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Atrium > Mathematics, Physics & Philosophy: a trig identity

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- CommentRowNumber1.
- CommentAuthorUrs
- CommentTime2 days ago
- (edited 2 days ago)
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Author: Urs
Format: MarkdownItexThis may just show my ignorance, but I am wondering how to see the following: $\,$ For $k \in \mathbb{N}_{\gt 0}$ and $a \in \mathbb{N}$, the expression $$ S_k(a) \,\coloneqq\, \sum_{n = 0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} (n + a)^2 } $$ is independent of $a$ if $k$ is even: $$ k\;\text{is even} \;\;\;\Rightarrow\;\;\; \underset{a}{\forall} \big( \; S_k(a) = S_k(0) \; \big) \,. $$ In a roundabout way this follows from the modular transformation law of conformal characters of the $\mathrm{U}(1)$ WZW model. But what's an elementary arithmetic way to see this? (If the exponent had another factor of "2", then it would follow easily, also for odd $k$, by modular arithmetic, now in the other sense of "modular". But the question is for the exponent as given.)

This may just show my ignorance, but I am wondering how to see the following:

$<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mspace width="0.16667em"/></mrow><annotation encoding="application/x-tex">\,</annotation></semantics></math>$

For $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi><mo>∈</mo><msub><mi>ℕ</mi> <mrow><mo>&gt;</mo><mn>0</mn></mrow></msub></mrow><annotation encoding="application/x-tex">k \in \mathbb{N}_{\gt 0}</annotation></semantics></math>$ and $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>a</mi><mo>∈</mo><mi>ℕ</mi></mrow><annotation encoding="application/x-tex">a \in \mathbb{N}</annotation></semantics></math>$ , the expression
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msub><mi>S</mi> <mi>k</mi></msub><mo stretchy="false">(</mo><mi>a</mi><mo stretchy="false">)</mo><mspace width="0.16667em"/><mo>≔</mo><mspace width="0.16667em"/><munderover><mo lspace="0.16667em" rspace="0.16667em">∑</mo> <mrow><mi>n</mi><mo>=</mo><mn>0</mn></mrow> <mrow><mi>k</mi><mo>−</mo><mn>1</mn></mrow></munderover><msup><mi>e</mi> <mrow><mstyle displaystyle="false"><mfrac><mrow><mi>π</mi><mi mathvariant="normal">i</mi></mrow><mi>k</mi></mfrac></mstyle><mo stretchy="false">(</mo><mi>n</mi><mo>+</mo><mi>a</mi><msup><mo stretchy="false">)</mo> <mn>2</mn></msup></mrow></msup></mrow><annotation encoding="application/x-tex"> S_k(a) \,\coloneqq\, \sum_{n = 0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} (n + a)^2 } </annotation></semantics></math>$
is independent of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math>$ if $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>$ is even:
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>k</mi><mspace width="0.27778em"/><mtext>is even</mtext><mspace width="0.27778em"/><mspace width="0.27778em"/><mspace width="0.27778em"/><mo>⇒</mo><mspace width="0.27778em"/><mspace width="0.27778em"/><mspace width="0.27778em"/><munder><mo>∀</mo><mi>a</mi></munder><mo maxsize="1.2em" minsize="1.2em">(</mo><mspace width="0.27778em"/><msub><mi>S</mi> <mi>k</mi></msub><mo stretchy="false">(</mo><mi>a</mi><mo stretchy="false">)</mo><mo>=</mo><msub><mi>S</mi> <mi>k</mi></msub><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mspace width="0.27778em"/><mo maxsize="1.2em" minsize="1.2em">)</mo><mspace width="0.16667em"/><mo>.</mo></mrow><annotation encoding="application/x-tex"> k\;\text{is even} \;\;\;\Rightarrow\;\;\; \underset{a}{\forall} \big( \; S_k(a) = S_k(0) \; \big) \,. </annotation></semantics></math>$
In a roundabout way this follows from the modular transformation law of conformal characters of the $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi mathvariant="normal">U</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\mathrm{U}(1)</annotation></semantics></math>$ WZW model.

But what’s an elementary arithmetic way to see this?

(If the exponent had another factor of “2”, then it would follow easily, also for odd $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>$ , by modular arithmetic, now in the other sense of “modular”. But the question is for the exponent as given.)
- CommentRowNumber2.
- CommentAuthorDavid_Corfield
- CommentTime2 days ago
- PermaLink
Author: David_Corfield
Format: MarkdownItexI guess the squares repeat with period $k$. If $k$ is even then $(k + a)^2 - a^2 = 2a k + k^2$ is $0$ mod $2 k$.

I guess the squares repeat with period $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>$ . If $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>$ is even then $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><mi>k</mi><mo>+</mo><mi>a</mi><msup><mo stretchy="false">)</mo> <mn>2</mn></msup><mo>−</mo><msup><mi>a</mi> <mn>2</mn></msup><mo>=</mo><mn>2</mn><mi>a</mi><mi>k</mi><mo>+</mo><msup><mi>k</mi> <mn>2</mn></msup></mrow><annotation encoding="application/x-tex">(k + a)^2 - a^2 = 2a k + k^2</annotation></semantics></math>$ is $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mn>0</mn></mrow><annotation encoding="application/x-tex">0</annotation></semantics></math>$ mod $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mn>2</mn><mi>k</mi></mrow><annotation encoding="application/x-tex">2 k</annotation></semantics></math>$ .
- CommentRowNumber3.
- CommentAuthorUrs
- CommentTime2 days ago
- PermaLink
Author: Urs
Format: MarkdownItexI am prepared to find that I am missing the obvious -- but could you expand on how this answers the question?

I am prepared to find that I am missing the obvious – but could you expand on how this answers the question?
- CommentRowNumber4.
- CommentAuthorUrs
- CommentTime2 days ago
- PermaLink
Author: Urs
Format: MarkdownItexOh, I think I see what you mean. Right.

Oh, I think I see what you mean. Right.
- CommentRowNumber5.
- CommentAuthorUrs
- CommentTime2 days ago
- PermaLink
Author: Urs
Format: MarkdownItexRight, so the summands are $k$-periodic after all, because $$ e^{ \tfrac{\pi \mathrm{i}}{k} (n + k)^2 } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} (n^2 + 2 n k + k^2) } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \underset{ 1\;if\;k\;even }{ \underbrace{ e^{ \pi \mathrm{i}(2n + k) } } } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } $$ and so if we shift $n \mapsto n + a$ in the exponent, then we may just relabel to get rid of the $a$. Thanks! :-)

Right, so the summands are $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>$ -periodic after all, because
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mi>e</mi> <mrow><mstyle displaystyle="false"><mfrac><mrow><mi>π</mi><mi mathvariant="normal">i</mi></mrow><mi>k</mi></mfrac></mstyle><mo stretchy="false">(</mo><mi>n</mi><mo>+</mo><mi>k</mi><msup><mo stretchy="false">)</mo> <mn>2</mn></msup></mrow></msup><mspace width="0.27778em"/><mo>=</mo><mspace width="0.27778em"/><msup><mi>e</mi> <mrow><mstyle displaystyle="false"><mfrac><mrow><mi>π</mi><mi mathvariant="normal">i</mi></mrow><mi>k</mi></mfrac></mstyle><mo stretchy="false">(</mo><msup><mi>n</mi> <mn>2</mn></msup><mo>+</mo><mn>2</mn><mi>n</mi><mi>k</mi><mo>+</mo><msup><mi>k</mi> <mn>2</mn></msup><mo stretchy="false">)</mo></mrow></msup><mspace width="0.27778em"/><mo>=</mo><mspace width="0.27778em"/><msup><mi>e</mi> <mrow><mstyle displaystyle="false"><mfrac><mrow><mi>π</mi><mi mathvariant="normal">i</mi></mrow><mi>k</mi></mfrac></mstyle><msup><mi>n</mi> <mn>2</mn></msup></mrow></msup><munder><munder><mrow><msup><mi>e</mi> <mrow><mi>π</mi><mi mathvariant="normal">i</mi><mo stretchy="false">(</mo><mn>2</mn><mi>n</mi><mo>+</mo><mi>k</mi><mo stretchy="false">)</mo></mrow></msup></mrow><mo>⏟</mo></munder><mrow><mn>1</mn><mspace width="0.27778em"/><mi>if</mi><mspace width="0.27778em"/><mi>k</mi><mspace width="0.27778em"/><mi>even</mi></mrow></munder><mspace width="0.27778em"/><mo>=</mo><mspace width="0.27778em"/><msup><mi>e</mi> <mrow><mstyle displaystyle="false"><mfrac><mrow><mi>π</mi><mi mathvariant="normal">i</mi></mrow><mi>k</mi></mfrac></mstyle><msup><mi>n</mi> <mn>2</mn></msup></mrow></msup></mrow><annotation encoding="application/x-tex"> e^{ \tfrac{\pi \mathrm{i}}{k} (n + k)^2 } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} (n^2 + 2 n k + k^2) } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \underset{ 1\;if\;k\;even }{ \underbrace{ e^{ \pi \mathrm{i}(2n + k) } } } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } </annotation></semantics></math>$
and so if we shift $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>n</mi><mo>↦</mo><mi>n</mi><mo>+</mo><mi>a</mi></mrow><annotation encoding="application/x-tex">n \mapsto n + a</annotation></semantics></math>$ in the exponent, then we may just relabel to get rid of the $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math>$ .

Thanks! :-)
- CommentRowNumber6.
- CommentAuthorDavid_Corfield
- CommentTime2 days ago
- PermaLink
Author: David_Corfield
Format: MarkdownItexSorry for being cryptic. A snatched moment when supposed to be supporting my wife in her clothes shopping for our daughter's wedding!

Sorry for being cryptic. A snatched moment when supposed to be supporting my wife in her clothes shopping for our daughter’s wedding!

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Atrium > Mathematics, Physics & Philosophy: a trig identity