Simplify the expression

xoxo1415 · **Joined:** Wed, 17 Aug 2011 23:43:31 UTC **Posts:** 11

1/1+a^n + 1/1+a^-n

My attempt:
=1/1+a^n + 1+a^n/1
=1+{1+a^n(1+a^n)}/1+a^n
=1+1+2a^n+a^n^2/1+a^n
=2+2a^n+a^n^2/1+a^n
let a^n=m
=2+2m+2m^2/1+m
=2(1+m+m^2)/1+m

I don't know what to do next, the answer is supposed to be 1.

Shadow · **Posted:** Thu, 18 Aug 2011 00:08:44 UTC

xoxo1415 wrote:

1/1+a^n + 1/1+a^-n

My attempt:
=1/1+a^n + 1+a^n/1
=1+{1+a^n(1+a^n)}/1+a^n
=1+1+2a^n+a^n^2/1+a^n
=2+2a^n+a^n^2/1+a^n
let a^n=m
=2+2m+2m^2/1+m
=2(1+m+m^2)/1+m

I don't know what to do next, the answer is supposed to be 1.

The answer cannot be 1, you have ${1\over 1}+a^n+{1\over 1}+a^{-n}=2+a^n+a^{-n}$ . If you want a simple fraction, then you get:

$${2a^n+a^{2n}+1\over a^n}={(a^n+1)^2\over a^n}$ , if you want something all within one set of parentheses, then you can transform this further:

$(a^{n/2}+a^{-n/2})^2$

If you want this in terms of a simpler function, this is $4\cosh^2({n\log a\over 2})$ , but none of these are equal to 1.

xoxo1415 · **Joined:** Wed, 17 Aug 2011 23:43:31 UTC **Posts:** 11

Shadow wrote:

xoxo1415 wrote:

1/1+a^n + 1/1+a^-n

My attempt:
=1/1+a^n + 1+a^n/1
=1+{1+a^n(1+a^n)}/1+a^n
=1+1+2a^n+a^n^2/1+a^n
=2+2a^n+a^n^2/1+a^n
let a^n=m
=2+2m+2m^2/1+m
=2(1+m+m^2)/1+m

I don't know what to do next, the answer is supposed to be 1.

The answer cannot be 1, you have ${1\over 1}+a^n+{1\over 1}+a^{-n}=2+a^n+a^{-n}$ . If you want a simple fraction, then you get:

$${2a^n+a^{2n}+1\over a^n}={(a^n+1)^2\over a^n}$ , if you want something all within one set of parentheses, then you can transform this further:

$(a^{n/2}+a^{-n/2})^2$

If you want this in terms of a simpler function, this is $4\cosh^2({n\log a\over 2})$ , but none of these are equal to 1.

I'm sorry I think I didn't type my question properly or you misunderstood it, this is how the question goes :

{1/(1+a^n)} + {1/(1+a^-n)}

Shadow · **Posted:** Thu, 18 Aug 2011 00:27:09 UTC

xoxo1415 wrote:

Shadow wrote:

xoxo1415 wrote:

1/1+a^n + 1/1+a^-n

My attempt:
=1/1+a^n + 1+a^n/1
=1+{1+a^n(1+a^n)}/1+a^n
=1+1+2a^n+a^n^2/1+a^n
=2+2a^n+a^n^2/1+a^n
let a^n=m
=2+2m+2m^2/1+m
=2(1+m+m^2)/1+m

I don't know what to do next, the answer is supposed to be 1.

The answer cannot be 1, you have ${1\over 1}+a^n+{1\over 1}+a^{-n}=2+a^n+a^{-n}$ . If you want a simple fraction, then you get:

$${2a^n+a^{2n}+1\over a^n}={(a^n+1)^2\over a^n}$ , if you want something all within one set of parentheses, then you can transform this further:

$(a^{n/2}+a^{-n/2})^2$

If you want this in terms of a simpler function, this is $4\cosh^2({n\log a\over 2})$ , but none of these are equal to 1.

I'm sorry I think I didn't type my question properly or you misunderstood it, this is how the question goes :

{1/(1+a^n)} + {1/(1+a^-n)}

Ah yes, if you put parentheses around it that way, then yes, you get 1, simply get a common denominator:

${1\over 1+a^n}+{1\over 1+a^{-n}}={1+a^{-n}+1+a^n\over 1+a^n+1+a^{-n}}=1$

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

1/(1+x) + 1/(1 + 1/x) = 1/(1+x) + x/(x+1) = (1+x)/(1+x)=1

x=a^n (is a red herring)

Shadow · **Posted:** Thu, 18 Aug 2011 00:41:27 UTC

mathematic wrote:

1/(1+x) + 1/(1 + 1/x) = 1/(1+x) + x/(x+1) = (1+x)/(1+x)=1

x=a^n (is a red herring)

What do you mean a red herring, all you did was rename the variable, if the whole thing was 1, it's obvious that it CANNOT depend on a or n by definition of being a constant.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

Shadow wrote:

mathematic wrote:

1/(1+x) + 1/(1 + 1/x) = 1/(1+x) + x/(x+1) = (1+x)/(1+x)=1

x=a^n (is a red herring)

What do you mean a red herring, all you did was rename the variable, if the whole thing was 1, it's obvious that it CANNOT depend on a or n by definition of being a constant.

(red herring) There was no reason to put it in the form a^n, a simple x will do.

Shadow · **Posted:** Thu, 18 Aug 2011 22:33:59 UTC

mathematic wrote:

Shadow wrote:

mathematic wrote:

1/(1+x) + 1/(1 + 1/x) = 1/(1+x) + x/(x+1) = (1+x)/(1+x)=1

x=a^n (is a red herring)

What do you mean a red herring, all you did was rename the variable, if the whole thing was 1, it's obvious that it CANNOT depend on a or n by definition of being a constant.

(red herring) There was no reason to put it in the form a^n, a simple x will do.

I disagree, if it were put in the other form it would be just what it is, the fact that they used $a^{\pm n}$ means they probably wanted to get the student practice at seeing the more general principles in more specific situations, which is arguably as important as being able to see the more general principles. For example, of course you can say:

"Solve x^2-2x=1

" and people will think "quadratic formula"

However, saying "Solve $\sin^4 x-2\sin^2 x=1$ " is perhaps a more reasonable question to get in some applications, and indeed it is just a disguised quadratic, but what use is the formula to you if you cannot see the general in the specific, something which may be one of the most central ideas in mathematics as a subject. It's the reason we can tell that some students just see math as a bunch of symbol pushing and cannot solve problems once they've just been even slightly tweaked to be in different forms which are still essentially the same as the more obvious versions.

Soroban · **Posted:** Fri, 19 Aug 2011 00:20:03 UTC

Hello, xoxo1415!

Quote:

$$\text{Simplify: }\:\frac{1}{1+a^n} + \frac{1}{1+a^{-n}}$

Multiply the second fraction by $\dfrac{a^n}{a^n}.$

. . $$\frac{1}{1+a^n}\;+\;\frac{a^n}{a^n}\cdot\frac{1}{1+a^{-n}} \;=\;\frac{1}{1+a^n}\;+\;\frac{a^n}{a^n + 1} \;=\;\frac{1}{1+a^n}\;+\;\frac{a^n}{1+a^n} \;=\;\frac{1+a^n}{1+a^n} \;=\;1$

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

Shadow wrote:

mathematic wrote:

Shadow wrote:

mathematic wrote:

1/(1+x) + 1/(1 + 1/x) = 1/(1+x) + x/(x+1) = (1+x)/(1+x)=1

x=a^n (is a red herring)

What do you mean a red herring, all you did was rename the variable, if the whole thing was 1, it's obvious that it CANNOT depend on a or n by definition of being a constant.

(red herring) There was no reason to put it in the form a^n, a simple x will do.

I disagree, if it were put in the other form it would be just what it is, the fact that they used $a^{\pm n}$ means they probably wanted to get the student practice at seeing the more general principles in more specific situations, which is arguably as important as being able to see the more general principles. For example, of course you can say:

"Solve x^2-2x=1

" and people will think "quadratic formula"

However, saying "Solve $\sin^4 x-2\sin^2 x=1$ " is perhaps a more reasonable question to get in some applications, and indeed it is just a disguised quadratic, but what use is the formula to you if you cannot see the general in the specific, something which may be one of the most central ideas in mathematics as a subject. It's the reason we can tell that some students just see math as a bunch of symbol pushing and cannot solve problems once they've just been even slightly tweaked to be in different forms which are still essentially the same as the more obvious versions.

You assume that the original question was part of an exercise given to a student. You may be right. I just did not make any such assumption.

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