If x = 0 , then x − x 2 + x 3 − x 4 + ⋯ = 0 .
Which of the following is the value of y − y 2 + y 3 − y 4 + ⋯ , where y is some real number?
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There is missing information about infinite sum. Therefore it is possible "might be" equal to 0 if x=1 and it is even number of elements.
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This is supposed to be a series, not a finite sum.
What if S=x-x2+x3-x4...=x-(x2-x3)-(x3-x4)-(x5-x6), which is S=1 when x=1
If x=-2 then S=2. It's not restricted to positive values. IMHO Brilliant got this one wrong
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then, in that case we can 3 answers(2,3,4)
Nope. If ∣ y ∣ > 1 then the series will diverge.
That formula is correct only when series is converging
This is probably wrong...this is not always a converging series....
Relevant wiki: Geometric Progression Sum
The sum to infinity of a geometric series is: 1 − c o m m o n r a t i o f i r s t t e r m
The first term of this series is x and the common ratio is -x.
So assuming the series converges, S= 1 + x x . Rearranging, we get:
S+Sx=x
S=x-Sx
S=x(1-S)
x= 1 − S S
In order for the series to converge, x must be less than 1 and greater than -1. x= 1 − S S does not satisfy this criterion for S=1,2,3,4. Therefore, none of the given answers are possible.
What are all the values that such a series can take?
Given expression is x-x^2+x^3-x^4... = x(1-x)+x^3(1-x).... =(1-x)(x+x^3+x^5....) Clearly for any values of x, the result is either positive or negative and the second factor tells us that it is a big one certainly not among 1, 2, 3 or 4 YOU may use integers like 2 or 3 to test it! or even 1! The result is a big negative or positive integer (Possibly to infinity) The result when we use 1 is 0 which is also not present here in the options!
Series when x=1 actually equals 1/2 as it goes to infinity
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Can you explain that?
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It is Grandi's series; it doesn't actually converge on 1/2 but we define it to be equal to 1/2. So we want to know the value of the sum of the infinite series 1 - 1 + 1 - 1 + 1 - 1 + ...
We can see by putting parenthesis around the 1s with negative signs that the series will sum to 0: (1 - 1) + (1 - 1) + (1 - 1) + ... = 0 + 0 + 0 + ...
But 1 - 1 + 1 is also equal to 1 - (1 - 1) and when we do that the series actually sums to 1: 1 - (1 - 1) - (1 - 1) - (1 - 1) ... = 1 - 0 - 0 - 0 ...
We can set S equal to the sum of the series: S = 1 - 1 + 1 - 1 + ... If we take the negative of S we get: -S = -1 + 1 - 1 + 1 - ...
By taking the negative, we see that -S is equal to (S-1), because -S is the original series S with a -1 in front of it. Setting -S = S - 1, we get 1/2 as the answer.
But realize that the series will never actually converge on 1/2, it is a divergent series, and if you picked a random place such as the millionth number in the series, it would be either a 1 or 0. But as it goes to infinity, we define it to be 1/2, due to the argument above. I hope this helped.
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@Dylan Wedel – Interesting. But is this series not also an instance of the infinity-infinity indetermination?
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@Rubén Martínez – This series has been debated by mathematicians for over three centuries but the general consensus is that the series value is defined to be 1/2. Basically the argument goes like this:
We can sum 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... and see that it approaches 2 by taking the partial sums: 1, 1 + 1/2 = 1.5, 1 + 1/2 + 1/4 = 1.75, 1 + 1/2 + 1/4 + 1/8 = 1.875, 1.9375, 1.96875 ... = 2
We can also sum this another way by taking the averages of the partial sums: 1, (1 + 1.5)/2 = 1.25, (1 + 1.5 + 1.75)/3 = 1.417, (1 + 1.5 + 1.75 + 1.875)/4 = 1.53125, 1.6125, 1.671875 ... and see this also approaches 2.
If we do this for Grandi's series we get 1, 0, 1, 0, 1 ... as the partial sums. This is where it gets interesting. Taking the averages of the partial sums, we get 1, (1 + 0)/2 = 1/2, (1 + 0 + 1)/3 = 2/3, (1 + 0 + 1 + 0)/4 = 1/2, (1 + 0 + 1 + 0 + 1)/5 = 3/5 ... or
1, 1/2, 2/3, 1/2, 3/5, 1/2, 4/7, 1/2, 5/9, 1/2, 6/11 ... and looking at the odd places, we can see a convergence on 1/2 as the series goes to infinity: 1, 2/3, 3/5, 4/7, 5/9, 6/11, 7/13 ... = 1/2.
This argument along with the -S = S - 1 argument is why most modern mathematicians agree that if Grandi's series should be assigned a value, it should be 1/2. But there are people who disagree, creating compelling arguments against this consensus, like 20th century philosopher James F. Thomson who devised Thomson's lamp.
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@Dylan Wedel – I disagree. The Grandi's Series does not have a well defined value. A series has a value only if it converges.
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@Agnishom Chattopadhyay – Like I said, there is some disagreement in the mathematical community. It's not conclusive; we either give it the value of 1/2 or we don't
@Agnishom Chattopadhyay – There are a lot of ways to assign values to divergent series, so rather than argue whether Grandi's series has a value in some natural sense, I would argue the only problem with Dylan's post is he omitted a short phrase regarding in which sense its value exists.
I.e. There is absolutely no disagreement in the mathematics community regarding the correctness of the following two statements:
In fact, this is just what Dylan showed, even if he didn't state his conclusion like this.
As a sidenote, I should point out that even though the usual definition for the value of an infinite series may seem natural to you, it has decidedly non-intuitive implications (just ask one of the "uninitiated" whether . 9 ˉ = 1 to see an example of this). From this standpoint, I believe Cesaro summability is equally natural as Riemann summability (though this is something we could legitimately argue ;-) )
For x=1, the sum is neither 0, nor 1/2, nor 1. It is not a convergent series. If you sum an odd number of terms the outcome is 1, if you sum an even number of terms the outcome is 0. For the ... it is not defined whether you should pick an even or an odd number of terms, so there is no unambiguous answer.
The series 1-1+1-1+1-... is known as Grandi's series ( https://en.wikipedia.org/wiki/Grandi%27s_series ). A method of making it summable is Cesàro summation, in which case you'd get 1/2. Basically this means you take the average value of the partial sums 1, 0, 1, 0, 1, 0, ..., which would be 1, 1/2, 2/3, 1/2, 3/5, 1/2, ... This is a convergent series, converging on 1/2. But note: this is no longer classical summation. Saying the series equals 1/2 therefore is false.
I don't seem to be able to follow your argument. If I put in x=0.5, I do not get a huge outcome, I get 1/2*(1/2+1/8+1/32+...), which is a real number less than 0.5. Are you sure there are no outcomes between this outcome and infinity? And if not, why is 1, 2, 3, or 4 not an option?
Your outcome of 0 for x=1 is incorrect. Taking an odd number of terms from the original expression would yield 1, and only an even number of terms yields 0. Your equals signs therefore are not really correct, they assume a specific grouping of the terms, in which you always combine an odd term with the consecutive even term, causing you to miss out on the outcome for x=1 being undefined.
If ∣ x ∣ ≥ 1 , the series does not converge.
If 0 < x < 1 , because it is an alternating series with terms decreasing in magnitude, each partial sum is at most x, thus the sum of the series is also at most x, and thus less than 1.
If − 1 < x < 0 , each partial sum is negative, and thus the sum is also negative.
If we factor out the lower-order term of each pair of x-terms that are being subtracted from one another, then the original equation:
x − x 2 + x 3 − x 4 + x 5 − x 6 + . . . = ?
Thought of like this:
( x − x 2 ) + ( x 3 − x 4 ) + ( x 5 − x 6 ) + . . . = ?
Can be rewritten as:
x ( 1 − x ) + x 3 ( 1 − x ) + x 5 ( 1 − x ) + . . .
And a ( 1 − x ) can be further factored out, giving:
( 1 − x ) ( x + x 3 + x 5 + . . . )
From here, we can easily try a few different options for x and rule out some possibilities:
Let A = ( 1 − x ) , B = ( x + x 3 + x 5 + . . . )
if x < 0 : A > 0 , B < 0 , thus A B < 0
if x = 0 : A = 1 , B = 0 , thus A B = 0
if 0 < x < 1 : 0 < A < 1 , 0 < B < 1 , thus 0 < A B < 1
if x > 1 : A < 0 , B > 0 , thus A B < 0
All of these conditions cover every possible real-number value for x , yet we can see that none of the offered solutions (1, 2, 3, or 4) are possible.
Because all terms with an even power are negative, and all terms with an odd power are positive, any negative number will result in a negative value, so we can eliminate those as an option. For all numbers greater than 1, the value will continually fluctuate between increasingly larger positive and negative values as each term is added. For 1, it will continually fluctuate between 1 and 0. Neither of these options result in a defined limit. So if a solution exists, we are limited to values of x between 0 and 1.
For any number between 0 and 1, the absolute value of each term will be less than the last. In other words, the value of the sequence will change by less and less as each term is added, and thus the value of the sequence has a defined limit. But, looking at the first two terms, the second term will subtract from the given value of x, and since the sequence will change by less and less, we know that, for this range of values, the value of the sequence can never exceed the value of x. Therefore, it will never be one of the positive integers listed.
Let y = x - x^2 + x^3 - x^4 + ...
Factor out an x
y = x (1 - x + x^2 - x^3 + x^4 - ...)
Note that the part in parentheses, from the second term on, is just - y. Therefore,
y = x (1 - y)
x = y / (1 - y)
This does, however, introduce some false solutions, so you have to check them.
if y=1, this is undefined
if y=2, x = -1, but plugged back into the original formula gives y = -1 -1 -1 -1 -1 ... so that doesn't work.
Similarly, if y=3 and y=4, x= -3/2 and -4/3, which also race to negative infinity.
Curiously, y = -1, x = -1/2 works just fine.
Relevant wiki: Geometric Progression Sum
Let S= x − x 2 + x 3 . . . . .
We know that by Geometric Progression for infinite addititon,
S = 1 − r a where a is 1st term & r is common ratio(r<1).
x − x 2 + x 3 − . . . . . . = 1 − x x , which is never greater than 1.
Thus, None of these are possible.
For what values of x does the series converge?
Providing that ∣ x ∣ < 1 , the sum of the power series x − x 2 + x 3 − x 4 + ⋯ is x ( 1 − x + x 2 − x 3 + ⋯ ) = 1 + x x . Let this number be denoted by c . Then x = 1 − c c . Since ∣ x ∣ < 1 , we find c < 2 1 . So, none of the offered numbers couldn't satisfy the condition c < 2 1 , and is not possible.
1 is a solution if you add an even number of terms together. So perhaps you must state that it is an odd number of terms.
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For x = 1 we've got a summable divergent series. For example, Cesàro summation assigns divergent series 1 − 1 + 1 − 1 + ⋯ + 1 − 1 + 1 − 1 + ⋯ the value 2 1 .
This is an infinite geometric sequence whose first term is t 1 = x and whose ratio is r = − x . The sum of an infinite geometric sequence converges to S = 1 − r t 1 when − 1 < r < 1 or in this case S = 1 + x x when − 1 < x < 1 .
When − 1 < x < 0 , 1 + x x < 0 (because a negative divided by a positive is a negative) so S < 0 , which isn't any of the options.
When x = 0 , 1 + x x = 0 so S = 0 , which was given but still isn't any of the options.
When 0 < x < 1 , 0 < 1 + x x < 1 (because a positive divided by a greater positive must be a value between 0 and 1) so 0 < S < 1 , which isn't any of the options, either.
Therefore, none of the options are possible.
Let S = x − x 2 + x 3 − . . .
cases:
x > = 1 this series does not converge
x ∈ ( − 1 , 1 ) this series converges as shown below
x < = − 1 this series diverges to − ∞
when S converges the same terms can be rewritten as equal to x ( 1 − S ) , so
S = 1 + x x < 1 for x ∈ ( − 1 , 1 )
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Let S = x − x 2 + x 3 − . . .
Then ( − x ) S = − x 2 + x 3 − . . .
Subtracting the two, we see that S ( x + 1 ) = x (for a b s ( x ) strictly less than 1)
Thus, S = x + 1 x = 1 - x + 1 1
For a b s ( x ) < 1 , x + 1 1 is strictly greater than 0.5.
Hence, 1 - x + 1 1 is strictly less than 0.5. None of the options are possible.