Meaning of biochemical pi

Meaning of biochemical pi

Amino acids, proteins, nucleic acids, properties of amino acids, amphotericity and isoelectric point. PI isoelectric point: in a solution with a certain pH, amino acids dissociate into cation and anion with the same tendency and degree. The net charge is zero and the solution is electrically neutral. At this time, the pH of the solution is called the isoelectric point of the amino acid
Isoelectric point
Definition: the pH of solution when the net charge of protein or amphoteric electrolyte (such as amino acid) is zero, and the mobility of protein or amphoteric electrolyte in electric field is zero. The symbol is pi.
PI represents the phosphate group, which is the farthest phosphate group from adenosine (a) in ATP molecule
Application of factorization
The hydraulic press (cylinder) has m hollow steel columns, each of which is h m in height, d m in outer diameter and d m in inner diameter. The mass of each cubic meter of steel is g tons. The total mass of M columns w? Is calculated, where H (D-D) = - 10.8, m (D + D) = 5.6, g = 7.8 and π = 3.14
1/4m【πD²-πd²】hg
=1/4×3.14m（D+d）（D-d）hg
=1/4×3.14×5.6×10.8×7.8
=370.319 tons
Find the limit of X * sin (1 / x) when x → 0
Some people say that the result is 0, but others think that the result is 1 (because when x → 0, sin (1 / x) can be replaced by 1 / x). Which result is right?
Because when x → 0, sin (1 / x) has no limit
Because 1 / X tends to infinity (it can't be replaced by 1 / x, only when 1 / X is very small), then the frequency of sine function changes very fast between positive and negative, and there is no limit, but sin (1 / x) must be between positive and negative, and it is a finite value
So I think it should be 0
Zero
First of all, why is 1 wrong? When x → 0, 1 / X →∞, sin (1 / x) = 1 / X does not hold
Sin (T) = t is true if x approaches 0, and 1 / X in the formula is equivalent to t in the formula, so the statement in the original question is not true.
Why is the answer 0
Sine function sin (1 / x), when x → 0, sin (1 / x) can take any value in [- 1, 1], it is a bounded function, and 1 / X is an infinitesimal, according to the theorem infinitesimal and... Expansion
Zero
First of all, why is 1 wrong? When x → 0, 1 / X →∞, sin (1 / x) = 1 / X does not hold
Sin (T) = t is true if x approaches 0, and 1 / X in the formula is equivalent to t in the formula, so the statement in the original question is not true.
Why is the answer 0
Sine function sin (1 / x), when x → 0, sin (1 / x) can take any value in [- 1, 1], is a bounded function, and 1 / X is an infinitesimal. According to the theorem, the product of infinitesimal and bounded function is still an infinitesimal.
The answer is 0
Because when x → 0, sin (1 / x) can be replaced by 1 / x, which is wrong
limx→0[x*sin(1/x)]=limx→0[x*sin(1/x)]=0
Because when x → 0, sin (1 / x) is a bounded function
The product of bounded function and 0 is infinitesimal 0
0 ≤ | xsin (1 / x) | ≤ | x | - - - > 0
Application of prior factorization
On the circular steel plate with radius r, punch out four small circles with radius r, and calculate the remaining area when r = 7.8cm and R = 1.1cm (π = 3.14, calculated by factorization)
The area of the remaining part
=π[R^2-4r^2]
=π(R+2r)(R-2r)
=3.14*(7.8+2*1.1)(7.8-2*1.1)
=3.14*10*5.6
=175.84
πR^-4πr^=π（7.8*7.8-4*1.1^）=175.84
1. Let f '(x0) exist and find the limit of [f (x0 + △ x) - f (x0-2 △ x)] / 2 △ x when △ x → 0
2. Let f (x) be continuous at x = 2, and if x → 2, the limit of F (x) / (X-2) is equal to 2, find f '(2)
1. Let f '(x0) exist and find the limit of [f (x0 + △ x) - f (x0-2 △ x)] / 2 △ x when △ x → 0
The original formula = [f (x0 + △ x) - f (x0)] + [f (x0) - f (x0-2 △ x) / 2 △ x
=[f(x0+△x)-f(x0)]/2△x-[f(x0)-f(x0-2△x)/-2△x
=f'（x0）/2-f'（x0）
=-f'（x0）/2
2. Let f (x) be continuous at x = 2, and if x → 2, the limit of F (x) / (X-2) is equal to 2, find f '(2)
When x → 2, X-2 → 0, and the above limit exists, so f (2) = 0
So f '(2) = [f (x) - f (2)] / (X-2) = 2
(square of 1-x) (square of 1-y) + 4xy factorization
The original formula = 1-x & sup2; - Y & sup2; + X & sup2; Y & sup2; + 4xy
=(x²y²+2xy+1)-(x²-2xy+y²)
=(xy+1)²-(x-y)²
=(xy+x-y+1)(xy-x+y+1)
Why sin (x0 + Δ x) - sinx0 = 2cos (x0 + Δ X / 2) sin (Δ X / 2)?
According to sin α - sin β = 2cos [(α + β) / 2] · sin [(α - β) / 2], it can be concluded that:
sin(x0+Δx)-sinx0=2cos(x0+Δx/2)sin(Δx/2)
Use the formula X & # 178; + (a + b) x + AB = (x + a) (x + b) to factorize A & # 178; - a-6
a²-a-6
=(a-3)(a+2)
Prove how the | cos x-cos x0 | in the limit of cosx at x0 gets 2 * | sin ((x + x0) / 2) | * | sin ((x-x0) / 2)|
This is obtained by using the sum difference product formula
The sum difference product formula: cos α - cos β = - 2Sin [(α + β) / 2] · sin [(α - β) / 2]
We obtain that | cos x-cos x0 | = 2| sin ((x + x0) / 2) | * | sin ((x-x0) / 2)|
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(X & # 178; - Y & # 178;) (x + y) - (X-Y) & # 179; factorization by formula method
The original formula = (x + y) &# 178; (X-Y) - (X-Y) &# 179;
=(x-y)[(x+y)²-(x-y)²]
=(x-y)(x+y+x-y)(x+y-x+y)
=4xy(x-y)
(X+Y)^2(X-Y)-(X-Y)(X^2+XY+Y^2)
=(X-Y)(X^2+2XY+Y^2-X^2-XY-Y^2)
=(X-Y)XY
﹙x²－y²﹚﹙x＋y﹚－﹙x－y﹚³ =（X+Y）(X-Y)(X+Y)-(X-Y)3=(X-Y)((X+Y)2-(X-Y)2)=4XY(X-Y)