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**Yusuke00**- Replies: 2

Ok so i have the next exercise.

I didn't had any problems solving a) and b)

What about c?

Can anyone tell me how to resolve it?

I googled it and found out that's Si(x)(whatever that is) and i don't know about what's that since i'm just last grade of high school.

Any help?

Indeed,good point you are right.

http://www.mathsisfun.com/data/function-grapher.php?func1=sqrt%28x%29&func2=2

I know the problem is to prove that x does not have roots on (-1,0) because it's easy to see on the other cases.Ideas?

In my opinion the first one is always positive for any x real but i don't really know how to prove it.

Not any opinions/ideas yet?

It's N* sorry math. I don't really know how to write N* or R*+.How you do that?

Done yeah cool now i found out how it works. hehe

**Yusuke00**- Replies: 13

Hey guys.I have a little question or just a personal wonder about them.

Question is: for any ?

What about ?

Test:

integral of 1/(sin^4x+cos^4x)dx

f:[0,3]->R f(x)=max{3-x,2x+[x]} . Show that f is integrable on[0,3] and calculated integral of f(x)

integral of sqrt(x-x^2)

ok i understand....thank you!

**Yusuke00**- Replies: 16

integral of 1/(x^4+x^2+1):D:D

It means that f(x) =y,y is part of R.

But it's very helpful,i like it.Thank you very much ^_^.

For a>=1 there are real solutions.For a<1 there are not.That function grapher is for R not for C.

f(x1)=0,f(x2)=0,f(x3)=0,f(x4)=0;

Sorry that's how i know to note them.

Yes i had the same ecuations,but i've been tired and just couldn't focus yesterday.Got them now thank you.

By the way,any clues on 1st problem?

Ok i got those ecuations either.W/E i'm stupid.Thank you.

bobbym wrote:

Hi;

f(1)=9

f(2)=-5

f(3)=0

f(4)=35

Could you describe your solution step by step please? For f(1)=8 not 9 sorry.

Ok guys a second problem

f(1)=8

f(2)=-5

f(3)=0

f(4)=35

f=?

The thing is i know how to solve the problem but i cannot really solve the system.

f(3)=0 -> x1=3 solution ->f = (X-3)(aX^2+bX+c)==>

==>f=aX^3+(b-3a)X^2+(c-3b)X-3c

so f(3)=0 ==>-3c=0 =>c=0

f(1)=8 ==> a+b=-4

f(2)=-5 ==>-4a-2b=-5==>4a+2b=5

f(4)=35 ==> 16a+4b=35

Can anyone solve the system or at least tell me where i'm wrong ? I'm getting very mad here.

**Yusuke00**- Replies: 14

Hello guys long time no see,got a new exercise:

On R[X] f=X^4-aX^3-aX+1

Proove that if |a|<1 then |x1|=|x2|=|x3|=|x4|=1;

Good luck.

How do you write mathematical here?

I mean with all the symbols and etc.

Interesting proof,i thought of a different one.