document.write( "Question 1189036: Can the following expression ((X+5)^0.5) be multiplied out like the following expression (x+5)^2. Thank you for the consideration.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #820265 by greenestamps(13198)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "I very much think the other tutor completely missed the point of your question....

\n" ); document.write( "Yes, you can \"multiply out\" the expression ((x+5)^(0.5)) using binomial expansion.

\n" ); document.write( "Many students will be familiar with binomial expansion when the power is an integer; few students will be familiar with it when the power is not an integer.

\n" ); document.write( "Because the exponent is not an integer, the expansion does not terminate; it has an infinite number of terms. The first few terms will give good approximations.

\n" ); document.write( "The key to understanding binomial expansion with non-integer powers is to look at how binomial coefficients are calculated.

\n" ); document.write( "For all n, including non-integer values of n...

\n" ); document.write( "\"C%28n%2C0%29+=+1\"

\n" ); document.write( "For all other binomial coefficients C(n,r), the coefficient can be calculated as a fraction with \"r\" numbers in both numerator and denominator, starting with n and decrementing by 1 in the numerator, and starting with r and decrementing by 1 in the denominator. That gives us...

\n" ); document.write( "\"C%28n%2C1%29+=+n%2F1+=+n\"
\n" ); document.write( "\"C%28n%2C2%29+=+%28%28n%29%28n-1%29%29%2F%282%2A1%29\"
\n" ); document.write( "\"C%28n%2C3%29+=+%28%28n%29%28n-1%29%28n-2%29%29%2F%283%2A2%2A1%29\"
\n" ); document.write( "etc....

\n" ); document.write( "For example for n=3, we get what (I hope) are the familiar coefficients for n=3:

\n" ); document.write( "\"C%283%2C0%29+=+1\"
\n" ); document.write( "\"C%283%2C1%29+=+%283%2F1%29+=+3\"
\n" ); document.write( "\"C%283%2C2%29+=+%28%283%2A2%29%2F%282%2A1%29%29+=+6%2F2+=+3\"
\n" ); document.write( "\"C%283%2C3%29+=+%28%283%2A2%2A1%29%2F%283%2A2%2A1%29%29+=+6%2F6+=+1\"

\n" ); document.write( "And using the same method for calculating the coefficients for n=0.5, we get the unfamiliar coefficients for a binomial expansion to the 0.5 power:

\n" ); document.write( "\"C%280.5%2C0%29+=+1\"
\n" ); document.write( "\"C%280.5%2C1%29+=+%280.5%2F1%29+=+1%2F2\"
\n" ); document.write( "\"C%280.5%2C2%29+=+%28%28%280.5%29%28-0.5%29%29%2F%282%2A1%29%29+=+-1%2F8\"
\n" ); document.write( "\"C%280.5%2C3%29+=+%28%28%280.5%29%28-0.5%29%28-1.5%29%29%2F%283%2A2%2A1%29%29+=+1%2F16\"

\n" ); document.write( "Now we can write out the first few terms of the binomial expansion of (x+5)^(-0.5). We have the binomial coefficients; and from one term to the next the power of x decreases by 1 and the power of 5 increases by 1:

\n" ); document.write( "

\n" ); document.write( "Evaluating that expression for values of x that should produce whole number results, we find
\r\n" );
document.write( "\r\n" );
document.write( "   x  x+5  (x+5)^(0.5)  f(x)  (4 decimal places)\r\n" );
document.write( "  ------------------------------\r\n" );
document.write( "   4   9        3       3.1035\r\n" );
document.write( "  11  16        4       4.0042\r\n" );
document.write( "  20  25        5       5.0006\r\n" );
document.write( "  31  36        6       6.0001\r\n" );
document.write( "  44  49        7       7.0000\r\n" );
document.write( "  59  64        8       8.0000

\n" ); document.write( "For small values of x, where this approximation is not very accurate, using 2 or 3 more terms of the binomial expansion would improve the accuracy.

\n" ); document.write( " \n" ); document.write( "
\n" );