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What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?
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Dhamnekar Winod said:View attachment 11392
View attachment 11393
What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?
But, how to use this information to solve the given problem 1 or lemma 7.1?Klaas van Aarsen said:If $x>0$ then the inequality in (1.9) must be true, because the left side is then slightly less than n(x), and the right side is slightly more than n(x).
Let's consider the derivatives of the expressions in (1.8).
$$\frac d{dx}(1-\Re(x)) = -\Re'(x) = -n(x)$$
Let's first find $n'(x)$.
We have:
$$n'(x)=\frac d{dx} \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} = \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} \cdot -x =-xn(x)$$
Then we have for instance:
$$\frac d{dx}(x^{-1}n(x)) = -x^{-2}n(x) + x^{-1}n'(x) = -x^{-2}n(x)+x^{-1}\cdot -x n(x) = -(1+x^{-2})n(x)$$
So we see that the derivatives of the expressions in (1.8) are indeed the negatives of the expressions in (1.9).
We have that $1-\Re(x)$ is in between 2 expressions, so its integration must also be between the integrations of the those 2 expressions.
Qed.
To prove the more general formula, we need to repeat these steps for the additional terms.
Write (7.1) in the same form as (1.8) with the series on the left and also on the right.Dhamnekar Winod said:But, how to use this information to solve the given problem 1 or lemma 7.1?
Thanks for your answer. But sorry for not understanding it. I want to know your way of answering this problem.Klaas van Aarsen said:Write (7.1) in the same form as (1.8) with the series on the left and also on the right.
Take the derivatives to find an expression that is in the same form as (1.9).
Then the proof follows in the same fashion.