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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