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seraphim [82]
3 years ago
7

If i is raised to an odd power, then it can not simplify to be

Mathematics
2 answers:
Tomtit [17]3 years ago
6 0
A real number

ok so
any even power is i^(2n) where n is a whole number
if n=1
i^2=(√-1)^2=-1

if i^3, then it is equal to (i^2)(i)=(-1)(i)=-i
if i^4, then it is equal to (i^2)(i^2)=(-1)(-1)=1

therefor
i^1=i
i^2=-1
i^3=-i
i^4=1
then it repeats

look at the odd powers
i and -i
they are complex

therefor

if i is raised to an odd power, it cannot be simplified to be a real number
oksano4ka [1.4K]3 years ago
3 0

Answer:

-1

Step-by-step explanation:

If i is raised to an odd power, then it can not simplify to be

<u><em>-1</em></u>

-i

i

Odyssey

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a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter
andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|, converges if 0< b_{n+1} \leq b_n for all n, and \lim_{n \to \infty} b_n = 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

<em>Part a</em>

An Alternating series is an infinite series given on these three possible general forms given by:

\sum_{n=0}^{\infty} (-1)^{n} b_n

\sum_{n=0}^{\infty} (-1)^{n+1} b_n

\sum_{n=0}^{\infty} (-1)^{n-1} b_n

For all a_n >0, \forall n

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

<em>Part b</em>

An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|  converges if 0< b_{n+1} \leq b_n for all n and \lim_{n \to \infty} b_n =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

<em>Theorem (Alternating series test)</em>

If a sequence of positive terms {bn} is monotonically decreasing and

<em>\lim_{n \to \infty} b_n = 0<em>, then the alternating series \sum (-1)^{n-1} b_n converges if:</em></em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

then <em>\sum_{n=1}^{\infty}(-1)^{n-1} b_n  converges</em>

<em>Proof</em>

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

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<em>Theorem (Alternating series sum estimation)</em>

<em>If  \sum (-1)^{n-1} b_n  is the sum of an alternating series that satisfies</em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

Then then \mid s - s_n \mid \leq b_{n+1}

<em>Proof</em>

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\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}

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