Mixing
Prove the set of functions $T$ is a vector space.
$begingroup$
Show that $$T={t(x)|t(x)=a(x^2-1)+bln x+ccot x}$$ is a vector space.
My attempt:
To prove a vector space is to prove for $x_1,x_2in T$, $x_1+x_2in T$. So I calculated $$t(x_1)+t(x_2)=a({x_1}^2+{x_2}^2-2)+b ln x_1x_2+c(cot x_1+cot x_2)$$ But clearly it is not in T. Where did I miss?
linear-algebra
edited Jan 21 at 20:25
greedoid
46.2k1160117
asked Jan 21 at 17:06
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
Show that $$T={t(x)|t(x)=a(x^2-1)+bln x+ccot x}$$ is a vector space.
My attempt:
To prove a vector space is to prove for $x_1,x_2in T$, $x_1+x_2in T$. So I calculated $$t(x_1)+t(x_2)=a({x_1}^2+{x_2}^2-2)+b ln x_1x_2+c(cot x_1+cot x_2)$$ But clearly it is not in T. Where did I miss?
linear-algebra
edited Jan 21 at 20:25
greedoid
46.2k1160117
asked Jan 21 at 17:06
Yibei HeYibei He
3139
$endgroup$
2
$begingroup$
You need to compute $(t_1+t_2)(x)$, not $t(x_1)+t(x_2)$!
$endgroup$
– Mindlack
Jan 21 at 17:10
$begingroup$
Objects in $T$ are functions, say $f$ and $g$, then to be closed under addition, you need to show that $f+g$ is in $T$ as well. What you have done is taken inputs to the functions as the objects of $T$, which is incorrect.
$endgroup$
– Anurag A
Jan 21 at 20:08
add a comment |
$begingroup$
Show that $$T={t(x)|t(x)=a(x^2-1)+bln x+ccot x}$$ is a vector space.
My attempt:
To prove a vector space is to prove for $x_1,x_2in T$, $x_1+x_2in T$. So I calculated $$t(x_1)+t(x_2)=a({x_1}^2+{x_2}^2-2)+b ln x_1x_2+c(cot x_1+cot x_2)$$ But clearly it is not in T. Where did I miss?
linear-algebra
edited Jan 21 at 20:25
greedoid
46.2k1160117
asked Jan 21 at 17:06
Yibei HeYibei He
3139
$endgroup$
Show that $$T={t(x)|t(x)=a(x^2-1)+bln x+ccot x}$$ is a vector space.
My attempt:
To prove a vector space is to prove for $x_1,x_2in T$, $x_1+x_2in T$. So I calculated $$t(x_1)+t(x_2)=a({x_1}^2+{x_2}^2-2)+b ln x_1x_2+c(cot x_1+cot x_2)$$ But clearly it is not in T. Where did I miss?
linear-algebra
linear-algebra
edited Jan 21 at 20:25
greedoid
46.2k1160117
asked Jan 21 at 17:06
Yibei HeYibei He
3139
edited Jan 21 at 20:25
greedoid
46.2k1160117
asked Jan 21 at 17:06
Yibei HeYibei He
3139
edited Jan 21 at 20:25
greedoid
46.2k1160117
edited Jan 21 at 20:25
greedoid
46.2k1160117
edited Jan 21 at 20:25
greedoid
46.2k1160117
46.2k1160117
asked Jan 21 at 17:06
Yibei HeYibei He
3139
asked Jan 21 at 17:06
Yibei HeYibei He
3139
asked Jan 21 at 17:06
Yibei HeYibei He
3139
3139
2
$begingroup$
You need to compute $(t_1+t_2)(x)$, not $t(x_1)+t(x_2)$!
$endgroup$
– Mindlack
Jan 21 at 17:10
$begingroup$
Objects in $T$ are functions, say $f$ and $g$, then to be closed under addition, you need to show that $f+g$ is in $T$ as well. What you have done is taken inputs to the functions as the objects of $T$, which is incorrect.
$endgroup$
– Anurag A
Jan 21 at 20:08
add a comment |
2
$begingroup$
You need to compute $(t_1+t_2)(x)$, not $t(x_1)+t(x_2)$!
$endgroup$
– Mindlack
Jan 21 at 17:10
$begingroup$
Objects in $T$ are functions, say $f$ and $g$, then to be closed under addition, you need to show that $f+g$ is in $T$ as well. What you have done is taken inputs to the functions as the objects of $T$, which is incorrect.
$endgroup$
– Anurag A
Jan 21 at 20:08
2
2
$begingroup$
You need to compute $(t_1+t_2)(x)$, not $t(x_1)+t(x_2)$!
$endgroup$
– Mindlack
Jan 21 at 17:10
$begingroup$
You need to compute $(t_1+t_2)(x)$, not $t(x_1)+t(x_2)$!
$endgroup$
– Mindlack
Jan 21 at 17:10
$begingroup$
Objects in $T$ are functions, say $f$ and $g$, then to be closed under addition, you need to show that $f+g$ is in $T$ as well. What you have done is taken inputs to the functions as the objects of $T$, which is incorrect.
$endgroup$
– Anurag A
Jan 21 at 20:08
$begingroup$
Objects in $T$ are functions, say $f$ and $g$, then to be closed under addition, you need to show that $f+g$ is in $T$ as well. What you have done is taken inputs to the functions as the objects of $T$, which is incorrect.
$endgroup$
– Anurag A
Jan 21 at 20:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Ther must be $x_1= x_2=x$ and $$t_1(x)= a_1(x^2-1)+b_1ln x +c_1cot x$$ $$t_2(x)= a_2(x^2-1)+b_2ln x +c_2cot x$$ so
$$t_1(x)+t_2(x)= underbrace{(a_1+a_2)}_{a}(x^2-1)+underbrace{(b_1+b_2)}_{b}ln x +underbrace{(c_1+c_2)}_{c}cot x$$
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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$begingroup$
Ther must be $x_1= x_2=x$ and $$t_1(x)= a_1(x^2-1)+b_1ln x +c_1cot x$$ $$t_2(x)= a_2(x^2-1)+b_2ln x +c_2cot x$$ so
$$t_1(x)+t_2(x)= underbrace{(a_1+a_2)}_{a}(x^2-1)+underbrace{(b_1+b_2)}_{b}ln x +underbrace{(c_1+c_2)}_{c}cot x$$
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
$endgroup$
add a comment |
$begingroup$
Ther must be $x_1= x_2=x$ and $$t_1(x)= a_1(x^2-1)+b_1ln x +c_1cot x$$ $$t_2(x)= a_2(x^2-1)+b_2ln x +c_2cot x$$ so
$$t_1(x)+t_2(x)= underbrace{(a_1+a_2)}_{a}(x^2-1)+underbrace{(b_1+b_2)}_{b}ln x +underbrace{(c_1+c_2)}_{c}cot x$$
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
$endgroup$
add a comment |
$begingroup$
Ther must be $x_1= x_2=x$ and $$t_1(x)= a_1(x^2-1)+b_1ln x +c_1cot x$$ $$t_2(x)= a_2(x^2-1)+b_2ln x +c_2cot x$$ so
$$t_1(x)+t_2(x)= underbrace{(a_1+a_2)}_{a}(x^2-1)+underbrace{(b_1+b_2)}_{b}ln x +underbrace{(c_1+c_2)}_{c}cot x$$
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
$endgroup$
Ther must be $x_1= x_2=x$ and $$t_1(x)= a_1(x^2-1)+b_1ln x +c_1cot x$$ $$t_2(x)= a_2(x^2-1)+b_2ln x +c_2cot x$$ so
$$t_1(x)+t_2(x)= underbrace{(a_1+a_2)}_{a}(x^2-1)+underbrace{(b_1+b_2)}_{b}ln x +underbrace{(c_1+c_2)}_{c}cot x$$
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
answered Jan 21 at 17:12
greedoidgreedoid
46.2k1160117
46.2k1160117
add a comment |
add a comment |
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$begingroup$
You need to compute $(t_1+t_2)(x)$, not $t(x_1)+t(x_2)$!
$endgroup$
– Mindlack
Jan 21 at 17:10
$begingroup$
Objects in $T$ are functions, say $f$ and $g$, then to be closed under addition, you need to show that $f+g$ is in $T$ as well. What you have done is taken inputs to the functions as the objects of $T$, which is incorrect.
$endgroup$
– Anurag A
Jan 21 at 20:08