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My way of solving infinite series

To understand this concept you must understand a variable number, a variable number is a number dedicated to or denotes a variable in a series, this number can have more than one variable but can't share a variable with other numbers, it also can't be a decimal. Ex1:

\[\sum_{x=1}^∞ a \frac{b}{c}\]

In the example x=1, right? 1 is the variable number and X is its dedicated variable(seems simple but trust me, it can be any number that's not ∞ or a decimal) this is what most of these problems look like:

\[\sum_{x=5}^∞ 2\frac{4\times x^{2}}{10} = \frac{a}{b}\] Find a+b

\[x=5\] \[x^{2}=25\] \[25\times 4 =100\] \[\frac{100}{10}=100+10=\boxed{110}\]

As you saw converging is not done in this method,the way around this is a dummy answer:

\[\frac{100}{10}=\boxed{10}\]

I hope you consider this a lesson for my future problems, take care!

Note by Ark3 Graptor
7 months, 4 weeks ago

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I am afraid you are missing the point about what a series is. All your method seems to be doing is finding the first term in a sum, and not the sum itself.

By putting \(x=1\) in the expression \[\sum_{x=1}^\infty 2^{-x}\] your method gets \(\tfrac12\), and not the correct answer of \(1\). If (as you do) you only consider the first term of a series, you will continue to get finite answers to series that have no finite limit.

Incidentally, your notion of a variable number is what is normally called a dummy variable. It is a place-holder variable used to indicate how to perform summations, integrals or other calculations. For example. in \[ \sum_{x=1}^\infty \frac{1}{x^2} \] \(x\) is an integer-valued dummy variable which can take any value greater than \(1\), while in \[ \int_0^3 \sin x\,dx \] \(x\) is a real-valued dummy variable which can take any value between \(0\) and \(3\). For that matter (just to show that dummy variables are not confined to sums and integrals), in \[ \bigcup_{j=1}^5 A_j \; = \; A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \] \(j\) is an integer-valued dummy variable taking values between \(1\) and \(5\). Mark Hennings · 7 months, 3 weeks ago

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@Mark Hennings Thank you for your input, but one thing that bugged me in your comment was the part where the answer was 1/2 and not 1, that is where an error was, you have to not treat it like a fraction if its not a fraction, for example: \[\sum_{x=1}^\infty 4\times x^{2}\] There are the three formulas and we have to pick the correct one \[\sum_{x=1}^\infty 4\times x^{2}=a,a\frac{b}{c} or \frac{a}{b}\]

The correct formula for this series is the first formula

The rest made sense, thanks again Ark3 Graptor · 7 months, 3 weeks ago

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@Ark3 Graptor I am sorry, but this is nonsense. There is no correct formula for the series you mention, which diverges.

Your method would pick \(x=1\) and get the answer of \(4\) here, I suppose. The whole point of a dummy variable here is that it indicates the range of values over which the sum is to be performed. Thus, as well as \(x=1\) giving \(4\), we have to add the contribution from \(x=2\), which is \(16\), then the contribution from \(x=3\), which is \(36\) and so on, including the contribution from all positive values of \(x\). The expression \[ \sum_{x=1}^\infty 4x^2 \] is mathematical shorthand for the sum \[ 4 + 16 + 36 + 64 + 100 + 144 + \cdots \] which does not converge to any value. Mark Hennings · 7 months, 3 weeks ago

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@Mark Hennings I'm going to have to think about this and do some testing, I will decide later if you are right or I have more of an argument Ark3 Graptor · 7 months, 3 weeks ago

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@Ark3 Graptor Sorry,That was my bad excision of certain things Ark3 Graptor · 7 months, 3 weeks ago

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@Ark3 Graptor I am not quite sure what you mean by that.

I am keen that you take some time to try to understand convergence properly. Isham Singh has given you a link to a suitable Wiki, which would be a good point to start. Mark Hennings · 7 months, 3 weeks ago

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\[\sum_{a=1}^{\infty}\sum_{b=2}^{\infty}\sum_{c=3}^{\infty} \dots \sum_{x=24}^{\infty}\sum_{y=25}^{\infty}\sum_{z=26}^{\infty} \dfrac{a+b^2+c^3+\dots+x^{24}+y^{25}+z^{26}}{a^{26}+b^{25}+c^{24}+\dots + x^{3}+y^2+z} = \dfrac{\alpha}{\gamma}\]

This works for all \(a,b,c,d,\dots , x,y,z\) who either are not in decimal and not infinity. Take it as a challenge dude @Ark3 Graptor ! Nihar Mahajan · 7 months, 4 weeks ago

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@Nihar Mahajan no, its not a=1,b=2,c=3 etc. its this: a can equal 1,2,3,4,5....any other number, just not a=.1 or a=\[\infty\] Ark3 Graptor · 7 months, 3 weeks ago

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@Ark3 Graptor Are you trying to argue the definition of a summation? This is not philosophy, I think you are in the wrong place. Julian Poon · 7 months, 3 weeks ago

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Please read up on how to use the notation before posting. Using notation in a way that doesn't make sense might make you look smart among people who don't know what the notation means but it makes you look stupid among those who know. Julian Poon · 7 months, 3 weeks ago

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@Julian Poon You are absolutely right. Aditya Kumar · 7 months, 3 weeks ago

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@Ark3 Graptor This might help. Ishan Singh · 7 months, 4 weeks ago

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What's the proof that your hypothesis is correct? Aditya Kumar · 7 months, 4 weeks ago

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The series you are talking about does not converge for \(x\geq1\). Aditya Kumar · 7 months, 4 weeks ago

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@Aditya Kumar The variable number changes that so x=1, did you forget to read that part? Ark3 Graptor · 7 months, 4 weeks ago

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@Ark3 Graptor If you don't mind, can you explain the concept of variable number to me with a few more examples? Aditya Kumar · 7 months, 4 weeks ago

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@Aditya Kumar Ah, the variable number is designed for people to understand the true meaning and thought on a series, infinite or finite. This is what more than one variable numbers look like(the amount of sums can be infinite)

\[\sum_{x=1}^\infty \sum_{y=2}^\infty \sum_{z=3}^\infty \frac{x+y+z}{z-y+x} = \frac{a}{b}=\frac{6}{2}\] The true meaning of a infinite series is to have no limit and for the number to not be changed Ark3 Graptor · 7 months, 4 weeks ago

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@Ark3 Graptor If you don't mind can I talk to you on any other private platform(other than fb)? I need to ask you some questions. Aditya Kumar · 7 months, 4 weeks ago

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@Aditya Kumar And converging isn't necessarily in this problem the only form of that is the dummy answer I edited in this morning Ark3 Graptor · 7 months, 4 weeks ago

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