Question on Lagrange Error Bound

<p>Estimate the error using the MacLaurin Series generated by e^x to approximate the value of e.</p>

<p>I used the langrange error bound formula: |f^{(n+1)}(c)(x-a)^{n+1}/(n+1)!|</p>

<p>But how does that equal e^c |x^{n+1}|/(n+1)!</p>

<p>Where does the e^c in the second expression come from?</p>

<p>the piece, f^<a href=“c”>(n+1)</a>, in the lagrange equation is simply some derivative of the function. since you use e^x, all derivatives are e^x, no matter the value of n. hence, e^c.</p>

<p>c is just some value between a and x, by the way.</p>

<p>Some examples of the Lagrange error bound</p>

<p><a href=“http://www.stewartcalculus.com/data/CALCULUS%206E%20Early%20Transcendentals/upfiles/topics/6et_at_02_frtts.pdf[/url]”>http://www.stewartcalculus.com/data/CALCULUS%206E%20Early%20Transcendentals/upfiles/topics/6et_at_02_frtts.pdf&lt;/a&gt;&lt;/p&gt;

<p>Look at pages 4 and 5</p>

<p>f^<a href=“c”>(n+1)</a> is the degree of the function or the (n+1)th derivative. For example, a 4th degree function is f^<a href=“x”>4</a> or the fourth derivative.</p>