math problem help needed!

[darkforce]

f(x) | (x+1) 余数是 1
f(x) | (x-1) 余数是 3

 

[猫罐头]

这回计算对了得1

 

[darkforce]

~~~请教一下怎么算的？？

 

[猫罐头]

最后有两个式子

f^2(x)/(x-1)(x+1)=3; 3/(x+1)(x-1)=(?)=f(x)/(x-1)(x+1)

2式子一乘,同项相削就行了

 

[darkforce]

最初由 Sceleton JACK 发布
最后有两个式子

f^2(x)/(x-1)(x+1)=3; 3/(x+1)(x-1)=(?)=f(x)/(x-1)(x+1)

2式子一乘,同项相削就行了

好难啊

 

[猫罐头]

最初由 darkforce 发布


好难啊

那怎么办?心领神会吧

 

[白米饭]

最初由 Sceleton JACK 发布
最后有两个式子

f^2(x)/(x-1)(x+1)=3; 3/(x+1)(x-1)=(?)=f(x)/(x-1)(x+1)

2式子一乘,同项相削就行了

你的2个等式是怎样算的？
那里求出余数了？？

 

[darkforce]

f(x)/(x+1)不等于1

 

[白米饭]

First, we know f(x) = q(x)g(x) + r(x)
Now, we have g(x) = (x+1)(x-1), find r(x)?

And we also know f(-1) = 1, f(1) = 3, sub in the first equation
f(-1) = q(-1)(-1+1)(-1-1) + r(-1) = 1
f(1) = q(1)(1+1)(1-1) + r(1) = 3

then becomes r(-1) =1 or r(1) = 3, since (x+1)(x-1) is a polynomial with degree of 2
then r(x) is a polynomial with dregree of one or zero(constant)
so, assume r(x) = ax + b, then
1 = a(-1) + b
3 = a(1) + b

...... problem solve

 

[garion]

How did you get 1??

If f(x)=a(x^2-1)+1, then plug in x=1 and x=-1, both should give you 1, contradictory to what's given in the problem f(1)=3!!

Since you are given two initial conditions, and you have f(x)=m(x+1)+1=n(x-1)+3, you should be able to uniquely solve this equation set and get f(x)=x+2. As of the remainder divided by x^2-1, it's unclear to me how it can be defined given x is an arbitrary real number.

最初由 Sceleton JACK 发布
这回计算对了得1

 

[garion]

okay, got it. this sounds right. that basically gives you x+2 as the remainder.

最初由 白米饭 发布
First, we know f(x) = q(x)g(x) + r(x)
Now, we have g(x) = (x+1)(x-1), find r(x)?

And we also know f(-1) = 1, f(1) = 3, sub in the first equation
f(-1) = q(-1)(-1+1)(-1-1) + r(-1) = 1
f(1) = q(1)(1+1)(1-1) + r(1) = 3

then becomes r(-1) =1 or r(1) = 3, since (x+1)(x-1) is a polynomial with degree of 2
then r(x) is a polynomial with dregree of one or zero(constant)
so, assume r(x) = ax + b, then
1 = a(-1) + b
3 = a(1) + b

...... problem solve

 

[darkyun]

the remainder should be a constant.

 

[garion]

no. see the three posts directly above.

最初由 darkyun 发布
the remainder should be a constant.

 

[白米饭]

This is the only trick in this problem:

(x+1)(x-1) is a polynomial with degree of 2
then r(x) is a polynomial with dregree of one or zero(constant)

 

[猫罐头]

了解了，真是犯了个大错，贻笑大方了。不好意思楼主。

不过，这样一来，题就简单了，成了个纯算的题了。

let f(x)/(x-1) + 1=a; f(x)/(x-1) + 3=b

然后将f(x),x换成以a,b 为未知数的多项分式方程，再最后带如上式，解出a,b.

再带入。。。就什么都解了。

我以解出a,b了；但由于计算量较大。建议各位想一个简洁的方法。

我这个是下下策。