<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>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>