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What is the limit of $phi_{t_k}(x_k,y_k) = 1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$ when $y_k...
$begingroup$
Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.
sequences-and-series limits
asked Jan 9 at 16:27
geo17geo17
967
$endgroup$
add a comment |
$begingroup$
Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.
sequences-and-series limits
asked Jan 9 at 16:27
geo17geo17
967
$endgroup$
$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25
add a comment |
$begingroup$
Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.
sequences-and-series limits
asked Jan 9 at 16:27
geo17geo17
967
$endgroup$
Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.
sequences-and-series limits
sequences-and-series limits
asked Jan 9 at 16:27
geo17geo17
967
asked Jan 9 at 16:27
geo17geo17
967
asked Jan 9 at 16:27
geo17geo17
967
asked Jan 9 at 16:27
geo17geo17
967
asked Jan 9 at 16:27
geo17geo17
967
967
$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25
add a comment |
$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25
$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25
$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25
add a comment |
1 Answer
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$begingroup$
First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
add a comment |
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1 Answer
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$begingroup$
First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
add a comment |
$begingroup$
First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
add a comment |
$begingroup$
First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 9 at 19:58
Mostafa AyazMostafa Ayaz
15.5k3939
15.5k3939
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
add a comment |
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53
add a comment |
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$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25