An advanced algebraic proof question, find an algebraic proof question and answer

Using the integral median theorem: 0qmqux e (t 178;） dt ＝ x ＊ e＾（ξ178；) is between 0 and x. ＝ x ＊ e＾（ x＆＃178；), where 0,1 when t gt; At 0, let f(t) e (t 178;） f ＆＃39；（t） ＝2t e＾（t＆＃178；） gt；", f(t) is strictly single-incremented, so that in the above equation is the only 1 47;x＆＃178；） ln【∫ 0，x］ e＾（t＆＃178；） dt ＆＃47； x】 lim（x－＆gt；＋∞lim（x－＆gt；＋∞1＆＃47；x＆＃178；） ln【∫ 0，x］ e＾（t＆＃178；） dt ＆＃47； x】＝ lim（x－＆gt；＋∞1＆＃47；x＆＃178；） ln（ ∫0koswx］ e＾（t＆＃178；） dt ） ln x 】＝lim（x－＆gt；＋∞e＾（x＆＃178；） 47； ∫0，x］ e＾（t＆＃178；） dt － 1＆＃47；x 】＆47；(2x) Lopida's law lim(x gt; ＋∞x e＾（x＆＃178；） 0，x］ e＾（t＆＃178；） dt 】＆47； （2 x＆＃178； ∫0，x］ e＾（t＆＃178；） dt ）＝lim（x－＆gt；＋∞2 x＆＃178； e＾（x＆＃178；） 47； 【2 x＆＃178； e＾（x＆＃178；） 4 x ∫ 040x］ e＾（t＆＃178；） dt 】＝62c。

Twice Lopida's Law 1

Left end =

0 1+x[1] 1+x[1]² 1+x[1]ⁿ0 1+x[2] 1+x[2]² 1+x[2]ⁿ0 1+x[n] 1+x[n]² 1+x[n]ⁿ1 x[1] x[1]² x[1]ⁿ

1 x[2] x[2]² x[2]ⁿ

1 x[n] x[n]² x[n]ⁿ

0 x[1] x[1]² x[1]ⁿ

0 x[2] x[2]² x[2]ⁿ

0 x[n] x[n]² x[n]ⁿ

1 x[1] x[1]² x[1]ⁿ

1 x[2] x[2]² x[2]ⁿ

1 x[n] x[n]² x[n]ⁿ

x[1] x[1]² x[1]ⁿ

x[2] x[2]² x[2]ⁿ

x[n] x[n]² x[n]ⁿ

1 x[1] x[1]² x[1]ⁿ

1 x[2] x[2]² x[2]ⁿ

1 x[n] x[n]² x[n]ⁿ

2·x[1]·x[2]·.x[n]·

1 x[1] .x[1]ⁿ⁻

1 x[2] .x[2]ⁿ⁻

1 x[n] .x[n]ⁿ⁻

1 x[1] x[1]² x[1]ⁿ

1 x[2] x[2]² x[2]ⁿ

1 x[n] x[n]² x[n]ⁿ

2∏ x[i] ·x[k]-x[j]) x[i]-1) ·x[k]-x[j])

x[k]-x[j]) 2 x[i] -x[i]-1)) right end. The penultimate equal sign uses the product of the vandermonde determinant.

There are two ways to prove it.

Algebraic reasoning. Because.

n(n+1)=n^2+n

So left = (1 2 + 2 2 +.n^2)+(1+2+3+..n) 1 2 + 2 2 ...n^2=n(n+1)(2n+1)/61+2+3+..n=n(n+1)/2

So it's a paragraph. Left.

n(n+1)(2n+1)/6+n(n+1)/2n(n+1)(n+2)/3

Mathematical induction.

When n=1, 1*2=2, 1*(1+1)*(1+2) 3=2, the equation is broken.

Let the equation hold when n=m (m is a positive integer), i.e., 1*2+2*3+.m(m+1)=m(m+1)(m+2)/3

Then when n=m+1, left.

1*2+2*3+..m(m+1)]+m+1)(m+2)m(m+1)(m+2)/3+(m+1)(m+2)(m+1)(m+2)(m+3)/3

The equation also holds.

It is proved by the proposition.

I used it. n(n+1)=[n+2)(n+1)n-(n+1)n(n-1)]/3

That is, to build dates.

The loss is based on the original formula = (3*2*1-2*1*0) 3....n+2)(n+1)n-(n+1)n(n-1)] 3 intermediate terms can be eliminated before they are removed.

n+2)(n+1)n/3

The first thing to remember is the formula.

1+2+3+..n=1/2n*(n+1)

1^2+2^2+3^2+..n^2=1/6n*(n+1)*(2n+1)

Then you can come to the answer to the delay.

The original formula can be turned into a dust line mode:

Principle n*(n+1)=n 2+n

1^2+2^2+3^2+..n^2)+(1+2+3+..n) Then press the tape and follow the formula.

That's it.

Divide it into two determinants, the first one puts a as the final determinant multiplied by a cubic multiplication, and the latter raises b

Prove it by counter-evidence.

Assuming that (x n-1) is not divisible by f(x n), then there is q(x)<>0 and k<>0 such that.

f(x^n)=(x^n-1)q(x)+k

x-1)[x^(n-1)+x^(n-2)+…1] q(x)+k, i.e., p(x)=[x(n-1)+x(n-2)+....1] q(x)<>0, and k<>0 make it.

f(x n) = (x-1)p(x)+k holds. This contradicts the known (x-1) divisible f(x n).

Therefore, the assumption is not valid.

then (x n-1) is divisible by f(x n).

See image below

If you have any questions, please feel free to ask @

It's about validating by definition.

1.Symmetry.

For any f,g c[a,b], f,g) = f(x)g(x) dx = g(x)f(x) dx = (g,f)

2.Bilinear.

For any f,g,h c[a,b], f,g+h) = f(x)(g(x)+h(x)) dx

f(x)g(x) dx+∫ f(x)h(x) dx = (f,g)+(f,h).

For any real number c, (f,c·g) = f(x)(c·g(x)) dx = c· f(x)g(x) dx = c· (f,g).

Thus (,) is linear for the second component, and by symmetry, it is also linear for the first component.

3.Positive certainty.

For any f c[a,b], f,f) = f(x) dx 0(f is a real function, so f(x) 0).).

is not constant zero for f, and there is t a,b], so that f(t) ≠0.

Consecuted by f(x), there exists an interval [r,s] (r < s) containing t, such that |f(x)| f(t)|2 is established on [r,s].

then (f,f) = f(x) dx f(x) dx f(t)|/2)² dx = (s-r)f(t)²/4 > 0.

Therefore (f,f) = 0 if and only if f is constant equal to zero on [a,b] (i.e., a zero element in linear space c[a,b]).

It is only necessary to prove that a has n linearly independent eigenvectors, and according to the knowledge of the higher generation, the eigenvectors corresponding to different eigenvalues are linearly independent, so only the sum of the eigenvectors corresponding to different eigenvalues needs to be n.

If 2009 is the eigenvalue, the number of linearly independent eigenvectors for the Fung response is equal to the dimension of the solution space of (a-2009e)x=0, i.e., n-rank(a-2009e);

The same is true for 2010 and 2011, so the sum of the number of eigenvectors with 2009, 2010, 2011 as eigenvalues is 3n-rank(a-2009e)-rank(a-2010e)-rank(a-2011e)=n, so it can be diagonalized. It should be noted that even if one of the numbers is not an eigenvalue, it does not affect the result, because it is equivalent to seeing the "eigenvalue" of the empty space.

If you promote, rank(a-k1e)+rank(a-k2e)+.rank(a-kme)=(m-1)n, a similar method can be used to prove diagonalization.

1. A is positively definite, then there is a nonsingular array g such that a=g tg, so det(xa-b)=det(xg tg-b)=det(g t)det(xe-g (-t)bg (-1))det(g), so det(xa-b)=0 is equivalent to det(xe-g (-t)bg (-1)))=0, when the eigenroots are all greater than -1, that is, the eigenvalues of g (-t)bg (-1) are all greater than -1, so e+g (-t)bg (-1). ) is a positive definite matrix, so a+b=g t(e+g (-t)bg (-1))g is a positive definite matrix. Otherwise, just go backwards.

2. Obviously, ker(t2) is contained in ker(t1t2), and any x is located in ker(t1t2), that is, t1t2x=0, so t2x belongs to ker(t1), and obviously t2x is located in im(t2) at the same time. If ker(t1) intersects im(t2) as 0, then t2x=0, so ker(t1t2) is included in ker(t2). Conversely, if ker(t1t2) is included in ker(t2), it is proved that ker(t1) and im(t2) are intersected as 0.

Let y be at both ker(t1) and im(t2), i.e., t1y=0, and x, so that y=t2x, then t1t2x=t1y=0, i.e., x at ker(t1t2)=ker(t2), then t2x=0, then y=t2x=0. And so the conclusion was established.

The sub-questions (1) and (2) of question 1 are easy to prove, and can be verified directly with the definitions of subspace and invariant subspace. Below is sub-question (3).

Proof of 2 questions.

Proof that if x,y belong to t-1(0), then t(ax+by)=at(x)+bt(y)=a*0+b*0=0, that is, ax+by belongs to t-1(0), so t-1(0) is a subspace of v.

Let x,y belong to t(v), then there is a, b belongs to v so that t(a)=x,t(b)=y, so k1x+k2y=k1t(a)+k2t(b)=t(k1a+k2b) belong to t(v), so t(v) is a subspace of v.

2) t(t-1(0))=0 belongs to t-1(0), so t-1(0) is an invariant subspace of t, and t(t(v)) belongs to t(v), so t(v) is also an invariant subspace of t. These two spaces are mundane invariant subspaces of t.

3) The conclusion of the rank-zeroing degree theorem and 1) is used to prove it.

The eigenequation is x 2=x, so the eigenvalue is 0 or 1

2) t(a-t(a))=t(a)-t 2(a)=t(a)-t(a)=0, so it is included in t-1(0).

On the other hand, if any given b belongs to t-1(0), then b=b-t(b), then t-1(0) is included in.

Proven. 3) Use 3 of 1 question), transfer t-1(0) to t(v)=

Counter-proof: If it is not true, then a, b belong to v so that a-t(a)=t(b)≠0, so t 2(b)=t(b)=t(a-t(a))=0 contradiction!

Therefore there must be t-1(0) and t(v)=, so that the proposition is proved.