document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=380351>Question 380351</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please would someone help,\r<BR>\n" );
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document.write( "Obtain the Maclaurin series expansion about the point 0 for the function ln(x + 1) as<BR>\n" );
document.write( "ln(x + 1) = x -x^2/2+x^3/3 - ...
+ (-1)^n+1 x^n/n+ ... 
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document.write( "Note that we cannot find a Maclaurin expansion of the function ln x since ln x does not exist at x = 0 and so cannot be differentiated at x = 0.\r<BR>\n" );
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document.write( "Thanks in advance </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #269924 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=269924\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is equivalent to finding the power series of ln x centered around x = 1. Note that all derivatives of ln x at x = 1 are equal to 1 or -1. Since we have the power series\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln+%28x%29+=+%28x-1%29+-+%28%28x-1%29%5E2%29%2F2+%2B+%28%28x-1%29%5E3%29%2F3+-+... ALIGN=MIDDLE  ALT=\"ln+%28x%29+=+%28x-1%29+-+%28%28x-1%29%5E2%29%2F2+%2B+%28%28x-1%29%5E3%29%2F3+-+...\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Adding one to all the x terms produces the given result. </SPAN>\n" );
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