If the root of the equation ln (x + 1) + 2x-1 = 0 is x = m, then the range of M is

If the root of the equation ln (x + 1) + 2x-1 = 0 is x = m, then the range of M is

Let f (x) = ln (x + 1) + 2x-1; x > - 1
F '(x) = 1 / (1 + x) + 2 = 0; X = - 1.5
So the function increases monotonically in x > - 1;
f(0)=-1;f(1)=1+ln2>0
So from the zero point theorem, we can see that there is only one root in the interval (0,1)

If the range of function y = ln (x2 + 2x + m2) is r, then the range of real number m is r______ ．

The value range of ∵ function y = ln (x2 + 2x + m2) is r, ∵ f (x) can be obtained by taking all positive numbers, △ ≥ 0 ∵ △ ≥ 0, 4-4m2 ≥ 0, the solution is - 1 ≤ m ≤ 1, so the answer is [- 1, 1]

The second derivative of y = e ^ X / X

y' = (e^x)'/x + e^x (1/x)'
=e^x (1/x -1/x²)
y'' = (e^x)' (1/x -1/x²) + e^x (1/x -1/x²)'
= e^x [1/x -1/x² - 1/x² + 1/(2x³)]
=e^x=[1/x -2/x²+ 1/(2x³)]

The second derivative of y = (x ^ 2) (e ^ x)

y‘=2x(e^x)+(x^2)(e^x)
y''=2e^x+2x(e^x)+(x^2)(e^x)=(x^2+2x+2)(e^x)

Hilbert transformation for sin (α T + θ)
The specific method of Hilbert transformation for sin (α T + θ)

So h [sin (α T + θ)] = f [sin (α T + θ)] · [- J · SGN (ω)] = J π [δ (ω + α) - δ (ω - α)] [- J · SGN (ω)] = π [δ (ω + α) - δ (ω - α)] · SGN (ω)

What is the Hilbert transformation of M (T) = cos (20000 π T) + cos (4000 π T)?
Is there a formula for this? What is the formula?
thank you!

If a function is f (T), its Hilbert transformation is: 1 / π {∫ [f (U) / (T-U)] Du}, where: π is the circumference, and the integral interval in the braces is from negative infinity to positive infinity. Except for some special functions, the integral can not be obtained. The residue theorem is often used in the process of integral calculation

Hilbert transformation of sin (WT)

I don't know what you want to achieve, using Matlab's own Hilbert function can achieve transformation

If the complex Z satisfies Z + Z module = 2 + 8i, find the square of Z module?

It is known that M = {Z | Z module is 1, Z is complex}, n = {Z | (Z + 1 + I) module minus the square of (z-1-i) module is 4a, Z is complex}
And M intersection n is not an empty set, the value range of a is obtained

Let z = x + Yi
|z+1+i|^2-|z+-1-i|^2
=(x+1)^2+(y+1)^2-(x-1)^2-(y-1)^2
=x^2+2x+1+y^2+2y+1-x^2+2x-1-y^2+2y-1
=4x+4y
If | Z | = 1, let z = cost + isint, x = cost, y = Sint
=4(cost+sint)
=4√2sin(t+π/2)∈[-4√2,4√2]
In other words, as long as 4A ∈ [- 4 √ 2,4 √ 2], there must be a t0 such that Z0 ∈ m ∩ n
Therefore,
a∈[-√2,√2]
If you don't understand, please ask

Change the following sentences into plural
The leaf is turning yellow.
Which is your pen?
What is this?
Who is he?
Look at that book.
There is an apple on the desk.
It is a lovely dog.
Show me your book.
Better say it today! Thank you!

The leaves are turning yellow. Which are your pens? What are these? Who are they? Look at those books. There are apples on the desk. These are lovely dogs. Show us your books.