leehuan's All-Levels-Of-Maths SOS thread (3 Viewers)

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leehuan

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MX2 problem. Required - c and d but all parts are given

 

leehuan

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I will admit this is just a bit of confusion in my head and laziness to sort it out.

 

leehuan

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Only the case for (-1,0) but the whole question is again given (with a previous part)

 

Paradoxica

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Only the case for (-1,0) but the whole question is again given (with a previous part)



This is probably the identity in question required to prove that statement given the nature of the question.
 
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Paradoxica

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The most straightforward way, however, would be to re-write the polynomial as a sum of perfect squares with zeroes that occur at different x-values.



as none of the terms all simultaneously share the same zero point over the reals, it is impossible for this expression to ever be zero, and hence it is strictly positive over the reals.
 
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seanieg89

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Alternatively alternatively:

P(x)=(x^5-1)/(x-1) for x =/= 1.

But x^5 is increasing, so the numerator has the same sign as the denominator and P(x) > 0 apart from possibly at x=1.

(And P(1)=5 so we are positive here too.)

Edit: Haha, Integrand beat me to it.
 

leehuan

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Lol thanks everyone. I ended up doing this:

Let x=-b, 0<b<1
RTP:1-b+b^2-b^3+b^4

1>b implies 1-b>0
b>0 allows multiplying by b^3: b^4-b^3>0
b>0 also implies b^2>0
Sum them up.
________________

Getting confused with what I learnt. Someone please break down surjective functions for me, and injective just means one-to-one both x and y (therefore inverse function exists for the maximum domain) right?
 

seanieg89

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Lol thanks everyone. I ended up doing this:

Let x=-b, 0<b<1
RTP:1-b+b^2-b^3+b^4

1>b implies 1-b>0
b>0 allows multiplying by b^3: b^4-b^3>0
b>0 also implies b^2>0
Sum them up.
________________

Getting confused with what I learnt. Someone please break down surjective functions for me, and injective just means one-to-one both x and y (therefore inverse function exists for the maximum domain) right?
When you specify a function f: X->Y you are also specifying the domain X and the co-domain Y.

With this in mind:

f is said to be injective if any two distinct elements of X get sent to distinct elements of Y by f.

f is said to be surjective if for every y in the codomain Y there exists an x in X with f(x)=y. Ie f "hits everything" in the codomain.

f is said to be bijective if it is both injective and surjective.
 
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