5

L. Counkler tho Mquenco d,01,01, whro to 2,Aml, to " 2% 4 uln U)an Prove thac, for n 2 2, It n not (ruo (hat Aaw" Suppae X i8 4 Rl quud % P(X) P(X) In An ...

Question

L. Counkler tho Mquenco d,01,01, whro to 2,Aml, to " 2% 4 uln U)an Prove thac, for n 2 2, It n not (ruo (hat Aaw" Suppae X i8 4 Rl quud % P(X) P(X) In An alyvbralc ekmuro opralor UIl X. Prove thal, for uy wts A,B € x, YMA)OM(B)) X(A)nxw):

L. Counkler tho Mquenco d,01,01, whro to 2,Aml, to " 2% 4 uln U)an Prove thac, for n 2 2, It n not (ruo (hat Aaw" Suppae X i8 4 Rl quud % P(X) P(X) In An alyvbralc ekmuro opralor UIl X. Prove thal, for uy wts A,B € x, YMA)OM(B)) X(A)nxw):



Answers

Show that $a_{n-2}=a_{n}-2 \nabla a_{n}+\nabla^{2} a_{n}$.

You hear? The question is para magnetism of or a magnetism two plus is due to is due to presidents. Often presidents often or the electron in molecular orbital so so Or two plus, iron has 15 electrons. And after writing the molecular or vital structure of this, we will find out The para magnetism of 40 plus. Iron is due to present some odd electron in the Bye, based on why are white are vital. This is our correct answer. So let's save the options. So we see obscenity is correct.

Right. So for this one you're assuming that n is odd. So let's and be an odd integer. And our pride of nation. We can write in us too. Okay. Last one for something. Did you Okay. Right. Note that. Okay. is equal to N -1/2. Just manipulating the equation for us. We don't have the floor of an X squared or four. Safe water. We know what and is and is odd. So two K plus one squared over four Which is equal to four K square. That's for Okay. Last one go for. Yeah. And so this would be K squared. That's okay. Plus 1/4. We can separate them. So we have K squared the flow of case where but the floor. Okay. Plus That's Love one for case where is an integer? So in this case correct? So he's gay the flow of uh 1/4 0. So we left it case square plus k. Um If we go back to the note part which we said case and -1 over to you can substitute back that back in So and managed one of the two squared plus And -1/2. Okay so I can take out and then minus 1/2 and I'll be left with And -1/2. Last one and This is equal to and -1/2 china's and minus one Plus two for 2. So in -1 plus two. Last one. Okay And ah so we just we just multiplying this sort of the top becomes And squared -1 In the bottom two times 2 is four. And this besides that we're looking for.

In this question When you calm down the function F thanks is even if the I am from the minus X must equal to the f x in discussion Here were given the function f x ego Thio Thio an expert t o n plus the HBO and managed to expert Thio and managed to and plus so on until the H two expert you lost zero here we noticed that another power for the next year they are on even number Ondas even power here and now the next time we need to test for the f of the minus X So we need to block in the minus x inside we have the about you. I'm Joanne minus x about you and plus H E minus two minus expected U minus two plus on up to the age you minus X about you plus nine zero. And because I'm the even power, it will turn everything to the punch deal So h u and expect you and that's about you and managed to expert you and managed to That's so, um, up to the HQ expert you that zero we see this one exactly equal Judah and from the X. So we already proved that the after the minus X it could you FX And by the definition, here, dysfunction F X is even.

Okay, so here we want. So the goal and one is two proved that a on minus K can be expressed in terms of a sit and backwards difference. This is Ben. Backwards difference squared of is then so on to the backwards difference to the K power. Uh, a N It's really use induction. So let's let puk e So this idea that 80 and minus K can be expressed as these backwards differences. Okay, So take the bass case where K is equal to one. Well, since the backwards difference of ace of N is equal to a set and minus a said n plus one, it follows that ace of and plus one is equal to a seven minus the backwards difference of a seven. Okay, so therefore p one is true. Yes, that holds. Let's move to the induction step. So let's p k be true. Then we need to show that p of K plus one is also true. Okay, So, consider then. Is that an minus? So k o b k plus one. But it is equal to a some n minus. Okay. Minus one. It was equal to Aye said and minus one minus K. All right. And since p f. K is true, this is gonna be equal to some function written as a and minus one and the backwards difference of M s one. We have to be backwards difference. Like a power a minus one. Okay, then, by definition of the first and second and Capel's first differences, this could be rewritten as a seven, minus the backwards difference of a n the backwards difference of a n minus t backwards, different squared of a n only to be backwards difference to the K a n minus t backwards difference Kate plus one n. Okay, so therefore P of K plus one is also true. And so by the principle of mathematical deduction, that sort of ran out of space that previous page. But so by induction, you know that PN must be true for all positive integers end positive into gear can It concludes the proof


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