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For How Many Integers \$a\$ Is \$frac2^10 Cdot 3 ^8 Cdot 5^6a^4\$ an Integer?

Hint: Factor \$a\$ into primes; it's evident that these primes must be \$2\$, \$3\$, and \$5\$ (why?). Then \$a = 2^d cdot 3^b cdot 5^c\$, so that\$\$a^4 = 2^4d cdot 3^4b cdot 5^4c\$\$What conditions on \$d, b, c\$ are necessary and sufficient to make this a divisor of \$2^10 cdot 3^8 cdot 5^6\$?

1. Boundedness of the Laplace Transform on \$(C[0, 1], |cdot|_2)\$.

\$|Tf(s)| leq (int_0^1|f(x)|dx)leq (int_0^1|f(x)|^2dx)^1/2=|f|\$ and hence \$|Tf| leq |f|\$, \$T\$ is bounded with \$|T|leq 1 \$.\$T\$ is one-to-one: If \$Tf=0\$ then \$int_0^1e^-sxf(x)dx=0\$ for all \$s in ,\$. Repeatedly differentiating and putting \$s=0\$ we get \$int_0^1x^nf(x)dxdx=0\$ for all \$n geq 0\$. [At each step you have to apply DCT to justify differentiating inside the integral sign]. A standard application of Weierstrass Approximation Thoerem now show that \$f=0\$, so \$T\$ is one-to-one.\$T\$ is not onto. It is easy to see \$Tf\$ is differentiable so non-differentiable continuous are not in the range.

2. Solve recurrence relation \$T(n) = ncdot T(n/2)^2 \$

We have \$\$4nT(n) = left(4fracn2T(n/2)

ight)^2\$\$ so that \$\$F(n) = F(n/2)^2 \$\$ where \$F(n) = 4nT(n).\$ Then take logs \$\$ln(F(n)) = 2ln(F(n/2)) \$\$so that we have \$\$ G(n) = 2G(n/2)\$\$ where \$G(n) = ln(F(n)).\$ The solution to this is obvious: when \$n\$ doubles, so does \$G. \$ Thus it is a linear function \$\$ G(n) = an.\$\$Unwinding the transformations gives \$F(n) = e^an\$ and \$T(n) = frace^an4n.\$As for your question about whether the master method works, I am only familiar with that as a theorem about asymptotics, but when you take logs of the initial equation it immediately becomes a recurrence of that type and you can conclude that the log of the solution is \$Theta(n). \$.

3. Why is \$frac 1cdot2cdot3cdotâ€¦cdot n(n1)(n2)â€¦(2n)le frac 1 n1\$

Hint: \$binom2nk\$ is unimodal by moving k with maximum at \$binom2nn\$. See also that \$n1leq 2n=binom2n1\$

4. \$int_1^infty operatornamesech x cdot ln x dx\$

It is convergent for sure since:\$\$ 0leq int_1^inftyfrac2log xe^xe^-x,dxleq int_1^infty2(x-1)e^-x,dx = frac2e.\$\$ The upper bound can be improved up to \$fracsqrtpie\$ if we exploit \$log xleqsqrtx-1\$ for any \$xgeq 1\$.

5. Show that the set of points that are nearer \$a\$ than \$b\$ with respect to \$lVert cdot

Vert_2\$ is convex

Let \$phi(x) = |x-a|^2-|x-b|^2= |a|^2-|b|^22 langle b-a, x

angle\$. \$phi\$ is affine (that is, linear plus a constant) hence convex, and so \$ x | phi(x) 0\$ is convex, we see that the sets are open half-spaces separated by the hyperplane \$ x | phi(x) = 0\$.

6. \$W\$ is binary R.V. and \$Y= Wcdot Y_1 (1-W)Y_0\$, where \$Y_1, Y_0\$ are also R.V.s. Does \$E[Wcdot Y]= E[Wcdot Y_1]\$?

Since \$W\$ only takes the values \$0\$ and \$1\$, it follows that \$W^2=W\$ and \$W(1-W)=0\$. Therefore \$\$ WY=W^2Y_1W(1-W)Y_0=WY_1 \$\$ hence \$\$ mathbbE[WY]=mathbbE[WY_\$\$ Note that \$W\$ does not need to be independent of \$Y_0\$ and \$Y_1\$.

7. Solve summation \$sum_i=1^n lfloor ecdot i

floor \$

Following are three possible ideas, the first two are not that satisfactory. The third one is a modification of the second ideas which might work. I hope they can inspire others to create something that is useful.As a Fourier seriesFirst, we can rewrite \$lfloor x

floor\$ as a Fourier series.\$\$lfloor x

floor = x - x = x - frac12 sum_m=1^infty fracsin(2pi m x)pi mtag*1\$\$Since the discontinuities of \$lfloor x

floor\$ is contained inside \$mathbbZ subset mathbbQ\$. the floor function is continuous at irrational \$x\$. As a result, RHS of \$(*1)\$ converges pointwisely to LHS for irrational \$x\$.Substitute \$x\$ by \$ek\$ for \$k = 1, ldots, n\$ and sum over \$k\$, we obtain.\$\$sum_k=1^n lfloor ek

floor = frace2n(n1) - fracn2 underbracefrac12pisum_m=1^infty fraccos(pi m e) - cos(pi m e(2n1))msin(pi m e)_I \$\$ In principle, if we can approximate the series \$I\$ on RHS accurate enough, we can round the RHS to nearest integer and it will give us the value of LHS. The problem is when we approximate \$I\$ by its partial sums, the \$sin(m pi e)\$ factor in denominator make it very hard to figure out the correct number of terms to keep! Recursive evaluationIf we did not insist for a closed formula, it is possible to evaluate the sum in a recursive manner. For \$alpha in (1,infty)setminus mathbbQ\$ and \$n in mathbbZ\$, define \$displaystyle;S(alpha,n) stackreldef= sum_k=1^nlfloor alpha k

floor\$. The sum we want is \$S(e,n)\$. There are two branches in the recursion:Case I - \$alpha > 2\$.Rewrite \$alpha\$ as \$beta m\$ where \$beta in (1,2)\$ and \$m = lfloor alpha - 1

floor\$, we have \$\$S(alpha,n) = sum_k=1^n left( mk lfloor beta k

floor

ight) = fracm2n(n1) S(beta,n)\$\$Case II - \$alpha 2 quadtext and quad overbracebeginalign alpha &leftarrow [ 1 a_1; a_2, ldots, a_k ] n &leftarrow leftlfloorfracn [ a_0 - 1; a_2, ldots, a_k]

ight

floor endalign^alpha n\$ and then set \$k = 3ell\$. For \$n approx 10^4000\$, \$k approx 4011\$ should be enough.On my PC, I can compute \$S(e,10^4000)\$ using maxima in less than a minute. However, I've to admit I have no way to verify I got the right answer.

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