Is it possible to bring a rod of length $ sqrt {3} l $ into a side cube of length l?

A cube of side length l may be suitable for a rod of maximum length $ sqrt {3} l $. But can such a rod be brought into a cube to be fixed in this position?

(Not even sure it's a mathematical question 🙂 and any help with the tags would be appreciated.)

Prove $ forall k geq 4, log (1 + x_k) -x_k leq {-1 over6k} $ where $ x_k = {(- 1) ^ k over sqrt k} $

The question:

Prove $ forall k geq 4, log (1 + x_k) -x_k leq {-1 over6k} $ or $ x_k = {(- 1) ^ k over sqrt k} $.

The inequality above is valid if and only if $$ begin {align} & log (1 + x_k) leq x_k- {1 on 6k} \
& Leftrightarrow 1 + x_k leq exp (x_k- {1 over6k})
end {align} $$

Using $$ x + 1 leq e ^ x $$we have $$ x_k + 1 leq e ^ {x_k} $$

Calculate the maximum value of $ sqrt { dfrac {3yz} {3yz + x}} + sqrt { dfrac {3zx} {3zx + 4y}} + sqrt { dfrac {xy} {xy + 3z}} where $ x + 2y + 3z = 2 $.

$ x $, $ y $ and $ z $ are positive such as $ x + 2y + 3z = $ 2. Calculate the maximum value of $$ large sqrt { dfrac {3yz} {3yz + x}} + sqrt { dfrac {3zx} {3zx + 4y}} + sqrt { dfrac {xy} {xy + 3z}} $$

This problem is brought to you by a recent contest. There should be different, more creative answers than the one I provided. Never mind…

inequality – Calculates the minimum value of $ sum_ {cyc} frac {a ^ 2} {b + c} $ where $ a, b, c> 0 $ and $ sum_ {cyc} sqrt {a ^ 2 + b ^ 2} = 1 $.

$ a $, $ b $ and $ c $ are positive such as $ sqrt {a ^ 2 + b ^ 2} + sqrt {b ^ 2 + c ^ 2} + sqrt {c ^ 2 + a ^ 2} = $ 1. Calculate the minimum value of $$ large frac {a ^ 2} {b + c} + frac {b ^ 2} {c + a} + frac {c ^ 2} {a + b} $$

I do not have any ideas yet to solve the problem. I will probably do it in the near future, but for now, I can not.

abstract algebra – Calculating the product of ideals $ left (4, frac {3+ sqrt {-71}} {2} right) $ and $ left (3, frac {1+ sqrt {- 71 }} {2} right) $

Let $ frak {a} = left (4, frac {3 + sqrt {-71}} {2} right) $ and $ frak {b} = left (3, frac {1 + sqrt {-71}} {2} right) $ to be the ideals in $ mathbb {Z}[sqrt{-71}]$. I would like to show that $ frak {a} frak {b} = left (12, frac {-5 + sqrt {-71}} {2} right) $. Clearly, $ frak {a} frak {b} $ contains the ideal $ left (12, frac {-5 + sqrt {-71}} {2} right) $. How about the conversation? Actually, I do not know how to write an arbitrary element $ frak {a} frak {b} $ explicitly since $ mathbb {Z}[sqrt{-71}]$ is not a PID. Thank you!

algebra precalculus – If $ x = frac { sqrt {111} -1} {2} $, calculate $ (2x ^ {5} + 2x ^ {4} – 53x ^ {3} – 57x + 54) ^ { 2004} $.

I already have two solutions to this problem, it's for high school students with an advanced level. I would like to know if there are better or more creative approaches to the problem. Here are my solutions:

  1. (1st solution): Note that
    $$ x ^ {2} = ( frac { sqrt {111} -1} {2}) ^ {2} = 28 – frac { sqrt {111}} {2} $$
    $$ x ^ {3} = x cdot x ^ {2} = left ( frac { sqrt {111}} {2} – frac {1} {2} right) left (28 – frac { sqrt {111}} {2} right) = 14 sqrt {111} – 111/4 – 14 + frac { sqrt {111}} {4} = frac {57 sqrt {111} } {4} – frac {167} {4} $$
    $$ x ^ {4} = (x ^ {2}) ^ {2} = frac {111 + 4 cdot 784} {4} – 28 sqrt {111} $$
    $$ x ^ {5} = x ^ {4} cdot x = left ( frac {111 + 4 cdot 784} {4} – 28 sqrt {111} right) left ( frac { sqrt {111} -1} {2} right) = left ( frac {3247} {4} – 28 sqrt {111} right) left ( frac { sqrt {111} -1} { 2} right) $$
    $$ = sqrt {111} frac {3359} {8} – frac {15679} {8} $$

So we have
$$ 2x ^ {5} + 2x ^ {4} – 53x ^ {3} – 57x + 54 = $$
$$ ( sqrt {111} frac {3359} {4} – frac {15679} {4}) + ( frac {111 + 4 cdot 784} {2} – 56 sqrt {111}) – 53 ( frac {57 sqrt {111}} {4} – frac {167} {4}) – 57 ( frac { sqrt {111} -1} {2} + 54 $
$$ = sqrt {111} ( frac {3359} {4} – frac {3359} {4}) – frac {15679} {4} + frac {111 + 4 cdot 784} {2} + frac {53 cdot 167} {4} + frac {57} {2} + 54 $$
$$ = – frac {15679} {4} + frac {222 + 8 cdot 784} {4} + frac {53 cdot 167} {4} + frac {114} {4} + frac {216} {4} $$
$$ = – frac {15679} {4} + frac {5600 + 894} {4} + frac {5300 + 3180 + 371} {4} + frac {114} {4} + frac {216} } {4} $$
$$ = – frac {15679} {4} + frac {6494} {4} + frac {8851} {4} + frac {114} {4} + frac {216} {4} $$
$$ = -4/4 = -1, $$
and the answer is $ (-1) ^ {2004} = 1. $ The above solution requires tedious calculations. You will find below an alternative solution.

  1. (2nd solution): Note that $ x = frac { sqrt {111} -1} {2} $ is equivalent with
    $$ (2x + 1) ^ {2} = 111 $$
    $$ 4x ^ {2} + 4x + 1 = 111 $$
    $$ 4x ^ {2} + 4x – 110 = 0 $$
    $$ (2x ^ {2} + 2x – 55) = 0 : : …….. : : (1) $$
    Multiply $ (1) $ with $ x ^ {3} $ we have
    $$ (2x ^ {5} + 2x ^ {4} – 55x ^ {3}) = 0 $$
    Multiply $ (1) $ with $ x $ we have
    $$ (2x ^ {3} + 2x ^ {2} – 55x) = 0 $$
    Sum both and we get:
    $$ 2x ^ {5} + 2x ^ {4} – 53 x ^ {3} + 2x ^ {2} – 55x = 0 : : …….. : : (2) $ $
    and now we have the first 3 terms of the form we want to calculate. Subtract $ (2) $ with $ (1) $ get:
    $$ 2x ^ {5} + 2x ^ {4} – 53 x ^ {3} – 57x + 55 = 0 $$
    $$ 2x ^ {5} + 2x ^ {4} – 53 x ^ {3} – 57x + 54 = -1 $$
    So the answer is $ (- 1) ^ {2004} = 1. $

real analysis – $ x_1 = sqrt {2} $ and we have $ x_ {n + 1} = ( sqrt {2}) ^ {x_n} $. Discover the limit.

The limit of this sequence, if it exists, will be a fixed point of $ g (x) = sqrt {2} ^ x $. So, if you solve the equation $ x = sqrt {2} ^ x $you have all the possible limits (the solutions are 2 and 4). In the end, you have to use the fixed point theorem to establish convergence when you start with $ x_1 = sqrt {2} $.

Why is the $ x ^ 2 $ integral of $ 0 to $ 5 not equal to the $ sqrt x $ integral of $ 0 to $ 25?

My calculator says the first is 41.67 and the last is 83.34. Why is that? Should not they be equal?

Show that: $ displaystyle sum_ {cyc} frac { sqrt {x ^ 2 + y ^ 2}} {x ^ 2 + y ^ 2 + xy} ≥ sqrt {2} $

CA watch:

$ displaystyle sum_ {cyc} frac { sqrt {x ^ 2 + y ^ 2}} {x ^ 2 + y ^ 2 + xy} ≥ sqrt {2} $

Or $ a, b, c $ in R $

I'm trying to use the incumbent inequality

Please give me some ideas or tips

Number Theory – Show that $ sqrt {d} $ has a length of period 1 ssi $ d = a ^ 2 + 1 $

Since I know if $ d $ is an integer that $ sqrt {d} =[alpha_0,bar{alpha_1},…bar{alpha_n},bar{2alpha_0}]$. I want to show that $ sqrt {d} $ has a length of period 1 if and only if $ d = a ^ 2 + 1 $, for a natural number a.

This is what I think I should do. Could someone confirm it to me?

($ Leftarrow $) Assumed $ d = a ^ 2 + 1 $then $ sqrt {d} = sqrt {a ^ 2 + 1} $.

$ (a-1) ^ 2 leq a ^ 2 + 1, Rightarrow (a-1) leq sqrt {a ^ 2 + 1} $

So the soil of $ sqrt {a ^ 2 + 1} $ is $ a-1 = alpha_0 $.
By subtracting this and inverting, we obtain:

$ tfrac {1} { sqrt {a ^ 2 + 1} – (a-1)} = tfrac { sqrt {a ^ 2 + 1} + (a-1)} {2a} $, which has floor 1.

So we know $ alpha_1 = $ 1, subtracting this and inverting, we obtain:

$ tfrac {2a} { sqrt {a ^ 2 + 1} – (a-1)} = tfac {2a ( sqrt {a ^ 2 + 1} + (a-1)} {2a} = sqrt {a ^ 2 + 1} + (a-1) $who has the floor $ 2a-2 $, subtracting and inverting gives,

$ tfrac {1} { sqrt {a ^ 2 + 1} – (a-1) $So, we are back at the beginning and the continuous fraction has a period length of 1.

$ ( Rightarrow) $ assume $ sqrt {d} $ for the length of period 1. Then $ sqrt {d} =[alpha_0,bar{alpha_1}]=[alpha_0,bar{alpha_1},bar{2alpha_0}]$.

So we can write:

$ sqrt {d} = alpha_0 + tfrac {1} { alpha_1 + tfrac {1} {2 alpha_0 + tfrac {1} {…}}} = alpha_0 + tfrac {1} { alpha_1 + tfrac {1} {2 sqrt {d}}} $.

I have not yet managed to solve this problem, but $ d = a ^ 2 + 1 $, do you come down to rearrange what I just wrote?