Felix didn't use induction. I saw his answer and realised what my friend was trying to get me to do... We just started FP1 so he was trying to play with it!
(2n−1)=n2[br]n=1:1=1check![br][br](2k−1)=k2[br][br]k+1:2(k+1)−1=(k+1)2[br]2(k+1)−1=2(k+1)−1[br]k2+(2(k+1)−1)=(k+1)2[br]2(k+1)−1=(k+1)2 Not even sure if that makes sense right now. Brain = fried.
I know it's not induction. That's why I was questioning it.
I honestly cannot see what you are trying to do....That top line isn't an identity.
Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?! What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!
A nice little pattern I noticed is this: if I have an n*n box - or a perfect square xxxx xxxx xxxx xxxx
in this case where n = 4, x = 16
where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'-squares' so the sum of the x-coordinate value for a will always equal the sum for a's y coordinate etc...)
abcd abcc abbb aaaa
So we have produced 7a + 5b + 3b + 1a = 16x or, 16 (a square number) = 7 + 5 + 3 + 1...
Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol
Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have
tan(sin(t))=tan(t)=t
and
ln(1+sin(t))=ln(1+t)=t
we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of
cos(x)21
Which is 1.
I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.
Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have
tan(sin(t))=tan(t)=t
and
ln(1+sin(t))=ln(1+t)=t
we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of
cos(x)21
Which is 0.
I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.
Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?! What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!
A nice little pattern I noticed is this: if I have an n*n box - or a perfect square xxxx xxxx xxxx xxxx
in this case where n = 4, x = 16
where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'-squares' so the sum of the x-coordinate value for a will always equal the sum for a's y coordinate etc...)
abcd abcc abbb aaaa
So we have produced 7a + 5b + 3b + 1a = 16x or, 16 (a square number) = 7 + 5 + 3 + 1...
Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol
(n+1)^2=n^2+2n+1; In essence, to step from the nth square number to the n+1th, you must add the nth odd number. You can see this must be so as you say, the nth square number is a square, n*n. To get to the (n+1)th, you need to add a column of n, a row of n, and then one more (ie the 3 a's going downwards, the 3 a's at the bottom, then the one in the bottom left). Fairly clear, from a certain viewpoint. Otherwise, it's not entirely obvious that the sum of odd numbers should be square.
There are several ways to prove this, most straight forward is to use the explicit formula for the fibonacci numbers and it is just an exercise in summing geometric series.
Using standard results from analysis we have that h is increasing and that
h(x)>x2
This is the best inequality we can have, as h can be made as close to x2 as we like.
Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings. Use IBP and set x - \frac{ab}{a+b} = v and arccos(...) = u Our uv term (arcos1-arccos1) can be safely set as 0 if I am to make any geometrical sense of this. After a bit of algebra bashing we obtain;
We now work on the RHS. We set y=b−a2x−a−b ∫−114(a+b)(1−y2)(b2−a2)(y+1)dy Taking out the constants from the integral, we see our integral reduces down to 4(a+b)b2−a2∫−111−y21dy As y/(1-y^2) is obviously 0 over the interval as it is an odd function. This is now a well known integral; ∫−111−y21dy=arcsin(1)−arcsin(−1)=π And thus I=π4(a+b)(b2−a2)
Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).
I'll take a look at the tribonacci when I have some more time at a computer.
T0=0,T1=T2=1 and Tn+3=Tn+2+Tn+1+Tn for n≥0.
Find k=1∑nTk2.
This is harder than I first thought. I have now failed with 3 methods, all failing because the reccurance relation has an odd number of terms (I think they would work with all ones with an even number of terms), the sums just don't all cancel as 1 is always left out.
I do have a method that will certainly work, but it involves solving teh recurrance relation to get it into an explict form, squaring that and then summing all the geometric series. I don't want to do this ans I doubt it was the intention.
Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings. Use IBP and set x - \frac{ab}{a+b} = v and arccos(...) = u Our uv term (arcos1-arccos1) can be safely set as 0 if I am to make any geometrical sense of this. After a bit of algebra bashing we obtain;
We now work on the RHS. We set y=b−a2x−a−b ∫−114(a+b)(1−y2)(b2−a2)(y+1)dy Taking out the constants from the integral, we see our integral reduces down to 4(a+b)b2−a2∫−111−y21dy As y/(1-y^2) is obviously 0 over the interval as it is an odd function. This is now a well known integral; ∫−111−y21dy=arcsin(1)−arcsin(−1)=π And thus I=π4(a+b)(b2−a2)
Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).
I'll take a look at the tribonacci when I have some more time at a computer.
I used a different second substitution so missed out on the nice geometric link.
Problem 351 ** ∫0∞(x−2x3+2⋅4x5−2⋅4⋅6x7+...)(1+22x2+22⋅42x4+22⋅42⋅62x6+...)dx