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irina [24]
3 years ago
13

Theorem: the sum of any two even integers equals 4k for some integer k. "proof: suppose m and n are any two even integers. by de

finition of even, m = 2k for some integer k and n = 2k for some integer k. by substitution, m + n = 2k + 2k = 4k. this is what was to be shown." what's the mistakes in this proof?
Mathematics
1 answer:
icang [17]3 years ago
3 0

Remark

The proof is only true if m and n are equal. Make it more general.

m = 2k

n = 2v

m + n = 2k + 2v = 2(k + v).

k and v can be equal but many times they are not. From that simple equation you cannot do anything for sure but divide by 2.

There are 4 combinations

m is divisible by 4 and n is not. The result will not be divisible by 4.

m is not divisible by 4 but n is. The result will not be divisible by 4.

But are divisible by 4 then the sum will be as well. Here's the really odd result

If both are even and not divisible by 4 then their sum is divisible by 4

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3 years ago
SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).
Ann [662]

Answer:

Required series is:

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

Step-by-step explanation:

Given that

                           f'(x) = -\frac{1}{1 + x^{2}} ---(1)

We know that:

                  \frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}} ---(2)

Comparing (1) and (2)

                           f'(x)=-(tan^{-1}x) ---- (3)

Using power series expansion for tan^{-1}x

f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx

= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]

=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}

=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

as

                 tan^{-1}(0)=0 \implies C=0

Hence,

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

7 0
3 years ago
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Step-by-step explanation:

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Answer:29.3

Step-by-step explanation: 88/3

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67 to the 2nd power, plz tell me if the answer is a prime number<br><br> Who wants to t a l k
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Answer:

No, it is not prime.

Step-by-step explanation:

67^2 = 4489

A number cannot be prime if it is divisible by a number besides 1 and itself. 4489 is divisible by 67; no perfect square can be prime.

8 0
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