9630 ax-pre-lttri 9631 ax-pre-lttrn 9632 ax-pre-ltadd 9633 ax-pre-mulgt0 9634 This theorem depends on definitions: df-bi 190 df-or 377 df-an 378 df-3or 1008 df-3an 1009 df-tru 1455 df-ex 1672 df-nf 1676 df-sb 1806 df-eu 2323 df-mo 2324 df-clab 2458 df-cleq 2464 df-clel 2467 df-nfc 2601 df-ne 2643 df-nel 2644 df-ral 2761 df-rex 2762 df-reu 2763 df-rmo 2764 df-rab 2765 df-v 3033 df-sbc 3256 df-csb 3350 df-dif 3393 df-un 3395 df-in 3397 df-ss 3404 df-pss 3406 df-nul 3723 df-if 3873 df-pw 3944 df-sn 3960 df-pr 3962 df-tp 3964 df-op 3966 df-uni 4191 df-int 4227 df-iun 4271 df-iin 4272 df-br 4396 df-opab 4455 df-mpt 4456 df-tr 4491 df-eprel 4750 df-id 4754 df-po 4760 df-so 4761 df-fr 4798 df-se 4799 df-we 4800 df-xp 4845 df-rel 4846 df-cnv 4847 df-co 4848 df-dm 4849 df-rn 4850 df-res 4851 df-ima 4852 df-pred 5387 df-ord 5433 df-on 5434 df-lim 5435 df-suc 5436 df-iota 5553 df-fun 5591 df-fn 5592 df-f 5593 df-f1 5594 df-fo 5595 df-f1o 5596 df-fv 5597 df-isom 5598 df-riota 6270 df-ov 6311 df-oprab 6312 df-mpt2 6313 df-om 6712 df-1st 6812 df-2nd 6813 df-supp 6934 df-wrecs 7046 df-recs 7108 df-rdg 7146 df-1o 7200 df-oadd 7204 df-er 7381 df-en 7588 df-dom 7589 df-sdom 7590 df-fin 7591 df-fsupp 7902 df-oi 8043 df-card 8391 df-pnf 9695 df-mnf 9696 df-xr 9697 df-ltxr 9698 df-le 9699 df-sub 9882 df-neg 9883 df-nn 10632 df-2 10690 df-n0 10894 df-z 10962 df-uz 11183 df-fz 11811 df-fzo 11943 df-seq 12252 df-hash 12554 df-ndx 15202 df-slot 15203 df-base 15204 df-sets 15205 df-ress 15206 df-plusg 15281 df-0g 15418 df-gsum 15419 df-mre 15570 df-mrc 15571 df-acs 15573 df-mgm 16566 df-sgrp 16605 df-mnd 16615 df-submnd 16661 df-mulg 16754 df-cntz 17049 df-cmn 17510 This theorem is referenced by: gsummpt1n0 17675 dprdfid 17728 evlslem3 18814 evlslem1 18815 coe1tmmul2 18946 coe1tmmul 18947 uvcresum 19428 frlmup2 19434 mamulid 19543 mamurid 19544 coe1mul3 23127 tayl0 23396 jensen 23993 linc1 40726
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# toricGraver -- calculates the Graver basis of the toric ideal; invokes "graver" from 4ti2
## Synopsis
• Usage:
toricGraver(A) or toricGraver(A,R)
• Inputs:
• A, , whose columns parametrize the toric variety. The toric ideal IA is the kernel of the map defined by A
• R, a ring, polynomial ring in which the toric ideal IA should live
• Outputs:
• B, , whose rows give binomials that form the Graver basis of the toric ideal of A, or
• I, an ideal, whose generators form the Graver basis for the toric ideal
## Description
The Graver basis for any toric ideal IA contains (properly) the union of all reduced Groebner basis of IA. Any element in the Graver basis of the ideal is called a primitive binomial.
```i1 : A = matrix "1,1,1,1; 1,2,3,4" o1 = | 1 1 1 1 | | 1 2 3 4 | 2 4 o1 : Matrix ZZ <--- ZZ``` ```i2 : toricGraver(A) o2 = | 1 -2 1 0 | | 2 -3 0 1 | | 1 -1 -1 1 | | 0 1 -2 1 | | 1 0 -3 2 | 5 4 o2 : Matrix ZZ <--- ZZ```
If we prefer to store the ideal instead, we may use:
```i3 : R = QQ[a..d] o3 = R o3 : PolynomialRing``` ```i4 : toricGraver(A,R) 2 3 2 2 3 2 o4 = ideal (- b + a*c, - b + a d, - b*c + a*d, - c + b*d, - c + a*d ) o4 : Ideal of R```
Note that this last ideal equals the toric ideal IA since every Graver basis element is actually in IA.
## Ways to use toricGraver :
• toricGraver(Matrix)
• toricGraver(Matrix,Ring)
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# IMO Level 1- Mathematics Olympiad (SOF) Class 2: Questions 34 - 41 of 578
Access detailed explanations (illustrated with images and videos) to 578 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. Unlimited Access for Unlimited Time! View Sample Explanation or View Features.
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## Question number: 34
Edit
MCQ▾
### Question
Which fish has the largest value?
### Choices
Choice (4)Response
a.
P
b.
Q
c.
R
d.
All of the above
## Question number: 35
Edit
MCQ▾
### Question
Which of the following is the greatest number in the given figure?
### Choices
Choice (4)Response
a.
667
b.
967
c.
887
d.
567
## Question number: 36
Edit
MCQ▾
### Question
How many cubes are shown in the figure?
### Choices
Choice (4)Response
a.
218
b.
211
c.
201
d.
210
## Question number: 37
Edit
MCQ▾
### Question
The place value of digit in is ________.
### Choices
Choice (4)Response
a.
33
b.
30
c.
3
d.
300
## Question number: 38
Edit
MCQ▾
### Question
Which is the largest numbers in the group?
### Choices
Choice (4)Response
a.
978
b.
878
c.
776
d.
474
## Question number: 39
Edit
MCQ▾
### Question
Study the number chain. Which number box is incorrectly placed?
### Choices
Choice (4)Response
a.
218
b.
400
c.
310
d.
475
## Question number: 40
Edit
MCQ▾
### Question
When written in words, 800 is same as________.
### Choices
Choice (4)Response
a.
Eight hundred ninety
b.
Eight hundred and nine
c.
Eight hundred
d.
Eighty and nine
## Question number: 41
Edit
MCQ▾
### Question
Arrange the number stars to make the greatest possible three-digit number.
### Choices
Choice (4)Response
a.
789
b.
879
c.
978
d.
987
Developed by:
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Next: Exercises Up: Basic MPI Previous: Exercises
## Dividing the Pie
We are now going to analyze the program you should have run already, cpi. The source of this program is distributed with MPICH2, and it lives in the directory /N/hpc/mpich2/src/mpich2-0.94b1/examples. The file name is cpi.c. For your convenience and because it is an open software program, I also quote it below:
gustav@bh1 $pwd /N/hpc/mpich2/src/mpich2-0.94b1/examples gustav@bh1$ cat cpi.c
#include "mpi.h"
#include <stdio.h>
#include <math.h>
double f(double);
double f(double a)
{
return (4.0 / (1.0 + a*a));
}
int main(int argc,char *argv[])
{
int done = 0, n, myid, numprocs, i;
double PI25DT = 3.141592653589793238462643;
double mypi, pi, h, sum, x;
double startwtime = 0.0, endwtime;
int namelen;
char processor_name[MPI_MAX_PROCESSOR_NAME];
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&numprocs);
MPI_Comm_rank(MPI_COMM_WORLD,&myid);
MPI_Get_processor_name(processor_name,&namelen);
fprintf(stdout,"Process %d of %d is on %s\n",
myid, numprocs, processor_name);
fflush(stdout);
n = 10000; /* default # of rectangles */
if (myid == 0)
startwtime = MPI_Wtime();
MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);
h = 1.0 / (double) n;
sum = 0.0;
/* A slightly better approach starts from large i and works back */
for (i = myid + 1; i <= n; i += numprocs)
{
x = h * ((double)i - 0.5);
sum += f(x);
}
mypi = h * sum;
MPI_Reduce(&mypi, &pi, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD);
if (myid == 0) {
endwtime = MPI_Wtime();
printf("pi is approximately %.16f, Error is %.16f\n",
pi, fabs(pi - PI25DT));
printf("wall clock time = %f\n", endwtime-startwtime);
fflush(stdout);
}
MPI_Finalize();
return 0;
}
gustav@bh1 $ This program calculates by using the formula (5.1) The calculation is numerical, i.e., the segment [0,1] is divided into 10,000 equal fragments. Within each fragment we evaluate for the middle of the fragment (this is the height of the rectangle) and multiply it by the width of the fragment to obtain the area of the rectangle. Then we add all the areas together and this is our integral, that is also the approximate value of . The program begins with the usual incantations to MPI_Init, MPI_Comm_size, MPI_Comm_rank and MPI_Get_processor_name. Then each process writes on standard output its rank number (which is called myid here), the size of the communicator (which is called numprocs here) and the name of the processor it runs on. We have not talked about it so far, but this writing on standard output is far from trivial. Every process writes on its own standard output. In some early versions of MPI, e.g., in LAM MPI, these writes were simply lost. But MPICH2 engine collects all standard outputs from its processes and combines them together on the so called console node. This is always the node whose rank is zero, the node on which we run mpdboot. The output you see on the PBS output file is the output from the MPICH2 console node. Now we hardwire the number of division of segment [0,1] into n and the console node begins the timing of the program by calling function MPI_Wtime. This function measures wall-clock time down to a microsecond, sic! It is very accurate indeed. Now the value of n is broadcast to all processes. This is completely unnecessary in this particular program, because all nodes know what n is from the very beginning, but this program has been modified from its interactive predecessor, which used to read n from standard input. It was the console node that used to do the reading. Now every process calculates the width of the rectangles: h = 1.0 / (double) n; Every process initializes its own local variable sum to zero, and then calls on the rectangle that corresponds to its own rank number. Then it jumps to the rectangle that corresponds to its rank number plus the number of processes in the pool, and so on. The values of are added for all rectangles and only at the end they are multiplied by the width of the rectangle. This, of course, saves on unnecessary multiplications. Now we encounter a new operation, MPI_Reduce. This operation takes as its input the content of mypi for each process in the pool (the first argument). We tell it that there is one item (the third argument) of type MPI_DOUBLE (the fourth argument) in there. We are then telling it that it should perform summation over all instances of mypi (the fifth argument, MPI_SUM) and that it should write the result of the operation on pi (the second argument) on the root process (the sixth argument), which is this case is the console process, i.e., process of rank zero. The whole operation is performed within the MPI_COMM_WORLD communicator. At the end of this operation only the root (console) process has the right answer in its location that corresponds to pi. But this is fine. In the next clause we instruct the console process to measure the wall clock time again, then print the value of on standard output together with the estimate of the accuracy of the computation. Then the console process also writes how long it took to carry out the computation. Here is the output of the program. gustav@bh1$ pwd
/N/B/gustav/PBS
gustav@bh1 $cat mpi_out Local MPD console on bc89 Process 0 of 8 is on bc89 Process 1 of 8 is on bc40 Process 2 of 8 is on bc42 Process 3 of 8 is on bc41 Process 4 of 8 is on bc44 Process 5 of 8 is on bc43 Process 6 of 8 is on bc88 Process 7 of 8 is on bc46 pi is approximately 3.1415926544231247, Error is 0.0000000008333316 wall clock time = 0.009541 gustav@bh1$
MPI_SUM is not the only operation you can use with MPI_Reduce. MPI provides a number of predefined reduction operations that can be used in this context, and you can define your own operations too. The predefined ones are:
MPI_MAX MPI_MIN MPI_SUM MPI_PROD MPI_LAND MPI_BAND MPI_LOR MPI_BOR MPI_LXOR MPI_BXOR MPI_MAXLOC MPI_MINLOC
Next: Exercises Up: Basic MPI Previous: Exercises
Zdzislaw Meglicki
2004-04-29
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# chemistry
How much water should be added to 55.0 mL of 0.201 M HCl to dilute the solution to 0.155 M?
i used M1V1 = M2V2 and i got 71.3mL
its wrong so im confused
1. 👍 0
2. 👎 0
3. 👁 128
1. 71.3 mL is the final volume, not the amount of water added to achieve that volume. Does that help?
1. 👍 0
2. 👎 0
posted by GK
2. yeaa thanks!!
1. 👍 0
2. 👎 0
posted by lyne
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# S1 question: What the hell is it talking about?!?!?!? watch
Announcements
1. Hi, first I would like to say that 'EDEXCEL are \$h!t'.
Now, I have an S1 resit tomorrow morning. I got 58 last year so I will definitely need to resit. I'm currently going through the Heinemann's Revise for Statstics 1 book and there's a question on (for those who have it) page 60. It's questions (a) and (b), with (a) as a proof question and then (b) as working out the general probability distribution for X. Anyways, here it goes:
When a certain type of cell is subjected to radiation, the cell may die, survive as a single cell, or divide into two cells with probabilities 1/2, 1/3, 1/6 respectively.
Two cells are independently subjected to radiation. The random variable X represents the total number of cells in existence after this experiment.
(a) Show that P(X=2) = 5/18
(b) Find the probability distribution of X
I have tried all sorts of things wth (a). I did manage to get 5/18 once but I'm pretty sure it wasn't how you work it out. Stupid textbook never gives answers for proof questions. I implemented my method to (b) and it's not right.
Here are the answers for (b):
x: 0 1 2 3 4
p(x): 9/36 12/36 10/36 4/46 1/36
So can anyone help me then because I absolutely have NO IDEA WHAT IS GOING ON. This is why I hate Statistics because some questions blab on and on and I have no-idea what it's talking about. Why can't the questions be straightforward? The exam boards should realise that not everyone is quick at realising stuff and it's about getting the correct answer, not about getting it as quick as possible. I would probably get the answer after looking at it for a few hours but I still need to do a past paper or two before I finish tonight.
2. For a): The probability that there will be 2 cells left at the end is the sum of the probabilities that they both survive and that one dies and the other divides. So:
Probability that they both survive = (1/3)^2 = 1/9
Probability that one dies and the other divides = (1/2)*(1/6) + (1/2)*(1/6)
= 1/6
1/9 + 1/6 = 5/18
Ben
3. (Original post by Ben.S.)
For a): The probability that there will be 2 cells left at the end is the sum of the probabilities that they both survive and that one dies and the other divides. So:
Probability that they both survive = (1/3)^2 = 1/9
Probability that one dies and the other divides = (1/2)*(1/6) + (1/2)*(1/6)
= 1/6
1/9 + 1/6 = 5/18
Ben
Oh I see! I thought that the probability that they either remain as one cell or divde already means that they will stay alive! They should word these questions better! Thanks man!
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C2 Maths AS aqa 2016 (unofficial mark scheme new)Watch
Announcements
2 years ago
#281
Anybody, for the sector question i thought you could write it as 19.9 = 2 multiplide by area of sector? isnt this right im sure it is, since the sector multiplied would get you area of triangle?
0
2 years ago
#282
(Original post by beanigger)
UPDATED THE MARK SCHEME SORRY FOR THE MISTAKES
i know the marks dont add up to 75 but i cant recall all the correct marks for each question
if there is anything else wrong please let me know
thanks
hey where the updated version ? and well mistakes ok cant expect to b perfect did better than i would have
0
2 years ago
#283
(Original post by Alexgood123)
I believe that question 5a) said to expand the binomial equation and show it in its simplest form which cant you divide the expansion in your answer by 8 to show it in its simplest form.
It just said expand this in the form " " with p q and r being integers or something.
0
2 years ago
#284
(Original post by SunDun111)
Anybody, for the sector question i thought you could write it as 19.9 = 2 multiplide by area of sector? isnt this right im sure it is, since the sector multiplied would get you area of triangle?
Thats correct; because it is area of sector=area of triangle-area of sector. So you rearrange and get area of sector x 2=area of triangle. theta x r^2=19.9
0
2 years ago
#285
(Original post by Parhomus)
Thats correct; because it is area of sector=area of triangle-area of sector. So you rearrange and get area of sector x 2=area of triangle. theta x r^2=19.9
I didnt show the rearrangement in my answer though because of time restrictions but thanks
0
2 years ago
#286
(Original post by SunDun111)
I didnt show the rearrangement in my answer though because of time restrictions but thanks
Yeah, I considered doing it but I wasn't sure, so I showed it anyways. I think you get the marks for the working regardless because it is implied.
0
2 years ago
#287
(Original post by Porkieee.ee)
18/2(56+17(-2))
9(56-34)
9(22)
=198
No, if you are going to do it that way, your first term is u4, which is 22, not 28. If you use this, you get the answer 90, which is the same as above.
0
2 years ago
#288
(Original post by beanigger)
Unoffical Mark scheme for C2 AQA 2016
It would help if you could correct the amount of marks in each question if it is wrong please
Questions:
1)a) integrate something to get -36x-1 + ax2 /2 [3]
b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a = - 2 [2]
2)a) Draw the graph of (0.2)x and state where itcrosses axis (0,1)[2]
b) use logarithms to solve (0.2)x = 4 , x = -0.861 [2]
c) describe transformation of (0.2)x onto (5)x reflection in y axis [1]
3)a) differentiate something to get 3x-1/2 -1 [2]
b) find y co-ordinate of the maximum point M M(9,6) [3]
c) equation of normal to curve at P, P was (4,5) y=-2x+13 [2]
d) the normal to curve at P is translated by (k,0) find value of k, k = 5.5 [3]
4)a) show that a+10d=8 [2]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series D=-2 and A=28, u12 = 6 [3]
c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]
5)a) show that thetha (in radians) was 0.568 [2]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given -> area of sector = area of shaded area
you had to realise that the area of sector = area of triangle - area of sector
re-arrange to get 2(area of sector) = 19.9 and solve for r to get 5.829
perimeter of shaded = (9-r) + (8-r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.8 [6]
6)a)trapezium rule to get 65.6 [4]
b) describe transformation of graph y=(x2 +9)1/2 onto y = 5 + (x2 +9)1/2
Translation by (0, 5) [2]
c) describe transformation of graph y=(x2 +9)1/2 onto y=3(x2 +1)1/2 Stretch by scale factor 1/3 in x direction [2]
7)a) expand (1+2x)5 work out p q r, i think p = -10, q= 40 r = - 80 (correct me if im wrong) [4]
b)find the coefficient of x10 in the expansion of (1+2x)^5 * (2+x)^7
it was something like -1648 [5]
8)a)find the value of tan(x) tan(x) = -5/4 [2]
b) values of tanx in the range of 0 to 360 degrees 45, 129, 225, 309 [3]
b) re-arrange 16+9sin2x / 5 - 3cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]
9)re-arrange something into the form (c)1/2 / d^2 to get 3y y = 1/2 ( m - 12n) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4]
3c was 3 marks, not 2
Q4 a,b,c was 3, 4 and 3 marks
Q5 and Q6 are the wrong way around - Q5 is the trapezium rule a4, b2, c2 marks and Q6 the triangle question - a3, b2,c6 marks
Q7 a, b are 3 and 5 marks
The rest of the marks are correct
0
2 years ago
#289
(Original post by SunDun111)
Anybody, for the sector question i thought you could write it as 19.9 = 2 multiplide by area of sector? isnt this right im sure it is, since the sector multiplied would get you area of triangle?
Yeah that'd work, as it's just area of sector = area of triangle / 2 rearranged
0
2 years ago
#290
(Original post by Alexgood123)
I believe that question 5a) said to expand the binomial equation and show it in its simplest form which cant you divide the expansion in your answer by 8 to show it in its simplest form.
It wasn't an equation, just the expansion of (1-2x)^5. It didn't ask for it in its simplest form, but as it's not an equation, you can't divide by 8 anyway
0
2 years ago
#291
(Original post by Eisobdxhsonw)
For question 6 part C I got stretch in the X by a scale factor of 1/9
Ill show you how I got it
First I rearranged the y=(x^2+9)^1/2 to make it easier to compare
Y=(x^2+9)^1/2
Y=(9(x^2/9 +1))^1/2
Y=3(x^2/9+1)^1/2
Comparing that with y=3(x^2+1)^1/2 you would have to stretch it in the X by a scale factor of 1/9???
I know this doesn't look too nice with all the powers and divides but if you write it down it makes more sense
I was just wondering if anyone could confirm my method or tell me where I went wrong as I believe this could have been one of those transformations with multiple answers??
I might be wrong, but aren't you doing this the wrong way around? We want to transform (x^2 +9)^1/2 into 3(x^2+1)^1/2. For it to become 3(x^2+1)^1/2, it must first become (9(x^2 + 1)^1/2 = (9x^2+9)^1/2 = ((3x)^2+9)^1/2. This means x has transformed into 3x, so it is a stretch parallel to the x axis, sf 1/3)
0
2 years ago
#292
(Original post by Bosssman)
Yeah that'd work, as it's just area of sector = area of triangle / 2 rearranged
kinda nervous cos i didnt right that statement down tho, but i did show it in my working i think
0
Thread starter 2 years ago
#293
(Original post by smithj)
3c was 3 marks, not 2
Q4 a,b,c was 3, 4 and 3 marks
Q5 and Q6 are the wrong way around - Q5 is the trapezium rule a4, b2, c2 marks and Q6 the triangle question - a3, b2,c6 marks
Q7 a, b are 3 and 5 marks
The rest of the marks are correct
thanks, its been updated
0
2 years ago
#294
(Original post by beanigger)
Unoffical Mark scheme for C2 AQA 2016
7)a) expand (1+2x)5 work out p q r, i think p = -10, q= 40 r = - 80 (correct me if im wrong) [3]
b)find the coefficient of x10 in the expansion of (1+2x)^5 * (2+x)^7
it was something like -1648 [5]
Not sure if I read it wrong but I thought it was (2-x)^7 rather than positive
0
Thread starter 2 years ago
#295
(Original post by KiranG23)
Not sure if I read it wrong but I thought it was (2-x)^7 rather than positive
yeah ur right my bad ill change it
0
2 years ago
#296
If i'm lucky i got 63, but probably go a less so i'm worried I wont get an A, what are the grade boundary predictions? Hopefully i can pull up my grade with stats but thats usually my worst one!
0
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#### Math - 2018-19
G.9 The student will verify and use properties of quadrilaterals to solve problems, including practical problems.
### BIG IDEAS
• I can lay out a football or baseball field, create floor plans, and plan building construction with engineering and architecture.
• I will develop analytical skills through comparing, contrasting, and sorting while improving my ability to look at things through multiple perspectives making me a better problem solver.
### UNDERSTANDING THE STANDARD
2016 VDOE Curriculum Framework - G.9 Understanding
· Deductive or inductive reasoning is used in mathematical proofs. In this course, deductive reasoning and logic are used in direct proofs. Direct proofs are presented in different formats (typically two-column or paragraph) and employ definitions, postulates, theorems, and algebraic justifications including coordinate methods.
· Quadrilaterals have a hierarchical nature based on the relationships between their sides, angles, and diagonals.
· Properties of quadrilaterals can be used to identify the quadrilateral and to determine the measures of sides and angles.
· Given coordinate representations of quadrilaterals, the distance, slope, and midpoint formulas may be used to verify that quadrilaterals have specific properties.
· The angle relationships formed when parallel lines are intersected by a transversal can be used to prove properties of quadrilaterals.
· Congruent triangles can be used to prove properties of quadrilaterals.
· A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Properties of a parallelogram include the following:
opposite sides are congruent;
opposite angles are congruent;
consecutive angles are supplementary; and
diagonals bisect each other.
· A rectangle is a quadrilateral with four right angles. Properties of rectangle include the following:
opposite sides are parallel and congruent; and
diagonals are congruent and bisect each other.
· A rhombus is a quadrilateral with four congruent sides. Properties of a rhombus include the following:
all sides are congruent;
opposite sides are parallel;
opposite angles are congruent;
diagonals are perpendicular bisectors of each other;
diagonals bisect opposite angles; and
diagonals divide the rhombus into four congruent right triangles.
· A square is a quadrilateral that is a regular polygon with four congruent sides and four right angles. Properties of a square include the following:
opposite sides are parallel;
diagonals are congruent;
diagonals are perpendicular bisectors of each other; and
diagonals divide the square into four congruent
45°-45°-90° triangles.
· A trapezoid is a quadrilateral with exactly one pair of parallel sides. The parallel sides of a trapezoid are called bases. The nonparallel sides of a trapezoid are called legs.
· An isosceles trapezoid has the following properties:
nonparallel sides are congruent;
diagonals are congruent; and
base angles are congruent.
· The construction of the perpendicular bisector of a line segment can be justified using the properties of quadrilaterals.
· The construction of the perpendicular to a given line from a point on, or not on, the line can be justified using the properties of quadrilaterals.
· The construction of the perpendicular to a given line from a point on the line can be justified using the properties of quadrilaterals.
· The construction of a bisector of a given angle can be justified using the properties of quadrilaterals.
· The construction of a square inscribed in a circle can be justified using the properties of squares.
### ESSENTIALS
The student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to
· G.91 Solve problems, including practical problems, using the properties specific to parallelograms, rectangles, rhombi, squares, isosceles trapezoids, and trapezoids.
· G.92 Prove that quadrilaterals have specific properties, using coordinate and algebraic methods, such as the distance formula, slope, and midpoint formula.
· G.93 Prove the properties of quadrilaterals, using direct proofs.
### KEY VOCABULARY
verify, characteristic, quadrilateral, properties of quadrilaterals, solve, parallelogram, rectangle, rhombus, square, isosceles trapezoid, trapezoid, coordinate method, algebraic method, distance formula, slope formula, midpoint formula, deductive reasoning, sides, angles, inscribed, circle, diagonal, mid-segment, base angles, opposite angles, right angle, base, segment, characteristics, identify, measure
Updated: Aug 23, 2018
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# Peak Focus Optimized Performance Formula
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Peak Focus Optimized Performance Formula - We developed this e-book to provide you and your team with an easy way to not only measure your level of performance, but to easily identify opportunities to elevate your level of performance - and then do something about it!
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### Peak Focus Optimized Performance Formula
1. 1. Peak Focus Optimized Performance Formula SM __________________________________________________________________ “Without continual growth and progress, such words asimprovement, achievement, and success have no meaning.” - Benjamin Franklin 2
3. 3. Peak Focus Optimized Performance Formula SM __________________________________________________________________ “Continuous improvement is better than delayed perfection.” - Mark TwainTABLE OF CONTENTS PageIntroduction 5The Formula 6(Ability x Activity x Aim)Passion = Performance level The Components - (the really short version) 6Why the Trio Walked an Icy Mile 7The Components - (the longer version) 8How to Use This Formula 10What if One of the Scores is 0? 12The Real Magic 13The Formula Worksheet: TASK 14Questions to Uncover Potential ImprovementThe Formula Worksheet: PERSONAL ASSESSMENT 16Questions to Uncover Potential ImprovementThe Formula Worksheet: TEAM / COMPANY 18Questions to Uncover Potential ImprovementNext Steps 21A Quick Note on Exponents 22About Peak Focus 23 4
4. 4. Peak Focus Optimized Performance Formula SM __________________________________________________________________ Introduction “When you improve a little each day, eventually big things occur…. Don’t look for the big, quick improvement. Seek the small improvement one day at a time. That’s the only way it happens – and when it happens, it lasts.” - John WoodenAchieving optimized performance is not a destination but a continuous journey. Bydeliberately elevating our level of performance in everything we do, we get one step closerto achieving optimized performance, a crucial step in reaching the greatest level of successpossible.You may be familiar with the philosophy known as Kaizen, created in Japan following WorldWar II. The word Kaizen means “continuous improvement”. It comes from the Japanesewords “kai” which means “change” or “to correct” and “zen” which means “good”.Kaizen is based on making changes anywhere that improvements can be made. Westernphilosophy may be summarized as, “if it ain’t broke, don’t fix it.” The Kaizen philosophy is to“do it better, make it better, and improve it even if it isn’t broken, because if we don’t, wecan’t compete with those who do.”Kaizen is a long-term approach that systematically seeks to achieve small, incremental andconstant changes.The Peak Focus Optimized Performance Formula was developed to provide you with an easyway to not only measure your level of performance, but to easily identify opportunities toelevate your level of performance - and then do something about it!This book will walk you through the four components that compose the formula: Ability Activity Aim PassionWe believe that these four components make up your performance level. And with agoal of optimizing your performance level, it’s important to increase the level of thesefour components to net a greater outcome. We will demonstrate how to achieve small,incremental and constant changes that can produce dramatic results.Optimized performance really is a special feeling - one you’ll never forget. 5
5. 5. Peak Focus Optimized Performance Formula SM __________________________________________________________________ Peak Focus Optimized Performance Formula(Ability x Activity x Aim)Passion = Performance level The Components - (the really short version)Ability The combination of hard and soft skills for any person, team or organization. (What do I know and what can I do?)Activity The actual use of hard and soft skills. (What am I doing?)Aim The focus, or reason to use hard and soft skills. (Why, exactly, am I doing this?)Passion The fire in the belly for the combination of using an ability through an activity toward a particular aim. (How fired up am I to be doing this? Flip side: What five other things would I prefer to be doing?) 6
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# Online Interactive parallelogram
## Interactive parallelogram calculator
Move the points in the grafic or define the point coordinates in the numeric field.
Scale:
Number of digits =
Point coordinates x, y
A =
B =
C =
Axes ranges
x-min=
x-max=
y-min=
y-max=
## Parallelogram properties
The opposite sides are the same. The opposite sides are parallel. The diagonals halve each other. Opposite angles are equal. Two adjacent angles together result 180 °.
### Parallelogram circumference
With the side length a = AB = CD und b = BC = DA result the circumference:
$U=2a+2b$
### Parallelogram area
With the side length a and b and the angle α is the area:
$F=a\cdot b\cdot \mathrm{sin}\alpha$
### Length of the parallelogram diagonals
With the side lengths a and b the length of the diagonals is given by:
${d}_{1}=\sqrt{{a}^{2}+{b}^{2}-2ab\mathrm{cos}\left(\alpha \right)}$
${d}_{2}=\sqrt{{a}^{2}+{b}^{2}+2ab\mathrm{cos}\left(\alpha \right)}$
### Screenshot of the Image
Print or save the image via right mouse click.
## More Calculators
Here is a list of of further useful graphical calculators:
Index Trigonometric calculations Triangle Circle Ellipse Rectangle
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https://mathoverflow.net/questions/467727/reverse-pinskers-inequality-for-smooth-density-classes/467735
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Reverse Pinsker's inequality for smooth density classes
Suppose we are given a class of probability density functions $$\mathcal{F}$$ so that for every $$f \in \mathcal{F}$$ we have $$\alpha \leq f \leq \beta$$ for some positive $$\alpha, \beta \in \mathbb{R}_+$$ and the density have the same support set $$B$$.
Here it is proved that in this setting the $$L_2$$ norm between density functions, i.e., $$\|f-g\|^2_2 = \int_B (f(x) - g(x))^2 \mu(dx)$$ is equivalent to the KL divergence between $$f$$ and $$g$$, i.e., for $$D_{\operatorname{KL}}(f \| g) = \int_B f(x) \log \frac{f(x)}{g(x)} \mu(dx)$$ we have $$D_{\operatorname{KL}}(f \| g) \asymp \|f-g\|^2_2$$ where $$\asymp$$ denotes equiality up to constant factors (which may depend on $$\alpha, \beta$$ but we consider those fixed).
Here a question was asked regarding the comparison of $$L_1$$ and $$L_2$$, and was argued that for such $$\mathcal{F}$$ classes of functions, it is possible that $$\|f-g\|_1 = \int_B |f(x) - g(x)| \mu(dx) \ll \|f-g\|_2$$ in general.
By Pinsker's inequality we know that $$\|f-g\|_1 \lesssim \sqrt{D_{\operatorname{KL}}(f \| g)} \asymp \|f-g\|_2$$ where $$\lesssim$$ indicates inequality up to absolute constant factors.
My question is as follows. Suppose we impose smoothness assumptions on $$\mathcal{F}$$. For instance suppose $$\mathcal{F}$$ consists of Lipschitz continuous functions with Lipschitz constant $$L$$. Does a reverse Pinsker's inequality hold in this case i.e. is it possible that $$\|f-g\|_1 \gtrsim \sqrt{D_{\operatorname{KL}}(f \| g)} \asymp \|f-g\|_2\;?$$
The counterexample given here fails to work in the Lipschtiz $$\mathcal{F}$$ case, and is not clear how it can be recreated because the functions need to be smooth.
• I've generally seen reverse pinsker inequalities only under assumptions that $c \leq f(x)/g(x)\leq c'$ are bounded for all $x$, see for example this. One can get something sort-of like a reverse pinsker inequality via bounds $D_{\mathsf{KL}(T(f)||T(g))} \leq C_T \lVert f -g\rVert_1^2$, where $T(f)$ is the markov kernel $T$ acting on the random variable with density $f$, and $C_T$ is a constant that depends on $T$, but not $f,g$. See for example this. Not relevant to your exact question, but hopefully useful. Mar 25 at 22:13
E.g., suppose that both $$f$$ and $$g$$ are supported on $$B=[0,1]$$, with $$f=1$$ on $$B$$ and $$g(x)=(1-h+x)\,1(0\le x for $$h\in(0,1/2)$$ and $$x\in B$$.
Then $$f$$ and $$g$$ are $$1$$-Lipschitz on $$B$$, $$\|f-g\|_1=h^2$$, and $$D_{\operatorname{KL}}(f\|g) =-\int_0^h dx\,\ln(1-h+x)-\int_{1-h}^1 dx\,\ln(x+h) \\ =-\int_0^h dx\,\ln(1-(h-x)^2) \ge\int_0^h dx\,(h-x)^2=h^3/3,$$ so that $$\|f-g\|_1\not\gtrsim\sqrt{D_{\operatorname{KL}}(f \| g)}$$ if $$h$$ is small enough.
• Awesome counterexample thanks! Do you have any thoughts on whether the exponents of $h$ can be made even more discrepant? In this example for instance $\|f-g\|_1 = h^2$ while $\|f-g\|_2 = h$. Mar 25 at 20:32
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78. geography
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79. math
A cone just fits inside a cylinder with volume 300cm^3. What is the volume of the cone?
80. math
A cone just fits inside a cylinder with volume 300cm^3. What is the volume of the cone?
81. science
1. how does a region's climate determine plant and animal life there? 2. how does climate affect our life style?
82. calculus, again
Hey can you help me find the exact value of x in the equation 5^2x=3^(x+1)? Thanks a lot, this website is cool and helpful!
83. AP Calculus BC Resources
Hi there, I was hoping if any one was familiar with any sites that have printable integration/differentiation packets? I'm hoping to use them for everyday practice. I skipped Calculus AB and need a little help. I love packets so if there are any out there
84. Trig
Which of the following is not a fifth root of -1? a.) cos 36°+ i sin 36° b.) cos 252°+ i sin 252° c.) cos 108°+ i sin 108° d.) cos 180°+ i sin 180° e.) cos 244°+ i sin 244° f.) cos 324°+ i sin 324°
85. chemistry
A 33.4 g sample of a nonelectrolyte was dissolved is 680. g of water. The solution's freezing point was -2.89°C. What is the molar mass of the compound?
86. English
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A monochromatic laser beam, with a wavelength of 532 nm passes through a diffraction grating. If the second order diffraction maximum occurs at an angle of 35.0 degrees to the straight through position, calculate the separation of adjacent slits on the
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Calculate the time it takes for a gamma ray to travel from the Sun to the Earth Distance (mean) Earth to Sun=1.5 e11 m
89. Marh
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90. Maths
A bag contains 2 green balls and 1 red ball Two balls are randomly selected without replacement. What is the probability of selecting one of each colour
91. Maths
Find the rule for the volume of a cylinder in terms of r only if the height is equal to its circumference. The surface area of a cylinder is 2pi square units. Find a rule for h in terms of r.
92. Maths
A page of text is analysed, and of the 150 words on it, 30 are nouns, 10 of which start with a vowel. Of the words that are not nouns, 85 of them do not start with vowels. a. If the word on the page is chosen at random, what is the probability that it is a
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97. finance
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98. algebra
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# How to Multiply and Divide Fractions
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• I've been getting weird mail and notes popping up in random places in my house. Some contain creepy messages and two contain math problems I can't figure out. I wrote them down in my notebook. The notes are always gone when I wake up! What do I do? This guy even took my favorite toy! A robot I made myself!
• I'll help you solve the problems! Maybe we can find the mystery man!
• 8/45 24/8
• The first problem is 4/5 times 9/12. The next is 3/4 divided by 2/8.
• Luck for us I can solve those problems! I'll show you how and maybe we can find the mystery man!
• 4/5 times 2/9 is 8/45. To multiply fractions you just multiply the numerator by the numerator and the denominator by the denominator. 4* 2=8 and 9*5=45. That makes 8/45. 3/4 divided by 2/8 is 24/8, or 3. To divide fractions, you just flip the second fraction upside down and multiply. 3/4 times 8/2 is 24/8.
• Thank you! Now I can multiply and divide fractions! Thanks you so much!
• You're welcome! Do you have any idea what 8/45 and 24/8 means!
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# Placement of Loads on Influence Lines
Objectives of the materials covered: For a given influence line, the student should be
able to determine where to place either concentrated or uniform live loads on simply
supported beams to insure their maximum effect on reactions, internal shears and
moments.
Placement of concentrated loads for maximum reactions: Once an influence line has
been generated, it can be used to determine where to position live loads, when the loads
can be positioned anywhere on the structure. Unlike “dead” or stationary loads, which are
bridge spans must be carefully positioned to produce the maximum possible effect. An
influence line assists the engineer in determining where to position these loads to produce
the maximum effect.
Load placement for maximum left reaction: For example, to determine the maximum
possible value for the left reaction on a simple beam, the influence line for this reaction is
first generated. This was accomplished in the previous section, and is repeated below:
Figure 1
The influence line summarizes the effect a concentrated load placed anywhere on the
span would have on the reaction at A. In this case it is seen that loads placed on the left
end of the structure will influence the reaction more than would loads placed farther to
the right, because the height of the influence line is greater on the left end.
L/4
Figure 2
it becomes obvious that the loads should be reversed in direction, such that the 20 kip
load can be positioned over the reaction at A, with the 10 kip load positioned L/4 away
from the reaction. The final positioning of the loads on the beam should then be as shown
below.
1.0
0.75
Figure 3
This decision is evident since placing the largest load over the highest part of the
influence line will cause the greatest influence on the reaction. This leaves the smaller
load to be placed as close to the highest point on the diagram as possible, while not
changing the fixed distance between the two wheel loads (the distance between axles on
the truck.)
Note that you might say “No fair! This is a one-way bridge and you have the trucks going
the wrong way.” But this accounts for the probability that the other bridge might be shut
down at some time during its life, and require this bridge to be temporarily converted to
two-way traffic, thus reversing the direction of travel shown.
Placement of uniform loads for maximum reaction: Influence lines can also be used to
determine critical placement of uniform live loads by placing the loads over the highest
sections. Thus the proper placement of a uniformly distributed load of unlimited length
can be determined from the influence line of Figure 1. It tells us where to place the
uniform load for maximum effect, namely across the entire length of the beam where the
influence line is positive.
W
Figure 4
## Placement of mixed loads for maximum shear:
To induce the maximum positive shear at the quarter-point of a simply supported doubly-
overhanging beam, the influence line generated earlier is shown here:
0.75
0.25
0.25
0.50
Figure 5
For maximum positive shear, the loads are situated such that the highest loads are placed
over the highest positive ordinates on the influence line, as follows:
L/4
w w
0.75
0.25
0.25
0.50
Figure 6
For maximum negative shear, the loads are situated such that the highest loads are placed
over the highest negative ordinates on the influence line, as follows:
0.75 L/4
0.25 w w
0.25
0.50
Figure 7
Note that in both cases, the concentrated loads are placed such that the larger load is over
the higher ordinate, and the uniform live loads are placed only in areas of positive (or
negative) values.
## To induce the maximum positive moment at the quarter-point of a simply supported
doubly-overhanging beam, the influence line generated earlier is shown here:
For maximum positive moment, the loads are situated such that the highest loads are
placed only over the highest positive ordinates on the influence line, and not in the
negative regions, as follows:
L/4
w
Note that the large wheel load is placed over the maximum positive ordinate of the
influence line, while the smaller wheel load is placed to its right. This is because the
influence line has a smaller slope, and descends more gradually to the right of the peak,
than to the left of the peak, thus making that the higher ordinate at L/4 from the peak.
## To induce the maximum negative moment at the quarter-point of a simply supported
doubly-overhanging beam, the influence line generated earlier is shown here:
w w
Again, the large wheel load is placed at the maximum negative ordinate, with the smaller
wheel load placed at L/4 away. The uniform live loads are placed continuously
throughout all negative regions.
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Convert Psi-second to kg/(m·s) (Dynamic viscosity)
## Psi-second into kg/(m·s)
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Psi-second+to+kg+m+s.php
# Convert Psi-second to kg/(m·s):
1. Choose the right category from the selection list, in this case 'Dynamic viscosity'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Psi-second'.
4. Finally choose the unit you want the value to be converted to, in this case 'kg/(m·s)'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '678 Psi-second'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Dynamic viscosity'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '31 Psi-second to kg/(m·s)' or '18 Psi-second into kg/(m·s)' or '81 Psi-second -> kg/(m·s)' or '46 Psi-second = kg/(m·s)'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(80 * 28) Psi-second'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '678 Psi-second + 2034 kg/(m·s)' or '47mm x 59cm x 43dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 6.560 999 940 294 9×1029. For this form of presentation, the number will be segmented into an exponent, here 29, and the actual number, here 6.560 999 940 294 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 6.560 999 940 294 9E+29. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 656 099 994 029 490 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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# Álgebras de Lie by Miguel A. Rodríguez
Posted by
By Miguel A. Rodríguez
Best abstract books
An introductory course in commutative algebra
The authors supply a concise creation to themes in commutative algebra, with an emphasis on labored examples and functions. Their remedy combines based algebraic thought with purposes to quantity idea, difficulties in classical Greek geometry, and the idea of finite fields, which has vital makes use of in different branches of technological know-how.
Analysis in Integer and Fractional Dimensions
This publication presents a radical and self-contained examine of interdependence and complexity in settings of sensible research, harmonic research and stochastic research. It specializes in "dimension" as a uncomplicated counter of levels of freedom, resulting in designated family members among combinatorial measurements and numerous indices originating from the classical inequalities of Khintchin, Littlewood and Grothendieck.
Abstract Algebra: A Concrete Introduction
This can be a new textual content for the summary Algebra direction. the writer has written this article with a special, but historic, procedure: solvability by means of radicals. This procedure is dependent upon a fields-first association. despite the fact that, professors wishing to begin their direction with staff conception will locate that the desk of Contents is extremely versatile, and incorporates a beneficiant volume of workforce assurance.
Basic Modern Algebra with Applications
The booklet is essentially meant as a textbook on glossy algebra for undergraduate arithmetic scholars. it's also valuable should you have an interest in supplementary examining at the next point. The textual content is designed in any such method that it encourages self reliant considering and motivates scholars in the direction of extra research.
Extra resources for Álgebras de Lie
Sample text
Adem´ as adx (N ) ⊂ k, luego adx act´ ua trivialmente en el cociente. Esto solo puede ser si el cociente es trivial, es decir, si k = N (k). Veamos algunas propiedades elementales de las sub´algebras de Cartan. 4 Sea g un ´ algebra de Lie de dimensi´ on finita y h una sub´ algebra nilpotente de g. adh adh Entonces, h es una sub´ algebra de Cartan si y solo si h = g0 donde g0 es la componente nula de la descomposici´ on de Fitting de g respecto a ad h (es decir el subespacio propio generalizado de autovalor 0 de ad h) (ver [4], p´ agina 80).
Versi´on 4. 3. 49) y sea π una representaci´ on de dimensi´ on finita en un espacio V . Sean H = π(h), E = π(e), F = π(f ). Los endomorfismos {H, E, F } verifican las mismas relaciones de conmutaci´on anteriores. 51) De esta manera obtenemos una sucesi´ on de vectores en V linealmente independientes (porque corresponden a distintos autovalores de H) y que necesariamente debe acabar, pues V es de dimensi´on finita. 52) Sea v = E p v. 54) Aplicamos ahora el operador F . Tenemos una sucesi´on de autovectores v0 = v, v1 = F v, .
17) ˜ = (adx )l g. Entonces: Sea g0 = χ(ad ˜ x )g, g ˜ (como espacios vectoriales, se trata de la descomposici´ 1. g = g0 ⊕ g on de Fitting) 2. g0 es una sub´ algebra de Cartan, dim g0 = l ˜] = [x, g ˜] = g ˜ 3. [g0 , g Veamos la relaci´ on entre los espacios de ra´ıces. Sea g un ´algebra de Lie de dimensi´on finita, h una sub´algebra nilpotente y V un m´odulo de g, con representaci´on π. Supongamos (como ya hemos dicho) que el cuerpo K es algebraicamente cerrado (o que los polinomios cracter´ısticos involucrados tienen todas sus ra´ıces en el cuerpo).
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Originally Posted by Existing Light
...I'm using a formula that Ralph Lambrecht posted on a pinhole thread here, which is PinholeDiameter=sqrt( (2.44)(wavelength)(Focal Length) )
I found on a seperate website (cant remember where) that 5500Angstroms is the wavelength of average daylight, so I guess I'll go with that. That same website said 1Angstrom = 1nm, which is 0.00055mm, right?
To convert 7.5inches to millimeters, I multiplied 7.5 by 2.54 to get 19.05cm. I multiplied that by 10 to get 190.5mm.
i plugged the wavelength and focal length in millimeters in to the formula. Assuming I did my math right, the diameter should be about 0.5mm. Is that right?...
Looks all right to me!
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# Calculate Interest on Overdraft Against Fixed Deposit
Before opening a Fixed Deposit (FD) account or taking an overdraft against an existing FD, it's essential to calculate the potential returns and interest earnings.
To calculate the interest on your Fixed Deposit, you need to multiply the principal amount (say Rs. 5,00,000) by the interest rate (let’s take 6%) and then divide the result by 100. Following this formula and figures, the interest on your Fixed Deposit amounts to Rs. 30,000.
For the overdraft facility, you can only avail up to 90% of the principal amount - Rs. 4,50,000 in this case.
The interest rate is determined by adding the base interest rate (6%) with the spread (2%). This calculation results in an overdraft interest rate of 8%.
To determine the interest payable on a selected amount utilised from the overdraft ( say you need Rs.1,00,000), you multiply the overdraft amount by the overdraft interest rate and divide the result by 100. Thus, the interest payable on the amount (of Rs.1,00,000) used from the overdraft comes to Rs. 8,000.
Consequently, when utilising Rs. 1,00,000 from the overdraft facility against your Fixed Deposit, the interest payable would amount to Rs. 8,000.
For overdraft against FD, a calculator can help carry out the calculations easily. You can also understand the interest earned on your FD and the interest payable. Remember that calculators will only give you the numbers without factoring in for other provisions and circumstances, the actual returns and interest earnings may vary accordingly.
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# Full text: Musschenbroek, Petrus: Physicae experimentales, et geometricae de magnete, tuborum capillarium vitreorumque speculorum attractione, magnitudine terrae, cohaerentia corporum firmorum dissertationes
CORPORUM FIRMORUM. erecta ſit perpendicularis P A, ſitque P quædam particula attracta ab
ducatur recta P E, in recta P A ducatur P F = P E, ex F ducatur
F K parallela ad A D, quæ repræſentet vires, quibus punctum E at-
trahit particulam P, ſitque I K L curva linea, quam punctum K
perpetuo tangit: Occurrat eadem circuli plano in L. in P A capia-
tur P H æqualis P D, & erigatur perpendiculum H I cur væ oc-
currens in I. demonſtravit Newtonus Lib. 1. prop. 90. Princ. Philoſ. corpuſculi P attractionem eſſe, ut area A H I L. eſt ducta in alti-
tudinem A P.
Vocetur P F x. F K y, & ſit F K, vel vis quâ punctum E
attrahit corpus P, reciproce ut aliqua potentia ipſius P F, ſit
hæc n. tum æquatio curvæ erit y = {1/X n } & area A H I K L uti
{1/PA n - 1 } - {1/PH n - 1 } unde attractio corpuſculi P in circulum erit
ut {1/PA n - 2 } - {PA/PH n - 1 }.
Hæc Cl. Chynæus in Philoſophical Principles of Natural reli-
gion eleganter explicuit §. 44. Si n = i, & P A = 0. tum radius cir-
culi attrahentis productus coincidet cum Aſy mptoto P O, in quo
caſu area A H I L erit infinita, cum curva ſit Hyperbola vulgaris; & P A = 0, ſive evaneſcente intervallo inter particulam & circu-
lum attrahentem, erit attractio = P A X A H I L = 0 X ? = i.
Si n = 1, & P A = ? , hoc eſt quando planum attrahens A D
poſitum eſt ad concurſum Hyperbolæ, cum ſuâ Aſymptoto P H,
tum arcus D H, cujus centrum eſt P, & cujus radius eſt P D = P A
= ? , coincidet cum A D, & ideo A L coincidet cum H I, unde
P A X A H I L = ? X 0 = i.
Si n = i. & P A = a tum A H vocetur y, & P H = x = a + y,
unde corpuſculum P a circulo attrahetur vi = P A X A H I L =
{y-yy/2a} + {y 3 /3aa} - {y 4 /4a 3 } & c.
Si n = 2 & P A = 0. tum area A H I L erit plus quam infinita,
unde attractio erit A P X A H I L = 0 multiplicato per pluſquam
infinitum, unde liquet attractionem in hoc caſu, poſito P A = 0.
## Note to user
Dear user,
In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser.
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Maximum likelihood estimation tutorial
Dave Harris on Maximum Likelihood Estimation R-bloggers
Maximum likelihood Saylor. R is well-suited for programming your own maximum likelihood routines. Indeed, important that we store the results from the estimation into an object. The, PyMC Tutorial #1: Bayesian Parameter Estimation for Bernoulli Distribution to estimate the parameter of a Bernoulli distribution. Maximum Likelihood Estimation.
tutorialsmle.html [Auton Lab]
TAM Tutorials edmeasurementsurveys.com. PyMC Tutorial #1: Bayesian Parameter Estimation for Bernoulli Distribution to estimate the parameter of a Bernoulli distribution. Maximum Likelihood Estimation, Maximum-Likelihood Estimation (MLE) is a statistical technique for estimating model parameters. It basically sets out to answer the question: what model parameters.
In this tutorial, you will learn the (OLS) while logistic regression is estimated using Maximum Likelihood Estimation (MLE) approach. Maximum Likelihood Luque-Fernandez, MA; Schomaker, M; Rachet, B; Schnitzer, ME (2018) Targeted maximum likelihood estimation for a binary treat-ment: A tutorial. Statistics in medicine.
Maximum likelihood estimation or otherwise noted as MLE is a popular mechanism which is used to estimate the model parameters of a regression model. Other than Things we will look at today Maximum Likelihood Estimation ML for Bernoulli Random Variables Maximizing a Multinomial Likelihood: Lagrange Multipliers
The Trinity Tutorial by Avi Kak 1.4: Maximum Likelihood (ML) Estimation of О We seek that value for О which maximizes the likelihood shown on the previous slide. Maximum Likelihood Estimation S. Purcell. Contents and Keywords. Introduction . probability models parameters conditional probability binomial probability distribution
Maximum Likelihood Estimation. The only restriction is that they are not freely available for use as teaching materials in classes or tutorials outside degree Things we will look at today Maximum Likelihood Estimation ML for Bernoulli Random Variables Maximizing a Multinomial Likelihood: Lagrange Multipliers
Maximum-Likelihood Estimation (MLE) is a statistical technique for estimating model parameters. It basically sets out to answer the question: what model parameters Things we will look at today Maximum Likelihood Estimation ML for Bernoulli Random Variables Maximizing a Multinomial Likelihood: Lagrange Multipliers
A Tutorial on Restricted Maximum Likelihood Estimation in Linear Regression and Linear Mixed-E ects Model Xiuming Zhang zhangxiuming@u.nus.edu A*STAR-NUS Clinical Targeted maximum likelihood estimation is a semiparametric double The reader should gain sufficient understanding of TMLE from this introductory tutorial to be
Topic 14: Maximum Likelihood Estimation November, 2009 As before, we begin with a sample X= (X 1;:::;X n) of random variables chosen according to one of a family Topic 14: Maximum Likelihood Estimation November, 2009 As before, we begin with a sample X= (X 1;:::;X n) of random variables chosen according to one of a family
TAM Tutorials edmeasurementsurveys.com
tutorials [Auton Lab]. Journal of Mathematical Psychology 47 (2003) 90–100. Tutorial Tutorial on maximum likelihood estimation In Jae Myung* Department of Psychology, Ohio State, 25/09/2018 · Slides and notebooks for my tutorial at Newton-based maximum likelihood estimation in classifiers maximum-likelihood-estimation maximum-a-posteriori.
tutorialsmle.html [Auton Lab]. This article covers the topic of Maximum Likelihood Estimation (MLE) - how to derive it, where it can be used, and a case study to solidify the concept in R., Luque-Fernandez, MA; Schomaker, M; Rachet, B; Schnitzer, ME (2018) Targeted maximum likelihood estimation for a binary treat-ment: A tutorial. Statistics in medicine..
Neural Machine Translation Tutorial ACL 2016
Dave Harris on Maximum Likelihood Estimation R-bloggers. 12/09/2017В В· In this post I want to talk about regression and the maximum likelihood estimate. Instead of going the usual way of deriving the least square (LS) estimate Targeted maximum likelihood estimation is a semiparametric double The reader should gain sufficient understanding of TMLE from this introductory tutorial to be.
Maximum Likelihood Estimation. Gaussian Bayes Classifiers. Cross-Validation. The most recent version is going to be in the tutorial project in Auton CVS. 25/09/2018В В· Slides and notebooks for my tutorial at Newton-based maximum likelihood estimation in classifiers maximum-likelihood-estimation maximum-a-posteriori
The above example gives us the idea behind the maximum likelihood estimation. Here, we introduce this method formally. To do so, we first define the likelihood function. Maximum Likelihood Estimation. The only restriction is that they are not freely available for use as teaching materials in classes or tutorials outside degree
Maximum-Likelihood Estimation (MLE) is a statistical technique for estimating model parameters. It basically sets out to answer the question: what model parameters Targeted Maximum Likelihood Estimation for a binary treatment: A tutorial. Statistics in Medicine. 2017 - migariane/SIM-TMLE-tutorial
Maximum likelihood 1 Maximum likelihood In statistics, maximum-likelihood estimation (MLE) is a method of estimating the parameters of a statistical See worked out examples of how maximum likelihood functions are used Tutorials Statistics Formulas The basic idea behind maximum likelihood estimation is that
Stat 411 { Lecture Notes 03 Likelihood and Maximum Likelihood Estimationy Ryan Martin www.math.uic.edu/~rgmartin Version: August 19, 2013 1 Introduction Maximum Likelihood Estimation. Gaussian Bayes Classifiers. Cross-Validation. The most recent version is going to be in the tutorial project in Auton CVS.
Maximum Likelihood Estimation. The only restriction is that they are not freely available for use as teaching materials in classes or tutorials outside degree Topic 14: Maximum Likelihood Estimation November, 2009 As before, we begin with a sample X= (X 1;:::;X n) of random variables chosen according to one of a family
Tutorial 3 - Maximum Likelihood Estimation & Canonical Link (last updated January 30, 2009) 1. Find the canonical link for (a) Normal distribution with unknown mean Lesson 4.2 Likelihood function and maximum likelihood. demonstrating maximum likelihood estimation and confidence intervals for binomial data.
Things we will look at today Maximum Likelihood Estimation ML for Bernoulli Random Variables Maximizing a Multinomial Likelihood: Lagrange Multipliers Tutorial 3 - Maximum Likelihood Estimation & Canonical Link (last updated January 30, 2009) 1. Find the canonical link for (a) Normal distribution with unknown mean
Jack Frost Hoodie . MissAmeriko 08/28 I've been wondering what I should use to make a Jack Frost jacket and yours looks Thanks for the tutorial. Read more . 0 Jack frost hoodie tutorial Nova Scotia Temukan dan simpan ide tentang Jack frost di Pinterest. Lihat ide lainnya tentang Dreamworks, Jack Frost's hoodie - TUTORIAL by ~LiLavi on deviantART
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https://www.clutchprep.com/chemistry/practice-problems/83149/a-balloon-filled-with-1-92-g-of-helium-has-a-volume-of-12-5-l-what-is-the-balloo
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# Problem: A balloon filled with 1.92 g of helium has a volume of 12.5 L. What is the balloon’s volume after 0.850 g of helium has leaked out through a small hole (assume constant pressure and temperature)?
###### FREE Expert Solution
Remember the ideal gas law: PV = nRT
Constant pressure and temperature, R is also a constant
80% (384 ratings)
###### Problem Details
A balloon filled with 1.92 g of helium has a volume of 12.5 L. What is the balloon’s volume after 0.850 g of helium has leaked out through a small hole (assume constant pressure and temperature)?
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# SAT Flashcards: How to Use Them for SAT Prep?
Last Updated on April 20, 2023
Many SAT students wonder whether they should invest time in making flashcards to use as study aids while preparing for the exam. In this blog, we’ll examine the top 6 reasons why I advise our SAT students to make and frequently use flashcards. I’ll also give some illustrations of how to use your SAT prep flashcards most efficiently.
I advise our SAT students to make and frequently use flashcards.
## Reason #1: Use Flashcards for Both Math and Verbal Learning
You’ve probably heard that the best way to memorize critical math formulas is with SAT math flashcards. As you probably know, a great SAT math score requires the memorization of numerous formulas. Flashcards help us internalize essential formulas and allow us to quickly see what we know and what we do not.
For example, in Geometry, you’ll need to memorize formulas such as:
• Area of rectangle = base x height
• Pythagorean theorem : a^2 + b^2 = c^2
In Algebra, you’ll need to know procedures such as:
• The Distributive Law: x(x + c) = x^2 + xc
• The Zero Products Property: If (x – a)(x – b) = 0 then x = a or x = b
These are just a few types of math facts you can commit to flashcards.
In addition to being great skill builder tools for learning math formulas and processes, flashcards can significantly help you learn verbal concepts, especially grammar and usage rules.
For example, you could review:
• Commonly tested homophones, such as their/there/they’re or its/it’s
• Examples of the usage of the relative pronouns who, that, and which
• Appropriate use of semicolons and commas
TTP PRO TIP:
Flashcards are an excellent tool for committing math and verbal facts to memory.
Next, let’s discuss the versatility of flashcards as a learning tool.
## Reason #2: Tailor Flashcards to Your Unique Needs
Flashcards are an extremely flexible learning tool because you can tailor them to your unique needs. For example, you can select which format and type of flashcard you prefer: physical cards, such as index cards, or cards made with a digital phone app. Additionally, you can further personalize your paper flashcards by color-coding them or making stacks of those you’ve already mastered and those you haven’t. Digital flashcard apps allow you to practice your cards via quizzes or games, making study more enjoyable and motivating.
Regardless of which route you take, you can also customize the content of your flashcards. In addition to including math formulas on your cards, you might put practice problems on the cards. Furthermore, for verbal, you might put practice sentences that illustrate the grammar concepts that have given you the most trouble.
Flashcards also allow you to switch up your test prep study routine. They are excellent for independent study, but you can also involve your family or friends. For example, when I was preparing for the SAT, I had my brother or mom and dad quiz me on my flashcards. Studying with your friends or family is also a great way to keep up a high level of motivation for your SAT.
TTP PRO TIP:
Use physical or digital flashcards and personalize them to your unique needs.
Next, let’s discuss how flashcards make it easier to study.
## Reason #3: Use Flashcards Anytime, Anywhere
With SAT flashcards, you can fit in “quick-hit” study sessions whenever and wherever you like. You will normally sit at a desk or table to complete your regular SAT studying. However, it never hurts to knock out an extra 10 or 15 minutes of SAT studying while you’re on the go.
You can also study while waiting in line for an appointment, during your lunch or study hall, or on the way to school if you take the bus. Using these moments to study your flashcards can add 20 minutes or more to your daily study time. That extra time would add up to 20 x 7 = 140 minutes each week, or 980 minutes per month, which is almost 16 hours. That’s a lot of extra study time!
TTP PRO TIP:
You can get in some extra study time by using your flashcards no matter where you are.
Let’s now consider how flashcards can help organize you.
## Reason #4: Flashcards Organize Your SAT Prep
With SAT flashcards, it’s easy to keep track of the verbal and math topics you’ve mastered and those you haven’t. Simply separate your cards into piles marked “mastered” and “not mastered.” If you use this surefire strategy, you’ll always know exactly what you need to study at any given time.
While you will be spending most of your time on the “not mastered” pile, occasionally look back at the “mastered” pile because you need to ensure that nothing has slipped through the cracks. When looking at that pile, if you discover that you’ve forgotten a topic or concept, return those cards to the “not mastered” pile.
TTP PRO TIP:
Divide your SAT flashcards into two piles — “mastered” and “not mastered” — and devote more time to reviewing the “not mastered” pile.
Let’s now learn how we can adapt flashcards to our study needs.
When you begin your SAT studying, I’m sure you will have some pretty “standard” flashcards to help you with some initial concepts. You can use flashcards for math and verbal. Let’s first run through an example of a math flashcard.
### Math Flashcard Example
For math, for example, you may have a flashcard on the difference of two squares, as follows:
Front of Card:
Back of Card:
Certainly, this generic flashcard can be helpful, but it is even more useful if it includes a specific problem that has given you issues during your review. For instance, suppose you incorrectly answered a question involving the factored form of x^100 – y^100, which is a particular application of the difference of squares formula. So, rather than just reviewing the solution and moving on, you can add the problem you missed to the flashcard. Thus, by studying that flashcard, you can ensure that you will never make the same mistake again.
Your updated difference of squares flashcard would now be this:
New Front of Card:
New Back of Card:
You now have a super-effective flashcard because it addresses both the general formula and your specific knowledge gap related to the difference of two squares. So, don’t be afraid to update and add more detail to your flashcards. Next, let’s look at a verbal flashcard.
### Verbal Flashcard Example
Your flashcards can specify particular verbal concepts that need extra attention. For example, consider the difference between the quantity words “number” and “amount.”
Front of Card:
Back of Card:
As an example, suppose you incorrectly answer a grammar/usage question with the answer “amount of people” instead of “number of people.” You could add this detail to the generic definition of the quantity words “number” and “amount.” Perhaps you could add a sentence that correctly uses the quantity words. You could adjust the back of your flashcard as follows:
New Back of Card:
By adding your own example, you will invest more in the flashcard’s information and, consequently, will be more likely to remember the difference between the two “quantity” words.
TTP PRO TIP:
Use your flashcards to include specific examples of questions that you have answered incorrectly.
Let’s now focus on how flashcards can be a source of motivation.
## Reason #6: Flashcards Help to Motivate You
To engage in useful SAT studying, you need to be an active participant in your prep. Too often, I hear from students who spend hours studying and do not retain anything. Don’t let that be you!
Using SAT flashcards is one way to be an “active learner.” When you’re using your flashcards, change things up from time to time. For instance, don’t always study your flashcards in the same order.
Also, try to gamify your studying by seeing how many flashcards you get right in a row. Or see how many your friends or family can get right. Turn your studying into a friendly competition! You must make every effort to maintain motivation and keep the process engaging because studying for the SAT is a journey.
TTP PRO TIP:
To make your SAT prep more challenging, more engaging, and more effective, “gamify” your flashcard study.
## Key Takeaways
Flashcards are an easy, convenient way to review or memorize key information while you’re studying for the SAT.
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Question:
# 1cm 1cm 2cm The magnetic field B shown in figure increases at the rate AB/At=1 mT/s. The resistance of the wire is 2m22/cm. Dete
1cm 1cm 2cm The magnetic field B shown in figure increases at the rate AB/At=1 mT/s. The resistance of the wire is 2m22/cm. Determine the direction of the current and the magnitude of the current in the circuit. 3cm 2cm 3cm
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# Ramsey's theorem [closed]
I'm reading introduction to combinatorics and encountered an exercise I couldn't answer
Let S be a set of six points in the plane, with no three of the points collinear. Color either red or blue each of the 15 line segments determined by the points of S. Show that there is at least two triangles determined by points of S which are either a red triangle or a blue triangle.(Both may be red, or both may be blue and one may be red the other bule)
Thanks if you can give me any help.
-
## closed as off-topic by Chris Eagle, William, Thomas Andrews, Sasha, AtaraxiaAug 3 '13 at 4:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." – Chris Eagle, William, Thomas Andrews, Sasha, Ataraxia
If this question can be reworded to fit the rules in the help center, please edit the question.
Since you're new, I'd like to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are. That way, people won't tell you stuff you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help if you show that you've tried the problem yourself. Also, some may consider your post rude because it is phrased as a command, not a request for help, so please consider rewriting it. – Zev Chonoles Jul 29 '13 at 0:42
It seems as though you might be confused about the question. When they say that "there is at least one triangles determined by points of S that is either a red triangle or a blue triangle", they mean that in any such graph, one of them must exist, but the other doesn't have to. For example, if every line is blue, you can't find a red triangle, but you can definitely find a blue one with any three points. So, you can't make sure that the first triangle exist, because it might not. But if it does not, then we can find the other one. – Omnomnomnom Jul 29 '13 at 0:55
Did you see the answers so far? Have you tried something along those lines? – Omnomnomnom Jul 29 '13 at 1:01
@yuan I haven't shown that any triangle exists yet. All I've shown is that we have a point, and three lines of the same color coming out of that point. – Omnomnomnom Jul 29 '13 at 1:19
– Andrés E. Caicedo Jul 29 '13 at 2:54
## 3 Answers
To the modified version of your question: we can always find a second triangle, but not a third. The proof is as follows:
We start by acknowledging the first triangle (let's say that it's red), as guaranteed from the previous proof. Choose a point not in that triangle. Following the steps of the original proof, either we have a new triangle, or there are three blue edges from this point to each point on the red triangle.
Now look at the three points outside of the red triangle, supposing that each has three blue edges from each point on the red triangle. If any two of these outside points are connected with a blue edge, we have a blue triangle. Otherwise, the outside points form another red triangle. Either way, we have found a second triangle.
To disprove the guaranteed existence of a third triangle, consider the graph in which these edges are blue and the rest are red. We have two red triangles, but no more.
-
Good answer to the OP's real question. I have often seen the proof that there is one triangle and a statement that there is another, but no proof for the second. – Ross Millikan Jul 29 '13 at 2:43
@RossMillikan Thank you. For the record, the original question was the one-triangle statement. OP then edited the post to ask this follow-up. – Omnomnomnom Jul 29 '13 at 2:49
Hint: consider any point on the graph. This point has $5$ protruding line segments, each of which are red or blue. We note that there must be at least $3$ line segments of the same color.
We can show that the points on the end of these $3$ segments form a red or a blue triangle. How?
Hint: Suppose that the three segments we found were red. Let's look at the points on the end of those segments, and the connection from each to the other. If any two are connected with a red line, then we have a red triangle. Draw this to check.
Now, what if none of the two points at the end of these $3$ segments are connected with a red line? Can we say anything about either a red or blue triangle?
Some helpful pictures: We looked at a point $A$, and noticed that whatever colors come out of $A$, we have to have $3$ that are the same
We then looked at the three points on the end. If there's a red line between any $2$ of them, then we have a red triangle. What happens if there isn't?
ANSWER:
There must be a blue triangle
-
if there isn't exist a red line then there is a blue triangle – yuan Jul 29 '13 at 1:29
Exactly. So, we've shown that: 1. you always have three points of the same color (let's say red). 2. If you have three points of the same color, you must have either a red triangle or a blue triangle. 3. Therefore, any graph on six points colored red and blue will have either a red triangle or a blue triangle. – Omnomnomnom Jul 29 '13 at 1:32
ok,my question is that if there exist a second triangle(maybe blue,maybe red)? – yuan Jul 29 '13 at 1:34
I'm not sure, but I think there must be an example of a graph with only the one triangle. – Omnomnomnom Jul 29 '13 at 1:51
To be precisely,the question I post is to proof that the second triangle exists. If you think there be an example of a graph with only one triangle, can you draw it? – yuan Jul 29 '13 at 1:58
This is really a graph theory problem, so there is no need to worry about planar embeddings or collinearity. What you mean is that any 2-coloring of the edges in $K_6$ (complete graph on six vertices) contains a monochromatic triangle.
The classic proof of this is as follows: pick some vertex $v$. There are five vertices connected to it, so three of these edges will have the same color (say, red). The three corresponding vertices will form a monochromatic triangle (why?).
The moral of this exercise is essentially to see how to apply the pigeonhole principle. Whenever you've got only two colors and a lot of edges, look for large collections of monochromatic edges. This is a common theme of Ramsey theory on finite graphs.
Here is another exercise with a similar solution: show that if you 2-color the planar lattice $\mathbb{N}^2$, you must have a monochromatic rectangle. Edit: 2-color the vertices of the lattice, I mean.
Edit: I have removed a crucial step in my proof, since Omnomnomnom has given a very good hint along the lines of "figure out the missing step".
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very much appreciated. I feel that homework questions require a delicate touch. – Omnomnomnom Jul 29 '13 at 0:58
For the plane lattice question, we can't we color all horizontal edges red and all vertical edges blue? – Andrew Salmon Jul 29 '13 at 1:35
Andrew, for the plane lattice problem, you need to show that for -any- coloring, you can find some monochromatic rectangle. So you don't get to set the colors. – Andrew Poelstra Jul 29 '13 at 2:36
@AndrewPoelstra Right, but I was wondering how coloring horizontal edges red and vertical edges blue fails to generate a counterexample (how does this contain a monochromatic rectangle)...but maybe I misunderstand the question and you're supposed to color the vertices. – Andrew Salmon Jul 29 '13 at 14:48
Oh, you're right! You -were- supposed to color the vertices, but I had a dumb moment and thought that the distinction wouldn't matter. (Because the edges look like a vertex lattice if you zoom out enough, right? ;)). I will edit my original answer. – Andrew Poelstra Jul 29 '13 at 15:18
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https://socratic.org/questions/how-do-you-find-a-third-degree-polynomial-given-roots-4-and-4i
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Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team
# How do you find a third degree polynomial given roots -4 and 4i?
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#### Explanation
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#### Explanation:
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Dec 27, 2017
${x}^{3} + 4 {x}^{2} + 16 x + 64 = 0$
#### Explanation:
Since the question ask for a third degree polynomial, I am going to assume that you want a polynomial with real coefficients, with $- 4 i$ as the third root, being the complex conjugate of $4 i$.
Each zero $a$ corresponds to a factor $\left(x - a\right)$, so we can write:
$f \left(x\right) = \left(x + 4\right) \left(x - 4 i\right) \left(x + 4 i\right)$
$\textcolor{w h i t e}{f \left(x\right)} = \left(x + 4\right) \left({x}^{2} - {\left(4 i\right)}^{2}\right)$
$\textcolor{w h i t e}{f \left(x\right)} = \left(x + 4\right) \left({x}^{2} + 16\right)$
$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} + 4 {x}^{2} + 16 x + 64$
So we can write a cubic equation:
${x}^{3} + 4 {x}^{2} + 16 x + 64 = 0$ with the desired roots.
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