<p>A power series can be added, differentiated and integrated just like any polynomial.</p>
<p>To determine the coefficients \(c_n\), follow the steps below:</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">First evaluate the series at \(x = 0\). We have&nbsp;\(\mathrm{f}(0) = c_0\)</span></div></li><li><div class="list-item-content"><span class="list-item-text">Then differentiate \(\mathrm{f}(x)\) with respect to \(x\):</span></div></li></ol>
<p style="text-align: center">\[\mathrm{f}'(x) = c_1 + 2c_2x + 3c_3x^2 + \dots\]</p>
<p>Therefore, when&nbsp;\(x = 0\), \(c_1 = \mathrm{f}'(0)\)</p>
<ol class="i-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Differentiating again, \(\mathrm{f}''(x) = 2c_2 + 3(2)c_3x + \dots\)</span></div></li></ol>
<p style="text-align: center">\[\therefore \mathrm{f}''(0) = 2c_2 \implies c_2 = \frac{\mathrm{f}''(0)}{2}\]</p>
<ol class="i-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Continuing in this manner, we find that&nbsp;\(\mathrm{f}^{(n)}(0) = n!c_n\)</span></div></li></ol>
<p style="text-align: center">\[\therefore c_n = \frac{\mathrm{f}^{(n)}(0)}{n!} \text{ where}\hspace{0.5em} 0! = 1 \text{ and}\hspace{0.5em} \mathrm{f}^{(0)}(x) = \mathrm{f}(x)\]</p>
<ol class="i-list" start="5"><li><div class="list-item-content"><span class="list-item-text">So if&nbsp;\(\mathrm{f}(x) = \sum_{n=0}^{\infty} c_nx^n, \quad |x| &lt; r\)</span></div></li></ol>
<p>Then \[\mathrm{f}(x) = \mathrm{f}(0) + \frac{\mathrm{f}'(0)}{1!}x + \frac{\mathrm{f}''(0)}{2!}x^2 + \dots + \frac{\mathrm{f}^{(n)}(0)}{n!}x^n + \dots\]</p>

A power series can be added, differentiated and integrated just like any polynomial.

To determine the coefficients \(c_n\), follow the steps below:

First evaluate the series at \(x = 0\). We have \(\mathrm{f}(0) = c_0\)

Then differentiate \(\mathrm{f}(x)\) with respect to \(x\):

\[\mathrm{f}'(x) = c_1 + 2c_2x + 3c_3x^2 + \dots\]

Therefore, when \(x = 0\), \(c_1 = \mathrm{f}'(0)\)

Differentiating again, \(\mathrm{f}''(x) = 2c_2 + 3(2)c_3x + \dots\)

\[\therefore \mathrm{f}''(0) = 2c_2 \implies c_2 = \frac{\mathrm{f}''(0)}{2}\]

Continuing in this manner, we find that \(\mathrm{f}^{(n)}(0) = n!c_n\)

\[\therefore c_n = \frac{\mathrm{f}^{(n)}(0)}{n!} \text{ where } 0! = 1 \text{ and } \mathrm{f}^{(0)}(x) = \mathrm{f}(x)\]

So if \(\mathrm{f}(x) = \sum_{n=0}^{\infty} c_nx^n, \quad |x| < r\)

Then \[\mathrm{f}(x) = \mathrm{f}(0) + \frac{\mathrm{f}'(0)}{1!}x + \frac{\mathrm{f}''(0)}{2!}x^2 + \dots + \frac{\mathrm{f}^{(n)}(0)}{n!}x^n + \dots\]

Steps to calculate the associated Maclaurin's series

Steps to calculate the associated Maclaurin's series