How would you solve a negative number raised to a fraction a/b if b is odd and a is evem?
Ignoring imaginary numbers
i.e $(-1)^\frac23$ Calculator returns an error
$(-1)^\frac 13 (-1)^\frac 13$ = -1.-1 = 1 (By law of indices)
or
$(-1^\frac13 )^2$ = 1
or
$(-1^2)^\frac13$ = 1
What about for other cases of a and b?
In Stephen Abbott's Understanding Analysis, when proving that the set of rational numbers is dense in the set of real numbers, Abbott picks an integer $m$ such that $$m - 1 \leqslant na < m,$$ where $n$ and $a$ are a natural number and a real number respectively.
Intuitively it makes sense that such an $m$ exists, but it seems to me that Abbott is taking liberties here. How do I know from the fact that the set of reals is a complete ordered field that such an $m$ exists? (Abbott doesn't discuss the real number axioms but I know them from elsewhere.)
We are given $(3^n-1)/2$ coins, and among those coins there is just one counterfeit coin. All the other coins weigh the same, but the counterfeit coin weighs slightly heavier or lighter (we don't know which is the case) than a normal coin.
Using a balance scale only, is it possible to identify the counterfeit coin, weighing not more than $n$ times?
My attempt : Using $-1, 0, +1$ instead of $0, 1, 2$, we may write each integer from $-(3^n-1)/2$ to $(3^n-1)/2$ in base 3, with digits among $-1, 0, +1$, in a unique manner. Label the coins from $1$ to $(3^n-1)/2$, and for each coin, labelled $k$, say, we designate one number in $\left\{-k,+k\right\}$ as being 'heavy' and the other one as being 'light'.
My claim is that it is possible to do this so that the resulting designation satisfies the following property :
For each $0\leq i\leq n-1$, the number of heavy numbers whose $3^i$-digit is -1 equals the number of heavy numbers whose $3^i$-digit is +1.
If this is the case, than the following weighing strategy works :
On $i$th weighing, put all the coins whose 'heavy' representation has $-1$ on its $3^i$-digit on the left. Put all the coins whose 'heavy' representation has $+1$ on its $3^i$-digit on the right.
Assuming the counterfeit coin is heavy, our $i$th weighing shows us the $3^i$-digit of the coin. After $n$ weighings, we may deduce which coin is counterfeit.
If the counterfeit coin is light, a similar reasoning also works, and it is clear from our construction the final answer does not depend on whether the counterfeit coin is heavier or lighter than a normal coin.
I believe that this strategy makes sense, but I was not capable proving that such a designation is possible (see the quote box). Am I on the right track? Is such a dsignation possible? Are there any more beautiful solutions to this innocent-looking puzzle?
Answer
Note that we are only asked to find the counterfeit coin (which we know exists), but strictly speaking need not find out if it is heavy or light.
Thus we might lay one coin aside and use the following theorem, interpreting "all coins weigh the same" as "the coin put aside is counterfeit".
If you think this is a loophole to the problem, the theorem allows us one extra weighing in that last case and we can use it to compare the counterfeit coin against one of the known good coins.
Theorem. Given $(3^n-3)/2$ coins, it is possible with $n$ weighings to arrive at one of the following results: "Coin $i$ is a heavy counterfeit" with $1\le i\le \frac{3^n-3}{2}$, or "Coin $i$ is a light counterfeit" with $1\le i\le \frac{3^n-3}{2}$, or to arrive at the result "all coins weigh the same" already after $n-1$ weighings.
Proof.
For $n=1$ this is clear: Given $0$ coins, it takes $0$ weighings to declare that they "all" weigh the same.
For $n>1$, (in your mind) glue the coins into groups of three (note that $3\mid \frac{3^n-3}{2}$). This leaves us with $\frac{3^{n-1}-1}2=\frac{3^{n-1}-3}2+1$ big coins. Ignoring the extra coin, we can either determine that the first $\frac{3^{n-1}-3}2$ big coins weigh the same with $n-2$ weighings, or determine with $n-1$ weighings which of the first the first $\frac{3^{n-1}-3}2$ big coins is counterfeit and how.
In the first case, "un-glue" the last coin into three coins $a,b,c$; also un-glue another coin into $d,e,f$. We know that $d,e,f$ are not counterfeit! Compare $a+b$ against $c+d$. In case of equilibrium, we know that all coins weigh the same and we used $n-1$ weighings for that.
If $a+b>c+d$, either $a$ or $b$ is too heavy or $c$ is too light. Compare $a$ against $b$; in case of equilibrium, $c$ is too light, and otherwise the heavier of the two coins is found. That is, we have determined the counterfeit coin and its type in $n$ weighings.
In the second case, we have found a counterfeit big coin and know if it is too heavy or too light. Un-glue it into three smaller coins $a,b,c$ and compare $a$ against $b$. In case of equilibrium, $c$ is counterfeit and in the same way the big coin was. And otherwise, we see which of $a,b$ is too light or too heavy, respectively. That is, we have determined the counterfeit coin and its type in $n$ weighings.
The theorem now follows by induction. $\square$
Sum of $$\lim_{n\rightarrow \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots \cdots \cdots +\frac{n}{n^2+n}\right)$$
$\bf{My\; Try::}$ I have solved it using Squeeze theorem
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+n}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+1}$$
So we get $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r} = 1$$
My question is can we solve above limit without using Squeeze Theorem,
If yes then plz explian me, Thanks
Answer
Using harmonic numbers $$S_n=\sum^{n}_{r=1}\frac{n}{n^2+r}=n \left(H_{n^2+n}-H_{n^2}\right)$$ Now, using the expansion $$H_p=\gamma+\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p ^4}\right)$$ and applying to each term you should arrive to $$S_n=n\left(\frac{-6 n^3-6 n^2+2 n+1}{12 n^4 (n+1)^2}+\log \left(1+\frac{1}{n}\right)+O\left(\frac{1}{n ^2}\right)\right)$$ Taylor again, $$S_n=1-\frac{1}{2 n}+O\left(\frac{1}{n ^2}\right)$$
Edit
May be of interest
$$T_n=\sum^{n}_{r=1}\frac{n}{n^k+r}=n \left(H_{n^k+n}-H_{n^k}\right)$$ leads to $$T_n=\frac{ \left(n^{2-k}-6 n^{k+1}-6 n^2+2 n\right)}{12 n^k\left(n^k+n\right)^2}+\log \left(1+\frac 1 {n^{k-1}}\right)$$ which makes $$n^{k-1}T_n=1-\frac{1}{2 n^{k-1}}+O\left(\frac{1}{n^k}\right)$$
If $90! = (90)(89)(88)...(2)(1)$, then what is the exponent of the highest power of $2$ which will divide $90!$ ?
How would I apply one of the easiest method from Here?
I need help on applying the link to this question.
I do not understand which one explains my case, and how I can solve using the method.
I would appreciate if someone showed how it is applied to this
I recently came across this problem in probability
Roll a fair six-sided die until you roll a number that is less than one of your previous rolls. To three decimal places, what is the expected value of the number of rolls made?
This is all what the question has to offer, any ideas?
This proves: Similar matrices have the same characteristic polynomial. (Lay P277 Theorem 4)
I prefer https://math.stackexchange.com/a/8407/53259, but this proves that they have the same eigenvalues.
Are they equivalent? What about in general, even for matrices which are NOT similar?
Answer
