Solving inequality system The square of M-1 is less than the square of 2m-1 The absolute value of 00 < 2m-1 < 2 Find the value range of M

Solving inequality system The square of M-1 is less than the square of 2m-1 The absolute value of 00 < 2m-1 < 2 Find the value range of M

It can be seen from the condition that (m-1) ^ 2 - (2m-1) ^ 2 < 0
That is (m-1-2m + 1) (m-1 + 2m-1) < 0
m(3m-2)>0
Then M > 2 / 3 or m < 0
(2m-1) ^ 2 < 4
So - 1 / 2 to sum up, - 1 / 2
Question:
What is: - 2 < 1-2m < 2m
Answer:
-1 / 22 / 3 or m < 0 gives - 1 / 2
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Approximate calculation of differential How to calculate the approximate values of Radix 65 and 997

By using the differential approximation, f (X. + x) ≈ f (X.) + F '(X.)} x, where │x │

Differential approximation The normal section of the water pipe wall is a ring. Suppose its inner diameter is R and its wall thickness is h, the approximate value of the ring area is calculated by differential calculus (H is quite small) This is the limit of the solution, the key to this problem is to use differential calculation, can you give a differential approximate calculation method to solve this problem? Please give the solution process

Area: S = 1 / 2 * π * (R + H) ^ 2-1 / 2 * π * H ^ 2 = 1 / 2 * π (R * (R + 2H)) = 1 / 2 * π * (R ^ 2 + 2RH) = 1 / 2 * π R ^, because the value of the latter formula can be ignored when h - > 0

Differential operation rules, compound functions for differential For example, y = x · sin2x

y=x·sin2x
dy=（sin2x+2xcos2x）dx

Approximate value ln0.98 by differential approximation formula

Let f (x) = LNX, X1 = 1, Δ x = - 0.02
∵f'(1)=1/1=1
∴ln0.98 ≈f(1)+f'(1)Δx=0+1*(-0.02)=-0.02
Primary school students' Math group,

Differential of y = 3 ^ cosx

-ln3(3^cosx)sinxdx

Differential of y = e ^ (- x) cosx

dy=de^(-x)cosx
=cosxde^(-x)+e^(-x)dcosx
=cosxe^(-x)d(-x)+e^(-x)(-sinxdx)
=-cosxe^(-x)dx-sinxe^(-x)dx
=-(cosx+sinx)e^(-x)dx

Differential of y = xsin2x + cosx

dy=d(xsin2x)+dcosx
=sin2xdx+xdsin2x-sinxdx
=sin2xdx+xcos2xd2x-sinxdx
=sin2xdx+2xcos2xdx-sinxdx
=(sin2x+2xcos2x-sinx)dx

Approximate value by differential: √ (1.05)

Let f (x) = (1 + x) ^ 0.5 be expanded at zero Taylor
f(x)=f(0)+f'(0)x+0.5*f"(0)x^2+o(x^2)≈1+0.5x-0.5*0.25x^2
x=0.05
f(x)≈1.02496875
Error analysis:
If x ^ 2 = 0.0025, the order of magnitude of the error is thousandth
The actual value is 1.024695 and the actual error is 0.00027
If you want higher accuracy, just keep a few more terms when Taylor expansion
above

What is the differential approximation formula in higher numbers?

f(x)=ln(1+x)
df(x)=dx/(1+x)
When x is very small, f (x) - f (0) ≈ f '(0) * x = x / (1 + 0) = X
The formula is summarized as follows
ln(1+x))≈x
Take x = 0.01
ln(1.01)≈ln(1+0)+0.01=0.01