How do solve X^0=1?
x^0=x^(n-n)Where n any positive numberFrom law of indices x^(m-n) = (x^m)/(x^n)Thereforex^(n-n) = (x^n)/(x^n)= 1
Factorial help: Solve for n: (n+2)! / (n)! = 110?
Notice that (n + 2)! = (n + 2)(n + 1)(n!), and so: (n + 2)!/n! = (n + 2)(n + 1). Hence, we can re-write the equation as: (n + 2)(n + 1) = 110. Expanding the left side: n^2 + 3n + 2 = 110. Then, setting the right side equal to zero by subtracting 110 from both sides gives: n^2 + 3n - 108 = 0. The left side factors to (n + 12)(n - 9), so the equation now becomes: (n + 12)(n - 9) = 0. Then, by the zero-product property, the solutions to (n + 2)(n + 1) = 110 are: n + 12 = 0 and n - 9 = 0 ==> n = -12 and n = 9. --- Now, we cannot take the factorial of a negative number, which is what happens with n = -12; thus, n = -12 is an extraneous solution and must be discarded. n = 9 has no such issues with the original equation, and so n = 9 is a solution. Therefore, n = 9.
How do I solve [math]\int_{1}^{n} \frac{2+\sqrt{n}}{n^2 -n+1}[/math]?
I assume this integral is in fact [math]\int_{1}^{n} \frac{2 + \sqrt{n}}{n^2 - n + 1} \mathrm{d}n[/math], which is only sensible (non-circular) in the presence of capture-avoiding substitution. For this reason and to avoid rewriting bounds, I’ll first solve for the indefinite integral and use that, noting that our bound of integration exclude [math]x = 0[/math].To make this integral easier to evaluate, let [math]x = n^{1/2}[/math], in which case we have [math]\frac{\mathrm{d}x}{\mathrm{d}n} = \frac{1}{2} n^{-1/2} \Rightarrow \mathrm{d}n = 2 \mathrm{d}x n^{1/2}[/math] for [math]n \neq 0[/math].Performing this substitution, again for [math]n \neq 0[/math], in the integral gives:[math]\int \frac{2 + \sqrt{n}}{n^2 - n + 1} \mathrm{d}n = \int \frac{2 x + 4 x^2}{x^4 - x^2 + 1} \mathrm{d}x[/math]Now note that [math]\frac{2 x + 4 x^2}{x^4 - x^2 + 1}[/math] decomposes as [math]-\frac{1 + 2 x}{\sqrt{3} (-1 + \sqrt{3} x - x^2)} - \frac{1 + 2 x}{\sqrt{3} (1 + \sqrt{3} x + x^2)}[/math]. We’ll integrate this expression termwise.Both of these terms are of the form [math]\int \frac{m x + n}{ax^2 + bx + c} \mathrm{d}x[/math] for which [math]4ac - b^2 > 0[/math], so each of their solutions is of the form:[math]\frac{m}{2a} \ln | ax^2 + bx + c | + \frac{2an-bm}{a\sqrt{4ac-b^2}} \arctan \frac{2ax + b}{\sqrt{4ac - b^2}} + C[/math]After substituting these and simplifying (which I did with a computer algebra system), you get:[math]3^{-1} \left( 2\sqrt{3} \arctan\left(\frac{\sqrt{3}}{1-2x^2}\right) + (3 + i \sqrt{3}) \sqrt{-2 -2 i \sqrt{3}} \arctan\left(\frac{1}{2} \left(1 - i \sqrt{3}\right) x\right) + \left(3 - i\sqrt{3} \right) \sqrt{-2 + 2i\sqrt{3}} \arctan\left(\frac{1}{2}\left(1 + i \sqrt{3}\right)\right) \right) + C[/math]Solving the integral is now a question of evaluating this expression at the bounds [math]n=1 \Rightarrow x = n^{1/2} = \sqrt{1} = 1[/math] and [math]n = x^2[/math], which I unfortunately don’t have the CAS horsepower or manual time for right now. Good luck!
Solve n /n-4 +n =12-4n /n-4?
Your typing leaves things ambiguous. Do you mean n/(n-4) + n = 12 - (4n/(n-4))? Or do you mean n/(n-4) + n = (12 - 4n) / (n-4)? In any case, go multiply both sides by (n-4) to clear away the fractions, then multiply all of the resulting terms out, combine like terms, and solve the quadratic.
Can anyone solve this chemistry problem?
Carbon has two naturally occurring isotopes: C-12 (natural abundance is 98.93%) and C-13 (natural abundance is 1.07%). How many C-13 atoms are present, on average, in a 24000-atom sample of carbon? Please and thank you!
How does one solve this series with the comparison test [math] \sum_ {n=6} ^ {\infty} \frac {n} {(n^ {4} - 0.9) ^\frac{3}{7}}[/math]?
*A2A[math]\dfrac{n}{(n^4-0.9)^{\frac 37}}<\dfrac{1}{n^{\frac57}}\tag*{}[/math][math]a_n=\dfrac{n}{(n^4-0.9)^{\frac37}},b=\dfrac1{n^{\frac57}}\\==================\\\lim_\limits{n\to\infty}\dfrac{a_n}{b_n}=\lim_\limits{n\to\infty}\dfrac{n}{(n^4-0.9)^{\frac37}}\cdot n^{\frac57}\\=\lim_\limits{n\to\infty}\dfrac{n^{\frac{12}7}}{n^{\frac{12}7}}=1\tag*{}[/math]Since the limit is positive and finite, hence the series diverges via the Limit Comparison Test. This is because [math]\sum \dfrac1{n^{\frac57}}[/math] diverges (p-series test)
How do I prove that [math]\frac{1}{n}\left[(n+1)(n+2)\ldots (n+n)\right]^{\frac{1}{n}}[/math] converges to [math]\frac{4}{e}?[/math]
I finally have the answer to this question after a lot of hard work. As I wrote in my comment on Nikhil's answer:(1/n)*[(n+1)(n+2).....(n+n)]^(1/n) = [(1+1/n)*(1+2/n).....(1+n/n)]^(1/n)lim n-inf [(1+1/n)*(1+2/n).....(1+n/n)]^(1/n) = lim n-inf exp{ (1/n) log ((1+1/n)*(1+2/n).....(1+n/n))}Now log(1+k/n)= k/n - (1/2)*(k/n)^2+(1/3)*(k/n)^3.............and log (a1.a2......an) = log(a1)+log(a2)+......log(an)log ((1+1/n)*(1+2/n).....(1+n/n) = k=1:n Σ (j=1:n(Σ ( j)^k/ k)*(-1)^(k)lim n- inf exp{ (1/n) log ((1+1/n)*(1+2/n).....(1+n/n))}=exp( Σ coefficient of highest powers of n in the above power series) = exp( 1/2-1/6+1/12-1/20-1/30+ 1/42+-------)Now we need to calculate S= 1/2-1/6+1/12-1/20+1/30 - 1/42+---S= 1/2-1/6+1/12-1/20+1/30 -1/42 +--- = (1/2+1/12+1/42+ ...............) - (1/6+1/20+ 1/42................) = S1 - S2 S1= 1/2+1/12+1/42............... = 1-1/2+ 1/3-1/4+ 1/5-1/6 +............. = ln2 S2 = 1/6+1/20+ 1/42.............. = 1/2-1/3+1/4-1/5+1/6-1/7 ......... = 1- (1/2-1/3+1/4-1/5+.........) = 1- ln2 S= S1-S2 = ln2-(1-ln2)= 2ln2-1= ln4 -1 lim n-inf [(1+1/n)*(1+2/n).....(1+n/n)]^(1/n)= exp( 1/2-1/6+1/12-1/20-1/30+ 1/42+-------)= exp(ln4 -1 ) = exp(ln4)*exp(-1)= 4/e PS: I have cut on a few steps here and there. If there's a doubt and it is pointed out, I would try my best to clarify it.
What is the solution of this question? If n is a whole number such that n* (n+28) is a prime number, find the prime number. Explain your answer.
Understanding what a prime number is will help you in solving this question.Prime numbers are numbers divisible by itself and 1.This implies that if n is any number greater than 1, that automatically rules out the final result as a prime number, given that the n>1 will also be a factor of the final result.From the above, n can only be 1. This means that the answer you're looking for is 29.
Does the series tan(1/n), n from 1 to infinity, converge or diverge?
Note that tan(x) > 0 on (0, 2π) on (0, π/2). For n > 1, 0 < 1/n < 1. Since 1 < π/2, tan(1/n) > 0 for all n > 1, the limit comparison test applies. Note that when x ≈ 0: tan(x) ≈ x. So when n is large: tan(1/n) ≈ 1/n. This implies that 1/n is a good candidate for the limit comparison test. With the limit comparison test with a_n = tan(1/n) and b_n = 1/n: lim (n-->infinity) (a_n)/(b_n) => lim (n-->infinity) tan(1/n)/(1/n) = lim (n-->infinity) n*tan(1/n) = lim (n-->infinity) [n*sin(1/n)/cos(1/n)] = lim (n-->infinity) n*sin(1/n) * lim (n-->infinity) 1/cos(1/n) = lim (x-->0+) sin(x)/x * lim (x-->0+) 1/cos(x) (with sub x = 1/n) = 1 * 1 = 1. This means that tan(1/n) and 1/n diverge/converge together. However, since 1/n is the divergent harmonic series, the series in question diverges. I hope this helps!
Find the solution to the following differential equation subject to the initial conditions give?
You have: (x^2)*y" - x*y' + y = 2x*ln(x) This is an inhomogeneous Cauchy-Euler equation. Let: x = exp(t) or t = ln(x) dx/dt = exp(t) = x then, dy/dt = dx/dt * dy/dx = x*dy/dx and d²y/dt² = d/dt(dy/dt) = d/dt(x*dy/dx) = dx/dt*dy/dx + x*d/dt(dy/dx) = exp(t)*dy/dx + x*d²y/dx²*dx/dt = x*dy/dx + (x^2)*d²y/dx² = dy/dt + (x^2)*d²y/dx² (x^2)*d²y/dx² = d²y/dt² - dy/dt Plugging these into the differential equation yields: (d²y/dt² - dy/dt) - dy/dt + y = 2*exp(t)*ln(exp(t)) d²y/dt² - 2dy/dt + y = 2t*exp(t) This is now a non-homogeneous, 2nd-order, linear equation with constant coefficients. First solve for the solution to the homogeneous problem: d²y/dt² - 2dy/dt + y = 0 This has the characteristic equation: k^2 - 2k + 1 = 0 k = 1 (repeated roots) so the solution to the homogeneous equation is: y(t) = A*exp(t) + B*t*exp(t) Now solve for the particular solution using the method of undetermined coefficients. An initial guess at the form of the particular solution would be: y_p = a*exp(t) + b*t*exp(t) However, both of these terms appear in the homogeneous solution. We can get new, linearly independent, candidate particular solutions by multiplying the first guess by t^2, then: y_p = a*(t^2)*exp(t) + b*(t^3)*exp(t) dy_p/dt = 2a*t*exp(t) + (a + 3b)*(t^2)*exp(t) + b*(t^3)*exp(t) d²y_p/dt² = 2a*exp(t) + (4a + 6b)*t*exp(t) + (a + 6b)*(t^2)*exp(t) + b*(t^3)*exp(t) Plugging these into the differential equation and simplifying: 2a*exp(t) + 6b*t*exp(t) = 2t*exp(t) This implies that a = 0 and b = 1/3, so the complete solution for y(t) is: y(t) = A*exp(t) + B*t*exp(t) + ((t^3)/3)*exp(t) Now back substitute for t: y(x) = A*exp(ln(x)) + B*ln(x)*exp(ln(x) + ((ln(x))^3)/3*exp(ln(x)) y(x) = A*x + B*x*ln(x) + (x/3)*(ln(x))^3 Now use the initial conditions to solve for A and B: y'(x) = (A + B) + B*ln(x) + (ln(x))^2 +(1/3)*(ln(x))^3 y(1) = 2 = A + 0 + 0, so A = 2 y'(1) = 1 = (2+B) + 0 + 0 + 0, so B = -1 The solution to the initial value problem is then: y(x) = 2*x - x*ln(x) + (x/3)*(ln(x))^3