Proving the Summation Problem: P(x) and the Limit of |e^(x-1)-1| for x>0

Since all the coefficients are positive, P'(1) is positive. So ##|P'(1)|<=1##In summary, if ##|P(x)|<=|e^{x-1}-1|## for all x> 0 where ##P(x)=\sum\limits_{r=0}^na_rx^r##, then it can be shown that ##|\sum\limits_{r=0}^nra_r|<=1##. This can be proved by considering that ##P(1)=0## and that the derivative of ##e^{x-1}-1## at x=1 is 1, which shows that ##|P'(1)|<=1##.
  • #1
AdityaDev
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Homework Statement


If ##|P(x)|<=|e^{x-1}-1|## for all x> 0 where ##P(x)=\sum\limits_{r=0}^na_rx^r## then prove that ##|\sum\limits_{r=0}^nra_r|<=1##

Homework Equations



None

The Attempt at a Solution



##P(1)=a_0+a_1+...##
If the constants are positive, then ##P(1)<=|e^0-1|##
So P(1)<=0
so ##a_0+a_1+a_2+... <=1##
But how do I prove that ##0a_0+1.a_1+2.a_2+...<=1##
 
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  • #2
AdityaDev said:

Homework Statement


If ##|P(x)|<=|e^{x-1}-1|## for all x> 0 where ##P(x)=\sum\limits_{r=0}^na_rx^r## then prove that ##|\sum\limits_{r=0}^nra_r|<=1##

Homework Equations



None

The Attempt at a Solution



##P(1)=a_0+a_1+...##
If the constants are positive, then ##P(1)<=|e^0-1|##
So P(1)<=0
so ##a_0+a_1+a_2+... <=1##
But how do I prove that ##0a_0+1.a_1+2.a_2+...<=1##

Uh, ##P(1)=0##. Say why that's true. Then you want to show ##|P'(1)| \le 1##. Think about considering what the derivative of ##e^{x-1}-1## at ##x=1## might tell you about that.
 
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  • #3
|P(1)|<=0. So the only possibility for P(1) is zero.
P'(1)=a1+2a2+3a3+...
Derivative of ##e^{x-1}-1## at x=1 is 1. So P'(1) lies between 1 and -1.
 

Related to Proving the Summation Problem: P(x) and the Limit of |e^(x-1)-1| for x>0

What is the summation problem?

The summation problem is a mathematical concept that involves finding the sum of a sequence of numbers. It is typically denoted by the symbol Σ and is commonly encountered in calculus and other areas of mathematics.

What is P(x)?

P(x) is a function that represents the sum of a sequence of numbers. It is often used in conjunction with the summation problem to find the value of the sequence at a specific point or to determine the limit of the sequence.

What is the limit of |e^(x-1)-1| for x>0?

The limit of |e^(x-1)-1| for x>0 is equal to 0. This means that as x approaches positive infinity, the value of the expression e^(x-1)-1 gets closer and closer to 0.

How is the summation problem related to the limit of |e^(x-1)-1| for x>0?

The summation problem and the limit of |e^(x-1)-1| for x>0 are closely related because the limit of the expression is often used to determine the value of the summation problem at a specific point. Additionally, the limit can also be used to determine whether a series of numbers converges or diverges.

Why is proving the summation problem and its limit important?

Proving the summation problem and its limit is important because it allows us to understand the behavior of sequences of numbers and to accurately calculate their values. It also has many practical applications in fields such as physics, engineering, and economics, where the use of series and summations is common.

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