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]]]
``````

Then because it does

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

Give only one of the 3 previous cases?

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?

\$ 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?