text:
#### Closure Operators for Setoid Algebras

Fix a signature 𝑆, let 𝒦 be a class of 𝑆-algebras, and define

• H 𝒦 = algebras isomorphic to a homomorphic image of a members of 𝒦;
• S 𝒦 = algebras isomorphic to a subalgebra of a member of 𝒦;
• P 𝒦 = algebras isomorphic to a product of members of 𝒦.
```{-# OPTIONS --without-K --exact-split --safe #-}

open import Base.Algebras.Basic using ( 𝓞 ; 𝓥 ; Signature )

module Setoid.Varieties.Closure {𝑆 : Signature 𝓞 𝓥} where

-- imports from Agda and the Agda Standard Library -------------------------------------------
open import Agda.Primitive         using ( _⊔_ ; lsuc )      renaming ( Set to Type ; lzero to ℓ₀)
open import Data.Product           using ( _,_ ; Σ-syntax )  renaming ( _×_ to _∧_ )
open import Data.Unit.Polymorphic  using ( ⊤ ; tt )
open import Function.Base          using ( id )
open import Function.Bundles       using ()                  renaming ( Func to _⟶_ )
open import Level                  using ( Level ; Lift ; lift ; lower )
open import Relation.Binary        using ( Setoid )
open import Relation.Unary         using ( Pred ; _∈_ ; _⊆_ )

-- Imports from the Agda Universal Algebra Library ---------------------------------------------
open import Setoid.Algebras.Basic                   {𝑆 = 𝑆}  using ( Algebra ; ov ; Lift-Alg )
open import Setoid.Algebras.Products                {𝑆 = 𝑆}  using ( ⨅ )
open import Setoid.Homomorphisms.Basic              {𝑆 = 𝑆}  using ( IsHom )
open import Setoid.Homomorphisms.Isomorphisms       {𝑆 = 𝑆}  using ( _≅_ ; ≅-trans ; ≅-sym ; Lift-≅ ; ⨅≅⨅ℓρ )
open import Setoid.Homomorphisms.HomomorphicImages  {𝑆 = 𝑆}  using ( _IsHomImageOf_ ; IdHomImage ; HomImage-≅ )
using ( HomImage-≅' ; Lift-HomImage-lemma )
open import Setoid.Subalgebras.Subalgebras          {𝑆 = 𝑆}  using ( _≤_ ; _≤c_ )
open import Setoid.Subalgebras.Properties           {𝑆 = 𝑆}  using ( ≤-reflexive ; ≤-trans ; ≅-trans-≤ )
using ( ≤-trans-≅ ; Lift-≤-Lift ; ≤-Lift )

open _⟶_ renaming ( f to _⟨\$⟩_ )

module _ {α ρᵃ β ρᵇ : Level} where

private a = α ⊔ ρᵃ ; b = β ⊔ ρᵇ

Level-closure : ∀ ℓ → Pred(Algebra α ρᵃ) (a ⊔ ov ℓ) → Pred(Algebra β ρᵇ) (b ⊔ ov(a ⊔ ℓ))
Level-closure ℓ 𝒦 𝑩 = Σ[ 𝑨 ∈ Algebra α ρᵃ ]  (𝑨 ∈ 𝒦)  ∧  𝑨 ≅ 𝑩

module _ {α ρᵃ β ρᵇ : Level} where

Lift-closed : ∀ ℓ → {𝒦 : Pred(Algebra α ρᵃ) _}{𝑨 : Algebra α ρᵃ} → 𝑨 ∈ 𝒦
→            Lift-Alg 𝑨 β ρᵇ ∈ (Level-closure ℓ 𝒦)
Lift-closed _ {𝑨 = 𝑨} kA = 𝑨 , (kA , Lift-≅)

module _  {α ρᵃ β ρᵇ : Level} where

private a = α ⊔ ρᵃ ; b = β ⊔ ρᵇ

H : ∀ ℓ → Pred(Algebra α ρᵃ) (a ⊔ ov ℓ) → Pred(Algebra β ρᵇ) (b ⊔ ov(a ⊔ ℓ))
H _ 𝒦 𝑩 = Σ[ 𝑨 ∈ Algebra α ρᵃ ] 𝑨 ∈ 𝒦 ∧ 𝑩 IsHomImageOf 𝑨

S : ∀ ℓ → Pred(Algebra α ρᵃ) (a ⊔ ov ℓ) → Pred(Algebra β ρᵇ) (b ⊔ ov(a ⊔ ℓ))
S _ 𝒦 𝑩 = Σ[ 𝑨 ∈ Algebra α ρᵃ ] 𝑨 ∈ 𝒦 ∧ 𝑩 ≤ 𝑨

P : ∀ ℓ ι → Pred(Algebra α ρᵃ) (a ⊔ ov ℓ) → Pred(Algebra β ρᵇ) (b ⊔ ov(a ⊔ ℓ ⊔ ι))
P ℓ ι 𝒦 𝑩 = Σ[ I ∈ Type ι ] (Σ[ 𝒜 ∈ (I → Algebra α ρᵃ) ] (∀ i → 𝒜 i ∈ 𝒦) ∧ (𝑩 ≅ ⨅ 𝒜))

module _  {α ρᵃ β ρᵇ : Level} where

private a = α ⊔ ρᵃ ; b = β ⊔ ρᵇ

SP : ∀ ℓ ι → Pred(Algebra α ρᵃ) (a ⊔ ov ℓ) → Pred(Algebra β ρᵇ) (b ⊔ ov(a ⊔ ℓ ⊔ ι))
SP ℓ ι 𝒦 = S{α}{ρᵃ} (a ⊔ ℓ ⊔ ι) (P ℓ ι 𝒦)

module _  {α ρᵃ β ρᵇ γ ρᶜ δ ρᵈ : Level} where

private a = α ⊔ ρᵃ ; b = β ⊔ ρᵇ ; c = γ ⊔ ρᶜ ; d = δ ⊔ ρᵈ

V : ∀ ℓ ι → Pred(Algebra α ρᵃ) (a ⊔ ov ℓ) →  Pred(Algebra δ ρᵈ) (d ⊔ ov(a ⊔ b ⊔ c ⊔ ℓ ⊔ ι))
V ℓ ι 𝒦 = H{γ}{ρᶜ}{δ}{ρᵈ} (a ⊔ b ⊔ ℓ ⊔ ι) (S{β}{ρᵇ} (a ⊔ ℓ ⊔ ι) (P ℓ ι 𝒦))

```

Thus, if 𝒦 is a class of 𝑆-algebras, then the variety generated by 𝒦 is denoted by `V 𝒦` and defined to be the smallest class that contains 𝒦 and is closed under `H`, `S`, and `P`.

With the closure operator V representing closure under HSP, we represent formally what it means to be a variety of algebras as follows.

```module _ {α ρᵃ ℓ ι : Level} where

is-variety : (𝒱 : Pred (Algebra α ρᵃ)(α ⊔ ρᵃ ⊔ ov ℓ)) → Type (ov (α ⊔ ρᵃ ⊔ ℓ ⊔ ι))
is-variety 𝒱 = V{β = α}{ρᵇ = ρᵃ}{γ = α}{ρᶜ = ρᵃ} ℓ ι 𝒱 ⊆ 𝒱

variety : Type (ov (α ⊔ ρᵃ ⊔ ov ℓ ⊔ ι))
variety = Σ[ 𝒱 ∈ (Pred (Algebra α ρᵃ) (α ⊔ ρᵃ ⊔ ov ℓ)) ] is-variety 𝒱

```

#### Closure properties of S

`S` is a closure operator. The fact that S is expansive won’t be needed, so we omit the proof, but we will make use of monotonicity and idempotence of `S`.

```module _ {α ρᵃ : Level} where

private a = α ⊔ ρᵃ

S-mono : ∀{ℓ} → {𝒦 𝒦' : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)}
→       𝒦 ⊆ 𝒦' → S{β = α}{ρᵃ} ℓ 𝒦 ⊆ S ℓ 𝒦'

S-mono kk {𝑩} (𝑨 , (kA , B≤A)) = 𝑨 , ((kk kA) , B≤A)

```

We say `S` is idempotent provided \af{S} (\af{S} \ab{𝒦}) \as{=} \af{S} \ab{𝒦}. Of course, this is proved by establishing two inclusions, but one of them is trivial, so only the other need be formalized, which we do as follows.

``` S-idem : ∀{β ρᵇ γ ρᶜ ℓ} → {𝒦 : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)}
→       S{β = γ}{ρᶜ} (a ⊔ ℓ) (S{β = β}{ρᵇ} ℓ 𝒦) ⊆ S{β = γ}{ρᶜ} ℓ 𝒦

S-idem (𝑨 , (𝑩 , sB , A≤B) , x≤A) = 𝑩 , (sB , ≤-trans x≤A A≤B)

```

#### Closure properties of P

`P` is a closure operator. This is proved by checking that `P` is monotone, expansive, and idempotent. The meaning of these terms will be clear from the definitions of the types that follow.

``` H-expa : ∀{ℓ} → {𝒦 : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)} → 𝒦 ⊆ H ℓ 𝒦
H-expa {ℓ} {𝒦}{𝑨} kA = 𝑨 , kA , IdHomImage

S-expa : ∀{ℓ} → {𝒦 : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)} → 𝒦 ⊆ S ℓ 𝒦
S-expa {ℓ}{𝒦}{𝑨} kA = 𝑨 , (kA , ≤-reflexive)

P-mono : ∀{ℓ ι} → {𝒦 𝒦' : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)}
→       𝒦 ⊆ 𝒦' → P{β = α}{ρᵃ} ℓ ι 𝒦 ⊆ P ℓ ι 𝒦'

P-mono {ℓ}{ι}{𝒦}{𝒦'} kk {𝑩} (I , 𝒜 , (kA , B≅⨅A)) = I , (𝒜 , ((λ i → kk (kA i)) , B≅⨅A))

open _≅_
open IsHom

P-expa : ∀{ℓ ι} → {𝒦 : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)} → 𝒦 ⊆ P ℓ ι 𝒦
P-expa {ℓ}{ι}{𝒦}{𝑨} kA = ⊤ , (λ x → 𝑨) , ((λ i → kA) , Goal)
where
open Algebra 𝑨 using () renaming (Domain to A)
open Algebra (⨅ (λ _ → 𝑨)) using () renaming (Domain to ⨅A)
open Setoid A using ( refl )
open Setoid ⨅A using () renaming ( refl to refl⨅ )

to⨅ : A ⟶ ⨅A
(to⨅ ⟨\$⟩ x) = λ _ → x
cong to⨅ xy = λ _ → xy
to⨅IsHom : IsHom 𝑨 (⨅ (λ _ → 𝑨)) to⨅
compatible to⨅IsHom =  refl⨅

from⨅ : ⨅A ⟶ A
(from⨅ ⟨\$⟩ x) = x tt
cong from⨅ xy = xy tt
from⨅IsHom : IsHom (⨅ (λ _ → 𝑨)) 𝑨 from⨅
compatible from⨅IsHom = refl

Goal : 𝑨 ≅ ⨅ (λ x → 𝑨)
to Goal = to⨅ , to⨅IsHom
from Goal = from⨅ , from⨅IsHom
to∼from Goal = λ _ _ → refl
from∼to Goal = λ _ → refl

V-expa : ∀ ℓ ι → {𝒦 : Pred (Algebra α ρᵃ)(a ⊔ ov ℓ)} → 𝒦 ⊆ V ℓ ι 𝒦
V-expa ℓ ι {𝒦} {𝑨} x = H-expa {a ⊔ ℓ ⊔ ι} (S-expa {a ⊔ ℓ ⊔ ι} (P-expa {ℓ}{ι} x) )

```

We sometimes want to go back and forth between our two representations of subalgebras of algebras in a class. The tools `subalgebra→S` and `S→subalgebra` are made for that purpose.

```module _ {α ρᵃ β ρᵇ ℓ ι : Level}{𝒦 : Pred (Algebra α ρᵃ)(α ⊔ ρᵃ ⊔ ov ℓ)}
{𝑨 : Algebra α ρᵃ}{𝑩 : Algebra β ρᵇ} where

S-≅ : 𝑨 ∈ S ℓ 𝒦 → 𝑨 ≅ 𝑩 → 𝑩 ∈ S{α ⊔ β}{ρᵃ ⊔ ρᵇ}(α ⊔ ρᵃ ⊔ ℓ) (Level-closure ℓ 𝒦)
S-≅ (𝑨' , kA' , A≤A') A≅B = lA' , (lklA' , B≤lA')
where
lA' : Algebra (α ⊔ β) (ρᵃ ⊔ ρᵇ)
lA' = Lift-Alg 𝑨' β ρᵇ
lklA' : lA' ∈ Level-closure ℓ 𝒦
lklA' = Lift-closed ℓ kA'
subgoal : 𝑨 ≤ lA'
subgoal = ≤-trans-≅ A≤A' Lift-≅
B≤lA' : 𝑩 ≤ lA'
B≤lA' = ≅-trans-≤ (≅-sym A≅B) subgoal

V-≅ : 𝑨 ∈ V ℓ ι 𝒦 → 𝑨 ≅ 𝑩 → 𝑩 ∈ V{β = α}{ρᵃ} ℓ ι 𝒦
V-≅ (𝑨' , spA' , AimgA') A≅B = 𝑨' , spA' , HomImage-≅ AimgA' A≅B

module _ {α ρᵃ ℓ : Level}
(𝒦 : Pred(Algebra α ρᵃ) (α ⊔ ρᵃ ⊔ ov ℓ))
(𝑨 : Algebra (α ⊔ ρᵃ ⊔ ℓ) (α ⊔ ρᵃ ⊔ ℓ)) where

private ι = ov(α ⊔ ρᵃ ⊔ ℓ)

V-≅-lc : Lift-Alg 𝑨 ι ι ∈ V{β = ι}{ι} ℓ ι 𝒦
→       𝑨 ∈ V{γ = ι}{ι} ℓ ι 𝒦

V-≅-lc (𝑨' , spA' , lAimgA') = 𝑨' , (spA' , AimgA')
where
AimgA' : 𝑨 IsHomImageOf 𝑨'
AimgA' = Lift-HomImage-lemma lAimgA'

```

The remaining theorems in this file are as yet unused, but may be useful later and/or for reference.

```module _ {α ρᵃ ℓ ι : Level}{𝒦 : Pred (Algebra α ρᵃ)(α ⊔ ρᵃ ⊔ ov ℓ)} where

-- For reference, some useful type levels:
classP : Pred (Algebra α ρᵃ) (ov(α ⊔ ρᵃ ⊔ ℓ ⊔ ι))
classP = P{β = α}{ρᵃ} ℓ ι 𝒦

classSP : Pred (Algebra α ρᵃ) (ov(α ⊔ ρᵃ ⊔ ℓ ⊔ ι))
classSP = S{β = α}{ρᵃ} (α ⊔ ρᵃ ⊔ ℓ ⊔ ι) (P{β = α}{ρᵃ} ℓ ι 𝒦)

classHSP : Pred (Algebra α ρᵃ) (ov(α ⊔ ρᵃ ⊔ ℓ ⊔ ι))
classHSP = H{β = α}{ρᵃ}(α ⊔ ρᵃ ⊔ ℓ ⊔ ι) (S{β = α}{ρᵃ}(α ⊔ ρᵃ ⊔ ℓ ⊔ ι) (P{β = α}{ρᵃ}ℓ ι 𝒦))

classS : ∀{β ρᵇ} → Pred (Algebra β ρᵇ) (β ⊔ ρᵇ ⊔ ov(α ⊔ ρᵃ ⊔ ℓ))
classS = S ℓ 𝒦
classK : ∀{β ρᵇ} → Pred (Algebra β ρᵇ) (β ⊔ ρᵇ ⊔ ov(α ⊔ ρᵃ ⊔ ℓ))
classK = Level-closure{α}{ρᵃ} ℓ 𝒦

module _ {α ρᵃ β ρᵇ γ ρᶜ ℓ : Level}{𝒦 : Pred (Algebra α ρᵃ)(α ⊔ ρᵃ ⊔ ov ℓ)} where
private a = α ⊔ ρᵃ ; b = β ⊔ ρᵇ ; c = γ ⊔ ρᶜ

LevelClosure-S : Pred (Algebra (α ⊔ γ) (ρᵃ ⊔ ρᶜ)) (c ⊔ ov(a ⊔ b ⊔ ℓ))
LevelClosure-S = Level-closure{β}{ρᵇ} (a ⊔ ℓ) (S ℓ 𝒦)

S-LevelClosure : Pred (Algebra (α ⊔ γ) (ρᵃ ⊔ ρᶜ)) (ov(a ⊔ c ⊔ ℓ))
S-LevelClosure = S{α ⊔ γ}{ρᵃ ⊔ ρᶜ}(a ⊔ ℓ) (Level-closure ℓ 𝒦)

S-Lift-lemma : LevelClosure-S ⊆ S-LevelClosure
S-Lift-lemma {𝑪} (𝑩 , (𝑨 , (kA , B≤A)) , B≅C) = Lift-Alg 𝑨 γ ρᶜ
, (Lift-closed{β = γ}{ρᶜ} ℓ kA)
, C≤lA
where
B≤lA : 𝑩 ≤ Lift-Alg 𝑨 γ ρᶜ
B≤lA = ≤-Lift B≤A
C≤lA : 𝑪 ≤ Lift-Alg 𝑨 γ ρᶜ
C≤lA = ≅-trans-≤ (≅-sym B≅C) B≤lA

module _ {α ρᵃ : Level} where

P-Lift-closed : ∀ ℓ ι → {𝒦 : Pred (Algebra α ρᵃ)(α ⊔ ρᵃ ⊔ ov ℓ)}{𝑨 : Algebra α ρᵃ}
→              𝑨 ∈ P{β = α}{ρᵃ} ℓ ι 𝒦
→              {γ ρᶜ : Level} → Lift-Alg 𝑨 γ ρᶜ ∈ P (α ⊔ ρᵃ ⊔ ℓ) ι (Level-closure ℓ 𝒦)
P-Lift-closed ℓ ι {𝒦}{𝑨} (I , 𝒜 , kA , A≅⨅𝒜) {γ}{ρᶜ} = I
, (λ x → Lift-Alg (𝒜 x) γ ρᶜ)
, goal1 , goal2

where
goal1 : (i : I) → Lift-Alg (𝒜 i) γ ρᶜ ∈ Level-closure ℓ 𝒦
goal1 i = Lift-closed ℓ (kA i)
goal2 : Lift-Alg 𝑨 γ ρᶜ ≅ ⨅ (λ x → Lift-Alg (𝒜 x) γ ρᶜ)
goal2 = ≅-trans (≅-sym Lift-≅) (≅-trans A≅⨅𝒜 (⨅≅⨅ℓρ{ℓ = γ}{ρ = ρᶜ}))

```


text:
article one of the constitution illustrates how the national government's power is

# article one of the constitution illustrates how the national government's power is

S
9.9k points
Which of the following is a true statement?Reserved powers are listed in the Constitution and belong to the state governments.
Which statement best describes why the ruling of Gibbons v. Ogden is significant?It clarified the meaning of the commerce clause to say that national government has the right to regulate interstate commerce.
Which of the following quotes best reflects the principle of popular sovereignty in the Declaration of Independence?"He has refused for a long time, after such dissolutions, to cause others to be elected."
Implied powers are those that areinterpreted from the language of the Constitution as belonging to the national government
The act of collaborating to write and send the Declaration of Independence reflected the principle ofpopular sovereignty
Imagine the state of California passes a law to tax all banking activities within the state. Which of the following could happen?The national bank of the United States, or the Federal Reserve, could challenge the California law based on the McCulloch v. Maryland decision.
Which of the following is a true statement about the Articles of Confederation?It created a strong legislative branch, a weak executive branch, and no judicial branch.
The Constitution gives the legislative branch the power toestablish postal service
How does the U.S. Constitution reflect the principle of democracy?Most of the people can elect leaders and petition the government.
Which of the following quotes best reflects the principle of natural rights in the Declaration of Independence?"For depriving us in many cases, of the benefits of Trial by Jury."
The Declaration of Independence reflected the idea of social contract because itsaid a new government must form due to the king's actions
The power to declare war isgranted to the national government, not states
The District of Columbia v. Heller case clarified the meaning of theSecond Amendment
The executive branch has the responsibility ofenforcing laws passed by the legislative branch
How did early American leaders think the Articles of Confederation would protect natural rights?They believed that keeping the federal government weak would protect the rights of the states and people.
In the U.S. Constitution, the principle "separation of powers" isevident because the first three articles describe the three branches of government
The "necessary and proper clause" is a phrase in the U.S. Constitution designed to protect the principle ofrule of law
The grievances listed in the Declaration of Independence werepresented as evidence for the king's breaking of the social contract
Which of the following makes a true statement?Reserved powers are those that are exclusive to the states, such as ratifying amendments.
The power to create and collect taxes is an example ofa concurrent power
Article Six of the Constitution illustrates how the national government's power isderived from the states and the people
Which of the following statements accurately describes the relationship between the principle of "separation of powers" and the Articles of Confederation?Because the Articles of Confederation created a central government that contained only a legislative branch, it does not reflect the principle of "separation of powers".
Under the Articles of Confederation, the principle of "republicanism" helps explain whystate governments
The framers of the Constitution believed that they created a governmentlimited to the powers described in the document
Article One of the Constitution describes the powers of thelegislative branch, such as declaring war
Under the Constitution, the principle of "federalism" is the reason that power and responsibility areshared by state and national governments
The Articles of Confederation reflected the principle of federalism because itdivided power between state and national governments
Article Two of the Constitution describes the powers of theexecutive branch, such as enforcing laws
Articles I, II, and III of the Constitution describe thepowers of the branches of federal government
How is the principle of "rule of law" evident in the Constitution?It creates a government that can make, enforce, and review its own laws.
Every Bill which shall have passed the House of Representatives and the Senate, shall, before it become a Law, be presented to the President of the United States: If he approve he shall sign it, but if not he shall return it, with his Objections to that House in which it shall have originated..."-U.S. Constitution This quote directly reflects the principle ofchecks and balances
The first three articles of the U.S. Constitutionexplain the powers delegated to the national government
article one
studybud

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text:
# NEET Physics Current Electricity Questions Solved

In the figure given the value of X resistance will be, when the p.d. between B and D is zero

(1) 4 ohm

(2) 6 ohm

(3) 8 ohm

(4) 9 ohm

(3) Potential difference between B and D is zero, it means Wheatstone bridge is in balanced condition

So $\frac{P}{Q}=\frac{R}{S}$

$\frac{21}{3+\frac{8X}{\left(8+X\right)}}=\frac{18}{6}$

$X=8\Omega$

Difficulty Level:

• 14%
• 15%
• 68%
• 5%
Crack NEET with Online Course - Free Trial (Offer Valid Till September 21, 2019)


text:
I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my short series on a couple of easily stated but remarkably difficult geometry problems.

Part 1: The world’s second hardest easy geometry problem.

Part 2: The world’s hardest easy geometry problem.

# Langley’s Adventitious Angles (Part 1)

Math With Bad Drawings had an interesting post about solving for $x$ in the following picture (this picture is taken from http://thinkzone.wlonk.com/MathFun/Triangle.htm):

I had never heard of this problem before, but it’s apparently well known and is called Langley’s Adventitious Angles. See Math With Bad Drawings, Wikipedia, and Math Pages for more information about the solution of this problem. Math Pages has a nice discussion about mathematical aspects of this problem, including connections to the Laws of Sines and Cosines and to various trig identities.

I’d encourage you to try to solve for $x$ without clicking on any of these links… a certain trick out of the patented Bag of Tricks is required to solve this problem using only geometry (as opposed to the Law of Cosines and the Law of Sines). I have a story that I tell my students about the patented Bag of Tricks: Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students. In the same post, Math With Bad Drawings has a nice discussion about pedagogical aspects of this problem concerning when a “trick” becomes a “technique”.

I recommend this problem for advanced geometry students who need to be challenged; even bright students will be stumped concerning coming up with the requisite trick on their own. Indeed, the problem still remains quite challenging even after the trick is shown.

# Inverse Functions: Arccosine and SSS (Part 21)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve $\triangle ABC$ if $a = 16$, $b = 20$, and $c = 25$.

When solving for the three angles, it’s best to start with the biggest angle (that is, the angle opposite the biggest side). To see why, let’s see what happens if we first use the Law of Cosines to solve for one of the two smaller angles, say $\alpha$:

$a^2 = b^2 + c^2 - 2 b c \cos \alpha$

$256 = 400 + 625 - 1000 \cos \alpha$

$-769 = -1000 \cos \alpha$

$0.769 = \cos \alpha$

$\alpha \approx 39.746^\circ$

So far, so good. Now let’s try using the Law of Sines to solve for $\gamma$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 39.746^\circ}{16} \approx \displaystyle \frac{\sin \gamma}{25}$

$0.99883 \approx \sin \gamma$

Uh oh… there are two possible solutions for $\gamma$ since, hypothetically, $\gamma$ could be in either the first or second quadrant! So we have no way of knowing, using only the Law of Sines, whether $\gamma \approx 87.223^\circ$ or if $\gamma \approx 180^\circ - 87.223^\circ = 92.777^\circ$.

For this reason, it would have been far better to solve for the biggest angle first. For the present example, the biggest answer is $\gamma$ since that’s the angle opposite the longest side.

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.223^\circ$.

We now use the Law of Sines to solve for either $\alpha$ or $\beta$ (pretending that we didn’t do the work above). Let’s solve for $\alpha$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin \alpha}{16} \approx \displaystyle \frac{\sin 87.223}{25}$

$\sin \alpha \approx 0.63949$

This equation also has two solutions in the interval $[0^\circ, 180^\circ]$, namely, $\alpha \approx 39.736^\circ$ and $\alpha \approx 180^\circ - 39.736^\circ = 140.264^\circ$. However, we know full well that the answer can’t be larger than $\gamma$ since that’s already known to be the largest angle. So there’s no need to overthink the matter — the answer from blindly using arcsine on a calculator is going to be the answer for $\alpha$.

Naturally, the easiest way of finding $\beta$ is by computing $180^\circ - \alpha - \gamma$.

# Inverse Functions: Arccosine and SSS (Part 19)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve $\triangle ABC if$latex a = 16$, $b = 20$, and $c = 25$. To solve for, say, the angle $\gamma$, we employ the Law of Cosines: $c^2 = a^2 + b^2 - 2 a b \cos \gamma$ $625 = 256 + 400 - 640 \cos \gamma$ $-31 =-640 \cos \gamma$ $0.0484375 = \cos \gamma$ Using a calculator, we find that $\gamma \approx 87.2^\circ$. And the good news is that there is no need to overthink this… this is guaranteed to be the angle since the range of $y = \cos^{-1} x$ is $[0,\pi]$, or $[0^\circ, 180^\circ]$ in degrees. So the equation $\cos x = \hbox{something}$ is guaranteed to have a unique solution between $0^\circ$ and $180^\circ$. (But there are infinitely many solutions on $\mathbb{R}$. And since an angle in a triangle must lie between $0^\circ$ and $180^\circ$, the practical upshot is that just plugging into a calculator blindly is perfectly OK for this problem. This is in stark contrast to the Law of Sines, for which some attention must be paid for solutions in the interval $[0^\circ,90^\circ]$ and also the interval $[90^\circ, 180^\circ]$. From this point forward, the Law of Cosines could be employed again to find either $\alpha$ or $\beta$. Indeed, this would be my preference since the sides $a$, $b$, and $c$ are exactly. However, my experience is that students prefer the simplicity of the Law of Sines to solve for one of these angles, using the now known pair of $c$ (exactly known) and $\gamma$ (approximately known with a calculator). # Inverse functions: Arcsine and SSA (Part 17) In the last few posts, we studied the SSA case of solving for a triangle, when two sides and an non-included angle are given. (Some mathematics instructors happily prefer the angle-side-side acronym to bluntly describe the complications that arise from this possibly ambiguous case. I personally prefer not to use this acronym.) A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:https://meangreenmath.files.wordpress.com/2014/10/philistines.png Suppose that $a$, $c$, and the nonincluded angle $\alpha$ are given, and we are supposed to solve for $b$, $\beta$, and $\gamma$. As we’ve seen in this series, there are four distinct cases — and handling these cases requires accurately solving equation like $\sin \gamma = \hbox{something}$ on the interval $[0^\circ, 180^\circ]$. Case 1. $b < c \sin \alpha$. In this case, there are no solutions. When the Law of Sines is employed and we reach the step $\sin \gamma = \hbox{something}$ the $\hbox{something}$ is greater than 1, which is impossible. Case 2. $b = c \sin \alpha$. This rarely arises in practice (except by careful writers of textbooks). In this case, there is exactly one solution. When the Law of Sines is employed, we obtain $\sin \gamma = 1$ We conclude that $\gamma = 90^\circ$, so that $\triangle ABC$ is a right triangle. Case 3. $c \sin \alpha < b < c$. This is the ambiguous case that yields two solutions. The Law of Sines yields $\sin \gamma = \hbox{something}$ so that there are two possible choices for $\gamma$, $\hbox{some angle}$ and $180^\circ - \hbox{some angle}$. Case 4. $b > c$. This yields one solution. Similar to Case 3, the Law of Sines yields $\sin \gamma = \hbox{something}$ so that there are two possible choices for $\gamma$, $\hbox{some angle}$ and $180^\circ - \hbox{some angle}$. However, when the second larger value of $\gamma$ is attempted, we end up with a negative angle for $\beta$, which is impossible (unlike Case 3). Many mathematics students prefer to memorize rules like those listed above. However, I try to encourage my students not to blindly use rules when solving the SSA case, as it’s just too easy to make a mistake in identifying the proper case. Instead, I encourage them to use the Law of Sines and to remember that the equation $\sin \gamma = t$ has two solutions in $[0^\circ, 180^\circ]$ as long as$0 < t < 1\$:

$\gamma = \sin^{-1} t \qquad \hbox{and} \qquad \gamma = 180^\circ - \sin^{-1} t$

If they can remember this fact, then students can just follow their noses when applying the Law of Sines, identifying impossible and ambiguous cases when the occasions arise.

# Inverse functions: Arcsine and SSA (Part 16)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has two different solutions:

Solve $\triangle ABC$ if $a = 8$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

This time, the red circle intersects the dashed black line at two different points. So there will be two different solutions for this case. In other words, the phrasing of the question is somewhat deceptive. Usually when the question asks, “Solve the triangle…”, it’s presumed that there is only one solution. In this case, however, there are two different solutions.

These two different solutions appear when using the Law of Sines:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 30^\circ}{8} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{8} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{5}{8} = \sin \gamma$

At this point, the natural inclination of a student is to pop out the calculator and find $\sin^{-1} \frac{1}{3}$.

This is incorrect logic that, as discussed extensively in earlier in this series of posts, there are two angles between $0^\circ$ and $180^\circ$ with a sine of $5/8$:

$\sin^{-1} \frac{5}{8} \qquad \hbox{and} \qquad \pi - \sin^{-1} \frac{5}{8}$,

or, in degrees,

$\gamma \approx 38.68^\circ \qquad \hbox{and} \qquad \gamma \approx 141.32^\circ$

So we have two different cases to check. Unlike the previous posts in this series, it’s really, really important that we list both of these cases.

Case 1: $\gamma \approx 38.68^\circ$. We begin by solving for $\beta$:

$\beta = 180^\circ - \alpha - \gamma \approx 111.32^\circ$

Then we can use the Law of Sines to find $b$. In this case, it’s best to use the pair $\alpha - a$ instead of $\gamma - c$ since the values of $\alpha$ and $a$ are both known exactly.

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}$

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 111.32^\circ}{b}$

$b = \displaystyle \frac{8 \sin 111.32^\circ}{\sin 30^\circ}$

$b \approx 14.9$

This triangle with $\gamma \approx 38.68^\circ$, $\beta \approx 111.32^\circ$, and $b \approx 14.9$ corresponds to the bigger of the two triangles in the above picture, or the rightmost of the two places where the dotted circle intersects the black dotted line.

Case 2: $\gamma \approx 141.32^\circ$. We again begin by solving for $\beta$:

$\beta = 180^\circ - \alpha - \gamma \approx 8.68^\circ$

Unlike yesterday’s example, this is possible. So we have to continue the calculation to find $b$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}$

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 8.68^\circ}{b}$

$b = \displaystyle \frac{8 \sin 8.68^\circ}{\sin 30^\circ}$

$b \approx 2.4$

This second triangle with $\gamma \approx 141.32^\circ$, $\beta \approx 8.68^\circ$, and $b \approx 2.4$ corresponds to the thinner of the two triangles in the above picture, or the leftmost of the two places where the dotted circle intersects the black dotted line.

# Inverse functions: Arcsine and SSA (Part 15)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve $\triangle ABC$ if $a = 15$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case. We also note that the circle would have intersected the black dashed line had the dashed line been extended to the left. This will become algebraically clear in the solution below.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{15} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1}{3} = \sin \gamma$

At this point, the natural inclination of a student is to pop out the calculator and find $\sin^{-1} \frac{1}{3}$.

This is incorrect logic that, as discussed extensively in yesterday’s post, there are two angles between $0^\circ$ and $180^\circ$ with a sine of $1/3$:

$\sin^{-1} \frac{1}{3} \qquad \hbox{and} \qquad \pi - \sin^{-1} \frac{1}{3}$,

or, in degrees,

$\gamma \approx 19.47^\circ \qquad \hbox{and} \qquad \gamma \approx 160.53^\circ$

So we have two different cases to check.

Case 1: $\gamma \approx 19.47^\circ$. We begin by solving for $\beta$:

$\beta = 180^\circ - \alpha - \gamma \approx 130.53^\circ$

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find $b$. In this case, it’s best to use the pair $\alpha - a$ instead of $\gamma - c$ since the values of $\alpha$ and $a$ are both known exactly.

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}$

$\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin 130.53^\circ}{b}$

$b = \displaystyle \frac{15 \sin 130.53^\circ}{\sin 30^\circ}$

$b \approx 22.8$

Case 2: $\gamma \approx 130.53^\circ$. We again begin by solving for $\beta$:

$\beta = 180^\circ - \alpha - \gamma \approx -10.53^\circ$

Oops. That’s clearly impossible. So there is only one possible triangle, and the missing pieces are $\gamma \approx 19.47^\circ$, $\beta \approx 130.53^\circ$, and $b \approx 22.8$. Judging from the above (correctly drawn) picture, these numbers certainly look plausible.

It turns out that Case 2 will always fail in SSA will always fail as long as the side opposite the given angle is longer than the other given side (in this case, $a > c$). However, I prefer that my students not memorize this rule. Instead, I’d prefer that they list the two possible values of $\gamma$ and then run through the logical consequences, stopping when an impossibility is reached. As we’ll see in tomorrow’s post, it’s perfectly possible for Case 2 to produce a second valid solution with the proper choice for the length of the side opposite the given angle.

# Inverse functions: Arcsine and SSA (Part 14)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve $\triangle ABC$ if $a = 15$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case. We also note that the circle would have intersected the black dashed line had the dashed line been extended to the left. This will become algebraically clear in the solution below.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{15} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1}{3} = \sin \gamma$

At this point, the natural inclination of a student is to pop out the calculator and find $\sin^{-1} \frac{1}{3}$.

This is incorrect logic that, as we’ll see tomorrow, nevertheless leads to the correct conclusion. This is incorrect logic because there are two angles between $0^\circ$ and $180^\circ$ with a sine of $1/3$. There is one solution in the first quadrant (the unique answer specified by arcsine), and there is another answer in the second quadrant — which is between $90^\circ$ and $180^\circ$ and hence not a permissible value of arcsine. Let me demonstrate this in three different ways.

First, let’s look at the graph of $y = \sin x$ (where, for convenience, the units of the $x-$axis are in degrees). This graph intersects the line $y = \frac{1}{3}$ in two different places between $0^\circ$ and $180^\circ$. This does not violate the way that arcsine was defined — arcsine was defined using the restricted domain $[-\pi/2,\pi/2]$, or $[-90^\circ, 90^\circ]$ in degrees.

Second, let’s look at drawing angles in standard position. The angle in the second quadrant is clearly the reflection of the angle in the first quadrant through the $y-$axis.

Third, let’s use a trigonometric identity to calculate $\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right)$:

$\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right) = \sin \pi \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) - \cos \pi \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)$

$=0 \cdot \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) + 1 \cdot \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)$

$= \displaystyle \frac{1}{3}$

Fourth, and perhaps most convincingly for modern students (to my great frustration), let’s use a calculator:

All this to say, blinding computing $\sin^{-1} \frac{1}{3}$ uses incorrect logic when solving this problem.

Tomorrow, we’ll examine what happens when we try to solve the triangle using these two different solutions for $\gamma$.

# Inverse functions: Arcsine and SSA (Part 13)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve $\triangle ABC$ if $a = 5$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case.

Of course, the reason that the dashed circle and line intersect at exactly one point is because $a = c \sin \alpha$, so that the triangle is a right triangle.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle 1 = \sin \gamma$

$90^\circ = \gamma$

The jump to the last step is only possible because there’s exactly one angle between $0^\circ$ and $90^\circ$ whose sine is equal to $1$. In the next couple posts in this series, we’ll see what happens when we get a step where $0 < \sin \gamma < 1$.

Anyway, for the problem at hand, from this point forward it’s easy to solve for the remaining pieces. We begin by finding $\beta$:

$\beta = 180^\circ - \alpha - \gamma = 60^\circ$

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find $b$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}$

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 60^\circ}{b}$

$\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sqrt{3}/2}{b}$

$b = 5\sqrt{3}$

In the next few posts of this series, I’ll consider the other SSA cases — including the case where two solutions are possible.

# Inverse functions: Arcsine and SSA (Part 12)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has no solution:

Solve $\triangle ABC$ if $a = 3$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

The red dashed circle with center $B$ illustrates the dilemma: “side” $BC$ is simply too short to reach the horizontal dashed line to make the vertex $C$, dangling limply from the vertex $B$.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 30^\circ}{3} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{3} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{5}{3} = \sin \gamma$

Since $\sin \gamma$ must like between $0$ and $1$ (said another way, $\sin^{-1} \frac{5}{3}$ is undefined), we know that this triangle cannot be solved.

In the next few posts of this series, I’ll consider the other SSA cases — including the case where two solutions are possible.


text:
시간 제한메모리 제한제출정답맞힌 사람정답 비율
3 초 512 MB3510829.630%

## 문제

San Bytecisco is a beautifully situated coastal town. It consists of n small, yet densely populated islands, numbered from 1 to n. Certain islands are connected with bridges, used for (bidirectional) road traffic. Each pair of islands can be connected with at most one bridge. The islands are connected in such a way that every island can be reached from every other by using the bridges only.

Byteasar and Bytie are going for a bike trip in San Bytecisco. The will start their ride at the island no. 1. They intend to visit every island, while passing along every bridge once and ending the trip where it began, i.e., the island no. 1. Being quite seasoned riders, they expect some serious trouble from... the wind! After all, it is very windy along the coast, and especially so on the bridges between the islands. Obviously, depending on its speed and direction, the wind makes it hard to cross the bridge in different extent for either direction. For simplicity we will assume for every bridge and direction of crossing, the opposing wind speed is constant.

Help Byteasar and Bytie to find a route as they desire that will in addition be the least tiresome. Byteasar and Bytie agreed on the maximum opposing wind speed as a measure of a route's tiresomeness.

## 입력

In the first line of the standard input there are two integers separated by a single space: n and m (2 ≤ n ≤ 1,000, 1 ≤ m ≤ 2,000), denoting the number of islands and the number of bridges in San Bytecisco respectively. The islands are numbered from 1 to n, while the bridges from 1 to m. The following m lines specify the bridges. The line no. (i+1) contains four integers ai, bi, li, pi (1 ≤ ai,bi ≤ n, ai≠bi, 1 ≤ li,pi ≤ 1,000), separated by single spaces. These denote that the bridge no. i connects the islands no. ai and bi. The opposing wind speeds are li when one goes moves from ai to bi, and pi if one goes from bi to ai.

## 출력

If there is no route satisfying the requirements of the daring two riders, the first and only line of the standard output should hold the word NIE (no in Polish). Otherwise, the output should have two lines, specifying the least tiresome route over San Bytecisco. The first line should hold the maximum opposing wind speed for that route, i.e., the number we wish to minimize. The second line should hold m integers, separated by single spaces, giving the numbers of successive bridges one crosses on the least tiresome route.

Should there be more than one least tiresome route, your program can pick one arbitrarily.

## 예제 입력 1

4 4
1 2 2 4
2 3 3 4
3 4 4 4
4 1 5 4

## 예제 출력 1

4
4 3 2 1


text:
# Practical Course ‘Specification and Verification’

## Overview

 Module Bachelor-Praktikum (Practical Course for BSc students), IN0012Master-Praktikum (Practical Course for MSc students, IN2106 Prerequisites Basic knowledge of Isabelle (e.g. through ‘Functional Data Structures’, IN2347, ‘Semantics’, IN2055) Language German, English Organisation Simon Roßkopf, Prof. Tobias NipkowContact Simon Roßkopf for all course matters, or your advisor (once you have been assigned one).

## Content

Participants will work on a project by themselves using the interactive theorem prover Isabelle. The practical course will run throughout the semester.

## Application

The application will be through the Matching Platform. There will be no kick-off meeting; if you have questions or want to participate contact Simon Roßkopf via email in advance (i.e. before the matching starts) and indicate what prior experience you have with Isabelle (e.g. through one of the above-mentioned lectures) and possibly what particular topics you are interested in.

Note that prior experience with Isabelle is mandatory

## Topics

Topics still subject to change!

### Verification of an Interesting Algorithm or Data Structure

You are welcome to propose an algorithm or data structure and discuss the realizability with your advisor. Some examples of algorithms and data structures that were verified in past lab courses: Knuth-Morris-Pratt, A*, Kruskal, Finger Trees, Skew Binomial Queues, Dijkstra’s Algorithm, Conversion Between Regular Expressions and Finite Automata.

Ideas: String Search Algorithms (Boyer-Moore), Hopcroft’s Minimization Algorithm, B-trees

Advisor: Simon Roßkopf, Lukas Stevens

### NP Zoo

#### Polynomial Reductions

This topic aims to formalize some parts of the zoo of NP-complete problems and the polynomial-time reductions between them. See Figure 6 of [1] for an overview of a plethora of such problems and their interrelations.

#### Cook-Levin theorem

The Cook-Levin theorem is a fundamental result in Computational Complexity Theory. The student would join an ongoing effort to formalise the Cook-Levin theorem. This will involve the following steps:

• Extending the existing refinement approaches in Isabelle/HOL to refine programs specified in Isabelle/HOL’s functional programming language to an already formalised simple While-language.
• Extending an existing formalisation of a While-language to include the notion of a cost of computation steps, and thus a notion of running time.
• Proving running time upper bounds on an already verified reduction from While-programs to SAT.

Further reading: [1]

### Real-time double-ended queues

Hood and Melville showed how to implement functionally a queue with constant time enqueue and dequeue operations. A verification is described here and was covered in the FDS course 2021. Chuang and Goldberg generalised the work of Hood and Melville to double-ended queues. These are queues where you can enqueue and dequeue at both ends. The aim of this project is to verify the implementation by Chuang and Goldberg.

Advisor: Tobias Nipkow

### Verification of an approximation algorithm for a graph theoretic problem

Many basic graph theoretic problems are either NP-hard or cannot be solved in better than polynomial time. This makes solving those problems prohibitive if not impossible for real-world graphs. Approximation algorithms circumvent that by using less resources than exact algorithms, at the expense of providing only approximate solutions. In this project the student would formally verify that 1) the approximate solutions of those algorithms meet a certain quality citerion, 2) the upper bounds on their runtimes are correct. A particularly interesting algorithm is the algorithm described below for approximating the diameter of an undirected graph due to Aingworth et al..

Further Reading: [1]

Advisor: Mohammad Abdulaziz, Lukas Stevens

### Verified Translations of AI Planning Problems:

AI planning is a discipline aiming to build computer programs that act rationally to achieve a certain goal. One established way of solving planning problems is via translating them to SAT [1]. In this project you would extend a recent verification [2] of the forall-step SAT encoding of classical AI planning. Possible directions: translating first-order logic planning problems to propositional planning problems, adding symmetry breaking formulae [3], adding formulae that represent invariants in the state space [4].

Further reading: [1] [2] [3] [4]

Advisor: Mohammad Abdulaziz

### Interval Analysis Based Verification of Neural Networks:

Deep neural networks are currently in heavy use in many applications, some of which are safety critical. One problem with them is that there are not theoretical foundations, let alone formal guarantees, on machine learning algorithms that train these deep neural networks. One approach to resolve that situation is by verifying that the trained neural networks themselves satisfy certain properties, e.g. that the outputs remain within certain bounds. One successful approach to perform this is via interval analysis [1,2]. In this project you would formally verify an interval analysis based method for verifying neural networks.

Further reading: [1] [2]

Advisor: Mohammad Abdulaziz, Tobias Nipkow

### Graph Theory and Algorithms:

There is an ongoing effort to build a unified library of graph theory and algorithms in Isabelle 1. In this project you would participate in the development of this graph library by either porting results from other theorem provers, or by formalising new results into that library. Some interesting directions would be: algorithms for computing strongly connected components in a digraph, connecting graph theoretic concepts, like matchings and weighted matchings, to linear programming. You can also suggest an interesting graph theoretic concept or algorithm, if you have one.

Further reading: [1]

Advisor: Mohammad Abdulaziz

### Number Theoretic Transform

The number theoretic transform (NTT) is an extension of the classic discrete Fourier transform (DFT) to finite fields. Similar to the DFT, it is a convolution to quickly compute a multiplication eg. of large integers or large degree polynomials. This has applications in computer arithmetic and cryptography, especially in lattice-based cryptography. The goal of this project would be to define the number theoretic transform and verify its correctness (eg. along these lines).

Advisor: Katharina Kreuzer

### Algebraic lattices

The goal of this project is to formalise some basic concepts from algebraic lattice theory. In algebra, lattices are discrete subgroups of $$\mathbb{R}^n$$, for example $$\mathbb{Z}^n$$ is a lattice (see for example). The application of algebraic lattices is huge: from chemistry where they describe the underlying structures of crystals to wallpaper groups and number theory in mathematics and lattice-based cryptography in computer science. The latter is of special interest, since the most known quantum-resistant cryptographic schemes rely on lattices. Example topics include the formalisation of lattices, their bases, transition between bases, the fundamental domain, Minkowski’s theorem or others. Another interesting topic is to formalise the basic lattice problems, such as the shortest vector problem (SVP), the closest vector problem (CVP) or similar

Advisor: Katharina Kreuzer

### Unification Hints

Unification hints are a controlled mechanism to extend unification procedures with new inference rules: if the standard unification procedure fails, the set of collected unification hints is matched against the goal for an appropriate alternative.

This mechanism is particularly useful for interactive theorem provers such as Isabelle, where users can provide such hints to guide the unifier to find solutions for seemingly trivial equalities which might not hold by pure unification but by using special domain knowledge. Beyond extending the strength of the unifier, unification hints can be used to build mechanisms such as type classes and reification tactics.

The goal of this project is to extend the existing implementation of unification hints in Isabelle. Currently, the first-order unifier and higher-order pattern matcher are able to work with unification hints. The first step of the project is to extend the full higher-order unification procedure to work with unification hints as well.

The second step of the project is to prove the usefulness of unification hints in the Isabelle/HOL environment. First, a non-trivial reification tactic should be implemented. Finally, a mechanism to integrate type classes in the user space of Isabelle (that is outside the Isabelle kernel) should be designed and implemented.

The project requires good functional programming experience as the code must be written in Isabelle/ML as well as basic experience with Isabelle/HOL.

Advisor: Kevin Kappelmann


text:
# Blumenthal and Kolmogorov 0-1 law

Blumenthal's 0-1 law see theorem 5.8/5.9 tells us that an event in the germ $\sigma-$ algebra has either probability zero or one with respect to a measure induced by a Brownian motion starting in some point $x.$ (Theorem 5.8)

Conversely, one can show that this is also the case for events in the tail $\sigma-$ algebra. This is not the standard Kolmogorov 0-1 law, but I want to call it that way, as it is also name in the used in the link. (Theorem 5.9)

Now, as you can see in the proof of theorem 5.9, that the two sigma algebras (tail and germ) can be mapped onto each other. Thus, the Kolmogorov 0-1 law for Brownian motion is in some sense the same as Blumenthal's 0-1 law.

My question is now the following: I am almost sure, that the probability $P^x(A)$ does not depend on $x$ for events in the tail sigma algebra, but afaik it is in general dependend on $x$ in Blumenthal's 0-1 law. Now, since the two sigma algebras agree (according to the proof given in the link), this does not make sense. Either both should depend on $x$ or they are both independent.

Does anybody know where my reasoning is wrong? I have been thinking about this for quite a time now but since I am not a probability theorist, I thought that somebody who is familiar with these things can easily help me out here.

• "The two $\sigma$-fields agree" - careful with that word "agree". What's constructed is a map from one to the other. It doesn't preserve the measure $P^x$, unless $x=0$. Nov 9 '15 at 7:52

The issue, informally, is this. Suppose $W_t$ is a Brownian motion started at some $x \in \mathbb{R}^d$. Let $X_{t} = t W_{1/t}$ as in the linked notes. Then $X_t$ is a Brownian motion, but started at 0, not $x$. (By the strong law of large numbers, we have $\lim_{t \to 0} X_t = \lim_{s \to \infty} \frac{1}{s} W_s = 0$ almost surely.)
To see the issue more formally, it's helpful to work with an explicit sample space $\Omega$. Let's take $\Omega = C((0, \infty), \mathbb{R}^d)$, to be the set of all continuous $\omega : (0, \infty) \to \mathbb{R}^d$, equipped with the Borel $\sigma$-algebra coming from the topology of uniform convergence on compact sets. For each $t > 0$, let $W_t : \Omega \to \mathbb{R}^d$ be simply $W_t(\omega) = \omega(t)$. And for each $x$, let $P^x$ be the unique probability measure on $\Omega$ under which $\{W_t\}$ is a Brownian motion started at $x$.
Now consider the map $\Phi : \Omega \to \Omega$ given by $(\Phi\omega)(t) = t \omega(1/t)$. I agree with your claim that $\Phi$ maps the tail $\sigma$-algebra to the germ $\sigma$-algebra and vice versa. But it sounds like you are thinking that $\Phi$ also preserves all the measures $P^x$, and that is not true. What is true is that $P^x(\Phi^{-1}(A)) = P^0(A)$ for all events $A$.
The Kolmogorov 0-1 law says that for any $x$, the tail $\sigma$-algebra is $P^x$-almost trivial. So a consequence is that the germ $\sigma$-algebra is $P^0$-almost trivial. In other words, Kolmogorov for $P^x$ directly implies Blumenthal for $P^0$ only. If you want to prove Blumenthal for $P^x$, you need one more (easy) step; say, that $P^x$ is the pushforward of $P^0$ under translation by $x$, and this translation preserves the germ $\sigma$-algebra.
• @user82546: The tail and germ $\sigma$-algebras of $\Omega$ do not coincide as sets! What they show is that the tail $\sigma$-algebra of $W_s$ coincides with the germ $\sigma$-algebra of $X_s$. Nov 10 '15 at 0:43


text:
Poster Open Access

# THE SUBTERAHERTZ SOLAR CYCLE: POLAR AND EQUATORIAL RADII DERIVED FROM SST AND ALMA

Menezes, Fabian

### Citation Style Language JSON Export

{
"publisher": "Zenodo",
"DOI": "10.5281/zenodo.4563673",
"language": "eng",
"title": "THE SUBTERAHERTZ SOLAR CYCLE: POLAR AND EQUATORIAL RADII DERIVED FROM SST AND ALMA",
"issued": {
"date-parts": [
[
2021,
2,
26
]
]
},
"abstract": "
At subterahertz frequencies – i.e., millimeter and submillimeter wavelengths – there is a gap of mea- surements of the solar radius as well as other parameters of the solar atmosphere. As the observational wavelength changes, the radius varies because the altitude of the dominant electromagnetic radiation is produced at different heights in the solar atmosphere. Moreover, radius variations throughout long time series are indicative of changes in the solar atmosphere that may be related to the solar cycle. Therefore, the solar radius is an important parameter for the calibration of solar atmospheric models enabling a better understanding of the atmospheric structure. In this work we use data from the So- lar Submillimeter-wave Telescope (SST) and from the Atacama Large Millimeter/submillimeter Array (ALMA), at the frequencies of 100, 212, 230, and 405 GHz, to measure the equatorial and polar radii of the Sun. The radii measured with extensive data from the SST agree with the radius-vs-frequency trend present in the literature. The radii derived from ALMA maps at 230 GHz also agree with the radius-vs-frequency trend, whereas the 100-GHz radii are slightly above the values reported by other authors. In addition, we analyze the equatorial and polar radius behavior over the years, by determin- ing the correlation coefficient between solar activity and subterahertz radii time series at 212 and 405 GHz (SST). The variation of the SST-derived radii over 13 years are correlated to the solar activity when considering equatorial regions of the solar atmosphere, and anticorrelated when considering polar regions. The ALMA derived radii time series for 100 and 230 GHz show very similar behaviors with those of SST.
",
"author": [
{
"family": "Menezes, Fabian"
}
],
"type": "graphic",
"id": "4563673"
}
36
37
views


text:
# aid with rewriting absolute values and converting log form to exponents

• Jul 15th 2013, 06:22 PM
deltemis
aid with rewriting absolute values and converting log form to exponents
my first problem is to rewrite |x-2|+|3-x| given that 0<x<2

What I've done so far is that given the values possible for x, the x-2 portion is negative making it:
(-x+2)
with the same reasoning the 3-x portion is always positive keeping it as is so I come up with:
(-x+2)+(3-x)

and once I simplify I get 5-2x, is this correct or did I do a step wrong?

the second problem I have is rewriting ln(sqrt(e)) = 1/2 to exponential form

What I did here was rewrite the sqrt(e) to e^1/2, then bring down the 1/2 via the ln(a^b) = b ln (a) rule and I'm stuck at
1/2 ln(e) = 1/2

I'm not sure what I do now to get it to exponential form, the ln and sqrt e get me confused, any help on what I'm supposed to do?

Thanks
• Jul 15th 2013, 07:37 PM
Prove It
Re: aid with rewriting absolute values and converting log form to exponents
Yes the first one is correct.

As for the second, you seem to be misunderstanding what the question is asking of you. You are not asked to SOLVE an equation, you are asked to SHOW that the two sides of the equation are in fact the same thing.

\displaystyle \begin{align*} \ln{ \left( \sqrt{e} \right) } &= \ln {\left( e^{\frac{1}{2}} \right) } \\ &= \frac{1}{2}\ln{(e)} \\ &= \frac{1}{2} \cdot 1 \\ &= \frac{1}{2} \end{align*}
• Jul 15th 2013, 07:39 PM


text:
### Increased computational accuracy in multi-compartmental cable models (Lindsay et al. 2005)

Accession:129149
Compartmental models of dendrites are the most widely used tool for investigating their electrical behaviour. Traditional models assign a single potential to a compartment. This potential is associated with the membrane potential at the centre of the segment represented by the compartment. All input to that segment, independent of its location on the segment, is assumed to act at the centre of the segment with the potential of the compartment. By contrast, the compartmental model introduced in this article assigns a potential to each end of a segment, and takes into account the location of input to a segment on the model solution by partitioning the effect of this input between the axial currents at the proximal and distal boundaries of segments. For a given neuron, the new and traditional approaches to compartmental modelling use the same number of locations at which the membrane potential is to be determined, and lead to ordinary differential equations that are structurally identical. However, the solution achieved by the new approach gives an order of magnitude better accuracy and precision than that achieved by the latter in the presence of point process input.
Reference:
1 . Lindsay AE, Lindsay KA, Rosenberg JR (2005) Increased computational accuracy in multi-compartmental cable models by a novel approach for precise point process localization. J Comput Neurosci 19:21-38 [PubMed]
Model Information (Click on a link to find other models with that property)
 Model Type: Neuron or other electrically excitable cell; Brain Region(s)/Organism: Cell Type(s): Channel(s): I Na,t; I K; Gap Junctions: Receptor(s): Gene(s): Transmitter(s): Simulation Environment: NEURON; C or C++ program; Model Concept(s): Methods; Implementer(s):
Search NeuronDB for information about:  I Na,t; I K;
 / LindsayEtAl2005 readme.txt 03-192.pdf AnalyseResults.c BitsAndPieces.c CellData.dat CompareSpikeTrain.c Ed04.tex ExactSolution.dat GammaCode Gen.tex Gen1.tex Gen2.tex Gen3.tex Gen4.tex Gen5.tex Gen6.tex GenCom.c GenCom1.c GenCom2.c GenComExactSoln.c GenerateInput.c GenerateInputText.c GenRan.ran GetNodeNumbers.c Info100.dat Info20.dat Info200.dat Info30.dat Info300.dat Info40.dat Info400.dat Info50.dat Info500.dat Info60.dat Info70.dat Info80.dat Info90.dat InputCurrents.dat InputDendrite.dat JaySpikeTrain.c JayTest1.dat JayTest100.dat KenSpikeTrain.c KenTest1.dat * KenTest10.dat KenTest100.dat * KenTest10p.dat KenTest1p.dat * KenTest2.dat KenTest2p.dat KenTest3.dat KenTest3p.dat KenTest4.dat KenTest4p.dat KenTest5.dat KenTest5p.dat KenTest6.dat KenTest6p.dat KenTest7.dat KenTest7p.dat KenTest8.dat KenTest8p.dat KenTest9.dat KenTest9p.dat LU.c Mean50.dat Mean500.dat mosinit.hoc NC.pdf NC.tex NC1.tex NC2.tex NC3.tex NC4.tex NC5.tex NC6.tex NCFig2.eps * NCFig3.eps * NCFig4.eps * NCFig5a.eps * NCFig5b.eps * NCFig6.eps * NCPics.tex NeuronDriver.hoc NewComExactSoln.c NewComp.pdf NewComp.ps NewComp.tex NewComp.toc NewComp1.tex NewComp2.tex NewComp3.tex NewComp4.tex NewComp5.tex NewComp6.tex NewCompFig1.eps NewCompFig2.eps * NewCompFig3.eps * NewCompFig4.eps * NewCompFig5a.eps * NewCompFig5b.eps * NewCompFig6.eps * NewCompPics.tex NewComSpikeTrain.c NewRes.dat NewRes60.dat NewRes70.dat NewRes80.dat NewSynRes40.dat NewTestCell.d3 NResults.res OldComExactSoln.c out.res principles_01.tex rand Ratio.dat RelErr.dat ReviewOfSpines.pdf SpikeTimes.dat TestCell.d3 TestCell1.d3 TestCell2.d3 TestCell3.d3 TestCell4.d3 testcellnew2.hoc TestCGS.c TestGen1.c TestSim.hoc TestSim020.hoc TestSim030.hoc TestSim040.hoc TestSim050.hoc TestSim060.hoc TestSim070.hoc TestSim080.hoc TestSim090.hoc TestSim1.hoc TestSim100.hoc TestSim200.hoc TestSim300.hoc TestSim400.hoc TestSim500 TestSim500.hoc
\section{Simulation exercises}
Two different simulation exercises are used to compare the
performance of the new compartmental model with that of a
traditional compartmental model. The first simulation exercise
assesses the performance of each type of compartmental model with
respect to an analytical solution of the continuum model. In these
simulations the NEURON simulator (Hines and Carnevale,
\cite{Hines97}) is used to characterise the behaviour of a
traditional compartmental model. The second simulation exercise
compares the behaviour of a traditional compartmental model and
the new compartmental model with respect to the spike train
activity generated in each model by large scale synaptic activity.
In these simulations, a traditional compartmental model developed
by the authors is used. The behaviour of this model in the first
simulation exercise is indistinguishable from that of the NEURON
simulator. Finally, all solutions of the compartmental models are
advanced in one microsecond time steps to ensure that errors in
temporal integration make no significant contribution to the error
in the estimated membrane potential.

\subsection{First simulation exercise}\label{sim1}
The performance of the traditional compartmental model is first
compared against that of the new compartmental model by assessing
the accuracy with which both models determine the time course of
the somal potential of the model neuron (Figure \ref{TestNeuron})
when subjected to large scale exogenous point input. The
comparison relies on the fact that the model neuron has the
property that the effect at its soma of an exogenous input at a
given electrotonic distance from its soma is identical to the
effect at the soma of that input acting on a uniform dendritic
cylinder at the same electrotonic distance from the soma.

Each simulation distributes 75 point inputs at random over the
dendritic tree of the model neuron, where each input has strength
$2\times10^{-5}\,\mu$A. These inputs are then mapped to positions
on the Rall equivalent cylinder at the same electrotonic distance
from the soma (assumed to be a sphere of diameter $40\,\mu$m). The
time course of the potential at the soma of the equivalent
cylinder due to the combined effect of these inputs is determined
analytically and taken to be the reference potential with respect
to which error in both compartmental models is assessed. The
potential at the soma in response to this stimulus regimen is
obtained for the traditional and new compartmental models. The
difference between a computed potential and its exact value is
determined at one millisecond intervals in the first 10
milliseconds of the simulation, and each difference is divided by
the exact potential at that time to get a relative measure of
error at these times. The entire simulation procedure is now
repeated 2000 times for each of 13 different levels of spatial
discretisation (number of compartments).

\subsubsection{Results}
The results for the first simulation exercise of the traditional
and new compartmental models are set out in Table \ref{simex1}.
This table shows the common logarithms of the mean value of the
modulus of the relative error and the standard deviation of that
error estimated at ten milliseconds after the initiation of the
stimulus and based on 2000 simulations for each level of spatial
discretisation (number of compartments). Similar results, not
shown, hold at all time points at which the errors were estimated.

\pagebreak[4]

\begin{table}[!h]
$\begin{array}{c|cc|cc} \hline \mbox{\begin{tabular}{c} Compartments \\[-5pt] (log_{10}(\mbox{Compartments})) \end{tabular}} & \multicolumn{2}{|c|}{\mbox{\begin{tabular}{cc} NEURON & New Model \\[-5pt] \multicolumn{2}{c}{\log_{10}(Mean)} \end{tabular}}} & \multicolumn{2}{|c}{\mbox{\begin{tabular}{cc} NEURON & New Model \\[-5pt] \multicolumn{2}{c}{\log_{10}(Standard Dev.)} \end{tabular}}}\\ \hline \phantom{0}17 \quad (1.2305) & -2.41151 & -2.71945 & -2.62290 & -3.19338 \\ \phantom{0}21 \quad (1.3222) & -2.47233 & -2.77674 & -2.69851 & -3.24583 \\ \phantom{0}34 \quad (1.5314) & -2.94299 & -3.41196 & -3.06731 & -3.88820 \\ \phantom{0}41 \quad (1.6127) & -3.04729 & -3.62138 & -3.17081 & -4.14997 \\ \phantom{0}54 \quad (1.7323) & -3.21258 & -3.89150 & -3.34889 & -4.41251 \\ \phantom{0}61 \quad (1.7853) & -3.24692 & -3.91268 & -3.37653 & -4.45051 \\ \phantom{0}75 \quad (1.8750) & -3.35180 & -4.12056 & -3.46881 & -4.65463 \\ \phantom{0}82 \quad (1.9138) & -3.39846 & -4.23567 & -3.51591 & -4.76498 \\ \phantom{0}93 \quad (1.9684) & -3.45602 & -4.30636 & -3.57633 & -4.82045 \\ 193 \quad (2.2855) & -3.77417 & -4.94731 & -3.89829 & -5.47886 \\ 293 \quad (2.4668) & -3.94409 & -5.31876 & -4.07811 & -5.84771 \\ 390 \quad (2.5910) & -4.08234 & -5.57349 & -4.20025 & -6.10791 \\ 495 \quad (2.6946) & -4.15996 & -5.78252 & -4.28525 & -6.32790 \\ \hline \end{array}$
\centering
\parbox{5in}{\caption{\label{simex1} The results of 2000 simulations
for each of 13 different compartmental models based on the new and
traditional (NEURON) representations of a compartment. The common
logarithms of the mean value of the modulus of the relative error
and the standard deviation of that error are estimated at ten
milliseconds after the initiation of the stimulus.}}
\end{table}

The left hand panel of Figure \ref{mean} shows regression lines of
the common logarithms of the modulus of the mean relative error
for the traditional (dashed line) and new (solid line)
compartmental models on the logarithm of the number of
compartments used to represent the model neuron. These lines are
based on the data in Table \ref{simex1} and have equations
$$\label{mean1} \begin{array}{rcl} \log_{10}(\mbox{Mean Relative Error: NEURON}) & = & -1.09-1.17\log_{10}(\mbox{Compartments})\,, \$5pt] \log_{10}(\mbox{Mean Relative Error: New Model}) & = & -0.17-2.10\log_{10}(\mbox{Compartments}) \end{array}$$ in which the regressions are achieved with respective adjusted R^2 values of 97.4\% and 99.5\%. In view of their very high R^2 values, a number of conclusions can be drawn from these results. For a fixed number of compartments, the error in the new model is always less than that of the traditional model. The regression equations (\ref{mean1}) support the argument and subsequent assertion made in Section \ref{assertion} that the error in a traditional compartmental model is approximately O(1/n), whereas that in the new compartmental model is approximately O(1/n^2). In practice, this means that the accuracy achieved by a traditional compartmental model using 500/100 compartments is achieved in the new compartmental model by the use of approximately 100/40 compartments. \begin{figure}[!h] \centerline{\includegraphics[ ]{NewCompFig5a.eps} \quad\includegraphics[ ]{NewCompFig5b.eps}} \centering \parbox{5.8in}{\caption{\label{mean} The left panel shows the regression lines of the mean relative errors in the new compartmental model (solid line) and that of a traditional compartmental model (NEURON - dashed line) against number of compartments. All errors are measured ten milliseconds after initiation of the stimulus. The right panel shows the regression lines for the standard deviations of the mean relative errors for the new compartmental model (solid line) and for a traditional compartmental model (NEURON - dashed line).}} \end{figure} %\begin{figure}[!h] %\centerline{\begin{mfpic}[56][24]{0.4}{3}{-7.5}{1} %\headlen7pt %\pen{1pt} %\dotspace=4pt %\dotsize=1.5pt %% %% x-axis %\tlabel[br](3.0,0.9){\textsf {\log_{10}(\mbox{No. Compartments})}} %\lines{(1.0,0),(3.0,0)} %\lines{(1.5,0),(1.5,-0.2)} %\lines{(2.0,0),(2.0,-0.2)} %\lines{(2.5,0),(2.5,-0.2)} %\lines{(3.0,0),(3.0,-0.2)} %\tlabel[bc](1.0,0.3){\textsf{1.0}} %\tlabel[bc](1.5,0.3){\textsf{1.5}} %\tlabel[bc](2.0,0.3){\textsf{2.0}} %\tlabel[bc](2.5,0.3){\textsf{2.5}} %\tlabel[bc](3.0,0.3){\textsf{3.0}} %% y-axis %\tlabel[bc](0.5,-6){\rotatebox{90}{\textsf{\log_{10}(\mbox{Mean relative error})}}} %\lines{(1,0),(1,-7)} %\lines{(1.0,-1.0),(1.05,-1.0)} %\lines{(1.0,-2.0),(1.05,-2.0)} %\lines{(1.0,-3.0),(1.05,-3.0)} %\lines{(1.0,-4.0),(1.05,-4.0)} %\lines{(1.0,-5.0),(1.05,-5.0)} %\lines{(1.0,-6.0),(1.05,-6.0)} %\lines{(1.0,-7.0),(1.05,-7.0)} %\tlabel[cr](0.95,-0.0){\textsf{0.0}} %\tlabel[cr](0.95,-1.0){\textsf{-1.0}} %\tlabel[cr](0.95,-2.0){\textsf{-2.0}} %\tlabel[cr](0.95,-3.0){\textsf{-3.0}} %\tlabel[cr](0.95,-4.0){\textsf{-4.0}} %\tlabel[cr](0.95,-5.0){\textsf{-5.0}} %\tlabel[cr](0.95,-6.0){\textsf{-6.0}} %\tlabel[cr](0.95,-7.0){\textsf{-7.0}} %% %% Mean values at t=10 %\dashed\lines{(1.2,-2.494),(3.0,-4.60)} %\lines{(1.2,-2.686),(3.0,-6.466)} %\end{mfpic} %\begin{mfpic}[56][24]{0}{3}{-7.5}{1} %\headlen7pt %\pen{1pt} %\dotspace=4pt %\dotsize=1.5pt %% %% x-axis %\tlabel[br](3.0,0.9){\textsf{\log_{10}(\mbox{No, Compartments})}} %\lines{(1.0,0),(3.0,0)} %\lines{(1.5,0),(1.5,-0.2)} %\lines{(2.0,0),(2.0,-0.2)} %\lines{(2.5,0),(2.5,-0.2)} %\lines{(3.0,0),(3.0,-0.2)} %\tlabel[bc](1.0,0.3){\textsf{1.0}} %\tlabel[bc](1.5,0.3){\textsf{1.5}} %\tlabel[bc](2.0,0.3){\textsf{2.0}} %\tlabel[bc](2.5,0.3){\textsf{2.5}} %\tlabel[bc](3.0,0.3){\textsf{3.0}} %% y-axis %\tlabel[bc](0.5,-6){\rotatebox{90}{\textsf{\log_{10}(\mbox{Standard Dev.})}}} %\lines{(1,0),(1,-7)} %\lines{(1.0,-1.0),(1.05,-1.0)} %\lines{(1.0,-2.0),(1.05,-2.0)} %\lines{(1.0,-3.0),(1.05,-3.0)} %\lines{(1.0,-4.0),(1.05,-4.0)} %\lines{(1.0,-5.0),(1.05,-5.0)} %\lines{(1.0,-6.0),(1.05,-6.0)} %\lines{(1.0,-7.0),(1.05,-7.0)} %\tlabel[cr](0.95,-0.0){\textsf{0.0}} %\tlabel[cr](0.95,-1.0){\textsf{-1.0}} %\tlabel[cr](0.95,-2.0){\textsf{-2.0}} %\tlabel[cr](0.95,-3.0){\textsf{-3.0}} %\tlabel[cr](0.95,-4.0){\textsf{-4.0}} %\tlabel[cr](0.95,-5.0){\textsf{-5.0}} %\tlabel[cr](0.95,-6.0){\textsf{-6.0}} %\tlabel[cr](0.95,-7.0){\textsf{-7.0}} %% %% Standard deviations at t=10 %\dashed\lines{(1.2,-2.664),(3.0,-4.680)} %\lines{(1.2,-3.169),(3.0,-7.021)} %\end{mfpic}} %\centering %\parbox{5.8in}{\caption{\label{mean} The left panel shows the %regression lines of the mean relative errors in the new %compartmental model (solid line) and that of a traditional %compartmental model (NEURON - dashed line) against number of %compartments. All errors are measured ten milliseconds after %initiation of the stimulus. The right panel shows the regression %lines for the standard deviations of the mean relative errors for %the new compartmental model (solid line) and for a traditional %compartmental model (NEURON - dashed line).}} %\end{figure} \pagebreak[4] The standard deviation (SD) of the modulus of the relative error can be regarded as an indicator of the reliability of a single application of the model. The right hand panel of Figure \ref{mean} shows regression lines of the common logarithms of the standard deviation of the modulus of the relative error for the traditional (dashed line) and new (solid line) compartmental models on the logarithm of the number of compartments used to represent the model neuron. These lines are based on the data in Table \ref{simex1} and have equations $$\label{mean1} \begin{array}{rcl} \log_{10}(\mbox{SD of Relative Error: NEURON}) & = & -1.32-1.12\log_{10}(\mbox{Compartments})\,, \\[5pt] \log_{10}(\mbox{SD of Relative Error: New Model}) & = & -0.60-2.14\log_{10}(\mbox{Compartments}) \end{array}$$ in which the regressions are achieved with respective adjusted R^2 values of 98.7\% and 99.4\%. These lines show that the new compartmental model is more reliable than a traditional compartmental model. For example, a traditional compartmental model requires at least 100 compartments to give a standard deviation of the modulus of the relative error that is smaller than that of the new compartmental model using 40 compartments. \subsection{Second simulation exercise}\label{sim2} In the second simulation exercise 100 synapses are distributed at random over the dendritic tree of the model neuron illustrated in Figure \ref{TestNeuron}. Each synapse is activated independently of all other synapses, has a maximum conductance of 3\times10^{-5}\,\mbox{mS} and a time constant of 0.5 msec. Activation times for each synapse follow Poisson statistics with a mean rate of 30 pre-synaptic spikes per second. By contrast with the first simulation exercise in which the somal membrane was passive, the behaviour of the somal membrane in the second simulation exercise is active and obeys Hodgkin-Huxley kinetics. This simulation exercise is based on 12 different levels of spatial discretisation (number of compartments) in which each simulation of the traditional and new compartmental models use identical synaptic firing times and identical numbers of compartments. \subsubsection{Results} Table \ref{spikerate} gives the spike rate of soma-generated action potentials based on 11 seconds of activity, the first second of which is ignored. \begin{table}[!h] \[ \begin{array}{c|c|c} \hline \mbox{\begin{tabular}{c} Compartments \\[-5pt] (log_{10}(\mbox{Compartments})) \end{tabular}} & \mbox{\begin{tabular}{c} Traditional Model \\[-5pt] Mean Firing Rate \end{tabular}} &\mbox{\begin{tabular}{c} New Model \\[-5pt] Mean Firing Rate \end{tabular}}\\ \hline \phantom{0}34 \quad (1.5314) & 31.5 & 27.6 \\ \phantom{0}41 \quad (1.6127) & 30.3 & 27.9 \\ \phantom{0}54 \quad (1.7323) & 30.5 & 27.5 \\ \phantom{0}61 \quad (1.7853) & 29.8 & 27.2 \\ \phantom{0}75 \quad (1.8750) & 29.2 & 27.0 \\ \phantom{0}82 \quad (1.9138) & 28.5 & 27.0 \\ \phantom{0}93 \quad (1.9684) & 28.3 & 26.8 \\ 193 \quad (2.2855) & 26.5 & 26.5 \\ 293 \quad (2.4668) & 25.9 & 26.2 \\ 390 \quad (2.5910) & 26.2 & 26.2 \\ 495 \quad (2.6946) & 26.7 & 26.2 \\ 992 \quad (2.9965) & 26.0 & 26.1 \\ \hline \end{array}$
\centering
\parbox{5in}{\caption{\label{simex2} The results of the second
simulation exercise for a traditional compartmental model and the
new compartmental model in which 10 second records of spike train
activity are obtained for both models for 12 different levels of
spatial discretisation (number of compartments).}}
\end{table}

Figure \ref{spikerate} illustrates the data set out in Table
\ref{simex2} in which the spike rates for the traditional model
(dashed line) and new model (solid line) are plotted against the
common logarithm of the number of compartments used in each
simulation. As the number of compartments used in each model is
increased, the spike rates generated by both models approach a
common limit. However, the spike rate of the traditional model
approaches the limit in a monotonic fashion, and achieves the
limiting value with fewer compartments. For example, the spike
rate obtained using the traditional model with 100 compartments is
achieved with only 40 compartments in the new model. The spike
rate obtained using the traditional model with 500 compartments is
achieved in the new model with only 100 compartments. These
differences in the number of compartments required to achieve the
same level of accuracy in both models are identical to those
observed in the first simulation exercise.

\begin{figure}[!h]
\centering
\includegraphics[ ]{NewCompFig6.eps}
\vskip5pt
\parbox{5.5in}{\caption{\label{spikerate} The spike rate plotted against
the common logarithm of the number of compartments for a
traditional compartmental model (dashed line) and the new
compartmental model (solid line). The dotted line shows the
expected spike rate.}}
\end{figure}

%\begin{figure}[!h]
%\centerline{\begin{mfpic}[75][20]{0.3}{3}{-1}{8.5}
%\pen{1pt}
%\dotspace=4pt
%\dotsize=1.5pt
%%
%% x-axis
%\tlabel[tr](3.0,-1){\textsf {$\log_{10}(\mbox{No. Compartments})$}}
%\lines{(1.0,0),(3.0,0.0)}
%\lines{(1.0,0),(1.0,-0.3)}
%\lines{(1.5,0),(1.5,-0.3)}
%\lines{(2.0,0),(2.0,-0.3)}
%\lines{(2.5,0),(2.5,-0.3)}
%\lines{(3.0,0),(3.0,-0.3)}
%\tlabel[tc](1.0,-0.5){\textsf{1.0}}
%\tlabel[tc](1.5,-0.5){\textsf{1.5}}
%\tlabel[tc](2.0,-0.5){\textsf{2.0}}
%\tlabel[tc](2.5,-0.5){\textsf{2.5}}
%\tlabel[tc](3.0,-0.5){\textsf{3.0}}
%%
%% Expected spike rate
%\dotted\lines{(1.5,2.6),(3.2,2.6)}
%%
%% Traditional model (Modulo spike rate of 25)
%\dashed\lines{
%(1.531,8.0),(1.613,6.8),(1.732,7.0),(1.785,6.3),
%(1.875,5.7),(1.914,5.0),(1.968,4.8),(2.286,3.0),
%(2.467,2.4),(2.591,2.7),(2.696,3.2),(2.997,2.5)}
%%
%% New model (Modulo spike rate of 25)
%\lines{
%(1.531,4.1),(1.613,4.4),(1.732,4.0),(1.785,3.7),
%(1.875,3.6),(1.914,3.5),(1.968,3.3),(2.286,3.0),
%(2.467,2.7),(2.591,2.7),(2.695,2.7),(3.000,2.6)}
%% y-axis
%\lines{(1,0),(1,0.5)}
%\dashed\lines{(1,0.5),(1,2.0)}
%\lines{(1,2.0),(1,8.5)}
%\lines{(1.0,0.0),(0.95,0.0)}
%\lines{(1.0,2.5),(0.95,2.5)}
%\lines{(1.0,4.5),(0.95,4.5)}
%\lines{(1.0,6.5),(0.95,6.5)}
%\lines{(1.0,8.5),(0.95,8.5)}
%\tlabel[cr](0.9,0.0){\textsf{0.0}}
%\tlabel[cr](0.9,2.5){\textsf{26.0}}
%\tlabel[cr](0.9,4.5){\textsf{28.0}}
%\tlabel[cr](0.9,6.5){\textsf{30.0}}
%\tlabel[cr](0.9,8.5){\textsf{32.0}}
%\tlabel[tc](0.5,6.5){\rotatebox{90}{\textsf{Spikes per second}}}
%\end{mfpic}}
%\centering
%\vskip5pt
%\parbox{5.5in}{\caption{\label{spikerate} The spike rate plotted against
%the common logarithm of the number of compartments for a
%traditional compartmental model (dashed line) and the new
%compartmental model (solid line). The dotted line shows the
%expected spike rate.}}
%\end{figure}

\section{Concluding remarks}
This investigation has demonstrated that it is possible to achieve
a significant increase in the accuracy and precision of
compartmental models once the actual placement of input is
reflected in the structure of the compartmental model. This
finding is relevant to recent physiological studies that have
demonstrated the extreme accuracy of the timing of events in spike
trains (\emph{e.g.}, Fellous \emph{et al}., \cite{Fellous01}). To
investigate this phenomenon it is essential that the numerical
solution of the mathematical model accounting for this behaviour
is sufficiently accurate to allow one to distinguish between
different biophysical models. The simulation exercises demonstrate
that the new compartmental model is better able to achieve this
aim than a traditional compartmental model using an identical
number of compartments.

\section*{Acknowledgement}
A.E. Lindsay would like to thank the Wellcome Trust for the award
of Vacation Scholarship (VS/03/GLA/8/SL/TH/FH).


text:
# Solve the linear equations by considering y as a function of x, that is, y = y(x). displaystylefrac{{{left.{d}{y}right.}}}{{{left.{d}{x}right.}}}-{y}={4}{e}^{x},{y}{left({0}right)}={4}

Question
Solve the linear equations by considering y as a function of x, that is, y = y(x).
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}-{y}={4}{e}^{x},{y}{\left({0}\right)}={4}$$

2021-02-26
Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}={\left.{d}{x}\right.}$$
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
$$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=\int{\left.{d}{x}\right.}$$
which is$$\displaystyle \ln{{\left|{{y}}\right|}}={x}+{c}$$
By taking exponents, we obtain
$$\displaystyle{\left|{{y}}\right|}={e}^{{{x}+{c}}}={e}^{x}\cdot{e}^{c}$$
Hence,we obtain
$$\displaystyle{y}={C}{e}^{x}$$
where $$\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={e}^{x}$$ is the compleentary solution
Next, we need to find the particular solution $$y_p$$.
Therefore, we consider $$uy_c$$, and try to find u, a function of x, that will make this work.
Let's assume that $$uy_c$$. is a solution of the given equation. Hence, it satisfies the given equation. Substituting $$uy_c$$, and its derivative in the equation gives
$$\displaystyle{\left({u}{y}_{{c}}\right)}'-{u}{y}_{{c}}={4}{e}^{x}$$
$$\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}-{u}{y}_{{c}}={4}{e}^{x}$$
$$\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}-{y}_{{c}}\right)}}}_{{\text{=0 }\ }}={4}{e}^{x}$$
since y_c is a solution
Therefore
$$\displaystyle{u}'{y}_{{c}}={4}{e}^{x}\Rightarrow{u}'=\frac{{{4}{e}^{x}}}{{{y}_{{c}}}}$$
which gives
$$\displaystyle{u}=\int\frac{{{e}^{x}}}{{{y}_{{c}}}}{\left.{d}{x}\right.}$$
Now, we can find the function u :
$$\displaystyle{u}=\int\frac{{{4}{e}^{x}}}{{{e}^{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle={4}\int{\left.{d}{x}\right.}$$
$$=4x+c$$
Since we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,
$$u=4x$$
Recall that $$y_p = uy_c$$. Therefore,
$$\displaystyle{y}_{{p}}={4}{x}{e}^{x}$$
The general solution is
$$y=Cy_c+y_p$$
$$\displaystyle={e}^{x}{\left({C}+{4}{x}\right)}$$
Integrating Factor technique
This equation is linear with $$\displaystyle{P}{\left({x}\right)}=-{1}{\quad\text{and}\quad}{Q}{\left({x}\right)}={4}{e}^{x}$$
Hence,
$$\displaystyle{h}=\int{P}{\left.{d}{x}\right.}=-\int{\left.{d}{x}\right.}=-{x}$$
So, an integrating factor is
$$\displaystyle{e}^{h}={e}^{{-{x}}}$$
and the general solution is
$$\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle={e}^{x}{\left({c}+\int{4}{e}^{x}\cdot{e}^{{-{x}}}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle={e}^{x}{\left({c}+{4}\int{\left.{d}{x}\right.}\right)}$$
$$\displaystyle={e}^{x}{\left({c}+{4}{x}\right)}$$
Now, we can use the given initial condition. Substitute 0 for x and 4 for y to obtain the numeric value of c.
$$\displaystyle{4}={y}{\left({0}\right)}={e}^{0}{\left({c}+{4}\cdot{0}\right)}\Rightarrow{4}={c}$$
Hence, the solution is
$$\displaystyle{y}={e}^{x}{\left({c}+{4}{x}\right)}$$

### Relevant Questions

Solve the linear equations by considering y as a function of x, that is, y = y(x).
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+\frac{y}{{x}}=\frac{{ \cos{{x}}}}{{x}},{y}{\left(\frac{\pi}{{2}}\right)}=\frac{4}{\pi},{x}>{0}$$
Solve the linear equations by considering y as a function of x, that is, y = y(x).
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{1}{{x}}\right)}{y}={x}$$
Solve the linear equations by considering y as a function of x, that is, y = y(x).
\displaystyle{y}'+{y} \tan{{x}}= \sec{{x}},\)
\displaystyle{y}{\left(\pi\right)}={1}\)
Solve the linear equations by considering y as a function of x, that is
$$y=y(x)$$
$$\displaystyle{y}'-{2}{x}{y}={e}^{{{x}^{2}}}$$
Solve the linear equations by considering y as a function of x, that is, y = y(x).
$$y'+2y=4$$
Solve the linear equations by considering y as a function of x, that is,
$$y = y(x)$$
$$3xy'+y=12x$$
Solve the linear equations by considering y as a function of x, that is, $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}.{\left\lbrace{x}\right\rbrace}{\left\lbrace{y}\right\rbrace}'+{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}$$
Solve the linear equations by considering y as a function of x, that is, $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}.{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}}-{\left\lbrace{2}\right\rbrace}{\frac{{{x}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}{\left\lbrace{y}\right\rbrace}={\left\lbrace{x}\right\rbrace}^{{{2}}}$$
Solve the correct answer linear equations by considering y as a function of x, that is, $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}.{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}}+{\left\lbrace{y}\right\rbrace}={\cos{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}$$
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.


text:
# Different approach for checking if vectors are linearly (in)dependent

The usual method I see for checking if vectors are LD is to set them as column vectors and perform RREF. However, why can't one just set them as row vectors and perform RREF? If there are no rows of all 0s, then they are LI, otherwise they are LD.

• Of course. Why can't you do that? It's perfectly fine. – Teresa Lisbon Sep 26 '16 at 4:10
• Idk I guess I just never saw it anywhere so I thought there was something obvious I was missing. – sir_thursday Sep 26 '16 at 4:11
• It's not obvious, the above fact. Just follows from row rank = column rank. – Teresa Lisbon Sep 26 '16 at 4:13
• Actually sticking them in the rows will not only tell you if the rows are linearly independent, but will get you a (generally) simpler set of basis vectors for the space regardless of whether the vectors are linearly independent or not. So I will usually do it that way (in the rare circumstance where I do it by hand). – user137731 Sep 26 '16 at 4:13


text:
Math Help - Brownian motion estimate

1. Brownian motion estimate

Let $W_t$ be a one-dimensional Brownian motion. How do you prove that one can find constants $c, \lambda >0$ such that:
$\mathbb{P}\Big(\sup_{1\leq s\leq 2^t}\frac{|W_s|}{\sqrt{s}}\leq 1\Big)\leq ce^{-\lambda t},\quad t\geq 1.$
One usual approach to prove such inequalities, using stopping times along the lines of the proof of the Maximal Theorem, doesn't seem to quite work here.

By advance, thank you very much for your help.

2. a trivial solution is available, since $P(anything) \leq 1$. Values of $c , \lambda$ can be chosen from that.

3. Well, since we need lambda to be > 0, e.g. the bound depends explicitly on time and goes to zero at infinity, this won't quite work here...