document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1024132>Question 1024132</A>: <I><SPAN STYLE=\"word-break: break-all;\"> procedure to solve cube root of 35 with out calculator </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #639651 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=639651\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "There are a couple of ways to go about it.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You can just approximate.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3 is too small, because 3^3 = 27<BR>\n" );
document.write( "4 is too large, because 4^3 = 64\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Interpolate: 64 - 27 is 37 and 35 - 27 is 8, so use 3 8/37 or a little less than 3 and 1/4.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Multiply out 3.25^3 and you'll find out that is a little small.  So pick something a little larger and try again.  Keep it up until you get as close as you like.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You can also use Newton-Raphson:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You want to solve the equation:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \  x^3\ -\ 35\ =\ 0\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Newton-Raphson says that if you make a guess, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x_o\">, then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x_1\"> is a closer guess if\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_o\ -\ \frac{f(x_o)}{f'(x_o)}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So for the cube root of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large a\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_o\ -\ \frac{x_o^3\ -\ a}{3x_o^2}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_o\ -\ \frac{1}{3}\left(x_o\ -\ \frac{a}{x_o^2}\right)\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So, since <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large a = 35\"> for your problem and 3 is an integer lower bound on the desired value, you can use 3 as your first guess:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ 3\ -\ \frac{1}{3}\left(3\ -\ \frac{35}{9}\right)\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Doing the arithmetic:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{89}{27}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "If you calculate\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{89}{27}\right)^3\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You will find that you are a little on the low side, so perform Newton-Raphson again using <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x_1\ =\ \frac{89}{27}\"> as the initial guess.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \left(\frac{89}{27}\right)\ -\ \frac{1}{3}\left(\left(\frac{89}{27}\right)\ -\ \frac{35}{\left(\frac{89}{27}\right)^2}\right)\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Yes, the arithmetic is as ugly as a mud fence, but it is all rational numbers so you can do it old school without a calculator.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The good news is that, given an appropriate first guess, Newton-Raphson generally converges rapidly for well-behaved functions.  In this particular case, if you do the calculation above, your approximation will be correct to the third decimal place; that is your answer will be on the high side by only 0.0002 or so.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \\r\"> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );