Severe Weather 4-in x 4-in x 8-ft #2 Pressure Treated Lumber. Item #639138 Model #11687. Get Pricing and Availability. Use Current Location. Check Other Stores closed #2 grade southern yellow pine. Severe Weather Ground Contact pressure treated exterior wood protected with Copper Azole (CA-C) provides lasting support and protection for all. Base 8 Octal form base 10 Decimal form (common used) base 16 Hexadecimal form (hexa) Examples. 3424 (base 10) = 00 (base 2) 12623 (base 10) = 314f (base 16). Tagr 4.0 it's just what I needed to keep my music organized. I find it really useful and easy to use. And with its new fixed Discogs feature, you can even tag albums that have no ID3 tag at all, just by searching the artist and the album and selecting the one you want from the large data collection from Discogs.
To find a missing number in a Sequence, first we must have a Rule
Sequence
A Sequence is a set of things (usually numbers) that are in order.
Each number in the sequence is called a term (or sometimes 'element' or 'member'), read Sequences and Series for a more in-depth discussion.
Finding Missing Numbers
To find a missing number, first find a Rule behind the Sequence.
Sometimes we can just look at the numbers and see a pattern:
Example: 1, 4, 9, 16, ?
Answer: they are Squares (12=1, 22=4, 32=9, 42=16, ..)
Rule: xn = n2
Sequence: 1, 4, 9, 16, 25, 36, 49, ..
Did you see how we wrote that rule using 'x' and 'n' ?
xn means 'term number n', so term 3 is written x3
And we can calculate term 3 using:
x3 = 32 = 9
We can use a Rule to find any term. For example, the 25th term can be found by 'plugging in' 25 wherever n is.
x25 = 252 = 625
Tagr 4 8 16
How about another example:
Example: 3, 5, 8, 13, 21, ?
After 3 and 5 all the rest are the sum of the two numbers before,
That is 3 + 5 = 8, 5 + 8 = 13 etc, which is part of the Fibonacci Sequence:
3, 5, 8, 13, 21, 34, 55, 89, ..
Pluraleyes 3 5 5. Which has this Rule:
Rule: xn = xn-1 + xn-2
Now what does xn-1 mean? It means 'the previous term' as term number n-1 is 1 less than term number n.
And xn-2 means the term before that one.
Let's try that Rule for the 6th term:
x6 = x6-1 + x6-2
x6 = x5 + x4
So term 6 equals term 5 plus term 4. We already know term 5 is 21 and term 4 is 13, so:
x6 = 21 + 13 = 34
Many Rules
One of the troubles with finding 'the next number' in a sequence is that mathematics is so powerful we can find more than one Rule that works.
What is the next number in the sequence 1, 2, 4, 7, ?
Here are three solutions (there can be more!):
Solution 1: Add 1, then add 2, 3, 4, ..
So, 1+1=2, 2+2=4, 4+3=7, 7+4=11, etc..
Rule: xn = n(n-1)/2 + 1
Sequence: 1, 2, 4, 7, 11, 16, 22, ..
(That rule looks a bit complicated, but it works)
Solution 2: After 1 and 2, add the two previous numbers, plus 1:
Rule: xn = xn-1 + xn-2 + 1
Sequence: 1, 2, 4, 7, 12, 20, 33, ..
Solution 3: After 1, 2 and 4, add the three previous numbers
Rule: xn = xn-1 + xn-2 + xn-3
Sequence: 1, 2, 4, 7, 13, 24, 44, ..
So, we have three perfectly reasonable solutions, and they create totally different sequences.
Which is right? They are all right.
And there are other solutions .... it may be a list of the winners' numbers .. so the next number could be .. anything! |
Simplest Rule
Tagr 4 8 12
When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.
Finding Differences
Sometimes it helps to find the differences between each pair of numbers .. this can often reveal an underlying pattern.
Here is a simple case:
Tagr 4 8 14
The differences are always 2, so we can guess that '2n' is part of the answer.
Let us try 2n: Posterino 3 7.
The last row shows that we are always wrong by 5, so just add 5 and we are done:
Rule: xn = 2n + 5
OK, we could have worked out '2n+5' by just playing around with the numbers a bit, but we want a systematic way to do it, for when the sequences get more complicated.
Second Differences
In the sequence {1, 2, 4, 7, 11, 16, 22, ..} we need to find the differences ..
.. and then find the differences of those (called second differences), like this:
The second differences in this case are 1.
With second differences we multiply by n22
In our case the difference is 1, so let us try just n22:
n: | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
Terms (xn): | 1 | 2 | 4 | 7 | 11 |
n22: | 0.5 | 2 | 4.5 | 8 | 12.5 |
Wrong by: | 0.5 | 0 | -0.5 | -1 | -1.5 |
We are close, but seem to be drifting by 0.5, so let us try: n22 − n2
Wrong by 1 now, so let us add 1:
n22 − n2 + 1 | 1 | 2 | 4 | 7 | 11 |
---|---|---|---|---|---|
Wrong by: | 0 | 0 | 0 | 0 | 0 |
We did it!
The formula n22 − n2 + 1 can be simplified to n(n-1)/2 + 1
So by 'trial-and-error' we discovered a rule that works:
Rule: xn = n(n-1)/2 + 1
Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ..
Other Types of Sequences
Tagr 4 8 15
Read Sequences and Series to learn about:
And there are also:
And many more!
In truth there are too many types of sequences to mention here, but if there is a special one you would like me to add just let me know.
In mathematics, the infinite series1/2 + 1/4 + 1/8 + 1/16 + ··· is an elementary example of a geometric series that converges absolutely.
There are many different expressions that can be shown to be equivalent to the problem, such as the form: 2−1 + 2−2 + 2−3 + ..
The sum of this series can be denoted in summation notation as:
- 12+14+18+116+⋯=∑n=1∞(12)n=121−12=1.{displaystyle {frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+{frac {1}{16}}+cdots =sum _{n=1}^{infty }left({frac {1}{2}}right)^{n}={frac {frac {1}{2}}{1-{frac {1}{2}}}}=1.}
Proof[edit]
As with any infinite series, the infinite sum
- 12+14+18+116+⋯{displaystyle {frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+{frac {1}{16}}+cdots }
is defined to mean the limit of the sum of the first n terms
- sn=12+14+18+116+⋯+12n−1+12n{displaystyle s_{n}={frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+{frac {1}{16}}+cdots +{frac {1}{2^{n-1}}}+{frac {1}{2^{n}}}}
as n approaches infinity.
Multiplying sn by 2 reveals a useful relationship:
- 2sn=22+24+28+216+⋯+22n=1+[12+14+18+⋯+12n−1]=1+[sn−12n].{displaystyle 2s_{n}={frac {2}{2}}+{frac {2}{4}}+{frac {2}{8}}+{frac {2}{16}}+cdots +{frac {2}{2^{n}}}=1+left[{frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+cdots +{frac {1}{2^{n-1}}}right]=1+left[s_{n}-{frac {1}{2^{n}}}right].}
Subtracting sn from both sides,
- sn=1−12n.{displaystyle s_{n}=1-{frac {1}{2^{n}}}.}
As n approaches infinity, sntends to 1.
History[edit]
Zeno's paradox[edit]
This series was used as a representation of many of Zeno's paradoxes, one of which, Achilles and the Tortoise, is shown here.[1] In the paradox, the warrior Achilles was to race against a tortoise. The track is 100 meters long. Achilles could run at 10 m/s, while the tortoise only 5. The tortoise, with a 10-meter advantage, Zeno argued, would win. Achilles would have to move 10 meters to catch up to the tortoise, but by then, the tortoise would already have moved another five meters. Achilles would then have to move 5 meters, where the tortoise would move 2.5 meters, and so on. Zeno argued that the tortoise would always remain ahead of Achilles.
The Eye of Horus[edit]
The parts of the Eye of Horus were once thought to represent the first six summands of the series.[2]
In a myriad ages it will not be exhausted[edit]
'Zhuangzi', also known as 'South China Classic', written by Zhuang Zhou. In the miscellaneous chapters 'All Under Heaven', he said: 'Take a chi long stick and remove half every day, in a myriad ages it will not be exhausted.'
See also[edit]
References[edit]
- ^Wachsmuth, Bet G. 'Description of Zeno's paradoxes'. Archived from the original on 2014-12-31. Retrieved 2014-12-29.
- ^Stewart, Ian (2009). Professor Stewart's Hoard of Mathematical Treasures. Profile Books. pp. 76–80. ISBN978 1 84668 292 6.
.. it may be a list of the winners' numbers .. so the next number could be .. anything! |
Simplest Rule
Tagr 4 8 12
When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.
Finding Differences
Sometimes it helps to find the differences between each pair of numbers .. this can often reveal an underlying pattern.
Here is a simple case:
Tagr 4 8 14
The differences are always 2, so we can guess that '2n' is part of the answer.
Let us try 2n: Posterino 3 7.
The last row shows that we are always wrong by 5, so just add 5 and we are done:
Rule: xn = 2n + 5
OK, we could have worked out '2n+5' by just playing around with the numbers a bit, but we want a systematic way to do it, for when the sequences get more complicated.
Second Differences
In the sequence {1, 2, 4, 7, 11, 16, 22, ..} we need to find the differences ..
.. and then find the differences of those (called second differences), like this:
The second differences in this case are 1.
With second differences we multiply by n22
In our case the difference is 1, so let us try just n22:
n: | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
Terms (xn): | 1 | 2 | 4 | 7 | 11 |
n22: | 0.5 | 2 | 4.5 | 8 | 12.5 |
Wrong by: | 0.5 | 0 | -0.5 | -1 | -1.5 |
We are close, but seem to be drifting by 0.5, so let us try: n22 − n2
Wrong by 1 now, so let us add 1:
n22 − n2 + 1 | 1 | 2 | 4 | 7 | 11 |
---|---|---|---|---|---|
Wrong by: | 0 | 0 | 0 | 0 | 0 |
We did it!
The formula n22 − n2 + 1 can be simplified to n(n-1)/2 + 1
So by 'trial-and-error' we discovered a rule that works:
Rule: xn = n(n-1)/2 + 1
Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ..
Other Types of Sequences
Tagr 4 8 15
Read Sequences and Series to learn about:
And there are also:
And many more!
In truth there are too many types of sequences to mention here, but if there is a special one you would like me to add just let me know.
In mathematics, the infinite series1/2 + 1/4 + 1/8 + 1/16 + ··· is an elementary example of a geometric series that converges absolutely.
There are many different expressions that can be shown to be equivalent to the problem, such as the form: 2−1 + 2−2 + 2−3 + ..
The sum of this series can be denoted in summation notation as:
- 12+14+18+116+⋯=∑n=1∞(12)n=121−12=1.{displaystyle {frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+{frac {1}{16}}+cdots =sum _{n=1}^{infty }left({frac {1}{2}}right)^{n}={frac {frac {1}{2}}{1-{frac {1}{2}}}}=1.}
Proof[edit]
As with any infinite series, the infinite sum
- 12+14+18+116+⋯{displaystyle {frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+{frac {1}{16}}+cdots }
is defined to mean the limit of the sum of the first n terms
- sn=12+14+18+116+⋯+12n−1+12n{displaystyle s_{n}={frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+{frac {1}{16}}+cdots +{frac {1}{2^{n-1}}}+{frac {1}{2^{n}}}}
as n approaches infinity.
Multiplying sn by 2 reveals a useful relationship:
- 2sn=22+24+28+216+⋯+22n=1+[12+14+18+⋯+12n−1]=1+[sn−12n].{displaystyle 2s_{n}={frac {2}{2}}+{frac {2}{4}}+{frac {2}{8}}+{frac {2}{16}}+cdots +{frac {2}{2^{n}}}=1+left[{frac {1}{2}}+{frac {1}{4}}+{frac {1}{8}}+cdots +{frac {1}{2^{n-1}}}right]=1+left[s_{n}-{frac {1}{2^{n}}}right].}
Subtracting sn from both sides,
- sn=1−12n.{displaystyle s_{n}=1-{frac {1}{2^{n}}}.}
As n approaches infinity, sntends to 1.
History[edit]
Zeno's paradox[edit]
This series was used as a representation of many of Zeno's paradoxes, one of which, Achilles and the Tortoise, is shown here.[1] In the paradox, the warrior Achilles was to race against a tortoise. The track is 100 meters long. Achilles could run at 10 m/s, while the tortoise only 5. The tortoise, with a 10-meter advantage, Zeno argued, would win. Achilles would have to move 10 meters to catch up to the tortoise, but by then, the tortoise would already have moved another five meters. Achilles would then have to move 5 meters, where the tortoise would move 2.5 meters, and so on. Zeno argued that the tortoise would always remain ahead of Achilles.
The Eye of Horus[edit]
The parts of the Eye of Horus were once thought to represent the first six summands of the series.[2]
In a myriad ages it will not be exhausted[edit]
'Zhuangzi', also known as 'South China Classic', written by Zhuang Zhou. In the miscellaneous chapters 'All Under Heaven', he said: 'Take a chi long stick and remove half every day, in a myriad ages it will not be exhausted.'
See also[edit]
References[edit]
- ^Wachsmuth, Bet G. 'Description of Zeno's paradoxes'. Archived from the original on 2014-12-31. Retrieved 2014-12-29.
- ^Stewart, Ian (2009). Professor Stewart's Hoard of Mathematical Treasures. Profile Books. pp. 76–80. ISBN978 1 84668 292 6.