Nov
26
comment Can you construct a field with 6 elements?
Why must $\langle F, +\rangle$ be a cyclic group?
Nov
24
comment Find $x$ such that $x+x^2+x^3=x^9-x^7$
$x^9-x^7-x^3-x^2-x=(x^3-x-1)(x^5+x^2+1)x$
Nov
24
comment Does L'Hôpital's work the other way?
-1 The question is not formulated well. L'Hôpital's rule does not say that this two limites are equal. It says they are equal if some preconditions are fulfilled. Also the limes of the undefinite integrals does not make sense.
Nov
23
comment Solving $(z+1)^5 = z^5$
Why follows $$ \pars{1 + {1 \over z_{n}}}^{5} = \expo{2n\pi\ic}\,,\qquad n = 1, 2, 3, 4 $$ from $$ \pars{1 + {1 \over z}}^{5} = 1 $$
Nov
21
comment Can any one tell me the books for power series?
A book is Abramuwitz & Stegun: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables You can use WolframAlpha which is documented here. Look up wikipedia for square root. Note the comment of @André Nicolas
Nov
16
comment Is $ n^2-14n+24 $ a prime number?
-1 because it is incomplete
Nov
14
comment which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$
+1 for $\stackrel{?}{\lt\gt}$
Nov
12
comment Finding Prime Number
I think the answer you accepted is blunder
Nov
12
comment Finding Prime Number
Your arguments dont work: 1) If two expressions are not divisible by $6p$ this does not mean that the sum of these expressions is not divisible by $6p$. 2) if $a^p \equiv -1$ then $a^p+1$ is divisible by $p$.
Nov
12
comment The largest value of $f(n) $
The answer you have accepted is wrong
Nov
12
comment The largest value of $f(n) $
I think almost all statements in this answer are wrong: From $2$ it does not follow that $f$ is increasing. Take $p=10$. $f(99)=99+f(9)=99+(9+f(0))=108$ and $f(101)=101+f(10)=101+f(1)=101+(1+f(0))=102$. It is wrong that $f(x)=x/p$ if $x$ is a multiple of $p$. Again, take $p=10$: $f(100)=f(10)=f(1)=1+f(0)=1$. If $a_k$ should be the max-value this is also wrong. For $p=3$ and $k=2$ the maxvalue is $f(8)=8+f(2)=8+2+f(0)=10$
Nov
12
comment Which is larger? $20!$ or $2^{40}$?
because you added the braces after you read my comment. I assumed that $16 \cdot 16 \cdot 16$ was a lower bound for $16 \cdot 17 \cdots$ and $4 \cdot 4 \cdot 4 \cdots 4 \cdot 4$ is a lower bound for $4 \cdot 5 \cdot 6 \cdots 15$.
Nov
11
comment Which is larger? $20!$ or $2^{40}$?
+1, but $$1\cdot1\cdot1\cdot4\cdot4\cdots 4\cdot 4\cdot16\cdot16\cdot 16 =2^{24} \cdot 2^{12} \lt 2^{40}$$ Better $$1\cdot1\cdot1\cdot4\cdot4\cdots 4\cdot 4\cdot16 \cdots \cdot 16 = 2^{44} \gt 2^{40}$$
Nov
8
comment Is $ n^2-14n+24 $ a prime number?
Whats about $(-p)*(-1)$ and $(-1)*(-p)$ ?
Nov
6
comment How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$
What is the linear fractional transform $g$? Can you add it to your post?
Nov
6
comment How prove $\frac{2}{3}<\frac{3x^6+15x^2+2}{2x^6+15x^4+3}\le\frac{3}{2}$
are you familiar with the standard technics to solve such an inequation?
Nov
4
comment Can my MSE reputation be any positive integer?
if one downvotes something one gets -1.
Nov
2
comment Factorize the polynomial $x^3+y^3+z^3-3xyz$
@Aaron San: added notes about integral domains
Oct
29
comment Factorize the polynomial $x^3+y^3+z^3-3xyz$
-1 $$2\,y^2+5\,x\,y+2\,x^2 = \left(y+2\,x\right)\,\left(2\,y+x\right)$$ a symmetric polynomial can have factors that are not symmteric
Oct
28
comment Minimum number of weighings necessary
@Ross Millikan: I already took into account that the weighing stops when the scale is balanced (therefore there are 63 different outcomes after 5 weighings and not $3^5$). Also I think that the fact that the result of the 1st, 2nd, 3rd, 4th or 5th weighing can be equality does not mean that there are 5 different possible splits that will balance (nor $2^0+2^1+2^2+2^3+2^4$) but only that the split you did for the 1st,2nd,... weighing is the unique solution. If you add my name preceded by a '@' to your answer I will get a notification.