document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1103836>Question 1103836</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the Maclaurin series (i.e., Taylor series about c=0) and its intervals of convergence.<BR>\n" );
document.write( "a) f (x) = ln(1+ x)<BR>\n" );
document.write( "b) f (x) = e^-x\r<BR>\n" );
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document.write( "Hello, I hope you can explain this question to me. thank you so much. God bless you  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #718551 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=718551\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A Maclaurin series is a Taylor Series about x = 0<BR>\n" );
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document.write( "The Maclaurin series is constructed this way<BR>\n" );
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document.write( "f^0(0) + f^1(0)x + f^2(0)x^2/2! + f^3(0)x^3/3! + .... + f^n(0)x^n/n!, where n = 0, 1, 2, 3,....+infinity n represents the nth derivative which are the coefficients in this expansion<BR>\n" );
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document.write( "a) f(0) = ln(1) = 0<BR>\n" );
document.write( "f^1(x) = 1/(1+x), f^1(0) = 1/(1+0) = 1<BR>\n" );
document.write( "f^2(x) = -1/(1+0)^2, f^2(0) = -1/(1+0))^2 = -1<BR>\n" );
document.write( "f^3(x) = 2/(1+0)^3, f^3(0) = 2/(1+0)^3 = 2<BR>\n" );
document.write( "f^4(x) = -6/(1+0)^4, f^4(0) = -6/(1+0)^4 = -6<BR>\n" );
document.write( "Our expansion looks like<BR>\n" );
document.write( "0 + x - x^2/2 + x^3/3 - X^4/4 and the pattern emerges for the series representation,<BR>\n" );
document.write( "ln(1+x) = -summation for n = 1 to +infinity of (-1)^n * x^n / n<BR>\n" );
document.write( "using ratio test this series converses for x < |1|<BR>\n" );
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document.write( "b) f(0) = e^(-0) = 1<BR>\n" );
document.write( "f^1(x) = -e^-x, f^1(0) = -1<BR>\n" );
document.write( "f^2(x) = e^-x, f^2(0) = 1<BR>\n" );
document.write( "f^3(x) = -e^-x, f^3(0) = -1<BR>\n" );
document.write( "e^-x = summation for n = 0 to +infinity of (-x)^n / n!<BR>\n" );
document.write( "using ratio test this series converges everywhere<BR>\n" );
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