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1.5: Induction


You are familiar, no doubt, with proofs by induction. It is our goal in this section to discuss the proofs by induction that you know so well, put them in a different light, and then generalize that notion of induction to a setting that will allow us to use induction to prove things about terms and formulas rather than just the natural numbers.

Just to remind you of the general form of a proof by induction on the natural numbers, let us state and prove a familiar theorem, assuming for the moment that the set of natural numbers is ({1, 2, 3, ldots }).

Theorem 1.4.1. For every natural number (n),

[1 + 2 + cdots + n = frac{n left( n + 1 ight)}{2}.]

Proof. If (n = 1), simple computation shows that the equality holds. For the inductive case, fix (k geq 1) and assume that

[1 + 2 + cdots + k = frac{k left( k + 1 ight)}{2}.]

If we add (k + 1) to both sides of this equation, we get

[1 + 2 + cdots + k + left( k + 1 ight) = frac{k left( k + 1 ight)}{2} + left( k + 1 ight),]

and simplifying the right-hand side of this equation shows that

[1 + 2 + cdots + left( k + 1 ight) = frac{left( k + 1 ight) left( left( k + 1 ight) + 1 ight)}{2},]

finishing the inductive step, and the proof.

As you look at the proof of this theorem, you notice that there is a base case, when (n = 1), and an inductive case. In the inductive step of the proof, we prove the implication

If the formula holds for (k), then the formula holds for (k + 1).

We prove this implication by assuming the antecedent, that the theorem holds for a (fixed, but unknown) number (k), and from that assumption proving the consequent, that the theorem hods for the next number, (k + 1). Notice that this is not the same as assuming the theorem that we are trying to prove. The theorem is a universal statement - it claims that a certain formula holds for every natural number.

Looking at this from a slightly different angle, what we have done is to construct a set of numbers with a certain property. If we let (S) stand for the set of numbers for which our theorem holds, in our proof by induction we show the following facts about S:

  1. The number 1 is an element of (S). We prove this explicitly in the base case of the proof.
  2. If the number (k) is an element of (S), then the number (k + 1) is an element of (S). This is the content of the inductive step of the proof.

But now, notice that we know that the collection of natural numbers can be defined as the smallest set such that:

  1. The number 1 is a natural number.
  2. If (k) is a natural number, then (k + 1) is a natural number.

So (S), the collection of numbers for which the theorem holds, is identical with the set of natural numbers, thus the theorem holds for every natural number (n), as needed. (If you caught the slight lie here, just substitute "superset" where appropriate.)

So what makes a proof by induction work is the fact that the natural numbers can be defined recursively. There is a base case, consisting of the smallest natural number ("1 is a natural number"), and there is a recursive case, showing how to construct bigger natural numbers from smaller ones ("If (k) is a natural number, then (k + 1) is a natural number").

Now, let us look at Definition 1.3.3, the definition of a formula. Notice that the five clauses of the definition can be separated into two groups. The first two clauses, the atomic formulas, are explicitly defined: For example, the first case says that anything that is of the form (= t_1 t_2) is a formula if (t_1) and (t_2) are terms. These first two clauses form the base case of the definition. The last three clauses are the recursive case, showing how if (alpha) and (eta) are formulas, they can be used to build more complex formulas, such as (left( alpha lor eta ight)) or (left( forall v ight) left( alpha ight)).

Now since the collection of formulas is defined recursively, we can use an inductive-style proof when we want to prove that something is true about every formula. The inductive proof will consist of two parts, a base case and an inductive case. In the base case of the proof we will verify that the theorem is true about every atomic formula - about every string that is known to be a formula from the base case of the definition. In the inductive step of the proof, we assume that the theorem is true about simple formulas ((alpha) and (eta)), and use that assumption to prove that the theorem holds a more complicated formula (phi) that is generated by a recursive clause of the definition. This method of proof is called induction on the complexity of the formula, or induction on the structure of the formula.

There are (at least) two ways to think about the word "simple" in the last paragraph. One way in which a formula (alpha) might be simpler than a complicated formula (phi) is if (alpha) is a subformula of (phi). The following theorem, although mildly interesting in its own right, is included here mostly so that you can see an example of a proof by induction in this setting:

Theorem 1.4.2. Suppose that (phi) is a formula in the language (mathcal{L}). Then the number of left parentheses occurring in (phi) is equal to the number of right parentheses occurring in (phi).

Proof. We will present this proof in a fair bit of detail, in order to emphasize the proof technique. As you become accustomed to proving theorems by induction on complexity, not so much detail is needed.'

Base Case. We begin our inductive proof with the base case, as you would expect. Our theorem makes an assertion about all formulas, and the simplest formulas are the atomic formulas. They constitute our base case. Suppose that (phi) is an atomic formula. There are two varieties of atomic formulas: Either (phi) begins with an equals sign followed by two terms, or (phi) begins with a relation symbol followed by several terms. As there are no parentheses in any term (we are using the official definition of term, here), there are no parentheses in (phi). Thus, there are as many left parentheses as right parentheses in (phi), and we have established the theorem if (phi) is an atomic formula.

Inductive Case. The inductive step of a proof by induction on complexity of a formula takes the following form: Assume that (phi) is a formula by virtue of clause (3), (4), or (5) of Definition 1.3.3. Also assume that the statement of the theorem is true when applied to the formulas (alpha) and (eta). With those assumptions we will prove that the statement of the theorem is true when applied to the formula (phi). Thus, as every formula is a formula either by virtue of being an atomic formula or by application of clause (3), (4), or (5) of the definition, we will have shown that the statement of the theorem is true when applied to any formula, which has been our goal.

So, assume that (alpha) and (eta) are formulas that contain equal numbers of left and right parentheses. Suppose that there are (k) left parentheses and (k) right parentheses in (alpha) and (l) left parentheses and (l) right parentheses in (eta).

If (phi) is a formula by virtue of clause (3) of the definition, then (phi : equiv left( eg alpha ight)). We observe that there are (k + 1) left parentheses and (k + 1) right parentheses in (phi), and thus (phi) has an equal number of left and right parentheses, as needed.

If (phi) is a formula because of clause (4), then (phi : equiv left( alpha lor eta ight)), and (phi) contains (k + l + 1) left and right parentheses, an equal number of each type.

Finally, if (phi : equiv left( forall v ight) left( alpha ight)), then (phi) contains (k + 2) left parentheses and (k + 2) right parentheses, as needed.

This concludes the possibilities for the inductive case of the proof, so we have established that in every formula, the number of left parentheses is equal to the number of right parentheses.

A second way in which we might structure a proof by induction on the structure of the formula is to say that (alpha) is simpler than (phi) if the number of connectives/quantifiers in (alpha) is less than the number in (phi). In this case one could argue that the induction argument is really an ordinary induction on the natural numbers. Here is an outline of how such a proof might proceed:

Proof. We argue by induction on the structure of (phi).

Base Case. Assume (phi) has 0 connectives/quantifiers. This means that (phi) is an atomic formula. {Insert argument establishing the theorem for atomic formulas.}

Inductive Case. Assume that (phi) has (k + 1) connectives/quantifiers. Then either (phi : equiv eg alpha), or (phi : equiv alpha lor eta) or (phi : equiv left( forall x ight) alpha), and we can assume that the theorem holds for every formula that has (k) or fewer connectives/quantifiers. We now argue that the theorem holds for the formula (phi). {Insert argument for the three inductive cases.}

Between the base case and the inductive case we have established that the theorem holds for (phi) no matter how many connectives/quantifiers the formula (phi) contains, so by induction on the structure of (phi), we have established that the theorem holds for all formulas (phi).

This might be a bit confusing on first glance, but the power of this proof technique will become very evident as you work through the following exercises and when we discuss the semantics of our language.

Notice also that the definition of a term (Definition 1.3.1) is also a recursive definition, so we can use induction on the complexity of a term to prove that a theorem holds for every term.

Exercises

  1. Prove, by ordinary induction on the natural numbers, that
    [1^2 + 2^2 + cdots + n^2 = frac{n left( n + 1 ight) left( 2n + 1 ight)}{6}.]
  2. Prove, by induction, that the sum of the interior angles in a convex (n)-gon is (left( n - 2 ight) 180^ ext{o}). (A convex (n)-gon is a polygon with (n) sides, where the interior angles are all less than (180^ ext{o}).)
  3. Prove by induction that if (A) is a set consisting of (n) elements, then (A) has (2^n) subsets.
  4. Suppose that (mathcal{L}) is ({0, f, g}), where 0 is a constant symbol, (f) is a binary function symbol, and (g) is a 4-ary function symbol. Use induction on complexity to show that every (mathcal{L})-term has an odd number of symbols.
  5. If (mathcal{L}) is ({0, <}), where 0 is a constant symbol and (<) is a binary relation symbol, show that the number of symbols in any formula is divisible by 3.
  6. If (s) and (t) are strings, we say that (s) is an initial segment of (t) if there is a nonempty string (u) such that (t : equiv su), where (su) is the string (s) followed by the string (u). For example, (KUMQ) is an initial segment of (KUMQUAT) and (+24) is an initial segment of (+24 u - v). Prove, by induction on the complexity of (s), that if (s) and (t) are terms, then (s) is not an initial segment of (t). [Suggestion: The base case, when (s) is either a variable or a constant symbol, should be easy. Then suppose that (s) is an initial segment of (t) and (s : equiv f t_1 t_2 ldots t_n), where you know that each (t_i) is not an initial segment of any other term. Look for a contradiction.]
  7. A language is said to satisfy unique readability for terms if, for each term (t), (t) is in exactly one of the following categories:
    (a) Variable
    (b) Constant symbol
    (c) Complex term
    and furthermore, if (t) is a complex term, then there is a unique function symbol (f) and a unique sequence of terms (t_1, t_2, ldots, t_n) such that (t : equiv f t_1 t_2 ldots t_n). Prove that our languages satisfy unique readability for terms. [Suggestion: You mostly have to worry about uniqueness - for example, suppose that (t) is (c), a constant symbol. How do you know that (t) is not also a complex term? Suppose that (t) is (f t_1 t_2 ldots t_n). How do you show that the (f) and the (t_i)'s are unique? You may find Exercise 6 useful.]
  8. To say that a language satisfies unique readability for formulas is to say that every formula (phi) is in exactly one of the following categories:
    (a) Equality (if (phi : equiv = t_1 t_2))
    (b) Other atomic (if (phi : equiv R t_1 t_2 ldots t_n) for an (n)-ary relation symbol (R))
    (c) Negation
    (d) Disjunction
    (e) Quantified
    Also, it must be that if (phi) is both (= t_1 t_2) and (= t_3 t_4), then (t_1) is identical to (t_3) and (t_2) is identical to (t_4), and similarly for other atomic formulas. Furthermore, if (for example) (phi) is a negation (left( eg alpha ight)), then it must be the case that there is not another formula (eta) such that (phi) is also (left( eg eta ight)), and similarly for disjunctions and quantified formulas. You will want to look at, and use, Exercise 7. You may have to prove an analog of Exercise 6, in which it may be helpful to think about the parentheses in an initial segment of a formula, in order to prove that no formula is an initial segment of another formula.
  9. Take the proof of Theorem 1.4.2 and write it out in the way that you would present it as part of a homework assignment. Thus, you should cut out all of the inessential motivation and present only what is needed to make the proof work.

HINT: Let $f(n)=4^+5^<2n-1>$ you want to show that $f(n)$ is divisible by $21$ for all $nge 1$.

Presumably the base case, proving that $f(1)$ is divisible by $21$, causes no trouble. For your induction step you want to show that if $f(n)$ is divisible by $21$ for some $nge 1$, then $f(n+1)$ is divisible by $21$. Note that this will be true if and only if $f(n+1)-f(n)$ is divisible by $21$, so you ought to look at

Now see if you can rearrange $(1)$ to get something that is demonstrably a multiple of $21$.

Ostensibly, it can be easily established using congruence.

But as induction is required,it can be done as follows:

So using induction,we can show that $21mid f(n)$ if $nge 1$

So if $(a^2-b)mid(a+b^2), (a^2-b)mid f(n)$ if $nge 1$

Here $a=5,b=4implies a^2-b=25-4=21$ and $a+b^2=5+4^2=21$

$5^ <2n-1>= 5 cdot 25^$. Hence, $5^ <2n-1>equiv 5 cdot (21+4)^ pmod <21>equiv 5 cdot 4^ pmod<21>$

Hence, $4^ + 5^ <2n-1>equiv 4^ + 5 cdot 4^ pmod <21>equiv 21 cdot 4^ pmod <21>equiv 0 pmod<21>$

For a proof by induction, note that if $f(n) = 4^ + 5^<2n-1>$, then $f(n+1) - 4f(n) = 4^ + 5^ <2n+1>- 4^ - 4 cdot 5^ <2n-1>= 21 cdot 5^<2n-1>$

Hence, $f(n+1) = 4f(n) + 21 cdot 5^<2n-1>$

Hint $ $ Times $5$ it's $ m, 25^n!+20cdot 4^nequiv, 4^n!- 4^nequiv, 0 !pmod<21> $ (easily proved by induction if need be)

Remark $ $ More generally $ m: a^3equiv -1:Rightarrow: (a^2)^!+a^<2n-1>! =, a^<2n-1>(a^3+1)equiv 0$

Another way of doing this is to reverse engineer a recurrence relation of the form $f(n+1)=pf(n)+qf(n-1)$ - not so different from what Marvis has done, but illustrating some different techniques and possibilities.

Such a recurrence is linear, so given solutions $f(n)$ and $g(n)$, $af(n)+bg(n)$ is also a solution, and that gives enough scope to fit two initial conditions (eg the values of f(0) and f(1)).

This kind of recurrence has solutions of the form $f(n)=alpha^n$, which can be used as a basis for the solution space, and substituting in to the recurrence we find that $alpha^2-palpha-q = 0$. Since the solution is determined by $f(0)$ and $f(1)$ it is easy to prove that the two solutions to the quadratic form a basis. The only complication is if the quadratic has equal roots, which doesn't apply in this case.

We are given a function which combines multiples of $4^n$ and $25^n$ so we want the solutions of the quadratic to be 4 and 25. Hence $p=29, q=-100$.

More importantly, provided $p$ and $q$ and $f(n)$ are integers, any integer which is a divisor of $f(n-1)$ and $f(n)$ will also be a divisor of $f(n+1)$.

The recurrence can therefore be used to give a simple inductive step. NB this method might be disallowed in some contexts as insufficiently elementary.


So you need to prove $ sum_^n (5k-4) = n(5n-3)/2 $ by induction.

You start by checking that the formula works for some $n$ ($n=1$ for example) $ n=1 Rightarrow sum_^1 (5k-4) = (5-4) = 1 = (5-3)/2 $ which is true.

Then your goal is to prove that the formula holding for $n$ implies that it holds for $n+1$. Start by writing the $n+1$ case using the $n$ case $ sum_^ (5k-4) = sum_^ (5k-4) + 5(n+1) - 4 $ and substitute the formula for the $n$ case (known to hold for some $n$) $ sum_^ (5k-4) = n(5n-3)/2 + 5(n+1) - 4 .$

The proof is complete if the last form can be shown to be equal to $(n+1)(5(n+1)-3)/2$. (Because you have shown that it holds for $n=1$ and that if it holds for some $n$ it holds also for $n+1$, therefore it holds for all $n geq 1$.)

First, show that this is true for $n=1$:

Second, assume that this is true for $n$:

Third, prove that this is true for $n+1$:

$left(sumlimits_^5i-4 ight)+5(n+1)-4=dfrac<2>+5(n+1)-4$ assumption used here

You have the induction assumption $1+2+ldots+(5(n-1)-4)=frac<(n-1)(5(n-1)-3)><2>$ Add $5n-4$ to that. And you should get $P(n-1)Rightarrow P(n)$ and this completes the induction

After some algebra. $(5k^2+7k+2)/2==(5k^2+7k+2)/2$ which is true.

A Proof Without Using Induction:

I will outline a simplified version that is more drawn out but probably answers your question more clearly. Your goal is to prove that the statement $P(n)$, that is, $ P(n) : 1+6+11+cdots+(5n-4)=frac <2>$ holds for all $ngeq 1$.

Base step: As you noted, check the $n=1$ case for the base step. Using $n=1$, we have that $ 1=frac<1(5(1)-3)><2>, $ which is correct. Thus, the base step checks out.

Inductive step: Fix $kgeq 1$ and assume that $ P(k) : 1+6+11+cdots+(5k-4)=frac <2>$ holds. It remains to show that $ P(k+1) : 1+6+11+cdots+(5k-4)+[5(k+1)-4]=frac<(k+1)[5(k+1)-3]> <2>$ follows. Starting with the left-hand side of $P(k+1)$, egin 1+cdots+(5k-4)+[5(k+1)-4] &leq frac<2>+[5(k+1)-4] ag[1em] &=frac<2> ag[1em] &=frac<5k^2-3k+10k+10-8><2> ag[1em] &=frac<5k^2+7k+2><2> ag[1em] &=frac<(k+1)(5k+2)><2> ag[1em] &=frac<(k+1)[5(k+1)-3]><2> ag end one arrives at the right-hand side of $P(k+1)$, thus completing the inductive step.


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How an induction cooktop works

Induction cooktops use magnetic induction technology to heat the cookware. An alternating current is made to pass through a copper element embedded underneath the cooktop. It creates a magnetic field. This magnetic field permeates the base of ferromagnetic cookware. It results in the generation of a resistive electric current inside the cookware.

The resistive electric current is also known as eddy currents. These currents generate heat and heat produced in this way cooks the food.

It is the other way round here. The source of heat is the cookware and not the cooktop. There is a certain condition under which the induction process works. The cookware should be made of a material which is ferromagnetic and magnetically conductive. Cast iron, iron, enameled or glazed iron, magnetic stainless steels materials work on induction.

Copper, aluminum, glass, and non-magnetic stainless steel will not work unless a converter plate is kept underneath between the cooktop and the cookware. Copper and aluminum clad with iron or stainless steel can be used. A method to check the compatibility of cookware on the induction is to place a magnet and if it sticks to the base of the cookware, it will work very well on induction.

Induction cooktops are favored by millions due to its excellent advantages including efficiency, less wastage, performance, safety, quick-cooking and easy to clean and maintain.

With all these advantages, many consumers have experienced noises while cooking on an induction cooktop. Let us find out details about these noises.


Docetaxel Based Chemotherapy Plus or Minus Induction Chemotherapy to Decrease Events in Head and Neck Cancer (DeCIDE) (DeCIDE)


Condition or disease Intervention/treatment Phase
Cancer of the Pharynx Cancer of the Larynx Cancer of the Nasal Cavity Paranasal Sinus Neoplasms Cancer of the Oral Cavity Drug: docetaxel Drug: cisplatin Drug: hydroxyurea Drug: fluorouracil Procedure: chemotherapy Procedure: radiotherapy Phase 3

Phase III trial of induction therapy with docetaxel followed by chemoradiotherapy versus chemoradiotherapy alone in patients with nodal stage N2 or N3 head and neck cancer

  • To determine the effect on overall survival when induction chemotherapy is administered prior to chemoradiotherapy in patients with N2 or N3 disease.
  • To determine the effect of induction chemotherapy when administered prior to chemoradiotherapy on distant failure-free survival, failure pattern, progression free survival and quality of life.

Arm A - Induction + chemoradiotherapy

Arm B - Chemoradiotherapy alone

  • Induction therapy: Two 21-day cycles of chemotherapy consisting of docetaxel (day 1), cisplatin (day 1), and 5-fluorouracil (days 1-5). Total duration of 6 weeks.
  • Chemoradiotherapy: Five 14-day cycles of docetaxel (day 1), 5-fluorouracil (day 0-4), and hydroxyurea (days 0-4) with twice daily radiation (days 1-5). Total duration of 10 weeks.
  • All patients will undergo surgical evaluation after chemoradiation for possible neck dissection.
  • Upon completion of treatment, patients will be monitored every three months during the first year, every six months during the second and third years, and annually thereafter, up to seven years.
  • Patients will be followed for Quality of Life (QOL) during the course of treatment, as well as annually thereafter, up to five years.

Layout table for study information
Study Type : Interventional (Clinical Trial)
Actual Enrollment : 285 participants
Allocation: Randomized
Intervention Model: Parallel Assignment
Masking: None (Open Label)
Primary Purpose: Treatment
Official Title: A Phase III Randomized Trial of Docetaxel Based Induction Chemotherapy in Patients With N2/N3 Locally Advanced Head and Neck Cancer
Study Start Date : November 2004
Actual Primary Completion Date : December 2016
Actual Study Completion Date : December 2016

Resource links provided by the National Library of Medicine

Induction therapy: Two 21-day cycles of chemotherapy consisting of docetaxel (75 mg/m2, day 1), cisplatin (75 mg/m2, day 1), and 5-fluorouracil (750 mg/m2/day, days 1-5). Total duration of 6 weeks.

Chemoradiotherapy: Five 14-day cycles of docetaxel (25 mg/m2, day 1), 5-fluorouracil (600 mg/m2/day, day 0-4), and hydroxyurea (500 mg PO q 12 hours x 6 days (11 doses)) with twice daily radiation (150 cGy given bid, days 1-5). Total duration of 10 weeks.

    Distant Failure-free Survival (DFFS): Time From Randomization to Distant Recurrence or Death From Any Cause [ Time Frame: Up to 6 years ]
Information from the National Library of Medicine

Choosing to participate in a study is an important personal decision. Talk with your doctor and family members or friends about deciding to join a study. To learn more about this study, you or your doctor may contact the study research staff using the contacts provided below. For general information, Learn About Clinical Studies.

Layout table for eligibility information
Ages Eligible for Study: 18 Years and older (Adult, Older Adult)
Sexes Eligible for Study: All
Accepts Healthy Volunteers: No
  • Age 18 years or older
  • Histologically or cytologically confirmed diagnosis of squamous cell or poorly differentiated carcinomas of the head and neck (excluding lip), or lymphoepithelioma
  • No prior chemotherapy or radiotherapy
  • Prior surgical therapy will consist only of incisional or excisional biopsy, and organ sparing procedures such as debulking of airway-compromising tumors or neck dissection in a patient with an existing primary tumor
  • Karnofsky performance status of >= 70%
  • Intact organ and bone marrow function
  • Obtained informed consent
  • Demonstration of metastatic disease (i.e. M1 disease).
  • Patients with a history of severe allergic reaction to docetaxel or other drugs formulated with polysorbate 80. History of allergic reactions attributed to compounds of similar chemical or biologic composition to cisplatin, 5-fluorouracil, or hydroxyurea
  • Other coexisting malignancies or malignancies diagnosed within the previous 3 years with the exception of basal cell carcinoma, cervical cancer in situ, and other treated malignancies with no evidence of disease for at least 3 years.
  • Prior surgical therapy other than incisional or excisional biopsy and organ-sparing procedures such as debulking of airway-compromising tumors or neck dissection in a patient with an unknown primary tumor. Any non-biopsy procedure must have taken place less than 3 months from initiating protocol treatment.
  • Incomplete healing from previous surgery
  • Pregnancy or breast feeding (men and women of child-bearing potential are eligible but must consent to using effective contraception during therapy and for at least 3 months after completing therapy)
  • Uncontrolled intercurrent illness including, but not limited to, ongoing or active infection, symptomatic congestive heart failure (CHF), unstable angina pectoris, cardiac arrhythmia, or psychiatric illness/social situations that would limit compliance with study requirements
  • Patients with clinically significant pulmonary dysfunction, cardiomyopathy, or any history of clinically significant CHF are excluded. The exclusion of patients with active coronary artery disease will be at the discretion of the attending physician.
  • Uncontrolled active infection unless curable with treatment of their cancer.
Information from the National Library of Medicine

To learn more about this study, you or your doctor may contact the study research staff using the contact information provided by the sponsor.

Please refer to this study by its ClinicalTrials.gov identifier (NCT number): NCT00117572

Show 26 study locations Hide 26 study locations

How is Health and Safety Induction Training Done?

A health and safety induction training is provided for new employees. Safety induction trainings can help new employees become familiar with their work activities, responsibilities, their colleagues or teammates, and operating policies, procedures, and rules. The topics covered in a health and safety induction training will vary depending on the employee’s role.

Below are some of the common safety induction topics that can be covered during a safety induction training session:

  • Hazards and risks in your workplace
  • Special equipment, such as personal protective equipment (PPE), which may require additional training
  • Safe work practices
  • Work health and safety legislation
  • Emergency procedures
  • First aid and other emergency contacts.

1.5: Induction

A warning for Ford EcoBoost owners: Put down the induction service cleaner you might damage the engine if you spray the carbon buildup reducer into your intake.

A Ford service technician posted two videos earlier this year addressing the issue. Essentially he discovered that aging EcoBoost engines are experiencing misfires caused by carbon buildup on the intake valves. When he called Ford for a fix, he was told that Ford is recommending changing the cylinder heads because forced-induction service (sometimes accomplished by spraying a special cleaning fluid into the intake) will ruin the turbos.

Mike Levine, Ford's truck communications manager, concurs. He said a forced-induction service is "not part of the official maintenance guidelines for EcoBoost engines."

How Does Carbon Buildup Happen?

Over time, direct-injection engines like the EcoBoost can run extraordinarily rich at startup, especially in cold weather. Combine this richness with a truck making a short commute — meaning the engine doesn't reach its optimal temperature — and carbon deposits begin to attach themselves to the cylinders, including valves and heads. Eventually these carbon buildups rob the engine of performance and fuel economy.

For naturally aspirated engines, carbon deposits can be prevented by following the manufacturer's recommended service interval for air filters and spark plugs, and doing periodic injector clearings.

If these don't prevent the carbon buildup, or if you really want to clean your engine, a forced-induction service is normally the way to go — unless you have a Ford EcoBoost engine.

What Is Forced-Induction Service?

Forced-induction service involves spraying chemicals into the engine to clean it out. There are pros and cons to using these chemicals. Benefits may include a smoother idle, better fuel economy and more power (although these improvements could also be due to regular maintenance of fuel filters and spark plugs). The downside to using them includes harming the engine through improper use, and the facts that amateurs should not use them and automakers don't recommend them. If used incorrectly, these chemicals cause problems like catalytic converter failure, carbon dioxide sensor failure, damaged actuators inside the air-intake manifold and other issues.

Other solutions used by dealers and mechanics include things such as TerraClean. This is a decarbonizing machine that connects to the fuel system and sprays a chemical mixture that decarbonizes the engine. It is used in many automotive shops.

Why Don't They Work for Ford's EcoBoost Engine?

We talked with an automotive expert who confirmed what the video explains: Forced-induction chemicals cause a reaction that increases the temperature around the turbochargers leading to premature failure. A single turbo can cost thousands to replace.

The reality is that while the new EcoBoost is an impressive engine, carbon buildup will happen and for now, the only fix seems to be replacing the cylinder heads.


Introduction

Kenaf, Hibiscus cannabinus, is an annual plant belonging to Malvaceae, which includes cotton and okra. Kenaf grows quickly and has a fiber-enriched woody base thus, it is regarded as an important fiber crop. The properties of kenaf enable it to be applied to biocomposite materials as a filler, especially in automotive parts (Holbery and Houston 2006). Conventional breeding can improve the property of kenaf, however, establishment of complete genetic transformation procedure is essential to dramatically improve the physical properties of kenaf as a structural material in short term. Among the several ways of genetic transformation established in plants, agrobacterium-mediated transformation is the most popular, and, in general, it utilizes the regeneration of plants from the callus that can be induced by the combination of the plant growth regulators auxins and cytokinins. There are several reports on regeneration of shoots from kenaf using various explants (Herath et al. 2004 McLean et al. 1992 Samanthi et al. 2013 Srivatanakul et al. 2000 Zapata et al. 1999). Using the kenaf shoot apex or shoot, shoots are efficiently regenerated without the callus state or adding plant growth regulators, but the addition of N 6 -benzyladenine (BA) and thidiazuron, cytokinin-like plant growth regulators, can increase the number of regenerating shoots (Herath et al. 2004 Srivatanakul et al. 2000 Zapata et al. 1999). McLean et al. (1992) showed efficient induction of the callus from kenaf internodal stems by supplementing 1-naphthaleneacetic acid (NAA)/BA or 2,4-dichlorophenoxyacetic acid (2,4-D)/kinetin as the auxin/cytokinin combination, but the regeneration efficiency was lower than 15%. Samanthi et al. (2013) showed the induction of the kenaf callus from an intact cotyledon with a maximum of 80% efficiency by adding indole-3-butyric acid (IBA) as the auxin and BA, and the callus regenerated shoots with 68.7% efficiency. Since the callus induction and shoot regeneration has not been achieved 100% efficiency yet, there are still possibilities to improve the regeneration efficiency. Moreover, information regarding the flowering and seed setting is lacking in kenaf. Here, we report the complete regeneration of kenaf including improved callus induction and shoot regeneration, flowering, and seed setting methods.


All off these machines may be custom configured for your specific use. For example: This machine has a separate transformer unit to isolate excess heat from the electronics.

SP-25ABD an Industrial 100% duty cycle 25KW Machine with built in timer

Specifications:

★ : 440V 3-phase, 50 or 60 HZ
★ Max input current : 33A,
★ Duty cycle : 100%
★ Output Oscillation Frequency : 30-80KHZ
★ Solid State Power
★ Max input power : 25KW,
★ Minimum water cooling required: 22,000btu, 3 gal/min (0.4Mpa, 10 L/min)
★ Cable length: 6.5 ft / 2 m
★ Dimensions: Main unit: 24 x 12 x 10.5" (60 X 26 X 50cm) - 62lb / 28kg
Transformer: 18.5 x 10.5 x 14" (47 X 27 X 35cm) - 57lb / 26kg


Watch the video: ΦΥΣΙΚΗ Γ ΛΥΚΕΙΟΥ, Φαινόμενο της Επαγωγής ΝΕΑ ΥΛΗ (November 2021).