The elements in matrix are all variables. How to define them in MATLAB,

The elements in matrix are all variables. How to define them in MATLAB,

We can consider the following definition methods: > > Syms a b c d%%%% define symbolic variable > > A = [a B; C D]%%%%% generate matrix A = [a, b] [C, D] > > subs (a, {a, B, C, D}, {1, 2, 3, 4})%%%%%%%%%% variable assignment ans = 1, 2, 3, 4 supplementary answer

Can matlab symbolic variables be defined in matrix form?
I want to define K symbolic variables
for i=1:100
syms (['x',num2str(i)]);
end
This can define x1, X2... X100, but I can't use X (I) when I use it later, so I want to store these 100 variables in a matrix A, let a (1) = X1... A (100) = X100, so that I can do a loop a (I), but how to realize a (I) =? It's definitely not good to fill in X (I) here, even worse to fill in Xi

>> for i=1:100
A(i)=sym(['x' num2str(i)]);
end

If A1 = 3 and a (n + 1) = the square of an, the general term formula is obtained

a1=3>0
Suppose that when n = K (K ∈ n +), AK > 0, then a (K + 1) = AK ^ 2 > 0
K is any positive integer, so for any positive integer n, an is constant > 0
a(n+1)=an^2
log3[a(n+1)]=log3(an^2)=2log3(an)
Log3 [a (n + 1)] / log3 (an) = 2
log3(a1)=log3(3)=1
The sequence {log3 (an)} is an equal ratio sequence with 1 as the first term and 2 as the common ratio
log3(an) =1×2^(n-1)=2^(n-1)
an=3^[2^(n-1)]
The general formula of sequence {an} is an = 3 ^ [2 ^ (n-1)]

A square needs 5400 pieces of cement square bricks with a side length of 4 decimeters; how many pieces of cement square bricks with a side length of 6 decimeters?

Suppose that if the side length is 6 decimeters, X bricks are needed, 6 × 6x = 4 × 4 × 5400, & nbsp; 36x = 86400, & nbsp; & nbsp; & nbsp; X = 2400; a: 2400 bricks are needed

1+3+5+7+9+11+13=___ 2．

1 + 3 + 5 + 7 + 9 + 11 + 13 = (1 + 13) × 7 / 2 = 14 × 7 / 2 = 49 = 72, so the answer is: 7

The measurable function sequence xn, YN, where xn exists everywhere and is finite in X. it is proved that the upper limit of XN + yn on N tending to infinity is equal to the upper limit of X + yn on N tending to infinity

Do you want to say that the limit of XN exists almost everywhere and is x?
From the properties of upper limit, it is easy to know that there is a subsequence NK such that limk (xnk + ynk) limit exists and is equal to the upper limit of XN + yn tending to infinity in N. because xnk limit exists, ynk limit also exists and is less than or equal to the upper limit of YN, so the left

Given that the center of the circle is on the x-axis, the radius is 5 and the chord length with a (5,4) as the midpoint is 25, then the equation of the circle is ()
A. (x-3) 2 + y2 = 25B. (X-7) 2 + y2 = 25C. (x ± 3) 2 + y2 = 25d. (x-3) 2 + y2 = 25 or (X-7) 2 + y2 = 25

Let the center of the circle be C (a, 0), the radius of the circle be r = 5, and the length of the chord be BD = 25. According to the vertical diameter theorem, we can get that AC is perpendicular to the chord BD, a (5, 4), a (5-a) 2 + 42 + 5 = 25, a = 3 or a = 7, then the equation of the circle is (x-3) 2 + y2 = 25 or (X-7) 2 + y2 = 25

Given the function y = x + 16 / x + 2, X ∈ (- 2, positive infinity), find the minimum value of this function
Given the function f (x) = x + 16 / x + 2, X ∈ [6, positive infinity), find the minimum value of this function

When x ∈ (- 2, positive infinity)
y=x+16/(x+2)
=x+2+16/(x+2)-2
≥ 2 √ 16 (x + 2) / (x + 2) - 2. Inequality properties
=8-2
=6
Minimum = 6
When x ∈ [6, positive infinity)
f(x)≥6+16/(6+2)=6+2=8
Minimum = 8

When can the limit be solved first? For example, when x tends to 0, (1 + xsinx cosx) / (1 + xsinx + cosx) = (1 + xsinx cosx) / 2

Part is weaker than the whole, part is weaker than the whole, this is the principle: most limit problems can be like this: 1. For X → 0 type, use Infinitesimal Substitution Method: such as SiNx TaNx; 1 - e ^ x, etc.; if the global limit is normal, such as not 0 or infinite, it is generally the correct result, if not

A rectangular vegetable field is twice as long as it is wide, and its perimeter is 54 meters. How many meters are the length and width of this vegetable field?

Set the width as X meters
X+2X=54÷2
3X=27
X=9
Length: 54 △ 2-9 = 18m