solving equations – Should I reduce all cases when $ sqrt {x y} = sqrt x sqrt and $?

My understanding is that Reduce gives all conditions (using or) where the entry is true.

Now, $ sqrt {xy} = sqrt x sqrt and $, where $ x, y $ are real, under the following three conditions / cases

$$
begin {align *}
x geq 0, y geq0 \
x geq0, and leq0 \
x leq0, y geq 0 \
end {align *}
$$

but not when $ x <0, and <0 $

This is verified by doing

Clear all[x,y]
Assuming[Element[{x,y},Reals]&& x> = 0 && and <= 0, simplify[ Sqrt[x*y] - Sqrt[x]* Sqrt[y]]]Assuming[Element[{x,y},Reals]&& X<= 0&&y>= 0, simplify[ Sqrt[x*y] - Sqrt[x]* Sqrt[y]]]Assuming[Element[{x,y},Reals]&& X<= 0&&y>= 0, simplify[ Sqrt[x*y] - Sqrt[x]* Sqrt[y]]]Assuming[Element[{x,y},Reals]&& x <= 0 && and <= 0, simplify[ Sqrt[x*y] - Sqrt[x]* Sqrt[y]]]

Mathematical graphics

Then because it does

    Reduce[ Sqrt[x*y] - Sqrt[x]* Sqrt[y]== 0, {x, y}, Real]

Give only one of the 3 previous cases?

Mathematical graphics

It is my understanding of Reduce wrong or should Reduce Have you given the other two cases?

V 12 in the windows.

calculation – A quick question about the integral of $ frac { sqrt {x}} { sqrt {x} -3} $

Consider the function $$ int frac { sqrt {x}} { sqrt {x} -3} $$
Using the substitution we obtain $ begin {align *}
u & = sqrt {x} -3 \
x & = (u + 3) ^ 2
& dx = 2 (u + 3) du
end {align *} $

$ (u + 3) ^ 2 = $$ u ^ 2 + 6u + 9 $, connecting all this we get
$$ int frac {u ^ 2 + 6u + 9} {u} du $$
This started a question with even number in my textbook, I used an online calculator that gave me
$$ int frac {2 (u ^ 2 + 6u + 9)} {u} du $$
and I do not know where that $ 2 $ He came from. I think it's from $ dx $ but that would also mean having $ (u + 3) $ there, or could have come from $ du = frac {1} {2 sqrt {x}} $ but then where did that $ sqrt {x} $ to go?

complex analysis: calculate the integral $ int_0 ^ 1 frac {x ^ 3} { sqrt {x (1-x)}} dx $ using the remainder

I am trying to calculate the following integral:

$$ int_0 ^ 1 frac {x ^ 3} { sqrt {x (1-x)}} dx $$

So what I'm trying to do is use a dog bone outline, let me call D, so in the lower part I have the negative sign, but the orientation is reversed in the upper part, because the only singularity I have is in x = $ infty $ I have that

$$ int_D f (z) dz = 2 int_0 ^ 1f (x) dx + ( int _ { gamma_1} + int _ { gamma_2}) f ( xi) d xi = 2 pi i Res (f, infty) $$ and then calculating the residue in infinity I found this $ 3 pi / 8 $, but the answer is actually $ 35 pi / 128 $. Can someone help me figure out what to do?

The outline of the dog bone that I imagine.

$ K: = int _ { pi -2} ^ { pi 2} {1+ sqrt {cos (x)}} dx $ integral $ K.M.N $ Who is the biggest and who is the smallest?

Set
$$ M: = int _ { pi -2} ^ { pi 2} frac {((1 + x) ^ 2)} {(1 + x ^ 2)} dx $$
$$ N: = int _ { pi -2} ^ { pi 2} frac {(1 + x)} {(e ^ {x})} dx $$
$$ K: = int _ { pi -2} ^ { pi 2} {1+ sqrt {cos (x)}} dx $$
integral $ K.M.N $ Who is the biggest and who is the smallest?

integration: calculate $ lim limits_ {n to infty} frac {1} {n} sum_ {i, j = 1} ^ n frac {1} { sqrt {i ^ 2 + j ^ 2 $

Calculate $ lim limits_ {n to infty} frac {1} {n} sum_ {i, j = 1} ^ n frac {1} { sqrt {i ^ 2 + j ^ 2}} $.
I'm not looking for a solution using double integrals. I tried to convert this into a Riemann sum, but I could not make any progress.
I thought about using the compression theorem, but I can not find any useful inequality, I just tried to use AM-GM in the denominator, but it did not help.

inequality – Test $ x + y + z = 3, , x ^ {, 2} + y ^ {, 2} + z ^ {, 2} = 9 , therefore , y- x leqq 2 sqrt {3} $

Try out
$$ x + y + z = 3, , x ^ {, 2} + y ^ {, 2} + z ^ {, 2} = 9 , therefore , y- x leqq 2 sqrt {3} $$
I have a solution, and I hope to see the best ones, thanks for your interest! We have
$$ (, x + y + z ,) ^ {, 2} + (, – , x + y + z ,) ^ {, 2} + (, x- y + z ,) ^ {, 2} + (, x + y- z ,) ^ {, 2} = 4 (, x ^ {, 2} + y ^ {, 2} + z ^ { , 2} ,) = 36 $$
OR
$$ left (z + (, y- x ,) right) ^ {, 2} + left (z- (, y- x ,) right) ^ {, 2} + (, 3- 2 , z ,) ^ {, 2} = 27 $$
OR
$$ 3 , z ^ {, 2} – 6 , z + (, y- x ,) ^ {, 2} = 9 $$
OR
$$ (, y- x ,) ^ {, 2} = – , 3 (, z- 1 ,) ^ {, 2} + 12 leqq 12 , therefore , y- x leqq | , y- x , | leqq 2 sqrt {3} $$
Q.E.D. The condition of equality occurs when $ z = 1 , therefore , x + y = 2 , therefore , x = 1- sqrt {3}, , y = 1+ sqrt {3} $.

precalculus algebra: solve the equation on reals: $ sqrt {5x ^ 2 + 27x + 25} – 5 sqrt {x + 1} = sqrt {x ^ 2 – 4} $.

Solve the equation over reais: $ sqrt {5x ^ 2 + 27x + 25} – 5 sqrt {x + 1} = sqrt {x ^ 2 – 4} $.

This problem is an adaptation of a recent competition. And I can not solve it.

The solutions are $ sqrt {5} + 1 $ Y $ dfrac {13 + sqrt {65}} {8} $, as Wolfram Alpha says.

Actual analysis: How to prove that $ left ( frac {(- 1) ^ n} { sqrt {n}} right) _ {n geq 1} $ is a null sequence?

How to prove that $ left ( frac {(- 1) ^ n} { sqrt {n}} right) _ {n geq 1} $ Is it a null sequence?

My attempt for induction:

If I prove that the denominator grows faster than the numerator, I can conclude that it is a null sequence, right? So I have to prove that $ sqrt {n} geq (-1) ^ n $ for $ forall n in mathbb {N}: n geq 1 $

Base case with $ n_0 = 1 $: $ sqrt {1} = 1 geq (-1) ^ 1 = -1 checkmark $

Induction hypothesis: $ exists n in mathbb {N}: sqrt {n} geq (-1) ^ n $

Induction claim: $ Longrightarrow sqrt {n + 1} geq (-1) ^ {n + 1} $

Inductive step: $$ begin {gather} sqrt {n + 1} geq (-1) ^ {n + 1} | cdot (-1) \ – sqrt {n + 1} leq (-1) ^ {n + 2} end {meets} $$ This is true, since $ – sqrt {n + 1} $ can not be $ geq -1 $.

Is that a valid test?

Example of $ sum a_n $ convergent but $ sum sqrt {a_ {2 ^ k}} $ divergent?

The problem is this:

Show that the convergence of $ sum a_n $ implies the convergence of $$ sum frac { sqrt {a_n}} {n}, $$ Yes $ a_n geq 0. $

I am using the Cauchy method:

$ sum frac { sqrt {a_n}} {n} $ converge $ iff $ $ sum2 ^ k frac { sqrt {a_ {2 ^ k}}} {2 ^ k} = sum sqrt {a_ {2 ^ k}} $ converge

Yes $ sum a_n $ converge $ implies $ $ sum sqrt {a_ {2 ^ k}} $ converge is true, I'm finished.

As $ sum a_n $ converge $ implies $ $ a_n a 0 $, at least we have $ sqrt {a_ {2 ^ k}} a 0 $. But the thing is, close. $ 0 $, squarerooting makes it bigger. So I can not use the comparison test.

But we are choosing $ a_ {2 ^ k} $ With great distance between them, it will decay much faster. I like it $ 2 ^ k $ Faster. So I'm pretty sure it will converge. However, I can not find $ N $ such that $ n> N $ it implies $ sqrt {a_ {2 ^ n}} leq a_n. $ Actually, I think there would be a counterexample (so the title is).

How do I proceed from here? And is there a counterexample?

calculation: help is needed with this integral $ int frac {6x ^ {3} + 7x ^ 2-12x + 1} { sqrt {x ^ 2 + 4x + 6}} dx $

$$ int frac {6x ^ {3} + 7x ^ 2-12x + 1} { sqrt {x ^ 2 + 4x + 6}} dx $$

I received a response from Achille Hui from this thread: How do you evaluate this integral $ int frac {6x ^ {3} + 7x ^ 2-12x + 1} { sqrt {x ^ 2 + 4x + 6}} dx $

There are still some points that I do not understand from your answer.

He says set $ B (y) = ay ^ 2 + for + c $ , RHS (* 1) becomes $ (6-3a) and ^ 3 + … $ Pin up $ a $ to $ 2 $
$ B (y) = 2y ^ 2 + for + c $ , RHS (* 1) becomes $ (- 2b-29) and ^ 2 + … $ Pin up $ b $ to $ frac {-29} {2} $

Repeat this procedure once again, we find that c must be set to 24.

I do not understand how he fixes these numbers? Do you leave them equal 0 to get the final result?