How Does Differentiation Relate to Polynomial Inequalities?

[phoenixXL]

Homework Statement

Suppose p(x)\ =\ a_0\ +\ a_1x\ +\ a_2x^2\ +\ ...\ + a_nx^n.

Now if |p(x)|\ <=\ |e^{x-1}\ -\ 1| for all x\ >=\ 0 then

Prove |a_1\ +\ 2a_2\ +\ ...\ + na_n|\ <=\ 1.

2. Relevant Graph( |e^{x-1}\ -\ 1| )

The Attempt at a Solution

From the graph we can conclude that
p(x) should pass through (1,0)
=> a_1\ +\ a_2\ +\ ...\ + a_n\ =\ 0

Further, I'm not able to apply any other condition given to simplify the expression. Their is of course something to do with the derivative as I found this question in a book of differentiation.
Any help would be highly appreciated.

Thanks

 

[ehild]

What is the derivative of p(x)?

Have you copied the problem correctly?



ehild

 

[phoenixXL]

Have you copied the problem correctly?

Sorry, there isn't x in the proof
I have corrected the problem.

p'(x)\ =\ a_1\ +\ 2a_2x\ +\ 3a_3x^2\ +\ ...\ +\ na_nx^{n-1}

 

[benorin]

Perhaps it's p'(1) =\ a_1\ +\ 2a_2\ +\ 3a_3\ +\ ...\ +\ na_n

 

[micromass]

p'(x)\ =\ a_1\ +\ 2a_2\ +\ 3a_3\ +\ ...\ +\ na_n

That is not the correct derivative of $p(x)$. The derivative should have some variables $x$.

 

[benorin]

From the graph we can conclude that
p(x) should pass through (1,0)
=> a_1\ +\ a_2\ +\ ...\ + a_n\ =\ 0

wrong. note that p passing through (1,0) means that p(1)=a_0+a_1+\cdots +a_n=0\Rightarrow a_0=-(a_1+\cdots +a_n)

 

[phoenixXL]

wrong. note that p passing through (1,0) means that p(1)=a_0+a_1+\cdots +a_n=0\Rightarrow a_0=-(a_1+\cdots +a_n)

right I messed with it. Additionally I don't know why I'm not able to edit the first post.

That is not the correct derivative of $p(x)$. The derivative should have some variables $x$.

I corrected it. Thanks.

 

[phoenixXL]

If I could just prove that the inequality of the funciton
i.e. |p(x)|\ <=\ |e^{x-1}\ −\ 1|
also holds true for the derivative
i.e. (|p(x)|)'\ <=\ (|e^{x-1}\ −\ 1|)'
then I could just substitute x = 1 and prove the question.

But I just couldn't formulate how to deduce from the given conditions to that part.
Any Ideas ?

 

[phoenixXL]

.../

 

[vela]

As you noted, p(1) = 0, so you can say that
$$\lvert p(x) - p(1) \rvert \le \lvert e^{x-1}-1 \rvert.$$ Now consider the definition of a derivative as a limit of a difference quotient and write down the expression for p'(1). Do you see the connection now?

 

[phoenixXL]

Do you see the connection now?

Right.

p&#039;(1)\ =\ lim_{h\rightarrow0}\frac{p(1+h)-p(1)}{h}\\<br /> \implies a_1+2a_2+3a_3+...+na_n\ =\ lim_{h\rightarrow0}\frac{p(1+h)}{h},\ as\ p(1)\ =\ 0<br />
I was able to solve it.
Thank you for your time