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What else can I help you with?
Does the series n factorial times 1 divided by e to the n converge or diverge?
Use the ratio test: Let an = n!/enlim n→∞ |an+1/an|= lim n→∞ |[(n + 1)!/en+1]/(n!/en)= lim n→∞ |[(n + 1)!/en+1](en/n!)= lim n→∞ |[(n + 1)n!en]/(enn!e)= (1/e) lim n→∞ (n + 1) = ∞, so the given series diverges.
You can approximate e by substituting large values for n into the expression Use the caret to enter exponents?
(1+1/n)^n
What is the integral of x to the power n?
For n not equal to -1, it is 1/(n+1)*xn+1 while for n = -1, it is ln(|x|), the logarithm to base e.
Can a sequence of numbers be both geometric and arithmetic?
Yes, it can both arithmetic and geometric.The formula for an arithmetic sequence is: a(n)=a(1)+d(n-1)The formula for a geometric sequence is: a(n)=a(1)*r^(n-1)Now, when d is zero and r is one, a sequence is both geometric and arithmetic. This is because it becomes a(n)=a(1)1 =a(1). Note that a(n) is often written anIt can easily observed that this makes the sequence a constant.Example:a(1)=a(2)=(i) for i= 3,4,5...if a(1)=3 then for a geometric sequence a(n)=3+0(n-1)=3,3,3,3,3,3,3and the geometric sequence a(n)=3r0 =3 also so the sequence is 3,3,3,3...In fact, we could do this for any constant sequence such as 1,1,1,1,1,1,1...or e,e,e,e,e,e,e,e...In general, let k be a constant, the sequence an =a1 (r)1 (n-1)(0) with a1 =kis the constant sequence k, k, k,... and is both geometric and arithmetic.
ζ(s):= n=1 1 ns equals?
E=ζ(s):= n=1 1 ns
What is slovene's formula?
it's a random sampling technique formula to estimate sampling size n=N/1+N(e)2 n- sampling size N-total population e-level of confidence
What is the latitude and longitude of Somalia?
Latitude: 1°30′S to 12°N Longitude: 41°E to 51°30′E Mogadishu, Somalia, Latitude = 2.04º N (2º2'23" N) Longitude = 45.34º E (45º20'24" E). They are 6.0442° N, 45.7194° E. 10 N, 49 E
How can you prove e-infinity equals 0?
I think you mean e to the (- infinity) power. The proof would be a limit proof. The limit (as n-->infinity) of [( en) ] = 0 You should have some other limits in class that you have proven. Show that your limit is less than one of those given for all values of n then you have your proof. For instance, if you already know that lim (as n-->infinity) of [(1/n) ] = 0 then for n = 1, 1/e1 < 1/1 true for n = 2, 1/e2 < 1/2 true Then assume that it is true for n = k so for n = K, 1/eK < 1/K assumed true therefore: eK > K multiply both by e e(k+1) > ke but we know ke > k+1 because e>1 SO: e(k+1) > ke > k+1 now take the reciprocal (reverses the inequalities) 1/e(k+1) < 1/ke < 1/ (k+1) by transitive prop of inequalities eliminate the middle term so that 1/e(k+1) < 1/ (k+1) this proves the case for n=K+1 and therefore will be true for all values of n since k was never a specified value. And if: 1/e(k+1) < 1/ (k+1) by one of the properties of limits, since the lim of 1/n is zero, then the lim of 1/en is also zero when n --> infinity.
You can approximate e by substituting large values for n into the expression?
(1+(1/n))^n you have to enter this way...payattention to parenthesis
Write a program to generate CRC?
/*written by: Bibhakar Jha;objective : to implement CRC in c programming language;*///Program code to add CRC check bit#include#include#define N strlen(g)char t[28],cs[28],g[]="10001000000100001";int a,e,c;void xor(){for(c = 1;c < N; c++)cs[c] = (( cs[c] == g[c])?'0':'1');}void crc(){for(e=0;e
What are the release dates for M-E-N- - 2010 1-3?
M-E-N- - 2010 1-3 was released on: USA: 23 March 2010
What are the release dates for M-E-N- - 2010 1-4?
M-E-N- - 2010 1-4 was released on: USA: 10 April 2010
