document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1027731>Question 1027731</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much more would $10,000 earn in 20 years in an account compounded continuously than an account compounded annually if the interest rate on both accounts is 5%? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #642934 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=642934\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> compounded annually, the formula is:\r<BR>\n" );
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document.write( "f = p * (1+r)^n\r<BR>\n" );
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document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r i the interest rate per time period (per year).<BR>\n" );
document.write( "n is the number of time periods (years).\r<BR>\n" );
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document.write( "at 5%, with a p of 10,000, for 20 years, the formula becomes:\r<BR>\n" );
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document.write( "f = 10,000 * (1.05)^20 = 26532.98\r<BR>\n" );
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document.write( "compounded continuously, the formula is:\r<BR>\n" );
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document.write( "f = p * e^(rt)\r<BR>\n" );
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document.write( "f =  future value<BR>\n" );
document.write( "p = present value<BR>\n" );
document.write( "r = interest rate per time period (per year).<BR>\n" );
document.write( "n = number of time periods (years).\r<BR>\n" );
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document.write( "with p = 10,000 and r = .05 and t = 20, the formula becomes:\r<BR>\n" );
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document.write( "f = 10,000 * e^(.05*20) = 27182.82\r<BR>\n" );
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document.write( "continuous compounding is the most compounding you can do.\r<BR>\n" );
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document.write( "the more compounding periods per year, the closer you get to continuous compounding.\r<BR>\n" );
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document.write( "for example, if you compounded daily, then:\r<BR>\n" );
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document.write( "assume 365 days per year.\r<BR>\n" );
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document.write( "p = 10,000<BR>\n" );
document.write( "r = .05/365 = .000136986301 per day.<BR>\n" );
document.write( "n = 20 * 365 = 7300 days.\r<BR>\n" );
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document.write( "formula becomes:\r<BR>\n" );
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document.write( "f = 10,000 * (1.000136986301)^7300 = 27180.95669\r<BR>\n" );
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document.write( "that's a lot closer.\r<BR>\n" );
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document.write( "if you compounded per hour, you would get much closer.\r<BR>\n" );
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document.write( "the limit is continuous compounding.\r<BR>\n" );
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document.write( "you can't get a higher value than continuous compounding.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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