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Is it the same $a^alpha + b^alpha = x$ as $(a^alpha)^frac1alpha + (b^alpha)^frac1alpha = x^frac1alpha$
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I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?
linear-algebra linear-transformations exponentiation
asked Jul 18 at 16:32
JoeDi
102
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I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?
linear-algebra linear-transformations exponentiation
asked Jul 18 at 16:32
JoeDi
102
1
No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32
The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33
Short answer : No.
– Alex Silva
Jul 18 at 16:41
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?
linear-algebra linear-transformations exponentiation
asked Jul 18 at 16:32
JoeDi
102
I want to ask if it is possible to transform equation $a^alpha + b^alpha = x$ into $a + b = x^frac1alpha$ by elevating each parameter by $frac1alpha$? Or if we elevate each parameter by $frac1alpha$ it must be $(a^alpha + b^alpha)^frac1alpha = x^frac1alpha$?
linear-algebra linear-transformations exponentiation
asked Jul 18 at 16:32
JoeDi
102
asked Jul 18 at 16:32
JoeDi
102
asked Jul 18 at 16:32
JoeDi
102
asked Jul 18 at 16:32
JoeDi
102
102
1
No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32
The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33
Short answer : No.
– Alex Silva
Jul 18 at 16:41
add a comment |Â
1
No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32
The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33
Short answer : No.
– Alex Silva
Jul 18 at 16:41
1
1
No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32
No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32
The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33
The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33
Short answer : No.
– Alex Silva
Jul 18 at 16:41
Short answer : No.
– Alex Silva
Jul 18 at 16:41
add a comment |Â
1 Answer
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No it isn't in general since
$$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$
For general $a$ and $b$ it is trivially true only for $alpha=1$.
answered Jul 18 at 16:35
gimusi
65.4k73584
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
No it isn't in general since
$$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$
For general $a$ and $b$ it is trivially true only for $alpha=1$.
answered Jul 18 at 16:35
gimusi
65.4k73584
add a comment |Â
up vote
1
down vote
accepted
No it isn't in general since
$$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$
For general $a$ and $b$ it is trivially true only for $alpha=1$.
answered Jul 18 at 16:35
gimusi
65.4k73584
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
No it isn't in general since
$$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$
For general $a$ and $b$ it is trivially true only for $alpha=1$.
answered Jul 18 at 16:35
gimusi
65.4k73584
No it isn't in general since
$$x=a^alpha + b^alpha implies x^frac1alpha=(a^alpha + b^alpha)^frac1alpha$$
For general $a$ and $b$ it is trivially true only for $alpha=1$.
answered Jul 18 at 16:35
gimusi
65.4k73584
answered Jul 18 at 16:35
gimusi
65.4k73584
answered Jul 18 at 16:35
gimusi
65.4k73584
answered Jul 18 at 16:35
gimusi
65.4k73584
65.4k73584
add a comment |Â
add a comment |Â
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1
No, yes$$.
– Lord Shark the Unknown
Jul 18 at 16:32
The answer to the title question is "no"
– Wojowu
Jul 18 at 16:33
Short answer : No.
– Alex Silva
Jul 18 at 16:41