Suppose `p(x)=a_0+a_1x+a_2x^2++a_n x^ndot` If `|p(x)|lt=e^(x-1)-1|` for all `xgeq0,` prove that `|a_1+2a_2++n a_n|lt=1.`

Explanation

`"Given "p(x)=a_(0)+a_(1)x+ax^(2)+...+an^(x^(n))`
`therefore" "f'(x)=0+a_(1)+2a_(2)x+...+na_(n)x^(n-1)`
`"or "p'(1)=a_(1)+2a_(2)+...+na_(n)" (1)"`
`"Now, "|p(x)|le|e^(x-1)-1|`
`therefore" "|p(1)|le0" "(because|e^(1-1)-1|=|e^(0)-1|=|1-1|=0)`
`"or "p(1)=0" "[therefore|p(1)|ge0]`
As `|p(x)|le|e^(x-1)-1|`,we get
`|p(1+h)|le|e^(h)-1|AAh gt-1, h ne0`
`"or "|p(1+h)-p(1)|le|e^(h-1)|`
`"or "|(p(1+h)-p(1))/(h)|le|(e^(h)-1)/(h)|`
Taking limit as `hrarr0`, we get
`underset(hrarr0)lim|(p(1+h)-p(1))/(h)|leunderset(hrarr0)lim|(e^(h)-1)/(h)|`
`"or "|p'(1)|le1`
`"or "|a_(1)+2a_(2)+...+na_(n)|le|"`