bcddd214

06-24-2011, 01:12 PM

Brain fart,,,

doesn't ?(1+2/y^6) evaluate to 1+4/y^4?

doesn't ?(1+2/y^6) evaluate to 1+4/y^4?

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bcddd214

06-24-2011, 01:12 PM

Brain fart,,,

doesn't ?(1+2/y^6) evaluate to 1+4/y^4?

doesn't ?(1+2/y^6) evaluate to 1+4/y^4?

soroban

06-24-2011, 02:50 PM

Hello, bcddd214!

\text{Doesn't }\sqrt{1+\frac{2}{y^6}}\,\text{ evaluate to: }\,1 +\frac{4}{y^4}\:?

Are you trying for a new world record?

Let's look at what you've done . . .

\sqrt{1 + \frac{2}{y^6}} \;=\;\sqrt{1} + \sqrt{\frac{2}{y^6}} .?

. . . . . . . =\;1 + \frac{\sqrt{2}}{\sqrt{y^6}}

Now: .\sqrt{2} \,=\,4 .?

and: .\sqrt[2]{y^6} \:=\:y^{6-2} \:=\:y^4 .?

\text{So you have: }\:1 + \frac{4}{y^4}

Was that your reasoning?

\text{Doesn't }\sqrt{1+\frac{2}{y^6}}\,\text{ evaluate to: }\,1 +\frac{4}{y^4}\:?

Are you trying for a new world record?

Let's look at what you've done . . .

\sqrt{1 + \frac{2}{y^6}} \;=\;\sqrt{1} + \sqrt{\frac{2}{y^6}} .?

. . . . . . . =\;1 + \frac{\sqrt{2}}{\sqrt{y^6}}

Now: .\sqrt{2} \,=\,4 .?

and: .\sqrt[2]{y^6} \:=\:y^{6-2} \:=\:y^4 .?

\text{So you have: }\:1 + \frac{4}{y^4}

Was that your reasoning?

Subhotosh Khan

06-24-2011, 03:07 PM

Brain fart,,,

doesn't ?(1+2/y^6) evaluate to 1+4/y^4? No

Without knowing exactly where did "that" come from - I have to say it cannot be reduced further.

doesn't ?(1+2/y^6) evaluate to 1+4/y^4? No

Without knowing exactly where did "that" come from - I have to say it cannot be reduced further.

Denis

06-24-2011, 03:24 PM

Brain fart,,,

doesn't ?(1+2/y^6) evaluate to 1+4/y^4?

Next time, check these silly things yourself:

assign a value to y, then CHECK!

Say we use y=1:

sqrt(1 + 2/y^6) = sqrt(3)

1 + 4/y^4 = 5

Does sqrt(3) = 5 ?

doesn't ?(1+2/y^6) evaluate to 1+4/y^4?

Next time, check these silly things yourself:

assign a value to y, then CHECK!

Say we use y=1:

sqrt(1 + 2/y^6) = sqrt(3)

1 + 4/y^4 = 5

Does sqrt(3) = 5 ?

bcddd214

06-24-2011, 03:51 PM

I apologize

doesn't ?(1+(4/y^6)) evaluate to 1+2/y^4?.

doesn't ?(1+(4/y^6)) evaluate to 1+2/y^4?.

JeffM

06-24-2011, 03:59 PM

I apologize

doesn't ?(1+(4/y^6)) evaluate to 1+2/y^4?.

No.

doesn't ?(1+(4/y^6)) evaluate to 1+2/y^4?.

No.

mmm4444bot

06-24-2011, 04:45 PM

doesn't ?(1+(4/y^6)) evaluate to 1+2/y^4?

No.

Actually, the correct answer to bcddd214's question above is "yes". :lol:

No.

Actually, the correct answer to bcddd214's question above is "yes". :lol:

mmm4444bot

06-24-2011, 04:49 PM

doesn't ?(1+ 4/y^6) evaluate to 1 + 2/y^4?

Please note that I removed the unneccesary grouping symbols from the first expression above.

We cannot evaluate sqrt(1 + 4/y^6) without knowing the value of y. I think that you're trying to ask about simplifying, instead.

So, yes, the expression sqrt(1 + 4/y^6) does not simplify to 1 + 2/y^4.

Please note that I removed the unneccesary grouping symbols from the first expression above.

We cannot evaluate sqrt(1 + 4/y^6) without knowing the value of y. I think that you're trying to ask about simplifying, instead.

So, yes, the expression sqrt(1 + 4/y^6) does not simplify to 1 + 2/y^4.

bcddd214

06-24-2011, 08:18 PM

doesn't ?(1+ 4/y^6) evaluate to 1 + 2/y^4?

Please note that I removed the unneccesary grouping symbols from the first expression above.

We cannot evaluate sqrt(1 + 4/y^6) without knowing the value of y. I think that you're trying to ask about simplifying, instead.

So, yes, the expression sqrt(1 + 4/y^6) does not simplify to 1 + 2/y^4.

Thank you.

The reason I 'thought' it did was because it looked pretty BUT, Maple said the same thing as you.

I plugged in evalf (sqrt(1+(4/y^6))); and it spit out the identical equation hinting that it could not be done.

I was just double checking the calculators answer to confirm there was no user error.

Thank you!

Please note that I removed the unneccesary grouping symbols from the first expression above.

We cannot evaluate sqrt(1 + 4/y^6) without knowing the value of y. I think that you're trying to ask about simplifying, instead.

So, yes, the expression sqrt(1 + 4/y^6) does not simplify to 1 + 2/y^4.

Thank you.

The reason I 'thought' it did was because it looked pretty BUT, Maple said the same thing as you.

I plugged in evalf (sqrt(1+(4/y^6))); and it spit out the identical equation hinting that it could not be done.

I was just double checking the calculators answer to confirm there was no user error.

Thank you!

JeffM

06-24-2011, 09:37 PM

doesn't ?(1+(4/y^6)) evaluate to 1+2/y^4?

No.

Actually, the correct answer to bcddd214's question above is "yes". :lol:

You are the mathematician, but may a wannabe historian present a possible counter-example a la mode du bon Denis.

Let's try y = 2.

(1 + (4/y[sup:3ud5rmel]6[/sup:3ud5rmel]))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = (1 + (4/2[sup:3ud5rmel]6[/sup:3ud5rmel]))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = (1 + (4/64))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = (1 + (1/16))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = sqrt(17) / 4.

1 + (2/y[sup:3ud5rmel]4[/sup:3ud5rmel]) = 1 + (2/16) = 1 + (1/8) = 9/8.

But (9/8)[sup:3ud5rmel]2[/sup:3ud5rmel] = 81/64 = 5.0625/4.

My problem is seeing how 5.06252[sup:3ud5rmel]2[/sup:3ud5rmel] > 25 = 17.

Of course, I admittedly am frequently in error, and arithmetic is not my forte, but I would love to know how you demonstrate that (1 + (4/y^6))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = 1 + (2/y[sup:3ud5rmel]4[/sup:3ud5rmel]) :P

PS I have this horrible feeling I am going to look extremely foolish before this is all over. Such as, you were teasing me and I was too literal-minded to twig, or else that devil Alzheimers has indeed caught up with me (which my newly acquired 18-year old wife keeps assuring me is the case as she hands me the power of attorney papers to sign).

No.

Actually, the correct answer to bcddd214's question above is "yes". :lol:

You are the mathematician, but may a wannabe historian present a possible counter-example a la mode du bon Denis.

Let's try y = 2.

(1 + (4/y[sup:3ud5rmel]6[/sup:3ud5rmel]))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = (1 + (4/2[sup:3ud5rmel]6[/sup:3ud5rmel]))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = (1 + (4/64))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = (1 + (1/16))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = sqrt(17) / 4.

1 + (2/y[sup:3ud5rmel]4[/sup:3ud5rmel]) = 1 + (2/16) = 1 + (1/8) = 9/8.

But (9/8)[sup:3ud5rmel]2[/sup:3ud5rmel] = 81/64 = 5.0625/4.

My problem is seeing how 5.06252[sup:3ud5rmel]2[/sup:3ud5rmel] > 25 = 17.

Of course, I admittedly am frequently in error, and arithmetic is not my forte, but I would love to know how you demonstrate that (1 + (4/y^6))[sup:3ud5rmel]1/2[/sup:3ud5rmel] = 1 + (2/y[sup:3ud5rmel]4[/sup:3ud5rmel]) :P

PS I have this horrible feeling I am going to look extremely foolish before this is all over. Such as, you were teasing me and I was too literal-minded to twig, or else that devil Alzheimers has indeed caught up with me (which my newly acquired 18-year old wife keeps assuring me is the case as she hands me the power of attorney papers to sign).

mmm4444bot

06-24-2011, 10:14 PM

Yeah, I was joshin' Jeff.

bcddd214 did not ask, "Does [this] evaluate to [that]".

They asked, "Does not [this] evaluate to [that]".

It does not, so the answer is "yes".

(I hope you're smiling right now.)

bcddd214 did not ask, "Does [this] evaluate to [that]".

They asked, "Does not [this] evaluate to [that]".

It does not, so the answer is "yes".

(I hope you're smiling right now.)

mmm4444bot

06-24-2011, 10:22 PM

The reason I 'thought' it did was because it looked pretty

Just as in life, looks can be deceiving. Don't be fooled by every pretty expression that crosses your path.

I plugged in evalf(sqrt(1+(4/y^6))); and it spit out the identical equation hinting that it could not be done.

That's because EVALF() is Maple's command for "evaluate the expression to a floating point number". You cannot get a Real number for this expression without first knowing a number that symbol y represents.

If you do not have any values for the symbol y, then the expression remains symbolic.

Another note about terminology: we don't call sqrt(1 + 4/y^6) an equation; it's called an expression. Equations always contain an equals sign. 8-)

Just as in life, looks can be deceiving. Don't be fooled by every pretty expression that crosses your path.

I plugged in evalf(sqrt(1+(4/y^6))); and it spit out the identical equation hinting that it could not be done.

That's because EVALF() is Maple's command for "evaluate the expression to a floating point number". You cannot get a Real number for this expression without first knowing a number that symbol y represents.

If you do not have any values for the symbol y, then the expression remains symbolic.

Another note about terminology: we don't call sqrt(1 + 4/y^6) an equation; it's called an expression. Equations always contain an equals sign. 8-)

JeffM

06-24-2011, 11:43 PM

Yeah, I was joshin' Jeff.

bcddd214 did not ask, "Does [this] evaluate to [that]".

They asked, "Does not [this] evaluate to [that]".

It does not, so the answer is "yes".

(I hope you're smiling right now.)

You got me right between the eyes.

bcddd214 did not ask, "Does [this] evaluate to [that]".

They asked, "Does not [this] evaluate to [that]".

It does not, so the answer is "yes".

(I hope you're smiling right now.)

You got me right between the eyes.

Denis

06-25-2011, 03:45 AM

You got me right between the eyes.

...you asked for it: you were leading with your nose 8-)

...you asked for it: you were leading with your nose 8-)

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