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- #1

- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

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so

[tex]\displaystyle \begin{align*} x^5 + \frac{1}{x^5} &= \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 + \frac{1}{ \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 } \end{align*}[/tex]

which probably simplifies quite a lot

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My solution:

Hence:

\(\displaystyle x+\frac{1}{x}=\pm3\)

Also, we find:

\(\displaystyle \left(x+\frac{1}{x} \right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x} \right)\)

\(\displaystyle \pm27=x^3+\frac{1}{x^3}\pm9\)

\(\displaystyle x^3+\frac{1}{x^3}=\pm18\)

And finally:

\(\displaystyle \left(x+\frac{1}{x} \right)^5=x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3} \right)+10\left(x+\frac{1}{x} \right)\)

Hence:

\(\displaystyle x^5+\frac{1}{x^5}=\pm(243-5\cdot18-10\cdot3)=\pm123\)

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- #4

- Feb 14, 2012

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Hi

so

[tex]\displaystyle \begin{align*} x^5 + \frac{1}{x^5} &= \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 + \frac{1}{ \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 } \end{align*}[/tex]

which probably simplifies quite a lot

After reading your step-by-step solution, I really think your answer is correct but just it needed more simplification to it! And thanks for participating!

Well done,My solution:

Hence:

\(\displaystyle x+\frac{1}{x}=\pm3\)

Also, we find:

\(\displaystyle \left(x+\frac{1}{x} \right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x} \right)\)

\(\displaystyle \pm27=x^3+\frac{1}{x^3}\pm9\)

\(\displaystyle x^3+\frac{1}{x^3}=\pm18\)

And finally:

\(\displaystyle \left(x+\frac{1}{x} \right)^5=x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3} \right)+10\left(x+\frac{1}{x} \right)\)

Hence:

\(\displaystyle x^5+\frac{1}{x^5}=\pm(243-5\cdot18-10\cdot3)=\pm123\)

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- #5

- Mar 31, 2013

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Good solution by MarkFL

Here is another

Then f(0) = 2

Now

f(n)f(m) = (x^n + x^(-n)) ( x^m + x^(-m) = x^(n+m) + x^(n-m) + x^-(n-m) + x^(-(n+m))

= x^(n+m) + x^(-(n+m)) + x^(n-m) + x^-(n-m)

= f(n+m) + f(n-m)

so we got f(n)f(m) = f(n+m) + f(n-m)

n= 1,m= 1

give f(2) + f(0) = f(1)^2 = 9 so f(1) = 3 or – 3

for f(1) = 3

now n = 2, m= 1 we get

f(2)f(1) = f(3) + f(1) or f(3) = f(2) f(1) – f(1) = 7 * 3 – 3 = 18

n =3 , m= 2 give

f(3)f(2) = f(5) + f(1) or f(5) = f(3) f(2) – f(1) = 18 * 7 – 3 = 126 -3 = 123

for f(1) = - 3 we can compute similarly

Last edited:

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- Feb 14, 2012

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HiGood solution by MarkFL

Here is another

Then f(0) = 2

Now

f(n)f(m) = (x^n + x^(-n)) ( x^m + x^(-m) = x^(n+m) + x^(n-m) + x^-(n-m) + x^(-(n+m))

= x^(n+m) + x^(-(n+m)) + x^(n-m) + x^-(n-m)

= f(n+m) + f(n-m)

so we got f(n)f(m) = f(n+m) + f(n-m)

n= 1,m= 1

give f(2) + f(0) = f(1)^2 = 9 so f(1) = 3 or – 3

for f(1) = 3

now n = 2, m= 1 we get

f(2)f(1) = f(3) + f(1) or f(3) = f(2) f(1) – f(1) = 7 * 3 – 3 = 18

n =3 , m= 2 give

f(3)f(2) = f(5) + f(1) or f(5) = f(3) f(2) – f(1) = 18 * 7 – 3 = 126 -3 = 123

for f(1) = - 3 we can compute similarly

Thanks for participating and your method has given me a completely new insight on how to tackle the problem using another route...well done,

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- #8

- Feb 14, 2012

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My solution:I still would like to see it!

i.) \(\displaystyle \sec^2 y=7\tan y\)

ii.)\(\displaystyle \sin 2y=\frac{2}{7}\),

iii.) \(\displaystyle \sin y \cos y=\frac{1}{7}\),

iv.) \(\displaystyle \sin y+\cos y=\frac{3}{\sqrt{7}}\),

v.) \(\displaystyle \sin^4 y+\cos^4 y=\frac{47}{49}\),

vi.) \(\displaystyle \sqrt{\tan y}\cos y= \pm \frac{1}{\sqrt{7}}\),

Next, we have to evaluate \(\displaystyle x^5+\frac{1}{x^5}\), i.e.

\(\displaystyle x^5+\frac{1}{x^5}=\frac{x^{10}+1}{x^5}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{(x^{2})^5+1}{(x^2)^4(x)}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{(\tan y)^5+1}{(\tan y)^4(\sqrt{\tan y})}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{\sin^5 y+\cos^5 y}{\cos^3 y \sin^2 y(\sqrt{\tan y})}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{(\sin y+\cos y)(\sin^4 y+\cos^4 y-\sin y \cos y(\sin^2 y + \cos^2 y- \sin y \cos y))}{(\cos y \sin y)^2(\sqrt{\tan y})\cos y}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\pm \frac{(\frac{3}{\sqrt{7}})(\frac{47}{49}-\frac{1}{7}(1- \frac{1}{7}))}{(\frac{1}{7})^2(\frac{1}{\sqrt{7}})}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\pm 123\)

I know, I know, this looks like a mess...but this is my thought and I must post it to show how much I appreciate to the fact that there are so many members answered to my challenge problems!

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- #9

I just want to say it speaks highly of your character to be willing to post your solution, even if you feel it is not as straightforward as that given by another. But, I think others benefit by seeing different approaches, so you have done a good thing for MHB.

I know I have posted solutions to problems, only to have someone else come along and post a much more direct and simple solution, and the feeling of "D'oh! Why didn't I think of that?" can momentarily be very powerful!

But, in the end, we can console ourselves with the knowledge that at least we were able to solve the problem, and we can learn from the solutions given by others.

Thank you for sharing your solution.

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- #10

- Feb 14, 2012

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Thank you for your kind words, Mark! I agree with all that you said and you're always a joy to read!I just want to say it speaks highly of your character to be willing to post your solution, even if you feel it is not as straightforward as that given by another. But, I think others benefit by seeing different approaches, so you have done a good thing for MHB.

I know I have posted solutions to problems, only to have someone else come along and post a much more direct and simple solution, and the feeling of "D'oh! Why didn't I think of that?" can momentarily be very powerful!

But, in the end, we can console ourselves with the knowledge that at least we were able to solve the problem, and we can learn from the solutions given by others.

Thank you for sharing your solution.

- Sep 16, 2013

- 337

Extremely well said, Mark^^

For me personally, when an problem I find interesting gets posted on here - which thankfully happens a lot - I enjoy looking at all of the different solutions. And sometimes, like in this case, I find the less obvious / more circumspect ones the most interesting...

Well done Anemone! And thanks for sharing!!

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- #12

- Feb 14, 2012

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HiExtremely well said, Mark^^

For me personally, when an problem I find interesting gets posted on here - which thankfully happens a lot - I enjoy looking at all of the different solutions. And sometimes, like in this case, I find the less obvious / more circumspect ones the most interesting...

Well done Anemone! And thanks for sharing!!

That's extremely nice of you to say...and as a matter of fact, I know that I am the one who owes thanks to those members who have taught me how to solve the math problems on which I was stuck, and to those who have participated in my challenge problems, and to those who have read the challenge problems that I posted!