Cover Sheet A S 0 9 6 0 0 5 5 5 5 S: Fill & Download for Free

There are a lot of answers here that are realistic. This is not one of them. As a single guy with no dependents and a job in tech, I can get away with personal finance murder and nobody's the wiser. If you're interested in an answer opposite the conservative (read: responsible) end of the spectrum, read on.~~~~~~~~~~~~~~~~~~~~~What is the minimal salary you need? $0 / yr. Just pay for the thing in cash.Okay, that probably wasn't your question. What is a doable salary? In my case, I got my 2012 model S when I was somewhat fresh out of college with a salary of just above $100k. I had $30k saved up. The car + options I wanted (tech package, P85, etc...) would cost around $102k. The banks would only lend me up to $30k. I had about 4 months to come up with the money.So I:Had a base of 30Borrowed 30 (@ 4% or so)Took out a personal line of credit for 10 (@ 11% or so)Opened a new credit card line (10) and eventually did a cash advance (@ 24%)Made 1/3 of my annual post-tax salary, of which I could keep everything after expenses. Let's say it was around ~10Turned off tax withholding** by claiming 20+ (an ungodly number) allowances on my W-4. This gave me another ~5.Stopped contributing to my 401k. This was another ~3Borrowed ~4.5 from my dad (@ 0%)Stopped paying my other credit cards for a virtual loan.Ensured that I would have cashflows after the purchase to cover interest until I could pay down the debt.And voila!Then I ruthlessly targeted my debt after I made the purchase in order to lower the aggregate interest rate of my virtual loan (2 thru 9) until I only had the car loan subcomponent outstanding.Why go through this exercise?A) My dad had placed a reservation in 2009 and our spot in line was forfeit if he couldn't cough up the cash or we didn't co-purchase. He had just moved and was illiquid at the time. Our spot in line was #1920 (vin #1917).B) The car is amazing. I'd been following Tesla on /. since people started reporting about it and the company. I really believed in the mission, the corporate strategy, and everything around the car. I also told myself in college that my next car (and all others after it) would be electric.C) I love a good challenge.Was it worth it? Hell yes. I got a cool car, but more importantly, my attitude towards finances (and goal oriented thinking) experienced a fundamental shift for the better.So I'd say the ultimate answer is: it depends on you. What's your risk/reward, and what does your personal balance sheet look like? How creative do you want to get? Do the math, and don't let your thinking be limited by just one number!** I did the Math ahead of time and my 2012 taxes would come out to basically $0 assuming I didn't do anything else crazy.

For starters you can readDeepak Mehta (दीपक मेहता)'s answer to What are some great Excel power user hacks and tips?The other answers are also wonderful. I will try to cover those not already covered. Forgive me, if there are any repetitions.The first mistake every manager will point out when you give him your balance sheet is how alignment is all wrong or how the numbers are incomprehensible because you forgot to use commas.So first things first, you have to format it appropriately, so that it looks presentable. These few basic ones do help a lot :1)Alt + O C A / ALT +HOIThis will auto-fit the contents in the cell by increasing the cell width2)Alt +HWThis will wrap the contents of the cell. Especially useful when you don’t want to increase the cell width3)Alt + HMCThis is used when you want to merge adjacent cells. Select the cells you want to merge and press Alt + HMC .Note this will keep the text centre aligned.4)Alt + HAC/HAR/HAL/HAT/HAM/HABThis is used to change the alignment - Centre/Right/Left /Top/Middle/bottom.5)Alt + HFCThis will give you the shortcut to change font color. Use the arrow keys to move between color options.6)Alt + HH/HHNThis helps in selecting the cell color. ALT + HHN will remove the color.7)ALT + HBThese will be useful for giving borders to the cell.Alt + HBA – will give border to the whole cell.Alt + HBN – Remove bordersAlt + HBU – Top & bottom double underlined8)Alt + HKThis will present your numbers with a comma separator.9)Alt + H +0/9This is to either increase (0) the decimals/reduce decimals (9)10)Ctrl VP/VNThis will give you the page break view and VN will give you the normal view. Especially useful when you are printing balance sheets and schedules.A few more basic1)Ctrl+ [This will lead you the first cell which is used in the formula. Suppose the formula in the cell is “=Details!B2+Details!B3” then it will take you to worksheet “Details Cell B2”Press F5 to get back to the original cell.Also read Anantha Krishna’s comment to Ankit Ruparel's answer to What are some useful Microsoft Excel tricks for Chartered Accountants?2)Ctrl plus +/-+ Insert row/column or move cells right/down- Delete row/ column or move cells left/up3)Ctrl + Shift + right arrow keyThis will enable you to select entire words when you don’t want to select the whole line.4)Ctrl + *This will enable you to select the whole table (provided there are no row/column breaks) without the need to continuously press shift+ arrow keys5)ALT + AM will remove duplicatesSelect the cells in the column or entire column and then press these keys to get the option of removing duplicates6)Alt + OHH/OHUAlt OHH is used to hide worksheet within the file and ALT OHU is used to unhide hidden sheets.A little advanced functions1)Filters=>Alt + HSF/DFF/HSCSelect cells and press either of these. You will be able to add filter to the selected cells.Press Alt + HSC when you want to clear the filter you have applied.=>Alt + down arrow keyWhen used in table where filter is applied it gives you a drop menu, from which you can select criteria’s you want to result to display. Press Up/down arrow key along with Spacebar to select multiple criteria’s.2)Alt +TUDThis is a useful formula especially when you are dealing with inter-linked sheets. Using this formula you can trace the cells which have been given reference to selected cells. When link are between multiple sheets it will show as below. When you double click on the arrow a dialog box as below shall appear. To remove this arrow press ALT + TUA.3)Break LinksOften we are given files which contain links to other work papers or files. For example formula links contain file references which are not provided to us. In such cases, make sure you break links. This can be done using the Edit Links option in your Data tab and then selecting break links option.4)ProofingGo to file menu in your MS excel 2010 and above then select options. Select proofing tab and go to autocorrect options. Then, in the replace text column type the Short text for example “ ABC” and in the with text column type the text you want to replace to replace it with as in our examples “ ABC enterprises Private Limited “OrReplace : BS 16With : Balance sheet as on 31st March, 2016So you are saved the trouble of typing this whole thing every time5)Pivot TablePivot is one of the best sources of analyzing the contents.viz. you can use this tool to check region wise sales for different months to identify any unreasonable fluctuations or to check month wise salary cost for employees to check if there were any increments given or whether the employee is new joinee or whether he has resigned. All in all a wonderful tool for analysis6)DelimitingDelimiting is a greatly under-valued functionality, as suggested by Ankit Ruparel is very useful when your entire data comes in a single column. Besides you can also use the same to convert date columns as per format required by you. Select the relevant cells and then go to data tab and select text to columns and then select delimiter as other.Then select “DMY” in date column or as per date format in your cells and voila. This is useful when you are given sales/purchases register and the date columns are in other formats like 30.05.2016 making it to difficult to analyze using filterThat’s all I can think of right now. I will add in case I recall a few more.Hope that helps.Ciao!!!

As Alon Amit explained, this is stated as a conjecture on a recent paper. This, however, isn’t a terribly difficult problem to solve, so it was open only because not enough people had thought about it.I’ll outline the key points of the proof, but let me first explain why it should intuitively be true:First, if we define [math]A[/math][math] + B = \{ x+y | x \in [/math][math]A[/math][math] \text{ and } y \in B \}[/math], what should we expect from [math]K_1 + K_2[/math] where [math]K_1[/math] and [math]K_2[/math] are (one-dimensional) Cantor sets?For any [math]A[/math][math], B \subset \R[/math], to find [math]A+B[/math], we can start with the Cartesian product [math]A[/math][math] \times B[/math], and project it into the real line at [math]45[/math] degrees, i.e. [math]f(x,y) = x + y[/math].Intuitively speaking, when we’re projecting something that’s [math]n[/math]-dimensional into a [math]d[/math]-dimensional subspace, we expect the projection to be [math]n[/math]-dimensional, if [math]n<d[/math] or [math]d[/math]-dimensional if [math]n \geq d[/math]. Think of shadows in our three-dimensional space. The shadow of a one-dimensional curve will still be one-dimensional, but the shadow of two-dimensional surfaces or three-dimensional objects will be two-dimensional (i.e. will contain a disk) unless in very denegerate cases (think a paper sheet perfectly aligned with the light source).The same reasoning works with non-integer (Hausdorff) dimensions. When we project a set with Hausdorff dimension [math]D[/math] into the real line, in general, the image of the projection will have dimension [math]D[/math] if [math]D<1[/math] and will contains intervals if [math]D>1[/math], unless in very degenerate cases.But the Hausdorff dimension of [math]K_1 \times K_2[/math] is generally equal to (and always [math]\geq[/math]) [math]dim(K_1) + dim(K_2)[/math]. I won’t prove it here in details, but it’s intuitively reasonable that, when we take Cartesian products, we add the dimensions - the Cartesian product of a [math]n[/math]-dimensional space and a [math]m[/math]-dimensional space is a [math](m+n)[/math]-dimensional space.That means that, as proven by J.-C. Yoccoz and C.G.T.A. Moreira, in general, if [math]dim(K_1) + dim(K_2) > 1[/math] then [math]K_1 + K_2[/math] contains intervals, unless in very degenerate cases.So, if we consider [math]C[/math] the classical middle-third Cantor set, and [math]K = \{x^2 | x \in C\}[/math], what we want to prove is that [math][0,1] \subset K + K + K + K[/math]. By what we described above, if [math]dim(K)[/math] is big enough, we should expect [math]K \times K \times K \times K[/math] to be thick enough, hence [math]K+K+K+K[/math] to be so thick as to cover the whole interval.What’s the Hausdorff dimension of [math]K[/math]? I claim that [math]dim(K) = dim(C) = \frac{\log 2}{\log 3} \approx 0.63[/math]. Why? The function [math]f(x) = x^2[/math] that maps [math]C[/math] onto [math]K[/math], in any interval [math](\epsilon, 1][/math], is a smooth function such that both [math]f[/math] and [math]f^{-1}[/math] have bounded derivatives, which means that there exists a constant [math]C[/math] such that, for any interval [math]I \subset (\epsilon,1][/math], the length of [math]f(I)[/math] is between [math]1/C[/math] and [math]C[/math] times the length of [math]I[/math]. Whenever that happens - i.e. [math]f[/math] doesn’t distort distances too much - [math]f[/math] preserves the Hausdorff dimension, that is [math]dim(f(A)) = dim(A)[/math] for any set [math]A[/math][math][/math].Putting all of this together, [math]dim(K^4) = 4 dim(K) = \frac{4 \log 2}{\log 3} > 2.52[/math], i.e. this set is way too “thick” for its projection not to contain all the intervals that it wouldn’t obviously miss.Now, working with Hausdorff dimensions leads to this “almost all” types of results, which don’t help much with specific explicit cases. So, it’s time to introduce another concept, and another theorem, that was actually proven a few years before Yoccoz-Moreira, and was a step in the direction that culminated on their result.The concept we need is called the thickness of a Cantor set. To define it, let’s look at any intermediate step in the construction of the set. At that stage, we’ll have a collection of intervals separated by gaps, some generated at previous stages, and some generated at the current stage. For any gap generated at the current stage, take the length of the adjacent intervals divided by the length of the gap. For instance, if these are the gaps we created at the second stage of the construction of our Cantor setthe ratios we’re interested at are [math]\frac{5}{3}, \frac{4}{3}, \frac{6}{2}, \frac{3}{2}[/math]. Now, we take the minimum of these ratios, in this case, [math]\frac{4}{3}[/math]. This will be our thickness at stage [math]2[/math]. Now we take the thickness of the Cantor set as the infimum of the thicknesses at all stages. The intuition is that, if the thickness is greater than 1, we’re always removing less than what we’re leaving, that “the gaps are smaller than the intervals”. That means that, if we slide two thick Cantor sets across each other, the “intervals” won’t fit in the “gaps”, and there’s always be an intersection point, i.e. their sum will contain intervals.And this is what S. Astels proved, that, if the product of the thickness of [math]K_1[/math] and [math]K_2[/math] is greater than or equal to [math]1[/math] then [math]K_1 + K_2[/math] will always contain intervals. [1] More than that, if there’s no obvious gap, i.e., if we cannot fit the whole Cantor set in a gap of the other, their sum will be the whole sum of the intervals they “live in”.So now, the only thing we need to prove is that the set [math]K+K[/math] is thick enough. Let’s do this explicitly.The first stage of [math]K[/math] is given by the intervals [math][[/math][math]0[/math][math], 1/9], [4/9, 1][/math], which means the first stage of [math]K+K[/math] is given by the intervals [math][[/math][math]0[/math][math], 2/9][/math] and [math][4/9,2][/math]. Or, in terms of intervals and gaps, intervals of length [math]2/9[/math] and [math]14/9[/math] separated by a gap of length [math]2/9[/math]. That means that the thickness at the first stage is exactly [math]1[/math].Now let’s look at the second stage of [math]K[/math]. It’s intervals (in bold), and gaps are:1/81, 3/81, 5/81, 27/81, 13/81, 15/81, 17/81(the gap in italic is from the previous stage)A bit of calculation will show that, for K+K, we’ll have2/81, 2/81, 14/81, 18/81, 22/81, 6/81, 21/81, 15/81, 17/81Again, the thickness at that stage is exactly 1, and is given by the ratio of the first interval and the first gap. All the other gaps are comfortably smaller than the adjacent gaps (Remember that we should ignore the gap in italic, coming from a previous stage).It’s easy to see that the first interval will always have length [math]\frac{2}{9^n}[/math], as well as the first gap, and it’s pretty straighforward* to see that the remaining gaps will continue to be smaller than the adjacent intervals. So the thickness of [math]K+K[/math] at every stage will be exactly [math]1[/math], which means that the thickness of [math]K+K[/math] is [math]1[/math], and, by Astels theorem, [math](K+K) + (K+K)[/math] is the whole interval [math][0,4][/math], Q.E.D. [math]\blacksquare[/math]* Let me elaborate on this and prove that claim that all the other gaps will be strictly smaller than the adjacent gaps.We proved that this is true for stages [math]1[/math] and [math]2[/math]. Now assume it’s true for all stages [math]\leq N[/math].The first observation is that the stage [math]N+1[/math] of [math]K \cap [0,1/9][/math] is just a scaled-down version of the stage [math]N[/math] of [math]K[/math], and the same holds for [math]K+K \cap [0,2/9][/math] and [math]K+K[/math].So that part is covered by simple induction. Now let’s look at what’s happenning at [math]K \cap [4/9,1] [/math]. Each interval of the [math]N[/math]-th stage of [math]K[/math] is of the form [math][k^2/9^N, (k+1)^2/9^N] [/math], and now the [math](N+1)[/math]-th stage will split it into interval-gap-interval, with lengths [math]\frac{6k+1}{9^{N+1}}, \frac{6k+3}{9^{N+1}}, \frac{6k+5}{9^{N+1}}[/math].From now on, I’ll just multiply everything by [math]9^{N+1}[/math], since we’re only interested in ratios. So what we have is that every gap of length [math]L[/math]’s adjacent intervals have length [math]L-2[/math] and [math]L+2[/math].When we now look at [math]K+K[/math], what happens? Let’s start with the leftmost point of that sequence of intervals (i.e. [math]k^2/9^N[/math]), and slide “the other copy of K” to the right, i.e. consider [math]\{x\} + K[/math], as we increase [math]x[/math].It’s obvious that the whole interval of length [math]L-2[/math] will be in [math]K+K[/math], but, more than that, the first point where we risk having a gap is this(These are valid pictures because [math]L-2[/math] is always greater than 9 in that region). That means that, in [math]K+K[/math], the corresponding interval has length at least [math]L+3[/math], and the gap has length at most [math]L-5[/math]. So we have a gap of length [math]L-5[/math] surrounded by intervals of length at least [math]L+2 > L-5[/math], as we wanted.That covers [math]K+K \cap [4/9,1][/math]. For [math]K+K \cap [1,2][/math] we can apply the same reasoning, but going in the reverse direction.And things are even easier, because, instead of moving the smaller and most dissimilar intervals, we’re moving the biggest and most similar, and it’s obvious that it will either cut at least 2 from the gap and/or add at least 2 to the smallest interval.Putting all of this together, we’ve seen that all the gaps will be strictly smaller than the adjacent gaps, except for the very first, that equals the first interval. [math]\blacksquare[/math]Footnotes[1] https://www.ams.org/journals/era/1999-05-15/S1079-6762-99-00068-2/S1079-6762-99-00068-2.pdf