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There is a very extreme method, you can assume a function with four unknown coefficients, and replace the number given above to determine all coefficients.
Like F(x)=aX^3+bX^2+cX+d, then when x=1, f(1)=1.
x=2 , f(2)=5 , x=4, f(4)=20, x=6, f(6)=100
So,there are 2, 3, base cases induction right? That will be very kool and convenient!
Fibonacci Number(f_(n-1)+f_n=f_(n+1)) , f_0=0 , f_1=1, prove that f_n=f_(k+1)*f_(n-k)+f_k*f(n-k-1)
I want to use induction to prove that , so I need to assume when n=k is true , then move on to prove k+1 is true , but in this case , I need to assume k and k+1 are true , but can I assume k and k+1 are true , it's a little wierd assuming too much .
Let m , n and k be positive integers. Prove that v_p(mn)=v_p(m)+v_p(n) , (we define v_p(n) as the greatest ineger r such that p^r divides n.
My proof is :
Let v_p(mn)=p^r that divides mn , if (p^r, m)=1 , then p^r must divides n , then v_p(m)=0 , v_p(n)=p^r , so v_p(mn)=v_p(m)+v_p(n), vice versa
If (p^r, m) not equal to 1 , and so is (p^r, n) , there exists i , k such that p^r=p^i*p^k (p^i,m)=p^i , (p^k,n)=p^k,then r=i+k
v_p(mn)=v_p(m)+v_p(n). ,
CHeck my proof , I am unsure of the second part.
I don't understand why this proof holds , the logic is so strange
Prove:Every positive integer can be written as a product of primes
Here is the proof
Since an empty product is equal to 1 , we can write 1 as the empty product ofprimes , Let n>=2 , suppose that every postive integer less than n is a product of primes . If n is prime , we are done. If n is composite , then n=dm , where 1<d<m<n .By the induction hypothesis , d and m are both products of primes ,and so n=dm is a product of primes .
What I don't understand is that ,why induction can be used like that , It seems all proof is based on hypothesis , I wonder that will holds.
you need to draw out all the line , first ignore the < or > , draw 3x-y=6 , x=1 ,y=3 . 3x-y<=6 divides the plane in two sides , only one of them satisfies the need. Notice that (0,0) is above the line, assume x=0. y=0, then the inequality 3x0-0<6 , it satisfies, so it's the upper area , then find the intersection of the three , done .
What? how can the segment CF besect any angle? Name the angle gamma ,I don't really understand ,lol
Yeah , I am working on that now,don't know if I can solve it though
Can anyone draw a picture of it , I am confused since my English is poor
Is it Linear ? Impossible to tell from only two statist
y=1/(x+1) , hehe
Eliminate the x or y and solve the equation
slope same , intercept different
I don't know the answer either ,lol
I divide two sides by 2^n, then A(n-1)/2^(n-1) + An/2^n=1
Denote An/2^n as another sequence Bn , then Bn+B(n-1)=1
you can add a minor to it
B2+B1=1
-B3-B2=-1
.......
IF N is even the last one will be Bn+B(n-1)=1 , add altogether Bn+B1=1 , thus Bn=1-B1
An=2^n - 2^(n-1)*A1
If N is odd , the last one will be -Bn-B(n-1)=-1 , add altogether ,-Bn+B1=0 , thus Bn=B1
An=2^(n-1)*A1
That's what I did ~lol
Oh ,yeah, I can assume x, y in C , instead I assume them in other wrong places , It's really confusing at the beginning ~
and another one , Let a,b,c,d, be integers such that ad-bc=1 , for integer u, v define
m=au+bv , n=cu+dv . Prove that (m,n) = (u,v)
here is the question Prove that the intersection of a family of subgroups of a group G is a subgroup of G.
Hmmm. when x-->0 , f(x)--->c , x=0, f(0)=c ,The same as mathsyperson , suggest y=x+K
Wow , there is such wonderful thing? I totally miss it. (quickly write down the forumla)
tranform it into cos10cos20cos40=sin10cos10cos20cos40/sin10=sin20cos20cos40/2sin10=sin40cos40/4sin10=sin80/8sin10=(1/8)*(sin80/sin10)=1/8*cot10. LOL , obviously it's not equal to 1/8~
slope? you mean the k in y=kx+b ? if two k in two lines have this kind of relation (use m to denote the diff k) km=-1 then two lines are perpendicular, if k=m then paralell. I think this is what you are looking for right?
I don't understand why the intersection of groups is still a group ?
A steel chain, whose mass measured m per unit , is placed on ground. Now grab one end of the chain and lift it vertically ,making it rise at a constant speed v , when the end of the chain is at height x, determine the grabing force F at this moment.
I did F=m(x*g+1/2*v^2) but the correct answer is F=m(x*g+v^2)
Here is what I did
there must be a acceleration process , so assume the end of the chain went through a distant L (limL--->0) before it reaches the speed v . when at the height x , it will require the force of it's weight to keep the speed v , so the F=xmg+gL
We go back to the acceleration process , the work the force does equal to TL=1/2*L*m*v^2
T=gL , after tranformation gL=1/2*m*v^2
substitute the gL , F=m(x*g+1/2*v^2) , but the real answer doesn't contain that 1/2, so I wanna know what had gone wrong?
Correct~ anyone attempt the second question?
Oh , Thank you anyway. I got to gain more exp lol