The New Phi Function and some of it's Applications

In Mathematics, we see a large number of functions, each having their own properties. Some of these are very interesting and contribute greatly to the intensive research in the ﬁeld of Mathematics. This paper deals with one such function (which we have termed as the phi function) which emerges from a chain of inequalities, established from the basic concepts of diﬀerential calculus. This paper establishes several inequalities which relate functions and their integrals. There is another important expression (from the point of view of notations),which links a class of divergent inﬁnite series to the phi function. Finally we will dive into a brief overview of the phi − form of plane trigonometric functions and derive the trigonometric identity sin 2 ( θ ) + cos 2 ( θ ) = 1, thus marking their importance. Throughout the paper, we will be analyzing functions in R + such that the functions are always greater than 0. We will also consider that the functions are continuous and diﬀerentiable in the intervals under consideration.


Abstract
In Mathematics, we see a large number of functions, each having their own properties. Some of these are very interesting and contribute greatly to the intensive research in the field of Mathematics. This paper deals with one such function (which we have termed as the phi function) which emerges from a chain of inequalities, established from the basic concepts of differential calculus. This paper establishes several inequalities which relate functions and their integrals. There is another important expression (from the point of view of notations),which links a class of divergent infinite series to the phi function. Finally we will dive into a brief overview of the phi − f orm of plane trigonometric functions and derive the trigonometric identity sin 2 (θ) + cos 2 (θ) = 1, thus marking their importance. Throughout the paper, we will be analyzing functions in R + such that the functions are always greater than 0. We will also consider that the functions are continuous and differentiable in the intervals under consideration.

Background
Proof. By the product rule, ).
Proof. By the quotient rule, d dx ( f ′ i (x) (f i (x)) n m=1 fm(x) . The last step was a change in the index from k to m.

Let us examine the function
[applying P roposition 2.1].
Similar to 3.1.1, let's define another function φ 2 (x) whose value is equal to Observation : We can now formulate a general form of the φ function that we were giving instances of, in (0.1) and (0.2). That is: Proof. Let us examine the differential equation d dx (

More Generalization
3.2.1. It does apparently seem that the function, φ Ω (x), which we defined in 3.1, is generalized.
But we can establish a more generalized expression for φ Ω (x): Let us examine the differential equation d [applying P roposition 2.1] Let us define another function, φ ′ 2 (x) [similar to that we defined in 2.1], whose value is equal to . Therefore, Remark: Here, φ ′ 2 (x) does not mean the derivative of the function φ 2 (x). In this context, φ ′ 2 (x) depicts a function that is different from φ 2 (x), as defined in 0.1.

3.2.2.
Without taking further instances into consideration, let us compute now the differen- [applying P roposition 1.1] Thus, we computed the new φ Ω (x), which is equal to . Thus, for Ω ∈ Z + : Remark: From the above expression, we can argue that φ Ω (x) function is actually an operator which acts on several other functions.

Rewriting the Differential Equation
For establish the inequalities, we shall first write the differential equation in section 3.2.2 in a different form, using the Proposition 2.3 (mainly transforming the second term): From, 3.2.2, we got the differential equation- Also, from Proposition 2.3, we know that: Thus, applying Proposition 2.3 to the differential equation, we get:

Establishing Inequalities
With reference to the differential equation 0.6, we can now establish our first inequality.
Theorem 0.1. For all Ω ∈ Z + and x ∈ R, if for all i ∈ [1, n], f i (x) is an increasing and positive function, then the following inequality holds: Equality occurs when the functions under consideration are all constant functions.
Before we move on to the next as well as important theorem, let us form an inequality bridge between integrals and the φ Ω (x) function: Let there be a function f(x) which is continuous on the interval (a,b). Now let us compute the ) Ω−1 } As mentioned earlier, Ω can take up values like 2, 3, ... that is, Ω ≥ 2. Now we are fully equipped to analyze the following theorem. But before introducing the theorem, let us introduce another theorem with which we will relate Theorem 0.3. and subsequently Theorem 0.4.
Consequently, the average value of f over [a,b] lies between m and M. [2] Similar to the above theorem, the next theorem gives a lower bound as well as an upper bound for linear functions under certain conditions. as well as defined at x=a and x=b with c, d ∈ Z + , the following inequality holds: where a and b are positive integers with b > a > 2.
Proof. Applying the concept introduced before the theorem, we can say that Ω . Therefore, in order to prove the first part of the theorem (the case of lower bound), we have to prove c 2 (b + a) + d > 2c : Since a > 2 and b > a,therefore, a + b > 4. Also since c is a positive integer, we can write the equality as (a + b)c > 4c. Moreover, since d is also a positive integer, we can argue that: (a + b)c + 2d > 4c. Dividing 2 on both sides of the inequality, we get c 2 (a + b) + d > 2c as expected. Now,i order to prove the second half of the theorem, we first need to compute From these 2 expressions, we can observe that the terms (b − a) and c 2 (b + a) + d are common in both expressions. Thus in order to prove that second half of the theorem, we need to show that 3c > 1 is is true because c is itself a positive integer and therefore, 3 2 is common and in both the left hand side and right hand side of the inequality and positive as well, then it is sufficient to check if 2m n < m 2n (b + a) + w p is true. We know that a > 2. Thus, (a + b) > 4. Dividing by 2 on both sides, we get (a+b) 2 > 2. Multiplying by m n on both sides, we get m 2n (b + a) > 2m n . Since w p is also greater than 0, so we can also state that m 2n (b + a) + w p > 2m n . Thus, the statement subject to verification is indeed true! Since from our analysis it is clear that m 2n (b + a) + w p > 2m n , let us also verify and check whether the inequality 3 is also true or not: Again, since the term (b − a) is common in both the expressions, it will be sufficient to verify For the sake of simplification, let us do some operations on the left hand side: This simplification makes the fact clear that the inequality is true only when 3 m n > 1, that is when 3m > n. Thus we can now state our more generalized theorem 0.3. in the form of theorem 0.4.: Theorem 0.4. For a linear function f (x) = m n x + w p , which is continuous and increasing on the interval (a,b) as well as defined at x = a and x = b and where m, n, w, p ∈ Z + , the following inequality holds true if 3m > n: where a and b are positive integers with b > a > 2.
Remark: It can be observed that the non-generalized theorem 4.3. is a special case of the generalized theorem 4.3 where n and p are both equal to 1 and that automatically satisfies the condition 3m > n.
Comparing Theorems 0.2. and 0.4.: In this section, we will attempt to establish a comparison between the lower and upper bounds of the integral of the class linear functions established in both these theorems. We will first try to compare the lower bounds, 2(f (b) − f (a)) and m(b − a). For he purpose of this context, we can argue that m = f (a). Therefore, the comparison is between 2(f (b) − f (a)) and f (a)(b − a). Now let us move on to comparing the upper bounds. Interestingly, the scenario will be different from that of the lower bounds. a))( m n (b + a) + 2 w p ). Similar to above, since we are considering a class of continuous and increasing linear functions, s we can also argue that M = f (b). Therefore, , let us check that whether the difference is positive and negative: . Since we know that 3m > n from the conditions of the theorem, therefore ( 3m n − 1) is positive. But no conclusion can be drawn for ( 2 + e 2 + 2b 3 cd + 2bde + 2b 2 ce + c 2 a 4 + d 2 a 2 + e 2 + 2a 3 cd + 2ade + 2a 2 ce + c 2 a 2 b 2 + ab 2 cd + b 2 ce + a 2 bcd + abd 2 + bde + a 2 ce + ade + e 2 } = 4 3 (b − a)(bc + ac + cd){c 2 b 4 + c 2 a 4 + d 2 b 2 + d 2 a 2 + 3e 2 + 2b 3 cd + 2a 3 cd + 3bde + 3ade + 3b 2 ce + 3a 2 ce + a 2 b 2 c 2 + ab 2 cd + a 2 bcd + abd 2 }.
Now let us investigate Theorem 0.4. in more depth by relating it to Theorem 0.6.
Theorem 0.6. If f 1 (x) and f 2 (x) are two functions of x, neither of which is a constant, such that if either one of them experiences an increase in any finite integral included in the interval of integration the other experiences a decrease, and conversely, then If on the other hand the two functions experience increments in the same intervals and decrements in the same intervals, the inequality sign above is reversed. [3] Now we will attempt to modify Theorem 0.6. and relate the concept with Theorems 0.4.and 0.5.
Corollary 0.1. If f 1 (x) is an increasing linear function in the interval from a to b and f 2 (x) is a function that is decreasing in the same interval of integration then, for b > a > 2, Proof. From our assumptions about f 1 (x) and f 2 (x), we can apply Theorem 0.6. to find that: After careful observation of Theorems 0.3. and 0.5., what comes to our minds is that there must be a general theorem that summarizes all of these. Let us introduce the general theorem, establishing a link between the integrals and the φ Ω (x) function: is increasing on the interval (a,b) as well as defined at x=a and x=b, the following inequality holds true:

Proof. By Induction
Let P (n) be the proposition that Step: To show that for n ≥ 1, P(n) =⇒ P(n+1) is true.
Assume P(n) to be true for the purpose of induction, a (a n x n + a n−1 x n−1 + ... + a 0 )dx. From our consideration of P(n) to be true, we can argue that: {(a n b n + a n−1 b n−1 + ...a 0 ) n+1 − (a n a n + a n−1 a n−1 + ...

More Applications of the φ Ω (x) Function
Before introducing the next applications of the function, let us first go through some preliminary steps mentioned in 5.1 and 5.2:

The Notation
In this section we shall be introducing a new notation for the convenience of identifying the functions being considered inside the φ Ω (x) function: = Ω(e x −e −x ) Ω−2 (e x +e −x ) 2 Ω−1 Remark : In the above example, we considered only one function, that is sinh(x). And this is why, for this example, n=1. Now, let us consider another example with 2 hyperbolic trigonometric functions sinh(x) and tanh(x). Therefore, After few more steps of manipulation, we shall get: 5.1.1. Applying the new notation: We shall now attempt to write the general form of a diverging series using the new notation introduced in 5.1-Let us focus on logarithmic functions. We shall select a class of functions of the form ln(x k ) for k ∈ Z + and examine their relation with the φ Ω (x) function.

Conclusion
From all the analysis we did, we can suggest that the new phi function leads the path to the derivation of various useful inequalities, lets us write expressions using the new notation and the phi-trigonometric functions can be used in several theorems which can in-turn lead to many new and important theorems as well as occasionally ease our computation process. Besides these, there are several important results that are linked to inequalities and the φ Ω (x) function which will present in my next research paper.