FP2: Method of Differences - Where have I gone wrong?

Ah I see! But just to make clear, is n and r the same thing? If not, what's the difference?

r represents what we are putting in r=1,2,3,4....,n-1,n
For each term! Here it is represnting the general term and r takes all the values listed in the sum and n is one of them


Posted from TSR Mobile

Ah I see! Thanks!

The other question I'm stuck on is finding (sigma) 2r+3/r(r+1) times by 1/3^r

I've converted 2r+3/r(r+1) into partial fractions but I don't know what to convert 1/3^r to..

I'm not quite sure what the question is - would you mind posting a screenshot or something?

Of course!

You shouldn't be "converting" the 3^r to anything!

On the face of it, that question looks quite horrible. THEREFORE you expect that applying partial fractions to the bit you can convert will simplify the sum to something that is going to be easy to evaluate! What do you get for your partial fraction expansion?

I got 3/r - 1/r+1

but even if I did (sigma)2r+3/r(r+1) times by 1/3^r and do n=1, n=2 etc nothing really cancels out..

I got 3/r - 1/r+1

but even if I did (sigma)2r+3/r(r+1) times by 1/3^r and do n=1, n=2 etc nothing really cancels out..

Of course!

You end up with $\displaystyle \sum_{r=1}^n \left(\frac{3}{r} - \frac{1}{r+1}\right) \cdot 3^{-r} = \sum_{r=1}^n \left(\frac{3^{1-r}}{r} - \frac{3^{-r}}{r+1}\right)$

Which, if you write it out term by term, you see you get

$\displaystyle \left(1 - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{27}\right) + \left(\frac{1}{27} - \frac{1}{108}\right) + \cdots$

Can you take it from there?

Ah yes, I'll give that a shot now!

In the meantime, how would you go about drawing 3x+1/x-3 I found the asymptotes to be y=3 and x=3 but I don't know how to draw the shape of it..

I'm assuming you're talking about $\frac{3x+1}{x-3}$ (use brackets to clear up ambiguity) - I would identify the asymptotes, as you have rightly done, then looked for x-intercepts ($x = -\frac{1}{3}$), y-intercepts, looked at what happens when x approaches 3 from the left or the right, maximum and minmum might be overkill but comes in useful every now and then, all of that combined should be enough to plot the graph.

Yeah I've found the asymptotes and intercepts but I can't figure out what the shape of it would be still.. :/

Basically, you need to be staring at your graph with the asympotes drawn with dotted lines and your important points marked. Now, look at your vertical asymptote, what happens to the function as x gets closer to 3 from the left, that is, 2.95, 2.99, 2.999 - does the function get more negative, more positive, etc... In this case the function becomes increasingly negative (because you're dividing a positive number by a tiiiiiny negative number), so you know the curve should move downwards to infinity, but it can never touch the asymptote. Then, on the other side of the asymptote, your curve should start from the top of the graph instead because it suddenly became positive and it should move downwards gradually before levelling out at your y-asymptote. Can you understand?

Ah yes, I think that made sense! If it was mod (3x+1)/x-3 would this be the correct graph?

Not quuiiiiite. You're taking the modulus of a huge negative number, which would give you a huge positive number near x=3.

Remember, modding a function just means reflecting what's under the x-axis in the x-axis. So your curve should curve upwards to x=3.

But isn't that what I've done? I've flipped everything under the x axis so it's now above the x-axis

But when you flip something that's far below the x-axis, it should go far above the x-axis once you're reflected it.

Yeah, isn't that what I've done or is it not obvious enough?

Ah, I see what you mean! But the graph still isn't right though, I don't see why it touches the x-axis and then shoots back up; it should look more like this: