# Application of Derivatives

1. Jul 9, 2014

### phoenixXL

1. The problem statement, all variables and given/known data
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( $|e^{x-1}\ -\ 1|$ )

3. 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

Last edited: Jul 10, 2014
2. Jul 10, 2014

### ehild

What is the derivative of p(x)?

Have you copied the problem correctly?

ehild

Last edited: Jul 10, 2014
3. Jul 10, 2014

### phoenixXL

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

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

Last edited: Jul 10, 2014
4. Jul 10, 2014

### benorin

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

5. Jul 10, 2014

### micromass

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

6. Jul 10, 2014

### benorin

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

Last edited: Jul 10, 2014
7. Jul 10, 2014

### phoenixXL

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

I corrected it. Thanks.

8. Jul 11, 2014

### 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 ?

9. Jul 13, 2014

### phoenixXL

.../

10. Jul 13, 2014

### vela

Staff Emeritus
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?

11. Jul 14, 2014

### phoenixXL

Right.

$$p'(1)\ =\ lim_{h\rightarrow0}\frac{p(1+h)-p(1)}{h}\\ \implies a_1+2a_2+3a_3+......+na_n\ =\ lim_{h\rightarrow0}\frac{p(1+h)}{h},\ as\ p(1)\ =\ 0$$
I was able to solve it.