A question in Advanced Mathematics 1, ask 1 High School Mathematics

x→∞"Indicates that x tends to infinity. It's not an actual number, it's just a concept, because any specific number you give has a larger number.

It can also be understood in this way: a number line marked with an origin is the "end" of the number line straight up. It is also a point that does not exist.

x xo means that x is infinitely close to may not exist, but f(x) the limit exists when x xo.

x stands for x infinity or infinitesimal size. represents infinity (infinity or infinitesimal small) x - represents negative infinity (- represents infinitesimal infinitesim).

x + stands for positive infinity (+ stands for infinity).

x xo (x tends to x zero?) Illustrated in terms of positive and negative.

When xxo, x tends to x zero from the right.

xo is a constant and x is a variable. xo can be any number on the number axis but relatively"x", equivalent to"0"

For example, if you set the number line can be xo=5 and x=9

It can also be said that (relative to x) xo=0 and x=4

x xo means that x tends to xo from 9, that is, infinitely tends to 5, but cannot be less than 5

"x" means that x is infinitely close, but not equal to

The limits don't have to be all "x"."x→1""x→-1000"...Both.

For example, if x tends to 1, after finding the limit, you can substitute 1 into x, which is a method, and sometimes you can directly substitute it inside!

I'm still very good at advanced mathematics, and I can't express it very accurately in words, anyway, when I took the self-exam, I got a high score in advanced mathematics!

x means that x tends to plus or minus infinity, which is equivalent to defining a set of spaces, that is, the range of x values tends to be negative infinitesimal at the left end of the number line, and the right end tends to be positive infinity, as shown in the following figure (x is the part of a short solid line, the leftmost end is , and the rightmost end is ).

—m)…0…(m)——

It's that x tends to infinity, that is, infinity, I don't know if you're asking about this.

Solution: Because.

s=a1 (1-q),sn=s(1-q n),sn-2s=-s(1+q n),lim(sn-2s)=1,slim(1+q n)=1, infinite proportional series, 0<|q|<1,lim(q^n)=0,∴ s=-1,a1/(1-q)=-1,q=a1+1.

0<|a1+1|<1, the value range of the first term a1 of -2 is (-2,-1) (1,0).

If the clearing point is changed to (x,y), then pa 2+pb 2 answer judgment (x 1) 2 (x 1) 2 2y 2 2 (x 2 y and this 2) 2, so only the minimum value of x 2 y 2 is required. Since xy is on the circle (x-3) 2+(y-4) 2=4, the geometric meaning of x 2 y 2 is the minimum distance from the circle 5 2 3, so the minimum value of pa 2 + pb 2 is 2 3 2 8

This question can be decided by drawing a picture. First, draw an image of f(x), which is also in the range [0,1].

At x=1 2, zero. It can be seen that f(x)=1 2 corresponds to two solutions, and it happens to be at the symmetrical distribution position at point (,0), and f(f(x))=1 2 requires that there are two different solutions for f(x) in it, and these two different connections correspond to four different solutions, so it can be imagined that f(f(f(f(x))) has eight different solutions. So we make a bold guess that the number of solutions is a proportional series of numbers with 2 as the base.

That is, to the 2nd power, prove it by mathematical induction:

f(x)=2, suppose there are 2 n different solutions for f[n](x), in f[n+1](x), each of these 2 n solutions corresponds to two different solutions, so there are 2*2 n, that is, 2 (n+1).

Answer: 2n Analysis: Use the combination of numbers and shapes.

It can be seen that the value range of f(n-1)(x) is the defined domain of fn(x)!

From the image, we know that f1 (x) has two segments [0,1], so f2 (x) has four segments, and so on, you can get the answer!

In a series where terms are positive, the first n terms of the series and s n satisfy s n = 1 2 (a n + 1 a n).

1) Seek a, a, a; (2) the general formula for conjecturing the sequence of numbers from (1); (3) Find s n

Solution: a = s = (1 2) (a +1 a) = (1 2) (a +1) a

Therefore 2a = a +1, a =1

s₂=a₁+a₂=1+a₂=(1/2)(a₂+1/a₂)=(1/2)(a₂²+1)/a₂

Therefore there is 2a +2a =a +1,, a +2a -1=0, a =(-2+ 8) 2=-1+ 2

s₃=a₁+a₂+a₃=1+(-1+√2)+a₃=√2+a₃=(1/2)(a₃²+1)/a₃

2(√2)a₃+2a₃²=a₃²+1,， a₃²+2(√2)a₃-1=0，∴a₃=(-2√2+√12)/2=-√2+√3.

a₁=1；a₂=√2-1， a₃=√3-√2, .a‹n›=√n-√(n-1)

Therefore s n = 1 + ( 2-1) + (3 - 2) + (4 - 3) +n-2)+√n-3)]+n-1)-√n-2)]+n-√(n-1)]=√n

(1) s1=1 2(a1+1 a1) gets 2a1=a1+1 a1 gets a1=1 a1 so a1=1

s2=1 2(a2+1 a2) This gives a2= -1;

and so on a3=1;

(1) Substituting n for 1 yields a1=1

Then a2 = root number 2-1, a3 = root number 3 - root number 2

2) Conjecture an=root-n-n-n-n>-n

3) sn=root number n+1

Solution: (1) A1=S1=1 2(A1+1 A1) is solved by a1=1

From a2=s2-s1=1 2(a2+1 a2)-1 2(a1+1 a1) (a2) 2+2(a2)=1 a2=1 a2=(root number 2)-1.

From a3=s3-s2=1 2(a3+1 a3)-1 2(a2+1 a2) (a3) 2+2 (root number 2)(a3) = 3 solution a3 = (root number 3) - (root number 2).

2) From (1) conjecture, an=(root n)-(root number(n-1)).

3) So sn=1+((root2)-1)+(root3)-(root2))+rootn)-(root(n-1))))=rootn.

Yes, draw the two function images within the agreed coordinate system.

It can be seen that when x<1 and x > another intersection value, y=a x is greater than y=-x+1;The rest is less than.

In this problem, only the part of x>0 is taken, which is the right half axis of the y-axis, then y=a x is greater than y=-x+1

Let f(x)=a x-(-x+1)=a x+x-1f(x)derivative=a x*lna+1

Because 00, a x <1

A x*LNA<0 is indeterminate in size.

If a=1 e, then LNA=-1 x>0 a x<1 -10

If a=1 e 2 then LNA=-2 x approaches 1, -2 so the monotonicity is uncertain.

And because f(0)=0, y=a x is not necessarily greater than y=-x+1

Look at the tangent slope of y=a x at x=0.

Slope k = a x·ln(a).

k(0)=ln(a)

When LNA-1 i.e., A1E.

In line with the topic. In summary, 1 e a 1

Two properties of the cross product are used here: a a=0 and a b=-b a (almost obvious).

emm, no problem, I don't know where your question is, speculate:

axb=-bxa

axa=0 and above are vectors.

It's OK to understand the algorithms (I don't know if they're called algorithms, just those commutative laws, associative laws, and distributive laws).

From the inscription, it can be seen that the sum of the areas of the two parts is 3 2 which is integrated with the x-axis, which is exactly a rectangular area, and the area is xf(x). As shown below: