Mixing
Regarding whether two functions are even or odd
$begingroup$
I am solving the integral
$$int_frac{-π}{2}^frac{π}{2} x cos x+ tan x^5 ,dx$$
In the solution, both $x cos x$ and $tan x^5$ are said to be odd functions.
Are these functions even or odd?
I noticed:
$$left(frac{-π}{2} right)cos left(frac{-π}{2} right)=-0$$ and $$left(tan left(frac{-π}{2} right) right)^5 =-∞$$
If these are odd functions then $0=-0$ and $∞=-∞$, the latter of which is making me have doubts.
calculus integration even-and-odd-functions
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
asked Jan 21 at 11:53
AashishAashish
688
$endgroup$
add a comment |
$begingroup$
I am solving the integral
$$int_frac{-π}{2}^frac{π}{2} x cos x+ tan x^5 ,dx$$
In the solution, both $x cos x$ and $tan x^5$ are said to be odd functions.
Are these functions even or odd?
I noticed:
$$left(frac{-π}{2} right)cos left(frac{-π}{2} right)=-0$$ and $$left(tan left(frac{-π}{2} right) right)^5 =-∞$$
If these are odd functions then $0=-0$ and $∞=-∞$, the latter of which is making me have doubts.
calculus integration even-and-odd-functions
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
asked Jan 21 at 11:53
AashishAashish
688
$endgroup$
1
$begingroup$
the fact that $left(tan left(frac{-π}{2} right) right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $left(tan left(frac{+π}{2} right) right)^5 =∞$
$endgroup$
– Calvin Khor
Jan 21 at 12:19
add a comment |
$begingroup$
I am solving the integral
$$int_frac{-π}{2}^frac{π}{2} x cos x+ tan x^5 ,dx$$
In the solution, both $x cos x$ and $tan x^5$ are said to be odd functions.
Are these functions even or odd?
I noticed:
$$left(frac{-π}{2} right)cos left(frac{-π}{2} right)=-0$$ and $$left(tan left(frac{-π}{2} right) right)^5 =-∞$$
If these are odd functions then $0=-0$ and $∞=-∞$, the latter of which is making me have doubts.
calculus integration even-and-odd-functions
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
asked Jan 21 at 11:53
AashishAashish
688
$endgroup$
I am solving the integral
$$int_frac{-π}{2}^frac{π}{2} x cos x+ tan x^5 ,dx$$
In the solution, both $x cos x$ and $tan x^5$ are said to be odd functions.
Are these functions even or odd?
I noticed:
$$left(frac{-π}{2} right)cos left(frac{-π}{2} right)=-0$$ and $$left(tan left(frac{-π}{2} right) right)^5 =-∞$$
If these are odd functions then $0=-0$ and $∞=-∞$, the latter of which is making me have doubts.
calculus integration even-and-odd-functions
calculus integration even-and-odd-functions
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
asked Jan 21 at 11:53
AashishAashish
688
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
asked Jan 21 at 11:53
AashishAashish
688
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
edited Jan 21 at 13:34
Eevee Trainer
7,42821338
7,42821338
asked Jan 21 at 11:53
AashishAashish
688
asked Jan 21 at 11:53
AashishAashish
688
asked Jan 21 at 11:53
AashishAashish
688
688
1
$begingroup$
the fact that $left(tan left(frac{-π}{2} right) right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $left(tan left(frac{+π}{2} right) right)^5 =∞$
$endgroup$
– Calvin Khor
Jan 21 at 12:19
add a comment |
1
$begingroup$
the fact that $left(tan left(frac{-π}{2} right) right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $left(tan left(frac{+π}{2} right) right)^5 =∞$
$endgroup$
– Calvin Khor
Jan 21 at 12:19
1
1
$begingroup$
the fact that $left(tan left(frac{-π}{2} right) right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $left(tan left(frac{+π}{2} right) right)^5 =∞$
$endgroup$
– Calvin Khor
Jan 21 at 12:19
$begingroup$
the fact that $left(tan left(frac{-π}{2} right) right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $left(tan left(frac{+π}{2} right) right)^5 =∞$
$endgroup$
– Calvin Khor
Jan 21 at 12:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.
Let $f(x) = x cdot cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
$$begin{align}
f(-x) &=-x cdot cos(-x) \
&= -x cdot cos(x) \
&= -f(x) \
end{align}$$
Thus, $f$ is odd, as $f(-x) = -f(x)$.
Let $g(x) = tan^5 (x)$. Recall tangent is odd, and thus
$$begin{align}
g(-x) &= tan^5 (-x)\
&= (-1)^5 tan^5 (x) \
&= - tan^5 (x) \
&= - g(x)
end{align}$$
(If instead you meant $tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.
It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 neq infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $tan(x)$ is not defined for $x = kpi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
$endgroup$
add a comment |
$begingroup$
To check, we have $$f(x)= xcos(x)= xcos(-x)= -(-xcos(-x))= -f(-x)$$ So $f(x)=xcos(x)$ is an odd function. Next, $$g(x)=(tan(x))^{5}= bigg(frac{sin(x)}{cos(x)}bigg)^{5} = bigg(frac{-sin(-x)}{cos(-x)}bigg)^{5}=(-1)^{5}bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -g(-x).$$ So we have the both are odd functions.
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.
Let $f(x) = x cdot cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
$$begin{align}
f(-x) &=-x cdot cos(-x) \
&= -x cdot cos(x) \
&= -f(x) \
end{align}$$
Thus, $f$ is odd, as $f(-x) = -f(x)$.
Let $g(x) = tan^5 (x)$. Recall tangent is odd, and thus
$$begin{align}
g(-x) &= tan^5 (-x)\
&= (-1)^5 tan^5 (x) \
&= - tan^5 (x) \
&= - g(x)
end{align}$$
(If instead you meant $tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.
It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 neq infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $tan(x)$ is not defined for $x = kpi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
$endgroup$
add a comment |
$begingroup$
Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.
Let $f(x) = x cdot cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
$$begin{align}
f(-x) &=-x cdot cos(-x) \
&= -x cdot cos(x) \
&= -f(x) \
end{align}$$
Thus, $f$ is odd, as $f(-x) = -f(x)$.
Let $g(x) = tan^5 (x)$. Recall tangent is odd, and thus
$$begin{align}
g(-x) &= tan^5 (-x)\
&= (-1)^5 tan^5 (x) \
&= - tan^5 (x) \
&= - g(x)
end{align}$$
(If instead you meant $tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.
It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 neq infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $tan(x)$ is not defined for $x = kpi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
$endgroup$
add a comment |
$begingroup$
Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.
Let $f(x) = x cdot cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
$$begin{align}
f(-x) &=-x cdot cos(-x) \
&= -x cdot cos(x) \
&= -f(x) \
end{align}$$
Thus, $f$ is odd, as $f(-x) = -f(x)$.
Let $g(x) = tan^5 (x)$. Recall tangent is odd, and thus
$$begin{align}
g(-x) &= tan^5 (-x)\
&= (-1)^5 tan^5 (x) \
&= - tan^5 (x) \
&= - g(x)
end{align}$$
(If instead you meant $tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.
It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 neq infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $tan(x)$ is not defined for $x = kpi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
$endgroup$
Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.
Let $f(x) = x cdot cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
$$begin{align}
f(-x) &=-x cdot cos(-x) \
&= -x cdot cos(x) \
&= -f(x) \
end{align}$$
Thus, $f$ is odd, as $f(-x) = -f(x)$.
Let $g(x) = tan^5 (x)$. Recall tangent is odd, and thus
$$begin{align}
g(-x) &= tan^5 (-x)\
&= (-1)^5 tan^5 (x) \
&= - tan^5 (x) \
&= - g(x)
end{align}$$
(If instead you meant $tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.
It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 neq infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $tan(x)$ is not defined for $x = kpi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
edited Jan 21 at 12:18
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
answered Jan 21 at 12:09
Eevee TrainerEevee Trainer
7,42821338
7,42821338
add a comment |
add a comment |
$begingroup$
To check, we have $$f(x)= xcos(x)= xcos(-x)= -(-xcos(-x))= -f(-x)$$ So $f(x)=xcos(x)$ is an odd function. Next, $$g(x)=(tan(x))^{5}= bigg(frac{sin(x)}{cos(x)}bigg)^{5} = bigg(frac{-sin(-x)}{cos(-x)}bigg)^{5}=(-1)^{5}bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -g(-x).$$ So we have the both are odd functions.
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
$endgroup$
add a comment |
$begingroup$
To check, we have $$f(x)= xcos(x)= xcos(-x)= -(-xcos(-x))= -f(-x)$$ So $f(x)=xcos(x)$ is an odd function. Next, $$g(x)=(tan(x))^{5}= bigg(frac{sin(x)}{cos(x)}bigg)^{5} = bigg(frac{-sin(-x)}{cos(-x)}bigg)^{5}=(-1)^{5}bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -g(-x).$$ So we have the both are odd functions.
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
$endgroup$
add a comment |
$begingroup$
To check, we have $$f(x)= xcos(x)= xcos(-x)= -(-xcos(-x))= -f(-x)$$ So $f(x)=xcos(x)$ is an odd function. Next, $$g(x)=(tan(x))^{5}= bigg(frac{sin(x)}{cos(x)}bigg)^{5} = bigg(frac{-sin(-x)}{cos(-x)}bigg)^{5}=(-1)^{5}bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -g(-x).$$ So we have the both are odd functions.
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
$endgroup$
To check, we have $$f(x)= xcos(x)= xcos(-x)= -(-xcos(-x))= -f(-x)$$ So $f(x)=xcos(x)$ is an odd function. Next, $$g(x)=(tan(x))^{5}= bigg(frac{sin(x)}{cos(x)}bigg)^{5} = bigg(frac{-sin(-x)}{cos(-x)}bigg)^{5}=(-1)^{5}bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -bigg(frac{sin(-x)}{cos(-x)}bigg)^{5}= -g(-x).$$ So we have the both are odd functions.
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
answered Jan 21 at 12:14
abdullah azzamabdullah azzam
294
294
add a comment |
add a comment |
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$begingroup$
the fact that $left(tan left(frac{-π}{2} right) right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $left(tan left(frac{+π}{2} right) right)^5 =∞$
$endgroup$
– Calvin Khor
Jan 21 at 12:19