how do you apply taylor expansion for small numbers?
931519
This question is extremely confused/confusing!  For one thing, the speed of light (exceedable only by warp drive) is wavelength TIMES frequency, not wavelength OVER frequency, as implied by two of your lines.  Any Starfleet Academy student should know that.\n\nIgnoring that, I THINK that the nub of your question is this: presumably v/c is small in the original question that you must have been asked.  Call that small value x.  Then, for small x, a BINOMIAL SERIES expansion (NOT a Taylor expansion) tells you that \n(1 + x)^(-1) is approximately 1 - x, ignoring terms of order x^2 and higher.\n\nLive long and prosper.
476547
First, make the substitution y = mx.  That will give you\n\n[x - 5mx]/[x^2 + (mx)^2].\n\nNow, factor out the x term and you get\n\nx[1 - m]/(x^2)[1 + m^2] = [1 - m]/x[1 + m^2]\n\nAs you can see from this expression, changing the value of m will change the value of the limit.  For a limit to exist, it must be unique.  Therefore, since the limit depends upon the value of m, it cannot be unique and so it does not exist.
490983
You would need more and more terms of the series to get an accurate result for the larger values of the argument.  Sum the positive and negative ones separately, from least to greatest absolute value, before adding those two subtotals together.\n\nHow does the Pade approximation do?  For example see\n\nhttp://math.fullerton.edu/mathews/n2003/pade/PadeApproximationMod/Links/PadeApproximationMod_lnk_15.html\n\nP subscript{5, 5} (x) for arcsin(x) = the fraction given by:\n\nnumerator: (922320 x - 717780 x^3 + 69049 x^5)\ndenominator: (15 (61488 - 58100 x^2 + 9675 x^4))\n\nIf you are using C, why not just call the\n\ndouble asin(double x);\n\nfunction in <math.h> ? :-)
895128
If by 'convert' to a linear function you mean approximate by a linear function, then a possible solution is given by the Taylor's expansion of the exponential function.\n\nConsider a point x0, then the linear approximation of y around x0 is given by:\n\ny(x) = e^(x0+a) + e^(x0+a) * (x+a-x0) ,\n\nwhich is linear in x.  Obviously this is only an approximation as the Taylor expansion of the exponential function will involve a very large number of higher order (i.e. quadratic, cubic etc) terms in the right hand side of the equation.
545599
subract 32 then divide by 1.8 (or mulitply by 5/9 --- same thing)
753811
well the way i was tought .. is that u make a table\n\nn | f^n(x) | c = 0\n0 |funct   |value at 0\n1| first deriv | value at 0 ( v1 ) \n2|second driv | value at 0 (v2)\n\nu do that till u hit number four and the fourth deriv\n\nand since e^0 cos(4*0) is 1 then u start your taylor series by 1+  v1/1! x + v2/2! x^2 + v3/3! x^3 + v4/4! x^4\n\nand there it is just replace the v's with the values u get from the table
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)
105048
Directly, there is no solution.  If you use a series expansion around x=0, \nsin(x)=sum((-1)^n(x^(2n+1))/(factorial(2n+1)))\nDivide this series by x, then integrate.  After the 3rd term, the error term goes to the ten millionths digit, so I'd say you're pretty safe :)\n\nsin(x)/x = 1 - (x^2)/6 + (x^4)/120 - (x^6)/5040 + .....\nIntegrated is:  x - (x^3)/18 + (x^5)/600 - (x^7)/32580 +\nor \nsum((-1)^n)(x^(2n+1))/(2n+1)*factorial(2n+1))
1148067
You elevate the number to the power 2 like this:\n\n                                    x^2\n\nYou should still do the normal product but it is easier to thing you et the square of a certain number and most calculators already have the elevate to the second power function integrated\n\nCynthia
782057
look at a few examples and it will make sense\n\nfor example \n\n02 (where the 2 is small)   -> _ O    (the _ is where a number goes)  (2)
891988
Multiply that by about 0.458
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
590233
Hint:\nDo a simple substitution u = x-5\nWhich means x = u + 5\n\nYou should get\n\n(2/5)(x-5)^(5/2) + 2(x-5)^(3/2) + C
974331
= ln x / x^5\nln x -> inf\nx^5 -> inf\ndifferentiate both sides (Lagrange)\n\n=> 1/x  / 5x^4\n= 1/ (5x^5)\n=> 0 \n\nOr ln x increases much slower as x gets bigger, whereas x^5 increases faster, (exponentially) as it gets bigger. polynomial wins over logarithm. \n\nThe limit is 0.
126603
The above method will not work, as the data will overflow. The factorial of 5000 is a very large number, and will not fit in any atomic data type.\n\nI will not give the answer outright here, but I will let you know the steps required.\n\nThe most obvious thing to do is create an arry of "int", where each member of the array holds a single digit. As far as how many digits you require, you may use Stirling's asymptotic approximation to figure out the approximate value of 5000!. You can get the formula for the approximation here:\n\nhttp://en.wikipedia.org/wiki/Stirling's_approximation\n\nMake sure that you add on a few extra places for safety.\n\nNow, your array should start off will every element being zero, except for the last one, which is 1. This corresponds to the factorial of 1.\n\nNow, inside of a loop from 2 to 5000, you basically need to do a long multiplication of the current array by the loop value. At each position, you need to do exactly as you learned in grade school - do the muliplication, add the carry, and store the last digit (mod 10 of result), and compute the carry (div 10 of result) and continue.\n\nTry for some increasingly larger numbers, and for small numbers that are less than 69!, you can verify the first few digits from your calculator. For larger numbers, you can again use Stirling's approximation to find out if you are in the right neighbourhood.\n\nI know that I have left some of the details a bit fuzzy here, but there is enough information to get you started in the right direction.\n\nHope this helps and good luck!
685997
u should write it like\n7.52*10^-4