Melody

Thanks Heureka,

you are asked for tn  when n = 345

n is a function of k where k is 0 or a positive integer.

in this case k = the floor (whole part) of  345/4   

345/4 = 86 with 1 left over.

k is 86 and 1 is left over

So   

n=4k+1

so tn = 2

think about it and I think you will understand.

Please proofread your question.

It is not right.

Do you repost because you are too lazy to finish it yourself? 

I answered this yesterday.

No it is not

Let a be the smallest integer satisfying the inequality x^2 - 15 < 2x, and let b be the largest integer satisfying the same inequality. What is b-a?

\(x^2 - 15 < 2x\\ x^2 - 15 -2x <0\\ x^2 -2x - 15 <0\\\)

This is a concave up parabola

It will be less than 0 between the roots

the roots are

\(x = {2 \pm \sqrt{4+60} \over 2}\\ x = {2 \pm 8 \over 2}\\ x=1\pm4 \)

the equation is satified for all x values between -3 and 5  BUT  NOT inclusive of -3 and 5)

So the smallest integer is a=-2   and the largest is b=4

4 - - 2 = 6

This is just asking fo whole numbers between -2 and 4     (NOT including -2 or 4)

\(Let\;\;f(x)=\lfloor x\lfloor x \rfloor \rfloor \quad for \;\;x\ge 0\\ (a)\;\; find\;all\;x\ge0\;\;such\;\;that\;\; f(x)=1\)

If  0<=x<1 then  f(x)=0

If  1<=x<2 then   f(x)= floor of (x*1) = 1

If x>=2      then    f(x)>1

so      \(if\;\; 1\le x<2 \;\;then \;\;f(x)=1\)

(b)   

\(Let\;\;f(x)=\lfloor x\lfloor x \rfloor \rfloor \quad for \;\;x\ge 0\\ (a)\;\; find\;all\;x\ge0\;\;such\;\;that\;\; f(x)=3\)

\( let\;\; 0\le\delta <1\\ If\;\;x=2+\delta \;\;then\\ \lfloor x\lfloor x \rfloor \rfloor=\lfloor (2+\delta) 2) \rfloor=\lfloor (4+2\delta) \rfloor=4\;\;or\;\;5 \)

So if  x is 1 and a bit then f(x)=1

If x=2 and a bit then f(x)=4 or 5

f(x) cannot equal 3

You can think about the rest yourself.

The floor function means round down to the nearest integer

The ceiling function means round up to the nearest integer.

For example:

\( \lfloor -4 -.5 \rfloor =\lfloor -4.5 \rfloor =-5\\ \lceil 4 +.5 \rceil =\lceil 4.5 \rceil =5\\ \)

Can you take it from there?